Chapter 1 Problems 1-1 through 1-6 are for student research. No standard solutions are provided. From Fig. 1-2, cost of grinding to  0.0005 in is 270%. Cost of turning to  0.003 in is 60%. Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans. ______________________________________________________________________________ 1-7 1-8 CA = CB, 10 + 0.8 P = 60 + 0.8 P  0.005 P 2 P 2 = 50/0.005  P = 100 parts Ans. ______________________________________________________________________________ 1-9 Max. load = 1.10 P Min. area = (0.95)2A Min. strength = 0.85 S To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be nd  1.10 0.85  0.95  2 1.43 Ans. ______________________________________________________________________________ 1-10 (a) X1 + X2: x1  x2  X 1  e1  X 2  e2 error e  x1  x2    X 1  X 2  e1  e2 (b) X1  X2: Ans. x1  x2  X 1  e1   X 2  e2  e  x1  x2    X 1  X 2  e1  e2 (c) X1 X2: Ans. x1 x2  X 1  e1   X 2  e2  e  x1 x2  X 1 X 2  X 1e2  X 2 e1  e1e2  e e   X 1e2  X 2e1  X 1 X 2  1  2   X1 X 2  Shigley’s MED, 11th edition Ans. Chapter 1 Solutions, Page 1/12 (d) X1/X2: x1 X 1  e1 X 1  1  e1 X 1      x2 X 2  e2 X 2  1  e2 X 2  1  e2  e2 1  1  X2  X2  then  1  e1 X 1   e1   e2  e1 e2     1   1  1  X1   X2  X1 X 2  1  e2 X 2   x1 X 1 X 1  e1 e2       Ans. x2 X 2 X 2  X 1 X 2  ______________________________________________________________________________ Thus, 1-11 (a) e x1 = 7 = 2.645 751 311 1 X1 = 2.64 (3 correct digits) x2 = 8 = 2.828 427 124 7 X2 = 2.82 (3 correct digits) x1 + x2 = 5.474 178 435 8 e1 = x1  X1 = 0.005 751 311 1 e2 = x2  X2 = 0.008 427 124 7 e = e1 + e2 = 0.014 178 435 8 Sum = x1 + x2 = X1 + X2 + e = 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks (b) X1 = 2.65, X2 = 2.83 (3 digit significant numbers) e1 = x1  X1 =  0.004 248 688 9 e2 = x2  X2 =  0.001 572 875 3 e = e1 + e2 =  0.005 821 564 2 Sum = x1 + x2 = X1 + X2 + e = 2.65 +2.83  0.001 572 875 3 = 5.474 178 435 8 Checks ______________________________________________________________________________ S  nd 1-12  32  1000  25  10  d3 2.5 Table A-17: d = 1 14 in 3   d 1.006 in Ans. Ans. 25  103  S n  4.79  32  1000  Ans.   1.25 Factor of safety: ______________________________________________________________________________ 3 Shigley’s MED, 11th edition Chapter 1 Solutions, Page 2/12 1-13 (a) x 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 fx f x2 120 7200 70 4900 240 19200 450 40500 800 80000 1320 145200 720 86400 1300 169000 1120 156800 750 112500 320 51200 510 86700 360 64800 190 36100 0 0 210 44100 8480 1 104 600 f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 69  x  Eq. (1-6) 1 N k fx i i i 1  8480 122.9 kcycles 69 Eq. (1-7) k sx  fx 2 i i  N x2 i 1 N  1  1 104 600  69(122.9) 2    69  1   z115  (b) Eq. (1-5) 1/ 2  30.3 kcycles Ans. x  x x  x 115  122.9  115    0.2607 ˆ x sx 30.3 Interpolating from Table (A-10) 0.2600 0.2607 0.2700 0.3974 x 0.3936  N(0.2607) = 69 (0.3971) = 27.4  27 Shigley’s MED, 11th edition x = 0.3971 Ans. Chapter 1 Solutions, Page 3/12 From the data, the number of instances less than 115 kcycles is 2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal) ____________________________________________________________________________ 1-14 x 174 182 190 198 206 214 222  x  Eq. (1-6) 6 9 44 67 53 12 6 fx 1044 1638 8360 13266 10918 2568 1332 f x2 181656 298116 1588400 2626668 2249108 549552 295704 197 39126 7789204 f 1 N k fx i i i 1 k fx 2 i i  39 126 198.61 kpsi 197  N x2 12  7 789 204  197(198.61) 2  sx    9.68 kpsi Ans. N1 197  1   Eq. (1-7) ______________________________________________________________________________ i 1 1-15 L 122.9 kcycles and sL  30.3 kcycles z10  Eq. (1-5) Thus, x   x x10  L x10  122.9   sL 30.3 ˆ x10 = 122.9 + 30.3 z10 = L10 From Table A-10, for 10 percent failure, z10 = 1.282. Thus, L10 = 122.9 + 30.3(1.282) = 84.1 kcycles Ans. ___________________________________________________________________________ Shigley’s MED, 11th edition Chapter 1 Solutions, Page 4/12 1-16 x f 93 95 97 99 101 103 105 107 109 111  19 25 38 17 12 10 5 4 4 2 136 x  Eq. (1-6) 1 N k fx i i fx 2 i i 13 364 / 136  98.26471 = 98.26 kpsi  N x2 i 1 N1 Eq. (1-7) fx2 164331 225625 357542 166617 122412 106090 55125 45796 47524 24642 1315704 i 1 k sx  fx 1767 2375 3686 1683 1212 1030 525 428 436 222 13364  1 315 704  136(98.26471) 2    136  1   12 4.30 kpsi Note, for accuracy in the calculation given above, x needs to be of more significant figures than the rounded value. For a normal distribution, from Eq. (1-5), and a yield strength exceeded by 99 percent (R = 0.99, pf = 0.01), x  x x  x x  98.26 z0.01   0.01  0.01 sx 4.30 ˆ x Solving for the yield strength gives x0.01 = 98.26 + 4.30 z0.01 From Table A-10, z0.01 =  2.326. Thus x0.01 = 98.26 + 4.30( 2.326) = 88.3 kpsi Ans. ______________________________________________________________________________ n 1-17  Ri Eq. (1-9): R = i 1 = 0.98(0.96)0.94 = 0.88 Overall reliability = 88 percent Ans. Shigley’s MED, 11th edition Chapter 1 Solutions, Page 5/12 ______________________________________________________________________________ 1-18 Obtain the coefficients of variance for strength and stress ˆ S 23.5 CS  sy   0.07532 S sy 312 C  ˆ ˆ 145  T   0.09667  T 1 500 For R = 0.99, from Table A-10, z =  2.326. Eq. (1-12): n  1  1   1  z 2CS2   1  z 2C2  1  z 2CS2 2 2 2 2 1  1   1    2.326   0.07532    1    2.326   0.09667        1.3229 1.32 2 2 1    2.326   0.07532  From the given equation for stress,  max  S sy n  Ans. 16T d3 Solving for d gives 1/ 3 1/ 3  16 T n   16(1500)1.3229   d    0.0319 m  31.9 mm Ans. 6   S    (312)10  sy   ______________________________________________________________________________ 1-19 Obtain the coefficients of variance for stress and strength ˆ ˆ 5 C    P   0.09231  P 65 CS  (a) 6.59 ˆ S ˆ S y    0.06901 S Sy 95.5 n 1.2 z  Eq. (1-11): Shigley’s MED, 11th edition nd  1 nd2CS2  C2  1.2  1 1.22  0.069012   0.092312   1.6127 Chapter 1 Solutions, Page 6/12 n  Sy   (b) Interpolating Table A-10, 1.61 0.0537 1.6127  1.62 0.0526  R = 1  0.0534 = 0.9466 Ans. Sy P / ( d / 4) 2   d 2S y 4P  d   = 0.0534 4  65  1.2 4 Pn   1.020 in Ans.  Sy   95.5  n 1.5 1.5  1 z  1.52  0.069012   0.092312 3.6 3.605 3.7 0.000159  0.000108 R = 1  0.00015645 = 0.9998   3.605   = 0.00015645 Ans. 4  65  1.5 4Pn   1.140 in Ans.  Sy   95.5  d  ______________________________________________________________________________ 1-20  max   max   a   b  90  383  473 MPa From footnote 9 of text, ˆ max   ˆ2a  ˆ2b  C max  CS y  n  1/ 2  (8.42  22.32 )1/ 2  23.83 MPa ˆ max ˆ 23.83   max   0.0504 473  max  max ˆ S y S y Sy  max Shigley’s MED, 11th edition   ˆ S y Sy  42.7  0.0772 553 553  1.169  1.17 Ans. 473 Chapter 1 Solutions, Page 7/12 z  Eq. (1-11): From Table A-10, nd  1 nd2 CS2  C2  1.169  1 1.1692  0.0772 2   0.0504 2   1.635 ( 1.635) = 0.05105 R = 1  0.05105 = 0.94895 = 94.9 percent Ans. ______________________________________________________________________________ 1-21 (a) a = 1.500  0.001 in b = 2.000  0.003 in c = 3.000  0.004 in d = 6.520  0.010 in w d  a  b  c = 6.520  1.5  2  3 = 0.020 in tw  tall = 0.001 + 0.003 + 0.004 +0.010 = 0.018 w = 0.020  0.018 in Ans. (b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore, d = 6.520 + 0.008 = 6.528 in Ans. ______________________________________________________________________________ 1-22 V = xyz, and x = a   a, y = b   b, z = c   c, V abc V  a a   b b   c c  abc bca acb abc abc bca cab abc The higher order terms in  are negligible. Thus, V bca  acb  abc and, V bca  acb  abc a b c a b c        Ans. V abc a b c a b c For the numerical values given, V 1.500  1.875 3.000 8.4375 in 3 V 0.002 0.003 0.004    0.004267 V 1.500 1.875 3.000 V = 8.4375  0.0360 in3 Shigley’s MED, 11th edition  V 0.004267  8.4375  0.0360 in 3 Ans. Chapter 1 Solutions, Page 8/12 8.4735 8.473551.. V  8.4015 in, whereas, exact is 8.401551.. in This answer yields ______________________________________________________________________________ V  1-23 wmax = 0.05 in, wmin = 0.004 in 0.05  0.004 w= 0.027 in 2 Thus,  w = 0.05  0.027 = 0.023 in, and then, w = 0.027  0.023 in. w=a  b  c 0.027 a  0.042  1.5 a 1.569 in tw = Thus, t all  0.023 = ta + 0.002 + 0.005 a = 1.569  0.016 in  ta = 0.016 in Ans. ______________________________________________________________________________ 1-24 Do Di  2d 3.734  2  0.139  4.012 in t Do  tall 0.028  2  0.004  0.036 in Do = 4.012  0.036 in Ans. ______________________________________________________________________________ 1-25 From O-Rings, Inc. (oringsusa.com), Di = 9.19  0.13 mm, d = 2.62  0.08 mm Do Di  2d 9.19  2  2.62  14.43 mm t Do  tall 0.13  2  0.08  0.29 mm Do = 14.43  0.29 mm Ans. ______________________________________________________________________________ 1-26 From O-Rings, Inc. (oringsusa.com), Di = 34.52  0.30 mm, d = 3.53  0.10 mm Do Di  2d 34.52  2  3.53 41.58 mm Shigley’s MED, 11th edition Chapter 1 Solutions, Page 9/12 t Do  tall 0.30  2  0.10  0.50 mm Do = 41.58  0.50 mm Ans. ______________________________________________________________________________ 1-27 From O-Rings, Inc. (oringsusa.com), Di = 5.237  0.035 in, d = 0.103  0.003 in Do Di  2d 5.237  2  0.103 5.443 in tDo  tall 0.035  2  0.003 0.041 in Do = 5.443  0.041 in Ans. ______________________________________________________________________________ 1-28 From O-Rings, Inc. (oringsusa.com), Di = 1.100  0.012 in, d = 0.210  0.005 in Do Di  2d 1.100  2  0.210  1.520 in t Do  tall 0.012  2  0.005  0.022 in Do = 1.520  0.022 in Ans. ______________________________________________________________________________ 1-29 From Table A-2, (a)  = 150/6.89 = 21.8 kpsi Ans. (b) F = 2 /4.45 = 0.449 kip = 449 lbf Ans. (c) M = 150/0.113 = 1330 lbf  in = 1.33 kip  in (d) A = 1500/ 25.42 = 2.33 in2 (e) I = 750/2.544 = 18.0 in4 (f) E = 145/6.89 = 21.0 Mpsi (g) v = 75/1.61 = 46.6 mi/h Ans. Ans. Ans. Ans. Ans. (h) V = 1000/946 = 1.06 qt Ans. ______________________________________________________________________________ 1-30 From Table A-2, Shigley’s MED, 11th edition Chapter 1 Solutions, Page 10/12 (a) l = 5(0.305) = 1.53 m Ans. (b)  = 90(6.89) = 620 MPa (c) p = 25(6.89) = 172 kPa (d) Z =12(16.4) = 197 cm3 Ans. Ans. Ans. (e) w = 0.208(175) = 36.4 N/m Ans. (f)  = 0.001 89(25.4) = 0.048 0 mm (g) v = 1 200(0.0051) = 6.12 m/s Ans. Ans. (h)  = 0.002 15(1) = 0.002 15 mm/mm (i) V = 1830(25.43) = 30.0 (106) mm3 Ans. Ans. ______________________________________________________________________________ 1-31 (a)  = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi (b)  = F /A = 9440/23.8 = 397 psi Ans. Ans. (c) y =Fl3/3EI = 270(31.5)3/[3(30)106(0.154)] = 0.609 in Ans. (d)  = Tl /GJ = 9 740(9.85)/[11.3(106)( /32)1.004] = 8.648(102) rad = 4.95 Ans. ______________________________________________________________________________ 1-32 (a)  =F / wt = 1000/[25(5)] = 8 MPa Ans. (b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4 Ans. (c) I = d4/64 =  (25.4)4/64 = 20.4(103) mm4 Ans. (d)  =16T / d 3 = 16(25)103/[ (12.7)3] = 62.2 MPa Ans. ______________________________________________________________________________ 1-33 (a)  =F /A = 2 700/[ (0.750)2/4] = 6110 psi = 6.11 kpsi Ans. (b)  = 32Fa/ d 3 = 32(180)31.5/[ (1.25)3] = 29 570 psi = 29.6 kpsi (c) Z = (do4  di4)/(32 do) =  (1.504  1.004)/[32(1.50)] = 0.266 in3 Shigley’s MED, 11th edition Ans. Ans. Chapter 1 Solutions, Page 11/12 (d) k = (d 4G)/(8D 3 N) = 0.062 54(11.3)106/[8(0.760)3 32] = 1.53 lbf/in Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 1 Solutions, Page 12/12 Chapter 8 ______________________________________________________________________________ 8-1 (a) Thread depth= 2.5 mm Ans. Width = 2.5 mm Ans. dm = 25 - 1.25 - 1.25 = 22.5 mm dr = 25 - 5 = 20 mm l = p = 5 mm Ans. (b) Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. dm = 22.5 mm dr = 20 mm l = p = 5 mm Ans. ______________________________________________________________________________ 8-2 From Table 8-1, d r  d  1.226 869 p d m  d  0.649 519 p d  1.226 869 p  d  0.649 519 p  d  0.938 194 p d  2 d 2  (d  0.938 194 p) 2 Ans. 4 4 ______________________________________________________________________________ At  8-3  From Eq. (c) of Sec. 8-2, tan   f PR  F 1  f tan  Pd Fd m tan   f TR  R m  2 2 1  f tan  1  f tan  T Fl / (2 ) 1  f tan  e 0   tan  TR Fd m / 2 tan   f tan   f Shigley’s MED, 11th edition Ans. Chap. 8 Solutions, Page 1/68 Using f = 0.08, form a table and plot the efficiency curve. e , deg. 0 0 0 0.678 20 0.796 30 0.838 40 0.8517 45 0.8519 ______________________________________________________________________________ 8-4 Given F = 5 kN, l = 5 mm, and dm = d  p/2 = 25  5/2 = 22.5 mm, the torque required to raise the load is found using Eqs. (8-1) and (8-6) TR  5  22.5   5    0.09  22.5  5  0.06  45  15.85 N  m    2 22.5 0.09 5 2        Ans. The torque required to lower the load, from Eqs. (8-2) and (8-6) is TL  5  22.5     0.09  22.5  5  5  0.06  45  7.83 N  m   2 22.5 0.09 5 2         Ans. Since TL is positive, the thread is self-locking. From Eq.(8-4) the efficiency is 5 5  0.251 Ans. 2 15.85  ______________________________________________________________________________ e 8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom segment of the screws must be in compression. Whereas, tension specimens and their grips must be in tension. Both screws must be of the same-hand threads. ______________________________________________________________________________ 8-6 Screws rotate at an angular rate of n 1720  28.67 rev/min 60 (a) The lead is 0.25 in, so the linear speed of the press head is Shigley’s MED, 11th edition Chap. 8 Solutions, Page 2/68 (b) F = 2500 lbf/screw V = 28.67(0.25) = 7.17 in/min Ans. d m  2  0.25 / 2  1.875 in sec   1 / cos(29o / 2)  1.033 Eq. (8-5): TR  2500(1.875)  0.25   (0.05)(1.875)(1.033)     221.0 lbf · in 2   (1.875)  0.05(0.25)(1.033)  Eq. (8-6): Tc  2500(0.08)(3.5 / 2)  350 lbf · in Ttotal  350  221.0  571 lbf · in/screw 571(2) Tmotor   20.04 lbf · in 60(0.95) Tn 20.04(1720) H    0.547 hp Ans. 63 025 63 025 ______________________________________________________________________________ 8-7 AISI 1006 CD steel. Table A-20: Sy = 41 kpsi. (a) The handle has maximum bending moment where it enters the screw body. M = (3.5  0.375) F = 3.125 F 32M 32(3.125) F Sy      41 000 3 d  (0.375)3 F = 67.9 lbf Ans. (b) Using Fig. 8-3 for the Acme thread, p = l = 1/6 = 0.1667 in d m  d  p / 2  0.75  1/12  0.6667 in , d r  d  p  0.75  1/ 6  0.5833 in  = 29/2 = 14.5, sec 14.5 = 1.033 From Eqs. (8-5) and (8-6), Fd  l   fd m sec   Ff c d c Ttotal  m   2   d m  fl sec   2 F  0.6667   0.1667    0.15  0.6667 1.033   F  0.15 1  0.1542 F   2 2    0.6667   0.15  0.1667 1.033   From part (a), Ttotal = 3.5 F = 3.5(67.9) = 237.7 lbf٠in 237.7 F  1542 lbf Ans. 0.1542 (c) Using Eqs. (8-11), (8-8), (8-7), and (8-12): Bending stress in first thread, with the force on the first thread being 0.38F and nt = 1, 6(0.38 F ) 6(0.38)(1542) x    11 510 psi  11.5 kpsi  d r nt p  (0.5833)(1)(0.1667)  Shigley’s MED, 11th edition Chap. 8 Solutions, Page 3/68 Axial stress in body of screw, 4F 4(1542) y     5770 psi  5.77 kpsi 2  dr  (0.5833) 2 Torsion in body of screw: 16T 16(237.7)  yz    6100 psi  6.10 kpsi 3  d r  (0.5833)3 The tangential shear stress given by Eq. (8-12) with one thread carrying 0.38 T: 4  0.38T  4  0.38  237.7  zx   2    2030 psi  2.03 kpsi 2  d r 1 p   0.5833 1 0.1667 From Eq. (5-14), 1/2 2 1 2 2 2 11.5   5.77     5.77  0    0  11.5   6  6.10   6  2.03   2  18.9 kpsi Sy 41   2.2 Ans. ny    18.9 (d) The column has one end fixed and the other end pivoted in the swivel joint of the anvil striker, so from Table 4-2, C = 1.2. We will use the root diameter of the screw body to check buckling, neglecting the effect of the threads. A = (0.58332)/4 = 0.267 in2, Sy = 41 kpsi, E = 30(106) psi, L = 8 in,   k I  d 4 / 64   d / 4  0.5833 / 4  0.1458 in , L/k = 8/0.1458 = 54.9 A d2 / 4 From Eq. (4-45),  2 2 1.2  30 106     131.7  41 000   Since 54.9 < 131.7, the J.B. Johnson formula is applicable. From Eq. (4-48), the critical clamping force for buckling is 2   Sy l  1  Pc r  A  S y       2 k  CE  2 3     41 10   1   (54.9)   0.267 41103     9995 lbf Ans 6 2 1.2 30 10          P 9995  6.5 Ans. n  cr  F 1542 It is confirmed that the weak link is the yielding of the handle. ______________________________________________________________________________ 8-8 T = 8(3.5) = 28 lbf  in 1/ 2 2  l   2 CE        k 1  S y  dm  1/2 3 1   0.6667 in 4 12 Shigley’s MED, 11th edition Chap. 8 Solutions, Page 4/68 l = 1 = 0.1667 in, 6  = 290 = 14.50, 2 sec 14.50 = 1.033 From Eqs. (8-5) and (8-6) Ttotal  0.6667 F 2  0.1667    0.15  0.6667 1.033  0.15 1 F  0.1542 F   0.6667 0.15 0.1667 1.033 2          28  182 lbf Ans. 0.1542 _____________________________________________________________________________ F 8-9 dm = 1.5  0.25/2 = 1.375 in, l = 2(0.25) = 0.5 in From Eq. (8-1) and Eq. (8-6), 2.2 103  (1.375)  0.5   (0.10)(1.375)  2.2 103  (0.15)(2.25) TR    (1.375)  0.10(0.5)   2 2    330  371  701 lbf · in Since n = V/l = 2/0.5 = 4 rev/s = 240 rev/min so the power is 701 240  Tn   2.67 hp Ans. 63 025 63 025 ______________________________________________________________________________ H 8-10 dm = 40  4 = 36 mm, l = p = 8 mm From Eqs. (8-1) and (8-6) T     H  T  F  36 F  8   (0.14)(36)  0.09(100) F  2   (36)  0.14(8)  2 (3.831  4.5) F  8.33F N · m (F in kN) 2 n  2 (1)  2 rad/s T 3000 H   477 N · m 2  477  57.3 kN Ans. 8.33 Shigley’s MED, 11th edition Chap. 8 Solutions, Page 5/68 Fl 57.3(8)   0.153 Ans. 2 T 2 (477) ______________________________________________________________________________ e 8-11 (a) Table A-31, nut height H = 12.8 mm. L ≥ l + H = 2(15) + 12.8 = 42.8 mm. Rounding up, L = 45 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm From Table 8-7, ld = L  LT = 45 34 = 11 mm, lt = l  ld = 2(15)  11 = 19 mm, Ad =  (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17) kb  153.9 115  207 Ad At E   874.6 MN/m Ad lt  At ld 153.9 19   115 11 Ans. (c) From Eq. (8-22), with l = 2(15) = 30 mm km  0.5774  207 14 0.5774 Ed   3 116.5 MN/m  0.5774l  0.5d   0.5774  30   0.5 14   2 ln  5  2 ln 5   0.5774l  2.5d   0.5774  30   2.5 14   Ans. 8-12 (a) Table A-31, nut height H = 12.8 mm. Table A-33, washer thickness t = 3.5 mm. Thus, the grip is l = 2(15) + 3.5 = 33.5 mm. L ≥ l + H = 33.5 + 12.8 = 46.3 mm. Rounding up L = 50 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm From Table 8-7, ld = L  LT = 50 34 = 16 mm, lt = l  ld = 33.5  16 = 17.5 mm, Ad =  (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17) kb  153.9 115  207 Ad At E   808.2 MN/m Ad lt  At ld 153.9 17.5   115 16  Ans. (c) Shigley’s MED, 11th edition Chap. 8 Solutions, Page 6/68 From Eq. (8-22) km  0.5774  207 14 0.5774 Ed   2 969 MN/m  0.5774l  0.5d   0.5774  33.5  0.5 14   2 ln  5  2 ln 5   0.5774l  2.5d   0.5774  33.5   2.5 14   Ans. ______________________________________________________________________________ 8-13 (a) Table 8-7, l = h + d /2 = 15 + 14/2 = 22 mm. L ≥ h + 1.5d = 36 mm. Rounding up L = 40 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm From Table 8-7, ld = L  LT = 40 34 = 6 mm, lt = l  ld = 22  6 = 16 mm Ad =  (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17) kb  153.9 115  207 Ad At E   1 162.2 MN/m Ad lt  At ld 153.9 16   115  6  Ans. (c) From Eq. (8-22), with l = 22 mm 0.5774  207 14 0.5774 Ed   3 624.4 MN/m Ans.  0.5774l  0.5d   0.5774  22   0.5 14   2 ln  5  2 ln 5   0.5774l  2.5d   0.5774  22   2.5 14   ______________________________________________________________________________ 8-14 (a) From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 2 + 1 + 7/16 = 3 7/16 in. Rounding up, L = 3.5 in Ans. km  (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L  LT = 3.5  1.25 = 2.25 in, lt = l  ld = 3  2.25 = 0.75 in Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) kb  (c) 0.1963  0.1419  30 Ad At E   1.79 Mlbf/in Ad lt  At ld 0.1963  0.75   0.1419  2.25  Shigley’s MED, 11th edition Ans. Chap. 8 Solutions, Page 7/68 Top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20) k1  0.5774  30  0.5 1.155 1.5   0.75  0.5  0.75  0.5  ln  1.155 1.5   0.75  0.5  0.75  0.5   22.65 Mlbf/in Lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.75 + 2(1) tan 30 = 1.905 in, E = 30 Mpsi. Eq. (8-20)  k2 = 210.7 Mlbf/in Cast iron: t = 1 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20)  k3 = 12.27 Mlbf/in From Eq. (8-18), km = (1/k1 + 1/k2 +1/k3)1 = (1/22.65 + 1/210.7 + 1/12.27)1 = 7.67 Mlbf/in Ans. 8-15 (a) From Table A-32, the washer thickness is 0.095 in. Thus, l = 2 + 1 + 2(0.095) = 3.19 in. From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 3.19 + 7/16 = 3.63 in. Rounding up, L = 3.75 in Ans. (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L  LT = 3.75  1.25 = 2.5 in, lt = l  ld = 3.19  2.5 = 0.69 in Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) kb  0.1963  0.1419  30 Ad At E   1.705 Mlbf/in Ad lt  At ld 0.1963  0.69   0.1419  2.5  Shigley’s MED, 11th edition Ans. Chap. 8 Solutions, Page 8/68 (c) Each steel washer frustum: t = 0.095 in, d = 0.531 in (Table A-32), D = 0.75 in, E = 30 Mpsi. From Eq. (8-20) k1  0.5774  30  0.531 1.155  0.095   0.75  0.531  0.75  0.531 ln  1.155  0.095   0.75  0.531  0.75  0.531  89.20 Mlbf/in Top plate, top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30 = 0.860 in, E = 30 Mpsi. Eq. (8-20)  k2 = 28.99 Mlbf/in Top plate, lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.860 + 2(1) tan 30 = 2.015 in, E = 30 Mpsi. Eq. (8-20)  k3 = 234.08 Mlbf/in Cast iron: t = 1 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30 = 0.860 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20)  k4 = 15.99 Mlbf/in From Eq. (8-18) km = (2/k1 + 1/k2 +1/k3+1/k4)1 = (2/89.20 + 1/28.99 + 1/234.08 + 1/15.99)1 = 8.08 Mlbf/in Ans. ______________________________________________________________________________ 8-16 (a) From Table 8-7, l = h + d /2 = 2 + 0.5/2 = 2.25 in. L ≥ h + 1.5 d = 2 + 1.5(0.5) = 2.75 in Ans. (b) From Table 8-7, LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in ld = L  LT = 2.75  1.25 = 1.5 in, lt = l  ld = 2.25  1.5 = 0.75 in Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) Shigley’s MED, 11th edition Chap. 8 Solutions, Page 9/68 kb  0.1963  0.1419  30 Ad At E   2.321 Mlbf/in Ad lt  At ld 0.1963  0.75   0.1419 1.5  Ans. (c) Top steel frustum: t = 1.125 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20) k1  0.5774  30  0.5 1.155 1.125   0.75  0.5   0.75  0.5  ln  1.155 1.125   0.75  0.5  0.75  0.5   24.48 Mlbf/in Lower steel frustum: t = 0.875 in, d = 0.5 in, D = 0.75 + 2(0.25) tan 30 = 1.039 in, E = 30 Mpsi. Eq. (8-20)  k2 = 49.36 Mlbf/in Cast iron: t = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20)  k3 = 23.49 Mlbf/in From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3)1 = (1/24.48 + 1/49.36 + 1/23.49)1 = 9.645 Mlbf/in Ans. ______________________________________________________________________________ 8-17 a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in. Rounding up, L = 4.75 in Ans. (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L  LT = 4.75  1.25 = 3.5 in, lt = l  ld = 4.19  3.5 = 0.69 in Shigley’s MED, 11th edition Chap. 8 Solutions, Page 10/68 Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) kb  0.1963  0.1419  30 Ad At E   1.322 Mlbf/in Ad lt  At ld 0.1963  0.69   0.1419  3.5  Ans. (c) Upper and lower halves are the same. For the upper half, Steel frustum: t = 0.095 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi. From Eq. (8-20) k1  0.5774  30  0.531 1.155  0.095   0.75  0.531  0.75  0.531 ln  1.155  0.095   0.75  0.531  0.75  0.531  89.20 Mlbf/in Aluminum: t = 2 in, d = 0.5 in, D =0.75 + 2(0.095) tan 30 = 0.860 in, and E = 10.3 Mpsi. Eq. (8-20)  k2 = 9.24 Mlbf/in For the top half, km = (1/k1 + 1/k2)1 = (1/89.20 + 1/9.24)1 = 8.373 Mlbf/in Since the bottom half is the same, the overall stiffness is given by km = (1/ km + 1/ km )1 = km /2 = 8.373/2 = 4.19 Mlbf/in Ans ______________________________________________________________________________ 8-18 (a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in. Rounding up, L = 4.75 in Ans. (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L  LT = 4.75  1.25 = 3.5 in, lt = l  ld = 4.19  3.5 = 0.69 in Shigley’s MED, 11th edition Chap. 8 Solutions, Page 11/68 Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) kb  0.1963  0.1419  30 Ad At E   1.322 Mlbf/in Ad lt  At ld 0.1963  0.69   0.1419  3.5  Ans. (c) Upper aluminum frustum: t = [4 + 2(0.095)] /2 = 2.095 in, d = 0.5 in, D = 0.75 in, and E = 10.3 Mpsi. From Eq. (8-20) 0.5774 10.3  0.5  7.23 Mlbf/in k1  1.155  2.095   0.75  0.5  0.75  0.5  ln 1.155  2.095   0.75  0.5  0.75  0.5  Lower aluminum frustum: t = 4  2.095 = 1.905 in, d = 0.5 in, D = 0.75 +4(0.095) tan 30 = 0.969 in, and E = 10.3 Mpsi. Eq. (8-20)  k2 = 11.34 Mlbf/in Steel washers frustum: t = 2(0.095) = 0.190 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi. Eq. (8-20)  k3 = 53.91 Mlbf/in From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3)1 = (1/7.23 + 1/11.34 + 1/53.91)1 = 4.08 Mlbf/in Ans. ______________________________________________________________________________ 8-19 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 50 + 8.4 = 58.4 mm. Rounding up, L = 60 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L  LT = 60  26 = 34 mm, lt = l  l = 50  34 = 16 mm. Ad =  (102) / 4 = 78.54 mm2. From Table 8-1, Shigley’s MED, 11th edition Chap. 8 Solutions, Page 12/68 At = 58 mm2. From Eq. (8-17) 78.54  58.0  207 Ad At E kb    292.1 MN/m Ad lt  At ld 78.54 16   58.0  34  Ans. (c) Upper and lower frustums are the same. For the upper half, Aluminum: t = 10 mm, d = 10 mm, D = 15 mm, and from Table 8-8, E = 71 GPa. From Eq. (8-20) 0.5774  7110 k1   1576 MN/m 1.155 10   15  10 15  10  ln 1.155 10   15  10  15  10  Steel: t = 15 mm, d = 10 mm, D = 15 + 2(10) tan 30 = 26.55 mm, and E = 207 GPa. From Eq. (8-20) k2  0.5774  207 10 1.155 15   26.55  10   26.55  10  ln  1.155 15   26.55  10   26.55  10   11 440 MN/m For the top half, km = (1/k1 + 1/k2)1 = (1/1576 + 1/11 440)1 = 1385 MN/m Since the bottom half is the same, the overall stiffness is given by km = (1/ km + 1/ km )1 = km /2 = 1385/2 = 692.5 MN/m Ans. 8-20 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 60 + 8.4 = 68.4 mm. Shigley’s MED, 11th edition Chap. 8 Solutions, Page 13/68 Rounding up, L = 70 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L  LT = 70  26 = 44 mm, lt = l  ld = 60  44 = 16 mm. Ad =  (102) / 4 = 78.54 mm2. From Table 8-1, At = 58 mm2. From Eq. (8-17) kb  78.54  58.0  207 Ad At E   247.6 MN/m Ad lt  At ld 78.54 16   58.0  44  Ans. (c) Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq. (8-20) 0.5774 10.3  71 k1   1576 MN/m 1.155  2.095   15  10  15  10  ln 1.155  2.095   15  10  15  10  Lower aluminum frustum: t = 20 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq. (8-20)  k2 = 1 201 MN/m Top steel frustum: t = 20 mm, d = 10 mm, D = 15 + 2(10) tan 30 = 26.55 mm, and E = 207 GPa. Eq. (8-20)  k3 = 9 781 MN/m Lower steel frustum: t = 10 mm, d = 10 mm, D = 15 + 2(20) tan 30 = 38.09 mm, and E = 207 GPa. Eq. (8-20)  k4 = 29 070 MN/m Shigley’s MED, 11th edition Chap. 8 Solutions, Page 14/68 From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3+1/k4)1 = (1/1 576 + 1/1 201 + 1/9 781 +1/29 070)1 = 623.5 MN/m Ans. ______________________________________________________________________________ 8-21 (a) From Table 8-7, l = h + d /2 = 10 + 30 + 10/2 = 45 mm. L ≥ h + 1.5 d = 10 + 30 + 1.5(10) = 55 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L  LT = 55  26 = 29 mm, lt = l  ld = 45  29 = 16 mm. Ad =  (102) / 4 = 78.54 mm2. From Table 8-1, At = 58 mm2. From Eq. (8-17) 78.54  58.0  207 Ad At E   320.9 MN/m kb  Ans. Ad lt  At ld 78.54 16   58.0  29  (c) Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq. (8-20) 0.5774 10.3  71  1576 MN/m k1  1.155  2.095   15  10  15  10  ln 1.155  2.095   15  10  15  10  Lower aluminum frustum: t = 5 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq. (8-20)  k2 = 2 300 MN/m Top steel frustum: t = 12.5 mm, d = 10 mm, D = 15 + 2(10) tan 30 = 26.55 mm, and E = 207 GPa. Eq. (8-20)  k3 = 12 759 MN/m Lower steel frustum: t = 17.5 mm, d = 10 mm, D = 15 + 2(5) tan 30 = 20.77 mm, and E = 207 GPa. Eq. (8-20)  k4 = 6 806 MN/m From Eq. (8-18) Shigley’s MED, 11th edition Chap. 8 Solutions, Page 15/68 km = (1/k1 + 1/k2 +1/k3+1/k4)1 = (1/1 576 + 1/2 300 + 1/12 759 +1/6 806)1 = 772.4 MN/m Ans. ______________________________________________________________________________ kb 8-22 Equation (f ), Sec. 8-7: C  kb  k m Ad At E Eq. (8-17): kb  Ad lt  At ld Eq. (8-22): 0.5774  207  d km   0.5774  40   0.5d  2 ln 5   0.5774  40   2.5d  See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are mm, mm 2, MN/m): d 10 12 14 16 20 24 30 d 10 12 14 16 20 24 30 l 40 40 40 40 40 40 40 At 58 84.3 115 157 245 353 561 ld 24 25 21 17 14 11 4 lt 16 15 19 23 26 29 36 Ad 78.53982 113.0973 153.938 201.0619 314.1593 452.3893 706.8583 kb 356.0129 518.8172 686.2578 895.9182 1373.719 1944.24 2964.343 H 8.4 10.8 12.8 14.8 18 21.5 25.6 L> 48.4 50.8 52.8 54.8 58 61.5 65.6 km 1751.566 2235.192 2761.721 3330.796 4595.515 6027.684 8487.533 L 50 55 55 55 60 65 70 LT 26 30 34 38 46 54 66 C 0.16892 0.188386 0.199032 0.211966 0.230133 0.243886 0.258852 Use a M14  2 bolt, with length 55 mm. Ans. _____________________________________________________________________________ kb 8-23 Equation (f ), Sec. 8-7: C  kb  k m Ad At E Eq. (8-17): kb  Ad lt  At ld Shigley’s MED, 11th edition Chap. 8 Solutions, Page 16/68 For upper frustum, Eq. (8-20), with D = 1.5 d and t = 1.5 in: k1  0.5774  30  d  1.155 1.5   0.5d   2.5d   ln     1.155 1.5   2.5d   0.5d    0.5774  30  d  1.733  0.5d   ln 5   1.733  2.5d   Lower steel frustum, with D = 1.5d + 2(1) tan 30 = 1.5d + 1.155, and t = 0.5 in: 0.5774  30  d k2   1.733  0.5d  2.5d  1.155   ln    1.733  2.5d  0.5d  1.155   For cast iron frustum, let E = 14. 5 Mpsi, and D = 1.5 d, and t = 1 in: 0.5774 14.5  d k3   1.155  0.5d   ln 5   1.155  2.5d   Overall, km = (1/k1 +1/k2 +1/k3)1 See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are in, in 2, Mlbf/in): Shigley’s MED, 11th edition Chap. 8 Solutions, Page 17/68 d 0.375 0.4375 0.5 0.5625 0.625 0.75 0.875 At 0.0775 0.1063 0.1419 0.182 0.226 0.334 0.462 d 0.375 0.4375 0.5 0.5625 0.625 0.75 0.875 ld 2.5 2.375 2.25 2.125 2.25 2 1.75 Ad H L> 0.110447 0.328125 3.328125 0.15033 0.375 3.375 0.19635 0.4375 3.4375 0.248505 0.484375 3.484375 0.306796 0.546875 3.546875 0.441786 0.640625 3.640625 0.60132 0.75 3.75 lt 0.5 0.625 0.75 0.875 0.75 1 1.25 L 3.5 3.5 3.5 3.5 3.75 3.75 3.75 LT 1 1.125 1.25 1.375 1.5 1.75 2 l 3 3 3 3 3 3 3 kb k1 k2 k3 km C 1.031389 15.94599 178.7801 8.461979 5.362481 0.161309 1.383882 19.21506 194.465 10.30557 6.484256 0.175884 1.791626 22.65332 210.6084 12.26874 7.668728 0.189383 2.245705 26.25931 227.2109 14.35052 8.915294 0.20121 2.816255 30.03179 244.2728 16.55009 10.22344 0.215976 3.988786 38.07191 279.7762 21.29991 13.02271 0.234476 5.341985 46.7663 317.1203 26.51374 16.06359 0.24956 Ans. Use a 169 12 UNC  3.5 in long bolt ______________________________________________________________________________ kb 8-24 Equation (f ), Sec. 8-7: C  kb  k m Eq. (8-17): kb  Shigley’s MED, 11th edition Ad At E Ad lt  At ld Chap. 8 Solutions, Page 18/68 Top frustum, Eq. (8-20), with E = 10.3Mpsi, D = 1.5 d, and t = l /2: 0.5774 10.3  d k1   1.155 l / 2  0.5d  ln 5   1.155 l / 2  2.5d  Middle frustum, with E = 10.3 Mpsi, D = 1.5d + 2(l  0.5) tan 30, and t = 0.5  l /2 0.5774 10.3  d k2  1.155  0.5  0.5l   0.5d  2  l  0.5  tan 30 0   2.5d  2  l  0.5  tan 30 0  ln 1.155  0.5  0.5l   2.5d  2  l  0.5  tan 300  0.5d  2  l  0.5  tan 300      Lower frustum, with E = 30Mpsi, D = 1.5 d, t = l  0.5 0.5774  30  d k3   1.155  l  0.5   0.5d   ln 5     1.155  l  0.5   2.5d   See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are in, in 2, Mlbf/in) Size 1 2 3 4 5 6 8 10 d 0.073 0.086 0.099 0.112 0.125 0.138 0.164 0.19 At 0.00263 0.0037 0.00487 0.00604 0.00796 0.00909 0.014 0.0175 Ad 0.004185 0.005809 0.007698 0.009852 0.012272 0.014957 0.021124 0.028353 L> 0.6095 0.629 0.6485 0.668 0.6875 0.707 0.746 0.785 L 0.75 0.75 0.75 0.75 0.75 0.75 0.75 1 LT 0.396 0.422 0.448 0.474 0.5 0.526 0.578 0.63 l 0.5365 0.543 0.5495 0.556 0.5625 0.569 0.582 0.595 ld 0.354 0.328 0.302 0.276 0.25 0.224 0.172 0.37 Size 1 2 3 4 5 6 8 10 d 0.073 0.086 0.099 0.112 0.125 0.138 0.164 0.19 lt 0.1825 0.215 0.2475 0.28 0.3125 0.345 0.41 0.225 kb 0.194841 0.261839 0.333134 0.403377 0.503097 0.566787 0.801537 1.15799 k1 1.084468 1.321595 1.570439 1.830494 2.101297 2.382414 2.974009 3.602349 k2 1.954599 2.449694 2.993366 3.587564 4.234381 4.936066 6.513824 8.342138 k3 7.09432 8.357692 9.621064 10.88444 12.14781 13.41118 15.93792 18.46467 km 0.635049 0.778497 0.930427 1.090613 1.258846 1.434931 1.809923 2.214214 C 0.23478 0.251687 0.263647 0.27 0.285535 0.28315 0.306931 0.343393 Use a 256 UNC  0.75 in long bolt. Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chap. 8 Solutions, Page 19/68 8-25 For half of joint, Eq. (8-20): t = 20 mm, d = 14 mm, D = 21 mm, and E = 207 GPa k1  0.5774  207 14 1.155  20   21  14   21  14  ln  1.155  20   21  14   21  14   5523 MN/m km = (1/k1 + 1/k1)1 = k1/2 = 5523/2 = 2762 MN/m Ans. From Eq. (8-22) with l = 40 mm km  0.5774  207 14  0.5774  40   0.5 14   2 ln 5   0.5774  40   2.5 14    2762 MN/m Ans. which agrees with the earlier calculation. For Eq. (8-23), from Table 8-8, A = 0.787 15, B = 0.628 73 km = 207(14)(0.78 715) exp [0.628 73(14)/40] = 2843 MN/m Ans. This is 2.9% higher than the earlier calculations. ______________________________________________________________________________ 8-26 (a) Grip, l = 10 in. Nut height, H = 41/64 in (Table A-31). L ≥ l + H = 10 + 41/64 = 10.641 in. Let L = 10.75 in. Table 8-7, LT = 2d + 0.5 = 2(0.75) + 0.5 = 2 in, ld = L  LT = 10.75  2 = 8.75 in, lt = l  ld = 10  8.75 = 1.25 in Ad = (0.752)/4 = 0.4418 in2, At = 0.373 in2 (Table 8-2) Eq. (8-17), 0.4418  0.373  30 Ad At E   1.296 Mlbf/in kb  Ans. Ad lt  At ld 0.4418 1.25   0.373  8.75  Eq. (4-4), 2 2 Am Em  / 4  1.125  0.75  30 km  Ans.   1.657 Mlbf/in l 10 Eq. (f), Sec. 8-7, Shigley’s MED, 11th edition C = kb/(kb + km) = 1.296/(1.296 + 1.657) = 0.439 Ans. Chap. 8 Solutions, Page 20/68 (b) Let: Nt = no. of turns, p = pitch of thread (in), N = no. of threads per in = 1/p. Then,  = b + m = Nt p = Nt / N (1) But, b = Fi / kb, and, m = Fi / km. Substituting these into Eq. (1) and solving for Fi gives Fi  kb k m N t kb  k m N  2 1.296 1.657 106 1/ 3 Ans.  15 150 lbf 1.296  1.657 16 ______________________________________________________________________________  8-27 Proof for the turn-of-nut equation is given in the solution of Prob. 8-26, Eq. (2), where Nt =  / 360. The relationship between the turn-of-nut method and the torque-wrench method is as follows.  k  km  Nt   b  Fi N  kb k m  T  KFd i (turn-of-nut) (torque-wrench) Eliminate Fi  k  km  NT  Nt   b Ans.   360  kb km  Kd ______________________________________________________________________________ 8-28 (a) From Ex. 8-4, Fi = 14.4 kip, kb = 5.21(106) lbf/in, km = 8.95(106) lbf/in Eq. (8-27): T = kFid = 0.2(14.4)(103)(5/8) = 1800 lbf · in Ans. From Prob. 8-27, Shigley’s MED, 11th edition Chap. 8 Solutions, Page 21/68  5.21  8.95   k  km  Nt   b (14.4)(103 )11  Fi N   6  5.218.95 10   kb km   0.0481 turns  17.3 Ans. Bolt group is (1.5) / (5/8) = 2.4 diameters. Answer is much lower than RB&W recommendations. ______________________________________________________________________________ 8-29 C = kb / (kb + km) = 3/(3+12) = 0.2, P = Ptotal/ N = 80/6 = 13.33 kips/bolt Table 8-2, At = 0.141 9 in2; Table 8-9, Sp = 120 kpsi; Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(0.141 9)(120) = 12.77 kips (a) From Eq. (8-28), the factor of safety for yielding is 120  0.141 9   1.10 CP  Fi 0.2 13.33  12.77 (b) From Eq. (8-29), the overload factor is np  nL  S p At  S p At  Fi CP  Ans. 120  0.141 9   12.77  1.60 0.2 13.33 Ans. (c) From Eq. (8-30), the joint separation factor of safety is Fi 12.77   1.20 Ans. P 1  C  13.33 1  0.2  ______________________________________________________________________________ n0  8-30 1/2  13 UNC Grade 8 bolt, K = 0.20 (a) Proof strength, Table 8-9, Sp = 120 kpsi Table 8-2, At = 0.141 9 in2 Maximum, Fi = Sp At = 120(0.141 9) = 17.0 kips Ans. (b) From Prob. 8-29, C = 0.2, P = 13.33 kips Joint separation, Eq. (8-30) with n0 = 1 Minimum Fi = P (1  C) = 13.33(1  0.2) = 10.66 kips Ans. (c) Fi = (17.0 + 10.66)/2 = 13.8 kips Eq. (8-27), T = KFi d = 0.2(13.8)103(0.5)/12 = 115 lbf  ft Ans. ______________________________________________________________________________ 8-31 (a) Table 8-1, At = 20.1 mm2. Table 8-11, Sp = 380 MPa. Eq. (8-31), Fi = 0.75 Fp = 0.75 At Sp = 0.75(20.1)380(103) = 5.73 kN kb 1 Eq. (f ), Sec. 8-7, C   0.278 kb  km 1  2.6 Eq. (8-28) with np = 1, Shigley’s MED, 11th edition Chap. 8 Solutions, Page 22/68 P 0.25S p At S p At  Fi  C C Ptotal = NP = 8(6.869) = 55.0 kN  0.25  20.1 380 10 3  Ans. 0.278  6.869 kN (b) Eq. (8-30) with n0 = 1, F 5.73 P i   7.94 kN 1  C 1  0.278 Ptotal = NP = 8(7.94) = 63.5 kN Ans. Bolt stress would exceed proof strength ______________________________________________________________________________ 8-32 (a) Table 8-2, At = 0.141 9 in2. Table 8-9, Sp = 120 kpsi. Eq. (8-31), Fi = 0.75 Fp = 0.75 At Sp = 0.75(0.141 9)120 = 12.77 kips Eq. (f ), Sec. 8-7, C Eq. (8-28) with np = 1,  S A  Fi Ptotal  N  p t C  kb 4   0.25 kb  k m 4  12  0.25 NS p At  C  80  0.25  PtotalC   4.70 0.25S p At 0.25 120  0.141 9 Round to N = 5 bolts Ans. N (b) Eq. (8-30) with n0 = 1,  F  Ptotal  N  i   1 C  P 1  C  80 1  0.25    4.70 N  total 12.77 Fi Round to N = 5 bolts Ans. ______________________________________________________________________________ 8-33 Bolts: From Table A-31, the nut height is H = 10.8 mm. L ≥ l +H = 40 + 10.8 = 50.8 mm. Round up to L = 55 mm Ans. Eq. (8-14): LT = 2d + 6 = 2(12) + 6 = 30 mm Table 8-7: ld = L  LT = 55  30 = 25 mm, lt = l ld = 40  25 = 15 mm Ad = (122)/4 = 113.1 mm2, Table 8-1: At = 84.3 mm2 Shigley’s MED, 11th edition Chap. 8 Solutions, Page 23/68 Eq. (8-17): kb  113.1 84.3  207 Ad At E   518.8 MN/m Ad lt  At ld 113.1 15   84.3  25  Members: Steel cyl. head: t = 20 mm, d = 12 mm, D = 18 mm, E = 207 GPa. Eq. (8-20), k1  0.5774  207 12 1.155  20   18  12  18  12  ln  1.155  20   18  12  18  12   4470 MN/m Cast iron: t = 20 mm, d = 12 mm, D = 18 mm, E = 100 GPa (from Table 8-8). The only difference from k1 is the material k2 = (100/207)(4470) = 2159 MN/m Eq. (8-18): km = (1/4470 + 1/2159)1 = 1456 MN/m C = kb / (kb + km) = 518.8/(518.8+1456) = 0.263 Table 8-11: Sp = 650 MPa For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(84.3)(650)103 = 41.1 kN The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 100 mm. The external load per bolt is P = Ptotal /N. Thus P = [6 (1002)/4](103)/10 = 4.712 kN/bolt Yielding factor of safety, Eq. (8-28): np  S p At CP  Fi  650  84.310 3  1.29 0.263  4.712   41.10 Ans. Overload factor of safety, Eq. (8-29): nL  S p At  Fi CP  650  84.3103  41.10  11.1 0.263  4.712  Ans. Separation factor of safety, Eq. (8-30): Shigley’s MED, 11th edition Chap. 8 Solutions, Page 24/68 Fi 41.10   11.8 Ans. P 1  C  4.712 1  0.263 ______________________________________________________________________________ n0  8-34 Bolts: Grip, l = 1/2 + 5/8 = 1.125 in. From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 1.125 + 7/16 = 1.563 in. Round up to L = 1.75 in Ans. Eq. (8-13): LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in Table 8-7: ld = L  LT = 1.75  1.25 = 0.5 in, lt = l ld = 1.125  0.5 = 0.625 in Ad =  (0.52)/4 = 0.196 3 in2, Table 8-2: At = 0.141 9 in2 Eq. (8-17): 0.196 3  0.141 9  30 Ad At E kb    4.316 Mlbf/in Ad lt  At ld 0.196 3  0.625   0.141 9  0.5  Members: Steel cyl. head: t = 0.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. Eq. (8-20), k1  0.5774  30  0.5 1.155  0.5   0.75  0.5  0.75  0.5  ln  1.155  0.5   0.75  0.5  0.75  0.5   33.30 Mlbf/in Cast iron: Has two frusta. Midpoint of complete joint is at (1/2 + 5/8)/2 = 0.5625 in. Upper frustum, t = 0.5625 0.5 = 0.0625 in, d = 0.5 in, D = 0.75 + 2(0.5) tan 30 = 1.327 in, E = 14.5 Mpsi (from Table 8-8) Eq. (8-20)  k2 = 292.7 Mlbf/in Lower frustum, t = 0.5625 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi Eq. (8-20)  k3 = 15.26 Mlbf/in Eq. (8-18): km = (1/33.30 + 1/292.7 + 1/15.26)1 = 10.10 Mlbf/in C = kb / (kb + km) = 4.316/(4.316+10.10) = 0.299 Table 8-9: Sp = 85 kpsi For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.141 9)(85) = 9.05 kips The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 3.5 in. The external load per bolt is P = Ptotal /N. Thus, P = [1 500 (3.52)/4](103)/10 = 1.443 kips/bolt Yielding factor of safety, Eq. (8-28): S p At 85  0.141 9    1.27 np  CP  Fi 0.299 1.443  9.05 Overload factor of safety, Eq. (8-29): Shigley’s MED, 11th edition Ans. Chap. 8 Solutions, Page 25/68 nL  S p At  Fi  CP Separation factor of safety, Eq. (8-30): 85  0.141 9   9.05 0.299 1.443  6.98 Ans. Fi 9.05   8.95 Ans. P 1  C  1.443 1  0.299  ______________________________________________________________________________ 8-35 Bolts: Grip: l = 20 + 25 = 45 mm. From Table A-31, the nut height is H = 8.4 mm. L ≥ l +H = 45 + 8.4 = 53.4 mm. Round up to L = 55 mm Ans. Eq. (8-14): LT = 2d + 6 = 2(10) + 6 = 26 mm Table 8-7: ld = L  LT = 55  26 = 29 mm, lt = l ld = 45  29 = 16 mm Ad = (102)/4 = 78.5 mm2, Table 8-1: At = 58.0 mm2 Eq. (8-17): 78.5  58.0  207 Ad At E kb    320.8 MN/m Ad lt  At ld 78.5 16   58.0  29  n0  Members: Steel cyl. head: t = 20 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20), 0.5774  207 10 k1   3503 MN/m 1.155  20   15  10  15  10  ln 1.155  20   15  10  15  10  Cast iron: Has two frusta. Midpoint of complete joint is at (20 + 25)/2 = 22.5 mm Upper frustum, t = 22.5  20 = 2.5 mm, d = 10 mm, D = 15 + 2(20) tan 30 = 38.09 mm, E = 100 GPa (from Table 8-8), Eq. (8-20)  k2 = 45 880 MN/m Lower frustum, t = 22.5 mm, d = 10 mm, D = 15 mm, E = 100 GPa Eq. (8-20)  k3 = 1632 MN/m Eq. (8-18): km = (1/3503 + 1/45 880 + 1/1632)1 = 1087 MN/m C = kb / (kb + km) = 320.8/(320.8+1087) = 0.228 Table 8-11: Sp = 830 MPa For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(58.0)(830)103 = 36.1 kN The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 0.8 m. The external load per bolt is P = Ptotal /N. Thus, P = [550 (0.82)/4]/36 = 7.679 kN/bolt Shigley’s MED, 11th edition Chap. 8 Solutions, Page 26/68 Yielding factor of safety, Eq. (8-28): 830  58.0 103   1.27 np  CP  Fi 0.228  7.679   36.1 S p At Ans. Overload factor of safety, Eq. (8-29): S p At  Fi 830  58.0 103  36.1 nL    6.88 Ans. CP 0.228  7.679  Separation factor of safety, Eq. (8-30): Fi 36.1   6.09 n0  Ans. P 1  C  7.679 1  0.228  ______________________________________________________________________________ 8-36 Bolts: Grip, l = 3/8 + 1/2 = 0.875 in. From Table A-31, the nut height is H = 3/8 in. L ≥ l + H = 0.875 + 3/8 = 1.25 in. Let L = 1.25 in Ans. Eq. (8-13): LT = 2d + 0.25 = 2(7/16) + 0.25 = 1.125 in Table 8-7: ld = L  LT = 1.25  1.125 = 0.125 in, lt = l ld = 0.875  0.125 = 0.75 in Ad =  (7/16)2/4 = 0.150 3 in2, Table 8-2: At = 0.106 3 in2 Eq. (8-17), 0.150 3  0.106 3  30 Ad At E kb    3.804 Mlbf/in Ad lt  At ld 0.150 3  0.75   0.106 3  0.125  Members: Steel cyl. head: t = 0.375 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi. Eq. (8-20), 0.5774  30  0.4375  31.40 Mlbf/in k1  1.155  0.375   0.65625  0.4375   0.65625  0.4375  ln 1.155  0.375   0.65625  0.4375   0.65625  0.4375  Cast iron: Has two frusta. Midpoint of complete joint is at (3/8 + 1/2)/2 = 0.4375 in. Upper frustum, t = 0.4375 0.375 = 0.0625 in, d = 0.4375 in, D = 0.65625 + 2(0.375) tan 30 = 1.089 in, E = 14.5 Mpsi (from Table 8-8) Eq. (8-20)  k2 = 195.5 Mlbf/in Lower frustum, t = 0.4375 in, d = 0.4375 in, D = 0.65625 in, E = 14.5 Mpsi Eq. (8-20)  k3 = 14.08 Mlbf/in Eq. (8-18): km = (1/31.40 + 1/195.5 + 1/14.08)1 = 9.261 Mlbf/in C = kb / (kb + km) = 3.804/(3.804 + 9.261) = 0.291 Table 8-9: Sp = 120 kpsi Shigley’s MED, 11th edition Chap. 8 Solutions, Page 27/68 For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.106 3)(120) = 9.57 kips The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 3.25 in. The external load per bolt is P = Ptotal /N. Thus, P = [1 200 (3.252)/4](103)/8 = 1.244 kips/bolt Yielding factor of safety, Eq. (8-28): S p At 120  0.106 3   1.28 np  Ans. CP  Fi 0.2911.244   9.57 Overload factor of safety, Eq. (8-29): S p At  Fi 120  0.106 3  9.57 nL    8.80 Ans. CP 0.2911.244  Separation factor of safety, Eq. (8-30): Fi 9.57   10.9 n0  Ans. P 1  C  1.244 1  0.291 ______________________________________________________________________________ 8-37 From Table 8-7, h = t1 = 20 mm For t2 > d, l = h + d /2 = 20 + 12/2 = 26 mm L ≥ h + 1.5 d = 20 + 1.5(12) = 38 mm. Round up to L = 40 mm LT = 2d + 6 = 2(12) + 6 = 30 mm ld = L  LT = 40  20 = 10 mm lt = l  ld = 26  10 = 16 mm From Table 8-1, At = 84.3 mm2. Ad =  (122)/4 = 113.1 mm2 Eq. (8-17), 113.1 84.3 207 Ad At E kb    744.0 MN/m Ad lt  At ld 113.1 16   84.3 10  Similar to Fig. 8-23, we have three frusta. Top frusta, steel: t = l / 2 = 13 mm, d = 12 mm, D = 18 mm, E = 207 GPa. Eq. (8-20) k1  0.5774  207 12 1.155 13  18  12  18  12  ln  1.155 13  18  12 18  12   5 316 MN/m Middle frusta, steel: t = 20  13 = 7 mm, d = 12 mm, D = 18 + 2(13  7) tan 30 = 24.93 mm, E = 207 GPa. Eq. (8-20)  k2 = 15 660 MN/m Lower frusta, cast iron: t = 26  20 = 6 mm, d = 12 mm, D = 18 mm, E = 100 GPa (see Table 8-8). Eq. (8-20)  k3 = 3 887 MN/m Eq. (8-18), km = (1/5 316 + 1/15 660 + 1/3 887)1 = 1 964 MN/m Shigley’s MED, 11th edition Chap. 8 Solutions, Page 28/68 C = kb / (kb + km) = 744.0/(744.0 + 1 964) = 0.275 Table 8-11: Sp = 650 MPa. For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(84.3)(650)103 = 41.1 kN The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 100 mm. The external load per bolt is P = Ptotal /N. Thus P = [6 (1002)/4](103)/10 = 4.712 kN/bolt Yielding factor of safety, Eq. (8-28) S p At 650  84.310 3   1.29 np  CP  Fi 0.275  4.712   41.1 Overload factor of safety, Eq. (8-29) S p At  Fi 650  84.310 3  41.1 nL    10.7 CP 0.275  4.712  Ans. Ans. Separation factor of safety, Eq. (8-30) Fi 41.1   12.0 n0  Ans. P 1  C  4.712 1  0.275 ______________________________________________________________________________ 8-38 From Table 8-7, h = t1 = 0.5 in For t2 > d, l = h + d /2 = 0.5 + 0.5/2 = 0.75 in L ≥ h + 1.5 d = 0.5 + 1.5(0.5) = 1.25 in. Let L = 1.25 in LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in. All threaded. From Table 8-1, At = 0.141 9 in2. The bolt stiffness is kb = At E / l = 0.141 9(30)/0.75 = 5.676 Mlbf/in Similar to Fig. 8-23, we have three frusta. Top frusta, steel: t = l / 2 = 0.375 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi 0.5774  30  0.5 k1   38.45 Mlbf/in 1.155  0.375   0.75  0.5  0.75  0.5  ln 1.155  0.375   0.75  0.5  0.75  0.5  Middle frusta, steel: t = 0.5  0.375 = 0.125 in, d = 0.5 in, D = 0.75 + 2(0.75  0.5) tan 30 = 1.039 in, E = 30 Mpsi. Eq. (8-20)  k2 = 184.3 Mlbf/in Lower frusta, cast iron: t = 0.75  0.5 = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi. Eq. (8-20)  k3 = 23.49 Mlbf/in Eq. (8-18), km = (1/38.45 + 1/184.3 + 1/23.49)1 = 13.51 Mlbf/in C = kb / (kb + km) = 5.676 / (5.676 + 13.51) = 0.296 Table 8-9, Sp = 85 kpsi. For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.141 9)(85) = 9.05 kips Shigley’s MED, 11th edition Chap. 8 Solutions, Page 29/68 The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 3.5 in. The external load per bolt is P = Ptotal /N. Thus P = [1 500 (3.52)/4](103)/10 = 1.443 kips/bolt Yielding factor of safety, Eq. (8-28) S p At 85  0.141 9    1.27 np  Ans. CP  Fi 0.296 1.443  9.05 Overload factor of safety, Eq. (8-29) S p At  Fi 85  0.141 9   9.05 nL    7.05 CP 0.296 1.443 Ans. Separation factor of safety, Eq. (8-30) Fi 9.05   8.91 n0  Ans. P 1  C  1.443 1  0.296  ______________________________________________________________________________ 8-39 From Table 8-7, h = t1 = 20 mm For t2 > d, l = h + d /2 = 20 + 10/2 = 25 mm L ≥ h + 1.5 d = 20 + 1.5(10) = 35 mm. Let L = 35 mm LT = 2d + 6 = 2(10) + 6 = 26 mm ld = L  LT = 35  26 = 9 mm lt = l  ld = 25  9 = 16 mm From Table 8-1, At = 58.0 mm2. Ad =  (102)/4 = 78.5 mm2 Eq. (8-17), 78.5  58.0  207 Ad At E kb    530.1 MN/m Ad lt  At ld 78.5 16   58.0  9  Similar to Fig. 8-23, we have three frusta. Top frusta, steel: t = l / 2 = 12.5 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20) 0.5774  207 10 k1   4 163 MN/m 1.155 12.5   15  10  15  10  ln 1.155 12.5   15  10  15  10  Middle frusta, steel: t = 20  12.5 = 7.5 mm, d = 10 mm, D = 15 + 2(12.5  7.5) tan 30 = 20.77 mm, E = 207 GPa. Eq. (8-20)  k2 = 10 975 MN/m Lower frusta, cast iron: t = 25  20 = 5 mm, d = 10 mm, D = 15 mm, E = 100 GPa (see Table 8-8). Eq. (8-20)  k3 = 3 239 MN/m Eq. (8-18), km = (1/4 163 + 1/10 975 + 1/3 239)1 = 1 562 MN/m C = kb / (kb + km) = 530.1/(530.1 + 1 562) = 0.253 Table 8-11: Sp = 830 MPa. For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(58.0)(830)103 = 36.1 kN The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 0.8 m. The external load per bolt is P = Ptotal /N. Thus Shigley’s MED, 11th edition Chap. 8 Solutions, Page 30/68 P = [550 (0.82)/4]/36 = 7.679 kN/bolt Yielding factor of safety, Eq. (8-28) S p At 830  58.0 10 3   1.27 np  Ans. CP  Fi 0.253  7.679   36.1 Overload factor of safety, Eq. (8-29) S p At  Fi 830  58.0 103  36.1   6.20 nL  CP 0.253  7.679  Ans. Separation factor of safety, Eq. (8-30) Fi 36.1   6.29 n0  Ans. P 1  C  7.679 1  0.253 ______________________________________________________________________________ 8-40 From Table 8-7, h = t1 = 0.375 in For t2 > d, l = h + d /2 = 0.375 + 0.4375/2 = 0.59375 in L ≥ h + 1.5 d = 0.375 + 1.5(0.4375) = 1.031 in. Round up to L = 1.25 in LT = 2d + 0.25 = 2(0.4375) + 0.25 = 1.125 in ld = L  LT = 1.25  1.125 = 0.125 lt = l  ld = 0.59375  0.125 = 0.46875 in Ad =  (7/16)2/4 = 0.150 3 in2, Table 8-2: At = 0.106 3 in2 Eq. (8-17), 0.150 3  0.106 3  30 Ad At E kb    5.724 Mlbf/in Ad lt  At ld 0.150 3  0.46875   0.106 3  0.125  Similar to Fig. 8-23, we have three frusta. Top frusta, steel: t = l / 2 = 0.296875 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi 0.5774  30  0.4375  35.52 Mlbf/in k1  1.155  0.296875   0.656255  0.4375   0.75  0.656255  ln 1.155  0.296875   0.75  0.656255   0.75  0.656255  Middle frusta, steel: t = 0.375  0.296875 = 0.078125 in, d = 0.4375 in, D = 0.65625 + 2(0.59375  0.375) tan 30 = 0.9088 in, E = 30 Mpsi. Eq. (8-20)  k2 = 215.8 Mlbf/in Lower frusta, cast iron: t = 0.59375  0.375 = 0.21875 in, d = 0.4375 in, D = 0.65625 in, E = 14.5 Mpsi. Eq. (8-20)  k3 = 20.55 Mlbf/in Eq. (8-18), km = (1/35.52 + 1/215.8 + 1/20.55)1 = 12.28 Mlbf/in C = kb / (kb + km) = 5.724/(5.724 + 12.28) = 0.318 Table 8-9, Sp = 120 kpsi. For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.106 3)(120) = 9.57 kips Shigley’s MED, 11th edition Chap. 8 Solutions, Page 31/68 The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 3.25 in. The external load per bolt is P = Ptotal /N. Thus P = [1 200 (3.252)/4](103)/8 = 1.244 kips/bolt Yielding factor of safety, Eq. (8-28) S p At 120  0.106 3   1.28 np  CP  Fi 0.318 1.244   9.57 Overload factor of safety, Eq. (8-29) S p At  Fi 120  0.106 3  9.57 nL    8.05 CP 0.318 1.244  Ans. Ans. Separation factor of safety, Eq. (8-30) Fi 9.57   11.3 n0  Ans. P 1  C  1.244 1  0.318  ______________________________________________________________________________ 8-41 This is a design problem and there is no closed-form solution path or a unique solution. What is presented here is one possible iterative approach. We will demonstrate this with an example. 1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on members, and combining using Eq. (8-18), yields km = 1 141 MN/m (see Prob. 8-33 for method of calculation. 2. Look up the nut height in Table A-31. For the example, H = 8.4 mm. From this, L is rounded up from the calculation of l + H = 40 + 8.4 = 48.4 mm to 50 mm. Next, calculations are made for LT = 2(10) + 6 = 26 mm, ld = 50  26 = 24 mm, lt = 40  24 = 16 mm. From step 1, Ad = (102)/4 = 78.54 mm2. Next, from Table 8-1, At = 78.54 mm2. From Eq. (8-17), kb = 356 MN/m. Finally, from Eq. (f), Sec. 8-7, C = 0.238. 3. From Prob. 8-33, the bolt circle diameter is E = 200 mm. Substituting this for Db in Eq. (8-34), the number of bolts are N  Db    200  4d 4 10  Rounding this up gives N = 16.  15.7 4. Next, select a grade bolt. Based on the solution to Prob. 8-33, the strength of ISO 9.8 was so high to give very large factors of safety for overload and separation. Try ISO 4.6 with Sp = 225 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi = 9.79 kN. Shigley’s MED, 11th edition Chap. 8 Solutions, Page 32/68 5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-33, pg = 6 MPa, and Ag =  (1002)/4. This gives P = 3.39 kN/bolt. 6. Using Eqs. (8-28) to (8-30) yield np = 1.23, nL = 4.05, and n0 = 3.79. Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables used from the text. The results for four bolt sizes are shown below. The dimension of each term is consistent with the example given above. d 8 10 12 14 km 854 1141 1456 1950 H 6.8 8.4 10.8 12.8 L 50 50 55 55 LT 22 26 30 34 ld 28 24 25 21 lt 12 16 15 19 d 8 10 12 14 C 0.215 0.238 0.263 0.276 N 20 16 13* 12 Sp 225 225 225 225 Fi 6.18 9.79 14.23 19.41 P 2.71 3.39 4.17 4.52 np 1.22 1.23 1.24 1.25 Ad At kb 50.26 36.6 233.9 78.54 58 356 113.1 84.3 518.8 153.9 115 686.3 nL 3.53 4.05 4.33 5.19 n0 2.90 3.79 4.63 5.94 *Rounded down from13.08997, so spacing is slightly greater than four diameters. Any one of the solutions is acceptable. A decision-maker might be cost such as N  cost/bolt, and/or N  cost per hole, etc. ________________________________________________________________________ Shigley’s MED, 11th edition Chap. 8 Solutions, Page 33/68 8-42 This is a design problem and there is no closed-form solution path or a unique solution. What is presented here is one possible iterative approach. We will demonstrate this with an example. 1. Select the diameter, d. For this example, let d = 0.5 in. Using Eq. (8-20) on three frusta (see Prob. 8-34 solution), and combining using Eq. (8-19), yields km = 10.10 Mlbf/in. 2. Look up the nut height in Table A-31. For the example, H = 0.4375 in. From this, L is rounded up from the calculation of l + H = 1.125 + 0.4375 = 1.5625 in to 1.75 in. Next, calculations are made for LT = 2(0.5) + 0.25 = 1.25 in, ld = 1.75  1.25 = 0.5 in, lt = 1.125  0.5 = 0.625 in. From step 1, Ad = (0.52)/4 = 0.1963 in2. Next, from Table 8-1, At = 0.141 9 in2. From Eq. (8-17), kb = 4.316 Mlbf/in. Finally, from Eq. (f), Sec. 8-7, C = 0.299. 3. From Prob. 8-34, the bolt circle diameter is E = 6 in. Substituting this for Db in Eq. (834), for the number of bolts N  Db    6 4d 4  0.5  Rounding this up gives N = 10.  9.425 4. Next, select a grade bolt. Based on the solution to Prob. 8-34, the strength of SAE grade 5 was adequate. Use this with Sp = 85 kpsi. From Eqs. (8-31) and (8-32) for a nonpermanent connection, Fi = 9.046 kips. 5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-34, pg = 1 500 psi, and Ag =  (3.52)/4 . This gives P = 1.660 kips/bolt. 6. Using Eqs. (8-28) to (8-30) yield np = 1.26, nL = 6.07, and n0 = 7.78. d 0.375 0.4375 0.5 0.5625 km 6.75 9.17 10.10 11.98 H 0.3281 0.375 0.4375 0.4844 d 0.375 0.4375 0.5 0.5625 C 0.261 0.273 0.299 0.308 N 13 11 10 9 L LT ld 1.5 1 0.5 1.5 1.125 0.375 1.75 1.25 0.5 1.75 1.375 0.375 Sp 85 85 85 85 Fi 4.941 6.777 9.046 11.6 P 1.277 1.509 1.660 1.844 lt 0.625 0.75 0.625 0.75 np 1.25 1.26 1.26 1.27 Ad At 0.1104 0.0775 0.1503 0.1063 0.1963 0.1419 0.2485 0.182 nL 4.95 5.48 6.07 6.81 kb 2.383 3.141 4.316 5.329 n0 5.24 6.18 7.78 9.09 Any one of the solutions is acceptable. A decision-maker might be cost such as N  cost/bolt, and/or N  cost per hole, etc. ________________________________________________________________________ Shigley’s MED, 11th edition Chap. 8 Solutions, Page 34/68 8-43 This is a design problem and there is no closed-form solution path or a unique solution. What is presented here is one possible iterative approach. We will demonstrate this with an example. 1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on three frusta (see Prob. 8-35 solution), and combining using Eq. (8-19), yields km = 1 087 MN/m. 2. Look up the nut height in Table A-31. For the example, H = 8.4 mm. From this, L is rounded up from the calculation of l + H = 45 + 8.4 = 53.4 mm to 55 mm. Next, calculations are made for LT = 2(10) + 6 = 26 mm, ld = 55  26 = 29 mm, lt = 45  29 = 16 mm. From step 1, Ad = (102)/4 = 78.54 mm2. Next, from Table 8-1, At = 58.0 mm2. From Eq. (8-17), kb = 320.9 MN/m. Finally, from Eq. (f), Sec. 8-7, C = 0.228. 3. From Prob. 8-35, the bolt circle diameter is E = 1000 mm. Substituting this for Db in Eq. (8-34), for the number of bolts  Db  1000   78.5 4d 4 10  Rounding this up gives N = 79. A rather large number, since the bolt circle diameter, E is so large. Try larger bolts. N  4. Next, select a grade bolt. Based on the solution to Prob. 8-35, the strength of ISO 9.8 was so high to give very large factors of safety for overload and separation. Try ISO 5.8 with Sp = 380 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi = 16.53 kN. 5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-35, pg = 0.550 MPa, and Ag =  (8002)/4 . This gives P = 4.024 kN/bolt. 6. Using Eqs. (8-28) to (8-30) yield np = 1.26, nL = 6.01, and n0 = 5.32. Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables used from the text. The results for three bolt sizes are shown below. The dimension of each term is consistent with the example given above. d 10 20 36 km 1087 3055 6725 H 8.4 18 31 L 55 65 80 d 10 20 36 C 0.228 0.308 0.361 N 79 40 22 Sp 380 380 380 Shigley’s MED, 11th edition LT 26 46 78 ld 29 19 2 lt 16 26 43 Fi P np 16.53 4.024 1.26 69.83 7.948 1.29 232.8 14.45 1.3 Ad 78.54 314.2 1018 At 58 245 817 nL 6.01 9.5 14.9 n0 5.32 12.7 25.2 kb 320.9 1242 3791 Chap. 8 Solutions, Page 35/68 A large range is presented here. Any one of the solutions is acceptable. A decision-maker might be cost such as N  cost/bolt, and/or N  cost per hole, etc. ________________________________________________________________________ 8-44 This is a design problem and there is no closed-form solution path or a unique solution. What is presented here is one possible iterative approach. We will demonstrate this with an example. 1. Select the diameter, d. For this example, let d = 0.375 in. Using Eq. (8-20) on three frusta (see Prob. 8-36 solution), and combining using Eq. (8-19), yields km = 7.42 Mlbf/in. 2. Look up the nut height in Table A-31. For the example, H = 0.3281 in. From this, L ≥ l + H = 0.875 + 0.3281 = 1.2031 in. Rounding up, L = 1.25. Next, calculations are made for LT = 2(0.375) + 0.25 = 1 in, ld = 1.25  1 = 0.25 in, lt = 0.875  0.25 = 0.625 in. From step 1, Ad = (0.3752)/4 = 0.1104 in2. Next, from Table 8-1, At = 0.0775 in2. From Eq. (8-17), kb = 2.905 Mlbf/in. Finally, from Eq. (f), Sec. 8-7, C = 0.263. 3. From Prob. 8-36, the bolt circle diameter is E = 6 in. Substituting this for Db in Eq. (834), for the number of bolts N  Db   6 4d 4  0.375  Rounding this up gives N = 13.  12.6 4. Next, select a grade bolt. Based on the solution to Prob. 8-36, the strength of SAE grade 8 seemed high for overload and separation. Try SAE grade 5 with Sp = 85 kpsi. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi = 4.941 kips. 5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-36, pg = 1 200 psi, and Ag =  (3.252)/4. This gives P = 0.881 kips/bolt. 6. Using Eqs. (8-28) to (8-30) yield np = 1.27, nL = 6.65, and n0 = 7.81. Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables used from the text. For this solution we only looked at one bolt size, 83 16 , but evaluated changing the bolt grade. The results for four bolt grades are shown below. The dimension of each term is consistent with the example given above. d km H L 0.375 7.42 0.3281 1.25 Shigley’s MED, 11th edition LT 1 ld lt Ad At kb 0.25 0.625 0.1104 0.0775 2.905 Chap. 8 Solutions, Page 36/68 d 0.375 0.375 0.375 0.375 C 0.281 0.281 0.281 0.281 SAE grade 1 2 4 5 N 13 13 13 13 Sp 33 55 65 85 Fi 1.918 3.197 3.778 4.941 P 0.881 0.881 0.881 0.881 np 1.18 1.24 1.25 1.27 nL 2.58 4.30 5.08 6.65 n0 3.03 5.05 5.97 7.81 Note that changing the bolt grade only affects Sp, Fi , np, nL, and n0. Any one of the solutions is acceptable, especially the lowest grade bolt. ________________________________________________________________________ 8-45 (a) Fb  RFb,max sin  Half of the external moment is contributed by the line load in the interval 0 ≤  ≤   M   FbR 2 sin  d  0 2 M   Fb,max R 2 2 2  0 Fb,max R 2 sin 2  d M  R2 from which Fb,max  Fmax   2  1 FbR sin  d  M 2 M R sin  d  (cos 1 - cos  2 ) 2  R 1 R Noting 1 = 75, 2 = 105, Fmax  (b) 12 000 (cos 75 - cos105 )  494 lbf  (8 / 2) Fmax  Fb,max R  Fmax  Ans. M 2M  2  ( R)    2  N RN R 2(12 000)  500 lbf (8 / 2)(12) Ans. (c) F = Fmax sin  M = 2 Fmax R [(1) sin2 90 + 2 sin2 60 + 2 sin2 30 + (1) sin2 (0)] = 6FmaxR from which, Fmax  Shigley’s MED, 11th edition M 12 000   500 lbf 6R 6(8 / 2) Ans. Chap. 8 Solutions, Page 37/68 The simple general equation resulted from part (b) 2M RN ________________________________________________________________________ Fmax  8-46 (a) From Table 8-11, Sp = 600 MPa. From Table 8-1, At = 353 mm2. Eq. (8-31): Table 8-15: Fi  0.9 At S p  0.9  353  600  10 3   190.6 kN K = 0.18 Eq. (8-27): T = K Fi d = 0.18(190.6)(24) = 823 Nm Ans. (b) Washers: t = 4.6 mm, d = 24 mm, D = 1.5(24) = 36 mm, E = 207 GPa. Eq. (8-20), k1  0.5774  207  24 1.155  4.6   36  24   36  24  ln  1.155  4.6   36  24   36  24   31 990 MN/m Cast iron: t = 20 mm, d = 24 mm, D = 36 + 2(4.6) tan 30 = 41.31 mm, E = 135 GPa. Eq. (8-20)  k2 = 10 785 MN/m Steel joist: t = 20 mm, d = 24 mm, D = 41.31 mm, E = 207 GPa. Eq. (8-20)  k3 = 16 537 MN/m Eq. (8-18): km = (2 / 31 990 + 1 / 10 785 +1 / 16 537)1 = 4 636 MN/m Bolt: l = 2(4.6) + 2(20) = 49.2 mm. Nut, Table A-31, H = 21.5 mm. L > 49.2 + 21.5 = 70.7 mm. From Table A-17, use L = 80 mm. From Eq. (8-14) LT = 2(24) + 6 = 54 mm, ld = 80  54 = 26 mm, lt = 49.2  26 = 23.2 mm From Table (8-1), At = 353 mm2, Ad =  (242) / 4 = 452.4 mm2 Eq. (8-17): kb  452.4  353  207 Ad At E   1680 MN/m Ad lt  At ld 452.4  23.2   353  26  C = kb / (kb + km) = 1680 / (1680 + 4636) = 0.266, Sp = 600 MPa, Fi = 190.6 kN, P = Ptotal / N = 18/4 = 4.5 kN Shigley’s MED, 11th edition Chap. 8 Solutions, Page 38/68 Yield: From Eq. (8-28) np  S p At CP  Fi  600  35310 3 0.266  4.5   190.6  1.10 Ans. Load factor: From Eq. (8-29) nL  S p At  Fi CP  600  353103  190.6 0.266  4.5   17.7 Ans. Separation: From Eq. (8-30) n0  Fi 190.6   57.7 P 1  C  4.5 1  0.266  Ans. As was stated in the text, bolts are typically preloaded such that the yielding factor of safety is not much greater than unity which is the case for this problem. However, the other load factors indicate that the bolts are oversized for the external load. ______________________________________________________________________________ 8-47 (a) ISO M 20  2.5 grade 8.8 coarse pitch bolts, lubricated. Table 8-2, At = 245 mm2 Table 8-11, Sp = 600 MPa Fi = 0.90 At Sp = 0.90(245)600(103) = 132.3 kN Table 8-15, K = 0.18 Eq. (8-27), T = KFi d = 0.18(132.3)20 = 476 N  m Ans. (b) Table A-31, H = 18 mm, L ≥ LG + H = 48 + 18 = 66 mm. Round up to L = 80 mm per Table A-17. LT  2d  6  2(20)  6  46 mm ld  L - LT  80  46  34 mm lt  l - ld  48  34  14 mm Ad =  (202) /4 = 314.2 mm2, kb  Ad At E 314.2(245)(207)   1251.9 MN/m Ad lt  Alt d 314.2(14)  245(34) Members: Since all members are steel use Eq. (8-22) with E = 207 MPa, l = 48 mm, d = 20mm Shigley’s MED, 11th edition Chap. 8 Solutions, Page 39/68 km  0.5774  207  20 0.5774 Ed   4236 MN/m  0.5774l  0.5d   0.5774  48  0.5  20   2 ln  5  2 ln 5   0.5774l  2.5d   0.5774  48   2.5  20   kb 1251.9   0.228 kb  km 1251.9  4236 P = Ptotal / N = 40/2 = 20 kN, C Yield: From Eq. (8-28) S p At 600  245 103   1.07 Ans. np  CP  Fi 0.228  20   132.3 Load factor: From Eq. (8-29) nL  S p At  Fi CP  600  245 103  132.3 0.228  20   3.22 Ans. Separation: From Eq. (8-30) Fi 132.3   8.57 Ans. P 1  C  20 1  0.228 ______________________________________________________________________________ n0  8-48 From Prob. 8-29 solution, Pmax =13.33 kips, C = 0.2, Fi = 12.77 kips, At = 0.141 9 in2 F 12.77 i  i   90.0 kpsi At 0.141 9 CP 0.2 13.33 Eq. (8-39), a    9.39 kpsi 2 At 2  0.141 9  Eq. (8-41),  m   a   i  9.39  90.0  99.39 kpsi (a) Goodman Eq. (8-45) for grade 8 bolts, Se = 23.2 kpsi (Table 8-17), Sut = 150 kpsi (Table 8-9) S  S   i  23.2 150  90.0    0.856 n f  e ut Ans.  a  Sut  Se  9.39 150  23.2  (b) Gerber Eq. (8-46) 1  nf  Sut Sut2  4 Se  S e   i   Sut2  2 i Se   2 a Se   1 150 150 2  4  23.2  23.2  90.0   150 2  2  90.0  23.2   1.32  2  9.39  23.2  Ans. (c) ASME-elliptic Eq. (8-47) with Sp = 120 kpsi (Table 8-9) Shigley’s MED, 11th edition Chap. 8 Solutions, Page 40/68 nf    Se S p S p2  Se2   i2   i Se 2 2  a  S p  Se   23.2 120 120 2  23.22  90 2  90  23.2    1.30  9.39 120 2  23.2 2   Ans. ______________________________________________________________________________ 8-49 Attention to the Instructor. Part (d) requires the determination of the endurance strength, Se, of a class 5.8 bolt. Table 8-17 does not provide this and the student will be required to estimate it by other means [see the solution of part (d)]. Per bolt, Pbmax = 60/8 = 7.5 kN, Pbmin = 20/8 = 2.5 kN kb 1 C   0.278 kb  km 1  2.6 (a) Table 8-1, At = 20.1 mm2; Table 8-11, Sp = 380 MPa Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(20.1)380(103) = 5.73 kN S p At 380  20.110 3   0.98 np  Ans. Yield, Eq. (8-28), CP  Fi 0.278  7.5   5.73 S p At  Fi 380  20.1103  5.73  0.915 0.278  7.5  Ans. (b) Overload, Eq. (8-29), nL  (c) Separation, Eq. (8-30), n0  (d) Goodman, Eq. (8-35), C  Pb max  Pb min  0.278  7.5  2.5 103   34.6 MPa a  2 At 2  20.1 CP  Fi 5.73   1.06 P 1  C  7.5 1  0.278  Ans. C  Pb max  Pb min  Fi 0.278  7.5  2.5 103 5.73 10 Eq. (8-36),  m     At 2 At 2  20.1 20.1 3   354.2 MPa Table 8-11, Sut = 520 MPa, i = Fi /At = 5.73(103)/20.1 = 285 MPa We have a problem for Se. Table 8-17 does not list Se for class 5.8 bolts. Here, we will estimate Se using the methods of Chapter 6. Estimate S e from the, Eq. (6-10): Se  0.5Sut  0.5  520   260 MPa . Table 6-2: Eq. (6-18): a = 3.04, b =  0.217 k a  aSutb  3.04  5200.217   0.783 Eq. (6-19): kb = 1 Eq. (6-25): kc = 0.85 The fatigue stress-concentration factor, from Table 8-16, is Kf = 2.2. For simple axial loading and infinite-life it is acceptable to reduce the endurance limit by Kf and use the nominal stresses in the stress/strength/design factor equations. Thus, from Eq. (6-17), Se = ka kb kc S e / Kf = 0.783(1)0.85(260) / 2.2 = 78.7 MPa Shigley’s MED, 11th edition Chap. 8 Solutions, Page 41/68 Eq. (8-38), nf  78.7  520  285  Se  Sut   i    0.789 Sut a  Se  m   i  520  34.6   78.7  354.2  285  Ans. It is obvious from the various answers obtained, the bolted assembly is undersized. This can be rectified by a one or more of the following: more bolts, larger bolts, higher class bolts. ______________________________________________________________________________ 8-50 Per bolt, Pbmax = Pmax /N = 80/10 = 8 kips, Pbmin = Pmin /N = 20/10 = 2 kips C = kb / (kb + km) = 4/(4 + 12) = 0.25 (a) Table 8-2, At = 0.141 9 in2, Table 8-9, Sp = 120 kpsi and Sut = 150 kpsi Table 8-17, Se = 23.2 kpsi Eqs. (8-31) and (8-32), Fi = 0.75 At Sp  i = Fi /At = 0.75 Sp = 0.75(120) =90 kpsi Eq. (8-35), a  Eq. (8-36), m  Eq. (8-38), C  Pb max  Pb min  0.25  8  2    5.29 kpsi 2 At 2  0.141 9  0.25  8  2  C  Pb max  Pb min  i   90  98.81 kpsi 2 At 2  0.141 9  23.2 150  90  Se  Sut   i    1.39 Ans. Sut a  Se  m   i  150  5.29   23.2  98.81  90  ______________________________________________________________________________ 8-51 Given: (8) M8  1.5 class 9.8 bolts, kb = 1.5 MN/mm, km = 3.9 MN/mm, reversing load with Pmax = 50 kN. Per bolt, P = Pmax/N = 50/8 = 6.25 kN Table 8-1, At = 36.6 mm2, Table 8-11, Sut = 900 MPa, Sp = 650 MPa, Table 8-17, Se = 140 MPa Eq. (f), Sec. 8-7: C = kb / (kb + km) = 1.5/(1.5 + 3.9) = 0.278 Eqs. (8-31) and (8-32): Fi = 0.75 At Sp = 0.75(36.6)650(103) = 17.84 kN i = Fi / At = 17.84(103)/36.6 = 487.5 MPa Eq. (8-39): a = CP / (2At) = 0.278(6.25)103/[2(36.6)] = 23.74 MPa (a) Goodman: Eq. (8-45): S  S   i  140  900  487.5  n f  e ut   2.34 Ans.  a  Sut  Se  23.74  900  140  (b) Gerber: Eq. (8-46): 1  nf  Sut Sut2  4 Se  S e   i   Sut2  2 i S e   2 a Se  nf  1 900 900 2  4 140 140  487.5   900 2  2  487.5    3.52 Ans.  2  23.74 140   (c) Morrow: Use Goodman Eq. (8-45) replacing Sut with  f . Equation (6-44) gives  Shigley’s MED, 11th edition Chap. 8 Solutions, Page 42/68  f = Sut + 345 = 900 + 345 = 1 245MPa nf  Se  f   i   a  f  Se   140 1245  487.5  23.74 1245  140   3.23 Ans. ______________________________________________________________________________ 8-52 From Prob. 8-33, C = 0.263, Pmax = 4.712 kN / bolt, Fi = 41.1 kN, Sp = 650 MPa, and At = 84.3 mm2 i = 0.75 Sp = 0.75(650) = 487.5 MPa 3 CP 0.263  4.712 10 a    7.350 MPa 2 At 2  84.3 Eq. (8-39): CP Fi   7.350  487.5  494.9 MPa 2 At At (a) Goodman: From Table 8-11, Sut = 900 MPa, and from Table 8-17, Se = 140 MPa S  S   i  140  900  487.5  Eq. (8-45):   7.55 n f  e ut Ans.  a  Sut  Se  7.350  900  140  (b) Gerber: Eq. (8-46): 1  nf  Sut Sut2  4 Se  Se   i   Sut2  2 i Se   2 a Se  m  Eq. (8-40)  1 900 9002  4 140 140  487.5   900 2  2  487.5 140    2  7.350 140   11.4 Ans. (c) ASME-elliptic: Eq. (8-47): nf    Se S p S p2  Se2   i2   i S e 2 2  a  S p  Se   140 650 6502  140 2  487.52  487.5 140    9.73 2 2   7.350  650  140  Ans. ______________________________________________________________________________ 8-53 From Prob. 8-34, C = 0.299, Pmax = 1.443 kips/bolt, Fi = 9.05 kips, Sp = 85 kpsi, and At = 0.141 9 in2  i  0.75S p  0.75  85   63.75 kpsi Shigley’s MED, 11th edition Chap. 8 Solutions, Page 43/68 Eq. (8-37): a  CP 0.299 1.443   1.520 kpsi 2 At 2  0.141 9  Eq. (8-38) m  CP   i  1.520  63.75  65.27 kpsi 2 At (a) Goodman: From Table 8-9, Sut = 120 kpsi, and from Table 8-17, Se = 18.8 kpsi S  S   i  18.8 120  63.75  Eq. (8-45): n f  e ut Ans.   5.01  a  Sut  Se  1.520 120  18.8  (b) Gerber: Eq. (8-46): 1  nf  Sut Sut2  4 Se  Se   i   Sut2  2 i S e    2 a S e  1 120 1202  4 18.6 18.6  63.75   120 2  2  63.75 18.6   2 1.520 18.6   7.45 Ans. (c) ASME-elliptic: Eq. (8-47): nf    Se S p S p2  S e2   i2   i Se 2 2  a  S p  Se   18.6 85 852  18.6 2  63.752  63.75 18.6   6.22 2 2   1.520  85  18.6  Ans. ______________________________________________________________________________ 8-54 From Prob. 8-35, C = 0.228, Pmax = 7.679 kN/bolt, Fi = 36.1 kN, Sp = 830 MPa, and At = 58.0 mm2 i = 0.75 Sp = 0.75(830) = 622.5 MPa 3 CP 0.228  7.679 10 a    15.09 MPa Eq. (8-37): 2 At 2  58.0  CP   i  15.09  622.5  637.6 MPa 2 At (a) Goodman: From Table 8-11, Sut = 1040 MPa, and from Table 8-17, Se = 162 MPa Eq. (8-38) Shigley’s MED, 11th edition m  Chap. 8 Solutions, Page 44/68 Eq. (8-45): nf  Se  Sut   i  162 1040  622.5    3.73  a  Sut  S e  15.09 1040  162  Ans. (b) Gerber: Eq. (8-46): 1  nf  Sut Sut2  4 Se  Se   i   Sut2  2 i Se    2 a Se  1 1040 10402  4 162 162  622.5   1040 2  2  622.5 162    2 15.09 162   5.74 Ans. (c) ASME-elliptic: Eq. (8-47): nf    Se S p S p2  Se2   i2   i S e  a  S p2  Se2   162 830 8302  162 2  622.52  622.5 162    5.62 2 2   15.09 830  162  Ans. ______________________________________________________________________________ 8-55 From Prob. 8-36, C = 0.291, Pmax = 1.244 kips/bolt, Fi = 9.57 kips, Sp = 120 kpsi, and At = 0.106 3 in2  i  0.75S p  0.75 120   90 kpsi Eq. (8-37): a  CP 0.2911.244    1.703 kpsi 2 At 2  0.106 3 Eq. (8-38) m  CP   i  1.703  90  91.70 kpsi 2 At (a) Goodman: From Table 8-9, Sut = 150 kpsi, and from Table 8-17, Se = 23.2 kpsi S S i  23.2 150  90  Eq. (8-45): n f  e ut Ans.   4.72  a  Sut  Se  1.703 150  23.2  (b) Gerber: Eq. (8-46): Shigley’s MED, 11th edition Chap. 8 Solutions, Page 45/68 nf   1  S S 2  4 S  S     S 2  2 S  ut ut e e i ut i e  2 a Se  1 150 1502  4  23.2  23.2  90   150 2  2  90  23.2   2 1.703 23.2   7.28 Ans. (c) ASME-elliptic: Eq. (8-47): nf    Se S p S p2  Se2   i2   i Se 2 2  a  S p  Se   23.2 120 1202  23.22  90 2  90  23.2    7.24  1.703 1202  18.62   Ans. ______________________________________________________________________________ 8-56 From Prob. 8-52, C = 0.263, Se = 140 MPa, Sut = 900 MPa, At = 84.4 mm2, i = 487.5 MPa, and Pmax = 4.712 kN. Pmin = Pmax / 2 = 4.712/2 = 2.356 kN a  Eq. (8-35): Eq. (8-36): m  C  Pmax  Pmin  0.263  4.712  2.356 103   3.675 MPa 2 At 2  84.3 C  Pmax  Pmin  2 At  i 0.263  4.712  2.356 103   487.5  498.5 MPa 2  84.3 Eq. (8-38): 140  900  487.5  Se  Sut   i    11.9 Ans. Sut a  Se  m   i  900  3.675   140  498.5  487.5  ______________________________________________________________________________ nf  Shigley’s MED, 11th edition Chap. 8 Solutions, Page 46/68 8-57 From Prob. 8-53, C = 0.299, Se = 18.8 kpsi, Sut = 120 kpsi, At = 0.141 9 in2, i = 63.75 kpsi, and Pmax = 1.443 kips Pmin = Pmax / 2 = 1.443/2 = 0.722 kips a  Eq. (8-35): Eq. (8-36): m   C  Pmax  Pmin  2 At  0.299 1.443  0.722  2  0.141 9   0.760 kpsi C  Pmax  Pmin  i 2 At 0.299 1.443  0.722   63.75  66.03 kpsi 2  0.141 9  Eq. (8-38): Se  Sut   i  18.8 120  63.75  Ans.   7.89 Sut a  Se  m   i  120  0.760   18.8  66.03  63.75  ______________________________________________________________________________ nf  8-58 From Prob. 8-54, C = 0.228, Se = 162 MPa, Sut = 1040 MPa, At = 58.0 mm2, i = 622.5 MPa, and Pmax = 7.679 kN. Pmin = Pmax / 2 = 7.679/2 = 3.840 kN a  Eq. (8-35): Eq. (8-36): m  C  Pmax  Pmin  0.228  7.679  3.840 103   7.546 MPa 2 At 2  58.0  C  Pmax  Pmin  2 At i 0.228  7.679  3.840 103   622.5  645.1 MPa 2  58.0  Eq. (8-38): Shigley’s MED, 11th edition Chap. 8 Solutions, Page 47/68 162 1040  622.5  Se  Sut   i    5.88 Ans. Sut a  Se  m   i  1040  7.546   162  645.1  622.5  ______________________________________________________________________________ nf  8-59 From Prob. 8-55, C = 0.291, Se = 23.2 kpsi, Sut = 150 kpsi, At = 0.106 3 in2, i = 90 kpsi, and Pmax = 1.244 kips Pmin = Pmax / 2 = 1.244/2 = 0.622 kips a  Eq. (8-35): Eq. (8-36): m   C  Pmax  Pmin  2 At  0.2911.244  0.622  2  0.106 3  0.851 kpsi C  Pmax  Pmin  i 2 At 0.2911.244  0.622   90  92.55 kpsi 2  0.106 3 Eq. (8-38): 23.2 150  90  Se  Sut   i    7.45 Ans. Sut a  Se  m   i  150  0.851  23.2  92.55  90  ______________________________________________________________________________ nf  8-60 Let the repeatedly-applied load be designated as P. From Table A-22, Sut = 93.7 kpsi. Referring to the Figure of Prob. 3-136, the following notation will be used for the radii of Section AA. ri = 1.5 in, ro = 2.5 in, rc = 2.0 in From Table 3-4, with R = 0.5 in, R2 0.52 rn    1.968 246 in 2 rc  rc2  R 2 2 2  22  0.52  e co ci A        rc  rn  2.0  1.968 246  0.031 754 in ro - rn  2.5  1.968 246  0.531 754 in rn - ri  1.968 246  1.5  0.468 246 in  (12 ) / 4  0.7854 in 2 If P is the maximum load Shigley’s MED, 11th edition Chap. 8 Solutions, Page 48/68 M  Prc  2 P 2(0.468)  P rc  P   i  1  c i   1    26.29 P A eri  0.785 4  0.031 754(1.5)  26.294 P   13.15P a  m  i  2 2 (a) Eye: Section AA, Table 6-2: a = 11.0 kpsi, b =  0.650 Eq. (6-18): k a  11.0(93.7) 0.650  0.575 Eq. (6-23): de = 0.370 d Eq. (6-19): Eq. (6-25): Eq. (6-10): Eq. (6-18): 0.107  0.37  kb    0.978   0.30  kc = 0.85 Se  0.5Sut  0.5  93.7   46.85 kpsi Se = 0.575(0.978)0.85(46.85) = 22.4 kpsi From Eq. 6-48, for Gerber, 2 2   2 m S e   1  Sut   a  1  1   nf     2   m  Se  Sut a      With m = a, 2 2    2Se   1 1 Sut2  93.7 2 2(22.4)   1.616   1  1   1  1   nf       P 2  a Se   93.7    Sut   2 13.15P(22.4)    where P is in kips. Thread: Die cut. Table 8-17 gives Se = 18.6 kpsi for rolled threads. Use Table 8-16 to find Se for die cut threads Se = 18.6(3.0/3.8) = 14.7 kpsi 2 Table 8-2, At = 0.663 in ,  = P/At = P /0.663 = 1.51 P, a = m = /2 = 0.755 P From Eq. 6-48, Gerber 2 2    2S e   1 1 Sut2  93.72 2(14.7)   19.01   1  1   1  1   nf      P 2  a Se   93.7    Sut   2 0.755P(14.7)    Comparing 19.01/P with 1.616/P, we conclude that the eye is weaker in fatigue. Ans. (b) Strengthening steps can include heat treatment, cold forming, cross section change (a round section is a poor cross section for a curved bar in bending because the bulk of the material is located where the stress is small). Ans. Shigley’s MED, 11th edition Chap. 8 Solutions, Page 49/68 (c) For nf = 2 1.616 103   808 lbf, max. load Ans. 2 ______________________________________________________________________________ P  8-61 Member, Eq. (8-22) with E =16 Mpsi, d = 0.75 in, and l = 1.5 in km  0.5774 16  0.75 0.5774 Ed   13.32 Mlbf/in  0.5774l  0.5d   0.5774 1.5   0.5  0.75   2 ln  5  2 ln 5   0.5774l  2.5d   0.5774 1.5   2.5  0.75   Bolt, Eq. (8-13), LT = 2d + 0.25 = 2(0.75) + 0.25 = 1.75 in l = 1.5 in ld = L  LT = 2.5  1.75 = 0.75 in lt = l  ld = 1.5  0.75 = 0.75 in Table 8-2, Eq. (8-17), At = 0.373 in2 Ad = (0.752)/4 = 0.442 in2 kb  0.442  0.373 30 Ad At E   8.09 Mlbf/in Ad lt  At ld 0.442  0.75   0.373  0.75  C kb 8.09   0.378 kb  km 8.09  13.32 Eq. (8-35), a  C  Pmax  Pmin  0.378  6  4    1.013 kpsi 2 At 2  0.373 m  C  Pmax  Pmin  Fi 0.378  6  4  25     72.09 kpsi At 2 At 2  0.373 0.373 Eq.(8-36), (a) From Table 8-9, Sp = 85 kpsi, and Eq. (8-51), the yielding factor of safety is np  Shigley’s MED, 11th edition Sp m a  85  1.16 72.09  1.013 Ans. Chap. 8 Solutions, Page 50/68 (b) From Eq. (8-29), the overload factor of safety is nL  S p At  Fi CPmax  85  0.373  25  2.96 0.378  6  Ans. (c) From Eq. (8-30), the factor of safety based on joint separation is n0  Fi 25   6.70 Pmax 1  C  6 1  0.378  Ans. (d) From Table 8-17, Se = 18.6 kpsi; Table 8-9, Sut = 120 kps; the preload stress is i = Fi / At = 25/0.373 = 67.0 kpsi; and from Eq. (8-38) Se  Sut   i  18.6 120  67.0    4.56 Ans. Sut a  S e  m   i  120 1.013  18.6  72.09  67.0  ______________________________________________________________________________ nf  8-62 (a) Table 8-2, At = 0.1419 in2 Table 8-9, Sp = 120 kpsi, Sut = 150 kpsi Table 8-17, Se = 23.2 kpsi Eqs. (8-31) and (8-32), i = 0.75 Sp = 0.75(120) = 90 kpsi kb 4   0.2 kb  k m 4  16 CP 0.2 P a    0.705P kpsi 2 At 2(0.141 9) C  Eq. (8-45) for the Goodman criterion, S S  i  23.2(150  90) 11.4    2 n f  e ut  a  Sut  Se  0.705 P(150  23.2) P  P  5.70 kips Ans. (b) Fi = 0.75At Sp = 0.75(0.141 9)120 = 12.77 kips Yield, Eq. (8-28), S p At 120  0.141 9    1.22 np  Ans. CP  Fi 0.2  5.70   12.77 Load factor, Eq. (8-29), S A - Fi 120(0.141 9)  12.77 nL  p t   3.74 Ans. CP 0.2(5.70) Separation load factor, Eq. (8-30) Shigley’s MED, 11th edition Chap. 8 Solutions, Page 51/68 12.77 Fi   2.80 Ans. P(1 - C ) 5.70(1  0.2) ______________________________________________________________________________ n0  8-63 Table 8-2, At = 0.969 in2 (coarse), At = 1.073 in2 (fine) Table 8-9, Sp = 74 kpsi, Sut = 105 kpsi Table 8-17, Se = 16.3 kpsi Coarse thread, Fi = 0.75 At Sp = 0.75(0.969)74 = 53.78 kips i = 0.75 Sp = 0.75(74) = 55.5 kpsi a  CP 0.30 P   0.155P kpsi 2 At 2(0.969) Gerber, Eq. (8-46), nf  1  S S 2  4 S  S     S 2  2 S  ut ut e e i ut i e  2 a Se  1 105 1052  4 16.316.3  55.5   1052  2  55.5 16.3  64.28  P 2  0.155 P 16.3  With nf = 2, 64.28 P   32.14 kip Ans. 2 Fine thread, Fi = 0.75 At Sp = 0.75(1.073)74 = 59.55kips i = 0.75 Sp = 0.75(74) = 55.5 kpsi CP 0.32 P a    0.149 P kpsi 2 At 2(1.073)  The only thing that changes in Eq. (8-46) is a. Thus, 0.155 64.28 66.87 nf    2  P  33.43 kips Ans. 0.149 P P Percent improvement, 33.43  32.14 (100) ฀ 4% Ans. 32.14 ______________________________________________________________________________ 8-64 For an M 30 × 3.5 ISO 8.8 bolt with P = 65 kN/bolt and C = 0.28 Table 8-1, Table 8-11, Table 8-17, At = 561 mm2 Sp = 600 MPa, Sut = 830 MPa Se = 129 MPa Shigley’s MED, 11th edition Chap. 8 Solutions, Page 52/68 Eq. (8-31), Fi = 0.75Fp = 0.75 At Sp = 0.75(5610600(103) = 252.45 kN i = 0.75 Sp = 0.75(600) = 450 MPa Eq. (8-39), CP 0.28  65 10 a    16.22 MPa 2 At 2  561 3 Gerber, Eq. (8-46), nf   1  S S 2  4 S  S     S 2  2 S  ut ut e e i ut i e  2 a S e  1 830 830 2  4 129 129  450   830 2  2  450 129   2 16.22 129   4.75 Ans. The yielding factor of safety, from Eq. (8-28) is 600  561103   1.24 np  CP  Fi 0.28  65   252.45 S p At Ans. From Eq. (8-29), the load factor is nL  S p At  Fi CP 600  561103  252.45   4.62 0.28  65  Ans. The separation factor, from Eq. (8-30) is Fi 252.45 Ans.   5.39 P 1  C  65 1  0.28  ______________________________________________________________________________ 8-65 (a) Table 8-2, At = 0.077 5 in2 Table 8-9, Sp = 85 kpsi, Sut = 120 kpsi Table 8-17, Se = 18.6 kpsi Unthreaded grip, n0  Ad E  (0.375)2 (30) kb    0.245 Mlbf/in per bolt l 4(13.5) Am   [( D  2t ) 2 - D 2 ]   (4.752 - 4 2 )  5.154 in 2 4 4 Am E 5.154(30)  1  km      2.148 Mlbf/in/bolt. l 12 6 Shigley’s MED, 11th edition Ans. Ans. Chap. 8 Solutions, Page 53/68 (b) Fi = 0.75 At Sp = 0.75(0.0775) (85) = 4.94 kip  i  0.75S p  0.75(85)  63.75 kpsi 2000    (4) 2   4189 lbf/bolt  6 4  kb 0.245 C    0.102 kb  k m 0.245  2.148 CP 0.102(4.189)   2.77 kpsi a  2 At 2(0.0775) P  pA  From Eq. (8-46) for Gerber fatigue criterion, nf  1  S S 2  4 S  S     S 2  2 S  ut ut e e i ut i e  2 a S e  1 120 1202  4 18.6 18.6  63.75   120 2  2  63.75 18.6   4.09  2  2.77 18.6  (c) Pressure causing joint separation from Eq. (8-30)  Ans. Fi 1 P(1  C ) Fi 4.94   5.50 kip P  1  C 1  0.102 P 5.50  p  6  2.63 kpsi Ans. A  (42 ) / 4 ______________________________________________________________________________ n0  8-66 From the solution of Prob. 8-64, At = 0.077 5 in2, Sut = 120 kpsi, Se = 18.6 kpsi, C = 0.102, i = 63.75 kpsi Pmax = pmaxA = 2  (42)/4 = 25.13 kpsi, Pmin = pminA = 1.2  (42)/4 = 15.08 kpsi, C  Pmax  Pmin  0.102  25.13  15.08  Eq. (8-35),   6.61 kpsi a  2 At 2  0.077 5  Eq. (8-36), m  C  Pmax  Pmin  0.102  25.13  15.08  i   63.75  90.21 kpsi 2 At 2  0.077 5  Eq. (8-38), Se  Sut   i  18.6 120  63.75  nf    0.814 Sut a  Se  m   i  120  6.61  18.6  90.21  63.75  Ans. This predicts a fatigue failure. ______________________________________________________________________________ 8-67 Members: Sy = 57 kpsi, Ssy = 0.577(57) = 32.89 kpsi. Shigley’s MED, 11th edition Chap. 8 Solutions, Page 54/68 Bolts: SAE grade 5, Sy = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear in bolts,   (0.252 )   0.0982 in 2 As  2   4   AS 0.0982(53.08) Fs  s sy   2.61 kips n 2 Bearing on bolts, Ab = 2(0.25)0.25 = 0.125 in2 AS 0.125(92) Fb  b yc   5.75 kips n 2 Bearing on member, 0.125(57) Fb   3.56 kips 2 Tension of members, At = (1.25  0.25) (0.25) = 0.25 in2 0.25(57)  7.13 kip 2 F  min(2.61, 5.75, 3.56, 7.13)  2.61 kip Ft  Ans. The shear in the bolts controls the design. ______________________________________________________________________________ 8-68 Members, Table A-20, Sy = 42 kpsi Bolts, Table 8-9, Sy = 130 kpsi, Ssy = 0.577(130) = 75.01 kpsi Shear of bolts,    5 /16 2  2 As  2    0.1534 in 4    n Fs 5   32.6 kpsi As 0.1534 S sy   75.01  2.30 32.6 Ans. Bearing on bolts, Ab = 2(0.25) (5/16) = 0.1563 in2 5 b    32.0 kpsi 0.1563 Shigley’s MED, 11th edition Chap. 8 Solutions, Page 55/68 n Sy b  130  4.06 32.0 Bearing on members, n Tension of members, Sy b  Ans. 42  1.31 32 Ans. At = [2.375  2(5/16)] (1/4) = 0.4375 in2 t  5  11.4 kpsi 0.4375 Sy 42  3.68 Ans.  t 11.4 ______________________________________________________________________________ n  8-69 Members: Table A-20, Sy = 490 MPa, Ssy = 0.577(490) = 282.7 MPa Bolts: Table 8-11, ISO class 5.8, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Shear in bolts,   (202 )   628.3 mm 2 As  2    4  AS 628.3(242.3)10 3 Fs  s sy   60.9 kN n 2.5 Bearing on bolts, Ab = 2(20)20 = 800 mm2 AS 800(420)10 3 Fb  b yc   134 kN n 2.5 Bearing on member, 800(490)10 3 Fb   157 kN 2.5 Tension of members, At = (80  20) (20) = 1 200 mm2 1 200(490)10 3 Ft   235 kN 2.5 F  min(60.9, 134, 157, 235)  60.9 kN Ans. The shear in the bolts controls the design. ______________________________________________________________________________ Shigley’s MED, 11th edition Chap. 8 Solutions, Page 56/68 8-70 Members: Table A-20, Sy = 320 MPa Bolts: Table 8-11, ISO class 5.8, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Shear of bolts, As =  (202)/4 = 314.2 mm2 90 103  s   95.48 MPa 3  314.2  n S sy s  242.3  2.54 95.48 Ans. Bearing on bolt, Ab = 3(20)15 = 900 mm2 90 103  b    100 MPa 900 S 420 n y   4.2 Ans. b 100 Bearing on members, S 320 n y   3.2 Ans. b 100 Tension on members, 90 103  F t    46.15 MPa A 15[190  3  20 ] S 320 n y   6.93 Ans. 46.15 t ______________________________________________________________________________ 8-71 Members: Sy = 57 kpsi Bolts: Sy = 100 kpsi, Ssy = 0.577(100) = 57.7 kpsi Shear of bolts,   1/ 4 2  2 A  3   0.1473 in 4   F 5 s    33.94 kpsi As 0.1473 n Bearing on bolts, Shigley’s MED, 11th edition S sy s  57.7  1.70 33.94 Ans. Ab = 3(1/4) (5/16) = 0.2344 in2 Chap. 8 Solutions, Page 57/68 b   n Bearing on members, Sy b 5 F   21.3 kpsi Ab 0.2344  100  4.69 21.3 Ans. Ab = 0.2344 in2 (From bearing on bolts calculation) b =  21.3 kpsi (From bearing on bolts calculation) n Sy b  57  2.68 21.3 Ans. Tension in members, failure across two bolts, At  5  2.375  2 1 / 4    0.5859 in 2 16 F 5   8.534 kpsi At 0.5859 Sy 57 n   6.68 Ans.  t 8.534 ______________________________________________________________________________ t  8-72 By symmetry, the reactions at each support is 1.6 kN. The free-body diagram for the left member is M M B 0 1.6(250)  50RA  0  RA  8 kN A 0 200(1.6)  50RB  0  RB  6.4 kN Members: Table A-20, Sy = 370 MPa Bolts: Table 8-11, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa  Bolt shear, As  (122 )  113.1 mm 2 4 Shigley’s MED, 11th edition Chap. 8 Solutions, Page 58/68 Fmax 8(103 )     70.73 MPa As 113.1 S 242.3 n  sy   3.43  70.73 Ab = td = 10(12) = 120 mm2 8(103 ) b    66.67 MPa 120 S 370  5.55 n y  b 66.67 Bearing on member, Strength of member. The bending moments at the hole locations are: In the left member at A, MA = 1.6(200) = 320 N · m. In the right member at B, MB = 8(50) = 400 N · m. The bending moment is greater at B 1 I B  [10(503 )  10(123 )]  102.7(103 ) mm 4 12 400(25) M c B  A  (103 )  97.37 MPa IA 102.7(103 ) S 370 n y   3.80  A 97.37 At the center, call it point C, MC = 1.6(350) = 560 N · m 1 (10)(503 )  104.2(103 ) mm 4 IC  12 560(25) M c (103 )  134.4 MPa C  C  3 104.2(10 ) IC S 370 n y   2.75  3.80 more critical at C  C 134.4 n  min(3.04, 3.80, 2.75)  2.72 Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chap. 8 Solutions, Page 59/68 8-73 The free-body diagram of the bracket, assuming the upper bolt takes all the shear and tensile load is Fs = 2500 lbf P 2500  3 7  1071 lbf Table A-31, H = 7/16 = 0.4375 in. Grip, l = 2(1/2) = 1 in. L ≥ l + H = 1.4375 in. Use 1.5 in bolts. Eq. (8-13), LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in Table 8-7, ld = L  LT = 1.5  1.25 = 0.25 in lt = l  ld = 1  0.25 = 0.75 in Table 8-2, At = 0.141 9 in2 Ad =  (0.52) /4 = 0.196 3 in2 0.196 3  0.141 9  30 Ad At E Eq. (8-17), kb    4.574 Mlbf/in Ad lt  At ld 0.196 3  0.75   0.141 9  0.25  Eq. (8-22), 0.5774  30  0.5 0.5774 Ed km    16.65 Mlbf/in  0.5774 l  0.5d   0.5774 1  0.5  0.5   2 ln  5  2 ln  5   0.5774 l  2.5d   0.5774 1  2.5  0.5   kb 4.574 C   0.216 kb  km 4.574  16.65 Table 8-9, Sp = 65 kpsi Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(0.141 9)65 = 6.918 kips i = 0.75 Sp = 0.75(65) = 48.75 kips Eq. (a): Direct shear, CP  Fi 0.216 1.071  6.918   50.38 kpsi At 0.141 9 F 3 s  s   21.14 kpsi At 0.141 9 b  Shigley’s MED, 11th edition Chap. 8 Solutions, Page 60/68 von Mises stress, Eq. (5-15),     b2  3 s2  1/2  50.382  3  21.142   1/2  62.3 kpsi Sp 65   1.04 Ans.   62.3 ______________________________________________________________________________ n 8-74 2P(200)  14(50) 14(50) P  1.75 kN per bolt 2(200) Fs  7 kN/bolt S p  380 MPa At  245 mm 2 , Ad   (202 )  314.2 mm 2 4 Fi  0.75(245)(380)(103 )  69.83 kN  i  0.75  380   285 MPa CP  Fi  0.25(1.75)  69.83  3   (10 )  287 MPa 245 At   3 7(10 ) F   s   22.3 MPa 314.2 Ad    [287 2  3(22.32 )]1/ 2  290 MPa b  Sp 380   1.31 Ans.   290 ______________________________________________________________________________ n 8-75 Using the result of Prob. 5-80 for lubricated assembly (replace 0.2 with 0.18 per Table 8-15) 2 f T Fx  0.18d With a design factor of nd gives T  or T/d = 716. Also, Shigley’s MED, 11th edition 0.18nd Fxd 0.18(3)(1000)d   716d 2 f 2 (0.12) Chap. 8 Solutions, Page 61/68 T  K (0.75S p At ) d  0.18(0.75)(85 000) At  11 475 At Form a table Size 1 4 - 28 5 16 3 8 At T/d = 11 475At n 0.0364 417.70 1.75 - 24 0.058 665.55 2.8  24 0.0878 1007.50 4.23 where the factor of safety in the last column of the table comes from n Select a 3" 8 2 f (T / d ) 2 (0.12)(T / d )   0.0042(T / d ) 0.18Fx 0.18(1000) - 24 UNF cap screw. The setting is given by T = (11 475At )d = 1007.5(0.375) = 378 lbf · in Given the coarse scale on a torque wrench, specify a torque wrench setting of 400 lbf · in. Check the factor of safety 2 f T 2 (0.12)(400)   4.47 0.18Fx d 0.18(1000)(0.375) ______________________________________________________________________________ n 8-76 Bolts, from Table 8-11, Sy = 420 MPa Channel, From Table A-20, Sy = 170 MPa. From Table A-7, t = 6.4 mm Cantilever, from Table A-20, Sy = 190 MPa F A = F B = F C = F / 3 Shigley’s MED, 11th edition Chap. 8 Solutions, Page 62/68 M = (50 + 26 + 125) F = 201 F FA  FC  201F  2.01 F 2  50  1  (1) FC  FC  FC    2.01 F  2.343F 3  Shear on Bolts: The shoulder bolt shear area, As = (102) / 4 = 78.54 mm2 Max. force, Ssy = 0.577(420) = 242.3 KPa  max  FC S sy  As n From Eq. (1), FC = 2.343 F. Thus F Ssy  As  242.3  78.54  3   10  4.06 kN n  2.343  2.0  2.343  Bearing on bolt: The bearing area is Ab = td = 6.4(10) = 64 mm2. Similar to shear F S y  Ab  420  64  3   10  5.74 kN n  2.343  2.0  2.343  Bearing on channel: Ab = 64 mm2, Sy = 170 MPa. S  A  170  64  3 F  y  b   10  2.32 kN n  2.343  2.0  2.343  Bearing on cantilever: Ab = 12(10) = 120 mm2, Sy = 190 MPa. F S y  Ab  190  120  3   10  4.87 kN n  2.343  2.0  2.343  Bending of cantilever: At C I 1 12   503  103   1.24 105  mm 4 12  max  Shigley’s MED, 11th edition Sy n  Mc 151Fc  I I  F Sy  I    n  151 c  Chap. 8 Solutions, Page 63/68 5 190 1.24 10   3   10  3.12 kN F 2.0  151 25     So F = 2.32 kN based on bearing on channel. Ans. ______________________________________________________________________________ 8-77 Bolts, from Table 8-11, Sy = 420 MPa Bracket, from Table A-20, Sy = 210 MPa 12  4 kN; M  12(200)  2400 N · m 3 2400  37.5 kN FA  FB  64 FA  FB  (4) 2  (37.5) 2  37.7 kN FO  4 kN F  Bolt shear: The shoulder bolt shear area, As = (122) / 4 = 113.1 mm2 Ssy = 0.577(420) = 242.3 KPa 37.7(10)3  333 MPa 113 S 242.3 n  sy  Ans.  0.728  333 Bearing on bolts: Ab  12(8)  96 mm 2 37.7(10)3 b    393 MPa 96 S 420 n  yc  Ans.  1.07 b 393   Bearing on member:  b  393 MPa S 210 n  yc   0.534 b 393 Shigley’s MED, 11th edition Ans. Chap. 8 Solutions, Page 64/68 Bending stress in plate:  bd 3  bh3 bd 3   2  a 2bd  12 12  12  3 3 3  8(12)  8(136) 8(12)    2  (32) 2 (8)(12)  12 12  12  6 4  1.48(10) mm Ans. 2400(68) Mc   (10)3  110 MPa  6 1.48(10) I Sy 210   1.91 n  Ans.  110 I  Failure is predicted for bolt shear and bearing on member. ______________________________________________________________________________ 8-78 FA  1208  125  1083 lbf, Bolt shear:  3625  FA  FB     1208 lbf  3  FB 1208  125  1333 lbf As = ( / 4)(0.3752) = 0.1104 in2  max  Fmax 1333   12 070 psi As 0.1104 From Table 8-10, Sy = 100 kpsi, Ssy = 0.577(100) = 57.7 kpsi n S sy  max  57.7  4.78 12.07 Ans. Bearing on bolt: Bearing area is Ab = td = 0.375 (0.375) = 0.1406 in2. Shigley’s MED, 11th edition Chap. 8 Solutions, Page 65/68 1333 F   9 481 psi Ab 0.1406 b   Sy n b  100  10.55 9.481 Ans. Bearing on member: From Table A-20, Sy = 54 kpsi. Bearing stress same as bolt Sy n b  54  5.70 9.481 Ans. Bending of member: At B, M = 250(13) = 3250 lbfin I 3 1 3 3  3  4  2       0.2484 in 12  8    8   Mc 3250 1   13 080 psi I 0.2484  Sy 54  4.13 Ans.  13.08 ______________________________________________________________________________ n  8-79 The direct shear load per bolt is F = 2000/6 = 333.3 lbf. The moment is taken only by the four outside bolts. This moment is M = 2000(5) = 10 000 lbf · in. 10 000 Thus F   1000 lbf and the resultant bolt load is 2(5) F  (333.3) 2  (1000) 2  1054 lbf Bolt strength, Table 8-9, Sy = 100 kpsi; Channel and Plate strength, Sy = 42 kpsi Shear of bolt: As =  (0.5)2/4 = 0.1963 in2 S (0.577)(100) n  sy   10.7  1.054 / 0.1963 Ans. Bearing on bolt: Channel thickness is t = 3/16 in, Ab = 0.5(3/16) = 0.09375 in2 n Shigley’s MED, 11th edition 100  8.89 1.054 / 0.09375 Ans. Chap. 8 Solutions, Page 66/68 Bearing on channel: Bearing on plate: n 42  3.74 1.054 / 0.09375 Ab = 0.5(0.25) = 0.125 in2 n 42  4.98 1.054 / 0.125 Ans. Ans. Strength of plate: 0.25(7.5)3 0.25(0.5)3  12 12 3  0.25(0.5)   2   0.25  0.5  (2.5) 2   7.219 in 4 12   M  5000 lbf · in per plate Mc 5000(3.75)     2597 psi I 7.219 42  16.2 Ans. n 2.597 ______________________________________________________________________________ I  A = 58 mm2. From Table A-9, beam 14, R1 = R2 = F/2 = 10/2 = 5 kN, M1 = M2 = Fl/8 = 10(0.4)103/8 = 500 N٠m.  0.050  M 1  500  4 F1    F1  5000 N  2  R 5000 F2  1   1250 N 4 4 Bolt A has the maximum force of FA = F1 + F2 = 5000 + 1250 = 6250 N F 6250 Ans.  max   A  A   108 MPa 58 A ______________________________________________________________________________ 8-81 Given: 250 hp at 600 rev/min, r1 = 2.5 in, r2 = 5 in. 63 025  250   26.26 103  lbf  in Eq. (3-42): T  600 F1 F2 r 5 (a)   F2  2 F1  F1  2 F1 2.5 r1 r2 r1 8-80 4 F1r1  4 F2 r2  T  4 F1 (2.5)  4  2 F1  5  26.26 103  Yields F1 = 525.2 lbf and F2 = 1050.4 lbf  max  4 F2 d2  d From Table A-17 select d = Shigley’s MED, 11th edition 4 F2  max 5 16 in  4 1050.4   0.258 in   20 103 Ans. Chap. 8 Solutions, Page 67/68 (b) F   max d 2 4 nFr2 = T  20 103    4   5 16 2  1534 lbf n (1534) 5 = 26.26 (103)  n = 3.4 bolts Use four bolts without need of the inner bolts Ans. ______________________________________________________________________________ 8-82 to 8-84 Specifying bolts, screws, dowels and rivets is the way a student learns about such components. However, choosing an array a priori is based on experience. Here is a chance for students to build some experience. ______________________________________________________________________________ 8-85 Assume tensile forces to be proportional to deflections. Thus, F1 F2   F2  9 F1 (1) 1 9 MA = 0 = 2 F1 (1) + 2 F2 (9)  1000 (10) Substitute Eq. (1) in, 2 F1 (1) + 2 (9 F1 ) (9)  1000 (10) = 0 Yields F1 = 60.98 lbf, F2 = 548.8 lbf. Combining the preload, 5000  548   27.7 103 psi  27.7 kpsi 0.2 The shear forces are F3 =1000/4 = 250 lbf.   The shear stress is, F 250  3  1250 psi  1.25 kpsi A 0.2 The maximum tensile stress is,  2 2 27.7    27.7  2     2    Ans.   1.25  27.8 kpsi 2 2 2  2  ______________________________________________________________________________  max  Shigley’s MED, 11th edition Chap. 8 Solutions, Page 68/68 Chapter 2 2-1 From Tables A-20, A-21, A-22, and A-24c, (a) UNS G10200 HR: Sut = 380 (55) MPa (kpsi), Syt = 210 (30) MPa (kpsi) Ans. (b) SAE 1050 CD: Sut = 690 (100) MPa (kpsi), Syt = 580 (84) MPa (kpsi) Ans. (c) AISI 1141 Q&T at 540C (1000F): Sut = 896 (130) MPa (kpsi), Syt = 765 (111) MPa (kpsi) Ans. (d) 2024-T4: Sut = 446 (64.8) MPa (kpsi), Syt = 296 (43.0) MPa (kpsi) Ans. (e) Ti-6Al-4V annealed: Sut = 900 (130) MPa (kpsi), Syt = 830 (120) MPa (kpsi) Ans. ______________________________________________________________________________ 2-2 (a) Maximize yield strength: Q&T at 425C (800F) Ans. (b) Maximize elongation: Q&T at 650C (1200F) Ans. ______________________________________________________________________________ 2-3 Conversion of kN/m3 to kg/ m3 multiply by 1(103) / 9.81 = 102 AISI 1018 CD steel: Tables A-20 and A-5 3 S y 370  10   47.4 kN m/kg Ans.  76.5  102  2011-T6 aluminum: Tables A-22 and A-5 3 S y 169  10   62.3 kN m/kg Ans.  26.6  102  Ti-6Al-4V titanium: Tables A-24c and A-5 3 S y 830  10   187 kN m/kg Ans.  43.4  102  ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the ultimate strength in tension 3 Sut 42.5  6.89   10   40.7 kN m/kg Ans  70.6  102  ______________________________________________________________________________ 2-4 AISI 1018 CD steel: Table A-5 6 E 30.0  10   106  106  in  0.282 2011-T6 aluminum: Table A-5 6 E 10.4  10   106  106  in  0.098 Shigley’s MED, 11th edition Ans. Ans. Chapter 2 Solutions, Page 1/22 Ti-6Al-6V titanium: Table A-5 6 E 16.5  10   103  10 6  in Ans.  0.160 No. 40 cast iron: Table A-5 6 E 14.5  10   55.8  106  in Ans.  0.260 ______________________________________________________________________________ E  2G 2G 2-5 Using values for E and G from Table A-5, 30.0  2  11.5 v 0.304 Ans. 2  11.5  Steel: The percent difference from the value in Table A-5 is 2G (1  v) E  v 0.304  0.292  0.0411  4.11 percent 0.292 10.4  2  3.90 v 0.333 2  3.90 Ans. Ans. Aluminum: The percent difference from the value in Table A-5 is 0 percent Ans. Beryllium copper: 18.0  2  7.0  v 0.286 2  7.0  Ans. The percent difference from the value in Table A-5 is 0.286  0.285  0.00351  0.351 percent 0.285 14.5  2  6.0  v 0.208 2  6.0  Ans. Ans. Gray cast iron: The percent difference from the value in Table A-5 is 0.208  0.211   0.0142   1.42 percent Ans. 0.211 ______________________________________________________________________________ 2-6 (a) A0 =  (0.503)2/4 = 0.1987 in2,  = Pi / A0 Shigley’s MED, 11th edition Chapter 2 Solutions, Page 2/22 For data in the elastic range, from Eq. (2–2),  = (l – l0) / l0 =  l / l0 =  l / 2 A  A  0 A For data in the plastic range, from Eq. (2–8), On the next two pages, the data and plots are presented. Figure (a) shows the linear part of the curve from data points 1 to 7. Figure (b) shows data points 1 to 12. Figure (c) shows the complete range. Note: The exact value of A0 is used without rounding off. (b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsi Ans. From Fig. (b) the equation for the dotted offset line is found to be  = 30.5(106)  61 000 (1) The equation for the line between data points 8 and 9 is  = 7.60(105) + 42 900 (2) Solving Eqs. (1) and (2) simultaneously yields  = 45.6 kpsi which is the 0.2 percent offset yield strength. Thus, Sy = 45.6 kpsi Ans. The ultimate strength from Figure (c) is Su = 85.6 kpsi Ans. The reduction in area is given by Eq. (2-25) as R A0  Af A0  0.1987  0.1077 0.458 45.8 % 0.1987 Data Point Pi l, Ai   1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 0 1000 2000 3000 4000 7000 8400 8800 9200 8800 9200 9100 13200 15200 17000 16400 14800 0.0004 0.0006 0.001 0.0013 0.0023 0.0028 0.0036 0.0089 0.1984 0.1978 0.1963 0.1924 0.1875 0.1563 0.1307 0.1077 0.00020 0.00030 0.00050 0.00065 0.00115 0.00140 0.00180 0.00445 0.00158 0.00461 0.01229 0.03281 0.05980 0.27136 0.52037 0.84506 0 5032 10065 15097 20130 35227 42272 44285 46298 44285 46298 45795 66428 76492 85551 82531 74479 Shigley’s MED, 11th edition Ans. Chapter 2 Solutions, Page 3/22 Stress (psi) 45000 40000 f(x) = 30451576.81 x − 10.57 35000 30000 25000 20000 15000 10000 5000 0 0.000 0.001 0.001 0.002 Linear () Strain Stress (psi) (a) Linear range 50000 45000 40000 35000 30000 25000 20000 15000 10000 5000 0 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Strain (b) Offset yield 90000 80000 70000 Stress (psi) 60000 50000 40000 30000 20000 10000 0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Strain (c) Complete range Shigley’s MED, 11th edition Chapter 2 Solutions, Page 4/22 (c) The material is ductile since there is a large amount of deformation beyond yield. (d) The closest material to the values of Sy, Sut, and R is SAE 1045 HR with Sy = 45 kpsi, Sut = 82 kpsi, and R = 40 %. Ans. ______________________________________________________________________________ 2-7 To plot  vs.  , the following equations are applied to the data. P   A Eq. (2-4) l  ln for 0 l 0.0028 in (0 P 8 400 lbf ) l0  ln Eq. (2–9) A0  A0 A for l  0.0028 in (P  8400 lbf )  (0.503) 2 0.1987 in 2 4 where The results are summarized in the following table and plot. The last 5 points of data are used to plot log  vs log  . The curve fit gives m = 0.2306 log 0 = 5.1852  0 = 153.2 kpsi Ans. The true strain corresponding to 20% cold work is given by Eq. (2–28) as  1   1   ln   ln   0.2231  1 W   1  0.2  Eq. (2-30): S y  0 m 153.2(0.2231) 0.2306 108.4 kpsi Ans. Eq. (2-32), with Su 85.6 from Prob. 2-6, S 85.6 Su  u  107 kpsi 1  W 1  0.2 Shigley’s MED, 11th edition Ans. Chapter 2 Solutions, Page 5/22 l P 0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800 0 0.000 4 0.000 6 0.001 0 0.001 3 0.002 3 0.002 8 Shigley’s MED, 11th edition A 0.198 7 0.198 7 0.198 7 0.198 7 0.198 7 0.198 7 0.198 7 0.198 4 0.197 8 0.196 3 0.192 4 0.187 5 0.156 3 0.130 7 0.107 7  0 0.000 2 0.000 3 0.000 5 0.000 65 0.001 15 0.001 4 0.001 51 0.004 54 0.012 15 0.032 22 0.058 02 0.240 02 0.418 89 0.612 45    0 5 032.71 10 065.4 15 098.1 20 130.9 35 229 42 274.8 44 354.8 46 511.6 46 357.6 68 607.1 81 066.7 108 765 125 478 137 419 log  -3.699 -3.523 -3.301 -3.187 -2.940 -2.854 -2.821 -2.343 -1.915 -1.492 -1.236 -0.620 -0.378 -0.213 log  3.702 4.003 4.179 4.304 4.547 4.626 4.647 4.668 4.666 4.836 4.909 5.036 5.099 5.138 Chapter 2 Solutions, Page 6/22 ______________________________________________________________________________ 2-8 Tangent modulus at  = 0 is E  5000  0  25  106  psi  0.2  10 3   0 Ans. At  = 20 kpsi Shigley’s MED, 11th edition Chapter 2 Solutions, Page 7/22  26  19   103  E20   1.5  1  10 3   (10-3) 0 0.20 0.44 0.80 1.0 1.5 2.0 2.8 3.4 4.0 5.0 14.0  106  psi Ans.  (kpsi) 0 5 10 16 19 26 32 40 46 49 54 ______________________________________________________________________________ 2-9 W = 0.20, (a) Before cold working: Annealed AISI 1018 steel. Table A-22, Sy = 32 kpsi, Su = 49.5  kpsi, 0 = 90.0 kpsi, m = 0.25, f = 1.05 m u 0.25 After cold working: Eq. (2-23),  1   1  20 ln   ln   0.223  u  1 W   1  0.2  Eq. (2-28): Eq. (2-30): S y  020m 90  0.223 Eq. (2-32): S 49.5 Su   u  61.9 kpsi 1  W 1  0.20 Su 49.5  1.55 Sy 32 0.25 61.8 kpsi Ans. 93% increase Ans. Ans. 25% increase Ans. Su 61.9  1.002 S y 61.8 (b) Before: After: Ans. Lost most of its ductility. ______________________________________________________________________________ 2-10 W = 0.20, (a) Before cold working: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, Su = 61.5 kpsi, 0  = 110 kpsi, m = 0.24, f = 0.85  After cold working: Eq. (2-23), m  u 0.24 Shigley’s MED, 11th edition Chapter 2 Solutions, Page 8/22 Eq. (2-28):  1 20 ln   1 W   1   ln   0.223  u   1  0.2  Eq. (2-30): S y  020m 110  0.223 Eq. (2-32): S 61.5 Su   u  76.9 kpsi 1  W 1  0.20 Su 61.5  2.20 Sy 28 0.24 76.7 kpsi Ans. 174% increase Ans. Ans. 25% increase Ans. Su 76.9  1.002 S y 76.7 (b) Before: After: Ans. Lost most of its ductility. ______________________________________________________________________________ 2-11 W = 0.20, (a) Before cold working: 2024-T4 aluminum alloy. Table A-22, Sy = 43.0 kpsi, Su = 64.8  kpsi, 0 = 100 kpsi, m = 0.15, f = 0.18  After cold working: Eq. (2-23), m  u 0.15  1   1  20 ln   ln   0.223   f  1 W   1  0.2  Eq. (2-28): Material fractures. Ans. ______________________________________________________________________________ 2-12 For HB = 275, Eq. (2-36), Su = 3.4(275) = 935 MPa Ans. ______________________________________________________________________________ 2-13 Gray cast iron, HB = 200. Eq. (2-37), Su = 0.23(200)  12.5 = 33.5 kpsi Ans. From Table A-24, this is probably ASTM No. 30 Gray cast iron Ans. ______________________________________________________________________________ 2-14 Eq. (2-36), 0.5HB = 100  HB = 200 Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 2 Solutions, Page 9/22 2-15 For the data given, converting HB to Su using Eq. (2-36) Su (kpsi) 115 116 116 117 117.5 117.5 117.5 118 118 119.5 HB 230 232 232 234 235 235 235 236 236 239 Su = Eq. (1-6) Su  S u N 1172 Su2 (kpsi) 13225 13456 13456 13689 13806.25 13806.25 13806.25 13924 13924 14280.25 Su2 = 137373 1172  117.2 117 kpsi Ans. 10 Eq. (1-7), 10 S 2 u  NSu2 137373  10  117.2  sSu   1.27 kpsi Ans. N1 9 ______________________________________________________________________________ 2 i 1 2-16 For the data given, converting HB to Su using Eq. (2-37) HB 230 232 232 234 235 235 235 236 236 239 Su = Su (kpsi) 40.4 40.86 40.86 41.32 41.55 41.55 41.55 41.78 41.78 42.47 Su2 (kpsi) 1632.16 1669.54 1669.54 1707.342 1726.403 1726.403 1726.403 1745.568 1745.568 1803.701 414.12 Su2 = 17152.63 Eq. (1-6) Shigley’s MED, 11th edition Chapter 2 Solutions, Page 10/22 Su  S u N  414.12 41.4 kpsi Ans. 10 Eq. (1-7), 10 S 2 u  NSu2 17152.63  10  41.4  sSu   1.20 Ans. N1 9 ______________________________________________________________________________ 2 i 1 uR  2-17 (a) Eq. (2-16) 45.62 34.7 in lbf / in 3 2(30) Ans. (b) A0 = (0.5032)/4 = 0.19871 in2 l P 0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800 (A0 / A) – 1 A 0 0.000 4 0.000 6 0.001 0 0.001 3 0.002 3 0.002 8 0.003 6 0.008 9 0.196 3 0.192 4 0.187 5 0.156 3 0.130 7 0.107 7 0.012 28 0.032 80 0.059 79 0.271 34 0.520 35 0.845 03  0 0.000 2 0.000 3 0.000 5 0.000 65 0.001 15 0.001 4 0.001 8 0.004 45 0.012 28 0.032 80 0.059 79 0.271 34 0.520 35 0.845 03  = P/A0 0 5 032 10 070 15 100 20 130 35 230 42 270 44 290 46 300 45 800 66 430 76 500 85 550 82 530 74 480 From the figures on the next page, 5 1 uT  Ai  (43 000)(0.001 5)  45 000(0.004 45  0.001 5) 2 i 1 1   45 000  76 500  (0.059 8  0.004 45) 2  81 000  0.4  0.059 8   80 000  0.845  0.4    66.7 103 in lbf/in 3 Shigley’s MED, 11th edition Ans. Chapter 2 Solutions, Page 11/22 Shigley’s MED, 11th edition Chapter 2 Solutions, Page 12/22 2-18, 2-19 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________ 2-20 Appropriate tables: Young’s modulus and Density (Table A-5); 1020 HR and CD (Table A20); 1040 and 4140 (Table A-21); Aluminum (Table A-24); Titanium (Table A-24c) Appropriate equations:  For diameter, F F  S y A   / 4 d 2  d 4F  Sy Weight/length = A, Cost/length = $/in = ($/lbf) Weight/length, Deflection/length =  /L = F/(AE) With F = 100 kips = 100(103) lbf, Young's Yield Material Modulus Density Strength Cost/lbf Diameter units Mpsi lbf/in3 kpsi $/lbf in 1020 HR 1020 CD 1040 4140 Al Ti 30 30 30 30 10.4 16.5 0.282 0.282 0.282 0.282 0.098 0.16 30 57 80 165 50 120 0.27 0.30 0.35 0.80 1.10 7.00 2.060 1.495 1.262 0.878 1.596 1.030 Weight/ length lbf/in Cost/ length $/in 0.9400 0.4947 0.3525 0.1709 0.1960 0.1333 0.25 0.15 0.12 0.14 0.22 $0.93 Deflection/ length in/in 1.000E-03 1.900E-03 2.667E-03 5.500E-03 4.808E-03 7.273E-03 The selected materials with minimum values are shaded in the table above. Ans. ______________________________________________________________________________ 2-21 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three would favor steel, cast iron, or maybe a less common ferrous material. The expectation would likely be hot-rolled steel. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 7.95 lbf, the unit weight is determined to be w W 7.95 lbf  0.281 lbf/in 3 2 Al [ (1 in) / 4](36 in) which agrees well with the unit weight of 0.282 lbf/in3 reported in Table A-5 for carbon steel. Nickel steel and stainless steel have similar unit weights, but surface finish and darker coloring do not favor their selection. To select a likely specification from Table A20, perform a Brinell hardness test, then use Eq. (2-36) to estimate an ultimate strength of Shigley’s MED, 11th edition Chapter 2 Solutions, Page 13/22 Su 0.5H B 0.5(200) 100 kpsi . Assuming the material is hot-rolled due to the rough surface finish, appropriate choices from Table A-20 would be one of the higher carbon steels, such as hot-rolled AISI 1050, 1060, or 1080. Ans. ______________________________________________________________________________ 2-22 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous material like aluminum. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 2.90 lbf, the unit weight is determined to be W 2.9 lbf w  0.103 lbf/in 3 2 Al [ (1 in) / 4](36 in) which agrees reasonably well with the unit weight of 0.098 lbf/in3 reported in Table A-5 for aluminum. No other materials come close to this unit weight, so the material is likely aluminum. Ans. ______________________________________________________________________________ 2-23 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous copper-based material such as copper, brass, or bronze. To further distinguish the material, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 9 lbf, the unit weight is determined to be w W 9.0 lbf  0.318 lbf/in 3 Al [ (1 in) 2 / 4](36 in) which agrees reasonably well with the unit weight of 0.322 lbf/in3 reported in Table A-5 for copper. Brass is not far off (0.309 lbf/in3), so the deflection test could be used to gain additional insight. From the measured deflection and utilizing the deflection equation for an end-loaded cantilever beam from Table A-9, Young’s modulus is determined to be 100  24  Fl 3 E  17.7 Mpsi 3Iy 3   (1) 4 64  (17 / 32) 3 which agrees better with the modulus for copper (17.2 Mpsi) than with brass (15.4 Mpsi). The conclusion is that the material is likely copper. Ans. ______________________________________________________________________________ 2-24 This problem is for student research. No standard solution is provided. ______________________________________________________________________________ 2-25 For strength,  = F/A = S  A = F/S For mass, m = Al = (F/S) l Shigley’s MED, 11th edition Chapter 2 Solutions, Page 14/22 Thus, f 3(M ) =  /S , and maximize S/ ( = 1) In Fig. (2-27), draw lines parallel to S/ The higher strength aluminum alloys have the greatest potential, as determined by comparing each material’s bubble to the S/ guidelines. Ans. ______________________________________________________________________________ 2-26 For stiffness, k = AE/l  A = kl/E For mass, m = Al = (kl/E) l =kl2  /E Thus, f 3(M) =  /E , and maximize E/ ( = 1) In Fig. (2-24), draw lines parallel to E/ Shigley’s MED, 11th edition Chapter 2 Solutions, Page 15/22 From the list of materials given, tungsten carbide (WC) is best, closely followed by aluminum alloys. They are close enough that other factors, like cost or availability, would likely dictate the best choice. Polycarbonate polymer is clearly not a good choice compared to the other candidate materials. Ans. ______________________________________________________________________________ 2-27 For strength,  = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b)]. The section modulus is strictly a function of the dimensions of the cross section and has the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a number. 4   For example, for a circular cross section, C = Fl S CA3/2 Shigley’s MED, 11th edition  1 . Then, for strength, Eq. (1) is  Fl  A    CS  2/3 (2) Chapter 2 Solutions, Page 16/22 For mass, Thus,  Fl  m  Al     CS  2/3 F l    C f 3(M) =  /S 2/3, and maximize S 2/3/ 2/3    l 5/3  2/3  S  ( = 2/3) In Fig. (2-27), draw lines parallel to S 2/3/ From the list of materials given, a higher strength aluminum alloy has the greatest potential, followed closely by high carbon heat-treated steel. Tungsten carbide is clearly not a good choice compared to the other candidate materials. .Ans. ______________________________________________________________________________ 2-28 Equation (2-41) applies to a circular cross section. However, for any cross section shape it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 = cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12 (a constant). Thus, Eq. (2-42) becomes Shigley’s MED, 11th edition Chapter 2 Solutions, Page 17/22  kl 3  A    3CE  and Eq. (2-44) becomes 1/2  k  m  Al     3C  1/2    l 5/2  1/2  E  E1/2  M f3  M   1/2  . From Fig. (2-24) E , or maximize Thus, minimize From the list of materials given, aluminum alloys are clearly the best followed by steels and tungsten carbide. Polycarbonate polymer is not a good choice compared to the other candidate materials. Ans. ______________________________________________________________________________ 2-29 For stiffness, k = AE/l  A = kl/E For mass, m = Al = (kl/E) l =kl2  /E So, f 3(M) =  /E, and maximize E/ . Thus,  = 1. Ans. ______________________________________________________________________________ 2-30 For strength,  = F/A = S  A = F/S Shigley’s MED, 11th edition Chapter 2 Solutions, Page 18/22 For mass, m = Al = (F/S) l Ans. So, f 3(M ) =  /S, and maximize S/ . Thus,  = 1. ______________________________________________________________________________ 2-31 Equation (2-41) applies to a circular cross section. However, for any cross section shape it can be shown that I = CA 2, where C is a constant. For the circular cross section, C = (4)1 [see Eq. (2-41)]. Another example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 = cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12, a constant. Thus, Eq. (2-42) becomes  kl 3  A    3CE  and Eq. (2-44) becomes 1/2  k  m  Al     3C  1/2    l 5/2  1/2  E  1/2 E  M f3  M   1/2  . Thus,  = 1/2. Ans. E , or maximize So, minimize ______________________________________________________________________________ 2-32 For strength,  = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b)]. The section modulus is strictly a function of the dimensions of the cross section and has the units in3 (ips) or m3 (SI). The area of the cross section has the units in2 or m2. Thus, for a given cross section, Z =C (A)3/2, where C is a number. For example, for a circular cross 4   section, Z = d /(32)and the area is A = d /4. This leads to C = 3 2 1 . So, with 3/2 Z =C (A) , for strength, Eq. (1) is Fl S CA3/2 For mass,  Fl  A    CS    Fl  m  Al     CS  2/3 F l    C 2/3 2/3 (2)    l 5/3  2/3  S  So, f 3(M) =  /S 2/3, and maximize S 2/3/. Thus,  = 2/3. Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 2 Solutions, Page 19/22 2-33 For stiffness, k=AE/l, or, A = kl/E. Thus, m = Al = (kl/E )l = kl 2  /E. Then, M = E / and  = 1. From Fig. 2-24, lines parallel to E / for ductile materials include steel, titanium, molybdenum, aluminum alloys, and composites. For strength, S = F/A, or, A = F/S. Thus, m = Al = F/Sl = Fl  /S. Then, M = S/ and  = 1. From Fig. 2-27, lines parallel to S/ give for ductile materials, steel, aluminum alloys, nickel alloys, titanium, and composites. Common to both stiffness and strength are steel, titanium, aluminum alloys, and composites. Ans. ______________________________________________________________________________ s 2-34 See Prob. 1-13 solution for x = 122.9 kcycles and x = 30.3 kcycles. Also, in that solution it is observed that the number of instances less than 115 kcycles predicted by the normal distribution is 27; whereas, the data indicates the number to be 31. ˆ From Eq. (1-4), the probability density function (PDF), with   x and   sx , is  1  x  x 2   1  x  122.9  2  1 1 f ( x)  exp    exp         sx 2 30.3 2  2  sx    2  30.3   (1) The discrete PDF is given by f /(Nw), where N = 69 and w = 10 kcycles. From the Eq. (1) and the data of Prob. 1-13, the following plots are obtained. Shigley’s MED, 11th edition Chapter 2 Solutions, Page 20/22 Range midpoint (kcycles) x 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 Frequency f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 Observed PDF f /(Nw) 0.002898551 0.001449275 0.004347826 0.007246377 0.011594203 0.017391304 0.008695652 0.014492754 0.011594203 0.007246377 0.002898551 0.004347826 0.002898551 0.001449275 0 0.001449275 Normal PDF f (x) 0.001526493 0.002868043 0.004832507 0.007302224 0.009895407 0.012025636 0.013106245 0.012809861 0.011228104 0.008826008 0.006221829 0.003933396 0.002230043 0.001133847 0.000517001 0.00021141 Plots of the PDF’s are shown below. 0.02 0.02 0.02 0.01 0.01 Normal Distribution Histogram 0.01 0.01 0.01 0 0 0 0 20 40 60 80 100 120 140 160 180 200 220 Shigley’s MED, 11th edition Chapter 2 Solutions, Page 21/22 It can be seen that the data is not perfectly normal and is skewed to the left indicating that the number of instances below 115 kcycles for the data (31) would be higher than the hypothetical normal distribution (27). Shigley’s MED, 11th edition Chapter 2 Solutions, Page 22/22 Chapter 10 10-1 From Eqs. (10-4) and (10-5) KW  K B  4C  1 0.615 4C  2   4C  4 C 4C  3 Plot 100(KW  KB)/ KW vs. C for 4  C  12 obtaining 1.4 1.3 1.2 100(KW-KB)/KW 1.1 1 0.9 0.8 0.7 4 6 8 10 12 C We see the maximum and minimum occur at C = 4 and 12 respectively where Maximum = 1.36 % Ans., and Minimum = 0.743 % Ans. ______________________________________________________________________________ A = Sdm dim(Auscu) = [dim (S) dim(d m)]uscu = kpsiinm 10-2 dim(ASI) = [dim (S) dim(d m)]SI = MPammm ASI  MPa mm m m m  m Auscu  6.894757  25.4  Auscu ฀ 6.895  25.4  Auscu kpsi in Ans. For music wire, from Table 10-4: Auscu = 201 kpsiinm, m = 0.145; what is ASI? ASI = 6.895(25.4)0.145 (201) = 2215 MPammm Ans. ______________________________________________________________________________ 10-3 Given: Music wire, d = 2.5 mm, OD = 31 mm, plain ground ends, Nt = 14 coils. Shigley’s MED, 11th edition Chapter 10 Solutions, Page 1/47 (a) Table 10-1: Na = Nt  1 = 14  1 = 13 coils D = OD  d = 31 2.5 = 28.5 mm C = D/d = 28.5/2.5 = 11.4 Table 10-5: d = 2.5/25.4 = 0.098 in Eq. (10-9): k (b) Table 10-1: G = 81.0(103) MPa  2.54  81103 d 4G   1.314 N / mm 8 D 3 N a 8  28.53 13 Ls = d Nt = 2.5(14) = 35 mm A = 2211 MPammm Table 10-4: m = 0.145, Eq. (10-14): Sut  Table 10-6: Ssy = 0.45(1936) = 871.2 MPa Eq. (10-5): KB  Eq. (10-7): Fs  (c) L0   L0 cr  Ans. A 2211   1936 MPa m d 2.50.145 4C  2 4 11.4   2   1.117 4C  3 4 11.4   3  d 3 S sy 8K B D    2.53  871.2 8 1.117  28.5  167.9 N Fs 167.9  Ls   35  162.8 mm k 1.314 Ans. Ans. 2.63  28.5   149.9 mm . Spring needs to be supported. Ans. 0.5 ______________________________________________________________________________ (d) 10-4 Given: Design load, F1 = 130 N. Referring to Prob. 10-3 solution, C = 11.4, Na = 13 coils, Ssy = 871.2 MPa, Fs = 167.9 N, L0 = 162.8 mm and (L0)cr = 149.9 mm. Eq. (10-18): 4 ≤ C ≤ 12 C = 11.4 O.K. Eq. (10-19): 3 ≤ Na ≤ 15 Na = 13 O.K. F 167.9 Eq. (10-17):   s  1   1  0.29 F1 130 Shigley’s MED, 11th edition Chapter 10 Solutions, Page 2/47 Eq. (10-20):   0.15,   0.29 From Eq. (10-7) for static service O.K . 8(130)(28.5)  8F1D   1.117  674 MPa 3   (2.5)3  d  S 871.2  1.29 n  sy  1 674 1  K B  Eq. (10-21): ns ≥ 1.2, n = 1.29 O.K.  167.9   167.9    674    870.5 MPa  130   130  S sy /  s  871.2 / 870.5 ฀ 1  s  1  Ssy/s ≥ (ns )d : Not solid-safe (but was the basis of the design). Not O.K. L0 ≤ (L0)cr: 162.8  149.9 Not O.K. Design is unsatisfactory. Operate over a rod? Ans. ______________________________________________________________________________ 10-5 Given: Oil-tempered wire, d = 0.2 in, D = 2 in, Nt = 12 coils, L0 = 5 in, squared ends. (a) Table 10-1: Ls = d (Nt + 1) = 0.2(12 + 1) = 2.6 in (b) Table 10-1: Table 10-5: Na = Nt  2 = 12  2 = 10 coils G = 11.2 Mpsi Eq. (10-9): k Ans. 0.24 11.2 106 d 4G   28 lbf/in 8D3 N a 8  23 10 Fs = k ys = k (L0  Ls ) = 28(5  2.6) = 67.2 lbf (c) Eq. (10-1): C = D/d = 2/0.2 = 10 4C  2 4 10   2   1.135 4C  3 4 10   3 Eq. (10-5): KB  Eq. (10-7):  s  KB Table 10-4: m = 0.187, A = 147 kpsiinm Shigley’s MED, 11th edition Ans. 8  67.2  2 8 Fs D  1.135  48.56 103  psi 3 3 d   0.2  Chapter 10 Solutions, Page 3/47 147 A   198.6 kpsi m d 0.20.187 Eq. (10-14): Sut  Table 10-6: Ssy = 0.50 Sut = 0.50(198.6) = 99.3 kpsi S sy 99.3  2.04 Ans.  s 48.56 ______________________________________________________________________________ ns  10-6  Given: Oil-tempered wire, d = 4 mm, C = 10, plain ends, L0 = 80 mm, and at F = 50 N, y = 15 mm. (a) k = F/y = 50/15 = 3.333 N/mm (b) D = Cd = 10(4) = 40 mm Ans. OD = D + d = 40 + 4 = 44 mm Ans. (c) From Table 10-5, G = 77.2 GPa Eq. (10-9): Table 10-1: 4 3 d 4G 4  77.2 10 Na    11.6 coils 8kD3 8  3.333 403 Nt = Na = 11.6 coils Ans. (d) Table 10-1: Ls = d (Nt + 1) = 4(11.6 + 1) = 50.4 mm (e) Table 10-4: m = 0.187, A = 1855 MPammm Ans. A 1855   1431 MPa d m 40.187 Eq. (10-14): Sut  Table 10-6: Ssy = 0.50 Sut = 0.50(1431) = 715.5 MPa ys = L0  Ls = 80  50.4 = 29.6 mm Fs = k ys = 3.333(29.6) = 98.66 N 4C  2 4(10)  2   1.135 4C  3 4(10)  3 Eq. (10-5): KB  Eq. (10-7):  s  KB Shigley’s MED, 11th edition 8  98.66  40 8 Fs D  1.135  178.2 MPa 3 d   43  Chapter 10 Solutions, Page 4/47 S sy 715.5  4.02 Ans.  s 178.2 ______________________________________________________________________________ ns  10-7  Static service spring with: HD steel wire, d = 0.080 in, OD = 0.880 in, Nt = 8 coils, plain and ground ends. Preliminaries Table 10-5: A = 140 kpsi · inm, m = 0.190 140 A Eq. (10-14): Sut  m   226.2 kpsi d 0.0800.190 Table 10-6: Ssy = 0.45(226.2) = 101.8 kpsi Then, D = OD  d = 0.880  0.080 = 0.8 in Eq. (10-1): C = D/d = 0.8/0.08 = 10 4C  2 4(10)  2 KB  Eq. (10-5):   1.135 4C  3 4(10)  3 Table 10-1: Na = Nt  1 = 8  1 = 7 coils Ls = dNt = 0.08(8) = 0.64 in Eq. (10-7) For solid-safe, ns = 1.2 : 3 3  d 3S sy / ns   0.08  101.8 10  / 1.2  Fs    18.78 lbf 8K B D 8(1.135)(0.8) Eq. (10-9): k  0.084 11.5 106 d 4G   16.43 lbf/in 8D 3 N a 8  0.83  7 ys  Fs 18.78   1.14 in k 16.43 (a) L0 = ys + Ls = 1.14 + 0.64 = 1.78 in Ans. L 1.78 (b) Table 10-1: p  0   0.223 in Ans. Nt 8 (c) From above: Fs = 18.78 lbf Ans. (d) From above: k = 16.43 lbf/in Ans. 2.63D 2.63(0.8) (e) Table 10-2 and Eq. (10-13): ( L0 ) c r    4.21 in  0.5 Since L0 < (L0)cr, buckling is unlikely Ans. ______________________________________________________________________________ 10-8 Given: Design load, F1 = 16.5 lbf. Referring to Prob. 10-7 solution, C = 10, Na = 7 coils, Ssy = 101.8 kpsi, Fs = 18.78 lbf, ys = 1.14 in, L0 = 1.78 in, and (L0)cr = 4.21 in. Eq. (10-18): 4 ≤ C ≤ 12 Shigley’s MED, 11th edition C = 10 O.K. Chapter 10 Solutions, Page 5/47 Eq. (10-19): 3 ≤ Na ≤ 15 Na = 7 O.K. Fs 18.78 1   1  0.14 F1 16.5 Eq. (10-20):   0.15,   0.14 not O.K . , but probably acceptable. From Eq. (10-7) for static service Eq. (10-17):    8F D  8(16.5)(0.8)  74.5 103  psi  74.5 kpsi  1  K B  1 3   1.135 3 d (0.080)     n Eq. (10-21): Ssy 1  101.8  1.37 74.5 ns ≥ 1.2, n = 1.37 O.K.  18.78   18.78    74.5    84.8 kpsi  16.5   16.5  ns  S sy /  s  101.8 / 84.8  1.20  s  1  Eq. (10-21): ns ≥ 1.2, ns = 1.2 It is solid-safe (basis of design). O.K. Eq. (10-13) and Table 10-2: L0 ≤ (L0)cr 1.78 in  4.21 in O.K. ______________________________________________________________________________ 10-9 Given: A228 music wire, squared and ground ends, d = 0.007 in, OD = 0.038 in, L0 = 0.58 in, Nt = 38 coils. Eq. (10-1): Eq. (10-5): Table 10-1: Table 10-5: Eq. (10-9): Table 10-1: Eq. (10-7): Table 10-4: D = OD  d = 0.038  0.007 = 0.031 in C = D/d = 0.031/0.007 = 4.429 4C  2 4  4.429   2   1.340 KB  4C  3 4  4.429   3 Na = Nt  2 = 38  2 = 36 coils (high) G = 12.0 Mpsi 0.007 4 12.0 106 d 4G k   3.358 lbf/in 8D3 N a 8  0.0313  36 Ls = dNt = 0.007(38) = 0.266 in ys = L0  Ls = 0.58  0.266 = 0.314 in Fs = kys = 3.358(0.314) = 1.054 lbf 8 1.054  0.031 8F D  s  K B s 3  1.340  325.1103  psi d   0.0073  (1) A = 201 kpsiinm, m = 0.145 Shigley’s MED, 11th edition Chapter 10 Solutions, Page 6/47 Eq. (10-14): Table 10-6: 201 A   412.7 kpsi m d 0.007 0.145 Ssy = 0.45 Sut = 0.45(412.7) = 185.7 kpsi Sut  s > Ssy, that is, 325.1 > 185.7 kpsi, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving  Ssy / ns   d 3 185.7 103  / 1.2   0.0073  ys   8 K B kD The free length should be wound to 8 1.340  3.358  0.031 L0 = Ls + ys = 0.266 + 0.149 = 0.415 in  0.149 in Ans. This only addresses the solid-safe criteria. There are additional problems. ______________________________________________________________________________ 10-10 Given: B159 phosphor-bronze, squared and ground. ends, d = 0.014 in, OD = 0.128 in, L0 = 0.50 in, Nt = 16 coils. Eq. (10-1): Eq. (10-5): Table 10-1: Table 10-5: Eq. (10-9): Table 10-1: Eq. (10-7): Table 10-4: Eq. (10-14): Table 10-6: D = OD  d = 0.128  0.014 = 0.114 in C = D/d = 0.114/0.014 = 8.143 4C  2 4  8.143   2   1.169 KB  4C  3 4  8.143  3 Na = Nt  2 = 16  2 = 14 coils G = 6 Mpsi 0.014 4  6 106 d 4G k   1.389 lbf/in 8D3 N a 8  0.1143 14 Ls = dNt = 0.014(16) = 0.224 in ys = L0  Ls = 0.50  0.224 = 0.276 in Fs = kys = 1.389(0.276) = 0.3834 lbf 8  0.3834  0.114 8F D  s  K B s 3  1.169  47.42 103  psi 3 d   0.014  (1) A = 145 kpsiinm, m = 0 145 A  145 kpsi Sut  m  d 0.0140 Ssy = 0.35 Sut = 0.35(135) = 47.25 kpsi s > Ssy, that is, 47.42 > 47.25 kpsi, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving  Ssy / ns   d 3 47.25 103  / 1.2   0.0143  ys   8 K B kD The free length should be wound to Shigley’s MED, 11th edition 8 1.169 1.389  0.114   0.229 in Chapter 10 Solutions, Page 7/47 L0 = Ls + ys = 0.224 + 0.229 = 0.453 in Ans. ______________________________________________________________________________ 10-11 Given: A313 stainless steel, squared and ground ends, d = 0.050 in, OD = 0.250 in, L0 = 0.68 in, Nt = 11.2 coils. Eq. (10-1): Eq. (10-5): Table 10-1: Table 10-5: Eq. (10-9): Table 10-1: Eq. (10-7): Table 10-4: Eq. (10-14): Table 10-6: D = OD  d = 0.250  0.050 = 0.200 in C = D/d = 0.200/0.050 = 4 4C  2 4  4   2   1.385 KB  4C  3 4  4   3 Na = Nt  2 = 11.2  2 = 9.2 coils G = 10 Mpsi 0.050 4 10 106 d 4G k   106.1 lbf/in 8D3 N a 8  0.23  9.2 Ls = dNt = 0.050(11.2) = 0.56 in ys = L0  Ls = 0.68  0.56 = 0.12 in Fs = kys = 106.1(0.12) = 12.73 lbf 8 12.73 0.2 8F D  s  K B s 3  1.385  71.8 103  psi d   0.0503  A = 169 kpsiinm, m = 0.146 A 169  261.7 kpsi Sut  m  d 0.0500.146 Ssy = 0.35 Sut = 0.35(261.7) = 91.6 kpsi S sy 91.6  1.28 Spring is solid-safe (ns > 1.2) Ans.  s 71.8 ______________________________________________________________________________ ns   10-12 Given: A227 hard-drawn wire, squared and ground ends, d = 0.148 in, OD = 2.12 in, L0 = 2.5 in, Nt = 5.75 coils. Eq. (10-1): Eq. (10-5): Table 10-1: Table 10-5: Eq. (10-9): Table 10-1: D = OD  d = 2.12  0.148 = 1.972 in C = D/d = 1.972/0.148 = 13.32 (high) 4C  2 4 13.32   2   1.099 KB  4C  3 4 13.32   3 Na = Nt  2 = 5.75  2 = 3.75 coils G = 11.4 Mpsi 0.1484 11.4 106 d 4G k   23.77 lbf/in 8D3 N a 8 1.9723  3.75 Ls = dNt = 0.148(5.75) = 0.851 in ys = L0  Ls = 2.5  0.851 = 1.649 in Fs = kys = 23.77(1.649) = 39.20 lbf Shigley’s MED, 11th edition Chapter 10 Solutions, Page 8/47 8  39.20 1.972 8 Fs D  1.099  66.7 103  psi 3 3 d   0.148  Eq. (10-7):  s  KB Table 10-4: A = 140 kpsiinm, m = 0.190 140 A  201.3 kpsi Sut  m  d 0.1480.190 Ssy = 0.35 Sut = 0.45(201.3) = 90.6 kpsi Eq. (10-14): Table 10-6: S sy 90.6  1.36 Spring is solid-safe (ns > 1.2) Ans.  s 66.7 ______________________________________________________________________________ ns   10-13 Given: A229 OQ&T steel, squared and ground ends, d = 0.138 in, OD = 0.92 in, L0 = 2.86 in, Nt = 12 coils. D = OD  d = 0.92  0.138 = 0.782 in Eq. (10-1): Eq. (10-5): Table 10-1: C = D/d = 0.782/0.138 = 5.667 4C  2 4  5.667   2   1.254 KB  4C  3 4  5.667   3 Na = Nt  2 = 12  2 = 10 coils A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels, G = 11.5 Mpsi. 0.1384 11.5 106 d 4G Eq. (10-9): k   109.0 lbf/in 8D3 N a 8  0.7823 10 Table 10-1: Eq. (10-7): Table 10-4: Eq. (10-14): Table 10-6: Ls = dNt = 0.138(12) = 1.656 in ys = L0  Ls = 2.86  1.656 = 1.204 in Fs = kys = 109.0(1.204) = 131.2 lbf 8 131.2  0.782 8F D  s  K B s 3  1.254  124.7 103  psi d   0.1383  (1) A = 147 kpsiinm, m = 0.187 A 147  212.9 kpsi Sut  m  d 0.1380.187 Ssy = 0.50 Sut = 0.50(212.9) = 106.5 kpsi s > Ssy, that is, 124.7 > 106.5 kpsi, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving  Ssy / ns   d 3 106.5 103  / 1.2   0.1383  ys  8 K B kD  8 1.254 109.0  0.782   0.857 in The free length should be wound to Shigley’s MED, 11th edition Chapter 10 Solutions, Page 9/47 L0 = Ls + ys = 1.656 + 0.857 = 2.51 in Ans. ______________________________________________________________________________ 10-14 Given: A232 chrome-vanadium steel, squared and ground ends, d = 0.185 in, OD = 2.75 in, L0 = 7.5 in, Nt = 8 coils. Eq. (10-1): Eq. (10-5): D = OD  d = 2.75  0.185 = 2.565 in C = D/d = 2.565/0.185 = 13.86 (high)  4 13.86 2    1.095 4C  2  KB  4C  3 4 13.86   3 Table 10-1: Na = Nt  2 = 8  2 = 6 coils Table 10-5: G = 11.2 Mpsi. 0.1854 11.2 106 d 4G k   16.20 lbf/in 8D3 N a 8  2.5653  6 Eq. (10-9): Table 10-1: Eq. (10-7): Ls = dNt = 0.185(8) = 1.48 in ys = L0  Ls = 7.5  1.48 = 6.02 in Fs = kys = 16.20(6.02) = 97.5 lbf 8  97.5  2.565 8F D  s  K B s 3  1.095  110.1 103  psi d   0.1853  (1) A = 169 kpsiinm, m = 0.168 A 169 Eq. (10-14): Sut  m   224.4 kpsi d 0.1850.168 Table 10-6: Ssy = 0.50 Sut = 0.50(224.4) = 112.2 kpsi S sy 112.2   1.02 Spring is not solid-safe (ns < 1.2) ns   s 110.1 Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving 3 3 S sy / ns   d 3 112.2 10  / 1.2    0.185     5.109 in ys  8 K B kD 8 1.095 16.20  2.565  The free length should be wound to Table 10-4: L0 = Ls + ys = 1.48 + 5.109 = 6.59 in Ans. ______________________________________________________________________________ 10-15 Given: A313 stainless steel, squared and ground ends, d = 0.25 mm, OD = 0.95 mm, L0 = 12.1 mm, Nt = 38 coils. D = OD  d = 0.95  0.25 = 0.7 mm Eq. (10-1): C = D/d = 0.7/0.25 = 2.8 (low) 4C  2 4  2.8   2 Eq. (10-5):   1.610 KB  4C  3 4  2.8   3 Shigley’s MED, 11th edition Chapter 10 Solutions, Page 10/47 Table 10-1: Na = Nt  2 = 38  2 = 36 coils Table 10-5: G = 69.0(103) MPa. 0.254  69.0 103 d 4G k   2.728 N/mm 8D3 N a 8  0.7 3  36 Eq. (10-9): Table 10-1: Eq. (10-7): (high) Ls = dNt = 0.25(38) = 9.5 mm ys = L0  Ls = 12.1  9.5 = 2.6 mm Fs = kys = 2.728(2.6) = 7.093 N 8  7.093 0.7 8F D  s  K B s 3  1.610  1303 MPa d   0.253  (1) Table 10-4 (dia. less than table): A = 1867 MPammm, m = 0.146 1867 A Eq. (10-14): Sut  m   2286 MPa d 0.250.146 Table 10-6: Ssy = 0.35 Sut = 0.35(2286) = 734 MPa s > Ssy, that is, 1303 > 734 MPa, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving  Ssy / ns   d 3  734 / 1.2    0.253    1.22 mm 8 K B kD 8 1.610  2.728  0.7  The free length should be wound to ys  L0 = Ls + ys = 9.5 + 1.22 = 10.72 mm Ans. This only addresses the solid-safe criteria. There are additional problems. ______________________________________________________________________________ 10-16 Given: A228 music wire, squared and ground ends, d = 1.2 mm, OD = 6.5 mm, L0 = 15.7 mm, Nt = 10.2 coils. D = OD  d = 6.5  1.2 = 5.3 mm Eq. (10-1): C = D/d = 5.3/1.2 = 4.417 4C  2 4  4.417   2 Eq. (10-5):   1.368 KB  4C  3 4  4.417   3 Table (10-1): Na = Nt  2 = 10.2  2 = 8.2 coils Table 10-5 (d = 1.2/25.4 = 0.0472 in): G = 81.7(103) MPa. 1.24  81.7 103 d 4G Eq. (10-9): k   17.35 N/mm 8D3 N a 8  5.33  8.2 Table 10-1: Ls = dNt = 1.2(10.2) = 12.24 mm ys = L0  Ls = 15.7  12.24 = 3.46 mm Fs = kys = 17.35(3.46) = 60.03 N Shigley’s MED, 11th edition Chapter 10 Solutions, Page 11/47 8  60.03 5.3 8 Fs D  1.368  641.4 MPa 3 d  1.23  Eq. (10-7):  s  KB Table 10-4: A = 2211 MPammm, m = 0.145 2211 A Sut  m  0.145  2153 MPa d 1.2 Ssy = 0.45 Sut = 0.45(2153) = 969 MPa Eq. (10-14): Table 10-6: S sy (1) 969  1.51 Spring is solid-safe (ns > 1.2) Ans.  s 641.4 ______________________________________________________________________________ ns   10-17 Given: A229 OQ&T steel, squared and ground ends, d = 3.5 mm, OD = 50.6 mm, L0 = 75.5 mm, Nt = 5.5 coils. Eq. (10-1): Eq. (10-5): Table 10-1: D = OD  d = 50.6  3.5 = 47.1 mm C = D/d = 47.1/3.5 = 13.46 (high) 4C  2 4 13.46   2   1.098 KB  4C  3 4 13.46   3 Na = Nt  2 = 5.5  2 = 3.5 coils A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels, G = 79.3(103) MPa. 3.54  79.3103 d 4G Eq. (10-9): k   4.067 N/mm 8D3 N a 8  47.13  3.5 Table 10-1: Eq. (10-7): Ls = dNt = 3.5(5.5) = 19.25 mm ys = L0  Ls = 75.5  19.25 = 56.25 mm Fs = kys = 4.067(56.25) = 228.8 N 8  228.8  47.1 8F D  s  K B s 3  1.098  702.8 MPa d   3.53  (1) A = 1855 MPammm, m = 0.187 A 1855 Eq. (10-14): Sut  m   1468 MPa d 3.50.187 Table 10-6: Ssy = 0.50 Sut = 0.50(1468) = 734 MPa S sy 734 ns    1.04 Spring is not solid-safe (ns < 1.2)  s 702.8 Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving S sy / ns   d 3  734 / 1.2    3.53     48.96 mm ys  8 K B kD 8 1.098  4.067  47.1 The free length should be wound to Table 10-4: Shigley’s MED, 11th edition Chapter 10 Solutions, Page 12/47 L0 = Ls + ys = 19.25 + 48.96 = 68.2 mm Ans. ______________________________________________________________________________ 10-18 Given: B159 phosphor-bronze, squared and ground ends, d = 3.8 mm, OD = 31.4 mm, L0 = 71.4 mm, Nt = 12.8 coils. Eq. (10-1): Eq. (10-5): D = OD  d = 31.4  3.8 = 27.6 mm C = D/d = 27.6/3.8 = 7.263 4C  2 4  7.263   2   1.192 KB  4C  3 4  7.263  3 Table 10-1: Na = Nt  2 = 12.8  2 = 10.8 coils Table 10-5: G = 41.4(103) MPa. 3.84  41.4 103 d 4G k   4.752 N/mm 8 D 3 N a 8  27.63 10.8 Eq. (10-9): Table 10-1: Eq. (10-7): Ls = dNt = 3.8(12.8) = 48.64 mm ys = L0  Ls = 71.4  48.64 = 22.76 mm Fs = kys = 4.752(22.76) = 108.2 N 8 108.2  27.6 8F D  s  K B s 3  1.192  165.2 MPa d   3.83  (1) Table 10-4 (d = 3.8/25.4 = 0.150 in): A = 932 MPammm, m = 0.064 A 932 Eq. (10-14): Sut  m   855.7 MPa d 3.80.064 Table 10-6: Ssy = 0.35 Sut = 0.35(855.7) = 299.5 MPa S sy 299.5   1.81 Spring is solid-safe (ns > 1.2) Ans. ns   s 165.2 ______________________________________________________________________________ 10-19 Given: A232 chrome-vanadium steel, squared and ground ends, d = 4.5 mm, OD = 69.2 mm, L0 = 215.6 mm, Nt = 8.2 coils. Eq. (10-1): Eq. (10-5): D = OD  d = 69.2  4.5 = 64.7 mm C = D/d = 64.7/4.5 = 14.38 (high) 4C  2 4 14.38   2   1.092 KB  4C  3 4 14.38   3 Table 10-1: Na = Nt  2 = 8.2  2 = 6.2 coils Table 10-5: G = 77.2(103) MPa. 4.54  77.2 103 d 4G k   2.357 N/mm 8D3 N a 8  64.73  6.2 Eq. (10-9): Table 10-1: Ls = dNt = 4.5(8.2) = 36.9 mm Shigley’s MED, 11th edition Chapter 10 Solutions, Page 13/47 Eq. (10-7): Table 10-4: Eq. (10-14): Table 10-6: ys = L0  Ls = 215.6  36.9 = 178.7 mm Fs = kys = 2.357(178.7) = 421.2 N 8  421.2  64.7 8F D  s  K B s 3  1.092  832 MPa d   4.53  (1) A = 2005 MPammm, m = 0.168 2005 A  1557 MPa Sut  m  d 4.50.168 Ssy = 0.50 Sut = 0.50(1557) = 779 MPa s > Ssy, that is, 832 > 779 MPa, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving  Ssy / ns   d 3  779 / 1.2    4.53    139.5 mm 8 K B kD 8 1.092  2.357  64.7  The free length should be wound to ys  L0 = Ls + ys = 36.9 + 139.5 = 176.4 mm Ans. This only addresses the solid-safe criteria. There are additional problems. ______________________________________________________________________________ 10-20 Given: A227 HD steel. From the figure: L0 = 4.75 in, OD = 2 in, and d = 0.135 in. Thus D = OD  d = 2  0.135 = 1.865 in (a) By counting, Nt = 12.5 coils. Since the ends are squared along 1/4 turn on each end, N a  12.5  0.5  12 turns Ans. p  4.75 / 12  0.396 in Ans. The solid stack is 13 wire diameters Ls = 13(0.135) = 1.755 in Ans. (b) From Table 10-5, G = 11.4 Mpsi 0.1354 (11.4) 106  d 4G k    6.08 lbf/in 8D 3 N a 8 1.8653  (12) (c) Fs = k(L0 - Ls ) = 6.08(4.75  1.755) = 18.2 lbf Ans. Ans. (d) C = D/d = 1.865/0.135 = 13.81 Shigley’s MED, 11th edition Chapter 10 Solutions, Page 14/47 4(13.81)  2  1.096 4(13.81)  3 8F D 8(18.2)(1.865)  38.5 103  psi  38.5 kpsi  s  K B s 3  1.096 3 d   0.135  KB  Ans. ______________________________________________________________________________ 10-21 Given: Plain end, hard drawn steel, 12 gauge W & M wire, OD = 0.75 in, Nt = 20 coils, L0 = 3.75 in. Table A-28: d = 0.1055 in (a) D = OD  d = 0.75  0.1055 = 0.6445 in. C = D / d = 0.6445/0.1055 = 6.109 Ans. (b) Table 10-1: Na = Nt = 20 coils, p = (L0  d)/ Na = (3.75  0.1055)/20 = 0.1822 in/coil Ans. (c) Table 10-1: Ls = d (Nt + 1) = 0.1055 (20 + 1) = 2.2155 in, ys = L0  Ls = 3.75  2.2155 = 1.5345 in Ans. (d) Eq. (10-8): 0.10554 11.5 106 1.5345  d 4Gys   50.36 lbf Fs  Ans. 1     3 1 8D N 1  2  8  0.6445 3  20  1  2  2C   2  6.109   (e) Eq. (10-5): K B  Eq. (10-7):  s  KB 4C  2 4  6.109   2   1.233 4C  3 4  6.109   3 8  50.36  0.6445 8 Fs D  1.233  86.8 103  psi  86.8 kpsi 3 3 d   0.1055  (f) Table 10-4 and Eq. (10-14): Sut  Ans. 140 A   214.6 kpsi 0.190 m d  0.1055  Table 10-6: Ssy = 0.45 Sut = 0.45(214.6) = 96.57 kpsi. S sy 96.57 ns    1.11 Ans. s 86.8 (g) Exact, k = Fs / ys = 50.36/1.5345 = 32.82 lbf/in Ans. 4 4 0.1055 11.5 106 d G Approximate using Eq. (10-9): k    33.26 lbf/in 3 8D3 N 8  0.6445  20 Ans. Approximate is 1.34 percent higher than the exact. Ans. ______________________________________________________________________________ 10-22 Given: Squared and ground, oil tempered steel, d = 3 mm, OD = 30 mm, Nt = 32 coils, L0 = 240 mm. Table A-28: d = 0.1055 in (a) D = OD  d = 30  3 = 27 mm. C = D / d = 27/3 = 9 Ans. (b) Table 10-1: Na = Nt  2 = 32  2 = 30 coils, p = (L0  2d)/ Na = [240  2(3)] / 30 = 7.8 mm/coil Ans. (c) Table 10-1: Ls = d Nt = 3 (32) = 96 mm, ys = L0  Ls = 240  96 = 144 mm Ans. Shigley’s MED, 11th edition Chapter 10 Solutions, Page 15/47 (d) Eq. (10-8):  0.003 77.2 109  0.144 d 4Gys   189.45 N Fs  Ans. 1     3 1 3 8D N 1  2  8  0.027   30  1  2  2C   2  9   4C  2 4  9   2 (e) Eq. (10-5): K B    1.152 4C  3 4  9   3 Eq. (10-7): 8 189.45  0.027 8F D  s  K B s 3  1.152 106   555.8 MPa Ans.  3 d   0.003 (f) Table 10-4 and Eq. (10-14): Sut  4 1855 A   1510.5 MPa 0.187 m d  3 Table 10-5: Ssy = 0.45 Sut = 0.45(1510.5) = 679.7 MPa. S sy 679.7 ns    1.22 Ans.  s 555.8 (g) Exact, k = Fs / ys = 189.45/144 = 1.316 N/mm Ans. Approximate using Eq. (10-9): 4  0.003 77.2 109  3 d 4G  k 10   1.324 N/mm 3 8D3 N 8  0.027  30 Ans. Approximate is 0.61 percent higher than the exact. Ans. ______________________________________________________________________________ 10-23 y = 50 mm, F = 90 N, k = F / y = 90/50 = 1.8 N/mm Ans. ys = 60 mm, Fs = k ys = 1.8(60) = 108 N. A 2065 Sut  m  0.263 Eq. (10-14), Table 10-4, assume 2.5  d  5 mm: d d 2065 722.75 (1) Table 10-6 (includes KB): S ys  0.35Sut  0.35 0.263  0.263 d d Eq. (10-7) (with ns = 1.2 but without KB): 8n F D 8n F C 8 1.2 108 10  3300  max  s s3  s s2   2 MPa (2) d d d d2 722.75 3300 3300  d 1.737   d  2.40 mm Equate Eqs. (1) and (2), 0.263  2 d d 722.75 Since this is less than 2.5 mm, return to Eq. (10-14), Table 10-4, for 0.3  d  2.5 mm: 653.45 A 1867  (1) Sut  m  0.146 S ys  0.146 d d d Again, equate Eqs. (1) and (2), 653.45 3300 3300  2  d 1.854   d  2.40 mm Ans. 0.146 d d 653.45 The final factor of safety is Shigley’s MED, 11th edition Chapter 10 Solutions, Page 16/47 653.45 / d 0.146 653.45  2.40  ns     1.20 Ans. s 8 108 10  8 Fs C   2   d  OD =Cd + d = (C + 1) d = 11(2.40) = 26.4 mm Ans. ID = (C  1) d = 9(2.40) = 21.6 mm Ans. k = 1.8 N/mm Ans. (found earlier) Table 10-5, G = 69.0 GPa, Eq. (10-9): 2.4 103  69 109  d 4G dG Na     11.5 coils 8 D3k 8C 3 k 8 103 1.8 103  1.854 S ys Table 10-1: Nt = Na + 2 = 13.5 coils Ans. Ls = d (Nt + 1) = 2.4(13.5 + 1) = 34.8 mm Ans. L0 = Ls + ys = 34.8 + 60 = 94.8 mm Ans. 10  2.4  D Eq. (10-13):   2.63  2.63  0.666 L0 94.8 Stable if supported between fixed-fixed ends. Otherwise would need to be supported by hole or rod. ______________________________________________________________________________ 10-24 Phosphor-bronze, closed ends, C = 10, at y = 2 in F = 15 lbf, ys = 3 in, ns = 1.2. k = F / y = 15/2 = 7.5 lbf/in, Fs = k ys = 7.5 (3) = 22.5 lbf. Eq. (10-14) with Table 10-4 assuming 0.022  d  0.075 in, A = 121 kpsi٠inm and m = 0.028. Then, from Table 10-5, Sys = 0.45 Sut: 121 54.45 (1) S ys  0.45 0.028  0.028 d d 4C  2 4 10   2   1.135 4C  3 4 10   3 Eq. (10-5): K B  Eq. (10-7) with ns = 1.2: 8  22.5 10 8F C 0.7804  max  ns K B s 2  1.2 1.135  103    2 d d d2 (2) Where max is in kpsi. Equating (1) and (2) gives 54.45 0.7804  d 0.028 d2  d 1.972  0.7804 54.45  d  0.116 in Returning to Table 10-4, use A =110 kpsi٠inm and m = 0.028 for 0.075  d  0.3 in, S ys  0.45 110 49.5 0.7804 0.7804  0.064   d 1.936   d  0.117 in 0.064 2 d d d 49.5 Table A-17, select the preferred size of: Shigley’s MED, 11th edition d = 0.12 in Ans. Chapter 10 Solutions, Page 17/47 Check ns, ns  S ys s  49.5 /  d 0.064   FC  K B  8 s 2 103   d   49.5 103   d 1.936 49.5 103    0.12  (3) 8 K B Fs C 1.936 ns  8 1.135  22.5 10   1.26 Ans. OD = Cd + d = (C + 1) d = (10 + 1)0.12 = 1.32 in Ans. ID = (C  1) d = (10  1)0.12 = 1.08 in Found earlier, k = 7.5 lbf/in Ans. Table 10-5, G = 6 Mpsi. Eq. (10-9): N a  Table 10-1: 0.12  6 106 dG   12 coils 8C 3 k 8 103  7.5 Nt = Na + 2 = 14 coils Ans. Ls = d (Nt + 1) = 0.12(14 + 1) = 1.8 in L0 = Ls + ys = 1.8 + 3 = 4.8 in Ans. Ans. Eq. (10-13):  = 2.63 D / L0 = 2.63 (10)0.12/4.8 = 0.658 Table 10-2: Stable if supported between fixed-fixed ends. Otherwise would need to be supported by hole or rod. ______________________________________________________________________________ 10-25 From Prob. 10-24, d = 0.12 in and from Eq. (3), 1.936 49.5 103   d 1.936 49.5 103    0.12  C   10.46 Ans. 8ns K B Fs 8 1.2 1.135  22.5  OD = (C + 1) d = 11.46(0.12) = 1.375 in Ans. ID = (C  1) d = 9.46(0.12) = 1.135 in Ans. Table 10-5, G = 6 Mpsi, Eq. (10-9): N a  Table 10-1: 0.12  6 106 dG   10.5 coils 8C 3 k 8 10.463  7.5 Nt = Na + 2 = 12.5 coils Ans. Ls = d (Nt + 1) = 0.12(12.5 + 1) = 1.62 in Ans. L0 = Ls + ys = 1.62 + 3 = 4.62 in Ans. Eq. (10-13):  = 2.63 D / L0 = 2.63 (10.46)0.12/4.62 = 0.715 Shigley’s MED, 11th edition Chapter 10 Solutions, Page 18/47 Table 10-2: Stable if supported between fixed-fixed ends, or one end on flat surface and other end hinged. Otherwise would need to be supported by hole or rod. ______________________________________________________________________________ 10-26 For the wire diameter analyzed, G = 11.75 Mpsi per Table 10-5. Use squared and ground ends. The following is a spread-sheet study using Fig. 10-3 for parts (a) and (b). For Na, k = Fmax /y = 20/2 = 10 lbf/in. For s, F = Fs = 20(1 + ) = 20(1 + 0.15) = 23 lbf. Source Eq. (10-1) Eq. (10-9) Table 10-1 Table 10-1 1.15y + Ls Eq. (10-13) Table 10-4 Table 10-4 Eq. (10-14) Table 10-6 Eq. (10-5) Eq. (10-7) Eq. (10-3) Eq. (10-22) (a) Spring over a Rod (b) Spring in a Hole Parameter Values Source Parameter Values d 0.075 0.080 0.085 d 0.075 0.080 ID 0.800 0.800 0.800 OD 0.950 0.950 D 0.875 0.880 0.885 D 0.875 0.870 C 11.667 11.000 10.412 Eq. (10-1) C 11.667 10.875 Na 6.937 8.828 11.061 Eq. (10-9) Na 6.937 9.136 Nt 8.937 10.828 13.061 Table 10-1 Nt 8.937 11.136 Ls 0.670 0.866 1.110 Table 10-1 Ls 0.670 0.891 L0 2.970 3.166 3.410 1.15y + Ls L0 2.970 3.191 (L0)cr 4.603 4.629 4.655 Eq. (10-13) (L0)cr 4.603 4.576 A 201.000 201.000 201.000 Table 10-4 A 201.000 201.000 m 0.145 0.145 0.145 Table 10-4 m 0.145 0.145 Sut 292.626 289.900 287.363 Eq. (10-14) Sut 292.626 289.900 Ssy 131.681 130.455 129.313 Table 10-6 Ssy 131.681 130.455 KB 1.115 1.122 1.129 Eq. (10-5) KB 1.115 1.123 135.335 112.948 95.293 Eq. (10-7) 135.335 111.787 s s ns 0.973 1.155 1.357 Eq. (10-3) ns 0.973 1.167 fom 0.282 0.391 0.536 Eq. (10-22) fom 0.282 0.398 For ns ≥ 1.2, the optimal size is d = 0.085 in for both cases. 0.085 0.950 0.865 10.176 11.846 13.846 1.177 3.477 4.550 201.000 0.145 287.363 129.313 1.133 93.434 1.384 0.555 ______________________________________________________________________________ 10-27 In Prob. 10-26, there is an advantage of first selecting d as one can select from the available sizes (Table A-28). Selecting C first requires a calculation of d where then a size must be selected from Table A-28. Consider part (a) of the problem. It is required that ID = D  d = 0.800 in. (1) From Eq. (10-1), D = Cd. Substituting this into the first equation yields d 0.800 C 1 (2) Starting with C = 10, from Eq. (2) we find that d = 0.089 in. From Table A-28, the closest diameter is d = 0.090 in. Substituting this back into Eq. (1) gives D = 0.890 in, with C = 0.890/0.090 = 9.889, which are acceptable. From this point the solution is the same as Prob. 10-26. For part (b), use OD = D + d = 0.950 in. (3) Shigley’s MED, 11th edition Chapter 10 Solutions, Page 19/47 and, d 0.800 C 1 (a) Spring over a rod Parameter Values C 10.000 10.5 Eq. (2) d 0.089 0.084 Table A-28 d 0.090 0.085 Eq. (1) D 0.890 0.885 Eq. (10-1) C 9.889 10.412 Eq. (10-9) Na 13.669 11.061 Table 10-1 Nt 15.669 13.061 Table 10-1 Ls 1.410 1.110 1.15y + Ls L0 3.710 3.410 Eq. (10-13) (L0)cr 4.681 4.655 Table 10-4 A 201.000 201.000 Table 10-4 m 0.145 0.145 Eq. (10-14) Sut 284.991 287.363 Table 10-6 Ssy 128.246 129.313 Eq. (10-5) KB 1.135 1.128 Eq. (10-7) s 81.167 95.223 ns = Ssy/s ns 1.580 1.358 Eq. (10-22) fom -0.725 -0.536 Source (4) (b) Spring in a Hole Source Parameter Values C 10.000 Eq. (4) d 0.086 Table A-28 d 0.085 Eq. (3) D 0.865 Eq. (10-1) C 10.176 Eq. (10-9) Na 11.846 Table 10-1 Nt 13.846 Table 10-1 Ls 1.177 1.15y + Ls L0 3.477 Eq. (10-13) (L0)cr 4.550 Table 10-4 A 201.000 Table 10-4 m 0.145 Eq. (10-14) Sut 287.363 Table 10-6 Ssy 129.313 Eq. (10-5) KB 1.135 Eq. (10-7) s 93.643 ns = Ssy/s ns 1.381 Eq. (10-22) fom -0.555 Again, for ns  1.2, the optimal size is = 0.085 in. Although this approach used less iterations than in Prob. 10-26, this was due to the initial values picked and not the approach. ______________________________________________________________________________ 10-28 One approach is to select A227 HD steel for its low cost. Try L0 = 48 mm, then for y = 48  37.5 = 10.5 mm when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286 N/mm. For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm. Starting with d = 2 mm, D = ID + d = 11.25 + 2 = 13.25 mm C = D/d = 13.25/2 = 6.625 (acceptable) Table 10-5 (d = 2/25.4 = 0.0787 in): G = 79.3 GPa 24 (79.3)103 d 4G Na    15.9 coils Eq. (10-9): 8kD 3 8(4.286)13.253 Assume squared and closed. Table 10-1: Nt = Na + 2 = 15.9 + 2 = 17.9 coils Ls = dNt = 2(17.9) =35.8 mm Shigley’s MED, 11th edition Chapter 10 Solutions, Page 20/47 Eq. (10-5): ys = L0  Ls = 48  35.8 = 12.2 mm Fs = kys = 4.286(12.2) = 52.29 N 4C  2 4  6.625   2   1.213 KB  4C  3 4  6.625   3  8(52.29)13.25  8Fs D 1.213     267.5 MPa d3   23    Eq. (10-7):  s  KB Table 10-4: A = 1783 MPa · mmm, m = 0.190 A 1783 Sut  m  0.190  1563 MPa d 2 Ssy = 0.45Sut = 0.45(1563) = 703.3 MPa Eq. (10-14): Table 10-6: ns  S sy s  703.3  2.63 1.2 267.5 O.K . No other diameters in the given range work. So specify A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L0 = 48 mm. Ans. ______________________________________________________________________________ 10-29 Select A227 HD steel for its low cost. Try L0 = 48 mm, then for y = 48  37.5 = 10.5 mm when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286 N/mm. For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm. and, D  d = 11.25 (1) D =Cd (2) Starting with C = 8, gives D = 8d. Substitute into Eq. (1) resulting in d = 1.607 mm. Selecting the nearest diameter in the given range, d = 1.6 mm. From this point, the calculations are shown in the third column of the spreadsheet output shown. We see that for d = 1.6 mm, the spring is not solid safe. Iterating on C we find that C = 6.5 provides acceptable results with the specifications A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L0 = 48 mm. Ans. Shigley’s MED, 11th edition Chapter 10 Solutions, Page 21/47 Source Eq. (2) Table A-28 Eq. (1) Eq. (10-1) Eq. (10-9) Table 10-1 Table 10-1 L0 Ls Fs = kys Table 10-4 Table 10-4 Eq. (10-14) Table 10-6 Eq. (10-5) Eq. (10-7) ns = Ssy/s C d d D C Na Nt Ls ys Fs A m Sut Ssy KB s ns Parameter Values 8.000 7 6.500 1.607 1.875 2.045 1.600 1.800 2.000 12.850 13.050 13.250 8.031 7.250 6.625 7.206 10.924 15.908 9.206 12.924 17.908 14.730 23.264 35.815 33.270 24.736 12.185 142.594 106.020 52.224 1783.000 1783.000 1783.000 0.190 0.190 0.190 1630.679 1594.592 1562.988 733.806 717.566 703.345 1.172 1.200 1.217 1335.568 724.943 268.145 0.549 0.990 2.623 The only difference between selecting C first rather than d as was done in Prob. 10-28, is that once d is calculated, the closest wire size must be selected. Iterating on d uses available wire sizes from the beginning. ______________________________________________________________________________ 10-30 A stock spring catalog may have over two hundred pages of compression springs with up to 80 springs per page listed. • Students should be made aware that such catalogs exist. • Many springs are selected from catalogs rather than designed. • The wire size you want may not be listed. • Catalogs may also be available on disk or the web through search routines. • It is better to familiarize yourself with vendor resources rather than invent them yourself. • Sample catalog pages can be given to students for study. ______________________________________________________________________________ 10-31 Given: ID = 0.6 in, C = 10, L0 = 5 in, Ls = 5  3 = 2 in, sq. & grd ends, unpeened, HD A227 wire. (a) With ID = D  d = 0.6 in and C = D/d = 10 10 d  d = 0.6  d = 0.0667 in Ans., and D = 0.667 in. (b) Table 10-1: Ls = dNt = 2 in  Nt = 2/0.0667 = 30 coils Ans. (c) Table 10-1: Table 10-5: Eq. (10-9): Shigley’s MED, 11th edition Na = Nt  2 = 30  2 = 28 coils G = 11.5 Mpsi 0.0667 4 11.5 106 d 4G   3.424 lbf/in k 8D3 N a 8  0.667 3  28 Ans. Chapter 10 Solutions, Page 22/47 (d) Table 10-4: Eq. (10-14): Table 10-6: Eq. (10-5): A = 140 kpsiinm, m = 0.190 140 A  234.2 kpsi Sut  m  d 0.0667 0.190 Ssy = 0.45 Sut = 0.45 (234.2) = 105.4 kpsi Fs = kys = 3.424(3) = 10.27 lbf 4C  2 4 10   2   1.135 KB  4C  3 4 10   3 Eq. (10-7):  s  KB 8 10.27  0.667 8Fs D 1.135  d3   0.06673   66.72 103  psi  66.72 kpsi S sy 105.4  1.58 Ans.  s 66.72 (e) a = m = 0.5s = 0.5(66.72) = 33.36 kpsi, r = a / m = 1. Using the Gerber fatigue failure criterion with Zimmerli data, ns  Eq. (10-30):  Ssu = 0.67 Sut = 0.67(234.2) = 156.9 kpsi The Gerber ordinate intercept for the Zimmerli data is obtained using Eqs. (10-28) and (10-29b). S sa 35 S se    39.9 kpsi 2 2 1   S sm / S su  1   55 /156.9  The Gerber fatigue criterion from Eq. (6-48), adapted for shear, 2 2   2 m S se   1  Ssu   a  1  1   nf     2   m  S se  Ssu a      2  2  2  33.36  39.9   1  156.9  33.36    1  1     2  33.36  39.9  156.9  33.36        1.13 Ans. ______________________________________________________________________________ 10-32 Given: OD  0.9 in, C = 8, L0 = 3 in, Ls = 1 in, ys = 3  1 = 2 in, sq. ends, unpeened, music wire. (a) Try OD = D + d = 0.9 in, C = D/d = 8  D = 8d  9d = 0.9  d = 0.1 Ans. D = 8(0.1) = 0.8 in (b) Table 10-1: Ls = d (Nt + 1)  Nt = Ls / d  1 = 1/0.1 1 = 9 coils Ans. Table 10-1: Shigley’s MED, 11th edition Na = Nt  2 = 9  2 = 7 coils Chapter 10 Solutions, Page 23/47 (c) Table 10-5: Eq. (10-9): (d) G = 11.75 Mpsi k 0.14 11.75 10 6 d 4G   40.98 lbf/in 8D3 N a 8  0.83  7 Ans. Fs = kys = 40.98(2) = 81.96 lbf 4C  2 4  8   2   1.172 4C  3 4  8   3 Eq. (10-5): KB  Eq. (10-7):  s  KB Table 10-4: A = 201 kpsiinm, m = 0.145 Eq. (10-14): Sut  Table 10-6: Ssy = 0.45 Sut = 0.45(280.7) = 126.3 kpsi ns  8  81.96  0.8 8 Fs D  1.172  195.7 103  psi  195.7 kpsi 3 3 d   0.1  201 A  0.145  280.7 kpsi m d 0.1 S sy s  126.3  0.645 195.7 Ans. (e) a = m = s /2 = 195.7/2 = 97.85 kpsi. Using the Gerber fatigue failure criterion with Zimmerli data, Eq. (10-30): Ssu = 0.67 Sut = 0.67(280.7) = 188.1 kpsi The Gerber ordinate intercept for the Zimmerli data is obtained using Eqs. (10-28) and (10-29b). S sa 35 S se    38.3 kpsi 2 2 1   S sm / S su  1   55 / 188.1 The Gerber fatigue criterion from Eq. (6-48), adapted for shear, 2 2   2 m S se   1  S su   a  nf   1  1     S su a   2   m  S se     2  2  2  97.85  38.3   1  188.1  97.85    1  1     2  97.85  38.3  188.1 97.85        0.38 Ans. Obviously, the spring is severely under designed and will fail statically and in fatigue. Increasing C would improve matters. Try C = 12. This yields ns = 1.83 and nf = 1.00. Shigley’s MED, 11th edition Chapter 10 Solutions, Page 24/47 ______________________________________________________________________________ 10-33 Given: Fmax = 300 lbf, Fmin = 150 lbf, y = 1 in, OD = 2.1  0.2 = 1.9 in, C = 7, unpeened, squared & ground, oil-tempered wire. (a) D = OD  d = 1.9  d (1) C = D/d = 7  D = 7d (2) Substitute Eq. (2) into (1) 7d = 1.9  d  d = 1.9/8 = 0.2375 in (b) From Eq. (2): D = 7d = 7(0.2375) = 1.663 in (c) k (d) Table 10-5: G = 11.6 Mpsi Ans. F 300  150   150 lbf/in y 1 Ans. 4 6 d 4G 0.2375 11.6 10   6.69 coils 8D 3k 8 1.6633 150 Eq. (10-9): Na  Table 10-1: Nt = Na + 2 = 8.69 coils Ans. Table 10-6: A = 147 kpsiinm, m = 0.187 147 A  192.3 kpsi Sut  m  d 0.23750.187 Ssy = 0.5 Sut = 0.5(192.3) = 96.15 kpsi Eq. (10-5): KB  Eq. (10-7):  s  KB (e) Table 10-4: Eq. (10-14): Fs  Ans. 4C  2 4  7   2   1.2 4C  3 4  7   3 8 Fs D  S sy d3  d 3 S sy 8K B D    0.23753  96.15 103  8 1.2 1.663  253.5 lbf ys = Fs / k = 253.5/150 = 1.69 in Table 10-1: Shigley’s MED, 11th edition Ls = Nt d = 8.46(0.2375) = 2.01 in Chapter 10 Solutions, Page 25/47 L0 = Ls + ys = 2.01 + 1.69 = 3.70 in Ans. ______________________________________________________________________________ 10-34 For a coil radius given by: R  R1  R2  R1  2 N The torsion of a section is T = PR where dL = R d U T 1 1 2 N 3  T dL  PR d  P P GJ GJ 0 3 P 2 N  R2  R1     d  R1  2 N GJ 0   P  P  GJ  2 N 4 R2  R1    1   2 N       R1    2 N  4   R2  R1     0  PN 2GJ ( R2 R  R) 4 2  R14    PN ( R1  R2 )  R12  R22  2GJ PN 16 J  d4  p  ( R1  R2 )  R12  R22  4 Gd 32 P d 4G  k  Ans.  P 16 N ( R1  R2 )  R12  R22   1 ______________________________________________________________________________ 10-35 Given: Fmin = 4 lbf, Fmax = 18 lbf, k = 9.5 lbf/in, OD  2.5 in, nf = 1.5. For a food service machinery application select A313 Stainless wire. Table 10-5: G = 10(106) psi Note that for 0.013 ≤ d ≤ 0.10 in A = 169, m = 0.146 0.10 < d ≤ 0.20 in A = 128, m = 0.263 18  4 18  4 Fa   7 lbf , Fm   11 lbf , r  7 / 11 2 2 169 Try, d  0.080 in, Sut   244.4 kpsi (0.08)0.146 Ssu = 0.67Sut = 163.7 kpsi, Ssy = 0.35Sut = 85.5 kpsi The Gerber ordinate intercept for the Zimmerli data is obtained using Eqs. (10-28) and (10-29b). S se  S sa 35   39.5 kpsi 2 1  (S sm / S su ) 1  (55 / 163.7) 2 Shigley’s MED, 11th edition Chapter 10 Solutions, Page 26/47 Let r = a /m = 7/11. The Gerber fatigue criterion from Eq. (6-48), adapted for shear, 2   2S se   1 S su2 r  1  1   nf   2  m S se  S su r      Solving for m gives, 2 2 2    2  39.5     2 S se   1 163.7   7 / 11  1 S su2 r  m  1  1   1  1      S su r   2 1.5  39.5  2 n f S se  163.7  7 / 11          36.70 kpsi But, 8 F C 4C  2  8 FmC  m  KB m2    4C  3   d 2  d 8 11103 8F Let  = m = 36.70 kpsi, and   m2   4.377 kpsi From Eq. (10-23), 2 d   0.08  2  2    3 2   C     4  4  4  2  36.70   4.377 4  4.377  2  2  36.70   4.377  3  36.70     6.98   4  4.377    4  4.377  D = Cd = 6.98(0.08) = 0.558 in 4C  2 4(6.98)  2   1.201 4C  3 4(6.98)  3  8(7)(0.558) 3   8F D   a  K B  a 3   1.201  (10 )   23.3 kpsi 3  d    (0.08 )  KB  The Gerber fatigue criterion from Eq. (6-48), adapted for shear, 2 2   2 m S se   1  S su   a  nf   1  1     2   m  S se   S su a     2 2 1  163.7  23.3    11  39.5      1  1   2      2  36.6  39.5    7  163.7      1.50 checks 4 Gd 10(106 )(0.08) 4 Na    31.02 coils 8kD3 8(9.5)(0.558)3 Shigley’s MED, 11th edition Chapter 10 Solutions, Page 27/47 Nt = 31.02 + 2 = 33 coils Ls = dNt = 0.08(33) = 2.64 in ymax = Fmax / k = 18 / 9.5 = 1.895 in ys = (1 + ) ymax = (1 + 0.15)(1.895) = 2.179 in L0 = Ls + ymax = 2.64 + 2.179 = 4.819 in (L0)cr = 2.63 D /  = 2.63(0.558) / 0.5 = 2.935 in s = (1 + )( Fmax /Fa) a = 1.15(18/7)23.3 = 68.8 kpsi ns = Ssy / s = 85.5/68.9 = 1.24 kg 9.5(386)   109 Hz 2 2  d DN a  (0.08 )(0.558)(31.02)(0.283) These steps are easily implemented on a spreadsheet, as shown below, for different diameters. f  2 2 d m A d1 d2 d3 d4 0.080 0.092 0.106 0.121 0.146 0.146 0.263 0.263 169.000 169.000 128.000 128.000 Sut 244.363 239.618 231.257 223.311 Ssu 163.723 160.544 154.942 149.618 Ssy 85.527 83.866 80.940 78.159 Ssa 35.000 35.000 35.000 35.000 Sse 39.452 39.654 40.046 40.469 m   C D 36.667 36.667 4.377 6.977 0.558 36.667 36.667 3.346 9.603 0.879 36.667 36.667 2.517 13.244 1.397 36.667 36.667 1.929 17.702 2.133 KB 1.201 1.141 1.100 1.074 a 23.333 23.333 23.333 23.333 nf 1.500 1.500 1.500 1.500 Na 30.993 13.594 5.975 2.858 Nt 32.993 15.594 7.975 4.858 Ls 2.639 1.427 0.841 0.585 ys 2.179 2.179 2.179 2.179 Shigley’s MED, 11th edition Chapter 10 Solutions, Page 28/47 L0 4.818 3.606 3.020 2.764 (L0)cr 2.936 4.622 7.350 11.220 69.000 69.000 69.000 69.000 1.240 1.215 1.173 1.133 s ns f, (Hz) 108.895 114.578 118.863 121.775 The shaded areas depict conditions outside the recommended design conditions. Thus, one spring is satisfactory. The specifications are: A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, L0 = 3.606 in, and Nt = 15.59 turns Ans. ______________________________________________________________________________ 10-36 The steps are the same as in Prob. 10-35 except that the Gerber-Zimmerli criterion is replaced with the Goodman-Zimmerli relationship of Eq. (10-29a) : S se  S sa 1   S sm S su  The problem then proceeds as in Prob. 10-30. The results for the wire sizes are shown below (see solution to Prob. 10-35 for additional details). Iteration of d for the first trial d d1 0.080 d2 0.0915 d3 0.1055 d4 0.1205 d1 0.080 d2 0.0915 d3 0.1055 d4 0.1205 m A 0.146 169.000 0.146 169.000 0.263 128.000 0.263 128.000 KB a 1.151 29.008 1.108 29.040 1.078 29.090 1.058 29.127 Sut Ssu 244.363 163.723 239.618 160.544 231.257 154.942 223.311 149.618 nf Na 1.500 14.191 1.500 6.456 1.500 2.899 1.500 1.404 Ssy Ssa 85.527 35.000 83.866 35.000 80.940 35.000 78.159 35.000 Nt Ls 16.191 1.295 8.456 0.774 4.899 0.517 3.404 0.410 Sse m   52.706 45.585 53.239 45.635 54.261 45.712 55.345 45.771 ys L0 2.179 3.474 2.179 2.953 2.179 2.696 2.179 2.589 45.585 4.377 45.635 3.346 45.712 2.517 45.771 1.929 (L0)cr 3.809 85.782 5.924 85.876 9.354 86.022 14.219 86.133 C D 9.052 0.724 12.309 1.126 16.856 1.778 22.433 2.703 ns f, (Hz) d s 0.997 0.977 0.941 0.907 141.284 146.853 151.271 154.326 Without checking all of the design conditions, it is obvious that none of the wire sizes satisfy ns ≥ 1.2. Also, the Gerber line is closer to the yield line than the Goodman. Setting nf = 1.5 for Goodman makes it impossible to reach the yield line (ns < 1) . The table below uses nf = 2. Shigley’s MED, 11th edition Chapter 10 Solutions, Page 29/47 Iteration of d for the second trial d m d1 0.080 0.146 d2 0.0915 0.146 d3 0.1055 0.263 d4 0.1205 0.263 A Sut 169.000 244.363 169.000 239.618 128.000 231.257 128.000 223.311 a Ssu Ssy 163.723 85.527 160.544 83.866 154.942 80.940 Ssa Sse 35.000 52.706 35.000 53.239 m   34.188 34.188 4.377 6.395 0.512 C D d1 0.080 1.221 d2 0.0915 1.154 d3 0.1055 1.108 d4 0.1205 1.079 nf 21.756 2.000 21.780 2.000 21.817 2.000 21.845 2.000 149.618 78.159 Na Nt 40.243 42.243 17.286 19.286 7.475 9.475 3.539 5.539 35.000 54.261 35.000 55.345 Ls ys 3.379 2.179 1.765 2.179 1.000 2.179 0.667 2.179 34.226 34.226 34.284 34.284 34.329 34.329 L0 (L0)cr 5.558 2.691 3.944 4.266 3.179 6.821 2.846 10.449 3.346 8.864 0.811 2.517 12.292 1.297 1.929 16.485 1.986 s d KB ns f, (Hz) 64.336 1.329 99.816 64.407 64.517 64.600 1.302 1.255 1.210 105.759 110.312 113.408 The satisfactory spring has design specifications of: A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.811 + 0.092 = 0.903 in, L0 = 3.944 in, and .Nt = 19.3 turns. Ans. ______________________________________________________________________________ 10-37 This is the same as Prob. 10-35 since Ssa = 35 kpsi. Therefore, the specifications are: A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, L0 = 3.606 in, and Nt = 15.59 turns Ans. ______________________________________________________________________________ 10-38 For the Gerber-Zimmerli fatigue-failure criterion, Ssu = 0.67Sut , S sa , S se  1  (S sm / S su ) 2  1 S su2 r  m  1  1  2 n f Sse   2  2S se      S su r    See the process used in Prob. 10-36. The last 2 columns of diameters of Ex. 10-5 are presented below with additional calculations. Shigley’s MED, 11th edition Chapter 10 Solutions, Page 30/47 d 0.105 0.112 d 0.105 0.112 Sut 278.691 276.096 Nt 10.915 8.190 Ssu 186.723 184.984 Ls 1.146 0.917 L0 3.446 3.217 Sse 38.325 38.394 m   38.508 38.502 (L0)cr 6.630 8.160 38.508 38.502 KB 1.111 1.095 2.887 2.538 a 23.105 23.101 C 12.004 13.851 nf 1.500 1.500 D 1.260 1.551 s 70.855 70.844 ID 1.155 1.439 ns 1.770 1.754 OD 1.365 1.663 fn Na 8.915 6.190 fom 105.433 106.922 -0.973 -1.022 There are only slight changes in the results. ______________________________________________________________________________ 10-39 As in Prob. 10-38, the basic change is Ssa. S sa 1  (S sm / S su ) Recalculate m using Eq. (6-41) for shear. That is, For Goodman, using Eq. (10-29a):    nf   a  m   S se S su  1  S se  S se S su S se S su   a S su   m S se  m  rS su  S se  Where r = a / m. Thus, m  S se S su n f  rS su  S se  See the process used in Prob. 10-36. Calculations for the last 2 diameters of Ex. 10-5 are given below. Shigley’s MED, 11th edition Chapter 10 Solutions, Page 31/47 d 0.105 0.112 d 0.105 0.112 Sut 278.691 276.096 Nt 11.153 8.353 Ssu 186.723 184.984 Ls 1.171 0.936 L0 3.471 3.236 Sse 49.614 49.810 m   38.207 38.201 (L0)cr 6.572 8.090 38.207 38.201 KB 1.112 1.096 2.887 2.538 a 22.924 22.920 C 11.899 13.732 nf 1.500 1.500 D 1.249 1.538 s 70.301 70.289 ID 1.144 1.426 ns 1.784 1.768 OD 1.354 1.650 fn Na 9.153 6.353 fom 104.509 106.000 -0.986 -1.034 There are only slight differences in the results. ______________________________________________________________________________ 10-40 Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi · inm , m = 0.190, rel cost = 1. 140 Try d  0.067 in, Sut   234.0 kpsi (0.067) 0.190 Table 10-6: Ssy = 0.45Sut = 105.3 kpsi Table 10-7: Sy = 0.75Sut = 175.5 kpsi Eq. (10-34) with D/d = C and C1 = C S F  A  max2 [( K ) A (16C )  4]  y ny d  d 2S y 4C 2  C  1 (16C )  4  n y Fmax 4C (C  1)   d 2S y  4C 2  C  1  (C  1)   1  4ny Fmax    2 2    d Sy 1 1   d Sy C 2  1   1 C    2  0   4  4n y Fmax 4  4n y Fmax   Shigley’s MED, 11th edition Chapter 10 Solutions, Page 32/47 2   2   d 2S y   d 2S y 1   d Sy     2  take positive root C   16n F   2 16n y Fmax 4 n F y max  y max    1   (0.067 2 )(175.5)(103 )   2 16(1.5)(18) 2    (0.067) 2 (175.5)(103 )   (0.067) 2 (175.5)(103)  2       4.590  16(1.5)(18) 4(1.5)(18)    D  Cd  4.59  0.067   0.3075 in Fi   d 3 i 8D  d3  33 500 C  3    1000  4    8D  exp(0.105C ) 6.5    Use the lowest Fi in the preferred range. This results in the best fom. Fi   (0.067)3  33 500 4.590  3     1000  4      6.505 lbf 8(0.3075)  exp[0.105(4.590)] 6.5   For simplicity, we will round up to the next integer or half integer. Therefore, use Fi = 7 lbf 18  7 k   22 lbf/in 0.5 d 4G (0.067) 4 (11.5)(10 6 ) Na    45.28 turns 8kD3 8(22)(0.3075)3 G 11.5 Nb  N a   45.28   44.88 turns E 28.6 L0  (2C  1  N b )d  [2(4.590)  1  44.88](0.067)  3.555 in L18 lbf  3.555  0.5  4.055 in Body: K B   max  4C  2 4(4.590)  2   1.326 4C  3 4(4.590)  3 8K B Fmax D 8(1.326)(18)(0.3075) 3 (10 )  62.1 kpsi  d3  (0.067)3 (n y ) body  S sy  max  105.3  1.70 62.1 r2  2d  2(0.067)  0.134 in, (K )B  C2  4C2  1 4(4)  1   1.25 4C2  4 4(4)  4 Shigley’s MED, 11th edition 2r2 2(0.134)  4 d 0.067 Chapter 10 Solutions, Page 33/47  8(18)(0.3075)  3  8Fmax D   1.25   (10 )  58.58 kpsi 3  3  d    (0.067)  S 105.3 (ny ) B  sy   1.80 B 58.58  B  (K ) B  fom  (1)  2d 2 ( N b  2) D 4   2 (0.067) 2 (44.88  2)(0.3075) 4  0.160 Several diameters, evaluated using a spreadsheet, are shown below. d Sut Ssy Sy C D Fi (calc) Fi (rd) k Na Nb L0 L18 lbf KB max (ny)body B (ny)B (ny)A fom 0.067 233.97 7 105.29 0 175.48 3 4.589 0.307 6.505 7.0 22.000 45.29 44.89 3.556 4.056 1.326 62.118 1.695 58.576 1.797 1.500 0.160 0.072 230.79 9 103.86 0 173.10 0 5.412 0.390 5.773 6.0 24.000 27.20 26.80 2.637 3.137 1.268 60.686 1.711 59.820 1.736 1.500 0.144 0.076 228.441 102.798 171.331 6.099 0.463 5.257 5.5 25.000 19.27 18.86 2.285 2.785 1.234 59.707 1.722 60.495 1.699 1.500 -0.138 0.081 225.69 2 101.56 1 169.26 9 6.993 0.566 4.675 5.0 26.000 13.10 12.69 2.080 2.580 1.200 58.636 1.732 61.067 1.663 1.500 0.135 0.085 223.63 4 100.63 5 167.72 6 7.738 0.658 4.251 4.5 27.000 9.77 9.36 2.026 2.526 1.179 57.875 1.739 61.367 1.640 1.500 0.133 0.09 0.095 0.104 221.21 218.95 215.22 9 8 4 99.548 98.531 96.851 165.91 4 8.708 0.784 3.764 4.0 28.000 7.00 6.59 2.071 2.571 1.157 57.019 1.746 61.598 1.616 1.500 0.135 164.21 8 9.721 0.923 3.320 3.5 29.000 5.13 4.72 2.201 2.701 1.139 56.249 1.752 61.712 1.597 1.500 0.138 161.41 8 11.650 1.212 2.621 3.0 30.000 3.15 2.75 2.605 3.105 1.115 55.031 1.760 61.712 1.569 1.500 0.154 Except for the 0.067 in wire, all springs satisfy the requirements of length and number of coils. The 0.085 in wire has the highest fom. ______________________________________________________________________________ 10-41 Given: Nb = 84 coils, Fi = 16 lbf, OQ&T steel, OD = 1.5 in, d = 0.162 in. D = OD  d = 1.5  0.162 = 1.338 in (a) Eq. (10-39): L0 = 2(D  d) + (Nb + 1)d = 2(1.338  0.162) + (84 + 1)(0.162) = 16.12 in or Shigley’s MED, 11th edition Ans. 2d + L0 = 2(0.162) + 16.12 = 16.45 in overall Chapter 10 Solutions, Page 34/47 D 1.338   8.26 d 0.162 4C  2 4(8.26)  2 KB    1.166 4C  3 4(8.26)  3 8(16)(1.338)  8F D   i  K B  i 3   1.166  14 950 psi  (0.162)3  d  (c) From Table 10-5 use: G = 11.4(106) psi and E = 28.5(106) psi (b) C  11.4 G  84   84.4 turns 28.5 E (0.162) 4 (11.4)(106 ) d 4G k    4.855 lbf/in 8D 3 N a 8(1.338)3 (84.4) (d) Table 10-4: A = 147 psi · inm , m = 0.187 147 Sut   207.1 kpsi (0.162)0.187 S y  0.75(207.1)  155.3 kpsi S sy  0.50(207.1)  103.5 kpsi Ans. N a  Nb  Body  d 3S sy F   KBD  (0.162)3 (103.5)(103 )  8(1.166)(1.338) Ans.  110.8 lbf Torsional stress on hook point B 2r2 2(0.25  0.162 / 2)   4.086 d 0.162 4C2  1 4(4.086)  1   1.243 (K )B  4C2  4 4(4.086)  4  (0.162)3 (103.5)(103) F   103.9 lbf 8(1.243)(1.338) C2  Normal stress on hook point A 2r1 1.338   8.26 d 0.162 4C12  C1  1 4(8.26) 2  8.26  1   1.099 (K ) A  4C1(C1  1) 4(8.26)(8.26  1) C1  Shigley’s MED, 11th edition Chapter 10 Solutions, Page 35/47 4  16( K ) A D S yt    F   3  d 2   d 155.3(103 ) F   85.8 lbf 16(1.099)(1.338) /  (0.162)3   4 /  (0.162)2   min(110.8, 103.9, 85.8)  85.8 lbf Ans.   (e) Eq. (10-48): 85.8  16 F  Fi   14.4 in Ans. k 4.855 ______________________________________________________________________________ y  10-42 Fmin = 9 lbf, Fmax = 18 lbf 18  9 18  9 Fa   4.5 lbf , Fm   13.5 lbf 2 2 A313 stainless: 0.013 ≤ d ≤ 0.1 A = 169 kpsi · inm , 0.1 ≤ d ≤ 0.2 A = 128 kpsi · inm , E = 28 Mpsi, G = 10 Gpsi Try d = 0.081 in and refer to the discussion following Ex. 10-7 169 Sut   243.9 kpsi (0.081)0.146 S su  0.67 Sut  163.4 kpsi S sy  0.35Sut  85.4 kpsi S y  0.55Sut  134.2 kpsi Table 10-8: m = 0.146 m = 0.263 Sr = 0.45Sut = 109.8 kpsi 109.8 / 2 Sr / 2   57.8 kpsi Se  2 1  [ Sr / (2Sut )] 1  [(109.8 / 2) / 243.9]2 r   a /  m  Fa / Fm  4.5 / 13.5 For Gerber, Eq. (6-48), solving for (m)A gives 2   2Se   1 Sut2 r  1  1    m  A   2 n f Se   Sut r     2  2    2  57.8  1 243.9  4.5 / 13.5   1  1      63.3 kpsi 2 2  57.8   243.9  4.5 / 13.5      Hook bending  16C 4  13.5  (4C 2  C  1)16C  ( m ) A  Fm ( K ) A 4      (1)  d 2  d 2   d 2  4C (C  1)   Shigley’s MED, 11th edition Chapter 10 Solutions, Page 36/47 Let    a  A 103  d 2  13.5 Equation (1) reduces to C2   16 63.3 103    0.081 C 13.5  8 16 2  96.65 0 The useable root for C is 2 2  1  96.647  1      96.647     C      8     96.647  8   4 8  4 8 8 8         4.91 F 4.5 63.3  21.1 kpsi  a  A  a  m  A  13.5 Fm n  f A 2 2   2 m Se   1  Sut    a        1  1    Sut a   2   m   Se      2 2   2(63.3)(57.8)   1  243.9   21.1       2.00     1  1   2  63.3   57.8   243.9(21.1)      checks D = Cd = 0.398 in Using Eq. (10-4) for i C  3   d 3 i  d 3  33 500  Fi    1000  4    8D 8D  exp(0.105C ) 6.5    Use the lowest Fi in the preferred range.  (0.081)3  33 500 4.91  3     1000  4  Fi    8(0.398)  exp[0.105(4.91)] 6.5     8.55 lbf For simplicity we will round Fi up to next 1/4 integer. Let Fi = 8.75 lbf. 18  9  36 lbf/in 0.25 d 4G (0.081) 4 (10)(106 ) Na    23.7 turns 8kD3 8(36)(0.398)3 G 10  23.7   23.3 turns Nb  N a  E 28 L0  (2C  1  N b )d  [2(4.91)  1  23.3](0.081)  2.602 in k  Lmax  L0  ( Fmax  Fi ) / k  2.602  (18  8.75) / 36  2.859 in Shigley’s MED, 11th edition Chapter 10 Solutions, Page 37/47  a  A  Body: Fa 4.5 63.3  21.1 kpsi  m  A  13.5 Fm 4C  2 4(4.91)  2   1.300 4C  3 4(4.91)  3 8(1.300)(4.5)(0.398) 3 (10 )  11.16 kpsi  a body   (0.081)3 F 13.5 (11.16)  33.48 kpsi  m body  m  a body  Fa 4.5 KB  The repeating allowable stress from Table 10-8 is Ssr = 0.30Sut = 0.30(243.9) = 73.17 kpsi The Gerber intercept is given by Eq. (10-42) as S se  From Eq. (6-48), 73.17 / 2  38.5 kpsi 1  [(73.17 / 2) / 163.4]2 2 2   2 m S se   1  S su    a   (n f ) body      1  1    2   m   S se   S su a      2 2   2(33.47)(38.5)   1  163.4   11.16       2.53     1  1   2  33.47   38.5   163.4(11.16)      Let r2 = 2d = 2(0.081) = 0.162 2r 4(4)  1 C2  2  4, ( K ) B   1.25 4(4)  4 d (K )B 1.25 ( a ) B  a  (11.16)  10.73 kpsi 1.30 KB (K )B 1.25 m  ( m ) B  (33.48)  32.19 kpsi 1.30 KB Table 10-8: (Ssr )B = 0.28Sut = 0.28(243.9) = 68.3 kpsi 68.3 / 2  35.7 kpsi (S se ) B  1  [(68.3 / 2) / 163.4]2 2 2   2(32.18)(35.7)   1  163.4   10.73   (n f ) B       1  1      2.51 2  32.18   35.7    163.4(10.73)     Yield Bending: Shigley’s MED, 11th edition Chapter 10 Solutions, Page 38/47  4 Fmax  (4C 2  C  1)  1 2  d  C 1  2  4(18)  4(4.91)  4.91  1   1 (10-3 )  84.4 kpsi 2  4.91  1  (0.081 )   134.2 (n y ) A   1.59 84.4 ( A ) max  Body:  i  ( Fi / Fa ) a  (8.75 / 4.5)(11.16)  21.7 kpsi r   a /( m   i )  11.16 / (33.47  21.7)  0.948 r 0.948 (S sy   i )  (85.4  21.7)  31.0 kpsi r 1 0.948  1 (S ) 31.0  sa y   2.78 11.16 a (S sa ) y  (n y ) body Hook shear: S sy  0.3Sut  0.3(243.9)  73.2 kpsi  max  ( a ) B  ( m ) B  10.73  32.18  42.9 kpsi 73.2  1.71 (n y ) B  42.9 7.6 2d 2 ( N b  2) D 7.6 2 (0.081) 2 (23.3  2)(0.398)   1.239 fom   4 4 A tabulation of several wire sizes follow 0.081 d 0.085 0.092 0.098 0.105 0.120 Sut 243.920 242.210 239.427 237.229 234.851 230.317 Ssu 163.427 162.281 160.416 158.943 157.350 154.312 Ssy Sr 85.372 84.773 83.800 83.030 82.198 80.611 109.764 108.994 107.742 106.753 105.683 103.643 Se 57.809 (m)A C 63.331 62.887 62.164 61.594 60.976 59.799 96.695 105.734 122.443 137.659 156.443 200.389 4.916 5.497 6.563 7.527 8.713 11.477 (a)A 21.110 20.962 20.721 20.531 20.325 19.933 (nf)A D OD Fi (calc) 2.000 0.398 0.479 2.000 0.467 0.552 2.000 0.604 0.696 2.000 0.738 0.836 2.000 0.915 1.020 2.000 1.377 1.497 8.537 7.842 6.769 5.960 5.117 3.618 Fi (rd) 8.750 8.750 8.750 8.750 8.750 8.750  Shigley’s MED, 11th edition 57.403 56.744 56.223 55.659 54.585 Chapter 10 Solutions, Page 39/47 k 36.000 36.000 36.000 36.000 36.000 36.000 Na 23.677 17.767 11.301 7.979 5.512 2.756 Nb 23.320 17.410 10.944 7.622 5.155 2.399 L0 2.604 2.329 2.122 2.124 2.266 2.922 Lmax 2.861 2.586 2.379 2.381 2.523 3.179 KB 1.300 1.263 1.215 1.184 1.157 1.117 (a)body 11.162 11.015 10.796 10.638 10.478 10.197 (m)body 33.486 33.044 32.388 31.913 31.433 30.591 Ssr 73.176 72.663 71.828 71.169 70.455 69.095 Sse 38.519 38.249 37.809 37.462 37.087 36.371 (nf)body 2.526 2.542 2.564 2.578 2.591 2.611 (KB)B 4.000 4.000 4.000 4.000 4.000 4.000 (a)B 10.732 10.898 11.106 11.226 11.320 11.416 (m)B 32.196 32.695 33.319 33.679 33.960 34.248 (Ssr)B 68.298 67.819 67.040 66.424 65.758 64.489 (Sse)B 35.708 35.458 35.050 34.728 34.380 33.717 (nf)B 2.512 2.457 2.383 2.336 2.293 2.230 Sy (A)max 134.156 133.215 131.685 130.476 129.168 126.674 84.441 83.849 82.886 82.125 81.302 79.732 1.589 1.589 1.589 1.589 1.589 1.589 r 21.704 0.947 21.417 0.947 20.992 0.947 20.684 0.947 20.373 0.947 19.828 0.947 (Ssa)y 30.974 30.822 30.555 30.330 30.077 29.570 2.775 2.798 2.830 2.851 2.871 2.900 (Ssy)B 73.176 72.663 71.828 71.169 70.455 69.095 (B)max 42.928 43.594 44.426 44.905 45.280 45.664 (ny)B fom 1.705 -1.240 1.667 -1.229 1.617 -1.240 1.585 -1.278 1.556 -1.353 1.513 -1.636 (ny)A i (ny)body optimal fom The shaded areas show the conditions not satisfied. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 10 Solutions, Page 40/47 10-43 For the hook, M = FR sin, F  1 EI   /2 0 ∂M/∂F = R sin F  R sin   R d  2  FR3 2 EI The total deflection of the body and the two hooks   FR 3  8FD3 N b 8FD 3 N b  F (D / 2)3    2   d 4G d 4G E ( / 64)(d 4 )  2 EI  8FD 3  G  8FD3 N a  4  Nb    d G  E d 4G G Q.E.D.  N a  Nb  E ______________________________________________________________________________   10-44 Table 10-5 (d = 4 mm = 0.1575 in): E = 196.5 GPa Table 10-4 for A227: Eq. (10-14): Eq. (10-57): A = 1783 MPa · mmm, m = 0.190 1783 A Sut  m  0.190  1370 MPa d 4 Sy = all = 0.78 Sut = 0.78(1370) = 1069 MPa D = OD  d = 32  4 = 28 mm C = D/d = 28/4 = 7 Eq. (10-43): Ki  2 4C 2  C  1 4  7   7  1   1.119 4C (C  1) 4(7)(7  1) 32 Fr d3 At yield, Fr = My ,  = Sy. Thus, Eq. (10-44):   Ki My   d 3S y 32 Ki Count the turns when M = 0 N  2.5  Shigley’s MED, 11th edition    43 1069 103  32(1.119)  6.00 N · m My k Chapter 10 Solutions, Page 41/47 where from Eq. (10-51): k d 4E 10.8 DN Thus, N  2.5  Solving for N gives N   My 4 d E / (10.8DN ) 2.5 1  [10.8DM y / (d 4 E )] 2.5   1  10.8(28)(6.00) / 4 4 (196.5)  This means (2.5  2.413)(360) or 31.3 from closed.  2.413 turns Ans. Treating the hand force as in the middle of the grip, 87.5 r  112.5  87.5   68.75 mm 2 6.00 103  My   87.3 N Ans. Fmax  r 68.75 ______________________________________________________________________________ 10-45 The spring material and condition are unknown. Given d = 0.081 in and OD = 0.500, (a) D = 0.500  0.081 = 0.419 in Using E = 28.6 Mpsi for an estimate k  (0.081) 4 (28.6)(106 ) d 4E   24.7 lbf · in/turn 10.8DN 10.8(0.419)(11) for each spring. The moment corresponding to a force of 8 lbf Fr = (8/2)(3.3125) = 13.25 lbf · in/spring The fraction windup turn is n Fr 13.25   0.536 turns k 24.7 The arm swings through an arc of slightly less than 180, say 165. This uses up 165/360 or 0.458 turns. So n = 0.536  0.458 = 0.078 turns are left (or 0.078(360) = 28.1 ). The original configuration of the spring was Shigley’s MED, 11th edition Chapter 10 Solutions, Page 42/47 Ans. (b) D 0.419   5.17 d 0.081 4C 2  C  1 4(5.17) 2  5.17  1   1.168 Ki  4C  C  1 4(5.17)(5.17  1) C    Ki  32(13.25)  32M  1.168   297 103  psi  297 kpsi 3 3 d   (0.081)  Ans. To achieve this stress level, the spring had to have set removed. ______________________________________________________________________________ 10-46 (a) Consider half and double results Straight section: M = 3FR, M  3R F Upper 180 section: M  F [ R  R(1  cos  )] M  FR(2  cos  ),  R(2  cos  ) F Lower section: M = FR sin , Considering bending only: Shigley’s MED, 11th edition M  R sin  F Chapter 10 Solutions, Page 43/47  U 2  l /2  9 FR 2 dx   FR 2 (2  cos  ) 2 R d   0 F EI  0  2F  9 2       R l  R 3  4  4sin  0    R 3     EI  2 2   4  2 2 2 FR  19 9  FR  R  l  (19 R  18l )  EI  4 2  2 EI     /2 0 F ( R sin  ) 2 R d   The spring rate is k F   2 EI R (19 R  18 l ) 2 Ans. (b) Given: A227 HD wire, d = 2 mm, R = 6 mm, and l = 25 mm. Table 10-5 (d = 2 mm = 0.0787 in): k E = 197.2 GPa 2 197.2 109   0.0024  /  64  0.006 19  0.006   18  0.025   2  10.65 103  N/m  10.65 N/mm Ans. (c) The maximum stress will occur at the bottom of the top hook where the bendingmoment is 3FR and the axial fore is F. Using curved beam theory for bending, i  Eq. (3-65): Axial: a  Combining, Mci 3FRci  2 Aeri  d / 4  e  R  d / 2  F F  A d2 / 4  max   i   a  F 4F d2   3Rci  1  S y   e  R  d / 2   d 2Sy   3Rci 4  1  e  R  d / 2  (1) Ans. For the clip in part (b), Eq. (10-14) and Table 10-4: Eq. (10-57): Shigley’s MED, 11th edition Sut = A/dm = 1783/20.190 = 1563 MPa Sy = 0.78 Sut = 0.78(1563) = 1219 MPa Chapter 10 Solutions, Page 44/47 Table 3-4: rn  12  2 6  62  12   5.95804 mm e = rc  rn = 6  5.95804 = 0.04196 mm ci = rn  (R  d /2) = 5.95804  (6  2/2) = 0.95804 mm Eq. (1):   0.0022 1219 106   46.0 N Ans.  3  6  0.95804   1 4  0.04196 6 1     ______________________________________________________________________________ F 10-47 (a) M  x F M   Fx, M  Fl  FR 1  cos   ,  0 xl M  l  R 1  cos   F 0   / 2  /2 2 1 l    ( ) Fx x dx F l  R 1  cos    Rd   0 0 EI F  4l 3  3R  2 l 2  4   2  l R   3  8  R 2  12 EI F     The spring rate is k F F  12 EI 4l  3R  2 l  4   2  l R   3  8  R 2  Shigley’s MED, 11th edition 3 2 Ans. Chapter 10 Solutions, Page 45/47 (b) Given: A313 stainless wire, d = 0.063 in, R = 0.625 in, and l = 0.5 in. Table 10-5: E = 28 Mpsi I k  64 d4   64  0.063   7.733 10  in 4 7 4 12  28 106  7.733107 4  0.53   3  0.625   2  0.52   4   2  0.5  0.625    3  8   0.6252   36.3 lbf/in (c) Table 10-4: Ans. A = 169 kpsiinm, m = 0.146 Eq. (10-14): Sut = A/ d m = 169/0.0630.146 = 253.0 kpsi Eq. (10-57): Sy = 0.61 Sut = 0.61(253.0) = 154.4 kpsi One can use curved beam theory as in the solution for Prob. 10-41. However, the equations developed in Sec. 10-12 are equally valid. C = D/d = 2(0.625 + 0.063/2)/0.063 = 20.8 Eq. (10-43): Ki  2 4C 2  C  1 4  20.8   20.8  1   1.037 4C  C  1 4  20.8  20.8  1 Eq. (10-44), setting  = Sy: Ki Solving for F yields 32 Fr  Sy d3  F = 3.25 lbf 1.037 32 F  0.5  0.625    0.063 3   154.4 103  Ans. Try solving part (c) of this problem using curved beam theory. You should obtain the same answer. ______________________________________________________________________________ 10-48 (a) M =  Fx  M Fx Fx   2 I / c I / c bh / 6 Constant stress, Shigley’s MED, 11th edition Chapter 10 Solutions, Page 46/47 bh 2 Fx  6  6 Fx b h  (1) Ans. At x = l, 6 Fl b ho  (b)  h  ho x / l Ans. M =  Fx,  M / F = x l y 0  l l  Fx   x  M  M / F  1 12 Fl 3/ 2 1/ 2 dx   1 3 dx x dx  EI E 0 12 bho  x / l 3/ 2 bho3 E 0 2 12 Fl 3/ 2 3/ 2 8 Fl 3 l  3 bho E 3 bho3 E F bho3 E  3 Ans. y 8l ______________________________________________________________________________ k 10-49 Computer programs will vary. ______________________________________________________________________________ 10-50 Computer programs will vary. Shigley’s MED, 11th edition Chapter 10 Solutions, Page 47/47 Chapter 4 For a torsion bar, kT = T/ = Fl/, and so  = Fl/kT. For a cantilever, kl = F/ , = F/kl. For the assembly, k = F/y, or, y = F/k = l +  Thus F Fl 2 F y   k kT kl Solving for k kk 1  2l T k 2 Ans. l 1 kl l  kT  kT kl ______________________________________________________________________________ 4-1 For a torsion bar, kT = T/ = Fl/, and so  = Fl/kT. For each cantilever, kl = F/l, l = F/kl, and,L = F/kL. For the assembly, k = F/y, or, y = F/k = l + l +L. Thus F Fl 2 F F y    k kT kl k L Solving for k k L kl kT 1  k 2 Ans. 2 l 1 1 kl k Ll  kT k L  kT kl   kT kl kL ______________________________________________________________________________ 4-2 4-3 (a) For a torsion bar, k =T/ =GJ/l. Two springs in parallel, with J =di 4/32, and d1 = d2 = d, J G J G   d4 d4  k  1  2  G 1  2  x l  x 32  x l  x   1  1 Gd 4    32  x lx Deflection equation,   Ans. (1) T1 x T2  l  x   JG JG T2  l  x  (2) x From statics, T1 + T2 = T = 1500. Substitute Eq. (2) results in T1  Shigley’s MED, 11th edition Chapter 4 Solutions, Page 1/96 lx T2    T2  1500  x  T2  1500  x l Ans. (3) lx Ans. (4) l 1   1 3 (b) From Eq. (1), k   0.54 11.5 106      28.2 10  lbf  in/rad 32  5 10  5  10  5 From Eq. (4), T1  1500  750 lbf  in Ans. 10 5 From Eq. (3), T2  1500  750 lbf  in Ans. 10 16Ti 16 1500    30.6 103  psi  30.6 kpsi Ans. From either section,   3 3  di   0.5  Substitute into Eq. (2) resulting in T1  1500 Ans. ______________________________________________________________________________ 4-4 Deflection to be the same as Prob. 4-3 where T1 = 750 lbfin, l1 = l / 2 = 5 in, and d1 = 0.5 in 1=2= T1  4   32 Or, d14G  T2  6   32 d 24G   32 T1  15 103  d14  0.54  G  4T1 6T2  4  60 103  4 d1 d2 (1) (2) T2  10 103  d 24 Equal stress,  1   2  750  5 (3) T T 16T1 16T2   13  23 3 3 d1 d 2  d1  d 2 (4) Divide Eq. (4) by the first two equations of Eq.(1) results in T1 T2 3 d1 d3  2  d 2  1.5d1 (5) 4T1 6T2 d14 d 24 Statics, T1 + T2 = 1500 (6) Substitute in Eqs. (2) and (3), with Eq. (5) gives 15 103  d14  10 103  1.5d1   1500 4 Solving for d1 and substituting it back into Eq. (5) gives d1 = 0.388 8 in, d2 = 0.583 2 in Ans. Shigley’s MED, 11th edition Chapter 4 Solutions, Page 2/96 From Eqs. (2) and (3), T1 = 15(103)(0.388 8)4 = 343 lbfin T2 = 10(103)(0.583 2)4 = 1 157 lbfin Ans. Ans. 343  4  T1l1   0.053 18 rad J1G   / 32   0.388 84 11.5 106  Deflection of T is 1  Spring constant is k The stress in d1 is 1  16  343 16T1   29.7 103  psi  29.7 kpsi 3  d1   0.388 8 3 Ans. The stress in d1 is 2  16 1 157  16T2   29.7 103  psi  29.7 kpsi 3 3  d 2   0.583 2  Ans. T 1  1500  28.2 103  lbf  in 0.053 18 Ans. ______________________________________________________________________________ 4-5 (a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the taper be  where tan  = (r2  r1)/2. Thus, the radius in the taper as a function of x is r = r1 + x tan , and the area is A =  (r1 + x tan )2. The deflection of the tapered portion is l l 1 F F dx F    dx  2   E 0  r1  x tan    E  r1  x tan   tan  AE 0   1 1 1 F  1 F        E  r1 tan  tan   r1  l tan     E tan   r1 r2   F r2  r1 F l tan  Fl    E tan  r1r2  E tan  r1r2  r1r2 E  4 Fl  d1d 2 E l 0 Ans. (b) For section 1, 1  4 Fl 4(1000)(2) Fl    3.40(10 4 ) in 2 2 6 AE  d1 E  (0.5 )(30)(10 ) For the tapered section, 4 Fl 4 1000(2)    2.26(10 4 ) in 6  d1d 2 E  (0.5)(0.75)(30)(10 ) For section 2, Shigley’s MED, 11th edition Ans. Ans. Chapter 4 Solutions, Page 3/96 4 Fl 4(1000)(2) Fl    1.51(10 4 ) in Ans. 2 2 6 AE  d1 E  (0.75 )(30)(10 ) ______________________________________________________________________________ 2  4-6 (a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the taper be  where tan  = (r2  r1)/2. Thus, the radius in the taper as a function of x is r = r1 + x tan , and the polar second area moment is J = ( /2) (r1 + x tan )4. The angular deflection of the tapered portion is l l T dx 2T 1 2T 1    dx   4  GJ 3  G  r1  x tan  3 tan   G 0  r1  x tan   0  l 0  2 T 1 1 2 T  1 1    3     3 3 G  r1 tan  tan   r1  l tan    3 G tan   r13 r23  2 2 2 2 T  l  r23  r13 2 Tl  r1  r1r2  r2  T r23  r13      3 G tan  r13r23 3 G  r2  r1  r13 r23 3 G r13r23  2 2 32 Tl  d1  d1d 2  d 2  3 G d13d 23 Ans. (b) The deflections, in degrees, are: Section 1, Tl GJ 1  32(1500)(2)  180   180  32Tl  180        2.44 deg 4 4 6      d1 G     (0.5 )11.5(10 )    Ans. Tapered section,  32 Tl (d12  d1d 2  d 2 2 )  180    Gd13 d 23 3    2 2 32 (1500)(2) 0.5  (0.5)(0.75)  0.75   180      1.14 deg 3 11.5(106 )(0.53 )(.753 )    Ans. Section 2, 32(1500)(2)  180  Tl  180  32Tl  180  Ans.        0.481 deg 4 GJ     d 2 G     (0.754 )11.5(106 )    ______________________________________________________________________________ 2  4-7 The area and the elastic modulus remain constant. However, the force changes with respect to x. From Table A-5, the unit weight of steel is  = 0.282 lbf/in3 and the elastic modulus is E = 30 Mpsi. Starting from the top of the cable (i.e. x = 0, at the top). F = (A)(lx) Shigley’s MED, 11th edition Chapter 4 Solutions, Page 4/96 c   l o 1     l 2 0.282 500(12)  Fdx w l   (l  x )dx   lx  x 2     0.169 in AE E 0 E 2  0 2E 2(30)106 2 l From the weight at the bottom of the cable, 4(5000) 500(12)  Wl 4Wl    5.093 in AE  d 2 E  (0.52 )30(106 )    c  W  0.169  5.093  5.262 in Ans. W  The percentage of total elongation due to the cable’s own weight 0.169 (100)  3.21% Ans. 5.262 ______________________________________________________________________________ 4-8 Fy = 0 = R 1  F  R 1 = F MA = 0 = M1  Fa  M1 = Fa VAB = F, MAB =F (x  a ), VBC = MBC = 0 Section AB:  F  x2 1  AB  F x  a dx      ax   C1  EI EI  2  AB = 0 at x = 0  C1 = 0 y AB  F  x2 F  x3 x2     ax  dx    a   C2 EI   2 EI  6 2  (1) (2) yAB = 0 at x = 0  C2 = 0, and Fx 2 y AB  Ans.  x  3a  6 EI Section BC:  BC  1 EI   0  dx  0  C 3 From Eq. (1), at x = a (with C1 = 0),    BC    F  a2 Fa 2 a ( a )    = C3. Thus,   EI  2 2 EI  Fa 2 2 EI Shigley’s MED, 11th edition Chapter 4 Solutions, Page 5/96 yBC   Fa 2 Fa 2 dx   x  C4 2 EI  2 EI (3) F  a3 a2  Fa3 From Eq. (2), at x = a (with C2 = 0), y  . Thus, from Eq. (3)  a    EI  6 2  3EI  Fa 2 Fa 3 a  C4   2 EI 3EI yBC   C4   Fa 3 6 EI Fa 2 Fa 3 Fa 2 x   a  3x  2 EI 6 EI 6 EI Substitute into Eq. (3), obtaining Ans. The maximum deflection occurs at x= l, Fa 2 Ans.  a  3l  6 EI ______________________________________________________________________________ ymax  4-9 MC = 0 = F (l /2)  R1 l  R1 = F /2 Fy = 0 = F /2 + R 2  F  R 2 = F /2 Break at 0  x  l /2: VAB = R 1 = F /2, MAB = R 1 x = Fx /2 Break at l /2  x  l : VBC = R 1  F =  R 2 =  F /2, MBC = R 1 x  F ( x  l / 2) = F (l  x) /2 Section AB:  AB  1 EI Fx F x2 dx   C1 2 EI 4 2 l F   2 C  0 From symmetry, AB = 0 at x = l /2  1 4 EI  AB  F x 2 Fl 2 F    4x2  l 2  EI 4 16 EI 16 EI Shigley’s MED, 11th edition  C1   Fl 2 . Thus, 16 EI (1) Chapter 4 Solutions, Page 6/96 y AB  F 16 EI yAB = 0 at x = 0 y AB    4x 2  l 2  dx  F  4 x3 2   l x   C2  16 EI  3   C2 = 0, and, Fx 4 x 2  3l 2   48 EI (2) yBC is not given, because with symmetry, Eq. (2) can be used in this region. The maximum deflection occurs at x =l /2, ymax l F  2  Fl 3 2  l   2 4 3 l         48EI   2  48EI  Ans. ______________________________________________________________________________ 4-10 From Table A-6, for each angle, I1-1 = 207 cm4. Thus, I = 2(207) (104) = 4.14(106) mm4 From Table A-9, use beam 2 with F = 2500 N, a = 2000 mm, and l = 3000 mm; and beam 3 with w = 1 N/mm and l = 3000 mm. Fa 2 wl4 (a  3l )  6 EI 8 EI 2500(2000)2 (1)(3000) 4    2000 3(3000)   6(207)103 (4.14)106 8(207)(103 )(4.14)(106 )  25.4 mm Ans. ymax  M O   Fa  ( wl 2 / 2) =  2500(2000)  [1(30002)/2] =  9.5(106) Nmm From Table A-6, from centroid to upper surface is y = 29 mm. From centroid to bottom surface is y = 29.0  100=  71 mm. The maximum stress is compressive at the bottom of the beam at the wall. This stress is 9.5(106 )(71) My   163 MPa Ans. I 4.14(106 ) ______________________________________________________________________________  max   Shigley’s MED, 11th edition Chapter 4 Solutions, Page 7/96 4-11 14 10 (450)  (300)  465 lbf 20 20 6 10 (450)  (300)  285 lbf RC  20 20 RO  M1 = 465(6)12 = 33.48(103) lbfin M2 = 33.48(103) +15(4)12 = 34.20(103) lbfin M max 34.2  15  Z  2.28 in 3 Z Z For deflections, use beams 5 and 6 of Table A-9 2 F1a[l  (l / 2)]  l  l  F2l 3 2 y x 10ft     a  2l   6 EIl 2  48 EI  2   max  0.5  450(72)(120) 300(2403 ) 2 2 2    120 72 240   48(30)(106 ) I 6(30)(106 ) I (240) I  12.60 in 4  I / 2  6.30 in 4 Select two 5 in-6.7 lbf/ft channels from Table A-7, I = 2(7.49) = 14.98 in4, Z =2(3.00) = 6.00 in3 12.60  1  ymidspan      0.421 in 14.98  2  34.2  5.70 kpsi  max  6.00 ______________________________________________________________________________ 4-12 I  (1.54 )  0.2485 in 4 64 From Table A-9 by superposition of beams 6 and 7, at x = a = 15 in, with b = 24 in and l = 39 in Fba 2 wa y [a  b 2  l 2 ]  (2la 2  a 3  l 3 ) 6 EIl 24 EI yA  340(24)15 152  242  392  6(30)106 (0.2485)39  (150 /12)(15)  2(39)(152 )  153  393   0.0978 in  6 24(30)10 (0.2485) Ans. At x = l /2 = 19.5 in 2 2 3  Fa[l  (l / 2)]  l  l  w(l / 2)   l   l  2 3 y    a  2l    2l       l  6 EIl 2  24 EI   2   2   2   Shigley’s MED, 11th edition Chapter 4 Solutions, Page 8/96 y 340(15)(19.5) 19.52  152  39 2  6 6(30)(10 )(0.2485)(39) (150 / 12)(19.5)  2(39)(19.52 )  19.53  393   0.1027 in  24(30)(106 )(0.2485)  Ans. 0.1027  0.0978 (100)  5.01% Ans. 0.0978 ______________________________________________________________________________ % difference  4-13 I  1 (6)(323 )  16.384 103 mm 4 12   From Table A-9-10, beam 10 Fa 2 yC   (l  a ) 3EI Fax 2 y AB  l  x2   6 EIl dy AB Fa 2  (l  3 x 2 ) dx 6 EIl dy AB  A dx Fal 2 Fal A   6 EIl 6 EI Fa 2l yO   A a   6 EI At x = 0, With both loads, Fa 2l Fa 2  (l  a) 6 EI 3EI 400(3002 ) Fa 2 (3l  2a)    3(500)  2(300)  3.72 mm Ans. 6 EI 6(207)103 (16.384)103 At midspan, 2 2 Fa(l / 2)  2  l   3 Fal 2 3 400(300)(500 2 ) yE    1.11 mm Ans. l      6 EIl  24 207 103 16.384 103   2   24 EI _____________________________________________________________________________  4 4-14 I  (2  1.54 )  0.5369 in 4 64 yO   From Table A-5, E = 10.4 Mpsi From Table A-9, beams 1 and 2, by superposition Shigley’s MED, 11th edition Chapter 4 Solutions, Page 9/96 200  4(12) 300  2(12)  FBl 3 FAa 2 yB     (a  3l )   2(12)  3(4)(12) 3EI 6 EI 3(10.4)106 (0.5369) 6(10.4)106 (0.5369) yB  1.94 in Ans. ______________________________________________________________________________ 3 2 4-15 From Table A-7, I = 2(1.85) = 3.70 in4 From Table A-5, E = 30.0 Mpsi From Table A-9, beams 1 and 3, by superposition 5  2(5 / 12)  (60 )  0.182 in Ans. Fl 3 ( w  wc )l 4 150(603 )    yA   6 3EI 8 EI 3(30)10 (3.70) 8(30)106 (3.70) ______________________________________________________________________________  4 4-16 I  d 64 From Table A-5, E  207(103 ) MPa From Table A-9, beams 5 and 9, with FC = FA = F, by superposition FB l 3 Fa 1   FB l 3  2 Fa(4a 2  3l 2 )  yB    (4a 2  3l 2 )  I  48 EI 24 EI 48 EyB 4 I 1 550(10003 )  2  375  (250) 4(2502 )  3(10002 )  3 48(207)10  2       53.624 103 mm4 d 4 64  I  4 64  (53.624)103  32.3 mm Ans. ______________________________________________________________________________ 4-17 From Table A-9, beams 8 (region BC for this beam with a = 0) and 10 (with a = a), by superposition MA 3 Fax 2 x  3lx 2  2l 2 x  l  x2 6 EIl 6 EIl 1  M A x 3  3lx 2  2l 2 x  Fax l 2  x 2  Ans.   6 EIl d M F (x  l)     A x3  3lx 2  2l 2 x   ( x  l )  [( x  l ) 2  a (3x  l )] 6 EI   x l  dx  6 EIl y AB  yBC           M Al F (x  l) (x  l)  [( x  l ) 2  a (3 x  l )] 6 EI 6 EI (x  l)   M Al  F  ( x  l ) 2  a (3 x  l )  Ans. 6 EI ______________________________________________________________________________   Shigley’s MED, 11th edition  Chapter 4 Solutions, Page 10/96 4-18 Note to the instructor: Beams with discontinuous loading are better solved using singularity functions. This eliminates matching the slopes and displacements at the discontinuity as is done in this solution. a wa   M C  0  R1l  wa  l  a  2   R1  2l  2l  a  Ans. wa wa 2  Fy  0  2l  2l  a   R2  wa  R2  2l Ans. wa w VAB  R1  wx =  2l  a   wx =  2l  a  x   a 2  Ans. 2l 2l 2 wa VBC   R2   Ans. 2l  x2  w  M AB   VAB dx   2l  ax    a 2 x   C1 2l   2  M AB  0 at x  0  C1  0  M AB  wx  2al  a 2  lx  Ans. 2l  wa 2 wa 2 dx   x  C2 2l 2l wa 2 wa 2 M BC  0 at x  l  C2   M BC  (l  x) Ans. 2 2l M 1 wx 1 w  2 1 2 2 1 3  2al  a 2  lx  dx   AB   AB dx    alx  a x  lx   C3    2 3  EI EI 2l EI  2l   M BC   VBC dx    y AB    AB dx  1 w 2 1 2 2 1 3   alx  a x  lx   C3  dx   2 3  EI  2l   1 w 1 3 1 2 3 1 4   alx  a x  lx   C3 x  C4   6 12  EI  2l  3   0 at x  0  C4  0  y AB  BC M 1   BC dx  EI EI  wa 2 1  wa 2  1 2  2l (l  x) dx  EI  2l  lx  2 x   C5   AB   BC at x  a  1 EI  wa 3 w  2 1 4 1 3  1  wa 2  1 2          C5 ala a la C la a C C 3 5 3      2l  2 3  2  6    EI  2l   Shigley’s MED, 11th edition (1) Chapter 4 Solutions, Page 11/96   1  wa 2  1 2 1  wa 2  1 2 1 3  lx  x  C dx  5    lx  x   C5 x  C6     2  6  EI  2l  EI  2l  2   2 2 wa l  0 at x  l  C6    C5l 6  1  wa 2  1 2 1 3 1 3    lx  x  l   C5 ( x  l )   EI  2l  2 6 3   yBC    BC dx  yBC yBC y AB  yBC at x  a  1 5 1 4 w 1 wa 2  1 2 1 3 1 3  3 ala  a  la  C a  3    la  a  l   C5 (a  l ) 2l  3 6 12 2l  2 6 3   C3 a  wa 2  3la 2  4l 3   C5 (a  l ) 24l Substituting (1) into (2) yields C5  (2) wa 2 a 2  4l 2  . Substituting this back into (2) gives  24l wa 2 C3  4al  a 2  4l 2  . Thus,  24l w y AB  4alx 3  2a 2 x 3  lx 4  4a 3lx  a 4 x  4a 2l 2 x   24 EIl wx  2  y AB  2ax 2 (2l  a )  lx 3  a 2  2l  a   Ans.  24 EIl  w y BC  6a 2lx 2  2a 2 x 3  a 4 x  4a 2l 2 x  a 4l  Ans.  24 EIl This result is sufficient for yBC. However, this can be shown to be equivalent to w w yBC  4alx 3  2a 2 x 3  lx 4  4a 2l 2 x  4a 3lx  a 4 x   ( x  a )4  24 EIl 24 EI w yBC  y AB  ( x  a) 4 Ans. 24 EI by expanding this or by solving the problem using singularity functions. ______________________________________________________________________________ 4-19 The beam can be broken up into a uniform load w downward from points A to C and a uniform load w upward from points A to B. Using the results of Prob. 4-18, with b = a for A to C and a = a for A to B, results in wx  wx  2 2 y AB  2bx 2 (2l  b)  lx 3  b 2  2l  b    2ax 2 (2l  a )  lx 3  a 2  2l  a    24 EIl   24 EIl  wx  2 2 2bx 2 (2l  b)  b 2  2l  b   2ax 2 (2l  a )  a 2  2l  a    24 EIl  w  2 2bx 3 (2l  b)  lx 4  b 2 x  2l  b    24 EIl  yBC   4alx 3  2a 2 x 3  lx 4  4a 2l 2 x  4a 3lx  a 4 x   l ( x  a ) 4  Shigley’s MED, 11th edition Ans. Ans. Chapter 4 Solutions, Page 12/96 w  4blx3  2b 2 x3  lx 4  4b 2l 2 x  4b3lx  b 4 x  l ( x  b) 4  24 EIl w  4alx 3  2a 2 x3  lx 4  4a 2l 2 x  4a 3lx  a 4 x  l ( x  a ) 4   24 EIl w ( x  b) 4  ( x  a) 4   y AB  Ans. 24 EI ______________________________________________________________________________ 4-20 Note to the instructor: See the note in the solution for Problem 4-18. wa 2 wa  Fy  0  RB  2l  wa  RB  2l  2l  a  Ans. For region BC, isolate right-hand element of length (l + a  x) wa 2 VAB   RA   , VBC  w  l  a  x  Ans. 2l wa 2 w 2 M AB   RA x   x, M BC    l  a  x  Ans. 2l 2 wa 2 2 EI AB   M AB dx   x  C1 4l wa 2 3 EIy AB   x  C1 x  C2 12l wa 2 3 yAB = 0 at x = 0  C2 = 0  EIy AB   x  C1 x 12l wa 2l yAB = 0 at x = l  C1   12 w a 2 3 w a 2l wa 2 x 2 wa 2 x 2 EIy AB   x  x l  x 2   y AB    l  x2  Ans. 12l 12 12l 12 EIl w 3 EI BC   M BC dx    l  a  x   C3 6 w 4 EIyBC    l  a  x   C3 x  C4 24 wa 4 wa 4 yBC = 0 at x = l    C3l  C4  0  C4   C3l (1) 24 24 wa 2l wa 2l wa 3 wa 2 AB = BC at x = l      C3  C3   l  a  4 12 6 6 wa 2 2  a  4l  l  a   . Substitute back into yBC Substitute C3 into Eq. (1) gives C4  24   1  w wa 2 wa 4 wa 2l 4 yBC  x l  a     l  a   l  a  x   EI  24 6 24 6  yCD  w  4 Ans.  l  a  x   4a 2  l  x  l  a   a 4   24 EI ______________________________________________________________________________  Shigley’s MED, 11th edition Chapter 4 Solutions, Page 13/96 4-21 Table A-9, beam 7, w l 100(10) R1  R2    500 lbf  2 2 wx 100 x  2(10) x 2  x3  103  2lx 2  x 3  l 3   y AB   6 24 EI 24  30 10  0.05   2.7778 10 6  x  20 x 2  x3  1000  Slope:  AB  d y AB w  6lx 2  4 x3  l 3   dx 24 EI w wl 3 2 3 3    6 l l 4 l l   x l 24 EI 24 EI 100 103  w l3 x l  x  l   x  10   2.7778 103   x  10  6 x l 24 EI 24(30)10 (0.05) At x = l,  AB yBC   AB  From Prob. 4-20, 2 wa 2 100  4  RA    80 lbf  2l 2(10) y AB RB  100  4  wa  2l  a    2(10)  4  480 lbf  2l 2(10) 100  42  x wa 2 x 2 2  10 2  x 2   8.8889 10 6  x 100  x 2  l x    6 12 EIl 12  30 10  0.05  w  4  l  a  x   4a 2  l  x  l  a   a 4   24 EI 100 10  4  x 4  4  4 2  10  x 10  4   44   6  24  30 10  0.05  yBC   4  2.7778 10 6  14  x   896 x  9216    Superposition, RA  500  80  420 lbf  RB  500  480  980 lbf  y AB  2.7778 10 6 yBC  2.7778 103  x  20 x  x  1000   8.8889 10  x 100  x    x 10   2.7778 10  14  x   896 x  9216  2 3 6 6 4 2 Ans. Ans. Ans. The deflection equations can be simplified further. However, they are sufficient for plotting. Using a spreadsheet, x y 0 0.5 1 1.5 2 2.5 3 3.5 0.000000 -0.000939 -0.001845 -0.002690 -0.003449 -0.004102 -0.004632 -0.005027 x y 4 4.5 5 5.5 6 6.5 7 7.5 -0.005280 -0.005387 -0.005347 -0.005167 -0.004853 -0.004421 -0.003885 -0.003268 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 14/96 x y x y 8 8.5 9 9.5 10 10.5 11 11.5 -0.002596 -0.001897 -0.001205 -0.000559 0.000000 0.000439 0.000775 0.001036 12 12.5 0.001244 0.001419 0.002 13 0.001575 13.5 14 0.001722 0.001867 Beam Deflection, Prob. 4-21 0.001 0.000 -0.001 y (in) -0.002 -0.003 -0.004 -0.005 -0.006 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 x (in) ______________________________________________________________________________ 4-22 (a) Useful relations k F 48EI  3 y l 3 kl 3 1800  36  I   0.05832 in 4 6 48 E 48(30)10 From I = bh 3/12, and b = 10 h, then I = 5 h 4/6, or, h 4 6 I 4 6(0.05832)   0.514 in 5 5 h is close to 1/2 in and 9/16 in, while b is close to 5.14 in. Changing the height drastically changes the spring rate, so changing the base will make finding a close solution easier. Trial and error was applied to find the combination of values from Table A-17 that yielded the closet desired spring rate. h (in) 1/2 1/2 1/2 9/16 9/16 Shigley’s MED, 11th edition b (in) 5 5½ 5¾ 5 4 b/h 10 11 11.5 8.89 7.11 k (lbf/in) 1608 1768 1849 2289 1831 Chapter 4 Solutions, Page 15/96 h = ½ in, b = 5 ½ in should be selected because it results in a close spring rate and b/h is still reasonably close to 10. (b) I  5.5(0.5)3 / 12  0.05729 in 4 4 I 4(60)103 (0.05729) Mc ( Fl / 4)c   F   1528 lbf  I I lc  36  (0.25) (1528)  36  Fl 3 y   0.864 in Ans. 48 EI 48(30)106 (0.05729) ______________________________________________________________________________ 3 4-23 From the solutions to Prob. 3-79, T1  60 lbf and T2  400 lbf I d4 64   (1.25) 4 64  0.1198 in 4 From Table A-9, beam 6, Fb x Fb x  z A   1 1 ( x 2  b12  l 2 )  2 2 ( x 2  b2 2  l 2 )  6 EIl  6 EIl  x 10in (575)(30)(10) 102  302  402   6 6(30)10 (0.1198)(40) 460(12)(10)  102  122  40 2   0.0332 in  6 6(30)10 (0.1198)(40)  Ans. dz Fb x  d Fb x      1 1 ( x 2  b12  l 2 )  2 2 ( x 2  b2 2  l 2 )    6 EIl  dx  x 10in   x 10in  dx  6 EIl Fb  Fb     1 1 (3x 2  b12  l 2 )  2 2 (3 x 2  b2 2  l 2 )  6 EIl  6 EIl  x 10in  A  y     (575)(30) 3 102   30 2  40 2  6  6(30)10 (0.1198)(40)  460(12) 3 102   122  402   6  6(30)10 (0.1198)(40)  Ans.  6.02(104 ) rad ______________________________________________________________________________ 4-24 From the solutions to Prob. 3-80, T1  2880 N and T2  432 N I d4 64   (30) 4 Shigley’s MED, 11th edition 64    39.76 103 mm 4 Chapter 4 Solutions, Page 16/96 The load in between the supports supplies an angle to the overhanging end of the beam. That angle is found by taking the derivative of the deflection from that load. From Table A-9, beams 6 (subscript 1) and 10 (subscript 2), y A   BC C  a2   beam6   y A beam10 (1)  d  F1a1  l  x  2    Fa  x  a12  2lx      1 1  6lx  3 x 2  a12  2l 2      x l   x l  6 EIl  dx  6 EIl  BC C   F1a1 2 l  a12   6 EIl Equation (1) is thus  F1a1 2 Fa2 l  a12  a2  2 2 (l  a2 )  6 EIl 3EI 2070(3002 ) 3312(230) 2 2    510 230 300     3(207)103 (39.76)103  510  300  6(207)103 (39.76)103 (510)  7.99 mm Ans. yA  The slope at A, relative to the z axis is  A  z  F1a1 2  d  F (x  l)  ( x  l ) 2  a2 (3x  l )    (l  a12 )    2  6 EIl   x  l  a2  dx  6 EI F1a1 2 F l  a12   2 3( x  l ) 2  3a2 ( x  l )  a2 (3x  l )   x  l  a2 6 EIl 6 EI Fa F  1 1 (l 2  a12 )  2  3a2 2  2la2  6 EIl 6 EI 3312(230)  5102  2302   3 3 6(207)10 (39.76)10 (510) 2070 3(3002 )  2(510)(300)   3 3  6(207)10 (39.76)10  0.0304 rad Ans. ______________________________________________________________________________  4-25 From the solutions to Prob. 3-81, T1  392.16 lbf and T2  58.82 lbf I d4  64 From Table A-9, beam 6, Shigley’s MED, 11th edition  (1) 4 64  0.049 09 in 4 Chapter 4 Solutions, Page 17/96 ( 350)(14)(8) Fb x  y A   1 1  x 2  b12  l 2   82  14 2  22 2   0.0452 in Ans.   6  6 EIl  x 8in 6(30)10 (0.049 09)(22) ( 450.98)(6)(8) F b x   z A   2 2 ( x 2  b2 2  l 2 )  82  62  22 2   0.0428 in Ans.  6 6 EIl 6(30)10 (0.049 09)(22)   x 8in The displacement magnitude is   y A2  z A2  0.04522  0.04282  0.0622 in Ans. d y  d Fb x Fb     1 1  x 2  b12  l 2     1 1 (3a12  b12  l 2 )    x  a1 6 EIl  d x  x  a1  dx  6 EIl  A  z    (350)(14) 3  82   14 2  22 2   0.00242 rad 6  6(30)10 (0.04909)(22)  Ans.  dz  d  F b x Fb      2 2 ( x 2  b22  l 2 )    2 2  3a12  b22  l 2     x  a1 6 EIl  dx  6 EIl  d x  x  a1  A  y     (450.98)(6) 3  82   62  222   0.00356 rad 6   6(30)10 (0.04909)(22) Ans. The slope magnitude is  A  0.00242 2   0.00356   0.00430 rad Ans. 2 ______________________________________________________________________________ 4-26 From the solutions to Prob. 3-82, T1  250 N and T2  37.5 N I d4 64   (20)4 64  7 854 mm 4 345sin 45o  (550)(300)   F1 y b1 x 2 2 2  ( x  b1  l )  3002  550 2  8502  yA     3 6(207)10 (7 854)(850)  6 EIl  x 300mm  1.60 mm Ans. Fb x F b x  z A   1z 1 ( x 2  b12  l 2 )  2 2 ( x 2  b2 2  l 2 )  6 EIl  6 EIl  x 300mm  345 cos 45  (550)(300) o   300 6(207)10 (7 854)(850) 2 3   5502  850 2  287.5(150)(300) 3002  150 2  8502   0.650 mm  3 6(207)10 (7 854)(850) The displacement magnitude is   Shigley’s MED, 11th edition Ans. y A2  z A2  1.602   0.650   1.73 mm 2 Ans. Chapter 4 Solutions, Page 18/96 F1 y b1  d  F1 y b1 x 2   d y x  b12  l 2       (3a12  b12  l 2 )    d x  x a1  dx  6 EIl   x a1 6 EIl  A  z      345sin 45o  (550) 3  3002   5502  8502   0.00243 rad  6(207)10 (7 854)(850)  Ans. 3 dz  d F b x Fb x      1z 1  x 2  b12  l 2   2 2  x 2  b2 2  l 2    6 EIl   x  a1  dx  6 EIl  d x  x  a1  A  y    F1z b1 Fb 3a12  b12  l 2   2 2  3a12  b22  l 2   6 EIl 6 EIl o  345cos 45  (550) 3 3002  5502  8502      6(207)103 (7 854)(850)    287.5(150) 3  3002   1502  8502   1.91 10 4 rad  6(207)103 (7 854)(850)  Ans. The slope magnitude is  A  0.002432  0.0001912  0.00244 rad Ans. ______________________________________________________________________________ 4-27 From the solutions to Prob. 3-83, FB  750 lbf I d4 64   (1.25) 4 64  0.1198 in 4 From Table A-9, beams 6 (subscript 1) and 10 (subscript 2) F2 y a2 x 2  F1 y b1 x 2  yA   x  b12  l 2   l  x2    6 EIl  6 EIl  x 16in  300 cos 20  (14)(16)  o 6(30)106 (0.119 8)(30)  0.0805 in 16 2  14  30   2  750sin 20  (9)(16) o 2 6(30)106 (0.119 8)(30)  30 2  162  Ans. F ax F b x  z A   1z 1  x 2  b12  l 2   2 z 2  l 2  x 2  6 EIl  6 EIl  x 16in  300sin 20  (14)(16) o  16 6(30)10 (0.119 8)(30) 6  0.1169 in 2  14  30 2   750 cos 20  (9)(16)  o  30 6(30)10 (0.119 8)(30) 6 2  162  Ans. The displacement magnitude is   Shigley’s MED, 11th edition 2 y A2  z A2  0.08052   0.1169   0.142 in 2 Ans. Chapter 4 Solutions, Page 19/96 F2 y a2 x 2   d y  d  F1 y b1 x 2   x  b12  l 2   l  x2      6 EIl  d x  x  a1  dx  6 EIl   x  a1  A  z   F1 y b1 F2 y a2  3a  b  l   6EIl  l  3a  6 EIl  300 cos 20  (14) 3 16  14  30      6(30)10 (0.119 8)(30)   2 1 2 1 2 2 2 1 o 2 2 2 6  750sin 20  (9) o  302  3 162    8.06 105  rad  6(30)10 (0.119 8)(30)  6 Ans. dz F ax  d F b x      1z 1  x 2  b12  l 2   2 z 2  l 2  x 2     6 EIl   x  a1  dx  6 EIl  d x  x  a1  A  y    F1z b1 F a 3a12  b12  l 2   2 z 2  l 2  3a12   6 EIl 6 EIl o 750 cos 20o  (9) 300sin 20  (14)   2 2 2   302  3 162    3 16   14  30   6 6   6(30)10 (0.119 8)(30) 6(30)10 (0.119 8)(30)    0.00115 rad Ans. The slope magnitude is  A  8.06 10 5    0.001152  0.00115 rad Ans. ______________________________________________________________________________ 2 4-28 From the solutions to Prob. 3-84, FB = 22.8 (103) N 4  d 4   50    306.8 103  mm 4 I 64 64 From Table A-9, beam 6, F2 y b2 x 2  F1 y b1 x 2  yA   ( x  b12  l 2 )  ( x  b2 2  l 2 )  6 EIl  6 EIl  x  400mm 11103  sin 20o  (650)(400)  4002  6502  10502    3 3 6(207)10 (306.8)10 (1050)  22.8 103  sin 25o  (300)(400)    4002  3002  10502  6(207)103 (306.8)103 (1050)  3.735 mm Ans. Shigley’s MED, 11th edition Chapter 4 Solutions, Page 20/96 F bx F b x  z A   1z 1 ( x 2  b12  l 2 )  2 z 2 ( x 2  b2 2  l 2 )  6 EIl  6 EIl  x  400mm 11103  cos 20o  (650)(400)    4002  6502  10502  6(207)103 (306.8)103 (1050)  22.8 103  cos 25o  (300)(400)   400 2  3002  1050 2   1.791 mm  3 3 6(207)10 (306.8)10 (1050) The displacement magnitude is   y A2  z A2   3.735  2  1.7912  4.14 mm Ans. Ans. d y F bx  d F b x     1z 1  x 2  b12  l 2   2 z 2  x 2  b2 2  l 2     6 EIl   x  a1  d x  x  a1  dx  6 EIl  A  z   F1 y b1 F2 y b2  3a  b  l   6EIl  3a  b  l  6 EIl 1110  sin 20  (650)  3  400   650   6(207)10 (306.8)10 (1050)   2 1 3 2 1 2 2 1 2 2 o 2 3 2 3 2  1050 2   22.8 103  sin 25o  (300)  3  4002   3002  1050 2    3 3  6(207)10 (306.8)10 (1050)   0.00507 rad Ans. dz  d F b x F bx      1z 1  x 2  b12  l 2   2 z 2  x 2  b2 2  l 2     6 EIl   x  a1  dx  6 EIl  d x  x  a1  A  y    F1z b1 F b 3a12  b12  l 2   2 z 2  3a12  b22  l 2   6 EIl 6 EIl 3 o 1110  cos 20  (650)  3  400 2   650 2  1050 2    3 3  6(207)10 (306.8)10 (1050)    22.8 103  cos 25o  (300)  3  4002   300 2  1050 2    3 3  6(207)10 (306.8)10 (1050)  Ans.  0.00489 rad The slope magnitude is  A   0.00507    0.00489  2 2  0.00704 rad Ans. ______________________________________________________________________________ 4-29 From the solutions to Prob. 3-79, T1 = 60 lbf and T2 = 400 lbf , and Prob. 4-23, I = 0.119 8 in4. From Table A-9, beam 6, Shigley’s MED, 11th edition Chapter 4 Solutions, Page 21/96 dz  d  F1z b1 x 2 F bx  x  b12  l 2   2 z 2  x 2  b22  l 2         6 EIl  x0  dx  6 EIl  d x  x 0 575(30) F b F b   1z 1  b12  l 2   2 z 2  b2 2  l 2     302  402  6 EIl 6 EIl 6(30)106 (0.119 8)(40) O  y     460(12) 12 2  40 2   0.00468 rad  6 6(30)10 (0.119 8)(40) Ans.  d  F1z a1  l  x  2 F a l  x  2   dz x  a12  2lx   2 z 2 x  a22  2lx          6 EIl  dx  6 EIl  d x  x l   x l F a F a     1z 1  6lx  2l 2  3 x 2  a12   2 z 2  6lx  2l 2  3x 2  a22   6 EIl  6 EIl  x l C  y    F1z a1 2 F a l  a12   2 z 2  l 2  a22   6 EIl 6 EIl 2 575(10)  40  102  460(28)  402  282     0.00219 rad Ans. 6(30)106 (0.119 8)(40) 6(30)106 (0.119 8)(40) ______________________________________________________________________________  4-30 From the solutions to Prob. 3-80, T1 = 2 880 N and T2 = 432 N, and Prob. 4-24, I = 39.76 (103) mm4. From Table A-9, beams 6 and 10 d y Fa x  d Fb x  O  z       1 1 ( x 2  b12  l 2 )  2 2 (l 2  x 2 )   6 EIl   x 0  d x  x 0  dx  6 EIl Fa Fb Fal  Fb    1 1 (3x 2  b12  l 2 )  2 2 (l 2  3x 2 )   1 1 (b12  l 2 )  2 2 6 EIl 6 EI  6 EIl  x 0 6 EIl 2 070(300)(510) 3 312(280) 2802  5102     3 3 6(207)10 (39.76)10 (510) 6(207)103 (39.76)103  0.0131 rad Ans. d y Fa x  d  F1a1 (l  x ) 2  ( x  a12  2lx)  2 2 (l 2  x 2 )      6 EIl   x l  d x  x l  dx  6 EIl C  z   Fa Fa Fal  Fa    1 1 (6lx  2l 2  3 x 2  a12 )  2 2 (l 2  3 x 2 )   1 1 (l 2  a12 )  2 2 6 EIl 3EI  6 EIl  x l 6 EIl 2 070(300)(510) 3 312(230) (510 2  230 2 )   3 3 6(207)10 (39.76)10 (510) 3(207)103 (39.76)103  0.0191 rad Ans. ______________________________________________________________________________ 4-31 From the solutions to Prob. 3-81, T1 = 392.19 lbf and T2 = 58.82 lbf , and Prob. 4-25, I = 0.0490 9 in4. From Table A-9, beam 6 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 22/96 F b d y    d  F1 y b1 x 2 x  b12  l 2     1 y 1 (b12  l 2 )      d x  x  0  dx  6 EIl   x 0 6 EIl 350(14)  142  22 2   0.00726 rad Ans.  6 6(30)10 (0.04909)(22) O  z   dz F b  d  F2 z b2 x 2  x  b22  l 2      2 z 2  b22  l 2       6 EIl   x 0  dx  6 EIl  d x  x 0 450.98(6)   62  222  6(30)106 (0.04909)(22)  0.00624 rad Ans. O  y    The slope magnitude is O  0.007262   0.00624   0.00957 rad Ans. 2  d  F1 y a1 (l  x) 2   d y x  a12  2lx         d x  x l  dx  6 EIl   x l F1 y a1 2  F1 y a1   6lx  2l 2  3x 2  a12    (l  a12 )   6 EIl  x l 6 EIl 350(8) 222  82   0.00605 rad Ans.   6 6(30)10 (0.0491)(22) C  z   dz  d  F2 z a2 (l  x) 2  x  a22  2lx           x l  dx  6 EIl  d x  x l C  y    F a F a     2 z 2  6lx  2l 2  3 x 2  a22     2 z 2  l 2  a22  6 EIl  6 EIl  x l 450.98(16) 222  162   0.00846 rad   6 6(30)10 (0.04909)(22) The slope magnitude is C   0.00605  2 Ans.  0.008462  0.0104 rad Ans. ______________________________________________________________________________ 4-32 From the solutions to Prob. 3-82, T1 =250 N and T1 =37.5 N, and Prob. 4-26, I = 7 854 mm4. From Table A-9, beam 6 F b   d y  d  F1 y b1 x 2 x  b12  l 2     1 y 1 (b12  l 2 )      d x  x  0  dx  6 EIl   x 0 6 EIl O  z    345sin 45o  (550)  5502  8502   0.00680 rad  3 6(207)10 (7 854)(850) Shigley’s MED, 11th edition Ans. Chapter 4 Solutions, Page 23/96 dz  d  F1z b1 x 2 F bx  x  b12  l 2   2 z 2  x 2  b22  l 2        6 EIl  x0  dx  6 EIl  d x  x0 O  y    345 cos 45o  (550) F1z b1 2 2 F2 z b2 2 2  b1  l   b2  l    5502  8502     3 6 EIl 6 EIl 6(207)10 (7 854)(850) 287.5(150)  1502  8502   0.00316 rad Ans. 6(207)103 (7 854)(850) The slope magnitude is O  0.00680 2  0.00316 2  0.00750 rad Ans.  d  F1 y a1 (l  x ) 2   F a  d y x  a12  2lx      1 y 1  6lx  2l 2  3 x 2  a12        d x  x l  dx  6 EIl   x l  6 EIl  x l C  z    345sin 45o  (300) (l  a )  8502  3002   0.00558 rad   3 6 EIl 6(207)10 (7 854)(850) F1 y a1 2 2 1 Ans. dz  d  F1z a1 (l  x) 2 F a (l  x) 2  x  a12  2lx   2 z 2 x  a22  2lx          6 EIl   x l  dx  6 EIl  d x  x l C  y    345cos 45o  (300) F1z a1 2 F2 z a2 2 2 2   l  a1   6EIl  l  a2    6(207)103 (7 854)(850) 8502  3002  6 EIl 287.5(700)  8502  7002   6.04 105  rad Ans.  3 6(207)10 (7 854)(850) The slope magnitude is C   0.00558 2  6.04 105    0.00558 rad Ans. 2 ________________________________________________________________________ 4-33 From the solutions to Prob. 3-83, FB = 750 lbf, and Prob. 4-27, I = 0.119 8 in4. From Table A-9, beams 6 and 10  d F b x F ax  d y O  z       1 y 1  x 2  b12  l 2   2 y 2  l 2  x 2   6 EIl  d x  x 0  dx  6 EIl   x 0 F a F b F al F b    1 y 1  3 x 2  b12  l 2   2 y 2  l 2  3x 2   1 y 1  b12  l 2   2 y 2 6 EIl 6 EI  6 EIl  x 0 6 EIl  300 cos 20o  (14) 750sin 20o  (9)(30) 2 2  14  30   6(30)106 (0.119 8)  0.00751 rad 6(30)106 (0.119 8)(30) Shigley’s MED, 11th edition Ans. Chapter 4 Solutions, Page 24/96 dz  d  F1z b1 x 2 F ax  x  b12  l 2   2 z 2  l 2  x 2         6 EIl   x 0  dx  6 EIl  d x  x 0 O  y    F a F b F al F b     1z 1  3 x 2  b12  l 2   2 z 2  l 2  3 x 2     1z 1  b12  l 2   2 z 2 6 EIl 6 EIl 6 EI  6 EIl  x0 300sin 20o  (14)  750 cos 20o  (9)(30) 2 2  14  30   6(30)106 (0.119 8)  0.0104 rad 6(30)106 (0.119 8)(30) Ans. The slope magnitude is O  0.007512  0.01042  0.0128 rad Ans. F ax    dy   d  F1 y a1 (l  x) 2 x  a12  2lx   2 y 2  l 2  x 2        6 EIl  dx  x l  dx  6 EIl   x l F a F a F al F a    1 y 1  6lx  2l 2  3x 2  a12   2 y 2  l 2  3 x 2    1 y 1 (l 2  a12 )  2 y 2 6 EIl 3EI  6 EIl  x l 6 EIl C  z   750sin 20o  (9)(30)  300 cos 20o  (16) 2 2   30  16   3(30)106 (0.119 8)  0.0109 rad 6(30)106 (0.119 8)(30) Ans. dz F a x  d  F1z a1 (l  x) 2  x  a12  2lx   2 z 2  l 2  x 2         6 EIl   x l  dx  6 EIl  d x  x l C  y    F a F a F al F a     1z 1  6lx  2l 2  3x 2  a12   2 z 2  l 2  3 x 2    1z 1  l 2  a12   2 z 2 6 EIl 6 EIl 3EI  6 EIl  x l  300sin 20o  (16)  750 cos 20o  (9)(30) 2 2  30 1 6    3(30)106 (0.119 8)  0.0193 rad 6(30)106 (0.119 8)(30)  0.0109    0.0193 2 The slope magnitude is C  2 Ans.  0.0222 rad Ans. ______________________________________________________________________________ 4-34 From the solutions to Prob. 3-84, FB = 22.8 kN, and Prob. 4-28, I = 306.8 (103) mm4. From Table A-9, beam 6 F2 y b2 x 2  d  F1 y b1 x 2   d y x  b12  l 2   x  b22  l 2        6 EIl  d x  x  0  dx  6 EIl   x 0 O  z   11103  sin 20o  (650)  b l  b l     6502  10502     3 3 6 EIl 6 EIl 6(207)10 (306.8)10 (1050) F1 y b1 2 1 2 F2 y b2 2 2 2  22.8 103  sin 25o  (300)    300 2  10502   0.0115 rad  3 3 6(207)10 (306.8)10 (1050) Shigley’s MED, 11th edition Ans. Chapter 4 Solutions, Page 25/96 dz F bx  d  F1z b1 x 2  x  b12  l 2   2 z 2  x 2  b22  l 2         6 EIl   x 0  dx  6 EIl  d x  x0 F b F b   1z 1  b12  l 2   2 z 2  b22  l 2  6 EIl 6 EIl 11103  cos 20o  (650)     6502  10502  6(207)103 (306.8)103 (1050) O  y     22.8 103  cos 25o  (300)  300 2  1050 2   0.00427 rad   3 3 6(207)10 (306.8)10 (1050) The slope magnitude is O   0.0115   0.00427  2 2 Ans.  0.0123 rad Ans. F2 y a2 (l  x) 2   d y  d  F1 y a1 (l  x) 2 x  a12  2lx   x  a22  2lx         6 EIl  d x  x l  dx  6 EIl   x l F2 y a2  F1 y a1  (6lx  2l 2  3x 2  a12 )  6lx  2l 2  3x 2  a22     6 EIl  6 EIl  x l C  z   11103  sin 20o  (400)     l  a   6EIl  l  a   6(207)10 10502  4002  3 6 EIl (306.8)103 (1050) F1 y a1 2 2 1 F2 y a2 2 2 2  22.8 103  sin 25o  (750)  10502  750 2   0.0133 rad    3 3 6(207)10 (306.8)10 (1050) Ans. dz  d  F1z a1 (l  x) 2 F a (l  x) 2  x  a12  2lx   2 z 2 x  a22  2lx          6 EIl   x l  dx  6 EIl  d x  x l C  y    F a F a     1z 1  6lx  2l 2  3x 2  a12   2 z 2  6lx  2l 2  3 x 2  a22  6 EIl  6 EIl  x l 11103  cos 20o  (400) F1z a1 2 F2 z a2 2 2 2   10502  4002  l  a1   l  a2        3 3 6 EIl 6 EIl 6(207)10 (306.8)10 (1050)  22.8 103  cos 25o  (750)   10502  7502   0.0112 rad  3 3 6(207)10 (306.8)10 (1050) Ans. The slope magnitude is C  0.01332  0.01122  0.0174 rad Ans. ______________________________________________________________________________ 4-35 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-29, I = 0.119 8 in4, and it was found that the greater angle occurs at the bearing at O where (O)y =  0.00468 rad. Since is inversely proportional to I, Shigley’s MED, 11th edition Chapter 4 Solutions, Page 26/96  new Inew =  old Iold  4 Inew =  d new /64 =  old Iold / new 1/ 4 or, d new  64  old   I old     new   The absolute sign is used as the old slope may be negative. 1/ 4  64 0.00468  d new   0.119 8   1.82 in Ans.   0.00105  ______________________________________________________________________________ 4-36 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-30, I = 39.76 (103) mm4, and it was found that the greater angle occurs at the bearing at C where (C)y =  0.0191 rad. See the solution to Prob. 4-35 for the development of the equation 1/ 4 d new  64  old   I old     new   1/ 4  64 0.0191  d new   39.76 103    62.0 mm Ans.   0.00105  ______________________________________________________________________________ 4-37 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-31, I = 0.0491 in4, and the maximum slope is C = 0.0104 rad. See the solution to Prob. 4-35 for the development of the equation 1/ 4 d new  64  old  I old      new   1/ 4  64 0.0104  d new   0.0491  1.77 in Ans.   0.00105  ______________________________________________________________________________ 4-38 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-32, I = 7 854 mm4, and the maximum slope is O = 0.00750 rad. See the solution to Prob. 4-35 for the development of the equation Shigley’s MED, 11th edition Chapter 4 Solutions, Page 27/96 1/ 4 d new  64  old   I old     new   1/ 4  64 0.00750  d new   7 854   32.7 mm Ans.   0.00105  ______________________________________________________________________________ 4-39 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-33, I = 0.119 8 in4, and the maximum slope  = 0.0222 rad. See the solution to Prob. 4-35 for the development of the equation 1/ 4 d new  64  old  I old      new   1/ 4  64 0.0222  d new   0.119 8   2.68 in Ans.   0.00105  ______________________________________________________________________________ 4-40 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-34, I = 306.8 (103) mm4, and the maximum slope is C = 0.0174 rad. See the solution to Prob. 4-35 for the development of the equation 1/ 4 d new  64  old  I old      new   1/ 4  64 0.0174  d new   306.8 103    100.9 mm Ans.   0.00105  ______________________________________________________________________________ 4-41 IAB =  14/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = (0.25)(1.5)3/12 = 0.07031 in4, ICD =  (3/4)4/64 = 0.01553 in4. For Eq. (3-41), b/c = 1.5/0.25 = 6   = 0.299. The deflection can be broken down into several parts 1. The vertical deflection of B due to force and moment acting on B (y1). 2. The vertical deflection due to the slope at B, B1, due to the force and moment acting on B (y2 = CD B1 = 2B1). Shigley’s MED, 11th edition Chapter 4 Solutions, Page 28/96 3. The vertical deflection due to the rotation at B, B2, due to the torsion acting at B (y3 = BC B1 = 5B1). 4. The vertical deflection of C due to the force acting on C (y4). 5. The rotation at C, C, due to the torsion acting at C (y3 = CD C = 2C). 6. The vertical deflection of D due to the force acting on D (y5). 1. From Table A-9, beams 1 and 4 with F =  200 lbf and MB = 2(200) = 400 lbfin 400  6 2  200  63  y1     0.01467 in 3  30 10 6  0.04909  2  30 106  0.04909  2. From Table A-9, beams 1 and 4  d  Fx 2 M x2   Fx M x  B1     x  3l   B      3x  6l   B  EI  x l 2 EI   x l  6 EI  dx  6 EI 6  l     200  6   2  400    0.004074 rad   Fl  2M B   6  2 EI  2  30 10  0.04909  y 2 = 2(0.004072) = 0.00815 in 3. The torsion at B is TB = 5(200) = 1000 lbfin. From Eq. (4-5)  TL  1000  6  B2    0.005314 rad   6  JG  AB 0.09818 11.5 10 y 3 = 5(0.005314) = 0.02657 in 4. For bending of BC, from Table A-9, beam 1 y4   200  53  3  30 10 6  0.07031  0.00395 in 5. For twist of BC, from Eq. (3-41), with T = 2(200) = 400 lbfin C  400  5   0.02482 rad 0.299 1.5  0.253 11.5 10 6 y 5 = 2(0.02482) = 0.04964 in 6. For bending of CD, from Table A-9, beam 1 y6   200  23  3  30 106  0.01553  Shigley’s MED, 11th edition  0.00114 in Chapter 4 Solutions, Page 29/96 Summing the deflections results in 6 y D   yi  0.01467  0.00815  0.02657  0.00395  0.04964  0.00114  0.1041 in Ans. i 1 This problem is solved more easily using Castigliano’s theorem. See Prob. 4-78. ______________________________________________________________________________ 4-42 The deflection of D in the x direction due to Fz is from: 1. The deflection due to the slope at B, B1, due to the force and moment acting on B (x1 = BC B1 = 5B1). 2. The deflection due to the moment acting on C (x2). 1. For AB, IAB =  14/64 = 0.04909 in4. From Table A-9, beams 1 and 4 M B x 2   M x  d  Fx 2  Fx   x l 3    3x  6l   B      EI  x l 2 EI   x l  6 EI  dx  6 EI  B1   6  l    100  6   2  200    0.002037 rad   Fl  2M B   6  2 EI  2  30 10  0.04909  x 1 = 5( 0.002037) =  0.01019 in 2. For BC, IBC = (1.5)(0.25)3/12 = 0.001953 in4. From Table A-9, beam 4 2  100  5 M Cl 2   0.04267 in x2  2 EI 2  30 106  0.001953 The deflection of D in the x direction due to Fx is from: 3. The elongation of AB due to the tension. For AB, the area is A =  12/4 = 0.7854 in2 150  6   Fl  x3    3.82 10 5  in   6  AE  AB 0.7854  30 10 4. The deflection due to the slope at B, B2, due to the moment acting on B (x1 = BC B2 = 5B2). With IAB = 0.04907 in4, B2  5  150  6 M Bl   0.003056 rad EI 30 106  0.04909 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 30/96 x4 = 5( 0.003056) =  0.01528 in 5. The deflection at C due to the bending force acting on C. With IBC = 0.001953 in4 150  53   Fl 3     0.10667 in x5     6 3 3 30 10 0.001953 EI       BC 6. The elongation of CD due to the tension. For CD, the area is A =  (0.752)/4 = 0.4418 in2 150  2   Fl  x6    2.26 10 5  in   6  AE CD 0.4418  30 10 Summing the deflections results in 6 xD   xi  0.01019  0.04267  3.82 105  i 1  0.01528  0.10667  2.26 105   0.1749 in Ans. ______________________________________________________________________________ 4-43 JOA = JBC =  (1.54)/32 = 0.4970 in4, JAB =  (14)/32 = 0.09817 in4, IAB =  (14)/64 = 0.04909 in4, and ICD =  (0.754)/64 = 0.01553 in4. T  lOA l AB lBC   Tl   Tl   Tl               GJ OA  GJ  AB  GJ  BC G  J OA J AB J BC  250(12)  2 9 2      0.0260 rad 6  11.5(10 )  0.4970 0.09817 0.4970  Simplified Tl 250(12)(13) s   GJ 11.5 106   0.09817   Ans.  s  0.0345 rad Ans. Simplified is 0.0345/0.0260 = 1.33 times greater Ans. yD  Fy lOC 3 3EI AB   s  lCD   Fy lCD 3 3EI CD  250 133  3(30)106  0.04909   0.0345(12)  250 123  3(30)106  0.01553  yD  0.847 in Ans. ______________________________________________________________________________ 4-44 Reverse the deflection equation of beam 7 of Table A-9. Using units in lbf, inches Shigley’s MED, 11th edition Chapter 4 Solutions, Page 31/96 y  3000 / 12  x 2 25 x 2  x3   25 12  3 wx 2lx 2  x 3  l 3          24 EI 24  30 106  485    7.159 1010  x  27 106   600 x 2  x 3   Ans. The maximum height occurs at x = 25(12)/2 = 150 in ymax  7.159 10 10 150  27 106   600 150 2   1503   1.812 in Ans. ______________________________________________________________________________ 4-45 From Table A-9, beam 6, Fbx 2 yL  x  b2  l 2   6 EIl Fb 3 yL  x  b2 x  l 2 x   6 EIl dyL Fb  3x2  b2  l 2  dx 6 EIl dyL dx Let   dyL dx  Fb  b 2  l 2  6 EIl x 0 and set I  x 0 dL   d L4 64 . Thus, 32 Fb  b 2  l 2  1/ 4 Ans. 3 El For the other end view, observe beam 6 of Table A-9 from the back of the page, noting that a and b interchange as do x and –x dR  32 Fa  l 2  a 2  3 El 1/4 Ans. For a uniform diameter shaft the necessary diameter is the larger of d L and d R . ______________________________________________________________________________ 4-46 The maximum slope will occur at the left bearing. Incorporating a design factor into the solution for d L of Prob. 4-45, Shigley’s MED, 11th edition Chapter 4 Solutions, Page 32/96  32nFb  l 2  b 2    d  3 El   d 4 32(1.28)(3000)(200)  300 2  200 2  3 (207)103 (300)(0.001) d  38.1 mm I 1/4   38.14  Ans.  103.4 103  mm 4 64 From Table A-9, beam 6, the maximum deflection will occur in BC where dyBC /dx = 0  d  Fa  l  x  2 x  a 2  2lx   0  3 x 2  6lx   a 2  2l 2   0   dx  6 EIl  3 x 2  6  300  x  1002  2  3002    0  x 2  600 x  63333  0 x 1 600  600 2  4(1)63 333   463.3, 136.7 mm  2 x = 136.7 mm is acceptable.  Fa  l  x  2  ymax   x  a 2  2lx     6 EIl  x 136.7 mm 3 103 100  300  136.7  136.7 2  1002  2  300 136.7   0.0678 mm Ans. 6  207 10 103.4 10  300  ______________________________________________________________________________  3 3 4-47 I =  (1.254)/64 = 0.1198 in4. From Table A-9, beam 6 2  F a (l  x) 2  F b x    1 1 ( x  a12  2lx)    2 2 ( x 2  b2 2  l 2 )   6 EIl   6 EIl  2 2  150(5)(20  8)   2 2   8  5  2(20)(8)  6   6(30)10  0.1198  (20)    250(10)(8) 82  102  202    6  6(30)10  0.1198  (20)  2 1/ 2     0.0120 in Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 4 Solutions, Page 33/96 4-48 I =  (1.254)/64 = 0.1198 in4. For both forces use beam 6 of Table A-9. For F1 = 150 lbf: 0x5 150 15  x Fb x y  1 1  x 2  b12  l 2   x 2  152  202   6 6 EIl 6  30 10  0.1198  20   5.217 10 6  x  x 2  175  5  x  20 y F1a1  l  x  6 EIl x 2 (1)  a12  2lx   150  5  20  x   x 2  52  2  20  x  6  30 10  0.1198  20  6  1.739 10 6   20  x   x 2  40 x  25  (2) For F2 = 250 lbf: 0  x  10 250 10  x Fb x z  2 2  x 2  b22  l 2   x 2  10 2  20 2   6 6 EIl 6  30 10  0.1198  20   5.797 106  x  x 2  300  (3) 10  x  20 F a l  x  2 250 10  20  x   x 2  102  2  20  x  z 2 2 x  a22  2lx    6 EIl 6  30 106  0.1198  20    5.797 106   20  x   x 2  40 x  100  (4) Plot Eqs. (1) to (4) for each 0.1 in using a spreadsheet. There are 201 data points, too numerous to tabulate here but the plot is shown below, where the maximum deflection of  = 0.01255 in occurs at x = 9.9 in. Ans. 0.015 0.01 0.005 y (in) 0 Displacement (in) z (in) -0.005 Total (in) -0.01 -0.015 0 5 10 15 20 x (in) ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 4 Solutions, Page 34/96 4-49 The larger slope will occur at the left end. From Table A-9, beam 8 MBx 2 ( x  3a 2  6al  2l 2 ) 6 EIl dy AB M B  (3 x 2  3a 2  6al  2l 2 ) dx 6 EIl y AB  With I =  d 4/64, the slope at the left bearing is dy AB dx x 0  A  MB (3a 2  6al  2l 2 ) 4 6 E  d / 64  l Solving for d 32 M B 32(1000) 3(42 )  6(4)(10)  2 102   3a 2  6al  2l 2   4 d4  6  3 E Al 3 (30)10 (0.002)(10)  Ans.  0.461 in ______________________________________________________________________________ 4-50 From Table A-5, E = 10.4 Mpsi MO = 0 = 18 FBC  6(100)  FBC = 33.33 lbf The cross sectional area of rod BC is A =  (0.52)/4 = 0.1963 in2. The deflection at point B will be equal to the elongation of the rod BC. 33.33(12)  FL  yB    6.79 105  in   6  AE  BC  0.1963 30 10  Ans. ______________________________________________________________________________ 4-51 MO = 0 = 6 FAC  11(100)  FAC = 183.3 lbf The deflection at point A in the negative y direction is equal to the elongation of the rod AC. From Table A-5, Es = 30 Mpsi. 183.3 12   FL  yA     3.735 104  in   2 6   0.5  / 4  30 10   AE  AC   By similar triangles the deflection at B due to the elongation of the rod AC is y A y B1  6 18  y B1  3 y A  3( 3.735)10 4  0.00112 in From Table A-5, Ea = 10.4 Mpsi The bar can then be treated as a simply supported beam with an overhang AB. From Table A-9, beam 10, Shigley’s MED, 11th edition Chapter 4 Solutions, Page 35/96  dy  Fa 2  d  F (x  l) Fa 2  2   yB 2  BD  BC  ( l  a )  7 ( x  l )  a (3 x  l )  (l  a)         x l  a 3EI  dx  6 EI  dx x l  a  3EI F Fa 2 7 Fa Fa 2 2 3( x  l )  3a( x  l )  a (3 x  l )  |x l  a  (l  a )   (2l  3a )  (l  a ) 7 6 EI  3EI 6 EI 3EI 100  52  7 100  5  (6  5)  2(6)  3(5)  6(10.4)106  0.25(23 ) / 12  3(10.4)106  0.25(23 ) / 12     0.01438 in yB = yB1 + yB2 =  0.00112  0.01438 =  0.0155 in Ans. ______________________________________________________________________________ 4-52 From Table A-5, EA = 71.7 GPa, ES = 207 GPa. MO = 0 = 450 FCD – 650(4000)  FCD = 5777.8 N 5 777.8  220   FL   0.2172 mm yD       / 4  0.0062  207 109  AE CD The deflection of B due to yD is (yB)1 = (650/450) ( 0.2172) =  0.3137 mm Treat beam OADB as simply-supported at O and D. Use beam 10 of Table A-9 and use the equation for yC, 2 Fa 2  l  a  4000  0.2   0.450  0.2    yB 2   3EI 3  71.7 109  0.012  0.0503 / 12   3.868 10 3  m  3.868 mm Superposition: yB = (yB)1 + (yB)2 =  0.3137  3.868 =  4.18 mm Ans. ______________________________________________________________________________ 4-53 From Table A-5, EA = 71.7 MPa, ES = 207 MPa MO = 0 = 450 FCD – 150(4000)  FCD = 1333 N 1333  220   FL  yD     0.0501 mm    / 4  0.0062  207 109  AE CD The deflection of B due to yD is (yB)1 = (650/450) ( 0.0501) =  0.0724 mm Treat beam OADB as simply-supported at O and D. Find slope at B for beam 6 of Table A-9, Fa   x 3  3lx 2   2l 2  a 2  x  la 2  yBC   6 EIl  dy Fa  3 x 2  6lx  2l 2  a 2   BC  BC  dx 6 EIl For OADB, Shigley’s MED, 11th edition Chapter 4 Solutions, Page 36/96 4000  0.15   3  0.450   6  0.450  0.450  2  0.450   0.152    D  9 3 6  71.7 10  0.012   0.050 / 12  0.450 2 2  0.004 463 rad Deflection of B due to slope at D is (yB)2 = 0.004 463(200) = 0.8926 mm Superposition: yB = (yB)1 + (yB)2 =  0.0725 + 0.8926 = 0.820 mm Ans. ______________________________________________________________________________ 4-54 From Table A-5, EA = 10.4 Mpsi, ES = 30 Mpsi MO = 0 = 18 FCD – 6(100)  FCD = 33.33 lbf 33.33 12   FL   6.792 105  in yB      2 6  / 4  0.5  30 10  AE  BC The deflection of A due to yB is (yA)1 = (6/18) ( 6.792)105 =  2.264 (105) in Treat beam OAB as simply-supported at O and D. Use beam 6 of Table A-9  y A 2 100 12  6  62  122  182  Fbx 2 2 2   5.538 10 3  in x b l     6 3 6 EIl 6 10.4 10  0.25  2 / 12   18 Superposition: yA = (yA)1 + (yA)2 =  2.264 (105)  5.538 (103) =  5.56 (103) in Ans. ______________________________________________________________________________ 4-55 From Table A-5, ES = 30 Mpsi MO = 0 = 18 FAC – 11(100)  FAC = 183.3 lbf 183.3 12   FL   3.734 104  in yA    Ans.   2 6 AE  / 4 0.5 30 10       AC _____________________________________________________________________________________________________________________ 4-56 From Table A-5, EA = 71.7 MPa, ES = 207 MPa MO = 0 = 450 FCD – 650(4000)  FCD = 5777.8 N 5 777.8  220   FL  yD     0.2172 mm    / 4  0.0062  207 109  AE CD The deflection of A due to yD is (yA)1 = (150/300) ( 0.2172) =  0.1086 mm Treat beam OADB as simply-supported at O and D. Use beam 10 of Table A-9 4000  0.2  0.15  0.450 2  0.150 2  Fax 2 2  y A 2   l  x   6 71.7 109  0.012 0.0503 / 12  0.450 6 EIl        0.8926 103  m  0.8926 mm Superposition: yA = (yA)1 + (yA)2 =  0.1086 + 0.8926 = 0.784 mm Ans. ______________________________________________________________________________ 4-57 From Table A-5, EA = 71.7 MPa, ES = 207 MPa Shigley’s MED, 11th edition Chapter 4 Solutions, Page 37/96 MO = 0 = 450 FCD – 150(4000)  FCD = 1333 N 1333  220   FL  yD     0.0501 mm    / 4  0.0062  207 109  AE CD The deflection of A due to yD is (yA)1 = (150/300) ( 0.0501) =  0.02505 mm Treat beam OADB as simply-supported at O and D. Use beam 6 of Table A-9 4000  0.3 0.15  0.152  0.32  0.452  Fbx 2 2 2  y A 2   x  b  l   6 71.7 109  0.012 0.0503 / 12  0.450 6 EIl        0.6695 103  m  0.6695 mm Superposition: yA = (yA)1 + (yA)2 =  0.02505  0.6695 =  0.695 mm Ans. ______________________________________________________________________________ 4-58 From Table A-5, E = 207 GPa, and G = 79.3 GPa. FlOC l AB 2 Fl AC l AB 2 Fl AB 3 Fl AB 3  Tl   Tl  yB   l l       AB   AB 3EI AB G  dOC 4 / 32  G  d AC 4 / 32  3E  d 34 / 64   GJ OC  GJ  AC  l 32 Fl AB 2  lOC 2l AB   AC 4   4 4   Gd OC Gd AC 3Ed AB  The spring rate is k = F/ yB. Thus  32l 2 k   AB    lOC l 2l AB    AC 4   4 4   GdOC Gd AC 3Ed AB   1 1  32  2002     2  200  200 200      3 4 3 4 3 4   79.3 10 18 79.3 10 12 3  207 10 8      8.10 N/mm Ans. _____________________________________________________________________________ 4-59 For the beam deflection, use beam 5 of Table A-9. F R1  R2  2 F F 1  , and  2  2k1 2k 2 y AB  1  1   2 l x Fx (4 x 2  3l 3 ) 48 EI  1 k2  k1  x  y AB  F   x (4 x 2  3l 3 )  48 EI  2k1 2k1k2l  Shigley’s MED, 11th edition Ans. Chapter 4 Solutions, Page 38/96 For BC, since Table A-9 does not have an equation (because of symmetry) an equation will need to be developed as the problem is no longer symmetric. This can be done easily using beam 6 of Table A-9 with a = l /2 F  l / 2  l  x   2 l 2   F Fk 2  Fk1  yBC  x  x   2lx  2k1 2k1k 2l 4 EIl    1 k2  k1  l  x  4 x 2  l 2  8lx  Ans.  F   x   48 EI  2k1 2k1k2l  ______________________________________________________________________________ 4-60 Fa F , and R2  (l  a ) l l Fa F , and  2  (l  a ) 1  lk1 lk 2 R1  y AB  1  1   2 l x Fax 2 (l  x 2 ) 6 EIl  a  x ax 2  k a  k1  l  a    y AB  F   (l  x 2 )  Ans. 2  2 6 EIl  k1l k1k2l    F (x  l) ( x  l ) 2  a (3 x  l )  yBC  1  1 2 x  l 6 EI   a  x (x  l) ( x  l ) 2  a (3 x  l )   yBC  F   Ans.  k a  k1  l  a    2  2 6 EI  k1l k1k2l  ______________________________________________________________________________ 4-61 Let the load be at x ≥ l/2. The maximum deflection will be in Section AB (Table A-9, beam 6) y AB  Fbx 2 x  b2  l 2   6 EIl dy AB Fb   3x 2  b2  l 2   0 dx 6 EIl x  l 2  b2 l2 , xmax   0.577l 3 3 3x 2  b2  l 2  0 Ans. For x  l/2, xmin  l  0.577l  0.423l Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 4 Solutions, Page 39/96 4-62 M O  1(3000)(1500)  2500(2000)  9.5 106  N·mm RO  1(3000)  2500  5 500 N From Prob. 4-10, I  4.14(106 ) mm 4 x2 1  2500 x - 2 000 2 dy x3 2 6 2 EI  9.5 10  x  2 750 x   1250 x  2000  C1 6 dx M  9.5 106   5500 x  dy  0 at x  0  C1  0 dx dy x3 2 EI  9.5 106  x  2 750 x 2   1250 x  2000 dx 6 x4 3 EIy  4.75 106  x 2  916.67 x 3   416.67 x  2000  C2 24 y  0 at x  0  C2  0 , and therefore y 1  3 114 106  x 2  22 103  x 3  x 4  10 103  x  2000    24 EI yB   1 114 106  3000 2  22 103  30003 3 6  24  207 10  4.14 10 3 30004  10 103   3000  2000     25.4 mm Ans. MO = 9.5 (106) Nm. The maximum stress is compressive at the bottom of the beam where y = 29.0  100 =  71 mm 9.5 106  (71) My  max     163 106  Pa  163MPa Ans. 4.14(106 ) I The solutions are the same as Prob. 4-10. ______________________________________________________________________________ 4-63 See Prob. 4-11 for reactions: RO = 465 lbf and RC = 285 lbf. Using lbf and inch units M = 465 x  450 x  721  300 x  1201 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 40/96 dy 2 2  232.5 x 2  225 x  72  150 x  120  C1 dx EIy = 77.5 x3  75 x  723  50 x  1203  C1x EI y = 0 at x = 0  C2 = 0 y = 0 at x = 240 in 0 = 77.5(2403)  75(240 72)3  50(240  120)3 + C1 x  and, EIy = 77.5 x3  75 x  723  50 x  1203 2.622(106) x C1 =  2.622(106) lbfin2 Substituting y =  0.5 in at x = 120 in gives 30(106) I ( 0.5) = 77.5 (1203)  75(120  72)3  50(120  120)3 2.622(106)(120) I = 12.60 in4 Select two 5 in  6.7 lbf/ft channels; from Table A-7, I = 2(7.49) = 14.98 in4 12.60  1  Ans.     0.421 in 14.98  2  The maximum moment occurs at x = 120 in where Mmax = 34.2(103) lbfin ymidspan  Mc 34.2(103 )(2.5)   5 710 psi O.K. I 14.98 The solutions are the same as Prob. 4-11. ______________________________________________________________________________  max  4-64 I =  (1.54)/64 = 0.2485 in4, and w = 150/12 = 12.5 lbf/in. 1 24 RO  12.5  39  (340)  453.0 lbf 2 39 12.5 2 1 M  453.0 x  x  340 x  15 2 dy 12.5 3 2 EI  226.5 x 2  x  170 x  15  C1 dx 6 3 EIy  75.5 x 3  0.5208 x 4  56.67 x  15  C1 x  C2 y  0 at x  0  C2  0 y  0 at x  39 in y  C1  6.385(10 4 ) lbf  in 2 Thus, 1  3 75.5 x 3  0.5208 x 4  56.67 x  15  6.385 10 4  x    EI Evaluating at x = 15 in, Shigley’s MED, 11th edition Chapter 4 Solutions, Page 41/96 1  75.5 153   0.5208 154   56.67 15  15 3  6.385 104  (15)   30(10 )(0.2485)   0.0978 in Ans. yA  6 1 75.5 19.53   0.5208 19.54   56.67 19.5  15 3  6.385 10 4  (19.5)   30(10 )(0.2485)    0.1027 in Ans. ymidspan  6 5 % difference Ans. The solutions are the same as Prob. 4-12. ______________________________________________________________________________ 3 14 100 7 14 100 4-65 I = 0.05 in4, RA   420 lbf  and RB   980 lbf  10 10 M = 420 x  50 x2 + 980  x  10 1 EI dy 2  210 x 2  16.667 x 3  490 x  10  C1 dx 3 EIy  70 x 3  4.167 x 4  163.3 x  10  C1 x  C 2 y = 0 at x = 0  C2 = 0 y = 0 at x = 10 in  C1 =  2 833 lbfin2. Thus, y 1 70 x 3  4.167 x 4  163.3 x  10 3  2833 x  6  30 10  0.05  3 Ans.  6.667 107  70 x 3  4.167 x 4  163.3 x  10  2833 x    The tabular results and plot are exactly the same as Prob. 4-21. ______________________________________________________________________________ 4-66 RA = RB = 400 N, and I = 6(323) /12 = 16 384 mm4. First half of beam, M =  400 x + 400  x  300 1 dy 2 EI  200 x 2  200 x  300  C1 dx From symmetry, dy/dx = 0 at x = 550 mm   0 =  200(5502) + 200(550 – 300) 2 + C1 C1 = 48(106) N·mm2 EIy =  66.67 x3 + 66.67  x  300 3 + 48(106) x + C2 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 42/96 y = 0 at x = 300 mm C2 =  12.60(109) N·mm3.  The term (EI)1 = [207(103)16 384] 1 = 2.949 (1010 ) Thus y = 2.949 (1010) [ 66.67 x3 + 66.67  x  300 3 + 48(106) x  12.60(109)] yO =  3.72 mm Ans. yx = 550 mm =2.949 (1010) [ 66.67 (5503) + 66.67 (550  300)3 + 48(106) 550  12.60(109)] = 1.11 mm Ans. The solutions are the same as Prob. 4-13. ______________________________________________________________________________ 4-67 1  M A  Fa  l 1  M A  0  M A  R2l  F (l  a)  R2  l  Fl  Fa  M A  M B  0  R1l  Fa  M A M  R1 x  M A  R2 x  l  R1  1 1 dy 1 2  R1 x 2  M A x  R2 x  l  C1 2 dx 2 1 1 1 3 EIy  R1 x 3  M A x 2  R2 x  l  C1 x  C2 6 2 6 EI y = 0 at x = 0  y = 0 at x = l  EIy  y C2 = 0 1 1 C1   R1l 2  M Al . Thus, 6 2 1 1 1 1 3  1  R1 x3  M A x 2  R2 x  l    R1l 2  M Al  x 6 2 6 2  6  1  3  M A  Fa  x 3  3M A x 2l   Fl  Fa  M A  x  l   Fal 2  2M Al 2  x   6 EIl Ans. In regions, 1  M A  Fa  x 3  3M A x 2l   Fal 2  2 M Al 2  x   6 EIl  x  M A  x 2  3lx  2l 2   Fa  l 2  x 2   Ans.   6 EIl  y AB  Shigley’s MED, 11th edition Chapter 4 Solutions, Page 43/96 1  3  M A  Fa  x3  3M A x 2l   Fl  Fa  M A  x  l    Fal 2  2M Al 2  x   6 EIl 1 3 3 M A  x 3  3 x 2l   x  l   2 xl 2   F   ax 3   l  a  x  l   axl 2       6 EIl 1 2   M A  x  l  l 2  Fl  x  l   x  l   a  3 x  l     6 EIl  x  l   M l  F  x  l 2  a 3x  l  Ans.     A  6 EI yBC        The solutions reduce to the same as Prob. 4-17. ______________________________________________________________________________ w b  a  1    R1  4-68  M D  0  R1l  w  b  a  l  b   b  a    2l  b  a  2 2l   w w 2 2 M  R1 x  xa  xb 2 2 dy 1 w w 3 3 EI  R1 x 2  xa  x  b  C1 dx 2 6 6 1 w w 4 4 EIy  R1 x 3  xa  x  b  C1 x  C2 6 24 24 y = 0 at x = 0 y = 0 at x = l  C2 = 0 1 1 w w 4 4 C1    R1l 3   l  a    l  b   24 24 l 6  y 1  1 w b  a w w 4  2l  b  a  x3  x  a  x  b  2l 24 24 EI  6 4 1  1 w b  a  w w 4 4   x   2l  b  a  l 3   l  a    l  b    2l 24 24 l 6     w 4 2  b  a  2l  b  a  x3  l x  a  l x  b 24 EIl 4  4 4  x  2  b  a  2l  b  a  l 2   l  a    l  b     Ans. The above answer is sufficient. In regions, w 4 4 2  b  a  2l  b  a  x 3  x  2  b  a  2l  b  a  l 2   l  a    l  b   y AB     24 EIl wx  4 4  2  b  a  2l  b  a  x 2  2  b  a  2l  b  a  l 2   l  a    l  b     24 EIl  Shigley’s MED, 11th edition Chapter 4 Solutions, Page 44/96 yBC   w 4 2  b  a  2l  b  a  x3  l  x  a  24 EIl  4 4  x  2  b  a  2l  b  a  l 2   l  a    l  b     yCD   w 4 4 2  b  a  2l  b  a  x3  l  x  a   l  x  b  24 EIl  4 4  x  2  b  a  2l  b  a  l 2   l  a    l  b     These equations can be shown to be equivalent to the results found in Prob. 4-19. ______________________________________________________________________________ 4-69 I1 =  (1.3754)/64 = 0.1755 in4, I2 =  (1.754)/64 = 0.4604 in4, R1 = 0.5(180)(10) = 900 lbf Since the loading and geometry are symmetric, we will only write the equations for the first half of the beam 2 For 0  x  8 in M  900 x  90 x  3 At x = 3, M = 2700 lbfin Writing an equation for M / I, as seen in the figure, the magnitude and slope reduce since I 2 > I 1. To reduce the magnitude at x = 3 in, we add the term,  2700(1/I 1  1/ I 2) x  3 0. The slope of 900 at x = 3 in is also reduced. We account for this with a ramp function,  x  31 . Thus, 1 1 1 1 M 900 x 90 0 1 x3   2700    x  3  900    x  3  I I1 I2  I1 I 2   I1 I 2  0 1  5128 x  9520 x  3  3173 x  3  195.5 x  3 E 2 2 dy 1 2 3  2564 x 2  9520 x  3  1587 x  3  65.17 x  3  C1 dx Boundary Condition: dy  0 at x  8 in dx 0  2564  8   9520 8  3   1587  8  3   65.17 8  3   C1  2 2 3 3 4 C1 =  68.67 (103) lbf/in2 2 Ey  854.7 x 3  4760 x  3  529 x  3  16.29 x  3  68.67(103 ) x  C2 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 45/96 y = 0 at x = 0 C2 = 0  Thus, for 0  x  8 in 1  2 3 4 y 854.7 x 3  4760 x  3  529 x  3  16.29 x  3  68.7(103 ) x  6   30(10 ) Ans. Using a spreadsheet, the following graph represents the deflection equation found above Beam Deflection 0 -0.002 0 1 2 3 4 5 6 7 8 -0.004 y (in) -0.006 -0.008 -0.01 -0.012 x (in) The maximum is ymax  0.0102 in at x  8 in Ans. ______________________________________________________________________________ 4-70 The force and moment reactions at the left support are F and Fl respectively. The bending moment equation is M = Fx  Fl Plots for M and M /I are shown. M /I can be expressed using singularity functions M F Fl Fl l  x  x I 2 I1 2 I1 4 I1 2 0 F l  x 2 I1 2 1 where the step down and increase in slope at x = l /2 are given by the last two terms. Integrate E dy F 2 Fl Fl l  x  x x 2 I1 4 I1 2 dx 4 I1 Shigley’s MED, 11th edition 1  F l x 4 I1 2 2  C1 Chapter 4 Solutions, Page 46/96 dy/dx = 0 at x = 0  C1 = 0 2 3 F 3 Fl 2 Fl l F l Ey  x  x  x  x  C2 12 I1 4 I1 8I1 2 12 I1 2 y = 0 at x = 0  C2 = 0 2 3 F  3 l l  2 2 x y  2 x  6lx  3l x   24 EI1  2 2  3 2  5 Fl 3 F  l l y x l / 2  Ans.  2    6l    3l (0)  2(0)    24 EI1   2  96 EI1 2  2 3 F  l  3Fl 3 3  l  2 y x l  Ans.  2  l   6l  l   3l  l    2  x      24 EI1  2   16 EI1  2  The answers are identical to Ex. 4-10. ______________________________________________________________________________ 4-71 Place a fictitious force, Q, at the center. The reaction, R1 = wl / 2 + Q / 2 M x wx2  wl Q  M   x  2 Q 2  2 2 Integrating for half the beam and doubling the results ymax  1  2  EI l/2  0  M M  Q   2  dx    Q  0 EI l /2  0  wl  wx2   x  x   dx  2  2   2    Note, after differentiating with respect to Q, it can be set to zero l/2 l/2 w  x 3l x 4  5w   ymax x l x dx 0   2EI  3  4   384 EI Ans. 0 ______________________________________________________________________________ w  2 EI 2 4-72 Place a fictitious force Q pointing downwards at the end. Use the variable x originating at the free end and positive to the left wx2 M M  Qx   x 2 Q l l  1 l  M   1  wx 2  w ymax    M  x3dx     x  dx   dx     2 EI 0  EI 0   Q   Q  0 EI 0  2  wl 4 Ans. 8EI ______________________________________________________________________________  Shigley’s MED, 11th edition Chapter 4 Solutions, Page 47/96 4-73 From Table A-7, I1-1 = 1.85 in4. Thus, I = 2(1.85) = 3.70 in4 First treat the end force as a variable, F. Adding weight of channels of 2(5)/12 = 0.833 lbf/in. Using the variable x as shown in the figure M  F x  M  x F A  1 EI  60 0 5.833 2 x   F x  2.917 x 2 2 M M 1 dx F EI  60 0 ( F x  2.917 x 2 )( x ) d x  60 1 ( Fx 3 / 3  2.917 x 4 / 4) 0 EI (150)(603 ) / 3  (2.917)(60 4 ) / 4   0.182 in in the direction of the 150 lbf force 30(106 )(3.70)  y A   0.182 in Ans. ______________________________________________________________________________ 4-74 P a RA   Q 2 l P l  a  RB   Q l 2 M a P a  M    Q x  x Q l 2 l  l l M a P a   M    Q x  P x    xl  x Q l 2 2 2 l   M l  x  (l  a) M  Q l  a  x   l  a  x Q We observe that for section BD, the only force is Q, which is zero, so there is no contribution to the deflection at D from the strain energy in section BD. 0 x l 2 l  Pa  1 l  a M  Pa 2 Pa   1  l /2 Pa 2 yD   M dx x dx x2  x  x  dx         l 0 0 /2 Q Q  0 EI  l 2l 2    2l  EI The first two integrals can be combined from 0 to l, 3 2 1  Pal 3 Pa  3  l   Pa  2  l    Pal 2     yD  l l Ans.           EI  6l l   2   4   2    16 EI (b) Table A-9, beam 5 with F = P, dy P P Slope:   12 x 2  3l 2      4x2  l 2  dx 48EI 16 EI Shigley’s MED, 11th edition Chapter 4 Solutions, Page 48/96 Pl 2 Pl 2 . By symmetry,  C  16 EI 16 EI 2 Pal yC  a C  Ans. This agrees with part (a) 16 EI ______________________________________________________________________________ At x = 0,  A   4-75 The energy includes torsion in AC, torsion in CO, and bending in AB. Neglecting transverse shear in AB M  Fx, M x F In AC and CO, T T  Fl AB ,  l AB F The total energy is  T 2l   T 2l  U       2GJ  AC  2GJ CO l AB  0 M2 dx 2 EI AB The deflection at the tip is    k U Tl AC T TlCO T    F GJ AC F GJ CO F l AB  0 Tl l Tl l 1 M M dx  AC AB  CO AB  EI 3 F GJ AC GJ CO EI AB l AB  Fx dx 2 0 3 2 2 3 Tl AC l AB TlCO l AB Fl AC l AB FlCO l AB Fl AB Fl AB      4 4 4 GJ AC GJ CO 3EI AB G  d AC / 32  G  dCO / 32  3E  d AB / 64  2  l AC l 32 Fl AB 2l AB   CO4   4 4    Gd AC Gd CO 3Ed AB  F    2 32l AB  l AC l 2l AB   CO4   4 4   Gd AC GdCO 3Ed AB  1 1   2  200  200 200    8.10 N/mm Ans.    32  2002   79.3 103 184 79.3 103 124 3  207 103 84   ______________________________________________________________________________  4-76 I1 =  (1.3754)/64 = 0.1755 in4, I2 =  (1.754)/64 = 0.4604 in4 Place a fictitious force Q pointing downwards at the midspan of the beam, x = 8 in Shigley’s MED, 11th edition Chapter 4 Solutions, Page 49/96 R1  1 1 (10)180  Q  900  0.5Q 2 2 For 0  x  3 in M   900  0.5Q  x M  0.5 x Q M  0.5 x Q For 3  x  13 in M   900  0.5Q  x  90( x  3) 2 By symmetry it is equivalent to use twice the integral from 0 to 8 3 8  8 M M  1 1  2 2    2 dx   900 x dx  900 x  90  x  3  x dx    EI 2 3   0 EI Q Q 0 EI1 0 3 8 300 x 3 1  1 9    300 x3  90( x 4  2 x 3  x 2 )   EI1 0 EI 2  4 2 3 120.2 103  8100 1 8100 3 3 145.5 10   25.3110       30 106 0.1755  30 106 0.4604 EI1 EI 2       0.0102 in Ans. ______________________________________________________________________________ 4-77 I =  (0.54)/64 = 3.068 (103) in4, J = 2 I = 6.136 (103) in4, A = (0.52)/4 = 0.1963 in2. Consider x to be in the direction of OA, y vertically upward, and z in the direction of AB. Resolve the force F into components in the x and y directions obtaining 0.6 F in the horizontal direction and 0.8 F in the negative vertical direction. The 0.6 F force creates strain energy in the form of bending in AB and OA, and tension in OA. The 0.8 F force creates strain energy in the form of bending in AB and OA, and torsion in OA. Use the dummy variable x to originate at the end where the loads are applied on each segment, 0.6 F: AB OA M  0.6 F x M  4.2 F Fa  0.6 F M  0.6 x F M  4.2 F Fa  0.6 F 0.8 F: AB M  0.8 F x M  0.8 x F OA M  0.8 F x M  0.8 x F Shigley’s MED, 11th edition Chapter 4 Solutions, Page 50/96 T  5.6 F Once the derivatives are taken the value of F = 15 lbf can be substituted in. The deflection of B in the direction of F is* T  5.6 F U  Fa L  Fa  TL  T 1 M M dx        F  AE OA F  JG OA F EI F 0.6 15 15 5.6 15 15 0.6     5.6  6  0.1963  30 10 6.136 103 11.5 10 6   B  F  7 15 15  4.22  15 2 0.6   x d x dx   30 106  3.068 10 3  0 30 106  3.068 103  0  7 15 15 15 2 2  0.8 x  d x   0.8 x  d x 3  6 6 3  30 10  3.068 10  0 30 10  3.068 10  0  1.38 105   0.1000  6.71103   0.0431  0.0119  0.1173  0.279 in Ans. *Note. Be careful, this is not the actual deflection of point B. For this, fictitious forces must be placed on B in the x, y, and z directions. Determine the energy due to each, take derivatives, and then substitute the values of Fx = 9 lbf, Fy =  12 lbf, and Fz = 0. This can be done separately and then added by vector addition. The actual deflections of B are found to be B = 0.0831 i  0.2862 j  0.00770 k in From this, the deflection of B in the direction of F is  B  F  0.6  0.0831  0.8  0.2862   0.279 in which agrees with our result. ______________________________________________________________________________ 4-78 Strain energy. AB: Bending and torsion, BC: Bending and torsion, CD: Bending. IAB =  (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = 0.25(1.53)/12 = 0.07031 in4, ICD =  (0.754)/64 = 0.01553 in4. For the torsion of bar BC, Eq. (3-41) is in the form of  =TL/(JG), where the equivalent of J is Jeq = bc 3. With b/c = 1.5/0.25 = 6, JBC = bc 3 = 0.299(1.5)0.253 = 7.008 (103) in4. Use the dummy variable x to originate at the end where the loads are applied on each segment, M AB: Bending M  F x  2 F  x 2 F Shigley’s MED, 11th edition Chapter 4 Solutions, Page 51/96 Torsion BC: Bending Torsion CD: Bending T  5F M Fx T  2F M Fx T 5 F M x F T 2 F M x F U M 1 Tl T   M dx F F JG F EI 6 5F  6  2F 5 1 2    F  x  2 d x 5 2   6  3 6 6 0.09818 11.5 10 7.008 10 11.5 10  30 10  0.04909 0 D   5 2 1 1 F x 2d x  F x 2d x   6 6 30 10  0.07031 0 30 10  0.01553 0  1.329 10 4  F  2.482 10 4  F  1.141 10 4  F  1.98 10 5  F  5.72 10 6  F  5.207 104  F  5.207 104  200  0.104 in Ans. ______________________________________________________________________________ 4-79 AAB =  (12)/4 = 0.7854 in2, IAB =  (14)/64 = 0.04909 in4, IBC = 1.5 (0.253)/12 = 1.953 (103) in4, ACD =  (0.752)/4 = 0.4418 in2, IAB =  (0.754)/64 = 0.01553 in4. For (D )x let F = Fx =  150 lbf and Fz =  100 lbf . Use the dummy variable x to originate at the end where the loads are applied on each segment, M y CD: 0 M y  Fz x F Fa 1 Fa  F F M y M y  F x  2 Fz x BC: F Fa 0 Fa  Fz F M y 5 M y  5 F  2 Fz  Fz x AB: F Fa Fa  F 1 F Shigley’s MED, 11th edition Chapter 4 Solutions, Page 52/96  D  x 1 U  FL  Fa     F  AE CD F EI BC   F  2 1 EI AB 0.4418  30 10  6 5   F x  2F  x d x z 0 6   5F  2 F z 0 1   FL  Fa  Fz x  5  d x     AE  AB F 1 3 F  5   Fz  52  3  30 10 1.953 10   3  6 F  6 F 1   25 F  6   10 Fz  6   z  62  5  1 6    2 30 10  0.04909   0.7854  30 10 6  1.509 107  F  7.112 104  F  4.267 104  Fz  1.019 104  F  1.019 104  Fz  2.546 10 7  F  8.135 10 4  F  5.286 10 4  Fz Substituting F = Fx =  150 lbf and Fz =  100 lbf gives  D  x  8.135 104   150   5.286 10 4   100   0.1749 in Ans. ______________________________________________________________________________ 4-80 IOA = IBC =  (1.54)/64 = 0.2485 in4, JOA = JBC = 2 IOA = 0.4970 in4, IAB =  (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, ICD =  (0.754)/64 = 0.01553 in4 Let Fy = F, and use the dummy variable x to originate at the end where the loads are applied on each segment, OC: M Fx M  x, F DC: M Fx M x F T  12 F T  12 F U 1 M  TL  T dx    M  EI F F  JG OC F The terms involving the torsion and bending moments in OC must be split up because of the changing second-area moments.  D  y  Shigley’s MED, 11th edition Chapter 4 Solutions, Page 53/96  D  y  2 12 F  4  12 F  9  1   12 12 F x 2d x   6  6   6 0.4970 11.5 10 0.09818 11.5 10 30 10  0.2485 0 11 13 12 1 1 1  F x 2d x  F x 2d x  F x 2d x    6 6 6 30 10  0.04909 2 30 10  0.2485 11 30 10  0.01553 0  1.008 104  F  1.148 10 3  F  3.58 10 7  F  2.994 104  F  3.872 10 5  F  1.2363 10 3  F  2.824 103  F  2.824 10 3  250  0.706 in Ans. For the simplified shaft OC,  B  y  13 12 12 F 13 1 1 2 12   F x d x F x 2d x  6    6 6 0.09818 11.5 10 30 10  0.04909 0 30 10  0.01553 0  1.6580 103  F  4.973 104  F  1.2363 10 3  F  3.392 10 3  F  3.392 10 3  250 Ans.  0.848 in Simplified is 0.848/0.706 = 1.20 times greater Ans. ______________________________________________________________________________ 4-81 Place a fictitious force Q pointing downwards at point B. The reaction at C is RC = Q + (6/18)100 = Q + 33.33 This is the axial force in member BC. Isolating the beam, we find that the moment is not a function of Q, and thus does not contribute to the strain energy. Thus, only energy in the member BC needs to be considered. Let the axial force in BC be F, where F  Q  33.33 F 1 Q  FL  F   0  33.3312   1  6.79 105  in Ans.    2 6  AE Q    BC   Q 0   0.5  / 4  30 10  Q 0 ______________________________________________________________________________ B  4-82 U Q IOB = 0.25(23)/12 = 0.1667 in4 AAC =  (0.52)/4 = 0.1963 in2 MO = 0 = 6 RC  11(100)  18 Q Shigley’s MED, 11th edition Chapter 4 Solutions, Page 54/96 RC = 3Q + 183.3 MA = 0 = 6 RO  5(100)  12 Q  RO = 2Q + 83.33 Bending in OB. BD: Bending in BD is only due to Q which when set to zero after differentiation gives no contribution. AD: Using the variable x as shown in the figure above M   7  x  Q OA: Using the variable x as shown in the figure above M  100 x  Q  7  x  M  2 x Q M    2Q  83.33 x Axial in AC: F  3Q  183.3 F 3 Q  U   FL  F   1        Q Q 0  AE  Q  Q 0  EI B   183.3 12  1  3  6   0.1963  30 10 EI  1.121103   5 M M  dx  Q Q 0 6 2  100 x  7  x  d x  0 2 83.33 x dx 0 5 6   2   100 7 166.7 x x d x x dx    0  10.4 106  0.1667  0 1  1.121103   5.768 107  100 129.2   166.7  72    0.0155 in Ans. ______________________________________________________________________________ 4-83 Table A-5, EA = 71.7 GPa, ES = 207 GPa, EI = 71.7(109)0.012(0.053)/12 = 8962.5 N٠m2 F =  4000 lbf MO = 0 = 450 FAC + 650F  FAC =  1.444 F F FAC AC LAC F  yB 1  EA 1.444  4000  1.444  220  207 109   / 4  0.006 2  0.3135 mm Shigley’s MED, 11th edition Chapter 4 Solutions, Page 55/96 MA = 0 = – 450 RO + 200F  RO = 0.4444 F 0  x  0.450 m M  0.4444 Fx M / F  0.4444 x 450  x  0.650 m M / F   0.650  x  0.65 M 1  0.45 2 0.4444  Fx 2 dx   F  0.65  x  dx  dx    0 0.45  F EI  0 2  4000   0.4444  1 3 0.65  0.453   0.65  x     3.867 mm 0.45 8962.5  3 3  yB = (yB)1 + (yB)2 =  0.3135  3.867 =  4.18 mm Ans. ______________________________________________________________________________ 4-84 Table A-5, EA = 71.7 GPa, ES = 207 GPa, EI = 71.7(109)0.012(0.053)/12 = 8962.5 N٠m2 MO = 0 = 450 FCD + 650Q – 150(4000)  FCD = 1333  1.444 Q F FCD CD LCD Q Q 0  yB 1  EA 1333  1.444  220  207 109   / 4  0.0062  yB 2  1 EI M  F  0.650  x  l M  0.0724 mm Fy = 0 = RO – 4000 + 1333 – 1.444 Q + Q  RO = 2667 + 0.444 Q 0  x  0.150 m M   2667  0.444Q  x M / Q  0.444 x 150  x  0.450 m M   2667  0.444Q  x  4000  x  0.150    1333  0.444Q  x  600 0.450  x  0.650 m M / Q  0.444 x M   0.65  x  Q  yB 2  1 EI l M 0 M / Q   x 0.45 M 1  0.15 dx  2667 x  0.444 x  dx    1333 x  600  0.444 x  dx   0.15  Q EI  0 Q 0  2667  0.444  0.153  1333  0.444  600  0.444    0.453  0.153   0.452  0.152        3 3 2     1   7.996   8.922 104  m  0.8922 mm 8962.5 yB = (yB)1 + (yB)2 =  0.0724 + 0.8922 = 0.820 mm Ans. ______________________________________________________________________________ 4-85 Table A-5, EA = 10.4 Mpsi, ES = 30 Mpsi, EI = 10.4(106)0.25(23)/12 = 1.733 (106) lbf٠in3 MO = 0 = 18 FBC + 6F  FBC =  F / 3  1 EI Shigley’s MED, 11th edition Chapter 4 Solutions, Page 56/96  y A 1  FBC FBC LBC F AE  1/ 3  100 12   / 4  0.52  30 106 2  2.264 105  in Fy = 0 = RO +F + FBC 2 0  x  6 in M   Fx 3  RO =  2 F / 3 2 M / F   x 3 1 1 M  FBC x   F x M / F   x 3 3 6 12 l M 1 1  2 2 dx  2 / 3 Fx 2 dx    1/ 3  F x 2 d x   y A 2  0 M     0 0 EI F EI  F  4 3 1 3 100   96   5.540 103  in  6  12   6 EI  27 27  1.733 10  0  x  12in yA = (yA)1 + (yA)2 =  2.264 (105)  5.540 (103) =  5.56 (103) in Ans. ______________________________________________________________________________ 4-86 Table A-5, EA = 10.4 Mpsi, ES = 30 Mpsi MO = 0 =6(FAC +Q)  11(100) FAC FAC = 183.3  Q  1 Q  y A 1  FAC FAC LAC Q Q 0 AE 183.3  112   3.734 104  in 2 6  / 4  0.5  30 10 Treating beam OADB as a simply-supported beam pinned at O and A we see that the force Q does not induce any bending. Thus, yA = (yA)1 =  3.734 (104) in Ans. ______________________________________________________________________________ 4-87 Table A-5, EA = 71.7 GPa, k = P /  = 10 (103)/(103) = 10 (106) N/m, EI = 71.7 (109)0.005(0.033)/12 = 806.6 N٠m2 (a) OA: RO = F M M = RO x = Fx x x Shigley’s MED, 11th edition Chapter 4 Solutions, Page 57/96  yB 1  1 EI l M 0 1 M dx  EI F  0.250 0 Fx 2 dx  300 0.2503 103  1.937 mm  3  806.6  Ans. M x F Since AB is the same length as OA, the integration is identical to part (a). Thus (yB)2 =  1.937 mm Ans. (c) AC: Eq. (4-15): MO = 0 = 250 FAC + 500 F  FAC =  2F F FAC AC 2 F 2 2 FAC    4  300   0.12 mm Ans. U F    y B 3   U 2k k k F 10 106  (b) AB: M Fx (d) yB = (yB)1 + (yB)2 + (yB)3 = 2( 1.937)  0.12 =  3.994 mm Ans. ______________________________________________________________________________ 4-88 Table A-5, E = 30 Mpsi, G = 11.5 Mpsi. (EI)AB = 30(106) (/64)14 = 1.473 (106) lbf٠in2, (JG)AB = (/32)14 (11.5)106 = 1.129 (106) lbf٠in2, (EI)BD = 30(106) 0.25 (1.253)/12 = 1.221 (106) lbf٠in2. F = 200 lbf. M y (a) AB at xB: M y  500 xB 1  EI  AB  l AB 0 0 No contribution from F M z  xB F M z   F  300  xB  y D i  F Mz 6  200  300  63 M z 1 2   dxB  F x dx 300   B B F 3 1.473106 1.473 106  0  4.888 10 3  in Tx = 4F Shigley’s MED, 11th edition Tx 4 F  yD ii Tx LAB 4 200 4 6      0.01701 in  F  1.129 106   JG  AB Tx Chapter 4 Solutions, Page 58/96 (yD)1 = (yD)i + (yD)ii =  4.888 (103) +0.01701 = 0.0121 in (b) BC at xC has no F. Therefore, (yD)2 = 0 Ans. (c) BD at xD: M x  xD F M x  FxD  y D 3 1   EI  BD  4 0 2 D Fx dxD   3.494 103  in (d) yD = (yD)1 + (yD)2 + (yD)3 = 0.0121 + 0 + 0.0035 = 0.0156 in 200  43  3 1.221106 Ans. Ans. ______________________________________________________________________________ 4-89 Table A-5, E = 207 GPa, G = 79.3 GPa, (EI)AB = 207 (109)(/64) 0.0254 = 3.969 (103) m4, (JG)AB = (/32) 0.0254 (79.3)109 = 3.041 (103) m4 (a) AB: Bending: M y 0 M y  300 xB Q No contribution from Q. M z  xB Q M z   450  Q  xB 1 EI  y D i    M z dxB Q Q 0 l M 0 1  EI  AB z  0.150 0 450 xB2 dxB 450  0.1503  3  3.969  103 Torsion:  1.276 104  m  0.1275 mm Tx  Q  0.1  450  0.125  0.1Q  56.25  yD ii  Tx Tx LAB Q Q 0  JG  AB  56.25  0.1 0.150 3.041 10 3  Tx  0.1 Q  2.775 104  m  0.2775 mm (yD)1 = (yD)i + (yD)ii = 0.1275  0.2775 =  0.150 mm (b) BC: Break at xC has no Q in it. Thus, (yD)2 = 0 Ans. Ans. (c) BD: Axial has no Q. Thus no contribution. Bending only has Q and since Q = 0, no contribution. Therefore, (yD)3 = 0 Ans. Shigley’s MED, 11th edition Chapter 4 Solutions, Page 59/96 (d) yD = (yD)1 +(yD)2 +(yD)3 = 0.150 + 0 + 0 =  0.150 mm Ans. ______________________________________________________________________________ 4-90 Table A-5, E = 30 Mpsi, G = 11.5 Mpsi. (EI)AB = 30 (106) (/64) 14 = 1.473 (106) lbf٠in2. (a) AB: Break at xB:  z D i   1  EI  AB  l 0 My 500  63  M y   Q  500  xB M y Q M y Q  xB dxB Q 0 3 1.473106  0.0244 in No contribution of Q to Tx, Mz. Thus, (zD)1 = (zD)i =  0.0244 in Ans. (b) BC: Break xC shows no contributions from Q. Thus, (zD)2 = 0 Ans. (c) BD: Break xD shows no contributions from Q for Mx, My, or Tz. For axial, F Q F  1 but setting Q = 0 gives nothing. Thus, (zD)3 = 0 Q Ans. (d) zD = (zD)1 =  0.0244 in Ans. ______________________________________________________________________________ 4-91 Table A-5, E = 207 GPa, (EI)AB = 207 (109)(/64) 0.0254 = 3.969 (103) m4, (EA)BD = 207(109) (/4) 0.0252 = 101.6 (106) m2. F =  300 N. Shigley’s MED, 11th edition Chapter 4 Solutions, Page 60/96 (a) AB: Break at xB shows only contribution to My from F M y M y  FxB 1  zD 1  EI l M 0 y F  xB M y 1 dx  F  EI  AB  0.150 0 2 B Fx dxB   8.50 105  m  0.0850 mm 300  0.1503  3  3.969 103 Ans. (b) BC: Break at xB shows no contribution from F. Thus, (zD)2 = 0 Ans. (c) BD: Break at xD shows only contribution to axial force from F Fz  F  z D 3 Fz 1 F Fz LBD 300 1 0.100  F     2.95 107  m   2.95 10 4  mm 6 EA 101.6 10  BD   Fz (d) zD =  0.0850 + 0  2.95 (104) =  0.0853 mm Ans. Ans. ______________________________________________________________________________ 4-92 There is no bending in AB. Using the variable, rotating counterclockwise from B M  PR sin  Fr  P cos  F  P sin  MF  2 PR sin 2  P M  R sin  P Fr  cos  P F  sin  P A  6(4)  24 mm 2 , ro  40  12 (6)  43 mm, ri  40  12 (6)  37 mm, From Table 3-4, for a rectangular cross section 6 rn   39.92489 mm ln(43 / 37) From Eq. (4-33), the eccentricity is e = R  rn =40  39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 61/96 From Eq. (4-38)  M  M   2 0 AeE  P  2 4-93 2  2 PR  sin     1   MF   2 2 CFr R  Fr d   d       0 AE 0 P AG  P    d   2 PR sin 2  2 CPR  cos    d   d   d   d 0 0 0 0 AeE AE AE AG EC   PR  R  (10)(40)  40 (207 103 )(1.2)  1 2 1 2             4 AE  e G  4(24)(207 103 )  0.07511 79.3 103    0.0338 mm Ans. ______________________________________________________________________________  P  R sin     2 F R  F d     0 AE  P   2 2  2 Place a fictitious force Q pointing downwards at point A. Bending in AB is only due to Q which when set to zero after differentiation gives no contribution. For section BC use the variable, rotating counterclockwise from B M M  PR sin   Q  R  R sin    R 1  sin   Q Fr Fr   P  Q  cos   cos  Q F F   P  Q  sin   sin  Q MF   PR sin   QR 1  sin     P  Q  sin  MF  PR sin 2    PR sin  1  sin    2QR sin  1  sin   Q But after differentiation, we can set Q = 0. Thus, MF  PR sin  1  2sin   Q A  6(4)  24 mm 2 , ro  40  12 (6)  43 mm, ri  40  12 (6)  37 mm, From Table 3-4, for a rectangular cross section 6 rn   39.92489 mm ln(43 / 37) From Eq. (4-33), the eccentricity is e = R  rn =40  39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 From Eq. (4-38), Shigley’s MED, 11th edition Chapter 4 Solutions, Page 62/96     1   MF  2 F R  F  2 2 CFr R  Fr  d  d  d          d    0 0 AE 0 AE 0 Q AG  Q    Q  PR 2 2 PR 2 2 PR 2 d d          sin 1 sin sin sin  1  2sin   d   AeE 0 AE 0 AE 0 CPR 2  cos 2  d  0 AG 2  CE  R   PR  PR    PR  CPR PR        1     2   1  2   4 G   4  AeE 4 AE  4  AE 4 AG AE  4  e    2 M  M  AeE  Q 3    1.2  207 10   40   1 2      24  207 103  4  0.07511 4 79.3 103     Ans.  0.0766 mm ______________________________________________________________________________ 4-94 A = 3(2.25) 2.25(1.5) = 3.375 in2 (1  1.5)(3)(2.25)  (1  0.75  1.125)(1.5)(2.25) R  2.125 in 3.375  10  40  Section is equivalent to the “T” section of Table 3-4, 2.25(0.75)  0.75(2.25)  1.7960 in 2.25ln[(1  0.75) /1]  0.75ln[(1  3) / (1  0.75)] e  R  rn  2.125  1.7960  0.329 in rn  For the straight section 1 I z  (2.25)  33   2.25(3)(1.5  1.125) 2 12 2 1 2.25    3   (1.5)  2.25   1.5(2.25)  0.75   1.125   2    12  2.689 in 4 For 0  x  4 in M   Fx M   x, F V F V 1 F For    /2 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 63/96 Fr  cos  , F Fr  F cos  F  F sin  F  sin  F M  (4  2.125sin  ) F MF MF  F (4  2.125sin  ) F sin   2 F (4  2.365sin  ) sin  F M  F (4  2.125sin  ) Use Eqs. (4-31) and (4-24) (with C = 1) for the straight part, and Eq. (4-38) for the curved part, integrating from 0 to π/2, and double the results  /2 F (4)(1) 2 1 4 2 (4  2.125sin  ) 2 F d      Fx dx  E I 0 3.375(G / E ) 0 3.375(0.329)  / 2 2 F (4  2.125sin  ) sin  F sin 2  (2.125) d   d 0 0 3.375 3.375 2  / 2 (1) F cos  (2.125)  d 0 3.375(G / E )   /2 Substitute I = 2.689 in4, F = 6700 lbf, E = 30 (106) psi, G = 11.5 (106) psi 2  6700   43    4 1    16    17(1)  4.516        6  30 10   3  2.689  3.375(11.5 / 30) 3.375(0.329)   2   4   2.125    2  2.125          4 1 2.125          3.375  4  3.375   4   3.375 11.5 / 30   4    0.0226 in Ans. ______________________________________________________________________________ 4-95 Since R/h = 35/4.5 = 7.78 use Eq. (4-38), integrate from 0 to  , and double the results M  R 1  cos  M  FR 1  cos  F Fr  sin  Fr  F sin  F F  cos  F  F cos  F MF  F 2 Rcos 1  cos    MF  F  2 FRcos 1  cos  From Eq. (4-38), Shigley’s MED, 11th edition Chapter 4 Solutions, Page 64/96  FR 2 FR   (1  cos  ) 2 d  cos 2  d   2   0 0 AE  AeE   2 FR  1.2 FR  2  cos  1  cos   d  sin  d    AE 0 AG 0  E 2 FR  3 R 3   0.6   AE  2 e 2 G A = 4.5(3) = 13.5 mm2, E = 207 (103) N/mm2, G = 79.3 (103) N/mm2, and from Table 3-4, 4.5 h rn    34.95173 mm 37.25 ro ln ln 32.75 ri and e = R  rn = 35  34.95173 = 0.04827 mm. Thus, 2 F  35   3 35 3 207     0.6   0.08583 F 3  13.5  207 10  2 0.04827 2 79.3  1 where F is in N. For  = 1 mm, F   11.65 N Ans. 0.08583 Note: The first term in the equation for  dominates and this is from the bending moment. Try Eq. (4-41), and compare the results. ______________________________________________________________________________ 4-96 R/h = 20 > 10 so Eq. (4-41) can be used to determine deflections. Consider the horizontal reaction to be applied at B and subject to the constraint ( B ) H  0. M  FR (1  cos  )  HR sin  2 M   R sin  H 0    2 By symmetry, we may consider only half of the wire form and use twice the strain energy Eq. (4-41) then becomes, ( B ) H    /2 0 2 U  H EI  /2  M    M H  Rd  0 0  FR   2 (1  cos  )  HR sin   (  R sin  ) R d  0 F F  F 30   H 0  H    9.55 N 2 4 4   Shigley’s MED, 11th edition Ans. Chapter 4 Solutions, Page 65/96 The reaction at A is the same where H goes to the left. Substituting H into the moment equation we get, FR M R   [ (1  cos  )  2sin  ] 0    (1  cos  )  2 sin   2 F 2 2 2  /2 FR U 2  M  2 P   [ (1  cos  )  2sin  ]2 R d M Rd  2  0 EI  F  EI P 4 3  /2 FR  2  ( 2   2 cos 2   4sin 2   2 2 cos   4 sin   4 sin  cos  ) d 2 EI 0  FR 3          2  2     2    4    2 2  4  2  2 EI   2  4 4  M   (30)(403 ) (3 2  8  4) FR 3 (3 2  8  4)   0.224 mm EI 8 8 207 103    2 4  / 64  Ans. ______________________________________________________________________________ 4-97 The radius is sufficiently large compared to the wire diameter to use Eq. (4-41) for the curved beam portion. The shear and axial components will be negligible compared to bending. Place a fictitious force Q pointing to the left at point A. M M  PR sin   Q ( R sin   l )  R sin   l Q Note that the strain energy in the straight portion is zero since there is no real force in that section. From Eq. (4-41),   A     1  M  1 M Rd   EI  Q   Q 0 EI  /2 0 PR 2  EI   /2 0   /2 0 PR sin   R sin   l Rd 1(2.52 ) PR 2       R sin   l sin   d  EI  4 R  l    (2.5)  2  6 4  30 10    0.125  / 64   4 2  0.0689 in Ans. ______________________________________________________________________________ 4-98 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. M AB x P Straight portion: M AB  Px Curved portion: M BC  P  R (1  cos  )  l  M BC   R (1  cos  )  l  P From Eq. (4-41) with the addition of the bending strain energy in the straight portion of the wire, Shigley’s MED, 11th edition Chapter 4 Solutions, Page 66/96  /2 1  M BC  1  M AB  M AB dx      M BC Rd 0 EI 0 EI  P  P   P l 2 PR  /2 2 x dx  R (1  cos  )  l  d     EI 0 EI 0 3 Pl PR  / 2 2  R (1  2 cos   cos 2  )  2 Rl (1  cos  )  l 2 d    0 3EI EI Pl 3 PR  / 2 2  R cos 2    2 R 2  2 Rl  cos   ( R  l ) 2 d    3EI EI 0   Pl 3 PR   2  R   2 R 2  2 Rl   ( R  l )2     2 3EI EI  4  A   l P  EI  l3  3  2 2   R  R  2 R  2 Rl   R( R  l )  2 3 4   23  1  2 3 2  3  4 (2.5 )  2.5  2(2.5 )  2(2.5)(2)   2  2.5  2.5  2   6 4 30 10    0.125  / 64    0.106 in Ans. ______________________________________________________________________________ 4-99 R = 2.5 in, d = 0.125 in, l = 2 in, P = 1 lbf, E = 30 Mpsi. EI = 30 (106) (/64) 0.1254 = 359.53 lbf٠in2. Member ABC: 0  x  l / 2 M  Qx M x Q Since Q = 0, there is no contribution. M  Qx  P  x  l / 2  l/2 xl  1  EI  A 1   l M 0 M x Q l M  1  0   P  x  l / 2  xdx  dx   l /2  Q  Q  0 EI  l 3 2 P  x 3 x 2l  P  l 3   l / 2   l 3  l / 2  l            EI  3 4  l / 2 EI  3  3  4 4    3 3 5 1 2 5 Pl    2.32 103  in 48 EI 48  359.53  Member CD: Shigley’s MED, 11th edition Chapter 4 Solutions, Page 67/96 M  Q l  R 1  cos     P l / 2  R 1  cos    0   M  l  R 1  cos    Q  1  P l / 2  R 1  cos    l  R 1  cos    Rd EI 0  PR   2 2 l / 2   3l / 2  R 1  cos    R 2 1  cos    d    EI 0  PR  2 l / 2   3l / 2  R   3l / 2  R cos   R 2  2 R 2 cos   R 2 cos 2   d    0 EI PR   2  1  cos 2 l / 2   3l / 2  R  R 2   3l / 2  R cos   2 R 2 cos   R 2     0 EI  2   A  2     d  PR  2 l / 2   3l / 2  R   3 / 2  R 2    3l / 2  R  2 R 2  cos   1 / 2  R 2 cos 2 d  0 EI  1 2.5  PR 2 l / 2   3l / 2  R   3 / 2  R 2   22 / 2   3  2  / 2   2.5    3 / 2  2.52    359.53 EI        0.4123 in A = (A)1 +(A)2 = 2.32 (103) + 0.4123 = 0.4146 in Ans. ______________________________________________________________________________ 4-100 E = 30 Mpsi, EI = 30 (106) (/64) (0.125)4 = 359.53 lbf٠in2. 0  1   / 2 M   P  Q  R cos 1 0  2   M  QR cos  2 M  R cos 1 Q M  R cos  2 Q No contribution in 2 range since Q = 0. Thus 1 A  EI    /2 0 M 1  M Rd1 Q EI Q 0 PR 3   sin 21   2 EI  2 2  /2 0   /2 0 PR 3 PR cos 1 Rd1  2 EI 2  /2  1  cos 2 d 2 0   PR 3  1 2.53   0.0341 in   4 EI 4 359.53    1 1 Ans. ______________________________________________________________________________ 4-101 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. Shigley’s MED, 11th edition Chapter 4 Solutions, Page 68/96  Place a fictitious force, Q, at A vertically downward. The only load in the straight section is the axial force, Q. Since this will be zero, there is no contribution. In the curved section M  R 1  cos   Q M  PR sin   QR 1  cos   From Eq. (4-41)   / 2 1  M  A V   0 M EI  Q    PR 3 EI   /2 0   1 Rd     Q  0 EI  sin   sin  cos   d  1 2.53  2  30 106   0.1254  / 64    /2 0 PR sin   R 1  cos  Rd PR 3  1  PR 3 1    EI  2  2 EI  0.0217 in  Ans. _____________________________________________________________________________ 4-102 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. Place a fictitious force, Q, at A vertically downward. The load in the straight section is the axial force, Q, whereas the bending moment is only a function of P and is not a function of Q. When setting Q = 0, there is no axial or bending contribution. In the curved section M   R sin  Q M  P  R 1  cos    l   QR sin  From Eq. (4-41)   / 2 1  M   1  A V   0 M Rd   EI  Q    Q  0 EI   PR 2 EI  /2  0   /2 0 P  R 1  cos    l    R sin   Rd  R sin   R sin  cos   l sin   d   1 2.52  2  30 106   0.1254  / 64  PR 2  PR 2 1  R l R      R  2l    EI  2  2 EI  2.5  2  2    0.0565 in Since the deflection is negative,  is in the opposite direction of Q. Thus the deflection is   0.0565 in  Ans. _____________________________________________________________________________ 4-103 EI =30(106) (/64)(0.125)4 = 359.53 lbf٠in2. Shigley’s MED, 11th edition Chapter 4 Solutions, Page 69/96 AB: Only Q and since Q = 0, no contribution to (A)V. CD: M  QR sin   P  l / 2   R 1  cos    A V  1 EI   0 M M  R sin  Q M Rd Q Q 0 1  P  l / 2   R 1  cos    R sin  Rd EI 0   PR 2 PR 2    l / 2  cos   R cos    R / 2  sin 2     l  2R  0 EI EI 1 2.52    2  2  2.5    0.1217 in  Ans. 359.53   ______________________________________________________________________________ 4-104 EI =30(106) (/64)(0.125)4 = 359.53 lbf٠in2. AB: Only Q. Since Q = 0, no contribution. BC: M  R 1  cos   Q M  QR 1  cos    PR sin   A V  1 EI   /2 0 M M Rd Q  Q 0 1 EI  /2   PR sin   R 1  cos   d 0  /2 1 2  PR 3  PR 3  1  cos sin         1   EI  EI  2  2 0 3  2.53  3PR 3    0.0652 in  2 EI 2  359.53 Ans. ________________________________________________________________________ Shigley’s MED, 11th edition Chapter 4 Solutions, Page 70/96 4-105 Consider the force of the mass to be F, where F = 9.81(1) = 9.81 N. The load in AB is tension FAB FAB  F 1 F For the curved section, the radius is sufficiently large to use Eq. (4-41). There is no bending in section DE. For section BCD, let  be counterclockwise originating at D M M  FR sin   R sin  0   F Using Eqs. (4-29) and (4-41)  Fl  F  1  M  Fl  FR 3 AB    sin 2  d 1  0  M Rd  0 EI  F  AE EI  AE  AB F   403   Fl  FR 3 F  l  R 3  9.81  80         AE 2 EI E  A 2 I  207 103     2 2  / 4  2   24  / 64        6.067 mm Ans. _____________________________________________________________________________ 4-106 AOA = 2(0.25) = 0.5 in2, IOAB = 0.25(23)/12 = 0.1667 in4, IAC =  (0.54)/64 = 3.068 (10-3) in4 Applying a force F at point B, using statics (AC is a two-force member), the reaction forces at O and C are as shown. FOA OA: Axial FOA  3F 3 F Bending M OA  2 Fx AB: Bending M AB   F x M OA  2 x F M AB  x F AC: Isolating the upper curved section M AC  3FR  sin   cos   1 Shigley’s MED, 11th edition M AC  3R  sin   cos   1 F Chapter 4 Solutions, Page 71/96 10 1 F  Fl  1 OA    4 Fx 2 dx     AE F EI  OAB 0  EI OAB  OA 9 FR3   EI  AC   /2   sin   cos   1 0 3F 10  0.5 10.4 106   3  9 F 103  2 20 Fx 2 dx 0 d 4 F 103  F  203   3 10.4 106  0.1667  3 10.4 10 6  0.1667   /2 30 10  3.068 10 6 3  sin  2   2 sin  cos   2 sin   cos 2   2 cos   1 d 0     1.731105  F  7.69110 4  F  1.538 10 3  F  0.09778 F   1  2   2   4 2 4  0.0162 F  0.0162 100   1.62 in Ans. _____________________________________________________________________________ 4-107 AOA = 2(0.25) = 0.5 in2, IOAB = 0.25(23)/12 = 0.1667 in4, IAC =  (0.54)/64 = 3.068 (10-3) in4 Applying a vertical fictitious force, Q, at A, from statics the reactions are as shown. The fictitious force is transmitted through section OA and member AC. FOA 1 OA: FOA  3F  Q Q AC: M AC   3F  Q  R sin    3F  Q  R 1  cos   M AC  R  sin   cos   1 Q  Fl   FOA   1   / 2  M AC    Rd       M AC   Q  AE OA  Q   EI  AC 0  Q 0   3FlOA 3FR 3   AE OA  EI  AC 3 100 10  /2   sin   cos   1 2 d 0 3 100 103      1  2   2    0.462 in 4 2 10.4 10  0.5 30 10  3.068 10   4 6  6 3 Ans. _____________________________________________________________________________ 4-108 I =  (64)/64 = 63.62 mm4 0 /2 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 72/96 M  R sin  F T T  FR (1  cos  )  R (1  cos  ) F According to Castigliano’s theorem, a positive  U/ F will yield a deflection of A in the negative y direction. Thus the deflection in the positive y direction is M  FR sin  ( A ) y   U 1   F  EI   /2 0 F ( R sin  )2 R d  1 GJ   /2 0  F [ R (1  cos  )]2 R d   Integrating and substituting J  2 I and G  E /  2 1    FR3   FR 3  3              (1 ) 2 4 8 (3 8)     EI  4 4 EI  4  3 (250)(80)  [4  8  (3  8)(0.29)]   12.5 mm Ans. 4(200)103  63.62  _____________________________________________________________________________ ( A ) y   4-109 The force applied to the copper and steel wire assembly is Fc  Fs  400 lbf (1) Since the deflections are equal,  c   s  Fl   Fl       AE c  AE  s Fc l Fs l  2 6 3( / 4)(0.1019) (17.2)10 ( / 4)(0.1055) 2 (30)10 6 Yields, Fc = 1.6046 Fs. Substituting this into Eq. (1) gives 1.6046 Fs  Fs  2.6046Fs  400  Fs  153.6 lbf Fc  1.6046 Fs  246.5 lbf F 246.5 c  c   10 075 psi  10.1 kpsi Ans. Ac 3( / 4)(0.1019)2 F 153.6 s  s   17 571 psi  17.6 kpsi Ans. As ( / 4)(0.10552 )  Fl  153.6(100)(12)  0.703 in     2 6  AE  s ( / 4)(0.1055) (30)10 Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 4 Solutions, Page 73/96  b  0.75(65)  48.8 kpsi 4-110 (a) Bolt stress Ans.   Fb  6 b Ab  6(48.8)   (0.52 )  57.5 kips 4 F 57.43 Cylinder stress c   b   13.9 kpsi Ans. Ac ( / 4)(5.52  52 ) (b) Force from pressure  D2  (52 ) (500)  9817 lbf  9.82 kip P p 4 4 Fx = 0 Total bolt force Pb + Pc = 9.82 (1) Since  c   b , Pc l Pb l  2 2 ( / 4)(5.5  5 ) E 6( / 4)(0.52 ) E Pc = 3.5 Pb (2) Substituting this into Eq. (1) Pb + 3.5 Pb = 4.5 Pb = 9.82  Pb = 2.182 kip. From Eq. (2), Pc = 7.638 kip Using the results of (a) above, the total bolt and cylinder stresses are 2.182  b  48.8   50.7 kpsi Ans. 6( / 4)(0.52 ) 7.638  c  13.9   12.0 kpsi Ans. ( / 4)(5.52  52 ) _____________________________________________________________________________ 4-111 (1) Tc + T s = T c = s  Tc l Tl  s  JG c  JG s Substitute this into Eq. (1)  JG c T T  T  JG s s s  Tc   Ts   JG c T  JG  s s (2)  JG s T  JG s   JG c The percentage of the total torque carried by the shell is % Torque  100  JG  s  JG s   JG c Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 4 Solutions, Page 74/96 4-112 RO + RB = W (1) OA = AB  Fl   Fl       AE OA  AE  AB 400 RO 600 RB 3   RO  RB AE AE 2 Substitute this unto Eq. (1) 3 RB  RB  4 2  RB  1.6 kN (2) Ans. 3 RO  1.6  2.4 kN Ans. 2 2 400(400)  Fl  A    0.0223 mm Ans.   3  AE OA 10(60)(71.7)(10 ) ______________________________________________________________________________ From Eq. (2) 4-113 See figure in Prob. 4-112 solution. Procedure 1: 1. Let RB be the redundant reaction. 2. Statics. RO + RB = 4 000 N 3. Deflection of point B.  B   RO = 4 000  RB RB  600  AE  (1)  RB  4000  400   0 AE (2) 4. From Eq. (2), AE cancels and RB = 1 600 N Ans. and from Eq. (1), RO = 4 000  1 600 = 2 400 N Ans. 2 400(400)  Fl   0.0223 mm Ans.   3  AE OA 10(60)(71.7)(10 ) _____________________________________________________________________________ A   4-114 (a) Without the right-hand wall, the deflection of point C would be 5 103  8 2 103  5 Fl C     AE   / 4  0.752 10.4 106  / 4  0.52 10.4 106  0.01360 in  0.005 in  Hits wall Shigley’s MED, 11th edition Ans. Chapter 4 Solutions, Page 75/96 (b) Let RC be the reaction of the wall at C acting to the left (). Thus, the deflection of point C is now 5 103   RC  8  2 103   RC  5      C  2 6 2  / 4  0.75 10.4 10  / 4  0.5 10.4 106  0.01360  4 RC 5   8  2   0.005 6  2  10.4 10  0.75 0.5  or, 0.01360  4.190 10 6  RC  0.005 RC  2 053 lbf  2.05 kip  Ans.   Fx = 5 000 + RA  2 053 = 0  RA = 2 947 lbf,  RA = 2.95 kip  Ans. Deflection. AB is 2 947 lbf in tension. Thus 2 947  8  RA  8    5.13 10 3  in  Ans. AAB E   / 4  0.752 10.4 106 _____________________________________________________________________________  B   AB  4-115 Since OA = AB, TOA (4) TAB (6)  JG JG Statics. TOA + TAB = 200 (2)  Substitute Eq. (1) into Eq. (2), 3 5 TAB  TAB  TAB  200  2 2 From Eq. (1) 3 TOA  TAB 2 TAB  80 lbf  in (1) Ans. 3 3 TOA  TAB  80  120 lbf  in Ans. 2 2 80  6  180  TL  A    0.390 0   4 6  JG  AB  / 32  0.5 11.5 10   max  16T d3  AB   16 80    0.53   OA  16 120    0.53  Ans.  4890 psi  4.89 kpsi  3260 psi  3.26 kpsi Ans. Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 4 Solutions, Page 76/96 4-116 Since OA = AB, TOA (4) TAB (6)  4  / 32  0.5 G  / 32  0.754 G TOA  0.2963 TAB  (1) Statics. TOA + TAB = 200 (2) Substitute Eq. (1) into Eq. (2), 0.2963TAB  TAB  1.2963TAB  200 TAB  154.3 lbf  in  TOA  0.2963TAB  0.2963 154.3  45.7 lbf  in From Eq. (1) A  154.3  6  180  0.1480 4 6  / 32  0.75 11.5 10   max  16T d3  AB    OA  16 154.3   0.753  16  45.7    0.53  Ans. Ans. Ans.  1862 psi 1.86 kpsi  1862 psi 1.86 kpsi Ans. Ans. _____________________________________________________________________________ 4-117 Procedure 1 1. Arbitrarily, choose RC as a redundant reaction. 2. Statics. Fx = 0, 12(103)  6(103)  RO  RC = 0 RO = 6(103)  RC (1) 3. The deflection of point C. 12(103 )  6(103 )  RC  (20) 6(103 )  RC  (10) R (15) C     C 0 AE 4. The deflection equation simplifies to  45 RC + 60(103) = 0 From Eq. (1), AE AE  RC = 1 333 lbf = 1.33 kip RO = 6(103)  1 333 = 4 667 lbf = 4.67 kip Ans. Ans. FAB = FB + RC = 6 +1.333 = 7.333 kips compression Shigley’s MED, 11th edition Chapter 4 Solutions, Page 77/96 FAB 7.333   14.7 kpsi Ans. A (0.5)(1) Deflection of A. Since OA is in tension, Rl 4 667(20)  A   OA  O OA   0.00622 in Ans. AE (0.5)(1)(30)106 _____________________________________________________________________________  AB  4-118 Procedure 1 1. Choose RB as redundant reaction. 2. Statics. RC = wl  RB (1) 1 w l 2  RB  l  a  (2) 2 3. Deflection equation for point B. Superposition of beams 2 and 3 of Table A-9, MC  yB  RB  l  a  3 3EI  w l  a  2 24 EI  4l  l  a    l  a 2  6l 2   0   4. Solving for RB. w  2 2 RB  6l  4l  l  a    l  a    8 l  a    w  3l 2  2al  a 2  8 l  a  Ans. Substituting this into Eqs. (1) and (2) gives RC  wl  RB  w 5l 2  10al  a 2   8 l  a  Ans. w 1 2 wl  RB  l  a    l 2  2al  a 2  Ans. 2 8 _____________________________________________________________________________ 4-119 See figure in Prob. 4-118 solution. MC  Procedure 1 1. Choose RB as redundant reaction. 2. Statics. RC = wl  RB (1) 1 M C  wl 2  RB  l  a  2 Shigley’s MED, 11th edition (2) Chapter 4 Solutions, Page 78/96 3. Deflection equation for point B. Let the variable x start at point A and to the right. Using singularity functions, the bending moment as a function of x is 1 M   w x 2  RB x  a 2 yB  1 l 1 U M M dx   RB EI 0 RB l or, M  xa RB 1 l 1 1 1  1    w x 2  0  dx   w x 2  RB  x  a    x  a  dx  0    EI 0 2 EI a  2  1 1 a 3 3  R  w   l 4  a 4    l 3  a 3    B  l  a    a  a    0  2 4 3  3  Solving for RB gives RB  w 8l  a  3 w 2 2  3  l 4  a 4   4a  l 3  a 3      8  l  a   3l  2al  a  Ans. From Eqs. (1) and (2) RC  wl  RB  w 5l 2  10al  a 2   8 l  a  Ans. w 1 2 wl  RB  l  a    l 2  2al  a 2  Ans. 2 8 ______________________________________________________________________________ MC  4-120 Note: When setting up the equations for this problem, no rounding of numbers was made in the calculations. It turns out that the deflection equation is very sensitive to rounding. Procedure 2 1. Statics. R1 + R2 = wl 1 2 wl 2 2. Bending moment equation. R2l  M 1  Shigley’s MED, 11th edition (1) (2) Chapter 4 Solutions, Page 79/96 1 M  R1 x  w x 2  M 1 2 1 dy 1  R1 x 2  w x3  M 1 x  C1 EI dx 2 6 1 1 1 EIy  R1 x3  w x 4  M 1 x 2  C1 x  C2 6 24 2 (3) (4) EI = 30(106)(0.85) = 25.5(106) lbfin2. 3. Boundary condition 1. At x = 0, y =  R1/k1 =  R1/[1.5(106)]. Substitute into Eq. (4) with value of EI yields C2 =  17 R1. Boundary condition 2. At x = 0, dy /dx =  M1/k2 =  M1/[2.5(106)]. Substitute into Eq. (3) with value of EI yields C1 =  10.2 M1. Boundary condition 3. At x = l, y =  R2/k3 =  R1/[2.0(106)]. Substitute into Eq. (4) with value of EI yields 1 3 1 1 R1l  wl 4  M 1l 2  10.2 M 1l  17 R1 (5) 6 24 2 Equations (1), (2), and (5), written in matrix form with w = 500/12 lbf/in and l = 24 in, are 12.75 R2  1 0   1   24 1   0  2287 12.75 532.8     R1   1      3  R2    12  10   M  576   1   Solving, the simultaneous equations yields R1 = 554.59 lbf, R2 = 445.41.59 lbf, M1 = 1310.1 lbfin Ans. For the deflection at x = l /2 = 12 in, Eq. (4) gives y x 12in  1 1 500 4 1 1 3  554.59 12 12  1310.112 2   6  24 12 2 25.5 10   6 10.2 1310.112  17  554.59    5.51103  in Ans. ______________________________________________________________________________ 4-121 Cable area, A  Procedure 2  4 (0.52 )  0.1963 in 2 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 80/96 1. Statics. RA + FBE + FDF = 5(103) (1) 3 FDF + FBE = 10(103) (2) 2. Bending moment equation. 1 M  RA x  FBE x  16  5000 x  32 1 1 dy 1 2 2  RA x 2  FBE x  16  2500 x  32  C1 2 dx 2 1 1 2500 3 3 EIy  RA x 3  FBE x  16  x  32  C1 x  C2 6 6 3 EI 3. B.C. 1: At x = 0, y = 0  (3) (4) C2 = 0 B.C. 2: At x = 16 in, FBE (38)  Fl   6.453(106 ) FBE yB      6 AE 0.1963(30)10   BE Substituting into Eq. (4) and evaluating at x = 16 in 1 EIyB  30(106 )(1.2)(6.453)(106 ) FBE  RA 163   C1 (16) 6 Simplifying gives 682.7 RA + 232.3 FBE + 16 C1 = 0 (5) B.C. 2: At x = 48 in, FDF (38)  Fl  yD     6.453(106 ) FDF   6 0.1963(30)10  AE  DF Substituting into Eq. (4) and evaluating at x = 48 in, 1 1 2500 RA 483  FBE (48  16)3  (48  32)3  48C1 6 6 3 18 432 RA + 5 461 FBE + 232.3 FDF + 48 C1 = 3.413(106) EIyD  232.3FDF  Simplifying gives   (6) Equations (1), (2), (5) and (6) in matrix form are 1 1 0   RA   5000   1     1 3 0   FBE   10 000   0    0  682.7 232.3 0 16   FDF      6    3.413 10 C 18 432 5 461 232.3 48     1    Solve simultaneously or use software. The results are RA =  970.5 lbf, Shigley’s MED, 11th edition FBE = 3956 lbf, FDF = 2015 lbf, and C1 =  16 020 lbfin2. Chapter 4 Solutions, Page 81/96 3956 2015  20.2 kpsi,  DF   10.3 kpsi 0.1963 0.1963 EI = 30(106)(1.2) = 36(106) lbfin2  BE  y  Ans. 1 2500 3 3  970.5 3 3956   x  x  16  x  32  16 020 x   6 6 6 3 36 10     161.8 x  659.3 x  16   1 36 106 3 B: x = 16 in, yC  1  161.8  323   659.3  32  16 3  16 020  32   6   36 10   0.0865 in   36 10 6  1  161.8 163   16 020 16    0.0255 in 6   36 10  D: x = 48 in, 1 3  833.3 x  32  16 020 x yB  C: x = 32 in, yD  3 Ans. Ans.    161.8 483  659.3  48  16 3  833.3  48  32 3  16 020  48      0.0131 in Ans. ______________________________________________________________________________ 4-122 Beam: EI = 207(103)21(103) = 4.347(109) Nmm2. Rods: A = ( /4)82 = 50.27 mm2. Procedure 2 1. Statics. RC + FBE  FDF = 2 000 (1) RC + 2FBE = 6 000 (2) 2. Bending moment equation. M =  2 000 x + FBE x  75 1 + RC x  150 1 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 82/96 1 1 dy 2 2  1000 x 2  FBE x  75  RC x  150  C1 2 2 dx 1000 3 1 1 3 3 EIy   x  FBE x  75  RC x  150  C1 x  C2 3 6 6 EI (3) (4) 3. B.C 1. At x = 75 mm, FBE  50   Fl  yB     4.805 106  FBE   3 50.27 207 10 AE     BE Substituting into Eq. (4) at x = 75 mm, 4.347 109   4.805 10 6  FBE    1000  753   C1  75  C2 3 Simplifying gives 20.89 103  FBE  75C1  C2  140.6 10 6  (5) B.C 2. At x = 150 mm, y = 0. From Eq. (4),  or, 1000 1 3 1503   FBE 150  75   C1 150   C2  0  3 6 70.31103  FBE  150C1  C2  1.125 109  (6) B.C 3. At x = 225 mm, FDF  65   Fl   6.246 106  FDF yD     3  AE  DF 50.27  207 10 Substituting into Eq. (4) at x = 225 mm, 4.347 109   6.246 106  FDF    Simplifying gives 1000 1 3 2253   FBE  225  75   3 6 1 3  RC  225  150   C1  225  C2 6 70.31103  RC  562.5 103  FBE  27.15 103  FDF  225C1  C2  3.797 109  (7) Equations (1), (2), (5), (6), and (7) in matrix form are Shigley’s MED, 11th edition Chapter 4 Solutions, Page 83/96  2 103   1 1 0 0 1      RC   3 1 2 0 0 0  6 10      FBE       0 20.89 103  0 75 1   F   140.6 106       DF   3       9 0 70.3110  0 150 1 C1  1.125 10       70.31103  562.5 103  27.15 103  225 1   C2   9   3.797 10   Solve simultaneously or use software. The results are RC =  2378 N, FBE = 4189 N, FDF =  189.2 N Ans. 7 2 8 3 and C1 = 1.036 (10 ) Nmm , C2 =  7.243 (10 ) Nmm . The bolt stresses are BE = 4189/50.27 = 83.3 MPa, DF =  189/50.27=  3.8 MPa Ans. The deflections are From Eq. (4) yA  1  7.243 108   0.167 mm  4.347 109   Ans. For points B and D use the axial deflection equations*. 4189  50   Fl  yB     0.0201 mm   50.27  207 103  AE  BE Ans. 189  65   Fl  yD   Ans.  1.18 10 3  mm   3 AE 50.27 207 10     DF *Note. The terms in Eq. (4) are quite large, and due to rounding are not very accurate for calculating the very small deflections, especially for point D. ______________________________________________________________________________ 4-123 Everything in Ex. 4-15 is the same except kB = 40 MN/m. yB =  FB / kB =  FB / 40 (106) =  2.5 (108) FB Equation (7) is replaced by EI  2.5 108 FB   FA F 3 3  0.2   B  0   0.2C1  C2 6 6 With EI = 1.25 (104) N٠m2, the equation reduces to  1.3333 (103) FA + 3.125 (104) FB + 0.2 C1 + C2 Shigley’s MED, 11th edition Chapter 4 Solutions, Page 84/96 The remaining equations come from Ex. 4-15 1 0 1 1   4 0 0 3  4   4.7746 10  0 0 0   1.3333 103  3.125 104  0 0.2  3 4 4   7.1458 10  5.625 10  6.3662 10  0.35 0  FA   0  0     FB 0     1   F   18.75   C     1  C1 0      C 0      1  2   Solving for the unknowns gives  FA   2350 N   F   5484 N   B     F 3134 N  C    C   95.239 N  m 2   1   C2  17.628 N  m3  Equation (3): 1.25 10 4  y   Ans. 2350 3 5484 3 x  x  0.2  95.239 x  17.628 6 6 Which reduces to y =  0.03133 x3 + 0.07312 x  0.23 +7.619 (103) x  1.410 (103) At x = 0, y = yA =  1.410 (103) m =  1.41 mm Ans. At x = 0.2 m, y = yB =  0.03133(0.2)3 +7.619(103)0.2  1.410(103) =  1.37 (104) m =  0.137 mm Ans. At x = 0.35 m, y = yC =  0.03133(0.35)3 + 0.07312 (0.35  0.2)3 +7.619(103)0.35  1.410(103) = 1.60 (104) m =  0.160 mm Ans. ______________________________________________________________________________ 4-124 (a) The cross section at A does not rotate. Thus, for a single quadrant we have U 0 M A The bending moment at an angle  to the x axis is FR M 1 M  MA  1  cos   2 M A The rotation at A is  /2 U M 1 A   M Rd  0  M A EI 0 M A Shigley’s MED, 11th edition Chapter 4 Solutions, Page 85/96 Thus, 1 EI  /2 or,    M A  0 FR 1  cos    1 Rd  0 2   FR   FR  0 MA    2 2 2  FR  2  1   2   Substituting this into the equation for M gives 2 FR  M  cos    (1)  2  The maximum occurs at B where  =  /2 MA  M max  M B   FR Ans.  (b) Assume B is supported on a knife edge. The deflection of point D is  U/ F. We will deal with the quarter-ring segment and multiply the results by 4. From Eq. (1) M R  2   cos    F  2 Thus, U 4  D  F EI  /2  0 FR 3 M M Rd  F EI  /2  0 2 FR3   2  2  cos  d          EI  4    3 FR  2  8  Ans.  4 EI ______________________________________________________________________________  4-125 Pcr  I C 2 EI l2   D4  d 4    D4 1  K  4 where K  64 64 2 4  C E   D 1  K 4  Pcr  2   l  64  d D 1/4   64 Pcr l 2  D 3 Ans. 4   CE 1  K   ______________________________________________________________________________ 4-126 A   D 2 1  K 2  , I  D 4 1  K 4   4 64 The radius of gyration, k, is given by Shigley’s MED, 11th edition  64 D 4 1  K 2 1  K 2  , where K = d / D. Chapter 4 Solutions, Page 86/96 k2  I D2  1 K 2   A 16 From Eq. (4-46) S y2l 2 S y2l 2 Pcr S  S   / 4  D 2 1  K 2  y 4 2 k 2CE y 4 2  D 2 / 16 1  K 2  CE 4 Pcr   D 1  K 2 2 S y   D 1  K  S y  4 Pcr  2 2 4 S y2l 2 D 2 1  K 2   2 D 2 1  K 2  CE 4 S y2l 2 1  K 2   1  K 2  CE   4 S y2l 2 1  K 2  4 Pcr   D 2 2 2   S y 1  K   1  K  CE 1  K  S y  1/ 2 1/2   S yl 2 Pcr   2  Ans. 2 2 2   S y 1  K   CE 1  K   ______________________________________________________________________________ 0.9 4-127 (a) M A  0, (0.75)(800)  0.92  0.52 FBO (0.5)  0  FBO  1373 N Using nd = 4, design for Fcr = nd FBO = 4(1373) = 5492 N l  0.92  0.52  1.03 m, S y  165 MPa In-plane: 1/ 2 1/ 2  bh3 / 12  I  k      A  bh  l 1.03   142.7 k 0.007218  0.2887 h  0.2887(0.025)  0.007 218 m, C  1.0 1/ 2 2 9  l   2 (207)(10 )     157.4    6  k 1  165(10 )  Since (l / k )1  (l / k ) use Johnson formula. Try 25 mm x 12 mm, 2   165 106  1   6   (142.7)  29.1 kN Pcr  0.025(0.012) 165 10  9   2  1(207)10     This is significantly greater than the design load of 5492 N found earlier. Check out-ofplane.   Shigley’s MED, 11th edition   Chapter 4 Solutions, Page 87/96 Out-of-plane: k  0.2887(0.012)  0.003 464 in, l 1.03   297.3 k 0.003 464 Since (l / k )1  (l / k ) use Euler equation. Pcr  0.025(0.012) C  1.2 1.2 2  207 109  8321 N 297.32 This is greater than the design load of 5492 N found earlier. It is also significantly less than the in-plane Pcr found earlier, so the out-of-plane condition will dominate. Iterate the process to find the minimum h that gives Pcr greater than the design load. With h = 0.010, Pcr = 4815 N (too small) h = 0.011, Pcr = 6409 N (acceptable) Use 25 mm x 11 mm. If standard size is preferred, use 25 mm x 12 mm. Ans. P 1373   10.4 106 Pa  10.4 MPa dh 0.012(0.011) No, bearing stress is not significant. Ans. ______________________________________________________________________________   (b)  b   4-128 (a) M A  0  800  750   9 2 9  52 FBD  1373 N (b)   FBD  500  Ans. 1373 F   4.99 MPa A 25 11 Ans. (c) lBD  0.92  0.52  1.030 m In plane, Table 4-2  C=1 1/ 2  bh /12  k  I / A   bh    3  h / 12  0.025 / 12  0.007 217 m l / k = 1.030/0.007 217 = 142.7. 1/ 2 2  l   2 CE  Eq. (4-45):       k 1  S y   2 2 1 207 109     165 106    1/ 2  157.4 Since (l / k)1 > l / k, use Johnson formula, Eq. (4-48) Shigley’s MED, 11th edition Chapter 4 Solutions, Page 88/96 2 2   165 106     Sy l  1  1   6   Pcr  A  S y     0.25  0.011 165 10   142.7   9   2  1 207 10     2 k  CE     26 720 N  26.72 kN n Pcr 26.72   19.5 FBD 1.373 Ans. (d) Out of plane, Table 4-2, C = 1.2, k = 0.011/ 12 = 3.175 (103) m, l / k = 1.030/3.175 (103) = 324.4. Since l / k > (l / k)1, use the Euler formula, Eq. (4-44)  C 2 E  1.2 2  207 109  Pcr  A  0.025 0.011        6 407 N 2 324.42   l / k     n Pcr 6 407   4.67 FBD 1373 Ans. ______________________________________________________________________________ 4-129 Out of plane bending, C =1.2.   82  4 2 M A  0  400  1200  FBC    8   FBC  1386 N Ans. k I/A  bh 3 /12  /  bh   h / 12  0.010 / 12  2.887 103  m l / k = 0.8/[2.887(103)] = 277.1 1/2 2  l   2 CE  Eq. (4-45):       k 1  S y   2 2 1.2  207 109     200 106    1/ 2  156.6 Since l / k > (l / k)1, use the Euler formula, Eq. (4-44)  C 2 E  1.2 2  207 109  Pcr  A   0.02  0.01    6 386 N 2  277.12   l / k     Shigley’s MED, 11th edition Chapter 4 Solutions, Page 89/96 n Pcr 6 386   4.6 Ans. FBD 1386 ______________________________________________________________________________ 4-130 This is an open-ended design problem with no one distinct solution. ______________________________________________________________________________ F = 1500( /4)22 = 4712 lbf. From Table A-20, Sy = 37.5 kpsi Pcr = nd F = 2.5(4712) = 11 780 lbf 4-131 (a) Assume Euler with C = 1 1/4 1/ 4  64 11790  502   64 Pcr l 2   3    d  3  6   1 30 10     CE  Use d = 1.25 in. The radius of gyration, k = ( I / A)1/2 = d /4 = 0.3125 in l 50   160 k 0.3125  P l2 I  d 4  cr 2 64 C E  2 2 (1)30 106     126   37.5 103     2 6 4   30 10  / 64 1.25  14194 lbf Pcr  502  1.193 in 1/ 2 1/2 2  l   2 CE        k 1  S y  Since 14 194 lbf > 11 780 lbf, d = 1.25 in is satisfactory.  use Euler Ans. 1/4  64 11780 162    0.675 in, so use d = 0.750 in d  3 6   1 30 10   k = 0.750/4 = 0.1875 in l 16   85.33 use Johnson k 0.1875 (b) 2    37.5 103   1   3 Pcr   0.750  37.5 10     12 748 lbf 85.33 6 4  2  1 30 10     2 Use d = 0.75 in. (c) Shigley’s MED, 11th edition Chapter 4 Solutions, Page 90/96 n( a )  14194  3.01 4 712 Ans. 12 748 Ans.  2.71 4 712 ______________________________________________________________________________ n(b )  4-132 From Table A-20, Sy = 180 MPa 4F sin = 2 943 735.8 sin  In range of operation, F is maximum when  = 15 735.8 Fmax   2843 N per bar sin15o F Pcr = ndFmax = 3.50 (2 843) = 9 951 N l = 350 mm, h = 30 mm Try b = 5 mm. Out of plane, k = b / 12 = 5/ 12 = 1.443 mm l 350   242.6 k 1.443 2 9  l   2 1.4  207 10       180 106   k 1   1/ 2 Pcr  A C 2 E l / k  2  5(30)  178.3 1.4 2  207 103  242.6  2  use Euler  7 290 N Too low. Try b = 6 mm. k = 6/ 12 = 1.732 mm 350 l   202.1 k 1.732 1.4 2  207 103 C 2 E  6(30)  12 605 N Pcr  A 2 2 l / k   202.1 O.K. Use 25  6 mm bars Ans. The factor of safety is 12 605  4.43 Ans. 2843 ______________________________________________________________________________ n 4-133 P = 1 500 + 9 000 = 10 500 lbf Shigley’s MED, 11th edition Ans. Chapter 4 Solutions, Page 91/96 MA = 10 500 (4.5/2)  9 000 (4.5) +M = 0 M = 16 874 lbfin e = M / P = 16 874/10 500 = 1.607 in Ans. From Table A-8, A = 2.160 in2, and I = 2.059 in4. The stresses are determined using Eq. (4-55) k2  I 2.059   0.953 in 2 A 2.160 10 500  1.607  3 / 2   P  ec  1    17157 psi  17.16 kpsi Ans. 1  2    A k  2.160  0.953  ______________________________________________________________________________ c   4-134 This is a design problem which has no single distinct solution. ______________________________________________________________________________ 4-135 Loss of potential energy of weight = W (h + ) 1 Increase in potential energy of spring = k 2 2 1 W (h + ) = k 2 2 2 2 W W or,  2   h  0 . W = 30 lbf, k = 100 lbf/in, h = 2 in yields k k  2  0.6   1.2 = 0 Taking the positive root [see discussion after Eq. (b), Sec. 4-17] 1  max  0.6  ( 0.6) 2  4(1.2)   1.436 in Ans.  2 Fmax = k  max = 100 (1.436) = 143.6 lbf Ans. ______________________________________________________________________________ 4-136 The drop of weight W1 converts potential energy, W1 h, to kinetic energy Equating these provides the velocity of W1 at impact with W2. W1h  Shigley’s MED, 11th edition 1 W1 2 v1 2 g  v1  2 gh 1 W1 2 v1 . 2 g (1) Chapter 4 Solutions, Page 92/96 Since the collision is inelastic, momentum is conserved. That is, (m1 + m2) v2 = m1 v1, where v2 is the velocity of W1 + W2 after impact. Thus W1  W2 W v2  1 v1 g g  v2  W1 W1 v1  W1  W2 W1  W2 2 gh (2) The kinetic and potential energies of W1 + W2 are then converted to potential energy of the spring. Thus, 1 W1  W2 2 1 v2  W1  W2    k 2 2 2 g Substituting in Eq. (1) and rearranging results in W1  W2 W12 h  2 0 (3) k W1  W2 k Solving for the positive root [see discussion after Eq. (b), Sec. 4-17] 2 2 2 W1  W2  W12 h  1  W1  W2     2  4  8  2 k k W1  W2 k      W1 = 40 N, W2 = 400 N, h = 200 mm, k = 32 kN/m = 32 N/mm. (4) 2 1   40  400  40  400  402 200     29.06 mm   2  4 8      2   32  40  400 32   32    Ans. Fmax = k = 32(29.06) = 930 N Ans. ______________________________________________________________________________ 1 4-137 The initial potential energy of the k1 spring is Vi = k1a 2 . The movement of the weight 2 1 1 2 W the distance y gives a final potential of Vf = k1  a  y   k 2 y 2 . Equating the two 2 2 energies give 1 1 1 2 k1a 2  k1  a  y   k 2 y 2 2 2 2 Simplifying gives Shigley’s MED, 11th edition  k1  k2  y 2  2ak1 y  0 Chapter 4 Solutions, Page 93/96 2k1a . Without damping the weight will vibrate between these k1  k 2 2k1a two limits. The maximum displacement is thus y max = Ans. k1  k 2 With W = 5 lbf, k1 = 10 lbf/in, k2 = 20 lbf/in, and a = 0.25 in This has two roots, y = 0, 2  0.25 10  0.1667 in Ans. 10  20 ______________________________________________________________________________ ymax  Shigley’s MED, 11th edition Chapter 4 Solutions, Page 94/96 Chapter 3 3-1 M O  0 18 RB  6(100)  0 RB  33.3 lbf Ans. Fy  0 RO  RB  100  0 RO  66.7 lbf Ans. RC  RB  33.3 lbf Ans. ______________________________________________________________________________ 3-2 Body AB: Fx  0 RAx  RBx Fy  0 RAy  RBy M B  0 RAy (10)  RAx (10)  0 RAx  RAy Body OAC: M O  0 RAy (10)  100(30)  0 RAy  300 lbf Ans. Fx  0 ROx   RAx  300 lbf Fy  0 ROy  RAy  100  0 ROy  200 lbf Ans. Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 1/114 3-3 0.8  1.39 kN Ans. tan 30 0.8 RA   1.6 kN Ans. sin 30 RO  ______________________________________________________________________________ 3-4 Step 1: Find RA & RE 4.5  7.794 m tan 30 M A  0 h 9 RE  7.794(400 cos 30 ) 4.5(400sin 30 )  0 RE  400 N F x F y 0 Ans. RAx  400 cos 30  0 RAx  346.4 N 0 RAy  400  400sin 30  0 RAy  200 N RA  346.42  2002  400 N Ans. Step 2: Find components of RC and RD on link 4 M C 0 400(4.5)   7.794  1.9  RD  0 RD  305.4 N F F Ans. x 0   RCx 4  305.4 N y 0  ( RCy ) 4  400 N Shigley’s MED, 11th edition Chapter 3 Solutions, Page 2/114 Step 3: Find components of RC on link 2  Fx  0  RCx 2  305.4  346.4  0  RCx 2  41 N F y 0 R  Cy 2  200 N Ans. _____________________________________________________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 3/114 3-5 M C  0 1500 R1  300(5)  1200(9)  0 R1  8.2 kN Ans. Fy  0 8.2  9  5  R2  0 R2  5.8 kN M 1  8.2(300)  2460 N  m Ans. Ans. M 2  2460  0.8(900)  1740 N  m Ans. M 3  1740  5.8(300)  0 checks! _____________________________________________________________________________ 3-6 Fy  0 RO  500  40(6)  740 lbf Ans. M O  0 M O  500(8)  40(6)(17)  8080 lbf  in M 1  8080  740(8)  2160 lbf  in Ans. Ans. M 2  2160  240(6)  720 lbf  in Ans. 1 M 3  720  (240)(6)  0 checks! 2 ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 4/114 3-7 M B  0 2.2 R1  1(2)  1(4)  0 R1  0.91 kN Fy  0 Ans. 0.91  2  R2  4  0 R2  6.91 kN Ans. M 1  0.91(1.2)  1.09 kN  m M 2  1.09  2.91(1)  4 kN  m Ans. Ans. M 3  4  4(1)  0 checks! ______________________________________________________________________________ 3-8 Break at the hinge at B Beam OB: From symmetry, R1  VB  200 lbf Ans. Beam BD: M D  0 200(12)  R2 (10)  40(10)(5)  0 R2  440 lbf Ans. Fy  0 200  440  40(10)  R3  0 R3  160 lbf Ans. Shigley’s MED, 11th edition Chapter 3 Solutions, Page 5/114 M 1  200(4)  800 lbf  in Ans. M 2  800  200(4)  0 checks at hinge M 3  800  200(6)  400 lbf  in Ans. 1 M 4  400  (240)(6)  320 lbf  in Ans. 2 1 M 5  320  (160)(4)  0 checks! 2 ______________________________________________________________________________ 3-9 q  R1 x 1  9 x  300 1  5 x  1200 0 1  R2 x  1500 0 V  R1  9 x  300  5 x  1200  R2 x  1500 1 1 0 M  R1 x  9 x  300  5 x  1200  R2 x  1500 1 (1) 1 (2) At x = 1500+ V = M = 0. Applying Eqs. (1) and (2), R1  9  5  R2  0  R1  R2  14 1500 R1  9(1500  300)  5(1500  1200)  0 R2  14  8.2  5.8 kN 0  x  300 :  R1  8.2 kN Ans. Ans. V  8.2 kN, M  8.2 x N  m 300  x  1200 : V  8.2  9  0.8 kN M  8.2 x  9( x  300)  0.8 x  2700 N  m 1200  x  1500 : V  8.2  9  5  5.8 kN M  8.2 x  9( x  300)  5( x  1200)  5.8 x  8700 N  m Plots of V and M are the same as in Prob. 3-5. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 6/114 3-10 q  RO x 1 2  MO x V  RO  M O x 1  500 x  8 1 0  40 x  14  40 x  20 0 1  500 x  8  40 x  14  40 x  20 1 2 M  RO x  M O  500 x  8  20 x  14  20 x  20 0 1 (1) 2 (2) at x  20 in, V  M  0, Eqs. (1) and (2) give RO  500  40  20  14   0  RO (20)  M O  500(20  8)  20(20  14) 2  0 0  x  8:  RO  740 lbf Ans. M O  8080 lbf  in Ans. V  740 lbf, M  740 x  8080 lbf  in 8  x  14 : V  740  500  240 lbf M  740 x  8080  500( x  8)  240 x  4080 lbf  in 14  x  20 : V  740  500  40( x  14)  40 x  800 lbf M  740 x  8080  500( x  8)  20( x  14) 2  20 x 2  800 x  8000 lbf  in Plots of V and M are the same as in Prob. 3-6. ______________________________________________________________________________ 3-11 q  R1 x 1  2 x  1.2 1  R2 x  2.2 0 1 0  4 x  3.2 V  R1  2 x  1.2  R2 x  2.2  4 x  3.2 1 1 1 0 (1) 1 M  R1 x  2 x  1.2  R2 x  2.2  4 x  3.2 at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2), (2) R1  2  R2  4  0  R1  R2  6 (3) 3.2 R1  2(2)  R2 (1)  0  3.2 R1  R2  4 (4) Solving Eqs. (3) and (4) simultaneously, R1 = -0.91 kN, R2 = 6.91 kN Ans. 0  x  1.2 : V  0.91 kN, M  0.91x kN  m 1.2  x  2.2 : V  0.91  2  2.91 kN M  0.91x  2( x  1.2)  2.91x  2.4 kN  m 2.2  x  3.2 : V  0.91  2  6.91  4 kN M  0.91x  2( x 1.2)  6.91( x  2.2)  4 x  12.8 kN  m Plots of V and M are the same as in Prob. 3-7. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 7/114 3-12 q  R1 x 1  400 x  4 0 1  R2 x  10 1 0 0  40 x  10  40 x  20  R3 x  20 0 1 1 V  R1  400 x  4  R2 x  10  40 x  10  40 x  20  R3 x  20 1 1 2 0 2 M  R1 x  400 x  4  R2 x  10  20 x  10  20 x  20  R3 x  20 M  0 at x  8 in  8R1  400(8  4)  0  R1  200 lbf 1 (1) 1 (2) Ans. at x = 20+, V =M = 0. Applying Eqs. (1) and (2), 200  400  R2  40(10)  R3  0 R2  R3  600  200(20)  400(16)  R2 (10)  20(10) 2  0  R2  440 lbf Ans. R3  600  440  160 lbf 0  x  4: Ans. V  200 lbf, M  200 x lbf  in 4  x  10 : V  200  400  200 lbf, M  200 x  400( x  4)  200 x  1600 lbf  in 10  x  20 : V  200  400  440  40( x  10)  640  40 x lbf M  200 x  400( x  4)  440( x  10)  20  x  10   20 x 2  640 x  4800 lbf  in Plots of V and M are the same as in Prob. 3-8. ______________________________________________________________________________ 2 3-13 Solution depends upon the beam selected. ______________________________________________________________________________ 3-14 (a) Moment at center,  l  2a  xc  2 2 wl  l   wl  l  2 M c   l  a        a 2  2  2   2  4  At reaction, M r  wa 2 2 a = 2.25, l = 10 in, w = 100 lbf/in Mc  100(10)  10    2.25   125 lbf  in 2  4  100  2.252   253 lbf  in 2 (b) Optimal occurs when M c  M r Mr  Shigley’s MED, 11th edition Ans. Chapter 3 Solutions, Page 8/114 2 wl  l  wa a    a 2  al  0.25l 2  0   2 4 2  Taking the positive root 1 l 2  1  0.207 l l  l 2  4  0.25l 2       2 2 for l = 10 in, w = 100 lbf, a = 0.207(10) = 2.07 in M min  100 2  2.07 2  214 lbf  in  a  Ans. ______________________________________________________________________________ 3-15 (a) 20  10  5 kpsi 2 20  10 CD   15 kpsi 2 C R  152  82  17 kpsi  1  5  17  22 kpsi  2  5  17  12 kpsi 1 8  p  tan 1    14.04 cw 2  15   1  R  17 kpsi s  45  14.04  30.96 ccw (b) 9  16  12.5 kpsi 2 16  9 CD   3.5 kpsi 2 C R  52  3.52  6.10 kpsi  1  12.5  6.1  18.6 kpsi  2  12.5  6.1  6.4 kpsi Shigley’s MED, 11th edition Chapter 3 Solutions, Page 9/114 1  5   tan 1    27.5 ccw 2  3.5   1  R  6.10 kpsi p  s  45  27.5  17.5 cw (c) 24  10  17 kpsi 2 24  10 CD   7 kpsi 2 C R  7 2  62  9.22 kpsi  1  17  9.22  26.22 kpsi  2  17  9.22  7.78 kpsi 1  7     p  90  tan 1     69.7 ccw 2 6   1  R  9.22 kpsi s  69.7  45  24.7 ccw (d) 12  22  5 kpsi 2 12  22 CD   17 kpsi 2 C R  17 2  122  20.81 kpsi  1  5  20.81  25.81 kpsi  2  5  20.81  15.81 kpsi 1  17       1  R  20.81 kpsi  p  90  tan 1     72.39 cw 2 12 s  72.39  45  27.39 cw ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 10/114 3-16 (a) 8  7  0.5 MPa 2 87  7.5 MPa CD  2 C R  7.52  62  9.60 MPa  1  9.60  0.5  9.10 MPa  2  0.5  9.6  10.1 Mpa 1  7.5     p  90  tan 1     70.67 cw 2  6   1  R  9.60 MPa s  70.67  45  25.67 cw (b) 96  1.5 MPa 2 96  7.5 MPa CD  2 C R  7.52  32  8.078 MPa  1  1.5  8.078  9.58 MPa  2  1.5  8.078  6.58 MPa 1  3    p  tan 1    10.9 cw 2  7.5   1  R  8.078 MPa s  45  10.9  34.1 ccw Shigley’s MED, 11th edition Chapter 3 Solutions, Page 11/114 (c) 12  4  4 MPa 2 12  4  8 MPa CD  2 C R  82  7 2  10.63 MPa  1  4  10.63  14.63 MPa  2  4  10.63  6.63 MPa 1  8      1  R  10.63 MPa  p  90  tan 1     69.4 ccw 2 7 s  69.4  45  24.4 ccw (d) 65  0.5 MPa 2 65  5.5 MPa CD  2 C R  5.52  82  9.71 MPa  1  0.5  9.71  10.21 MPa  2  0.5  9.71  9.21 MPa 1  8   p  tan 1    27.75 ccw 2  5.5   1  R  9.71 MPa s  45  27.75  17.25 cw ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 12/114 3-17 (a) 12  6  9 kpsi 2 12  6 CD   3 kpsi 2 C R  32  42  5 kpsi  1  5  9  14 kpsi  2  9  5  4 kpsi 1  4 tan 1    26.6 ccw  3 2  1  R  5 kpsi p  s  45  26.6  18.4 ccw (b) 30  10  10 kpsi 2 30  10  20 kpsi CD  2 C R  202  102  22.36 kpsi  1  10  22.36  32.36 kpsi  2  10  22.36  12.36 kpsi 1  10   p  tan 1    13.28 ccw 2  20   1  R  22.36 kpsi s  45  13.28  31.72 cw Shigley’s MED, 11th edition Chapter 3 Solutions, Page 13/114 (c) 10  18  4 kpsi 2 10  18  14 kpsi CD  2 C R  142  9 2  16.64 kpsi  1  4  16.64  20.64 kpsi  2  4  16.64  12.64 kpsi 1  14    p  90  tan 1     73.63 cw 2  9   1  R  16.64 kpsi s  73.63  45  28.63 cw (d) 9  19  14 kpsi 2 19  9  5 kpsi CD  2 C R  52  82  9.434 kpsi  1  14  9.43  23.43 kpsi  2  14  9.43  4.57 kpsi 1  5     p  90  tan 1     61.0 cw 2 8   1  R  9.34 kpsi s  61  45  16 cw ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 14/114 3-18 (a) 80  30  55 MPa 2 80  30  25 MPa CD  2 C R  252  202  32.02 MPa  1  0 MPa  2  55  32.02  22.98  23.0 MPa  3  55  32.0  87.0 MPa 1 2  23  11.5 MPa, 2  2 3  32.0 MPa, 1 3  87  43.5 MPa 2 (b) 30  60  15 MPa 2 60  30  45 MPa CD  2 C R  452  302  54.1 MPa  1  15  54.1  39.1 MPa  2  0 MPa  3  15  54.1  69.1 MPa 39.1  69.1  54.1 MPa 2 39.1  19.6 MPa 1 2  2 69.1  34.6 MPa 2 3  2 1 3  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 15/114 (c) 40  0  20 MPa 2 40  0 CD   20 MPa 2 C R  202  202  28.3 MPa  1  20  28.3  48.3 MPa  2  20  28.3  8.3 MPa  3   z  30 MPa 1 3  48.3  30  39.1 MPa, 2  1 2  28.3 MPa, 2 3  30  8.3  10.9 MPa 2 (d) 50  25 MPa 2 50 CD   25 MPa 2 C R  252  302  39.1 MPa  1  25  39.1  64.1 MPa  2  25  39.1  14.1 MPa  3   z  20 MPa 64.1  20 20  14.1  42.1 MPa,  1 2  39.1 MPa,  2 3   2.95 MPa 2 2 ______________________________________________________________________________ 1 3  3-19 (a) Since there are no shear stresses on the stress element, the stress element already represents principal stresses.  1   x  10 kpsi  2  0 kpsi  3   y  4 kpsi 10  ( 4)  7 kpsi 2 10  1 2   5 kpsi 2 0  (4)  2 kpsi 2 3  2 1 3  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 16/114 (b) 0  10  5 kpsi 2 10  0  5 kpsi CD  2 C R  52  4 2  6.40 kpsi  1  5  6.40  11.40 kpsi  2  0 kpsi,  3  5  6.40  1.40 kpsi  1 3  R  6.40 kpsi, 1 2  11.40  5.70 kpsi, 2 3  1.40  0.70 kpsi 2 (c) 2  8  5 kpsi 2 82  3 kpsi CD  2 C R  32  42  5 kpsi  1  5  5  0 kpsi,  2  0 kpsi  3  5  5  10 kpsi 1 3  10  5 kpsi, 2  1 2  0 kpsi,  2 3  5 kpsi (d) 10  30  10 kpsi 2 10  30  20 kpsi CD  2 C R  202  102  22.36 kpsi  1  10  22.36  12.36 kpsi  2  0 kpsi  3  10  22.36  32.36 kpsi 12.36 32.36  6.18 kpsi,  2 3   16.18 kpsi 2 2 ______________________________________________________________________________  1 3  22.36 kpsi, Shigley’s MED, 11th edition 1 2  Chapter 3 Solutions, Page 17/114 3-20 From Eq. (3-15),  3  (6  18  12) 2   6(18)  (6)(12)  18(12)  92  62  ( 15)2     6(18)( 12)  2(9)(6)(15)  ( 6)(6) 2  18( 15) 2  ( 12)(9) 2   0  3  594  3186  0 Roots are: 21.04, 5.67, –26.71 kpsi Ans. 21.04  5.67  7.69 kpsi 2 5.67  26.71  16.19 kpsi 2 3  2 21.04  26.71  23.88 kpsi  max   1 3  2 1 2  Ans. _____________________________________________________________________________ 3-21 From Eq. (3-15)   2  3  (20  0  20) 2   20(0)  20(20)  0(20)  402  20 2  0 2         20(0)(20)  2(40) 20 2 (0)  20 20 2   3  40 2  2 000  48 000  0 Roots are: 60, 20, –40 kpsi 60  20  20 kpsi 2 20  40  30 kpsi 2 3  2 60  40  50 kpsi  max   1 3  2  2   0(0) 2  20(40) 2   0  Ans. 1 2  Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 18/114 3-22 From Eq. (3-15) 2 2  3  (10  40  40) 2  10(40)  10(40)  40(40)  20 2   40    20       10(40)(40)  2(20)( 40)( 20)  10( 40)  40( 20)  40(20)   0  3  90 2  0 2 Roots are: 90, 0, 0 MPa 2 2 Ans. 2 3  0  1 2   1 3   max  90  45 MPa 2 Ans. _____________________________________________________________________________ 3-23  15000 F   33 950 psi  34.0 kpsi A  4   0.752   60 FL L    33 950  0.0679 in AE E 30 106   0.0679 1    1130 106   1130 L 60 From Table A-5, v = 0.292  2  v1  0.292(1130)  330  Ans. Ans.  Ans. Ans. d   2 d  330 106  (0.75)  248 10 6  in Ans. _____________________________________________________________________________ 3-24  3000 F   6790 psi  6.79 kpsi A  4   0.752   60 FL L    6790  0.0392 in AE E 10.4 106   0.0392  653 106   653 L 60 From Table A-5, v = 0.333 1    2  v1  0.333(653)  217  Ans. Ans. Ans. Ans. d   2 d  217 10 6  (0.75)  163 10 6  in Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 19/114 3-25 d 0.0001d   0.0001 d d From Table A-5, v = 0.326, E = 119 GPa  0.0001 1  2   306.7 106  v 0.326 FL F and   , so  AE A E =  1 E  306.7 10 6  (119) 109   36.5 MPa L 2  F   A  36.5 10 6    0.03 2 Ans.  25 800 N  25.8 kN 4 Sy = 70 MPa >  , so elastic deformation assumption is valid. _____________________________________________________________________________ 3-26  FL L 8(12)    20 000  0.185 in AE E 10.4 106  Ans. _____________________________________________________________________________ 3-27  FL L 3    140 106   0.00586 m  5.86 mm AE E 71.7 109  Ans. _____________________________________________________________________________ 3-28  FL L 10(12)    15 000  0.173 in AE E 10.4 106  Ans. _____________________________________________________________________________ 3-29 With  z  0, solve the first two equations of Eq. (3-19) simulatenously. Place E on the left-hand side of both equations, and using Cramer’s rule, x  E x E y v 1 1 v v 1  E x  vE y 1  v2  E   x  v y  1  v2 Likewise, Shigley’s MED, 11th edition Chapter 3 Solutions, Page 20/114 y  E   y   x  1  v2 From Table A-5, E = 207 GPa and ν = 0.292. Thus, E   x  v y  x  2  207 109  0.0019  0.292  0.000 72   2 1 v 1  0.292 9 207 10   0.000 72  0.292  0.0019   10   382 MPa 6 Ans. 106   37.4 MPa Ans. 1  0.2922 _____________________________________________________________________________ y  3-30 With  z  0, solve the first two equations of Eq. (3-19) simulatenously. Place E on the left-hand side of both equations, and using Cramer’s rule, x  E x E y v 1 1 v v 1  E x  vE y 1  v2  E   x  v y  1  v2 Likewise, y  E   y   x  1  v2 From Table A-5, E = 71.7 GPa and ν = 0.333. Thus, E   x  v y  71.7 109   0.0019  0.333  0.000 72   106   134 MPa Ans.  x   2 2 1 v 1  0.333 9 71.7 10   0.000 72  0.333  0.0019   y  106   7.04 MPa Ans. 1  0.3332 _____________________________________________________________________________ 3-31 For plane strain, z = 0. From the third equation of Eq. (3-19), Ans.  z    x   y  First of Eq. (3-19), 1  x   x   y    x   y  E 1  1  2   x  1    y  E 1  1    x  y   E  Similarly,  Shigley’s MED, 11th edition  Ans. Chapter 3 Solutions, Page 21/114 1  1    y  x  Ans. E  ______________________________________________________________________________ 3-32 c ac F (a) R1  F M max  R1a  l l 6M 6 ac  bh 2 l Ans. FF  2  2 bh bh l 6ac y     bm b  hm h   lm l1   1( s)( s) 2 (s)  s 2 F (b) m  m F ( s )( s )  am a  cm c  2 Ans. For equal stress, the model load varies by the square of the scale factor. _____________________________________________________________________________ 3-33 wl w l  l  wl 2 R1  , M max x l /2  (a) l    2 2 2 2 8 6M 6 wl 2 3Wl  2  2  4bh 2 bh bh 8 (b)  4  bh 2 W 3 l Ans. Wm ( m /  )(bm / b)(hm / h) 2 1( s)( s )2    s2 W lm / l s Ans. w m lm wm s 2  s2    s Ans. wl w s For equal stress, the model load w varies linearly with the scale factor. _____________________________________________________________________________ 3-34 (a) Can solve by iteration or derive equations for the general case. Find maximum moment under wheel W3 . WT  W at centroid of W’s l  x3  d 3 RA  WT l Under wheel 3, M 3  RA x3  W1a13  W2 a23  For maximum,  l  x3  d3  W l dM 3 W  0   l  d3  2 x3  T dx3 l Substitute into M  M 3  Shigley’s MED, 11th edition  l  d3  4l T x3  W1a13  W2 a23  x3  l  d3 2 2 WT  W1a13  W2 a23 Chapter 3 Solutions, Page 22/114 This means the midpoint of d 3 intersects the midpoint of the beam. For wheel i, l  di xi  , 2 Mi  l  di   4l 2 i 1 WT   W j a ji j 1 Note for wheel 1: W j a ji  0 WT  104.4, W1  W2  W3  W4  104.4  26.1 kips 4 476 (1200  238) 2  238 in, M 1  (104.4)  20128 kip  in 2 4(1200) Wheel 2: d 2  238  84  154 in Wheel 1: d1  M2  (1200  154)2 (104.4)  26.1(84)  21 605 kip  in  M max 4(1200) Ans. Check if all of the wheels are on the rail. xmax  600  77  523 in Ans. (b) (c) See above sketch. (d) Inner axles _____________________________________________________________________________ 3-35 (a) Let a = total area of entire envelope Let b = area of side notch A  a  2b  40(2)(37.5)  25  34   2150 mm2 1 1 3 3  40 75   34 25 12 12 6 4 I  1.36 10  mm Ans. I  I a  2Ib  (b) Aa  0.375(1.875)  0.703 125 in 2 Dimensions in mm. Ab  0.375(1.75)  0.656 25 in 2 A  2(0.703125)  0.656 25  2.0625 in 2 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 23/114 2(0.703 125)(0.9375)  0.656 25(0.6875)  0.858 in Ans. 2.0625 0.375(1.875)3  0.206 in 4 Ia  12 1.75(0.375)3  0.007 69 in 4 Ib  12 I1  2 0.206  0.703 125(0.0795) 2   0.00769  0.656 25(0.1705) 2   0.448 in 4 y Ans. (c) Use two negative areas. Aa  625 mm 2 , Ab  5625 mm 2 , Ac  10 000 mm 2 A  10 000  5625  625  3750 mm 2 ; ya  6.25 mm, yb  50 mm, yc  50 mm 10 000(50)  5625(50)  625(6.25)  57.29 mm 3750 c1  100  57.29  42.71 mm Ans. y Ans. 50(12.5)3  8138 mm 4 12 75(75)3  2.637 106  mm 4 Ib  12 100(100)3  8.333 106  in 4 Ic  12 2 2 I1  8.333 106   10 000(7.29) 2    2.637 10 6   5625  7.29    8138  625  57.29  6.25       Ia  I1  4.29 106  in 4 Ans. (d) Shigley’s MED, 11th edition Chapter 3 Solutions, Page 24/114 Aa  4  0.875   3.5 in 2 Ab  2.5  0.875   2.1875 in 2 A  Aa  Ab  5.6875 in 2 y 2.9375  3.5   1.25(2.1875) 5.6875  2.288 in Ans. 1 1 3 2 3 2 (4)  0.875   3.5  2.9375  2.288    0.875  2.5   2.1875  2.288  1.25  12 12 4 I  5.20 in Ans. _____________________________________________________________________________ I 3-36 1 (20)(40)3  1.067 105  mm4 12 A  20(40)  800 mm 2 Mmax is at A. At the bottom of the section, Mc 450 000(20)  max    84.3 MPa Ans. I 1.067 105  I Due to V, max is between A and B at y = 0. 3 V 3  3000     max    5.63 MPa Ans. 2 A 2  800  __________________________________________________________________________ 3-37 1 (1)(2)3  0.6667 in 4 12 A  1(2)  2 in 2 I M O  0 8 RA  100(8)(12)  0 RA  1200 lbf RO  1200  100(8)  400 lbf M max is at A. At the top of the beam,  max  Mc 3200(0.5)   2400 psi I 0.6667 Ans. Due to V,  max is at A, at y = 0. 3 V 3  800      600 psi Ans. 2 A 2 2  _____________________________________________________________________________  max  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 25/114 3-38 1 (0.75)(2)3  0.5 in 4 12 A  (0.75)(2)  1.5 in 2 I M A  0 15 RB  1000(20)  0 RB  1333.3 lbf RA  3000  1333.3  1000  2666.7 lbf M max is at B. At the top of the beam,  max  Mc 5000(1)   10 000 psi I 0.5 Ans. Due to V,  max is between B and C at y = 0. 3 V 3  1000      1000 psi Ans. 2 A 2  1.5  _____________________________________________________________________________  max  3-39 I d4   (50) 4    306.796 103 mm 4 64 64 2 d  (50) 2 A   1963 mm 2 4 4 M B  0 6(300)(150)  200 RA  0 RA  1350 kN RB  6(300)  1350  450 kN M max is at A. At the top, Mc 30 000(25)   2.44 kN/mm 2  2.44 GPa I 30 6796 Due to V,  max is at A, at y = 0.  max  Ans. 4 V 4  750  2     0.509 kN/mm  509 MPa Ans. 3 A 3  1963  _____________________________________________________________________________  max  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 26/114 3-40 8 I wl 2 wl 2 c   max   w  max 8 8I cl 2 (a) l  48 in; Table A-8, I  0.537 in 4 M max  w 8 12  103   0.537  1 482   22.38 lbf/in Ans. (b) l  60 in, I  1 12  2   33   1 12 1.625   2.6253   2.051 in 4 w 8 12  103   2.051 1.5  602   36.5 lbf/in Ans. (c) l  60 in; Table A-6, I  2  0.703   1.406 in 4 y = 0.717 in, cmax = 1.783 in 8 12  103  1.406  w  21.0 lbf/in Ans. 1.783  60 2  (d) l  60 in, Table A-7, I  2.07 in 4 w 8 12  103   2.07  1.5  602   36.8 lbf/in Ans. _____________________________________________________________________________ 3-41 I   0.5   3.068 10  in , 64 4 3 4 A  4  0.5   0.1963 in 2 2 Model (c) 500(0.5) 500(0.75 / 2) M    218.75 lbf  in 2 2 Mc 218.75(0.25)   I 3.068 10 3    17 825 psi  17.8 kpsi  max  Ans. 4 V 4 500   3400 psi  3.4 kpsi 3 A 3 0.1963 Shigley’s MED, 11th edition Ans. Chapter 3 Solutions, Page 27/114 Model (d) M  500(0.625)  312.5 lbf  in  Mc 312.5(0.25)  I 3.068 103    25 464 psi  25.5 kpsi  max  Ans. 4 V 4 500   3400 psi  3.4 kpsi 3 A 3 0.1963 Ans. Model (e) M  500(0.4375)  218.75 lbf  in  Mc 218.75(0.25)  I 3.068 103    17 825 psi  17.8 kpsi  max  Ans. 4 V 4 500   3400 psi  3.4 kpsi 3 A 3 0.1963 Ans. ____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 28/114 3-42 I   12   1018 mm , A  12   113.1 mm  64 4 4 4 2 2 Model (c) 2000(6) 2000(9)   15 000 N  mm M 2 2 Mc 15 000(6)   1018 I   88.4 N/mm 2  88.4 MPa Ans.  max  4 V 4  2000  2     23.6 N/mm  23.6 MPa 3 A 3  113.1  Ans. Model (d) M  2000(12)  24 000 N  mm Mc 24 000(6)  I 1018   141.5 N/mm 2  141.5 MPa Ans. 4 V 4  2000  2  max      23.6 N/mm  23.6 MPa 3 A 3  113.1   Ans. Model (e) M  2000(7.5)  15000 N  mm Mc 15000(6)  I 1018   88.4 N/mm 2  88.4 MPa Ans. 4 V 4  2000  2  max      23.6 N/mm  23.6 MPa 3 A 3  113.1   Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 29/114 3-43 (a)   d (b)   d (c)   d Mc M  d / 2  32M   I  d 4 / 64  d 3 3 32 M   3 32(218.75)  0.420 in Ans.  (30 000) V V  A d2 / 4 4V  4(500)  0.206 in  (15000)  Ans. 4V 4 V  3 A 3  d 2 / 4  4 4V  3  4 4(500)  0.238 in 3  (15000) Ans. _____________________________________________________________________________ 3-44 p1  p2 1 x  l  terms for x  l  a a p  p2 1 2 V   F  p1 x  l  1 x  l  terms for x  l  a 2a p p  p2 2 3 M   Fx  1 x  l  1 x  l  terms for x  l  a 2 6a q  F x 1 0  p1 x  l  At x  (l  a)  , V  M  0, terms for x > l + a = 0 p1  p2 2 2F a  0  p1  p2  a 2a 2 pa p  p2 3 6 F (l  a )  F (l  a )  1  1 a 0  2 p1  p2  a2 2 6a  F  p1a  From (1) and (2) Shigley’s MED, 11th edition p1  2F (3l  2a), a2 p2  2F (3l  a) a2 (1) (2) (3) Chapter 3 Solutions, Page 30/114 From similar triangles b a  p2 p1  p2  b ap2 p1  p2 (4) Mmax occurs where V = 0 xmax  l  a  2b p1 p  p2 ( a  2b) 2  1 (a  2b)3 2 6a  p2 p1 p   Fl  F ( a  2b)  ( a  2b) 2  1 (a  2b)3 2 6a Normally Mmax =  Fl M max   F (l  a  2b)  The fractional increase in the magnitude is F (a  2b)   p1 2  (a  2b)2   p1  p2  6a  (a  2b)3  Fl (5) For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in From (3) From (4) p1  2(1500) 3 1.5  2(1.2)   14 375 lbf/in 1.22  p2  2(1500) 3 1.5  1.2  11 875 lbf/in 1.22  b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in Substituting into (5) yields  = 0.036 89 or 3.7% higher than -Fl _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 31/114 3-45 300(30) 40  1800  6900 lbf 2 30 300(30) 10 R2   1800  3900 lbf 2 30 3900 a  13 in 300 R1  MB = 1800(10) = 18 000 lbfin Mx = 27 in = (1/2)3900(13) = 25 350 lbfin 0.5(3)  2.5(3)  1.5 in 6 1 I1  (3)(13 )  0.25 in 4 12 1 I 2  (1)(33 )  2.25 in 4 12 Applying the parallel-axis theorem, y I z   0.25  3(1.5  0.5) 2    2.25  3(2.5  1.5) 2   8.5 in 4 (a) 18000( 1.5) At x  10 in, y  1.5 in,  x    3176 psi 8.5 18000(2.5) At x  10 in, y  2.5 in,  x    5294 psi 8.5 25350( 1.5) At x  27 in, y  1.5 in,  x    4474 psi 8.5 25350(2.5) At x  27 in, y  2.5 in,  x    7456 psi 8.5 Ans. Max tension  5294 psi Max compression  7456 psi Ans. (b) The maximum shear stress due to V is at B, at the neutral axis. Vmax  5100 lbf Q  yA  1.25(2.5)(1)  3.125 in 3  max V (c) VQ 5100(3.125)   1875 psi Ans. Ib 8.5(1) There are three potentially critical locations for the maximum shear stress, all at  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 32/114 x = 27 in: (i) at the top where the bending stress is maximum, (ii) at the neutral axis where the transverse shear is maximum, or (iii) in the web just above the flange where bending stress and shear stress are in their largest combination. For (i): The maximum bending stress was previously found to be  7456 psi, and the shear stress is zero. From Mohr’s circle,  7456  3728 psi  max  max  2 2 For (ii): The bending stress is zero, and the transverse shear stress was found previously to be 1875 psi. Thus, max = 1875 psi. For (iii): The bending stress, at y = – 0.5 in, is 18000(0.5) x    1059 psi 8.5 The transverse shear stress is Q  yA  (1)(3)(1)  3.0 in 3  VQ 5100(3.0)   1800 psi Ib 8.5(1) From Mohr’s circle, 2  1059  2   1800  1876 psi  2  The critical location is at x = 27 in, at the top surface, where max = 3728 psi. Ans. _____________________________________________________________________________ 3-46  max   (a) I = bh3/12 = 50(503)/(12) = 520.83 (103) mm4 Element A: 2 000  200  50 / 2  My A   A    19.2 MPa I 520.83 103  A  VQA , QA  0   A  0 Ib Shigley’s MED, 11th edition Ans. Chapter 3 Solutions, Page 33/114 Element B: B   B  MyB , yB  0   B  0 I QB  yB A   25 / 2  25  50   15.625 103  mm3 3 VQB 2000 15.625 10   1.20 MPa Ib 520.83 103  50 Element C: C   2 000  200  25 / 2  520.83 103  Ans.  9.60 MPa QC   25 / 2  25 / 4  25 / 2  50  11.719 103  mm3 C  2 000 11.719 103 520.83 103  50  0.90 MPa Ans. (b) Point A:  max  A 2  19.2  9.6 MPa 2 Ans. Point B:  max   B  1.20 MPa Ans. Point C: 2  max     C    C2  2  2  9.60  2     0.9  4.88 MPa  2  (c) Point A is critical. Shigley’s MED, 11th edition Ans. Ans. Chapter 3 Solutions, Page 34/114 (d) Transverse shear does not change with respect to the length of the beam. 2 000 L  50 / 2   1.20  L  25.0 mm Ans. 2 2  520.83 103 ______________________________________________________________________________ 3-47 I = (/64)404 = 125.66(103) mm4, J = 2I (a) Point A: 3 M A c A 50 10 10  40 / 2  A    79.6 MPa 125.66 I Ans. 3 800 10 40 / 2     Tr A  A A   63.7 MPa 2 125.66 103 J  A max  A  Point B: B  0 3 4 V 4 50 10  B    53.1 MPa 3 A 3   40 / 2 2 Ans. Point C: C  0 C   B  Ans. Tr  53.1  63.7  116.8 MPa J (b) Point A:  79.6  2 2  max  A     63.7  75.1 MPa  2  Ans. Point B: (max)B = 53.1 MPa Shigley’s MED, 11th edition Ans. Chapter 3 Solutions, Page 35/114 Point C: (max)C = 116.8 MPa (c) max = 116.8 MPa at point C Ans. Ans. ______________________________________________________________________________ (a) L = 10 in. Element A: 3-48 My (1000)(10)(0.5)  10 3  101.9 kpsi 4 I ( / 64)(1) VQ A  , Q  0  A  0 Ib  A   2  max  2    101.9  2   A    A2     (0)  50.9 kpsi  2   2  Ans. Element B: B   My , y0 I  4r Q  y A    3 B  B  0  2 3 4  0.5     r  4r    1/ 12 in 3   6 6  2  3 VQ (1000)(1/ 12) 103  1.698 kpsi  4 Ib ( / 64)(1) (1)  0  2  max     1.6982  1.698 kpsi 2 Ans. Element C: C   (1000)(10)(0.25) My  103  50.93 kpsi I ( / 64)(1) 4 Shigley’s MED, 11th edition   Chapter 3 Solutions, Page 36/114 r r r y1 y1 y1   Q   ydA   y (2 x )dy   y 2 r 2  y 2 dy 2   r 2  y2 3   3/ 2 r y1 2    r2  r2 3    3/ 2   r 2  y12  3/ 2   3/ 2 2 2 r  y12 3 For C, y1 = r /2 =0.25 in   Q  3/2 2 0.52  0.252   0.05413 in3  3 b  2 x  2 r 2  y12  2 0.52  0.252  0.866 in C  VQ (1000)(0.05413) 103  1.273 kpsi  4 Ib ( / 64)(1) (0.866)  max  50.93  2     (1.273)  25.50 kpsi  2    2 Ans. (b) Neglecting transverse shear stress: Element A: Since the transverse shear stress at point A is zero, there is no change.  max  50.9 kpsi Ans. % error  0% Ans. Element B: Since the only stress at point B is transverse shear stress, neglecting the transverse shear stress ignores the entire stress. 2 0  max     0 psi Ans. 2  1.698  0  % error    * (100)  100% Ans.  1.698  Element C: 2  max  50.93      25.47 kpsi  2  Ans.  25.50  25.47  % error    *(100)  0.12% Ans. 25.50   (c) Repeating the process with different beam lengths produces the results in the table. Shigley’s MED, 11th edition Chapter 3 Solutions, Page 37/114 Bending stress, kpsi) Transverse shear stress, kpsi) Max shear stress, max kpsi) Max shear stress, neglecting  max kpsi) % error 102 0 50.9 0 1.70 1.27 50.9 1.70 25.50 50.9 0 25.47 0 100 0.12 40.7 0 20.4 0 1.70 1.27 20.4 1.70 10.26 20.4 0 10.19 0 100 0.77 10.2 0 5.09 0 1.70 1.27 5.09 1.70 2.85 5.09 0 2.55 0 100 10.6 1.02 0 0.509 0 1.70 1.27 0.509 1.70 1.30 0.509 0 0.255 0 100 80.4 L = 10 in A B C L = 4 in A B C L = 1 in A B C L = 0.1in A B C Discussion: The transverse shear stress is only significant in determining the critical stress element as the length of the cantilever beam becomes smaller. As this length decreases, bending stress reduces greatly and transverse shear stress stays the same. This causes the critical element location to go from being at point A, on the surface, to point B, in the center. The maximum shear stress is on the outer surface at point A for all cases except L = 0.1 in, where it is at point B at the center. When the critical stress element is at point A, there is no error from neglecting transverse shear stress, since it is zero at that location. Neglecting the transverse shear stress has extreme significance at the stress element at the center at point B, but that location is probably only of practical significance for very short beam lengths. _____________________________________________________________________________ 3-49 c F l c M  Fx 0  x  a l 6M 6  c l  Fx  2  bh bh2 R1  h 6 Fcx lb max 0 xa Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 38/114 3-50 c From Problem 3-49, R1  F  V , 0  x  a l 3V 3 (c / l ) F 3 Fc  max    h 2 bh 2 bh 2 lb max Ans. 6 Fcx . lb max Sub in x = e and equate to h above. 3 Fc 6 Fce  lb max 2 lb max From Problem 3-49, h( x)  3 Fc max Ans. 2 8 lb max _____________________________________________________________________________ e 3-51 (a) x-z plane M O  0  1.5(0.5)  2(1.5) sin(30 )(2.25)  R2 z (3) R2 z  1.375 kN Ans. Fz  0  R1z  1.5  2(1.5) sin(30 )  1.375 R1z  1.625 kN Ans. x-y plane M O  0  2(1.5) cos(30 )(2.25)  R2 y (3) R2 y  1.949 kN Ans. Fy  0  R1 y  2(1.5) cos(30 )  1.949 R1 y  0.6491 kN Ans. (b) Shigley’s MED, 11th edition Chapter 3 Solutions, Page 39/114 (c) The transverse shear and bending moments for most points of interest can readily be taken straight from the diagrams. For 1.5 < x < 3, the bending moment equations are parabolic, and are obtained by integrating the linear expressions for shear. For convenience, use a coordinate shift of x = x – 1.5. Then, for 0 < x < 1.5, Vz  x  0.125 My    x  V dx  z 2 2  0.125x  C At x  0, M y  C  0.9375  M y  0.5  x   0.125 x  0.9375 2 1.949 x  0.6491  1.732 x  0.6491 1.125 1.732 2 Mz   x   0.6491x  C 2 2 At x  0, M z  C  0.9737  M z  0.8662  x   0.125 x  0.9375 By programming these bending moment equations, we can find My, Mz, and their vector combination at any point along the beam. The maximum combined bending moment is found to be at x = 1.79 m, where M = 1.433 kN·m. The table below shows values at key locations on the shear and bending moment diagrams. Vy   x (m) 0 0.5 1.5 1.625 1.875 3 Vz (kN) Vy (kN) –1.625 0.6491 –1.625 0.6491 –0.1250 0.6491 0 0.4327 0.2500 0 1.375 –1.949 My Mz V (kN) (kNm) (kNm) 1.750 0 0 1.750 –0.8125 0.3246 0.6610 0.9375 0.9737 0.4327 –0.9453 1.041 0.2500 –0.9141 1.095 2.385 0 0 M (kNm) 0 0.8749 1.352 1.406 1.427 0 (d) The bending stress is obtained from Eq. (3-27), M z yA M y z A x   Iz Iy The maximum tensile bending stress will be at point A in the cross section of Prob. 3-35 (a), where distances from the neutral axes for both bending moments will be maximum. At A, for Mz, yA = –37.5 mm, and for My, zA = –20 mm. 40(75)3 34(25)3 Iz    1.36(106 ) mm 4  1.36(106 ) m 4 12 12  25(40)3  25(6)3 Iy  2  2.67(105 ) mm 4  2.67(107 ) m 4  12  12  It is apparent the maximum bending moment, and thus the maximum stress, will be in the parabolic section of the bending moment diagrams. Programming Eq. (3-27) with the bending moment equations previously derived, the maximum tensile bending stress is Shigley’s MED, 11th edition Chapter 3 Solutions, Page 40/114 found at x = 1.77 m, where My = – 0.9408 kN·m, Mz = 1.075 kN·m, and x = 100.1 MPa. Ans. _____________________________________________________________________________ 3-52 (a) x-z plane 3 600 M O  0  (1000)(4)  (10)  M Oy 5 2 M Oy  1842.6 lbf  in Ans. 3 600 Fz  0  ROz  (1000)  5 2 ROz  175.7 lbf Ans. x-y plane 4 600 (10)  M Oz M O  0   (1000)(4)  5 2 M Oz  7442.5 lbf  in Ans. 4 600 Fy  0  ROy  (1000)  5 2 ROy  1224.3 lbf Ans. (b) (c) 1/ 2 V ( x)  V y ( x ) 2  Vz ( x) 2  1/ 2 M ( x)   M y ( x )2  M z ( x ) 2  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 41/114 x (m) 0 4 10 Vz (kN) –175.7 –175.7 424.3 Vy (kN) 1224.3 1224.3 424.3 V (kN) 1237 1237 600 My (kNm) Mz (kNm) M (kNm) –1842.6 –7442.6 7667 –2545.4 –2545.4 3600 0 0 0 (d) The maximum tensile bending stress will be at the outer corner of the cross section in the positive y, negative z quadrant, where y = 1.5 in and z = –1 in. 2(3)3 (1.625)(2.625)3 Iz    2.051 in 4 12 12 3 3(2) (2.625)(1.625) 3 Iy    1.601 in 4 12 12 At x = 0, using Eq. (3-27), M y M z x   z  y Iz Iy ( 7442.6)(1.5) ( 1842.6)( 1) x     6594 psi 2.051 1.601 Check at x = 4 in, ( 2545.4)(1.5) ( 2545.4)( 1) x     2706 psi 2.051 1.601 The critical location is at x = 0, where x = 6594 psi. Ans. _____________________________________________________________________________ 3-53 (a) Moments at A: My = 300(0.050) = 15 N٠m, Mz = 200(0.055) = 11 N٠m, Torque at A: Tx = 200(0.060) = 12 N٠m M  M y2  M z2  152  112  18.601 N  m  Mz   11   tan 1    36.25o  M   15   y o Critical point will be 90 from , i.e. 126.25o from the vertical y axis. (b) M c 32 M 32 18.601 6  bend    10  109.6 MPa I  d3   0.0123    tan 1   axial  Ans. 4  300  F 4F 106  2.65 MPa   2 A d   0.0122   total   bend   axial  109.6  2.65  112.3 MPa  16 12  Tx c 16Tx 10 6  35.4 MPa   3 3 J d   0.012  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 42/114 Ans. (c) Ans. 2 112.3  112.3  2  56.2 MPa, R     35.4  66.4 MPa 2  2   1  C  R  56.2  66.4  122.6 MPa,  2  0, (d) C  3  C  R  56.2  66.4  10.2 MPa  max  R  66.4 MPa Ans. ______________________________________________________________________________ 3-54 (a) Moments at A: My =  200(0.060) =  12 N٠m, Mz = 300(0.095) = 28.5 N٠m Torque at A: Tx =  300(0.050) =  15 N٠m M  M y2  M z2   Mz M  y   tan 1   12 2  28.52  30.92 N  m  1  28.5  o   tan    112.8  12   Critical point will be 90o from , i.e. 202.8o from the vertical y. axis. M c 32 M 32  30.92  6  bend    10  182.3 MPa  d3 I   0.0123   axial  Ans. 4  200  F 4F   106  1.77 MPa 2 2 A d   0.012   total   bend   axial  182.3  1.77  184.1 MPa  16 15  Tx c 16Tx   10 6  44.2 MPa 3 3 d J   0.012  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 43/114 Ans. (c) Ans. (d) 2 184.1  184.1  2 C  92.1 MPa, R     44.2  102.1 MPa 2  2   1  C  R  92.1  102.1  194.2 MPa,  2  0,  3  C  R  92.1  102.1  10 MPa  max  R  102.1 MPa Ans. ______________________________________________________________________________ 3-55 (a) Moments at A: My = 60(5) + 200(5.5) = 1400 lbf٠in, Mz =  (300 – 75)5.5 =  1238 lbf٠in, Torque at A: Tx =  75(5) =  375 lbf٠in M  M y2  M z2  14002   1238   1869 lbf  in 2  Mz   1238  o  tan 1     318.5  41.5 M  1400    y o Critical point will be 90 from , i.e. 48.5o from the vertical y axis. Ans. M c 32 M 32 1869  3  bend  10  45.13 kpsi   (b) I  d3   0.753    tan 1   axial  4  60  F 4F    136 psi 2 A d   0.752   total   bend   axial  45.126  0.136  45.3 kpsi  Tx c 16Tx 16  375  3   10  4.53 kpsi  d 3   0.753  J Shigley’s MED, 11th edition Chapter 3 Solutions, Page 44/114 Ans. (c) Ans. (d) 2 45.3  45.3  2 C  22.7 kpsi, R     4.53  23.1 kpsi 2  2   1  C  R  22.7  23.1  45.8 kpsi,  2  0,  3  C  R  22.7  23.1  0.45 kpsi  max  R  23.1 kpsi Ans. ______________________________________________________________________________ 3-56 Given: b = 3.6 in, c = 2.5 in, and l = 40 in. From Table A-5, G = 11.5 Mpsi. From Table A-20, Sy = 42 kpsi. (a) For the table for Eq. (3-40), b/c = 3.6/2.5 = 1.44 b/c_ _____ 0.208 1  1.44 0.231 1.5 0.231   1.5  1.44   0.12    0.231  0.12  0.231  0.208   0.228 0.231  0.208 1.5  1 Equation (3-40): 30 103  T  max    5 850 psi  5.85 kpsi Ans.  bc 2 0.228  3.6  2.52 (b) For the table for Eq. (3-41), b/c = 3.6/2.5 = 1.44 _____ b/c_ 0.141 1  1.44 0.196 1.5 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 45/114 0.196   1.5  1.44   0.12    0.196  0.12  0.196  0.141  0.189 0.196  0.141 1.5  1 Equation (3-41): 30 103  40 Tl    9.815 10 3  rad  0.562o Ans. 3 3 6  bc G 0.189  3.6  2.5 11.5 10 (c) Ssy = 0.5(42) = 21 kpsi. Yield factor of safety, S sy 21 ny    3.59 Ans.  max 5.85 ______________________________________________________________________________ 3-57 Given: 250 hp at 540 rev/min, allow = 15 kpsi. 63 025 63 025 Eq. (3-42): T H 250  29.178 103  lbf  in n 540 Eq. (3-41) with table where for square cross-section, b/c =1 29.178 103  T  max    15 103   b  2.107 in 2 3 0.208bc 0.208b 2 From Table A-17, use b = 14 in Ans. ______________________________________________________________________________ 3-58 Given: T = 50 kN-m. OD = 300 mm, t = 2 mm, and l = 2 m. J = (/32)(0.3004 – 0.2964) = 4.157(105) m4. From Table A-5, G = 79.3 GPa. (a) Eq. (3-37): 3 Tr 50 10  0.15 6  max   10  180.4 MPa Ans. J 4.157 10 5  (b) Eq. (3-45): 50 103  T  max   106  179.2 MPa 2 2 Amt 2  / 4  0.298  0.002  Answer is 0.67 percent lower than part (a). (c) Eq. (3-35): 50 103  2 TL    0.0303 rad  1.74o JG 4.157 10 5  79.3 109  Ans. Ans. (d) Eq. (3-46): 50 103    0.298  2 TLml   0.303 rad  1.74o   1l  4GAm2 t 4  79.3109  / 4  0.2982  2 0.002   Ans. Within the same accuracy, the answer is the same as part (c). ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 46/114 3-59 Given: Rectangular tube with inner dimensions 1.5 in  2.0 in, t = 1 8 in, 1035 CD steel, n = 540 rev/min, and Ssy =0.5 Sy. (a) Table A-20, Sy = 67 kpsi. Factor of safety against yield, ny = 2. For the table for Eq. (3-40), with b/c = 2/1.5 = 1.333, _____ b/c_ 0.208 1  1.333 0.231 1.5 0.231   1.5  1.333   0.3333    0.231  0.3333  0.231  0.208   0.2233 0.231  0.208 1.5  1 Eq. (3-40):  max  Eq. (3-42): 0.5S y 0.5  67 103 T T     T  16.83 103  lbf  in 2 2 2 ny  bc 0.2233  2 1.5 16.83 103  540 Tn H   144 hp 63 025 63 025 (b) Eq. (3-45):  max  0.5S y ny  Ans. T 2 Amt 0.5  67 103 T   T  14.46 103  lbf  in 1 1 1 2   2  1.5  8 2.0  8  8 14.46 103  540 H  124 hp Ans. 63 025 ______________________________________________________________________________ 3-60 Outer dimensions 20  30 mm, t = 1 mm, l = 1 m, 1018 CD steel, Ssy = 0.5 Sy. Table A-20, Sy = 370 MPa. Ssy = 0.5(370) = 185 MPa. Table A-5, G = 79.3 GPa. (a) Am = 19(29) = 551 mm2. Eq, (3-45): T  185 106   T  2(551)106  0.001185 10 6   204 N  m  Ans. 2 Amt (b) Eq. (3-46): T  2 19  29 10 3 1 TLml      1l   3  T  52.5 N  m Ans.  2 4GAm2 t  180  4  79.3109  551 1012  0.001    ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 47/114 3-61 (a) The area within the wall median line, Am, is Square: Am  (b  t )2 . From Eq. (3-45) Tsq  2 Amt all  2(b  t ) 2 t all Round: Am   (b  t ) 2 / 4 Trd  2 (b  t ) 2 t all / 4 Ratio of Torques Tsq 2(b  t )2 t all 4    1.27 Ans. 2 Trd  (b  t ) t all / 2  (b) Twist per unit length from Eq. (3-46) is 1  TLm  L L 2 A t L  m all2 m  all m  C m 2 4GAm t 4GAm t 2G Am Am Square: sq  C Round: rd  C 4(b  t ) (b  t )2  (b  t ) 2  (b  t ) / 4 C 4(b  t ) (b  t )2  sq  1 . Twists are the same. Ans.  rd _____________________________________________________________________________ 3-62 (a) The area enclosed by the section median line is Am = (1 – 0.0625)2 = 0.8789 in2 and the length of the section median line is Lm = 4(1  0.0625) = 3.75 in. From Eq. (3-45), T  2 Am t  2(0.8789)(0.0625)(12 000)  1318 lbf  in From Eq. (3-46),   1l  TLm l 4GAm2 t  (1318)(3.75)  36    4 11.5  10 (0.8789)  0.0625  6 2 Ans.  0.0801 rad  4.59 Ans. (b) The radius at the median line is rm = 0.125 + (0.5) (0.0625) = 0.15625 in. The area enclosed by the section median line is Am = (1  0.0625)2 – 4(0.15625)2 + 4(π /4) (0.15625)2 = 0.8579 in2. The length of the section median line is Lm = 4[1 – 0.0625 – 2(0.15625)] + 2π (0.15625) = 3.482 in. Shigley’s MED, 11th edition Chapter 3 Solutions, Page 48/114 From Eq. (3-45), T  2 Am t  2(0.8579)(0.0625)(12 000)  1287 lbf  in Ans. From Eq. (3-46), (1287)(3.482)  36  TLm l   1l    0.0762 rad  4.37 Ans. 2 6 2 4GAmt 4 11.5  10 (0.8579)  0.0625    _____________________________________________________________________________ 3-63 1  3Ti Ti   GLi ci3 T  T1  T2  T3  1G 3 1GLi ci3 3  Li ci3 i 1 3 Ans. From Eq. (3-47), G1c G and 1 are constant, therefore the largest shear stress occurs when c is a maximum.  max  G1cmax Ans. _____________________________________________________________________________ 3-64 (b) Solve part (b) first since the angle of twist per unit length is needed for part (a).  max   allow  12  6.89   82.7 MPa 1   max Gcmax     0.348 rad/m 79.3 10  (0.003) 82.7 106 9 (a) T1  T2  1GL1c13 3  2GL2 c23 3   3GL3c33   0.348(79.3) 109 (0.020)(0.0023 ) 3 0.348(79.3) 109  (0.030)(0.0033 ) 3 9 0.348(79.3) 10  (0)(03 ) Ans.  1.47 N  m Ans.  7.45 N  m Ans.   0 Ans. 3 3 T  T1  T2  T3  1.47  7.45  0  8.92 N  m Ans. _____________________________________________________________________________ T3  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 49/114 3-65 (b) Solve part (b) first since the angle of twist per unit length is needed for part (a).  12 000 1  max   8.35 103 rad/in Ans. 6 Gcmax 11.5 10 (0.125)    (a) T1  1GL1c13 3   2GL2 c23  8.35 103  11.5 106   0.75   0.06253  3 3  8.35 10  11.5 106  1  0.1253   5.86 lbf  in Ans.  62.52 lbf  in Ans. 3 6 3 3 3GL3c33  8.35  10  11.5  10   0.625   0.0625  T3    4.88 lbf  in Ans. 3 3 T  T1  T2  T3  5.86  62.52  4.88  73.3 lbf  in Ans. _____________________________________________________________________________ T2   3 3-66 (b) Solve part (b) first since the angle of twist per unit length is needed for part (a).  max   allow  12  6.89   82.7 MPa 1   max Gcmax     0.348 rad/m 79.3 10  (0.003) 82.7 106 9 (a) T1  T2  1GL1c13 3  2GL2 c23 3   3GL3c33   0.348(79.3) 109 (0.020)(0.0023 ) 3 0.348(79.3) 109  (0.030)(0.0033 ) 3 9 0.348(79.3) 10  (0.025)(0.0023 ) Ans.  1.47 N  m  7.45 N  m Ans. Ans. Ans.   1.84 N  m 3 3 T  T1  T2  T3  1.47  7.45  1.84  10.8 N  m Ans. _____________________________________________________________________________ T3  3-67 (a) From Eq. (3-40), with two 2-mm strips,  80  106   0.030   0.0022  T   3.08 N  m 3  1.8 / (b / c) 3  1.8 /  0.030 / 0.002   max bc 2 Tmax  2(3.08)  6.16 N  m Shigley’s MED, 11th edition Ans. Chapter 3 Solutions, Page 50/114 From the table for Eqs. (3-40) and (3-41), with b/c = 30/2 = 15, and has a value between 0.313 and 0.333. From Eq. (3-40), 1   0.321 3  1.8 / (30 / 2) From Eq. (3-41), 3.08(0.3) Tl    0.151 rad 3  bc G 0.321 0.030  0.0023  79.3 109  kt  T   6.16  40.8 N  m 0.151    Ans. Ans. From Eq. (3-40), with a single 4-mm strip,  80  106   0.030   0.0042  Tmax    11.9 N  m 3  1.8 / (b / c) 3  1.8 /  0.030 / 0.004   max bc 2 Ans. Interpolating from the table for Eqs. (3-40) and (3-41), with b/c = 30/4 = 7.5, 7.5  6  (0.307  0.299)  0.299  0.305 86 From Eq. (3-41) Tl 11.9(0.3)    0.0769 rad 3  bc G 0.305  0.030  0.0043  79.3 109  kt  T    11.9  155 N  m 0.0769   Ans. Ans. (b) From Eq. (3-47), with two 2-mm strips,       2 6 Lc 2  0.030  0.002  80  10 T   3.20 N  m 3 3 Tmax  2(3.20)  6.40 N  m Ans. 3Tl 3(3.20)(0.3)  3   0.151 rad Lc G  0.030  0.0023  79.3 109   kt  T   6.40 0.151  42.4 N  m Ans. Ans. From Eq. (3-47), with a single 4-mm strip, Tmax     2 6 Lc 2  0.030  0.004  80  10    12.8 N  m 3 3 Shigley’s MED, 11th edition Ans. Chapter 3 Solutions, Page 51/114  3Tl 3(12.8)(0.3)   0.0757 rad 3 Lc G  0.030  0.0043  79.3 109     kt  T   12.8 0.0757  169 N  m Ans. Ans. The results for the spring constants when using Eq. (3-47) are slightly larger than when using Eq. (3-40) and Eq. (3-41) because the strips are not infinitesimally thin (i.e. b/c does not equal infinity). The spring constants when considering one solid strip are significantly larger (almost four times larger) than when considering two thin strips because two thin strips would be able to slip along the center plane. _____________________________________________________________________________ 3-68 (a) Obtain the torque from the given power and speed using Eq. (3-44). H (40000) T  9.55  9.55  152.8 N  m n 2500 Tr 16T  max   3 J d 13  16 152.8      0.0223 m  22.3 mm  6    70  10    H (40000)  1528 N  m (b) T  9.55  9.55 n 250 13  16T  d     max    Ans. 13   16(1528)   d  0.0481 m  48.1 mm Ans.    70  106    _____________________________________________________________________________   3-69 (a) Obtain the torque from the given power and speed using Eq. (3-42). 63025H 63025(50)   1261 lbf  in T n 2500 Tr 16T  max   3 J d 13 13  16T   16 1261  d      0.685 in   max    (20 000)  63025H 63025(50)   12610 lbf  in (b) T  n 250 Ans. 13 16(12 610)  d    1.48 in Ans.   (20 000)  _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 52/114 3-70  max 16T  d3  max d 3  50  106    0.033    265 N  m 16 16 Tn 265(2000)   55.5 103 W  55.5 kW Ans. Eq. (3-44), H  9.55 9.55 _____________________________________________________________________________ 3-71   16T   3  T   d 3  110  106 0.0203  173 N  m 16 16 d     0.0204  79.3 109 15  4 Tl  d G  180   l   32T 32(173) JG l  1.89 m Ans. _____________________________________________________________________________  T      3-72       16T  T   d 3   30 000  0.753  2485 lbf  in 3 16 16 d Tl 32Tl 32(2485)(24)     0.167 rad  9.57 Ans. 4 JG  d G  0.754 11.5 106        _____________________________________________________________________________ 3-73 (a) Tsolid J max  d o4 max   16d o r Thollow J max  (d o4  d i4 ) max   16d o r     364 Tsolid  Thollow di4 %T  (100%)  4 (100%)  (100%)  65.6% Tsolid do 404 (b) Wsolid  kdo2 ,  Whollow  k do 2  di 2  Ans.     362 Wsolid  Whollow di2 %W  (100%)  2 (100%)  (100%)  81.0% Wsolid do 402 Ans. _____________________________________________________________________________ 3-74 4  4 J max  d 4 max J max   d   xd    max   Thollow  (a) Tsolid  r 16d r 16d 4 T T ( xd ) % T  solid hollow (100%)  (100%)  x 4 (100%) Ans. 4 Tsolid d Shigley’s MED, 11th edition Chapter 3 Solutions, Page 53/114  Whollow  k d 2   xd  (b) Wsolid  kd 2 2  2 %W  Wsolid  Whollow  xd  (100%)  x 2 (100%) (100%)  Wsolid d2 Ans. Plot %T and %W versus x. Percent Reduction in Torque and Weight Percent Reduction 100 80 60 Weight 40 Torque 20 0 difference 0 0.2 0.4 0.6 0.8 1 x The value of greatest difference in percent reduction of weight and torque is 25% and occurs at x  2 2 . _____________________________________________________________________________ 3-75 Tc (a)   J   120 10    13 4200  d 2  6  4 32   d 4   0.70d        2.8149 104 d 3  2.8149 104    6.17 102 m  61.7 mm d  6  120(10 )    From Table A-17, the next preferred size is d = 80 mm. Ans. di = 0.7d = 56 mm. The next preferred size smaller is di = 50 mm   Ans. (b) 4200  di 2  4200  0.050 2  Tc    30.8 MPa Ans. 4 J  32   d 4   d    32   0.080 4   0.050 4  i     _____________________________________________________________________________  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 54/114 3-76 T  9.55 H (1500)  9.55  1433 N  m n 10 13  16 1433      0.045 m  45 mm 16T  16T   3  dC    =   dC     80 106      From Table A-17, select 50 mm. Ans. 16  2 1433  117 106  Pa  117 MPa Ans. (a)  start  3   0.050  13   (b) Design activity _____________________________________________________________________________ 3-77 T 63 025 H 63 025(1)   7880 lbf  in n 8 16T  3  dC  13  16T  dC       13  16  7880   =    15 000    1.39 in From Table A-17, select 1.40 in. Ans. _____________________________________________________________________________ 3-78 For a square cross section with side length b, and a circular section with diameter d,   Asquare  Acircular  b 2  d 2  b d 4 2 From Eq. (3-40) with b = c,  max square 3 1.8  T  1.8  T  2  T  T  2 3 (4.8)  6.896 3   3 3  3  bc  b/c b  1  d    d For the circular cross section, 16T T  max circular  3  5.093 3 d d T  max square 6.896 d 3   1.354  max circular 5.093 T d3 The shear stress in the square cross section is 35.4% greater. Ans. (b) For the square cross section, from the table for Eq. (3-41), β = 0.141. From Eq. (3-41), Tl Tl Tl Tl square     11.50 4 4 3 4 d G  bc G  b G    0.141 d G  2  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 55/114 For the circular cross section, Tl Tl Tl  rd    10.19 4 4 GJ G  d 32  d G Tl  sq 11.50 d 4G   1.129  rd 10.19 Tl 4 d G The angle of twist in the square cross section is 12.9% greater. Ans. _____________________________________________________________________________ 3-79 (a) T1  0.15T2  T  0  (500  75)(4)  T 2 1700  4.25T2  0  T1  0.15  400   60 lbf (b) M O T2  400 lbf Ans. Ans.  0  575(10)  460(28)  RC (40) RC  178.25 lbf F 0 R O (c)  T1  5   1700  T2  0.15T2  5  Ans.  575  460  178.25 RO  293.25 lbf Ans. (d) The maximum bending moment is at x = 10 in, and is M = 2932.5 lbf·in. Since the shaft rotates, each stress element will experience both positive and negative bending stress as it moves from tension to compression. The torque transmitted through the shaft Shigley’s MED, 11th edition Chapter 3 Solutions, Page 56/114 from A to B is T = (500  75)(4) = 1700 lbf·in. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Mc 32 M 32  2932.5     15 294 psi = 15.3 kpsi Ans. d3  (1.25)3 I Tr 16T 16(1700)   3  4433 psi = 4.43 kpsi Ans.  (1.25)3 J d  (e) 2 x 2 2 15.3 2    15.3  1,  2    x    xy        4.43 2 2  2   2   1  16.5 kpsi Ans.  2  1.19 kpsi Ans. 2 2 2 2    15.3   max   x    xy    Ans.    4.43   8.84 kpsi  2   2  _____________________________________________________________________________ 3-80 (a) T2  0.15T1  T  0  1800  270  (200)  T 2 306 103   106.25T1  0   T1  (125)  306 103   125  0.15T1  T1  T1  2880 N Ans. T2  0.15  2880   432 N Ans. (b) M O  0  3312(230)  RC (510)  2070(810) RC  1794 N Ans. F y (c)  0  RO  3312  1794  2070 RO  3036 N Ans. Shigley’s MED, 11th edition Chapter 3 Solutions, Page 57/114 (d) The maximum bending moment is at x = 230 mm, and is M = –698.3 N·m. Since the shaft rotates, each stress element will experience both positive and negative bending stress as it moves from tension to compression. The torque transmitted through the shaft from A to B is T = (1800  270)(0.200) = 306 N·m. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Mc 32 M 32  698.3    263 103  Pa  263 MPa Ans. 3 3 d  (0.030) I 16(306) Tr 16T   3  57.7 106 Pa  57.7MPa Ans. 3  (0.030) J d    (e) 2 x 2 2 263 2    263  1,  2    x    xy        57.7  2 2  2   2   1  275 MPa Ans.  2  12.1 MPa Ans. 2 2 2 2    263   max   x    xy       57.7   144 MPa  2   2  Ans. _____________________________________________________________________________ 3-81 (a) T2  0.15T1  T  0   300  50  (4)  T  T1  (3)  1000   0.15T1  T1  (3) 1000  2.55T1  0 T1  392.16 lbf Ans. 2  T2  0.15  392.16   58.82 lbf Ans. (b) M Oy  0  450.98(16)  RC z (22) RC z  327.99 lbf Ans. F z  0  RO z  450.98  327.99 RO z  122.99 lbf Ans. M Oz  0 350(8)  RC y (22) RC y  127.27 lbf Ans. F y  0  RO y  350  127.27 RO y  222.73 lbf Ans. Shigley’s MED, 11th edition Chapter 3 Solutions, Page 58/114 (c) (d) Combine the bending moments from both planes at A and B to find the critical location. M A  (983.92) 2  ( 1781.84) 2  2035 lbf  in M B  (1967.84) 2  ( 763.65) 2  2111 lbf  in The critical location is at B. The torque transmitted through the shaft from A to B is T = (300  50)(4) = 1000 lbf·in. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Mc 32 M 32  2111    21 502 psi = 21.5 kpsi Ans. d3  (1)3 I Tr 16T 16(1000)   3  5093 psi = 5.09 kpsi Ans. J d  (1)3  (e) 2 x 2 2 21.5 2    21.5    x    xy     1,  2     5.09  2 2  2   2  Ans.  1  22.6 kpsi  2  1.14 kpsi 2 Ans. 2 2 2    21.5  Ans.  max   x    xy       5.09   11.9 kpsi  2   2  _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 59/114 3-82 (a) T2  0.15T1  T  0   300  45 (125)  T 2 31 875  127.5T1  0 (b)   T1  (150)  31 875   0.15T1  T1  (150) T1  250 N  mm Ans. T2  0.15  250   37.5 N  mm Ans. M Oy  0  345sin 45o (300)  287.5(700)  RC z (850) RC z  150.7 N F z  0  RO z  345 cos 45o  287.5  150.7 RO z  107.2 N M Oz y Ans.  0  345sin 45o (300)  RC y (850) RC y  86.10 N F Ans. Ans.  0  RO y  345 cos 45o  86.10 RO y  157.9 N Ans. (c) (d) From the bending moment diagrams, it is clear that the critical location is at A where both planes have the maximum bending moment. Combining the bending moments from the two planes, Shigley’s MED, 11th edition Chapter 3 Solutions, Page 60/114 M  47.37    32.16  2 2  57.26 N  m The torque transmitted through the shaft from A to B is T = (300  45)(0.125) = 31.88 N·m. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Mc 32 M 32  57.26      72.9 106  Pa  72.9 MPa Ans.  d 3  (0.020)3 I Tr 16T 16(31.88)   3  20.3 106  Pa  20.3 MPa Ans. J d  (0.020)3 (e) 2 x 2 2 72.9 2    72.9  1,  2    x    xy        20.3 2 2  2   2   1  78.2 MPa Ans.  2  5.27 MPa Ans. 2 2 2 2    72.9   max   x    xy    Ans.    20.3   41.7 MPa  2   2  _____________________________________________________________________________ 3-83 (a)  T  0   300(cos 20º )(10)  FB (cos 20º )(4) FB  750 lbf Ans. (b) M Oz  0  300(cos 20º )(16)  750(sin 20º )(39) RC y (30) RC y  183.1 lbf Ans. F y  0 RO y  300(cos 20º )  183.1  750(sin 20º ) RO y  208.5 lbf Ans. M Oy  0  300(sin 20º )(16)  RC z (30)  750(cos 20º )(39) RC z  861.5 lbf Ans. F z  0 RO z  300(sin 20º )  861.5  750(cos 20º ) RO z  259.3 lbf Ans. Shigley’s MED, 11th edition Chapter 3 Solutions, Page 61/114 (c) (d) Combine the bending moments from both planes at A and C to find the critical location. M A  (3336) 2  (4149) 2  5324 lbf  in M C  (2308) 2  (6343) 2  6750 lbf  in The critical location is at C. The torque transmitted through the shaft from A to B is T  300 cos  20º 10   2819 lbf  in . For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Mc 32 M 32  6750     35 203 psi = 35.2 kpsi Ans. I d3  (1.25)3 Tr 16T 16(2819)   3  7351 psi = 7.35 kpsi Ans.  (1.25)3 J d  (e) 2 x 2 2 35.2 2    35.2  1,  2    x    xy        7.35  2 2  2   2   1  36.7 kpsi Ans.  2  1.47 kpsi 2 Ans. 2 2 2    35.2  Ans.  max   x    xy       7.35   19.1 kpsi  2   2  _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 62/114 3-84 (a)  T  0   11 000(cos 20º )(300)  F (cos 25º )(150) B (b) FB  22 810 N M Oz  0  11 000(sin 20º )(400)  22 810(sin 25º )(750) RC y (1050) RC y  8319 N F y Ans.  0 RO y  11000(sin 20º )  22 810 sin(25º )  8319 RO y  5083 N M Ans. Oy Ans.  0 11 000(cos 20º )(400)  22 810(cos 25º )(750)  RC z (1050) RC z  10 830 N Ans. F z  0 RO z  11 000(cos 20º )  22 810(cos 25º )  10 830 RO z  494 N Ans. (c) (d) From the bending moment diagrams, it is clear that the critical location is at B where both planes have the maximum bending moment. Combining the bending moments from the two planes, M  2496    3249  2 2  4097 N  m The torque transmitted through the shaft from A to B is T  11 000 cos  20º  0.3  3101 N  m . For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Shigley’s MED, 11th edition Chapter 3 Solutions, Page 63/114 Mc 32 M 32  4097     333.9 10 6  Pa  333.9 MPa Ans. 3 3 d  (0.050) I Tr 16T 16(3101)   3  126.3 106 Pa  126.3 MPa Ans. J d  (0.050)3    (e) 2 x 2 2 333.9 2    333.9    x    xy     1,  2    126.3 2 2  2   2  Ans.  1  376 MPa  2  42.4 MPa Ans. 2 2 2 2    333.9  Ans.  max   x    xy      126.3   209 MPa  2   2  _____________________________________________________________________________ 3-85 (a)  M D  z  6.13Cx  3.8(92.8)  3.88(362.8)  0 C x  287.2 lbf Ans.  M C  z  6.13Dx  2.33(92.8)  3.88(362.8)  0 Dx  194.4 lbf Ans. 3.8 (808)  500.9 lbf Ans. 6.13 2.33 (808)  307.1 lbf Ans.  M C  x  0  Dz  6.13 (b) For DQC, let x, y, z correspond to the original  y, x, z axes.  M D  x  0  Cz  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 64/114 (c) The critical stress element is just to the right of Q, where the bending moment in both planes is a maximum, and where the torsional and axial loads exist. T  808(3.88)  3135 lbf  in M  669.22  1167 2  1345 lbf  in 16T 16(3135)  3  11 070 psi Ans. d  1.133  b   32M 32(1345)    9495 psi 3 d  1.133 Ans. a   362.8 F   362 psi A ( / 4) 1.132 Ans.     (d) The critical stress element will be where the bending stress and axial stress are both in compression.  max  9495  362  9857 psi 2  max  9857  2     11 070  12 118 psi  12.1 kpsi  2  Ans. 2 9857  9857  2 1 , 2      11 070 2 2    1  7189 psi  7.19 kpsi Ans.  2  17 046 psi  17.0 kpsi Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 65/114 3-86 (a)  M D  z  0 6.13Cx  3.8(46.6)  3.88(140)  0 C x  117.5 lbf Ans.  M C  z  0 6.13Dx  2.33(46.6)  3.88(140)  0 Dx  70.9 lbf Ans. 3.8 (406)  251.7 lbf Ans. 6.13 2.33 (406)  154.3 lbf Ans.  M C  x  0  Dz  6.13 (b) For DQC, let x, y, z correspond to the original  y, x, z axes.  M D  x  0  Cz  (c) The critical stress element is just to the right of Q, where the bending moment in both planes is a maximum, and where the torsional and axial loads exist. T  406(3.88)  1575 lbf  in M  273.82  586.32  647.1 lbf  in 16T 16(1575)  3  8021 psi Ans. d  13  b   32M 32(647.1)    6591 psi 3 d  13 Shigley’s MED, 11th edition   Ans. Chapter 3 Solutions, Page 66/114 a   F 140   178.3 psi A ( / 4) 12   Ans. (d) The critical stress element will be where the bending stress and axial stress are both in compression.  max  6591  178.3  6769 psi 2  max  6769  2     8021  8706 psi  8.71 kpsi  2  Ans. 2 6769  6769  2   1, 2    8021 2 2    1  5321 psi  5.32 kpsi Ans.  2  12 090 psi  12.1 kpsi Ans. _____________________________________________________________________________ 3-87  M B  z  5.62(362.8)  1.3(92.8)  3 Ay  0 Ay  639.4 lbf Ans. B y  276.6 lbf Ans.  M A  z  2.62(362.8)  1.3(92.8)  3By  0 5.62 (808)  1513.7 lbf 3 2.62  0  Bz  (808)  705.7 lbf 3  M B  y  0  M A  y  Az  Shigley’s MED, 11th edition Ans. Ans. Chapter 3 Solutions, Page 67/114 (b) (c) The critical stress element is just to the left of A, where the bending moment in both planes is a maximum, and where the torsional and axial loads exist. T  808(1.3)  1050 lbf  in 16(1050)   7847 psi Ans.  0.883   M  (829.8) 2  (2117) 2  2274 lbf  in 32M 32(2274) b     33 990 psi 3 d  0.883  a    92.8 F   153 psi A ( / 4) 0.882   Ans. Ans. (d) The critical stress will occur when the bending stress and axial stress are both in compression.  max  33 990  153  34 143 psi 2  max  34 143  2     7847  18 789 psi  18.8 kpsi 2   Ans. 2 34143  34143  2 1 , 2      7847 2 2    1  1717 psi  1.72 kpsi Ans.  2  35 860 psi  35.9 kpsi Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 68/114 3-88 Ft  T 100   1600 N c / 2 0.125 / 2   M A z  0 Fn  1600 tan 20  582.4 N TC  Ft  b 2   1600  0.250 2   200 N  m 450 RDy  582.4(325)  2667(75)  0 P TC 200   2667 N  a 2   0.150 2  RDy  865.1 N   M A  y  0  450 RDz  1600(325)  RDz  1156 N  Fy  0  RAy  865.1  582.4  2667  RAy  2384 N  Fz  0  RAz  1156  1600  RAz  444 N AB The maximum bending moment will either be at B or C. If this is not obvious, sketch the shear and bending moment diagrams. We will directly obtain the combined moments from each plane. M B  AB R A2 y  R A2z  0.075 23842  4442  181.9 N  m M C  CD RD2 y  RD2 z  0.125 865.12  11562  180.5 N  m The stresses at B and C are almost identical, but the maximum stresses occur at B. 32M B 32(181.9) B    68.6 106 Pa  68.6 MPa 3 3 d  0.030  B     Ans. 16TB 16(200)   37.7 106 Pa  37.7 MPa 3 3 d  0.030  max      2 B 2 68.6    68.6  2   B    B2      37.7  85.3 MPa 2 2  2   2  2 Ans. 2    68.6  2  max   B    B2     37.7  51.0 MPa Ans.  2   2  _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 69/114 3-89 Ft  T 100   1600 N c / 2 0.125 / 2 Fn  1600 tan 20  582.4 N TC  Ft  b 2   1600  0.250 2   200 N  m P TC 200   2667 N  a 2   0.150 2   RDy  420.6 N   M A  z  0  450 RDy  582.4(325)   M A  y  0  450 RDz  1600(325)  2667(75)  RDz  711.1 N  Fy  0  RAy  420.6  582.4  Fz  0  RAz  711.1  1600  2667  RAy  161.8 N  RAz  1778 N The maximum bending moment will either be at B or C. If this is not obvious, sketch shear and bending moment diagrams. We will directly obtain the combined moments from each plane. 2 M B  AB RA2 y  RA2z  0.075 161.82   1778   133.9 N  m M C  CD RD2 y  RD2 z  0.125 420.62  711.12  103.3 N  m The maximum stresses occur at B. Ans. 32M B 32(133.9) B    50.5 106 Pa  50.5 MPa 3 3 d  0.030  B     16TB 16(200)   37.7 106 Pa  37.7 MPa 3 3 d  0.030 Shigley’s MED, 11th edition     Chapter 3 Solutions, Page 70/114  max 2 B 2 50.5    50.5  2    B    B2      37.7  70.6 MPa 2 2 2 2     2 2    50.5  2  max   B    B2     37.7  45.4 MPa  2   2  Ans. Ans. _____________________________________________________________________________ 3-90 T 900   180 lbf c / 2 10 / 2 Fn  180 tan 20  65.5 lbf Ft  TC  Ft  b 2   180  5 2   450 lbf  in P TC 450   150 lbf  a 2  6 2   M A  z  0  20 RDy  65.5(14)  150(4)  RDy  75.9 lbf  RDz  126 lbf   M A  y  0  20 RDz  180(14)  Fy  0  RAy  75.9  65.5  150  Fz  0  RAz  126  180  RAy  140 lbf  RAz  54.0 lbf The maximum bending moment will either be at B or C. If this is not obvious, sketch shear and bending moment diagrams. We will directly obtain the combined moments from each plane. M B  AB RA2 y  R A2z  4 140 2  542  600 lbf  in M C  CD RD2 y  RD2 z  6 75.92  1262  883 lbf  in The maximum stresses occur at C. C  C  32M C d 3 16TC d 3   32(883)   1.3753 16(450)   1.3753 Shigley’s MED, 11th edition   Ans.  3460 psi  882 psi Chapter 3 Solutions, Page 71/114  max 2 C 2 3460    3460  2    C    C2      882  3670 psi 2 2 2 2     2 2    3460  2  max   C    C2     882  1940 psi  2   2  Ans. Ans. _____________________________________________________________________________ 3-91 (a) Rod AB experiences constant torsion throughout its length, and maximum bending moment at the wall. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be at the wall, at either the top (compression) or the bottom (tension) on the y axis. We will select the bottom element for this analysis. (b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces. Mc M  d / 2  32M 32  8  200  x      16 297 psi  16.3 kpsi 3 I  d 4 / 64  d 3  1  xz  Tr T  d / 2  16T 16  5  200      5093 psi  5.09 kpsi 3 J  d 4 / 32  d 3  1 (c) 2 x 2 16.3 2 2    16.3    x    xz        5.09  2 2  2   2   1  17.8 kpsi Ans. 1,  2   2  1.46 kpsi 2 Ans. 2 2 2    16.3  Ans.  max   x    xz       5.09   9.61 kpsi  2   2  _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 72/114 3-92 (a) Rod AB experiences constant torsion throughout its length, and maximum bending moments at the wall in both planes of bending. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface at the wall, with its critical location determined by the plane of the combined bending moments. My = – (100)(8) = – 800 lbf·in Mz = (175)(8) = 1400 lbf·in M tot  M y2  M z2   800  2  14002  1612 lbf  in  My  1  800    tan    29.7º  1400   Mz  The combined bending moment vector is at an angle of 29.7º CCW from the z axis. The critical bending stress location, and thus the critical stress element, will be ±90º from this vector, as shown. There are two equally critical stress elements, one in tension (119.7º CCW from the z axis) and the other in compression (60.3º CW from the z axis). We’ll continue the analysis with the element in tension. (b) Transverse shear is zero at the critical stress elements on the outer surfaces. M c M  d / 2  32 M tot 32 1612   x  tot  tot 4    16 420 psi  16.4 kpsi 3 I  d / 64  d3  1  = tan 1   Tr T  d / 2  16T 16  5 175      4456 psi  4.46 kpsi 3 J  d 4 / 32  d 3  1 (c) 2 x 2 16.4 2    16.4  1,  2    x   2       4.46  2 2  2   2   1  17.5 kpsi Ans.  2  1.13 kpsi 2 Ans. 2 2    16.4   max   x    2      4.46   9.33 kpsi  2   2  Ans. _____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 73/114 3-93 (a) Rod AB experiences constant torsion and constant axial tension throughout its length, and maximum bending moments at the wall from both planes of bending. Both torsional shear stress and bending stress will be maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface at the wall, with its critical location determined by the plane of the combined bending moments. My = – (100)(8) – (75)(5) = – 1175 lbf·in Mz = (–200)(8) = –1600 lbf·in M tot  M y2  M z2   1175    1600  2 2  1985 lbf  in  My  1  1175    tan    36.3º 1600 M   z   The combined bending moment vector is at an angle of 36.3º CW from the negative z axis. The critical bending stress location will be ±90º from this vector, as shown. Since there is an axial stress in tension, the critical stress element will be where the bending is also in tension. The critical stress element is therefore on the outer surface at the wall, at an angle of 36.3º CW from the y axis. (b) Transverse shear is zero at the critical stress element on the outer surface. M c M  d / 2  32 M tot 32 1985   x ,bend  tot  tot 4    20 220 psi  20.2 kpsi 3 I  d / 64 d3  1  = tan 1   x ,axial  Fx Fx 75    95.5 psi  0.1 kpsi , which is essentially negligible 2 A  d / 4  12 / 4  x   x ,axial   x ,bend  20 220  95.5  20 316 psi  20.3 kpsi  Tr 16T 16  5  200     5093 psi  5.09 kpsi 3 J d3  1 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 74/114 (c) 2 x 2 20.3 2    20.3  1,  2    x   2       5.09  2 2  2   2   1  21.5 kpsi Ans.  2  1.20 kpsi 2 Ans. 2 2    20.3   max   x    2   Ans.    5.09   11.4 kpsi  2   2  _____________________________________________________________________________ 3-94 T = (2)(200) = 400 lbf·in The maximum shear stress due to torsion occurs in the middle of the longest side of the rectangular cross section. From the table for Eq. (3-40), with b/c = 1.5/0.25 = 6,  = 0.299. From Eq. (3-40), T 400  max    14 270 psi  14.3 kpsi Ans. 2 2  bc  0.299 1.5  0.25 ____________________________________________________________________________ 3-95 (a) The cross section at A will experience bending, torsion, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be at either the top (compression) or the bottom (tension) on the y axis. We’ll select the bottom element for this analysis. (b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces. Mc M  d / 2  32M 32 11 250  x      28 011 psi  28.0 kpsi 3 I  d 4 / 64  d 3  1  xz  Tr T  d / 2  16T 16 12  250      15 279 psi  15.3 kpsi 3 J  d 4 / 32  d 3  1 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 75/114 (c) 2 x 2 28.0 2 2    28.0  1,  2    x    xz       15.3 2 2  2   2   1  34.7 kpsi Ans.  2  6.7 kpsi Ans. 2 2 2 2    28.0   max   x    xz    Ans.   15.3  20.7 kpsi  2   2  ____________________________________________________________________________ 3-96 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) = 3600 lbf·in Mz = (250)(11) = 2750 lbf·in M tot  M y2  M z2   3600    2750  2  M z  My   = tan 1  2  4530 lbf  in   2750    tan 1    37.4º   3600   The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 37.4º CCW from the z axis. (b) M c M  d / 2  32 M tot 32  4530   x ,bend  tot  tot 4    46 142 psi  46.1 kpsi 3 I  d / 64  d3  1  x ,axial  Fx Fx 300    382 psi  0.382 kpsi 2 A  d / 4  12 / 4  x   x ,axial   x ,bend  46 142  382  46 524 psi  46.5 kpsi  Tr 16T 16 12  250     15 279 psi  15.3 kpsi 3 J d3  1 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 76/114 (c) 2 x 2 46.5 2    46.5  1,  2    x   2      15.3 2 2  2   2   1  51.1 kpsi Ans.  2  4.58 kpsi Ans. 2 2 2    46.5   max   x    2     15.3  27.8 kpsi  2   2  Ans. ____________________________________________________________________________ 3-97 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) – (–100)(11) = 4700 lbf·in Mz = (250)(11) = 2750 lbf·in M tot  M y2  M z2   4700    2750  2  M z  My   = tan 1  2  5445 lbf  in   2750    tan 1    30.3º  4700    The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 30.3º CCW from the z axis. (b) M c M  d / 2  32 M tot 32  5445   x ,bend  tot  tot 4    55 462 psi  55.5 kpsi 3 I  d / 64  d3  1 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 77/114  x ,axial  Fx Fx 300    382 psi  0.382 kpsi 2 A  d / 4  12 / 4  x   x ,axial   x ,bend  55 462  382  55 844 psi  55.8 kpsi  Tr 16T 16 12  250     15 279 psi  15.3 kpsi 3 J d3  1 (c) 2 x 2 55.8 2    55.8    x   2      15.3 2 2  2   2   1  59.7 kpsi Ans. 1,  2   2  3.92 kpsi Ans. 2  max 2 2    55.8    x   2     15.3  31.8 kpsi  2   2  Ans. ____________________________________________________________________________ 3-98 (a) The cross section at A will experience bending, torsion, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface, where the stress concentration will also be applicable. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be at either the top (compression) or the bottom (tension) on the y axis. We’ll select the bottom element for this analysis. (b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces. r / d  0.125 /1  0.125 D / d  1.5 /1  1.5 Fig. A-15-8 K t ,torsion  1.39 Fig. A-15-9 K t ,bend  1.59  x  K t ,bend 32 11 250  Mc 32 M  K t ,bend  (1.59)  44 538 psi  44.5 kpsi 3 3 d I  1  xz  K t ,torsion 16 12  250  Tr 16T  K t ,torsion  (1.39)  21 238 psi  21.2 kpsi 3 d3 J  1 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 78/114 (c) 2 x 2 44.5 2 2    44.5  1,  2    x    xz        21.2  2 2  2   2   1  53.0 kpsi Ans.  2  8.48 kpsi Ans. 2 2 2 2    44.5   max   x    xz    Ans.    21.2   30.7 kpsi  2   2  ____________________________________________________________________________ 3-99 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface, where the stress concentration will also be applicable. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) = 3600 lbf·in Mz = (250)(11) = 2750 lbf·in M tot  M y2  M z2   3600    2750  2  M z  My   = tan 1  2  4530 lbf  in   2750    tan 1    37.4º  3600    The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 37.4º CCW from the z axis. (b) r / d  0.125 /1  0.125 D / d  1.5 /1  1.5 Fig. A-15-7 K t , axial  1.75 K t ,torsion  1.39 Shigley’s MED, 11th edition Fig. A-15-8 Chapter 3 Solutions, Page 79/114 Fig. A-15-9 K t ,bend  1.59  x ,bend  K t ,bend 32  4530  Mc 32 M  K t ,bend  (1.59)  73 366 psi  73.4 kpsi 3 3 d I  1  x ,axial  K t ,axial Fx 300  1.75   668 psi  0.668 kpsi 2 A  1 / 4  x   x ,axial   x ,bend  73 366  668  74 034 psi  74.0 kpsi   K t ,torsion 16 12  250  Tr 16T  K t ,torsion  (1.39)  21 238 psi  21.2 kpsi 3 3 d J  1 (c) 2 x 2 74.0 2    74.0    x   2       21.2  2 2  2   2   1  79.6 kpsi Ans. 1,  2   2  5.64 kpsi 2  max Ans. 2 2    74.0    x   2      21.2   42.6 kpsi  2   2  Ans. ____________________________________________________________________________ 3-100 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface, where the stress concentration is also applicable. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. Shigley’s MED, 11th edition Chapter 3 Solutions, Page 80/114 My = (300)(12) – (–100)(11) = 4700 lbf·in Mz = (250)(11) = 2750 lbf·in M tot  M y2  M z2   4700    2750  2 2  5445 lbf  in  M z  My    2750    tan 1    30.3º  4700    The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 30.3º CCW from the z axis. (b) r / d  0.125 /1  0.125  = tan 1  D / d  1.5 /1  1.5 Fig. A-15-7 K t , axial  1.75 K t ,torsion  1.39 Fig. A-15-8 K t ,bend  1.59 Fig. A-15-9  x ,bend  K t ,bend 32  5445  Mc 32 M  K t ,bend  (1.59)  88185 psi  88.2 kpsi 3 d3 I  1  x ,axial  K t ,axial Fx 300  1.75   668 psi  0.668 kpsi 2 A  1 / 4  x   x ,axial   x ,bend  88 185  668  88 853 psi  88.9 kpsi   K t ,torsion 16 12  250  Tr 16T  K t ,torsion  (1.39)  21 238 psi  21.2 kpsi 3 3 d J  1 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 81/114 (c) 2 x 2 88.9 2    88.9  1,  2    x   2       21.2  2 2  2   2  Ans.  1  93.7 kpsi  2  4.80 kpsi Ans. 2 2 2    88.9  Ans.  max   x    2      21.2   49.2 kpsi  2   2  ____________________________________________________________________________ 3-101 (a) (Eq. 3-42): 200  60  Tn   0.1904 hp Ans. 63 025 63 025 (b) The output torque To = (i /o) Ti. But i r1 = o r2. Thus, To = (r2 / r1) Ti. So, To = (2.5/1) 200 = 500 lbf٠in Ans. P (c)  Mx = 0, (FG cos 20o)(1) – 200 = 0 FG = 212.84 lbf  (MB)y = 0, 2(FG cos 20o) – 3.5RAz = 0 RAz = 2(212.84 cos 20o)/3.5 = 114.29 lbf  Fz = 0, RBz + 114.29 – 212.84 cos 20o = 0 RBz = 85.71 lbf  (MB)z = 0, 2 (212.84 sin 20o) + 3.5 (FA)y = 0 RAy = – 41.60 lbf  Fy = 0, RBy – 41.60 +212.84 sin 20o = 0  RBy = – 31.19 lbf 2 2 RA  RAy  RAz   41.60  2 2 RB  RBy  RBz   31.19  2  114.292  121.6 lbf 2  85.712  91.21 lbf Ans. Ans. (d) Shigley’s MED, 11th edition Chapter 3 Solutions, Page 82/114 (e) (My)max = 2( 85.71) =  171.42 lbf-in, (Mz)max = 2( 31.19) =  62.38 lbf-in  171.42    62.38  182.42 32 M 32 182.42  3   10  14.87 kpsi 3 d3   0.5 16T 16  200  3  3 10  8.15 kpsi 3 d   0.5 2 M max  2 lbf-in Ans. Ans. (f) 2  2 14.87    14.87  2   1, 2       2    8.15 2 2 2 2      18.47,  3.60 kpsi Ans. 2 2    14.87  2 Ans.  max      2     8.15  11.03 kpsi 2  2  ______________________________________________________________________________ 3-102 (a) (Eq. 3-42): 200  60  Tn P   0.1904 hp Ans. 63 025 63 025 (b) The output torque To = (i /o) Ti. But i r1 = o r2. Thus, To = (r2 / r1) Ti. So, To = (2.5/1) 200 = 500 lbf٠in Ans. (c)  Mx = 0 = (FG cos 20o)(1) – 200 FG = 212.84 lbf  (MB)y = 0 = – 2(FG cos 20o) – 3.5RCz RCz = – 2(212.84 cos 20o)/3.5 = – 114.29 lbf  Fz = 0 = RDz – 114.29 + 212.84 cos 20o RDz = – 85.71 lbf  (MB)z = 0 = – 2 (212.84 sin 20o) + 3.5 RCy  RCy = 41.60 lbf  Fy = 0 = RDy + 41.60 – 212.84 sin 20o  RDy = 31.19 lbf 2 2 RC  RCy  RCz  41.60 2   114.29   121.6 lbf Ans. 2 2 RD  RDy  RDz  31.19 2   85.71  91.21 lbf Ans. 2 2 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 83/114 (d) (e) (My)max = 2(85.71) = 171.42 lbf٠in, (Mz)max = 2(31.19) = 62.38 lbf٠in M max  171.42 2  62.382  182.42 lbf  in  32 M 32 182.42  3  10  14.87 kpsi 3 d3   0.5  16T 16  500  3  10  20.37 kpsi  d 3   0.5 3 Ans. Ans. (f) 2  2 14.87    14.87  2     2      20.37 2 2 2  2   29.1,  14.2 kpsi Ans.  1, 2  2 2    14.87  2 Ans.  max      2     20.37  21.7 kpsi 2  2  ______________________________________________________________________________ 3-103 (a) M = F(p / 4), c = p / 4, I = bh3 / 12, b =  dr nt, h = p / 2  F  p / 4    p / 4  Mc Fp 2 b      3 I bh3 /12 16   d r nt  p / 2  /12 6F Ans.  d r nt p F F 4F (b)  a     2  2 Ans. A  dr / 4  dr Tr T  d r / 2  16T t    Ans.  d r4 / 32  d r3 J b   Shigley’s MED, 11th edition Chapter 3 Solutions, Page 84/114 (c) The bending stress causes compression in the x direction. The axial stress causes compression in the y direction. The torsional stress shears across the y face in the negative z direction. (d) Analyze the stress element from part (c) using the equations developed in parts (a) and (b). d r  d  p  1.5  0.25  1.25 in  x  b   6 1500  6F    4584 psi =  4.584 kpsi  d r nt p  1.25  2  0.25  y  a   4 1500  4F = =  1222 psi =  1.222 kpsi 2  dr  1.252   yz   t   16  235  16T = =  612.8 psi =  0.6128 kpsi 3  dr  1.253  Use Eq. (3-15) for the three-dimensional stress element. 2 2  3   4.584  1.222   2   4.584  1.222    0.6128        4.584  0.6128    0    5.806  5.226  1.721  0 3    2 The roots are at 0.2543, – 4.584, and –1.476. Thus, the ordered principal stresses are 1 = 0.2543 kpsi, 2 = –1.476 kpsi, and 3 = – 4.584 kpsi. Ans. From Eq. (3-16), the principal shear stresses are 0.2543   1.476    Ans.  1/ 2  1 2   0.8652 kpsi 2 2    3  1.476    4.584  Ans.  2/3  2   1.554 kpsi 2 2 0.2543   4.584     1/3  1 3   2.419 kpsi Ans. 2 2 ____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 85/114 3-104 As shown in Fig. 3-32, the maximum stresses occur at the inside fiber where r = ri. Therefore, from Eq. (3-50)  t ,max ri2 pi  ro2   2 2 1  2  ro  ri  ri   r2  r2   pi  o2 i2   r r   o i  Ans. ri2 pi  ro2  1     pi Ans. ro2  ri2  ri2  ______________________________________________________________________________  r ,max  3-105 If pi = 0, Eq. (3-49) becomes t   po ro2  ri 2 ro2 po / r 2 ro2  ri 2 po ro2  ri 2  1   ro2  ri 2  r 2  The maximum tangential stress occurs at r = ri. So   t ,max   For σr, we have r  2 po ro2 Ans. ro2  ri 2  po ro2  ri 2 ro2 po / r 2 ro2  ri 2 po ro2  ri 2   2 2  2  1 ro  ri  r  So σr = 0 at r = ri. Thus at r = ro po ro2  ri 2  ro2   r ,max  2 2  2    po Ans. ro  ri  ro  ______________________________________________________________________________ 3-106 The force due to the pressure on half of the sphere is resisted by the stress that is distributed around the center plane of the sphere. All planes are the same, so ( t ) av   1   2  p  / 4  di2  di t  pd i 4t Ans. The radial stress on the inner surface of the shell is, 3 =  p Ans. ______________________________________________________________________________ 3-107 σt > σl > σr Shigley’s MED, 11th edition Chapter 3 Solutions, Page 86/114 τmax = (σt − σr)/2 at r = ri ro2 pi 1  ri 2 pi  ro2  ri 2 pi  ro2   1 1         2  ro2  ri 2  ri 2  ro2  ri 2  ri 2   ro2  ri 2  max   ro2  ri 2 32  2.752 (10 000)  1597 psi Ans.  pi   max  ro 2 32 ______________________________________________________________________________ 3-108 σt > σl > σr τmax = (σt − σr)/2 at r = ri r2 p  r2  r2 p 1  r 2 p  r 2  r 2 p  r 2   max   2i i 2 1  o2   2i i 2 1  o2    2i i 2  o2   2o i 2 2  ro  ri  ri  ro  ri  ri   ro  ri  ri  ro  ri  ri  ro ( max  pi )  max  100 (25  4)106  91.7 mm 25 106  t  ro  ri  100  91.7  8.3 mm Ans. ______________________________________________________________________________ 3-109 σt > σl > σr τmax = (σt − σr)/2 at r = ri ri 2 pi  ro2  ro2 pi 1  ri 2 pi  ro2  ri 2 pi  ro2   1 1             2  ro2  ri 2  ri 2  ro2  ri 2  ri 2   ro2  ri 2  ri 2  ro2  ri 2  max   4 2 (500)  4129 psi Ans. 42  3.752 ______________________________________________________________________________  3-110 From Eq. (3-49) with pi = 0, r2 p  r2   t   2o o 2 1  i 2  ro  ri  r  ro2 po  ri 2   r   2 2 1  2  ro  ri  r  σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro  max  ro2 po  ri 2  ri 2 po 1  ro2 po  ri 2  ro2 po  ri 2      2 2 1  2   2 2 1  2   2 2  2   2 2 2  ro  ri  ro  ro  ri  ro   ro  ri  ro  ro  ri po  ro2  ri 2 32  2.752 (10 000)  1900 psi Ans.   max 2.752 ri 2 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 87/114 ______________________________________________________________________________ 3-111 From Eq. (3-49) with pi = 0, r2 p  r2   t   2o o 2 1  i 2  ro  ri  r  r   ro2 po  ri 2  1   ro2  ri 2  r 2  σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro ro2 po  ri2  ri 2 po 1  ro2 po  ri 2  ro2 po  ri 2   1 1           2  ro2  ri 2  ro 2  ro2  ri 2  ro 2   ro2  ri 2  ro 2  ro2  ri 2  max     ri  ro  max ( max  po )  100 25 106   25  4 106  92.8 mm t  ro  ri  100  92.8  7.2 mm Ans. ______________________________________________________________________________ 3-112 From Eq. (3-49) with pi = 0, t   ro2 po  ri 2  1   ro2  ri 2  r 2  r   ro2 po  ri 2  1   ro2  ri 2  r 2  σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro r2 p  r2  r2 p 1  r 2 p  r 2  r 2 p  r 2   max    2o o 2 1  i 2   2o o 2 1  i 2    2o o 2  i 2   2i o 2 2  ro  ri  ro  ro  ri  ro   ro  ri  ro  ro  ri 3.752 (500)  2  3629 psi Ans. 4  3.752 ______________________________________________________________________________ 3-113 From Table A-20, Sy=490 MPa From Eq. (3-49) with pi = 0, ro2 po  ri 2   t   2 2 1  2  ro  ri  r  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 88/114 Maximum will occur at r = ri 0.8(490)  25  19  (r  r )  2r 2 p  82.8 MPa Ans.  t ,max   2 o o2  po   t ,max o2 i   ro  ri 2ro 2(252 ) ______________________________________________________________________________ 2 2 2 2 3-114 From Table A-20, Sy = 71 kpsi From Eq. (3-49) with pi = 0, ro2 po  ri 2   t   2 2 1  2  ro  ri  r  Maximum will occur at r = ri  t ,max  ro  ri   0.8(71)  1  0.75  2ro2 po  t ,max   2 2  po      12.4 kpsi Ans. ro  ri 2ro2 2(12 ) ______________________________________________________________________________ 2 2 2 2 3-115 From Table A-20, Sy=490 MPa From Eq. (3-50) t  ri 2 pi  ro2  1   ro2  ri 2  r 2  Maximum will occur at r = ri  t ,max 2 2 ri 2 pi  ro2  pi  ro  ri   2 2 1  2   ro  ri  ri  ro2  ri 2  t ,max (ro2  ri 2 ) 0.8(490)  (252  19 2 )    105 MPa Ans. ro2  ri 2 (252  192 ) ______________________________________________________________________________ 3-116 From Table A-20, Sy =71 MPa From Eq. (3-50)  t  pi  ri 2 pi  ro2  1   ro2  ri 2  r 2  Maximum will occur at r = ri r 2 p  r 2  p (r 2  r 2 )  t ,max  2i i 2 1  o2   i 2o 2i ro  ri  ri  ro  ri  t ,max (ro2  ri 2 ) 0.8(71)  (12  0.752 )  15.9 ksi Ans. ro2  ri 2 (12  0.752 ) ______________________________________________________________________________  Shigley’s MED, 11th edition pi   Chapter 3 Solutions, Page 89/114 3-117 The longitudinal stress will be due to the weight of the vessel above the maximum stress point. From Table A-5, the unit weight of steel is s = 0.282 lbf/in3. The area of the wall is Awall = ( /4)(3602  358.52) = 846. 5 in2 The volume of the wall and dome are Vwall = Awall h = 846.5 (720) = 609.5 (103) in3 Vdome = (2 /3)(1803  179.253) = 152.0 (103) in3 The weight of the structure on the wall area at the tank bottom is W = s Vtotal = 0.282(609.5 +152.0) (103) = 214.7(103) lbf 214.7 103  W l     254 psi Awall 846.5 The maximum pressure will occur at the bottom of the tank, pi = water h. From Eq. (3-50) with r  ri t   ro2  ri 2  ri 2 pi  ro2  p   1   i 2 2  ro2  ri 2  ri 2   ro  ri    1 ft 2    1802  179.252    62.4(55)   5708 ฀ 5710 psi Ans. 2   2 2   144 in    180  179.25    1 ft 2  ri 2 pi  ro2   23.8 psi Ans.  r  2 2 1  2    pi  62.4(55)  2  ro  ri  ri   144 in  Note: These stresses are very idealized as the floor of the tank will restrict the values calculated. Since 1  2  3, 1 = t = 5708 psi, 2 = r =  24 psi and3 = l =  254 psi. From Eq. (3-16), 5708  254  2981  2980 psi 2 5708  24 Ans.  2866  2870 psi 1 2  2 24  254  115 psi 2 3  2 ______________________________________________________________________________ 1 3  3-118 Stresses from additional pressure are, Eq. (3-52), Shigley’s MED, 11th edition Chapter 3 Solutions, Page 90/114  l 50psi  Eq. (3-50) (r)50 psi 50 179.252  180 2  179.252 =  50 psi  5963 psi 1802  179.252  11 975 psi 180 2  179.252 Adding these to the stresses found in Prob. 3-117 gives  t 50psi  50 t = 5708 + 11 975 = 17683 psi = 17.7 kpsi Ans. r =  23.8  50 =  73.8 psi Ans. l =  254 + 5963 = 5709 psi Ans. Note: These stresses are very idealized as the floor of the tank will restrict the values calculated. From Eq. (3-16) 17 683  73.8 1 3   8879 psi 2 17 683  5709 Ans.  5987 psi 1 2  2 5709  23.8  2866 psi 2 3  2 ______________________________________________________________________________ 3-119 (a) For shapes let  (3 +  )/8 = 1,  = 0.3, ro = 2, and ri = 1. Then, 1  3 1  3  0.3   0.5758 3  3  0.3 Thus, Eqs. (3-55) are 4 4  t  1  4  2  0.5758 r 2 , r  1 4  2  r2 r r These equations are plotted: Rotating Ring 10 8 6 4 2 0 1 1.2 1.4 1.6 tangential stress 1.8 2 radial stress (b) The tangential stress, t is maximum at r = ri and is given by Shigley’s MED, 11th edition Chapter 3 Solutions, Page 91/114 3     2 2 ri 2 ro2 1  3 2  ri    ri  ro  2  3  ri  8    t max   2    3      3    1  3   2 2   2   ri  2ro    3   8       2 1   ri 2   3   ro2  Ans. 4 r = 0 at r = ri and r = ro. (r)max occurs where dr / dr = 0. 2 2  d r  3    2ri ro 4 2 2   2    3  2r   0  r  ri ro  r  ri ro dr r 8    Thus,  3    2 2 ri 2 ro2 2  3    ri ro    2  Ans.  r max   2    ri  ro    ro  ri  ri ro  8   8   ______________________________________________________________________________  3-120 Since σt and σr are both positive and σt > σr  max   t  max 2 From Eq. (3-55), t is maximum at r = ri = 0.3125 in. The term 2  0.282  2  5000    3  0.292        82.42 lbf/in 60 8  386       0.31252 2.752 1  3(0.292) 2 2 2   82.42 0.3125  2.75   0.3125   3  0.292 0.31252    1260 psi  3     8 2  t max  max   1260  630 psi 2 Ans.  r 2r 2     Radial stress:   r  k  ri2  ro2  i 2o  r 2  r   Maxima:  ri2 ro2  d r  k  2 3  2r   0  r  dr   Shigley’s MED, 11th edition  r  ri ro  0.3125(2.75)  0.927 in Chapter 3 Solutions, Page 92/114     0.31252 2.752 2 2 2   0.927  r max  82.42 0.3125  2.75    0.927 2    490 psi Ans. ______________________________________________________________________________ 3-121  = 2 (2000)/60 = 209.4 rad/s,  = 3320 20 kg/m3,  = 0.24, ri = 0.01 m, ro = 0.125 m Using Eq. (3-55) 2  3  0.24   2 2 1  3(0.24)  0.012  (10) 6   0.01  (0.125)  (0.125)  3  0.24  8    1.85 MPa Ans. ______________________________________________________________________________  t  3320(209.4)2  3-122 Eq (3-55):  2 2 ri 2 ro2 r 2   2 2 ri 2 ro2  K r r K  t   i  o  2   (1), and  r   ri  ro  2  r 2  (2) 2 r r    2 Where K =  (3 +  )/8 and (1 + 3 )/(3 +  ) = [1 + 3(0.2)] / (3 + 0.2) = 1/2. It can be seen that t is always positive. Check for maxima.  r 2r 2  d t  K  2 i 3o  r   0  r 4  2ri 2 ro2 No roots, no maxima. Check at extremes. dr r   At r = ri,   t i  K  ri2  ro2  ro2   2 ri 2  ri 2   K   2ro   2 2    2 2 2 ro2   2 ro2  At r = ro,  t i  K  ri  ro  ri    K  2ri   2 2   Eq. (3) > (4). Thus, (t)max = (t)i . For Eq. (2), r = 0 at r = ri and r = ro. Check for maxima.  r 2r 2  d r (5)  K  2 i 3o  2r   0  r  ri ro dr r   Substitute Eq. (5) into (2), (3) (4)  r max  K  ri2  ro2  ri ro  ri ro   K  ro2  2ri ro  ri2   K  ro  ri  2 (6)  r2  Comparing Eqs. (3) and (6) it is clear that K  2ro2  i   K  ro2  ri 2  2ri ro  . Thus, the 2   3   2 2 maximum stress is given by Eq. (3) simplified as,  max   2    4ro  ri  Ans.  16  2 2 3 (b) Vol = (/4)(5 – 0.75 )(1/16) = 1.1996 in .  = 2(12 000)/60 = 1256.6 rad/s. 5 /16  6.749 104  lbf-s 2 /in 4  386 1.1996  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 93/114 2 0.375  2  3  0.2    5 360 psi Ans. Eq. (3):  t  max  6.749 104 1256.62    2  2.5   2   8  The factor of safety corresponding to fracture is: S 12 n u   2.24 Ans.  max 5.36 ______________________________________________________________________________ 3-123  = 2 (3500)/60 = 366.5 rad/s, mass of blade = m = V = (0.282 / 386) [1.25(30)(0.125)] = 3.425(103) lbfs2/in F = (m/2)  2r = [3.425(103)/2]( 366.52)(7.5) = 1725 lbf Anom = (1.25  0.5)(1/8) = 0.093 75 in2 nom = F/ Anom = 1725/0.093 75 = 18 400 psi Ans. Note: Stress concentration Fig. A-15-1 gives Kt = 2.25 which increases σmax and fatigue. ______________________________________________________________________________ 3-124  = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57), 207(109 )  (0.052  0.0252 )(0.0252  0)  9 3   10  3.105(10 ) 2(0.025)3  (0.052  0)  where p is in MPa and  is in mm.  p  (1) Maximum interference, 1  max  [50.042  50.000]  0.021 mm Ans. 2 Minimum interference, 1  min  [50.026  50.025]  0.0005 mm Ans. 2 From Eq. (1) pmax = 3.105(103)(0.021) = 65.2 MPa Ans. pmin = 3.105(103)(0.0005) = 1.55 MPa Ans. ______________________________________________________________________________ 3-125  = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in Eq. (3-57), 30(106 )  (2 2  12 )(12  0)  7 p    1.125(10 ) 3 2 2(1 )  (2  0)  Shigley’s MED, 11th edition (1) Chapter 3 Solutions, Page 94/114 where p is in psi and  is in inches. Maximum interference, 1  max  [2.0016  2.0000]  0.0008 in 2 Minimum interference, 1 2  min  [2.0010  2.0010]  0 Ans. Ans. From Eq. (1), pmax = 1.125(107)(0.0008) = 9 000 psi Ans. pmin = 1.125(107)(0) = 0 Ans. ______________________________________________________________________________ 3-126  = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57), 207(109 )  (0.052  0.0252 )(0.0252  0)  9 3   10  3.105(10 ) 2(0.025)3  (0.052  0)  where p is in MPa and  is in mm.  p  (1) Maximum interference, 1  max  [50.059  50.000]  0.0295 mm Ans. 2 Minimum interference, 1  min  [50.043  50.025]  0.009 mm Ans. 2 From Eq. (1) pmax = 3.105(103)(0.0295) = 91.6 MPa Ans. pmin = 3.105(103)(0.009) = 27.9 MPa Ans. ______________________________________________________________________________ 3-127  = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in Eq. (3-57), 30(106 )  (2 2  12 )(12  0)  7 p    1.125(10 ) 2(13 )  (22  0)  (1) where p is in psi and  is in inches. Maximum interference, 1  max  [2.0023  2.0000]  0.00115 in 2 Shigley’s MED, 11th edition Ans. Chapter 3 Solutions, Page 95/114 Minimum interference, 1 2  min  [2.0017  2.0010]  0.00035 Ans. From Eq. (1), pmax = 1.125(107)(0.00115) = 12 940 psi pmin = 1.125(107)(0.00035) = 3 938 Ans. Ans. ______________________________________________________________________________ 3-128  = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57), 207(109 )  (0.052  0.0252 )(0.0252  0)  9 3  10  3.105(10 ) 3  2 2(0.025)  (0.05  0)  where p is in MPa and  is in mm.  p  (1) Maximum interference, 1  max  [50.086  50.000]  0.043 mm Ans. 2 Minimum interference, 1  min  [50.070  50.025]  0.0225 mm Ans. 2 From Eq. (1) pmax = 3.105(103)(0.043) = 134 MPa Ans. pmin = 3.105(103)(0.0225) = 69.9 MPa Ans. ______________________________________________________________________________ 3-129  = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in Eq. (3-57), 30(106 )  (2 2  12 )(12  0)  7 p    1.125(10 ) 3 2 2(1 )  (2  0)  (1) where p is in psi and  is in inches. Maximum interference, 1  max  [2.0034  2.0000]  0.0017 in 2 Minimum interference, 1 2  min  [2.0028  2.0010]  0.0009 Shigley’s MED, 11th edition Ans. Ans. Chapter 3 Solutions, Page 96/114 From Eq. (1), pmax = 1.125(107)(0.0017) = 19 130 psi Ans. pmin = 1.125(107)(0.0009) = 10 130 Ans. ______________________________________________________________________________ 3-130 From Table A-5, Ei = Eo = 30 Mpsi, i o ri = 0, R = 1 in, ro = 1.5 in 1 The radial interference is    2.002  2.000   0.001 in Ans. 2 Eq. (3-57),   2 2 2 2 E  ro  R R  ri p 3 ro2  ri 2 2R    8333 psi  8.33 kpsi    30 10  0.001  1.5  1 1  0   2 1    1.5  0  6 2 3 2 2 2 Ans. The tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. R2  r 2 12  02 ( t )i r  R   p 2 i2  (8333) 2 2  8333 psi  8.33 kpsi Ans. R  ri 1 0 ro2  R 2 1.52  12   21 670 psi  21.7 kpsi Ans. (8333) rR ro2  R 2 1.52  12 ______________________________________________________________________________ ( t )o p 3-131 From Table A-5, Ei = 30 Mpsi, Eo =14.5 Mpsi, i o   ri = 0, R = 1 in, ro = 1.5 in 1 The radial interference is    2.002  2.000   0.001 in Ans. 2 Eq. (3-56), p p   1  r 2  R2  1  o   R   o2 2  Ei  Eo  ro  R  R 2  ri 2   2 2  i    R  ri  0.001   1.52  12   12  02  1 1    1 0.211 0.292   2 2 2 6  6  2 14.5 10  1.5  1  30 10  1  0      4599 psi Ans.   The tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. R2  r 2 12  0 2  4599 psi Ans. ( t )i r  R   p 2 i 2  (4599) 2 R  ri 1  02 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 97/114 ro2  R 2 1.52  12   11 960 psi Ans. (4599) r R ro2  R 2 1.52  12 ______________________________________________________________________________ ( t )o p 3-132 From Table A-5, Ei = Eo = 30 Mpsi, i o ri = 0, R = 0.5 in, ro = 1 in The minimum and maximum radial interferences are 1  min  1.002  1.002   0.000 in Ans. 2 1  max  1.003  1.001  0.001 in Ans. 2 Since the minimum interference is zero, the minimum pressure and tangential stresses are zero. Ans. The maximum pressure is obtained from Eq. (3-57).   2 2 2 2 E  ro  R R  ri p 3 ro2  ri 2 2R            30 106 0.001  12  0.52 0.52  0  p 2 0.53 12  0       22 500 psi  Ans The maximum tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. R 2  ri 2 0.52  02 ( t )i r  R   p 2 2  (22 500) 2  22 500 psi Ans. R  ri 0.5  0 2 ro2  R 2 12  0.52 (22 500)   37 500 psi Ans. r R ro2  R 2 12  0.52 ______________________________________________________________________________ ( t )o p 3-133 From Table A-5, Ei = 10.4 Mpsi, Eo =30 Mpsi, i o   ri = 0, R = 1 in, ro = 1.5 in The minimum and maximum radial interferences are 1  min  [2.003  2.002]  0.0005 in Ans. 2 1  max  [2.006  2.000]  0.003 in Ans. 2 Eq. (3-56), Shigley’s MED, 11th edition Chapter 3 Solutions, Page 98/114 p p   1  r 2  R2  1 R   o2  o   2  Ei  Eo  ro  R  R 2  ri 2   2 2  i    R  ri    1  1.52  12   12  02  1 1  0.292   0.333   2 2 2 6  6  2  30 10  1.5  1  10.4 10  1  0     p  6.229 10   psi p  6.229 10   p  6.229 10     6 6 min 6 max min max Ans.  6.229 10 6   0.0005   3114.6 psi  3.11 kpsi  6.229 106   0.003   18 687 psi  18.7 kpsi Ans. Ans. The tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. Minimum interference: R2  r 2 12  02 ( t )i min   pmin 2 i 2  (3.11) 2 2  3.11 kpsi Ans. R  ri 1 0 ( t )o min  pmin ro2  R 2 1.52  12  (3.11) 2 2  8.09 kpsi ro2  R 2 1.5  1 Ans. Maximum interference: R2  r 2 12  02 ( t )i max   pmax 2 i2  (18.7) 2  18.7 kpsi R  ri 1  02 Ans. ro2  R 2 1.52  12 (18.7)   48.6 kpsi Ans. max ro2  R 2 1.52  12 ______________________________________________________________________________ ( t )o  pmax 3-134 d  20 mm, ri  37.5 mm, ro  57.5 mm From Table 3-4, for R = 10 mm, rc  37.5  10  47.5 mm rn   10 2 2 47.5  47.52  102   46.96772 mm e  rc  rn  47.5  46.96772  0.53228 mm ci  rn  ri  46.9677  37.5  9.4677 mm co  ro  rn  57.5  46.9677  10.5323 mm A   d 2 / 4   (20) 2 / 4  314.16 mm 2 M  Frc  4000(47.5)  190 000 N  mm Using Eq. (3-65) for the bending stress, and combining with the axial stress, Shigley’s MED, 11th edition Chapter 3 Solutions, Page 99/114 4000 190 000(9.4677) F Mci     300 MPa A Aeri 314.16 314.16(0.53228)(37.5) i  Ans. 4000 190 000(10.5323) F Mco     195 MPa Ans. A Aero 314.16 314.16(0.53228)(57.5) ______________________________________________________________________________ o  3-135 d  0.75 in, ri  1.25 in, ro  2.0 in From Table 3-4, for R = 0.375 in, rc  1.25  0.375  1.625 in rn  0.3752  2 1.625  1.6252  0.3752   1.60307 in e  rc  rn  1.625  1.60307  0.02193 in ci  rn  ri  1.60307  1.25  0.35307 in co  ro  rn  2.0  1.60307  0.39693 in A   d 2 / 4   (0.75) 2 / 4  0.44179 in 2 M  Frc  750(1.625)  1218.8 lbf  in Using Eq. (3-65) for the bending stress, and combining with the axial stress, 750 1218.8(0.35307) F Mci     37 230 psi  37.2 kpsi A Aeri 0.44179 0.44179(0.02193)(1.25) i  Ans. 750 1218.8(0.39693) F Mco     23 269 psi  23.3 kpsi Ans. A Aero 0.44179 0.44179(0.02193)(2.0) ______________________________________________________________________________ o  3-136 d  6 mm, ri  10 mm, ro  16 mm From Table 3-4, for R = 3 mm, rc  10  3  13 mm rn   32 2 13  132  32   12.82456 mm e  rc  rn  13  12.82456  0.17544 mm ci  rn  ri  12.82456  10  2.82456 mm co  ro  rn  16  12.82456  3.17544 mm A   d 2 / 4   (6) 2 / 4  28.2743 mm 2 M  Frc  300(13)  3900 N  mm Using Eq. (3-65) for the bending stress, and combining with the axial stress, Shigley’s MED, 11th edition Chapter 3 Solutions, Page 100/114 300 3900(2.82456) F Mci     233 MPa A Aeri 28.2743 28.2743(0.17544)(10) i  Ans. 300 3900(3.17544) F Mco     145 MPa Ans. A Aero 28.2743 28.2743(0.17544)(16) ______________________________________________________________________________ o  3-137 d  6 mm, ri  10 mm, ro  16 mm From Table 3-4, for R = 3 mm, rc  10  3  13 mm rn   32 2 13  132  32   12.82456 mm e  rc  rn  13  12.82456  0.17544 mm ci  rn  ri  12.82456  10  2.82456 mm co  ro  rn  16  12.82456  3.17544 mm A   d 2 / 4   (6) 2 / 4  28.2743 mm 2 The angle  of the line of radius centers is  Rd /2   1  10  6 / 2    sin 1    sin    30  Rd R  10  6  10  M  F  R  d / 2  sin   300 10  6 / 2  sin 30  1950 N  mm Using Eq. (3-65) for the bending stress, and combining with the axial stress, F sin  Mci 300 sin 30 1950(2.82456)     116 MPa A Aeri 28.2743 28.2743(0.17544)(10) i  Ans. 1950(3.17544) F sin  Mco 300sin 30     72.7 MPa Ans. A Aero 28.2743 28.2743(0.17544)(16) Note that the shear stress due to the shear force is zero at the surface. ______________________________________________________________________________ o  3-138 d  0.25 in, ri  0.5 in, ro  0.75 in From Table 3-4, for R = 0.125 in, rc  0.5  0.125  0.625 in rn   0.1252 2 2 0.625  0.625  0.125 2   0.618686 in e  rc  rn  0.625  0.618686  0.006314 in ci  rn  ri  0.618686  0.5  0.118686 in co  ro  rn  0.75  0.618686  0.131314 in Shigley’s MED, 11th edition Chapter 3 Solutions, Page 101/114 A   d 2 / 4   (0.25) 2 / 4  0.049087 in 2 M  Frc  75(0.625)  46.875 lbf  in Using Eq. (3-65) for the bending stress, and combining with the axial stress, i  75 46.875(0.118686) F Mci     37 428 psi  37.4 kpsi A Aeri 0.049087 0.049087(0.006314)(0.5) Ans. 75 46.875(0.131314) F Mco     24 952 psi  25.0 kpsi Ans. A Aero 0.049087 0.049087(0.006314)(0.75) ______________________________________________________________________________ 3-139 d  0.25 in, ri  0.5 in, ro  0.75 in o  From Table 3-4, for R = 0.125 in, rc  0.5  0.125  0.625 in rn   0.1252 2 0.625  0.6252  0.1252   0.618686 in e  rc  rn  0.625  0.618686  0.006314 in ci  rn  ri  0.618686  0.5  0.118686 in co  ro  rn  0.75  0.618686  0.131314 in A   d 2 / 4   (0.25) 2 / 4  0.049087 in 2 The angle  of the line of radius centers is  Rd /2  1  0.5  0.25 / 2     sin 1    sin    30  Rd R  0.5  0.25  0.5  M  F  R  d / 2  sin   75  0.5  0.25 / 2  sin 30  23.44 lbf  in Using Eq. (3-65) for the bending stress, and combining with the axial stress, i  F sin  Mci 75sin 30 23.44(0.118686)     18 716 psi  18.7 kpsi A Aeri 0.049087 0.049087(0.006314)(0.5) Ans. F sin  Mco 75sin 30 23.44(0.131314)     12 478 psi  12.5 kpsi Ans. A Aero 0.049087 0.049087(0.006314)(0.75) Note that the shear stress due to the shear force is zero at the surface. ______________________________________________________________________________ o  3-140 (a)    3(4)0.5(0.1094)  8021 psi  8.02 kpsi Mc  I (0.75)  0.10943  /12 Ans. (b) ri = 0.125 in, ro = ri + h = 0.125 + 0.1094 = 0.2344 in From Table 3-4, Shigley’s MED, 11th edition Chapter 3 Solutions, Page 102/114 rc  0.125  (0.5)(0.1094)  0.1797 in 0.1094  0.174006 in ln(0.2344 / 0.125) e  rc  rn  0.1797  0.174006  0.005694 in rn  ci  rn  ri  0.174006  0.125  0.049006 in co  ro  rn  0.2344  0.174006  0.060394 in A  bh  0.75(0.1094)  0.08205 in 2 M  3(4)  12 lbf  in The negative sign on the bending moment is due to the sign convention shown in Fig. 3-35. Using Eq. (3-65), i  Mci 12(0.049006)   10 070 psi  10.1 kpsi Aeri 0.08205(0.005694)(0.125) o   Mco 12(0.060394)   6618 psi  6.62 kpsi Aero 0.08205(0.005694)(0.2344)  i 10.1   1.26  8.02  6.62 Ko  o   0.825  8.02 (c) K i  Ans. Ans. Ans. Ans. ______________________________________________________________________________ 3-141 (a)    3(4)0.5(0.1406)  4856 psi  4.86 kpsi Mc  I (0.75)  0.14063  / 12 Ans. (b) ri = 0.125 in, ro = ri + h = 0.125 + 0.1406 = 0.2656 in From Table 3-4, rc  0.125  (0.5)(0.1406)  0.1953 in 0.1406  0.186552 in ln(0.2656 / 0.125) e  rc  rn  0.1953  0.186552  0.008748 in rn  ci  rn  ri  0.186552  0.125  0.061552 in co  ro  rn  0.2656  0.186552  0.079048 in A  bh  0.75(0.1406)  0.10545 in 2 M  3(4)  12 lbf  in The negative sign on the bending moment is due to the sign convention shown in Fig. 3-35. Using Eq. (3-65), Shigley’s MED, 11th edition Chapter 3 Solutions, Page 103/114 i  Mci 12(0.061552)   6406 psi  6.41 kpsi Aeri 0.10545(0.008748)(0.125) o   Ans. Mco 12(0.079048)   3872 psi  3.87 kpsi Aero 0.10545(0.008748)(0.2656) Ans.  i 6.41   1.32 Ans.  4.86  3.87 Ko  o   0.80 Ans.  4.86 ______________________________________________________________________________ (c) K i  3-142 (a)    3(4)0.5(0.1094)  8021 psi  8.02 kpsi Mc  I (0.75)  0.10943  /12 Ans. (b) ri = 0.25 in, ro = ri + h = 0.25 + 0.1094 = 0.3594 in From Table 3-4, rc  0.25  (0.5)(0.1094)  0.3047 in 0.1094  0.301398 in ln(0.3594 / 0.25) e  rc  rn  0.3047  0.301398  0.003302 in rn  ci  rn  ri  0.301398  0.25  0.051398 in co  ro  rn  0.3594  0.301398  0.058002 in A  bh  0.75(0.1094)  0.08205 in 2 M  3(4)  12 lbf  in The negative sign on the bending moment is due to the sign convention shown in Fig. 3-35. Using Eq. (3-65), i  Mci 12(0.051398)   9106 psi  9.11 kpsi Aeri 0.08205(0.003302)(0.25) o   Ans. Mco 12(0.058002)   7148 psi  7.15 kpsi Aero 0.08205(0.003302)(0.3594) Ans.  i 9.11   1.14 Ans.  8.02  7.15 Ko  o   0.89 Ans.  8.02 ______________________________________________________________________________ (c) K i  3-143 ri = 25 mm, ro = ri + h = 25 + 87 = 112 mm, rc = 25 + 87/2 = 68.5 mm The radius of the neutral axis is found from Eq. (3-63), given below. Shigley’s MED, 11th edition Chapter 3 Solutions, Page 104/114 rn  A   dA / r  For a rectangular area with constant width b, the denominator is ro  bdr  r ri  r   b ln roi Applying this equation over each of the four rectangular areas,  dA  45   54.5   92   112   9  ln   31 ln   31 ln   9  ln   16.3769 r 45   25    82.5   92  A  2  20(9)  31(9.5)  949 mm 2 rn  A   dA / r   949  57.9475 mm 16.3769 e  rc  rn  68.5  57.9475  10.5525 mm ci  rn  ri  57.9475  25  32.9475 mm co  ro  rn  112  57.9475  54.0525 mm M = 150F2 = 150(3.2) = 480 kN∙mm We need to find the forces transmitted through the section in order to determine the axial stress. It is not immediately obvious which plane should be used for resolving the axial versus shear directions. It is convenient to use the plane containing the reaction force at the bushing, which assumes its contribution resolves entirely into shear force. To find the angle of this plane, find the resultant of F1 and F2. Fx  F1x  F2 x  2.4 cos 60  3.2 cos 0  4.40 kN Fy  F1 y  F2 y  2.4sin 60  3.2sin 0  2.08 kN F   4.402  2.082  12  4.87 kN This is the pin force on the lever which acts in a direction   tan 1 Fy Fx  tan 1 2.08  25.3 4.40 On the surface 25.3° from the horizontal, find the internal forces in the tangential and normal directions. Resolving F1 into components, Ft  2.4cos  60  25.3   1.97 kN Fn  2.4sin  60  25.3   1.37 kN The transverse shear stress is zero at the inner and outer surfaces. Using Eq. (3-65) for the bending stress, and combining with the axial stress due to Fn, Shigley’s MED, 11th edition Chapter 3 Solutions, Page 105/114 i  Fn Mci 1370  3200 150   (32.9475)     64.6 MPa A Aeri 949 949(10.5525)(25) Ans. Fn Mco 1370  3200 150   (54.0525)     21.7 MPa Ans. A Aero 949 949(10.5525)(112) ______________________________________________________________________________ o  3-144 ri = 2 in, ro = ri + h = 2 + 4 = 6 in, rc  2  0.5(4)  4 in A  (6  2  0.75)(0.75)  2.4375 in 2 Similar to Prob. 3-143, dA 3.625 6  0.75 ln  0.75ln  0.682 920 in r 2 4.375 2.4375 A rn    3.56923 in 0.682 920 ( / ) dA r   e  rc  rn  4  3.56923  0.43077 in ci  rn  ri  3.56923  2  1.56923 in co  ro  rn  6  3.56923  2.43077 in M  Frc  6000(4)  24 000 lbf  in Using Eq. (3-65) for the bending stress, and combining with the axial stress, i  6000 24 000(1.56923) F Mci     20 396 psi  20.4 kpsi A Aeri 2.4375 2.4375(0.43077)(2) Ans. 6000 24 000(2.43077) F Mco     6 799 psi  6.80 kpsi Ans. A Aero 2.4375 2.4375(0.43077)(6) ______________________________________________________________________________ o  3-145 ri = 12 in, ro = ri + h = 12 + 3 = 15 in, rc = 12 + 3/2 = 13.5 in   I  a 3b  (1.53 )(0.75)  1.988 in 4 4 4 A   ab   (1.5)(0.75)  3.534 M  20(3  1.5)  90 kip  in Since the radius is large compared to the cross section, assume Eq. 3-67 is applicable for the bending stress. Combining the bending stress and the axial stress, 20 90(1.5)(13.5) F Mci rc     82.1 kpsi Ans. A Iri 3.534 (1.988)(12) 20 90(1.5)(13.5) F Mc r o   o c    55.5 kpsi Ans. A Iro 3.534 1.988(15) ______________________________________________________________________________ 3-146 ri = 1.25 in, ro = ri + h = 1.25 + 0.5 + 1 + 0.5 = 3.25 in i  Shigley’s MED, 11th edition Chapter 3 Solutions, Page 106/114 rc = (ri + ro) / 2 = (1.25 + 3.25)/2 = 2.25 in Ans. ro  dA  For outer rectangle,     b ln ri  r     A  r2   , A   r2   For circle,  O   2 2    dA r   r r r   2 c  c  O O  dA      2 (rc  rc2  r 2 )  r O Combine the integrals subtracting the circle from the rectangle     3.25 dA  1.25 ln  2 2.25  2.252  0.52  0.840 904 in r 1.25 A  1.25(2)   (0.52 )  1.714 60 in 2 Ans. A 1.714 60 rn    2.0390 in Ans. 0.840904 dA r ( / )   e  rc  rn  2.25  2.0390  0.2110 in Ans. ci  rn  ri  2.0390  1.25  0.7890 in co  ro  rn  3.25  2.0390  1.2110 in M  2000(4.5  1.25  0.5  0.5)  13 500 lbf  in 2000 13 500(0.7890) F Mci i      20 720 psi = 20.7 kpsi Ans. A Aeri 1.7146 1.7146(0.2110)(1.25) 2000 13 500(1.2110) F Mco o      12 738 psi  12.7 kpsi Ans. A Aero 1.7146 1.7146(0.2110)(3.25) ______________________________________________________________________________ 3-147 Table A-5: Glass: EG = 46.2 GPa, G = 0.245, Steel: ES =207 GPa, S = 0.292 Eq. (3-68): a 3 2 2 3F 1  1  / E1  1  2  / E2 8 1/ d1  1/ d 2 2 9 2 9 3  5  1  0.245  /  46.2 10   1  0.292  / 207 10    8 1/ 0.030  1/  3  1.1168 104  m  0.11168 mm Eq. (3-69): pmax  3 5 3F   191.4 MPa 2 2 2 a 2  0.11168 Ans. (b) Eq. (3-70): Shigley’s MED, 11th edition Chapter 3 Solutions, Page 107/114   1  1 z   1  1   2   pmax 1  tan  1   G   2  a z / a  2 1   z / a          1 1 z     1  191.4  1    1  0.245   tan 2   z / 0.11168   2 1   z / 0.11168       0.11168    Eq. (3-71):  pmax 191.4 3   (2) 2 2 1   z / a  1   z / 0.11168  The maximum shear stress is given by  3 (3)  max  1 2 (c) Plots of Eqs. (1) – (3) as absolute dimensionless stresses,stress / pmax, are given. Note, at z = 0, 1 = 2 =  0.745 pmax. (1) 1.0 0.9 0.8  Stress/pmax 0.7 Ans. 1,2 0.6 0.5 0.4 max 0.3 0.2 0.1 0.0 0.0 0.2 0.4 0.6 0.8 1.0 z/a (d) max  0.323 pmax = 0.323(191.4) = 61.8 MPa at z = 0.05 mm Ans. (e) From Fig. 3-38, max = 0.3(191.4) = 57.4 MPa Ans. ______________________________________________________________________________ 3-148 From Eq. (3-68), Shigley’s MED, 11th edition Chapter 3 Solutions, Page 108/114 13 a  KF 1 3  2 1  2  E   1 3  3    F    2 1 d   8   Use  = 0.292, F in newtons, E in N/mm2 and d in mm, then 1/3  3  [(1  0.2922 ) / 207 000]  K     1/ 30  8    0.03685 From Eq. (3-69), pmax  3F 3F 3F 1/3 3F 1/3     352 F 1/3 MPa 2 1/3 2 2 2 2 a 2 ( KF ) 2 K 2 (0.03685) From Eq. (3-71), the maximum principal stress occurs on the surface where z = 0, and is equal to – pmax.  max   z   pmax  352 F 1/3 MPa Ans. From Fig. 3-38,  max  0.3 pmax  106 F 1/3 MPa Ans. ______________________________________________________________________________ 3-149 From Eq. (3-68), 1  12  E1  1  22  E2  3 F   a3  1 d1  1 d 2  8   3 10   1  0.292 a3   8  2   207 000   1  0.333   71 700   0.0990 mm 2 1 25  1 40 From Eq. (3-69), 3 10  3F   487.2 MPa pmax  2 2 a 2  0.09902  From Fig. 3-38, the maximum shear stress occurs at a depth of z = 0.48 a. z  0.48a  0.48  0.0990   0.0475 mm Ans. The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48.   1  1   2  487.2  1  0.48 tan 1 1/ 0.48   1  0.333   101.3 MPa  2 1  0.482    487.2 3   396.0 MPa 1  0.482 Shigley’s MED, 11th edition Chapter 3 Solutions, Page 109/114 From Eq. (3-72),  101.3   396.0   147.4 MPa Ans.    max  1 3  2 2 Note that if a closer examination of the applicability of the depth assumption from Fig. 338 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow for calculating and plotting the stresses versus the depth for specific values of . For = 0.333 for aluminum, the maximum shear stress occurs at a depth of z = 0.492a with max = 0.3025 pmax. This gives max = 0.3025 pmax = (0.3025)(487.2) = 147.38 MPa. Even though the depth assumption was a little off, it did not have significant effect on the the maximum shear stress. ______________________________________________________________________________ 3-150 From the solution to Prob. 3-149, a = 0.0990 mm and pmax = 487.2 MPa. Assuming applicability of Fig. 3-38, the maximum shear stress occurs at a depth of z = 0.48 a = 0.0475 mm. Ans. The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48.   1   2  487.2  1  0.48 tan 1 1/ 0.48   1  0.292    3  487.2  396.0 MPa 1  0.482 From Eq. (3-72),  92.09    396.0   152.0 MPa    max  1 3  2 2  1   92.09 MPa 2 1  0.482   Ans. Note that if a closer examination of the applicability of the depth assumption from Fig. 338 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow for calculating and plotting the stresses versus the depth for specific values of . For = 0.292 for steel, the maximum shear stress occurs at a depth of z = 0.478a with max = 0.3119 pmax. ______________________________________________________________________________ 3-151 From Eq. (3-68), 2 1  2  E F 3   a3   8  1 d1  1 d 2 2  3  20   2 1  0.292   207 000  a3  0.1258 mm  1 30  1   8  From Eq. (3-69), Shigley’s MED, 11th edition Chapter 3 Solutions, Page 110/114 pmax  3  20  3F   603.4 MPa 2 2 a 2 0.12582   From Fig. 3-38, the maximum shear stress occurs at a depth of z  0.48a  0.48  0.1258   0.0604 mm Ans. Also from Fig. 3-38, the maximum shear stress is  max  0.3 pmax  0.3(603.4)  181 MPa Ans. ______________________________________________________________________________ 3-152 Aluminum Plate-Ball interface: From Eq. (3-68), 2 2  3F  1  1  E1  1  2  E2 a   1 d1  1 d 2  8  3 2 6 2 6  3F  1  0.292   30  10   1  0.333  10.4  10   a   3.517 103  F 1/3 in  1 11   8  3 From Eq. (3-69), 3F 3F   3.860 104 F 1/3 psi pmax  2 2 2 a 2 3.517 103 F 1/3  By examination of Eqs. (3-70), (3-71), and (3-72), it can be seen that the only difference in the maximum shear stress for the plate and the ball will be due to Poisson’s ratio in Eq. (3-70). The larger Poisson’s ratio will create the greater maximum shear stress, so the aluminum plate will be the critical element in this interface. Applying the equations for the aluminum plate,   1  1  3.86 104  F 1/3  1  0.48 tan 1 1/ 0.48   1  0.333   8025 F 1/3 psi 2  2 1  0.48    3.86 104  F 1/3 3   3.137 104  F 1/3 psi 2 1  0.48  From Eq. (3-72),  max  1   3     8025F    3.137 10  F    1.167 10  F 2 1/3 4 1/3 4 2 Comparing this stress to the allowable stress, and solving for F, 1/3 psi 3  20 000    5.03 lbf F  4 1.167 10   Table-Ball interface: From Eq. (3-68), Shigley’s MED, 11th edition Chapter 3 Solutions, Page 111/114 2 6 2 6  3F  1  0.292   30  10    1  0.211  14.5  10   a   3.306 103  F 1/3 in  1 11   8  3 From Eq. (3-69), 3F 3F   4.369 10 4 F 1/3 psi pmax  2 2 3 1/3 2 a 2 3.306 10 F  The steel ball has a higher Poisson’s ratio than the cast iron table, so it will dominate.       1  4.369 10 4  F 1/3  1  0.48 tan 1 1/ 0.48  1  0.292   3   4.369 10 4  F 1/3 1  0.48 From Eq. (3-72),  max  1   3 2  1  8258 F 1/3 psi 2  2 1  0.48    3.55110 4  F 1/3 psi  8258F    3.55110  F    1.363 10  F 2 1/3 4 1/3 4 1/3 2 Comparing this stress to the allowable stress, and solving for F, psi 3  20 000    3.16 lbf F  4 1.363 10   The steel ball is critical, with F = 3.16 lbf. Ans. ______________________________________________________________________________ 3-153 v1 = 0.333, E1 = 10.4 Mpsi, l = 2 in, d1 = 1.25 in, v2 = 0.211, E2 = 14.5 Mpsi, d2 = –12 in. From Eq. 3-73, with b = KcF1/2 12  2 1  0.3332  10.4 106    1  0.2112  14.5 106       Kc     (2)  1 /1.25  1/  12     2.593 104  By examination of Eqs. (3-75), (3-76), and (3-77), it can be seen that the only difference in the maximum shear stress for the two materials will be due to Poisson’s ratio in Eq. (375). The larger Poisson’s ratio will create the greater maximum shear stress, so the aluminum roller will be the critical element in this interface. Instead of applying these equations, we will assume the Poisson’s ratio for aluminum of 0.333 is close enough to 0.3 to make Fig. 3-40 applicable. Shigley’s MED, 11th edition Chapter 3 Solutions, Page 112/114  max  0.3 pmax pmax  4000  13 300 psi 0.3 From Eq. (3-74), pmax = 2F / (bl ), so we have pmax 2F 2F 1 2    lK c F 1 2  lK c So,   lK c pmax  F   2   2   (2)(2.593) 10 4  (13 300)      2    117.4 lbf Ans. ______________________________________________________________________________ 2 3-154 v = 0.292, E = 30 Mpsi, l = 0.75 in, d1 = 2(0.47) = 0.94 in, d2 = 2(0.62) = 1.24 in. Eq. (3-73): 12  2(40) 2 1  0.2922  30 106      b   (0.75)  1/ 0.94  1/ 1.24   Eq. (3-74): pmax   1.052 103  in 2  40  2F   32 275 psi  32.3 kpsi  bl  1.052  103   0.75  Ans. From Fig. 3-40,  max  0.3 pmax  0.3  32 275 = 9682.5 psi  9.68 kpsi Ans. ______________________________________________________________________________ 3-155 Use Eqs. (3-73) through (3-77). 1/ 2  2 F (1  12 ) / E1  (1  v22 ) / E2  b  (1/ d1 )  (1/ d 2 )  l  1/ 2  2(600) (1  0.292 2 ) / (30(10 6 ))  (1  0.292 2 ) / (30(106 ))    1/ 5  1/    (2)  b  0.007 631 in Shigley’s MED, 11th edition Chapter 3 Solutions, Page 113/114 pmax  2F 2(600)   25 028 psi  bl  (0.007 631)(2)  z2 z     2  0.292  25 028  2  b b   Ans.  7102 psi  7.10 kpsi  x  2 pmax  1   1  0.786 2  0.786    z2  1  2 0.7862   1 2 2    z b    y   pmax 2  25 028  2  0.786   2 2    b  1   0.786  z 1      b2    4 646 psi  4.65 kpsi Ans.  pmax 25 028 z    19 677 psi  19.7 kpsi Ans. z2 1  0.7862 1 2 b    z 4 646   19 677    7 516 psi  7.52 kpsi  max  y Ans. 2 2 ______________________________________________________________________________ 3-156 Use Eqs. (3-73) through (3-77).  2 F 1  12  E1  1  2 2  E2   b  l  1/ d1  1/ d 2   12 12  2(2000) 1  0.2922   207 103   1  0.2112  100 103           (40)  1/150  1/    b  0.2583 mm pmax  2F 2(2000)   123.2 MPa  bl  (0.2583)(40)  z2 z     2  0.292 123.2  1  0.786 2  0.786 2  b b   Ans.  35.0 MPa   z2  1  2 0.7862   1 2 2    z b   2  123.2  2  0.786    y   pmax 2   2   b z  1   0.786     1 2  b    22.9 MPa Ans.  x  2 pmax  1  Shigley’s MED, 11th edition   Chapter 3 Solutions, Page 114/114 z   pmax  max   z2 1 2 b  y  z 2  123.2 1  0.786 2  96.9 MPa 22.9   96.9  2 Ans.  37.0 MPa Ans. ______________________________________________________________________________ 3-157 Use Eqs. (3-73) through (3-77).  2 F 1  12  E1  1  2 2  E2   b  l  1/ d1  1/ d 2   12 12  2(250) 1  0.2112  14.5 106    1  0.2112  14.5 106          (1.25)  1/ 3  1/    b  0.007 095 in pmax  2F 2(250)   17 946 psi  bl  (0.007 095)(1.25)  z2 z     2  0.21117 946  2  b b    3 680 psi  3.68 kpsi Ans.  x  2 pmax  1   1  0.786 2  0.786    z2  1  2 0.7862   1 2 2    z b   2  17 946  2  0.786    y   pmax 2   2   b z  1   0.786  1      b2   Ans.  3 332 psi  3.33 kpsi  pmax 17 946 z    14 109 psi  14.1 kpsi Ans. z2 1  0.7862 1 2 b    z 3 332   14 109    5 389 psi  5.39 kpsi  max  y Ans. 2 2 ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 3 Solutions, Page 115/114 Chapter 13 13-1 d P  17 / 8  2.125 in N 1120 dG  2 d P   2.125  4.375 in N3 544 NG  PdG  8  4.375  35 teeth Ans. C   2.125  4.375  / 2  3.25 in Ans. ______________________________________________________________________________ nG  1600 15 / 60   400 rev/min p   m  3 mm Ans. 13-2 C  3 15  60   2  112.5 mm Ans. Ans. ______________________________________________________________________________ NG  16  4   64 teeth 13-3 Ans. dG  NG m  64  6   384 mm d P  N P m  16  6   96 mm C   384  96  / 2  240 mm Ans. Ans. Ans. ______________________________________________________________________________ 13-4 Mesh: a  1/ P  1/ 3  0.3333 in Ans. b  1.25 / P  1.25 / 3  0.4167 in Ans. c  b  a  0.0834 in Ans. p   / P   / 3  1.047 in Ans. t  p / 2  1.047 / 2  0.523 in Ans. Pinion Base-Circle: d1  N1 / P  21/ 3  7 in d1b  7 cos 20  6.578 in Gear Base-Circle: Ans. d 2  N 2 / P  28 / 3  9.333 in d 2b  9.333cos 20  8.770 in Base pitch: Ans. pb  pc cos    / 3 cos 20  0.984 in Ans. mc  Lab / pb  1.53 / 0.984  1.55 Ans. Contact Ratio: See the following figure for a drawing of the gears and the arc lengths. Shigley’s MED, 11th edition Chapter 13 Solutions, Page 1/36 ______________________________________________________________________________ 13-5 (a)  14 / 6  2  32 / 6 2  AO        2   2   (b)   tan 1 14 / 32   23.63 Ans.   tan 1  32 /14   66.37 Ans. (c) d P  14 / 6  2.333 in 1/2  2.910 in Ans. d G  32 / 6  5.333 in Ans. (d) From Table 13-3, 0.3AO = 0.3(2.910) = 0.873 in and 10/P = 10/6 = 1.67 0.873 < 1.67  F  0.873 in Shigley’s MED, 11th edition Ans. Ans. Chapter 13 Solutions, Page 2/36 ____________________________________________________________________ 13-6 (a) pn   / Pn   / 4  0.7854 in pt  pn / cos  0.7854 / cos 30  0.9069 in px  pt / tan  0.9069 / tan 30  1.571 in (b) Eq. (13-7): (c) Pt  Pn cos  4 cos 30  3.464 teeth/in pnb  pn cos n  0.7854 cos 25  0.7380 in Ans. t  tan 1  tan n / cos   tan 1 (tan 25 / cos 30 )  28.3 Ans. (d) Table 13-4: a  1/ 4  0.250 in Ans. b  1.25 / 4  0.3125 in Ans. 20 dP   5.774 in Ans. 4 cos 30 36 dG   10.39 in Ans. 4 cos 30 ______________________________________________________________________________ 13-7 N P  19 teeth, N G  57 teeth, n  20 , mn  2.5 mm (a) pn   mn    2.5  7.854 mm Ans. pn 7.854   9.069 mm Ans. cos cos 30 pt 9.069 px    15.71 mm Ans. tan tan 30 mn 2.5 mt    2.887 mm Ans. cos cos 30 pt  (b) Shigley’s MED, 11th edition Chapter 13 Solutions, Page 3/36  tan 20   22.80   cos 30   t  tan 1  (c) a  mn  2.5 mm Ans. Ans. b  1.25mn  1.25  2.5   3.125 mm dP  Ans. N  Nmt  19  2.887  =54.85 mm Pt dG  57  2.887   164.6 mm Ans. Ans. ______________________________________________________________________________ 13-8 (a) Using Eq. (13-11) with k = 1, = 20º, and m = 2, NP    2k m  m 2  1  2m  sin 2  1  2m  sin 2   2  1  2  2   sin  20  2 1 2   2 2     14.16 teeth  1  2  2   sin 2 20 Round up for the minimum integer number of teeth. NP = 15 teeth Ans. (b) Repeating (a) with m = 3, NP = 14.98 teeth. Rounding up, NP = 15 teeth. Ans. (c) Repeating (a) with m = 4, NP = 15.44 teeth. Rounding up, NP = 16 teeth. Ans. (d) Repeating (a) with m = 5, NP = 15.74 teeth. Rounding up, NP = 16 teeth. Ans. Alternatively, a useful table can be generated to determine the largest gear that can mesh with a specified pinion, and thus also the maximum gear ratio with a specified pinion. The Max NG column was generated using Eq. (13-12) with k = 1, = 20º, and rounding up to the next integer. Min NP 13 14 15 16 17 18 Max NG 16 26 45 101 1309 unlimited Max m = Max NG / Min NP 1.23 1.86 3.00 6.31 77.00 unlimited With this table, we can readily see that gear ratios up to 3 can be obtained with a minimum NP of 15 teeth, and gear ratios up to 6.31 can be obtained with a minimum NP of 16 teeth. This is consistent with the results previously obtained. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 13 Solutions, Page 4/36 13-9 Repeating the process shown in the solution to Prob. 13-8, except with = 25º, we obtain the following results. (a) For m = 2, NP = 9.43 teeth. Rounding up, NP = 10 teeth. Ans. (b) For m = 3, NP = 9.92 teeth. Rounding up, NP = 10 teeth. Ans. (c) For m = 4, NP = 10.20 teeth. Rounding up, NP = 11 teeth. Ans. (d) For m = 5, NP = 10.38 teeth. Rounding up, NP = 11 teeth. Ans. For convenient reference, we will also generate the table from Eq. (13-12) for = 25º. Min NP Max NG Max m = Max NG / Min NP 9 13 1.44 10 32 3.20 11 249 22.64 12 unlimited unlimited ______________________________________________________________________________ 13-10 (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10). NP    2k 1  1  3sin 2  2 3sin  2 1 1  3sin 20 2  12.32   1  3sin 2 20  13 teeth  Ans. (b) The smallest pinion that will mesh with a gear ratio of mG = 2.5, from Eq. (13-11) is NP    2k m  m 2  1  2m  sin 2  2 1  2m  sin  2 1 2.5  1  2  2.5   sin 20 2  14.64  2.52  1  2  2.5   sin 2 20  15 teeth   Ans. The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is N P2 sin 2   4k 2 NG  4k  2 N P sin 2   152 sin 2 20  4 1 2 4 1  2 15  sin 2 20  45.49  45 teeth Ans. (c) The smallest pinion that will mesh with a rack, from Eq. (13-13), Shigley’s MED, 11th edition Chapter 13 Solutions, Page 5/36 2 1 2k  2 sin  sin 2 20  17.097  18 teeth Ans. ______________________________________________________________________________ NP  13-11 n  20 ,  30 From Eq. (13-19), t  tan 1  tan 20 / cos 30   22.80 (a) The smallest pinion tooth count that will run with itself, from Eq. (13-21) is NP    2k cos 1  1  3sin 2 t 3sin 2 t 2 1 cos 30 1  3sin 22.80 2   1  3sin 2 22.80  8.48  9 teeth  Ans. (b) The smallest pinion that will mesh with a gear ratio of m = 2.5, from Eq. (13-22) is 2 1 cos 30 NP  2.5  2.52  1  2  2.5   sin 2 22.80 2  1  2  2.5   sin 22.80  9.95  10 teeth Ans.   The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-23) is NG   N P2 sin 2 t  4k 2 cos 2  4k cos  2 N P sin 2 t 102 sin 2 22.80  4 1 cos 2 30 4 1 cos 2 30  2  20  sin 2 22.80  26.08  26 teeth Ans. (c) The smallest pinion that will mesh with a rack, from Eq. (13-24) is  2k cos 2 1 cos 30 NP   sin 2 t sin 2 22.80  11.53  12 teeth Ans. ______________________________________________________________________________ 13-12 From Eq. (13-19), Shigley’s MED, 11th edition   tan n  1  tan 20   22.796 t  tan    tan     cos   cos 30  1 Chapter 13 Solutions, Page 6/36 Program Eq. (13-23) on a computer using a spreadsheet or code, and increment NP. The first value of NP that can be doubled is NP = 10 teeth, where NG ≤ 26.01 teeth. So NG = 20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc. Use NP = 10 teeth, NG = 20 teeth Ans. Note that the given diametral pitch (tooth size) is not relevant to the interference problem. ______________________________________________________________________________   tan n  1  tan 20    27.236 tan     cos cos 45      Program Eq. (13-23) on a computer using a spreadsheet or code, and increment NP. The first value of NP that can be doubled is NP = 6 teeth, where NG ≤ 17.6 teeth. So NG = 12 teeth will work. Higher tooth counts will work also, for example 7:14, 8:16, etc. 13-13 From Eq. (13-19), t  tan 1  Use NP = 6 teeth, NG = 12 teeth Ans. ______________________________________________________________________________ 13-14 The smallest pinion that will operate with a rack without interference is given by Eq. (1313). 2k NP  sin 2  Setting k = 1 for full depth teeth, NP = 9 teeth, and solving for ,   sin 1 2 1 2k  sin 1  28.126 NP 9 Ans. ______________________________________________________________________________ 13-15 (a) (b) Eq. (13-3): pn   mn  3 mm Eq. (13-16): pt  pn / cos  3 / cos 25  10.40 mm Eq. (13-17): px  pt / tan  10.40 / tan 25  22.30 mm Eq. (13-3): mt  pt /   10.40 /   3.310 mm Eq. (13-19): Shigley’s MED, 11th edition t  tan 1 Ans. Ans. Ans. tan n tan 20  tan 1  21.88  cos cos 25  Ans. Ans. Chapter 13 Solutions, Page 7/36 (c) Eq. (13-2): dp = mt Np = 3.310 (18) = 59.58 mm Ans. Eq. (13-2): dG = mt NG = 3.310 (32) = 105.92 mm Ans. ______________________________________________________________________________ 13-16 (a) Sketches of the figures are shown to determine the axial forces by inspection. The axial force of gear 2 on shaft a is in the negative z-direction. The axial force of gear 3 on shaft b is in the positive z-direction. Ans. The axial force of gear 4 on shaft b is in the positive z-direction. The axial force of gear 5 on shaft c is in the negative z-direction. Ans. 12  16     700   77.78 rev/min ccw 48  36  (b) nc  n5  (c) d P 2  12 / 12 cos 30  1.155 in dG 3    48 / 12 cos 30   4.619 in  1.155  4.619  2.887 in 2  16 /  8 cos 25   2.207 in Cab  dP4 Ans.  Ans.  d G 5  36 / 8 cos 25  4.965 in Cbc  3.586 in Ans. ______________________________________________________________________________ 20  8   20  4     40  17   60  51 4 nd   00   47.06 rev/min cw Ans. 51 ______________________________________________________________________________ 13-17 e 6  18  20   3  3       10  38  48   36  304 3 n9  1200   11.84 rev/min cw Ans. 304 ______________________________________________________________________________ 13-18 e Shigley’s MED, 11th edition Chapter 13 Solutions, Page 8/36 13-19 (a) (b) 12 1   540   162 rev/min 40 1     Ans. cw about x, as viewed from the positive x axis     d P  12 /  8 cos 23   1.630 in nc    d G  40 / 8 cos 23  5.432 in d P  dG  3.531 in 2 Ans. N 32   8 in Ans. P 4 ______________________________________________________________________________ (c) d 13-20 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 45. For an exact ratio, we will choose to factor the train value into integers, such that N2 / N3 = 9 N4 / N5 = 5 (1) (2) Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the in-line requirement of the compound reverted configuration is N2 + N3 = N4 + N5 (3) With three equations and four unknowns, one free choice is available. It is necessary that all of the unknowns be integers. We will use a normalized approach to find the minimum free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity, thus N3 = 1. From (1), N2 = 9. From (3), N2 + N3 = 9 + 1 = 10 = N4 + N5 Substituting N4 = 5 N5 from (2) gives 10 = 5 N5 + N5 = 6 N5 N5 = 10 / 6 = 5 / 3 To eliminate this fraction, we need to multiply the original free choice by a multiple of 3. In addition, the smallest gear needs to have sufficient teeth to avoid interference. From Eq. (13-11) with k = 1, = 20°, and m = 9, the minimum number of teeth on the pinion to avoid interference is 17. Therefore, the smallest multiple of 3 greater than 17 is 18. Setting N3 = 18 and repeating the solution of equations (1), (2), and (3) yields Shigley’s MED, 11th edition Chapter 13 Solutions, Page 9/36 N2 = 162 teeth N3 = 18 teeth N4 = 150 teeth N5 = 30 teeth Ans. ______________________________________________________________________________ 13-21 The solution to Prob. 13-20 applies up to the point of determining the minimum number of teeth to avoid interference. From Eq. (13-11), with k = 1, = 25°, and m = 9, the minimum number of teeth on the pinion to avoid interference is 11. Therefore, the smallest multiple of 3 greater than 11 is 12. Setting N3 = 12 and repeating the solution of equations (1), (2), and (3) of Prob. 13-20 yields N2 = 108 teeth N3 = 12 teeth N4 = 100 teeth N5 = 20 teeth Ans. ______________________________________________________________________________ 13-22 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 30. For an exact ratio, we will choose to factor the train value into integers, such that N2 / N3 = 6 N4 / N5 = 5 (1) (2) Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the in-line requirement of the compound reverted configuration is N2 + N3 = N4 + N5 (3) With three equations and four unknowns, one free choice is available. It is necessary that all of the unknowns be integers. We will use a normalized approach to find the minimum free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity, thus N3 = 1. From (1), N2 = 6. From (3), N2 + N3 = 6 + 1 = 7 = N4 + N5 Substituting N4 = 5 N5 from (2) gives 7 = 5 N5 + N5 = 6 N5 N5 = 7 / 6 To eliminate this fraction, we need to multiply the original free choice by a multiple of 6. In addition, the smallest gear needs to have sufficient teeth to avoid interference. From Eq. (13-11) with k = 1, = 20°, and m = 6, the minimum number of teeth on the pinion to avoid interference is 16. Therefore, the smallest multiple of 6 greater than 16 is 18. Setting N3 = 18 and repeating the solution of equations (1), (2), and (3) yields Shigley’s MED, 11th edition Chapter 13 Solutions, Page 10/36 N2 = 108 teeth N3 = 18 teeth N4 = 105 teeth N5 = 21 teeth Ans. ______________________________________________________________________________ 13-23 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 45. For an approximate ratio, we will choose to factor the train value into two equal stages, such that N 2 / N3  N 4 / N5  45 If we choose identical pinions such that interference is avoided, both stages will be identical and the in-line geometry condition will automatically be satisfied. From Eq. (13-11) with k = 1, = 20°, and m  45 , the minimum number of teeth on the pinions to avoid interference is 17. Setting N3 = N5 = 17, we get N 2  N 4  17 45  114.04 teeth Rounding to the nearest integer, we obtain N2 = N4 = 114 teeth N3 = N5 = 17 teeth Ans. Checking, the overall train value is e = (114 / 17) (114 / 17) = 44.97. ______________________________________________________________________________ 13-24 H = 25 hp, i = 2500 rev/min Let ωo = 300 rev/min for minimal gear ratio to minimize gear size. o 300 N N 1    2 4 i 2500 8.333 N 3 N 5 Let N2 N4 1 1    N3 N5 8.333 2.887 From Eq. (13-11) with k = 1, = 20°, and m = 2.887, the minimum number of teeth on the pinions to avoid interference is 15. Let N2 = N4 = 15 teeth N3 = N5 = 2.887(15) = 43.31 teeth Try N3 = N5 = 43 teeth. Shigley’s MED, 11th edition Chapter 13 Solutions, Page 11/36  15   15  o       2500   304.2  43   43  Too big. Try N3 = N5 = 44.  15   15  o       2500   290.55 rev/min  44   44  N2 = N4 = 15 teeth, N3 = N5 = 44 teeth Ans. ______________________________________________________________________________ 13-25 (a) The planet gears act as keys and the wheel speeds are the same as that of the ring gear. Thus, nA  n3  900 16 / 48   300 rev/min Ans. (b) nF  n5  0, nL  n6 , e  1 n  300 1  6 0  300 300  n6  300 n6  600 rev/min Ans. (c) The wheel spins freely on icy surfaces, leaving no traction for the other wheel. The car is stalled. Ans. ______________________________________________________________________________ 13-26 (a) The motive power is divided equally among four wheels instead of two. (b) Locking the center differential causes 50 percent of the power to be applied to the rear wheels and 50 percent to the front wheels. If one of the rear wheels rests on a slippery surface such as ice, the other rear wheel has no traction. But the front wheels still provide traction, and so you have two-wheel drive. However, if the rear differential is locked, you have 3-wheel drive because the rear-wheel power is now distributed 50-50. ______________________________________________________________________________ 13-27 Let gear 2 be first, then nF = n2 = 0. Let gear 6 be last, then nL = n6 = –12 rev/min. 20  16  16 nL  nA    30  34  51 nF  nA 16  0  nA   12  nA 51 e Shigley’s MED, 11th edition Chapter 13 Solutions, Page 12/36 nA  12  17.49 rev/min 35 / 51 (negative indicates cw as viewed from the bottom of the figure)     Ans.    ______________________________________________________________________________ 13-28 Let gear 2 be first, then nF = n2 = 0 rev/min. Let gear 6 be last, then nL = n6 = 85 rev/min. 20  16  16 nL  nA    30  34  51 nF  nA 16  0  nA    85  nA  51  16   nA    n A  85  51   16  nA  1    85  51  85 nA   123.9 rev/min 16 1 51 e The positive sign indicates the same direction as n6.  nA  123.9 rev/min ccw Ans. ______________________________________________________________________________ 13-29  = 20o, P = 6 teeth/in. Since N  d, then, N2 + N3 = N4 + N5 (1) Hour hand moves 1/12 of minute hand. Thus,  5 /  2 = 121 . Now,  5 /  4 = N 4 / N 5 ,  3 /  2 = N 2 / N 3, and  4 =  3. Thus, 2 N 3 N 5   12  4  3 2 N 2 N 4 So, try N3 = 4 N2, and N5 = 3 N4. Substituting N5 = 3 N4 into Eq. (1) gives N2 + N3 = 4 N4  N4 = (N2 + N3)/4. Let N2 = 1. Then, N3 = 4N2 = 4(1) = 4, N4 = (N2 + N3)/4 = (1 + 4)/4 = 5/4, N5 = 3 N4 = 3(5/4) = 15/4. Teeth must be a multiple of 4. To avoid interference for the smaller pinion use Eq. (13-11) with m = N 3 / N 2 = 4: 2k N2  m  m 2  1  2m  sin 2  2 1  2m  sin     2 1 4  1  2  4   sin 20 2 o  42  1  2  4   sin 2 20o  15.4 teeth Use N2 = 16 teeth which is a multiple of 4. Then, N3 = 4(16) = 64 teeth, N4 = (5/4)16 = 20 teeth, and N5 = (15/4) 16 = 60 teeth. Thus, Shigley’s MED, 11th edition Chapter 13 Solutions, Page 13/36 N2 = 16 teeth, N3 = 64 teeth, N4 = 20 teeth, and N5 = 60 teeth Ans. ______________________________________________________________________________ 13-30 The geometry condition is d 5 / 2  d 2 / 2  d 3  d 4 . Since all the gears are meshed, they will all have the same diametral pitch. Applying d = N / P, N 5 / (2 P )  N 2 / (2 P )  N3 / P  N 4 / P N5  N 2  2 N3  2 N 4  12  2 16   2 12   68 teeth Ans. Let gear 2 be first, nF = n2 = 320 rev/min. Let gear 5 be last, nL = n5 = 0 rev/min. e 12  16  12  3 nL  nA      16  12  68  17 nF  nA 320  n A  nA   17  0  nA  3 3  320   68.57 rev/min 14  nA  68.57 rev/min cw Ans. The negative sign indicates opposite of n2. ______________________________________________________________________________ 13-31 Let nF = n2, then nL = n7 = 0. e nL  n5 20  16  36      0.5217  16  30  46  nF  n5 0  n5  0.5217 10  n5 0.5217 10  n5   n5 5.217  0.5217 n5  n5  0 n5 1  0.5217   5.217 5.217 1.5217 n5  nb  3.428 turns in same direction Ans. ______________________________________________________________________________ n5  13-32 (a)   2 n / 60 H  T   2 Tn / 60 Shigley’s MED, 11th edition (T in N  m, H in W) Chapter 13 Solutions, Page 14/36   60 H 103 So T So F32t  2 n  9550 H / n (H in kW, n in rev/min) 9550  75   398 N  m Ta  1800 mN 2 5 17  r2    42.5 mm 2 2 Ta 398   9.36 kN r2 42.5 F3b   Fb3  2  9.36   18.73 kN in the positive x-direction. (b) Ans. mN 4 5  51   127.5 mm 2 2 Tc 4  9.36 127.5   1193 N  m ccw r4  T4c  1193 N  m cw Ans. Note: The solution is independent of the pressure angle. ______________________________________________________________________________ N N 13-33 d  P 6 d 2  4 in, d 4  4 in, d5  6 in, d 6  24 in  24   24  36  e           1/ 6  24   36  144  nF  n2  1000 rev/min nL  n6  0 n  nA 0  nA 1   e L nF  nA 1000  n A 6 Shigley’s MED, 11th edition Chapter 13 Solutions, Page 15/36 n A  200 rev/min Noting that power equals torque times angular velocity, the input torque is T2   550 lbf  ft/s   60 s   1 rev  12 in  H 25 hp        1576 lbf  in hp n2 1000 rev/min    min   2 rad  ft  For 100 percent gear efficiency, the output power equals the input power, so Tarm  H 25 hp  550 lbf  ft/s   60 s   1 rev  12 in         7878 lbf  in hp n A 200 rev/min    min   2 rad  ft  Next, we’ll confirm the output torque as we work through the force analysis and complete the free body diagrams. Gear 2 1576  788 lbf 2 F32r  788 tan 20  287 lbf Wt  Gear 4 FA4  2W t  2  788   1576 lbf Gear 5 Shigley’s MED, 11th edition Chapter 13 Solutions, Page 16/36 Arm Tout  1576  9   1576  4   7880 lbf  in Ans. ______________________________________________________________________________ 13-34 (a)  = 20o, P = 6 teeth/in. Since N  d, then, N2 + N3 = N4 + N5 (1) e = 40 = 8  5. Then, let N 2 / N 3 = 8 and N 4 / N 5 = 5. Thus, N 2 = 8 N 3 and N 4 = 5 N 5. Substituting N 4 = 5 N 5 into Eq. (1) gives N2 + N3 = 6N5  N5 = (N2 + N3)/6. To avoid interference use Eq. (13-11) with m = N 2 / N 3 = 8: 2k N2  m  m 2  1  2m  sin 2  2 1  2m  sin    2 1   8  82  1  2  8   sin 2 20o  16.2 teeth 1  2  8   sin 2 20o Try N 3 = 17 teeth  N 2 = 8 N 3 = 8(17) = 136 teeth, N5 = (136 + 17)/6 = 25.5 teeth which is unacceptable. Next, try N 3 = 18 teeth  N 2 = 8 N 3 = 8(18) = 144 teeth, N5 = (144 + 18)/6 = 27 teeth, and N 4 = 5 N5 = 5 (27) = 135 teeth. Thus, N 2 = 144 teeth, N 3 = 18 teeth, N 4 = 135 teeth, and N5 = 27 teeth Ans. (b) Gear diameter is d = N / P (with P = 6 teeth/in), 0.5 in wall clearance on two sides, addendum of each gear is 1/P = 1/6 in, and each wall thickness is 0.75 in. Thus, N /2 N /2 N 1  2  0.5   2    2  0.75  Y 2 3  4 P P P P 144 18 / 2 135 / 2 1     2  0.5   2    2  0.75   39.6 in Ans. 6 6 6 6 ______________________________________________________________________________  13-35 (a)  = 25o, P = 6 teeth/in. Since N  d, then, N2 + N3 = N4 + N5 (1) e = 40 = 8  5. Let N 2 / N 3 = 8 and N 4 / N 5 = 5. Thus, N 2 = 8 N 3 and N 4 = 5 N 5. Substituting N 4 = 5 N 5 into Eq. (1) gives N2 + N3 = 6N5  N5 = (N2 + N3)/6. To avoid interference use Eq. (13-11) with m = N 2 / N 3 = 8: 2k N2  m  m 2  1  2m  sin 2  2 1  2m  sin    2 1   8  82  1  2  8  sin 2 25o  10.7 teeth 1  2  8   sin 2 25o Try N 3 = 11 teeth  N 2 = 8 N 3 = 8(11) = 88 teeth, N5 = (88 + 11)/6 = 16.5 teeth which is unacceptable. Next, try N 3 = 12 teeth  N 2 = 8 N 3 = 8(12) = 96 teeth, N5 = (96 + 12)/6 = 18 teeth, and N 4 = 5 N5 = 5 (18) = 90 teeth. Thus, N 2 = 96 teeth, N 3 = 12 teeth, N 4 = 90 teeth, and N5 = 18 teeth Ans.  Shigley’s MED, 11th edition Chapter 13 Solutions, Page 17/36 (b) Gear diameter is d = N / P (with P = 6 teeth/in), 0.5 in wall clearance on two sides, addendum of each gear is 1/P = 1/6 in, and each wall thickness is 0.75 in. Thus, N /2 N /2 N 1  2  0.5   2    2  0.75  Y 2 3  4 P P P P 96 12 / 2 90 / 2 1     2  0.5   2    2  0.75   27.3 in Ans. 6 6 6 6 ______________________________________________________________________________ 13-36 (a) n = 20o,  = 45o, P = 6 teeth/in. Since N  d, then, N2 + N3 = N4 + N5 (1) e = 40 = 8  5. Let N 2 / N 3 = 8 and N 4 / N 5 = 5. Thus N 2 = 8 N 3 and N 4 = 5 N 5. Substituting N 4 = 5 N 5 into Eq. (1) gives N2 + N3 = 6N5  N5 = (N2 + N3)/6. From Eq. (13-19), o  tan n  1  tan 20    20o tan t  tan 1    o   tan   tan 45  Eq. (13-22): 2k cos Np  m  m 2  1  2m  sin 2 t 2 1  2m  sin t   2 1 cos 45o   8  82  1  2  8  sin 2 20o  11.5 teeth 1  2 8   sin 2 20o Try N 3 = 12 teeth  N 2 = 8 N 3 = 8(12) = 96 teeth, N5 = (96 + 12)/6 = 18 teeth, and N 4 = 5 N5 = 5 (18) = 90 teeth. Thus, N 2 = 96 teeth, N 3 = 12 teeth, N 4 = 90 teeth, and N5 = 18 teeth Ans. (b) Gear diameter is d = N / P (with P = 6 teeth/in), 0.5 in wall clearance on two sides, addendum of each gear is 1/P = 1/6 in, and each wall thickness is 0.75 in. Thus, N /2 N /2 N 1  2  0.5   2    2  0.75  Y 2 3  4 P P P P 96 12 / 2 90 / 2 1     2  0.5   2    2  0.75   27.3 in Ans. 6 6 6 6 ______________________________________________________________________________  13-37 e  40, P = 6 teeth/in,  = 20o. (a) Minimum size is N 2 / N 3 = N 4 / N 5 = (13-11) with m = 6.325: N2    40 = 6.325. To avoid interference use Eq. 2k m  m 2  1  2m  sin 2  2 m   1 2 sin   2 1 6.325  1  2  6.325   sin 20 Shigley’s MED, 11th edition 2 o   6.2352  1  2  6.325   sin 2 20o  16.0 teeth Chapter 13 Solutions, Page 18/36 Let N3 = 16 teeth. N2 = 16 40 = 101.2. Use N2 = 101 teeth, e = (101/16)2 = 39.85 which is ok since it is between 38 and 42. Thus, N 2 = N 4 = 101 teeth, and N 3 = N5 = 16 teeth Ans. (b) Gear diameter is d = N / P (with P = 6 teeth/in), 0.5 in wall clearance on two sides, addendum of each gear is 1/P = 1/6 in, and each wall thickness is 0.75 in. Thus, N /2 N /2 N 1  2  0.5   2    2  0.75  Y 2 3  4 P P P P 101 116 / 2 101/ 2 1     2  0.5   2    2  0.75   29.4 in Ans. 6 6 6 6 Comparing this to Prob. 13-34 where Y = 39.6 in, we see a large reduction in gearbox size. ______________________________________________________________________________ 13-38 (a) P = 6 teeth/in,  = 20o. Since N  d, then, N2 N N N 30 60 20 80  N3  4  5  N 6  7   20    N6  2 2 2 2 2 2 2 2 Solving for N6 yields N6 = 15 teeth Ans. N N N / 2 N 7 / 2 2 60 20 30 / 2 80 / 2 2         22.83 in Ans. (b) Y  4  3  2 P P P P P 6 6 6 6 6 (c) Gear 2, n2 = 300 rev/min,  d n   N 2 / P  n2   30 / 6  300   392.7 ft/min Ans. Eq, (13-34): V2  2 2  12 12 12 H 4 Wt  33 000  33 000  336.1 lbf Ans. (d) Eq. (13-35): V 392.7 Ans, (e) Fr = Wt tan  = 336.1 tan20o = 122.3 lbf 63 025 H 63 025  4  (f)   70.0 lbf  in Ti  Ans. 12n 12  300  (g) e N 2 N 3 N 5 N 6 N 2 N 5 30  20     0.125 N 3 N 4 N 6 N 7 N 4 N 7 60  80  1  1  To  Ti    70.0  Ans.   560 lbf  ft e  0.125  Ans. (h) o = ei = 0.125 (300) = 37.5 rev/min (i) Assuming no losses, Po = Pi = 4 hp Ans. ______________________________________________________________________________ 13-39 Given: m = 12 mm, nP = 1800 rev/min cw, N2 = 18T, N3 = 32T, N4 = 18T, N5 = 48T Pitch Diameters: d2 = 18(12) = 216 mm, d3 = 32(12) = 384 mm, d4 = 18(12) = 216 mm, d5 = 48(12) = 576 mm Gear 2 Shigley’s MED, 11th edition Chapter 13 Solutions, Page 19/36 From Eq. (13-36), 60 000 150  60 000 H   7.368 kN Wt   dn   216 1800  d   216  Ta 2  Wt  2   7.368    795.7 N  m  2   2  W r  7.368 tan 20  2.682 kN Gears 3 and 4  384   216  Wt    7.368 2  2  t W  13.10 kN W r  13.10 tan 20  4.768 kN Ans. ______________________________________________________________________________ 13-40 Given: P = 5 teeth/in, N2 = 18T, N3 = 45T, n  20 , H = 32 hp, n2 = 1800 rev/min Gear 2 Tin  63025  32  1800  1120 lbf  in 18  3.600 in 5 45 dG   9.000 in 5 1120 W32t   622 lbf 3.6 / 2 W32r  622 tan 20  226 lbf dP  Fat2  W32t  622 lbf,  Fa 2  622 2  2262 Shigley’s MED, 11th edition  1/ 2 Far2  W32r  226 lbf  662 lbf Chapter 13 Solutions, Page 20/36 Each bearing on shaft a has the same radial load of RA = RB = 662/2 = 331 lbf. Gear 3 W23t  W32t  622 lbf W23r  W32r  226 lbf Fb 3  Fb 2  662 lbf RC  RD  662 / 2  331 lbf Each bearing on shaft b has the same radial load which is equal to the radial load of bearings A and B. Thus, all four bearings have the same radial load of 331 lbf. Ans. ______________________________________________________________________________ 13-41 Given: P = 4 teeth/in, n  20 , NP = 20T, n2 = 900 rev/min N P 20   5.000 in P 4 63025  30  2  Tin   4202 lbf  in 900 W32t  Tin /  d2 / 2   4202 /  5 / 2   1681 lbf d2  W32r  1681 tan 20  612 lbf The motor mount resists the equivalent forces and torque. The radial force due to torque is Shigley’s MED, 11th edition Chapter 13 Solutions, Page 21/36 Fr  4202  150 lbf 14  2  Forces reverse with rotational sense as torque reverses. The compressive loads at A and D are absorbed by the base plate, not the bolts. For W32t , the tensions in C and D are M AB  0 1681 4.875  15.25   2 F 15.25   0 F  1109 lbf If W32t reverses, 15.25 in changes to 13.25 in, 4.815 in changes to 2.875 in, and the forces change direction. For A and B, 1681 2.875   2 F1 13.25   0  F1  182.4 lbf For W32r , Shigley’s MED, 11th edition Chapter 13 Solutions, Page 22/36 M  612  4.875  11.25 / 2   6426 lbf  in a 14 / 2   11.25 / 2  F2  6426  179 lbf 4 8.98 2 2  8.98 in At C and D, the shear forces are: 2 2 2 2 FS 1  153  179  5.625 / 8.98    179  7 / 8.98   At A and B, the shear forces are: FS 2  153  179  5.625 / 8.98    179  7 / 8.98    145 lbf The shear forces are independent of the rotational sense. The bolt tensions and the shear forces for cw rotation are, For ccw rotation, ______________________________________________________________________________ 13-42 (a) N2 = N4 = 15 teeth, P N d  d N3 = N5 = 44 teeth N P 15  2.5 in Ans. 6 44 d3  d5   7.33 in Ans. 6 d2  d4  Shigley’s MED, 11th edition Chapter 13 Solutions, Page 23/36 (b) (c)  d 2 n2   2.5  2500   1636 ft/min Ans. 12  d n   2.5   2500 15 / 44   Vo  V4  V5  4 4   558 ft/min 12 12 Vi  V2  V3  Input gears: Wti  33000 12  H 33000  25    504.3 lbf  504 lbf Vi 1636 Wri  Wti tan   504.3 tan 20  184 lbf W 504.3 Wi  ti   537 lbf Ans. cos  cos 20 Output gears: H 33000  25   1478 lbf Wto  33000  Vo 558 Ans. Ans. Ans. Ans. Wro  Wto tan   1478 tan 20  538 lbf Ans. Wto 1478 Wo    1573 lbf Ans.  cos 20 cos 20 (d) d Ti  Wti  2  2   2.5    630 lbf  in   504.3    2  2 Ans. 2  44   44  To  Ti    630    5420 lbf  in Ans.  15   15  ______________________________________________________________________________ (e) 13-43 H  35 hp, ni  1200 rev/min,  =20 N 2  N 4  16 teeth, N 3  N 5  48 teeth, P  10 teeth/in N 16 (a) nintermediate  n3  n4  2 ni  1200   400 rev/min N3 48 (b) no  N2 N4 16  16  ni    1200   133.3 rev/min N3 N5 48  48  P N d  d Shigley’s MED, 11th edition Ans. N P 16  1.6 in 10 48 d3  d5   4.8 in 10 d2  d4  Ans. Ans. Ans. Chapter 13 Solutions, Page 24/36  d 2 n2  1.6 1200   502.7 ft/min Ans. 12 12  d n  1.6  400  Vo  V4  V5  4 4   167.6 ft/min Ans. 12 12 Vi  V2  V3  (c)  H 33000  35    2298 lbf lbf Vi 502.7 Wti  33000 Wri  Wti tan   2298 tan 20  836.4 lbf W 2298 Wi  ti   2445 lbf Ans. cos  cos 20 H 33000  35    6891 lbf Vo 167.6 Wto  33000 Ans. Ans. Ans. Wro  Wto tan   6891tan 20  2508 lbf Ans. Wto 6891 Wo    7333 lbf Ans.  cos 20 cos 20 (d) d   1.6  Ti  Wti  2   2298    1838 lbf  in  2   2  2 Ans. 2  48   48  (e) To  Ti    1838    16 540 lbf  in Ans.  16   16  ______________________________________________________________________________  2 13-44 (a) For o  , from Eq. (13-11), with m = 2, k = 1,   20 i 1 NP  2 1 2  1  2  2   sin 20 So N P min  15 (b) P 2   22  1  2  2   sin 2 20  14.16 Ans. N 15   1.875 teeth/in d 8 Ans. (c) To transmit the same power with no change in pitch diameters, the speed and transmitted force must remain the same. For A, with  = 20°, WtA = FA cos20° = 300 cos20° = 281.9 lbf For A, with  = 25°, same transmitted load, Shigley’s MED, 11th edition Chapter 13 Solutions, Page 25/36 FA = WtA/cos25° = 281.9/cos25° = 311.0 lbf Ans. Summing the torque about the shaft axis, d  d  WtA  A   WtB  B   2   2   d / 2   W  d A   281.9  20   704.75 lbf WtB  WtA A     d B / 2  tA  d B   8  WtB 704.75 FB    777.6 lbf Ans.  cos 25 cos 25 ______________________________________________________________________________ 13-45 (a) For o 5  , from Eq. (13-11), with m = 5, k = 1,   20 i 1 2 1 5  52  1  2  5   sin 2 25  10.4 NP  2  1  2  5  sin 25  So (b) N P min  11 m  Ans. d 300   27.3 mm/tooth N 11 Ans. (c) To transmit the same power with no change in pitch diameters, the speed and transmitted force must remain the same. For A, with  = 20°, WtA = FA cos20° = 11 cos20° = 10.33 kN For A, with  = 25°, same transmitted load, FA = WtA/cos25° = 10.33 / cos 25° = 11.40 kN Ans. Summing the torque about the shaft axis, d  d  WtA  A   WtB  B   2   2   d / 2   W  d A   11.40  600   22.80 kN WtB  WtA A     d B / 2  tA  d B   300  WtB 22.80   25.16 kN Ans.  cos 25 cos 25 ______________________________________________________________________________ FB  13-46 (a) Using Eq. (13-11) with k = 1, = 20º, and m = 2, Shigley’s MED, 11th edition Chapter 13 Solutions, Page 26/36 NP    2k m  m 2  1  2m  sin 2  2 1  2m  sin   2  1  2  2   sin  20  2 1 2   2 2     14.16 teeth  1  2  2   sin 2 20 Round up for the minimum integer number of teeth. NF = 15 teeth, NC = 30 teeth d 125   8.33 mm/tooth Ans. N 15 2 kW H  1000 W   rev   60 s    100 N  m T  191 rev/min  kW   2 rad   min  (b) m (c) (d) Ans. From Eq. (13-36), Wt  60 000  2  60 000 H   1.60 kN  1600 N  dn  125 191 Ans. Or, we could have obtained Wt directly from the torque and radius, Wt  100 T   1600 N d / 2 0.125 / 2 Wr  Wt tan   1600 tan 20  583 N Ans. Wt 1600 W   1700 N Ans. cos  cos 20 ______________________________________________________________________________ 13-47 (a) Using Eq. (13-11) with k = 1, = 20º, and m = 2, NP    2k m  m 2  1  2m  sin 2  2 1 2 sin   m    2  1  2  2   sin  20  2 1 2   2 2     14.16 teeth  1  2  2   sin 2 20 Round up for the minimum integer number of teeth. NC = 15 teeth, NF = 30 teeth Shigley’s MED, 11th edition Ans. Chapter 13 Solutions, Page 27/36 (b) P (c) T (d) N 30   3 teeth/in d 10 Ans.  550 lbf  ft/s   12 in   rev   60 s  1 hp      70 rev/min  hp   ft   2 rad   min  T  900 lbf  in Ans. H  From Eqs. (13-34) and (13-35),  dn  10  70   183.3 ft/min 12 H 33000 1  180 lbf Ans. Wt  33000  V 183.3 Wr  Wt tan   180 tan 20  65.5 lbf Ans. Wt 180 W   192 lbf Ans. cos  cos 20 ______________________________________________________________________________ N  d   1.30   13-48 (a) Eq. (13-14):   tan 1  P   tan 1  P   tan 1  Ans.   18.5 N d 3.88    G  G V 12   dn   2 1.30  600  (b) Eq. (13-34): V (c) Eq. (13-35): Wt  33 000 12  12  408.4 ft/min Ans. Eq. (13-38): H  10   33 000    808 lbf V  408.4  Wr  Wt tan  cos   808 tan 20 cos18.5  279 lbf Ans. Eq. (13-38): Wa  Wt tan  sin   808 tan 20 sin18.5  93.3 lbf Ans. Ans. The tangential and axial forces agree with Prob. 3-85, but the radial force given in Prob. 3-85 is shown here to be incorrect. Ans. ______________________________________________________________________________ 13-49   tan 1  2 / 4   26.565   tan 1  4 / 2   63.435 rav  2  1.5sin 26.5650  / 2  1.665 in Tin  63 025H / n  63025  2.5  / 240  656.5 lbf  in W t  T / rav  656.5 / 1.665  394.3 lbf   a  2  1.5 cos 26.565 / 2  2.671 in W r  394.3 tan 20 cos 26.565  128.4 lbf Shigley’s MED, 11th edition Chapter 13 Solutions, Page 28/36 W a  394.3 tan 20 sin 26.565  64.2 lbf W = 128.4i – 64.2j + 394.3k lbf RAG = –1.665i + 5.171j, RAB = 2.5j M 4  R AG  W + R AB  FB  T  0 Solving gives R AB  FB  2.5 FBz i  2.5 FBx k R AG  W  2039i  656.5 j  557.1k So,  2039i  656.5 j  557.1k    2.5 FBz i  2.5 FBxk  Tj  0 FBz  2039 / 2.5  815.6 lbf T  656.5 lbf  in FBx  557.1 / 2.5  222.8 lbf So, FB = 222.8i 815.6k lbf Ans. 1/2 2 2 FB   222.88    815.6      845.5 lbf FA = – (FB + W) = – (–222.8i – 815.6k + 128.4i – 64.2j + 394.3k) = 94.4i + 64.2j + 421.3k Ans.  FA (radial)  94.4 2  421.32  1/ 2  431.7 lbf FA (thrust)  64.2 lbf ______________________________________________________________________________ 13-50 d 2  18 / 10  1.8 in, d3  30 / 10  3.0 in  d2 / 2  1  0.9     tan    30.96 d / 2 1.5    3    tan 1    90    59.04 rav  3.0 / 2   0.5sin 59.04o  / 2  1.286 in 9  0.5cos 59.04 / 2  0.6911 in 16 t W  25 lbf W r  25 tan 20 cos 59.04  4.681 lbf DE    W a  25 tan 20 sin 59.04  7.803 lbf W = –4.681i – 7.803j +25k Shigley’s MED, 11th edition Chapter 13 Solutions, Page 29/36 RDG = 1.286i + 0.6911j RDC = –0.625j M D  R DG  W  R DC  FC  T  0 R DG  W  17.28i  32.15 j  6.800k R DC  FC  0.625 FCz i  0.625 FCx k 17.28i  32.15 j  6.800k    0.625 FCz i  0.625 FCxk   Tj  0 T  32.15 lbf  in Ans. FC  10.88i  27.65k lbf  FC  10.882  27.652 F  0  1/ 2 Ans.  29.7 lbf Ans. FD  6.20i  7.80 j  52.65k lbf 2 1/2 FD (radial)   6.20    52.65    53.0 lbf Ans.   a FD (thrust)  W  7.80 lbf Ans. ______________________________________________________________________________ 2 13-51 NOTE: The shaft forces exerted on the gears are not shown in the figures above. Pt  Pn cos  5 cos 30  4.330 teeth/in t  tan 1 dP  tan n tan 20  tan 1  22.80 cos cos 30 18  4.157 in 4.330 The forces on the shafts will be equal to the forces transmitted to the gears through the meshing teeth. Shigley’s MED, 11th edition Chapter 13 Solutions, Page 30/36 Pinion (Gear 2) W r  W t tan t  800 tan 22.80  336 lbf W a  W t tan  800 tan 30  462 lbf W  336i  462 j  800k lbf Ans. 1/2 W   336    462   800 2    2 Gear 3 2  983 lbf Ans. W  336i  462 j  800k lbf Ans. W  983 lbf Ans. 32 dG   7.390 in 4.330 TG  W t r  800  7.390   5912 lbf  in ______________________________________________________________________________  tan n 1 tan 20 t  tan  tan  22.80 13-52  cos cos 30 Pinion (Gear 2) W r  W t tan t  800 tan 22.80  336 lbf 1 W a  W t tan  800 tan 30  462 lbf W  336i  462 j  800k lbf Ans. 1/2 2 2 2 W   336    462    800      983 lbf Ans. Idler (Gear 3) From the diagram for the idler, noting that the radial and axial forces from gears 2 and 4 cancel each other, the force acting on the shaft is W  1600k lbf Ans. Output gear (Gear 4) W  336i  462 j  800k lbf 2 1/2 2 2 W   336    462    800     Ans.  983 lbf Ans. NOTE: For simplicity, the above figures only show the gear contact forces. Shigley’s MED, 11th edition Chapter 13 Solutions, Page 31/36 Also, notice that the idler shaft reaction contains a couple tending to turn the shaft endover-end. Also the idler teeth are bent both ways. Idlers are more severely loaded than other gears, belying their name. Thus, be cautious. ______________________________________________________________________________ 13-53 Gear 3: Pt  Pn cos  7 cos 30  6.062 teeth/in tan t  tan 20  0.4203, t  22.8  cos 30 54 d3   8.908 in 6.062 W t  500 lbf W a  500 tan 30  288.7 lbf W r  500 tan 22.8  210.2 lbf W3  210.2i  288.7 j  500k lbf Ans. Gear 4: 14 d4   2.309 in 6.062 8.908 W t  500  1929 lbf 2.309 W a  1929 tan 30  1114 lbf W r  1929 tan 22.8  811 lbf W4  811i  1114 j  1929k lbf Ans. ______________________________________________________________________________ 13-54 Pt  6 cos 30  5.196 teeth/in 42 d3   8.083 in 5.196 t  22.8 16 d2   3.079 in 5.196 63025  25  T2   916 lbf  in 1720 Shigley’s MED, 11th edition Chapter 13 Solutions, Page 32/36 T 916   595 lbf r 3.079 / 2 W a  595 tan 30  344 lbf W r  595 tan 22.8  250 lbf Wt  W  344i  250 j  595k lbf R DC  6i, (1) R DG  3i  4.04 j M D  R DC  FC  R DG  W  T  0 R DG  W  2404i  1785 j  2140k R DC  FC  6 FCz j  6 FCy k Substituting and solving Eq. (1) gives T  2404i lbf  in FCz  297.5 lbf FCy  365.7 lbf F  FD  FC  W  0 Substituting and solving gives FCx  344 lbf FDy  106.7 lbf FDz  297.5 lbf FC  344i  356.7 j  297.5k lbf Ans. FD  106.7 j  297.5k lbf Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 13 Solutions, Page 33/36 13-55 Since the transverse pressure angle is specified, we will assume the given module is also in terms of the transverse orientation. d 2  mN 2  4 16   64 mm d3  mN3  4  36   144 mm d 4  mN 4  4  28  112 mm 6 kW  1000 W   rev   60 s   35.81 N  m  1600 rev/min  kW   2 rad   min  35.81 T Wt    1119 N d 2 / 2 0.064 / 2 T H  Shigley’s MED, 11th edition Chapter 13 Solutions, Page 34/36 W r  W t tan t  1119 tan 20  407.3 N W a  W t tan  1119 tan15  299.8 N F2 a  1119i  407.3 j  299.8k N Ans. F3b  1119  407.3 i  1119  407.3 j  711.7i  711.7 j N Ans. F4c  407.3i  1119 j  299.8k N Ans. ______________________________________________________________________________ 13-56 14 36 N   2.021 in, d3   5.196 in  Pn cos 8 cos 30 8cos 30 15 45 d4   3.106 in, d 5   9.317 in  5 cos15 5cos15 d2  For gears 2 and 3: t  tan 1  tan n / cos   tan 1  tan 20 / cos 30   22.8 For gears 4 and 5: t  tan 1  tan 20 / cos15   20.6 F23t  T2 / r2  1200 /  2.021/ 2   1188 lbf 5.196  1987 lbf 3.106 F23r  F23t tan t  1188 tan 22.8  499 lbf F54t  1188 F54r  1986 tan 20.6  746 lbf F23a  F23t tan   1188 tan 30  686 lbf F54a  1986 tan15  532 lbf Next, designate the points of action on gears 4 and 3, respectively, as points G and H, as shown. Position vectors are Shigley’s MED, 11th edition Chapter 13 Solutions, Page 35/36 R CG  1.553 j  3k R CH  2.598 j  6.5k R CD  8.5k Force vectors are F54  1986i  748 j  532k F23  1188i  500 j  686k FC  FCx i  FCy j FD  FDx i  FDy j  FDz k Now, a summation of moments about bearing C gives M C  R CG  F54  R CH  F23  R CD  FD  0 The terms for this equation are found to be R CG  F54  1412i  5961j  3086k R CH  F23  5026i  7722 j  3086k R CD  FD  8.5 FDy i  8.5 FDx j When these terms are placed back into the moment equation, the k terms, representing the shaft torque, cancel. The i and j terms give 3614  425 lbf 8.5 13683 FDx   1610 lbf 8.5 Next, we sum the forces to zero. FDy   F  FC  F54  F23  FD  0 Substituting, gives F x C  i  FCy j   1987i  746 j  532k    1188i  499 j  686k     1610i  425 j  FDz k  0 Solving gives FCx  1987  1188  1610  1565 lbf FCy  746  499  425  672 lbf FDz  532  686  154 lbf FC = 1565i + 672j lbf Ans. FD = 1610i  425j + 154k lbf Ans. ______________________________________________________________________________ So, Shigley’s MED, 11th edition Chapter 13 Solutions, Page 36/36 13-57 VW  WW t  dW nW    0.100  600  60 60 H 2000    637 N VW    m/s L  px NW  25 1  25 mm   tan 1  tan 1 L  dW 25  4.550 lead angle  100  WW t W cos n sin   f cos  V   3.152 m/s VS  W  cos  cos 4.550 In ft/min: VS = 3.28(3.152) = 10.33 ft/s = 620 ft/min Use f = 0.043 from curve A of Fig. 13-38. Then, from the first of Eq. (13-43) W 637  5323 N cos14.5 sin 4.55  0.043cos 4.55    W y  W sin n  5323sin14.5  1333 N   W z  5323 cos14.5 cos 4.55  0.043sin 4.55   5119 N The force acting against the worm is W  637i  1333 j  5119k N Thus, A is the thrust bearing. Ans. R AG  0.05 j  0.10k m, R AB  0.20k m M A  R AG  W  R AB  FB  T  0 R AG  W  122.6i  63.7 j  31.85k N  m R AB  FB  0.2 FBy i  0.2 FBx j Substituting and solving gives T  31.85 N  m Ans. FBx  318.5 N, FBy  613 N So FB  318.5i  613 j N Shigley’s MED, 11th edition Ans. Chapter 13 Solutions, Page 37/36 1/2 2 2 FB   613   318.5     F  FA  W  R B  0 Or  691 N radial FA    W  FB     637i  1333 j  5119k  318.5i  613j  318.5i  1946 j  5119k Radial Ans. FAr  318.5i  1946 j N 2 1/2 FAr   318.5    1946    1972 N   a Thrust FA  5119 N ______________________________________________________________________________ 2 13-58 From Prob. 13-57, WG  637i  1333 j  5119k N pt  p x So dG  N G px   48  25    382 mm Bearing D takes the thrust load. M D  R DG  WG  R DC  FC  T  0 R DG  0.0725i  0.191j m R DC  0.1075i m R DG  WG  977.7i  371.1j  25.02k N  m R DC  FC  0.1075 FCz j  0.1075 FCy k N  m Putting it together and solving, T  977.7 N  m Ans. FC  233 j  3450k N, F  FC  WG  FD  0 FC  3460 N Ans. FD    FC  WG   637i  1566 j  1669k N Ans. Radial FDr  1566 j  1669k N Or  FDr  1566 2  1669 2  1/ 2  2289 N (total radial) FDt  637i N (thrust) ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 13 Solutions, Page 38/36 13-59  1.5  600   235.7 ft/min 12 33000  0.75  W x  WW t   105.0 lbf 235.7  pt  px   0.3927 in 8 L  0.3927  2   0.7854 in VW  0.7854  9.46  1.5  105.0 W  515.3 lbf  cos 20 sin 9.46  0.05 cos 9.46 W y  515.3sin 20  176.2 lbf W z  515.3 cos 20  cos 9.46   0.05sin 9.46   473.4 lbf   tan 1 So W  105i  176 j  473k lbf T  105  0.75  78.8 lbf  in Ans. Ans. ______________________________________________________________________________ 13-60 Computer programs will vary. Shigley’s MED, 11th edition Chapter 13 Solutions, Page 39/36 Chapter 9 Figure for Probs. 9-1 to 9-4 9-1 Given, b = 50 mm, d = 50 mm, h = 5 mm, allow = 140 MPa. F = 0.707 hlallow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans. ______________________________________________________________________________ 9-2 Given, b = 2 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. F = 0.707 hlallow = 0.707(5/16)[2(2)](25) = 22.1 kip Ans. ______________________________________________________________________________ 9-3 Given, b = 50 mm, d = 30 mm, h = 5 mm, allow = 140 MPa. F = 0.707 hlallow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans. ______________________________________________________________________________ 9-4 Given, b = 4 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. F = 0.707 hlallow = 0.707(5/16)[2(4)](25) = 44.2 kip Ans. ______________________________________________________________________________ 9-5 Prob. 9-1 with E7010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in = 2.923(4.45/25.4) = 0.512 kN/mm F = f l = 0.512[2(50)] = 51.2 kN Ans. ______________________________________________________________________________ 9-6 Prob. 9-2 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in Shigley’s MED, 11th edition Chapter 9 Solutions, Page 1/39 F = f l = 4.64[2(2)] = 18.6 kip Ans. ______________________________________________________________________________ 9-7 Prob. 9-3 with E7010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in = 2.923(4.45/25.4) = 0.512 kN/mm F = f l = 0.512[2(50)] = 51.2 kN Ans. ______________________________________________________________________________ 9-8 Prob. 9-4 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64[2(4)] = 37.1 kip Ans. ______________________________________________________________________________ 9-9 Table A-20: 1018 CD: Sut = 440 MPa, Sy = 370 MPa 1018 HR: Sut = 400 MPa, Sy = 220 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4:  all  min(0.30Sut , 0.40S y )  min[0.30(400), 0.40(220)]  min(120, 88)  88 MPa for both materials. Eq. (9-3): F = 0.707hlall = 0.707(5)[2(50)](88)(103) = 31.1 kN Ans. ______________________________________________________________________________ 9-10 Table A-20: 1020 CD: Sut = 68 kpsi, Sy = 57 kpsi 1020 HR: Sut = 55 kpsi, Sy = 30 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4:  all  min(0.30Sut , 0.40S y )  min[0.30(55), 0.40(30)]  min(16.5, 12.0)  12.0 kpsi for both materials. Eq. (9-3): F = 0.707hlall = 0.707(5/16)[2(2)](12.0) = 10.6 kip Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 9 Solutions, Page 2/39 9-11 Table A-20: 1035 HR: Sut = 500 MPa, Sy = 270 MPa 1035 CD: Sut = 550 MPa, Sy = 460 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4:  all  min(0.30Sut , 0.40S y )  min[0.30(500), 0.40(270)]  min(150, 108)  108 MPa for both materials. Eq. (9-3): F = 0.707hlall = 0.707(5)[2(50)](108)(103) = 38.2 kN Ans. ______________________________________________________________________________ 9-12 Table A-20: 1035 HR: Sut = 72 kpsi, Sy = 39.5 kpsi 1020 CD: Sut = 68 kpsi, Sy = 57 kpsi, 1020 HR: Sut = 55 kpsi, Sy = 30 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4:  all  min(0.30Sut , 0.40S y )  min[0.30(55), 0.40(30)]  min(16.5, 12.0)  12.0 kpsi for both materials. Eq. (9-3): F = 0.707hlall = 0.707(5/16)[2(4)](12.0) = 21.2 kip Ans. ______________________________________________________________________________ 9-13 2 100  103  2F    141 MPa Ans. Eq. (9-3): hl 5  2  50  50   ______________________________________________________________________________ 9-14 Eq. (9-3):  2  40  2F   22.6 kpsi hl  5 /16   2  2  2  Ans. ______________________________________________________________________________ 2 100  103  2F   177 MPa Ans. 9-15 Eq. (9-3):   hl 5  2  50  30   ______________________________________________________________________________ 9-16 Eq. (9-3):   2  40  2F   15.1 kpsi hl  5 /16   2  4  2  Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 9 Solutions, Page 3/39 9-17 A = (throat area)(length) = 0.707h (b + d) Ans. b b2 x b  d   0  d   b   x 2 2 b  d  y b  d   0 b  d d  2  x d2 2 b  d  Ans. Ans. For line b:  J u b 2  b  b3  b2  b3  2   b   x   y    b   bx  x 2  y 2  12   4   2  12  b2  4b3  b  d 2  6b 4  b  d   3b5  3bd 4 b3 b3 b4 d4  b     2 2 2 12 12  b  d   4 2  b  d  4  b  d  4  b  d   Similarly,  J u d  4d 3  b  d   6d 4  b  d   3d 5  3b 4 d 2 12  b  d  J u   J u b   J u  d 2 4b3  b  d   6b 4  b  d   3b5  3bd 4  4d 3  b  d   6d 4  b  d   3d 5  3b 4 d 2  2 12  b  d  2 4b3  b  d   6b 4  b  d   3b 4  b  d   3d 4  b  d   4d 3  b  d   6d 4  b  d  2  2 12  b  d  2 This reduces to b 4  4b3 d  d 4  4bd 3 1 12  b  d  Add and subtract 6b2d 2 to Eq. (1) giving b 4  4b3d  6b 2 d 2  4bd 3  d 4   6b 2 d 2  Ju  12  b  d  Ju   b  d   6b 2 d 2  12  b  d  4 which is the same as Table 9  1 Ans. ______________________________________________________________________________ 9-18 A = (throat area) (length) = 0.707h (2b + d) Ans. d b2 x  2b  d   0  d    b   x Ans. 2 2b  d d  d b   2 d d  y  2b  d   0  b   d    d  b   y   Ans.  2b  d 2 2 For lines b: Shigley’s MED, 11th edition Chapter 9 Solutions, Page 4/39  J u b 2  b   b3  b2   b3  2   2   b   x   y     2b   bx  x 2  y 2    4   2   6 12 2b 4 2b5 2b 4 2b5 b3 b3 bd 2 2b3 bd 2          6 2 2b  d  2b  d 2 2 3 2b  d  2b  d 2 2 Line d: d3 d3 b4 d  J u d   d x 2   12 12  2b  d  2 Ju = (Ju)b + (Ju)d bd 2 d 3 b4 d 2b 3 2b 4 2b5 Ju       3 2b  d  2b  d 2 2 12  2b  d 2 4 5 4 8b3  6bd 2  d 3 2b  2b  d   2b  b d   2 12  2b  d   4 4 8b3  6bd 2  d 3 2b  2b  d   b  2b  d   2 12  2b  d  b4 8b3  6bd 2  d 3 Ans.  12 2b  d ______________________________________________________________________________  9-19 b = d =50 mm, c = 150 mm, h = 5 mm, and allow = 140 MPa. (a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kN and  in MPa): F 103  V  2.829 F  y   A 1.414  5 50  Secondary shear, Table 9-1: Ju  d  3b 2  d 2  6  50 3  502   502  6  83.33 103  mm3 J = 0.707 h Ju = 0.707(5)(83.33)(103) = 294.6(103) mm4  x   y  Mry J  175 F 103   25  294.6 103   14.85 F  max   x2   y   y   F 14.852   2.829  14.85   23.1F (1) 2 Shigley’s MED, 11th edition 2 Chapter 9 Solutions, Page 5/39 F  allow 23.1 140  6.06 kN Ans. 23.1  (b) For E7010 from Table 9-6, allow = 21 kpsi = 21(6.89) = 145 MPa 1020 HR bar: Sut = 380 MPa, Sy = 210 MPa 1015 HR support: Sut = 340 MPa, Sy = 190 MPa Table 9-3, E7010 Electrode: Sut = 482 MPa, Sy = 393 MPa The support controls the design. Table 9-4: allow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76) = 76 MPa The allowable load, from Eq. (1) is  allow 76  3.29 kN Ans. 23.1 23.1 ______________________________________________________________________________ F 9-20  b = d =2 in, c = 6 in, h = 5/16 in, and allow = 25 kpsi. (a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and  in kpsi): V F   1.132 F A 1.414  5 /16  2  Secondary shear, Table 9-1: 2 2 d  3b2  d 2  2 3  2   2    5.333 in 3 Ju  6 6  y  J = 0.707 h Ju = 0.707(5/16)(5.333) = 1.178 in4  x   y  Mry J  7 F 1 1.178  5.942 F  max   x2   y   y   F 5.942 2  1.132  5.942   9.24 F (1) 2 F Shigley’s MED, 11th edition  allow 9.24  2 25  2.71 kip Ans. 9.24 Chapter 9 Solutions, Page 6/39 (b) For E7010 from Table 9-6, allow = 21 kpsi 1020 HR bar: Sut = 55 kpsi, Sy = 30 kpsi 1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi The support controls the design. Table 9-4: allow = min(0.30Sut, 0.40Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11) = 11 kpsi The allowable load, from Eq. (1) is  allow 11  1.19 kip Ans. 9.24 9.24 ______________________________________________________________________________ F 9-21  b =50 mm, c = 150 mm, d = 30 mm, h = 5 mm, and allow = 140 MPa. (a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kN and  in MPa):  y  F 103  V   2.829 F A 1.414  5 50  Secondary shear, Table 9-1: Ju  d  3b 2  d 2  6  50 3  302   502  6  43.33 103  mm3 J = 0.707 h Ju = 0.707(5)(43.33)(103) = 153.2(103) mm4  x  Mry J  175 F 103  15  153.2 103   17.13F Mrx 175 F 10   25    28.55 F J 153.2 103  3  y   max   x2   y   y   F 17.132   2.829  28.55   35.8 F (1) 2 Shigley’s MED, 11th edition 2 Chapter 9 Solutions, Page 7/39 F  allow 35.8 140  3.91 kN 35.8  Ans. (b) For E7010 from Table 9-6, allow = 21 kpsi = 21(6.89) = 145 MPa 1020 HR bar: Sut = 380 MPa, Sy = 210 MPa 1015 HR support: Sut = 340 MPa, Sy = 190 MPa Table 9-3, E7010 Electrode: Sut = 482 MPa, Sy = 393 MPa The support controls the design. Table 9-4: allow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76) = 76 MPa The allowable load, from Eq. (1) is  allow 76  2.12 kN Ans. 35.8 35.8 ______________________________________________________________________________ F 9-22  b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, and allow = 25 kpsi. (a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and  in kpsi): V F   0.5658 F A 1.414  5 /16  4  Secondary shear, Table 9-1:  y  Ju  d  3b 2  d 2  6  4 3  22   42  6  18.67 in 3 J = 0.707 h Ju = 0.707(5/16)(18.67) = 4.125 in4  x   y  Mry J  8 F 1 4.125  1.939 F Mrx 8 F  2    3.879 F J 4.125  max   x2   y   y   F 1.9392   0.5658  3.879   4.85 F (1) 2 Shigley’s MED, 11th edition 2 Chapter 9 Solutions, Page 8/39 F  allow 4.85  25  5.15 kip Ans. 4.85 (b) For E7010 from Table 9-6, allow = 21 kpsi 1020 HR bar: Sut = 55 kpsi, Sy = 30 kpsi 1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi The support controls the design. Table 9-4: allow = min(0.30Sut, 0.40Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11) = 11 kpsi The allowable load, from Eq. (1) is  allow 11  2.27 kip Ans. 4.85 4.85 ______________________________________________________________________________ F 9-23  A = (throat area)(length) = 0.707h (2b + d) Ans. d b2 x  2b  d   0  d    b   x Ans. 2 2b  d d  d b   2 d d  y  2b  d   0  b   d    d  b   y   Ans.  2b  d 2 2 2 d3 d2 d   2b    Ans.  6b  d  12  2  12 ______________________________________________________________________________ 9-24 A = (throat area)(length) = 0.707h (b + 2d) Ans. b  b  d  b 2 b x  b  2d   0  d    b   b  d   x   Ans. b  2d 2 2 d d2 y  b  2d   0  b    2 d   y Ans. 2 b  2d Iu  Shigley’s MED, 11th edition Chapter 9 Solutions, Page 9/39 2 I u  by 2  d3 d3 2d 3 d   2d   y     b y 2  2d 2 y  2d y 2 12 6 2 2  2 3 d  2d 2 y   b  2 d  y 2 Ans. 3 ______________________________________________________________________________  9-25 Given, b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, allow = 140 MPa. Primary shear (F in kN,  in MPa, A in mm2): F 103  V   1.414 F y   A 1.414  5  50  50  Secondary shear: b  d   3 Table 9-1:  50  50   3  166.7 103  mm3 6 6 J = 0.707 h Ju = 0.707(5)166.7(103) = 589.2(103) mm4 Ju  x   y  Mry J  175 F 103  (25) 589.2 103   7.425 F Maximum shear:  max   x2   y   y   F 7.4252  1.414  7.425   11.54 F 2  allow 2 140  12.1 kN Ans. 11.54 11.54 ______________________________________________________________________________ F 9-26  Given, b = 2 in, c = 6 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. Primary shear:  y  Secondary shear: Table 9-1: V F   0.5658 F A 1.414  5 /16  2  2  b  d   Shigley’s MED, 11th edition 2  2  3  10.67 in 3 6 6 J = 0.707 h Ju = 0.707(5/16)10.67 = 2.357 in4 Ju  x   y  Maximum shear: 3 Mry J  7 F (1)  2.970 F 2.357 Chapter 9 Solutions, Page 10/39  max   x2   y   y   F 2.9702   0.566  2.970   4.618 F 2  allow 2 25  5.41 kip Ans. 4.618 4.618 ______________________________________________________________________________ F 9-27  Given, b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, allow = 140 MPa. Primary shear (F in kN,  in MPa, A in mm2): F 103  V  1.768 F  y   A 1.414  5  50  30  Secondary shear: Table 9-1: Ju b  d   3  50  30   3  85.33 103  mm3 6 6 J = 0.707 h Ju = 0.707(5)85.33(103) = 301.6(103) mm4  x   y  Mry J  175 F 103  (15) 301.6 103   8.704 F 3 Mrx 175 F 10  (25)   14.51F J 301.6 103  Maximum shear:  max   x2   y   y   F 8.7042  1.768  14.51  18.46 F 2  allow 2 140  7.58 kN Ans. 18.46 18.46 ______________________________________________________________________________ F 9-28  Given, b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. Primary shear:  y  Secondary shear: Table 9-1: Ju b  d   3  4  2  3  36 in 3 6 6 J = 0.707 h Ju = 0.707(5/16)36 = 7.954 in4  x  Shigley’s MED, 11th edition V F   0.3772 F A 1.414  5 /16  4  2  Mry J  8 F (1)  1.006 F 7.954 Chapter 9 Solutions, Page 11/39  y  Maximum shear: Mrx 8 F (2)   2.012 F J 7.954  max   x2   y   y   F 1.006 2   0.3772  2.012   2.592 F 2  allow 2 25  9.65 kip Ans. 2.592 2.592 ______________________________________________________________________________ 9-29 Given: b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, allow = 140 MPa. F 103  sin 45o V Primary shear (F in kN):  x   y   F A 1.414  5  50  50  F  Secondary shear: M = 0.707 F (103)(175  25) = 106.05(103) F   Table 9-1: J u   b  d  / 6   50  50  / 6  166.7 103 mm3 3 3 J  0.707 h J u  0.707  5 166.7 103   589.3 103  mm 4 Mry  x   y  J  106.5 103  F  25  589.3 103   4.50 F Upper right end of weld:  max   x   x    y   y  2 2  2  F  4.5 F   7.778 F 2 140  18.0 kN Ans. 7.778 ______________________________________________________________________________ 9-30 Given: b = 2 in, c = 6 in, d = 2 in, h = 165 in, allow = 25 kpsi.  max   allow F  Primary shear (F in kip) :  x   y  0.707 F V   0.4 F A 1.414 5  2  2    16 Secondary shear: M = 0.707 F (7  1) = 4.242 F 3 3 Table 9-1: J u   b  d  / 6   2  2  / 6  10.667 in 3 J  0.707 h J u  0.707 Mry  x   y  Upper right end of weld:  max  J   10.667  2.357 in 5 16 4.242 F 1 2.357  x   x    y   y  2 2 4  1.80 F  2  0.4 F  1.80 F   3.111F 2 25  8.04 kip Ans. 3.111 ______________________________________________________________________________  max   allow Shigley’s MED, 11th edition  F Chapter 9 Solutions, Page 12/39 9-31 Given: b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, allow = 140 MPa. F 103  sin 45o V Primary shear (F in kN):  x   y    1.25 F A 1.414  5  50  30  Secondary shear: M = 0.707 F (103)(175  15) = 113.12 (103) F   Table 9-1: J u   b  d  / 6   50  30  / 6  85.33 103 mm3 3 3 J  0.707 h J u  0.707  5  85.33 103   301.65 103  mm 4  x  Mry  J 113.12 103  F 15  301.65 103   5.625 F 3 Mrx 113.12 10  F  25  y    9.375 F J 301.65 103  Upper right end of weld:  x   x    y   y  2 2 2  F 1.25  5.625   1.25  9.375    12.66 F   140  max   allow  F  11.1 kN Ans. 12.66 ______________________________________________________________________________ 9-32 Given: b = 4 in, c = 6 in, d = 2 in, h = 165 in, allow = 25 kpsi.  max  2 Primary shear (F in kip) :  x   y  0.707 F V   0.2667 F A 1.414 5  4  2    16 Secondary shear: M = 0.707 F (8  1) = 4.949 F 3 3 Table 9-1: J u   b  d  / 6   4  2  / 6  36 in 3 J  0.707 h J u  0.707   36  7.954 in 5 16 4 4.949 F 1  0.6222 F J 7.954 Mr 4.949 F  2   1.244 F  y  x  J 7.954 Upper right end of weld:  x  Mry   x   x    y   y  2 2 2  F  0.2667  0.6222    0.2667  1.244    1.753F   25  max   allow  F  14.3 kip Ans. 1.753 ______________________________________________________________________________  max  9-33 2 Given, b = 50 mm, d = 50 mm, h = 5 mm, E6010 electrode. Shigley’s MED, 11th edition Chapter 9 Solutions, Page 13/39 A = 0.707(5)(50 +50 + 50) = 530.3 mm2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa. Eq. (6-18) and Table 6-2: ka = 54.9(320)0.758 = 0.693 kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-10) and (6-17): Se = 0.693(1)(0.59)(1)(0.5)(320) = 65.4 MPa Electrode endurance: E6010, Table 9-3, Eq. (6-18) and Table 6-2: Sut = 427 MPa ka = 54.9(427)0.758 = 0.557 As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.557(1)(0.59)(1)(0.5)(427) = 70.2 MPa The members and electrode are basically of equal strength. We will use Se = 65.4 MPa. For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5)  A 65.4  530.3  F  allow   12.8 103  N  12.8 kN Ans. K fs 2.7 ______________________________________________________________________________ 9-34 Given, b = 2 in, d = 2 in, h = 5/16 in, E6010 electrode. A = 0.707(5/16)(2 +2 + 2) = 1.326 in2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi. Eq. (6-18) and Table 6-2: ka = 12.7(47)0.758 = 0.686 kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-10) and (6-17): Se = 0.686(1)(0.59)(1)(0.5)(47) = 9.51 kpsi Electrode endurance: E6010, Table 9-3, Sut = 62 kpsi Eq. (6-18) and Table 6-2: ka = 12.7(62)0.758 = 0.556 As before, kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.556(1)(0.59)(1)(0.5)(62) = 10.2 kpsi For a factor of safety of 1, with Kfs = 2.7 (Table 9-5), using Se = 9.51 kpsi  A 9.511.326  F  allow   4.67 kip Ans. K fs 2.7 ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 9 Solutions, Page 14/39 9-35 Given, b = 50 mm, d = 30 mm, h = 5 mm, E7010 electrode. A = 0.707(5)(50 +50 + 30) = 459.6 mm2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa. Eq. (6-18) and Table 6-2: ka = 54.9(320)0.758 = 0.693 kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-10) and (6-17): Se = 0.693(1)(0.59)(1)(0.5)(320) = 65.4 MPa Electrode endurance: E6010, Table 9-3, Eq. (6-18) and Table 6-2: Sut = 482 MPa ka = 54.9(482)0.758 = 0.508 As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.508(1)(0.59)(1)(0.5)(482) = 72.2 MPa For a factor of safety of 1,with Kfs = 2.7 (Table 9-5), and using Se =65.4 MPa  A 65.4  459.6  F  allow   11.1103  N  11.1 kN Ans. K fs 2.7 ______________________________________________________________________________ 9-36 Given, b = 4 in, d = 2 in, h = 5/16 in, E7010 electrode. A = 0.707(5/16)(4 +4 + 2) = 2.209 in2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi. Eq. (6-18) and Table 6-2: ka = 12.7(47)0.758 = 0.686 kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-10) and (6-17): Se = 0.686(1)(0.59)(1)(0.5)(47) = 9.51 kpsi Electrode endurance: E7010, Table 9-3, Eq. (6-18) and Table 6-2: Sut = 70 kpsi ka = 12.7(70)0.758 = 0.507 As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.507(1)(0.59)(1)(0.5)(70) = 10.5 kpsi For a factor of safety of 1, with Kfs = 2.7 (Table 9-5), and using Se = 9.51 kpsi Shigley’s MED, 11th edition Chapter 9 Solutions, Page 15/39 9.51 2.209   7.78 kip Ans. K fs 2.7 ______________________________________________________________________________ 9-37 Primary shear:  = 0 (why?) Secondary shear: F  allow A  Table 9-1: Ju = 2 r3 = 2 (1.5)3 = 21.21 in3 J = 0.707 h Ju = 0.707(1/4)(21.21) = 3.749 in4 2 welds:    Mr 8 F 1.5    1.600 F 2 J 2  3.749      allow  1.600 F  20  F  12.5 kip Ans. ______________________________________________________________________________ 9-38 Direct shear Vz = F, Torsion Tx = 8F, Bending My = 6F B: Direct shear: V F F   0.6014 F  V   z  z  A 1.414 hr 1.414 1 1.5  4 Torsion, Table 9-1: Ju = 2r3 = 2 (1.5)3 = 21.21 in3 J = 0.707h Ju = 0.707 ( 14 ) 21.21 = 3.749 in4 Tr  T   z  x  8 F 1.5   3.201F J 3.749  max   z   z  0.6014 F  3.201F  3.802 F  20 F  allow   5.26 kip Ans. 3.802 3.802 C: Direct shear, torsion, and bending. Bending, Table 9-2: Iu = r3 =  (1.5)3 = 10.603 in3, I = 0.707h Iu = 0.707( 14 ) 10.603 =1.874 in4.   M  max   Mc 6 F 1.5    4.803F I 1.874  V   T    M 2 2 2  allow   0.6014 F  2   3.201F    4.803F   5.803F 2 2 20  3.45 kip Ans. 5.803 5.803 ______________________________________________________________________________ F Shigley’s MED, 11th edition  Chapter 9 Solutions, Page 16/39 9-39 l = 2 + 4 + 4 = 10 in 2 1  4  0   4  2   1 in 10 2  4  4  2  4  0  1.6 in y 10 x M = FR = F (10  1) = 9 F 1  1   4  1.6  2 r1  2  2.4 in r2  12   2  1.6   1.077 in 2  2  1 r3  2  1.62  1.887 in 1  0.707  5 /16   23   0.1473 in 4 12 1  J G3   0.707  5 /16   43   1.178 in 4 12 J G1  J G2 3  J   J i  Ai rG2i i 1   0.1473  0.707  5 /16  2   2.4 2   1.178  0.707  5 / 16  4  1.077 2   1.178  0.707  5 / 16  4  1.887 2   9.220 in 4  1.6  o   28.07  4 1    tan 1  r  1.62   4  1  3.4 in 2 Primary shear ( in kpsi, F in kip) :   Secondary shear:    Shigley’s MED, 11th edition V F   0.4526 F A 0.707  5 /16 10  Mr 9 F  3.4    3.319 F J 9.220 Chapter 9 Solutions, Page 17/39  max   3.319F sin 28.07    3.319F cos 28.07 o 2 o  0.4526 F  2  3.724 F Ans. max = allow  3.724 F = 25  F = 6.71 kip ______________________________________________________________________________ 9-40 l = 30 + 50 + 50 = 130 mm 30 15   50  0   50  25   13.08 mm 130 30  50   50  25   50  0   21.15 mm y 130 x M = FR = F(200  13.08) = 186.92 F (M in Nm, F in kN) r1  15  13.08   50  21.15  2 2  28.92 mm r2  13.082   25  21.15   13.63 mm 2  25  13.08 r3  2  21.152  24.28 mm 1  0.707  5   303   7.954 103  mm 4 12 1  J G3   0.707  5   503   36.82 103  mm 4 12 J G1  J G2 3  J   J i  Ai rG2i i 1   7.954 103   0.707  5  30   28.922   36.82 103   0.707  5  50  13.632   36.82 103   0.707  5  50   24.282   307.3 103  mm 4  21.15  o   29.81  50  13.08    tan 1  r  21.152   50  13.08   42.55 mm 2 Shigley’s MED, 11th edition Chapter 9 Solutions, Page 18/39 Primary shear ( in MPa, F in kN) : F 103  V  2.176 F    A 0.707  5 130  Secondary shear: 3 Mr 186.92 F 10   42.55    25.88 F    J 307.3 103   max   25.88F sin 29.81    25.88F cos 29.81 o 2 o  2.176 F  2  27.79 F Ans. max = allow  27.79 F = 140  F = 5.04 kN ______________________________________________________________________________ 9-41 Weld Pattern 1. 2. Figure of merit  a2  J u a 3 / 12 a 2      fom 0.0833   lh ah 12h  h  a  3a 2  a 2  a 2  a2  fom    0.3333   6  2a  h 3h  h  3.  2a   6a 2 a 2  5a 2  0.2083  a 2  fom    12  a  a  2ah 24h  h  4. fom  5.  2a  fom  Rank______ 5 1 4 4  a2  1  8a 3  6a 3  a 3 a4  0.3056       3ah  12 2a  a   h  2 3  a2  1 8a 3   0.3333   6h 4a 24ah  h  1 2  a / 2   a2  a3 6. 3 fom    0.25   4ah  ah  h  ______________________________________________________________________________ 3 Shigley’s MED, 11th edition Chapter 9 Solutions, Page 19/39 9-42 Weld Pattern 1. 2. 3. 4.* 5. & 7. Figure of merit 3  a2  I u  a / 12   0.0833   fom   lh ah  h  a3 / 6    a2   0.0833   fom  2ah  h  aa 2 / 2    a2  fom   0.25   2ah  h  a 2 /12   6a  a  7 a 2   a2   fom    0.1944   3ah 36h  h  a a2 a x , y  2 a  2a 3 Rank______ 6 6 1 2 2 a3 2a 3 a 2 a Iu   2a   a  2a     3 3 3 3 6. & 8. 3  a2  I u  a / 3 1  a 2      0.1111  fom   3ah 9 h  lh  h  a 2 / 6   3a  a  1  a 2    a2  fom      0.1667   4ah 6 h   h    a / 2 a2  a2  fom    0.125   8h  ah  h  5 3 3 9. 4 *Note. Because this section is not symmetric with the vertical axis, out-of-plane deflection may occur unless special precautions are taken. See the topic of “shear center” in books with more advanced treatments of mechanics of materials. ______________________________________________________________________________ 9-43 Attachment and member (1018 HR), Sy = 220 MPa and Sut = 400 MPa. The member and attachment are weak compared to the properties of the lowest electrode. Decision Specify the E6010 electrode Controlling property, Table 9-4: all = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa For a static load, the parallel and transverse fillets are the same. Let the length of a bead be l = 75 mm, and n be the number of beads. Shigley’s MED, 11th edition Chapter 9 Solutions, Page 20/39  F   all n  0.707  hl 100 103  F   21.43 nh  0.707l all 0.707  75  88  where h is in millimeters. Make a table Number of beads, n 1 2 3 4 Decision Leg size, h (mm) 21.43 10.71 7.14 5.36  6 mm Specify h = 6 mm on all four sides. Weldment specification: Pattern: All-around square, four beads each side, 75 mm long Electrode: E6010 Leg size: h = 6 mm ______________________________________________________________________________ 9-44 Decision: Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an optimal pattern (see Prob. 9-41) and have thus reduced a synthesis problem to an analysis problem: Table 9-1, case 2, rotated 90: A = 1.414hd = 1.414(h)(75) = 106.05h mm2 Primary shear 12 103  113.2 V  y    A 106.05h h Secondary shear: d (3b 2  d 2 ) 6 75[3(752 )  752 ]   281.3 103  mm 3 6 J  0.707(h)(281.3) 103   198.8 103  h mm 4 Ju  With  = 45, Shigley’s MED, 11th edition Chapter 9 Solutions, Page 21/39  x  3 Mry 12 10  (187.5)(37.5) 424.4 Mr cos 45o      y J J h 198.8 103  h  max   x 2   y   y   2 1 684.9 424.42  (113.2  424.4) 2  h h Attachment and member (1018 HR): Sy = 220 MPa, Sut = 400 MPa Decision: Use E60XX electrode which is stronger  all  min[0.3(400), 0.4(220)]  88 MPa  max   all  h 684.9  88 MPa h 684.9  7.78 mm 88 Decision: Specify 8 mm leg size Weldment Specifications: Pattern: Parallel horizontal fillet welds Electrode: E6010 Type: Fillet Length of each bead: 75 mm Leg size: 8 mm ______________________________________________________________________________ 9-45 Problem 9-44 solves the problem using parallel horizontal fillet welds, each 75 mm long obtaining a leg size rounded up to 8 mm. For this problem, since the width of the plate is fixed and the length has not been determined, we will explore reducing the leg size by using two vertical beads 75 mm long and two horizontal beads such that the beads have a leg size of 6 mm. Decision: Use a rectangular weld bead pattern with a leg size of 6 mm (case 5 of Table 9-1 with b unknown and d = 75 mm). Materials: Attachment and member (1018 HR): Sy = 220 MPa, Sut = 400 MPa From Table 9-4, AISC welding code, all = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa Select a stronger electrode material from Table 9-3. Decision: Specify E6010 Solving for b: In Prob. 9-44, every term was linear in the unknown h. This made solving for h relatively easy. In this problem, the terms will not be linear in b, and so we will use an iterative solution with a spreadsheet. Throat area and other properties from Table 9-1: A = 1.414(6)(b + 75) = 8.484(b + 75) Shigley’s MED, 11th edition (1) Chapter 9 Solutions, Page 22/39 Ju  b  75   6 3 , J = 0.707 (6) Ju = 0.707(b +75)3 (2) Primary shear ( in MPa, h in mm):  y  12 103  V  A A (3) Secondary shear (See Prob. 9-44 solution for the definition of ) : Mr    J 3 Mry 12 10  150  b / 2  (37.5) Mr  x    cos    cos   3 J J 0.707  b  75  3 Mr Mrx 12 10  150  b / 2  (b / 2)   y    sin   sin   3 J J 0.707  b  75   max   y 2   x   y  2 (4) (5) (6) Enter Eqs. (1) to (6) into a spreadsheet and iterate for various values of b. A portion of the spreadsheet is shown below. b (mm) 41 42 43 44 A (mm2) 984.144 992.628 1001.112 1009.596 J (mm4) 1103553.5 1132340.4 1161623.6 1191407.4 'y (Mpa) "y (Mpa) "x (Mpa) 12.19334 12.08912 11.98667 11.88594 69.5254 67.9566 66.43718 64.96518 38.00722 38.05569 38.09065 38.11291 max (Mpa) 90.12492 88.63156 87.18485 < 88 Mpa 85.7828 We see that b  43 mm meets the strength goal. Weldment Specifications: Pattern: Horizontal parallel weld tracks 43 mm long, vertical parallel weld tracks 75 mm long Electrode: E6010 Leg size: 6 mm ______________________________________________________________________________ 9-46 Materials: Member and attachment (1018 HR): Table 9-4: Shigley’s MED, 11th edition S y  32 kpsi, Sut  58 kpsi  all  min[0.3(58), 0.4(32)]  12.8 kpsi Chapter 9 Solutions, Page 23/39 Decision: Use E6010 electrode. From Table 9-3: S y  50 kpsi, Sut  62 kpsi,  all  min[0.3(62), 0.4(50)]  20 kpsi Decision: Since 1018 HR is weaker than the E6010 electrode, use  all  12.8 kpsi Decision: Use an all-around square weld bead track. l1 = 6 + a = 6 + 6.25 = 12.25 in Throat area and other properties from Table 9-1: A  1.414h(b  d )  1.414( h)(6  6)  16.97 h Primary shear   3 V F 20 10 1179   y    psi A A 16.97 h h Secondary shear (b  d )3 (6  6)3   288 in 3 6 6 J  0.707 h(288)  203.6h in 4 Ju   x   y   max Mry    20 103 (6.25  3)(3)  2726 psi h 203.6h 1 4762   x2  ( y   y ) 2  27262  (1179  2726) 2  psi h h J Relate stress to strength  max   all  4762  12.8 103 h    h 4762  0.372 in 12.8 103   Decision: Specify 3 / 8 in leg size Specifications: Pattern: All-around square weld bead track Electrode: E6010 Type of weld: Fillet Weld bead length: 24 in Leg size: 3 / 8 in Attachment length: 12.25 in ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 9 Solutions, Page 24/39 9-47 This is a good analysis task to test a student’s understanding. (1) Solicit information related to a priori decisions. (2) Solicit design variables b and d. (3) Find h and round and output all parameters on a single screen. Allow return to Step 1 or Step 2. (4) When the iteration is complete, the final display can be the bulk of your adequacy assessment. Such a program can teach too. ______________________________________________________________________________ 9-48 The objective of this design task is to have the students teach themselves that the weld patterns of Table 9-2 can be added or subtracted to obtain the properties of a contemplated weld pattern. The instructor can control the level of complication. We have left the presentation of the drawing to you. Here is one possibility. Study the problem’s opportunities, and then present this (or your sketch) with the problem assignment. Use b1 as the design variable. Express properties as a function of b1. From Table 9-3, case 3: A  1.414h(b  b1 ) bd 2 b1d 2 (b  b1 ) d 2 Iu    2 2 2 I  0.707 hI u V F    A 1.414h(b  b1 ) Mc Fa ( d / 2)     I 0.707hI u Parametric study Let a  10 in, b  8 in, d  8 in, b1  2 in,  all  12.8 kpsi, l  2(8  2)  12 in Shigley’s MED, 11th edition Chapter 9 Solutions, Page 25/39 A  1.414h(8  2)  8.48h in 2 I u  (8  2)(82 / 2)  192 in 3 I  0.707( h)(192)  135.7 h in 4 10 000 1179   psi  8.48h h 10 000(10)(8 / 2) 2948     psi 135.7h h 1 3175  max  11792  29482   12 800 psi h h from which h  0.248 in. Do not round off the leg size – something to learn. Iu 192   64.5 in hl 0.248(12) A  8.48(0.248)  2.10 in 2 fom '  I  135.7(0.248)  33.65 in 4 h2 0.2482 l 12  0.369 in 3 2 2 I 33.65 eff    91.2 in vol 0.369 1179  4754 psi   0.248 2948  11 887 psi    0.248 3175  max   12 800 psi 0.248 vol  Now consider the case of uninterrupted welds, b1  0 A  1.414( h)(8  0)  11.31h I u  (8  0)(82 / 2)  256 in 3 I  0.707(256) h  181h in 4 10 000 884    11.31h h 10 000(10)(8 / 2) 2210     181h h 1 2380  max    all 8842  2210 2  h h  2380  0.186 in h  max   all 12 800 Do not round off h. Shigley’s MED, 11th edition Chapter 9 Solutions, Page 26/39 A  11.31(0.186)  2.10 in 2 I  181(0.186)  33.67 in 4 884 0.186 2  4753 psi, vol  16  0.277 in 3 0.186 2 2210  11882 psi    0.186 I 256 fom '  u   86.0 in hl 0.186(16) 33.67 I eff  2   121.7 in (h / 2)l (0.1862 / 2)16   Conclusions: To meet allowable stress limitations, I and A do not change, nor do τ and σ. To meet the shortened bead length, h is increased proportionately. However, volume of bead laid down increases as h2. The uninterrupted bead is superior. In this example, we did not round h and as a result we learned something. Our measures of merit are also sensitive to rounding. When the design decision is made, rounding to the next larger standard weld fillet size will decrease the merit. Had the weld bead gone around the corners, the situation would change. Here is a follow up task analyzing an alternative weld pattern. ______________________________________________________________________________ 9-49 From Table 9-2 For the box A  1.414 h(b  d ) Subtracting b1 from b and d1 from d A  1.414h  b  b1  d  d1  Iu  Length of bead d2 d3 b d2 1 1 (3b  d )  1  1   b  b1  d 2  d 3  d13 6 6 2 2 6   l  2(b  b1  d  d1 ) fom  I u / hl ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 9 Solutions, Page 27/39 9-50 Computer programs will vary. ______________________________________________________________________________ 9-51 all = 12 kpsi. Use Fig. 9-17(a) for general geometry, but employ beads. beads and then Horizontal parallel weld bead pattern b = 3 in, d = 6 in Table 9-2: A  1.414hb  1.414(h)(3)  4.24h in 2 bd 2 3(6) 2   54 in 3 2 2 I  0.707 hI u  0.707( h)(54)  38.2h in 4 10 2.358    kpsi 4.24h h Mc 10(10)(6 / 2) 7.853      kpsi I 38.2h h Iu   max   2   2  1 8.199 2.3582  7.8532  kpsi h h Equate the maximum and allowable shear stresses.  max   all  8.199  12 h from which h  0.683 in. It follows that I  38.2(0.683)  26.1 in 4 The volume of the weld metal is vol  h 2l (0.683) 2 (3  3)   1.40 in 3 2 2 The effectiveness, (eff)H, is (eff) H  Shigley’s MED, 11th edition 26.1 I   18.6 in vol 1.4 Ans. Chapter 9 Solutions, Page 28/39 Vertical parallel weld beads b  3 in d  6 in From Table 9-2, case 2 A  1.414hd  1.414(h)(6)  8.48h in 2 d 3 63   72 in 3 6 6 I  0.707 hI u  0.707( h)(72)  50.9h 10 1.179    psi 8.48h h Mc 10(10)(6 / 2) 5.894      psi I 50.9h h 1 6.011  max   2   2  1.1792  5.894 2  kpsi h h Iu  Equating  max to  all gives h  0.501 in. It follows that I  50.9(0.501)  25.5 in 4 h 2l 0.5012 vol  (6  6)  1.51 in 3  2 2 I 25.5   16.7 in (eff ) V  vol 1.51 Ans. ______________________________________________________________________________ 9-52 F = 0, T = 15 kipin. Table 9-1: Ju = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4 Tr 15 1   13.5 kpsi Ans. J 1.111 ______________________________________________________________________________  max  9-53 F = 2 kip, T = 0. Table 9-2: A = 1.414  h r = 1.414  (1/4)(1) = 1.111 in2 Iu =  r 3 =  (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4 Shigley’s MED, 11th edition Chapter 9 Solutions, Page 29/39   V 2   1.80 kpsi A 1.111    Mr 2  6 1   21.6 kpsi I 0.5553  max = ( 2 +  2)1/2 = (1.802 + 21.62)1/2 = 21.7 kpsi Ans. ______________________________________________________________________________ 9-54 F = 2 kip, T = 15 kipin. Bending: Table 9-2: A = 1.414  h r = 1.414  (1/4)(1) = 1.111 in2 Iu =  r 3 =  (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4   V 2   1.80 kpsi A 1.111  M  Mr 2  6 1   21.6 kpsi I 0.5553 Torsion: Table 9-1: Ju = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4  T  Tr 15 1   13.5 kpsi J 1.111  max   2    M    T  1.802  21.62  13.52  25.5 kpsi 2 2 Ans. ______________________________________________________________________________ 9-55 F = 2 kip, T = 15 kipin. Bending: Table 9-2: A = 1.414  h r = 1.414  h (1) = 4.442h in2 Iu =  r 3 =  (1)3 = 3.142 in3, I = 0.707 h (3.142) = 2.221h in4   Shigley’s MED, 11th edition V 2 0.4502   kpsi A 4.442h h Chapter 9 Solutions, Page 30/39   M Mr 2  6 1 5.403   kpsi I 2.221h h  Torsion: Ju = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707 h (6.283) = 4.442 in4 Table 9-1:  T  max  Tr 15 1 3.377   kpsi J 4.442h h 2 2 2 6.387 2 2  0.4502   5.403   3.377  kpsi   2     M    T         h  h   h   h   max   all 6.387  20 h  Should specify a 83 in weld.  h  0.319 in Ans. Ans. ______________________________________________________________________________ 9-56 h  9 mm, d  200 mm, b  25 mm From Table 9-2, case 2: A = 1.414(9)(200) = 2.545(103) mm2 Iu  d 3 2003   1.333 106 mm3 6 6   I = 0.707h Iu = 0.707(9)(1.333)(106) = 8.484(106) mm4   25 103 F     9.82 MPa A 2.545(103 ) M = 25(150) = 3750 Nm    Mc 3750(100)  103  44.20 MPa I 8.484(106 )    max   2   2  9.822  44.202  45.3 MPa Ans. ______________________________________________________________________________ 9-57 h = 0.25 in, b = 2.5 in, d = 5 in. Table 9-2, case 5: Shigley’s MED, 11th edition A = 0.707h (b +2d) = 0.707(0.25)[2.5 + 2(5)] = 2.209 in2 Chapter 9 Solutions, Page 31/39 y d2 52   2 in b  2d 2.5  2  5  2d 3  2d 2 y   b  2 d  y 2 3 2  53    2  52   2    2.5  2  5    22   33.33 in 3 3 Iu  I = 0.707 h Iu = 0.707(1/4)(33.33) = 5.891 in4 Primary shear:   F 2   0.905 kpsi A 2.209 Secondary shear (the critical location is at the bottom of the bracket): y = 5  2 = 3 in    My 2  5  3    5.093 kpsi I 5.891  max   2   2  0.9052  5.0932  5.173 kpsi  all 18   3.48 Ans.  max 5.173 ______________________________________________________________________________ n 9-58 The largest possible weld size is 1/16 in. This is a small weld and thus difficult to accomplish. The bracket’s load-carrying capability is not known. There are geometry problems associated with sheet metal folding, load-placement and location of the center of twist. This is not available to us. We will identify the strongest possible weldment. Use a rectangular, weld-all-around pattern – Table 9-2, case 6: Shigley’s MED, 11th edition Chapter 9 Solutions, Page 32/39 A  1.414 h(b  d )  1.414(1 / 16)(1  7.5)  0.7512 in 2 x  b / 2  0.5 in y  d / 2  7.5 / 2  3.75 in d2 7.52 Iu  (3b  d )  [3(1)  7.5]  98.44 in 3 6 6 I  0.707hI u  0.707(1 / 16)(98.44)  4.350 in 4 M  (3.75  0.5)W  4.25W V W   1.331W   A 0.7512 Mc 4.25W (7.5 / 2)   3.664W    I 4.350  max   2   2  W 1.3312  3.664 2  3.90W Material properties: The allowable stress given is low. Let’s demonstrate that. For the 1020 CD bracket, use HR properties of Sy = 30 kpsi and Sut = 55. The 1030 HR support, Sy = 37.5 kpsi and Sut = 68. The E6010 electrode has strengths of Sy = 50 and Sut = 62 kpsi. Allowable stresses: 1020 HR: all = min[0.3(55), 0.4(30)] = min(16.5, 12) = 12 kpsi 1020 HR: all = min[0.3(68), 0.4(37.5)] = min(20.4, 15) = 15 kpsi E6010: all = min[0.3(62), 0.4(50)] = min(18.6, 20) = 18.6 kpsi Since Table 9-6 gives 18.0 kpsi as the allowable shear stress, use this lower value. Therefore, the allowable shear stress is all = min(14.4, 12, 18.0) = 12 kpsi However, the allowable stress in the problem statement is 1.5 kpsi which is low from the weldment perspective. The load associated with this strength is  max   all  3.90W  1500 W  1500  385 lbf 3.90 If the welding can be accomplished (1/16 leg size is a small weld), the weld strength is 12 000 psi and the load associated with this strength is W = 12 000/3.90 = 3077 lbf. Can the bracket carry such a load? There are geometry problems associated with sheet metal folding. Load placement is important and the center of twist has not been identified. Also, the load-carrying capability of the top bend is unknown. Shigley’s MED, 11th edition Chapter 9 Solutions, Page 33/39 These uncertainties may require the use of a different weld pattern. Our solution provides the best weldment and thus insight for comparing a welded joint to one which employs screw fasteners. ______________________________________________________________________________ 9-59 F FB FBx FBy  100 lbf ,  all  3 kpsi  100(16 / 3)  533.3 lbf  533.3cos 60  266.7 lbf  533.3cos 30  462 lbf It follows that RAy  562 lbf and RAx  266.7 lbf, RA = 622 lbf M = 100(16) = 1600 lbf · in The OD of the tubes is 1 in. From Table 9-1, case 6: A  2 1.414( hr )   2(1.414)( h)(1 / 2)  4.442h in 2 J u  2 r 3  2 (1 / 2)3  0.7854 in 3 J  2(0.707)hJ u  1.414(0.7854)h  1.111h in 4 The weld only carries the torsional load between the handle and tube A. Consequently, the primary shear in the weld is zero, and the maximum shear stress is comprised entirely of the secondary shear. Shigley’s MED, 11th edition Chapter 9 Solutions, Page 34/39 Tc Mc 1600(0.5) 720.1    psi J J h 1.111h 720.1   3000 h  max      max   all h 720.1  0.240  1 / 4 in 3000 Decision: Use 1/4 in fillet welds Ans. ______________________________________________________________________________ 9-60 For the pattern in bending shown, find the centroid G of the weld group. x 75  6 150   325  9 150   225 mm  6 150    9 150  I 6 mm  2  I G  Ax 2  6mm  0.707  6  1503   2  2  0.707  6 150  225  75    31.02 10 6  mm 4 12   I 9mm  0.707  9  1503   2  2  0.707  9 150 175  75   22.67 106  mm 4 12   I = I 6 mm + I 9 mm = (31.02 + 22.67)(106) = 53.69(106) mm4 The critical location is at B. With  in MPa, and F in kN F 103  V     0.3143F A 2 0.707  6  9 150   3 Mc 200 F 10   225    0.8381F    I 53.69 106  Shigley’s MED, 11th edition Chapter 9 Solutions, Page 35/39  max   2   2  F 0.31432  0.83812  0.8951F Materials: 1015 HR (Table A-20): Sy = 190 MPa, E6010 Electrode(Table 9-3): Sy = 345 MPa Eq. (5-21): all = 0.577(190) = 109.6 MPa  all / n 109.6 / 2  61.2 kN Ans. 0.8951 0.8951 ______________________________________________________________________________ F 9-61  In the textbook, Fig. Problem 9-61b is a free-body diagram of the bracket. Forces and moments that act on the welds are equal, but of opposite sense. (a) M = 1200(0.366) = 439 lbf · in Ans. (b) Fy = 1200 sin 30 = 600 lbf Ans. (c) Fx = 1200 cos 30 = 1039 lbf Ans. (d) From Table 9-2, case 6: A  1.414(0.25)(0.25  2.5)  0.972 in 2 d2 2.52 Iu  (3b  d )  [3(0.25)  2.5]  3.39 in 3 6 6 The second area moment about an axis through G and parallel to z is I  0.707hI u  0.707(0.25)(3.39)  0.599 in 4 Ans. (e) Refer to Fig. Problem 9-61b. The shear stress due to Fy is 1  Fy A  600  617 psi 0.972 The shear stress along the throat due to Fx is 2  Fx 1039   1069 psi A 0.972 The resultant of 1 and 2 is in the throat plane     12   22  617 2  10692  1234 psi The bending of the throat gives    Shigley’s MED, 11th edition Mc 439(1.25)   916 psi I 0.599 Chapter 9 Solutions, Page 36/39 The maximum shear stress is  max   2   2  1234 2  9162  1537 psi (f) Materials: 1018 HR Member: E6010 Electrode: n S sy  max Ans. Sy = 32 kpsi, Sut = 58 kpsi (Table A-20) Sy = 50 kpsi (Table 9-3)  0.577 S y  max  0.577(32)  12.0 1.537 Ans. (g) Bending in the attachment near the base. The cross-sectional area is approximately equal to bh. A1 ฀ bh  0.25(2.5)  0.625 in 2 1039 F  1662 psi  xy  x  0.625 A1 0.25(2.5) 2 I bd 2    0.260 in 3 c 6 6 At location A, F M y  y  A1 I / c 600 439 y    2648 psi 0.625 0.260 The von Mises stress   is     y2  3 xy2  26482  3(1662) 2  3912 psi Thus, the factor of safety is, S 32 n y   8.18   3.912 Ans. The clip on the mooring line bears against the side of the 1/2-in hole. If the clip fills the hole F 1200     9600 psi td 0.25(0.50) S 32(103 )  3.33 Ans. n y   9600 Further investigation of this situation requires more detail than is included in the task statement. (h) In shear fatigue, the weakest constituent of the weld melt is 1018 HR with Sut = 58 kpsi, Eq. (6-10), gives Shigley’s MED, 11th edition Chapter 9 Solutions, Page 37/39 Se  0.5Sut  0.5(58)  29.0 kpsi ka = 12.7(58)-0.758 = 0.585 Eq. (6-18): For uniform shear stress on the throat, assume kb = 1. Eq.(6-25): kc = 0.59 From Eq. (6-17), the endurance strength in shear is Sse = 0.585(1)(0.59)(29.0) = 10.0 kpsi From Table 9-5, the shear stress-concentration factor is Kf s = 2.7. The loading is repeatedly-applied  1.537  a   m  K f s max  2.7  2.07 kpsi 2 2 Eq. (6-48): Gerber factor of safety nf, adjusted for shear, with Ssu = 0.67Sut 2 1  S su   a   1  nf    2   m  S se    2 S   1   m se    S su a   2 2   2(2.07)(10.0)   1  0.67(58)   2.07        1  1      4.55 Ans. 2  2.07   10.0    0.67(58)(2.07)     Attachment metal should be checked for bending fatigue. ______________________________________________________________________________ 9-62 (a) Use b = d = 4 in. Since h = 5/8 in, the primary shear is   F  0.2829 F 1.414(5 / 8)(4) The secondary shear calculations, for a moment arm of 14 in give 4[3(42 )  42 ]  42.67 in 3 6 J  0.707hJ u  0.707(5 / 8)42.67  18.85 in 4 Mry 14 F (2)  x   y    1.485F J 18.85 Ju  Thus, the maximum shear and allowable load are: Shigley’s MED, 11th edition Chapter 9 Solutions, Page 38/39  max  F 1.4852  (0.2829  1.485) 2  2.309 F  all 25   10.8 kip Ans. 2.309 2.309 The load for part (a) has increased by a factor of 10.8/2.71 = 3.99 F  Ans. (b) From Prob. 9-20b, all = 11 kpsi Fall   all 2.309  11  4.76 kip 2.309 The allowable load in part (b) has increased by a factor of 4.76/1.19 = 4 Ans. ______________________________________________________________________________ 9-63 Purchase the hook having the design shown in Fig. Problem 9-63b. Referring to text Fig. 9-29a, this design reduces peel stresses. ______________________________________________________________________________ 9-64 (a) l/2 l /2 A 1 l / 2 P cosh( x) dx  A1  cosh( x) dx  1 sinh( x)    l / 2 l / 2   l  4b sinh(l / 2) l / 2 A1 A1  [sinh(l / 2)  sinh(l / 2)]  [sinh(l / 2)  ( sinh(l / 2))]   (b) (c) 2 A1 sinh(l / 2)  P P [2sinh(l / 2)]  4bl sinh(l / 2) 2bl  P cosh( l / 2) P  (l / 2)   Ans. 4b sinh( l / 2) 4b tanh( l / 2) K    (l / 2) P l / 2  2bl      4b tanh(l / 2)  P  tanh( l / 2) Ans. Ans. For computer programming, it can be useful to express the hyperbolic tangent in terms of exponentials: l exp(l / 2)  exp(l / 2) K  Ans. 2 exp(l / 2)  exp(l / 2) ______________________________________________________________________________ 9-65 This is a computer programming exercise. All programs will vary. Shigley’s MED, 11th edition Chapter 9 Solutions, Page 39/39 Chapter 5 5-1 Sy = 350 MPa. MSS: 1  3 = Sy /n DE: n      A2   A B   B2  1/ 2 n Sy  1   3    x2   x y   y2  3 xy2  1/ 2 Sy  (a) MSS: 1 = 100 MPa,2 = 100 MPa,3 = 0 n DE: (b) MSS: Ans.    (100 2  100(100)  1002 )1/ 2  100 MPa, 350  3.5 100  0 350  3.5 100 Ans. 350  4.04 86.6 Ans. n 1 = 100 MPa,2 = 50 MPa,3 = 0 n DE: 350  3.5 100  0 Ans.    (100 2  100(50)  50 2 )1/ 2  86.6 MPa, n 2 100  100  2 (c)  A ,  B      (75)  140,  40 MPa 2 2    1  140,  2  0,  3  40 MPa 350  1.94 Ans. MSS: n  140  ( 40) DE:    100 2  3  752   1/ 2  164 MPa, n 350  2.13 164 Ans. 2 50  75  50  75  2     ( 50)  11.0, 114.0 MPa 2 2    1  0,  2  11.0,  3  114.0 MPa 350 MSS: n  3.07 Ans. 0  (114.0) DE:    [(50) 2  ( 50)(75)  (75) 2  3( 50) 2 ]1/2  109.0 MPa 350 n  3.21 Ans. 109.0 (d)  A ,  B  (e)  A ,  B  2 100  20 2  100  20       20   104.7, 15.3 MPa 2 2   Shigley’s MED, 11th edition Chapter 5 Solutions, Page 1/58  1  104.7,  2  15.3,  3  0 MPa 350  3.34 104.7  0 MSS: n DE: 2    1002  100(20)  20 2  3  20   Ans. 1/2  98.0 MPa   350 n  3.57 Ans. 98.0 ______________________________________________________________________________ 5-2 Sy = 350 MPa. MSS: 1  3 = Sy /n DE: n      A2   A B   B2  (a) MSS: DE: (b) MSS: DE: 1/ 2  Sy  1   3  Sy n  n  1  100 MPa,  3  0  n  n Sy  350  3.5 100  0 350  3.5 [100  (100)(100)  1002 ]1/ 2 2  1  100 ,  3  100 MPa  n Ans. Ans. 350  1.75 100  ( 100) Ans. 350  2.02 Ans. 1/2 1002  (100)( 100)   100  2    350 (c) MSS:  1  100 MPa,  3  0  n   3.5 Ans. 100  0 350 DE: n  4.04 Ans. 1/ 2 1002  (100)(50)  50 2  350  1  100,  3  50 MPa  n   2.33 Ans. (d) MSS: 100  ( 50) 350 DE: n  2.65 Ans. 1/2 1002  (100)( 50)   50  2    350 (e) MSS:  1  0,  3  100 MPa  n   3.5 Ans. 0  (100) 350 DE: n  4.04 Ans. 1/2  50 2  ( 50)(100)   100 2    ______________________________________________________________________________ 5-3 n From Table A-20, Sy = 37.5 kpsi Shigley’s MED, 11th edition Chapter 5 Solutions, Page 2/58 MSS: 1  3 = Sy /n n      A2   A B   B2  1/ 2 DE: n Sy  1   3    x2   x y   y2  3 xy2  1/ 2 Sy  (a) MSS:  1  25 kpsi,  3  0 DE: n  37.5  252  (25)(15)  152  1/ 2 n  1.72  1  15 kpsi,  3  15  (b) MSS: DE: n 37.5  1.5 25  0 n 37.5 1/ 2 152  (15)(15)   15 2    Ans. Ans. 37.5  1.25 15  (15)  1.44 Ans. 2 20  20      ( 10)2  24.1,  4.1 kpsi 2  2   1  24.1,  2  0,  3  4.1 kpsi 37.5 n  1.33 Ans. MSS: 24.1  ( 4.1) (c)  A ,  B      202  3  102   DE: Ans. 1/2  26.5 kpsi  n 37.5  1.42 26.5 Ans. 2 12  15  12  15  2     ( 9)  17.7, 14.7 kpsi 2 2    1  17.7,  2  0,  3  14.7 kpsi 37.5 MSS: n  1.16 Ans. 17.7  ( 14.7) (d)  A ,  B  1/ 2    (12)2  (12)(15)  152  3( 9) 2  DE: n 37.5  1.33 28.1  28.1 kpsi Ans. 2 24  24 2  24  24       15   9,  39 kpsi 2 2    1  0,  2  9,  3  39 kpsi 37.5 MSS: n  0.96 Ans. 0   39  (e)  A ,  B  Shigley’s MED, 11th edition Chapter 5 Solutions, Page 3/58 DE: 2 2 2     24    24  24    24   3  15   1/ 2  35.4 kpsi   37.5 n  1.06 Ans. 35.4 ______________________________________________________________________________ 5-4 From Table A-20, Sy = 47 kpsi. MSS: 1  3 = Sy /n n  Sy  1   3  Sy n  47 (a) MSS:  1  30 kpsi,  3  0  n   1.57 Ans. 30  0 47 n  1.57 Ans. DE: 2 [30  (30)(30)  302 ]1/ 2 47 (b) MSS:  1  30 ,  3  30 kpsi  n   0.78 Ans. 30  (30) 47 DE: n  0.90 Ans. 1/ 2 302  (30)(30)   30 2    47 (c) MSS:  1  30 kpsi,  3  0  n   1.57 Ans. 30  0 47 DE: n  1.81 Ans. 1/ 2 2 30  (30)(15)  152  47  1  0,  3  30 kpsi  n   1.57 Ans. (d) MSS: 0  (30) 47 DE: n  1.81 Ans. 1/ 2 2  30   ( 30)(15)   15 2    47  0.78 Ans. (e) MSS:  1  10,  3  50 kpsi  n  10  (50) 47 DE: n  0.84 Ans. 1/2 2  50   ( 50)(10)  10 2    ______________________________________________________________________________ DE:     A2   A B   B2  Shigley’s MED, 11th edition 1/ 2  Sy  n Chapter 5 Solutions, Page 4/58 5-5 Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. (a) MSS and DE: n OB 4.95"   3.51 Ans. OA 1.41" (b) MSS: n OD 3.91"   3.49 Ans. OC 1.12" DE: n OE 4.51"   4.03 Ans. OC 1.12" (c) MSS: n OG 2.50"   2.00 Ans. OF 1.25" DE: n  OH 2.86"   2.29 Ans. OF 1.25" (d) MSS: n  OJ 3.51" OK 3.65"   3.05 Ans. , DE: n    3.17 Ans. OI 1.15" OI 1.15" OM 3.54" ON 3.77"   3.34 Ans. , DE: n    3.56 Ans. OL 1.06" OL 1.06" ______________________________________________________________________________ (e) MSS: n  Shigley’s MED, 11th edition Chapter 5 Solutions, Page 5/58 5-6 Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. (a) A = 25 kpsi, B = 15 kpsi MSS: n DE: OB 4.37"   1.50 OA 2.92" Ans. OC 5.02"   1.72 Ans. OA 2.92" (b) A = 15 kpsi, B =  15 kpsi n MSS: n OE 2.66"   1.25 OD 2.12" Ans. n OF 3.05"   1.44 OD 2.12" Ans. DE: (c)  A ,  B  MSS: n  2 20  20      ( 10) 2  24.1,  4.1 kpsi 2  2  OH 3.25"   1.34 OG 2.43" Ans. DE: n  OI 3.46"   1.42 OG 2.43" Ans. 2 12  15  12  15  2 (d)  A ,  B      ( 9)  17.7, 14.7 MPa 2 2   MSS: n  OK 2.67"   1.16 OJ 2.30" (e)  A ,  B  Ans. DE: n  OL 3.06"   1.33 OJ 2.30" Ans. 2 24  24 2  24  24       15   9,  39 kpsi 2 2   ON 3.85" OP 4.23"   0.96 Ans. DE: n    1.06 Ans. OM 4.00" OM 4.00" ______________________________________________________________________________ MSS: n  Shigley’s MED, 11th edition Chapter 5 Solutions, Page 6/58 5-7 Sy = 295 MPa, A = 75 MPa, B =  35 MPa, (a) n   Sy 2 A   A B    2 1/ 2 B  295 752  75  35    35  2    1/ 2  3.03 Ans. (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n OB 2.50"   3.01 Ans. OA 0.83" ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 7/58 5-8 Sy = 295 MPa, A = 30 MPa, B =  100 MPa, (a) n   Sy 2 A   A B    2 1/ 2 B  295 1/ 2 302  30  100    100 2     2.50 Ans. (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n OB 2.50"   3.01 Ans. OA 0.83" ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 8/58 5-9 2 100  100  2 Sy = 295 MPa,  A ,  B      (25)  105.9,  5.9 MPa 2  2  Sy 295   2.71 (a) n  1/2 2 2 1/ 2  A   A B   B  105.92  105.9  5.9    5.9 2  Ans. (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n OB 2.87"   2.71 Ans. OA 1.06" ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 9/58 5-10 2 30  65  30  65  2 Sy = 295 MPa,  A ,  B      40  3.8, 91.2 MPa 2 2   Sy 295 (a) n    3.30 1/ 2 2 2 1/ 2 2  A   A B   B   3.8   3.8 91.2    91.2 2  Ans. (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n OB 3.00"   3.33 Ans. OA 0.90" ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 10/58 5-11 2 80  30 2  80  30  Sy = 295 MPa,  A ,  B       10   30.9, 80.9 MPa 2 2   Sy 295 (a) n    2.95 Ans. 1/ 2 2 2 1/ 2  A   A B   B  30.92  30.9  80.9    80.9 2  (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n OB 2.55"   2.93 Ans. OA 0.87" ______________________________________________________________________________ 5-12 Syt = 60 kpsi, Syc = 75 kpsi. Eq. (5-26) for yield is    n 1  3  S   yt S yc  (a)  1  25 kpsi,  3  0 (b)  1  15,  3  15 kpsi Shigley’s MED, 11th edition 1 1  25 0  n      2.40  60 75   Ans. 1   15 15  n    2.22  60 75  Ans. Chapter 5 Solutions, Page 11/58 2 20  20  (c)  A ,  B      ( 10) 2  24.1,  4.1 kpsi , 2  2   1  24.1,  2  0,  3  4.1 kpsi  (d)  A ,  B  Ans. 2 12  15  12  15  2     ( 9)  17.7, 14.7 kpsi 2 2    1  17.7,  2  0,  3  14.7 kpsi (e)  A ,  B  1  24.1 4.1   n   2.19 75   60 1   17.7 14.7  n    2.04 75   60 Ans. 2 24  24 2  24  24       15   9,  39 kpsi 2 2   1  0 39   1  0,  2  9,  3  39 kpsi  n    Ans.   1.92  60 75  ______________________________________________________________________________ 5-13 Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. (a)  A  25,  B  15 kpsi n OB 3.49"   2.39 Ans. OA 1.46" (b)  A  15,  B  15 kpsi n (c) OD 2.36"   2.23 Ans. OC 1.06" 2 20  20   A ,  B      ( 10) 2  24.1,  4.1 kpsi 2  2  n (d) OF 2.67"   2.19 Ans. OE 1.22" 2 12  15  12  15  2  A , B      ( 9)  17.7, 14.7 kpsi 2 2   Shigley’s MED, 11th edition Chapter 5 Solutions, Page 12/58 n (e) OH 2.34"   2.03 Ans. OG 1.15" 2 24  24 2  24  24   A , B       15   9,  39 kpsi 2 2   OJ 3.85" n   1.93 Ans. OI 2.00" ______________________________________________________________________________ 5-14 Since f > 0.05, and Syt  Syc, the Coulomb-Mohr theory for ductile materials will be used. (a) From Eq. (5-26), 1 1     150 50  n 1  3      1.23 Ans. S  S 235 285   yt yc   (b) Plots for Problems 5-14 to 5-18 are found here. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n OB 1.94"   1.23 Ans. OA 1.58" ______________________________________________________________________________ 5-15 (a) From Eq. (5-26), 1 1     50 150  n 1  3      1.35 Ans. S  S 235 285   yc   yt Shigley’s MED, 11th edition Chapter 5 Solutions, Page 13/58 (b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OD 2.14"   1.35 Ans. OC 1.58" ______________________________________________________________________________ n 5-16 2 125  125  2  A,  B      (75)  160,  35 kpsi 2 2   (a) From Eq. (5-26), 1 1  1  3   160 35  n        1.24 Ans. S  235 285   yt S yc  (b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OF 2.04"   1.24 Ans. OE 1.64" ______________________________________________________________________________ n 5-17 2 80  125  80  125  2    A,  B    50  47.7,  157.3 kpsi 2 2   (a) From Eq. (5-26), 1 1  1  3  157.3   0 n        1.81 Ans. S 285   235  yt S yc  (b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OH 2.99"   1.82 Ans. OG 1.64" ______________________________________________________________________________ n 5-18 2 125  80  125  80  2    A,  B    ( 75)  180.8, 24.2 kpsi 2 2   (a) From Eq. (5-26), Shigley’s MED, 11th edition Chapter 5 Solutions, Page 14/58 1 1  1  3  0   180.8 n        1.30 Ans. S  235 285   yt S yc  (b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OJ 2.37"   1.30 Ans. OI 1.83" ______________________________________________________________________________ n 5-19 Sut = 30 kpsi, Suc = 90 kpsi BCM: Eqs. (5-31), MM: Eqs. (5-32) (a) A = 25 kpsi, B = 15 kpsi BCM : Eq. (5-31a), n  MM: Eq. (5-32a), n  Sut A Sut   A (b) A = 15 kpsi, B = 15 kpsi, 30  1.2 Ans. 25 30  1.2 Ans. 25 1  15 15  BCM: Eq. (5-31a), n      1.5 Ans.  30 90  MM: A  0  B, and B /A   1, Eq. (5-32a), n  Sut A  30  2.0 Ans. 15  30  1.24 Ans. 24.14 2 20  20  (c)  A ,  B      ( 10) 2  24.14,  4.14 kpsi 2  2  1  24.14 4.14  BCM: Eq. (5-31b), n      1.18 Ans. 90   30 MM: A  0  B, and B /A   1, Eq. (5-32a), n  (d)  A ,  B  Sut A 2 15  10  15  10  2     ( 15)  17.03,  22.03 kpsi 2 2   Shigley’s MED, 11th edition Chapter 5 Solutions, Page 15/58 1  17.03 22.03  BCM: Eq. (5-31b), n      1.23 Ans. 90   30 MM: A  0  B, and B /A   1, Eq. (5-32b), 1 1   90  30 17.03 22.03    S  Sut   A  B  n   uc     1.60 Ans.   Suc Sut Suc  90  30  90    (e)  A ,  B  2 20  20  20  20  2     ( 15)  5,  35 kpsi 2 2   BCM: Eq. (5-31c), n   Suc B  90  2.57 Ans. 35 90  2.57 Ans. B 35 ______________________________________________________________________________ MM: Eq. (5-32c), n   5-20 Suc  Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. (a) A = 25, B = 15 kpsi BCM & MM: OB 1.74" n   1.19 Ans. OA 1.46" (b) A = 15, B =  15 kpsi OC 1.59"   1.5 Ans. OD 1.06" OE 2.12" MM: n    2.0 Ans. OC 1.06" BCM: n  2 20  20  (c)  A ,  B      ( 10)2 2  2   24.14,  4.14 kpsi OG 1.44" BCM: n    1.18 Ans. OF 1.22" OH 1.52" MM: n    1.25 Ans. OF 1.22" Shigley’s MED, 11th edition Chapter 5 Solutions, Page 16/58 2 15  10  15  10  2 (d)  A ,  B      ( 15)  17.03,  22.03 kpsi 2 2   OJ 1.72" BCM: n    1.24 Ans. OI 1.39" OK 2.24" MM: n    1.61 Ans. OI 1.39" 2 20  20  20  20  2 (e)  A ,  B      ( 15)  5,  35 kpsi 2 2   OM 4.55" BCM and MM: n    2.57 Ans. OL 1.77" ______________________________________________________________________________ 5-21 From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi BCM: Eqs. (5-31), MM: Eqs. (5-32) (a) A = 15, B = 10 kpsi. 31  2.07 Ans.  A 15 S 31 MM: Eq. (5-32a), n  ut   2.07 Ans.  A 15 (b), (c) The plot is shown below is for Probs. 5-21 to 5-25. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. BCM: Eq. (5-31a), n  Sut  BCM and MM: n OB 1.86"   2.07 Ans. OA 0.90" ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 17/58 5-22 Sut = 31 kpsi, Suc = 109 kpsi BCM: Eq. (5-31), MM: Eqs. (5-32) (a) A = 15, B =  50 kpsi, B /A  > 1 1  15 50  BCM: Eq. (5-31b), n      1.06 Ans.  31 109  1 1  109  3115 50    S  Sut   A  B  MM: Eq. (5-32b), n   uc     1.24 Ans.   Suc Sut Suc  109    109  31 (b), (c) The plot is shown in the solution to Prob. 5-21. OD 2.78"   1.07 Ans. OC 2.61" OE 3.25" MM: n    1.25 Ans. OC 2.61" ______________________________________________________________________________ BCM: n  5-23 From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi BCM: Eq. (5-31), MM: Eqs. (5-32)  A,  B  2 15  15      (10) 2  20,  5 kpsi 2  2 1  20 5  (a) BCM: Eq. (5-32b), n      1.45 Ans.  31 109  S 31 MM: Eq. (5-32a), n  ut   1.55 Ans.  A 20 (b), (c) The plot is shown in the solution to Prob. 5-21. OG 1.48"   1.44 Ans. OF 1.03" OH 1.60" MM: n    1.55 Ans. OF 1.03" ______________________________________________________________________________ BCM: n  5-24 From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi BCM: Eq. (5-31), MM: Eqs. (5-32)  A,  B  2 10  25  10  25  2     ( 10)  5,  30 kpsi 2 2   Shigley’s MED, 11th edition Chapter 5 Solutions, Page 18/58 109  3.63 Ans.  B 30 S 109 MM: Eq. (5-32c), n   uc    3.63 Ans. 30 B (b), (c) The plot is shown in the solution to Prob. 5-21. (a) BCM: Eq. (5-31c), n   Suc  OJ 5.53"   3.64 Ans. OI 1.52" ______________________________________________________________________________ BCM and MM: n  5-25 From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi BCM: Eq. (5-31), MM: Eqs. (5-32) 2 35  13  35  13  2    A,  B    ( 10)  15,  37 kpsi 2 2   1  15 37  (a) BCM: Eq. (5-31b), n      1.21 Ans.  31 109  1 1  109  3115 37    S  Sut   A  B  MM: Eq. (5-32b), n   uc     1.46 Ans.   Suc Sut Suc  109    109  31 (b), (c) The plot is shown in the solution to Prob. 5-21. OL 2.42"   1.21 Ans. OK 2.00" OM 2.91" MM: n    1.46 Ans. OK 2.00" ______________________________________________________________________________ BCM: n  5-26 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), (a) A = 15, B =  10 kpsi. 1  15 10  BCM: Eq. (5-31b), n      1.42 Ans.  36 35  Shigley’s MED, 11th edition Chapter 5 Solutions, Page 19/58 (b) The plot is shown below is for Probs. 5-26 to 5-30. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OB 1.28" n   1.42 Ans. OA 0.90" ______________________________________________________________________________ 5-27 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), (a) A = 15, B =  15 kpsi. 1  10 15  BCM: Eq. (5-31b), n      1.42 Ans.  36 35  (b) The plot is shown in the solution to Prob. 5-26. OD 1.28"   1.42 Ans. OC 0.90" ______________________________________________________________________________ n 5-28 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), 2 12  12  (a)  A ,  B      (8) 2  16,  4 kpsi 2  2 1  16 4  BCM: Eq. (5-31b), n      1.79 Ans.  36 35  (b) The plot is shown in the solution to Prob. 5-26. Shigley’s MED, 11th edition Chapter 5 Solutions, Page 20/58 OF 1.47"   1.79 Ans. OE 0.82" ______________________________________________________________________________ n 5-29 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), 2 10  15  10  15  2 (a)  A ,  B      10  2.2,  22.8 kpsi 2 2   BCM: Eq. (5-31c), n   35  1.54 Ans. 22.8 (b) The plot is shown in the solution to Prob. 5-26. OH 1.76"   1.53 Ans. OG 1.15" ______________________________________________________________________________ n 5-30 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), 2 15  8 2  15  8  (a)  A ,  B       8   20.2, 2.8 kpsi 2  2  BCM: Eq. (5-31a), n  36  1.78 Ans. 20.2 (b) The plot is shown in the solution to Prob. 5-26. OJ 1.82"   1.78 Ans. OI 1.02" ______________________________________________________________________________ n 5-31 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32). For this problem, MM reduces to the MNS theory. S 36 (a) A = 15, B =  10 kpsi. Eq. (5-32a), n  ut   2.4 Ans.  A 15 (b) The plot on the next page is for Probs. 5-31 to 5-35. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. Shigley’s MED, 11th edition Chapter 5 Solutions, Page 21/58 n OB 2.16"   2.43 Ans. OA 0.90" ______________________________________________________________________________ 5-32 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the MNS theory. (a) A = 10, B =  15 kpsi. Eq. (5-32b) is not valid and must use Eq, (5-32c), S 35 n   uc    2.33 Ans. B 15 (b) The plot is shown in the solution to Prob. 5-31. OD 2.10" n   2.33 Ans. OC 0.90" ______________________________________________________________________________ 5-33 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the MNS theory. 2 12  12  (a)  A ,  B      (8) 2  16,  4 kpsi 2  2 S 36 n  ut   2.25 Ans.  A 16 (b) The plot is shown in the solution to Prob. 5-31. OF 1.86"   2.27 Ans. OE 0.82" ______________________________________________________________________________ n 5-34 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the MNS theory. Shigley’s MED, 11th edition Chapter 5 Solutions, Page 22/58 2 10  15  10  15  2 (a)  A ,  B      10  2.2,  22.8 kpsi 2 2   S 35 n   uc    1.54 Ans. 22.8 B (b) The plot is shown in the solution to Prob. 5-31. OH 1.76"   1.53 Ans. OG 1.15" ______________________________________________________________________________ n 5-35 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the MNS theory. 2 15  8 2  15  8       8   20.2, 2.8 kpsi 2  2  S 36 n  ut   1.78 Ans.  A 20.2 (b) The plot is shown in the solution to Prob. 5-31. (a)  A ,  B  OJ 1.82"   1.78 Ans. OI 1.02" ______________________________________________________________________________ n 5-36 Given: AISI 1006 CD steel, F = 0.55 kN, P = 4.0 kN, and T = 25 Nm. From Table A-20, Sy =280 MPa. Apply the DE theory to stress elements A and B 4  4 103 4P A: x    22.6 106  Pa  22.6 MPa 2 2 d   0.015  3 16  25  16T 4 V 4  0.55 10      33.6 106  Pa  33.6 MPa    xy  3  2 3  d 3 A   0.015  3   / 4  0.015        22.62  3  33.62   1/ 2 n B: 280  4.49 62.4  62.4 MPa Ans. 3 3 32 Fl 4 P 32  0.55 10  0.1 4  4 10 x      189 106  Pa  189 MPa d3 d2   0.0153    0.0152   xy  16  25  16T   37.7 106  Pa  37.7 MPa 3 3 d   0.015  Shigley’s MED, 11th edition Chapter 5 Solutions, Page 23/58     x2  3 xy2  1/2  1892  3  37.7 2   1/ 2  200 MPa S y 280   1.4 Ans.   200 ______________________________________________________________________________ n 5-37 From Prob. 3-45, the critical location is at the top of the beam at x = 27 in from the left end, where there is only a bending stress of  =  7 456 psi. Thus,  = 7 456 psi and (Sy)min = n = 2(7 456) = 14 912 psi Choose (Sy)min = 15 kpsi Ans. ______________________________________________________________________________ 5-38 From Table A-20 for 1020 CD steel, Sy = 57 kpsi. From Eq. (3-42) T 63 025H n (1) where n is the shaft speed in rev/min. From Eq. (5-3), for the MSS theory,  max  Sy 2nd  16T  d3 (2) where nd is the design factor. Substituting Eq. (1) into Eq. (2) and solving for d gives 1/3  32  63 025  Hnd  d   n S y   (3) Substituting H = 20 hp, nd = 3, n = 1750 rev/min, and Sy = 57(103) psi results in 1/3  32  63 025  20  3  d min   Ans.  0.728 in 3   1750  57 10  ______________________________________________________________________________ 5-39 Given: d = 30 mm, AISI 1018 steel, H = 10 kW, n = 200 rev/min. Table A-20, Sy = 220 MPa Eq. (3-44): T = 9.55 H/n = 9.55(10)103/200 = 477.5 N٠m 16T 16  477.5  6  max  3  10  90.07 MPa 3 d   0.030  (a) Eq. (5-3): n  Shigley’s MED, 11th edition Sy 2 max  220  1.22 2  90.07  Ans. Chapter 5 Solutions, Page 24/58 2   3 max  3  90.07   156.0 MPa (b) From Eq. (5-13),  max Sy 220   1.41 Ans.   max 156 ______________________________________________________________________________ Eq. (5-19): n  5-40 Given: d = 20 mm, AISC 1035 HR steel, ns = 400 rev/min, ny = 1.5. Table A-20, Sy = 270 MPa S 270 (a) Eq. (5-30):  max  y   90 MPa 2n y 2 1.5  Substituting Eq. (3-44) in the equation for max gives  max 16T 16  H 9.55    3 3  d d  ns  H  d 3 ns max 16  9.55     0.023  400  90 106 16  9.55   5.92 103  W  5.92 kW   3  max  (b) From Eq. (5-13):  max Thus, H   d 3 ns max Sy ny  270 1.5   0.023  400 103.9 106   max  Ans. 270  103.9 MPa 1.5 3  6.84 103  W  6.84 kW Ans 16  9.55  16  9.55  ______________________________________________________________________________ 5-41  Table A-20 for AISI 1040 CD steel, Sy = 490 MPa. From Prob. 3-47, 2 2    79.6  2 A: x =79.6 MPa, xy = 63.7 MPa.  max      xy2     63.7  75.1 MPa. 2  2  B: max = zx = 53.1 MPa. C: max = zx = 116.8 MPa. Critical case is at point C. S 490 (a) MSS Theory: n y  y   2.1 Ans. 2 max 2 116.8  Sy 490  2.4 Ans. 3 max 3 116.8  ______________________________________________________________________________ (b) DE Theory: n  5-42  Table A-20 for AISI 1040 CD steel, Sy = 490 MPa From Prob. 3-53,  1  122.6 MPa,  2  0,  3  10.2 MPa,  max  R  66.4 MPa Sy 490 (a) MSS Theory: ny    3.69 Ans. 2 max 2  66.4  2 (b) DE Theory, Eq. (5-13):    122.62  122.6  10.2    10.2     Shigley’s MED, 11th edition 1/ 2  128 MPa Chapter 5 Solutions, Page 25/58 S y 490   3.83 Ans.   128 ______________________________________________________________________________ ny  5-43 Table A-20 for AISI 1020 CD steel, Sy = 390 MPa From Prob. 3-54,  1  194.2 MPa,  2  0,  3  10 MPa,  max  102.1 MPa Sy 490 (a) MSS Theory: ny    1.91 Ans. 2 max 2 102.1 2 1/ 2 (b) DE Theory, Eq. (5-13):    194.22  194.2  10    10    199.4 MPa   Sy 490 ny    1.96 Ans.   194.2 ______________________________________________________________________________ 5-44 Table A-20 for AISI 1035 CD steel, Sy = 67 kpsi From Prob. 3-55,  1  45.8 kpsi,  2  0,  3  0.45 kpsi,  max  23.1 kpsi Sy 67 (a) MSS Theory: ny    1.45 Ans. 2 max 2  23.1 2 1/ 2 (b) DE Theory, Eq. (5-13):     45.82  45.8  0.45    0.45    46.0 MPa   S y 67 ny    1.46 Ans.   46 ______________________________________________________________________________ 5-45 Table A-20 for AISI 1040 CD steel, Sy = 71 kpsi From Prob. 3-101,  1  18.47 kpsi,  2  3.60 kpsi,  max  11.03 kpsi Sy 71 (a) MSS Theory: ny    3.22 Ans. 2 max 2 11.03  2 1/2 (b) DE Theory, Eq. (5-13):    18.47 2  18.47  3.60    3.60    20.51 MPa   Sy 71 ny    3.46 Ans.   20.51 ______________________________________________________________________________ 5-46 Table A-20 for AISI 1040 CD steel, Sy = 71 kpsi From Prob. 3-102,  1  29.1 kpsi,  2  14.2 kpsi,  max  21.7 kpsi Sy 71 (a) MSS Theory: ny    1.64 Ans. 2 max 2  21.7  2 (b) DE Theory, Eq. (5-13):     29.12  29.1 14.2    14.2     Shigley’s MED, 11th edition 1/ 2  38.2 MPa Chapter 5 Solutions, Page 26/58 Sy 71   1.86 Ans.   38.2 ______________________________________________________________________________ ny  5-47 Table A-21 for AISI 4140 steel Q & T 400o F , Sy = 238 kpsi, F = 15 kip. MA = 0 = 3 RD  2 F  RD = 2 (15)/3 = 10 kip, Fy = 0 = RA + RD  F  RA = 15  10 = 5 kip Critical sections are at points B and C where the areas are minimal. B: dB = 1.1 in, MB = RA (1) = 5 kip٠in, VB = RA = 5 kip, TB = 7 kip٠in AB = (/4) 1.12 = 0.9503 in2, C: dC = 1.3 in, MC = RD (1) = 10 kip٠in, VC = RD = 10 kip, TC = 7 kip٠in AC = (/4) 1.32 = 1.327 in2, Critical locations are at the outer surfaces where bending stresses are maximum, and at the center planes where the transverse shear stresses are maximum. In both cases, there exists the torsional shear stresses. B: Outer surface: 32  5  16  7  32 M B 16TB B    38.26 kpsi,  B    26.78 kpsi 3 3 3 3 d d  1.1  1.1  B max 2 2    38.26  2   B    B2     26.78  32.9 kpsi 2 2      B   B2  3 B2  38.262  3  26.78   60.1 kpsi 2 Center plane:  B V  4 5 4VB   7.02 kpsi 3 AB 3  0.9503  B max   B   B V  26.78  7.02  33.8  B  3  B max  3  33.8   58.5 kpsi kpsi C: Outer surface: 16  7  32 M C 32 10  16TC C    46.36 kpsi,  C    16.23 kpsi 3 3 3 3 d d  1.3  1.3  C max 2 2    46.36  2   C    C2     16.23  28.30 kpsi  2   2   C   C2  3 C2  46.362  3 16.23  54.2 kpsi 2 Center plane:  C V  4 10  4VC   10.05 kpsi 3 AC 3 1.327   C max   C   C V  16.23  10.05  26.28  C  3  C max  3  26.28   45.5 kpsi Shigley’s MED, 11th edition kpsi Chapter 5 Solutions, Page 27/58 (a) MSS Theory: Critical location is at point B at the center plane: Sy 238 ny    3.52 Ans. 2 max 2  33.8  (b) DE Theory: Critical location is at point B at the outer surface: S y 238 ny    3.96 Ans.  B 60.1 ______________________________________________________________________________ 5-48 MO = 0 = 40 RC  30(575) +12(460) RC = 293.25 lbf Fy = 0 = RO + 293.25 +460  575 RO =  178.25 lbf Mmax = 2.9325 kip٠in 32 M max 32  2.9325     29.87 kpsi d3  13   16T 16 1.5    7.64 kpsi  d 3  13  2  29.87  2 (a)  max     7.64  16.78 kpsi 2   Sy 50 ny    1.49 2 max 2 16.78  Ans. (b) Eq. (5-15): ’ = [29.872 +3 (7.64)2]1/2 = 32.67 kpsi Sy 50 ny    1.53 Ans.   32.67 ______________________________________________________________________________ 5-49 Given: AISI 1010 HR, ny = 2, L = 0.5 m, F = 150 N, T = 25 N٠m Table A-20, Sy = 180 MPa. Mz = FL = 150 (0.5) = 75 N٠m 32 M z 32  75  2400     d3 d3 d3 16T 16  25  400  3  d d3 d3 6 2 2 2 402.63 S y 180 10     1200   400  2 (a)  max         3     3  d3 2n 2  2 2  d  d   4  402.63   d  6  180 10   Shigley’s MED, 11th edition 1/3  0.0208 m  20.8 mm Ans. Chapter 5 Solutions, Page 28/58 (b)      3 2  2 1/ 2 2  2400  2  400      3 3   3    d     d  1/2 6 795.14 S y 180 10     2 d3 ny 1/3  2  795.14     0.0207 m  20.7 mm d  Ans. 6  180 10   ______________________________________________________________________________ 5-50 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-79, in the plane of analysis 1 = 16.5 kpsi, 2 =  1.19 kpsi, and max = 8.84 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 16.5 kpsi, 2 = 0, and 3 =  1.19 kpsi MSS: From Eq. (5-3), n Sy 1   3  54  3.05 16.5   1.19  Ans. Note: Whenever the two principal stresses of a plane stress state are of opposite sign, the maximum shear stress found in the analysis is the true maximum shear stress. Thus, the factor of safety could have been found from Sy 54  3.05 Ans. 2 max 2  8.84  DE: The von Mises stress can be found from the principal stresses or from the stresses found in part (d) of Prob. 3-79. That is, n  Eqs. (5-13) and (5-19) S Sy 54  n y  1/2 1/2    A2   A B   B2  16.52  16.5  1.19    1.19  2     3.15 Ans. or, Eqs. (5-15) and (5-19) using the results of part (d) of Prob. 3-79 Sy Sy 54   1/ 2 1/2    2  3 2  15.32  3  4.432     3.15 Ans. ______________________________________________________________________________ n 5-51 From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-80, in the plane of analysis Shigley’s MED, 11th edition Chapter 5 Solutions, Page 29/58 1 = 275 MPa, 2 =  12.1 MPa, and max = 144 MPa The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 275 MPa, 2 = 0, and 3 =  12.1 MPa MSS: From Eq. (5-3), n  Sy 1   3  370  1.29 275   12.1 Ans. DE: From Eqs. (5-13) and (5-19) S Sy 370 n y   1/2 1/2    A2   A B   B2   2752  275  12.1   12.1 2     1.32 Ans. ______________________________________________________________________________ 5-52 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-81, in the plane of analysis 1 = 22.6 kpsi, 2 =  1.14 kpsi, and max = 11.9 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 22.6 kpsi, 2 = 0, and 3 =  1.14 kpsi MSS: From Eq. (5-3), n  Sy 1   3  54  2.27 22.6   1.14  Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54 n y   1/2 1/2 2 2    A   A B   B   22.6 2  22.6  1.14    1.14  2     2.33 Ans. ______________________________________________________________________________ 5-53 From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-82, in the plane of analysis 1 = 78.2 MPa, 2 =  5.27 MPa, and max = 41.7 MPa The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 78.2 MPa, 2 = 0, and 3 =  5.27 MPa Shigley’s MED, 11th edition Chapter 5 Solutions, Page 30/58 MSS: From Eq. (5-3), n  Sy 1   3  370  4.43 78.2   5.27  Ans. DE: From Eqs. (5-13) and (5-19) S Sy 370 n y   1/2 1/2 2 2    A   A B   B  78.22  78.2  5.27    5.27  2     4.57 Ans. ______________________________________________________________________________ 5-54 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-83, in the plane of analysis 1 = 36.7 kpsi, 2 =  1.47 kpsi, and max = 19.1 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 36.7 kpsi, 2 = 0, and 3 =  1.47 kpsi MSS: From Eq. (5-3), n  Sy 1   3  54  1.41 36.7   1.47  Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54 n y   1/2 1/2    A2   A B   B2  36.7 2  36.7  1.47    1.47  2     1.44 Ans. ______________________________________________________________________________ 5-55 From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-84, in the plane of analysis 1 = 376 MPa, 2 =  42.4 MPa, and max = 209 MPa The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 376 MPa, 2 = 0, and 3 =  42.4 MPa MSS: From Eq. (5-3), n  Shigley’s MED, 11th edition Sy 1   3  370  0.88 376   42.4  Ans. Chapter 5 Solutions, Page 31/58 DE: From Eqs. (5-13) and (5-19) S Sy 370 n y   1/2 1/2    A2   A B   B2  3762  376  42.4    42.4 2     0.93 Ans. ______________________________________________________________________________ 5-56 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-85, in the plane of analysis 1 = 7.19 kpsi, 2 =  17.0 kpsi, and max = 12.1 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 7.19 kpsi, 2 = 0, and 3 =  17.0 kpsi MSS: From Eq. (5-3), n  Sy  54  2.23 7.19   17.0  Ans. 1   3 DE: From Eqs. (5-13) and (5-19) S Sy 54 n y   1/2 1/2    A2   A B   B2  7.192  7.19  17.0    17.0  2     2.51 Ans. ______________________________________________________________________________ 5-57 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-87, in the plane of analysis 1 = 1.72 kpsi, 2 =  35.9 kpsi, and max = 18.8 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 1.72 kpsi, 2 = 0, and 3 =  35.9 kpsi MSS: From Eq. (5-3), n  Sy 1   3  54  1.44 1.72   35.9  Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54 n y   1/2 1/2    A2   A B   B2  1.722  1.72  35.9    35.9 2     1.47 Ans. ______________________________________________________________________________ 5-58 From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-88, Bending: B = 68.6 MPa, Torsion: B = 37.7 MPa Shigley’s MED, 11th edition Chapter 5 Solutions, Page 32/58 For a plane stress analysis it was found that max = 51.0 MPa. With combined bending and torsion, the plane stress analysis yields the true max. S 370 MSS: From Eq. (5-3), n  y   3.63 Ans. 2 max 2  51.0  DE: From Eqs. (5-15) and (5-19) S Sy 370  n y  1/ 2 1/ 2    B2  3 B2  68.62  3  37.7 2      3.91 Ans. ______________________________________________________________________________ 5-59 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-90, Bending: C = 3460 psi, Torsion: C = 882 kpsi For a plane stress analysis it was found that max = 1940 psi. With combined bending and torsion, the plane stress analysis yields the true max. 54 103  Sy MSS: From Eq. (5-3), n    13.9 Ans. 2 max 2 1940  DE: From Eqs. (5-15) and (5-19) 54 103  Sy Sy   n    C2  3 C2 1/ 2 34602  3  8822  1/2    14.3 Ans. ______________________________________________________________________________ 5-60 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-91, in the plane of analysis 1 = 17.8 kpsi, 2 =  1.46 kpsi, and max = 9.61 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 17.8 kpsi, 2 = 0, and 3 =  1.46 kpsi MSS: From Eq. (5-3), n  Shigley’s MED, 11th edition Sy 1   3  54  2.80 17.8   1.46  Ans. Chapter 5 Solutions, Page 33/58 DE: From Eqs. (5-13) and (5-19) S Sy 54 n y   1/2 1/2    A2   A B   B2  17.82  17.8  1.46    1.46  2     2.91 Ans. ______________________________________________________________________________ 5-61 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-92, in the plane of analysis 1 = 17.5 kpsi, 2 =  1.13 kpsi, and max = 9.33 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 17.5 kpsi, 2 = 0, and 3 =  1.13 kpsi MSS: From Eq. (5-3), n  Sy 1   3  54  2.90 17.5   1.13 Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54 n y   1/2 1/2 2 2    A   A B   B  17.52  17.5  1.13   1.13 2     2.98 Ans. ______________________________________________________________________________ 5-62 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-93, in the plane of analysis 1 = 21.5 kpsi, 2 =  1.20 kpsi, and max = 11.4 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 21.5 kpsi, 2 = 0, and 3 =  1.20 kpsi MSS: From Eq. (5-3), n  Sy 1   3  54  2.38 21.5   1.20  Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54 n y   1/2 1/2    A2   A B   B2   21.52  21.5  1.20    1.20  2     2.44 Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 34/58 5-63 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-94, the concern was failure due to twisting of the flat bar where it was found that max = 14.3 kpsi in the middle of the longest side of the rectangular cross section. The bar is also in bending, but the bending stress is zero where max exists. S 54 MSS: From Eq. (5-3), n  y   1.89 Ans. 2 max 2 14.3 DE: From Eqs. (5-15) and (5-19) Sy Sy 54 n    2.18 Ans. 1/ 2 1/ 2 2 2    3 max   3 14.3     ______________________________________________________________________________ 5-64 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-95, in the plane of analysis 1 = 34.7 kpsi, 2 =  6.7 kpsi, and max = 20.7 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 34.7 kpsi, 2 = 0, and 3 =  6.7 kpsi MSS: From Eq. (5-3), n  Sy 1   3  54  1.30 34.7   6.7  Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54  n y  1/2 1/2 2 2    A   A B   B  34.7 2  34.7  6.7    6.7 2     1.40 Ans. ______________________________________________________________________________ 5-65 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-96, in the plane of analysis 1 = 51.1 kpsi, 2 =  4.58 kpsi, and max = 27.8 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 51.1 kpsi, 2 = 0, and 3 =  4.58 kpsi MSS: From Eq. (5-3), n  Shigley’s MED, 11th edition Sy 1   3  54  0.97 51.1   4.58  Ans. Chapter 5 Solutions, Page 35/58 DE: From Eqs. (5-13) and (5-19) S Sy 54 n y   1/2 1/2    A2   A B   B2  51.12  51.1 4.58    4.58  2     1.01 Ans. ______________________________________________________________________________ 5-66 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-97, in the plane of analysis 1 = 59.7 kpsi, 2 =  3.92 kpsi, and max = 31.8 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are 1 = 59.7 kpsi, 2 = 0, and 3 =  3.92 kpsi MSS: From Eq. (5-3), n  Sy 1   3  54  0.85 59.7   3.92  Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54 n y   1/2 1/2 2 2    A   A B   B  59.7 2  59.7  3.92    3.92  2     0.87 Ans. ______________________________________________________________________________ 5-67 For Prob. 3-95, from Prob. 5-64 solution, with 1018 CD, DE theory yields, n = 1.40. From Table A-21, for 4140 Q&T @400F, Sy = 238 kpsi. From Prob. 3-98 solution which considered stress concentrations for Prob. 3-95 1 = 53.0 kpsi, 2 =  8.48 kpsi, and max = 30.7 kpsi DE: From Eqs. (5-13) and (5-19) S Sy 238  n y  1/2 1/2    A2   A B   B2  53.02  53.0  8.48    8.48 2     4.12 Ans. Using the 4140 versus the 1018 CD, the factor of safety increases by a factor of 4.12/1.40 = 2.94. Ans. ______________________________________________________________________________ 5-68 Design Decisions Required: Shigley’s MED, 11th edition Chapter 5 Solutions, Page 36/58     Material and condition Design factor Failure model Diameter of pin Using F = 416 lbf from Ex. 5-3,  max  32M d3 1  32M  3 d     max  Decision 1: Select the same material and condition of Ex. 5-3 (AISI 1035 steel, Sy = 81 kpsi) Decision 2: Since we prefer the pin to yield, set nd a little larger than 1. Further explanation will follow. Decision 3: Use the Distortion Energy static failure theory. Decision 4: Initially set nd = 1 S S  max  y  y  81 000 psi nd 1 1  32(416)(15)  3 d    0.922 in   (81 000)  Choose preferred size of d  1.000 in F n  (1)3 (81 000) 32(15)  530 lbf 530  1.27 416 Set design factor to nd  1.27 Adequacy Assessment:  max  Sy nd  81 000  63 800 psi 1.27 1  32(416)(15)  3 d    1.00 in (OK)   (63 800)  Shigley’s MED, 11th edition Chapter 5 Solutions, Page 37/58  (1)3 (81 000)  530 lbf 32(15) 530 n  1.27 (OK) 416 ______________________________________________________________________________ F 5-69 From Table A-20, for a thin walled cylinder made of AISI 1020 CD steel, Syt = 57 kpsi, Sut = 68 kpsi. Since r/t = 7.5/0.0625 = 120 > 10, the shell can be considered thin-wall. From the solution of Prob. 3-106 the principal stresses are 1   2  pd p(15)   60 p, 4t 4(0.0625) 3   p From Eq. (5-12) 1 2 1  2   1/ 2 ( 1   2 )2  ( 2   3 ) 2  ( 3   1 ) 2  1/ 2 (60 p  60 p ) 2  (60 p  p )2  (  p  60 p) 2   61 p For yield,  = Sy  61p = 57 (103)  p = 934 psi Ans. For rupture, 61 p  68  p  1.11 kpsi Ans. ________________________________________________________________________ 5-70 Given: AISI CD 1040 steel, ny = 2, OD = 50 mm, ID = 42 mm, L = 150 mm. Table A-20, Sy = 490 MPa At r = ri = 21 mm, Eq. (3-51) gives ro2  ri 2 252  212  t max  pi 2 2  pi 2 2  5.793 pi   1 25  21 ro  ri  r max   pi   3 Closed end, Eq. (3-52) gives p  21 p r2  l  2 i i 2  i2  2.397 pi   2 25  212 ro  ri 2 (a)  max  1   3 2  max   Sy 2n y (b) Eq. (5-12): Shigley’s MED, 11th edition 5.793 pi    pi  2   3.397 pi 3.397 pi  490 2  2  pi  36.1 MPa Ans. Chapter 5 Solutions, Page 38/58 1/ 2   1   2  2    2   3  2    3   1  2      2   1/ 2   5.793  2.397 2   2.397  12   1  5.7932    pi  5.883 pi 2   S 490   y 5.883 pi  pi  41.6 MPa Ans.   ny 2 ______________________________________________________________________________ 5-71 Given: AISI 1040 CD steel, OD = 50 mm, ID = 42 mm, L = 150 mm, pi = 40 MPa Table A-20, Sy = 490 MPa At r = ri = 21 mm, Eq. (3-51) gives r2  r2 252  212  t max  pi o2 i2  40 2 2  231.74 MPa   1 ro  ri 25  21  r max   pi  40 MPa   3 Closed end, Eq. (3-52) gives 40  21 pi ri 2 l  2 2  2  95.87 MPa   2 ro  ri 25  212 2 (a)  max  1   3 ny  2 Sy 2 max (b) Eq. (5-12): 231.74   40   135.87 MPa 2 490   1.80 Ans. 2 135.87   1/ 2   1   2  2    2   3  2    3   1  2      2   1/ 2   231.74  95.87 2   95.87  40 2   40  231.74 2    2    235.3 MPa Sy 490   2.08 Ans.   235.3 _____________________________________________________________________________ ny  5-72 For AISI 1020 HR steel, from Tables A-5 and A-20, w = 0.282 lbf/in3, Sy = 30 kpsi, and  = 0.292. Then,  = w/g = 0.282/386 lbfs2/in. For the problem, ri = 3 in, and ro = 5 in. Substituting into Eqs. (3-55), gives Shigley’s MED, 11th edition Chapter 5 Solutions, Page 39/58 t  9  25  1  3  0.292  2  0.282 2  3  0.292     r   9  25  2  386 8 3  0.292 r    225    3.006 104   2  34  2  0.5699r 2   F  r   2 r   225    r  3.006 104   2  34  2  r 2   G  r   2 r   (1) (2) For the distortion-energy theory, the von Mises stress will be     t2   t r   22  1/ 2 1/2   2  F 2 (r )  F (r )G (r )  G 2 ( r )  (3) Although it was noted that the maximum radial stress occurs at r = (rori )1/2 we are more interested as to where the von Mises stress is a maximum. One could take the derivative of Eq. (3) and set it to zero to find where the maximum occurs. However, it is much easier to plot / 2 for 3  r  5 in. Plotting Eqs. (1) through (3) results in 70 60 50 '/2 40 tan 30 radial 20 von Mises 10 0 3 3.5 4 4.5 5 r (in) It can be seen that there is no maxima, and the greatest value of the von Mises stress is the tangential stress at r = ri. Substituting r = 3 in into Eq. (1) and setting  = Sy gives 1/2     30 103     3.006 104  34  225  0.5699 32           32   1361 rad/s 60 60(1361)   13 000 rev/min Ans. 2 2 ________________________________________________________________________ n 5-73 Since r/t = 1.75/0.065 = 26.9 > 10, we can use thin-walled equations. From Eqs. (3-53) and (3-54), Shigley’s MED, 11th edition Chapter 5 Solutions, Page 40/58 di  3.5  2(0.065)  3.37 in p(di  t ) 2t 500(3.37  0.065) t   13 212 psi  13.2 kpsi 2(0.065) pd 500(3.37)  6481 psi  6.48 kpsi l  i  4t 4(0.065)  r   pi  500 psi  0.5 kpsi t  These are all principal stresses, thus, from Eq. (5-12),   1/ 2 1 2 2 2 13.2  6.48   6.48  (0.5)   0.5  13.2  2  11.87 kpsi Sy 46 n    11.87 n  3.88 Ans. ________________________________________________________________________   5-74 From Table A-20, S y  320 MPa With pi = 0, Eqs. (3-49) are  b2  r2 p  r2   t   2o o 2 1  i 2   c 1  2  ro  ri  r   r  (1)  b2  ro2 po  ri 2   r   2 2 1  2   c 1  2  ro  ri  r   r  For the distortion-energy theory, the von Mises stress is     t2   t r    2 1/ 2 r 1/ 2  b 2  2  b 2  b 2   b 2  2   c 1  2    1  2 1  2   1  2    r   r  r   r   1/ 2  b4   c 1  3 4  r   We see that the maximum von Mises stress occurs where r is a minimum at r = ri. Here, r = 0 and thus  =  t . Setting  t = Sy = 320 MPa at r = 0.1 m in Eq. (1) results in 2ro2 po 2  0.15  po Ans.  t r  r  2 2   3.6 po  320  po  88.9 MPa i ro  ri 0.152  0.12 ________________________________________________________________________ 2 Shigley’s MED, 11th edition Chapter 5 Solutions, Page 41/58 5-75 From Table A-24, Sut = 31 kpsi for grade 30 cast iron. From Table A-5,  = 0.211 and w = 0.260 lbf/in3. In Prob. 5-72, it was determined that the maximum stress was the tangential stress at the inner radius, where the radial stress is zero. Thus at the inner radius, Eq. (3-55) gives 1  3v 2  0.260 2  3.211   1  3(0.211) 2   3 v  2 2 2 2 ri    t   2    3    2ro  ri   2 5   3  3  v  386 3.211  8   8      0.01471 2  31 103  1452 rad/ sec Ans. n = 60(1452)/(2 ) = 13 870 rev/min ________________________________________________________________________ 5-76 From Table A-20, for AISI 1035 CD, Sy = 67 kpsi. From force and bending-moment equations, the ground reaction forces are found in two planes as shown. The maximum bending moment will be at B or C. Check which is larger. In the xy plane, M B  223(8)  1784 lbf  in and M C  127(6)  762 lbf  in. In the xz plane, M B  123(8)  984 lbf  in and M C  328(6)  1968 lbf  in. 1 M B  [(1784) 2  (984) 2 ] 2  2037 lbf  in 1 M C  [(762) 2  (1968) 2 ] 2  2110 lbf  in So point C governs. The torque transmitted between B and C is T = (300  50)(4) = 1000 lbf·in. The stresses are Shigley’s MED, 11th edition Chapter 5 Solutions, Page 42/58  xz  16T 16(1000) 5093   3 psi d d3 d3 x  32 M C 32(2110) 21 492   psi d3  d3 d3 For combined bending and torsion, the maximum shear stress is found from 1/2  max   x  2 2       xz   2   1/2  21.49  2  5.09 2     3   3   2d   d    11.89 kpsi d3 Max Shear Stress theory is chosen as a conservative failure theory. From Eq. (5-3) Sy 11.89 67   d  0.892 in Ans. 3 2n d 2  2 ________________________________________________________________________ 5-77 As in Prob. 5-76, we will assume this to be a statics problem. Since the proportions are unchanged, the bearing reactions will be the same as in Prob. 5-76 and the bending moment will still be a maximum at point C. Thus  max   xy plane: M C  127(3)  381 lbf  in xz plane: M C  328(3)  984 lbf  in So 1/ 2 M max   (381) 2  (984) 2   1055 lbf  in 32 M C 2(1055) 10 746 10.75 x    psi  kpsi 3 3 3 d d d d3 Since the torsional stress is unchanged,  xz  5.09 kpsi d3 For combined bending and torsion, the maximum shear stress is found from 1/2  max    2    x    xz2   2   1/2  10.75 2  5.09 2     3   3   2d   d    7.40 kpsi d3 Using the MSS theory, as was used in Prob. 5-76, gives S 7.40 67  max  y  3   d  0.762 in Ans. 2n d 2  2 ________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 43/58 5-78 For AISI 1018 HR, Table A-20 gives Sy = 32 kpsi. Transverse shear stress is a maximum at the neutral axis, and zero at the outer radius. Bending stress is a maximum at the outer radius, and zero at the neutral axis. Model (c): From Prob. 3-41, at outer radius,      17.8 kpsi Sy 32   1.80   17.8 At neutral axis, n    3 2  3  3.4   5.89 kpsi 2 Sy 32   5.43   5.89 The bending stress at the outer radius dominates. n = 1.80 n Ans. Model (d): Assume the bending stress at the outer radius will dominate, as in model (c). From Prob. 3-41,      25.5 kpsi n Sy 32   1.25   25.5 Ans. Model (e): From Prob. 3-41,      17.8 kpsi Sy 32   1.80 Ans.   17.8 Model (d) is the most conservative, thus safest, and requires the least modeling time. Model (c) is probably the most accurate, but model (e) yields the same results with less modeling effort. ________________________________________________________________________ n 5-79 For AISI 1018 HR, from Table A-20, Sy = 32 kpsi. Model (d) yields the largest bending moment, so designing to it is the most conservative approach. The bending moment is M = 312.5 lbfin. For this case, the principal stresses are 32 M 1  ,  2  3  0 d3 Using a conservative yielding failure theory use the MSS theory and Eq. (5-3) 1   3  Sy n S 32 M  y 3 d n  1/3   32 Mn  d    S  y   1/3  32  312.5  2.5  11 Thus, d   in Ans.   0.629 in  Use d  3  32 10 16     ________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 44/58 5-80 When the ring is set, the hoop tension in the ring is equal to the screw tension. t  ri 2 pi  ro2  1   ro2  ri 2  r 2  The differential hoop tension dF at r for the ring of width w, is dF  w t dr . Integration yields F  ro ri wr 2 p w t dr  2 i 2i ro  ri  ro ri ro  ro2  ro2  wri 2 pi   dr  r  1     wri pi (1) 2  ro2  ri 2  r r  r  i The screw equation is Fi  T 0.2d (2) From Eqs. (1) and (2) pi  F T  wri 0.2d wri dFx  fpi ri d Fx   2 0  fpi wri d  2 f T 0.2d 2 f Tw ri  d 0 0.2d wri Ans. ________________________________________________________________________ 5-81 T = 20 Nm, Sy = 450 MPa (a) From Prob. 5-80, T = 0.2 Fi d 20 T Fi    16.7 103  N  16.7 kN 3  0.2d 0.2  6 10     Ans. (b) From Prob. 5-80, F =wri pi 16.7 103  Fi F pi     111.3 106  Pa  111.3 MPa wri wri 12 10 3    25 / 2  103      Shigley’s MED, 11th edition Ans. Chapter 5 Solutions, Page 45/58   pi ri 2  ro2 ri 2 pi  ro2   t  2 2 1    ro  ri  r r r ro2  ri 2 (c) i   111.3 0.01252  0.0252 2 0.025  0.0125 2   185.5 MPa Ans.  r   pi  111.3 MPa (d)  max  1   3  t r 2 2 185.5  ( 111.3)   148.4 MPa 2  '  ( A2   A B   B2 )1/ 2 Ans. 1/ 2  185.52  (185.5)(111.3)  ( 111.3) 2   259.7 MPa Ans. (e) Maximum Shear Stress Theory n Sy 2 max  450  1.52 2 148.4  Ans. Distortion Energy theory Sy 450   1.73 Ans.   259.7 ________________________________________________________________________ n Shigley’s MED, 11th edition Chapter 5 Solutions, Page 46/58 5-82 The moment about the center caused by the force F is F re where re is the effective radius. This is balanced by the moment about the center caused by the tangential (hoop) stress. For the ring of width w ro Fre   r t w dr ri  re  w pi ri 2 ro2  ri 2  ro ri  ro2  r    dr r   r  w pi ri 2  ro2  ri 2  ro2 ln o  2 2  ri  F  ro  ri   2 From Prob. 5-80, F = wri pi. Therefore, re  ri  ro2  ri 2 r   ro2 ln o  2 2  ro  ri  2 ri  For the conditions of Prob. 5-80, ri = 12.5 mm and ro = 25 mm  252  12.52 12.5 25   252 ln Ans.   17.8 mm 2 2  25  12.5  2 12.5  ________________________________________________________________________ re  5-83 (a) The nominal radial interference is nom = (2.002  2.001) /2 = 0.0005 in. From Eq. (3-57), 2 2 2 2 E   ro  R  R  ri     p 2R3  ro2  ri 2   30 106  0.0005  1.52  12 12  0.6252      3072 psi Ans.  1.52  0.6252 2 13    Inner member: pi = 0, po = p = 3072 psi. At fit surface r =R = 1 in, Eq. (3-49):  12  0.6252  R 2  ri 2  t   p 2 2  3072  2  7 010 psi 2  R  ri  1  0.625  r =  p =  3072 psi Shigley’s MED, 11th edition Chapter 5 Solutions, Page 47/58 Eq. (5-13):     A2   A B   A2  1/ 2 1/2 2   7010    7010  3072    3072      6086 psi Ans. Outer member: pi = p = 3072 psi, po = 0. At fit surface r =R = 1 in, t  p Eq. (3-49):  1.52  12  ro2  R 2 3072   2 2   7 987 psi ro2  R 2  1.5  1  r =  p =  3072 psi Eq. (5-13):     A2   A B   A2  1/ 2  7987 2  7987  3072    3072   1/2  9888 psi Ans. (b) For a solid inner tube, 30 106  0.0005  1.52  12 12      4167 psi Ans. p 1.52 2 13    Inner member: t = r =  p =  4167 psi 1/2 2 2     4167    4167  4167    4167      4167 psi Ans. Outer member: pi = p = 4167 psi, po = 0. At fit surface r =R = 1 in,  1.52  12  ro2  R 2  4167  2 2   10834 psi t  p 2 ro  R 2  1.5  1  Eq. (3-49): r =  p =  4167 psi Eq. (5-13):     A2   A B   A2  1/ 2 1/ 2  10 8342  10 834  4167    4167    13 410 psi Ans. ________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 48/58 5-84 From Table A-5, E = 207 (103) MPa. The nominal radial interference is nom = (40  39.98) /2 = 0.01 mm. From Eq. (3-57), 2 2 2 2 E   ro  R  R  ri     p 2 R3  ro2  ri 2   207 103  0.01   32.52  202  20 2  10 2      26.64 MPa Ans.  32.52  102 2  203    Inner member: pi = 0, po = p = 26.64 MPa. At fit surface r =R = 20 mm, t   p Eq. (3-49):  202  102  R 2  ri 2    44.40 MPa 26.64  2 2  R 2  ri 2  20  10  r =  p =  26.64 MPa Eq. (5-13):     A2   A B   A2  1/ 2 1/ 2 2   44.40    44.40  26.64    26.64      38.71 MPa Ans. Outer member: pi = p = 26.64 MPa, po = 0. At fit surface r =R = 20 mm, t  p Eq. (3-49):  32.52  202  ro2  R 2 26.64   59.12 MPa  2 2  ro2  R 2  32.5  20  r =  p =  26.64 MPa Eq. (5-13):     A2   A B   A2  1/ 2 1/ 2  59.12 2  59.12  26.64    26.64    76.03 MPa Ans. ________________________________________________________________________ 5-85 From Table A-5, E = 207 (103) MPa. The nominal radial interference is nom = (40.008  39.972) /2 = 0.018 mm. From Eq. (3-57), Shigley’s MED, 11th edition Chapter 5 Solutions, Page 49/58 2 2 2 2 E   ro  R  R  ri     p 2R3  ro2  ri 2   207 103  0.018   32.52  20 2  202  10 2      47.94 MPa Ans.  32.52  10 2 2  203    Inner member: pi = 0, po = p = 47.94 MPa. At fit surface r =R = 20 mm, t   p Eq. (3-49):  202  102  R 2  ri 2 47.94    79.90 MPa  2 2  20 10 R 2  ri 2    r =  p =  47.94 MPa Eq. (5-13):     A2   A B   A2  1/ 2 1/ 2 2   79.90    79.90  47.94    47.94      69.66 MPa Ans. Outer member: pi = p = 47.94 MPa, po = 0. At fit surface r =R = 20 mm, t  p Eq. (3-49):  32.52  202  ro2  R 2   106.4 MPa 47.94  2 2  ro2  R 2  32.5  20  r =  p =  47.94 MPa Eq. (5-13):     A2   A B   A2  1/ 2  106.4 2  106.4  47.94    47.94   1/2  136.8 MPa Ans. ________________________________________________________________________ 5-86 From Table A-5, for carbon steel, Es = 30 kpsi, and s = 0.292. While for Eci = 14.5 Mpsi, and ci = 0.211. For ASTM grade 20 cast iron, from Table A-24, Sut = 22 kpsi. For midrange values,  = (2.001  2.0002)/2 = 0.0004 in. Eq. (3-50): Shigley’s MED, 11th edition Chapter 5 Solutions, Page 50/58 p   1 R  Eo  ro2  R 2  1  R 2  ri 2     2  2 2  i   o  2  ro  R  Ei  R  ri  0.0004   22  12   12  1 1 1 0.211 0.292        2 2 2 6 6 14.5 10   2  1  30 10   1    2613 psi At fit surface, with pi = p =2613 psi, and po = 0, from Eq. (3-50) t  p  22  12  ro2  R 2 2613   2 2   4355 psi ro2  R 2  2 1  r =  p =  2613 psi From Modified-Mohr theory, Eq. (5-32a), since A > 0 > B and  B /A  <1, 22  5.05 Ans.  A 4.355 ________________________________________________________________________ n 5-87 Sut  E = 207 GPa Eq. (3-57) can be written in terms of diameters, E d p 2 D3   3  (d o2  D 2 )( D 2  di2 )  207 10 (0.062)  (502  452 )(452  40 2 )      3 ( d o2  di2 ) (502  40 2 ) 2  45      p  15.80 MPa Outer member: From Eq. (3-50), Outer radius:  t o  452 (15.80) (2)  134.7 MPa 50 2  452  r  o  0 Inner radius:  t i  452 (15.80)  502  1    150.5 MPa 502  452  452   r i  15.80 MPa Bending (no slipping): Shigley’s MED, 11th edition I = ( /64)(504  404) = 181.1 (103) mm4 Chapter 5 Solutions, Page 51/58 At ro :   x o   Mc 75(0.05 / 2)   93.2(106 ) Pa  93.2 MPa 9 I 181.1 10 At ri :  x  i   675(0.045 / 2)  83.9 106 Pa  83.9 MPa 9 181.1 10       Torsion: J = 2I = 362.2 (103) mm4 At ro :    Tc 900(0.05 / 2)   62.1 106 Pa  62.1 MPa J 362.2 109 At ri :    900(0.045 / 2)  55.9 106 Pa  55.9 MPa 9 362.2 10 xy o xy i         Outer radius, is plane stress. Since the tangential stress is positive the von Mises stress will be a maximum with a negative bending stress. That is,  x  93.2 MPa,  y  134.7 MPa,  xy  62.1 MPa     x2   x y   y2  3 xy2  1/ 2 Eq. (5-15) 1/ 2 2 2   93.2    93.2 134.7   134.7 2  3  62.1    S y 415 no    1.84 Ans.   226  226 MPa Inner radius, 3D state of stress From Eq. (5-14) with yz = zx = 0 and x = + 83.9 MPa 1/ 2 1  (83.9  150.5) 2  (150.5  15.8) 2  ( 15.8  83.9) 2  6(55.9) 2   174 MPa    2 With x =  83.9 MPa 1/ 2 1  ( 83.9  150.5) 2  (150.5  15.8) 2  ( 15.8  83.9) 2  6(55.9) 2   230 MPa   2 Shigley’s MED, 11th edition Chapter 5 Solutions, Page 52/58 S y 415   1.80 Ans.   230 ________________________________________________________________________ 5-88 From the solution of Prob. 5-87, p = 15.80 MPa ni  Inner member: From Eq. (3-50), Outer radius:  t o   ro2  ri 2 452  402   p (15.80)  134.8 MPa o ro2  ri 2 452  40 2  r o   p  15.80 MPa Inner radius:  t i   2 452 2ro2   2 2 po   2 (15.80)  150.6 MPa ro  ri 45  402  r i  0 Bending (no slipping): I = ( /64)(504  404) = 181.1 (103) mm4 Mc 75(0.045 / 2)  83.9(106 ) Pa  83.9 MPa At ro :   x o     9 I 181.1 10  At ri :   x i    675(0.040 / 2)  74.5 106 Pa  74.5 MPa 9 181.1 10     Torsion: J = 2I = 362.2 (103) mm4 At ro :    Tc 900(0.045 / 2)   55.9 106 Pa  55.9 MPa 9 J 362.2 10 At ri :    900(0.040 / 2)  49.7 106 Pa  49.7 MPa 9 362.2 10 xy o xy i         Outer radius, 3D state of stress Shigley’s MED, 11th edition Chapter 5 Solutions, Page 53/58 From Eq. (5-14) with yz = zx = 0 and x = + 83.9 MPa   1/ 2 1 (83.9  134.8) 2  ( 134.8  15.8) 2  ( 15.8  83.9) 2  6(55.9) 2   213 MPa  2 With x =  83.9 MPa 1/ 2 1 ( 83.9  134.8) 2  ( 134.8  15.8) 2  ( 15.8  83.9) 2  6(55.9) 2   142 MPa    2 S y 415 no    1.95 Ans.   213 Inner radius, plane stress. Worst case is when x is positive  x  74.5 MPa,  y  150.6 MPa,  xy  49.7 MPa Eq. (5-15)     x2   x y   y2  3 xy2  1/ 2 2 1/2   74.52  74.5  150.6    150.6   3  49.7    216 MPa   S y 415 ni    1.92 Ans.   216 ______________________________________________________________________________ 2 5-89 For AISI 1040 HR, from Table A-20, Sy = 290 MPa. From Prob. 3-124, pmax = 65.2 MPa. From Eq. (3-50) at the inner radius R of the outer member, r 2  R2 502  252   108.7 MPa  t  p o2 65.2 ro  R 2 502  252  r   p  65.2 MPa These are principal stresses. From Eq. (5-13)  o   t2   t r   r2  1/ 2 1/ 2 2  108.7 2  108.7  65.2    65.2      152.2 MPa Sy 290   1.91 Ans.  o 152.2 ________________________________________________________________________ n 5-90 For AISI 1040 HR, from Table A-20, Sy = 42 kpsi. Shigley’s MED, 11th edition Chapter 5 Solutions, Page 54/58 From Prob. 3-125, pmax = 9 kpsi. From Eq. (3-50) at the inner radius R of the outer member, ro2  R 2 22  12 t  p 2  9 2 2  15 kpsi ro  R 2 2 1  r   p  9 kpsi These are principal stresses. From Eq. (5-13)  o   t2   t r   r2  1/ 2 2  152  15( 9)   9     1/ 2  21 kpsi S y 42   2 Ans.  o 21 ________________________________________________________________________ n 5-91 For AISI 1040 HR, from Table A-20, Sy = 290 MPa. From Prob. 3-126, pmax = 91.6 MPa. From Eq. (3-50) at the inner radius R of the outer member, r 2  R2 502  252  t  p o2 91.6   152.7 MPa ro  R 2 50 2  252  r   p  91.6 MPa These are principal stresses. From Eq. (5-13)  o   t2   t r   r2  1/2 2  152.7 2  152.7( 91.6)   91.6     1/ 2  213.8 MPa Sy 290   1.36 Ans.  o 213.8 ________________________________________________________________________ n 5-92 For AISI 1040 HR, from Table A-20, Sy = 42 kpsi. From Prob. 3-127, pmax = 12.94 kpsi. From Eq. (3-50) at the inner radius R of the outer member, ro2  R 2 22  12 t  p 2  12.94 2 2  21.57 kpsi ro  R 2 2 1  r   p  12.94 kpsi These are principal stresses. From Eq. (5-13)  o   t2   t r   r2  1/ 2 Shigley’s MED, 11th edition 2   21.57 2  21.57( 12.94)   12.94     1/ 2  30.20 kpsi Chapter 5 Solutions, Page 55/58 Sy 42   1.39 Ans.  o 30.2 ________________________________________________________________________ 5-93 For AISI 1040 HR, from Table A-20, Sy = 290 MPa. n From Prob. 3-128, pmax = 134 MPa. From Eq. (3-50) at the inner radius R of the outer member, r 2  R2 502  252  t  p o2 134   223.3 MPa ro  R 2 502  252  r   p  134 MPa These are principal stresses. From Eq. (5-13)  o   t2   t r   r2  1/ 2 1/ 2 2   223.32  223.3(134)   134      312.6 MPa Sy 290   0.93 Ans.  o 312.6 ________________________________________________________________________ n 5-94 For AISI 1040 HR, from Table A-20, Sy = 42 kpsi. From Prob. 3-129, pmax = 19.13 kpsi. From Eq. (3-50) at the inner radius R of the outer member, ro2  R 2 22  12 t  p 2  19.13 2 2  31.88 kpsi ro  R 2 2 1  r   p  19.13 kpsi These are principal stresses. From Eq. (5-13)  o   t2   t r   r2  1/2 2  31.882  31.88(19.13)   19.13    1/ 2  44.63 kpsi Sy 42   0.94 Ans.  o 44.63 ________________________________________________________________________ n 5-95    K   1 K 3   p   2 I cos    I sin cos sin  2 2   2 r 2 2 2  2 r 2 2   3    KI sin cos cos    2 2 2    2 r Shigley’s MED, 11th edition 1/2 Chapter 5 Solutions, Page 56/58 1/2    2  2 2 3 2 2 2 3    cos sin cos sin sin cos cos     2  2 2 2 2 2 2         KI  KI  cos 1  sin   cos  sin cos   2 2 2 2 2 2 r  2 r  KI 2 r Plane stress: The third principal stress is zero and KI KI     cos 1  sin  ,  2  cos 1  sin  ,  3  0 Ans. 1  2 2 2 2 2 r 2 r Plane strain: Equations for 1 and2 are still valid,. However, KI  Ans. cos 2 2 r ________________________________________________________________________  3    1   2   2 5-96 For  = 0 and plane strain, the principal stress equations of Prob. 5-95 give 1   2  (a) DE: Eq. (5-18) or, KI KI ,  3  2  2 1 2 r 2 r 1  2 2 2 1/ 2  1   1    1  2 1    2 1   1    S y   2 1  21 = Sy 1 For   , 3   1  1  2  3    1  S y    (a) MSS: Eq. (5-3) , with n =1   1  3S y 1  3 = Sy  1 3   Ans. 1  21 = Sy  1  3S y Ans. 2 3  3   1  S y  2S y   3   1 Radius of largest circle 1 2   R  1  1   1 2 3  6 ________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 57/58 5-97 Given: a = 16 mm, KIc = 80 MPa m and S y  950 MPa (a) Ignoring stress concentration F = SyA =950(100  16)(12) = 958(103) N = 958 kN Ans. (b) From Fig. 5-26: h/b = 1, a/b = 16/100 = 0.16,  = 1.3 K I    a Eq. (5-37) 80  1.3 F  (16)103 100(12) F = 329.4(103) N = 329.4 kN Ans. ________________________________________________________________________ 5-98 Given: a = 0.5 in, KIc = 72 kpsi in and Sy = 170 kpsi, Sut = 192 kpsi ro = 14/2 = 7 in, ri = (14  2)/2 = 6 in 0.5 a   0.5, ro  ri 7  6 Fig. 5-30: Eq. (5-37):   2.4 K Ic    a ri 6   0.857 ro 7  72  2.4   0.5    23.9 kpsi Eq. (3-50) at r = ro = 7 in: ri 2 pi 62 pi  t  2 2  2   23.9  2 2  2   pi  4.315 kpsi Ans. ro  ri 7 6 ________________________________________________________________________ Shigley’s MED, 11th edition Chapter 5 Solutions, Page 58/58 Chapter 7 7-1 Eq. (7-6) A  4  K f M a   3  K fsTa   4  (2.2)(70)   3  (1.8)(45)  338.4 N m 2 2 2 2 B  4  K f M m   3  K fsTm   4  (2.2)(55)   3  (1.8)(35)  265.5 N m 2 2 2 2 (a) DE-Goodman, Eq. (7-8):  16(2)  338.4 265.5          210  106  700  10 6       3 d = 27.27 (10 ) m = 27.27 mm Ans.  16n  A B d       Se Sut     1/3 1/3 (b) DE-Morrow, Eq. (7-10):  16(2)  338.4 265.5          210  106  1045  106       3 d = 26.68 (10 ) m = 26.68 mm Ans.  16n  A B   d        Se  f   1/3 1/3 (c) DE-Gerber, Eq. (7-12): 1/3 2 1/ 2      8nA    2 BSe     1 1  d       Se    ASut          6  8(2)(338.4)    2(265.5)  210  10  1 1  6  6     210  10    338.4  700  10   3 d = 25.85 (10 ) m = 25.85 mm Ans.         2 1/ 2     1/3      (d) DE-SWT, Eq. (7-14): 1/3 1/2   16  2  1/2   16n 2  338.4  2  338.4  265.5    d  A  AB     6       210  10   Se   d = 27.99 (103) m = 27.99 mm Ans. ________________________________________________________________________ Criterion d (mm) Compared to DE-Goodman 1/3 DE-Goodman DE-Morrow DE-Gerber Shigley’s MED, 11th edition 27.27 26.68 25.85 2.2% Lower 5.2% Lower Less conservative Less conservative Chapter 7 Solutions, Page 1/48 DE-SWT 27.99 2.6% Higher More conservative ______________________________________________________________________________ 7-2 Given: AISI 1050 CD steel, Ma = 650 lbf ۰ in, Mm = 500 lbf ۰ in, Ta = 400 lbf ۰ in, Tm = 300 lbf ۰ in, Se = 30 kpsi, Kf = 2.3, Kfs = 1.9, ny = 2.5. Table A-20, Sy = 77 kpsi, Sut = 100 kpsi. Eq. (7-6): A  4  K f M a   3  K fsTa   4  2.3  650    3  1.9  400   3266.9 lbf in 2 2 2 2 B  4  K f M m   3  K fsTm   4  2.3  500    3  1.9  300   2502.9 lbf in (a) DE-Goodman criterion, Eq. (7-8): 2  16n  A B   d        Se Sut   1/3  16n  A B   d          Se  f   2 2  16(2.5)  3266.9 2502.9           30  106  100  106       (b) DE-Morrow criterion, 1/3 2  f 1/3 0.119 in Ans. = 50 + 100 = 150 kpsi, Eq. (7-10):  16(2.5)  3266.9 2502.9           30  106  150  106       1/3 0.117 in Ans. (c) DE-Gerber criterion, Eq. (7-12): 1/3 2 1/ 2      8nA    2 BSe     d  1 1       Se    ASut       1/3  2 1/2    6     8  2.5  3266.9    2  2502.9  30  10       1  1    0.113 in 6 6    30  10    3266.9  100  10         (d) DE-SWT criterion, Eq. (7-14): 1/2   16n 2 d  A  AB      Se  Ans. 1/3 1/3  16  2.5   1/2   3266.92  3266.9  2502.9    0.123 in  Ans. 6     30  10  ______________________________________________________________________________ 7-3 This problem has to be done by successive trials, since Se is a function of shaft size. The material is SAE 2340 for which Sut = 175 kpsi, Sy = 160 kpsi, and HB ≥ 370. Shigley’s MED, 11th edition Chapter 7 Solutions, Page 2/48 Eq. (6-18): ka 2.00(175)  0.217 0.65 Trial #1: Choose dr = 0.75 in Eq. (6-19 ): Eq. (6-10): Eq. (6-17): kb 0.879(0.75)  0.107 0.91 Se 0.5Sut 0.5  175  87.5 kpsi Se = 0.65 (0.91)(87.5) = 51.8 kpsi d r d  2r 0.75 D  2 D / 20 0.65D d 0.75 D r  1.15 in 0.65 0.65 D 1.15 r  0.058 in 20 20 Fig. A-15-14: d d r  2r 0.75  2(0.058) 0.808 in d 0.808  1.08 dr 0.75 r 0.058  0.077 dr 0.75 Kt = 1.9 Fig. 6-26: r = 0.058 in, q = 0.90 Eq. (6-32): Kf = 1 + 0.90 (1.9 – 1) = 1.81 Fig. A-15-15: Kts = 1.5 Fig. 6-27: r = 0.058 in, qs = 0.92 Eq. (6-32): Kfs = 1 + 0.92 (1.5 – 1) = 1.46 For the DE-Goodman criteria, Eqs. (7-6) and (7-8), with d as dr, Ma = 600 lbf ۰ in, Tm = 400 lbf ۰ in, and Mm = Ta = 0,  16n  1 d r    S  e   16  2.5      0.847 in 2 1/ 2  4 K M   f a   1  Sut 2 1/2  3 K T   fs m       1/3   1 1 2 1/ 2 2 1/2      4 1.81  600  3 1.46  400         3  3   51.8 10 175 10      1/3   Ans. Trial #2: Choose dr = 0.85 in. kb 0.879(0.85)  0.107 0.894 Shigley’s MED, 11th edition Chapter 7 Solutions, Page 3/48 Se = 0.65 (0.894)(0.5)(175) = 50.8 kpsi d 0.85 D r  1.31 in 0.65 0.65 r = D / 20 = 1.31/20 = 0.0655 in Figs. A-15-14 and A-15-15: d d r  2r 0.85  2(0.0655) 0.981 in d 0.981  1.15 dr 0.85 r 0.0655  0.077 dr 0.85 With these ratios only slightly different from the previous iteration, we are at the limit of readability of the figures. We will keep the same values as before. K t 1.9, K ts 1.5, q 0.90,  K f 1.81, K fs 1.46 qs 0.92 Using Eq. (7-8) produces dr = 0.852 in. Further iteration produces no change. With dr = 0.835 in, d 0.852 D r  1.31 in 0.65 0.65 d 0.75 D 0.75(1.31) 0.983 in Ans. ______________________________________________________________________________ 7-4 F cos 20(d / 2) = TA, F = 2 TA / ( d cos 20) = 2(340) / (0.150 cos 20) = 4824 N. The maximum bending moment will be at point C, with MC = 4824(0.100) = 482.4 N∙m. Due to the rotation, the bending is completely reversed, while the torsion is constant. Thus, Ma = 482.4 N∙m, Tm = 340 N∙m, Mm = Ta = 0. For sharp fillet radii at the shoulders, from Table 7-1, Kt = 2.7, and Kts = 2.2. Examining Figs. 6-26 and 6-27 with Sut 560 MPa, conservatively estimate q = 0.8 and qs 0.9. These estimates can be checked once a specific fillet radius is determined. Eq. (6-32): K f 1  0.8(2.7  1) 2.4 K fs 1  0.9(2.2  1) 2.1 (a) We will choose to include fatigue stress concentration factors even for the static analysis to avoid localized yielding. Eq. (7-15): 2   32 K f M a  2  16 K fsTm       max   3   3 3    d    d   Shigley’s MED, 11th edition 1/ 2 Chapter 7 Solutions, Page 4/48 Eq. (7-16): Solving for d, Sy  d 3S y n    max 16  4  K M  2  3 K T  2  f a fs m    1/2 1/3  16n 1/2    4( K f M a ) 2  3( K fsTa ) 2   d    S y   16(2.5)     420  106    4  (2.4)(482.4)   3  (2.1)(340)  2 d = 0.0430 m = 43.0 mm  2 1/2     1/3 Ans. ka 3.04(560) 0.217 0.77 (b) Eq. (6-18): Assume kb = 0.85 for now. Check later once a diameter is known. Se = 0.77(0.85)(0.5)(560) = 183 MPa Selecting the DE-Goodman criteria, use Eqs. (7-6) and (7-8) with M m Ta 0.  16n  1 d    S  e  4 K M 2   f a   1/2 1  Sut 3 K T 2   fs m   1/2 1/3      16  2.5    1 1 2 1/2 2 1/2          4  2.4 482.4   3  2.1 340   6  6       183 10 560 10     0.0574 m 57.4 mm Ans.   1/3   With this diameter, we can refine our estimates for kb and q. Eq. (6-19 ): kb 1.51d  0.157 1.51 57.4   0.157 0.80 Assuming a sharp fillet radius, from Table 7-1, r = 0.02d = 0.02 (57.4) = 1.15 mm. Fig. (6-26): Fig. (6-27): q = 0.72 qs = 0.77 Iterating with these new estimates, Eq. (6-32): Eq. (6-18): Eq. (7-8): Kf = 1 + 0.72 (2.7 – 1) = 2.2 Kfs = 1 + 0.77(2.2 – 1) = 1.9 Se = 0.77(0.80)(0.5)(560) = 172.5 MPa d = 57 mm Ans. Shigley’s MED, 11th edition Chapter 7 Solutions, Page 5/48 Further iteration does not change the results. _____________________________________________________________________________ 7-5 Given: AISI 1045 CD steel, Se = 40 kpsi, Kf = 2.1, Kfs = 1.7, nf = 2.5. Table A-20, Sy = 77 kpsi, Sut = 91 kpsi. (MO)y = 0 =  18 RCz + 12(300) RCz = 200 lbf (MO)z = 0 =  18 RCy + 6(600) RCy = 200 lbf (MB)y =  6(200) =  1200 lbf ۰ in, (MB)z = 6(200) = 1200 lbf ۰ in MB    1200  2  12002 1697 lbf in Mm = 0, Ma = 1697 lbf ۰ in, Tm = Ta = 1800/2 = 900 lbf ۰ in (a) DE-Gerber criterion, Eqs. (7-6) and Eq. (7-12): A  4  2.1 1697    3 1.7  900   7604 lbf in 2 2 B  3 1.7  900   2650 lbf in 2 1/3  2 1/2    3     8  2.5  7604    2  2650  40  10      d  1 1   1.353 in 3  3    40  10    7604  91 10         (b) DE-Goodman criterion, Eq. (7-6) and Eq. (7-8): Ans. 1/3  16  2.5   7604 2650      1.408 in d   Ans.  40  103  91 103         ______________________________________________________________________________ 7-6 (a) Let Kfb = (Kf)bending, Kfa = (Kf)axial, Kfs = (Kfs)torsion. Eqs. (6-66) and (6-67): 2  32M a 4 Fa  16Ta   a   K fb  K  3  K fs fa 3 2  d d  d3        1/2 1/2 2  32M m 4Fm  16Tm     m   K fb  K  3  K fs  fa 3 2  d d   d 3      For the DE-Goodman criterion, substitute stresses into Eq. (6-41), 2 2 1/2 1    8 K M  K F d  48 K T     fb a fa a fs a     d 3  Se   nf    1/2 2 2 4  1     8K fb M m  K fa Fm d   48  K fsTm     Sut   (b) For the free-body diagram in the problem statement,  Fx 0  FCx 173 lbf Shigley’s MED, 11th edition 1 Ans. Chapter 7 Solutions, Page 6/48   M  0 7 F  3.75  173  3.5  58  F 121.68 lbf   M  0  7 F  3.5  500   F 250 lbf   T  0 3.75  500   T  T 1875 lbf in A y A z z C z C y C y C A x M m 0, M a M B 3.5 121.682  2502 973.1 lbf in Tm = 1875 lbf ۰ in, Ta = 0, Fm = 173 lbf, Fa = 0 From part (a), with d = 1.5 in,  2 1/2 1  8 2 973.1  0          3   1.53   40  10  nf   2 2 1 4    2.3  173 1.5   48  1.5  1875   3  90  10   (c) Without axial force, use Eq. (7-7):     2 1 16  1  4  2  973.1   3  3 n   1.5   40  10    1/ 2  2 1 3 1.5  1875    3 90  10     1     4.38 1/2      Ans.    4.38   Ans. 1/2 Axial force is too small to add any appreciable difference. ______________________________________________________________________________ 7-7 From Prob. 7-6, Ma = 973.1 lbf ۰ in, Tm = 1875 lbf ۰ in, Ta = 0. From Eqs. (7-6) and (7-8): 1/3   16  2.5   2 1/2 2 1/2  1 1    1.245 in Ans. d  4  2  973.1  3  1.5  1875        40  103   90  103     ______________________________________________________________________________ 7-8 From Chap. 13, Eq. (13-1), with N = number of gear teeth and d = gear pitch diameter, the gear pitch is P = N/d = 90/6 = 15 teeth/in. Table 7-2, yall = 0.005 in. Eq. (7-17), 1/4 1/4 1.5  0.007  nd yold d new dold 0.75 0.903 in Ans. yall 0.005 ______________________________________________________________________________  7-9    We have a design task of identifying bending moment and torsion diagrams which are preliminary to an industrial roller shaft design. Let point C represent the center of the span of the roller. FCy 30(8) 240 lbf FCz 0.4(240) 96 lbf Shigley’s MED, 11th edition Chapter 7 Solutions, Page 7/48 T FCz (2) 96(2) 192 lbf in T 192 FBz   128 lbf 1.5 1.5 FBy FBz tan 20 128 tan 20 46.6 lbf (a) xy-plane M O 240(5.75)  FAy (11.5)  46.6(14.25) 0 240(5.75)  46.6(14.25) FAy  62.3 lbf 11.5 M A FOy (11.5)  46.6(2.75)  240(5.75) 0 240(5.75)  46.6(2.75) FOy  131.1 lbf 11.5 Bending moment diagram: xz-plane M O 0 96(5.75)  FAz (11.5)  128(14.25) 96(5.75) 128(14.25) FAz  206.6 lbf 11.5 M A 0 FOz (11.5)  128(2.75)  96(5.75) Shigley’s MED, 11th edition Chapter 7 Solutions, Page 8/48 96(5.75)  128(2.75) 17.4 lbf 11.5 Bending moment diagram: FOz  M C  1002  ( 754)2 761 lbf in M A  ( 128) 2  ( 352) 2 375 lbf in Torque: The torque is constant from C to B, with a magnitude previously obtained of 192 lbf∙in. (b) xy-plane: 2 2 M xy  131.1x  15 x  1.75  15 x  9.75  62.3 x  11.5 Bending moment diagram: 1 Mmax = –516 lbf ∙ in and occurs at 6.12 in. M C 131.1(5.75)  15(5.75  1.75) 2 514 lbf in This is reduced from 754 lbf ∙ in found in part (a). The maximum occurs at x 6.12 in rather than C, but it is close enough. xz-plane: Shigley’s MED, 11th edition Chapter 7 Solutions, Page 9/48 2 2 M xz 17.4 x  6 x  1.75  6 x  9.75  206.6 x  11.5 1 Bending moment diagram: Let M net  M xy2  M xz2 Plot Mnet(x), 1.75 ≤ x ≤ 11.5 in Mmax = 516 lbf ∙ in at x = 6.25 in Torque: The torque rises from 0 to 192 lbf∙in linearly across the roller, then is constant to B. Ans. ______________________________________________________________________________ 7-10 This is a design problem, which can have many acceptable designs. See the solution for Prob. 7-22 for an example of the design process. ______________________________________________________________________________ 7-11 If students have access to finite element or beam analysis software, have them model the shaft to check deflections. If not, solve a simpler version of shaft for deflection. The 1 in diameter sections will not affect the deflection results much, so model the 1 in diameter as 1.25 in. Also, ignore the step in AB. From Prob. 7-9, integrate Mxy and Mxz. xy plane, with dy/dx = y'   131.1 2 62.3 3 3 2 x  5 x  1.75  5 x  9.75  x  11.5  C1 2 2 131.1 3 5 5 62.3 4 4 3 EIy  x   x  1.75  x  9.75  x  11.5  C1 x  C2  6 4 4 6 EIy   Shigley’s MED, 11th edition (1) Chapter 7 Solutions, Page 10/48 From (1), y 0 at x 0  C2 0 y 0 at x 11.5  C1 1908.4 lbf in 3 x = 0: x = 11.5: EIy' = 1908.4 EIy' = –2153.1 xz plane (treating z   )   17.4 2 206.6 3 3 2 EIz   x  2 x  1.75  2 x  9.75  x  11.5  C3 2 2 17.4 3 1 1 206.6 4 4 3 EIz  x  x  1.75  x  9.75  x  11.5  C3 x  C4 6 2 2 6 (2)   z 0 at x 0  C4 0 From (2), z 0 at x 11.5  C3 8.975 lbf in 3 x = 0: EIz' = 8.975 x = 11.5: EIz' = –683.5 At O: EI  1908.42  8.9752 1908.4 lbf in 3 At A: EI  ( 2153.1) 2  ( 683.5) 2 2259.0 lbf in 3  n (dictates size) 2259 0.000 628 rad 30  10    / 64   1.254  6 0.001 1.59 0.000 628 At gear mesh, B xy plane With I I1 in section OCA, yA  2153.1/ EI1 Since y'B/A is a cantilever, from Table A-9-1, with I I 2 in section AB Fx( x  2l ) 46.6 yB / A   (2.75)[2.75  2(2.75)]  176.2 / EI 2 2 EI 2 2 EI 2 2153.1 176.2  yB  yA  yB / A   6 4 6 30 10   / 64  1.25 30 10   / 64  0.8754         = –0.000 803 rad (magnitude greater than 0.0005 rad) Shigley’s MED, 11th edition Chapter 7 Solutions, Page 11/48 xz plane 683.5 z A  , EI1 zB  z B / A  128  2.752  2 EI 2  484 EI 2 484 683.5   0.000 751 rad 30 10   / 64  1.254 30 106   / 64  0.8754   6        B  ( 0.000 803) 2  ( 0.000 751) 2 0.00110 rad Crowned teeth must be used. Finite element results: O 5.47(10 4 ) rad Error in simplified model 3.0% 4  A 7.09(10 ) rad 11.4% 3  B 1.10(10 ) rad 0.0% The simplified model yielded reasonable results. Strength Sut 72 kpsi, S y 39.5 kpsi At the shoulder at A, x 10.75 in. From Prob. 7-9, M xy  209.3 lbf in, M xz  293.0 lbf in, T 192 lbf in M  (  209.3) 2  ( 293)2 360.0 lbf in Se 0.5(72) 36 kpsi ka 2.00(72)  0.217  0.791  0.107  1  kb  0.879   0.3  kc kd ke k f 1 Se 0.791(0.879)(36) 25.0 kpsi Fig. A-15-8: Fig. A-15-9: Fig. 6-26: D / d = 1.25, r / d = 0.03 Kts = 1.8 Kt = 2.3 q = 0.65 Shigley’s MED, 11th edition Chapter 7 Solutions, Page 12/48 Fig. 6-27: Eq. (6-32): qs = 0.70 K f 1  0.65(2.3  1) 1.85 K fs 1  0.70(1.8  1) 1.56 Using DE-Goodman, Eqs. (7-6) and (7-7) with M m Ta 0,   B  3 1.56  192    A  4  1.85  360   2 1/2 2 1/2 1332 lbf in 518.8 lbf in   13   1332 518.8    n  16  25  103  72  103     1 n = 3.25 Perform a similar analysis at the profile keyway under the gear. The main problem with the design is the undersized shaft overhang with excessive slope at the gear. The use of crowned-teeth in the gears will eliminate this problem. ______________________________________________________________________________ 7-12 through 7-21 These are design problems, which can have many acceptable designs. See the solution for Prob. 7-22 for an example of the design process. ______________________________________________________________________________ 7-22 (a) One possible shaft layout is shown in part (e). Both bearings and the gear will be located against shoulders. The gear and the motor will transmit the torque through the keys. The bearings can be lightly pressed onto the shaft. The left bearing will locate the shaft in the housing, while the right bearing will float in the housing. (b) From summing moments around the shaft axis, the tangential transmitted load through the gear will be Wt T / (d / 2) 2500 / (4 / 2) 1250 lbf The radial component of gear force is related by the pressure angle. Wr Wt tan  1250 tan 20 455 lbf  W  Wr2  Wt 2  1/2   4552  12502  1/2 1330 lbf Reactions RA and RB , and the load W are all in the same plane. From force and moment balance, RA 1330(2 /11) 242 lbf Shigley’s MED, 11th edition Chapter 7 Solutions, Page 13/48 RB 1330(9 /11) 1088 lbf M max RA (9) 242(9) 2178 lbf in Shear force, bending moment, and torque diagrams can now be obtained. (c) Potential critical locations occur at each stress concentration (shoulders and keyways). To be thorough, the stress at each potentially critical location should be evaluated. For now, we will choose the most likely critical location, by observation of the loading situation, to be in the keyway for the gear. At this point there is a large stress concentration, a large bending moment, and the torque is present. The other locations either have small bending moments, or no torque. The stress concentration for the keyway is highest at the ends. For simplicity, and to be conservative, we will use the maximum bending moment, even though it will have dropped off a little at the end of the keyway. (d) At the gear keyway, approximately 9 in from the left end of the shaft, the bending is completely reversed and the torque is steady. M a 2178 lbf in Tm 2500 lbf in M m Ta 0 From Table 7-1, estimate stress concentrations for the end-milled keyseat to be Kt = 2.14 and Kts = 3.0. For the relatively low strength steel specified (AISI 1020 CD), roughly estimate notch sensitivities of q = 0.75 and qs = 0.80, obtained by observation of Figs. 6Shigley’s MED, 11th edition Chapter 7 Solutions, Page 14/48 26 and 6-27, assuming a typical radius at the bottom of the keyseat of r / d = 0.02, and a shaft diameter of up to 3 inches. Eq. (6-32): K f 1  0.75(2.14  1) 1.9 K fs 1  0.8(3.0  1) 2.6 Eq. (6-18): ka 2.00(68)  0.217 0.801 Eq. (6-20) For estimating kb , guess d 2 in. kb (2 / 0.3) 0.107 0.816 Eq. (6-17) Se 0.801(0.816)(0.5)(68) 22.2 kpsi Selecting the DE-Goodman criteria for a conservative first design, Eq. (7-8) with Eq. (7-6): 1/2 2 1/ 2     3 K T  2     16n   4  K f M a   fs m     d     Se Sut            1/3    4 1.9 2178 2  1/2  3 2.6 2500 2  1/2         16(1.5)      d     3 3  22.2 10 68 10          d 1.60 in 1/3   Ans. With this diameter, the estimates for notch sensitivity and size factor were conservative, but close enough for a first iteration until deflections are checked. Check yielding with this diameter. Eq. (7-15): 2   32 K f M a  2  16 K fsTm       max   3   3 3 d     d     1/2 2   32(1.9)(2178)  2  16(2.6)(2500)       max   3   3 3    (1.60)    (1.60)    57 /17.4 3.3 Ans. n y S y /  max 1/2 17 374 psi 17.4 kpsi (e) Now estimate other diameters to provide typical shoulder supports for the gear and bearings. Also, estimate the gear and bearing widths. Shigley’s MED, 11th edition Chapter 7 Solutions, Page 15/48 (f) Entering this shaft geometry into beam analysis software (or Finite Element software), the following deflections are determined: Left bearing slope: 0.000 493 rad Right bearing slope:  0.000 788 rad Gear slope:  0.000 505 rad Right end of shaft slope:  0.000 788 rad Gear deflection:  0.000 927 in Right end of shaft deflection: 0.005 73 in Comparing these deflections to the recommendations in Table 7-2, everything is within typical range except the gear slope is a little high for an uncrowned gear. (g) To use a non-crowned gear, the gear slope is recommended to be less than 0.0005 rad. Since all other deflections are acceptable, we will target an increase in diameter only for the long section between the left bearing and the gear. Increasing this diameter from the proposed 1.60 in to 1.75 in, produces a gear slope of  0.000 353 rad. All other deflections are improved as well. ______________________________________________________________________________ 7-23 (a) Use the DE-Goodman fatigue criterion. The torque and moment loadings on the shaft are shown in the solution to Prob. 7-22. Candidate critical locations for strength:  Left seat keyway  Right bearing shoulder  Right keyway Table A-20 for 1030 HR: Sut 68 kpsi, S y 37.5 kpsi, H B 137 Eq. (6-10): Se 0.5(68) 34.0 kpsi Eq. (6-18): ka 2.00(68)  0.217 0.801 kc k d ke 1 Left keyway See Table 7-1 for keyway stress concentration factors, Shigley’s MED, 11th edition Chapter 7 Solutions, Page 16/48 K t 2.14   Profile keyway K ts 3.0  For an end-mill profile keyway cutter of 0.010 in radius, estimate notch sensitivities. Fig. 6-26: q 0.51 Fig. 6-27: qs 0.57 Eq. (6-32): K fs 1  qs ( Kts  1) 1  0.57(3.0  1) 2.1 K f 1  0.51(2.14  1) 1.6  0.107 Eq. (6-19):  1.875  kb    0.30  Eq. (6-17): Se 0.801(0.822)(34.0) 22.4 kpsi 0.822 A  4  (1.6)(2178)  6969.6 2 Eq. (7-6): B  3  (2.1)(2500)  9093.3 2 Eq. (7-7):  (1.8753 )  6969.6 9093.3    nf    22.4  103  68  103   16   nf = 2.9 1 Ans. Right bearing shoulder The text does not give minimum and maximum shoulder diameters for 03-series bearings (roller). Use D = 1.75 in. r 0.030 D 1.75  0.019,  1.11 d 1.574 d 1.574 Fig. A-15-9: Kt 2.4 Fig. A-15-8: Kts 1.6 Fig. 6-26: q 0.65 Fig. 6-27: qs 0.70 Eq. (6-32): K f 1  0.65(2.4  1) 1.91 K fs 1  0.70(1.6  1) 1.42  0.453  M 2178   493 lbf in  2  A  4  (1.9)(493) 1873.4 2 Eq. (7-6): Shigley’s MED, 11th edition Chapter 7 Solutions, Page 17/48 B  3  (1.42)(2500)  6148.8 2 Eq. (7-7):  (1.5743 )  1873.4 6148.8    nf    22.4  103  68  103   16   nf = 4.4 Ans. 1 Right keyway Use the same stress concentration factors as for the left keyway. There is no bending moment, so there is no alternating stress, so fatigue is not predicted at this location Yielding Check for yielding at the left keyway, where the completely reversed bending is maximum, and the steady torque is present. Using Eq. (7-15), with Mm = Ta = 0, 2   32 K f M a  2  16 K fsTm       max   3   3 3    d    d   1/ 2 2   32 1.6 2178  2  16  2.1  2500          3        1.875  3       1.875  3       1/ 2 8791 psi 8.79 kpsi S 37.5 ny  y  4.3   max 8.79 Ans. Check in smaller diameter at right end of shaft where only steady torsion exists.   16 K fsTm  2    3   max   3    d   1/2   16 2.1 2500  2       3  3       1.5      13 722 psi 13.7 kpsi S 37.5 ny  y  2.7   max 13.7 1/2 Ans. (b) One could take pains to model this shaft exactly, using finite element software. However, for the bearings and the gear, the shaft is basically of uniform diameter, 1.875 in. The reductions in diameter at the bearings will change the results insignificantly. Use E = 30 Mpsi for steel. To the left of the load, from Table A-9, case 6, Shigley’s MED, 11th edition Chapter 7 Solutions, Page 18/48  AB  d y AB Fb 1449(2)(3 x 2  2 2  112 )  (3 x 2  b 2  l 2 )  dx 6 EIl 6(30)(106 )( / 64)(1.8754 )(11) 2.4124(10 6 )(3 x 2  117) At x = 0 in:   2.823(10 4 ) rad 4 At x = 9 in:  3.040(10 ) rad To the right of the load, from Table A-9, case 6,  BC   d yBC Fa   3 x 2  6 xl  2l 2  a 2 dx 6 EIl  At x = l = 11 in: Fa 2 1449(9)(112  9 2 )  l  a2  4.342(10 4 ) rad 6 4 6 EIl 6(30)(10 )( / 64)(1.875 )(11)   Obtain allowable slopes from Table 7-2. Left bearing: n fs  Allowable slope 0.001  3.5 Actual slope 0.000 282 3 n fs  0.0008 1.8 0.000 434 2 Ans. Right bearing: Ans. Gear mesh slope: Table 7-2 recommends a minimum relative slope of 0.0005 rad. While we don’t know the slope on the next shaft, we know that it will need to have a larger diameter and be stiffer. At the moment we can say 0.0005 1.6 Ans. 0.000 304 ______________________________________________________________________________ n fs  7-24 The most likely critical locations for fatigue are at locations where the bending moment is high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-83, indicate decreasing moments in both planes to the left of A and to the right of C, with combined values at A and C of MA = 5324 lbf∙in and MC = 6750 lbf∙in. The torque is constant between A and B, with T = 2819 lbf∙in. The most likely critical locations are at the stress concentrations near A and C. The two shoulders near A can be eliminated since the shoulders near C Shigley’s MED, 11th edition Chapter 7 Solutions, Page 19/48 have the same geometry but a higher bending moment. We will consider the following potentially critical locations:  keyway at A  shoulder to the left of C  shoulder to the right of C Table A-20: Eq. (6-10): Sut = 64 kpsi, Sy = 54 kpsi Se 0.5(64) 32.0 kpsi Eq. (6-18): ka 2.00(64)  0.217 0.811 kc k d ke 1 Keyway at A Assuming r / d = 0.02 for typical end-milled keyway cutter (see discussion prior to Table 7-1), with d = 1.75 in, r = 0.02d = 0.035 in. Table 7-1: Kt = 2.14, Kts = 3.0 Fig. 6-26: q = 0.65 Fig. 6-27: qs = 0.71 K f 1  q  K t  1 1  0.65(2.14  1) 1.7 Eq. (6-32): K fs 1  qs ( K ts  1) 1  0.71(3.0  1) 2.4  0.107 Eq. (6-19):  1.75  kb    0.30  Eq. (6-17): Se 0.811(0.828)(32) 21.5 kpsi 0.828 We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. Using Eqs. (7-6) and (7-11) with Mm = Ta = 0, A  4  K f M a   4   1.7   5324   18102 lbf in 18.10 kip in 2 2 B  3  K fsTm   3   2.4   2819   11 718 lbf in 11.72 kip in 2 Shigley’s MED, 11th edition 2 Chapter 7 Solutions, Page 20/48 2 1/2   1 8 A    2 BSe     1   1      n  d 3 Se    ASut        2 1/2    8  18.10     2  11.72   21.5       1  1      1.753   21.5      18.10   64         n = 1.2 Shoulder to the left of C r / d = 0.0625 / 1.75 = 0.036, D / d = 2.5 / 1.75 = 1.43 Fig. A-15-9: Fig. A-15-8: Fig. 6-26: Fig. 6-27: Kt = 2.2 Kts = 1.8 q = 0.71 qs = 0.76 K f 1  q  K t  1 1  0.71(2.2  1) 1.9 Eq. (6-32): K fs 1  qs ( K ts  1) 1  0.76(1.8  1) 1.6  0.107 Eq. (6-19):  1.75  kb    0.30  Eq. (6-17): Se 0.811(0.828)(32) 21.5 kpsi 0.828 For convenience, we will use the full value of the bending moment at C, even though it will be slightly less at the shoulder. Using Eqs. (7-6) and (7-11) with Mm = Ta = 0, A  4  K f M a   4   1.9   6750   25 650 lbf in 25.65 kip in 2 2 B  3  K fsTm   3   1.6   2819   7812 lbf in 7.812 kip in 2 2 2 1/2     2 BSe     1   1      AS  ut        2 1/2        8  25.65  2 7.812 21.5          1 1        1.753   21.5      25.65  64         1 8A  3 n  d Se n = 0.87 Shoulder to the right of C r / d = 0.625 / 1.3 = 0.048, D / d = 1.75 / 1.3 = 1.35 Fig. A-15-9: Kt = 2.0 Shigley’s MED, 11th edition Chapter 7 Solutions, Page 21/48 Fig. A-15-8: Fig. 6-26: Fig. 6-27: Kts = 1.7 q = 0.71 qs = 0.76 K f 1  q  K t  1 1  0.71(2.0  1) 1.7 Eq. (6-32): K fs 1  qs ( K ts  1) 1  0.76(1.7  1) 1.5  0.107 Eq. (6-19):  1.3  kb  0.855   0.30  Se 0.811(0.855)(32) 22.2 kpsi Eq. (6-17): For convenience, we will use the full value of the bending moment at C, even though it will be slightly less at the shoulder. Using Eqs. (7-6) and (7-11) with Mm = Ta = 0, A  4  K f M a   4   1.7   6750   22 950 lbf in 22.95 kip in 2 2 B  3  K fsTm   3   1.5   2819   7324 lbf in 7.324 kip in 2 2 2 1/2   1 8 A    2 BSe     1   1      n  d 3 Se    ASut        2 1/2      2  7.324   22.2       1  1      1.33   22.2      22.95   64         8  22.95  n = 0.41 The critical location is at the shoulder to the right of C, where n = 0.41 and finite life is predicted. Ans. Though not explicitly called for in the problem statement, a static check for yielding is especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with Mm = Ta = 0, 2   32 K f M a  2  16 K fsTm       max   3   3 3    d    d   1/2 2   32 1.7 6750  2  16  1.5   2819          3     3 3        1.3   1.3      1/2 55 845 psi 55.8 kpsi S 54 n y  0.97   max 55.8 Shigley’s MED, 11th edition Chapter 7 Solutions, Page 22/48 This indicates localized yielding is predicted at the stress-concentration, though after localized cold-working it may not be a problem. The finite fatigue life is still likely to be the failure mode that will dictate whether this shaft is acceptable. It is interesting to note the impact of stress concentration on the acceptability of the proposed design. This problem is linked with several previous problems (see Table 1-2) in which the shaft was considered to have a constant diameter of 1.25 in. In each of the previous problems, the 1.25 in diameter was more than adequate for deflection, static, and fatigue considerations. In this problem, even though practically the entire shaft has diameters larger than 1.25 in, the stress concentrations significantly reduce the anticipated fatigue life. ______________________________________________________________________________ 7-25 For a shaft with significantly varying diameters over its length, we will choose to use shaft analysis software or finite element software to calculate the deflections. Entering the geometry from the shaft as defined in Prob. 7-24, and the loading as defined in Prob. 3-83, the following deflection magnitudes are determined: Location Left bearing O Right bearing C Left Gear A Right Gear B Slope (rad) 0.0064 0 0.0043 4 0.0026 0 0.0107 8 Deflectio n (in) 0.00000 0.00000 0.04839 0.07517 Comparing these values to the recommended limits in Table 7-2, we find that they are all out of the desired range. This is not unexpected since the stress analysis of Prob. 7-24 also indicated the shaft is undersized for infinite life. The slope at the right gear is the most excessive, so we will attempt to increase all diameters to bring it into compliance. Using Eq. (7-18) at the right gear, d new nd  dy / dx  old  d old  slope  all 1/4 (1)(0.01078)  0.0005 1/4 2.15 Multiplying all diameters by 2.15, we obtain the following deflections: Shigley’s MED, 11th edition Location Slope (rad) Left bearing O Right bearing C 0.00030 0.00020 Deflectio n (in) 0.00000 0.00000 Chapter 7 Solutions, Page 23/48 Left Gear A 0.00012 0.00225 Right Gear B 0.00050 0.00350 This brings the slope at the right gear just to the limit for an uncrowned gear, and all other slopes well below the recommended limits. For the gear deflections, the values are below recommended limits as long as the diametral pitch is less than 20. ______________________________________________________________________________ 7-26 The most likely critical locations for fatigue are at locations where the bending moment is high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-84, indicate both planes have a maximum bending moment at B. At this location, the combined bending moment from both planes is M = 4097 N∙m, and the torque is T = 3101 N∙m. The shoulder to the right of B will be eliminated since its diameter is only slightly smaller, and there is no torque. Comparing the shoulder to the left of B with the keyway at B, the primary difference between the two is the stress concentration, since they both have essentially the same bending moment, torque, and size. We will check the stress concentration factors for both to determine which is critical. Table A-20: Sut = 440 MPa, Sy = 370 MPa Keyway at A Assuming r / d = 0.02 for typical end-milled keyway cutter (see discussion prior to Table 7-1) with d = 50 mm, r = 0.02d = 1 mm. Table 7-1: Fig. 6-26: Fig. 6-27: Kt = 2.14, Kts = 3.0 q = 0.66 qs = 0.72 K f 1  q  K t  1 1  0.66(2.14  1) 1.8 Eq. (6-32): K fs 1  qs ( K ts  1) 1  0.72(3.0  1) 2.4 Shoulder to the left of B Shigley’s MED, 11th edition Chapter 7 Solutions, Page 24/48 r / d = 2 / 50 = 0.04, D / d = 75 / 50 = 1.5 Fig. A-15-9: Fig. A-15-8: Fig. 6-26: Fig. 6-27: Kt = 2.2 Kts = 1.8 q = 0.73 qs = 0.78 K f 1  q  K t  1 1  0.73(2.2  1) 1.9 Eq. (6-32): K fs 1  qs ( K ts  1) 1  0.78(1.8  1) 1.6 Examination of the stress concentration factors indicates the keyway will be the critical location. Eq. (6-10): Se 0.5(440) 220 MPa Eq. (6-18): ka 3.04(440) 0.217 0.811  0.107 Eq. (6-19): Eq. (6-17):  50  kb  0.818   7.62  kc k d ke 1 Se 0.811(0.818)(220) 150 MPa We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. Using Eqs. (7-6) and (7-11) with Mm = Ta = 0, A  4  K f M a   4   1.8   4097   14 750 N m 2 2 B  3  K fsTm   3   2.4   3101  12 890 N m 2 2 2 1/2        2 BS 1 8A  e  3 1   1      n  d Se    ASut        2 1/2    6    2 12 890 150 10     8  14 750           1  1  3 6  6       0.050   150   10     14 750   440   10        n = 0.23 Infinite life is not predicted. Ans. Though not explicitly called for in the problem statement, a static check for yielding is especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with Mm = Ta = 0, Shigley’s MED, 11th edition Chapter 7 Solutions, Page 25/48 2   32 K f M a  2  16 K fsTm       max   3   3 3    d    d   1/2 2   32 1.8 4097  2  16  2.4   3101         3        0.050  3       0.050  3       S 370 n y  0.46   max 798 1/ 2   7.98 108 Pa 798 MPa This indicates localized yielding is predicted at the stress-concentration. Even without the stress concentration effects, the static factor of safety turns out to be 0.93. Static failure is predicted, rendering this proposed shaft design unacceptable. This problem is linked with several previous problems (see Table 1-2) in which the shaft was considered to have a constant diameter of 50 mm. The results here are consistent with the previous problems, in which the 50 mm diameter was found to slightly undersized for static, and significantly undersized for fatigue. Though in the current problem much of the shaft has larger than 50 mm diameter, the added contribution of stress concentration limits the fatigue life. ______________________________________________________________________________ 7-27 For a shaft with significantly varying diameters over its length, we will choose to use shaft analysis software or finite element software to calculate the deflections. Entering the geometry from the shaft as defined in Prob. 7-26, and the loading as defined in Prob. 3-84, the following deflection magnitudes are determined: Location Slope (rad) 0.01445 0.01843 0.00358 0.00366 Left bearing O Right bearing C Left Gear A Right Gear B Deflection (mm) 0.000 0.000 3.761 3.676 Comparing these values to the recommended limits in Table 7-2, we find that they are all well out of the desired range. This is not unexpected since the stress analysis in Prob. 7-26 also indicated the shaft is undersized for infinite life. The transverse deflection at the left gear is the most excessive, so we will attempt to increase all diameters to bring it into compliance. Using Eq. (7-17) at the left gear, assuming from Table 7-2 an allowable deflection of yall = 0.01 in = 0.254 mm, d new nd yold  d old yall Shigley’s MED, 11th edition 1/4 (1)(3.761)  0.254 1/4 1.96 Chapter 7 Solutions, Page 26/48 Multiplying all diameters by 2, we obtain the following deflections: Location Slope Deflection (rad) (mm) Left bearing O 0.00090 0.000 Right bearing C 0.00115 0.000 Left Gear A 0.00022 0.235 Right Gear B 0.00023 0.230 This brings the deflection at the gears just within the limit for a spur gear (assuming P < 10 teeth/in), and all other deflections well below the recommended limits. ______________________________________________________________________________ 7-28 (a) Label the approximate locations of the effective centers of the bearings as A and B, the fan as C, and the gear as D, with axial dimensions as shown. Since there is only one gear, we can combine the radial and tangential gear forces into a single resultant force with an accompanying torque, and handle the statics problem in a single plane. From statics, the resultant reactions at the bearings can be found to be RA = 209.9 lbf and RB = 464.5 lbf. The bending moment and torque diagrams are shown, with the maximum bending moment at D of MD = 209.9(6.98) = 1459 lbf∙in and a torque transmitted from D to C of T = 633 (8/2) = 2532 lbf∙in. Due to the shaft rotation, the bending stress on any stress element will be completely reversed, while the torsional stress will be steady. Since we do not have any information about the fan, we will ignore any axial load that it would introduce. It would not likely contribute much compared to the bending anyway. Shigley’s MED, 11th edition Chapter 7 Solutions, Page 27/48 Potentially critical locations are identified as follows:  Keyway at C, where the torque is high, the diameter is small, and the keyway creates a stress concentration.  Keyway at D, where the bending moment is maximum, the torque is high, and the keyway creates a stress concentration.  Groove at E, where the diameter is smaller than at D, the bending moment is still high, and the groove creates a stress concentration. There is no torque here, though.  Shoulder at F, where the diameter is smaller than at D or E, the bending moment is still moderate, and the shoulder creates a stress concentration. There is no torque here, though.  The shoulder to the left of D can be eliminated since the change in diameter is very slight, so that the stress concentration will undoubtedly be much less than at D. Table A-20: Eq. (6-10): Sut = 68 kpsi, Sy = 57 kpsi Se 0.5(68) 34.0 kpsi Eq. (6-18): ka 2.00(68)  0.217 0.801 Keyway at C Shigley’s MED, 11th edition Chapter 7 Solutions, Page 28/48 Since there is only steady torsion here, only a static check needs to be performed. We’ll use the maximum shear stress theory.  Eq. (5-3): Tr 2532  1.00 / 2   12.9 kpsi J   1.004  / 32 ny  Sy / 2   57 / 2 2.21 12.9 Keyway at D Assuming r / d = 0.02 for typical end-milled keyway cutter (see discussion prior to Table 7-1), with d = 1.75 in, r = 0.02d = 0.035 in. Table 7-1: Fig. 6-26: Fig. 6-27: Kt = 2.14, Kts = 3.0 q = 0.66 qs = 0.72 K f 1  q  K t  1 1  0.66(2.14  1) 1.8 Eq. (6-32): K fs 1  qs ( K ts  1) 1  0.72(3.0  1) 2.4  0.107 Eq. (6-19):  1.75  kb    0.30  Eq. (6-17): Se 0.801(0.828)(34.0) 22.5 kpsi 0.828 We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. Using Eqs. (7-6) and (7-11) with Mm = Ta = 0, A  4  K f M a   4   1.8   1459   5252 lbf in 5.252 kip in 2 2 B  3  K fsTm   3   2.4   2532   10 525 lbf in 10.53 kip in 2 2 2 1/2   1 8 A    2 BSe     1   1      n  d 3 Se    ASut        2 1/2    8  5.252     2  10.53  22.5       1  1      1.753   22.5      5.252   68         n = 3.39 Ans. Groove at E Shigley’s MED, 11th edition Chapter 7 Solutions, Page 29/48 We will assume Figs. A-15-14 is applicable since the 2 in diameter to the right of the groove is relatively narrow and will likely not allow the stress flow to fully develop. (See Fig.7-9 for the stress flow concept.) r / d = 0.1 / 1.55 = 0.065, D / d = 1.75 / 1.55 = 1.13 Fig. A-15-14: Kt = 2.1 Fig. 6-26: q = 0.76 K f 1  q  K t  1 1  0.76(2.1  1) 1.8 Eq. (6-32):  0.107  1.55  kb  0.839   0.30  Eq. (6-19): Eq. (6-17): Se 0.801(0.839)(34) 22.8 kpsi Using Eqs. (7-6) and (7-11) with Mm = Ta = Tm = 0, A  4  K f M a   4   1.8   1115   4122 lbf in 4.122 kip in B=0 2 1/2   1 8 A    2 BSe     1   1      n  d 3 Se    ASut        1/2 8  4.122  1   0 2   1     1.553  22.8  2  2   n = 4.04  Ans. Shoulder at F Fig. A-15-9: Fig. 6-26: Eq. (6-32): r / d = 0.125 / 1.40 = 0.089, D / d = 2.0 / 1.40 = 1.43 Kt = 1.7 q = 0.78 K f 1  q  K t  1 1  0.78(1.7  1) 1.5  0.107 Eq. (6-19): Eq. (6-17):  1.40  kb  0.848   0.30  Se 0.801(0.848)(34) 23.4 kpsi Using Eqs. (7-6) and (7-11) with Mm = Ta = Tm = 0, A  4  K f M a   4   1.5   845   2535 lbf in 2.535 kip in B=0 2 Shigley’s MED, 11th edition 2 Chapter 7 Solutions, Page 30/48 2 1/2   1 8 A    2 BSe     1   1      n  d 3 Se    ASut        1/2 8  2.535  1   0  2   1      1.403   23.4   n = 4.97  Ans. (b) The deflection will not be much affected by the details of fillet radii, grooves, and keyways, so these can be ignored. Also, the slight diameter changes, as well as the narrow 2.0 in diameter section, can be neglected. We will model the shaft with the following three sections: Sectio n 1 2 3 Diamete r (in) 1.00 1.70 1.40 Lengt h (in) 2.90 7.77 2.20 The deflection problem can readily (though tediously) be solved with singularity functions. For examples, see Ex. 4-7, or the solution to Prob. 7-29. Alternatively, shaft analysis software or finite element software may be used. Using any of the methods, the results should be as follows: Location Left bearing A Right bearing B Fan C Gear D Slope (rad) 0.00029 0 0.00040 0 0.00029 0 0.00014 6 Deflectio n (in) 0.000000 0.000000 0.000404 0.000928 Comparing these values to the recommended limits in Table 7-2, we find that they are all within the recommended range. ______________________________________________________________________________ 7-29 Shaft analysis software or finite element software can be utilized if available. Here we will demonstrate how the problem can be simplified and solved using singularity functions. Shigley’s MED, 11th edition Chapter 7 Solutions, Page 31/48 Deflection: First we will ignore the steps near the bearings where the bending moments are low. Thus let the 30 mm dia. be 35 mm. Secondly, the 55 mm dia. is very thin, 10 mm. The full bending stresses will not develop at the outer fibers so full stiffness will not develop either. Thus, ignore this step and let the diameter be 45 mm. Statics: Left support: R1 7(315  140) / 315 3.889 kN Right support: R2 7(140) / 315 3.111 kN Determine the bending moment at each step. x(mm) 0 40 100 140 210 275 M(N ∙ m) 0 155.56 388.89 544.44 326.67 124.44 31 5 0 I35 = (/64)(0.0354) = 7.366(10-8) m4, I40 = 1.257(10-7) m4, I45 = 2.013(10-7) m4 Plot M/I as a function of x. x(m) 0 0.04 0.04 0.1 0.1 0.14 0.14 0.21 0.21 0.275 0.275 0.315 M/I (109 N/m3) 0 2.112 1.2375 3.094 1.932 2.705 2.705 1.623 2.6 0.99 1.6894 0 Step Slope 52.8 Slope –0.8745 30.942 –21.86 –1.162 19.325 –11.617 0 –15.457 –34.78 0.977 -24.769 -9.312 0.6994 -42.235 -17.47 The steps and the change of slopes are evaluated in the table. From these, the function M/I can be generated: Shigley’s MED, 11th edition Chapter 7 Solutions, Page 32/48 M / I   52.8 x  0.8745 x  0.04  21.86 x  0.04  1.162 x  0.1  1 1 0  11.617 x  0.1  34.78 x  0.14  0.977 x  0.21 0 1 1  9.312 x  0.21  0.6994 x  0.275 0 0 1  17.47 x  0.275  109  Integrate twice: E dy  1 2 1  26.4 x 2  0.8745 x  0.04  10.93 x  0.04  1.162 x  0.1 dx  2 2 1  5.81 x  0.1  17.39 x  0.14  0.977 x  0.21 2 1  C1  109 (1)  2  0.581 x  0.1  4.655 x  0.21  0.6994 x  0.275  8.735 x  0.275 Ey   8.8 x3  0.4373 x  0.04  3.643 x  0.04  3 3  1.937 x  0.1  5.797 x  0.14  0.4885 x  0.21 2 3 3  1.552 x  0.21  0.3497 x  0.275 2 2 2  2.912 x  0.275 3  C1x  C2  109  Boundary conditions: y = 0 at x = 0 yields C2 = 0; y = 0 at x = 0.315 m yields C1 = –0.295 25 N/m2. Equation (1) with C1 = –0.295 25 provides the slopes at the bearings and gear. The following table gives the results in the second column. The third column gives the results from a similar finite element model. The fourth column gives the results of a full model which models the 35 and 55 mm diameter steps. x (mm) 0 140 315  (rad) F.E. Model Full F.E. Model –0.001 4260 –0.000 1466 0.001 3120 –0.001 4270 –0.000 1467 0.001 3280 –0.001 4160 –0.000 1646 0.001 3150 The main discrepancy between the results is at the gear location (x = 140 mm). The larger value in the full model is caused by the stiffer 55 mm diameter step. As was stated earlier, this step is not as stiff as modeling implicates, so the exact answer is somewhere between the full model and the simplified model which in any event is a small value. As expected, modeling the 30 mm dia. as 35 mm does not affect the results much. It can be seen that the allowable slopes at the bearings are exceeded. Thus, either the load has to be reduced or the shaft “beefed” up. If the allowable slope is 0.001 rad, then the maximum load should be Fmax = (0.001/0.001 426)7 = 4.91 kN. With a design factor this would be reduced further. To increase the stiffness of the shaft, apply Eq. (7-18) to the most offending deflection (at Shigley’s MED, 11th edition Chapter 7 Solutions, Page 33/48 x = 0) to determine a multiplier to be used for all diameters. d new nd  dy / dx  old  d old  slope  all 1/4 (1)(0.0014260)  0.001 1/4 1.093 Form a table: Old d, mm New ideal d, mm Rounded up d, mm 20.00 30.00 35.00 40.00 45.00 55.00 21.86 32.79 38.26 43.72 49.19 60.12 22.00 34.00 40.00 44.00 50.00 62.00 Repeating the full finite element model results in x = 0:  = –9.30  10-4 rad x = 140 mm:  = –1.09  10-4 rad x = 315 mm:  = 8.65  10-4 rad This is well within our goal. Have the students try a goal of 0.0005 rad at the gears. Strength: Due to stress concentrations and reduced shaft diameters, there are a number of locations to look at. A table of nominal stresses is given below. Note that torsion is only to the right of the 7 kN load. Using  = 32M/(d3) and  = 16T/(d3), x (mm)  (MPa)  (MPa)  (MPa) 0 15 40 100 110 140 210 275 300 330 0 22.0 37.0 61.9 47.8 60.9 52.0 39.6 17.6 0 0 0 0 0 0 6 8.5 12.7 20.2 68.1 0 22.0 37.0 61.9 47.8 61.8 53.1 45.3 39.2 118.0 Table A-20 for AISI 1020 CD steel: Sut = 470 MPa, Sy = 390 MPa At x = 210 mm: ka 3.04(470)  0.217 0.800 Eq. (6-18): Eq. (6-19): Eq. (6-17): Fig. A-15-8: Fig. A-15-9: Fig. 6-26: Fig. 6-27: Eq. (6-32): kb (40 / 7.62)  0.107 0.837 Se = 0.800 (0.837)(0.5)(470) = 157 MPa D / d = 45 / 40 = 1.125, r / d = 2 / 40 = 0.05 Kts = 1.4 Kt = 1.9 q = 0.75 qs = 0.79 Kf = 1 + 0.75(1.9 –1) = 1.68 Kf s = 1 + 0.79(1.4 – 1) = 1.32 Shigley’s MED, 11th edition Chapter 7 Solutions, Page 34/48 Using DE-Goodman, from Eqs. (7-6) and (7-7), with Mm = Ta = 0, A  4   1.68   326.67   1097.6 2 B  3   1.32   107   244.6 2   0.043   1097.6 244.6    n   157  106  390  106   16   n 1.65 1 At x = 330 mm: The von Mises stress is the highest but it comes from the steady torque only. So we will do a check on yield, using Eqs. (7-15) and (7-16).   16(1.32)(107)  2    max  3    3    (0.02 )   1/2 155.7 MPa S 390 n y  2.5   max 155.7 Check the other locations. If worse-case is at x = 210 mm, the changes discussed for the slope criterion will improve the strength issue. ______________________________________________________________________________ 7-30 and 7-31 With these design tasks each student will travel different paths and almost all details will differ. The important points are  The student gets a blank piece of paper, a statement of function, and some constraints – explicit and implied. At this point in the course, this is a good experience.  It is a good preparation for the capstone design course.  The adequacy of their design must be demonstrated and possibly include a designer’s notebook.  Many of the fundaments of the course, based on this text and this course, are useful. The student will find them useful and notice that he/she is doing it.  Don’t let the students create a time sink for themselves. Tell them how far you want them to go. ______________________________________________________________________________ 7-32 This task was once given as a final exam problem. This problem is a learning experience. Following the task statement, the following guidance was added.   Take the first half hour, resisting the temptation of putting pencil to paper, and decide what the problem really is. Take another twenty minutes to list several possible remedies. Shigley’s MED, 11th edition Chapter 7 Solutions, Page 35/48  Pick one, and show your instructor how you would implement it. The students’ initial reaction is that he/she does not know much from the problem statement. Then, slowly the realization sets in that they do know some important things that the designer did not. They knew how it failed, where it failed, and that the design wasn’t good enough; it was close, though. Also, a fix at the bearing seat lead-in could transfer the problem to the shoulder fillet, and the problem may not be solved. To many students’ credit, they chose to keep the shaft geometry, and selected a new material to realize about twice the Brinell hardness. ______________________________________________________________________________ 7-33 In Eq. (7-22) set d4 d2 I , A 64 4 to obtain 2     d  gE       l   4  (1) or d 4l 2 2  gE (2) (a) From Eq. (1) and Table A-5 2 9     0.025  9.81(207)(10 )   883 rad/s    76.5  103   0.6   4  Ans. (b) From Eq. (1), we observe that the critical speed is linearly proportional to the diameter. Thus, to double the critical speed, we should double the diameter to d = 50 mm. Ans. (c) From Eq. (2), 2 d l  4 l gE  Since d / l is the same regardless of the scale, l constant 0.6(883) 529.8 529.8  1766 rad/s Ans. 0.3 Shigley’s MED, 11th edition Chapter 7 Solutions, Page 36/48 Thus the first critical speed doubles. ______________________________________________________________________________ 7-34 From Prob. 7-33,  883 rad/s A 4.909  10 4  m 2 , I 1.917  10  8  m 4 , E 207(109 ) Pa,  7.65  10 4  N/m3 w  Al 4.909  10  4  7.65  10 4  (0.6) 22.53 N One element: Eq. (7-24): 11  0.3(0.3)  0.62  0.32  0.32  6(207)  10  (1.917)  10 9 8  y1 w111 22.53(1.134) 10 6 y12 6.528  10 10   1.134  10 6  m/N  (0.6)  2.555  10  m 5   wy 22.53(2.555) 10  5 5.756 10  4  wy 2 22.53(6.528)  10  10  1.471 10 8  5.756  10 4  wy 1  g  9.81 620 rad/s wy 2 1.471 10 8  (30% low) Two elements: 11  22  12  21  0.45(0.15)  0.62  0.452  0.152  6(207)  10  (1.917)  10 9 8  (0.6) 6.379  10 7  m/N 0.15(0.15)(0.62  0.152  0.152 ) 4.961 10 7 m/N 9 8 6(207) 10 (1.917) 10 (0.6)             y1  y2 w111  w212 11.265(6.379) 10 7  11.265(4.961) 10 7 1.277 10 5 m y12  y22 1.632  10 10  wy 2(11.265)(1.277) 10 5 2.877 10 4  wy 2(11.265)(1.632)  10 9 2 Shigley’s MED, 11th edition     10  3.677  10  Chapter 7 Solutions, Page 37/48    2.877 10 4 1  9.81   3.677 10 9   876 rad/s   (0.8% low) Three elements: 11  33   22   0.5(0.1) 0.62  0.52  0.12  6(207) 10 (1.917) 10 8  (0.6) 0.3(0.3)  0.62  0.32  0.32  6(207)  10  (1.917)  10 9 12  32  13    9 8   (0.6)    6(207) 10 (1.917) 10 8 6(207)  10  (1.917)  10   (0.6) 0.1(0.1)  0.62  0.12  0.12    3.500 10 7 m/N 1.134  10 6  m/N 0.3(0.1) 0.62  0.32  0.12 9    5.460 10 7 m/N 2.380  10  7  m/N  (0.6) y 7.51  3.500  10   5.460  10   2.380  10   8.516  10  y 7.51  5.460  10   1.134  10   5.460  10   1.672  10  y 7.51  2.380  10   5.460  10   3.500  10   8.516  10  wy 7.51  8.516  10   1.672  10   8.516  10   2.535  10  wy 7.51  8.516  10     1.672  10     8.516  10    3.189  10  9 8 7 7 7 6 7 6 7 5 7 7 7 6 1 2 3 6 5 6 2    2.535 10 4 1  9.81   3.189 10 9 2 6 5 2 4 6 2 9   883 rad/s   The result is the same as in Prob. 7-33. The point was to show that convergence is rapid using a static deflection beam equation. The method works because:  If a deflection curve is chosen which meets the boundary conditions of momentfree and deflection-free ends, as in this problem, the strain energy is not very sensitive to the equation used.  Since the static bending equation is available, and meets the moment-free and deflection-free ends, it works. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 7 Solutions, Page 38/48 7-35 (a) For two bodies, Eq. (7-26) is ( m111  1/  2 ) m212 m1 21 ( m2 22  1/  2 ) 0 Expanding the determinant yields, 2  1   1   2   (m111  m2 22 )  2   m1m2 (11 22  12 21 ) 0    1  (1) 2 2 Eq. (1) has two roots 1/ 1 and 1/ 2 . Thus  1 1  1 1   2  2   2  2  0   1    2  or, 2 2 1  1   1  1   1   1  2    2  2      2   2  0     1 2      1   2  (2) Equate the third terms of Eqs. (1) and (2), which must be identical. 1 1 m1m2 (11 22  12 21 ) 12 22  1 12 m1m2 (11 22  12 21 ) 22 and it follows that 1 2  1 g2 w1w2 (11 22  12 21 ) Ans. (b) In Ex. 7-5, part (b), the first critical speed of the two-disk shaft (w1 = 35 lbf, w 2 = 55 lbf) is 1 = 124.8 rad/s. From part (a), using influence coefficients, 1 3862 2  466 rad/s Ans. 124.8 35(55)  2.061(3.534)  2.234 2   10  8  ______________________________________________________________________________ 7-36 In Eq. (7-22), for 1, the term I / A appears. For a hollow uniform diameter shaft, Shigley’s MED, 11th edition Chapter 7 Solutions, Page 39/48 1  2 2 2 2   d o4  di4  / 64 I 1  d o  di   do  di  1    d o2  d i2 2 2 2 2 A 16 d o  di 4   d o  di  / 4 This means that when a solid shaft is hollowed out, the critical speed increases beyond that of the solid shaft of the same size. Ans. By how much? (1/ 4) d o2  di2 (1/ 4) d o2 d   1  i   do  2 The possible values of d i are 0 d i d o , so the range of the critical speeds is 1 1  0 to about 1 1  1 or from 1 to 2 1 . For the specific case where the inner diameter is half of the outer diameter, the ratio of the critical speeds is 2 1   do   di  1     1   2  1.12  do   do    2 Ans. ______________________________________________________________________________ 7-37 All steps will be modeled using singularity functions with a spreadsheet. Programming both loads will enable the user to first set the left load to 1, the right load to 0 and calculate 11 and 21. Then set the left load to 0 and the right to 1 to get 12 and 22. The spreadsheet shows the 11 and 21 calculation. A table for M / I vs. x is easy to make. First, draw the bending-moment diagram as shown with the data. x M x M 0 0 1 2 3 0.875 1.75 1.625 9 10 11 0.875 0.75 0.625 Shigley’s MED, 11th edition 12 0.5 4 1.5 5 6 7 1.375 1.25 1.125 13 14 15 0.375 0.25 0.125 8 1 16 0 Chapter 7 Solutions, Page 40/48 2 M (lbf-in) 1.5 1 0.5 0 0 2 4 6 8 10 12 14 16 x (in) The second-area moments are: 0  x 1 in and 15  x 16 in, I1   2 4  / 64 0.7854 in 4 1 x 9 in , I 2   2.472 4  / 64 1.833 in 4 9 x 15 in , I 3   2.7634  / 64 2.861 in 4 Divide M by I at the key points x = 0, 1, 2, 9, 14, 15, and 16 in and plot x M/I x M/I 0 0 1 1 2 2 3 4 5 6 7 8 1.1141 0.4774 0.9547 0.9547 0.8865 0.8183 0.7501 0.6819 0.6137 0.5456 9 9 10 11 12 13 14 14 15 15 0.4774 0.3058 0.2621 0.2185 0.1748 0.1311 0.0874 0.0874 0.0437 0.1592 16 0 1.2 1 M/I (lbf/in3) 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 14 16 x (in) From this diagram, one can see where changes in value (steps) and slope occur. Using a Shigley’s MED, 11th edition Chapter 7 Solutions, Page 41/48 spreadsheet, one can form a table of these changes. An example of a step is, at x = 1 in, M/I goes from 0.875/0.7854 = 1.1141 lbf/in3 to 0.875/1.833 = 0.4774 lbf/in3, a step change of 0.4774  1.1141 =  0.6367 lbf/in3. A slope change also occurs at at x = 1 in. The slope for 0  x  1 in is 1.1141/1 = 1.1141 lbf/in2, which changes to (0.9547  0.4774)/1 = 0.4774 lbf/in2, a change of 0.4774  1.1141 =  0.6367 lbf/in2. Following this approach, a table is made of all the changes. The table shown indicates the column letters and row numbers for the spreadsheet. A x 1a 1b 2 2 9a 9b 14 14 15a 15b 16 1 2 3 4 5 6 7 8 9 10 11 12 B M 0.875 0.875 1.75 1.75 0.875 0.875 0.25 0.25 0.125 0.125 0 C M/I 1.114085 0.477358 0.954716 0.954716 0.477358 0.305854 0.087387 0.087387 0.043693 0.159155 0.000000 D step 0.000000 -0.636727 0.000000 0.000000 0.000000 -0.171504 0.000000 0.000000 0.000000 0.115461 0.000000 E Slope 1.114085 0.477358 0.477358 -0.068194 -0.068194 -0.043693 -0.043693 -0.043693 -0.043693 -0.159155 -0.159155 F  Slope 0.000000 -0.636727 0.000000 -0.545552 0.000000 0.024501 0.000000 0.000000 0.000000 -0.115461 0.000000 The equation for M / I in terms of the spreadsheet cell locations is: 0 1 M / I E2 ( x)  D3 x  1  F3 x  1  F5 x  2 0 1 1 0 D7 x  9  F7 x  9  D11 x  15  F11 x  15 1 Integrating twice gives the equation for Ey. Assume the shaft is steel. Boundary conditions y = 0 at x = 0 and at x = 16 inches provide integration constants (C1 =  4.906 lbf/in and C2 = 0). Substitution back into the deflection equation at x = 2 and 14 in provides the  ’s. The results are: 11 = 2.917(10–7) and 12 = 1.627(10–7). Repeat for F1 = 0 and F2 = 1, resulting in 21 = 1.627(10–7) and 22 = 2.231(10–7). This can be verified by finite element analysis. y1 y2 y12  wy  18(2.917)(10 7 )  32(1.627)(10 7 ) 1.046(10  5 )  18(1.627)(10 7 )  32(2.231)(10 7 ) 1.007(10 5 )  1.093(10 10 ), y22  1.014(10  10 )  5.105(10 4 ),  wy 2  5.212(10 9 ) Neglecting the shaft, Eq. (7-23) gives Shigley’s MED, 11th edition Chapter 7 Solutions, Page 42/48 1  386 5.105(10 4 )  6149 rad/s or 58 720 rev/min 5.212(10 9 ) Ans. Without the loads, we will model the shaft using 2 elements, one between 0  x  9 in, and one between 0  x  16 in. As an approximation, we will place their weights at x = 9/2 = 4.5 in, and x = 9 + (16  9)/2 = 12.5 in. From Table A-5, the weight density of steel is  = 0.282 lbf/in3. The weight of the left element is    w1   d 2l 0.282    22  1  2.472 2  8   11.7 lbf 4  4 The right element is   w2 0.282    2.7632  6   22  1  11.0 lbf  4 The spreadsheet can be easily modified to give 11 9.605  10 7  , 12  21 5.718  10  7  ,  22 5.472  10  7  y1 1.753  10 5  , y2 1.271 10  5  y12 3.072  10 10  , y22 1.615  10 10   wy 3.449  10  ,  wy 4 2 5.371 10 9   3.449  10  4    4980 rad/s 1  386  9  5.371 10   A finite element model of the exact shaft gives 1 = 5340 rad/s. The simple model is 6.8% low. Combination: Using Dunkerley’s equation, Eq. (7-32): 1 1 1    1 3870 rad/s Ans. 2 2 1 6149 4980 2 ______________________________________________________________________________ 7-38 We must not let the basis of the stress concentration factor, as presented, impose a viewpoint on the designer. Table A-16 shows Kts as a decreasing monotonic as a function of Shigley’s MED, 11th edition Chapter 7 Solutions, Page 43/48 a/D. All is not what it seems. Let us change the basis for data presentation to the full section rather than the net section.  K ts 0 K ts 0 32T  32T  K ts  K ts  3 3   AD D  Therefore K ts  K ts A Form a table: K ts has the following attributes:     It exhibits a minimum; It changes little over a wide range; Its minimum is a stationary point minimum at a / D  0.100; Our knowledge of the minima location is 0.075 ( a / D) 0.125 We can form a design rule: In torsion, the pin diameter should be about 1/10 of the shaft diameter, for greatest shaft capacity, that is, a  0.10 D. Ans. However, it is not catastrophic if one forgets the rule. ______________________________________________________________________________ 7-39 From the solution to Prob. 3-83, the torque to be transmitted through the key from the gear to the shaft is T = 2819 lbf∙in. From Prob. 7-24, the nominal shaft diameter supporting the gear is 1.00 in. From Table 7-6, a 0.25 in square key is appropriate for a 1.00 in shaft diameter. The force applied to the key is T 2819 F  5638 lbf r 1.00 / 2 Shigley’s MED, 11th edition Chapter 7 Solutions, Page 44/48 Selecting 1020 CD steel for the key, with Sy = 57 kpsi, and using the distortion-energy theory, Ssy = 0.577 Sy = (0.577)(57) = 32.9 kpsi. Failure by shear across the key: F F  A tl S S n  sy  sy F / tl    l 1.1 5638  nF  0.754 in tS sy 0.25  32 900  Failure by crushing:  F F n Sy  Sy   2 F /  tl  A  t / 2 l  l 2 Fn 2  5638   1.1  0.870 in tS y 0.25  57   103  Select ¼-in square key, 7/8 in long, 1020 CD steel. Ans. ______________________________________________________________________________ 7-40 From the solution to Prob. 3-84, the torque to be transmitted through the key from the gear to the shaft is T = 3101 N∙m. From Prob. 7-26, the nominal shaft diameter supporting the gear is 50 mm. To determine an appropriate key size for the shaft diameter, we can either convert to inches and use Table 7-6, or we can look up standard metric key sizes from the internet or a machine design handbook. It turns out that the recommended metric key for a 50 mm shaft is 14 x 9 mm. Since the problem statement specifies a square key, we will use a 14 x 14 mm key. For comparison, using Table 7-6 as a guide, for d = 50 mm = 1.97 in, a 0.5 in square key is appropriate. This is equivalent to 12.7 mm. A 14 x 14 mm size is conservative, but reasonable after rounding up to standard sizes. The force applied to the key is T 3101 F  124  103  N r 0.050 / 2 Selecting 1020 CD steel for the key, with Sy = 390 MPa, and using the distortion-energy theory, Ssy = 0.577 Sy = 0.577(390) = 225 MPa. Failure by shear across the key:  n F F  A tl S sy   S sy F /  tl  Shigley’s MED, 11th edition  1.1 124   103  nF l  0.0433 m 43.3 mm tS sy  0.014   225   10 6  Chapter 7 Solutions, Page 45/48 Failure by crushing:  n Sy   F F  A  t / 2 l Sy 2 F /  tl   2  124   103   1.1 2 Fn l  0.0500 m 50.0 mm tS y  0.014   390   106  Select 14 mm square key, 50 mm long, 1020 CD steel. Ans. ______________________________________________________________________________ 7-41 Choose basic size D = d = 15 mm. From Table 7-9, a locational clearance fit is designated as 15H7/h6. From Table A-11, the tolerance grades are D = 0.018 mm and d = 0.011 mm. From Table A-12, the fundamental deviation is F = 0 mm. Hole: Eq. (7-36) : Shaft: Eq. (7-37): 7-42 Dmax = D + D = 15 + 0.018 = 15.018 mm Ans. Dmin = D = 15.000 mm Ans. Ans. dmax = d + F = 15.000 + 0 = 15.000 mm dmin = d + F – d = 15.000 + 0 – 0.011 = 14.989 mm Ans. Choose basic size D = d = 1.75 in. From Table 7-9, a medium drive fit is designated as H7/s6. From Table A-13, the tolerance grades are D = 0.0010 in and d = 0.0006 in. From Table A-14, the fundamental deviation is F = 0.0017 in. Hole: Eq. (7-36 ): Dmax = D + D = 1.75 + 0.0010 = 1.7510 in Ans. Dmin = D = 1.7500 in Ans. Shaft: Eq. (7-38): dmin = d + F = 1.75 + 0.0017 = 1.7517 in Ans. dmax = d + F + d = 1.75 + 0.0017 + 0.0006 = 1.7523 in Ans. ______________________________________________________________________________ 7-43 Choose basic size D = d = 45 mm. From Table 7-9, a sliding fit is designated as H7/g6. From Table A-11, the tolerance grades are D = 0.025 mm and d = 0.016 mm. From Table A-12, the fundamental deviation is F = –0.009 mm. Hole: Eq. (7-36): Dmax = D + D = 45 + 0.025 = 45.025 mm Ans. Dmin = D = 45.000 mm Ans. Shaft: Shigley’s MED, 11th edition Chapter 7 Solutions, Page 46/48 dmax = d + F = 45.000 + (–0.009) = 44.991 mm Ans. dmin = d + F – d = 45.000 + (–0.009) – 0.016 = 44.975 mm Ans. ______________________________________________________________________________ Eq. (7-37): 7-44 Choose basic size D = d = 1.250 in. From Table 7-9, a close running fit is designated as H8/f7. From Table A-13, the tolerance grades are D = 0.0015 in and d = 0.0010 in. From Table A-14, the fundamental deviation is F = –0.0010 in. Hole: Eq. (7-36): Dmax = D + D = 1.250 + 0.0015 = 1.2515 in Ans. Dmin = D = 1.2500 in Ans. Shaft: Eq. (7-37): dmax = d + F = 1.250 + (–0.0010) = 1.2490 in Ans. dmin = d + F – d = 1.250 + (–0.0010) – 0.0010 = 1.2480 in Ans. ______________________________________________________________________________ 7-45 Choose basic size D = d = 35 mm. From Table 7-9, a locational interference fit is designated as H7/p6. From Table A-11, the tolerance grades are D = 0.025 mm and d = 0.016 mm. From Table A-12, the fundamental deviation is F = 0.026 mm. Hole: Eq. (7-36): Dmax = D + D = 35 + 0.025 = 35.025 mm Dmin = D = 35.000 mm The bearing bore specifications are within the hole specifications for a locational interference fit. Now find the necessary shaft sizes. Shaft: Eq. (7-38): dmin = d + F = 35 + 0.026 = 35.026 mm Ans. dmax = d + F + d = 35 + 0.026 + 0.016 = 35.042 mm Ans. ______________________________________________________________________________ 7-46 Choose basic size D = d = 1.5 in. From Table 7-9, a locational interference fit is designated as H7/p6. From Table A-13, the tolerance grades are D = 0.0010 in and d = 0.0006 in. From Table A-14, the fundamental deviation is F = 0.0010 in. Hole: Eq. (7-36): Dmax = D + D = 1.5000 + 0.0010 = 1.5010 in Dmin = D = 1.5000 in The bearing bore specifications exactly match the requirements for a locational interference fit. Now check the shaft. Shaft: Eq. (7-38): dmin = d + F = 1.5000 + 0.0010 = 1.5010 in dmax = d + F + d = 1.5000 + 0.0010 + 0.0006 = 1.5016 in Shigley’s MED, 11th edition Chapter 7 Solutions, Page 47/48 The shaft diameter of 1.5020 in is greater than the maximum allowable diameter of 1.5016 in, and therefore does not meet the specifications for the locational interference fit. Ans. ______________________________________________________________________________ 7-47 (a) Basic size is D = d = 35 mm. Table 7-9: H7/s6 is specified for medium drive fit. Table A-11: Tolerance grades are D = 0.025 mm and d = 0.016 mm.  0.043 mm. Table A-12: Fundamental deviation is F Dmax = D + D = 35 + 0.025 = 35.025 mm Eq. (7-36): Dmin = D = 35.000 mm Eq. (7-38): dmin = d + F = 35 + 0.043 = 35.043 mm Ans. dmax = d + F + d = 35 + 0.043 + 0.016 = 35.059 mm Ans. (b) Eq. (7-42):  min d min  Dmax 35.043  35.025 0.018 mm Eq. (7-43):  max d max  Dmin 35.059  35.000 0.059 mm pmax Eq. (7-40):   2 2 2 2 E max  d o  d d  di   2d 3  d o2  di2            207 109  0.059   602  352 352  0    115 MPa  602  0 2 353   pmin    2 2 2 2 E min  d o  d d  di     d o2  di2 2d 3    207 109  0.018   602  352 352  0    35.1 MPa  602  0 2 353       Ans.    Ans. (c) For the shaft:  t ,shaft  p  115 MPa Eq. (7-44):  r ,shaft  p  115 MPa Eq. (7-46): Eq. (5-13):     12   1 2   22  1/2  ( 115) 2  ( 115)(  115)  (  115) 2  n S y /   390 /115 3.4 1/2 115 MPa Ans. For the hub: Shigley’s MED, 11th edition Chapter 7 Solutions, Page 48/48  t ,hub  p Eq. (7-45):  602  352  d o2  d 2  234 MPa 115  2 2  d o2  d 2  60  35  Eq. (7-46):  r ,hub  p  115 MPa Eq. (5-13):     12   1 2   22  1/2  (234) 2  (234)( 115)  ( 115) 2  n S y /   600 / 308 1.9 1/ 2 308 MPa Ans. (d) A value for the static coefficient of friction for steel to steel can be obtained online or from a physics textbook as approximately f = 0.8. T ( / 2) f pmin ld 2 Eq. (7-49) ( / 2)(0.8)(35.1)  106  (0.050)(0.035) 2 2700 N m Ans. ______________________________________________________________________________ 7-48 Basic size D = 2.5 in, L = 3 in, OD = 4 in, ID = 2.5 in, f = 0.7. (a) Table 7-9: medium drive fit  H7/S6 Table A-13: D = 0.0012 in, d = 0.0007 in Table A-14: F = 0.0021 in Eq. (7-36): Dmax = D + D = 2.5 + 0.0012 = 2.5012 in Ans. Dmin = D = 2.5000 in Ans. Eq. (7-38): dmax = d +F + d = 2.5 + 0.0021 + 0.0007 = 2.5028 in Ans. dmin = d +F = 2.5 + 0.0021 = 2.5021 in Ans. (b) Eq. (7-42): min = dmin  Dmax = 2.5021  2.5012 = 0.0009 in Eq. (7-40): 2 2 2 2 6 2 2 2 2 E   d o  d   d  di   30  10  0.0009   4  2.5   2.5  0      3291 psi p 3 3 2d  d o2  d i2 42  0 2 2 2.5         Eq. (7-49):   Tmax  fpLd 2   0.7  3291 3 2.52 67 850 lbf in 67.85 kip in 2 2 (c) (i) Eq. (7-43): max = dmax  Dmin = 2.5028  2.5 = 0.0028 in Eq. (7-40): 30  106  0.0028   4 2  2.52   2.52  0 2     10 240 psi =10.24 kpsi p 3 4 2  02 2  2.5    (ii) Eq. (7-45): d2 d2 42  2.52 23.4 kpsi   t  hub  p o2 2 10.24 2 do  d 4  2.52 t  (iii)  p  10.24 kpsi Ans. Ans. hub     12   1 2   22  Shigley’s MED, 11th edition Ans. 1/2 2  23.42  23.4   10.2     10.2     1/2 29.8 kpsi Chapter 7 Solutions, Page 49/48 S y 100  3.36 Ans.   29.8 ______________________________________________________________________________ n Shigley’s MED, 11th edition Chapter 7 Solutions, Page 50/48 Chapter 6 6-1 Eq. (2-36): Eq. (6-10): Table 6-2: Eq. (6-18): Sut  3.4 H B  3.4(300)  1020 MPa Se  0.5Sut  0.5(1020)  510 MPa a  1.38, b  0.067 k a  aSutb  1.38(1020) 0.067  0.868 Eq. (6-19): kb  1.24 d 0.107  1.24(10) 0.107  0.969 Se  k a kb Se  (0.868)(0.969)(510)  429 MPa Ans. Eq. (6-17): ______________________________________________________________________________ 6-2 (a) Table A-20: Eq. (6-10): Sut = 80 kpsi Se  0.5(80)  40 kpsi Ans. (b) Table A-20: Eq. (6-10): Sut = 90 kpsi Se  0.5(90)  45 kpsi Ans. (c) Aluminum has no endurance limit. Ans. (d) Eq. (6-10): Sut > 200 kpsi, Se  100 kpsi Ans. ______________________________________________________________________________ 6-3 Sut  120 kpsi,  ar  70 kpsi Fig. 6-23: f  0.82 Eq. (6-10): Se  S e  0.5(120)  60 kpsi Eq. (6-13): ( f Sut ) 2  0.82(120)  a   161.4 kpsi Se 60 Eq. (6-14):  f Sut 1 b   log  3  Se 2 1/ b  1  0.82(120)     log    0.0716 3 60    1    70  0.0716 N   ar    Eq. (6-15):  117 000 cycles Ans.   161.4   a  ______________________________________________________________________________ 6-4 Sut  1600 MPa,  ar  900 MPa Fig. 6-23: Sut = 1600 MPa. Off the graph, so estimate f = 0.77. Eq. (6-10): Sut > 1400 MPa, so Se = 700 MPa Eq. (6-13): ( f Sut ) 2  0.77(1600)  a   2168.3 MPa Se 700 2 Shigley’s MED, 11th edition Chapter 6 Solutions, Page 1/58 Eq. (6-14):  f Sut 1 b   log  3  Se  1  0.77(1600)     log    0.081838 3 700    1 1/ b    900  0.081838 N   ar    Eq. (6-15):  46 400 cycles Ans.   2168.3   a  ______________________________________________________________________________ 6-5 Sut  230 kpsi, N  150 000 cycles Fig. 6-23, point is off the graph, so estimate: f = 0.77 Eq. (6-10): Sut > 200 kpsi, so Se  Se  100 kpsi Eq. (6-13): ( f Sut ) 2  0.77(230)  a   313.6 kpsi Se 100 Eq. (6-14):  f Sut 1 b   log  3  Se Eq. (6-12): S f  aN b  313.6(150 000) 0.08274  117.0 kpsi 2  1  0.77(230)     log    0.08274 3  100   Ans. ______________________________________________________________________________ 6-6 Sut  1100 MPa = 160 kpsi Fig. 6-23: f = 0.79 Eq. (6-10): Se  S e  0.5(1100)  550 MPa Eq. (6-13): ( f Sut ) 2  0.79(1100)  a   1373 MPa Se 550 Eq. (6-14):  f Sut 1 b   log  3  Se Eq. (6-12): S f  aN b  1373(150 000) 0.06622  624 MPa 2  1  0.79(1100)     log    0.06622 3 550    Ans. ______________________________________________________________________________ 6-7 Sut  150 kpsi, S yt  135 kpsi, N  500 cycles Fig. 6-23: f = 0.80 From Fig. 6-21, we note that below 103 cycles on the S-N diagram constitutes the lowcycle region. The stress-life approach is not very reliable in this region, but for a rough Shigley’s MED, 11th edition Chapter 6 Solutions, Page 2/58 response to this question, we can write an equation in log-log scale for the line between (100, Sut) and (103, fSut) as S f  Sut N  log f  /3 log  0.80   /3  150  500   123 kpsi Ans. The testing should be done at a completely reversed stress of 123 kpsi, which is below the yield strength, so it is possible. Ans. ______________________________________________________________________________ 6-8 d = 1.5 in, Sut = 110 kpsi Se  0.5(110)  55 kpsi Eq. (6-10): Table 6-2: a = 2.00, b =  0.217 Eq. (6-18): k a  aSut b  2.00(110) 0.217  0.721 Eq. (6-19): kb = 0.879 d 0.107= 0.879(1.5) 0.107 =0.842 Eq. (6-17): Se = kakb S e = 0.721(0.842)(55) = 33.4 kpsi Ans. ______________________________________________________________________________ 6-9 For AISI 4340 as-forged steel, Eq. (6-10): Table 6-2: Eq. (6-18): Se = 100 kpsi a = 12.7, b =  0.758 ka = 12.7(260)0.758 = 0.188 Eq. (6-19):  0.75  kb     0.30  0.107  0.907 Each of the other modifying factors is unity. Se = 0.188(0.907)(100) = 17.1 kpsi Ans. For AISI 1040: Se  0.5(113)  56.5 kpsi ka  12.7(113)0.758  0.353 kb  0.907 (same as 4340) Each of the other modifying factors is unity Se  0.353(0.907)(56.5)  18.1 kpsi Ans. Not only is AISI 1040 steel a contender, it has a superior endurance strength. Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 6 Solutions, Page 3/58 6-10 From Table A-20, Sut = 570 MPa, Sy = 310 MPa. From a free-body diagram analysis, the bearing reaction forces are found to be R1 = 3.25 kN and R2 = 9.75 kN. The shear-force and bendingmoment diagrams are shown. The critical location is at the section where the bending moment is maximum, on the outer surface where the bending stress is maximum. With a rotating shaft, the bending stress will be completely reversed. Mc 487 500(25 / 2)   317.8 MPa I ( / 64)(25) 4 Sy 310   0.98 Ans. ny   max 317.8  max   ar  (a) Yielding is predicted, on the outer surface. For some applications, this might not prevent the part from being used, so we will continue checking for fatigue. (b) Eq. (6-10): Se'  0.5S ut  0.5(570)  285 MPa Eq. (6-18): k a  aSutb  3.04(570) 0.217  0.767 Eq. (6-19): kb  1.24d 0.107  1.24(25) 0.107  0.879 Eq. (6-25): kc  1 Ans. Eq. (6-17): Se  ka kb kc S e'  (0.767)(0.879)(1)(285)  192 MPa (c) For completely reversed stress, the fatigue factor of safety can be assessed as the ratio of the endurance limit to the completely reversed stress. S 192 nf  e   0.60 Ans.  ar 317.8 Infinite life is not predicted. Use the S-N diagram to estimate the life. (d) Fig. 6-23, or Eq. (6-11): f = 0.87  f Sut  2  0.87(570)  2 Eq. (6-13): a Eq. (6-14):  f Sut  1 1  0.87(570)  b   log     log    0.1374 3 3  192   Se  192 Se 1 Eq. (6-15):  1280.8 1    b  317.8  0.1374  25444 N   ar      a   1280.8  N = 25000 cycles Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 6 Solutions, Page 4/58 6-11 From Table A-20, Sut = 400 MPa, Sy = 220 MPa. Free-body, shear-force, and bending-moment diagrams are shown. 32 M 32(45000)  max    458366 / d 3 3 3 d d The load is repeatedly applied and released, so from Eqs. (6-8) and (6-9),  m   a   max / 2  229183 / d 3 Be sure to confirm that the units are legitimate for stress in MPa and d in mm. For yielding, n y  1.5  Sy  max  220 458366 / d 3 d  14.62 mm Now check fatigue, opting for the linear Goodman criterion for simplicity of solving for the diameter. First, determine the adjusted endurance limit. Eq. (6-10): Se'  0.5Sut  0.5(400)  200 MPa Eq. (6-18): k a  aSutb  3.04(400) 0.217  0.828 Estimate the size factor from the diameter determined for yielding. It can be adjusted later. Eq. (6-19): kb  1.24 d 0.107  1.24(15) 0.107  0.93 Eq. (6-25): kc  1 Ans. Eq. (6-17): Se  ka kb kc S e'  (0.767)(0.879)(1)(285)  192 MPa (c) For completely reversed stress, the fatigue factor of safety can be assessed as the ratio of the endurance limit to the completely reversed stress. S 192 nf  e   0.60 Ans.  ar 317.8 Infinite life is not predicted. Use the S-N diagram to estimate the life. (d) Fig. 6-23, or Eq. (6-11): f = 0.87  f Sut  2  0.87(570)  Eq. (6-13): a Eq. (6-14):  f Sut 1 b   log  3  Se 192 Se 1 Eq. (6-15): 2  1280.8  1  0.87(570)     log    0.1374 3 192    1    b  317.8  0.1374 N   ar     25444   a   1280.8  N = 25000 cycles Ans. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 5/58 6-12 D = 1 in, d = 0.8 in, T = 1800 lbfin, and from Table A-20 for AISI 1020 CD, Sut = 68 kpsi, and Sy = 57 kpsi. r 0.1 D 1 (a) Fig. A-15-15:   0.125,   1.25, K ts  1.40 d 0.8 d 0.8 Get the notch sensitivity either from Fig. 6-27, or from the curve-fit Eqs. (6-33) and (6-36). Using the equations, a  0.190  2.51103   68   1.35 105   68   2.67 10 8  683   0.07335 2 qs  Eq. (6-32): 1 1  a r 1  0.812 0.07335 1 0.1 Kfs = 1 + qs (Kts  1) = 1 + 0.812(1.40  1) = 1.32 For a purely reversing torque of T = 1800 lbfin, Tr K fs 16T 1.32(16)(1800)    23 635 psi  23.6 kpsi d3  (0.8)3 J Se  0.5(68)  34 kpsi  a  K fs Eq. (6-10): Eq. (6-18): ka = 2.00(68)0.217 = 0.80 Eq. (6-19): kb = 0.879(0.8)0.107 = 0.90 Eq. (6-25): kc = 0.59 Eq. (6-17) (labeling for shear): Sse = 0.80(0.90)(0.59)(34) = 14.4 kpsi For purely reversing torsion, use Eq. (6-58) for the ultimate strength in shear. Eq. (6-58): Ssu = 0.67 Sut = 0.67(68) = 45.6 kpsi Fig. 6-23: f = 0.9 Adjusting the fatigue strength equations for shear,  0.9(45.6)  Eq. (6-13):  f S su  a Eq. (6-14):  f S su 1 b   log  3  S se S se 2 1 2 14.4  117.0 kpsi  1  0.9(45.6)     log    0.151 61 3  14.4   1    b  23.6  0.151 61  38.5 103 cycles Ans. Eq. (6-15): N  a    a 117.0     (b) Estimate the ultimate strength at the operating temperature. Eq. (6-26): ( ST S RT )750  0.98  3.5(10 4 )(750)  6.3(10 7 )750 2  0.89 Thus,  Sut 750   ST Shigley’s MED, 11th edition    S RT 750  Sut 70  0.89(68)  60.5 kpsi Chapter 6 Solutions, Page 6/58 Eq. (6-10): Eq. (6-17):  Se 750  0.5  Sut 750  0.5(60.5)  30.3 kpsi Sse = 0.80(0.90)(0.59)(30.3) = 12.9 kpsi  Note that we use kd = 1 since the ultimate strength has been adjusted for the operating temperature. S su  0.67  Sut 750  0.67  60.5  40.5 kpsi Eq. (6-58):  f S su  a 2 S se 0.9(40.5)  2  103.0 kpsi 12.9  f S su 1 b   log  3  S se  1  0.9(40.5)     log    0.150 37 3  12.9   1 1    b  23.6  0.150 37 N  a   Ans.  18.0 103 cycles   a   103.0  ______________________________________________________________________________ 6-13   L  0.6 m, Fa  2 kN, n  1.5, N  104 cycles, Sut  770 MPa, S y  420 MPa (Table A-20) First evaluate the fatigue strength. Se  0.5(770)  385 MPa k a  38.6(770) 0.650  0.51 Since the size is not yet known, assume a typical value of kb = 0.85 and check later. All other modifiers are equal to one. Eq. (6-17): Fig. 6-23: Se = 0.51(0.85)(385) = 167 MPa f = 0.83 Eq. (6-13): a Eq. (6-14): Eq. (6-12):  f Sut  Se 2  0.83(770)   2 167  2446 MPa  f Sut  1 1  0.83(770)  b   log     log    0.1943 3 3  167   Se  S f  aN b  2446(104 ) 0.1943  409 MPa Now evaluate the stress. M max  (2000 N)(0.6 m)  1200 N  m  a   max  Mc M  b / 2  6 M 6 1200  7200     3 Pa, with b in m. I b(b3 ) / 12 b3 b3 b Compare strength to stress and solve for the necessary b. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 7/58 Sf 409 106   1.5 7200 / b3 b = 0.0298 m Select b = 30 mm. n a  Since the size factor was guessed, go back and check it now. 1/ 2 Eq. (6-24): d e  0.808  hb   0.808b  0.808  30   24.2 mm 0.107  24.2  Eq. (6-19): kb    0.88   7.62  Our guess of 0.85 was slightly conservative, so we will accept the result of b = 30 mm. Checking yield,  max  Ans. 7200 106   267 MPa 0.0303 Sy 420  1.57  max 267 ______________________________________________________________________________ ny  6-14  Given: w =2.5 in, t = 3/8 in, d = 0.5 in, nd = 2. From Table A-20, for AISI 1020 CD, Sut = 68 kpsi and Sy = 57 kpsi. Eq. (6-10): Se  0.5(68)  34 kpsi Table 6-2: Eq. (6-20): Eq. (6-25): ka  2.00(68) 0.217  0.80 kb = 1 (axial loading) kc = 0.85 Eq. (6-17): Se = 0.80(1)(0.85)(34) = 23.1 kpsi Table A-15-1: d / w  0.5 / 2.5  0.2, K t  2.5 Get the notch sensitivity either from Fig. 6-26, or from the curve-fit Eqs. (6-33) and (6-35). The relatively large radius is off the graph of Fig. 6-26, so we will assume the curves continue according to the same trend and use the equations to estimate the notch sensitivity. a  0.246  3.08 103   68   1.51105   68   2.67 10 8  683   0.09799 2 1  0.836 a 1  0.09799 1 0.25 r K f  1  q ( K t  1)  1  0.836(2.5  1)  2.25 q Eq. (6-32): Shigley’s MED, 11th edition 1  Chapter 6 Solutions, Page 8/58 Fa 2.25 Fa =  3Fa A (3 / 8)(2.5  0.5) a  K f Since a finite life was not mentioned, we’ll assume infinite life is desired, so the completely reversed stress must stay below the endurance limit. 23.1 2  a 3Fa Fa  3.85 kips Ans. ______________________________________________________________________________ nf  6-15 Se  Given: D  2 in, d  1.8 in, r  0.1 in, M max  25 000 lbf  in, M min  0. From Table A-20, for AISI 1095 HR, Sut = 120 kpsi and Sy = 66 kpsi. Eq. (6-10): Se  0.5Sut  0.5 120   60 kpsi Eq. (6-18): k a  aSutb  2.00(120) 0.217  0.71 Eq. (6-23): d e  0.370d  0.370(1.8)  0.666 in Eq. (6-19): kb  0.879d e 0.107  0.879(0.666) 0.107  0.92 Eq. (6-25): kc  1 Eq. (6-17): Se  ka kb kc Se  (0.71)(0.92)(1)(60)  39.2 kpsi Fig. A-15-14: D / d  2 / 1.8  1.11, r / d  0.1/ 1.8  0.056  K t  2.1 Get the notch sensitivity either from Fig. 6-26, or from the curve-fit Eqs. (6-33) and (6-35). Using the equations, a  0.246  3.08 103  120   1.5110 5  120   2.67 108 1203   0.04770 2 1 q 1 Eq. (6-32): a r  1  0.87 0.04770 1 0.1 K f  1  q ( K t  1)  1  0.87(2.1  1)  1.96 I  ( / 64) d 4  ( / 64)(1.8) 4  0.5153 in 4 Mc 25 000(1.8 / 2)   43 664 psi  43.7 kpsi I 0.5153 0  max   min Shigley’s MED, 11th edition Chapter 6 Solutions, Page 9/58 Eqs. (6-8) and (6-9):  m  K f a  K f  43.7  0   42.8 kpsi  max   min  1.96   max   min  1.96  2 2 2  43.7  0  1 2     42.8 42.8  nf   a  m      S S 39.2 120   e ut   Eq. (6-41): n f  0.69  42.8 kpsi 1 Ans. A factor of safety less than unity indicates a finite life. Check for yielding. It is not necessary to include the stress concentration for static yielding of a ductile material. Sy 66  1.51 Ans.  max 43.7 ______________________________________________________________________________ ny  6-16  From a free-body diagram analysis, the bearing reaction forces are found to be 2.1 kN at the left bearing and 3.9 kN at the right bearing. The critical location will be at the shoulder fillet between the 35 mm and the 50 mm diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. The bending moment at this point is M = 2.1(200) = 420 kN∙mm. With a rotating shaft, the bending stress will be completely reversed.  ar  Mc 420 (35 / 2)   0.09978 kN/mm 2  99.8 MPa 4 I ( / 64)(35) This stress is far below the yield strength of 390 MPa, so yielding is not predicted. Find the stress concentration factor for the fatigue analysis. Fig. A-15-9: r/d = 3/35 = 0.086, D/d = 50/35 = 1.43, Kt =1.7 Get the notch sensitivity either from Fig. 6-26, or from the curve-fit Eqs. (6-33) and (6-35). Using the equations, with Sut = 470 MPa and r = 3 mm, a  1.24  2.25 103  (470)  1.60 10 6  (470) 2  4.11 10 10  (470)3  0.4933 1 q 1 Shigley’s MED, 11th edition a r  1  0.78 0.4933 1 3 Chapter 6 Solutions, Page 10/58 Eq. (6-32): K f  1  q ( K t  1)  1  0.78(1.7  1)  1.55 Eq. (6-10): Se'  0.5Sut  0.5(470)  235 MPa Eq. (6-18): k a  aSutb  3.04(470) 0.217  0.80 Eq. (6-19): kb  1.24 d 0.107  1.24(35) 0.107  0.85 Eq. (6-25): kc  1 Eq. (6-17): Se  ka kb kc S e'  (0.80)(0.85)(1)(235)  160 MPa 160  1.03 Infinite life is predicted. Ans. K f  ar 1.55  99.8 ______________________________________________________________________________ nf  6-17 Se  From a free-body diagram analysis, the bearing reaction forces are found to be RA = 2000 lbf and RB = 1500 lbf. The shearforce and bending-moment diagrams are shown. The critical location will be at the shoulder fillet between the 1-5/8 in and the 1-7/8 in diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. M = 16 000 – 500 (2.5) = 14 750 lbf ∙ in With a rotating shaft, the bending stress will be completely reversed. Mc 14 750(1.625 / 2)   35.0 kpsi I ( / 64)(1.625) 4 This stress is far below the yield strength of 71 kpsi, so yielding is not predicted.  ar  Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.95 Get the notch sensitivity either from Fig. 6-26, or from the curve-fit Eqs. (6-33) and (6-35). Using the equations, a  0.246  3.08 103   85   1.51105  85   2.67 108   85   0.07690 2 1 q 1  0.76 . 0.07690 1 0.0625 a r K f  1  q ( K t  1)  1  0.76(1.95  1)  1.72 1 Eq. (6-32):  3 Shigley’s MED, 11th edition Chapter 6 Solutions, Page 11/58 Eq. (6-10): Se'  0.5S ut  0.5(85)  42.5 kpsi Eq. (6-18): k a  aSutb  2.00(85) 0.217  0.76 Eq. (6-19): kb  0.879d 0.107  0.879(1.625) 0.107  0.835 Eq. (6-25): kc  1 Eq. (6-17): Se  ka kb kc S e'  (0.76)(0.835)(1)(42.5)  27.0 kpsi 27.0  0.45 Ans. K f  ar 1.72  35.0  Infinite life is not predicted. Use the S-N diagram to estimate the life. nf  Fig. 6-23: Se f = 0.87   0.87(85)   Eq. (6-13):  f Sut  a Eq. (6-14):  f Sut  1 1  0.87(85)  b   log     log    0.1459 3 3  27.0   Se  Se 2 2 27.0  202.5 1 1  K f  ar  b  (1.72)(35.0)  0.1459  4082 cycles N  Eq. (6-15):     a   202.5  N = 4100 cycles Ans. ______________________________________________________________________________ 6-18 From a free-body diagram analysis, the bearing reaction forces are found to be RA = 1600 lbf and RB = 2000 lbf. The shear-force and bending-moment diagrams are shown. The critical location will be at the shoulder fillet between the 1-5/8 in and the 1-7/8 in diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. M = 12 800 + 400 (2.5) = 13 800 lbf ∙ in With a rotating shaft, the bending stress will be completely reversed. Mc 13 800(1.625 / 2)  ar    32.8 kpsi I ( / 64)(1.625) 4 This stress is far below the yield strength of 71 kpsi, so yielding is not predicted. Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.95 Shigley’s MED, 11th edition Chapter 6 Solutions, Page 12/58 Get the notch sensitivity either from Fig. 6-26, or from the curve-fit Eqs. (6-33) and (6-35). Using the equations, a  0.246  3.08 103   85   1.51105  85   2.67 108   85   0.07690 2 Eq. (6-32): 1  0.76 0.07690 a 1 1 0.0625 r K f  1  q ( K t  1)  1  0.76(1.95  1)  1.72 Eq. (6-10): Se'  0.5S ut  0.5(85)  42.5 kpsi Eq. (6-18): k a  aSutb  2.00(85) 0.217  0.76 Eq. (6-19): kb  0.879d 0.107  0.879(1.625) 0.107  0.835 Eq. (6-25): kc  1 Eq. (6-17): Se  ka kb kc S e'  (0.76)(0.835)(1)(42.5)  27.0 kpsi q 1 3  27.0  0.48 Ans. K f  ar 1.72  32.8  Infinite life is not predicted. Use the S-N diagram to estimate the life. Fig. 6-23: f = 0.87 2 2  f Sut   0.87(85)  a Eq. (6-13):   202.5 Se 27.0 nf  Eq. (6-14): Se   f Sut  1 1  0.87(85)  b   log     log    0.1459 3 3  27.0   Se  1 1  K f  ar  b  (1.72)(32.8)  0.1459  6370 cycles N  Eq. (6-15):     a   202.5  N = 6400 cycles Ans. ______________________________________________________________________________ 6-19 Table A-20: Sut  120 kpsi, S y  66 kpsi N = (950 rev/min)(10 hr)(60 min/hr) = 570 000 cycles One approach is to guess a diameter and solve the problem as an iterative analysis problem. Alternatively, we can estimate the few modifying parameters that are dependent on the diameter and solve the stress equation for the diameter, then iterate to check the estimates. We’ll use the second approach since it should require only one iteration, since the estimates on the modifying parameters should be pretty close. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 13/58 First, we will evaluate the stress. From a free-body diagram analysis, the reaction forces at the bearings are R1 = 2 kips and R2 = 6 kips. The critical stress location is in the middle of the span at the shoulder, where the bending moment is high, the shaft diameter is smaller, and a stress concentration factor exists. If the critical location is not obvious, prepare a complete bending moment diagram and evaluate at any potentially critical locations. Evaluating at the critical shoulder, M  2 kip 10 in   20 kip  in  ar  Mc M  d / 2  32M 32  20  203.7     kpsi I d3  d 4 / 64  d 3 d3 Now we will get the notch sensitivity and stress concentration factor. The notch sensitivity depends on the fillet radius, which depends on the unknown diameter. For now, let us estimate a value of q = 0.85 from observation of Fig. 6-26, and check it later. Fig. A-15-9: D / d  1.4d / d  1.4, Eq. (6-32): r / d  0.1d / d  0.1, K t  1.65 K f  1  q ( K t  1)  1  0.85(1.65  1)  1.55 Now, evaluate the fatigue strength. Se'  0.5(120)  60 kpsi ka  2.00(120) 0.217  0.71 Since the diameter is not yet known, assume a typical value of kb = 0.85 and check later. All other modifiers are equal to one. Se = (0.71)(0.85)(60) = 36.2 kpsi Determine the desired fatigue strength from the S-N diagram. Fig. 6-23: f = 0.82 Eq. (6-13): a Eq. (6-14):  f Sut 1 b   log  3  Se Eq. (6-12): S f  aN b  267.5(570 000) 0.1448  39.3 kpsi  f Sut  Se 2  0.82(120)  2 36.2  267.5  1  0.82(120)     log    0.1448 3  36.2   Compare strength to stress and solve for the necessary d. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 14/58 nf  Sf K f  ar  39.3  1.6 1.55   203.7 / d 3  d = 2.34 in Since the size factor and notch sensitivity were guessed, go back and check them now. Eq. (6-19): kb  0.91d 0.157  0.91 2.34   0.80 This is a little lower than our initial guess. From Fig. 6-26 with r = d/10 = 0.234 in, we are off the graph, but it appears our guess for q of 0.85 is low. Assuming the trend of the graph continues, we’ll choose q = 0.91 and iterate the problem with the new values of kb and q. Intermediate results are Se = 34.1 kpsi, Sf = 37.2 kpsi, and Kf = 1.59. This gives 0.157 nf  Sf K f  ar  37.2  1.6 1.59   203.7 / d 3  d = 2.41 in Ans. A quick check of kb and q show that our estimates are still reasonable for this diameter. ______________________________________________________________________________ 6-20 Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  m  15 kpsi,  a  25 kpsi,  m   a  0 Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 2   252  3  0     1/ 2 2   0 2  3 15     2 2    max  3 max  max  1/2   1/ 2  25.00 kpsi  25.98 kpsi 1/2 2 2   a   m   3  a   m     1/2   252  3 152   36.06 kpsi Sy 60 ny    1.66 Ans.   max 36.06 (a) Goodman, Equation (6-41) nf  1  1.05 (25.00 / 40)  (25.98 / 80) Ans. (b) Gerber, Equation (6-48) 2  2   2(25.98)(40)   1  80   25.00   nf     1.31    1  1   2  25.98   40    80(25.00)     Shigley’s MED, 11th edition Ans. Chapter 6 Solutions, Page 15/58 (c) Morrow Estimate the fatigue strength coefficient. Eq. (6-44):  f  Sut  50  80  50  130 kpsi Eq. (6-46): n f  1  1.21 (25.00 / 40)  (25.98 / 130) Ans. ______________________________________________________________________________ 6-21 Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  m  20 kpsi,  a  10 kpsi,  m   a  0 Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/2 2  10 2  3  0     1/ 2 2   02  3  20     2 2    max  3 max  max  1/2    10.00 kpsi 1/ 2  34.64 kpsi 1/2 2 2   a   m   3  a   m     1/2  102  3 202   36.06 kpsi Sy 60 ny    1.66 Ans.   max 36.06 (a) Goodman, Equation (6-41) 1  1.46 (10.00 / 40)  (34.64 / 80) (b) Gerber, Equation (6-48) nf  Ans. 2   2  2(34.64)(40)   1  80   10.00   nf      1.74     1  1   2  34.64   40    80(10.00)     Ans. (c) Morrow Estimate the fatigue strength coefficient. Eq. (6-44):  f  Sut  50  80  50  130 kpsi 1  1.94 Ans. (10.00 / 40)  (34.64 / 130) ______________________________________________________________________________ Eq. (6-46): n f  6-22 Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  a  10 kpsi,  m  15 kpsi,  a  12 kpsi,  m  0 Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 Shigley’s MED, 11th edition 2  12 2  3 10     2   0 2  3 15     1/ 2 1/ 2  21.07 kpsi  25.98 kpsi Chapter 6 Solutions, Page 16/58 2 2    max  3 max  max  1/ 2 1/ 2 2 2   a   m   3  a   m     1/ 2 2 2  12  0   3 10  15    44.93 kpsi   Sy 60 ny    1.34 Ans.   max 44.93 (a) Goodman, Equation (6-41) nf  1  1.17 (21.07 / 40)  (25.98 / 80) Ans. (b) Gerber, Equation (6-48) 2  2  2(25.98)(40)   1  80   21.07   nf      1.47     1  1   2  25.98   40    80(21.07)     Ans. (c) Morrow Estimate the fatigue strength coefficient. Eq. (6-44):  f  Sut  50  80  50  130 kpsi 1  1.38 Ans. (21.07 / 40)  (25.98 / 130) ______________________________________________________________________________ Eq. (6-46): n f  6-23 Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  a  30 kpsi,  m   a   a  0 Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 1/2 2  02  3  30      51.96 kpsi  0 kpsi 2 2    max  3 max  max  1/ 2 1/ 2 2 2   a   m   3  a   m     1/ 2 2  3  30    51.96 kpsi   Sy 60 ny    1.15 Ans.   max 51.96 (a) through (c) With a mean stress of zero, the Goodman, Gerber, and Morrow criteria all simplify to the same simple comparison of the alternating stress to the endurance limit, nf  Shigley’s MED, 11th edition Se 40   0.77  a 51.96 Ans. Chapter 6 Solutions, Page 17/58 Since infinite life is not predicted, estimate a life from the S-N diagram. Since 'm = 0, the stress state is completely reversed and the S-N diagram is applicable for 'a. Fig. 6-23: f = 0.875 Eq. (6-13): ( f Sut ) 2  0.875(80)  a   122.5 Se 40 Eq. (6-14):  f Sut  1 1  0.875(80)  b   log     log    0.08101 3 3 40    Se  2 1 1/ b    51.96  0.08101 Eq. (6-15): N   ar     39 600 cycles Ans.   122.5   a  ______________________________________________________________________________ 6-24 Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  a  15 kpsi,  m  15 kpsi,  m   a  0 Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/2  m   m2  3 m2  1/2 1/2 2  02  3 15      25.98 kpsi 1/2 2  152  3  0     2 2    max  3 max  max  1/ 2  15.00 kpsi 1/ 2 2 2   a   m   3  a   m     2 1/ 2  15   3 15    30.00 kpsi   Sy 60 ny    2.00 Ans.   max 30 2 (a) Goodman, Eq. (6-41) nf  1  1.19 (25.98 / 40)  (15.00 / 80) Ans. (b) Gerber, Eq. (6-48) 2  2  2(15.00)(40)   1  80   25.98   nf      1.43     1  1   2  15.00   40    80(25.98)     Ans. (c) Morrow Estimate the fatigue strength coefficient. Eq. (6-44):  f  Sut  50  80  50  130 kpsi Eq. (6-46): n f  1  1.31 (25.98 / 40)  (15.00 /130) Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 6 Solutions, Page 18/58 6-25 Given: Fmax  28 kN, Fmin  28 kN . From Table A-20, for AISI 1040 CD, Sut  590 MPa, S y  490 MPa, Check for yielding F 28 000  max  max   147.4 N/mm 2  147.4 MPa A 10(25  6) Sy 490  3.32 Ans.  max 147.4 Determine the fatigue factor of safety based on infinite life ny   Eq. (6-10): Se'  0.5(590)  295 MPa Eq. (6-18): k a  aSutb  3.04(590) 0.217  0.76 Eq. (6-20): kb  1 (axial) Eq. (6-25): kc  0.85 Eq. (6-17): Se  ka kb kc S e'  (0.76)(1)(0.85)(295)  190.6 MPa Fig. 6-26: Fig. A-15-1: q = 0.83 d / w  0.24, K t  2.44 K f  1  q ( K t  1)  1  0.83(2.44  1)  2.20 a  K f 28 000   28 000  Fmax  Fmin  2.2  324.2 MPa 2A 2(10)(25  6) Fmax  Fmin 0 2A Note, since m = 0, the stress is completely reversing, and m  K f 190.6  0.59 Ans.  a 324.2 Since infinite life is not predicted, estimate the life from the S-N diagram. With m = 0, the stress state is completely reversed, and the S-N diagram is applicable for a. nf  Fig. 6-23: Se  f = 0.87 Eq. (6-13): ( f Sut ) 2  0.87(590)  a   1382 Se 190.6 Eq. (6-14):  f Sut  1 1  0.87(590)  b   log     log    0.1434 3 3  190.6   Se  Eq. (6-15):   N   ar   a  2 1/ b Shigley’s MED, 11th edition 1  324.2  0.1434   24 613 cycles   1382  Chapter 6 Solutions, Page 19/58 N = 25 000 cycles Ans. ________________________________________________________________________ 6-26 Sut  590 MPa, S y  490 MPa, Fmax  28 kN, Fmin  12 kN Check for yielding Fmax 28 000   147.4 N/mm 2  147.4 MPa A 10(25  6) Sy 490 ny    3.32 Ans.  max 147.4  max  Determine the fatigue factor of safety based on infinite life. From Prob. 6-25: Se  190.6 MPa, K f  2.2 a  K f 28 000  12 000  Fmax  Fmin  2.2  92.63 MPa 2A 2(10)(25  6) m  K f  28 000  12 000  Fmax  Fmin  2.2    231.6 MPa 2A  2(10)(25  6)  Goodman criteria, Equation (6-41): 1     92.63 231.6  nf   a  m       190.6 590   Se Sut  n f  1.14 1 Ans. Gerber criteria, Equation (6-48): 2 2   2 m Se   1  Sut   a  1  1   nf     2   m  Se   Sut a     2 2   2(231.6)(190.6)   1  590  92.63    1  1     2  231.6  190.6  590(92.63)      n f  1.42 Ans. Morrow criteria: Estimate the fatigue strength coefficient. Eq. (6-44):  f  Sut  345  590  345  935 MPa 1 Eq. (6-46): 1     92.63 231.6  nf   a  m       Se  f   190.6 935    n f  1.36 Shigley’s MED, 11th edition Ans. Chapter 6 Solutions, Page 20/58 The results are consistent with Fig. 6-36, where for a mean stress that is about half of the yield strength, the Goodman line should predict failure significantly before the other two. ______________________________________________________________________________ 6-27 Sut  590 MPa, S y  490 MPa From Prob. 6-25: Se  190.6 MPa, K f  2.2 (a) Fmax  28 kN, Fmin  0 kN Check for yielding  max  ny  Fmax 28 000   147.4 N/mm 2  147.4 MPa A 10(25  6) Sy  max a  K f  490  3.32 147.4 Ans. Fmax  Fmin 28 000  0  2.2  162.1 MPa 2A 2(10)(25  6)  28 000  0  Fmax  Fmin  2.2    162.1 MPa 2A  2(10)(25  6)  For the Goodman criteria, Eq. (6-41): m  K f 1 1     162.1 162.1   nf   a  m      0.89  190.6 590   Se Sut  Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress. Using the Goodman criterion, a 162.1 Eq. (6-58):  ar    223.5 MPa 1  ( m / Sut ) 1  (162.1 / 590) Fig. 6-23: f = 0.87 Eq. (6-13): ( f Sut ) 2  0.87(590)  a   1382 Se 190.6 Eq. (6-14):  f Sut  1 1  0.87(590)  b   log     log    0.1434 3 3  190.6   Se  Eq. (6-15):   N   ar   a  2 1/ b 1  223.5  0.1434   329 000 cycles   1382  Ans. (b) Fmax  28 kN, Fmin  12 kN Shigley’s MED, 11th edition Chapter 6 Solutions, Page 21/58 The maximum load is the same as in part (a), so  max  147.4 MPa n y  3.32 Ans. Factor of safety based on infinite life: a  K f Fmax  Fmin 28 000  12 000  2.2  92.63 MPa 2A 2(10)(25  6) m  K f  28 000  12 000  Fmax  Fmin  2.2    231.6 MPa 2A  2(10)(25  6)  1 Eq. (6-41): 1     92.63 231.6   nf   a  m      1.14  190.6 590   Se Sut  Ans. (c) Fmax  12 kN, Fmin  28 kN The compressive load is the largest, so check it for yielding.  min  ny  28 000 Fmin   147.4 MPa A 10(25  6) S yc  min  490  3.32 147.4 Ans. Factor of safety based on infinite life: a  K f 12 000   28 000  Fmax  Fmin  2.2  231.6 MPa 2A 2(10)(25  6) m  K f 12 000   28 000   Fmax  Fmin  2.2    92.63 MPa 2A  2(10)(25  6)  190.6  0.82 Ans.  a 231.6 Since infinite life is not predicted, estimate a life from the S-N diagram. For a negative mean stress, we shall assume the equivalent completely reversed stress is the same as the actual alternating stress, consistent with the horizontal fatigue line in Fig. 6-34. Get a and b from part (a). For m < 0, Eq. (6-42): nf  Se 1/ b  1    231.6  0.1434 Eq. (6-15): N   ar    Ans.  257 000 cycles   1382   a  ______________________________________________________________________________ 6-28 Eq. (2-36): Sut = 0.5(400) = 200 kpsi Shigley’s MED, 11th edition Chapter 6 Solutions, Page 22/58 Eq. (6-10): Se'  0.5(200)  100 kpsi Eq. (6-18): k a  aSutb  11.0(200) 0.650  0.35 Eq. (6-24): d e  0.37 d  0.37(0.375)  0.1388 in Eq. (6-19): kb  0.879d e 0.107  0.879(0.1388) 0.107  1.09 Since we have used the equivalent diameter method to get the size factor, and in doing so introduced greater uncertainties, we will choose not to use a size factor greater than one. Let kb = 1. Eq. (6-17): Se  (0.35)(1)(100)  35.0 kpsi 40  20 40  20 Fa   10 lb Fm   30 lb 2 2 32M a 32(10)(12) a    23.18 kpsi d3  (0.375)3 m  32M m 32(30)(12)   69.54 kpsi d3  (0.375)3 (a) Goodman criterion, Eq. (6-41): 1  a  m 23.18 69.54     nf Se Sut 35.0 200 n f  0.99 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress. Using the Goodman criterion, a 23.18 Eq. (6-58):  ar    35.54 kpsi 1  ( m / Sut ) 1  (69.54 / 200) Fig. 6-23: f = 0.78 Eq. (6-13): ( f Sut ) 2  0.78(200)  a   695.3 Se 35 Eq. (6-14):  f Sut  1 1  0.78(200)  b   log     log    0.2164 3 3  35.0   Se  Eq. (6-15):   N   ar   a  2 1/ b 1  35.54  0.2164   929 000 cycles   695.3  Ans. (b) Gerber criterion, Eq. (6-48): Shigley’s MED, 11th edition Chapter 6 Solutions, Page 23/58 2 2    2 m Se   1  Sut   a  nf   1  1     2   m  Se  Sut a      2 2   2(69.54)(35.0)   1  200  23.18    1  1     2  69.54  35.0   200(23.18)     Ans.  1.23 Infinite life is predicted ______________________________________________________________________________ 6-29 E  207.0 GPa 1 (a) I  (20)(43 )  106.7 mm 4 12 Fl 3 3EIy y  F 3 3EI l 9 3(207)(10 )(106.7)(1012 )(2)(10 3 ) Fmin   48.3 N 1403 (109 ) Fmax  3(207)(109 )(106.7)(10 12 )(6)(10 3 )  144.9 N 1403 (10 9 ) Ans. Ans. (b) Get the fatigue strength information. Eq. (2-36): Sut = =3.4HB = 3.4(490) = 1666 MPa From problem statement: Sy = 0.9Sut = 0.9(1666) = 1499 MPa Se  700 MPa Eq. (6-10): Eq. (6-18): Eq. (6-24): Eq. (6-19): Eq. (6-17): ka = 1.38(1666)-0.067 = 0.84 de = 0.808[20(4)]1/2 = 7.23 mm kb = 1.24(7.23)-0.107 = 1.00 Se = 0.84(1)(700) = 588 MPa This is a relatively thick curved beam, so use the method in Sect. 3-18 to find the stresses. The maximum bending moment will be to the centroid of the section as shown. M = 142F N∙mm, A = 4(20) = 80 mm2, h = 4 mm, ri = 4 mm, ro = ri + h = 8 mm, rc = ri + h/2 = 6 mm Table 3-4: rn  Shigley’s MED, 11th edition h 4   5.7708 mm ln(ro / ri ) ln(8 / 4) Chapter 6 Solutions, Page 24/58 e  rc  rn  6  5.7708  0.2292 mm ci  rn  ri  5.7708  4  1.7708 mm co  ro  rn  8  5.7708  2.2292 mm Get the stresses at the inner and outer surfaces from Eq. (3-76) with the axial stresses added. The signs have been set to account for tension and compression as appropriate. i   o  Mci F (142 F )(1.7708) F     3.441F MPa 80(0.2292)(4) 80 Aeri A Mco F (142 F )(2.2292) F     2.145 F MPa 80(0.2292)(8) 80 Aero A ( i ) min  3.441(144.9)  498.6 MPa ( i ) max  3.441(48.3)  166.2 MPa ( o ) min  2.145(48.3)  103.6 MPa ( o ) max  2.145(144.9)  310.8 MPa ( i ) a  ( i )m  166.2   498.6   166.2 MPa 2 166.2   498.6   332.4 MPa 2 310.8  103.6  103.6 MPa ( o ) a  2 310.8  103.6 ( o ) m   207.2 MPa 2 To check for yielding, we note that the largest stress is –498.6 MPa (compression) on the inner radius. This is considerably less than the estimated yield strength of 1499 MPa, so yielding is not predicted. Check for fatigue on both inner and outer radii since one has a compressive mean stress and the other has a tensile mean stress. Inner radius: S 588 Since m < 0, Eq. (6-42): n f  e   3.54  a 166.2 Outer radius: Since m > 0, using the Goodman line, Eq. (6-41), Shigley’s MED, 11th edition Chapter 6 Solutions, Page 25/58 1     103.6 207.2   nf   a  m      588 1666   Se Sut  n f  3.33 1 Infinite life is predicted at both inner and outer radii. The outer radius is critical, with a fatigue factor of safety of nf = 3.33. Ans. ______________________________________________________________________________ 6-30 From Table A-20, for AISI 1018 CD, Sut  64 kpsi, S y  54 kpsi Eq. (6-10): Se'  0.5(64)  32 kpsi Eq. (6-18): k a  2.00(64) 0.217  0.81 Eq. (6-19): kb  1 (axial) Eq. (6-25): kc  0.85 Eq. (6-17): Se  (0.81)(1)(0.85)(32)  22.0 kpsi Fillet: Fig. A-15-5: D / d  3.5 / 3  1.17, r / d  0.25 / 3  0.083, K t  1.85 Use Fig. 6-26 or Eqs. (6-33) and (6-35) for q. Estimate a little high since it is off the graph. q = 0.85 K f  1  q ( K t  1)  1  0.85(1.85  1)  1.72  max  Fmax 5   3.33 kpsi w2 h 3.0(0.5)  min  16  10.67 kpsi 3.0(0.5) a  K f  max   min 2   max   min 2  m  K f  ny  Sy  min   1.72 3.33  (10.67)  12.0 kpsi 2   3.33  (10.67)    6.31 kpsi   1.72  2    54  5.06 10.67  Does not yield. Since the mean stress is negative, use Eq. (6-42). nf  Se a  22.0  1.83 12.0 Hole: Shigley’s MED, 11th edition Chapter 6 Solutions, Page 26/58 Fig. A-15-1: d / w1  0.4 / 3.5  0.11  K t  2.68 Use Fig. 6-26 or Eqs. (6-33) and (6-35) for q. Estimate a little high since it is off the graph, q = 0.85 K f  1  0.85(2.68  1)  2.43  max  Fmax 5   3.226 kpsi h  w1  d  0.5(3.5  0.4)  min  Fmin 16   10.32 kpsi h  w1  d  0.5(3.5  0.4)  max   min a  K f 2   max   min 2  m  Kf  ny  Sy  min   2.43 3.226  ( 10.32)  16.5 kpsi 2   3.226  ( 10.32)    8.62 kpsi   2.43  2    54  5.23 10.32  does not yield Since the mean stress is negative, use Eq. (6-42). nf  Se a  22.0  1.33 16.5 Thus the design is controlled by the threat of fatigue at the hole with a minimum factor of safety of n f  1.33. Ans. ______________________________________________________________________________ 6-31 Sut  64 kpsi, S y  54 kpsi Eq. (6-10): Se'  0.5(64)  32 kpsi Eq. (6-18): k a  2.00(64) 0.217  0.81 Eq. (6-19): kb  1 (axial) Eq. (6-25): kc  0.85 Eq. (6-17): Se  (0.81)(1)(0.85)(32)  22.0 kpsi Fillet: Fig. A-15-5: D / d  2.5 / 1.5  1.67, r / d  0.25 /1.5  0.17, K t  2.1 Use Fig. 6-26 or Eqs. (6-33) and (6-35) for q. Estimate a little high since it is off the graph. q = 0.85 K f  1  q ( K t  1)  1  0.85(2.1  1)  1.94 Shigley’s MED, 11th edition Chapter 6 Solutions, Page 27/58  max  Fmax 16   21.3 kpsi w2 h 1.5(0.5) 4  5.33 kpsi 1.5(0.5)    min 21.3  ( 5.33)  1.94  25.8 kpsi  a  K f max 2 2  min    max   min 2  m  K f  ny  Sy  max    21.3  ( 5.33)    15.5 kpsi   1.94  2    54  2.54 21.3  Does not yield. Using Goodman criteria, Eq. (6-41), 1 1     25.8 15.5  nf   a  m       0.71  22.0 64   Se Sut  Hole: Fig. A-15-1: d / w1  0.4 / 2.5  0.16  K t  2.55 Use Fig. 6-26 or Eqs. (6-33) and (6-35) for q. Estimate a little high since it is off the graph. q = 0.85 K f  1  0.85(2.55  1)  2.32  max  Fmax 16   15.2 kpsi h  w1  d  0.5(2.5  0.4)  min  4 Fmin   3.81 kpsi h  w1  d  0.5(2.5  0.4)  max   min  15.2  ( 3.81)   2.32    22.1 kpsi 2 2       min   15.2  ( 3.81)   m  K f  max   13.2 kpsi   2.32  2 2     Sy 54 ny    3.55  Does not yield.  max 15.2 a  Kf Using Goodman criteria, Eq. (6-41), 1 1     22.1 13.2  nf   a  m       0.83  22.0 64   Se Sut  Shigley’s MED, 11th edition Chapter 6 Solutions, Page 28/58 Thus the design is controlled by the threat of fatigue at the fillet with a minimum factor of safety of n f  0.71 Ans. ______________________________________________________________________________ 6-32 Sut  64 kpsi, S y  54 kpsi From Prob. 6-30, the fatigue factor of safety at the hole is nf = 1.33. To match this at the fillet, S S 22.0 nf  e  a  e   16.5 kpsi n f 1.33 a where Se is unchanged from Prob. 6-30. The only aspect of a that is affected by the fillet radius is the fatigue stress concentration factor. Obtaining a in terms of Kf, a  K f  max   min 2  Kf 3.33  ( 10.67)  7.00 K f 2 Equating to the desired stress, and solving for Kf,  a  7.00 K f  16.5  K f  2.36 Assume since we are expecting to get a smaller fillet radius than the original, that q will be back on the graph of Fig. 6-26, so we will estimate q = 0.8. K f  1  0.80( K t  1)  2.36  K t  2.7 From Fig. A-15-5, with D / d = 3.5/3 = 1.17 and Kt = 2.6, find r / d. Choosing r / d = 0.03, and with d = w2 = 3.0, r  0.03w2  0.03  3.0   0.09 in At this small radius, our estimate for q is too high. From Fig. 6-26, with r = 0.09, q should be about 0.75. Iterating, we get Kt = 2.8. This is at a difficult range on Fig. A-155 to read the graph with any confidence, but we’ll estimate r / d = 0.02, giving r = 0.06 in. This is a very rough estimate, but it clearly demonstrates that the fillet radius can be relatively sharp to match the fatigue factor of safety of the hole. Ans. ______________________________________________________________________________ 6-33 S y  60 kpsi, S ut  110 kpsi Inner fiber where rc  3 / 4 in 3 3 ro    0.84375 4 16(2) 3 3 ri    0.65625 4 32 Table 3-4, Shigley’s MED, 11th edition Chapter 6 Solutions, Page 29/58 rn  3 / 16 h   0.74608 in 0.84375 ro ln ln 0.65625 ri e  rc  rn  0.75  0.74608  0.00392 in ci  rn  ri  0.74608  0.65625  0.08983  3  3  A       0.035156 in 2  16   16  Eq. (3-65), i  Mci T (0.08983)   993.3T Aeri (0.035156)(0.00392)(0.65625) where T is in lbf∙in and  i is in psi. 1 2  a  496.7T  m  (993.3)T  496.7T Eq. (6-10): Se'  0.5 110   55 kpsi Eq. (6-18): k a  2.00(110) 0.217  0.72 Eq. (6-24): d e  0.808  3 /16  3 /16   Eq. (6-19): kb  0.879  0.1515  Eq. (6-18): Se  (0.72)(1)(55)  39.6 kpsi 1/ 2 0.107  0.1515 in  1.08 (round to 1) For a compressive mean component, from Eq. (6-42),  a  S e / n f . Thus, 39.6 3 T  26.6 lbf  in 0.4967T  Outer fiber where rc  2.5 in 3  2.59375 32 3 ri  2.5   2.40625 32 3 / 16 rn   2.49883 2.59375 ln 2.40625 e  2.5  2.49883  0.00117 in co  2.59375  2.49883  0.09492 in ro  2.5  Shigley’s MED, 11th edition Chapter 6 Solutions, Page 30/58 o  Mco T (0.09492)   889.7T psi Aero (0.035156)(0.00117)(2.59375) 1 2 (a) Using Eq. (6-41), for Goodman, we have  m   a  (889.7T )  444.9T psi 1 1     0.4449T 0.4449T   nf   a  m     3 39.6 110 S S   ut   e T  21.8 lbf  in Ans. (b) For Morrow, estimate the fatigue strength coefficient from Eq. (6-44),  f  Sut  50  110  50  160 kpsi 1 Eq. (6-46): 1 a m   0.4449T 0.4449T     nf    3  S    160   39.6 f   e T  23.8 lbf  in Ans. (c) To guard against yield, use T of part (b) and the inner stress. Sy 60  2.54 Ans.  i 0.9933(23.8) ______________________________________________________________________________ ny  6-34  From Prob. 6-33, Se  39.6 kpsi, S y  60 kpsi, and Sut  110 kpsi (a) Assuming the beam is straight,  max  Mc M  h / 2  6 M 6T   2   910.2T 3 I bh /12 bh (3 / 16)3 Using Eq. (6-41), for Goodman, we have 1 1     0.4551T 0.4551T  nf   a  m      3 110   39.6  Se Sut  T  21.3 lbf  in Ans. (b) ) For Morrow, estimate the fatigue strength coefficient from Eq. (6-44),  f  Sut  50  110  50  160 kpsi 1 Eq. (6-46): 1 a m   0.4551T 0.4551T     nf    3  S    160   39.6 f   e T  23.3 lbf  in Shigley’s MED, 11th edition Ans. Chapter 6 Solutions, Page 31/58 Sy 60  2.83 Ans.  max 0.9102(23.3) ______________________________________________________________________________ (c) 6-35 ny   K f ,bend  1.4, K f ,axial  1.1, K f ,tors  2.0, S y  300 MPa, Sut  400 MPa, Se  160 MPa Bending: Axial:  m  0,  a  60 MPa  m  20 MPa,  a  0  m  35 MPa,  a  35 MPa Torsion: Eqs. (6-66) and (6-67):  a  1.4(60)  0  m   0  1.1(20) 2 2  3 2.0(35)   147.5 MPa 2  3 2.0(35)   123.2 MPa 2    a   m , Check for yielding, using the conservative  max ny  Sy   a   m 300  1.11 147.5  123.2 Yielding is not predicted. Ans. Using Goodman, Eq. (6-41), 1     147.5 123.2  nf   a  m      S S 160 400   ut   e n f  0.81 1 Ans. Finite life is predicted. To use the Walker criterion for estimating an equivalent completely reversed stress, estimate the material fitting parameter for steels with Eq. (657).   0.0002 Sut  0.8818  0.0002(400)  0.8818  0.8018 Eq. (6-61): Fig. 6-23:  ar   m   a   a  123.2  147.5  Off the chart, so use f = 0.9 1  1 0.8018 Eq. (6-13): ( f Sut ) 2  0.9(400)  a   810 Se 160 Eq. (6-14):  f Sut 1 b   log  3  Se Eq. (6-15):   N   ar   a  147.50.8018  166.4 MPa 2 1/ b Shigley’s MED, 11th edition  1  0.9(400)     log    0.1174 3  160   1  166.4  0.1174   716 000 cycles   810  Ans. Chapter 6 Solutions, Page 32/58 ______________________________________________________________________________ 6-36 K f ,bend  1.4, K f ,tors  2.0, S y  300 MPa, S ut  400 MPa, Se  160 MPa Bending:  max  150 MPa,  min  40 MPa,  m  55 MPa,  a  95 MPa Torsion:  m  90 MPa,  a  9 MPa Eqs. (6-66) and (6-67):  a  1.4(95)   m  1.4(55)   3 2.0(9)   136.6 MPa 2 2 2  3 2.0(90)   321.1 MPa 2 Using Goodman, Eq. (6-41), 1 1     136.6 321.1  nf   a  m       0.60 400   160  Se Sut  Ans.    a   m , Check for yielding, using the conservative  max ny  Sy  a   m  300  0.66 136.6  321.1 Ans. Since the conservative yield check indicates yielding, we will check more carefully with  obtained directly from the maximum stresses, using the distortion energy failure  max theory, without stress concentrations. Note that this is exactly the method used for static failure in Ch. 5.    max  max  2  3  max   2 150  2  3  90  9   227.8 MPa 2 Sy 300   1.32 Ans.  227.8  max Since yielding is not predicted, and infinite life is not predicted, we would like to estimate a life from the S-N diagram. ny  To use the Walker criterion for estimating an equivalent completely reversed stress, estimate the material fitting parameter for steels with Eq. (6-57).   0.0002 Sut  0.8818  0.0002(400)  0.8818  0.8018 Eq. (6-61): Fig. 6-23:  ar   m   a   a   321.1  136.6  Off the chart, so use f = 0.9 Eq. (6-13): ( f Sut ) 2  0.9(400)  a   810 Se 160 1  1 0.8018 136.60.8018  173.6 MPa 2 Shigley’s MED, 11th edition Chapter 6 Solutions, Page 33/58 Eq. (6-14):  f Sut 1 b   log  3  Se  1  0.9(400)     log    0.1174 3  160   1 1/ b    173.6  0.1174 N   ar    Ans. Eq. (6-15):  499 000 cycles   810   a  _____________________________________________________________________________ 6-37 Table A-20: S ut  64 kpsi, S y  54 kpsi From Prob. 3-79, the critical stress element experiences  = 15.3 kpsi and  = 4.43 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 15.3 kpsi, m = 0 kpsi, a = 0 kpsi, m = 4.43 kpsi. Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/2 1/ 2 2  15.32  3  0      15.3 kpsi 1/ 2 2   0 2  3  4.43     2 2    max  max  3 max  1/ 2  7.67 kpsi 2  15.32  3  4.43     1/ 2  17.11 kpsi Check for yielding, using the distortion energy failure theory. Sy 54 ny    3.16   max 17.11 Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5  64   32 kpsi Eq. (6-18): ka  2.00(64) 0.217  0.81 Eq. (6-19): kb  0.879(1.25) 0.107  0.86 Se  0.81(0.86)(32)  22.3 kpsi Eq. (6-17): Using Goodman, Eq. (6-41), 1 1     15.3 7.67  nf   a  m     Ans.   1.24  22.3 64   Se Sut  ______________________________________________________________________________ 6-38 Table A-20: S ut  440 MPa, S y  370 MPa From Prob. 3-80, the critical stress element experiences  = 263 MPa and  = 57.7 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 263 MPa, m = 0, a = 0 MPa, m = 57.7 MPa. Obtain von Mises stresses for the alternating, mean, and maximum stresses. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 34/58  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 1/ 2 2   2632  3  0      263 MPa 1/ 2 2  0 2  3  57.7     2 2    max  3 max  max  1/ 2  99.9 MPa 2   2632  3  57.7     1/ 2  281 MPa Check for yielding, using the distortion energy failure theory. Sy 370   1.32  281  max Obtain the modifying factors and endurance limit. ny  Eq. (6-10): Se  0.5  440   220 MPa Eq. (6-18): k a  3.04(440) 0.217  0.81 Eq. (6-19): kb  1.24(30) 0.107  0.86 Eq. (6-17): Se  0.81(0.86)(220)  153 MPa Using Goodman, Eq. (6-41): 1     263 99.9   nf   a  m     S S 153 440   e ut   n f  0.51 1 Infinite life is not predicted. Ans. ______________________________________________________________________________ 6-39 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-81, the critical stress element experiences  = 21.5 kpsi and  = 5.09 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 21.5 kpsi, m = 0 kpsi, a = 0 kpsi, m = 5.09 kpsi. Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 1/ 2 2   21.52  3  0     1/ 2 2   0 2  3  5.09     2 2    max  max  3 max  1/ 2  21.5 kpsi  8.82 kpsi 2   21.52  3  5.09     1/ 2  23.24 kpsi Check for yielding, using the distortion energy failure theory. ny  Shigley’s MED, 11th edition Sy 54   2.32   max 23.24 Chapter 6 Solutions, Page 35/58 Obtain the modifying factors and endurance limit. Se  0.5  64   32 kpsi Eq. (6-10): Eq. (6-18): ka  2.00(64) 0.217  0.81 Eq. (6-19): kb  0.879(1) 0.107  0.88 Eq. (6-17): Se  0.81(0.88)(32)  22.8 kpsi Using Goodman, Eq. (6-41), 1 1     21.5 8.82  nf   a  m     Ans.   0.93  22.8 64   Se Sut  ______________________________________________________________________________ 6-40 Table A-20: S ut  440 MPa, S y  370 MPa From Prob. 3-82, the critical stress element experiences  = 72.9 MPa and  = 20.3 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 72.9 MPa, m = 0 MPa, a = 0 MPa, m = 20.3 MPa. Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/2 1/ 2 2  72.92  3  0     1/ 2 2   0 2  3  20.3     2 2    max  max  3 max  1/ 2  72.9 MPa  35.2 MPa 2   72.9 2  3  20.3    1/ 2  80.9 MPa Check for yielding, using the distortion energy failure theory. Sy 370 ny    4.57  80.9  max Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5  440   220 MPa Eq. (6-18): ka  3.04(440) 0.217  0.81 Eq. (6-19): kb  1.24(20) 0.107  0.90 Eq. (6-17): Se  0.81(0.90)(220)  160.4 MPa Using Goodman, Eq. (6-41), 1 1     72.9 35.2  nf   a  m     Ans.   1.87  160.4 440   Se Sut  ______________________________________________________________________________ 6-41 Table A-20: S ut  64 kpsi, S y  54 kpsi Shigley’s MED, 11th edition Chapter 6 Solutions, Page 36/58 From Prob. 3-83, the critical stress element experiences  = 35.2 kpsi and  = 7.35 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 35.2 kpsi, m = 0 kpsi, a = 0 kpsi, m = 7.35 kpsi. Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/2 1/ 2 2  35.2 2  3  0      35.2 kpsi 1/ 2 2   0 2  3  7.35     2 2    max  3 max  max  1/ 2  12.7 kpsi 2  35.22  3  7.35     1/ 2  37.4 kpsi Check for yielding, using the distortion energy failure theory. ny  Sy 54   1.44  37.4  max Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): ka  2.00(64) 0.217  0.81 Eq. (6-19): kb  0.879(1.25) 0.107  0.86 Eq. (6-17): Se  0.81(0.86)(32)  22.3 kpsi Using Goodman, Eq. (6-41), 1 1     35.2 12.7  nf   a  m     Ans.   0.56  22.3 64   Se Sut  ______________________________________________________________________________ 6-42 Table A-20: S ut  440 MPa, S y  370 MPa From Prob. 3-84, the critical stress element experiences  = 333.9 MPa and  = 126.3 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 333.9 MPa, m = 0 MPa, a = 0 MPa, m = 126.3 MPa. Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 2  333.9 2  3  0     1/ 2 2   0 2  3 126.3     2 2    max  max  3 max  1/ 2 1/ 2  333.9 MPa  218.8 MPa 1/2 2  333.92  3 126.3      399.2 MPa Check for yielding, using the distortion energy failure theory. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 37/58 ny  Sy 370   0.93   max 399.2 The sample fails by yielding, infinite life is not predicted. Ans. The fatigue analysis will be continued only to obtain the requested fatigue factor of safety, though the yielding failure will dictate the life. Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5(440)  220 MPa Eq. (6-18): ka  3.04(440) 0.217  0.81 Eq. (6-19): kb  1.24(50) 0.107  0.82 Eq. (6-17): Se  0.81(0.82)(220)  146.1 MPa Using Goodman, Eq. (6-41), 1 1     333.9 218.8  nf   a  m     Ans.   0.36  146.1 440   Se Sut  ______________________________________________________________________________ 6-43 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-85, the critical stress element experiences completely reversed bending stress due to the rotation, and steady torsional and axial stresses.  a ,bend  9.495 kpsi,  a ,axial  0 kpsi,  a  0 kpsi,  m,bend  0 kpsi  m,axial  0.362 kpsi  m  11.07 kpsi Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 2 2   9.495   3  0     1/ 2  9.495 kpsi 2 2   0.362   3 11.07     2 2    max  max  3 max  1/ 2 1/ 2  19.18 kpsi 1/ 2 2 2   9.495  0.362   3 11.07      21.56 kpsi Check for yielding, using the distortion energy failure theory. Sy 54 ny    2.50   max 21.56 Obtain the modifying factors and endurance limit. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 38/58 Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): ka  2.00(64) 0.217  0.81 Eq. (6-19): kb  0.879(1.13) 0.107  0.87 Eq. (6-17): Se  0.81(0.87)(32)  22.6 kpsi Using Goodman, Eq. (6-41), 1 1     9.495 19.18  nf   a  m     Ans.   1.39 64   22.6  Se Sut  ______________________________________________________________________________ 6-44 Table A-20: S ut  64 kpsi, S y  54 kpsi From Prob. 3-87, the critical stress element experiences completely reversed bending stress due to the rotation, and steady torsional and axial stresses.  a ,bend  33.99 kpsi,  m ,bend  0 kpsi  a ,axial  0 kpsi,  a  0 kpsi,  m,axial  0.153 kpsi  m  7.847 kpsi Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 2 2   33.99   3  0     1/ 2  33.99 kpsi 2 2   0.153  3  7.847     2 2    max  max  3 max  1/ 2 1/ 2  13.59 kpsi 1/ 2 2 2   33.99  0.153   3  7.847      36.75 kpsi Check for yielding, using the distortion energy failure theory. ny  Sy 54   1.47   max 36.75 Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): ka  2.00(64) 0.217  0.81 Eq. (6-19): kb  0.879(0.88) 0.107  0.89 Eq. (6-17): Se  0.81(0.89)(32)  23.1 kpsi Using Goodman, Eq. (6-41), 1 1     33.99 13.59  nf   a  m       0.59 64   23.1  Se Sut  Shigley’s MED, 11th edition Ans. Chapter 6 Solutions, Page 39/58 ______________________________________________________________________________ 6-45 Table A-20: S ut  440 MPa, S y  370 MPa From Prob. 3-88, the critical stress element experiences  = 68.6 MPa and  = 37.7 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 68.6 MPa, m = 0 MPa, a = 0 MPa, m = 37.7 MPa. Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 1/ 2 2   68.62  3  0     1/ 2 2   0 2  3  37.7      68.6 MPa  65.3 MPa 2 1/ 2 2 2    max  3 max  max  1/ 2   68.62  3  37.7    94.7 MPa   Check for yielding, using the distortion energy failure theory. ny  Sy 370   3.91   max 94.7 Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5(440)  220 MPa Eq. (6-18): ka  3.04(440) 0.217  0.81 Eq. (6-19): kb  1.24(30) 0.107  0.86 Eq. (6-17): Se  0.81(0.86)(220)  153 MPa Using Goodman, Eq. (6-41), 1 1     68.6 65.3  nf   a  m     Ans.   1.68 S S 153 440   ut   e ______________________________________________________________________________ 6-46 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-90, the critical stress element experiences  = 3.46 kpsi and  = 0.882 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 3.46 kpsi, m = 0, a = 0 kpsi, m = 0.882 kpsi. Obtain von Mises stresses for the alternating, mean, and maximum stresses.  a   a2  3 a2  1/ 2  m   m2  3 m2  1/ 2 1/ 2 2  3.462  3  0     1/ 2 2  02  3  0.882     2 2    max  max  3 max  1/ 2 Shigley’s MED, 11th edition  3.46 kpsi  1.53 kpsi 1/ 2 2  3.46 2  3  0.882      3.78 kpsi Chapter 6 Solutions, Page 40/58 Check for yielding, using the distortion energy failure theory. ny  Sy 54   14.3   max 3.78 Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): k a  2.00(64) 0.217  0.81 Eq. (6-19): kb  0.879(1.375) 0.107  0.85 Eq. (6-17): Se  0.81(0.85)(32)  22.0 kpsi Using Goodman, 1 1     3.46 1.53  nf   a  m     Eq. (6-41):   22.0 64   Se Sut  Ans. n f  5.5 ______________________________________________________________________________ 6-47 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-91, the critical stress element experiences  = 16.3 kpsi and  = 5.09 kpsi. Since the load is applied and released repeatedly, this gives max = 16.3 kpsi, min = 0 kpsi, max = 5.09 kpsi, min = 0 kpsi. Consequently,m = a = 8.15 kpsi, m = a = 2.55 kpsi. For bending, from Eqs. (6-33) and (6-35), a  0.246  3.08 103   64   1.51105   64   2.67 10 8   64   0.10373 2 1  0.75 a 1  0.10373 1 0.1 r K f  1  q ( K t  1)  1  0.75(1.5  1)  1.38 q Eq. (6-32): 1 3  For torsion, from Eqs. (6-33) and (6-36), a  0.190  2.51103   64   1.35 105   64   2.67 108   64   0.07800 2 1  0.80 a 1  0.07800 1 0.1 r K fs  1  qs ( K ts  1)  1  0.80(2.1  1)  1.88 q Eq. (6-32): Shigley’s MED, 11th edition 1 3  Chapter 6 Solutions, Page 41/58 Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).  2  a  1.38  8.15   3 1.88  2.55   2 1/ 2  13.98 kpsi  m   a  13.98 kpsi    a   m , Check for yielding, using the conservative  max Sy 54   1.93 ny   a   m 13.98  13.98 Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): k a  aSutb  2.00(64) 0.217  0.81 Eq. (6-23): de  0.370d  0.370 1  0.370 in Eq. (6-19): kb  0.879d e 0.107  0.879(0.370) 0.107  0.98 Eq. (6-17): Se  (0.81)(0.98)(32)  25.4 kpsi Using Goodman, Eq. (6-41): 1     13.98 13.98  nf   a  m      64   25.4  Se Sut  n f  1.3 1 Ans. ______________________________________________________________________________ 6-48 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-92, the critical stress element experiences  = 16.4 kpsi and  = 4.46 kpsi. Since the load is applied and released repeatedly, this gives max = 16.4 kpsi, min = 0 kpsi, max = 4.46 kpsi, min = 0 kpsi. Consequently,m = a = 8.20 kpsi, m = a = 2.23 kpsi. For bending, from Eqs. (6-33) and (6-35), a  0.246  3.08 103   64   1.51105   64   2.67 10 8   64   0.10373 2 1  0.75 a 1  0.10373 1 0.1 r K f  1  q ( K t  1)  1  0.75(1.5  1)  1.38 q Eq. (6-32): 1 3  For torsion, from Eqs. (6-33) and (6-36), Shigley’s MED, 11th edition Chapter 6 Solutions, Page 42/58 a  0.190  2.51103   64   1.35 105   64   2.67 108   64   0.07800 2 1 1  0.80 a 1  0.07800 1 0.1 r K fs  1  qs ( K ts  1)  1  0.80(2.1  1)  1.88 q Eq. (6-32): 3  Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).  2  a  1.38  8.20    3 1.88  2.23   2 1/ 2  13.45 kpsi  m   a  13.45 kpsi    a   m , Check for yielding, using the conservative  max Sy 54   2.01 ny   a   m 13.45  13.45 Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): k a  aSutb  2.00(64) 0.217  0.81 Eq. (6-23): d e  0.370d  0.370(1)  0.370 in Eq. (6-19): kb  0.879d e 0.107  0.879(0.370) 0.107  0.98 Eq. (6-17): Se  (0.81)(0.98)(32)  25.4 kpsi Using Goodman, Eq. (6-41): 1     13.45 13.45  nf   a  m      S S 25.4 64   ut   e n f  1.35 1 Ans. ______________________________________________________________________________ 6-49 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-93, the critical stress element experiences repeatedly applied bending, axial, and torsional stresses of x,bend = 20.2 kpsi, x,axial = 0.1 kpsi, and  = 5.09 kpsi.. Since the axial stress is practically negligible compared to the bending stress, we will simply combine the two and not treat the axial stress separately for stress concentration factor and load factor. This gives max = 20.3 kpsi, min = 0 kpsi, max = 5.09 kpsi, min = 0 kpsi. Consequently,m = a = 10.15 kpsi, m = a = 2.55 kpsi. For bending, from Eqs. (6-33) and (6-35), Shigley’s MED, 11th edition Chapter 6 Solutions, Page 43/58 a  0.246  3.08 103   64   1.51105   64   2.67 10 8   64   0.10373 2 1 1  0.75 0.10373 a 1 1 0.1 r K f  1  q ( K t  1)  1  0.75(1.5  1)  1.38 q Eq. (6-32): 3  For torsion, from Eqs. (6-33) and (6-36), a  0.190  2.51103   64   1.35 105   64   2.67 108   64   0.07800 2 1 1  0.80 a 1  0.07800 1 0.1 r K fs  1  qs ( K ts  1)  1  0.80(2.1  1)  1.88 q Eq. (6-32): 3  Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).  2  a  1.38 10.15    3 1.88  2.55   2 1/ 2  16.28 kpsi  m   a  16.28 kpsi    a   m , Check for yielding, using the conservative  max Sy 54   1.66 ny   a   m 16.28  16.28 Obtain the modifying factors and endurance limit. Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): k a  aSutb  2.00(64) 0.217  0.81 Eq. (6-23): d e  0.370d  0.370(1)  0.370 in Eq. (6-19): kb  0.879d e 0.107  0.879(0.370) 0.107  0.98 Eq. (6-17): Se  (0.81)(0.98)(32)  25.4 kpsi Using Goodman, 1 1     16.28 16.28  nf   a  m     Eq. (6-41):  S S 25.4 64   ut   e n f  1.12 Ans. ____________________________________________________________________________ Shigley’s MED, 11th edition Chapter 6 Solutions, Page 44/58 6-50 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-94, the critical stress element on the neutral axis in the middle of the longest side of the rectangular cross section experiences a repeatedly applied shear stress of max = 14.3 kpsi, min = 0 kpsi. Thus, m = a = 7.15 kpsi. Since the stress is entirely shear, it is convenient to check for yielding using the standard Maximum Shear Stress theory. S y / 2 54 / 2   1.89 ny   max 14.3 Find the modifiers and endurance limit. Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): k a  aSutb  2.00(64) 0.217  0.81 The size factor for a rectangular cross section loaded in torsion is not readily available. An equivalent diameter based on the 95 percent stress area is not readily obtained, since the stress situation in this case is nonlinear, as described in Section 3-12. Noting that the maximum stress occurs at the middle of the longest side, or with a radius from the center of the cross section equal to half of the shortest side, we will simply choose an equivalent diameter equal to the length of the shortest side. d e  0.25 in Eq. (6-19): kb  0.879de 0.107  0.879(0.25) 0.107  1.02 We will round down to kb = 1. Eq. (6-25): kc  0.59 Eq. (6-17): S se  0.81(1)(0.59)(32)  15.3 kpsi Since the stress is entirely shear, we choose to use a load factor kc = 0.59, and convert the ultimate strength to a shear value rather than using the combination loading method of Sec. 6-16. From Eq. (6-58), Ssu = 0.67Su = 0.67 (64) = 42.9 kpsi. Using Goodman, 1 1     7.15 7.15  nf   a  m     Eq. (6-41): Ans.   1.58 S S 15.3 42.9   se su   ______________________________________________________________________________ 6-51 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-95, the critical stress element experiences  = 28.0 kpsi and  = 15.3 kpsi. Since the load is applied and released repeatedly, this gives max = 28.0 kpsi, min = 0 Shigley’s MED, 11th edition Chapter 6 Solutions, Page 45/58 kpsi, max = 15.3 kpsi, min = 0 kpsi. Consequently,m = a = 14.0 kpsi, m = a = 7.65 kpsi. From Table A-15-8 and A-15-9, D / d  1.5 /1  1.5, K t ,bend  1.60, r / d  0.125 / 1  0.125 K t ,tors  1.39 Figs. 6-26 and 6-27: qbend = 0.78, qtors = 0.82 Eq. (6-32): K f ,bend  1  qbend  K t ,bend  1  1  0.78 1.60  1  1.47 K f ,tors  1  qtors  K t ,tors  1  1  0.82 1.39  1  1.32 Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).  2  a  1.47 14.0   3 1.32  7.65    2 1/ 2  27.0 kpsi  m   a  27.0 kpsi    a   m , Check for yielding, using the conservative  max Sy 54 ny    1.00  a   m 27.0  27.0 Since stress concentrations are included in this quick yield check, the low factor of safety is acceptable. Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): k a  aSutb  2.00(64) 0.217  0.81 Eq. (6-23): de  0.370d  0.370 1  0.370 in Eq. (6-19): kb  0.879d e 0.107  0.879(0.370) 0.107  0.98 Eq. (6-17): Se  (0.81)(0.98)(0.5)(64)  25.4 kpsi For the Morrow criterion, estimate the fatigue strength coefficient for steel. Eq. (6-44):  f  Sut  50  64  50  114 kpsi 1 Eq. (6-46): 1   a  m   27.0 27.0  nf        S     25.4 114  f   e n f  0.77 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress, again using Morrow. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 46/58 Eq. (6-59): Fig. 6-23:  a 27.0   35.4 kpsi 1  ( m /  f ) 1  (27.0 / 114) Off the chart, so use f = 0.9  ar  Eq. (6-13): ( f Sut ) 2  0.9(64)  a   130.6 Se 25.4 Eq. (6-14):  f Sut 1 b   log  3  Se 2  1  0.9(64)     log    0.1185 3  25.4   1 1/ b   ar   35.4  0.1185 N  Ans. Eq. (6-15):   60856 cycles  61 000 cycles    130.6   a  ______________________________________________________________________________ 6-52 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-96, the critical stress element experiences x,bend = 46.1 kpsi, x,axial = 0.382 kpsi and  = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to demonstrate the process. Since the load is applied and released repeatedly, this gives max,bend = 46.1 kpsi, min,bend = 0 kpsi, max,axial = 0.382 kpsi, min,axial = 0 kpsi, max = 15.3 kpsi, min = 0 kpsi. Consequently,m,bend = a,bend = 23.05 kpsi, m,axial = a,axial = 0.191 kpsi, m = a = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9, D / d  1.5 /1  1.5, K t ,bend  1.60, r / d  0.125 / 1  0.125 K t ,tors  1.39, K t ,axial  1.75 Eqs. (6-33), (6-35), and (6-36), or Figs. 6-26 and 6-27: qbend = qaxial =0.78, qtors = 0.82 Eq. (6-32): K f ,bend  1  qbend  Kt ,bend  1  1  0.78 1.60  1  1.47 K f ,axial  1  qaxial  Kt ,axial  1  1  0.78 1.75  1  1.59 K f ,tors  1  qtors  Kt ,tors  1  1  0.82 1.39  1  1.32 Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).      1.47  23.05  1.59  0.191   3 1.32  7.65   2 2 1/ 2 2 2 1/2  a  1.47  23.05   1.59  0.191   3 1.32  7.65  m  38.4 kpsi  38.4 kpsi    a   m , Check for yielding, using the conservative  max Sy 54 ny    0.70  a   m 38.4  38.4 Shigley’s MED, 11th edition Chapter 6 Solutions, Page 47/58 Since the conservative yield check indicates yielding, we will check more carefully with  obtained directly from the maximum stresses, using the distortion energy failure  max theory, without stress concentrations. Note that this is exactly the method used for static failure in Ch. 5.    max ny     max,axial   3  max   2 max,bend Sy 54   1.01  53.5  max 2  46.1  0.382  2  3 15.3  53.5 kpsi 2 Ans. This shows that yielding is imminent, and further analysis of fatigue life should not be interpreted as a guarantee of more than one cycle of life. Eq. (6-10): Se  0.5(64)  32 kpsi Eq. (6-18): k a  aSutb  2.00(64) 0.217  0.81 Eq. (6-23): de  0.370d  0.370 1  0.370 in Eq. (6-19): kb  0.879d e 0.107  0.879(0.370) 0.107  0.98 Eq. (6-17): Se  (0.81)(0.98)(0.5)(64)  25.4 kpsi For the Morrow criterion, estimate the fatigue strength coefficient for steel. Eq. (6-44):  f  Sut  50  64  50  114 kpsi 1 Eq. (6-46): 1     38.4 38.4   nf   a  m      S    25.4 114  f   e n f  0.54 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress, again using Morrow. Eq. (6-59): Fig. 6-23:  a 38.4   57.9 kpsi 1  ( m /  f ) 1  (38.4 / 114) Off the chart, so use f = 0.9  ar  Eq. (6-13): ( f Sut ) 2  0.9(64)  a   130.6 Se 25.4 Eq. (6-14):  f Sut 1 b   log  3  Se 2 1/ b  1  0.9(64)     log    0.1185 3 25.4    1    57.9  0.1185 N   ar    Ans. Eq. (6-15):  960 cycles   130.6   a  ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 6 Solutions, Page 48/58 6-53 Table A-20: Sut  64 kpsi, S y  54 kpsi From Prob. 3-97, the critical stress element experiences x,bend = 55.5 kpsi, x,axial = 0.382 kpsi and  = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to demonstrate the process. Since the load is applied and released repeatedly, this gives max,bend = 55.5 kpsi, min,bend = 0 kpsi, max,axial = 0.382 kpsi, min,axial = 0 kpsi, max = 15.3 kpsi, min = 0 kpsi. Consequently,m,bend = a,bend = 27.75 kpsi, m,axial = a,axial = 0.191 kpsi, m = a = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9, D / d  1.5 /1  1.5, r / d  0.125 / 1  0.125 K t ,bend  1.60, K t ,tors  1.39, K t ,axial  1.75 Eqs. (6-33), (6-35), and (6-36), or Figs. 6-26 and 6-27: qbend = qaxial =0.78, qtors = 0.82 Eq. (6-32): K f ,bend  1  qbend  Kt ,bend  1  1  0.78 1.60  1  1.47 K f ,axial  1  qaxial  Kt ,axial  1  1  0.78 1.75  1  1.59 K f ,tors  1  qtors  Kt ,tors  1  1  0.82 1.39  1  1.32 Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).      1.47  27.75  1.59  0.191   3 1.32  7.65    2 2 1/2 2 2 1/2  a  1.47  27.75  1.59  0.191   3 1.32  7.65   m  44.66 kpsi  44.66 kpsi Since these stresses are relatively high compared to the yield strength, we will go ahead and check for yielding using the distortion energy failure theory.    max ny     max,axial   3  max   2 max,bend Sy 54   0.87   max 61.8 2  55.5  0.382  2  3 15.3  61.8 kpsi 2 Ans. This shows that yielding is predicted. Further analysis of fatigue life is just to be able to report the fatigue factor of safety, though the life will be dictated by the static yielding failure, i.e. N = 1/2 cycle. Ans. Eq. (6-10): Se  0.5  64   32 kpsi Eq. (6-18): k a  aSutb  2.00(64) 0.217  0.81 Eq. (6-23): de  0.370d  0.370 1  0.370 in Eq. (6-19): kb  0.879d e 0.107  0.879(0.370) 0.107  0.98 Eq. (6-17): Se  (0.81)(0.98)(0.5)(64)  25.4 kpsi For the Morrow criterion, estimate the fatigue strength coefficient for steel. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 49/58 Eq. (6-44):  f  Sut  50  64  50  114 kpsi 1 Eq. (6-46): 1   a  m   44.66 44.66  nf        S    114   25.4 f   e n f  0.47 Ans. ______________________________________________________________________________ 6-54 From Table A-20, for AISI 1040 CD, Sut = 85 kpsi and Sy = 71 kpsi. From the solution to Prob. 6-17 we find the completely reversed stress at the critical shoulder fillet to be ar = 35.0 kpsi, producing a = 35.0 kpsi and m = 0 kpsi. This problem adds a steady torque which creates torsional stresses of m  Tr 2500 1.625 / 2    2967 psi  2.97 kpsi, J  1.6254  / 32  a  0 kpsi From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt,bend =1.95, Kt,tors =1.60 Eqs. (6-33), (6-35) and (6-36), or Figs. 6-26 and 6-27: qbend = 0.76, qtors = 0.81 Eq. (6-32): K f ,bend  1  qbend  K t ,bend  1  1  0.76 1.95  1  1.72 K f ,tors  1  qtors  K t ,tors  1  1  0.811.60  1  1.49 Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).      1.72  0    3 1.49  2.97    2  a  1.72  35.0    3 1.49  0   2 1/2 2 1/ 2 2 m  60.2 kpsi  7.66 kpsi    a   m , Check for yielding, using the conservative  max Sy 71 ny    1.05  a   m 60.2  7.66 From the solution to Prob. 6-17, Se = 27.0 kpsi. Using Goodman, 1 Eq. (6-41):     60.2 7.66  nf   a  m       27.0 85   Se Sut  n f  0.43 1 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress. Choosing the Goodman criterion, Shigley’s MED, 11th edition Chapter 6 Solutions, Page 50/58 Eq. (6-58): Fig. 6-23: Eq. (6-13):  a 60.2   66.2 kpsi 1  ( m / Sut ) 1  (7.66 / 85) f = 0.867 2 2  f Sut   0.867(85)  a   201.1 Se 27.0  ar  Eq. (6-14):  f Sut 1 b   log  3  Se Eq. (6-15):   N   ar   a  1/ b  1  0.867(85)     log    0.1454 3  27.0   1  66.2  0.1454   2084 cycles   201.1  N = 2100 cycles Ans. ______________________________________________________________________________ 6-55 From the solution to Prob. 6-18 we find the completely reversed stress at the critical shoulder fillet to be rev = 32.8 kpsi, producing a = 32.8 kpsi and m = 0 kpsi. This problem adds a steady torque which creates torsional stresses of m  Tr 2200 1.625 / 2    2611 psi  2.61 kpsi, J  1.6254  / 32  a  0 kpsi From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt,bend =1.95, Kt,tors =1.60 Eqs. (6-33), (6-35) and (6-36), or Figs. 6-26 and 6-27: qbend = 0.76, qtors = 0.81 Eq. (6-32): K f ,bend  1  qbend  K t ,bend  1  1  0.76 1.95  1  1.72 K f ,tors  1  qtors  K t ,tors  1  1  0.811.60  1  1.49 Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).      1.72  0    3 1.49  2.61   2  a  1.72  32.8   3 1.49  0   2 m 2 1/ 2 2 1/2  56.4 kpsi  6.74 kpsi    a   m , Check for yielding, using the conservative  max Sy 71 ny    1.12  a   m 56.4  6.74 From the solution to Prob. 6-18, Se = 27.0 kpsi. Using Goodman, Shigley’s MED, 11th edition Chapter 6 Solutions, Page 51/58 1 1     56.4 6.74   nf   a  m    Eq. (6-41):   27.0 85   Se Sut  n f  0.46 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress. Choosing the Goodman criterion,  a 56.4   61.3 kpsi 1  ( m / Sut ) 1  (6.74 / 85) f = 0.867 2 2 f Sut  0.867(85)    a   201.1 Se 27.0  ar  Eq. (6-58): Fig. 6-23: Eq. (6-13): Eq. (6-14):  f Sut 1 b   log  3  Se Eq. (6-15):   N   ar   a  1/ b  1  0.867(85)     log    0.1454 3  27.0   1  61.3  0.1454   3536 cycles   201.1  N = 3500 cycles Ans. ______________________________________________________________________________ 6-56 Sut  55 kpsi, S y  30 kpsi, K ts  1.6, L  2 ft, Fmin  150 lbf , Fmax  500 lbf Eqs. (6-33) and (6-36), or Fig. 6-27: qs = 0.80 Eq. (6-32): K fs  1  qs  Kts  1  1  0.80 1.6  1  1.48 Tmax  500(2)  1000 lbf  in, Tmin  150(2)  300 lbf  in  max   min  m  16 K fsTmax d 3 16 K fsTmin d  max   min 3 2    a  max min 2  16(1.48)(1000)  11 251 psi  11.25 kpsi  (0.875)3 16(1.48)(300)  3375 psi  3.38 kpsi  (0.875)3 11.25  3.38   7.32 kpsi 2 11.25  3.38   3.94 kpsi 2  Since the stress is entirely shear, it is convenient to check for yielding using the standard Maximum Shear Stress theory. S y / 2 30 / 2 ny    1.33  max 11.25 Shigley’s MED, 11th edition Chapter 6 Solutions, Page 52/58 Find the modifiers and endurance limit. Se  0.5(55)  27.5 kpsi Eq. (6-10): Eq. (6-18): k a  11.0(55) 0.650  0.81 Eq. (6-23): d e  0.370(0.875)  0.324 in Eq. (6-19): kb  0.879(0.324) 0.107  0.99 Eq. (6-25): kc  0.59 Eq. (6-17): S se  0.81(0.99)(0.59)(27.5)  13.0 kpsi Since the stress is entirely shear, we will use a load factor kc = 0.59, and convert the ultimate strength to a shear value rather than using the combination loading method of Sec. 6-16. From Eq. (6-58), Ssu = 0.67Su = 0.67 (55) = 36.9 kpsi. (a) Goodman, Eq. (6-41): 1 1     3.94 7.32  nf   a  m       1.99  13.0 36.9   S se S su  Ans. (b) Gerber 2 2   2 m S se   1  S su   a  1  1   nf   Eq. (6-48):   2   m  S se   S su a     2 2   2(7.32)(13.0)   1  36.9   3.94   nf       1  1   2  7.32   13.0   36.9(3.94)      n f  2.49 Ans. ______________________________________________________________________________ 6-57 Sut  145 kpsi, S y  120 kpsi From Eqs. (6-33) and (6-35), or Fig. 6-26, with a notch radius of 0.1 in, q = 0.9. Thus, with Kt = 3 from the problem statement, K f  1  q ( K t  1)  1  0.9(3  1)  2.80 4 P 2.80(4)( P)   2.476 P d2  (1.2) 2 1  m   a  (2.476 P )  1.238 P 2 f P  D  d  0.3P  6  1.2    0.54 P Tmax  4 4  max   K f Shigley’s MED, 11th edition Chapter 6 Solutions, Page 53/58 From Eqs. (6-33) and (6-36), or Fig. 6-27, with a notch radius of 0.1 in, qs  0.92. Thus, with Kts = 1.8 from the problem statement, K fs  1  qs ( K ts  1)  1  0.92(1.8  1)  1.74 16 K fsT 16(1.74)(0.54 P)  2.769 P d  (1.2)3  2.769 P  a   m  max   1.385 P 2 2 Eqs. (6-66) and (6-67):  max  3   a  [ a 2  3 a2 ]1/2  [(1.238 P)2  3(1.385P) 2 ]1/ 2  2.70 P  m  [ m 2  3 m2 ]1/ 2  [(1.238P) 2  3(1.385P ) 2 ]1/2  2.70 P Eq. (6-10): Se  0.5(145)  72.5 kpsi Eq. (6-18): ka  2.00(145)0.217  0.68 Eq. (6-19): kb  0.879(1.2) 0.107  0.862 Eq. (6-17): Se  (0.68)(0.862)(72.5)  42.5 kpsi 1 Eq. (6-41): 1     2.70 P 2.70 P  nf   a  m      3 145   42.5  Se Sut  P  4.1 kips Ans. Yield (conservative): Sy 120 ny    5.4 Yielding is not predicted. Ans.  a   m (2.70)(4.1)  (2.70)(4.1) ______________________________________________________________________________ 6-58 From Prob. 6-57, K f  2.80, K f s  1.74, S e  42.5 kpsi 4 Pmax 4(18)  2.80  44.56 kpsi 2 d  (1.2 2 ) 4P 4(4.5)  min   K f min2  2.80  11.14 kpsi d  (1.2) 2  Dd   6  1.2  Tmax  f Pmax    0.3(18)    9.72 kip  in  4   4   Dd   6  1.2  Tmin  f Pmin    0.3(4.5)    2.43 kip  in  4   4  16Tmax 16(9.72)  max  K f s  1.74  49.85 kpsi 3 d  (1.2)3 16Tmin 16(2.43)  min  K f s  1.74  12.46 kpsi 3 d  (1.2)3  max   K f Shigley’s MED, 11th edition Chapter 6 Solutions, Page 54/58 44.56  ( 11.14)  16.71 kpsi 2 44.56  ( 11.14) m   27.85 kpsi 2 49.85  12.46 a   18.70 kpsi 2 49.85  12.46 m   31.16 kpsi 2 a  Eqs. (6-66) and (6-67):  a  [( a / 0.85)2  3 a2 ]1/2  [(16.71/ 0.85) 2  3(18.70) 2 ]1/ 2  37.89 kpsi  m  [ m 2  3 m2 ]1/2  [(27.85)2  3(31.16) 2 ]1/2  60.73 kpsi Goodman: Eq. (6-41): 1     37.89 60.73   nf   a  m     145   42.5  Se Sut  nf = 0.76 1 Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). Choosing the Goodman criterion, Eq. (6-58): Fig. 6-23:  a 37.89   65.2 kpsi 1  ( m / Sut ) 1  (60.73 / 145) f = 0.8  ar   f Sut  2 0.8(145)  Eq. (6-13): a Eq. (6-14):  f Sut 1 b   log  3  Se 42.5 Se 1/ b Eq. (6-15):   ar  N    a  2  316.6  1  0.8(145)     log    0.1454 3  42.5   1  65.2  0.1454   52 460 cycles   316.6  N = 52 500 cycles Ans. ______________________________________________________________________________ 6-59 For AISI 1020 CD, From Table A-20, Sy = 390 MPa, Sut = 470 MPa. Given: Se = 175 MPa. Shigley’s MED, 11th edition Chapter 6 Solutions, Page 55/58  m 1  First Loading: 360  160  260 MPa, 2  a 1  360  160  100 MPa 2 Goodman, Eq. (6-58):  a e1  Fig. 6-23:  a 1 1   m 1 / Sut 100  223.8 MPa  Se  finite life 1  260 / 470 Off the graph, so let f = 0.9.  2 0.9  470    1022.5 MPa a 175 0.9  470  1  0.127 767 b   log 3 175 1/0.127767  223.8   145 920 cycles N    1022.5  320   200  320   200   60 MPa,  260 MPa Second loading:  m 2   a  2  2 2   a e 2  (a) Miner’s method: n1 n2  1 N1 N 2 260  298.0 MPa 1  60 / 470  298.0  N2     1022.5   1/0.127767  15 520 cycles n2 80 000  1 145 920 15 520  n2  7000 cycles Ans. (b) Manson’s method: The number of cycles remaining after the first loading Nremaining =145 920  80 000 = 65 920 cycles Two data points: 0.9(470) MPa, 103 cycles 223.8 MPa, 65 920 cycles Shigley’s MED, 11th edition Chapter 6 Solutions, Page 56/58 0.9  470  223.8  a2 103  b2 a2  65 920  2 b 1.8901   0.015170  2 b log1.8901  0.151 997 log 0.015170 223.8 a2   1208.7 MPa 0.151 997  65 920  b2  1/ 0.151 997  298.0  n2   Ans.  10 000 cycles   1208.7  ______________________________________________________________________________ 6-60 Given: Se = 50 kpsi, Sut = 140 kpsi, f =0.8. Using Miner’s method, 2 0.8 140    250.88 kpsi a 50 0.8 140  1  0.116 749 b   log 3 50 1/ 0.116 749  95   1  95 kpsi, N1    4100 cycles   250.88  1/ 0.116 749  2  80 kpsi,  80  N2     250.88   3  65 kpsi,  65  N3     250.88   17 850 cycles 1/ 0.116 749  105 700 cycles 0.2 N 0.5 N 0.3 N    1  N  12 600 cycles Ans. 4100 17 850 105 700 ______________________________________________________________________________ 6-61 Given: Sut = 530 MPa, Se = 210 MPa, and f = 0.9. (a) Miner’s method 2 0.9  530    1083.47 MPa a  210 0.9  530  1  0.118 766 b   log 3 210 1/ 0.118 766  350   1  350 MPa, N1     1083.47  Shigley’s MED, 11th edition  13 550 cycles Chapter 6 Solutions, Page 57/58 1/ 0.118 766  260   2  260 MPa, N 2     1083.47  225   165 600 cycles 1/ 0.118 766   3  225 MPa, N 3     1083.47   559 400 cycles n1 n2 n3   1 N1 N 2 N 3 n3 5000 50 000    184 100 cycles 13 550 165 600 559 400 Ans. (b) Manson’s method: The life remaining after the first series of cycling is NR1 = 13 550  5000 = 8550 cycles. The two data points required to define Se,1 are [0.9(530), 103] and (350, 8550). 0.9  530  350  a2 103  b2 a2  8550  b2  a2   b2 1.3629   0.11696  2 b log 1.362 9   0.144 280 log  0.116 96  350  8550  0.144 280  1292.3 MPa 1/0.144 280  260   67 090 cycles N2     1292.3  N R 2  67 090  50 000  17 090 cycles 0.9  530  260 b3   a3 103  b3 a3 17 090  log 1.834 6  log  0.058 514  b3  1.834 6   0.058 514  2  0.213 785, a3  b 260 17 090  0.213 785  2088.7 MPa 1/0.213 785  225   33 610 cycles Ans. N3     2088.7  ______________________________________________________________________________ 6-62 Given: Se = 45 kpsi, Sut = 85 kpsi, f = 0.86, and a = 35 kpsi and m = 30 kpsi for 12 (103) cycles. Goodman equivalent reversing stress, Eq. (6-58): Shigley’s MED, 11th edition Chapter 6 Solutions, Page 58/58  ar  Initial cycling a 35   54.09 kpsi 1   m / Sut  1   30 / 85 2 0.86  85    116.00 kpsi a  45 0.86  85  1  0.070 235 b   log 3 45  1  54.09 kpsi, 1/ 0.070 235  54.09  N1     116.00   52 190 cycles (a) Miner’s method: The number of remaining cycles at 54.09 kpsi is Nremaining = 52 190  12 000 = 40 190 cycles. The new coefficients are b = b, and a =Sf /Nb = 54.09/(40 190)  0.070 235 = 113.89 kpsi. The new endurance limit is Se,1  aN eb  113.89 106   0.070 235  43.2 kpsi Ans. (b) Manson’s method: The number of remaining cycles at 54.09 kpsi is Nremaining = 52 190  12 000 = 40 190 cycles. At 103 cycles, Sf = 0.86(85) = 73.1 kpsi. The new coefficients are b = [log(73.1/54.09)]/log(103/40 190) =  0.081 540 and a = 1/ (Nremaining) b = 54.09/(40 190)  0.081 540 = 128.39 kpsi. The new endurance limit is Se,1  aN eb  128.39 106   0.081 540  41.6 kpsi Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 6 Solutions, Page 59/58 Chapter 11 11-1 Eq. (b), Sec. 11-3: L10 = 60 L R nR = 60(3000)500 = 90 (106) rev Eq. (11-3):  10  10  1800  60    L n 60   6.30 kN C10  FD  D D  2.75  Ans. 90  106     L R nR 60  ______________________________________________________________________________ 1/ a 11-2 1/3 3 For the deep-groove 02-series ball bearing with R = 0.90, the design life xD, in multiples of rating life, is L 60L D nD 60  25000  350 xD  D   525 Ans. LR L10 106 The design radial load is FD 1.2  2.5  3.0 kN 1/3 Eq. (11-9):   525   C10 3.0  1/1.483   0.02   4.459  0.02   ln  1/ 0.9    C10 = 24.3 kN Table 11-2: Ans. Choose an 02-35 mm bearing with C10 = 25.5 kN. Ans.   525 3 / 25.5 3  0.02  1.483      R exp    Ans.   0.920 4.459  0.02     Eq. (11-21): ______________________________________________________________________________ 11-3 For the angular-contact 02-series ball bearing as described, the rating life multiple is L 60L D nD 60  40 000  520 xD  D   1248 LR L10 106 The design radial load is FD 1.4  725  1015 lbf 4.52 kN Eq. (11-9): Shigley’s MED, 11th edition Chapter 11 Solutions, Page 1/31 1/3   1248   C10 1015  1/1.483   0.02   4.459  0.02   ln  1 / 0.9    10 930 lbf 48.6 kN Table 11-2: Select an 02-60 mm bearing with C10 = 55.9 kN. Ans.   1248 4.52 / 55.9 3  0.02  1.483      R exp    Ans.   0.945 4.439     Eq. (11-21): ______________________________________________________________________________ 11-4 For the straight-roller 03-series bearing selection, xD = 1248 rating lives from Prob. 11-3 solution. FD 1.4  2235  3129 lbf 13.92 kN  1248  C10 13.92    1  Table 11-3: 3/10 118 kN Select an 03-60 mm bearing with C10 = 123 kN. Ans. 1.483 10/3     1248  13.92 /123   0.02   R exp     0.917 Ans. 4.459  0.02     Eq. (11-21): ______________________________________________________________________________ 11-5 The combined reliability of the two bearings selected in Probs. 11-3 and 11-4 is R  0.945   0.917  0.867 Ans. We can choose a reliability goal of 0.90 0.95 for each bearing. We make the selections, find the existing reliabilities, multiply them together, and observe that the reliability goal is exceeded due to the roundup of capacity upon table entry. Another possibility is to use the reliability of one bearing, say R1. Then set the reliability goal of the second as R2  0.90 R1 or vice versa. This gives three pairs of selections to compare in terms of cost, geometry implications, etc. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 11 Solutions, Page 2/31 11-6 Establish a reliability goal of contact ball bearing, 0.90 0.95 for each bearing. For an 02-series angular 1/3   1248   C10 1015  1/1.483   0.02  4.439  ln  1/ 0.95    12822 lbf 57.1 kN Select an 02-65 mm angular-contact bearing with C10 = 63.7 kN.   RA exp    1248  4.52 / 63.7  3  0.02    4.439   1.483    0.962  For an 03-series straight roller bearing, 3/10   1248   C10 13.92  1/1.483   0.02  4.439  ln  1/ 0.95    136.5 kN Select an 03-65 mm straight-roller bearing with C10 = 138 kN.   RB exp    1248  13.92 /138  10/3  0.02    4.439   1.483    0.953  The overall reliability is R = (0.962)(0.953) = 0.917, which exceeds the goal. ______________________________________________________________________________ 11-7 Given: R = 96 percent,  f = 1.2, x0 = 0,  = 4.48, b = 1.5. From Prob. 11-1, FD = 2.75 kN, LR = L10 = 60 L R nR = 60(3000)500 = 90 (106) rev, and LD = 60 L D nD = 60(10)103(1800) = 1.08 (109) rev The dimensionless multiple of rating life is 9 LD 1.08  10  xD   12 LR 90  106  Eq. (11-10): 1/ a 1/3     xD 12 C10 a f FD   1.2 2.75 9.37 kN Ans.     1/ b 1/1.5   x0     x0   1  RD    4.48  1  0.96   ______________________________________________________________________________ 11-8 For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.95 and FR = 20 kN. Shigley’s MED, 11th edition Chapter 11 Solutions, Page 3/31 xD  LD 60L D nD 60  8000  950   456 LR L10 106 3/10   456   C10 20  145 kN Ans. 1/1.483   0.02  4.439  ln  1 / 0.95    ______________________________________________________________________________ 11-9 Both bearings need to be rated in terms of the same catalog rating system in order to compare them. Using a rating life of one million revolutions, both bearings can be rated in terms of a Basic Load Rating. Eq. (11-3): L  C A FA  A   LR  8.96 kN 1/ a 1/ a  L n 60  FA  A A   LR    3000   500   60   2.0   106   1/3 Bearing B already is rated at one million revolutions, so CB = 7.0 kN. Since CA > CB, bearing A can carry the larger load. Ans. ______________________________________________________________________________ 11-10 FD = 2 kN, LD = 109 rev, R = 0.90 1/ a 1/3 L   109  C10 FD  D  2  6  20 kN Ans.  10   LR  Eq. (11-3): ______________________________________________________________________________ 11-11 FD = 800 lbf, D = 12 000 hours, nD = 350 rev/min, R = 0.90  12 000  350   60    L n 60  C10 FD  D D  800   5050 lbf Ans LR  106    Eq. (11-3): ______________________________________________________________________________ 1/3 1/ a 11-12 FD = 4 kN, D = 8 000 hours, nD = 500 rev/min, R = 0.90  8 000  500   60    L n 60  C10 FD  D D  4   24.9 kN Ans LR  106    Eq. (11-3): ______________________________________________________________________________ 1/ a 1/3 11-13 FD = 650 lbf, nD = 400 rev/min, R = 0.95 D = (5 years)(40 h/week)(52 week/year) = 10 400 hours Assume an application factor of one. The multiple of rating life is Shigley’s MED, 11th edition Chapter 11 Solutions, Page 4/31 xD  LD  10 400   400   60   249.6 LR 106 1/3 Eq. (11-9):   249.6   C10  1  650   1/1.483   0.02  4.439  ln  1 / 0.95    4800 lbf Ans. ______________________________________________________________________________ 11-14 FD = 9 kN, LD = 108 rev, R = 0.99 Assume an application factor of one. The multiple of rating life is LD 108 xD   6 100 LR 10 1/3 Eq. (11-9):   100   C10  1  9   1/1.483   0.02  4.439  ln  1 / 0.99    69.2 kN Ans. ______________________________________________________________________________ 11-15 FD = 11 kips, D = 20 000 hours, nD = 200 rev/min, R = 0.99 Assume an application factor of one. Use the Weibull parameters for Manufacturer 2 in Table 11-6. The multiple of rating life is xD  LD  20 000   200   60   240 LR 106 1/3 Eq. (11-9):   240   C10  1  11  1/1.483   0.02  4.439  ln  1/ 0.99    113 kips Ans. ______________________________________________________________________________ 11-16 From the solution to Prob. 3-79, the ground reaction force carried by the bearing at C is RC = FD = 178 lbf. Use the Weibull parameters for Manufacturer 2 in Table 11-6. xD  Shigley’s MED, 11th edition LD 15000  1200   60   1080 LR 106 Chapter 11 Solutions, Page 5/31 Eq. (11-10):   xD C10 a f FD   1/ b  x0     x0   1  RD   1/ a 1/3   1080 C10 1.2  178   1/1.483  0.02   4.459  0.02   1  0.95   2590 lbf Ans. ______________________________________________________________________________ 11-17 From the solution to Prob. 3-80, the ground reaction force carried by the bearing at C is RC = FD = 1.794 kN. Use the Weibull parameters for Manufacturer 2 in Table 11-6. xD  Eq. (11-10): LD 15000  1200   60   1080 LR 106   xD C10 a f FD   1/ b  x0     x0   1  RD   1/ a 1/3   1080 C10 1.2  1.794    1/1.483  0.02   4.459  0.02   1  0.95   26.1 kN Ans. ______________________________________________________________________________ 11-18 From the solution to Prob. 3-81, RCz = –327.99 lbf, RCy = –127.27 lbf 1/2 2 2 RC FD    327.99     127.27   351.8 lbf   Use the Weibull parameters for Manufacturer 2 in Table 11-6. 15000  1200   60  L xD  D  1080 LR 106 Eq. (11-10):   xD C10 a f FD   1/ b  xo     xo   1  RD   1/ a 1/3   1080 C10 1.2  351.8    1/1.483  0.02   4.459  0.02   1  0.95   5110 lbf Ans. ______________________________________________________________________________ 11-19 From the solution to Prob. 3-82, RCz = –150.7 N, RCy = –86.10 N 1/2 2 2 RC FD    150.7     86.10   173.6 N   Use the Weibull parameters for Manufacturer 2 in Table 11-6. Shigley’s MED, 11th edition Chapter 11 Solutions, Page 6/31 xD  Eq. (11-10): LD 15000  1200   60   1080 LR 106   xD C10 a f FD   1/ b  x0     x0   1  RD   1/ a 1/3   1080 C10 1.2  173.6    1/1.483  0.02   4.459  0.02   1  0.95   2520 N Ans. ______________________________________________________________________________ 11-20 From the solution to Prob. 3-88, RAz = 444 N, RAy = 2384 N  RA FD  4442  23842  1/ 2 2425 N 2.425 kN Use the Weibull parameters for Manufacturer 2 in Table 11-6. The design speed is equal to the speed of shaft AD, d 125 nD  F ni   191 95.5 rev/min dC 250 xD  Eq. (11-10): LD 12 000  95.5   60   68.76 LR 106   xD C10 a f FD   1/ b  x0     x0   1  RD   1/ a 1/3   68.76 C10  1  2.425    1/1.483  0.02   4.459  0.02   1  0.95   11.7 kN Ans. ______________________________________________________________________________ 11-21 From the solution to Prob. 3-90, RAz = 54.0 lbf, RAy = 140 lbf  RA FD  54.02  1402  1/2 150.1 lbf Use the Weibull parameters for Manufacturer 2 in Table 11-6. The design speed is equal to the speed of shaft AD, d 10 nD  F ni   280  560 rev/min dC 5 xD  Shigley’s MED, 11th edition LD 14 000  560   60   470.4 LR 106 Chapter 11 Solutions, Page 7/31 Eq. (11-10):   xD C10 a f FD   1/ b  x0     x0   1  RD   1/ a 3/10   470.4 C10  1  150.1   1/1.483  0.02   4.459  0.02   1  0.98   1320 lbf Ans. ______________________________________________________________________________ 11-22 (a) Fa 3 kN, Fr 7 kN, nD 500 rev/min, V 1.2 From Table 11-2, with a 65 mm bore, C0 = 34.0 kN. Fa / C0 = 3 / 34 = 0.088 From Table 11-1, 0.28  e  3.0. Fa 3  0.357 VFr  1.2   7  Since this is greater than e, interpolating Table 11-1 with Fa / C0 = 0.088, we obtain X2 = 0.56 and Y2 = 1.53. Eq. (11-12): Fe  X iVFr  Yi Fa  0.56   1.2   7    1.53  3 9.29 kN Ans. Fe > Fr so use Fe. (b) Use Eq. (11-10) to determine the necessary rated load the bearing should have to carry the equivalent radial load for the desired life and reliability. Use the Weibull parameters for Manufacturer 2 in Table 11-6. 10 000  500   60  L xD  D  300 LR 106 Eq. (11-10):   xD C10 a f FD   1/ b  x0     x0   1  RD   1/ a   300 C10  1  9.29    1/1.483  0.02   4.459  0.02   1  0.95   73.4 kN 1/3 From Table 11-2, the 65 mm bearing is rated for 55.9 kN, which is less than the necessary rating to meet the specifications. This bearing should not be expected to meet the load, life, and reliability goals. Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 11 Solutions, Page 8/31 11-23 (a) Fa 2 kN, Fr 5 kN, nD 400 rev/min, V 1 From Table 11-2, 30 mm bore, C10 = 19.5 kN, C0 = 10.0 kN Fa / C0 = 2 / 10 = 0.2 From Table 11-1, 0.34  e  0.38. Fa 2  0.4 VFr  1  5  Since this is greater than e, interpolating Table 11-1, with Fa / C0 = 0.2, we obtain X2 = 0.56 and Y2 = 1.27. F  X iVFr  Yi Fa  0.56   1  5    1.27   2  5.34 kN Ans. Eq. (11-12): e Fe > Fr so use Fe. (b) Solve Eq. (11-10) for xD.  C xD  10  a f FD  a  1/ b   x0     x0   1  RD    3  19.5   0.02   4.459  0.02   1  0.99  1/1.483  xD     1  5.34       xD 10.66 xD  LD L D nD  60   LR 106 LD    10.66  10  444 h xD 106 6 Ans. nD  60   400   60  ______________________________________________________________________________ 9 11-24 Fr 8 kN, R 0.9, LD 10 rev Eq. (11-3): L  C10 FD  D   LR  1/ a  109  8  6   10  1/3 80 kN From Table 11-2, select the 85 mm bore. Ans. ______________________________________________________________________________ Fr 8 kN, Fa 2 kN, V 1, R 0.99 11-25 Use the Weibull parameters for Manufacturer 2 in Table 11-6. Shigley’s MED, 11th edition Chapter 11 Solutions, Page 9/31 xD  LD 10 000  400   60   240 LR 106 First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-12): Fe 0.56  1  8   1.63  2  7.74 kN Fe < Fr, so just use Fr as the design load. Eq. (11-10):   xD C10 a f FD   1/ b  xo     xo   1  RD   1/ a 1/3   240 C10  1  8   82.5 kN  1/1.483  0.02   4.459  0.02   1  0.99   From Table 11-2, try 85 mm bore with C10 = 83.2 kN, C0 = 53.0 kN Iterate the previous process: Fa / C0 = 2 / 53 = 0.038 Table 11-1: 0.22  e  0.24 Fa 2  0.25  e VFr 1 8  Interpolate Table 11-1 with Fa / C0 = 0.038 to obtain X2 = 0.56 and Y2 = 1.89. Eq. (11-12): Fe 0.56(1)8  1.89(2) 8.26 > Fr 1/3 Eq. (11-10): Table 11-2: Iterate again:   240 C10  1  8.26    85.2 kN 1/1.483  0.02   4.459  0.02   1  0.99   Move up to the 90 mm bore with C10 = 95.6 kN, C0 = 62.0 kN. Fa / C0 = 2 / 62 = 0.032 Again, 0.22  e  0.24 Fa 2  0.25  e VFr 1 8  Interpolate Table 11-1 with Fa / C0 = 0.032 to obtain X2 = 0.56 and Y2 = 1.95. Table 11-1: Eq. (11- 12): Eq. (11-10): Fe 0.56(1)8  1.95(2) 8.38 > Fr   240 C10  1  8.38   1/1.483  0.02   4.459  0.02   1  0.99   Shigley’s MED, 11th edition 1/3 86.4 kN Chapter 11 Solutions, Page 10/31 The 90 mm bore is acceptable. Ans. ______________________________________________________________________________ F 8 kN, Fa 3 kN, V 1.2, R 0.9, LD 108 rev 11-26 r First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-12): Fe 0.56  1.2   8   1.63  3 10.3 kN Fe  Fr Eq. (11-3): L  C10 Fe  D   LR  1/ a  108  10.3  6   10  1/3 47.8 kN From Table 11-2, try 60 mm with C10 = 47.5 kN, C0 = 28.0 kN Iterate the previous process: Fa / C0 = 3 / 28 = 0.107 0.28  e  0.30 Fa 3  0.313  e VFr 1.2  8  Interpolate Table 11-1 with Fa / C0 = 0.107 to obtain X2 = 0.56 and Y2 = 1.46 Table 11-1: Eq. (11-12): Fe 0.56  1.2   8   1.46  3 9.76 kN > Fr 1/3  108  C10 9.76  6  45.3 kN  10  Eq. (11-3): From Table 11-2, we have converged on the 60 mm bearing. Ans. ______________________________________________________________________________ Fr 10 kN, Fa 5 kN, V 1, R 0.95 11-27 Use the Weibull parameters for Manufacturer 2 in Table 11-6. xD  LD 12 000  300   60   216 LR 106 First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-12): Fe 0.56  1  10   1.63  5  13.75 kN Fe > Fr, so use Fe as the design load. Eq. (11-10):   xD C10 a f FD   1/ b  x0     x0   1  RD   Shigley’s MED, 11th edition 1/ a Chapter 11 Solutions, Page 11/31   216 C10  1  13.75    1/1.483  0.02   4.459  0.02   1  0.95   From Table 11-2, try 95 mm bore with C10 = 108 kN, C0 = 69.5 kN Iterate the previous process: 1/3 97.4 kN Fa / C0 = 5 / 69.5 = 0.072 0.27  e  0.28 Fa 5  0.5  e VFr 1 10  Interpolate Table 11-1 with Fa / C0 = 0.072 to obtain X2 = 0.56 and Y2 = 1.62  1.63 Table 11-1: Since this is where we started, we will converge back to the same bearing. The 95 mm bore meets the requirements. Ans. ______________________________________________________________________________ Fr 9 kN, Fa 3 kN, V 1.2, R 0.99 11-28 Use the Weibull parameters for Manufacturer 2 in Table 11-6. LD 108 xD   6 100 LR 10 First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Fe 0.56  1.2   9   1.63  3 10.9 kN Eq. (11-12): Fe > Fr, so use Fe as the design load. Eq. (11-10):   xD C10 a f FD   1/ b  x0     x0   1  RD   1/ a   100 C10  1  10.9    1/1.483  0.02   4.459  0.02   1  0.99   1/3 83.9 kN From Table 11-2, try 90 mm bore with C10 = 95.6 kN, C0 = 62.0 kN. Try this bearing. Iterate the previous process: Fa / C0 = 3 / 62 = 0.048 0.24  e  0.26 Fa 3  0.278  e VFr 1.2  9  Interpolate Table 11-1 with Fa / C0 = 0.048 to obtain X2 = 0.56 and Y2 = 1.79 Table 11-1: Shigley’s MED, 11th edition Chapter 11 Solutions, Page 12/31 Eq. (11-12): Fe 0.56  1.2   9   1.79  3  11.4 kN  Fr 11.4 83.9  87.7 kN 10.9 From Table 11-2, this converges back to the same bearing. The 90 mm bore meets the requirements. Ans. ______________________________________________________________________________ C10  11-29 (a) nD 1200 rev/min, LD 15 kh, R 0.95, a f 1.2 From Prob. 3-83, RCy = 183.1 lbf, RCz = –861.5 lbf. 1/ 2 2 RC FD  183.12    861.5   881 lbf   15000  1200   60  L xD  D  1080 LR 106 Eq. (11-10):   1080 C10 1.2  881   1/1.483  0.02  4.439  1  0.95   12800 lbf 12.8 kips Ans. 1/3 (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ 11-30 (a) nD 1200 rev/min, LD 15 kh, R 0.95, a f 1.2 From Prob. 3-83, ROy = –208.5 lbf, ROz = 259.3 lbf. 1/ 2 2 RC FD  259.32    208.5   333 lbf   15000  1200   60  L xD  D  1080 LR 106 1/3   1080 C10 1.2  333   1/1.483  0.02  4.439  1  0.95   Eq. (11-10): 4837 lbf 4.84 kips Ans. (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ 11-31 (a) nD 900 rev/min, LD 12 kh, R 0.98, a f 1.2 From Prob. 3-84, RCy = 8.319 kN, RCz = –10.830 kN. Shigley’s MED, 11th edition Chapter 11 Solutions, Page 13/31 1/2 2 RC FD  8.3192    10.830   13.7 kN   12 000  900   60  L xD  D  648 LR 106 1/3   648 C10 1.2  13.7    204 kN Ans. 1/1.483  0.02  4.439  1  0.98   Eq. (11-10): (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ 11-32 (a) nD 900 rev/min, LD 12 kh, R 0.98, a f 1.2 From Prob. 3-84, ROy = 5083 N, ROz = 494 N. RC FD  50832  494 2  xD  1/2 5106 N 5.1 kN LD 12 000  900   60   648 LR 106 1/3   648 C10 1.2  5.1   76.1 kN Ans. 1/1.483  0.02  4.439  1  0.98   Eq. (11-10): (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ 11-33 Assume concentrated forces as shown. Pz 8  28  224 lbf Py 8  35 280 lbf T 224  2  448 lbf in T x  448  1.5 F cos 20 0 448 F 318 lbf 1.5  0.940  M Oz 5.75Py  11.5RAy  14.25 F sin 20 0 5.75  280   11.5 RAy  14.25  318   0.342  0 RAy  5.24 lbf Shigley’s MED, 11th edition Chapter 11 Solutions, Page 14/31 M Oy  5.75Pz  11.5 RAz  14.25 F cos 20 0  5.75  224   11.5 RAz  14.25  318   0.940  0 2 2 RA    482     5.24     z z z  F RO  Pz  RA  F cos 20 0 RAz  482 lbf; 1/2 482 lbf ROz  224  482  318  0.940  0 ROz  40.9 lbf F y ROy  Py  RAy  F sin 20 0 ROy  280  5.24  318  0.342  0 ROy  166 lbf 2 2 RO    40.9     166     1/2 171 lbf So the reaction at A governs. Reliability Goal: 0.92 0.96 FD 1.2  482  578 lbf xD 35000  350   60  /106 735 1/3   735   C10 578  1/1.483   0.02   4.459  0.02   ln  1/ 0.96    6431 lbf 28.6 kN From Table 11-2, a 40 mm bore angular contact bearing is sufficient with a rating of 31.9 kN. Ans. ______________________________________________________________________________ 11-34 For a combined reliability goal of 0.95, use 0.95 0.975 for the individual bearings. xD  40 000  420   60  106 Shigley’s MED, 11th edition 1008 Chapter 11 Solutions, Page 15/31 The resultants of the given forces are RO = [(–387)2 + 4672]1/2 = 607 lbf RB = [3162 + (–1615)2]1/2 = 1646 lbf At O: 1/3 Eq. (11-9):   1008   C10 1.2  607   1/1.483   0.02   4.459  0.02   ln  1/ 0.975    9978 lbf  44.4 kN From Table 11-2, select an 02-55 mm angular-contact ball bearing with a basic load rating of 46.2 kN. Ans. At B: 3/10 Eq. (11-9):   1008   C10 1.2  1646   1/1.483   0.02   4.459  0.02   ln  1 / 0.975    20827 lbf 92.7 kN From Table 11-3, select an 02-75 mm or 03-55 mm cylindrical roller. Ans. ______________________________________________________________________________ 11-35 The reliability of the individual bearings is R  0.98 0.9899 From statics, T = (270  50) = (P1  P2)125 = (P1  0.15 P1)125 P1 = 310.6 N, P2 = 0.15 (310.6) = 46.6 N P1 + P2 = 357.2 N FAy  357.2sin 45  252.6 N  FAz M F M F z O y y O z Shigley’s MED, 11th edition  850REy  300(252.6)  0  REy   89.2 N  252.6  89.2  ROy  0  ROy   163.4 N   850 REz  700(320)  300(252.6)  0  REz   174.4 N   174.4  320  252.6  ROz  0  ROz 107 N Chapter 11 Solutions, Page 16/31 RO    163.4 RE    89.2  2 2  107 2  195 N    174.4  196 N 2 The radial loads are nearly the same at O and E. We can use the same bearing at both locations. xD  60 000  1500   60  106 5400 1/3 Eq. (11-9):   5400   C10 1 0.196   1/1.483   0.02  4.439  ln  1 / 0.9899     5.7 kN From Table 11-2, select an 02-12 mm deep-groove ball bearing with a basic load rating of 6.89 kN. Ans. ______________________________________________________________________________ 11-36 R  0.96 0.980 T 12(240 cos 20 ) 2706 lbf in 2706 F 498 lbf 6 cos 25 In xy-plane: RCy 181 lbf ROy 82.1  210  181 111.1 lbf In xz-plane: M Oy 16(226)  30(451)  42 RCz 0 RCz  236 lbf ROz 226  451  236 11 lbf RO  111.12  112   RC  1812  2362 xD   1/2 112 lbf Ans. 1/ 2 297 lbf Ans. 50 000  300   60  106 900 1/3  C10  O   900   1.2  112   1/1.483   0.02  4.439  ln  1/ 0.980    1860 lbf 8.28 kN Shigley’s MED, 11th edition Chapter 11 Solutions, Page 17/31 1/3  C10  C   900   1.2  297   1/1.483   0.02  4.439  ln  1/ 0.980    4932 lbf 21.9 kN Bearing at O: Choose a deep-groove 02-17 mm. Ans. Bearing at C: Choose a deep-groove 02-35 mm. Ans. ______________________________________________________________________________ 11-37 Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust. The shaft floats within the endplay of the second (roller) bearing. Since the thrust force here is larger than any radial load, the bearing absorbing the thrust (bearing A) is heavily loaded compared to bearing B. Bearing B is thus likely to be oversized and may not contribute measurably to the chance of failure. If this is the case, we may be able to obtain the desired combined reliability with bearing A having a reliability near 0.99 and bearing B having a reliability near 1. This would allow for bearing A to have a lower capacity than if it needed to achieve a reliability of 0.99 . To determine if this is the case, we will start with bearing B. Bearing B (straight roller bearing) 30 000  500   60  xD  900 106  Fr  362  67 2  1/ 2 76.1 lbf 0.339 kN Try a reliability of 1 to see if it is readily obtainable with the available bearings. 3/10 Eq. (11-9):   900   C10 1.2  0.339   1/1.483   0.02  4.439  ln  1/1.0    10.1 kN The smallest capacity bearing from Table 11-3 has a rated capacity of 16.8 kN. Therefore, we select the 02-25 mm straight cylindrical roller bearing. Ans. Bearing at A (angular-contact ball) With a reliability of 1 for bearing B, we can achieve the combined reliability goal of 0.99 if bearing A has a reliability of 0.99.  Fr  362  2122  1/ 2 215 lbf 0.957 kN Fa 555 lbf 2.47 kN Trial #1: Tentatively select an 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN. Shigley’s MED, 11th edition Chapter 11 Solutions, Page 18/31 Fa 2.47  0.0392 C0 63.0 xD  30 000  500   60  106 900 Table 11-1: Interpolating, X2 = 0.56, Y2 = 1.88 Eq. (11-12): Fe 0.56  0.957   1.88  2.47  5.18 kN 1/3   900   C10 1.2  5.18   1/1.483   0.02  4.439  ln  1/ 0.99    99.54 kN  90.4 kN Eq. (11-9): Trial #2: Tentatively select a 02-90 mm angular-contact ball with C10 = 106 kN and C0 = 73.5 kN. Fa 2.47  0.0336 C0 73.5 Table 11-1: Interpolating, X2 = 0.56, Y2 = 1.93 Fe 0.56  0.957   1.93  2.47  5.30 kN 1/3   900   C10 1.2  5.30   102 kN < 106 kN O.K. 1/1.483    0.02  4.439 ln 1 / 0.99       Select an 02-90 mm angular-contact ball bearing. Ans. ______________________________________________________________________________ 11-38 We have some data. Let’s estimate parameters b and θ from it. In Fig. 11-5, we will use line AB. In this case, B is to the right of A. 115  2000   60  For F = 18 kN,  x 1  106 13.8 This establishes point 1 on the R = 0.90 line. Shigley’s MED, 11th edition Chapter 11 Solutions, Page 19/31 The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter Weibull distribution, x0 = 0 and points A and B are related by [see Eq. (11-8)]: x A   ln  1/ 0.90   1/ b xB   ln  1/ 0.20   1/ b (1) and xB/xA is in the same ratio as 600/115. Eliminating θ, b ln  ln  1/ 0.20  / ln  1/ 0.90   ln  600 /115  1.65 Ans. Solving for θ in Eq. (1),  xA  ln  1/ RA   1/1.65  1  ln  1/ 0.90   1/1.65 3.91 Ans. Therefore, for the data at hand,  R exp     x     3.91  1.65    Check R at point B: xB = (600/115) = 5.217   5.217 1.65  R exp      0.20   3.91   Shigley’s MED, 11th edition Chapter 11 Solutions, Page 20/31 Note also, for point 2 on the R = 0.20 line, log  5.217   log  1 log  xm  2  log  13.8   xm  2 72 ______________________________________________________________________________ 11-39 This problem is rich in useful variations. Here is one. Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of (0.99)1/6 = 0.9983. Shaft a FAr  2392 1112  1/2   FBr  5022  10752 264 lbf 1.175 kN 1/ 2 1186 lbf 5.28 kN Thus the bearing at B controls. 10 000  1200   60  xD  720 106 0.02  4.439  ln  1/ 0.9983    720  C10 1.2  5.28     0.080 26  1/1.483 0.3 97.2 kN Select an 02-80 mm with C10 = 106 kN. Shaft b   393 FCr  8742  22742 FDr 2  657 2   1/2 1/2 0.080 26 Ans. 2436 lbf 766 lbf or or 10.84 kN 3.41 kN The bearing at C controls. 10 000  240   60  xD  144 106  144  C10 1.2  10.84     0.080 26  Select an 02-90 mm with C10 = 142 kN. 0.3 123 kN Ans. Shaft c FEr  11132  23852  Shigley’s MED, 11th edition 1/2 2632 lbf or 11.71 kN Chapter 11 Solutions, Page 21/31 FFr  417 2  8952  1/2 987 lbf or 4.39 kN The bearing at E controls. 10 000  80   60  xD  48 106  48  C10 1.2  11.71    0.080 26  0.3 95.7 kN Select an 02-80 mm with C10 = 106 kN. Ans. ______________________________________________________________________________ 4 11-40 (a) Eq. (1-9): (b) R  Ri  Ri  4 R  4 0.92 0.9794 97.94 percent Ans. i 1  Mx = 0 = (FG cos 20o)(1) – 200 FG = 212.84 lbf  (MB)y = 0 = 2(FG cos 20o) – 3.5RAz RAz = 2(212.84 cos 20o)/3.5 = 114.29 lbf  (MB)z = 0, 2 (212.84 sin 20o) + 3.5 (FA)y = 0  RAy = – 41.60 lbf 2 2 RA  RAy  RAz    41.60  2  114.292 121.6 lbf Ans. (c) L10 = 1 (106) revolutions, LD = 30(103)60(60) = 108(106) revolutions. 6 LD 108  10  xD   108 L10 1 106  Eq. (11-10) with af = 1, xD = 108, a = 3, FD = 121.6 lbf, RD = 0.9794, x0 = 0.02,  = 4.459, and b =1.483:   xD C10 a f FD  1/ b   x0     x0   1  RD   1/ a 1/3   108 1 121.6   826.6 lbf Ans. 1/1.483   0.02   4.459  0.02   1  0.9794   ______________________________________________________________________________ 11-41 For the output shaft: N1 / N2 = 2 /1  2 = (N1 / N2)2 = (30/60) 200 = 100 rev/min. Shigley’s MED, 11th edition Chapter 11 Solutions, Page 22/31 Service time in minutes:  52 weeks   5 days   8 hours   60 minutes  3 6 years      748.8  10  minutes hour   year   week   day   3 For bearings C and D: LD = 748.8(10 )100 = 74.88(106) revolutions 74.88  10 6  L xD  D  74.88 L10 1 106  (a) R = RA RB RC RD  RD = R / (RA RB RC) = 0.90 / [0.97(0.96)0.98] = 0.986 Ans. (b) Given Fr = 9 kN, Fa = 5 kN. Eq. (11-12) with V = 1 (inner ring rotates): F e = X i F r + Yi F a Iteration will be necessary. First try. From the middle of Table 11-1, X2 = 0.56, Y2 = 1.63 Fe = 0.56(9) + 1.63(5) = 13.19 kN Eq. (11-10): 1/ a 1/3     xD 74.88  C10 a f FD  1 13.19     1/ b 1/1.483   x0     x0   1  RD    0.02   4.459  0.02   1  0.986   86.06 kN Ans. From Table 11-2 we see that C10 exceeds the 83.2 kN rating for an 85 mm bore bearing. To see if this bearing is possible, we need to determine Xi and Yi. For the 85 mm bearing: C0 = 53.0 kN, Fa / C0 = 5/53 = 0.0943 Fa / C0 e 0084 0.28 0.0943 e 0.110 0.30 0.110  0.0943 0.3  e 0.3  e   0.604   e 0.29 0.110  0.084 0.3  0.28 0.02 Fa 5  0.556 VFr 1 9  . Since 0.556 > 0.29, use X2 = 0.56, Y2 = 1.50 (interpolated) Fe = 0.56(9) + 1.50(5) = 12.54 kN 12.54 C10  86.06 81.8 kN 13.19 Since 81.8 kN < 83.2 kN it is acceptable to use the bearing with the 85 mm bore. Ans. ______________________________________________________________________________ 11-42 From Table 11-3, for a 03-30 mm cylindrical roller bearing, C10 =36.9 kN = 36.9/4.45 = 8.292 kips. K C10a L10 8.29210/3  106  1.154  109  Eq. (a), Sec. 11-7: 9 K 1.154  10  L1  a  5.399  106  rev 10/3 F1 5 At 5 kips: Shigley’s MED, 11th edition Chapter 11 Solutions, Page 23/31 9 K 1.154  10  L2  a  1.127  106  rev 10/3 F2 8 At 8 kips: l1 l2 l2 300 000  1   1 6 L1 L2 5.399  10  1.127  106  Eq. (11-16):  300 000   1.06  106  rev l2 1.127  106  1  Ans. 6  5.399  10   ______________________________________________________________________________ K C10a L10 and 11-43 From Table 11-3, select a 25 mm trial bearing. Using the equations a Li K / Fi Size 25 mm C10a L10 C10 28.6 kN 28.610/3  106  7.154(1010) L1 7.154  1010  L2 7.154  1010  L3 7.154  1010  2010/3 3.294(106) 2510/3 1.566(106) 3010/3 852.7(103) 1    f1 f 2 f 3  0.3 0.5 0.2  l        6 6 3  3.294  10  1.566  10  852.7  10    L1 L2 L3  Using a spreadsheet following the method above yields 1 1.55  106  rev Size l 25 mm 1.55(106) rev 30 mm 3.63(106) rev 35 mm 6.82(106) rev greater than 5 (106) rev. Thus, 35 mm Ans. ______________________________________________________________________________ 11-44 Express Eq. (11-1) as F1a L1 C10a L10 K For a ball bearing, a = 3 and for an 02-30 mm angular contact bearing, C10 = 20.3 kN. K  20.3  106  8.365  109  3 At a load of 18 kN, life L1 is given by: 9 K 8.365  10  L1  a  1.434  106  rev 3 F1 18 For a load of 30 kN, life L2 is: Shigley’s MED, 11th edition Chapter 11 Solutions, Page 24/31 L2    0.310 10 rev   30 8.365 109 6 3 In this case, Eq. (6-68) – the Palmgren-Miner cycle-ratio summation rule – can be expressed as l1 l2  1 L1 L2 Substituting, 200 000 l2  1 6 1.434 10 0.310 106 l2   0.267  10  6   rev Ans. ______________________________________________________________________________ 11-45 Total life in revolutions Let: l = total turns f1 = fraction of turns at F1 f2 = fraction of turns at F2 From the solution of Prob. 11-44, L1 = 1.434(106) rev and L2 = 0.310(106) rev. Palmgren-Miner rule: l1 l2 fl f l   1  2 1 L1 L2 L1 L2 from which l 1 f1 / L1  f 2 / L2 l 1  0.40 / 1.434  10      0.60 /  0.310  10    451 585 rev 6 6 Ans. Total life in loading cycles 4 min at 2000 rev/min = 8000 rev/cycle 6 min at 2000 rev/min = 12 000 rev/cycle Shigley’s MED, 11th edition Chapter 11 Solutions, Page 25/31 Total rev/cycle = 8000 + 12 000 = 20 000 451 585 rev 22.58 cycles 20 000 rev/cycle Ans. Total life in hours  min   22.58 cycles   10   3.76 h Ans.  cycle   60 min/h  ______________________________________________________________________________ FrA 560 lbf 11-46 FrB 1095 lbf Fae 200 lbf xD  LD 40 000  400   60   10.67 LR 90  106  R  0.90 0.949 FiA  0.47 FrA 0.47  560   175.5 lbf KA 1.5 FiB  0.47 FrB 0.47  1095   343.1 lbf KB 1.5 Eq. (11-18): Eq. (11-18): FiA ?  FiB  Fae  175.5 lbf  343.1  200  543.1 lbf, so Eq. (11-19) applies. We will size bearing B first since its induced load will affect bearing A, but is not itself affected by the induced load from bearing A [see Eq. (11-19)]. From Eq. (11-19b), FeB = FrB = 1095 lbf. FRB Eq. (11-10):   10.67  1.4  1095   1/1.5  4.48  1  0.949     3/10 3607 lbf Ans. Select cone 32305, cup 32305, with 0.9843 in bore, and rated at 3910 lbf with K = 1.95. Ans. With bearing B selected, we re-evaluate the induced load from bearing B using the actual value for K. Shigley’s MED, 11th edition Chapter 11 Solutions, Page 26/31 FiB  0.47 FrB 0.47  1095   263.9 lbf KB 1.95 Eq. (11-18): Find the equivalent radial load for bearing A from Eq. (11-19), which still applies. Eq. (11-19a): FeA 0.4 FrA  K A  FiB  Fae  FeA 0.4  560   1.5  263.9  200  920 lbf FeA  FrA Eq. (11-10):   10.67  FRA 1.4  920    4.48  1  0.949  1/1.5    3/10 3030 lbf Tentatively select cone M86643, cup M86610, with 1 in bore, and rated at 3250 lbf with K = 1.07. Iterating with the new value for K, we get FeA = 702 lbf and FrA = 2312 lbf. Ans. By using a bearing with a lower K, the rated load decreased significantly, providing a higher than requested reliability. Further examination with different combinations of bearing choices could yield additional acceptable solutions. ______________________________________________________________________________ 11-47 The thrust load on shaft CD is from the axial component of the force transmitted through the bevel gear, and is directed toward bearing C. By observation of Fig. 11-14, direct mounted bearings would allow bearing C to carry the thrust load. Ans. From the solution to Prob. 3-85, the axial thrust load is Fae = 362.8 lbf, and the bearing radial forces are FCx = 287.2 lbf, FCz = 500.9 lbf, FDx = 194.4 lbf, and FDz = 307.1 lbf. Thus, the radial forces are FrC  287.2 2  500.9 2 577 lbf FrD  194.42  307.12 363 lbf The induced loads are 0.47 FrC 0.47  577  FiC   181 lbf KC 1.5 Eq. (11-18): FiD  0.47 FrD 0.47  363  114 lbf KD 1.5 Eq. (11-18): Check the condition on whether to apply Eq. (11-19) or Eq. (11-20), where bearings C and D are substituted, respectively, for labels A and B in the equations. FiC ? FiD  Fae Shigley’s MED, 11th edition Chapter 11 Solutions, Page 27/31 181 lbf  114  362.8 476.8 lbf, so Eq.(11-19) applies Eq. (11-19a): FeC 0.4 FrC  KC  FiD  Fae  0.4  577  1.5  114  362.8  946 lbf  FrC , so use FeC Assume for tapered roller bearings that the specifications for Manufacturer 1 in Table 116 are applicable. xD  LD 108  1.11 LR 90  106  R  0.90 0.949 FRC Eq. (11-10):   1.11  1 946    4.48  1  0.949  1/1.5    3/10 1130 lbf Ans. Eq. (11-19b): FeD FrD 363 lbf 3/10   1.11  433 lbf Ans. FRD 1 363  1/1.5  4.48  1  0.949     Eq. (11-10): ______________________________________________________________________________ 11-48 The thrust load on shaft AB is from the axial component of the force transmitted through the bevel gear, and is directed to the right. By observation of Fig. 11-14, indirect mounted bearings would allow bearing A to carry the thrust load. Ans. From the solution to Prob. 3-87, the axial thrust load is Fae = 92.8 lbf, and the bearing radial forces are FAy = 639.4 lbf, FAz = 1513.7 lbf, FBy = 276.6 lbf, and FBz = 705.7 lbf. Thus, the radial forces are FrA  639.42  1513.7 2 1643 lbf FrB  276.62  705.7 2 758 lbf The induced loads are 0.47 FrA 0.47  1643 FiA   515 lbf K 1.5 A Eq. (11-18): FiB  0.47 FrB 0.47  758   238 lbf KB 1.5 Eq. (11-18): Check the condition on whether to apply Eq. (11-19) or Eq. (11-20). FiA ? FiB  Fae Shigley’s MED, 11th edition Chapter 11 Solutions, Page 28/31 515 lbf  238  92.8 330.8 lbf, so Eq.(11-20) applies Notice that the induced load from bearing A is sufficiently large to cause a net axial force to the left, which must be supported by bearing B. Eq. (11-20a): FeB 0.4 FrB  K B  FiA  Fae  0.4  758  1.5  515  92.8  937 lbf  FrB , so use FeB Assume for tapered roller bearings that the specifications for Manufacturer 1 in Table 11-6 are applicable. 6 LD 500  10  xD   5.56 LR 90  106  R  0.90 0.949 Eq. (11-10):   5.56  FRB 1 937   1/1.5  4.48  1  0.949     3/10 1810 lbf Ans. Eq. (11-19b): FeA FrA 1643 lbf 3/10   5.56  3180 lbf Ans. FRA 1 1643   4.48  1  0.949  1/1.5    Eq. (11-10): ______________________________________________________________________________ 11-49 The lower bearing is compressed by the axial load, so it is designated as bearing A. FrA 25 kN FrB 12 kN Fae 5 kN FiA  0.47 FrA 0.47  25  7.83 kN KA 1.5 FiB  0.47 FrB 0.47  12   3.76 kN KB 1.5 Eq. (11-18): Eq. (11-18): Check the condition on whether to apply Eq. (11-19) or Eq. (11-20) FiA ? FiB  Fae 7.83 kN  3.76  5 8.76 kN, so Eq.(11-19) applies Eq. (11-19a): FeA 0.4 FrA  K A  FiB  Fae  Shigley’s MED, 11th edition Chapter 11 Solutions, Page 29/31 0.4  25   1.5  3.76  5  23.1 kN  FrA, so use FrA  60 min   8 hr   5 day   52 weeks  LD  250 rev/min      5 yrs    yr  hr   day   week      156 106 rev Assume for tapered roller bearings that the specifications for Manufacturer 1 in Table 11-6 are applicable. Eq. (11-3): L  FRA a f FD  D   LR  3/10        156 106 1.2  25    90 106 Eq. (11-19b): FeB FrB 12 kN 3/10  156  FRB 1.2  12   17.0 kN  90   Eq. (11-3): 3/10 35.4 kN Ans. Ans. ______________________________________________________________________________ 11-50 The left bearing is compressed by the axial load, so it is properly designated as bearing A. FrA 875 lbf FrB 625 lbf Fae 250 lbf Assume K = 1.5 for each bearing for the first iteration. Obtain the induced loads. FiA  0.47 FrA 0.47  875   274 lbf KA 1.5 FiB  0.47 FrB 0.47  625   196 lbf KB 1.5 Eq. (11-18): Eq. (11-18): Check the condition on whether to apply Eq. (11-19) or Eq. (11-20). FiA ? FiB  Fae 274 lbf  196  250 lbf, so Eq.(11-19) applies We will size bearing B first since its induced load will affect bearing A, but it is not affected by the induced load from bearing A [see Eq. (11-19)]. From Eq. (11-19b), FeB = FrB = 625 lbf. Shigley’s MED, 11th edition Chapter 11 Solutions, Page 30/31 Eq. (11-3): L  FRB a f FD  D   LR  3/10  90 000  150   60    1 625   90 106   3/10   FRB 1208 lbf Select cone 07100, cup 07196, with 1 in bore, and rated at 1570 lbf with K = 1.45. Ans. With bearing B selected, we re-evaluate the induced load from bearing B using the actual value for K. 0.47 FrB 0.47  625  FiB   203 lbf K 1.45 B Eq. (11-18): Find the equivalent radial load for bearing A from Eq. (11-19), which still applies. Eq. (11-19a): FeA 0.4 FrA  K A  FiB  Fae  0.4  875   1.5  203  250  1030 lbf FeA  FrA Eq. (11-3): L  FRA a f FD  D   LR  3/10  90 000  150   60    1 1030   90 106   3/10   FRA 1990 lbf Any of the bearings with 1-1/8 in bore are more than adequate. Select cone 15590, cup 15520, rated at 2480 lbf with K = 1.69. Iterating with the new value for K, we get FeA = 1120 lbf and FrA = 2160 lbf. The selected bearing is still adequate. Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 11 Solutions, Page 31/31 Shigley’s MED, 11th edition Chapter 11 Solutions, Page 32/31 Chapter 14 d  14-1 N 22   3.667 in P 6 Table 14-2: Y = 0.331  dn  (3.667)(1200) V    1152 ft/min 12 12 Eq. (13-34): 1200  1152 Kv  1.96 1200 Eq. (14-4b): H 15 W t  33 000  33 000  429.7 lbf V 1152 Eq. (13-35) : K W t P 1.96(429.7)(6)   v   7633 psi  7.63 kpsi FY 2(0.331) Eq. (14-7): Ans. ________________________________________________________________________ N 18   1.8 in P 10 Table 14-2: Y = 0.309  dn  (1.8)(600) V    282.7 ft/min 12 12 Eq. (13-34): 1200  282.7 Kv   1.236 1200 Eq. (14-4b): H 2 W t  33 000  33 000  233.5 lbf V 282.7 Eq. (13-35) : K W t P 1.236(233.5)(10)   v   9340 psi  9.34 kpsi FY 1.0(0.309) Eq. (14-7): d  14-2 Ans. ________________________________________________________________________ d  mN  1.25(18)  22.5 mm 14-3 Table 14-2: Eq. (14-6b): Eq. (13-36): Y = 0.309  dn  (22.5)(10 3 )(1800) V    2.121 m/s 60 60 6.1  2.121 Kv   1.348 6.1 60 000H 60 000(0.5) Wt    0.2358 kN  235.8 N  dn  (22.5)(1800) Shigley’s MED, 11th edition Chapter 14 Solutions, Page 1/40   K vW t 1.348(235.8)   68.6 MPa FmY 12(1.25)(0.309) Ans. Eq. (14-8): ________________________________________________________________________ d  mN  8(16)  128 mm 14-4 Table 14-2: Eq. (14-6b): Eq. (13-36): Y = 0.296  dn  (128)(10 3 )(150) V    1.0053 m/s 60 60 6.1  1.0053 Kv   1.165 6.1 60 000 H 60 000(6) Wt    5.968 kN  5968 N  dn  (128)(150) K vW t 1.165(5968)     32.6 MPa FmY 90(8)(0.296) Ans. Eq. (14-8): ________________________________________________________________________ d  mN  1(16)  16 mm 14-5 Table 14-2: Eq. (14-6b): Eq. (13-36): Y = 0.296  dn  (16)(10 3 )(400) V    0.335 m/s 60 60 6.1  0.335 Kv  1.055 6.1 60 000H 60 000(0.15) Wt    0.4476 kN  447.6 N  dn  (16)(400) F  Eq. (14-8): K vW t 1.055(447.6)   10.6 mm  mY 150(1)(0.296) From Table 13-2, use F = 11 mm or 12 mm, depending on availability. Ans. ________________________________________________________________________ d  mN  2(20)  40 mm 14-6 Table 14-2: Eq. (14-6b): Eq. (13-36): Y = 0.322  dn  (40)(10 3 )(200) V    0.419 m/s 60 60 6.1  0.419 Kv   1.069 6.1 60 000H 60 000(0.5) Wt    1.194 kN  1194 N  dn  (40)(200) Shigley’s MED, 11th edition Chapter 14 Solutions, Page 2/40 F  Eq. (14-8): K vW t 1.069(1194)   26.4 mm  mY 75(2.0)(0.322) From Table 13-2, use F = 28 mm. Ans. ________________________________________________________________________ N 24   4.8 in P 5 Table 14-2: Y = 0.337  dn  (4.8)(50)   62.83 ft/min V  12 12 Eq. (13-34): 1200  62.83 Kv   1.052 1200 Eq. (14-4b): H 6 W t  33 000  33 000  3151 lbf V 62.83 Eq. (13-35) : K W t P 1.052(3151)(5) F  v   2.46 in 3  Y 20(10 )(0.337) Eq. (14-7): d  14-7 Use F = 2.5 in Ans. ________________________________________________________________________ N 16   4.0 in P 4 Table 14-2: Y = 0.296  dn  (4.0)(400) V    418.9 ft/min 12 12 Eq. (13-34): 1200  418.9 Kv   1.349 1200 Eq. (14-4b): H 20 W t  33 000  33 000 1575.6 lbf V 418.9 Eq. (13-35) : K W t P 1.349(1575.6)(4) F  v   2.39 in 3  Y 12(10 )(0.296) Eq. (14-7): d  14-8 Use F = 2.5 in Ans. ________________________________________________________________________ 14-9 Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309. V  Eq. (13-34): Shigley’s MED, 11th edition  dn  (2.25)(600)   353.4 ft/min 12 12 Chapter 14 Solutions, Page 3/40 Eq. (14-4b): Eq. (13-35): 1200  353.4  1.295 1200 H 2.5 W t  33 000  33 000  233.4 lbf V 353.4 Kv  F  Eq. (14-7): K vW t P 1.295(233.4)(8)   0.783 in Y 10(103 )(0.309) Using coarse integer pitches from Table 13-2, the following table is formed. d 9.000 2 6.000 3 4.500 4 3.000 6 V Kv 1413.717 2.178 942.478 1.785 706.858 1.589 471.239 1.393 2.250 8 353.429 1.295 1.800 10 282.743 1.236 1.500 12 235.619 1.196 1.125 16 176.715 1.147 Wt F 58.356 0.082 87.535 0.152 116.713 0.240 175.06 0.473 9 233.42 0.782 6 291.78 1.167 2 350.13 1.627 9 466.85 2.773 2 Other considerations may dictate the selection. Good candidates are P = 8 (F = 7/8 in) and P =10 (F = 1.25 in). Ans. ________________________________________________________________________ 14-10 Try m = 2 mm which gives d = 2(18) = 36 mm and Y = 0.309. V  Eq. (14-6b): Eq. (13-36):  dn  (36)(10 3 )(900)   1.696 m/s 60 60 6.1  1.696 1.278 6.1 60 000H 60 000(1.5) Wt    0.884 kN  884 N  dn  (36)(900) Kv  F  1.278(884)  24.4 mm 75(2)(0.309) Eq. (14-8): Using the preferred module sizes from Table 13-2: m d 1.00 1.25 Shigley’s MED, 11th edition V 18.0 22.5 Kv Wt F 0.848 1.139 1768.388 86.917 1.060 1.174 1414.711 57.324 Chapter 14 Solutions, Page 4/40 1.50 2.00 3.00 4.00 5.00 6.00 8.00 10.00 12.00 16.00 20.00 25.00 32.00 40.00 50.00 27.0 36.0 54.0 72.0 90.0 108.0 144.0 180.0 216.0 288.0 360.0 450.0 576.0 720.0 900.0 1.272 1.696 2.545 3.393 4.241 5.089 6.786 8.482 10.179 13.572 16.965 21.206 27.143 33.929 42.412 1.209 1178.926 40.987 1.278 884.194 24.382 1.417 589.463 12.015 1.556 442.097 7.422 1.695 353.678 5.174 1.834 294.731 3.888 2.112 221.049 2.519 2.391 176.839 1.824 2.669 147.366 1.414 3.225 110.524 0.961 3.781 88.419 0.721 4.476 70.736 0.547 5.450 55.262 0.406 6.562 44.210 0.313 7.953 35.368 0.243 Other design considerations may dictate the size selection. For the present design, m = 2 mm (F = 25 mm) is a good selection. Ans. ________________________________________________________________________ 14-11 Given: Sc = 2.22 HB +200 MPa, HB = 200, Pinion NP = 16 teeth,  = 20o, m = 8 mm, F = 90 mm, nP = 150 rev/min, W t = 6 kN, gear ratio 4:1. NG = 16(4) = 64 teeth. Eq. (13-2): dP = m NP = 8(16) = 128 mm, dG = m NG = 8(64) = 512 mm Pinion and Gear steel, Eq. (14-13):   E  C p  2  2  1     V Eq. (14-12): Eq. (14-14):  207  109     2  2  1  0.292   1/ 2 189.8  103  Pa 2 d P nP 2  0.128 150  1.005 m/s 2  60  2  60  Kv  Eq. (14-6b): 1/2 6.1  1.005 1.165 6.1 rP sin  128sin 20o r1   21.89 mm, 2 2  K W t  1 1   C  C p  v      F cos   r1 r2   r2 4r1 87.56 mm 1/2  1.165  6  103  1 1   189.8  10      o  0.090 cos 20  0.02189 0.08756   1/2 3  412.3  106  Pa  412.3 MPa Sc = 2.22 HB +200 = 2.22(200) + 200 = 644 MPa Shigley’s MED, 11th edition Chapter 14 Solutions, Page 5/40 2 2  S   644  n  c   Ans.  2.44 412.3     C ______________________________________________________________________________ 14-12 NP 20 N 50   2.5 in, d G  G   6.25 in P 8 P 8  (2.5)(1200) V   785.4 ft/min 12 1200  785.4 Kv   1.655 1200 H 12 W t  33 000  33 000  504.2 lbf V 785.4 dP  Eq. (14-4b): Eq. (13-36): C p  2100 psi Table 14-8: [Note: Using Eq. (14-13) can result in wide variation in Cp due to wide variation in cast iron properties.] 2.5sin 20 6.25sin 20 r1   0.4275 in, r2  1.069 in 2 2 Eq. (14-12): Eq. (14-14):  K W t  1 1  C   Cp  v    F cos   r1 r2    1/ 2 1/ 2  1.655(504.2)  1 1    2100      1.5cos 20  0.4275 1.069     92.5(103 ) psi   92.5 kpsi Ans. ________________________________________________________________________ 14-13 16 48  1.333 in, d G   4 in 12 12  (1.333)(700) V   244.3 ft/min 12 dP  Eq. (14-4b): Eq. (13-36): 1200  244.3  1.204 1200 H 1.5 Wt  33 000  33 000  202.6 lbf V 244.3 Kv  C p  2100 psi Table 14-8: [Note: Using Eq. (14-13) can result in wide variation in Cp due to wide variation in cast iron properties.] Shigley’s MED, 11th edition Chapter 14 Solutions, Page 6/40 Eq. (14-12): Eq. (14-14): r1  1.333sin 20  0.228 in, 2 r2   1.204(202.6)  1 1   C   2100      F cos 20  0.228 0.684   4 sin 20  0.684 in 2 1/ 2   100(103 ) 2  2100   1.204(202.6)   1 1  F      0.669 in 3     100(10 )   cos 20   0.228 0.684  Use F = 0.75 in Ans. ________________________________________________________________________ d p 5(24) 120 mm, d G 5(48) 240 mm 14-14 V Eq. (14-6a):  (120)(10 3 )(50) 0.3142 m/s 60 Kv  3.05  0.3142 1.103 3.05 60 000H 60(103 ) H   3.183H  dn  (120)(50) where H is in kW and Wt is in kN Wt  C p  163 MPa [Note: Using Eq. (14-13) can result in wide variation in Table 14-8: Cp due to wide variation in cast iron properties]. 120sin 20 240sin 20 r1   20.52 mm, r2   41.04 mm 2 2 Eq. (14-12): Eq. (14-14):  1.103(3.183)  103  H  1 1   690   163     60 cos 20o  20.52 41.04    1/ 2 H 3.94 kW Ans. ________________________________________________________________________ 14-15 Eq. (14-6a): d P  4(20)  80 mm, dG  4(32)  128 mm  (80)(10 3 )(1000) V   4.189 m/s 60 3.05  4.189 Kv   2.373 3.05 Wt  Shigley’s MED, 11th edition 60(10)(103 )  2.387 kN  2387 N  (80)(1000) Chapter 14 Solutions, Page 7/40 C p  163 MPa Table 14-8: [Note: Using Eq. (14-13) can result in wide variation in Cp due to wide variation in cast iron properties.] 80sin 20 128sin 20 r1   13.68 mm, r2   21.89 mm 2 2 Eq. (14-12): 1/ 2  2.373(2387)  1 1   C   163        617 MPa Ans. 50 cos 20 13.68 21.89     Eq. (14-14): ________________________________________________________________________ 14-16 The pinion controls the design. Bending YP = 0.303, YG = 0.359 17 30  1.417 in, dG   2.500 in 12 12  d n  (1.417)(525) V  P   194.8 ft/min 12 12 1200  194.8 Kv   1.162 1200 Se  0.5Sut  0.5(76)  38.0 kpsi dP  Eq. (14-4b): Eq. (6-10): Eq. (6-18): Eq. (14-3): ka = aSutb = 2.00(76)–0.217 = 0.781 2.25 2.25 l    0.1875 in Pd 12 3Y 3(0.303) x P   0.0379 in 2P 2(12) t  4lx  4(0.1875)(0.0379)  0.1686 in Eq. (b), Sec. 14-1: d e  0.808 hb  0.808 0.875(0.1686)  0.310 in Eq. (6-24):  0.107  0.107  d   0.310  kb     0.996    0.3   0.3  Eq. (6-19): kc = kd = ke = 1 Account for one-way bending with kf = 1.66. (See Ex. 14-2.) Eq. (6-17): Se = 0.781(0.996)(1)(1)(1)(1.66)(38.0) = 49.07 kpsi For stress concentration, find the radius of the root fillet (See Ex. 14-2). rf  0.300 0.300   0.025 in P 12 From Fig. A-15-6, Shigley’s MED, 11th edition Chapter 14 Solutions, Page 8/40 r r 0.025  f   0.148 d t 0.1686 Approximate D/d = ∞ with D/d = 3; from Fig. A-15-6, Kt = 1.68. From Fig. 6-26, with Sut = 76 kpsi and r = 0.025 in, q = 0.62. Eq. (6-32): Kf = 1 + q( Kt – 1) = 1 + 0.62 (1.68 – 1) = 1.42 S 49.07  all  e  15.36 psi K f nd 1.42(2.25) FY  0.875(0.303)(15 360) W t  P all   292.0 lbf K vPd 1.162(12) W tV 292.0(194.8) H    1.72 hp Ans. 33 000 33 000 Wear  1 =  2 = 0.292, Eq. (14-13): Eq. (14-12): E1 = E2 = 30(106) psi           1 1   Cp     2 2 2   1   P 1  G     1  0.292         2   EG    30  10 6      EP    1/ 2  2285 psi dP 1.417 sin   sin 20  0.242 in 2 2 d 2.500 r2  G sin   sin 20  0.428 in 2 2 1 1 1 1     6.469 in  1 r1 r2 0.242 0.428 r1  (SC )108  0.4H B  10 kpsi  [0.4(149)  10](103 )  49 600 psi Eq. (6-79): From the discussion and unnumbered equation immediately following Eq. (6-80),  S  8  49 600  C ,all   C 10    33 067 psi n 2.25 From Eq. (14-14):   2  2    F cos     33 067   0.875cos 20  W t   C     22.6 lbf  C p    1 1    2285   1.162(6.469)    K   v   r1 r2   Shigley’s MED, 11th edition Chapter 14 Solutions, Page 9/40 W tV 22.6(194.8)   0.133 hp 33 000 33 000 Rating power (pinion controls): H  Ans. H1 = 1.72 hp H2 = 0.133 hp Hall = (min 1.72, 0.133) = 0.133 hp Ans. ________________________________________________________________________ 14-17 See Prob. 14-16 solution for equation numbers. Pinion controls: YP = 0.322, YG = 0.447 Bending dP = 20/3 = 6.667 in, dG = 100/3 = 33.333 in V   d P n / 12   (6.667)(870) / 12 1519 ft/min K v  (1200  1519) / 1200  2.266 Se  0.5(113)  56.5 kpsi ka  2.00(113)  0.217  0.717 l  2.25 / Pd  2.25 / 3  0.75 in x  3(0.322) / [2(3)]  0.161 in t  4(0.75)(0.161)  0.695 in d e  0.808 2.5(0.695)  1.065 in kb  (1.065 / 0.30)  0.107  0.873 kc  k d  k e  1 kf = 1.66 (See Ex. 14-2.) Se  0.717(0.873)(1)(1)(1)(1.66)(56.5)  58.7 kpsi rf  0.300 / 3  0.100 in r r 0.100  f   0.144 d t 0.695 Kt = 1.75, q = 0.85, Kf = 1.64 S 58.7  all  e   23.9 kpsi K f nd 1.64(1.5) FY  2.5(0.322)(23 900) W t  P all   2830 lbf K vPd 2.266(3) H  W tV / 33 000  2830(1519) / 33 000  130 hp Ans. Wear Shigley’s MED, 11th edition Chapter 14 Solutions, Page 10/40 Eq. (14-13):       1 Cp    2   1  0.292    2  30  106       1/ 2  2285 psi Eq. (14-12): r1 = (6.667/2) sin 20° = 1.140 in r2 = (33.333/2) sin 20° = 5.700 in Eq. (6-68): SC = [0.4(262) – 10](103) = 94 800 psi  C ,all   SC / nd   94 800 / 1.5   77 400 psi 2   F cos    1 W   C ,all     Cp  K v  1 / r1  1 / r2    2 1   77 400   2.5cos 20          2285   2.266   1 / 1.140  1 / 5.700   1130 lbf W tV 1130(1519) H    52.0 hp Ans. 33 000 33 000 For 108 cycles (revolutions of the pinion), the power based on wear is 52.0 hp. Rating power (pinion controls): t H1 130 hp H 2  52.0 hp H rated  min(130, 52.0)  52.0 hp Ans. ________________________________________________________________________ 14-18 See Prob. 14-16 solution for equation numbers. Given:  = 20°, n = 1145 rev/min, m = 6 mm, F = 75 mm, NP = 16 milled teeth, NG = 30T, Sut = 900 MPa, HB = 260, nd = 3, YP = 0.296, and YG = 0.359. Pinion bending d P  mN P  6(16)  96 mm d G  6(30)  180 mm  d Pn  (96)(10 3 )(1145) V    5.76 m/s 60 (60) 6.1  5.76 Kv   1.944 6.1 Shigley’s MED, 11th edition Chapter 14 Solutions, Page 11/40 Se  0.5(900)  450 MPa ka  3.04(900)  0.217  0.695 l  2.25m  2.25(6) 13.5 mm x  3Ym / 2  3(0.296)6 / 2  2.664 mm t  4lx  4(13.5)(2.664) 12.0 mm d e  0.808 75(12.0)  24.23 mm  0.107  24.23  kb    0.884   7.62  kc  k d  ke  1 kf = 1.66 (See Ex. 14-2) Se  0.695(0.884)(1)(1)(1)(1.66)(450)  458.9 MPa rf  0.300m  0.300(6) 1.8 mm r/d = rf /t = 1.8/12 = 0.15, Kt = 1.68, q = 0.86, Kf = 1.58 S 458.9  all  e   223.4 MPa K f nd 1.58  1.3 Wt  Eq. (14-8): Eq. (13-36): FYm all 75(0.296)(6)(223.4)   15 310 N Kv 1.944 W t dn 15.31 (96)(1145) H    88.1 kW 60 000 60 000 Ans. Wear: Pinion and gear Eq. (14-12): Eq. (14-13): Eq. (6-79): r1 = (96/2) sin 20 = 16.42 mm r2 = (180/2) sin 20 = 30.78 mm       1 Cp    2   1  0.292    2  207  103       190 MPa SC = 6.89[0.4(260) – 10] = 647.7 MPa  647.7  C ,all   SC / nd    568 MPa 1.3   W   C ,all   Cp    t Eq. (14-14): 1/ 2 Shigley’s MED, 11th edition 2  F cos   1   K v  1 / r1  1 / r2  Chapter 14 Solutions, Page 12/40 2 o 1   568   75cos 20          3469 N  190   1.944   1 / 16.42  1 / 30.78  W t dn 3.469 (96)(1145) H    20.0 kW 60 000 60 000 Eq. (13-36): Thus, wear controls the gearset power rating; H = 20.0 kW. Ans. ________________________________________________________________________ NP = 17 teeth, NG = 51 teeth N 17 dP    2.833 in P 6 51 dG   8.500 in 6 V   d P n / 12   (2.833)(1120) / 12  830.7 ft/min 14-19 Eq. (14-4b): Kv = (1200 + 830.7)/1200 = 1.692 S 90 000  all  y   45 000 psi nd 2 Table 14-2: YP = 0.303, YG = 0.410 Wt  Eq. (14-7): H  Eq. (13-35): FYP all 2(0.303)(45 000)   2686 lbf K vP 1.692(6) W tV 2686(830.7)   67.6 hp 33 000 33 000 Based on yielding in bending, the power is 67.6 hp. (a) Pinion fatigue Bending Eq. (2-21): Eq. (6-10): Sut = 0.5 HB = 0.5(232) = 116 kpsi Se  0.5Sut  0.5(116)  58 kpsi Eq. (6-18): ka  aSutb  2.00(116)  0.217  0.713 l  1 1.25 2.25 2.25     0.375 in Pd Pd Pd 6 x 3YP 3(0.303)   0.0758 in 2P 2(6) Table 13-1: Eq. (14-3): Shigley’s MED, 11th edition Chapter 14 Solutions, Page 13/40 Eq. (b), Sec. 14-1: t  4lx  4(0.375)(0.0758)  0.337 in d e  0.808 F t  0.808 2(0.337)  0.663 in Eq. (6-24):  0.107  d   0.663  kb       0.3   0.3  kc = kd = ke = 1 Eq. (6-19):  0.107  0.919 Account for one-way bending with kf = 1.66. (See Ex. 14-2.) Eq. (6-17): Se  0.713(0.919)(1)(1)(1)(1.66)(58)  63.1 kpsi For stress concentration, find the radius of the root fillet (See Ex. 14-2). 0.300 0.300 rf    0.050 in P 6 r r 0.05  f   0.148 0.338 d t Fig. A-15-6: Estimate D/d = ∞ by setting D/d = 3, Kt = 1.68. Fig. 6-26: q = 0.86 K f 1  q  K t  1 1  (0.86)(1.68  1) 1.58 Eq. (6-32): S 63.1  all  e   20.0 kpsi K f nd 1.58(2) FY  2(0.303)(20 000) W t  P all   1194 lbf K vPd 1.692(6) W tV 1194(830.7) H    30.1 hp 33 000 33 000 Ans. (b) Pinion fatigue Wear 1/ 2 Eq. (14-13): Eq. (14-12): Shigley’s MED, 11th edition   1 Cp   2 6   2 [(1 - 0.292 ) / 30(10 )]   2285 psi dP 2.833 sin   sin 20o  0.485 in 2 2 d 8.500 r2  G sin   sin 20o  1.454 in 2 2 1 1 1 1   2.750 in     r r 0.485 1.454 2   1 r1  Chapter 14 Solutions, Page 14/40 Eq. (6-79): (SC )108  0.4 H B  10 kpsi In terms of gear notation C = [0.4(232) – 10]103 = 82 800 psi We will introduce the design factor of nd = 2 and because it is a contact stress apply it nd  2 . [See unnumbered equation immediately to the load Wt by dividing by after Eq. (6-80)].  82 800   58 548 psi  C ,all   c   2 2 Solve Eq. (14-14) for Wt: 2 o   58 548   2 cos 20  W      265 lbf  2285   1.692(2.750)  W tV 265(830.7) H all    6.67 hp Ans. 33 000 33 000 For 108 cycles (turns of pinion), the allowable power is 6.67 hp. (c) Gear fatigue due to bending and wear t Bending x Eq. (14-3): Eq. (b), Sec. 14-1: Eq. (6-24): 3YG 3(0.4103)   0.1026 in 2P 2(6) t  4 lx  4(0.375)(0.1026)  0.392 in  d e  0.808 F t  0.808 2(0.392)  0.715 in  0.107  0.715  kb    0.911  0.30   Eq. (6-19): kc = kd = ke = 1 kf = 1.66. (See Ex. 14-2.) S  0.713(0.911)(1)(1)(1)(1.66)(58)  62.5 kpsi Eq. (6-17): e r r 0.050  f   0.128 d t 0.392 Approximate D/d = ∞ by setting D/d = 3 for Fig. A-15-6; Kt = 1.80. Fig. 6-26: q = 0.82 K 1  q  K t  1 1  (0.82)(1.80  1) 1.66 Eq. (6-32): f S 62.5  all  e   18.8 kpsi K f nd 1.66(2) Shigley’s MED, 11th edition Chapter 14 Solutions, Page 15/40 Wt  FYP all 2(0.4103)(18 800)   1520 lbf K vPd 1.692(6) W tV 1520(830.7)   38.3 hp 33 000 33 000 The gear is thus stronger than the pinion in bending. H all  Ans. Wear Since the material of the pinion and the gear are the same, and the contact stresses are the same, the allowable power transmission of both is the same. Thus, Hall = 6.67 hp for 108 revolutions of each. As yet, we have no way to establish SC for 108/3 revolutions. (d) Pinion bending: H1 = 30.1 hp Pinion wear: H2 = 6.67 hp Gear bending: H3 = 38.3 hp Gear wear: H4 = 6.67 hp Power rating of the gear set is thus Hrated = min(30.1, 6.67, 38.3, 6.67) = 6.67 hp Ans. ________________________________________________________________________ 14-20 Given: H = 5 hp, nP = 300 rev/min,  = 20o, HB = 200, Qv = 6, L = 108 cycles with R = 90 percent. (a) Let F = 2 in, dP = NP/P = 16/6 = 2.667 in, dG = 48/6 = 8 in d n  (2.667)(300) V  P P   209.5 ft/min 12 12 Eq. (13-34): H 5 W t  33 000  33 000  787.6 lbf V 209.5 Eq. (13-35): Assuming uniform loading, Ko = 1. Qv = 6, Eq. (14-28): B = 0.25(12  Qv) = 0.25(12  6)2/3 = 0.8255 A = 50 + 56(1  B) = 50 + 56(1  0.8255) = 59.77 B Eq. (14-27): A V   59.77  209.5  K v       A  59.77    0.8255  1.196 YP  0.296 Table 14-2: From Eq. (a), Sec. 14-10, with F = 2 in Eq. (14-31): Eq. (14-32): F Y  2 0.296  ( K s ) P 1.192   1.192   6  P    Cmc = 1 Shigley’s MED, 11th edition 0.0535 1.088 Chapter 14 Solutions, Page 16/40 F 2  0.0375  0.0125F   0.0375  0.0125(2)  0.0625 10d P 10(2.667) Eq.(4-33) assuming gears are centered between bearings: Cpm = 1 Fig. 14-11: Cma = 0.16 Eq. (14-35): Ce = 1 K 1  Cmc  C pf C pm  Cma Ce  1  1  0.0625  1  0.16  1  1.2225 Eq. (14-30): m Eq. (14-40) assuming constant gear thickness: KB = 1 8  0.0178  0.977 For the pinion, N = 108 cycles, and from Fig. 14-14: (YN ) P 1.3558(10 ) J P  0.27 Fig. 14-6: Table 14-10, for R = 0.90: KR = 0.85 Figs. 14-17 and 14-18: KT = Cf = 1 Eq. (14-22): mG = NG/NP = 48/16 = 3 cos t sin t mG cos 20o sin 20o  3  I      0.1205 2  mN  mG  1 2(1) 3  1  Eq. (14-23): Cp f  C p  2300 psi Table 14-8: Strength: Grade 1 steel with HBP = HBG = 200 Fig. 14-2: (St)P = 77.3(200) + 12 800 = 28 260 psi Fig. 14-5: (Sc)P = 322(200) + 29 100 = 93 500 psi Fig. 14-15: (ZN )P = 1.4488(108)–0.023 = 0.948 Sec. 14-12: HBP/HBG = 1  CH = 1 Pinion tooth bending Eq. (14-15): P K K  6   (1.2207)(1)  ( ) P  W t K o K vK s d m B  787.8(1)(1.196)(1.088)    F J  2   0.27  13 904 psi Ans. Eq. (14-17): S YN 28 260  0.977   all  t  Ans.   32 482 psi S F KT K R 1  1(0.85)  28 260(0.977) / [(1)(0.85)]  S Y / ( KT K R )  (S F ) P   t N   2.34   13 904   Eq. (14-41): Ans Pinion tooth wear Eq. (14-16): 1/ 2  K C  ( c ) P  C p  W t K o K vK s m f  dPF I  P    1.2207   1    2300  787.8(1)(1.196)(1.088)     2.667(2)   0.1205    Shigley’s MED, 11th edition 1/ 2 101 485 psi Ans. Chapter 14 Solutions, Page 17/40  all ,c  Eq. (14-18): Eq. (14-42): Sc Z N CH 93 500  0.948(1)     104 280 psi S H KT K R 1  1(0.85)  Ans.  S Z /  KT K R   93 500  0.948  / [1 0.85  ] (S H ) P   c N  1.03   c 101 485  P Ans. Ans. 2 143 521 101 485  F F (b) Only check wear. From Eq. (14-16) in part (a): Sc ZN / (KT KR) = 93 500 (0.948)/[1(0.85)] = 104 280 psi  S Z /  KT K R   104 280 2  SH  P   c N   c   P 143 521 / F c  P   143 521  F 2   2.75 in  104 280  p = /6 = 0.5235 in, 3 p = 1.57 in and 5 p = 2.62. Fails to satisfy requirements since both are less than 2.75 in. Try P = 5 teeth/in. 3 p = 3/5 = 1.885 in and 5 p = 5/5 = 3.142 in. dP = 16/5 = 3.2 in, V =  (3.2) 300/12 = 251.3 ft/min 5 W t 33 000 656.6 lbf 251.3  59.77  251.3  K v    59.77   0.8255 1.214 Keep Km = 1.2207 and check later. KR = 0.85, I = 0.1205, Cp = 2300 psi , (Sc)P = 93 500 psi, (ZN)P = 0.948, and Sc ZN / (KT KR) = 104.28(103) psi. 1/2 115.5  103    1.2207  1    c  P 2300  656.6  1 1.214  1     F  3.2 F  0.1205    S Z /  KT K R   104 280 F SH )P   c N 2   c 115.5  103   P  115.5  103    2.22 in F 2   104 280    Use F = 2.5 in Ans. Rechecking Km. F 2.5 Cp f   0.0375  0.0125 F   0.0375  0.0125(2.5)  0.0719 10d P 10(3.2) Cma = 0.17 K m 1  Cmc  C pf C pm  Cma Ce  1  1  0.0719  1  0.17  1  1.2419 c  P 3 3 1/2  1.2419  115.5  10  116.5  10     F F  1.2207  Shigley’s MED, 11th edition Chapter 14 Solutions, Page 18/40  116.5  103    2.24 in F 2   104 280    F = 2.5 in Acceptable. Ans. ________________________________________________________________________ 14-21 dP = 2.5(20) = 50 mm, dG = 2.5(36) = 90 mm d n  (50)(10 3 )(100) V  P P   0.2618 m/s 60 60 60(120) Wt   458.4 N  (50)(10 3 )(100) With no specific information given to indicate otherwise, assume KB = Ko = Y = ZR = 1 Eq. (14-28): Eq. (14-27): Table 14-2: Qv = 6, B = 0.25(12 – 6)2/3 = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77  59.77  200(0.2618)  Kv    59.77   YP = 0.322, YG = 0.3775 0.8255  1.099 Similar to Eq. (a) of Sec. 14-10 but for SI units: Ks   1  0.8433 mF Y kb  0.0535 ( K s ) P  0.8433  2.5(18) 0.322  0.0535 ( K s )G  0.8433  2.5(18) 0.3775  Cmc  Ce  C pm  1 1.003 0.0535 =1.007 use 1 use 1 18  0.025  0.011 10(50)  0.247  0.0167(0.709)  0.765(10 4 )(0.7092 )  0.259 F  18 / 25.4  0.709 in, C pf  Cma K H  1  1[0.011(1)  0.259(1)] 1.27 Fig. 14-14: (YN )P = 1.3558(108)–0.0178 = 0.977 (YN )G = 1.3558(108/1.8)–0.0178 = 0.987 Fig. 14-6: Eq. (14-38): (YJ )P = 0.33, (YJ )G = 0.38 YZ = 0.658 – 0.0759 ln(1 – 0.95) = 0.885 cos 20o sin 20o  1.8  ZI     0.103 2(1)  1.8  1  Eq. (14-23): Shigley’s MED, 11th edition Chapter 14 Solutions, Page 19/40 Table 14-8: Z E  191 MPa Strength Grade 1 steel, HBP = HBG = 200 Fig. 14-2: Fig. 14-5: Fig. 14-15: (St)P = (St)G = 0.533(200) + 88.3 = 194.9 MPa (Sc)P = (Sc)G = 2.22(200) + 200 = 644 MPa (ZN )P = 1.4488(108)–0.023 = 0.948 ( Z N )G 1.4488(108 / 1.8)  0.023  0.961 Fig. 14-12: H BP / H BG 1  ZW  CH  1 Pinion tooth bending Eq. (14-15):  1 KH KB  ( ) P   W t K o K vK s  bmt YJ  P  Eq. (14-41) for SI:  1   1.27(1)   458.4(1)(1.099)(1)     43.08 MPa  18(2.5)   0.33  S Y  194.9  0.977  (S F ) P   t N    1(0.885)   4.99 Ans.  Y Y 43.08    Z P  Ans. Gear tooth bending  1   1.27(1)  ( )G  458.4(1)(1.099)(1)    0.38   37.42 MPa 18(2.5)    194.9  0.987  ( S F )G   5.81 Ans. 37.42  1(0.885)  Ans. Pinion tooth wear Eq. (14-16):  K Z  ( c ) P   Z E W t K o K vK s H R   d w1b Z I  P  Eq. (14-42) for SI:  1.27   1  191 458.4(1)(1.099)(1)     501.8 MPa  50(18)   0.103  S Z Z  644  0.948(1)  (S H ) P   c N W     1.37 Ans.   c Y YZ  P 501.8  1(0.885)  Ans. Gear tooth wear  (K )  ( c )G   s G   (K s ) P  ( S H )G  Shigley’s MED, 11th edition 1/ 2  1 ( c ) P     1 644 0.961(1) 1.39 501.8 1(0.885) 1/ 2 (501.8)  501.8 MPa Ans. Ans. Chapter 14 Solutions, Page 20/40 ________________________________________________________________________ 14-22 Pt = Pn cos  = 6cos 30o = 5.196 teeth/in 16 48 dP   3.079 in, d G  (3.079)  9.238 in 5.196 16 d n  (3.079)(300) V  P P   241.8 ft/min 12 12 Eq. (13-34): H 5 W t  33 000  33 000  682.3 lbf V 241.8 Eq. (13-35): Eq. (14-28): B = 0.25(12  Qv) = 0.25(12  6)2/3 = 0.8255 A = 50 + 56(1  B) = 50 + 56(1  0.8255) = 59.77 B Eq. (14-27): A V   59.77  241.8  K v       A  59.77    From Prob. 14-20: YP  0.296, K R  0.85, 0.8255  1.210 ( K s ) P 1.088, K B  1, mG  3, (YN ) P  0.977 (St ) P  28 260 psi, CH 1, ( Sc ) P  93 500 psi ( Z N ) P  0.948, C p  2300 psi The pressure angle is:  tan 20  t  tan  1    22.80  cos 30  3.079 cos 22.8 1.419 in, 2 a  1 / Pn  1 / 6  0.167 in (rb ) P  ( rb )G  3(rb ) P  4.258 in Eq. (14-25): 1/ 2 1/ 2 2 2 2 2 Z    rP  a    rb  P     rG  a    rb  G    rP  rG  sin t     1/ 2 1/ 2 2 2   3.079    9.238    2 2    0.167   1.419      0.167   4.258      2    2   3.079 9.238     sin 22.8 2   2  0.9479  2.1852  2.3865  0.7466 Conditions O.K . for use pN  pn cos n  mN  Eq. (14-21): Shigley’s MED, 11th edition  cos 20  0.4920 in 6 pN 0.492   0.6937 0.95Z 0.95(0.7466) Chapter 14 Solutions, Page 21/40  sin 22.8 cos 22.8   3  cos t sin t mG     3  1   0.193 2mN mG  1  2(0.6937)   JP  0.45 I  Eq. (14-23): Fig. 14-7: Fig. 14-8: Correction is 0.94. Thus, JP = 0.45(0.94) = 0.423. Eq. (14-31): Cmc = 1 F 2 C pf   0.0375  0.0125 F   0.0375  0.0125(2) 0.0525 10d P 10(3.079) Eq. (14-32: Eq. (14-33) assuming gears are centered between bearings: Cpm = 1 Fig. 14-11: Cma = 0.16 Eq. (14-35): Ce = 1 K 1  Cmc  C pf C pm  Cma Ce  1  1  0.0525  1  0.16  1  1.213 Eq. (14-30): m Pinion tooth bending P K K  5.196  1.213(1) ( ) P  W t K o K vK s d m B  682.3(1)(1.21)(1.088)   F J  2  0.423  6692 psi Ans. Eq. (14-15): (S F ) P  Eq. (14-41): Pinion tooth wear Eq. (14-16): StYN /  KT K R  28 260(0.977) / [1(0.85)]   4.85 6692   P Ans. 1/ 2  K C  ( c ) P  C p  W t K o K vK s m f  dPF I  P  1/ 2   1.213   1    2300 682.3(1)(1.21)(1.088)     69 640 psi Ans.   3.079(2)   0.193     S Z /  KT K R   93 500(0.948) / [(1)(0.85)] (S H ) P   c N 1.50    69 640 c  P (Eq. 14-43): Ans. ______________________________________________________________________________ 14-23 Given: R = 0.99 at 108 cycles, HB = 232 through-hardening Grade 1, core and case, both gears. NP = 17T, NG = 51T, Table 14-2: YP = 0.303, YG = 0.4103 Fig. 14-6: JP = 0.292, JG = 0.396 dP = NP / P = 17 / 6 = 2.833 in, dG = 51 / 6 = 8.500 in. Pinion bending From Fig. 14-2: 0.99 Shigley’s MED, 11th edition (St )107  77.3H B  12 800  77.3(232)  12 800  30 734 psi Chapter 14 Solutions, Page 22/40 Fig. 14-14: YN = 1.6831(108)–0.0323 = 0.928 V   d P n / 12   (2.833)(1120 / 12)  830.7 ft/min KT  K R  1, S F  2, S H  2 30 734(0.928)  all   14 261 psi 2(1)(1) Qv  5, B  0.25(12  5) 2 / 3  0.9148 A  50  56(1  0.9148)  54.77  54.77  830.7  K v    54.77   0.9148  1.472 0.0535  2 0.303  K s 1.192  1.089  use 1  6   K m  Cm f 1  Cmc (C p f C p m  CmaCe ) Cmc 1 F  0.0375  0.0125F 10d 2   0.0375  0.0125(2)  0.0581 10(2.833) 1 C pf  C pm Cma  0.127  0.0158(2)  0.093(10 4 )(22 )  0.1586 Ce  1 Eq. (14-15): K m  1  1[0.0581(1)  0.1586(1)]  1.217 K B 1 FJ P all Wt  K o K vK s Pd K m K B 2(0.292)(14 261)  775 lbf 1(1.472)(1)(6)(1.217)(1) W tV 775(830.7) H    19.5 hp 33 000 33 000  Pinion wear Fig. 14-15: ZN = 2.466N–0.056 = 2.466(108)–0.056 = 0.879 mG  51 / 17  3 I  Eq. (14-23): Shigley’s MED, 11th edition cos 20o sin 20 o  3     1.205, 2  3  1 CH  1 Chapter 14 Solutions, Page 23/40 Fig. 14-5: 0.99 ( Sc )107  322 H B  29 100  322(232)  29 100  103 804 psi 103 804(0.879)  c,all   64 519 psi 2(1)(1) 2 Eq. (14-16):   Fd P I W t   c,all   C  KK KK C  p  o v s m f 2  64 519   2(2.833)(0.1205)    2300   1(1.472)(1)(1.2167)(1)   300 lbf 300(830.7) W tV H    7.55 hp 33 000 33 000 The pinion controls, therefore Hrated = 7.55 hp Ans. ________________________________________________________________________ 14-24 l = 2.25/ Pd, x = 3Y / 2Pd  2.25   3Y  3.674 t  4 lx  4  Y    Pd  Pd   2 Pd   3.674  F Y d e  0.808 F t  0.808 F   Y  1.5487 Pd  Pd   1.5487 F Y / P  d  kb     0.30   F Y 1 K s   1.192  kb  Pd  0.107 F Y  0.8389   Pd     0.0535 0.0535  Ans.   ________________________________________________________________________ 14-25 YP = 0.331, YG = 0.422, JP = 0.345, JG = 0.410, Ko = 1.25. The service conditions are adequately described by Ko. Set SF = SH = 1. dP = 22 / 4 = 5.500 in dG = 60 / 4 = 15.000 in  (5.5)(1145) V  1649 ft/min 12 Pinion bending 0.99 ( St )107  77.3H B  12 800  77.3(250)  12 800  32 125 psi YN  1.6831[3(109 )] 0.0323  0.832 Shigley’s MED, 11th edition Chapter 14 Solutions, Page 24/40 Eq. (14-17):   all  P  32 125(0.832) 26 728 psi 1(1)(1) B  0.25(12  6) 2 / 3  0.8255 A  50  56(1  0.8255)  59.77  59.77  1649  K v    59.77   K s  1, Cm  1 0.8255 1.534 F  0.0375  0.0125 F 10d 3.25   0.0375  0.0125(3.25)  0.0622 10(5.5) Cmc  Cma  0.127  0.0158(3.25)  0.093(10 4 )(3.252 )  0.178 Ce  1 K m  Cm f 1  (1)[0.0622(1)  0.178(1)] 1.240 K B  1, KT  1 W1t  26 728(3.25)(0.345)  3151 lbf 1.25(1.534)(1)(4)(1.240) H1  3151(1649)  157.5 hp 33 000 Eq. (14-15): Gear bending By similar reasoning, W2t  3861 lbf and H 2  192.9 hp Pinion wear mG  60 / 22  2.727 cos 20o sin 20o  2.727     0.1176 2  1  2.727  0.99 ( S c )107  322(250)  29 100  109 600 psi I  (Z N ) P  2.466[3(109 )] 0.056  0.727 (Z N )G  2.466[3(109 ) / 2.727] 0.056  0.769 ( c,all ) P  109 600(0.727)  79 679 psi 1(1)(1) 2   Fd P I W   c,all   C p  K o K vK s K mC f   t 3 Shigley’s MED, 11th edition Chapter 14 Solutions, Page 25/40 2  79 679   3.25(5.5)(0.1176)       1061 lbf  2300   1.25(1.534)(1)(1.24)(1)  1061(1649) H3   53.0 hp 33 000 Gear wear Similarly, W4t  1182 lbf , H 4  59.0 hp Rating H rated  min(H1, H 2 , H 3 , H 4 )  min(157.5, 192.9, 53, 59)  53 hp Ans. Note differing capacities. Can these be equalized? ________________________________________________________________________ 14-26 From Prob. 14-25: W1t  3151 lbf , W2t  3861 lbf , W3t  1061 lbf , W4t  1182 lbf 33 000 K o H 33 000(1.25)(40) Wt    1000 lbf V 1649 Pinion bending: The factor of safety, based on load and stress, is (S F ) P  W1t 3151   3.15 1000 1000 Ans. Gear bending based on load and stress ( S F )G  W2t 3861   3.86 1000 1000 Ans. Pinion wear based on load: based on stress: n3  W3t 1061   1.06 1000 1000 ( S H ) P  1.06  1.03 Ans. Gear wear based on load: based on stress: Shigley’s MED, 11th edition n4  W4t 1182   1.18 1000 1000 (S H )G  1.18  1.09 Ans. Chapter 14 Solutions, Page 26/40 Factors of safety are used to assess the relative threat of loss of function 3.15, 3.86, 1.06, 1.18 where the threat is from pinion wear. By comparison, the AGMA safety factors (SF)P, (SF)G, (SH)P, (SH)G are 3.15, 3.86, 1.03, 1.09 or 3.15, 3.86, 1.061/2, 1.181/2 and the threat is again from pinion wear. Depending on the magnitude of the numbers, using SF and SH as defined by AGMA, does not necessarily lead to the same conclusion concerning threat. Therefore be cautious. ________________________________________________________________________ 14-27 Solution summary from Prob. 14-25: n = 1145 rev/min, Ko = 1.25, Grade 1 materials, NP = 22T, NG = 60T, mG = 2.727, YP = 0.331,YG = 0.422, JP = 0.345, JG = 0.410, Pd = 4T /in, F = 3.25 in, Qv = 6, (Nc)P = 3(109), R = 0.99, Km = 1.240, KT = 1, KB = 1, dP = 5.500 in, dG = 15.000 in, V = 1649 ft/min, Kv = 1.534, (Ks )P = (Ks )G = 1, (YN )P = 0.832, (YN )G = 0.859, KR = 1 Pinion HB: 250 core, 390 case Gear HB: 250 core, 390 case Bending ( all ) P ( all )G W1t W2t  26 728 psi  27 546 psi  3151 lbf ,  3861 lbf , ( St ) P  32 125 psi ( St ) G  32 125 psi H 1  157.5 hp H 2  192.9 hp Wear   20o , I  0.1176, (Z N )G  0.769, (Z N ) P  0.727 CP  2300 psi ( Sc ) P  Sc  322(390)  29 100  154 680 psi 154 680(0.727) 112 450 psi 1(1)(1) 154 680(0.769) ( c,all )G   118 950 psi 1(1)(1) ( c,all ) P  2  112 450  W3t    (1061)  2113 lbf ,  79 679  H3  2113(1649)  105.6 hp 33 000 2   118 950 W   (1182)  2354 lbf ,  109 600(0.769)  t 4 H4  2354(1649)  117.6 hp 33 000 Rated power Shigley’s MED, 11th edition Chapter 14 Solutions, Page 27/40 Hrated = min(157.5, 192.9, 105.6, 117.6) = 105.6 hp Ans. Prob. 14-25: Hrated = min(157.5, 192.9, 53.0, 59.0) = 53 hp The rated power approximately doubled. ________________________________________________________________________ 14-28 The gear and the pinion are 9310 grade 1, carburized and case-hardened to obtain Brinell 285 core and Brinell 580–600 case. Table 14-3: 0.99 (St )107  55 000 psi Modification of St by (YN )P = 0.832 produces ( all ) P  45 657 psi, Similarly for (YN )G = 0.859 ( all )G  47 161 psi, W1t  4569 lbf , W2t  5668 lbf , From Table 14-8, 0.99 and H1  228 hp H 2  283 hp C p  2300 psi. (Sc )107 Also, from Table 14-6:  180 000 psi Modification of Sc by YN produces ( c,all ) P  130 525 psi ( c,all )G  138 069 psi and W3t  2489 lbf , W4t  2767 lbf , H 3  124.3 hp H 4  138.2 hp Rating Hrated = min(228, 283, 124, 138) = 124 hp Ans. ________________________________________________________________________ 14-29 Grade 2, 9310 carburized and case-hardened to 285 core and 580 case in Prob. 14-28. Shigley’s MED, 11th edition Chapter 14 Solutions, Page 28/40 Summary: Table 14-3: 0.99 (St )107  65 000 psi ( all ) P  53 959 psi ( all )G  55 736 psi and it follows that W1t  5400 lbf , W2t  6699 lbf , H1  270 hp H 2  335 hp C p  2300 psi. From Table 14-8, Also, from Table 14-6: Sc  225 000 psi ( c,all ) P  181 285 psi ( c,all )G 191 762 psi Consequently, W3t  4801 lbf , W4t  5337 lbf , H 3  240 hp H 4  267 hp Rating Hrated = min(270, 335, 240, 267) = 240 hp. Ans. ________________________________________________________________________ 14-30 Given: n = 1145 rev/min, Ko = 1.25, NP = 22T, NG = 60T, mG = 2.727, dP = 2.75 in, dG = 7.5 in, YP = 0.331,YG = 0.422, JP = 0.335, JG = 0.405, P = 8T /in, F = 1.625 in, HB = 250, case and core, both gears. Cm = 1, F/dP = 0.0591, Cf = 0.0419, Cpm = 1, Cma = 0.152, Ce = 1, Km = 1.1942, KT = 1, KB = 1, Ks = 1,V = 824 ft/min, (YN )P = 0.8318, (YN )G = 0.859, KR = 1, I = 0.117 58 (St )107  32 125 psi ( all ) P  26 668 psi ( all )G  27 546 psi 0.99 and it follows that W1t  879.3 lbf , W2t  1098 lbf , H1  21.97 hp H 2  27.4 hp For wear Shigley’s MED, 11th edition Chapter 14 Solutions, Page 29/40 W3t  304 lbf , W4t  340 lbf , H 3  7.59 hp H 4  8.50 hp Rating Hrated = min(21.97, 27.4, 7.59, 8.50) = 7.59 hp In Prob. 14-25, Hrated = 53 hp. Thus, 7.59 1  0.1432  , 53.0 6.98 not 1 8 Ans. The transmitted load rating is t Wrated  min(879.3, 1098, 304, 340)  304 lbf In Prob. 14-25 t Wrated  1061 lbf Thus 304 1 1  0.2865  , not Ans. 1061 3.49 4 ________________________________________________________________________ 14-31 SP = SH = 1, Pd = 4, JP = 0.345, JG = 0.410, Ko = 1.25 Bending Table 14-4: 0.99 (St )107  13 000 psi 13 000(1)  13 000 psi 1(1)(1)  all FJ P 13 000(3.25)(0.345)    1533 lbf K o K vK s Pd K m K B 1.25(1.534)(1)(4)(1.24)(1) 1533(1649)   76.6 hp 33 000  W1t J G / J P 1533(0.410) / 0.345 1822 lbf  H1J G / J P  76.6(0.410) / 0.345  91.0 hp ( all ) P  ( all )G  W1t H1 W2t H2 Wear Table 14-8: Table 14-7: C p  1960 psi 0.99 (Sc )107  75 000 psi  ( c,all ) P  ( c,all )G Shigley’s MED, 11th edition Chapter 14 Solutions, Page 30/40 2  ( )  Fd p I W3t   c,all P   C p  K o K vK s K mC f 2  75 000  3.25(5.5)(0.1176)  1295 lbf W    1960  1.25(1.534)(1)(1.24)(1) W4t  W3t  1295 lbf 1295(1649)  64.7 hp H 4  H3  33 000 t 3 Rating Hrated = min(76.7, 94.4, 64.7, 64.7) = 64.7 hp Ans. Notice that the balance between bending and wear power is improved due to CI’s more favorable Sc/St ratio. Also note that the life is 107 pinion revolutions which is (1/300) of 3(109). Longer life goals require power de-rating. ________________________________________________________________________ 14-32 From Table A-24a, Eav = 11.8(106) Mpsi For  = 14.5 and HB = 156 SC  1.4(81)  51 693 psi 2sin14.5 / [11.8(106 )] For  = 20 1.4(112)  52 008 psi 2sin 20 / [11.8(106 )] SC  0.32(156)  49.9 kpsi The first two calculations were approximately 4 percent higher. ________________________________________________________________________ SC  14-33 Programs will vary. ________________________________________________________________________ 14-34 (YN ) P  0.977, (YN )G  0.996 (St ) P  (St )G  82.3(250)  12 150  32 725 psi 32 725(0.977)  37 615 psi 1(0.85) 37 615(1.5)(0.423) W1t   1558 lbf 1(1.404)(1.043)(8.66)(1.208)(1) 1558(925) H1   43.7 hp 33 000 ( all ) P  Shigley’s MED, 11th edition Chapter 14 Solutions, Page 31/40 32 725(0.996)  38 346 psi 1(0.85) 38 346(1.5)(0.5346) W2t   2007 lbf 1(1.404)(1.043)(8.66)(1.208)(1) 2007(925) H2   56.3 hp 33 000 (Z N ) P  0.948, (Z N )G  0.973 ( all )G  Table 14-6: 0.99 (Sc )107  150 000 psi  0.948(1)  ( c,allow ) P 150 000   167 294 psi  1(0.85)  2  167 294   1.963(1.5)(0.195)  W3t       2074 lbf  2300   1(1.404)(1.043)  2074(925) H3   58.1 hp 33 000 0.973 ( c,allow )G  (167 294) 171 706 psi 0.948 2  171 706   1.963(1.5)(0.195)  t W4       2167 lbf  2300   1(1.404)(1.052)  2167(925) H4   60.7 hp 33 000 H rated  min(43.7, 56.3, 58.1, 60.7)  43.7 hp Ans. Pinion bending is controlling. ________________________________________________________________________ (YN ) P 1.6831(108 )  0.0323  0.928 14-35 (YN )G 1.6831(108 / 3.059)  0.0323  0.962 Table 14-3: St = 55 000 psi 55 000(0.928) ( all ) P   60 047 psi 1(0.85) 60 047(1.5)(0.423) W1t   2487 lbf 1(1.404)(1.043)(8.66)(1.208)(1) 2487(925) H1   69.7 hp 33 000 0.962 ( all )G  (60 047)  62 247 psi 0.928 Shigley’s MED, 11th edition Chapter 14 Solutions, Page 32/40 62 247  0.5346    (2487)  3258 lbf 60 047  0.423  3258 H2  (69.7)  91.3 hp 2487 W2t  Table 14-6: Sc = 180 000 psi (Z N ) P  2.466(108 )  0.056  0.8790 (Z N )G  2.466(108 / 3.059)  0.056  0.9358 ( c,all ) P  180 000(0.8790) 186 141 psi 1(0.85) 2  186 141   1.963(1.5)(0.195)  W3t       2568 lbf  2300   1(1.404)(1.043)  2568(925) H3   72.0 hp 33 000 0.9358 ( c,all )G  (186 141) 198 169 psi 0.8790 2  198 169   1.043  t W4      (2568)  2886 lbf  186 141   1.052  2886(925) H4   80.9 hp 33 000 H rated  min(69.7, 91.3, 72, 80.9)  69.7 hp Ans. Pinion bending controlling ________________________________________________________________________ (YN)P = 0.928, 14-36 Table 14-3: (YN )G = 0.962 (See Prob. 14-34) St = 65 000 psi 65 000(0.928) ( all ) P   70 965 psi 1(0.85) 70 965(1.5)(0.423) W1t   2939 lbf 1(1.404)(1.043)(8.66)(1.208) 2939(925) H1   82.4 hp 33 000 65 000(0.962) ( all )G   73 565 psi 1(0.85) 73 565  0.5346  W2t    (2939)  3850 lbf 70 965  0.423  3850 H2  (82.4)  108 hp 2939 Shigley’s MED, 11th edition Chapter 14 Solutions, Page 33/40 Table 14-6: Sc = 225 000 psi (Z N ) P  0.8790, ( c,all ) P  (Z N )G  0.9358 225 000(0.879)  232 676 psi 1(0.85) 2  232 676   1.963(1.5)(0.195)  W      4013 lbf  2300   1(1.404)(1.043)  4013(925) H3   112.5 hp 33 000 0.9358 ( c,all )G  (232 676)  247 711 psi 0.8790 2  247 711   1.043  t W4      (4013)  4509 lbf  232 676   1.052  4509(925) H4   126 hp 33 000 H rated  min(82.4, 108, 112.5, 126)  82.4 hp Ans. t 3 The bending of the pinion is the controlling factor. ________________________________________________________________________ P = 2 teeth/in, d = 8 in, N = dP = 8 (2) = 16 teeth     F 4 p 4   4   2  P 2  M x 0 10(300) cos 20  4 FB cos20 14-37 FB = 750 lbf W t FB cos 20 750cos 20 705 lbf n = 2400 / 2 = 1200 rev/min  dn  (8)(1200)   2513 ft/min V  12 12 We will obtain all of the needed factors, roughly in the order presented in the textbook. Fig. 14-2: St = 102(300) + 16 400 = 47 000 psi Fig. 14-5: Fig. 14-6: Sc = 349(300) + 34 300 = 139 000 psi J = 0.27 cos 20o sin 20o  2  I     0.107 2(1)  2  1 Eq. (14-23): C p  2300 psi Table 14-8: Assume a typical quality number of 6. Shigley’s MED, 11th edition Chapter 14 Solutions, Page 34/40 Eq. (14-28): B  0.25(12  Qv) 2 / 3  0.25(12  6) 2 /3  0.8255 Eq. (14-27): A V   59.77  2513  K v       A  59.77    A  50  56(1  B)  50  56(1  0.8255)  59.77 B 0.8255  1.65 To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296. From Eq. (a), Sec. 14-10, 0.0535 0.0535 F Y  2 0.296  K s  1.192   1.192   1.23   2  P    The load distribution factor is applicable for straddle-mounted gears, which is not the case here since the gear is mounted outboard of the bearings. Lacking anything better, we will use the load distribution factor as a rough estimate. Eq. (14-31): Eq. (14-32): Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30): Cmc = 1 (uncrowned teeth) 2 Cp f   0.0375  0.0125(2 )  0.1196 10(8) Cpm = 1.1 Cma = 0.23 (commercial enclosed gear unit) Ce = 1 K m  1  1[0.1196(1.1)  0.23(1)] 1.36 For the stress-cycle factors, we need the desired number of load cycles. Fig. 14-14: Fig. 14-15: Eq. 14-38: N = 15 000 h (1200 rev/min)(60 min/h) = 1.1 (109) rev YN = 0.9 ZN = 0.8 K R 0.658  0.0759 ln  1  R  0.658  0.0759 ln  1  0.95  0.885 With no specific information given to indicate otherwise, assume Ko = KB = KT = Cf = 1 Tooth bending Eq. (14-15):   W t K o K vK s Pd K m K B F J  2  705(1)(1.65)(1.23)   2 Eq. (14-41):   (1.36)(1)   2294 psi    0.27   S Y / ( KT K R )  SF   t N    Shigley’s MED, 11th edition Chapter 14 Solutions, Page 35/40  47 000(0.9) / [(1)(0.885)]  20.8 2294 Ans. Tooth wear Eq. (14-16):  K C   c  C p  W t K o K vK s m f  dPF I   1/ 2 1/ 2   1.36   1    2300  705(1)(1.65)(1.23)     8(2 )   0.107     43 750 psi Since gear B is a pinion, CH is not used in Eq. (14-42) (see Fig. 14-18), where SH  Sc Z N / ( KT K R ) c 139 000(0.8) / [(1)(0.885)]     2.9 Ans 43 750   ________________________________________________________________________ m = 18.75 mm/tooth, d = 300 mm N = d/m = 300 / 18.75 = 16 teeth F b 4 p 4   m  4  18.75  236 mm 14-38 M x 0 300(11) cos 20  150 FB cos25 FB = 22.81 kN W t FB cos 25 22.81cos 25 20.67 kN n = 1800 / 2 = 900 rev/min  dn  (0.300)(900) V   14.14 m/s 60 60 We will obtain all of the needed factors, roughly in the order presented in the textbook. Fig. 14-2: Fig. 14-5: Fig. 14-6: Eq. (14-23): St = 0.703(300) + 113 = 324 MPa Sc = 2.41(300) + 237 = 960 MPa J = YJ = 0.27 cos 20o sin 20o  5  I  ZI     0.134 2(1)  5  1 Table 14-8: Z E 191 MPa Assume a typical quality number of 6. B  0.25(12  Qv) 2 / 3  0.25(12  6)2 / 3  0.8255 Eq. (14-28): A  50  56(1  B)  50  56(1  0.8255)  59.77 Shigley’s MED, 11th edition Chapter 14 Solutions, Page 36/40 B Eq. (14-27):  59.77  200(14.14)   A  200V  K v       A 59.77     0.8255 1.69 To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296. Similar to Eq. (a) of Sec. 14-10 but for SI units: Ks   1  0.8433 mF Y kb  0.0535 K s  0.8433  18.75(236) 0.296  0.0535 1.28 Convert the diameter and facewidth to inches for use in the load-distribution factor equations. d = 300/25.4 = 11.81 in, F = 236/25.4 = 9.29 in Eq. (14-31): Eq. (14-32): Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30): Cmc = 1 (uncrowned teeth) 9.29 C pf   0.0375  0.0125(9.29)  0.1573 10(11.81) Cpm = 1.1 Cma = 0.27 (commercial enclosed gear unit) Ce = 1 K m  K H 1  1[0.1573(1.1)  0.27(1)] 1.44 For the stress-cycle factors, we need the desired number of load cycles. N = 12 000 h (900 rev/min)(60 min/h) = 6.48 (108) rev Fig. 14-14: Fig. 14-15: Eq. 14-38: YN = 0.9 ZN = 0.85 K R 0.658  0.0759 ln  1  R  0.658  0.0759 ln  1  0.98  0.955 With no specific information given to indicate otherwise, assume Ko = KB = KT = ZR = 1. Tooth bending   W t K o K vK s Eq. (14-15): 1 KH KB bmt YJ    (1.44)(1)  1  20 670(1)(1.69)(1.28)      53.9 MPa  236(18.75)   0.27  Eq. (14-41):  S Y / ( KT K R )  SF   t N    Shigley’s MED, 11th edition Chapter 14 Solutions, Page 37/40  324(0.9) / [(1)(0.955)]  5.66 53.9 Ans. Tooth wear Eq. (14-16):  K Z   c  Z E  W t K o K vK s H R  d w1b Z I   1/ 2   1.44   1    191  20 670(1)(1.69)(1.28)     300(236)   0.134     498 MPa 1/ 2 Since gear B is a pinion, CH is not used in Eq. (14-42) (see Fig. 14-18), where Sc Z N / ( KT K R ) c 960(0.85) / [(1)(0.955)]  1.72 Ans 498 ________________________________________________________________________ SH  14-39 From the solution to Prob. 13-46, n = 191 rev/min, Wt = 1600 N, d = 125 mm, N = 15 teeth, m = 8.33 mm/tooth. F b 4 p 4   m  4  8.33 105 mm V   dn  (0.125)(191)   1.25 m/s 60 60 We will obtain all of the needed factors, roughly in the order presented in the textbook. Table 14-3: Table 14-6: Fig. 14-6: Eq. (14-23): St = 65 kpsi = 448 MPa Sc = 225 kpsi = 1550 MPa J = YJ = 0.25 cos 20o sin 20o  2  I  ZI     0.107 2(1)  2  1 Z E 191 MPa Table 14-8: Assume a typical quality number of 6. Eq. (14-28): B  0.25(12  Qv) 2 / 3  0.25(12  6) 2 /3  0.8255 Eq. (14-27):  59.77  200(1.25)   A  200V  K v       A 59.77     A  50  56(1  B)  50  56(1  0.8255)  59.77 B Shigley’s MED, 11th edition 0.8255  1.21 Chapter 14 Solutions, Page 38/40 To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290. Similar to Eq. (a) of Sec. 14-10 but for SI units: Ks   1  0.8433 mF Y kb  0.0535 0.0535 K s  0.8433  8.33(105) 0.290  1.17 Convert the diameter and facewidth to inches for use in the load-distribution factor equations. d = 125/25.4 = 4.92 in, F = 105/25.4 = 4.13 in Eq. (14-31): Cmc = 1 (uncrowned teeth) 4.13 C pf   0.0375  0.0125(4.13)  0.0981 10(4.92) Eq. (14-32): Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30): Cpm = 1 Cma = 0.32 (open gearing) Ce = 1 K m  K H 1  1[0.0981(1)  0.32(1)] 1.42 For the stress-cycle factors, we need the desired number of load cycles. N = 12 000 h (191 rev/min)(60 min/h) = 1.4 (108) rev Fig. 14-14: YN = 0.95 Fig. 14-15: ZN = 0.88 K R 0.658  0.0759 ln  1  R  0.658  0.0759 ln  1  0.95  0.885 Eq. 14-38: With no specific information given to indicate otherwise, assume Ko = KB = KT = ZR = 1. Tooth bending 1 KH KB   W t K o K vK s bmt YJ Eq. (14-15):    (1.42)(1)  1  1600(1)(1.21)(1.17)     14.7 MPa  105(8.33)   0.25  Since gear is a pinion, CH is not used in Eq. (14-42), where  S Y / ( KT K R )  SF   t N    448(0.95) / [(1)(0.885)]   32.7 14.7 Ans. Tooth wear Eq. (14-16):  K Z   c  Z E  W t K o K vK s H R  d w1b Z I   Shigley’s MED, 11th edition 1/ 2 Chapter 14 Solutions, Page 39/40   1.42   1    191  1600(1)(1.21)(1.17)     125(105)   0.107     289 MPa Eq. (14-42): 1/ 2  S Z / ( KT K R )  SH   c N  c   1550(0.88) / [(1)(0.885)]     5.33 Ans 289   ________________________________________________________________________ 14-40 From the solution to Prob. 13-47, n = 2(70) = 140 rev/min, Wt = 180 lbf, d = 5 in N = 15 teeth, P = 3 teeth/in.     F 4 p 4   4   4.2 in  P  3  dn  (5)(140) V    183.3 ft/min 12 12 We will obtain all of the needed factors, roughly in the order presented in the textbook. Table 14-3: Table 14-6: Fig. 14-6: Eq. (14-23): St = 65 kpsi Sc = 225 kpsi J = 0.25 cos 20o sin 20o  2  I     0.107 2(1)  2  1 C p  2300 psi Table 14-8: Assume a typical quality number of 6. B  0.25(12  Qv) 2 / 3  0.25(12  6) 2 /3  0.8255 Eq. (14-28): A  50  56(1  B)  50  56(1  0.8255)  59.77 B Eq. (14-27): A V   59.77  183.3  K v       A  59.77    0.8255 1.18 To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290. From Eq. (a), Sec. 14-10, 0.0535 Eq. (14-31): Eq. (14-32): F Y   4.2 0.290  K s  1.192  1.192    3  P    Cmc = 1 (uncrowned teeth) 4.2 C pf   0.0375  0.0125(4.2)  0.099 10(5) Shigley’s MED, 11th edition 0.0535  1.17 Chapter 14 Solutions, Page 40/40 Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30): Cpm = 1 Cma = 0.32 (Open gearing) Ce = 1 K m  1  1[0.099(1)  0.32(1)]  1.42 For the stress-cycle factors, we need the desired number of load cycles. N = 14 000 h (140 rev/min)(60 min/h) = 1.2 (108) rev Fig. 14-14: YN = 0.95 Fig. 14-15: ZN = 0.88 K R 0.658  0.0759 ln  1  R  0.658  0.0759 ln  1  0.98  0.955 Eq. 14-38: With no specific information given to indicate otherwise, assume Ko = KB = KT = Cf = 1. Tooth bending P K K   W t K o K vK s d m B F J Eq. (14-15):  3   (1.42)(1)   180(1)(1.18)(1.17)  1010 psi   4.2   0.25  Eq. (14-41):  S Y / ( KT K R )  SF   t N    65 000(0.95) / [(1)(0.955)]   64.0 1010 Ans. Tooth wear Eq. (14-16): 1/ 2    1.42   1   K C   c  C p  W t K o K vK s m f   2300  180(1)(1.18)(1.17)    dPF I   5(4.2)   0.107      28 800 psi Since gear B is a pinion, CH is not used in Eq. (14-42), where 1/ 2  S Z / ( KT K R )  SH   c N  c    225 000(0.88) / [(1)(0.955)]     7.28 Ans 28 800   ________________________________________________________________________ Shigley’s MED, 11th edition Chapter 14 Solutions, Page 41/40 Chapter 15 15-1 Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, NC = 109 rev of pinion at R = 0.999, NP = 20 teeth, NG = 60 teeth, Qv = 6, Pd = 6 teeth/in, normal pressure angle = 20°, shaft angle = 90°, np = 900 rev/min, F = 1.25 in, SF = SH = 1, Ko = 1. Fig. 15-7: JP = 0.249, JG = 0.216 Mesh dP = 20/6 = 3.333 in, dG = 60/6 = 10.000 in Eq. (15-7): vt = (3.333)(900/12) = 785.3 ft/min Eq. (15-6): B = 0.25(12 – 6)2/3 = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77 0.8255 Eq. (15-5):  59.77  785.3  K v    59.77   Eq. (15-8): vt,max = [59.77 + (6 – 3)]2 = 3940 ft/min 1.374 Since 785.3 < 3904, Kv = 1.374 is valid. The size factor for bending is: Eq. (15-10): Ks = 0.4867 + 0.2132 / 6 = 0.5222 For one gear straddle-mounted, the load-distribution factor is: Eq. (15-11): Km = 1.10 + 0.0036 (1.25)2 = 1.106 For KL, the more conservative “critical” equation will be used here. Though it is acceptable, according to AGMA, to use the less conservative “general” equation for general applications. Eq. (15-15): (KL)P = 1.6831(109)–0.0323 = 0.862 (KL)G = 1.6831(109 / 3)–0.0323 = 0.893 Eq. (15-14): (CL)P = 3.4822(109)–0.0602 = 1 (CL)G = 3.4822(109 / 3)–0.0602 = 1.069 Eq. (15-19): KR = 0.50 – 0.25 log(1 – 0.999) = 1.25 (or Table 15-3) CR  K R  1.25  1.118 Shigley’s MED, 11th edition Chapter 15 Solutions, Page 1/20 Bending Fig. 15-13: 0.99 St  sat  44(300)  2100  15 300 psi ( all ) P  swt  Eq. (15-4): WPt  Eq. (15-3): sat K L S F KT K R  15 300(0.862) 10 551 psi 1(1)(1.25) ( all ) P FK x J P Pd K o K vK s K m 10 551(1.25)(1)(0.249)  690 lbf 6(1)(1.374)(0.5222)(1.106) 690(785.3) H1   16.4 hp 33 000  ( all )G  Eq. (15-4): 15 300(0.893)  10 930 psi 1(1)(1.25) 10 930(1.25)(1)(0.216)  620 lbf 6(1)(1.374)(0.5222)(1.106) 620(785.3) H2   14.8 hp Ans. 33 000 WGt  The gear controls the bending rating. ________________________________________________________________________ 15-2 Refer to Prob. 15-1 for the gearset specifications. Wear Fig. 15-12: sac = 341(300) + 23 620 = 125 920 psi For the pinion, CH = 1. From Prob. 15-1, CR = 1.118. Thus, from Eq. (15-2): sac (CL ) P CH S H KT C R 125 920(1)(1) ( c,all ) P   112 630 psi 1(1)(1.118) ( c,all ) P  For the gear, from Eq. (15-16), B1  0.008 98(300 / 300)  0.008 29  0.000 69 CH  1  0.000 69(3  1) 1.001 38 From Prob. 15-1, (CL)G = 1.0685. Equation (15-2) thus gives Shigley’s MED, 11th edition Chapter 15 Solutions, Page 2/20 sac (C L )G CH S H KT CR 125 920(1.0685)(1.001 38) ( c,all )G   120 511 psi 1(1)(1.118) ( c,all )G  For steel: C p  2290 psi Eq. (15-9): Cs  0.125(1.25)  0.4375  0.593 75 Fig. 15-6: I = 0.083 Eq. (15-12): Cxc = 2 Eq. (15-1):  ( )  Fd P I W   c,all P   C  K K K CC p   o v m s xc 2 t P 2  1.25(3.333)(0.083)  112 630       2290   1(1.374)(1.106)(0.5937)(2)   464 lbf 464(785.3) H3   11.0 hp 33 000 2  1.25(3.333)(0.083)  120 511  t WG      2290   1(1.374)(1.106)(0.593 75)(2)   531 lbf 531(785.3) H4   12.6 hp 33 000 The pinion controls wear: H = 11.0 hp Ans. The power rating of the mesh, considering the power ratings found in Prob. 15-1, is H = min(16.4, 14.8, 11.0, 12.6) = 11.0 hp Ans. ________________________________________________________________________ 15-3 AGMA 2003-B97 does not fully address cast iron gears. However, approximate comparisons can be useful. This problem is similar to Prob. 15-1, but not identical. We will organize the method. A follow-up could consist of completing Probs. 15-1 and 15-2 with identical pinions, and cast iron gears. Given: Uncrowned, straight teeth, Pd = 6 teeth/in, NP = 30 teeth, NG = 60 teeth, ASTM 30 cast iron, material Grade 1, shaft angle 90°, F = 1.25, nP = 900 rev/min, n = 20, one gear straddle-mounted, Ko = 1, SF = 2, S H  2. Fig. 15-7: JP = 0.268, JG = 0.228 Shigley’s MED, 11th edition Chapter 15 Solutions, Page 3/20 Mesh dP = 30/6 = 5.000 in, dG = 60/6 = 10.000 in vt =  (5)(900 / 12) = 1178 ft/min Set NL = 107 cycles for the pinion. For R = 0.99, Table 15-7: Table 15-5: Eq. (15-4): sat = 4500 psi sac = 50 000 psi s K 4500(1) swt  at L   2250 psi S F KT K R 2(1)(1) The velocity factor Kv represents stress augmentation due to mislocation of tooth profiles along the pitch surface and the resulting “falling” of teeth into engagement. Equation (5-67) shows that the induced bending moment in a cantilever (tooth) varies directly with E of the tooth material. If only the material varies (cast iron vs. steel) in the same geometry, I is the same. From the Lewis equation of Section 14-1, M K W tP    v I /c FY We expect the ratio CI/steel to be  CI ( K v)CI ECI    steel ( K v)steel Esteel In the case of ASTM class 30, from Table A-24(a) (ECI)av = (13 + 16.2)/2 = 14.7 kpsi ( K v) CI  Then, 14.7 ( K v)steel  0.7( K v)steel 30 Our modeling is rough, but it convinces us that (Kv)CI < (Kv)steel, but we are not sure of the value of (Kv)CI. We will use Kv for steel as a basis for a conservative rating. Eq. (15-6): B = 0.25(12 – 6)2/3 = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77 Eq. (15-5):  59.77  1178  K v    59.77   Pinion bending Shigley’s MED, 11th edition 0.8255 1.454 (all)P = swt = 2250 psi Chapter 15 Solutions, Page 4/20 From Prob. 15-1, Kx = 1, Km = 1.106, Ks = 0.5222 ( ) FK J WPt  all P x P Pd K o K vK s K m Eq. (15-3): 2250(1.25)(1)(0.268) 149.6 lbf 6(1)(1.454)(0.5222)(1.106) 149.6(1178) H1   5.34 hp 33 000  Gear bending JG  0.228   149.6   127.3 lbf  0.268  JP 127.3(1178) H2   4.54 hp 33 000 WGt  WPt The gear controls in bending fatigue. H = 4.54 hp Ans. ________________________________________________________________________ 15-4 Continuing Prob. 15-3, Table 15-5: sac = 50 000 psi 50 000 swt   c,all   35 355 psi 2 2 Eq. (15-1):   Fd P I W   c,all   C p  K o K vK mCsC xc   Fig. 15-6: I = 0.86 t From Probs. 15-1 and 15-2: Cs = 0.593 75, Ks = 0.5222, Km = 1.106, Cxc = 2 C p  1960 psi From Table 14-8: 2  1.25(5.000)(0.086)  35 355   W   91.6 lbf     1960   1(1.454)(1.106)(0.59375)(2)  91.6(1178) H3  H 4   3.27 hp 33 000 Ans. t Thus, Rating Side note: Based on results of Probs. 15-3 and 15-4, H = min(5.34, 4.54, 3.27, 3.27) = 3.27 hp The mesh is weakest in wear fatigue. ________________________________________________________________________ Shigley’s MED, 11th edition Chapter 15 Solutions, Page 5/20 15-5 Uncrowned, through-hardened to 180 Brinell (core and case), Grade 1, 109 rev of pinion at R = 0.999, NP = z1 = 22 teeth, NG = z2 = 24 teeth, Qv = 5, met = 4 mm, S  S F  1, n = 20, shaft angle 90°, n1 = 1800 rev/min, SF = 1, H F = b = 25 C  190 MPa . mm, Ko = KA = KT = K = 1 and p Fig. 15-7: JP = YJ1 = 0.23, JG = YJ2 = 0.205 Mesh dP = de1 = mz1 = 4(22) = 88 mm, Eq. (15-7): vet = 5.236(10–5)(88)(1800) = 8.29 m/s Eq. (15-6): B = 0.25(12 – 5)2/3 = 0.9148 A = 50 + 56(1 – 0.9148) = 54.77 dG = met z2 = 4(24) = 96 mm 0.9148 Eq. (15-5): Eq. (15-10):  54.77  200(8.29)  Kv    1.663   54.77   Ks = Yx = 0.4867 + 0.008 339(4) = 0.520 Eq. (15-11): with Kmb = 1.25 (neither member straddle-mounted), Km = KH = 1.25 + 5.6(10–6)(252) = 1.2535 From Fig. 15-8, (CL ) P  ( Z NT ) P  3.4822(109 )  0.0602 1.00 (CL )G  (Z NT )G  3.4822[109 (22 / 24)] 0.0602 1.0054 Eq. (15-12): Cxc = Zxc = 2 (uncrowned) Eq. (15-19): KR = YZ = 0.50 – 0.25 log (1 – 0.999) = 1.25 CR  Z Z  YZ  1.25  1.118 From Fig. 15-10, CH = Zw = 1 Eq. (15-9): Zx = 0.004 92(25) + 0.4375 = 0.560 Wear of Pinion Fig. 15-12: H lim = 2.35HB + 162.89 = 2.35(180) + 162.89 = 585.9 MPa Fig. 15-6: I = ZI = 0.066 Shigley’s MED, 11th edition Chapter 15 Solutions, Page 6/20 ( H ) P  ( H Eq. (15-2):  ) (Z NT ) P ZW S H K Z Z lim P 585.9(1)(1)  524.1 MPa 1(1)(1.118) 2   bd e1Z I W  H   C  1000 K K K Z Z A v H  x xc  p Eq. (15-1): t The constant 1000 expresses W in kN. t P  524.1  W    190  t  dnW 1 H3  60 000 t P Eq. (13-36): Wear of Gear 2   25(88)(0.066)  1000(1)(1.663)(1.2535)(0.56)(2)   0.473 kN    (88)(1800)(0.473)   3.92 kW 60 000 H lim = 585.9 MPa 585.9(1.0054) ( H )G   526.9 MPa 1(1)(1.118) ( )  526.9  WGt  WPt H G  0.473    0.476 kN ( H ) P  524.1   (88)(1800)(0.476)  3.95 kW H4  60 000 Thus in wear, the pinion controls the power rating; H = 3.92 kW Ans. We will rate the gear set after solving Prob. 15-6. ________________________________________________________________________ 15-6 Refer to Prob. 15-5 for terms not defined below. Bending of Pinion ( K L ) P  (YNT ) P 1.6831(109 )  0.0323  0.862 ( K L )G  (YNT )G  1.6831[109 (22 / 24)] 0.0323  0.864 Fig. 15-13: F lim = 0.30HB + 14.48 = 0.30(180) + 14.48 = 68.5 MPa Eq. (15-13): Kx = Y = 1 From Prob. 15-5: Shigley’s MED, 11th edition YZ = 1.25, vet = 8.29 m/s, Chapter 15 Solutions, Page 7/20 K A 1, K v 1.663, K 1, Yx  0.52, K H  1.2535, YJ 1  0.23 ( F ) P  Eq. (5-4): WPt  Eq. (5-3):  F limYNT 68.5(0.862)   47.2 MPa S F K YZ 1(1)(1.25) ( F ) P bmetY YJ 1 1000 K A K vYx K H  47.2(25)(4)(1)(0.23)  1.00 kN 1000(1)(1.663)(0.52)(1.2535)   88   1800   1.00  H1   8.29 kW 60 000 Bending of Gear   F lim  68.5 MPa 68.5(0.864)  47.3 MPa 1(1)(1.25) 47.3(25)(4)(1)(0.205) WGt   0.895 kN 1000(1)(1.663)(0.52)(1.2535)   88   1800   0.895  H2   7.42 kW 60 000 Rating of mesh is Hrating = min(8.29, 7.42, 3.92, 3.95) = 3.92 kW Ans. with pinion wear controlling. ________________________________________________________________________ ( F ) G  15-7 (a)     (S F ) P   all   (S F )G   all    P   G (sat K L / KT K R ) P ( s K / KT K R )G  t at L (W Pd K o K vK s K m / FK x J ) P (W Pd K o K vK s K m / FK x J )G t All terms cancel except for sat , KL , and J, (sat)P(KL)P JP = (sat )G(KL)G JG From which ( sat )G  (sat ) P ( K L ) P J P J  ( sat ) P P mG ( K L )G J G JG where  = – 0.0178 or  = – 0.0323 as appropriate. This equation is the same as Shigley’s MED, 11th edition Chapter 15 Solutions, Page 8/20 Eq. (14-44). Ans. (b) In bending    sat K L  FK x J FK x J W t   all     S F Pd K o K vK s K m  11  S F KT K R Pd K o K vK s K m 11 (1) In wear 1/ 2  sacCLCU   W t K o K vK mCsCxc   C    p Fd P I  S H KT CR  22   22 Squaring and solving for Wt gives   s 2 C 2C 2   Fd P I W t   2 ac 2L 2H 2     S H KT CRCP  22  K o K vK mCsCxc  22 (2) Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and C  KR recognizing that R and PddP = NP, we obtain (sac ) 22  S H2 (sat )11( K L )11 K x J11KT CsCxc SF CH2 N P K s I Cp (CL ) 22 For equal Wt in bending and wear S H2  SF  SF  2 SF 1 So we get (sac ) G  Cp (CL )G CH (sat ) P ( K L ) P J P K x KT CsCxc N P IK s Ans. (c)     (S H ) P  ( S H )G   c,all    c,all   c  P  c  G Substituting in the right-hand equality gives [sacCL / (CR KT )]P  C W t K K K C C / ( Fd I )  o v m s xc P  p P  [ sacCLCH / (CR KT )]G  C W t K K K C C / (Fd I )  o v m s xc P  p G Denominators cancel, leaving Shigley’s MED, 11th edition Chapter 15 Solutions, Page 9/20 (sac)P(CL)P = (sac)G(CL)GCH Solving for (sac)P gives, (sac ) P  (sac )G (CL )G CH (CL ) P (1)  C  3.4822 N L 0.0602 and  CL  G 3.4822  N L / mG  From Eq. (15-14), L P Thus,  0.0602 CH  sac  G mG0.0602CH  sac  P  sac  G  1 mG  Ans.  0.0602 . This equation is the transpose of Eq. (14-45). ________________________________________________________________________ 15-8 Given (HB)11 = 300 Brinell Eq. (15-23): (sat )P = 44(300) + 2100 = 15 300 psi J P  0.0323  0.249   0.0323 mG 15 300   17 023 psi 3 JG  0.216  17 023  2100 ( H B )21   339 Brinell Ans. 44 2290 15 300(0.862)(0.249)(1)(0.593 25)(2) ( sac )G  1.0685(1) 20(0.086)(0.5222)  141 160 psi 141 160  23 600 ( H B ) 22   345 Brinell Ans. 341 (sac ) P  (sac )G mG0.0602CH 141 160(30.0602 )  1 150 811 psi (sat )G  (sat ) P ( H B )12  Pinion 150 811  23 600  373 Brinell 341 Core (HB)11 = 300 Ans. Case (HB)12 = 373 A n s . (HB)21 = 339 (HB)22 = 345 _______________________________________________________________________ 15-9 Pinion core Shigley’s MED, 11th edition Chapter 15 Solutions, Page 10/20 (sat ) P  44  300   2 100 15 300 psi ( all ) P  Wt  Shigley’s MED, 11th edition 15 300  0.862   10 551 psi 1 1  1.25  10 551 1.25   0.249   689.7 lbf 6  1  1.374   0.5222   1.106  Chapter 15 Solutions, Page 11/20 Gear core (sat )G  44(339)  2100 17 016 psi 17 016(0.893) 12 156 psi 1(1)(1.25) 12 156(1.25)(0.216) Wt   689.3 lbf 6(1)(1.374)(0.5222)(1.106) ( all )G  Pinion case (sac ) P  341(373)  23 620 150 813 psi ( c,all ) P  150 813(1) 134 895 psi 1(1)(1.118)  134 895  W    2290  2 t   1.25(3.333)(0.086)  1(1.374)(1.106)(0.593 75)(2)   689.0 lbf   Gear case (sac )G  341(345)  23 620 141 265 psi ( c,all )G  141 265(1.0685)(1)  135 010 psi 1(1)(1.118)  135 010  W    2290  t 2   1.25(3.333)(0.086)  1(1.1374)(1.106)(0.593 75)(2)   690.1 lbf   All four transmitted loads are essentially equal, so the equations developed within Prob. 15-7 are effective. Ans. ________________________________________________________________________ 15-10 The catalog rating is 5.2 hp at 1200 rev/min for a straight bevel gearset. Also given: NP = 20 teeth, NG = 40 teeth, n = 20, F = 0.71 in, JP = 0.241, JG = 0.201, Pd = 10 teeth/in, through-hardened to 300 Brinell-General Industrial Service, and Qv = 5 uncrowned. Mesh d P  20 / 10  2.000 in, d G  40 / 10  4.000 in  d P nP  (2)(1200)   628.3 ft/min 12 12 K o  1, S F  1, S H  1 vt  Eq. (15-6): Eq. (15-5): B = 0.25(12 – 5)2/3 = 0.9148 A = 50 + 56(1 – 0.9148) = 54.77  54.77  628.3  K v    54.77   Shigley’s MED, 11th edition 0.9148 1.412 Chapter 15 Solutions, Page 12/20 Eq. (15-10): Eq. (15-11): Ks = 0.4867 + 0.2132/10 = 0.508 Km = 1.25 + 0.0036(0.71)2 = 1.252, where Kmb = 1.25 Eq. (15-15): (KL)P = 1.3558(3∙106)–0.0178 = 1.040 (KL)G = 1.3558 (3∙106/2)–0.0178 = 1.053 Eq. (15-14): (CL)P = 3.4822(3∙106)–0.0602 = 1.419 (CL)G = 3.4822(3∙106/2)–0.0602 = 1.479 Eq. (15-19): KR = 0.50 – 0.25 log(1 – 0.99) = 1.0 CR  K R  1.0 Bending Pinion: Eq. (15-23): (sat )P = 44(300) + 2100 = 15 300 psi ( swt ) P  Eq. (15-4): Wt  Eq. (15-3): 15 300(1.040) 15 912 psi 1(1)(1) (swt ) P FK x J P Pd K o K vK s K m 15 912(0.71)(1)(0.241)  303.2 lbf 10(1)(1.412)(0.508)(1.252) 303.2(628.3) H1   5.8 hp 33 000  Gear: Eq. (15-4): (sat )G = 15 300 psi 15 300(1.053) ( swt )G  16 111 psi 1(1)(1) Wt  16 111(0.71)(1)(0.201)  256.0 lbf 10(1)(1.412)(0.508)(1.252) H2  256.0(628.3)  4.9 hp 33 000 Eq. (15-3): Wear Pinion: (CH )G  1, I  0.078, C p  2290 psi, Cs  0.125(0.71)  0.4375  0.526 25 Eq. (15-22): Eq. (15-2): C xc  2 (sac)P = 341(300) + 23 620 = 125 920 psi 125 920(1.419)(1) ( c,all ) P  178 680 psi 1(1)(1) Shigley’s MED, 11th edition Chapter 15 Solutions, Page 13/20 2  ( )  Fd P I W   c,all P   C p  K o K vK mCsCxc t Eq. (15-1):  178 680     2290   362.4 lbf H3  2   0.71(2.000)(0.078)  1(1.412)(1.252)(0.526 25)(2)    362.4(628.3)  6.9 hp 33 000 Gear: (sac )G 125 920 psi ( c,all )  125 920(1.479)(1) 186 236 psi 1(1)(1) 2  0.71(2.000)(0.078)  186 236   W   393.7 lbf     2290   1(1.412)(1.252)(0.526 25)(2)  393.7(628.3) H4   7.5 hp 33 000 t Rating: H = min(5.8, 4.9, 6.9, 7.5) = 4.9 hp Pinion wear controls the power rating. The catalog rating of 5.2 hp is slightly higher than predicted here, but seems reasonable. Ans. ________________________________________________________________________ 15-11 From Ex. 15-1, the core hardness of both the pinion and gear is 180 Brinell. So (HB)11 and (HB)21 are 180 Brinell and the bending stress numbers are: (sat ) P  44(180)  2100  10 020 psi ( sat )G  10 020 psi The contact strength of the gear case, based upon the equation derived in Prob. 15-7, is Cp S H2 (sat ) P ( K L ) P K x J P KT CsC xc (sac )G  (CL )G CH S F N P IK s Substituting (sat)P from above and the values of the remaining terms from Ex. 15-1, Shigley’s MED, 11th edition Chapter 15 Solutions, Page 14/20 2290 1.52  10 020(1)(1)(0.216)(1)(0.575)(2)  (sac )G    1.32(1) 1.5  25(0.065)(0.529)   114 331 psi 114 331  23 620 ( H B ) 22   266 Brinell 341 The pinion contact strength is found using the relation from Prob. 15-7: (sac ) P  (sac )G mG0.0602CH  114 331(1) 0.0602 (1) 114 331 psi 114 331  23 600 ( H B )12   266 Brinell 341 Pinion Realization of hardnesses The response of students to this part of the question would be a function of the extent to which heat-treatment procedures were covered in their materials and manufacturing prerequisites, and how quantitative it was. The most important thing is to have the student think about it. The instructor can comment in class when students’ curiosity is heightened. Options that will surface may include: (a) Select a through-hardening steel which will meet or exceed core hardness in the hot-rolled condition, then heat-treating to gain the additional 86 points of Brinell hardness by bath-quenching, then tempering, then generating the teeth in the blank. (b) Flame or induction hardening are possibilities. (c) The hardness goal for the case is sufficiently modest that carburizing and case hardening may be too costly. In this case the material selection will be different. (d)The initial step in a nitriding process brings the core hardness to 33–38 Rockwell C-scale (about 300–350 Brinell), which is too much. ________________________________________________________________________ 15-12 Computer programs will vary. ________________________________________________________________________ 15-13 A design program would ask the user to make the a priori decisions, as indicated in Sec. 15-5 of the text. The decision set can be organized as follows: Shigley’s MED, 11th edition Chapter 15 Solutions, Page 15/20 A priori decisions: • Function: H, Ko, rpm, mG, temp., NL, R S  nd • Design factor: nd (SF = nd , H ) • Tooth system: Involute, Straight Teeth, Crowning, n • Straddling: Kmb • Tooth count: NP (NG = mGNP) Design decisions: • Pitch and Face: Pd , F • Quality number: Qv • Pinion hardness: (HB)1, (HB)3 • Gear hardness: (HB)2, (HB)4 First, gather all of the equations one needs, then arrange them before coding. Find the required hardnesses, express the consequences of the chosen hardnesses, and allow for revisions as appropriate. Shigley’s MED, 11th edition Chapter 15 Solutions, Page 16/20 Pinion Bending Load-induced stress (Allowable stress) Tabulated strength Associated hardness st  W t PK o K vK m K s  s11 FK x J P ( sat ) P  Factor of safety Note: st  W t PK o K vK m K s  s21 FK x J G (sat )G  s21S F KT K R ( K L )G Pinion Wear Gear Wear 1/ 2  W t K o K vCsC xc   c  Cp    s12 Fd P I   s S KC (sac ) P  12 H T R (CL ) P (CH ) P s22 = s12 (sac )G  s22 S H KT CR (CL )G (CH )G  (sat ) P  2100  44 Bhn    (sat ) P  5980  48  (sat )G  2100  44 Bhn    (sat )G  5980  48  ( sac ) P  23 620  341 Bhn    ( sac ) P  29 560  363.6  (sac ) P  23 620  341 Bhn    (sac ) P  29 560  363.6 (HB)11 (HB)21 (HB)12 (HB)22 44( H B )11  2100 (sat1) P    48( H B )11  5980 44( H B ) 21  2100 (sat1)G    48( H B ) 21  5980 341( H B )12  23 620 (sac1) P   363.6( H B )12  29 560 341( H B ) 22  23 620 (sac1)G   363.6( H B ) 22  29 560 Chosen hardness New tabulated strength s11S F KT K R (K L )P Gear Bending n11   all (s ) ( K )  at1 P L P s11KT K R  n21  (sat1)G ( K L )G s21KT K R  (s ) (C ) (C )  n12   ac1 P L P H P  s12 KT CR   S F  nd , S H  S F Shigley’s MED, 11th edition Chapter 15 Solutions, Page 17/20 2 n22  (s ) (C ) (C )    ac1 G L G H G  s22 KT CR   2 15-14 NW = 1, NG = 56, Pt = 8 teeth/in, d = 1.5 in, Ho = 1hp, n = 20, ta = 70F, Ka = 1.25, nd = 1, Fe = 2 in, A = 850 in2 (a) mG = NG/NW = 56, dG = NG/Pt = 56/8 = 7.0 in px = / 8 = 0.3927 in, C = 1.5 + 7 = 8.5 in Eq. (15-39): a = px / = 0.3927 / = 0.125 in Eq. (15-40): b = 0.3683 px = 0.1446 in Eq. (15-41): ht = 0.6866 px = 0.2696 in Eq. (15-42): do = 1.5 + 2(0.125) = 1.75 in Eq. (15-43): dr = 3 – 2(0.1446) = 2.711 in Eq. (15-44): Dt = 7 + 2(0.125) = 7.25 in Eq. (15-45): Dr = 7 – 2(0.1446) = 6.711 in Eq. (15-46): c = 0.1446 – 0.125 = 0.0196 in Eq. (15-47): Eq. (13-27): Eq. (13-28): Eq. (15-62): ( FW ) max  2 2  7   0.125   2.646 in VW   (1.5)(1725 /12)  677.4 ft/min  (7)(1725 / 56) VG   56.45 ft/min 12 L  px NW  0.3927 in  0.3927  o   tan  1    4.764  (1.5)   P 8 Pn  t   8.028 cos  cos 4.764  pn   0.3913 in Pn  (1.5)(1725) Vs  12 cos 4.764  679.8 ft/min (b) Eq. (15-38): f  0.103exp   0.110(679.8)0.450   0.012  0.0250 Eq. (15-54): cos n  f tan  cos 20  0.0250 tan 4.764 e   0.7563 cos n  f cot  cos 20  0.0250 cot 4.764 Shigley’s MED, 11th edition Ans. Chapter 15 Solutions, Page 18/20 WGt  Eq. (15-58): 33 000nd H o K a 33 000(1)(1)(1.25)   966 lbf VGe 56.45(0.7563) WWt  WGt Eq. (15-57): Ans. cos n sin   f cos  cos n cos   f sin   cos 20 sin 4.764  0.025cos 4.764   966    cos 20 cos 4.764  0.025sin 4.764   106.4 lbf Ans. (c) Eq. (15-33): Cs = 1190 – 477 log 7.0 = 787 Eq. (15-36): Cm  0.0107  562  56(56)  5145  0.767 Eq. (15-37): Cv  0.659 exp[ 0.0011(679.8)]  0.312 Eq. (15-38): (Wt)all = 787(7)0.8(2)(0.767)(0.312) = 1787 lbf Since WGt  (W t )all , Eq. (15-61): Eq. (15-63): the mesh will survive at least 25 000 h. 0.025(966)   29.5 lbf 0.025sin 4.764  cos 20 cos 4.764 29.5(679.8) Hf   0.608 hp 33 000 106.4(677.4) HW   2.18 hp 33 000 966(56.45) HG   1.65 hp 33 000 Wf  The mesh is sufficient Ans. Pn  Pt / cos   8 / cos 4.764o  8.028 pn   / 8.028  0.3913 in G  966  39 500 psi 0.3913(0.5)(0.125) The stress is high. At the rated horsepower, G  Shigley’s MED, 11th edition 1 39 500  23 940 psi 1.65 acceptable Chapter 15 Solutions, Page 19/20 (d) Eq. (15-52): Amin = 43.2(8.5)1.7 = 1642 in2 < 1700 in2 Eq. (15-49): Hloss = 33 000(1 – 0.7563)(2.18) = 17 530 ft · lbf/min Assuming a fan exists on the worm shaft, Eq. (15-50): 1725  0.13  0.568 ft · lbf/(min · in 2 · oF) 3939 17 530 ts  70   88.2o F Ans. 0.568(1700)  CR  Eq. (15-51): ________________________________________________________________________ Shigley’s MED, 11th edition Chapter 15 Solutions, Page 20/20 15-15 to 15-22 Problem statement values of 25 hp, 1125 rev/min, mG = 10, Ka = 1.25, nd = 1.1, n = 20°, ta = 70°F are not referenced in the table. The first four parameters listed in the table were selected as design decisions. 15-15 15-16 15-17 15-18 15-19 15-20 15-21 15-22 px 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75 dW 3.60 3.60 3.60 3.60 3.60 4.10 3.60 3.60 FG 2.40 1.68 1.43 1.69 2.40 2.25 2.4 2.4 A 2000 2000 2000 2000 2000 2000 2500 2600 FAN FAN HW 38.2 38.2 38.2 38.2 38.2 38.0 41.2 41.2 HG 36.2 36.2 36.2 36.2 36.2 36.1 37.7 37.7 Hf 1.87 1.47 1.97 1.97 1.97 1.85 3.59 3.59 NW 3 3 3 3 3 3 3 3 NG 30 30 30 30 30 30 30 30 KW 125 80 50 115 185 Cs 607 854 1000 Cm 0.759 0.759 0.759 Cv 0.236 0.236 0.236 VG 492 492 492 492 492 563 492 492 t WG 2430 2430 2430 2430 2430 2120 2524 2524 WWt f e (Pt)G Pn C-to-C ts L  G dG 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 5103 16.71 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 7290 16.71 Shigley’s MED, 11th edition 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 8565 16.71 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 7247 16.71 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 5103 16.71 1038 1284 0.0183 0.034 0.951 0.913 1.571 1.795 1.732 1.979 11.6 10.156 171 179.6 6.0 5.25 24.98 24.9 4158 5301 19.099 16.7 1284 0.034 0.913 1.795 1.979 10.156 179.6 5.25 24.9 5301 16.71 Chapter 15 Solutions, Page 21/20 Chapter 16 16-1 Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, 1 = 0, 2 = 120, and a = 90. From which, sina = sin90 = 1. Eq. (1 0.28 pa (0.040)(0.150) 120 0 sin  (0.150  0.125cos  ) d 1  2.993  10 4  pa N · m Mf  Eq. (16-3): MN  pa (0.040)(0.150)(0.125) 120 2 4 0 sin  d  9.478  10  pa N · m 1 c = 2(0.125 cos 30) = 0.2165 m F  9.478  10 4  pa  2.993  10 4  pa Eq. (16-4): 0.2165  2.995  10 3  pa pa = F/ [2.995(103)] = 2200/ [2.995(103)] = 734.5(103) Pa for cw rotation 2200  Eq. (16-7): 9.478  10 4  pa  2.993  10 4  pa 0.2165 pa = 381.9(103) Pa for ccw rotation A maximum pressure of 734.5 kPa occurs on the RH shoe for cw rotation. Ans. (b) RH shoe: Eq. (16-6): 0.28(734.5)103 (0.040)0.1502 (cos 0o  cos120o )  277.6 N · m 1 LH shoe: 381.9 TL  277.6 144.4 N · m Ans. 734.5 Ttotal = 277.6 + 144.4 = 422 N · m Ans. TR  Shigley’s MED, 11th edition Ans. Chapter 16 Solutions, Page 1/27 (c) RH shoe: Fx = 2200 sin 30° = 1100 N, Eqs. (16-8): 1  A   sin 2   2  0o 2 / 3 rad 120o Rx  Eqs. (16-9): Ry  Fy = 2200 cos 30° = 1905 N  0.375,  1  B    sin 2  2 4 0 734.5  103  0.040(0.150) 1 1.264 [0.375  0.28(1.264)]  1100   1007 N 734.5  103  0.04(0.150) R  [  1007  LH shoe: Eqs. (16-10): 2 [1.264  0.28(0.375)]  1905  4128 N 1  41282 ]1/ 2  4249 N Ans. Fx = 1100 N, Rx  Fy = 1905 N 381.9  103  0.040(0.150) Ry  1 [0.375  0.28(1.264)]  1100  570 N 381.9  103  0.040(0.150) 1 R   597 2  7512  1/ 2 [1.264  0.28(0.375)]  1905  751 N  959 N Ans. ______________________________________________________________________________ 16-2 Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, 1 = 15, 2 = 105, and a = 90. From which, sina = sin90 = 1. Eq. (16-2): 0.28 pa (0.040)(0.150) 105 Mf  sin  (0.150  0.125cos  ) d  2.177  10  4  pa  15  1 Shigley’s MED, 11th edition Chapter 16 Solutions, Page 2/27 MN  Eq. (16-3): pa (0.040)(0.150)(0.125) 105 2 sin  d  7.765  10 4  pa  15  1 c = 2(0.125) cos 30° = 0.2165 m F  7.765  10 4  pa  2.177  10 4  pa 0.2165 Eq. (16-4):  2.581 10 3  pa pa = 2200/ [2.581(10 3)] = 852.4 (103) Pa = 852.4 kPa on RH shoe for cw rotation RH shoe: TR  Eq. (16-6): LH shoe: 2200  Ans. 0.28(852.4)103 (0.040)(0.1502 )(cos15  cos105 )  263 N · m 1 7.765  10  4  pa  2.177  10  4  pa 0.2165 pa  479.1 103  Pa  479.1 kPa on LH shoe for cw rotation Ans. 0.28(479.1)103 (0.040)(0.150 2 )(cos15  cos105 )  148 N · m 1  263  148  411 N · m Ans. TL  Ttotal Comparing this result with that of Prob. 16-1, a 2.6% reduction in torque is obtained by using 25% less braking material. ______________________________________________________________________________ 16-3 Given: 1 = 0°, 2 = 120°, a = 90°, sin a = 1, a = R = 3.5 in, b = 1.25 in, f = 0.30, F = 225 lbf, r = 11/2 = 5.5 in, counter-clockwise rotation. LH shoe: Eq. (16-2), with 1 = 0:  f pabr 2 f pabr  a 2  Mf  sin   r  a cos   d  r (1  cos  )  sin  2  2  sin  a 1 sin  a  2  0.30 pa (1.25)5.5  3.5 2  5.5(1  cos120o )  sin 120   1 2   14.31 pa lbf · in Eq. (16-3), with 1 = 0:  Shigley’s MED, 11th edition Chapter 16 Solutions, Page 3/27  MN p bra 2 2 p bra   2 1  sin  d  a  a  sin 2 2    sin  a 1 sin  a  2 4   pa (1.25)5.5(3.5)  120    1    sin 2(120 )   1  2  180  4   30.41 pa lbf · in   180o   2  o c  2r cos    2(5.5) cos 30  9.526 in 2   30.41 pa  14.31 pa F  225   1.690 pa 9.526 pa  225 / 1.690 133.1 psi Eq. (16-6): f pabr 2 (cos 1  cos  2 ) 0.30(133.1)1.25(5.52 )  [1  ( 0.5)] sin  a 1  2265 lbf · in  2.265 kip · in Ans. TL  RH shoe: 30.41 pa  14.31 pa  4.694 pa 9.526 pa  225 / 4.694  47.93 psi 47.93 TR  2265  816 lbf ·in  0.816 kip·in 133.1 F  225  Ttotal = 2.27 + 0.82 = 3.09 kip  in Ans. ______________________________________________________________________________ 16-4 (a) Given: 1 = 10°, 2 = 75°, a = 75°, pa = 106 Pa, f = 0.24, b = 0.075 m (shoe width), a = 0.150 m, r = 0.200 m, d = 0.050 m, c = 0.165 m. Some of the terms needed are evaluated here:  2 2 2 2 1 2    A  r  sin  d  a  sin  cos  d  r   cos     a  sin    1  1 1 2  1 75 1   200   cos   10  150  sin 2    77.5 mm 2  10 75 /180 rad 2  1  2 B   sin  d    sin 2   0.528 1 2 4  10 /180 rad 75 2 C   sin  cos  d  0.4514 1 Now converting to Pascals and meters, we have from Eq. (16-2), Shigley’s MED, 11th edition Chapter 16 Solutions, Page 4/27 0.24  106  (0.075)(0.200) f pabr Mf  A (0.0775)  289 N · m sin  a sin 75 From Eq. (16-3), MN  pabra 106 (0.075)(0.200)(0.150) B  (0.528) 1230 N · m sin  a sin 75 Finally, using Eq. (16-4), we have F  MN  M f c  1230  289  5.70 kN 165 Ans. (b) Use Eq. (16-6) for the primary shoe. T   fpabr 2 (cos 1  cos  2 ) sin  a 0.24  106  (0.075)(0.200) 2 (cos 10  cos 75 ) sin 75  541 N · m For the secondary shoe, we must first find pa. Substituting 1230 289 pa and M f  6 pa into Eq. (16 - 7), 6 10 10 (1230 / 106 ) pa  (289 / 106 ) pa 5.70  , solving gives 165 MN  pa  619  103  Pa Then T  0.24  619  103   0.075  0.2002   cos 10  cos 75  sin 75  335 N · m so the braking capacity is Ttotal = 2(541) + 2(335) = 1750 N · m Ans. (c) Primary shoes: Shigley’s MED, 11th edition Chapter 16 Solutions, Page 5/27 pabr  C  f B   Fx sin  a 106 (0.075)0.200  [0.4514  0.24(0.528)](10  3 )  5.70   0.658 kN sin 75 p br Ry  a ( B  f C )  Fy sin  a 106 (0.075)0.200  [0.528  0.24(0.4514)]  10 3   0  9.88 kN sin 75 Rx  Secondary shoes: Rx   pabr (C  f B)  Fx sin  a 0.619  106  0.075(0.200) sin 75   0.143 kN p br Ry  a ( B  f C )  Fy sin  a  0.619  106  0.075(0.200)  4.03 kN sin 75 [0.4514  0.24(0.528)]  10 3   5.70 [0.528  0.24(0.4514)]  10  3   0 Note from figure that +y for secondary shoe is opposite to +y for primary shoe. Combining horizontal and vertical components, RH   0.658  0.143   0.801 kN RV  9.88  4.03  5.85 kN R  ( 0.801) 2  5.852  5.90 kN Ans. ______________________________________________________________________________ 16-5 Given: Face width b = 1.25 in, F = 90 lbf, f = 0.25. Preliminaries: 1 = 45°  tan1(6/8) = 8.13°, 2 = 98.13°, a = 90°, a = (62 + 82)1/2 = 10 in Eq. (16-2):  Mf  f pabr 2 0.25 pa (1.25)6 sin   r  a cos   d   sin  a 1 1 98.13  sin   6  10 cos   d 8.13  3.728 pa lbf · in Shigley’s MED, 11th edition Chapter 16 Solutions, Page 6/27 Eq. (16-3):  MN p bra 2 2 p (1.25)6(10)  a sin  d  a  sin  a 1 1 98.13  sin 2  d 8.13  69.405 pa lbf · in Eq. (16-4): Using Fc = MN  Mf , we obtain 90(20)  (69.405  3.728) pa  pa  27.4 psi Ans. Eq. (16-6): 0.25(27.4)1.25  6 2   cos8.13  cos 98.13  fpabr 2  cos 1  cos  2  T   sin  a 1  348.7 lbf · in Ans. ______________________________________________________________________________ 16-6 For 3ˆ f : f  f  3ˆ f  0.25  3(0.025)  0.325 From Prob. 16-5, with f = 0.25, M f = 3.728 pa. Thus, M f = (0.325/0.25) 3.728 pa = 4.846 pa. From Prob. 16-5, M N = 69.405 pa. Eq. (16-4): Using Fc = MN  Mf , we obtain 90(20)  (69.405  4.846) pa  pa  27.88 psi Ans. From Prob. 16-5, pa = 27.4 psi and T = 348.7 lbfin. Thus,  0.325   27.88  T    348.7  461.3 lbf ·in  0.25   27.4  Similarly, for Ans.  3ˆ f : f  f  3ˆ f  0.25  3(0.025)  0.175 M f  (0.175 / 0.25) 3.728 pa  2.610 pa 90(20) = (69.405  2.610) pa  pa = 26.95 psi  0.175   26.95  T    348.7  240.1 lbf · in Ans.  0.25   27.4  ______________________________________________________________________________ 16-7 Preliminaries: 2 = 180°  30°  tan1(3/12) = 136°, 1 = 20°  tan1(3/12) = 6°, a = 90, sina = 1, a = (32 + 122)1/2 = 12.37 in, r = 10 in, f = 0.30, b = 2 in, pa = 150 psi. Shigley’s MED, 11th edition Chapter 16 Solutions, Page 7/27 Eq. (16-2): Eq. (16-3): Mf  0.30(150)(2)(10) 136o 6 sin  (10  12.37 cos  ) d  12 800 lbf · in sin 90 MN  150(2)(10)(12.37) 136 2 6 sin  d  53 300 lbf · in sin 90 LH shoe: cL = 12 + 12 + 4 = 28 in Now note that Mf is cw and MN is ccw. Thus, Eq. (16-6): FL  53 300  12 800  1446 lbf 28 TL  0.30(150)(2)(10) 2 (cos 6  cos136 ) 15 420 lbf · in sin 90 RH shoe: M N  53 300 pa  355.3 pa , 150 M f  12 800 pa  85.3 pa 150 On this shoe, both MN and Mf are ccw. Also, cR = (24  2 tan 14°) cos 14° = 22.8 in Fact  FL sin14  361 lbf Ans. FR  FL / cos14 1491 lbf 1491  Thus, Then, TR  355.3  85.3 pa  pa  77.2 psi 22.8 0.30(77.2)(2)(10) 2 (cos 6  cos136 )  7940 lbf · in sin 90 Ttotal = 15 420 + 7940 = 23 400 lbf · in Ans. ______________________________________________________________________________ 16-8 Shigley’s MED, 11th edition Chapter 16 Solutions, Page 8/27 2 M f  2 ( fdN )(acos   r ) 0 where dN  pbr d 2  2 fpbr  (acos   r ) d  0 0 From which 2 2 0 0 a cos  d  r  d r 2 r (60 )( / 180) a    1.209r sin  2 sin 60 Ans. Eq. (16-15): a  4r sin 60  1.170r 2(60)( / 180)  sin[2(60)] Ans. a differs with a  by 100(1.170 1.209)/1.209 =  3.23 % Ans. ______________________________________________________________________________ (a) Counter-clockwise rotation, 2 =  / 4 rad, r = 13.5/2 = 6.75 in Eq. (16-15): 4r sin  2 4(6.75)sin( / 4) a    7.426 in 2 2  sin 2 2 2 / 4  sin(2 / 4) 16-9 e  2a  2(7.426) 14.85 in Ans. (b)  M F x R = tan1(3/14.85) = 11.4°  0  3F x  6.375P  0   F x  Rx   F x  2.125P R x  F x  2.125P F y  F x tan11.4o  0.428P  Fy   P  F y  R y R y  P  0.428P  1.428P Shigley’s MED, 11th edition Chapter 16 Solutions, Page 9/27 Left shoe lever. M  0  7.78S x  15.28F x 15.28 Sx  (2.125P)  4.174P 7.78 S y  f S x  0.30(4.174P)  1.252P  Fy  0  R y  S y  F y R R y   F y  S y   0.428P  1.252P   1.68P  Fx  0  R x  S x  F x R x  S x  F x  4.174P  2.125P  2.049P Shigley’s MED, 11th edition Chapter 16 Solutions, Page 10/27 (c) The direction of brake pulley rotation affects the sense of Sy, which has no effect on the brake shoe lever moment and hence, no effect on Sx or the brake torque. The brake shoe levers carry identical bending moments but the left lever carries a tension while the right carries compression (column loading). The right lever is designed and used as a left lever, producing interchangeable levers (identical levers). But do not infer from these identical loadings. ______________________________________________________________________________ 16-10 r = 13.5/2 = 6.75 in, b = 6 in, 2 = 45° =  / 4 rad. From Table 16-3 for a rigid, molded non-asbestos lining use a conservative estimate of pa = 100 psi, f = 0.33. Equation (16-16) gives the horizontal brake hinge pin reaction which corresponds to Sx in Prob. 16-9. Thus, p br 100(6)6.75 N  S x  a  2 2  sin 2 2   2   / 4   sin  2  45   2 2  5206 lbf   which, from Prob. 16-9 is 4.174 P. Therefore, 4.174 P = 5206  P = 1250 lbf = 1.25 kip Ans. Applying Eq. (16-18) for two shoes, where from Prob. 16-9, a = 7.426 in T  2a f N  2(7.426)0.33(5206)  25 520 lbf · in  25.52 kip · in Ans. ______________________________________________________________________________ 16-11 Given: D = 350 mm, b = 100 mm, pa = 620 kPa, f = 0.30,  = 270. Eq. (16-22): Shigley’s MED, 11th edition Chapter 16 Solutions, Page 11/27 P1  pabD 620(0.100)0.350   10.85 kN 2 2 Ans. f   0.30(270 )( / 180 ) 1.414 Eq. (16-19): P2 = P1 exp( f  ) = 10.85 exp( 1.414) = 2.64 kN Ans. T  ( P1  P2 )( D / 2)  (10.85  2.64)(0.350 / 2) 1.437 kN · m Ans. ______________________________________________________________________________ 16-12 Given: D = 12 in, f = 0.28, pa  Eq. (16-22): b = 3.25 in,  = 270°, P1 = 1800 lbf. 2 P1 2(1800)   92.3 psi bD 3.25(12) Ans. f   0.28(270o )( / 180 o )  1.319 P2  P1 exp( f  )  1800 exp( 1.319)  481 lbf T  ( P1  P2 )( D / 2)  (1800  481)(12 / 2)  7910 lbf · in  7.91 kip · in Ans. ______________________________________________________________________________ 16-13  MO = 0 = 100 P2  325 F  P2 = 325(300)/100 = 975 N  100    cos  1    51.32  160    270  51.32  218.7 f   0.30(218.7)   / 180  1.145 P1  P2 exp( f  )  975exp(1.145)  3064 N Ans. T   P1  P2  (D / 2)  (3064  975)(200 / 2)  209  103  N · mm  209 N · m Ans. Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 16 Solutions, Page 12/27 16-14 (a) D = 16 in, b = 3 in n = 200 rev/min f = 0.20, pa = 70 psi Eq. (16-22): p bD 70(3)(16) P1  a   1680 lbf 2 2 f   0.20(3 / 2)  0.942 Eq. (16-14): P2  P1 exp( f  )  1680 exp( 0.942)  655 lbf D 16  (1680  655) 2 2  8200 lbf · in Ans. Tn 8200(200) H    26.0 hp 63 025 63 025 3P 3(1680) P  1   504 lbf Ans. 10 10 T  ( P1  P2 ) Ans. (b) Force of belt on the drum: R = (16802 + 6552)1/2 = 1803 lbf Force of shaft on the drum: 1680 and 655 lbf TP1  1680(8)  13 440 lbf · in TP2  655(8)  5240 lbf · in Net torque on drum due to brake band: Shigley’s MED, 11th edition Chapter 16 Solutions, Page 13/27 T  TP1  TP2  13 440  5240  8200 lbf · in The radial load on the bearing pair is 1803 lbf. If the bearing is straddle mounted with the drum at center span, the bearing radial load is 1803/2 = 901 lbf. Shigley’s MED, 11th edition Chapter 16 Solutions, Page 14/27 (c) Eq. (16-21): 2P bD 2 P1 2(1680)    70 psi 3(16) 3(16) p  p  0 Ans. 2 P2 2(655)   27.3 psi Ans. 3(16) 3(16) ______________________________________________________________________________ p  270  16-15 Given:  = 270°, b = 2.125 in, f = 0.20, T =150 lbf · ft, D = 8.25 in, c2 = 2.25 in (see figure). Notice that the pivoting rocker is not located on the vertical centerline of the drum. (a) To have the band tighten for ccw rotation, it is necessary to have c1 < c2 . When friction is fully developed, P1 / P2  exp( f  )  exp[0.2(3 / 2)]  2.566 If friction is not fully developed, P1/P2 ≤ exp( f  ) To help visualize what is going on let’s add a force W parallel to P1, at a lever arm of c3. Now sum moments about the rocker pivot. M  0  c3W  c1P1  c2 P2 From which c2 P2  c1P1 c3 The device is self locking for ccw rotation if W is no longer needed, that is, W ≤ 0. It follows from the equation above W  P1 c  2 P2 c1 When friction is fully developed 2.566  2.25 / c1 2.25 c1   0.877 in 2.566 When P1/P2 is less than 2.566, friction is not fully developed. Suppose P1/P2 = 2.25, then Shigley’s MED, 11th edition Chapter 16 Solutions, Page 15/27 c1  2.25  1 in 2.25 We don’t want to be at the point of slip, and we need the band to tighten. c2  c1  c2 P1 / P2 When the developed friction is very small, P1/P2 → 1 and c1 → c2 Ans. (b) Rocker has c1 = 1 in P1 c 2.25  2   2.25 P2 c1 1 ln( P1 / P2 ) ln 2.25 f    0.172 3 / 2  Friction is not fully developed, no slip. T  ( P1  P2 ) P D D  P2  1  1  2  P2 2 Solve for P2 2T 2(150)(12)   349 lbf [( P1 / P2 )  1]D (2.25  1)(8.25) P1  2.25P2  2.25(349)  785 lbf 2P 2(785) p 1   89.6 psi Ans. bD 2.125(8.25) P2  (c) The torque ratio is 150(12)/100 or 18-fold. 349 P2   19.4 lbf 18 P1  2.25P2  2.25(19.4)  43.6 lbf 89.6 p   4.98 psi Ans. 18 Comment: As the torque opposed by the locked brake increases, P2 and P1 increase (although ratio is still 2.25), then p follows. The brake can self-destruct. Protection could be provided by a shear key. ______________________________________________________________________________ 16-16 Given: OD = 250 mm, ID = 175 mm, f = 0.30, F = 4 kN. Shigley’s MED, 11th edition Chapter 16 Solutions, Page 16/27 (a) From Eq. (16-23), 2  4000  2F pa    0.194 N/mm 2 194 kPa Ans.  d ( D  d )  (175)(250  175) Eq. (16-25): Ff 4000(0.30) T  (D  d )  (250  175)10  3  127.5 N · m Ans. 4 4 (b) From Eq. (16-26), pa  4F 4(4000)   0.159 N/mm 2  159 kPa 2 2  ( D  d )  (2502  1752 ) Ans. Eq. (16-27): 3   f pa ( D 3  d 3 )  (0.30)159  103   2503  1753   10  3  12 12  128 N · m Ans. ______________________________________________________________________________ T  16-17 Given: OD = 6.5 in, ID = 4 in, f = 0.24, pa = 120 psi. (a) Eq. (16-23): pd  (120)(4) F  a (D  d )  (6.5  4)  1885 lbf Ans. 2 2 Eq. (16-24) with N sliding planes:  fpa d 2  (0.24)(120)(4) T  (D  d 2 ) N  (6.52  42 )(6) 8 8  7125 lbf · in Ans. T  (b)  (0.24)(120d ) (6.52  d 2 )(6) 8 d, in 2 3 4 5 6 T, lbf · in 5191 6769 7125 5853 2545 (c) The torque-diameter curve exhibits a stationary point maximum in the range of diameter d. The clutch has nearly optimal proportions. ______________________________________________________________________________ 16-18 (a) Eq. (16-24) with N sliding planes: Shigley’s MED, 11th edition Chapter 16 Solutions, Page 17/27 T   f pa d ( D 2  d 2 ) N  f pa N 2  D d  d3  8 8 Differentiating with respect to d and equating to zero gives dT  f pa N 2   D  3d 2   0 dd 8 D d*  Ans. 3 d 2T 3 f pa N  f pa N  6 d  d 2 dd 8 4 which is negative for all positive d. We have a stationary point maximum. d*  (b) 6.5  3.75 in 3 Ans. Eq. (16-24): T*     (0.24)(120) 6.5 / 3  2 2 6.5  6.5 / 3  (6)  7173 lbf · in   8   (c) The table indicates a maximum within the range: 3 ≤ d ≤ 5 in d 0.45   0.80 D (d) Consider: Multiply through by D, 0.45D  d  0.80 D 0.45(6.5)  d  0.80(6.5) 2.925  d  5.2 in * 1 d  0.577    d * /D  3  D which lies within the common range of clutches. Yes. Ans. ______________________________________________________________________________ 16-19 Given: d = 11 in, l = 2.25 in, T = 1800 lbf · in, D = 12 in, f = 0.28.  0.5    tan  1   12.53  2.25  Shigley’s MED, 11th edition Chapter 16 Solutions, Page 18/27 Uniform wear Eq. (16-45):  f pa d 2  D  d2 8sin   (0.28) pa (11) 2 1800  12  112  128.2 pa  8sin12.53 1800 pa   14.04 psi Ans. 128.2 T  Eq. (16-44): F   pa d  (14.04)11 (D  d )  (12  11)  243 lbf 2 2 Ans. Uniform pressure Eq. (16-48):  f pa  D3  d 3  12sin   (0.28) pa 1800  123  113  134.1 pa  12sin12.53 1800 pa   13.42 psi Ans. 134.1 T  Eq. (16-47):  pa 2  (13.42) 2 (D  d 2 )  12  112   242 lbf Ans.  4 4 ______________________________________________________________________________ F  16-20 Uniform wear Eq. (16-34): Eq. (16-33): Thus, 1 T  ( 2  1) f pari  ro2  ri 2  2 F = (2  1) pari (ro  ri) (1 / 2)( 2  1) f pa ri  ro2  ri 2  T  f FD f ( 2  1) pa ri (ro  ri )( D) r  ri D / 2  d / 2 1 d  o    1   O.K . 2D 2D 4 D Ans. Uniform pressure Shigley’s MED, 11th edition Chapter 16 Solutions, Page 19/27 1 T  ( 2  1) f pa  ro3  ri3  3 1 F  ( 2  1) pa  ro2  ri 2  2 Eq. (16-38): Eq. (16-37): Thus, (1 / 3)( 2  1) f pa  ro3  ri 3  T 2  ( D / 2) 3  (d / 2) 3      f FD (1 / 2) f ( 2  1) pa  ro2  ri 2  D 3   ( D / 2) 2  (d / 2) 2 D   2( D / 2)3  1  (d / D)3  1  1  (d / D)3      O.K . Ans. 3( D / 2) 2 1  (d / D) 2  D 3  1  ( d / D) 2  ______________________________________________________________________________ 16-21   2 n / 60  2 500 / 60  52.4 rad/s H 2(103 ) T    38.2 N · m 52.4  Key: T 38.2   3.18 kN r 12 Average shear stress in key is 3.18(103 )    13.2 MPa Ans. 6(40) Average bearing stress is F 3.18(103 ) b      26.5 MPa Ab 3(40) Let one jaw carry the entire load. F  Ans. 1  26 45  rav      17.75 mm 2 2 2 T 38.2 F    2.15 kN rav 17.75 The bearing and shear stress estimates are b   2.15  103    22.6 MPa Ans. 10(22.5  13) 2.15(103 )    0.869 MPa Ans. 10  0.25 (17.75) 2  ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 16 Solutions, Page 20/27 Shigley’s MED, 11th edition Chapter 16 Solutions, Page 21/27 16-22 1  2 n / 60  2 (1600) / 60 167.6 rad/s 2  0 From Eq. (16-51), I1I 2 Tt1 2800(8)    133.7 lbf · in · s 2 I1  I 2 1  2 167.6  0 Eq. (16-52): E  I1I 2 133.7 2 (167.6  0) 2  1.877  106  lbf in  1  2   2  I1  I 2  2 H = E / 9336 = 1.877(106) / 9336 = 201 Btu In Btu, Eq. (16-53): Eq. (16-54): T  H 201   41.9 F C pW 0.12(40) Ans. ______________________________________________________________________________ 16-23 n1  n2 260  240   250 rev/min 2 2 Cs = ( 2   1) /  = (n2  n1) / n = (260  240) / 250 = 0.08 n Eq. (16-62): Ans.  = 2 (250) / 60 = 26.18 rad/s From Eq. (16-64): 6.75  103  E2  E1 I    123.1 N · m · s 2 2 2 Cs 0.08(26.18) I  m 2 d o  di2   8  m  8I 8(123.1)  2  233.9 kg 2 d  di 1.5  1.4 2 Table A-5, cast iron unit weight = 70.6 kN/m3 Volume: 2 o   = 70.6(103) / 9.81 = 7197 kg / m3. V = m /  = 233.9 / 7197 = 0.0325 m3 V   t  d o2  d i2  / 4   t  1.52  1.42  / 4  0.2278t Equating the expressions for volume and solving for t, 0.0325 t   0.143 m  143 mm Ans. 0.2278 ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 16 Solutions, Page 22/27 16-24 (a) The useful work performed in one revolution of the crank shaft is U = 320 (103) 200 (103) 0.15 = 9.6 (103) J Accounting for friction, the total work done in one revolution is U = 9.6(103) / (1  0.20) = 12.0(103) J Since 15% of the crank shaft stroke accounts for 7.5% of a crank shaft revolution, the energy fluctuation is E2  E1 = 9.6(103)  12.0(103)(0.075) = 8.70(103) J (b) For the flywheel, Since Eq. (16-64): Ans. n  6(90)  540 rev/min 2 n 2 (540)     56.5 rad/s 60 60 Cs = 0.10 E  E 8.70(103 ) I  2 21   27.25 N · m · s 2 2 Cs 0.10(56.5) Assuming all the mass is concentrated at the effective diameter, d, md 2 4 4I 4(27.25) m 2   75.7 kg Ans. d 1.22 ______________________________________________________________________________ I  mr 2  16-25 Use Ex. 16-6 and Table 16-6 data for one cylinder of a 3-cylinder engine. Cs  0.30 n  2400 rev/min or 251 rad/s 3(3368) Tm   804 lbf · in Ans. 4 E2  E1  3(3531)  10 590 in · lbf E  E 10 590 I  2 21   0.560 in · lbf · s 2 Ans. 2 Cs 0.30(251 ) ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 16 Solutions, Page 23/27 16-26 (a) (1) (T2 )1   F21rP   T2 T rP  2 rG n Ans. Equivalent energy (2) (1 / 2) I 222  (1 / 2)( I 2 )1  12  ( I 2 )1  I 22 I  22 2 2 n 1 2 2 2 I G  rG   mG   rG   rG       n4 IP  rP   mP   rP   rP  (3) From (2) (b) Ans. ( I 2 )1  Ie  I M  I P  n2I P  IL n2 IG n4I P   n2I P 2 2 n n Ans. Ans. ______________________________________________________________________________ 16-27 (a) Reflect IL, IG2 to the center shaft Shigley’s MED, 11th edition Chapter 16 Solutions, Page 24/27 Reflect the center shaft to the motor shaft I e  I M  I P  n 2I P  (b) For R = constant = nm, I P m2 I  2 I P  2L 2 2 n n mn I e  I M  I P  n2 I P  Ans. IP R2I P I   L2 2 4 n n R Ans. I e 2(1) 4(102 )(1)  0  0  2n(1)  3   0 0 n n5 (c) For R = 10, n n6  n2  200 = 0 From which n*  2.430 Ans. 10 m*   4.115 2.430 Ans. Notice that n*and m* are independent of IL. ______________________________________________________________________________ 16-28 From Prob. 16-27, IP R 2I P I   L2 2 4 n n R 1 100(1) 100 10  1  n 2 (1)  2   2 n n4 10 1 100 12  n 2  2  4 n n Ie  I M  I P  n2I P  Shigley’s MED, 11th edition Chapter 16 Solutions, Page 25/27 Optimizing the partitioning of a double reduction lowered the gear-train inertia to 20.9/112 = 0.187, or to 19% of that of a single reduction. This includes the two additional gears. ______________________________________________________________________________ 16-29 Figure 16-29 applies, t2  10 s, t1  0.5 s t2 - t1 10  0.5   19 t1 0.5 The load torque, as seen by the motor shaft (Rule 1, Prob. 16-26), is TL  1300(12)  1560 lbf · in 10 The rated motor torque Tr is Tr  63 025(3)  168.07 lbf · in 1125 For Eqs. (16-65): 2 (1125) 117.81 rad/s 60 2 s  (1200) 125.66 rad/s 60  Tr 168.07 a     21.41 lbf in s/rad s  r 125.66  117.81 T 168.07(125.66) b r s   2690.4 lbf · in s  r 125.66  117.81 r  Shigley’s MED, 11th edition Chapter 16 Solutions, Page 26/27 The linear portion of the squirrel-cage motor characteristic can now be expressed as TM = 21.41 + 2690.4 lbf · in Eq. (16-68):  1560  168.07  T2  168.07    1560  T2  19 One root is 168.07 which is for infinite time. The root for 10 s is desired. Use a successive substitution method 19.30 24.40 26.00 26.50 Continue until convergence to New T2 19.30 24.40 26.00 26.50 26.67 T2 = 26.771 lbf  in Eq. (16-69): a  t2  t1   21.41(10  0.5)   110.72 lbf · in · s 2 ln  T2 / Tr  ln(26.771 / 168.07) T  b   a T  b 26.771  2690.4 max  2   124.41 rad/s Ans.  21.41 a min 117.81 rad/s Ans. 124.41  117.81  121.11 rad/s   2 124.41  117.81 max  min   0.0545 Ans. Cs  (max  min ) / 2 (124.41  117.81) / 2 1 1 E1  I r2  (110.72)(117.81)2  768 352 in · lbf 2 2 1 2 1 E2  I 2  (110.72)(124.41)2  856 854 in · lbf 2 2 E  E2  E1  856 854  768 352  88 502 in · lbf I  Eq. (16-64): E  Cs I  2  0.0545(110.72)(121.11) 2  88 508 in · lbf, close enough Ans. During the punch Shigley’s MED, 11th edition Chapter 16 Solutions, Page 27/27 63 025H n TL (60 / 2 ) 1560(121.11)(60 / 2 ) H    28.6 hp 63 025 63 025 T  The gear train has to be sized for 28.6 hp under shock conditions since the flywheel is on the motor shaft. From Table A-18,    m 2 W 2 d o  di2  d o  di2 8 8g 8 gI 8(386)(110.72) W  2  2 d o  di d o2  di2 I   If a mean diameter of the flywheel rim of 30 in is acceptable, try a rim thickness of 4 in di  30  (4 / 2)  28 in d o  30  (4 / 2)  32 in 8(386)(110.72) W   189.1 lbf 322  282 Rim volume V is given by V  l 2 l d o  d i2   (322  282 )  188.5l  4 4 where l is the rim width as shown in Table A-18. The specific weight of cast iron is  = 0.260 lbf / in3, therefore the volume of cast iron is V  W 189.1   727.3 in 3  0.260 Equating the volumes, 188.5 l  727.3 727.3 l   3.86 in wide 188.5 Proportions can be varied. ______________________________________________________________________________ 16-30 Prob. 16-29 solution has I for the motor shaft flywheel as I = 110.72 lbf · in · s2 Shigley’s MED, 11th edition Chapter 16 Solutions, Page 28/27 A flywheel located on the crank shaft needs an inertia of 102 I (Prob. 16-26, rule 2) I = 102(110.72) = 11 072 lbf · in · s2 A 100-fold inertia increase. On the other hand, the gear train has to transmit 3 hp under shock conditions. Stating the problem is most of the solution. Satisfy yourself that on the crankshaft: TL  1300(12)  15 600 lbf · in Tr  10(168.07)  1680.7 lbf · in r  117.81 / 10  11.781 rad/s s 125.66 / 10 12.566 rad/s a   21.41(100)   2141 lbf · in · s/rad b  2690.35(10)  26903.5 lbf · in TM   2141c  26 903.5 lbf · in  15 600  1680.5  T2  1680.6    15 600  T2  19 The root is 10(26.67) = 266.7 lbf · in  121.11 / 10 12.111 rad/s Cs  0.0549 (same) max 121.11 / 10 12.111 rad/s Ans. min 117.81 / 10 11.781 rad/s Ans. E1, E2, E and peak power are the same. From Table A-18 6 8 gI 8(386)(11 072) 34.19  10  W  2   2 d o  di2 d o2  di2 d o  di2 Scaling will affect do and di , but the gear ratio changed I. Scale up the flywheel in the Prob. 16-29 solution by a factor of 2.5. Thickness becomes 4(2.5) = 10 in. d  30(2.5)  75 in d o  75  (10 / 2)  80 in di  75  (10 / 2)  70 in Shigley’s MED, 11th edition Chapter 16 Solutions, Page 29/27 W  34.19  106   3026 lbf 802  702 W 3026 V    11 638 in 3  0.260  V  l (80 2  702 )  1178 l 4 11 638 l   9.88 in 1178 Proportions can be varied. The weight has increased 3026/189.1 or about 16-fold while the moment of inertia I increased 100-fold. The gear train transmits a steady 3 hp. But the motor armature has its inertia magnified 100-fold, and during the punch there are deceleration stresses in the train. With no motor armature information, we cannot comment. ______________________________________________________________________________ 16-31 This can be the basis for a class discussion. Shigley’s MED, 11th edition Chapter 16 Solutions, Page 30/27 Chapter 17 17-1 Begin with Eq. (17-10), F1  Fc  Fi 2 exp( f  ) exp( f  )  1 Introduce Eq. (17-9):  exp( f  )  1   2 exp( f  )  2T  exp( f  ) F1  Fc  d   Fc     d  exp( f  )   exp( f  )  1   exp( f  )  1  exp( f  ) F1  Fc  F exp( f  )  1  1   exp( f  )  Fc   Now add and subtract  exp( f  )  1  exp( f  )   exp( f  )   exp( f  )  F1  Fc  Fc   F   Fc      exp( f  )  1   exp( f  )  1   exp( f  )  1   exp( f  )   exp( f  )   ( Fc  F )   Fc  Fc     exp( f  )  1   exp( f  )  1   exp( f  )  Fc  ( Fc  F )     exp( f  )  1  exp( f  )  1 ( F  F ) exp( f  )  Fc  c Q.E.D. exp( f  )  1 From Ex. 17-2: d = 3.037 rad, F = 664 lbf, exp( f ) = exp[0.80(3.037)] = 11.35, and Fc = 73.4 lbf. (73.4  664)11.35  73.4  802 lbf (11.35  1) F2  F1  F  802  664  138 lbf 802  138 Fi   73.4  396.6 lbf 2 1  F  Fc  1  802  73.4  f   ln  1 ln   0.80    d  F2  Fc  3.037  138  73.4  F1  Ans. ______________________________________________________________________________ 17-2 Given: F-1 Polyamide, b = 6 in, d = 2 in with n = 1750 rev/min, Hnom = 2 hp, C = 9(12) = 108 in, velocity ratio = 0.5, Ks = 1.25, nd = 1 Shigley’s MED, 11th edition Chapter 17 Solutions, Page 1/39 V =  d n / 12 =  (2)(1750) / 12 = 916.3 ft/min Eq. (17-1): D = d / vel ratio = 2 / 0.5 = 4 in  4 2 D d  d    2sin  1    2sin  1    3.123 rad 2C  2(108)  Table 17-2: t = 0.05 in, dmin = 1.0 in, Fa = 35 lbf/in,  = 0.035 lbf/in3, f = 0.5 w = 12 bt = 12(0.035)6(0.05) = 0.126 lbf/ft 2 2 w V  0.126  916.3  Fc        0.913 lbf g  60  32.17  60  (a) Eq. (e), Sec. 17-2: Ans. 63 025H nom K snd 63 025(2)(1.25)(1)   90.0 lbf · in n 1750 2T 2(90.0) F   F1  a  F2    90.0 lbf d 2 T  Table 17-4: Cp = 0.70 Eq. (17-12): (F1)a = bFaCpCv = 6(35)(0.70)(1) = 147 lbf Ans. F2 = (F1)a  [(F1)a  F2] = 147  90 = 57 lbf Ans. Do not use Eq. (17-9) because we do not yet know f   F1  a  F2 147  57  0.913  101.1 lbf 2 2 Eq. (i), Sec. 17-2: Using Eq. (17-7) solved for f  (see step 8, just after Table 17-8), 1  ( F )  Fc  1  147  0.913  f   ln  1 a ln   0.307    d  F2  Fc  3.123  57  0.913  Fi   Fc  Ans. The friction is thus underdeveloped. (b) The transmitted horsepower is, with F = (F1)a  F2 = 90 lbf, (F )V 90(916.3) H    2.5 hp Ans. 33 000 33 000 Eq. (j), Sec. 17-2: H 2.5 nf s   1 H nom K s 2(1.25)  D    2sin  1 Eq. (17-1): Shigley’s MED, 11th edition  4 2 D d    2sin  1    3.160 rad 2C  2(108)  Chapter 17 Solutions, Page 2/39 L = [4C2  (D  d)2]1/2 + (DD + dd)/2 Eq. (17-2): = [4(108)2  (4  2)2]1/2 + [4(3.160) + 2(3.123)]/2 = 225.4 in (c) Eq. (17-13): 3C 2 w 3(108 / 12) 2 (0.126) dip    0.151 in 2Fi 2(101.1) Ans. Ans. Comment: The solution of the problem is finished; however, a note concerning the design is presented here. The friction is under-developed. Narrowing the belt width to 5 in (if size is available) will increase f  . The limit of narrowing is bmin = 4.680 in, whence w  0.0983 lbf/ft Fc  0.713 lbf T  90 lbf · in (same) F  ( F1)a  F2  90 lbf Fi  68.9 lbf ( F1) a 114.7 lbf F2  24.7 lbf f   f  0.50 dip  0.173 in Longer life can be obtained with a 6-inch wide belt by reducing Fi to attain f   0.50. Prob. 17-8 develops an equation we can use here (F  Fc ) exp( f  )  Fc exp( f  )  1 F2  F1  F F  F2 Fi  1  Fc 2 1  F  Fc  f   ln  1   d  F2  Fc  F1  dip  3C 2 w 2Fi which in this case, d = 3.123 rad, exp(f ) = exp[0.5(3.123)] = 4.766, w = 0.126 lbf/ft, F = 90.0 lbf, Fc = 0.913 lbf, and gives  0.913  90  4.766  0.913  114.8 lbf F1  4.766  1 F2 = 114.8  90 = 24.8 lbf Fi = (114.8 + 24.8)/ 2  0.913 = 68.9 lbf Shigley’s MED, 11th edition Chapter 17 Solutions, Page 3/39 1  114.8  0.913  ln    0.50 3.123  24.8  0.913  2 3  108 / 12  0.126 dip   0.222 in 2(68.9) f  So, reducing Fi from 101.1 lbf to 68.9 lbf will bring the undeveloped friction up to 0.50, with a corresponding dip of 0.222 in. Having reduced F1 and F2, the endurance of the belt is improved. Power, service factor and design factor have remained intact. ______________________________________________________________________________ 17-3 Double the dimensions of Prob. 17-2. In Prob. 17-2, F-1 Polyamide was used with a thickness of 0.05 in. With what is available in Table 17-2 we will select the Polyamide A-2 belt with a thickness of 0.11 in. Also, let b = 12 in, d = 4 in with n = 1750 rev/min, Hnom = 2 hp, C = 18(12) = 216 in, velocity ratio = 0.5, Ks = 1.25, nd = 1. V =  d n / 12 =  (4)(1750) / 12 = 1833 ft/min Eq. (17-1): D = d / vel ratio = 4 / 0.5 = 8 in  8 4 D d  d    2sin  1    2sin  1    3.123 rad 2C  2(216)  Table 17-2: t = 0.11 in, dmin = 2.4 in, Fa = 60 lbf/in,  = 0.037 lbf/in3, f = 0.8 w = 12 bt = 12(0.037)12(0.11) = 0.586 lbf/ft 2 2 w V  0.586  1833  Fc        17.0 lbf g  60  32.17  60  (a) Eq. (e), Sec. 17-2: Ans. 63 025H nom K snd 63 025(2)(1.25)(1)   90.0 lbf · in n 1750 2T 2(90.0) F   F1  a  F2    45.0 lbf d 4 T  Table 17-4: Cp = 0.73 Eq. (17-12): (F1)a = bFaCpCv = 12(60)(0.73)(1) = 525.6 lbf F2 = (F1)a  [(F1)a  F2] = 525.6  45 = 480.6 lbf Eq. (i), Sec. 17-2: Eq. (17-9): Shigley’s MED, 11th edition Fi   F1  a  F2 2  Fc  Ans. Ans. 525.6  480.6  17.0  486.1 lbf 2 Ans. Chapter 17 Solutions, Page 4/39 1  ( F1) a  Fc  1  525.6  17.0  ln  ln   0.0297    d  F2  Fc  3.123  480.6  17.0  The friction is thus underdeveloped. f  (b) The transmitted horsepower is, with F = (F1)a  F2 = 45 lbf, (F )V 45(1833)   2.5 hp 33 000 33 000 H 2.5   1 H nom K s 2(1.25) H  nf s  D    2sin  1 Eq. (17-1): Eq. (17-2): Ans.  8 4  D d    2sin  1    3.160 rad 2C  2(216)  L = [4C2  (D  d)2]1/2 + (DD + dd)/2 = [4(216)2  (8  4)2]1/2 + [8(3.160) + 4(3.123)]/2 = 450.9 in dip  3C 2 w 3(216 / 12) 2 (0.586)   0.586 in 2 Fi 2(486.1) Ans. Ans. (c) Eq. (17-13): ______________________________________________________________________________ 17-4 As a design task, the decision set just before Ex. 17-2 is useful. A priori decisions: • Function: Hnom = 60 hp, n = 380 rev/min, C = 192 in, Ks = 1.1 • Design factor: nd = 1 • Initial tension: Catenary • Belt material. Table 17-2: Polyamide A-3, Fa = 100 lbf/in,  = 0.042 lbf/in3, f = 0.8 • Drive geometry: d = D = 48 in • Belt thickness: t = 0.13 in Design variable: Belt width. Use a method of trials. Initially, choose b = 6 in Shigley’s MED, 11th edition Chapter 17 Solutions, Page 5/39  dn  (48)(380)   4775 ft/min 12 12 w  12bt  12(0.042)(6)(0.13)  0.393 lbf/ft wV 2 0.393(4775 / 60) 2 Fc    77.4 lbf g 32.17 63 025H nom K snd 63 025(60)(1.1)(1)  10 946 lbf · in T  n 380 2T 2(10 946) F    456.1 lbf d 48 F1  ( F1)a  bFaC pCv  6(100)(1)(1)  600 lbf F2  F1  F  600  456.1 143.9 lbf V  Transmitted power H F (V ) 456.1(4775)   66 hp 33 000 33 000 F  F2 600  143.9 Fi  1  Fc   77.4  294.6 lbf 2 2 1 F  Fc 1  600  77.4  f   ln 1  ln   0.656  d F2  Fc   143.9  77.4  H  Eq. (17-2): L = [4(192)2  (48  48)2]1/2 + [48() + 48()] / 2 = 534.8 in Friction is not fully developed, so bmin is just a little smaller than 6 in (5.7 in). Not having a figure of merit, we choose the most narrow belt available (6 in). We can improve the design by reducing the initial tension, which reduces F1 and F2, thereby increasing belt life (see the result of Prob. 17-8). This will bring f  to 0.80 F1   F  Fc  exp  f    Fc exp  f    1 exp  f    exp(0.80 ) 12.345 Therefore (456.1  77.4)(12.345)  77.4  573.7 lbf 12.345  1 F2  F1  F  573.7  456.1 117.6 lbf F  F2 573.7  117.6 Fi  1  Fc   77.4  268.3 lbf 2 2 F1  These are small reductions since f  is close to f , but improvements nevertheless. Shigley’s MED, 11th edition Chapter 17 Solutions, Page 6/39 f  1 F  Fc 1  573.7  77.4  ln 1  ln   0.80  d F2  Fc   117.6  77.4  3C 2 w 3(192 / 12) 2 (0.393)   0.562 in 2 Fi 2(268.3) ______________________________________________________________________________ dip  Shigley’s MED, 11th edition Chapter 17 Solutions, Page 7/39 17-5 From the last equation given in the problem statement, exp  f    1 1   2T / [d (a0  a2 )b]   2T 1   exp  f    1  d ( a a ) b 0 2     2T   exp  f    exp  f    1  d (a0  a2 )b  b 1  2T  a0  a2  d   exp  f      exp  f      1  But 2T/d = 33 000Hd/V. Thus, 1  33 000 H d b  a0  a2  V   exp  f     Q.E.D.    exp  f    1  ______________________________________________________________________________ 17-6 Refer to Ex. 17-1 for the values used below. (a) The maximum torque prior to slip is, T  63 025H nom K s nd 63 025(15)(1.25)(1.1)   742.8 lbf · in n 1750 Ans. The corresponding initial tension, from Eq. (17-9), is, Fi  T  exp( f  )  1  742.8  11.17  1        148.1 lbf d  exp( f  )  1  6  11.17  1  Ans. (b) See Prob. 17-4 statement. The final relation can be written bmin    33 000 H a exp  f    1   FaC pCv  (12t / 32.174)(V / 60) 2  V [exp  f    1]   33 000(20.6)(11.17)  1 2  100(0.7)(1)   [12(0.042)(0.13)] / 32.174 (2749 / 60)  2749(11.17  1)   4.13 in Ans. This is the minimum belt width since the belt is at the point of slip. The design must Shigley’s MED, 11th edition Chapter 17 Solutions, Page 8/39 round up to an available width. Eq. (17-1): D d  1  18  6   d    2sin  1      2sin    2C   2(96)   3.016 511 rad D d  1  18  6   D    2sin  1      2sin    2C   2(96)   3.266 674 rad Eq. (17-2): L  [4(96) 2  (18  6)2 ]1/ 2   230.074 in 1 [18(3.266 674)  6(3.016 511)] 2 Ans. 2T 2(742.8)   247.6 lbf d 6 ( F1) a  bFaC pCv  F1  4.13(100)(0.70)(1)  289.1 lbf F2  F1  F  289.1  247.6  41.5 lbf w  12bt  12(0.042)4.13(0.130)  0.271 lbf/ft F  (c) 2 2 w V  0.271  2749       17.7 lbf g  60  32.17  60  F  F2 289.1  41.5 Fi  1  Fc   17.7 147.6 lbf 2 2 Fc  Transmitted belt power H F (V ) 247.6(2749)   20.6 hp 33 000 33 000 H 20.6 n fs    1.1 H nom K s 15(1.25) H  dip  Dip: 2 3C 2w 3(96 / 12)  0.271   0.176 in 2 Fi 2(147.6) (d) If you only change the belt width, the parameters in the following table change as shown. Shigley’s MED, 11th edition Chapter 17 Solutions, Page 9/39 Ex. 17-1 This Problem 6.00 4.13 0.393 0.271 25.6 17.7 420 289 172.4 41.5 270.6 147.6 0.33* 0.80** 0.139 0.176 *Friction underdeveloped **Friction fully developed ______________________________________________________________________________ 17-7 The transmitted power is the same. Fc Fi (F1) n-Fold b = 6 in b = 12 in Change 25.65 51.3 2 270.35 664.9 2.46 420 840 2 a F2 Ha nfs f dip 172.4 20.62 1.1 0.139 0.328 592.4 20.62 1.1 0.125 0.114 3.44 1 1 0.90 0.34 If we relax Fi to develop full friction (f = 0.80) and obtain longer life, then n-Fold b = 6 in b = 12 in Change Fc 25.6 51.3 2 Fi 148.1 148.1 1 F1 297.6 323.2 1.09 F2 50 75.6 1.51 f 0.80 0.80 1 dip 0.255 0.503 2 ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 17 Solutions, Page 10/39 17-8 Find the resultant of F1 and F2: D d 2C D d sin   2C 2 1 D  d  cos  1    2  2C    sin  1 2  1 D  d   R x  F1 cos   F2 cos   ( F1  F2 )  1     2  2C    D d R y  F1 sin   F2 sin   ( F1  F2 ) Ans. 2C Ans. From Ex. 17-2, d = 16 in, D = 36 in, C = 16(12) = 192 in, F1 = 940 lbf, F2 = 276 lbf  36  16  o   sin  1    2.9855 2(192)   2  1  36  16   x R  (940  276)  1      1214.4 lbf 2  2(192)      36  16  R y  (940  276)    34.6 lbf  2(192)  d  16  T  ( F1  F2 )    (940  276)    5312 lbf · in  2  2 ______________________________________________________________________________ 17-9 This is a good class project. Form four groups, each with a belt to design. Once each group agrees internally, all four should report their designs including the forces and torques on the line shaft. If you give them the pulley locations, they could design the line shaft. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 17 Solutions, Page 11/39 17-10 If you have the students implement a computer program, the design problem selections may differ, and the students will be able to explore them. For Ks = 1.25, nd = 1.1, d = 14 in and D = 28 in, a polyamide A-5 belt, 8 inches wide, will do (bmin = 6.58 in) ______________________________________________________________________________ 17-11 An efficiency of less than unity lowers the output for a given input. Since the object of the drive is the output, the efficiency must be incorporated such that the belt’s capacity is increased. The design power would thus be expressed as H nom K snd Ans. eff ______________________________________________________________________________ Hd  17-12 Some perspective on the size of Fc can be obtained from 2 w V  12bt  V  Fc       g  60  g  60  2 An approximate comparison of non-metal and metal belts is presented in the table below. Non-metal Metal 0.04 0.280 , l b f / i n 3 b, in t, in 5.00 1.000 0.20 0.005 The ratio w / wm is w 12(0.04)(5)(0.2)   29 wm 12(0.28)(1)(0.005) The second contribution to Fc is the belt peripheral velocity which tends to be low in metal belts used in instrument, printer, plotter and similar drives. The velocity ratio squared influences any Fc / (Fc)m ratio. It is common for engineers to treat Fc as negligible compared to other tensions in the belting problem. However, when developing a computer code, one should include Fc. ______________________________________________________________________________ 17-13 Eq. (17-8): Shigley’s MED, 11th edition Chapter 17 Solutions, Page 12/39 exp( f  )  1 exp( f  )  1  F1 exp( f  ) exp( f  ) Assuming negligible centrifugal force and setting F1 = ab from step 3, just before Ex. 173, F  F1  F2  ( F1  Fc ) F exp( f  ) a exp( f  )  1 (F )V H d  H nom K snd  33 000 33 000 H nom K s nd F  V bmin  Also, (1) 1  33 000 H d  exp( f  ) bmin    a V exp( f  )  1 Ans. Substituting into Eq. (1), ______________________________________________________________________________ 17-14 The decision set for the friction metal flat-belt drive is: A priori decisions • Function: Hnom = 1 hp, n = 1750 rev/min, VR = 2 , C 15 in, Ks = 1.2 , Np = 106 belt passes. • Design factor: nd = 1.05 • Belt material and properties: 301/302 stainless steel Table 17-8: Sy = 175 kpsi, E = 28 Mpsi,  = 0.285 • Drive geometry: d = 2 in, D = 4 in • Belt thickness: t = 0.003 in Design variables: • Belt width, b • Belt loop periphery Preliminaries H d  H nom K snd 1(1.2)(1.05) 1.26 hp 63 025(1.26) T   45.38 lbf · in 1750 A 15 in center-to-center distance corresponds to a belt loop periphery of 39.5 in. The 40 in loop available corresponds to a 15.254 in center distance. Shigley’s MED, 11th edition Chapter 17 Solutions, Page 13/39  4 2   d    2sin  1    3.010 rad  2(15.254)   4 2   D    2sin  1    3.273 rad  2(15.274)  For full friction development exp( f  d ) exp[0.35(3.010)]  2.868  dn  (2)(1750) V    916.3 ft/min 12 12 Sy 175 kpsi Eq. (17-15): S y  14.17  106  N p 0.407 14.17  106   106   0.407  51.212  103  psi From selection step 3, just before Ex. 17-3,    Et 28(106 )(0.003)  3 a Sf  t  51.212(10 )  (0.003)  (1   2 )d  (1  0.2852 )(2)     16.50 lbf/in of belt width ( F1) a  ab  16.50b For full friction development, from Prob. 17-13, bmin  F exp( f  d ) a exp( f  d )  1 F  2T 2(45.38)   45.38 lbf d 2 bmin  45.38  2.868     4.23 in 16.50  2.868  1  So Decision #1: b = 4.5 in F1  ( F1) a  ab  16.5(4.5)  74.25 lbf F2  F1  F  74.25  45.38  28.87 lbf F  F2 74.25  28.87 Fi  1   51.56 lbf 2 2 Existing friction Shigley’s MED, 11th edition Chapter 17 Solutions, Page 14/39 1  F1  1  74.25  ln    ln   0.314  d  F2  3.010  28.87  (F )V 45.38(916.3) Ht    1.26 hp 33 000 33 000 Ht 1.26 n fs    1.05 H nom K s 1(1.2) f  This is a non-trivial point. The methodology preserved the factor of safety corresponding to nd = 1.1 even as we rounded bmin up to b. Decision #2 was taken care of with the adjustment of the center-to-center distance to accommodate the belt loop. Use Eq. (17-2) as is and solve for C to assist in this. Remember to subsequently recalculate d and D . ______________________________________________________________________________ 17-15 Decision set: A priori decisions • Function: Hnom = 5 hp, N = 1125 rev/min, VR = 3, C  20 in, Ks = 1.25, Np = 106 belt passes • Design factor: nd = 1.1 • Belt material: BeCu, Sy = 170 kpsi, E = 17 Mpsi,  = 0.220 • Belt geometry: d = 3 in, D = 9 in • Belt thickness: t = 0.003 in Design decisions • Belt loop periphery • Belt width b Preliminaries: H d  H nom K snd  5(1.25)(1.1)  6.875 hp 63 025(6.875) T   385.2 lbf · in 1125 Decision #1: Choose a 60-in belt loop with a center-to-center distance of 20.3 in.  9 3   d    2sin  1    2.845 rad  2(20.3)   9 3   D    2sin  1    3.438 rad  2(20.3)  For full friction development: Shigley’s MED, 11th edition Chapter 17 Solutions, Page 15/39 exp( f  d )  exp[0.32(2.845)]  2.485  dn  (3)(1125) V    883.6 ft/min 12 12 S f  56.67 kpsi From selection step 3, just before Ex. 17-3,    Et 17(106 )(0.003)  3 a Sf  t   56.67(10 )  (0.003) 116.4 lbf/in (1   2 )d  (1  0.22 2 )(3)    2T 2(385.2) F    256.8 lbf d 3 F  exp( f  d )  256.8  2.485  bmin        3.69 in a  exp( f  d )  1  116.4  2.485  1  Decision #2: b = 4 in F1  ( F1) a  ab 116.4(4)  465.6 lbf F2  F1  F  465.6  256.8  208.8 lbf F  F2 465.6  208.8 Fi  1   337.3 lbf 2 2 Existing friction 1  F1  1  465.6  ln    ln    0.282  d  F2  2.845  208.8  (F )V 256.8(883.6) H    6.88 hp 33 000 33 000 H 6.88 n fs    1.1 5(1.25) 5(1.25) f  Fi can be reduced only to the point at which f   f  0.32. From Eq. (17-9) Fi  T  exp( f  d )  1  385.2  2.485  1        301.3 lbf d  exp( f  d )  1  3  2.485  1  Eq. (17-10):  2 exp( f  d )   2(2.485)  F1  Fi   429.7 lbf   301.3   2.485  1   exp( f  d )  1  F2  F1  F  429.7  256.8  172.9 lbf Shigley’s MED, 11th edition Chapter 17 Solutions, Page 16/39 f   f  0.32 and ______________________________________________________________________________ 17-16 This solution is the result of a series of five design tasks involving different belt thicknesses. The results are to be compared as a matter of perspective. These design tasks are accomplished in the same manner as in Probs. 17-14 and 17-15 solutions. The details will not be presented here, but the table is provided as a means of learning. Five groups of students could each be assigned a belt thickness. You can form a table from their results or use the table given here. CD 0.002 4.000 20.300 109.700 3.000 9.000 310.600 0.003 3.500 20.300 131.900 3.000 9.000 333.300 439.000 461.700 182.200 209.000 1.100 60.000 0.309 301.200 1.100 60.000 0.285 301.200 429.600 429.600 172.800 172.800 0.320 0.320 t, in 0.005 4.000 20.300 110.900 3.000 9.000 315.20 0 443.60 0 186.80 0 1.100 60.000 0.304 301.20 0 429.60 0 172.80 0 0.320 0.008 1.500 18.700 194.900 5.000 15.000 215.300 0.010 1.500 20.200 221.800 6.000 18.000 268.500 292.300 332.700 138.200 204.300 1.100 70.000 0.288 195.700 1.100 80.000 0.192 166.600 272.700 230.800 118.700 102.400 0.320 0.320 The first three thicknesses result in the same adjusted Fi, F1 and F2 (why?). We have no figure of merit, but the costs of the belt and pulleys are about the same for these three thicknesses. Since the same power is transmitted and the belts are widening, belt forces are lessening. ______________________________________________________________________________ 17-17 This is a design task. The decision variables would be belt length and belt section, which could be combined into one, such as B90. The number of belts is not an issue. We have no figure of merit, which is not practical in a text for this application. It is suggested that you gather sheave dimensions and costs and V-belt costs from a principal Shigley’s MED, 11th edition Chapter 17 Solutions, Page 17/39 vendor and construct a figure of merit based on the costs. Here is one trial. Preliminaries: For a single V-belt drive with Hnom = 3 hp, n = 3100 rev/min, D = 12 in, and d = 6.2 in, choose a B90 belt, Ks = 1.3 and nd = 1. From Table 17-10, select a circumference of 90 in. From Table 17-11, add 1.8 in giving Lp = 90 + 1.8 = 91.8 in Eq. (17-16b):     C  0.25   91.8  (12  6.2)   2     31.47 in 2     2 91.8 (12 6.2) 2(12 6.2)        2   12  6.2   d    2sin -1    2.9570 rad  2(31.47)  exp( f  d )  exp[0.5123(2.9570)]  4.5489  dn  (6.2)(3100) V    5031.8 ft/min 12 12 Table 17-13: Angle   d 180  180   (2.957 rad)   169.42     The footnote regression equation of Table 17-13 gives K1 without interpolation: K1 = 0.143 543 + 0.007 468(169.42°)  0.000 015 052(169.42°)2 = 0.9767 The design power is Hd = HnomKsnd = 3(1.3)(1) = 3.9 hp From Table 17-14 for B90, K2 = 1. From Table 17-12 take a marginal entry of Htab = 4, although extrapolation would give a slightly lower Htab. Eq. (17-17): Ha = K1K2Htab = 0.9767(1)(4) = 3.91 hp The allowable Fa is given by Fa  63 025H a 63 025(3.91)   25.6 lbf n(d / 2) 3100(6.2 / 2) The allowable torque Ta is Shigley’s MED, 11th edition Chapter 17 Solutions, Page 18/39 Ta  Fa d 25.6(6.2)   79.4 lbf · in 2 2 From Table 17-16, Kc = 0.965. Thus, Eq. (17-21) gives, 2 2  V   5031.8  Fc  K c    0.965    24.4 lbf  1000   1000  At incipient slip, Eq. (17-9) provides:  T   exp( f  )  1   79.4   4.5489  1  Fi        20.0 lbf    d   exp( f  )  1   6.2   4.5489  1  Eq. (17-10):  2 exp( f  )   2(4.5489)  F1  Fc  Fi   24.4  20   57.2 lbf   4.5489  1   exp( f  )  1  Thus, F2 = F1  Fa = 57.2  25.6 = 31.6 lbf n fs  Eq. (17-26): H a Nb (3.91)(1)   1.003 Hd 3.9 Ans. If we had extrapolated for Htab, the factor of safety would have been slightly less than one. Life Use Table 17-16 to find equivalent tensions T1 and T2 . Kb 576  57.2   150.1 lbf d 6.2 K 576 T2  F1  ( Fb ) 2  F1  b  57.2   105.2 lbf D 12 T1  F1  ( Fb )1  F1  From Table 17-17, K = 1193, b = 10.926, and from Eq. (17-27), the number of belt passes is:   K  b  K  b  NP           T1   T2   1   1193   10.926  1193   10.926         105.2    150.1   1  6.72(109 ) passes From Eq. (17-28) for NP > 109, Shigley’s MED, 11th edition Chapter 17 Solutions, Page 19/39 109 (91.8) t   720V 720(5031.8) t  25 340 h Ans. N P Lp Suppose nf s was too small. Compare these results with a 2-belt solution. H tab  4 hp/belt, Ta  39.6 lbf · in/belt, Fa  12.8 lbf/belt, H a  3.91 hp/belt NH NH 2(3.91) n fs  b a  b a   2.0 Hd H nom K s 3(1.3) Also, F1 = 40.8 lbf/belt, Fi  9.99 lbf/belt, ( Fb )1  92.9 lbf/belt, T1  133.7 lbf/belt, N P  2.39(1010 ) passes, F2 = 28.0 lbf/belt Fc  24.4 lbf/belt ( Fb ) 2  48 lbf/belt T2  88.8 lbf/belt t  605 600 h Initial tension of the drive: (Fi)drive = NbFi = 2(9.99) = 20 lbf ______________________________________________________________________________ 17-18 Given: two B85 V-belts with d = 5.4 in, D = 16 in, n = 1200 rev/min, and Ks = 1.25 Table 17-11: Lp = 85 + 1.8 = 86.8 in Eq. (17-17b):     C  0.25   86.8  (16  5.4)   2     26.05 in Ans. 2     2 86.8 (16 5.4) 2(16 5.4)        2  Eq. (17-1) in degrees:  16  5.4   d  180  2sin  1    156.5  2(26.05)  From Table 17-13 footnote: K1 = 0.143 543 + 0.007 468(156.5°)  0.000 015 052(156.5°)2 = 0.944 Table 17-14: Shigley’s MED, 11th edition K2 = 1 Chapter 17 Solutions, Page 20/39 V  Belt speed:  (5.4)(1200)  1696 ft/min 12 Use Table 17-12 to interpolate for Htab.  2.62  1.59  H tab 1.59    (1696  1000)  2.31 hp/belt  2000  1000  two belts: Eq. (1717) for H a  K1K 2 Nb H tab  0.944(1)(2)(2.31)  4.36 hp Assuming nd = 1, Hd = KsHnomnd = 1.25(1)Hnom For a factor of safety of one, Ha  Hd 4.36  1.25H nom 4.36 H nom   3.49 hp Ans. 1.25 ______________________________________________________________________________ 17-19 Given: Hnom = 60 hp, n = 400 rev/min, d = D = 26 in on 12 ft centers. From Table 17-15, a conservative service factor of KS = 1.4 is selected. Design task: specify V-belt and number of strands (belts). Tentative decision: Use D360 belts. Table 17-11: Lp = 360 + 3.3 = 363.3 in Eq. (17-16b):     C  0.25   363.3  (26  26)   2    140.8 in (nearly 144 in) 2     2 363.3  (26  26)  2(26  26)    2   d   ,  D   , exp[0.5123 ]  5.0,  dn  (26)(400) V    2722.7 ft/min 12 12 Table 17-13: For  = 180°, K1 = 1 Table 17-14: For D360, K2 = 1.10 Table 17-12: Htab = 16.94 hp by interpolation Shigley’s MED, 11th edition Chapter 17 Solutions, Page 21/39 Thus, Ha = K1K2Htab = 1(1.1)(16.94) = 18.63 hp / belt Eq. (17-19): Hd = HnomKs nd = 60(1.4)(1) = 84 hp Number of belts, Nb Nb  Hd 84   4.51 Ha 18.63 Round up to five belts. It is left to the reader to repeat the above for belts such as C360 and E360. 63 025H a 63 025(18.63)   225.8 lbf/belt n(d / 2) 400(26 / 2) (Fa )d 225.8(26) Ta    2935 lbf · in/belt 2 2 Fa  Eq. (17-21): 2 2  V   2722.7  Fc  3.498   3.498   25.9 lbf/belt   1000   1000  At fully developed friction, Eq. (17-9) gives Fi  Eq. (17-10): Life T d  exp( f  )  1  2935  5  1   exp( f  )  1   26  5  1   169.3 lbf/belt      2 exp( f  )   2(5)  F1  Fc  Fi   25.9  169.3   308.1 lbf/belt   5  1   exp( f  )  1  F2  F1  Fa  308.1  225.8  82.3 lbf/belt 18.63  5 H N nf s  a b  1.109 Ans. Hd 84 From Table 17-16, T1  T2  F1  Kb 5 680  308.1   526.6 lbf d 26 Eq. (17-27):   K  b  K  b  NP           T1   T2   Thus, NP > 109 passes Shigley’s MED, 11th edition 1  5.28  10 9  passes Ans. Chapter 17 Solutions, Page 22/39 t  Eq. (17-28): N P Lp 720V  109 (363.3) 720(2722.7) Thus, t > 185 320 h Ans. ______________________________________________________________________________ 17-20 Preliminaries: D  60 in, 14-in wide rim, Hnom = 50 hp, n = 875 rev/min, Ks = 1.2, nd = 1.1, mG = 875/170 = 5.147, d  60 / 5.147  11.65 in (a) From Table 17-9, an 11-in sheave exceeds C-section minimum diameter and precludes D- and E-section V-belts. Decision: Use d = 11 in, C270 belts Table 17-11: Lp = 270 + 2.9 = 272.9 in Eq. (17-16b):     C  0.25   272.9  (60  11)   2     76.78 in 2     2  272.9  2 (60  11)   2(60  11)   This fits in the range D  C  3( D  d )  60  C  3(60  11)  60 in  C  213 in 60  11  d    2sin  1  2.492 rad  142.8 2(76.78) 60  11  D    2sin  1  3.791 rad 2(76.78) exp(f d) = exp[0.5123(2.492)] = 3.5846 For the flat on flywheel, f = 0.13 [see discussion just before Eq. (17-18)], exp(f D) = exp[0.13(3.791)] = 1.637. The belt speed is V   dn  (11)(875)   2520 ft/min 12 12 Table 17-13: K1 = 0.143 543 + 0.007 468(142.8°)  0.000 015 052(142.8°)2 = 0.903 Table 17-14: K2 = 1.15 For interpolation of Table 17-12, let x be entry for d = 11.65 in and n = 2000 ft/min, and y Shigley’s MED, 11th edition Chapter 17 Solutions, Page 23/39 be entry for d = 11.65 in and n = 3000 ft/min. Then, x  6.74 7.17  6.74   11.65  11 12  11 x  7.01 hp at 2000 ft/min 8.11  y 8.84  8.11   11.65  11 12  11 y  8.58 hp at 3000 ft/min and Interpolating these for 2520 ft/min gives Eq. (17-17): 8.58  H tab 3000  2520   H tab  7.83 hp/belt 8.58  7.01 3000  2000 Ha = K1K2Htab = 0.903(1.15)(7.83) = 8.13 hp Eq. (17-19): Hd = HnomKsnd = 50(1.2)(1.1) = 66 hp H 66 Nb  d   8.1 belts H 8.13 a Eq. (17-20): Decision: Use 9 belts. On a per belt basis, 63 025H a 63 025(8.13)  106.5 lbf/belt n(d / 2) 875(11 / 2) F d 106.5(11) Ta  a   586.8 lbf · in per belt 2 2 Table 17-16: Kc = 1.716 2 2  V   2520  Fc  1.716    1.716    10.9 lbf/belt 1000 1000     Eq. (17-21): Fa  At fully developed friction, Eq. (17-9) gives Fi  T  exp( f  d )  1  586.9  3.5846  1   94.6 lbf/belt    d  exp( f  d )  1  11  3.5846  1  Eq. (17-10):  2 exp( f  d )   2(3.5846)  F1  Fc  Fi  158.8 lbf/belt  10.9  94.6   3.5846  1   exp( f  d )  1  F2  F1  Fa  158.8  106.7  52.1 lbf/belt NH 9(8.13) nf s  b a   1.11 O.K . Ans. Hd 66 Durability: Shigley’s MED, 11th edition Chapter 17 Solutions, Page 24/39  Fb  1  Kb / d 1600 / 11 145.5 lbf/belt  Fb  2  Kb / D  1600 / 60  26.7 lbf/belt T1  F1   Fb  1  158.8  145.5  304.3 lbf/belt T2  F1   Fb  2  158.8  26.7 185.5 lbf/belt Eq. (17-27) with Table 17-17: 1   K  b  K  b    2038   11.173  2038   11.173  NP                 185.5    T1    304.3    T2   9 9  1.68  10  passes  10 passes Ans. 1 Since NP is greater than 109 passes and is out of the range of Table 17-17, life from Eq. (17-27) is N L 109 (272.9) t  P p   150  103  h 720V 720(2520) Remember: (Fi)drive = 9(94.6) = 851.4 lbf Table 17-9: C-section belts are 7/8 in wide. Check sheave groove spacing to see if 14 in width is accommodating. (b) The fully developed friction torque on the flywheel using the flats of the V-belts, from Eq. (17-9), is  exp( f  )  1   1.637  1  Tflat  Fi D   94.6(60)   1371 lbf · in per belt   1.637  1   exp( f  )  1  The flywheel torque should be Tfly = mGTa = 5.147(586.9) = 3021 lbf · in per belt but it is not. There are applications, however, in which it will work. For example, make the flywheel controlling. Yes. Ans. ______________________________________________________________________________ 17-21 (a) S is the spliced-in string segment length De is the equatorial diameter D is the spliced string diameter  is the radial clearance S + De =  D = (De + 2) = De + 2 Shigley’s MED, 11th edition Chapter 17 Solutions, Page 25/39 From which   S 2 The radial clearance is thus independent of De.   12(6)  11.5 in 2 Ans. This is true whether the sphere is the earth, the moon or a marble. Thinking in terms of a radial or diametral increment removes the basic size from the problem. Shigley’s MED, 11th edition Chapter 17 Solutions, Page 26/39 (b) and (c) Table 17-9: For an E210 belt, the thickness is 1 in. 210  4.5 210 4.5      4.5 2   4.5  0.716 in   2 d P  di  The pitch diameter of the flywheel is DP  2  D  DP  D  2  60  2(0.716)  61.43 in We could make a table: Diametral Growth A 1.3 2  B 1.8  Section C D 2.9 3.3   E 4.5  The velocity ratio for the D-section belt of Prob. 17-20 is mG  D  2 60  3.3 /    5.55 d 11 Ans. for the V-flat drive as compared to ma = 60/11 = 5.455 for the VV drive. The pitch diameter of the pulley is still d = 11 in, so the new angle of wrap, d, is D  2  d Ans. 2C D  2  d  D    2sin  1 Ans. 2C Equations (17-16a) and (17-16b) are modified as follows  d    2sin  1 Shigley’s MED, 11th edition Chapter 17 Solutions, Page 27/39  ( D    d )2 Lp  2C  ( D  2  d )  2 4C    C p  0.25   Lp  ( D  2  d )  2   Ans. 2       L p  ( D  2  d )   2( D  2  d ) 2  Ans. 2    The changes are small, but if you are writing a computer code for a V-flat drive, remember that d and D changes are exponential. ______________________________________________________________________________ 17-22 This design task involves specifying a drive to couple an electric motor running at 1720 rev/min to a blower running at 240 rev/min, transmitting two horsepower with a center distance of at least 22 inches. Instead of focusing on the steps, we will display two different designs side-by-side for study. Parameters are in a “per belt” basis with per drive quantities shown along side, where helpful. Parameter Four A-90 Belts Two A-120 Belts mG 7.33 7.142 Ks 1.1 1.1 nd 1.1 1.1 K1 0.877 0.869 K2 1.05 1.15 d, in 3.0 4.2 D, in 22 30 d, rad 2.333 2.287 V, ft/min 1350.9 1891 3.304 3.2266 exp(fd ) Lp, in 91.3 101.3 C, in 24.1 31 Htab, uncorr. 0.783 1.662 NbHtab, uncorr. 3.13 3.326 Ta, lbf · in 26.45(105.8) 60.87(121.7) Fa, lbf 17.6(70.4) 29.0(58) Ha, hp 0.721(2.88) 1.667(3.33) nf s 1.192 1.372 F1, lbf 26.28(105.2) 44(88) F2, lbf 8.67(34.7) 15(30) (Fb)1, lbf 73.3(293.2) 52.4(109.8) (Fb)2, lbf 10(40) 7.33(14.7) Fc, lbf 1.024 2.0 Fi, lbf 16.45(65.8) 27.5(55) T1, lbf · in 99.2 96.4 T2, lbf · in 36.3 57.4 N, passes 1.61(109) 2.3(109) t>h 93 869 89 080 Shigley’s MED, 11th edition Chapter 17 Solutions, Page 28/39 Conclusions: • Smaller sheaves lead to more belts. • Larger sheaves lead to larger D and larger V. • Larger sheaves lead to larger tabulated power. • The discrete numbers of belts obscures some of the variation. The factors of safety exceed the design factor by differing amounts. ______________________________________________________________________________ 17-23 In Ex. 17-5 the selected chain was 140-3, making the pitch of this 140 chain14/8 = 1.75 in. Table 17-19 confirms. ______________________________________________________________________________ 17-24 (a) Eq. (17-32): H1  0.004 N11.08n10.9 p (3  0.07 p) H2  1000 K r N11.5 p 0.8 n11.5 Eq. (17-33): Equating and solving for n1 gives  0.25(106 ) K r N10.42  n1    p (2.2  0.07 p)   1/ 2.4 Ans. (b) For a No. 60 chain, p = 6/8 = 0.75 in, N1 = 17, Kr = 17 1/ 2.4  0.25(106 )(17)(17)0.42  n1    [2.2  0.07(0.75)]  0.75  1227 rev/min Ans. Table 17-20 confirms that this point occurs at 1200 ± 200 rev/min. (c) Life predictions using Eq. (17-40) are possible at speeds greater than 1227 rev/min. Ans. ______________________________________________________________________________ 17-25 Given: a double strand No. 60 roller chain with p = 0.75 in, N1 = 13 teeth at 300 rev/min, N2 = 52 teeth. (a) Table 17-20: Table 17-22: Table 17-23: Use Eq. (17-37): Htab = 6.20 hp K1 = 0.75 K2 = 1.7 Ks = 1 Ha = K1K2Htab = 0.75(1.7)(6.20) = 7.91 hp Ans. (b) Eqs. (17-35) and (17-36) with L/p = 82 Shigley’s MED, 11th edition Chapter 17 Solutions, Page 29/39 13  52  82   49.5 2 2 p 52  13    2 C   49.5  49.5  8     23.95 p 4 2      C  23.95(0.75) 17.96 in, round up to 18 in Ans. A (c) For 30 percent less power transmission, H  0.7(7.91)  5.54 hp 63 025(5.54) T   1164 lbf · in 300 Ans. Eq. (17-29): 0.75  3.13 in sin(180o /13) T 1164 F    744 lbf Ans. r 3.13 / 2 ______________________________________________________________________________ D  17-26 Given: No. 40-4 chain, N1 = 21 teeth for n = 2000 rev/min, N2 = 84 teeth, h = 20 (103) hr. (a) From Table 17-19, pitch is p = 0.500 in. With C = 20 in, Eq. (17-34): L 2C N1  N 2  N1  N 2     p p 2 4 2C / p 2(20) 21  84 (84  21) 2    135 pitches (or links) 0.5 2 4 2 (20 / 0.5) L = 135p = 135(0.500) = 67.5 in Ans. 2 (b) Table 17-20: No.40 chain, at 2000 rpm, Htab = 7.72 hp Table 17-22: Assuming for now the “post-extreme”, that is, failure by fatigue of the roller, 1.5 1.5  N1   21  K1       1.37  17   17  Table 17-23: Four strands, K2 = 3.3 Eq. (17-37): Ha = K1K2Htab = 1.37(3.3)7.72 = 34.9 hp Ans. Note that this table value does not account for the differences from the assumptions for chain length and life. (c) Eq. (17-32): Shigley’s MED, 11th edition Chapter 17 Solutions, Page 30/39 H1  0.004 N11.08n10.9 p (3  0.07 p)  0.004(21)1.08 (2000) 0.9 (0.5) 3  0.07(.5) 12.84 hp To compare with the tabulated value from part (b), adjust with K2 for the number of strands. It is not necessary to adjust with K1, since the number of teeth is included in Eq. (17-37). H a K 2 H1 3.3(12.84) 42.4 hp This is larger than the table value from part (a), probably indicating that H2 is the limiting power in the table. H2  1000K r N11.5 p 0.8 1000(17)(21)1.5 0.50.8  10.5 hp n11.5 (2000)1.5 Eq. (17-33): Adjusting for the number of strands, Ha = K2H2 = 3.3(10.5) = 34.7 hp Ans. This compares pretty well to the adjusted table value from part (b) of 34.9 hp. Note that this still doesn’t account for the desired chain length or life being different from the table assumptions. (d) Eq. (17-39): 0.4   N 1.5 L p   15000  0.4  0.8  1 H 2 1000  K r   p        n1   100   h     21 1.5 0.8  135  0.4  15000  0.4  1000  17   0.5      10.56 hp  100   20000     2000  Adjusting for the 4 strands, H a K 2 H 2 3.3(10.56) 34.8 hp Ans. This is only slightly different from the 34.7 hp from part (c). Note that accounting for the longer chain length increased the power rating slightly, while accounting for the longer life decreased the power rating slightly. N1 pn 21(0.5)(2000)   1750 ft/min 12 12 (e) Eq. (17-30): (f) From the standard relationship of power, torque, and speed, H=T V  T Ans. H 34.8 hp  550 ft-lbf/s  1       60 s/min    91.4 lbf-ft  2000 rev/min  hp  2 rad/rev   D (g) Eq. (17-29): Shigley’s MED, 11th edition Ans. p 0.5  3.355 in  sin(180 / N ) sin(180 / 21) Chapter 17 Solutions, Page 31/39 F1 T / ( D / 2) 91.4 lbf-ft  12 in/ft  /   3.355 / 2  in  654 lbf Ans. ______________________________________________________________________________ 17-27 This is our first design/selection task for chain drives. A possible decision set: A priori decisions • Function: Hnom, n1, space, life, Ks • Design factor: nd • Sprockets: Tooth counts N1 and N2, factors K1 and K2 Decision variables • Chain number • Strand count • Lubrication type • Chain length in pitches Function: Motor with Hnom = 25 hp at n = 700 rev/min; pump at n = 140 rev/min; mG = 700/140 = 5 Design Factor: nd = 1.1 Sprockets: Tooth count N2 = mGN1 = 5(17) = 85 teeth–odd and unavailable. Choose 84 teeth. Decision: N1 = 17, N2 = 84 Evaluate K1 and K2 Eq. (17-38): Eq. (17-37): Hd = HnomKsnd Ha = K1K2Htab Equate Hd to Ha and solve for Htab : KnH H tab  s d nom K1K 2 Table 17-22: K1 = 1 Table 17-23: K2 = 1, 1.7, 2.5, 3.3 for 1 through 4 strands H tab   1.5(1.1)(25) 41.25  (1) K 2 K2 Prepare a table to help with the design decisions: Strands 1 2 3 4 Shigley’s MED, 11th edition Chain H  tab K2 No. 1.0 41.3 100 1.7 24.3 80 2.5 16.5 80 3.3 12.5 60 Htab nf s Type 59.4 1.58 B 31.0 1.40 B 31.0 2.07 B 13.3 1.17 B Chapter 17 Solutions, Page 32/39 Design Decisions: We need a figure of merit to help with the choice. If the best was 4 strands of No. 60 chain, then Decision #1 and #2: Choose four strand No. 60 roller chain with nf s = 1.17. n fs  K1K 2 H tab 1(3.3)(13.3)   1.17 K s H nom 1.5(25) Decision #3: Choose Type B lubrication Analysis: Table 17-20: Table 17-19: Htab = 13.3 hp p = 0.75 in Try C = 30 in in Eq. (17-34): L 2C N1  N 2 ( N 2  N1 ) 2    p p 2 4 2C / p 17  84 (84  17) 2  2(30 / 0.75)   2 4 2 (30 / 0.75) 133.3 L = 0.75(133.3) = 100 in (no need to round) A N1  N 2 L 17  84 100      82.83 2 p 2 0.75 Eq. (17-36) with p = 0.75 in: Eq. (17-35): 2 p N 2  N1    2 C    A  A  8   4 2      2  0.75 84  17   2     82.83    82.83  8    30.0 in  4  2      Decision #4: Choose C = 30.0 in. ______________________________________________________________________________ 17-28 Follow the decision set outlined in Prob. 17-27 solution. We will form two tables, the first for a 15 000 h life goal, and a second for a 50 000 h life goal. The comparison is useful. Function: Hnom = 50 hp at n = 1800 rev/min, npump = 900 rev/min, mG = 1800/900 = 2, Shigley’s MED, 11th edition Chapter 17 Solutions, Page 33/39 Ks = 1.2, life = 15 000 h, then repeat with life = 50 000 h Design factor: nd = 1.1 Sprockets: N1 = 19 teeth, N2 = 38 teeth Table 17-22 (post extreme): 1.5 1.5  N1   19  K1        1.18  17   17  Table 17-23: K2 = 1, 1.7, 2.5, 3.3, 3.9, 4.6, 6.0 Decision variables for 15 000 h life goal: K snd H nom 1.2(1.1)(50) 55.9   K1K 2 1.18K 2 K2 KK H 1.18K 2 H tab  1 2 tab   0.0197K 2 H tab K s H nom 1.2(50)   H tab nf s (1) Form a table for a 15 000 h life goal using these equations. K2 1 1.7 2.5 3.3 3.9 4.6 6 H'tab 55.90 32.90 22.40 16.90 14.30 12.20 9.32 Chain # 120 120 120 120 80 60 60 Htab 21.6 21.6 21.6 21.6 15.6 12.4 12.4 nf s 0.423 0.923 1.064 1.404 1.106 1.126 1.416 Lub C' C' C' C' C' C' C' There are 4 possibilities where nf s ≥ 1.1 Decision variables for 50 000 h life goal From Eq. (17-40), the power-life tradeoff is:  ) 2.550 000 ( H tab ) 2.515 000  ( H tab  15 000      )2.5  H tab ( H tab  50 000  Substituting from (1), 1/ 2.5   0.618 H tab  55.9  34.5 H tab    0.618   K2  K2  The H  notation is only necessary because we constructed the first table, which we normally would not do. Shigley’s MED, 11th edition Chapter 17 Solutions, Page 34/39 nf s   ) K1K 2 H tab K K (0.618H tab  1 2  0.618[(0.0197) K 2 H tab ] K s H nom K s H nom  0.0122 K 2 H tab  Form a table for a 50 000 h life goal. K2 1 1.7 2.5 3.3 3.9 4.6 6 H''tab 34.50 20.30 13.80 10.50 8.85 7.60 5.80 Chain # 120 120 120 120 120 120 80 Htab 21.6 21.6 21.6 21.6 21.6 21.6 15.6 nf s 0.264 0.448 0.656 0.870 1.028 1.210 1.140 Lub C' C' C' C' C' C' C' There are two possibilities in the second table with nf s ≥ 1.1. (The tables allow for the identification of a longer life of the outcomes.) We need a figure of merit to help with the choice; costs of sprockets and chains are thus needed, but is more information than we have. Decision #1: #80 Chain (smaller installation) Ans. nf s = 0.0122K2Htab = 0.0122(8.0)(15.6) = 1.14 O.K. Decision #2: 8-Strand, No. 80 Ans. Decision #3: Type C Lubrication Ans. Decision #4: p = 1.0 in, C is in midrange of 40 pitches L 2C N1  N 2 ( N 2  N1) 2    p p 2 4 2C / p 19  38 (38  19) 2  2(40)   2 4 2 (40)  108.7  110 even integer Ans. Eq. (17-36): A Eq. (17-35): N1  N 2 L 19  38 110      81.5 2 p 2 1 C 1  38  19     ( 81.5)  ( 81.5) 2  8   p 4  2   2    40.64   C = p(C/p) = 1.0(40.64/1.0) = 40.64 in (for reference) Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 17 Solutions, Page 35/39 17-29 The objective of the problem is to explore factors of safety in wire rope. We will express strengths as tensions. (a) Monitor steel 2-in 6  19 rope, 480 ft long. Table 17-2: Minimum diameter of a sheave is 30d = 30(2) = 60 in, preferably 45(2) = 90 in. The hoist abuses the wire when it is bent around a sheave. Table 17-24 gives the nominal tensile strength as 106 kpsi. The ultimate load is   (2) 2  Fu  (Su )nom Anom  106    333 kip  4  Ans. The tensile loading of the wire is given by Eq. (17-47) a W  Ft    wl   1   g m  W  4(2)  8 kip, m 1 Table (17-24): wl = 1.60d 2 l = 1.60(22)(480) = 3072 lbf = 3.072 kip Therefore, 2   Ft  (8  3.072)  1    11.76 kip  32.2  Eq. (17-41): Ed A Fb  r w m D and for the 72-in drum Fb  Ans. 12(106 )(2 / 13)(0.38)(22 )(10 3 )  39 kip 72 Ans. For use in Eq. (17-44), from Fig. 17-21 ( p / Su )  0.0014 Su  240 kpsi, p. 920 0.0014(240)(2)(72) Ff   24.2 kip 2 Ans. (b) Factors of safety Static, no bending: n Shigley’s MED, 11th edition Fu 333   28.3 Ft 11.76 Ans. Chapter 17 Solutions, Page 36/39 Static, with bending: ns  Eq. (17-46): Fu  Fb 333  39   25.0 Ft 11.76 Fatigue without bending: F 24.2 nf  f   2.06 Ft 11.76 Ans. Ans. Fatigue, with bending: For a life of 0.1(106) cycles, from Fig. 17-21 ( p / Su )  4 / 1000  0.004 0.004(240)(2)(72) Ff   69.1 kip 2 Eq. (17-45): nf  69.1  39  2.56 11.76 Ans. If we were to use the endurance strength at 106 cycles (Ff = 24.2 kip) the factor of safety would be less than 1 indicating 106 cycle life impossible. Comments: • There are a number of factors of safety used in wire rope analysis. They are different, with different meanings. There is no substitute for knowing exactly which factor of safety is written or spoken. • Static performance of a rope in tension is impressive. • In this problem, at the drum, we have a finite life. • The remedy for fatigue is the use of smaller diameter ropes, with multiple ropes supporting the load. See Ex. 17-6 for the effectiveness of this approach. It will also be used in Prob. 17-30. • Remind students that wire ropes do not fail suddenly due to fatigue. The outer wires gradually show wear and breaks; such ropes should be retired. Periodic inspections prevent fatigue failures by parting of the rope. ______________________________________________________________________________ 17-30 Since this is a design task, a decision set is useful. A priori decisions • Function: load, height, acceleration, velocity, life goal • Design Factor: nd • Material: IPS, PS, MPS or other • Rope: Lay, number of strands, number of wires per strand Decision variables: • Nominal wire size: d • Number of load-supporting wires: m Shigley’s MED, 11th edition Chapter 17 Solutions, Page 37/39 From experience with Prob. 17-29, a 1-in diameter rope is not likely to have much of a life, so approach the problem with the d and m decisions open. Function: 5000 lbf load, 90 foot lift, acceleration = 4 ft/s2, velocity = 2 ft/s, life goal = 105 cycles Design Factor: nd = 2 Material: IPS Rope: Regular lay, 1-in plow-steel 6  19 hoisting Design variables Choose 30-in Dmin. Table 17-27: w = 1.60d 2 lbf/ft wl = 1.60d 2l = 1.60d 2(90) = 144d 2 lbf, each Eq. (17-47): a   5000 4  W   Ft    wl   1      144d 2   1   g  m 32.2  m   5620   162d 2 lbf, each wire m From Fig. 17-21 for 105 cycles, p/Su = 0.004. For plow steel, let Su = 240 kpsi. Eq. (17-44): Ff  ( p / Su )Su Dd 0.004(240 000)(30d )  14 400d lbf each wire 2 2 Eq. (17-41) and Table 17-27: 6 2 Er d w Am 12  10  0.067d  0.4d  Fb    10 720d 3 lbf, each wire D 30 Eq. (17-45): F f  Fb 14 400d  10 720d 3 Ft (5620 / m)  162d 2 We could use a computer program to build a table similar to that of Ex. 17-6. Alternatively, we could recognize that 162 d 2 is small compared to 5620 / m, and therefore eliminate the 162d 2 term. nf  nf   14 400d  10 720d 3 m  (14 400d  10 720d 3 ) 5620 / m 5620 Maximize nf , n f d 0  m [14 400  3(10 720)d 2 ] 5620 From which d*  Shigley’s MED, 11th edition 14 400  0.669 in 3(10 720) Chapter 17 Solutions, Page 38/39 Back-substituting nf  m [14 400(0.669)  10 720(0.6693 )] 1.14 m 5620 Thus nf = 1.14, 2.28, 3.42, 4.56 for m=1, 2, 3, 4 respectively. If we choose d = 0.50 in, then m = 2. 14 400(0.5)  10 720(0.53 ) nf   2.06 (5620 / 2)  162(0.5) 2 This exceeds nd = 2 Decision #1: d = 1/2 in Decision #2: m = 2 ropes supporting load. Rope should be inspected weekly for any signs of fatigue (broken outer wires). Comment: Table 17-25 gives n for freight elevators in terms of velocity.  d 2  Fu  (Su ) nom Anom  106 000   83 252d 2 lbf, each wire   4  2 F 83 452(0.5) n u   7.32 Ft (5620 / 2)  162(0.5) 2 By comparison, interpolation for 120 ft/min gives 7.08 - close. The category of construction hoists is not addressed in Table 17-25. We should investigate this before proceeding further. ______________________________________________________________________________ 17-31 Given: 2000 ft lift, 72 in drum, 6  19 MS rope, cage and load 8000 lbf, accel. = 2 ft/s2. (a) Table 17-24: (Su)nom = 106 kpsi; Su = 240 kpsi (plow steel); Fig. 17-21: (p/Su)106 = 0.0014 Eq. (17-44): Ff  Table 17-24: Eq. (17-47):  p / Su  Su dD 2  0.0014(240)d (72) 12.1 d kip 2 wl = 1.6d 2 2000(103) = 3.2d 2 kip  a Ft  (W  wl )  1   g  2    (8  3.2d 2 )  1    32.2   8.5  3.4d 2 kip Shigley’s MED, 11th edition Chapter 17 Solutions, Page 39/39 Note that bending is not included. F 12.1d nf  f  8.5  3.4d 2 Ft d, in 0.500 1.000 1.500 1.625 1.750 2.000 nf 0.650 1.020 1.124 1.125 ← maximum nf Ans. 1.120 1.095 (b) Try m = 4 strands 2  8  Ft    3.2d 2   1   32.2  4   2.12  3.4d 2 kip F f  12.1d kip 12.1d nf  2.12  3.4d 2 d, in 0.5000 0.5625 0.6520 0.7500 nf 2.037 2.130 2.193 2.250 ← maximum nf Ans. 0.8750 2.242 1.0000 2.192 Comparing tables, multiple ropes supporting the load increases the factor of safety, and reduces the corresponding wire rope diameter, a useful perspective. ______________________________________________________________________________ 17-32 ad b / m  cd 2 dn f (b / m  cd 2 )a  ad (2cd )  0 dd (b / m  cd 2 ) 2 nf  From which d*  nf *  Shigley’s MED, 11th edition b mc Ans. a b / (mc) a m  (b / m)  c [b / (mc)] 2 bc Ans. Chapter 17 Solutions, Page 40/39 These results agree closely with the Prob. 17-31 solution. The small differences are due to rounding in Prob. 17-31. ______________________________________________________________________________ 17-33 From Prob. 17-32 solution: n1  ad b / m  cd 2 Solve the above equation for m b m (1) ad / n1  cd 2   ad / n1   ad 2  (0)  b   a / n1   2cd  dm 0   2 dd   ad / n1   cd 2  a d*  Ans. 2 cn 1 From which Substituting this result for d into Eq. (1) gives 4bcn1 Ans. a2 ______________________________________________________________________________ m*  17-34 Table 17-27: Am  0.40d 2  0.40(22 )  1.6 in 2 Er 12 Mpsi, w 1.6d 2 1.6(2 2 )  6.4 lbf/ft wl  6.4(480)  3072 lbf   wl /  Aml   3072 /  1.6(480)12   0.333 lbf/in 3 Treat the rest of the system as rigid, so that all of the stretch is due to the load of 7000 lbf, the cage weighing 1000 lbf, and the wire’s weight. From the solution of Prob. 4-7, Wl l 2  AE 2 E (1000  7000)(480)(12) 0.333(480 2 )12 2   1.6(12)(106 ) 2(12)(106 ) 1   2.4  0.460  2.860 in Ans. ______________________________________________________________________________ 17-35 to 17-38 Computer programs will vary. Shigley’s MED, 11th edition Chapter 17 Solutions, Page 41/39 ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 17 Solutions, Page 42/39 Chapter 20 Refer to the first paragraph of Sec. 20–1, p. 970. iv Ans. ______________________________________________________________________________ 20–1 Refer to the first paragraph on p. 974, and the first paragraph in Sec. 20–4 on p. 981. The size dimension of a feature of size should be directly toleranced. Ans. ______________________________________________________________________________ 20–2 Refer to the last paragraph on p. 972. ii Ans. ______________________________________________________________________________ stu ed d vi y re aC s o ou urc rs e eH w er as o. co m 20–3 Refer to the first paragraph on p. 973. Feature of size Ans. ______________________________________________________________________________ 20–4 Refer to the first paragraph on p. 973. size, location, orientation, and form Ans. ______________________________________________________________________________ 20–5 Refer to the Form Controls sub-section on p. 985. straightness, flatness, circularity, and cylindricity Ans. ______________________________________________________________________________ 20–6 Refer to the Orientation Controls sub-section on p. 987. angularity, parallelism, and perpendicularity Ans. ______________________________________________________________________________ 20–7 Refer to the Location Controls sub-section on p. 990. position, concentricity, symmetry, position Ans. The position control is the most prominently used. Ans. ______________________________________________________________________________ Th is 20–8 sh ar Refer to the Basic Dimensions sub-section on p. 983. ii Ans. ______________________________________________________________________________ 20–9 20–10 All of the features with a +/– tolerance are features of size. Ans. These include boss, hole, plate thickness, plate width, and plate height. The combined thickness of the plate and the boss (70 +/– 0.1) technically satisfies the definition, though it is not a typical feature of size. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 20 Solutions, Page 1/8 This study source was downloaded by 100000755497394 from CourseHero.com on 04-22-2021 22:27:38 GMT -05:00 https://www.coursehero.com/file/59623485/Chapter20-Solutions-11edocx/ stu ed d vi y re aC s o ou urc rs e eH w er as o. co m 20–11 The datum features are the physical surfaces of the plate. The datums are the theoretically perfect planes corresponding to the datum feature simulators of the physical surfaces. The datum reference frame is the xyz coordinate axis corresponding to the three datum planes. ______________________________________________________________________________ 20–12 For the first two questions, refer to the first two paragraphs in the Position sub-section on p. 990. Also refer to step 4 of Ex. 20–1, p. 998. iv Ans. iii Ans. For the third question, refer to the first full paragraph on p. 977. v Ans. Th is ______________________________________________________________________________ ar 20–13 (a) The boss is a feature of size, so it is directly toleranced. The maximum and minimum diameters allowed are 50.3 and 49.7, respectively. Ans. sh (b) The position tolerance affects the location and orientation of the center axis of the boss, but does not affect the diameter of the boss. No effect. Ans. (c) The position tolerance always defines the allowed location and orientation of an axis, center line, or center plane of a feature of size. Refer to the Position sub-section on p. Shigley’s MED, 11th edition Chapter 20 Solutions, Page 2/8 This study source was downloaded by 100000755497394 from CourseHero.com on 04-22-2021 22:27:38 GMT -05:00 https://www.coursehero.com/file/59623485/Chapter20-Solutions-11edocx/ 990. For a strict definition of the axis, refer to the Actual Mating Envelopes sub-section on p. 979. The center axis of the boss, as determined by the related actual mating envelope, must be within the tolerance zone. Ans. (d) Note that the position tolerance is specified at maximum material condition. If the boss is produced at a diameter of 50.3, which is the maximum material condition, then the specified tolerance applies. The tolerance zone diameter is 0.2. Ans. (e) If the boss is produced at a diameter of 49.7, which is 0.6 less than the maximum material condition, the tolerance zone diameter is the sum of this 0.6 and the specified tolerance of 0.2. The tolerance zone diameter is 0.8. Ans. stu ed d vi y re aC s o ou urc rs e eH w er as o. co m (f) The part is stabilized with respect to datums A, B, and C, in that order, then the central axis of the tolerance zone is oriented perpendicular to datum A, and located from datums B and C by the basic dimensions. Ans. (g) The position control defines a tolerance zone that defines the allowed location and orientation of the axis of the boss. Refer to the first paragraph of the Position sub-section on p. 990. Depending on the amount of deviation from the maximum material condition, the tolerance varies from 0.2 at MMC to 0.8 at LMC. ii Ans. (h) Cylindricity is a characteristic of form. Since a specific cylindricity form control is not specified, Rule #1 applies as the default form control for a feature of size. Refer to the second full paragraph on p. 982. iii Ans. ______________________________________________________________________________ 20–14 (a) The hole is a feature of size, so it is directly toleranced. The maximum and minimum diameters allowed are 30.1 and 29.9, respectively. Ans. ar Th is (b) The position tolerance affects the location and orientation of the center axis of the hole, but does not affect the diameter of the boss. No effect. Ans. sh (c) The position tolerance always defines the allowed location and orientation of an axis, center line, or center plane of a feature of size. Refer to the Position sub-section on p. 990. For a strict definition of the axis, refer to the Actual Mating Envelopes sub-section on p. 979. The center axis of the hole, as determined by the related actual mating envelope, must be within the tolerance zone. Ans. Shigley’s MED, 11th edition Chapter 20 Solutions, Page 3/8 This study source was downloaded by 100000755497394 from CourseHero.com on 04-22-2021 22:27:38 GMT -05:00 https://www.coursehero.com/file/59623485/Chapter20-Solutions-11edocx/ (d) Note that the position tolerance is specified at maximum material condition, which for the hole is 29.9. If the hole is produced at a diameter of 30.1, which is 0.2 greater than the maximum material condition, the tolerance zone is the sum of this 0.2 and the specified tolerance of 0.3. The tolerance zone diameter is 0.5. Ans. (e) If the hole is produced at a diameter of 29.9, which is the maximum material condition, then the specified tolerance applies. The tolerance zone diameter is 0.3. Ans. (f) The part is stabilized with respect to datums A, B, and C, in that order, then the central axis of the tolerance zone is oriented perpendicular to datum A, and located from datums B and C by the basic dimensions. Ans. stu ed d vi y re aC s o ou urc rs e eH w er as o. co m (g) The position control defines a tolerance zone that defines the allowed location and orientation of the axis of the boss. Refer to the first paragraph of the Position sub-section on p. 990. Depending on the amount of deviation from the maximum material condition, the tolerance varies from 0.3 at MMC to 0.5 at LMC. ii Ans. (h) Cylindricity is a characteristic of form. Since a specific cylindricity form control is not specified, Rule #1 applies as the default form control for a feature of size. Refer to the second full paragraph on p. 982. iii Ans. ______________________________________________________________________________ 20–15 Refer to the Actual Mating Envelopes sub-section on p. 979. An actual mating envelope is determined by fitting the largest gauge pin (or an expanding mandrel) into the hole, just touching the high points of the hole. The center axis of the gauge pin is defined to be the center axis of the hole. Ans. ______________________________________________________________________________ is 20–16 Refer to the second full paragraph on p. 982. iii Ans. ______________________________________________________________________________ ar sh Th 20–17 Refer to the first full paragraph on p. 982. Shaft diameter at MMC = 20.2 Ans. Shaft diameter at LMC = 19.8 Ans. ______________________________________________________________________________ 20–18 Refer to the first full paragraph on p. 982. Hole diameter at MMC = 19.8 Ans. Hole diameter at LMC = 20.2 Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 20 Solutions, Page 4/8 This study source was downloaded by 100000755497394 from CourseHero.com on 04-22-2021 22:27:38 GMT -05:00 https://www.coursehero.com/file/59623485/Chapter20-Solutions-11edocx/ 20–19 Refer to the second full paragraph on p. 982. The surface of the hole is limited by an envelope of perfect form at MMC. There is no limiting envelope defined at LMC. The diameter of the limiting envelope is 19.8. Ans. ______________________________________________________________________________ 20–20 Refer to the second full paragraph on p. 982. The surface of the shaft is limited by an envelope of perfect form at MMC. There is no limiting envelope defined at LMC. The diameter of the limiting envelope is 20.2. Ans. ______________________________________________________________________________ 20–21 Refer to Sec. 20–6, p. 994. (a) Since no material condition is specified, the tolerance zone diameter is the specified 0.1 at all produced diameters of the boss. 0.1, 0.1, 0.1 Ans. stu ed d vi y re aC s o ou urc rs e eH w er as o. co m (b) Since the maximum material condition is specified, the tolerance zone diameter is the specified 0.1 when the boss is produced at the MMC of 25.2, and increases by the amount of deviation from the MMC. 0.5, 0.3, 0.1 Ans. (c) Since the least material condition is specified, the tolerance zone diameter is the specified 0.1 when the boss is produced at the LMC of 24.8, and increases by the amount of deviation from the LMC. 0.1, 0.3, 0.5 Ans. ______________________________________________________________________________ 20–22 Refer to Sec. 20–6, p. 994. (a) Since no material condition is specified, the tolerance zone diameter is the specified 0.3 at all produced diameters of the hole. 0.3, 0.3, 0.3 Ans. is (b) Since the maximum material condition is specified, the tolerance zone diameter is the specified 0.3 when the hole is produced at the MMC of 32.0, and increases by the amount of deviation from the MMC. 0.3, 0.5, 0.7 Ans. ar sh Th (c) Since the least material condition is specified, the tolerance zone diameter is the specified 0.3 when the hole is produced at the LMC of 32.4, and increases by the amount of deviation from the LMC. 0.7, 0.5, 0.3 Ans. ______________________________________________________________________________ 20–23 Refer to the Cylindricity sub-section on p. 987. cylindricity Ans. ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 20 Solutions, Page 5/8 This study source was downloaded by 100000755497394 from CourseHero.com on 04-22-2021 22:27:38 GMT -05:00 https://www.coursehero.com/file/59623485/Chapter20-Solutions-11edocx/ 20–24 Refer to the third paragraph on p. 989. profile of a surface Ans. ______________________________________________________________________________ 20–25 Refer to the Form Controls sub-section on p. 985. The form controls (straightness, flatness, circularity, and cylindricity) never reference datums. They only relate to the form of the individual feature, and are independent of the feature’s location. Ans. ______________________________________________________________________________ 20–26 (a) Refer to the first paragraph of Sec. 20–6 on p. 994 for the two primary modifiers, and the last paragraph on p. 995 for the default material condition. MMC, LMC, and RFS Ans. stu ed d vi y re aC s o ou urc rs e eH w er as o. co m (b) Refer to the last paragraph on p. 995. RFS Ans. (c) Refer to the last paragraph on p. 994 and the third paragraph on p. 995. MMC and LMC Ans. (d) Refer to the last sentence on p. 994. iv Ans. (e) Refer to the first paragraph of Sec. 20–6. iii Ans. (f) Refer to the first paragraph on p. 995. MMC Ans. (g) Refer to the first partial paragraph on p. 996. RFS Ans. is (h) Refer to the third paragraph on p. 995. LMC Ans. ______________________________________________________________________________ ar Th 20–27 (a) The countersink diameter is specified as 12 ± 0.2. The minimum countersink diameter is 11.8. Ans. sh (b) The countersink depth is specified as 3 ± 0.1. The maximum countersink depth is 3.1. Ans. (c) The bolt holes are specified as 6 +0.1/–0.0. The maximum material condition for a hole is the smallest hole diameter. The diameter of the bolt holes at MMC is 6.0. Ans. Shigley’s MED, 11th edition Chapter 20 Solutions, Page 6/8 This study source was downloaded by 100000755497394 from CourseHero.com on 04-22-2021 22:27:38 GMT -05:00 https://www.coursehero.com/file/59623485/Chapter20-Solutions-11edocx/ (d) Refer to the second paragraph on p. 973. The features of size are all hole diameters and counterbore diameters, the slot width, the base widths, the base thickness, and the center protrusion width. Ans. (e) The default surface profile tolerance in the note provides a tolerance zone around all surfaces of 0.5, centered on the surface. This gives a tolerance of 0.25 outside the base dimension on each edge, for a total of 0.5. Therefore, the minimum and maximum allowed base dimensions are 59.5 and 60.5. Ans. stu ed d vi y re aC s o ou urc rs e eH w er as o. co m (f) Datum feature A is the bottom surface. Its datum is a theoretical plane corresponding to a datum feature simulator (such as a machine table). Datum feature B is the surface of the hole. Its datum is the center axis corresponding to a datum feature simulator, such as the largest gauge pin just touching the high points of the hole, while being held perpendicular to datum A. Datum feature C is the surface of the slot. Its datum is the center plane corresponding to a datum feature simulator, such as the largest gauge block just touching the high points of the slot, while being held perpendicular to datum A. Ans. The datum reference frame is made of three mutually perpendicular planes, with the origin of the coordinate axes at the bottom center of the datum B hole. Ans. The part is stabilized for manufacture or inspection by applying the datums in the following sequence: Constrain the base to be in contact with a datum feature simulator (an essentially flat surface) corresponding to datum A; constrain the translation of the part on datum A with a gauge pin in the datum feature B hole; constrain the final rotation of the part (around datum B) with a gauge pin or block in the groove. Ans. The purpose of the fixture is to locate and orient the large bore. This bore is located with respect to the bottom surface (datum A), the pin hole (datum B), and the groove (datum C). The edges of the base will not be touching anything for alignment, and can consequently have much looser tolerances. It is usually easier and cheaper to precisely locate a part with respect to locating pins than with respect to the edges of the part. Ans. ar sh Th is (g) Since no material condition modifier is specified for the tolerance zone associated with the hole, its tolerance zone is RFS (regardless of feature size). The specified tolerance zone diameter is the specified 0.05 whether the hole diameter is 6.00, or 6.05, or anything between. 0.05, 0.05 Ans. (h) The top line of the position control of the bolt holes specifies the PLTZF tolerance of 0.3 at MMC (maximum material condition). For the bolt holes, when manufactured at the MMC of 6.0, the diameter of the PLTZF tolerance zone is 0.3. When manufactured at the LMC of 6.1, the diameter of the PLTZF tolerance zone is 0.4, i.e. the sum of the specified tolerance plus the “bonus” tolerance of 0.1 due to the deviation of the hole from its MMC. 0.3, 0.4 Ans. Shigley’s MED, 11th edition Chapter 20 Solutions, Page 7/8 This study source was downloaded by 100000755497394 from CourseHero.com on 04-22-2021 22:27:38 GMT -05:00 https://www.coursehero.com/file/59623485/Chapter20-Solutions-11edocx/ (i) The second line of the position control of the bolt holes specifies the FRTZF tolerance of 0.1 at MMC (maximum material condition). For the bolt holes, when manufactured at the MMC of 6.0, the diameter of the FRTZF tolerance zone is 0.1. When manufactured at the LMC of 6.1, the diameter of the FRTZF tolerance zone is 0.2, i.e. the sum of the specified tolerance plus the “bonus” tolerance of 0.1 due to the deviation of the hole from its MMC. 0.1, 0.2 Ans. (j) MMC ensures a minimum clearance for the bolts in their holes, but allows greater deviations from ideal if the holes are produced larger. Ans. stu ed d vi y re aC s o ou urc rs e eH w er as o. co m (k) Orientation and straightness of the axis are controlled by the position tolerance zone which is specified in the feature control frame directly under the diameter dimension for the bore. The position tolerance specifies a cylindrical tolerance zone that the axis of the hole must be within, thus controlling its orientation and straightness. Ans. Cylindricity is a control of the surface of the bore, so is not controlled by the position tolerance zone which controls the axis of the bore. Since there is not an explicit control of the cylindricity specified, it could be controlled either by the default profile of a surface control, specified with the note at the bottom of the drawing, or by Rule #1. In this case, Rule #1 has the tighter tolerance, so it controls the cylindricity of the bore. Rule #1 provides for the cylindricity of the surface of the bore to be within an envelope of a perfect cylinder at MMC diameter of 14.9. Ans. (l) All exterior surfaces except for the bottom are controlled only by the default surface profile tolerance, so as-cast is sufficient for them. Ans. If the edges of the base were used as datum features, these surfaces would need to be controlled more tightly in order to locate the critical holes from them. The edges of the base could not be left as-cast. Ans. ______________________________________________________________________________ ar sh Th is 20–28 (a) A datum feature is the physical surface that will be used to define the theoretical datum. The datum B tag is associated with a control frame that is pointing to the datum feature. The datum feature is best described by answer iv The outer surface of the protruding cylinder. Ans. (b) 10.1 Ans. (c) Maximum material condition for a hole is when it is at its smallest allowed diameter (maximum material), thus 10.0. Ans. (d) The tolerance block indicates zero tolerance at maximum material condition, plus a bonus of any amount that the hole is larger than maximum material condition. At a diameter of 10.1, the hole is 0.1 larger than maximum material condition. Thus the diameter of the tolerance zone is 0.1. Ans. (e) The tolerance block indicates a 0.3 diameter tolerance zone, regardless of size. Thus, the tolerance zone has a diameter of 0.3. Ans. Shigley’s MED, 11th edition Chapter 20 Solutions, Page 8/8 This study source was downloaded by 100000755497394 from CourseHero.com on 04-22-2021 22:27:38 GMT -05:00 https://www.coursehero.com/file/59623485/Chapter20-Solutions-11edocx/ ar sh Th is stu ed d vi y re aC s o ou urc rs e eH w er as o. co m (f) The universal profile tolerance of note at the bottom of the drawing applies to all basic dimensions that aren’t otherwise toleranced. Thus, a profile tolerance of 0.2 is centered on the basic dimension of 90. The diameter may vary between a minimum of 89.9 and a maximum of 90.1. Ans. (g) The tolerance zone is located ideally by basic dimensions. The non-ideal center axis of the hole must lie entirely within the tolerance zone. The center axis of the hole is defined by its actual mating envelope. This is best describe by answer i, the largest gauge pin that fits into the hole. Ans. (h) There is not an explicit cylindricity control given, but the surface of the hole, including its cylindricity, must stay within the ideal cylinder defined by the hole’s maximum material condition. This is best described by answer iii. The envelope of a perfect cylinder with diameter of 19.95. Ans. (i) The tolerance for the holes is defined with the modifier for maximum material condition. Therefore, the center axes of the bolt holes (defined by their actual mating envelope) must be perfectly located if the holes are manufactured at maximum material condition. If the holes are manufactured larger than maximum material condition, then the tolerance zone size increases and the holes do not need to be perfectly located. Ans. Shigley’s MED, 11th edition Chapter 20 Solutions, Page 9/8 This study source was downloaded by 100000755497394 from CourseHero.com on 04-22-2021 22:27:38 GMT -05:00 https://www.coursehero.com/file/59623485/Chapter20-Solutions-11edocx/ Powered by TCPDF (www.tcpdf.org)