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A priori design decisions: The design decision will be: d Material and condition: 1095 HR and from Table A-20 Sut = 120, Sy = 66 kpsi. Design factor: nf = per problem statement. Life: (1150)(3) = 3450 cycles Function: carry 10 000 lbf load | Chapter 7 7-1 Hb 490 Eq. 3-17 Eq. 7-8 Table 7-4 Eq. 7-18 Eq. 7-19 Eq. 7-17 Sut 490 kpsi 212 kpsi S e 107 kpsi a b ka -0 085 3 16 Se kakbSe 107 kpsi Ans. 7-2 a Sut 68 kpsi S e 68 kpsi Ans. b Sut 112 kpsi S e 112 kpsi Ans. c 2024T3 has no endurance limit Ans. d Eq. 3-17 S e 107 kpsi Ans. 7-3 Eq. 7-8 Eq. 7-11 Eq. 7-9 Eq. 7-13 Eq. 7-12 Eq. 7-15 a F C0 m 115 0-22 kpsi Se kpsi log b -------------------- 64 log 2 106 f 2 103 -0-08364 2 . a ---------------- kpsi Sf aNb 12 500 -0 08364 kpsi Ans. N 1 b 409 530 cycles Ans. a 7-4 From Sf aNb Substituting 1 Sut From which log Sf log a b log N log Sut log a b log 1 a Sut Chapter 7 181 Substituting 103 fSut and a Sut log fSut log Sut b log 103 From which 1 b 3 log f a Sf SutN log f 3 1 N 103 For 500 cycles as in Prob. 7-3 500Sf 500 log0 8991 3 kpsi Ans. 7-5 Read from graph 103 90 and 106 50 . From S aNb log S1 log a b log N1 log S2 log a b log N2 From which log S1 log N2 - log S2 log N1 log a 6 log N2 N1 _ log 90 log 106 - log 50 log 103 log106 103 a 10log a 102-2095 b og5 09 3 Sf ax 162-0-08509 1 03 N 106 in kpsi Ans. Check 103 Sf ax 162 103 -0 08509 90 kpsi 106 Sf ax 162 106 -0 08509 50 kpsi The end points agree. 7-6 Eq. 7-8 S e 710 MPa Table 7-4 a b Eq. 7-18 ka 710 -0 265 d 32 Eq. 7-19 kb b Eq. 7-17 Se kakbS e 243 MPa Ans. 182 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design 7-7 For AISI 4340 as forged steel Eq. 7-8 Se 107 kpsi Table 7-4 a b Eq. 7-18 ka 260 -0 995 -0-107 Eq. 7-19 kb . Each of the other Marin factors is unity. Se 107 kpsi For AISI 1040 Se 113 kpsi ka 113 -0 995 kb same as 4340 Each of the .