7_ch 09 Mechanical Design budynas_SM_ch09

7_ch 09 Mechanical Design budynas_SM_ch09 - A = 1 . 414 hd...

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Chapter 9 245 9-13 An optimal square space (3 " × 3 " ) weldment pattern is ±± or or ± . In Prob. 9-12, there was roundup of leg size to 3 / 8 in. Consider the member material to be structural A36 steel. Decision: Use a parallel horizontal weld bead pattern for welding optimization and convenience. Materials: Attachment (1018 HR): S y = 32 kpsi, S ut = 58 kpsi Member (A36): S y = 36 kpsi, S ut 58–80 kpsi; use 58 kpsi From Table 9-4 AISC welding code, τ all = min[0 . 3(58), 0 . 4(32)] = min(16 . 6, 12 . 8) = 12 . 8 kpsi Select a stronger electrode material from Table 9-3. Decision: Specify E6010 Throat area and other properties:
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Unformatted text preview: A = 1 . 414 hd = 1 . 414( h )(3) = 4 . 24 h in 2 ¯ x = b / 2 = 3 / 2 = 1 . 5 in ¯ y = d / 2 = 3 / 2 = 1 . 5 in J u = d (3 b 2 + d 2 ) 6 = 3[3(3 2 ) + 3 2 ] 6 = 18 in 3 J = . 707 h J u = . 707( h )(18) = 12 . 73 h in 4 Primary shear: τ ± x = V A = 3000 4 . 24 h = 707 . 5 h Secondary shear: τ ±± = Mr J τ ±± x = τ ±± cos 45 ◦ = Mr J cos 45 ◦ = Mr x J τ ±± x = 3000(6 + 1 . 5)(1 . 5) 12 . 73 h = 2651 h τ ±± y = τ ±± x = 2651 h r y x r x r ± ±± ± ±± ± ± y x x ± ±± y...
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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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