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. N1 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 N1 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 2C2 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 C2 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 ˆ2a ˆ2b 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 C2 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 bca acb abc abc bca cab abc The higher order terms in are negligible. Thus, V bca acb abc and, V bca acb abc 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(102) 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 41103 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 7110 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 256 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 5.218.95 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(103) = 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)103 = 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](103)/10 = 4.712 kN/bolt Yielding factor of safety, Eq. (8-28): np S p At CP Fi 650 84.310 3 1.29 0.263 4.712 41.10 Ans. Overload factor of safety, Eq. (8-29): nL S p At Fi CP 650 84.3103 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](103)/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)103 = 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 58.0 103 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 58.0 103 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](103)/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 0.2911.244 9.57 Overload factor of safety, Eq. (8-29): S p At Fi 120 0.106 3 9.57 nL 8.80 Ans. CP 0.2911.244 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)103 = 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](103)/10 = 4.712 kN/bolt Yielding factor of safety, Eq. (8-28) S p At 650 84.310 3 1.29 np CP Fi 0.275 4.712 41.1 Overload factor of safety, Eq. (8-29) S p At Fi 650 84.310 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](103)/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)103 = 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 58.0 103 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](103)/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 FbR 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 FbR 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 Nm 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 35310 3 0.266 4.5 190.6 1.10 Ans. Load factor: From Eq. (8-29) nL S p At Fi CP 600 353103 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(103) = 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 600 245 103 1.07 Ans. np CP Fi 0.228 20 132.3 Load factor: From Eq. (8-29) nL S p At Fi CP 600 245 103 132.3 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(103) = 5.73 kN S p At 380 20.110 3 0.98 np Ans. Yield, Eq. (8-28), CP Fi 0.278 7.5 5.73 S p At Fi 380 20.1103 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 5200.217 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(103) = 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 0.2911.244 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 0.2911.244 0.622 2 0.106 3 0.851 kpsi C Pmax Pmin i 2 At 0.2911.244 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 16.316.3 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(103) = 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 561103 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 561103 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)(103 ) 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 lbfin 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 540C (1000F): 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 425C (800F) Ans. (b) Maximize elongation: Q&T at 650C (1200F) 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 020m 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 020m 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. N1 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. N1 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 = kpsiinm 10-2 dim(ASI) = [dim (S) dim(d m)]SI = MPammm 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 kpsiinm, m = 0.145; what is ASI? ASI = 6.895(25.4)0.145 (201) = 2215 MPammm 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 81103 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 MPammm 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 kpsiinm 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 MPammm 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 1.340 325.1103 psi d 0.0073 (1) A = 201 kpsiinm, 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 kpsiinm, 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 kpsiinm, 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 kpsiinm, 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 kpsiinm, 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 kpsiinm, 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 MPammm, 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 MPammm, 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 79.3103 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 MPammm, 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 MPammm, 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 MPammm, 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 1.152 106 555.8 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): 2.4 103 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 1.135 103 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 2 103 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 kpsiinm, 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 = 0.5s = 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 kpsiinm, 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 kpsiinm, 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 11103 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 103 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 xl 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 7.733107 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 kpsiinm, 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 lx 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 lx T2 T2 1500 x T2 1500 x l Ans. (3) lx 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 lbfin, 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 lbfin T2 = 10(103)(0.583 2)4 = 1 157 lbfin 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)(lx) 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) Nmm 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) lbfin M2 = 33.48(103) +15(4)12 = 34.20(103) lbfin 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 103 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 103 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(104 ) 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 8.06 105 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 11103 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 11103 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 1110 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 1110 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 x0 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 x0 dx 6 EIl d x x0 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 6.04 105 rad Ans. 3 6(207)10 (7 854)(850) The slope magnitude is C 0.00558 2 6.04 105 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 x0 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 11103 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 x0 F b F b 1z 1 b12 l 2 2 z 2 b22 l 2 6 EIl 6 EIl 11103 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 11103 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 11103 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 = 2B1). 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 = 5B1). 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 = 2C). 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 lbfin 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 lbfin. 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 lbfin 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 = 5B1). 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 = 5B2). 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 3.82 105 i 1 0.01528 0.10667 2.26 105 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 1010 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: 0x5 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 5.797 106 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 5.797 106 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 d4 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 6.79 105 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 3.735 104 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 6.792 105 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)105 = 2.264 (105) 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 (105) 5.538 (103) = 5.56 (103) 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 3.734 104 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 103 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 103 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) Nm. 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 721 300 x 1201 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 723 50 x 1203 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 723 50 x 1203 2.622(106) x C1 = 2.622(106) lbfin2 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) lbfin 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 lbfin2. Thus, y 1 70 x 3 4.167 x 4 163.3 x 10 3 2833 x 6 30 10 0.05 3 Ans. 6.667 107 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 (1010 ) Thus y = 2.949 (1010) [ 66.67 x3 + 66.67 x 300 3 + 48(106) x 12.60(109)] yO = 3.72 mm Ans. yx = 550 mm =2.949 (1010) [ 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 xa xb 2 2 dy 1 w w 3 3 EI R1 x 2 xa x b C1 dx 2 6 6 1 w w 4 4 EIy R1 x 3 xa 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 lbfin 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 31 . Thus, 1 1 1 1 M 900 x 90 0 1 x3 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 xl 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 25.3110 30 106 0.1755 30 106 0.4604 EI1 EI 2 0.0102 in Ans. ______________________________________________________________________________ 4-77 I = (0.54)/64 = 3.068 (103) in4, J = 2 I = 6.136 (103) 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 6.136 103 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 3.068 103 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 1.38 105 0.1000 6.71103 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 (103) 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 5.207 104 F 5.207 104 200 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 (103) 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 1.509 107 F 7.112 104 F 4.267 104 Fz 1.019 104 F 1.019 104 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 8.135 104 150 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 1.008 104 F 1.148 10 3 F 3.58 10 7 F 2.994 104 F 3.872 10 5 F 1.2363 10 3 F 2.824 103 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 103 F 4.973 104 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 33.3312 1 6.79 105 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 1.121103 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 1.121103 5.768 107 100 129.2 166.7 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 8.922 104 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 2.264 105 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 5.540 103 in 6 12 6 EI 27 27 1.733 10 0 x 12in yA = (yA)1 + (yA)2 = 2.264 (105) 5.540 (103) = 5.56 (103) 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 183.3 112 3.734 104 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 (104) in Ans. ______________________________________________________________________________ 4-87 Table A-5, EA = 71.7 GPa, k = P / = 10 (103)/(103) = 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 1.473106 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 (103) +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 3.494 103 in (d) yD = (yD)1 + (yD)2 + (yD)3 = 0.0121 + 0 + 0.0035 = 0.0156 in 200 43 3 1.221106 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: 1.276 104 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 2.775 104 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 1.473106 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 8.50 105 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 2.95 107 m 2.95 10 4 mm 6 EA 101.6 10 BD Fz (d) zD = 0.0850 + 0 2.95 (104) = 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 xl 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 2.32 103 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 (103) + 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 Rd1 Q EI Q 0 PR 3 sin 21 2 EI 2 2 /2 0 /2 0 PR 3 PR cos 1 Rd1 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 1.731105 F 7.69110 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 xa 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 8l 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) lbfin2. 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 lbfin 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 1310.112 2 6 24 12 2 25.5 10 6 10.2 1310.112 17 554.59 5.51103 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(106 ) 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)(106 ) 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(106 ) 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 lbfin2. 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) lbfin2 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) Nmm2. 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 4.805 106 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 70.31103 FBE 150C1 C2 1.125 109 (6) B.C 3. At x = 225 mm, FDF 65 Fl 6.246 106 FDF yD 3 AE DF 50.27 207 10 Substituting into Eq. (4) at x = 225 mm, 4.347 109 6.246 106 FDF Simplifying gives 1000 1 3 2253 FBE 225 75 3 6 1 3 RC 225 150 C1 225 C2 6 70.31103 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 70.3110 0 150 1 C1 1.125 10 70.31103 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 ) Nmm , C2 = 7.243 (10 ) Nmm . 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 (108) FB Equation (7) is replaced by EI 2.5 108 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 (103) FA + 3.125 (104) 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 103 3.125 104 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 0.23 +7.619 (103) x 1.410 (103) At x = 0, y = yA = 1.410 (103) m = 1.41 mm Ans. At x = 0.2 m, y = yB = 0.03133(0.2)3 +7.619(103)0.2 1.410(103) = 1.37 (104) 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(103)0.35 1.410(103) = 1.60 (104) 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 (103) m, l / k = 1.030/3.175 (103) = 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 2.887 103 m l / k = 0.8/[2.887(103)] = 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 lbfin 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 wl 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 87 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) 96 1.5 MPa 2 96 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) 65 0.5 MPa 2 65 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 82 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 106 1130 L 60 From Table A-5, v = 0.292 2 v1 0.292(1130) 330 Ans. Ans. Ans. Ans. d 2 d 330 106 (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 653 106 653 L 60 From Table A-5, v = 0.333 1 2 v1 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 2 306.7 106 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. 106 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 72 106 134 MPa Ans. x 2 2 1 v 1 0.333 9 71.7 10 0.000 72 0.333 0.0019 y 106 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. FF 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 3.068 103 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 3.068 103 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 lbfin Mx = 27 in = (1/2)3900(13) = 25 350 lbfin 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 yA 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 yA (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 , y0 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) 103 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 103 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) 103 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 xa 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) (kNm) (kNm) 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 (kNm) 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(106 ) m 4 12 12 25(40)3 25(6)3 Iy 2 2.67(105 ) mm 4 2.67(107 ) 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 (kNm) Mz (kNm) M (kNm) –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 106 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 106 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(105) 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 106 179.2 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 79.3109 / 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)106 0.001185 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 79.3109 551 1012 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), G1c G and 1 are constant, therefore the largest shear stress occurs when c is a maximum. max G1cmax 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 8.35 103 rad/in Ans. 6 Gcmax 11.5 10 (0.125) (a) T1 1GL1c13 3 2GL2 c23 8.35 103 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 86 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 6.17 102 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 12 / 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 12 / 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 12 / 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 and3 = 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 dr / 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) 0.012 (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 6.749 104 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 6.749 104 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(103) lbfs2/in F = (m/2) 2r = [3.425(103)/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) rR 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 Rd /2 1 10 6 / 2 sin 1 sin 30 Rd 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 Rd /2 1 0.5 0.25 / 2 sin 1 sin 30 Rd 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 104 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 a3 1 d1 1 d 2 8 3 10 1 0.292 a3 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 a3 8 1 d1 1 d 2 2 3 20 2 1 0.292 207 000 a3 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 3.517 103 F 1/3 in 1 11 8 3 From Eq. (3-69), 3F 3F 3.860 104 F 1/3 psi pmax 2 2 2 a 2 3.517 103 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 3.306 103 F 1/3 in 1 11 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 3.55110 4 F 1/3 psi 8258F 3.55110 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 2.593 104 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 1.052 103 in 2 40 2F 32 275 psi 32.3 kpsi bl 1.052 103 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 0.21117 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 = ei = 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 hlallow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans. ______________________________________________________________________________ 9-2 Given, b = 2 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. F = 0.707 hlallow = 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 hlallow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans. ______________________________________________________________________________ 9-4 Given, b = 4 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. F = 0.707 hlallow = 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 = 0.707hlall = 0.707(5)[2(50)](88)(103) = 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 = 0.707hlall = 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 = 0.707hlall = 0.707(5)[2(50)](108)(103) = 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 = 0.707hlall = 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 x2 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 x2 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 x2 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 x2 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 x2 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 x2 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 x2 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 x2 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 9.511.326 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 11.1103 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 = 2r3 = 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 Nm, 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 x2 ( 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 kipin. 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 kipin. 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 kipin. 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 Nm 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 3115 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 3115 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 Nm. 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 @400F, 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 12 1 5.7932 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 lbfs2/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 3.006 104 2 34 2 0.5699r 2 F r 2 r 225 r 3.006 104 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 3.006 104 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 lbfin. 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 Nm, 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 / 2 103 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 362.2 109 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 and2 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 21 = 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 21 = 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)103 100(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 (103) 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, yA 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 yA 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 zB 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 Al 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 w111 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 w111 w212 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 ( m111 1/ 2 ) m212 m1 21 ( m2 22 1/ 2 ) 0 Expanding the determinant yields, 2 1 1 2 (m111 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 lbfin, 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 2.51103 68 1.35 105 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 lbfin, 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 106 267 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 3.08 103 68 1.51105 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 3.08 103 120 1.5110 5 120 2.67 108 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 2.25 103 (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 3.08 103 85 1.51105 85 2.67 108 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 3.08 103 85 1.51105 85 2.67 108 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)(1012 )(2)(10 3 ) Fmin 48.3 N 1403 (109 ) 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 3.08 103 64 1.51105 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 2.51103 64 1.35 105 64 2.67 108 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 3.08 103 64 1.51105 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 2.51103 64 1.35 105 64 2.67 108 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 3.08 103 64 1.51105 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 2.51103 64 1.35 105 64 2.67 108 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 0.811.60 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 0.811.60 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 Dd 6 1.2 Tmax f Pmax 0.3(18) 9.72 kip in 4 4 Dd 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 aN 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 aN 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) JP 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 = 70F, 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