Correct Answer - Option 2 : 16
Concept:
Maximum permissible shear stress can be calculated using:
\({τ _{max}} = \frac{T}{{{Z_p}}}\)
where Zp = polar moment of inertia, T = torque applied
Given:
T = 50 Nm, τallowable = 140 MPa
F0S = 2
\({τ _{max}} = \frac{{{τ _{allowable\;}}}}{{FOS}} = \frac{{140}}{{2}}=70~MPa\)
Polar moment of inertia for the circular road is:
\({Z_P} = \frac{{\pi {d^3}}}{{16}}\)
Now maximum shear stress is
\({τ _{max}} = \frac{T}{{{Z_p}}}\)
⇒ \({Z_p} = \frac{T}{{{τ _{max}}}}\)
\(\frac{{\pi {d^3}}}{{16}} = \frac{T}{{{τ _{max\;}}}}\)
\(d = \sqrt[3]{{\frac{{16T}}{{\pi {τ _{max}}}}}} = \sqrt[3]{{\frac{{16 \times 50}}{{\pi \times 70 \times {{10}^6}}}}}\)
d = 0.01537 m = 15.37 mm ≃ 16 mm