16_ch 09 Mechanical Design budynas_SM_ch09

# 16_ch 09 Mechanical Design budynas_SM_ch09 - weldment...

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254 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design I u = d 2 6 (3 b + d ) = 7 . 5 2 6 [3(1) + 7 . 5] = 98 . 4in 3 I = 0 . 707 hI u = 0 . 707(1 / 16)(98 . 4) = 4 . 35 in 4 M = (3 . 75 + 0 . 5) W = 4 . 25 W τ ± = V A = W 0 . 751 = 1 . 332 W τ ±± = Mc I = 4 . 25 W (7 . 5 / 2) 4 . 35 = 3 . 664 W τ max = ± τ ± 2 + τ ±± 2 = W ± 1 . 332 2 + 3 . 664 2 = 3 . 90 W Material properties: The allowable stress given is low. Let’s demonstrate that. For the A36 structural steel member, S y = 36 kpsi and S ut = 58 kpsi .Fo r the 1020 CD attachment, use HR properties of S y = 30 kpsi and S ut = 55. The E6010 electrode has strengths of S y = 50 and S ut = 62 kpsi. Allowable stresses: A36: τ all = min[0 . 3(58), 0 . 4(36)] = min(17 . 4, 14 . 4) = 14 . 4 kpsi 1020: τ all = min[0 . 3(55), 0 . 4(30)] τ all = min(16 . 5, 12) = 12 kpsi E6010: τ all = min[0 . 3(62), 0 . 4(50)] = min(18 . 6, 20) = 18 . 6 kpsi Since Table 9-6 gives 18.0 kpsi as the allowable shear stress, use this lower value. Therefore, the allowable shear stress is τ all = min(14 . 4, 12, 18 . 0)
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Unformatted text preview: weldment perspective. The load associated with this strength is τ max = τ all = 3 . 90 W = 900 W = 900 3 . 90 = 231 lbf If the welding can be accomplished (1 / 16 leg size is a small weld), the weld strength is 12 000 psi and the load W = 3047 lbf . Can the bracket carry such a load? There are geometry problems associated with sheet metal folding. Load placement is important and the center of twist has not been identiﬁed. Also, the load-carrying capability of the top bend is unknown. These uncertainties may require the use of a different weld pattern. Our solution pro-vides the best weldment and thus insight for comparing a welded joint to one which em-ploys screw fasteners....
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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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