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.6-36:Se = 29.3LN(1, 0.138) kpsika = 0.782LN(1, 0.11)kb = 0.955For kc use Eq.(6-74):kc = 0.328(58)0.125LN(1, 0.125)= 0.545LN(1, 0.125)SSe = 0.782[LN(1, 0.11)](0.955)[0.545LN(1, 0.125)][29.3LN(1, 0.138)]¯SSe = 0.782(0.955)(0.545)(29.3) = 11.9 kpsiCSe = (0.112 + 0.1252 + 0.1382)1/2 = 0.216Table A-16: d/D = 0, a/D = 0.1, A = 0.92, Kts = 1.68From Eqs.(6-78), (6-79), Table 6-151.68LN(1, 0.10)Kf s = "1 + 2/ 0.125 [(1.68 - 1)/1.68](5/58)= 1.403LN(1, 0.10)Table A-16:À AD4 À(0.92)(1.254)Jnet = = = 0.220132 32TacÄa = Kf sJnet2.4(1.25/2)= 1.403[LN(1, 0.10)]0.2201= 9.56LN(1, 0.10) kpsiFrom Eq.(5-43), p.242:ln (11.9/9.56) (1 + 0.102)/(1 + 0.2162)z =- =-0.85ln[(1 + 0.102)(1 + 0.2162)]Table A-10, pf = 0.1977R = 1 - pf = 1 - 0.1977 = 0.80 Ans.6-38 This is a very important task for the student to attempt before starting Part 3.It illustratesthe drawback of the deterministic factor of safety method.It also identifies the a priori de-cisions and their consequences.The range of force fluctuation in Prob.6-23 is -16 to +4 kip, or 20 kip.Repeatedly-applied Fa is 10 kip.The stochastic properties of this heat of AISI 1018 CD are given. budynas_SM_ch06.qxd 11/29/2006 17:41 Page 175FIRST PAGESChapter 6 175Function ConsequencesAxial Fa = 10 kipFatigue load CFa = 0Ckc = 0.125Overall reliability R e" 0.998; z =-3.09with twin fillets CKf = 0.11"R e" 0.998 e" 0.999Cold rolled or machined Cka = 0.058surfacesAmbient temperature Ckd = 0Use correlation method CÆ = 0.138Stress amplitude CKf = 0.11Cà a = 0.11CSe = (0.0582 + 0.1252 + 0.1382)1/2Significant strength Se= 0.195Choose the mean design factor which will meet the reliability goal0.1952 + 0.112Cn = = 0.2231 + 0.112n = exp -(-3.09) ln(1 + 0.2232) + ln 1 + 0.2232¯n = 2.02¯Review the number and quantitative consequences of the designer s a priori decisions toaccomplish this.The operative equation is the definition of the design factorSeÃa =n¯ ¯ ¯Se K Fa SefÃa = Ò! =¯n w2h n¯ ¯Solve for thickness h.To do so we need-0.265¯ ¯ka = 2.67Sut = 2.67(64)-0.265 = 0.887kb = 1-0.078¯ ¯kc = 1.23Sut = 1.23(64)-0.078 = 0.889¯ ¯kd = ke = 1¯Se = 0.887(1)(0.889)(1)(1)(0.506)(64) = 25.5 kpsiFig.A-15-5: D = 3.75 in, d = 2.5in, D/d = 3.75/2.5 = 1.5, r/d = 0.25/2.5 = 0.10 budynas_SM_ch06.qxd 11/29/2006 17:41 Page 176FIRST PAGES176 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design4" Kt = 2.12.1¯K = " = 1.857f1 + 2/ 0.25 [(2.1 - 1)/(2.1)](4/64)¯K nFa 1.857(2.02)(10)¯fh = = = 0.667 Ans.¯2.5(25.5)w2Se¯¯This thickness separates Se and Ãa so as to realize the reliability goal of 0.999 at eachshoulder.The design decision is to make t the next available thickness of 1018 CD steelstrap from the same heat.This eliminates machining to the desired thickness and the extracost of thicker work stock will be less than machining the fares.Ask your steel supplierwhat is available in this heat.6-39 3"1200 lbf411"21"4Fa = 1200 lbfSut = 80 kpsi(a) Strengthka = 2.67(80)-0.265LN(1, 0.058)= 0.836LN(1, 0.058)kb = 1kc = 1.23(80)-0.078LN(1, 0.125)= 0.874LN(1, 0.125)Sa = 0.506(80)LN(1, 0.138)= 40.5LN(1, 0.138) kpsiSe = 0.836[LN(1, 0.058)](1)[0.874LN(1, 0.125)][40.5LN(1, 0.138)]¯Se = 0.836(1)(0.874)(40.5) = 29.6 kpsiCSe = (0.0582 + 0.1252 + 0.1382)1/2 = 0.195Stress: Fig.A-15-1; d/w = 0.75/1.5 = 0.5, Kt = 2.17.From Eqs.(6-78), (6-79) andTable 6-152.17LN(1, 0.10)Kf = "1 + 2/ 0.375 [(2.17 - 1)/2.17](5/80)= 1.95LN(1, 0.10)Kf FaÃa = , Cà = 0.10(w - d)t¯K Fa 1.95(1.2)fÃa = = = 12.48 kpsi¯(w - d)t (1.5 - 0.75)(0.25) budynas_SM_ch06.qxd 11/29/2006 17:41 Page 177FIRST PAGESChapter 6 177¯ ¯Sa = Se = 29.6 kpsi22¯ln (Sa/Ãa) 1 + Cà 1 + CS¯z =-22ln 1 + Cà 1 + CSln (29.6/12.48) (1 + 0.102)/(1 + 0.1952)=- =-3.9ln (1 + 0.102)(1 + 0.1952)From Table A-20pf = 4.481(10-5)R = 1 - 4.481(10-5) = 0.999 955 Ans.(b) All computer programs will differ in detail.6-40 Each computer program will differ in detail.When the programs are working, the experi-¯ence should reinforce that the decision regarding n is independent of mean values offstrength, stress or associated geometry.The reliability goal can be realized by noting theimpact of all those a priori decisions.6-41 Such subprograms allow a simple call when the information is needed.The calling pro-gram is often named an executive routine (executives tend to delegate chores to others andonly want the answers).6-42 This task is similar to Prob.6-41.6-43 Again, a similar task.6-44 The results of Probs.6-41 to 6-44 will be the basis of a class computer aid for fatigue prob-lems.The codes should be made available to the class through the library of the computernetwork or main frame available to your students.6-45 Peterson s notch sensitivity q has very little statistical basis.This subroutine can be used toshow the variation in q, which is not apparent to those who embrace a deterministic q.6-46 An additional program which is useful
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