2slowtogofast
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A cylindrical torsion bar made of steel (E = 209 GPa) with diameter of 20mm. The allowable shear stress for this material is 200 MPa what is the max torque that can be safely applied to this bar.
\tau = Tr / J
\tau = shear stress
T = torque
r = radius
J= second polar moment of area
I was thinking of using this formula but I would not have used E = 209GPa in my solution is this wrong? If so could somebody point me in the right direction.
\tau = Tr / J
\tau = shear stress
T = torque
r = radius
J= second polar moment of area
I was thinking of using this formula but I would not have used E = 209GPa in my solution is this wrong? If so could somebody point me in the right direction.