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MET 304 Welded joints example-3-solution

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MET 304 Welded joints example-3-solution

  1. 1. Example(3): Determine the maximum stress in the 2200 lbreinforced weld of the bracket plate in fig. 2 in 2 in 4.5 in(7.6) . Assume that the load varies from zero tothe maximum value Solution: The throat area of each weld is: A= 0.707×0.25× (6 – 0.5) ×1.2 = 1.16 6 in 2 in For each weld, r1=2 in, and the total polar O second moment of area is:  l2   5.5 2 2J = 2A ×  12 + r1 2  = 2 x1.16 ×     12 + 2  = 15.1 in ¼ in weld    The torque is T = 2200 × 6.5 = 14300 lb-in Fig. (7.6)The distance r in equation (7.4) is r = 2.75 2 + 2 2 = 3.35 inHence, the maximum torsional stress, by equation (7.4), is 14300 × 3.35 s= = 3172.5 psi 15.1This stress is resolved into a vertical component 3172.5 × 2 sv = = 1894 psi 3.35And a horizontal component 3172.5 × 2.75 sh = = 2604.3 psi 3.35The direct stress has only a vertical component, F 2200 s vd = = = 948.3 psi 2A 2 ×1.16The total vertical stress is svt = 1894 +948.3 = 2842.3 psiTherefore the resultant stress is s = 2842.3 2 + 2604.3 2 = 3855 psiWith a concentration factor of k’ = 2.7 from table (7.2) the significant stress is smax = 3855 × 2.7 = 10408.5 psi

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