hwk2 - n0 Fi P1 C 190.6 4.5 1 0.266 57.7 Ans. As was stated...

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Unformatted text preview: n0 Fi P1 C 190.6 4.5 1 0.266 57.7 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 Table 8-2, Table 8-11, Table 8-15, Eq. (8-27), 2.5 grade 8.8 coarse pitch bolts, lubricated. A t = 245 mm2 S p = 600 MPa F i = 0.90 A t S p = 0.90(245)600(10 3) = 132.3 kN K = 0.18 T = KF i d = 0.18(132.3)20 = 476 N m Ans. (b) Table A-31, H = 18 mm, L Table A-17. LT ld lt L G + H = 48 + 18 = 66 mm. Round up to L = 80 mm per 2d 6 2(20) 6 46 mm L - LT 80 46 34 mm l - ld 48 34 14 mm Ad = (202) /4 = 314.2 mm2, kb Ad At E Ad lt Ald t 314.2(245)(207) 314.2(14) 245(34) 1251.9 MN/m Members: Since all members are steel use Eq. (8-22) with E = 207 MPa, l = 48 mm, d = 20mm km 0.5774 Ed 0.5774l 0.5d 2 ln 5 0.5774l 2.5d C kb 0.5774 207 20 0.5 20 2.5 20 0.5774 48 2 ln 5 0.5774 48 0.228 4236 MN/m P = P total 1251.9 kb km 1251.9 4236 / N = 40/2 = 20 kN, Yield: From Eq. (8-28) S p At np CP Fi 600 245 10 0.228 20 3 132.3 1.07 Ans. Chap. 8 Solutions - Rev. A, Page 41/69 Load factor: From Eq. (8-29) nL S p At CP Fi 600 245 10 3 132.3 0.228 20 3.22 Ans. Separation: From Eq. (8-30) Fi 132.3 Ans. 8.57 P1 C 20 1 0.228 ______________________________________________________________________________ n0 8-48 From Prob. 8-29 solution, P max =13.33 kips, C = 0.2, F i = 12.77 kips, A t = 0.141 9 in2 Fi 12.77 90.0 kpsi i At 0.141 9 CP 0.2 13.33 Eq. (8-39), 9.39 kpsi a 2 At 2 0.141 9 Eq. (8-41), 9.39 90.0 99.39 kpsi m a i (a) Goodman Eq. (8-45) for grade 8 bolts, S e = 23.2 kpsi (Table 8-17), S ut = 150 kpsi (Table 8-9) Se Sut 23.2 150 90.0 i 0.856 nf Ans. 9.39 150 23.2 Se a Sut (b) Gerber Eq. (8-46) 1 2 nf Sut Sut 4Se Se 2 a Se i 2 Sut 2 i Se 1502 2 90.0 23.2 1.32 Ans. 1 150 1502 4 23.2 23.2 90.0 2 9.39 23.2 (c) ASME-elliptic Eq. (8-47) with S p = 120 kpsi (Table 8-9) Se 2 2 nf S p S p Se2 i i Se 2 2 Se a Sp 23.2 9.39 1202 23.22 120 120 2 23.2 2 902 90 23.2 1.30 Ans. ______________________________________________________________________________ 8-49 Attention to the Instructor. Part (d) requires the determination of the endurance strength, S e , 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, P bmax = 60/8 = 7.5 kN, P bmin = 20/8 = 2.5 kN Chap. 8 Solutions - Rev. A, Page 42/69 Table 8-17, S e = 23.2 kpsi Eqs. (8-31) and (8-32), F i = 0.75 A t S p Eq. (8-35), Eq. (8-36), Eq. (8-38), C Pb max Pb min 2 At C Pb max Pb min 2 At i = F i /A t = 0.75 S p = 0.75(120) =90 kpsi 5.29 kpsi 90 98.81 kpsi a 0.25 8 2 2 0.141 9 i m 0.25 8 2 2 0.141 9 Se Sut 23.2 150 90 i 1.39 Ans. 150 5.29 23.2 98.81 90 Sut a Se m i ______________________________________________________________________________ nf 8-51 From Prob. 8-33, C = 0.263, P max = 4.712 kN / bolt, F i = 41.1 kN, S p = 650 MPa, and A t = 84.3 mm2 i = 0.75 S p = 0.75(650) = 487.5 MPa 3 CP 0.263 4.712 10 Eq. (8-39): 7.350 MPa a 2 At 2 84.3 Eq. (8-40) m CP 2 At Fi At 7.350 487.5 494.9 MPa (a) Goodman: From Table 8-11, S ut = 900 MPa, and from Table 8-17, S e = 140 MPa Se Sut 140 900 487.5 i Eq. (8-45): 7.55 nf Ans. 7.350 900 140 Se a Sut (b) Gerber: Eq. (8-46): 1 2 nf Sut Sut 2 a Se 4 Se Se i 2 Sut 2 i Se 1 900 9002 2 7.350 140 11.4 Ans. 4 140 140 487.5 9002 2 487.5 140 (c) ASME-elliptic: Eq. (8-47): Chap. 8 Solutions - Rev. A, Page 44/69 nf a Se 2 S p Se2 2 Sp Sp Se2 2 i i Se 140 7.350 6502 1402 650 6502 140 2 487.52 487.5 140 9.73 Ans. ______________________________________________________________________________ 8-52 From Prob. 8-34, C = 0.299, P max = 1.443 kips/bolt,F i = 9.05 kips, S p = 85 kpsi, and A t = 0.141 9 in2 0.75S p 0.75 85 63.75 kpsi i Eq. (8-37): a CP 2 At CP 2 At 0.299 1.443 2 0.141 9 1.520 kpsi Eq. (8-38) m i 1.520 63.75 65.27 kpsi (a) Goodman: From Table 8-9, S ut = 120 kpsi, and from Table 8-17, S e = 18.8 kpsi Se Sut 18.8 120 63.75 i Eq. (8-45): 5.01 nf Ans. 1.520 120 18.8 Se a Sut (b) Gerber: Eq. (8-46): 1 2 2 nf Sut Sut 4Se Se Sut 2 i Se i 2 a Se 1 120 1202 2 1.520 18.6 7.45 Ans. 1202 2 63.75 18.6 4 18.6 18.6 63.75 (c) ASME-elliptic: Eq. (8-47): nf a Se 2 S p Se2 2 Sp Sp Se2 2 i i Se 18.6 1.520 852 18.62 85 852 18.62 63.752 63.75 18.6 6.22 Ans. ______________________________________________________________________________ Chap. 8 Solutions - Rev. A, Page 45/69 Eq. (8-38) m CP 2 At i 1.703 90 91.70 kpsi (a) Goodman: From Table 8-9, S ut = 150 kpsi, and from Table 8-17, S e = 23.2 kpsi Se Sut 23.2 150 90 i Eq. (8-45): nf 4.72 Ans. Se 1.703 150 23.2 a Sut (b) Gerber: Eq. (8-46): 1 2 nf Sut Sut 2 a Se 4 Se Se i 2 Sut 2 i Se 1 150 1502 2 1.703 23.2 7.28 Ans. 4 23.2 23.2 90 1502 2 90 23.2 (c) ASME-elliptic: Eq. (8-47): nf a Se 2 S p S e2 2 Sp Sp Se2 2 i i Se 23.2 1.703 120 2 18.6 2 120 120 2 23.2 2 902 90 23.2 7.24 Ans. ______________________________________________________________________________ 8-55 From Prob. 8-51, C = 0.263, S e = 140 MPa, S ut = 900 MPa, 487.5 MPa, and P max = 4.712 kN. A t = 84.4 mm2, i = P min = P max / 2 = 4.712/2 = 2.356 kN Eq. (8-35): Eq. (8-36): C Pmax a Pmin 0.263 4.712 2.356 103 2 84.3 2 At 3.675 MPa Chap. 8 Solutions - Rev. A, Page 47/69 m C Pmax Pmin 2 At i 0.263 4.712 2.356 103 2 84.3 Eq. (8-38): 487.5 498.5 MPa Se Sut 140 900 487.5 i 11.9 Ans. Sut a Se m 900 3.675 140 498.5 487.5 i ______________________________________________________________________________ nf 8-56 From Prob. 8-52, C = 0.299, S e = 18.8 kpsi, S ut = 120 kpsi, A t = 0.141 9 in2, kpsi, and P max = 1.443 kips i = 63.75 P min = P max / 2 = 1.443/2 = 0.722 kips Eq. (8-35): Eq. (8-36): m a C Pmax Pmin 2 At 0.299 1.443 0.722 2 0.141 9 0.760 kpsi C Pmax Pmin 2 At i 0.299 1.443 0.722 2 0.141 9 63.75 66.03 kpsi Eq. (8-38): Se Sut 18.8 120 63.75 i 7.89 Ans. Sut a Se m 120 0.760 18.8 66.03 63.75 i ______________________________________________________________________________ nf 8-57 From Prob. 8-53, C = 0.228, S e = 162 MPa, S ut = 1040 MPa, A t = 58.0 mm2, MPa, and P max = 7.679 kN. i = 622.5 P min = P max / 2 = 7.679/2 = 3.840 kN Eq. (8-35): C Pmax a Pmin 0.228 7.679 3.840 103 2 58.0 2 At 7.546 MPa Chap. 8 Solutions - Rev. A, Page 48/69 Thread: Die cut. Table 8-17 gives S e = 18.6 kpsi for rolled threads. Use Table 8-16 to find S e for die cut threads S e = 18.6(3.0/3.8) = 14.7 kpsi Table 8-2, A t = 0.663 in2, = P/A t = P /0.663 = 1.51 P, a = m = /2 = 0.755 P From Table 6-7, Gerber nf 2 1 Sut 2 a Se 1 1 2Se Sut 2 1 93.7 2 2 0.755P(14.7) 1 1 2(14.7) 93.7 2 19.01 P Comparing 1910/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 material is located where the stress is small). Ans. (c) For n f = 2 779 lbf, max. load Ans. 2 ______________________________________________________________________________ 8-60 Member, Eq. (8-22) with E =16 Mpsi, d = 0.75 in, and l = 1.5 in km 0.5774 Ed 0.5774l 0.5d 2 ln 5 0.5774l 2.5d 0.5774 16 0.75 0.5 0.75 2.5 0.75 13.32 Mlbf/in P 1.557 103 0.5774 1.5 2 ln 5 0.5774 1.5 Bolt, Eq. (8-13), L T = 2d + 0.25 = 2(0.75) + 0.25 = 1.75 in l = 1.5 in ld = L lt = l Table 8-2, A t = 0.373 in2 A d = (0.752)/4 = 0.442 in2 Eq. (8-17), L T = 2.5 l d = 1.5 1.75 = 0.75 in 0.75 = 0.75 in Chap. 8 Solutions - Rev. A, Page 51/69 kb Ad At E Ad lt At ld kb kb km 0.442 0.373 30 0.442 0.75 0.373 0.75 0.378 8.09 Mlbf/in C 8.09 8.09 13.32 Eq. (8-35), a C Pmax Pmin 2 At C Pmax Pmin 2 At 0.378 6 4 2 0.373 Fi At 1.013 kpsi Eq.(8-36), m 0.378 6 4 2 0.373 25 0.373 72.09 kpsi (a) From Table 8-9, S p = 85 kpsi, and Eq. (8-51), the yielding factor of safety is np m Sp a 85 1.16 72.09 1.013 Ans. (b) From Eq. (8-29), the overload factor of safety is nL S p At CPmax Fi 85 0.373 25 0.378 6 2.96 Ans. (c) From Eq. (8-30), the factor of safety based on joint separation is n0 Fi 1C 25 6 1 0.378 6.70 Ans. Pmax (d) From Table 8-17, S e = 18.6 kpsi; Table 8-9, S ut = 120 kps; the preload stress is i = F i / A t = 25/0.373 = 67.0 kpsi; and from Eq. (8-38) Se Sut 18.6 120 67.0 i 4.56 Ans. Sut a Se m 120 1.013 18.6 72.09 67.0 i ______________________________________________________________________________ nf 8-61 (a) Table 8-2, A t = 0.1419 in2 Table 8-9, S p = 120 kpsi, S ut = 150 kpsi Table 8-17, S e = 23.2 kpsi Eqs. (8-31) and (8-32), i = 0.75 S p = 0.75(120) = 90 kpsi Chap. 8 Solutions - Rev. A, Page 52/69 ...
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