11_ch 09 Mechanical Design budynas_SM_ch09

# 11_ch 09 Mechanical Design budynas_SM_ch09 -...

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Chapter 9 249 vol = h 2 2 l = 0 . 248 2 2 12 = 0 . 369 in 3 I vol = 33 . 65 0 . 369 = 91 . 2 = eff τ ± = 1179 0 . 248 = 4754 psi τ ±± = 2948 0 . 248 = 11 887 psi τ max = 4127 0 . 248 . = 12 800 psi Now consider the case of uninterrupted welds, b 1 = 0 A = 1 . 414( h )(8 0) = 11 . 31 h I u = (8 0)(8 2 / 2) = 256 in 3 I = 0 . 707(256) h = 181 h in 4 τ ± = 10 000 11 . 31 h = 884 h τ ±± = 10 000(10)(8 / 2) 181 h = 2210 h τ max = 1 h ± 884 2 + 2210 2 = 2380 h = τ all h = τ max τ all = 2380 12 800 = 0 . 186 in Do not round off h . A = 11 . 31(0 . 186) = 2 . 10 in 2 I = 181(0 . 186) = 33 . 67 τ ± = 884 0 . 186 = 4753 psi, vol = 0 . 186 2 2 16 = 0 . 277 in 3 τ ±± = 2210 0 . 186 = 11 882 psi fom ± = I u hl = 256 0 . 186(16) = 86 . 0 eff = I ( h 2 / 2) l = 33 . 67 (0 . 186 2 / 2)16 = 121 . 7 Conclusions: To meet allowable stress limitations, I and A do not change, nor do τ and σ . To meet the shortened bead length,
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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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