Question

A solid circular shaft needs to be designed to transmit a torque of 50 Nm. If the allowable shear stress of the material is 140 MPa, assuming a factor of safety of 2, the minimum allowable design diameter in mm is

1. 8
2. 16
3. 24
4. 32

Answer 1

Correct Answer - Option 2 : 16

Concept:

Maximum permissible shear stress can be calculated using:

\({τ _{max}} = \frac{T}{{{Z_p}}}\)

where Z_p = polar moment of inertia, T = torque applied

Given:

T = 50 Nm, τ_allowable = 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