Maximum shear stress developed on the surface of a solid circular shaft under pure torsion is 240 MPa. If the shaft diameter is doubled, what is the maximum shear stress developed for the same torque

Concept:

Torsion equation:

\(\frac{T}{J} = \frac{τ }{r} = \frac{{G\theta }}{L}\)

For solid shaft:

\(J=\frac{\pi }{{32}}{D^4}\)

The maximum shear stress developed on the surface of the shaft due to twisting moment T:

\(τ = \frac{{16T}}{{\pi {d^3}}}\)

Calculation:

Given:

d2 = 2d1 and τ1 = 240 MPa.

For same torque:

\(τ \propto \frac{{1}}{{ {d^3}}}\)

\(\frac{τ_2}{τ_1}= {\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^3} = {\left( {\frac{{{1}}}{{{2}}}} \right)^3}=\frac{1}{8}\)

\(τ_2=\frac{1}{8} τ_1=\frac{1}{8} \times 240=30~MPa\)