MECH 420 Problems Chap 8, Fall 2012

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, dr = d − 1.226 869p dm = d − 0.649 519p ¯d = d − 1.226 869p + d − 0.649 519p 2 = d − 0.938 194p At = π ¯d2 4 = π 4 (d − 0.938 194p)2 Ans. 8-3 From Eq. (c) of Sec. 8-2, P = F tan λ + f 1 − f tan λ T = Pdm 2 = Fdm 2 tan λ + f 1 − f tan λ e = T0 T = Fl/(2π) Fdm/2 1 − f tan λ tan λ + f = tan λ1 − f tan λ tan λ + f Ans. Using f = 0.08, form a table and plot the efficiency curve. λ, deg. e 0 0 10 0.678 20 0.796 30 0.838 40 0.8517 45 0.8519 1 0 50 �, deg. e 5 mm 5 mm 2.5 2.5 2.5 mm 25 mm 5 mm Chapter 8 205 8-4 Given F = 6 kN, l = 5 mm, and dm = 22.5 mm, the torque required to raise the load is found using Eqs. (8-1) and (8-6) TR = 6(22.5) 2 �5 + π(0.08)(22.5) π(22.5) − 0.08(5) � + 6(0.05)(40) 2 = 10.23 + 6 = 16.23 N · m Ans. The torque required to lower the load, from Eqs. (8-2) and (8-6) is TL = 6(22.5) 2 � π(0.08)22.5 − 5 π(22.5) + 0.08(5) � + 6(0.05)(40) 2 = 0.622 + 6 = 6.622 N · m Ans. Since TL is positive, the thread is self-locking. The efficiency is Eq. (8-4): e = 6(5) 2π(16.23) = 0.294 Ans. 8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom seg- ment of the screws must be in compression. Where as 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 75 = 22.9 rev/min (a) The lead is 0.5 in, so the linear speed of the press head is V = 22.9(0.5) = 11.5 in/min Ans. (b) F = 2500 lbf/screw dm = 3 − 0.25 = 2.75 in sec α = 1/cos(29/2) = 1.033 Eq. (8-5): TR = 2500(2.75) 2 �0.5 + π(0.05)(2.75)(1.033) π(2.75) − 0.5(0.05)(1.033) � = 377.6 lbf · in Eq. (8-6): Tc = 2500(0.06)(5/2) = 375 lbf · in Ttotal = 377.6 + 375 = 753 lbf · in/screw Tmotor = 753(2) 75(0.95) = 21.1 lbf · in H = Tn 63 025 = 21.1(1720) 63 025 = 0.58 hp Ans. 206 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 8-7 The force F is perpendicular to the paper. L = 3 − 1 8 − 1 4 − 7 32 = 2.406 in T = 2.406F M = � L − 7 32 � F = � 2.406 − 7 32 � F = 2.188F Sy = 41 kpsi σ = Sy = 32M πd3 = 32(2.188)F π(0.1875)3 = 41 000 F = 12.13 lbf T = 2.406(12.13) = 29.2 lbf · in Ans. (b) Eq. (8-5), 2α = 60◦, l = 1/14 = 0.0714 in, f = 0.075, sec α = 1.155, p = 1/14 in dm = 7 16 − 0.649 519 � 1 14 � = 0.3911 in TR = Fclamp(0.3911) 2 �Num Den � Num = 0.0714 + π(0.075)(0.3911)(1.155) Den = π(0.3911) − 0.075(0.0714)(1.155) T = 0.028 45Fclamp Fclamp = T 0.028 45 = 29.2 0.028 45 = 1030 lbf Ans. (c) The column has one end fixed and the other end pivoted. Base decision on the mean diameter column. Input: C = 1.2, D = 0.391 in, Sy = 41 kpsi, E = 30(106) psi, L = 4.1875 in, k = D/4 = 0.097 75 in, L/k = 42.8. For this J. B. Johnson column, the critical load represents the limiting clamping force for bucking. Thus, Fclamp = Pcr = 4663 lbf. (d) This is a subject for class discussion. 8-8 T = 6(2.75) = 16.5 lbf · in dm = 5 8 − 1 12 = 0.5417 in l = 1 6 = 0.1667 in, α = 29◦ 2 = 14.5◦, sec 14.5◦ = 1.033 1 4 " 3 16 D. " 7 16 " 2.406" 3" Chapter 8 207 Eq. (8-5): T = 0.5417(F/2) �0.1667 + π(0.15)(0.5417)(1.033) π(0.5417) − 0.15(0.1667)(1.033) � = 0.0696F Eq. (8-6): Tc = 0.15(7/16)(F/2) = 0.032 81F Ttotal = (0.0696 + 0.0328)F = 0.1024F F = 16.5 0.1024 = 161 lbf Ans. 8-9 dm = 40 − 3 = 37 mm, l = 2(6) = 12 mm From Eq. (8-1) and Eq. (8-6) TR = 10(37) 2 �12 + π(0.10)(37) π(37) − 0.10(12) � + 10(0.15)(60) 2 = 38.0 + 45 = 83.0 N · m Since n = V/l = 48/12 = 4 rev/s ω = 2πn = 2π(4) = 8π rad/s so the power is H = Tω = 83.0(8π) = 2086 W Ans. 8-10 (a) dm = 36 − 3 = 33 mm, l = p = 6 mm From Eqs. (8-1) and (8-6) T = 33F 2 �6 + π(0.14)(33) π(33) − 0.14(6) � + 0.09(90)F 2 = (3.292 + 4.050)F = 7.34F N · m ω = 2πn = 2π(1) = 2π rad/s H = Tω T = H ω = 3000 2π = 477 N · m F = 477 7.34 = 65.0 kN Ans. (b) e = Fl 2πT = 65.0(6) 2π(477) = 0.130 Ans. 8-11 (a) LT = 2D + 1 4 = 2(0.5) + 0.25 = 1.25 in Ans. (b) From Table A-32 the washer thickness is 0.109 in. Thus, l = 0.5 + 0.5 + 0.109 = 1.109 in Ans. (c) From Table A-31, H = 7 16 = 0.4375 in Ans. 208 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (d) l + H = 1.109 + 0.4375 = 1.5465 in This would be rounded to 1.75 in per Table A-17. The bolt is long enough. Ans. (e) ld = L − LT = 1.75 − 1.25 = 0.500 in Ans. lt = l − ld = 1.109 − 0.500 = 0.609 in Ans. These lengths are needed to estimate bolt spring rate kb. Note: In an analysis problem, you need not know the fastener’s length at the outset, although you can certainly check, if appropriate. 8-12 (a) LT = 2D + 6 = 2(14) + 6 = 34 mm Ans. (b) From Table A-33, the maximum washer thickness is 3.5 mm. Thus, the grip is, l = 14 + 14 + 3.5 = 31.5 mm Ans. (c) From Table A-31, H = 12.8 mm (d) l + H = 31.5 + 12.8 = 44.3 mm Adding one or two threads and rounding up to L = 50 mm. The bolt is long enough. Ans. (e) ld = L − LT = 50 − 34 = 16 mm Ans. lt = l − ld = 31.5 − 16 = 15.5 mm Ans. These lengths are needed to estimate the bolt spring rate kb. 8-13 (a) LT = 2D + 1 4 = 2(0.5) + 0.25 = 1.25 in Ans. (b) l′ > h + d 2 = t1 + d 2 = 0.875 + 0.5 2 = 1.125 in Ans. (c) L > h + 1.5d = t1 + 1.5d = 0.875 + 1.5(0.5) = 1.625 in From Table A-17, this rounds to 1.75 in. The cap screw is long enough. Ans. (d) ld = L − LT = 1.75 − 1.25 = 0.500 in Ans. lt = l′ − ld = 1.125 − 0.5 = 0.625 in Ans. 8-14 (a) LT = 2(12) + 6 = 30 mm Ans. (b) l′ = h + d 2 = t1 + d 2 = 20 + 12 2 = 26 mm Ans. (c) L > h + 1.5d = t1 + 1.5d = 20 + 1.5(12) = 38 mm This rounds to 40 mm (Table A-17). The fastener is long enough. Ans. (d) ld = L − LT = 40 − 30 = 10 mm Ans. lT = l′ − ld = 26 − 10 = 16 mm Ans. Chapter 8 209 8-15 (a) Ad = 0.7854(0.75)2 = 0.442 in2 Atube = 0.7854(1.1252 − 0.752) = 0.552 in2 kb = Ad E grip = 0.442(30)(106) 13 = 1.02(106) lbf/in Ans. km = AtubeE 13 = 0.552(30)(106) 13 = 1.27(106) lbf/in Ans. C = 1.02 1.02 + 1.27 = 0.445 Ans. (b) δ = 1 16 · 1 3 = 1 48 = 0.020 83 in |δb| = �|P|l AE � b = (13 − 0.020 83) 0.442(30)(106) |P| = 9.79(10−7)|P| in |δm| = �|P|l AE � m = |P|(13) 0.552(30)(106) = 7.85(10−7)|P| in |δb| + |δm| = δ = 0.020 83 9.79(10−7)|P| + 7.85(10−7)|P| = 0.020 83 Fi = |P| = 0.020 83 9.79(10−7) + 7.85(10−7) = 11 810 lbf Ans. (c) At opening load P0 9.79(10−7)P0 = 0.020 83 P0 = 0.020 83 9.79(10−7) = 21 280 lbf Ans. As a check use Fi = (1 − C)P0 P0 = Fi 1 − C = 11 810 1 − 0.445 = 21 280 lbf 8-16 The movement is known at one location when the nut is free to turn δ = pt = t/N Letting Nt represent the turn of the nut from snug tight, Nt = θ/360◦ and δ = Nt/N. The elongation of the bolt δb is δb = Fi kb The advance of the nut along the bolt is the algebraic sum of |δb| and |δm| Original bolt Nut advance A A � �m �b Equilibrium Grip 210 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design |δb| + |δm| = Nt N Fi kb + Fi km = Nt N Nt = N Fi � 1 kb + 1 km � = �kb + km kbkm � Fi N = θ 360◦ Ans. As a check invert Prob. 8-15. What Turn-of-Nut will induce Fi = 11 808 lbf? Nt = 16(11 808) � 1 1.02(106) + 1 1.27(106) � = 0.334 turns .= 1/3 turn (checks) The relationship between the Turn-of-Nut method and the Torque Wrench method is as follows. Nt = �kb + km kbkm � Fi N (Turn-of-Nut) T = K Fid (Torque Wrench) Eliminate Fi Nt = �kb + km kbkm � NT Kd = θ 360◦ Ans. 8-17 (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. 8-16, t = N Fi � 1 kb + 1 km � = 16(14.4)(103) � 1 5.21(106) + 1 8.95(106) � = 0.132 turns = 47.5◦ Ans. Bolt group is (1.5)/(5/8) = 2.4 diameters. Answer is lower than RB&W recommendations. (b) From Ex. 8-5, Fi = 14.4 kip, kb = 6.78 Mlbf/in, and km = 17.4 Mlbf/in T = 0.2(14.4)(103)(5/8) = 1800 lbf · in Ans. t = 11(14.4)(103) � 1 6.78(106) + 1 17.4(106) � = 0.0325 = 11.7◦ Ans. Again lower than RB&W. 8-18 From Eq. (8-22) for the conical frusta, with d/l = 0.5 km Ed ���� (d/l)=0.5 = 0.5774π 2 ln {5[0.5774 + 0.5(0.5)]/[0.5774 + 2.5(0.5)]} = 1.11 Chapter 8 211 Eq. (8-23), from the Wileman et al. finite element study, using the general expression, km Ed ���� (d/l)=0.5 = 0.789 52 exp[0.629 14(0.5)] = 1.08 8-19 For cast iron, from Table 8-8: A = 0.778 71, B = 0.616 16, E = 14.5 Mpsi km = 14.5(106)(0.625)(0.778 71) exp � 0.616 160.625 1.5 � = 9.12(106) lbf/in This member’s spring rate applies to both members. We need km for the upper member which represents half of the joint. kci = 2km = 2[9.12(106)] = 18.24(106) lbf/in For steel from Table 8-8: A = 0.787 15, B = 0.628 73, E = 30 Mpsi km = 30(106)(0.625)(0.787 15) exp � 0.628 730.625 1.5 � = 19.18(106) lbf/in ksteel = 2km = 2(19.18)(106) = 38.36(106) lbf/in For springs in series 1 km = 1 kci + 1 ksteel = 1 18.24(106) + 1 38.36(106) km = 12.4(106) lbf/in Ans. 8-20 The external tensile load per bolt is P = 1 10 �π 4 � (150)2(6)(10−3) = 10.6 kN Also, l = 40 mm and from Table A-31, for d = 12 mm, H = 10.8 mm. No washer is specified. LT = 2D + 6 = 2(12) + 6 = 30 mm l + H = 40 + 10.8 = 50.8 mm Table A-17: L = 60 mm ld = 60 − 30 = 30 mm lt = 45 − 30 = 15 mm Ad = π(12)2 4 = 113 mm2 Table 8-1: At = 84.3 mm2 Eq. (8-17): kb = 113(84.3)(207) 113(15) + 84.3(30) = 466.8 MN/m Steel: Using Eq. (8-23) for A = 0.787 15, B = 0.628 73 and E = 207 GPa 212 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Eq. (8-23): km = 207(12)(0.787 15) exp[(0.628 73)(12/40)] = 2361 MN/m ks = 2km = 4722 MN/m Cast iron: A = 0.778 71, B = 0.616 16, E = 100 GPa km = 100(12)(0.778 71) exp[(0.616 16)(12/40)] = 1124 MN/m kci = 2km = 2248 MN/m 1 km = 1 ks + 1 kci ⇒ km = 1523 MN/m C = 466.8 466.8 + 1523 = 0.2346 Table 8-1: At = 84.3 mm2, Table 8-11, Sp = 600 MPa Eqs. (8-30) and (8-31): Fi = 0.75(84.3)(600)(10−3) = 37.9 kN Eq. (8-28): n = Sp At − Fi C P = 600(10−3)(84.3) − 37.9 0.2346(10.6) = 5.1 Ans. 8-21 Computer programs will vary. 8-22 D3 = 150 mm, A = 100 mm, B = 200 mm, C = 300 mm, D = 20 mm, E = 25 mm. ISO 8.8 bolts: d = 12 mm, p = 1.75 mm, coarse pitch of p = 6 MPa. P = 1 10 �π 4 � (1502)(6)(10−3) = 10.6 kN/bolt l = D + E = 20 + 25 = 45 mm LT = 2D + 6 = 2(12) + 6 = 30 mm Table A-31: H = 10.8 mm l + H = 45 + 10.8 = 55.8 mm Table A-17: L = 60 mm ld = 60 − 30 = 30 mm, lt = 45 − 30 = 15 mm, Ad = π(122/4) = 113 mm2 Table 8-1: At = 84.3 mm2 2.5 dw D1 22.5 25 45 20 Chapter 8 213 Eq. (8-17): kb = 113(84.3)(207) 113(15) + 84.3(30) = 466.8 MN/m There are three frusta: dm = 1.5(12) = 18 mm D1 = (20 tan 30◦)2 + dw = (20 tan 30◦)2 + 18 = 41.09 mm Upper Frustum: t = 20 mm, E = 207 GPa, D = 1.5(12) = 18 mm Eq. (8-20): k1 = 4470 MN/m Central Frustum: t = 2.5 mm, D = 41.09 mm, E = 100 GPa (Table A-5) ⇒ k2 = 52 230 MN/m Lower Frustum: t = 22.5 mm, E = 100 GPa, D = 18 mm ⇒ k3 = 2074 MN/m From Eq. (8-18): km = [(1/4470) + (1/52 230) + (1/2074)]−1 = 1379 MN/m Eq. (e), p. 421: C = 466.8 466.8 + 1379 = 0.253 Eqs. (8-30) and (8-31): Fi = K Fp = K At Sp = 0.75(84.3)(600)(10−3) = 37.9 kN Eq. (8-28): n = Sp At − Fi C P = 600(10−3)(84.3) − 37.9 0.253(10.6) = 4.73 Ans. 8-23 P = 1 8 �π 4 � (1202)(6)(10−3) = 8.48 kN From Fig. 8-21, t1 = h = 20 mm and t2 = 25 mm l = 20 + 12/2 = 26 mm t = 0 (no washer), LT = 2(12) + 6 = 30 mm L > h + 1.5d = 20 + 1.5(12) = 38 mm Use 40 mm cap screws. ld = 40 − 30 = 10 mm lt = l − ld = 26 − 10 = 16 mm Ad = 113 mm2, At = 84.3 mm2 Eq. (8-17): kb = 113(84.3)(207) 113(16) + 84.3(10) = 744 MN/m Ans. dw = 1.5(12) = 18 mm D = 18 + 2(6)(tan 30) = 24.9 mm l � 26 t2 � 25 h � 20 13 13 7 6 D 12 214 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design From Eq. (8-20): Top frustum: D = 18, t = 13, E = 207 GPa ⇒ k1 = 5316 MN/m Mid-frustum: t = 7, E = 207 GPa, D = 24.9 mm ⇒ k2 = 15 620 MN/m Bottom frustum: D = 18, t = 6, E = 100 GPa ⇒ k3 = 3887 MN/m km = 1 (1/5316) + (1/55 620) + (1/3887) = 2158 MN/m Ans. C = 744 744 + 2158 = 0.256 Ans. From Prob. 8-22, Fi = 37.9 kN n = Sp At − Fi C P = 600(0.0843) − 37.9 0.256(8.48) = 5.84 Ans. 8-24 Calculation of bolt stiffness: H = 7/16 in LT = 2(1/2) + 1/4 = 1 1/4 in l = 1/2 + 5/8 + 0.095 = 1.22 in L > 1.125 + 7/16 + 0.095 = 1.66 in Use L = 1.75 in ld = L − LT = 1.75 − 1.25 = 0.500 in lt = 1.125 + 0.095 − 0.500 = 0.72 in Ad = π(0.502)/4 = 0.1963 in2 At = 0.1419 in2 (UNC) kt = At E lt = 0.1419(30) 0.72 = 5.9125 Mlbf/in kd = Ad E ld = 0.1963(30) 0.500 = 11.778 Mlbf/in kb = 1 (1/5.9125) + (1/11.778) = 3.936 Mlbf/in Ans. 5 8 1 2 0.095 3 4 1.454 1.327 3 4 0.860 1.22 0.61 Chapter 8 215 Member stiffness for four frusta and joint constant C using Eqs. (8-20) and (e). Top frustum: D = 0.75, t = 0.5, d = 0.5, E = 30 ⇒ k1 = 33.30 Mlbf/in 2nd frustum: D = 1.327, t = 0.11, d = 0.5, E = 14.5 ⇒ k2 = 173.8 Mlbf/in 3rd frustum: D = 0.860, t = 0.515, E = 14.5 ⇒ k3 = 21.47 Mlbf/in Fourth frustum: D = 0.75, t = 0.095, d = 0.5, E = 30 ⇒ k4 = 97.27 Mlbf/in km = � 4 � i=1 1/ki �−1 = 10.79 Mlbf/in Ans. C = 3.94/(3.94 + 10.79) = 0.267 Ans. 8-25 kb = At E l = 0.1419(30) 0.845 = 5.04 Mlbf/in Ans. From Fig. 8-21, h = 1 2 + 0.095 = 0.595 in l = h + d 2 = 0.595 + 0.5 2 = 0.845 D1 = 0.75 + 0.845 tan 30◦ = 1.238 in l/2 = 0.845/2 = 0.4225 in From Eq. (8-20): Frustum 1: D = 0.75, t = 0.4225 in, d = 0.5 in, E = 30 Mpsi ⇒ k1 = 36.14 Mlbf/in Frustum 2: D = 1.018 in, t = 0.1725 in, E = 70 Mpsi, d = 0.5 in ⇒ k2 = 134.6 Mlbf/in Frustum 3: D = 0.75, t = 0.25 in, d = 0.5 in, E = 14.5 Mpsi ⇒ k3 = 23.49 Mlbf/in km = 1 (1/36.14) + (1/134.6) + (1/23.49) = 12.87 Mlbf/in Ans. C = 5.04 5.04 + 12.87 = 0.281 Ans. 0.095" 0.1725" 0.25" 0.595" 0.5" 0.625" 0.4225" 0.845" 0.75" 1.018" 1.238" Steel Cast iron 216 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 8-26 Refer to Prob. 8-24 and its solution.Additional information: A = 3.5 in, Ds = 4.25 in, static pressure 1500 psi, Db = 6 in,C (joint constant) = 0.267, ten SAE grade 5 bolts. P = 1 10 π(4.252) 4 (1500) = 2128 lbf From Tables 8-2 and 8-9, At = 0.1419 in2 Sp = 85 000 psi Fi = 0.75(0.1419)(85) = 9.046 kip From Eq. (8-28), n = Sp At − Fi C P = 85(0.1419) − 9.046 0.267(2.128) = 5.31 Ans. 8-27 From Fig. 8-21, t1 = 0.25 in h = 0.25 + 0.065 = 0.315 in l = h + (d/2) = 0.315 + (3/16) = 0.5025 in D1 = 1.5(0.375) + 0.577(0.5025) = 0.8524 in D2 = 1.5(0.375) = 0.5625 in l/2 = 0.5025/2 = 0.251 25 in Frustum 1: Washer E = 30 Mpsi, t = 0.065 in, D = 0.5625 in k = 78.57 Mlbf/in (by computer) Frustum 2: Cap portion E = 14 Mpsi, t = 0.186 25 in D = 0.5625 + 2(0.065)(0.577) = 0.6375 in k = 23.46 Mlbf/in (by computer) Frustum 3: Frame and Cap E = 14 Mpsi, t = 0.251 25 in, D = 0.5625 in k = 14.31 Mlbf/in (by computer) km = 1 (1/78.57) + (1/23.46) + (1/14.31) = 7.99 Mlbf/in Ans. 0.8524" 0.5625" 0.25125" 0.8524" 0.6375" 0.18625" 0.5625" 0.6375" 0.065" Chapter 8 217 For the bolt, LT = 2(3/8) + (1/4) = 1 in. So the bolt is threaded all the way. Since At = 0.0775 in2 kb = 0.0775(30) 0.5025 = 4.63 Mlbf/in Ans. 8-28 (a) F′ b = RF′ b,max sin θ Half of the external moment is contributed by the line load in the interval 0 ≤ θ ≤ π. M 2 = � π 0 F′ bR2 sin θ dθ = � π 0 F′ b, maxR2 sin2 θ dθ M 2 = π 2 F′ b, maxR2 from which F′ b,max = M π R2 Fmax = � φ2 φ1 F′ bR sin θ dθ = M π R2 � φ2 φ1 R sin θ dθ = M π R(cos φ1 − cos φ2) Noting φ1 = 75◦, φ2 = 105◦ Fmax = 12 000 π(8/2)(cos 75◦ − cos 105◦) = 494 lbf Ans. (b) Fmax = F′ b, maxR�φ = M π R2(R) �2π N � = 2M RN Fmax = 2(12 000) (8/2)(12) = 500 lbf Ans. (c) F = Fmax sin θ M = 2FmaxR[(1) sin2 90◦ + 2 sin2 60◦ + 2 sin2 30◦ + (1) sin2(0)] = 6FmaxR from which Fmax = M 6R = 12 000 6(8/2) = 500 lbf Ans. The simple general equation resulted from part (b) Fmax = 2M RN 8-29 (a) Table 8-11: Sp = 600 MPa Eq. (8-30): Fi = 0.9At Sp = 0.9(245)(600)(10−3) = 132.3 kN Table (8-15): K = 0.18 Eq. (8-27) T = 0.18(132.3)(20) = 476 N · m Ans. 218 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) Washers:t = 3.4 mm, d = 20 mm, D = 30 mm, E = 207 GPa ⇒ k1 = 42 175 MN/m Cast iron: t = 20 mm, d = 20 mm, D = 30 + 2(3.4) tan 30◦ = 33.93 mm, E = 135 GPa ⇒ k2 = 7885 MN/m Steel: t = 20 mm, d = 20 mm, D = 33.93 mm, E = 207 GPa ⇒ k3 = 12 090 MN/m km = (2/42 175 + 1/7885 + 1/12 090)−1 = 3892 MN/m Bolt: l = 46.8 mm. Nut: H = 18 mm. L > 46.8 + 18 = 64.8 mm. Use L = 80 mm. LT = 2(20) + 6 = 46 mm, ld = 80 − 46 = 34 mm, lt = 46.8 − 34 = 12.8 mm, At = 245 mm2, Ad = π202/4 = 314.2 mm2 kb = Ad At E Adlt + Atld = 314.2(245)(207) 314.2(12.8) + 245(34) = 1290 MN/m C = 1290/(1290 + 3892) = 0.2489, Sp = 600 MPa, Fi = 132.3 kN n = Sp At − Fi C(P/N) = 600(0.245) − 132.3 0.2489(15/4) = 15.7 Ans. Bolts are a bit oversized for the load. 8-30 (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 Ad = π(20)2/4 = 314.2 mm2 Fp = 245(0.600) = 147 kN Fi = 0.90Fp = 0.90(147) = 132.3 kN T = 0.18(132.3)(20) = 476 N · m Ans. (b) L ≥ l + H = 48 + 18 = 66 mm. Therefore, set 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 14 Not to scale 80 48 grip 46 34 Chapter 8 219 kb = Ad At E Adlt + Atld = 314.2(245)(207) 314.2(14) + 245(34) = 1251.9 MN/m Use Wileman et al. Eq. (8-23) A = 0.787 15, B = 0.628 73 km Ed = A exp � Bd LG � = 0.787 15 exp � 0.628 73 �20 48 �� = 1.0229 km = 1.0229(207)(20) = 4235 MN/m C = 1251.9 1251.9 + 4235 = 0.228 Bolts carry 0.228 of the external load; members carry 0.772 of the external load. Ans. Thus, the actual loads are Fb = C P + Fi = 0.228(20) + 132.3 = 136.9 kN Fm = (1 − C)P − Fi = (1 − 0.228)20 − 132.3 = −116.9 kN 8-31 Given pmax = 6 MPa, pmin = 0 and from Prob. 8-20 solution, C = 0.2346, Fi = 37.9 kN, At = 84.3 mm2. For 6 MPa, P = 10.6 kN per bolt σi = Fi At = 37.9(103) 84.3 = 450 MPa Eq. (8-35): σa = C P 2At = 0.2346(10.6)(103) 2(84.3) = 14.75 MPa σm = σa + σi = 14.75 + 450 = 464.8 MPa (a) Goodman Eq. (8-40) for 8.8 bolts with Se = 129 MPa, Sut = 830 MPa Sa = Se(Sut − σi) Sut + Se = 129(830 − 450) 830 + 129 = 51.12 MPa n f = Sa σa = 51.12 14.75 = 3.47 Ans. 24 24 30 30 220 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) Gerber Eq. (8-42) Sa = 1 2Se � Sut � S2ut + 4Se(Se + σi) − S2 ut − 2σi Se � = 1 2(129) � 830 � 8302 + 4(129)(129 + 450) − 8302 − 2(450)(129) � = 76.99 MPa n f = 76.99 14.75 = 5.22 Ans. (c) ASME-elliptic Eq. (8-43) with Sp = 600 MPa Sa = Se S2p + S2e � Sp � S2p + S2e − σ 2 i − σi Se � = 129 6002 + 1292 � 600 � 6002 + 1292 − 4502 − 450(129) � = 65.87 MPa n f = 65.87 14.75 = 4.47 Ans. 8-32 P = pA N = π D2 p 4N = π(0.92)(550) 4(36) = 9.72 kN/bolt Table 8-11: Sp = 830 MPa, Sut = 1040 MPa, Sy = 940 MPa Table 8-1: At = 58 mm2 Ad = π(102)/4 = 78.5 mm2 l = D + E = 20 + 25 = 45 mm LT = 2(10) + 6 = 26 mm Table A-31: H = 8.4 mm L ≥ l + H = 45 + 8.4 = 53.4 mm Choose L = 60 mm from Table A-17 ld = L − LT = 60 − 26 = 34 mm lt = l − ld = 45 − 34 = 11 mm kb = Ad At E Adlt + Atld = 78.5(58)(207) 78.5(11) + 58(34) = 332.4 MN/m 2.5 22.5 22.5 25 20 15 10 Chapter 8 221 Frustum 1: Top, E = 207, t = 20 mm, d = 10 mm, D = 15 mm k1 = 0.5774π(207)(10) ln ��1.155(20) + 15 − 10 1.155(20) + 15 + 10 � �15 + 10 15 − 10 �� = 3503 MN/m Frustum 2: Middle, E = 96 GPa, D = 38.09 mm, t = 2.5 mm, d = 10 mm k2 = 0.5774π(96)(10) ln ��1.155(2.5) + 38.09 − 10 1.155(2.5) + 38.09 + 10 � �38.09 + 10 38.09 − 10 �� = 44 044 MN/m could be neglected due to its small influence on km. Frustum 3: Bottom, E = 96 GPa, t = 22.5 mm, d = 10 mm, D = 15 mm k3 = 0.5774π(96)(10) ln ��1.155(22.5) + 15 − 10 1.155(22.5) + 15 + 10 � �15 + 10 15 − 10 �� = 1567 MN/m km = 1 (1/3503) + (1/44 044) + (1/1567) = 1057 MN/m C = 332.4 332.4 + 1057 = 0.239 Fi = 0.75At Sp = 0.75(58)(830)(10−3) = 36.1 kN Table 8-17: Se = 162 MPa σi = Fi At = 36.1(103) 58 = 622 MPa (a) Goodman Eq. (8-40) Sa = Se(Sut − σi) Sut + Se = 162(1040 − 622) 1040 + 162 = 56.34 MPa n f = 56.34 20 = 2.82 Ans. (b) Gerber Eq. (8-42) Sa = 1 2Se � Sut � S2ut + 4Se(Se + σi) − S2 ut − 2σi Se � = 1 2(162) � 1040 � 10402 + 4(162)(162 + 622) − 10402 − 2(622)(162) � = 86.8 MPa 222 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design σa = C P 2At = 0.239(9.72)(103) 2(58) = 20 MPa n f = Sa σa = 86.8 20 = 4.34 Ans. (c) ASME elliptic Sa = Se S2p + S2e � Sp � S2p + S2e − σ 2 i − σi Se � = 162 8302 + 1622 � 830 � 8302 + 1622 − 6222 − 622(162) � = 84.90 MPa n f = 84.90 20 = 4.24 Ans. 8-33 Let the repeatedly-applied load be designated as P. From Table A-22, Sut = 93.7 kpsi. Referring to the Figure of Prob. 3-74, the following notation will be used for the radii of Section AA. ri = 1 in, ro = 2 in, rc = 1.5 in From Table 4-5, with R = 0.5 in rn = 0.52 2 � 1.5 − √ 1.52 − 0.52� = 1.457 107 in e = rc − rn = 1.5 − 1.457 107 = 0.042 893 in co = ro − rn = 2 − 1.457 109 = 0.542 893 in ci = rn − ri = 1.457 107 − 1 = 0.457 107 in A = π(12)/4 = 0.7854 in2 If P is the maximum load M = Prc = 1.5P σi = P A � 1 + rcci eri � = P 0.7854 � 1 + 1.5(0.457) 0.0429(1) � = 21.62P σa = σm = σi 2 = 21.62P 2 = 10.81P (a) Eye: Section AA ka = 14.4(93.7)−0.718 = 0.553 de = 0.37d = 0.37(1) = 0.37 in kb = �0.37 0.30 �−0.107 = 0.978 kc = 0.85 S′ e = 0.5(93.7) = 46.85 kpsi Se = 0.553(0.978)(0.85)(46.85) = 21.5 kpsi Chapter 8 223 Since no stress concentration exists, use a load line slope of 1. From Table 7-10 for Gerber Sa = 93.72 2(21.5) ⎡ ⎣−1 + � 1 + �2(21.5) 93.7 �2 ⎤ ⎦ = 20.47 kpsi Note the mere 5 percent degrading of Se in Sa n f = Sa σa = 20.47(103) 10.81P = 1894 P Thread: Die cut. Table 8-17 gives 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 Table 8-2: At = 0.663 in2 σ = P/At = P/0.663 = 1.51P σa = σm = σ/2 = 1.51P/2 = 0.755P From Table 7-10, Gerber Sa = 1202 2(14.7) ⎡ ⎣−1 + � 1 + �2(14.7) 120 �2 ⎤ ⎦ = 14.5 kpsi n f = Sa σa = 14 500 0.755P = 19 200 P Comparing 1894/P with 19 200/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 is a poor cross section for a curved bar in bending because the bulk of the mate- rial is located where the stress is small). Ans. (c) For n f = 2 P = 1894 2 = 947 lbf, max. load Ans. 8-34 (a) L ≥ 1.5 + 2(0.134) + 41 64 = 2.41 in. Use L = 21 2 in Ans. (b) Four frusta: Two washers and two members 1.125" D1 0.134" 1.280" 0.75" 224 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Washer: E = 30 Mpsi, t = 0.134 in, D = 1.125 in, d = 0.75 in Eq. (8-20): k1 = 153.3 Mlbf/in Member: E = 16 Mpsi, t = 0.75 in, D = 1.280 in, d = 0.75 in Eq. (8-20): k2 = 35.5 Mlbf/in km = 1 (2/153.3) + (2/35.5) = 14.41 Mlbf/in Ans. Bolt: LT = 2(3/4) + 1/4 = 13/4 in l = 2(0.134) + 2(0.75) = 1.768 in ld = L − LT = 2.50 − 1.75 = 0.75 in lt = l − ld = 1.768 − 0.75 = 1.018 in At = 0.373 in2 (Table 8-2) Ad = π(0.75)2/4 = 0.442 in2 kb = Ad At E Adlt + Atld = 0.442(0.373)(30) 0.442(1.018) + 0.373(0.75) = 6.78 Mlbf/in Ans. C = 6.78 6.78 + 14.41 = 0.320 Ans. (c) From Eq. (8-40), Goodman with Se = 18.6 kpsi, Sut = 120 kpsi Sa = 18.6[120 − (25/0.373)] 120 + 18.6 = 7.11 kpsi The stress components are σa = C P 2At = 0.320(6) 2(0.373) = 2.574 kpsi σm = σa + Fi At = 2.574 + 25 0.373 = 69.6 kpsi n f = Sa σa = 7.11 2.574 = 2.76 Ans. (d) Eq. (8-42) for Gerber Sa = 1 2(18.6) � 120 � 1202 + 4(18.6) � 18.6 + 25 0.373 � − 1202 − 2 � 25 0.373 � 18.6 � = 10.78 kpsi n f = 10.78 2.574 = 4.19 Ans. (e) nproof = 85 2.654 + 69.8 = 1.17 Ans. Chapter 8 225 8-35 (a) Table 8-2: At = 0.1419 in2 Table 8-9: Sp = 85 kpsi, Sut = 120 kpsi Table 8-17: Se = 18.6 kpsi Fi = 0.75At Sp = 0.75(0.1419)(85) = 9.046 kip C = 4.94 4.94 + 15.97 = 0.236 σa = C P 2At = 0.236P 2(0.1419) = 0.832P kpsi Eq. (8-40) for Goodman criterion Sa = 18.6(120 − 9.046/0.1419) 120 + 18.6 = 7.55 kpsi n f = Sa σa = 7.55 0.832P = 2 ⇒ P = 4.54 kip Ans. (b) Eq. (8-42) for Gerber criterion Sa = 1 2(18.6) � 120 � 1202 + 4(18.6) � 18.6 + 9.046 0.1419 � − 1202 − 2 � 9.046 0.1419 � 18.6 � = 11.32 kpsi n f = Sa σa = 11.32 0.832P = 2 From which P = 11.32 2(0.832) = 6.80 kip Ans. (c) σa = 0.832P = 0.832(6.80) = 5.66 kpsi σm = Sa + σa = 11.32 + 63.75 = 75.07 kpsi Load factor, Eq. (8-28) n = Sp At − Fi C P = 85(0.1419) − 9.046 0.236(6.80) = 1.88 Ans. Separation load factor, Eq. (8-29) n = Fi (1 − C)P = 9.046 6.80(1 − 0.236) = 1.74 Ans. 8-36 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 226 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Coarse thread, UNC Fi = 0.75(0.969)(74) = 53.78 kip σi = Fi At = 53.78 0.969 = 55.5 kpsi σa = C P 2At = 0.30P 2(0.969) = 0.155P kpsi Eq. (8-42): Sa = 1 2(16.3) � 105 � 1052 + 4(16.3)(16.3 + 55.5) − 1052 − 2(55.5)(16.3) � = 9.96 kpsi n f = Sa σa = 9.96 0.155P = 2 From which P = 9.96 0.155(2) = 32.13 kip Ans. Fine thread, UNF Fi = 0.75(1.073)(74) = 59.55 kip σi = 59.55 1.073 = 55.5 kpsi σa = 0.32P 2(1.073) = 0.149P kpsi Sa = 9.96 (as before) n f = Sa σa = 9.96 0.149P = 2 From which P = 9.96 0.149(2) = 33.42 kip Ans. Percent improvement 33.42 − 32.13 32.13 (100) .= 4% Ans. 8-37 For a M 30 × 3.5 ISO 8.8 bolt with P = 80 kN/bolt and C = 0.33 Table 8-1: At = 561 mm2 Table 8-11: Sp = 600 MPa Sut = 830 MPa Table 8-17: Se = 129 MPa Chapter 8 227 Fi = 0.75(561)(10−3)(600) = 252.45 kN σi = 252.45(10−3) 561 = 450 MPa σa = C P 2At = 0.33(80)(103) 2(561) = 23.53 MPa Eq. (8-42): Sa = 1 2(129) � 830 � 8302 + 4(129)(129 + 450) − 8302 − 2(450)(129) � = 77.0 MPa Fatigue factor of safety n f = Sa σa = 77.0 23.53 = 3.27 Ans. Load factor from Eq. (8-28), n = Sp At − Fi C P = 600(10−3)(561) − 252.45 0.33(80) = 3.19 Ans. Separation load factor from Eq. (8-29), n = Fi (1 − C)P = 252.45 (1 − 0.33)(80) = 4.71 Ans. 8-38 (a) Table 8-2: At = 0.0775 in2 Table 8-9: Sp = 85 kpsi, Sut = 120 kpsi Table 8-17: Se = 18.6 kpsi Unthreaded grip kb = Ad E l = π(0.375)2(30) 4(13.5) = 0.245 Mlbf/in per bolt Ans. Am = π 4 [(D + 2t)2 − D2] = π 4 (4.752 − 42) = 5.154 in2 km = Am E l = 5.154(30) 12 �1 6 � = 2.148 Mlbf/in/bolt. Ans. (b) Fi = 0.75(0.0775)(85) = 4.94 kip σi = 0.75(85) = 63.75 kpsi P = pA = 2000 6 �π 4 (4)2 � = 4189 lbf/bolt C = 0.245 0.245 + 2.148 = 0.102 σa = C P 2At = 0.102(4.189) 2(0.0775) = 2.77 kpsi 228 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Eq. (8-40) for Goodman Sa = 18.6(120 − 63.75) 120 + 18.6 = 7.55 kpsi n f = Sa σa = 7.55 2.77 = 2.73 Ans. (c) From Eq. (8-42) for Gerber fatigue criterion, Sa = 1 2(18.6) � 120 � 1202 + 4(18.6)(18.6 + 63.75) − 1202 − 2(63.75)(18.6) � = 11.32 kpsi n f = Sa σa = 11.32 2.77 = 4.09 Ans. (d) Pressure causing joint separation from Eq. (8-29) n = Fi (1 − C)P = 1 P = Fi 1 − C = 4.94 1 − 0.102 = 5.50 kip p = P A = 5500 π(42)/46 = 2626 psi Ans. 8-39 This analysis is important should the initial bolt tension fail. Members: Sy = 71 kpsi, Ssy = 0.577(71) = 41.0 kpsi. Bolts: SAE grade 8, Sy = 130 kpsi, Ssy = 0.577(130) = 75.01 kpsi Shear in bolts As = 2 �π(0.3752) 4 � = 0.221 in2 Fs = AsSsy n = 0.221(75.01) 3 = 5.53 kip Bearing on bolts Ab = 2(0.375)(0.25) = 0.188 in2 Fb = AbSyc n = 0.188(130) 2 = 12.2 kip Bearing on member Fb = 0.188(71) 2.5 = 5.34 kip Tension of members At = (1.25 − 0.375)(0.25) = 0.219 in2 Ft = 0.219(71) 3 = 5.18 kip F = min(5.53, 12.2, 5.34, 5.18) = 5.18 kip Ans. The tension in the members controls the design. Chapter 8 229 8-40 Members: Sy = 32 kpsi Bolts: Sy = 92 kpsi, Ssy = (0.577)92 = 53.08 kpsi Shear of bolts As = 2 �π(0.375)2 4 � = 0.221 in2 τ = Fs As = 4 0.221 = 18.1 kpsi n = Ssy τ = 53.08 18.1 = 2.93 Ans. Bearing on bolts Ab = 2(0.25)(0.375) = 0.188 in2 σb = −4 0.188 = −21.3 kpsi n = Sy |σb| = 92 |−21.3| = 4.32 Ans. Bearing on members n = Syc |σb| = 32 |−21.3| = 1.50 Ans. Tension of members At = (2.375 − 0.75)(1/4) = 0.406 in2 σt = 4 0.406 = 9.85 kpsi n = Sy At = 32 9.85 = 3.25 Ans. 8-41 Members: Sy = 71 kpsi Bolts: Sy = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear of bolts F = Ssy A/n Fs = 53.08(2)(π/4)(7/8)2 1.8 = 35.46 kip Bearing on bolts Fb = 2(7/8)(3/4)(92) 2.2 = 54.89 kip Bearing on members Fb = 2(7/8)(3/4)(71) 2.4 = 38.83 kip 230 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Tension in members Ft = (3 − 0.875)(3/4)(71) 2.6 = 43.52 kip F = min(35.46, 54.89, 38.83, 43.52) = 35.46 kip Ans. 8-42 Members: Sy = 47 kpsi Bolts: Sy = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear of bolts Ad = π(0.75)2 4 = 0.442 in2 τs = 20 3(0.442) = 15.08 kpsi n = Ssy τs = 53.08 15.08 = 3.52 Ans. Bearing on bolt σb = − 20 3(3/4)(5/8) = −14.22 kpsi n = −Sy σb = − � 92 −14.22 � = 6.47 Ans. Bearing on members σb = − F Ab = − 20 3(3/4)(5/8) = −14.22 kpsi n = −Sy σb = − 47 14.22 = 3.31 Ans. Tension on members σt = F A = 20 (5/8)[7.5 − 3(3/4)] = 6.10 kpsi n = Sy σt = 47 6.10 = 7.71 Ans. 8-43 Members: Sy = 57 kpsi Bolts: Sy = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear of bolts As = 3 �π(3/8)2 4 � = 0.3313 in2 τs = F A = 5.4 0.3313 = 16.3 kpsi n = Ssy τs = 53.08 16.3 = 3.26 Ans. Chapter 8 231 Bearing on bolt Ab = 3 �3 8 � � 5 16 � = 0.3516 in2 σb = − F Ab = − 5.4 0.3516 = −15.36 kpsi n = −Sy σb = − � 92 −15.36 � = 5.99 Ans. Bearing on members Ab = 0.3516 in2 (From bearing on bolt calculations) σb = −15.36 kpsi (From bearing on bolt calculations) n = −Sy σb = − � 57 −15.36 � = 3.71 Ans. Tension in members Failure across two bolts A = 5 16 � 23 8 − 2 �3 8 �� = 0.5078 in2 σ = F A = 5.4 0.5078 = 10.63 kpsi n = Sy σt = 57 10.63 = 5.36 Ans. 8-44 By symmetry, R1 = R2 = 1.4 kN � MB = 0 1.4(250) − 50RA = 0 ⇒ RA = 7 kN � MA = 0 200(1.4) − 50RB = 0 ⇒ RB = 5.6 kN Members: Sy = 370 MPa Bolts: Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Bolt shear: As = π 4 (102) = 78.54 mm2 τ = 7(103) 78.54 = 89.13 MPa n = Ssy τ = 242.3 89.13 = 2.72 A B 1.4 kN 200 50 RB RA C R1 350 350 R2 2.8 kN 232 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Bearing on member: Ab = td = 10(10) = 100 mm2 σb = −7(103) 100 = −70 MPa n = −Sy σ = −3