A 65 mm diameter solid shaft is to be welded to a flat plate and is required to carry a torque of 1750 N-m. If fillet joint is used for welding what will be the minimum size of the weld when working shear stress is 66 MPa.
"= (T*d\/2)J"
where T = torsion
d = diameter of the shaft
J = polar moment of inertia
"= (T*d\/2)\/ (\\Pi*t*d^3\/4)""=2T\/ (\\Pi*t*d^2)"
t = s sin 45= 0.707s
"shear stress = 2T\/ (\\Pi * 0.0707s *d^2)"
"66* 10^6 = 2.83*1750\/(\\Pi * s *0.065^2)"
"1\/s = 176.88 M"
"s = 5.653 *10^-3M"
Size of the weld = 5.653 mm.
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