32_ch 05 Mechanical Design budynas_SM_ch05

# 32_ch 05 Mechanical Design budynas_SM_ch05 - 9 2[9 − −...

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146 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-46 σ t = p d 2 t = 6000 N (1, 0 . 083 33)(0 . 75) 2(0 . 125) = 18 N (1, 0 . 083 33) kpsi σ l = p d 4 t = 6000 N (1, 0 . 083 33)(0 . 75) 4(0 . 125) = 9 N (1, 0 . 083 33) kpsi σ r =− p =− 6000 N (1, 0 . 083 33) kpsi These three stresses are principal stresses whose variability is due to the loading. From Eq. (5-12), we ﬁnd the von Mises stress to be σ ± = ± (18
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Unformatted text preview: 9) 2 + [9 − ( − 6)] 2 + ( − 6 − 18) 2 2 ² 1 / 2 = 21 . 0 kpsi ˆ σ σ ± = C p ¯ σ ± = . 083 33(21 . 0) = 1 . 75 kpsi z = − ¯ S − ¯ σ ± ( ˆ σ 2 S + ˆ σ 2 σ ± ) 1 / 2 = 50 − 21 . (4 . 1 2 + 1 . 75 2 ) 1 / 2 = − 6 . 5 The reliability is very high R = 1 − ± (6 . 5) = 1 − 4 . 02(10 − 11 ) . = 1 Ans....
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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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