Question

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.

Answer 1

$= (T*d/2)J$

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.