Mechanical Engineering Design (SOLUTIONS MANUAL) by Shigley

By Shigley

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30 2 For the lognormal distribution, from Eq. 02 0 90 92 94 96 98 100 102 x (kpsi) 104 106 108 110 112 The normal and lognormal are almost the same. However the data is quite skewed and perhaps a Weibull distribution should be explored. For a method of establishing the Weibull parameters see Shigley, J. , and C. R. , 1989, Sec. 4-12. 6 Eq. 4 kpsi Eq. (2-29) σˆ x = (θ − x0 )[ (1 + 2/b) − 2 Ans. 64 kpsi Ans. 031 Ans. 55 kpsi Ans. 38 kpsi Ans. 098 Ans. 846 Ans. 2 kpsi Ans. 82 kpsi Ans. qxd 34 7/21/03 3:28 PM Page 34 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design From Prob.

Eq. (3-14) Su = Eq. 9 kpsi Ans. 41 (250) = 853 MPa Eq. (3-17) 3-18 Ans. For the data given, H B = 2530 H B2 = 640 226 2530 = 253 H¯ B = 10 Eq. 2 kpsi Ans. 92 kpsi Ans. 3-19 From Prob. 887 Eq. 894 kpsi 3-20 . 5 in · lbf/in3 uR = 2(30) (a) Ans. Ans. Ans. 005 ␧ ␴ 90000 80000 70000 60000 50000 40000 A4 A5 30000 20000 A3 10000 0 . 4) . 7(103 )in · lbf/in3 Ans. 732 kN RB 60° 90° R B = 2 sin 30 = 1 kN 2 2 kN 30° 60° RB Ans. Ans. 04Њ ⇒ R A = 1100 N Ans. ⇒ R O = 377 N Ans. 078 kN Ans. 4 kN Ans. 5(400 sin 30) = 0 R E = 400 N Ans.

Note since τ is not uniform, the offset yield does not apply, so we are using the elastic limit as an approximation. 781) kpsi 3-16 50 x Ans. 5 kpsi σ0 = 110 kpsi, From Eq. 24 Eq. 2 Eq. 7 kpsi Ans. Eq. (3-14) Su = Eq. 9 kpsi Ans. 41 (250) = 853 MPa Eq. (3-17) 3-18 Ans. For the data given, H B = 2530 H B2 = 640 226 2530 = 253 H¯ B = 10 Eq. 2 kpsi Ans. 92 kpsi Ans. 3-19 From Prob. 887 Eq. 894 kpsi 3-20 . 5 in · lbf/in3 uR = 2(30) (a) Ans. Ans. Ans. 005 ␧ ␴ 90000 80000 70000 60000 50000 40000 A4 A5 30000 20000 A3 10000 0 .

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