6_ch 02 Mechanical Design budynas_SM_ch02

6_ch 02 Mechanical Design budynas_SM_ch02 -...

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Chapter 2 11 2-15 For the data given, ± H B = 2530 ± H 2 B = 640 226 ¯ H B = 2530 10 = 253 ˆ σ HB = ² 640 226 (2530) 2 / 10 9 = 3 . 887 Eq. (2-17) ¯ S u = 0 . 495(253) = 125 . 2 kpsi Ans. ¯ σ su = 0 . 495(3 . 887) = 1 . 92 kpsi Ans. 2-16 From Prob. 2-15, ¯ H B = 253 and ˆ σ HB = 3 . 887 Eq. (2-18) ¯ S u = 0 . 23(253) 12 . 5 = 45 . 7 kpsi Ans. ˆ σ = 0 . 23(3 . 887) = 0 . 894 kpsi Ans. 2-17 (a) u R . = 45 . 5 2 2(30) = 34 . 5in · lbf/in 3 Ans. (b) P ± LA A 0 / A 1 ε σ = P / A 0 00 0 0 1 000 0.0004 0.0002 5 032.39 2 000 0.0006 0.0003 10 064.78 3 000 0.0010 0.0005 15 097.17 4 000 0.0013 0.000 65 20 129.55 7 000 0.0023 0.00115 35 226.72 8 400 0.0028 0.0014 42 272.06
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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