a welded joint, as shown in fig. is subjected to an eccentric load of 2500 n. find the size of the weld, if the maximum shear stress in the weld is not to exceed 50 n/mm^2
Weld area :
A=PσA = \frac{P}{\sigma }A=σP
A=250050=50mm2A = \frac{2500}{50}=50 mm^2A=502500=50mm2
A=t×lA = t \times lA=t×l
Throat thickness t=5010=5mmt =\frac{50}{10}=5mmt=1050=5mm
Size of weld s=tks = \frac{t}{k}s=kt
s=50.7=7.14mm≈8mms = \frac{5}{0.7} =7.14 mm \approx 8 mms=0.75=7.14mm≈8mm
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