Question

A hollow shaft of 20 mm diameter and 16 mm inside diameter is subjected to a torque of 40 Nm. The shear stress at outside of the shaft will be:
1. 53.12 N/mm²
2. 43.13 N/mm²
3. 62.52 N/mm²
4. 34.50 N/mm²

Answer 1

Correct Answer - Option 2 : 43.13 N/mm²

Concept:

For hollow shaft shear stress will be given as:

\({\bf{\tau }} = \frac{{16{\bf{T}}}}{{{\bf{\pi }}{{\bf{D}}^3}\left( {1 - {{\bf{k}}^4}} \right)}}\)

where, τ = permissible shear stress, T = Resisting torque, D = outside diameter of the hollow shaft, k = ratio of inside and outside diameter of the hollow shaft \( = \frac{d}{D}\)

Calculation:

Given:

D = 20 mm, d = 16 mm, T = 40 Nm = 40 × 10³ Nmm, \(k = \frac{{16}}{{20}} = 0.4\)

\({\rm{\tau }} = \frac{{16{\rm{T}}}}{{{\rm{\pi }}{{\rm{D}}^3}\left( {1 - {{\rm{k}}^4}} \right)}} = \frac{{16 \times 40 \times {{10}^3}}}{{\pi \times {{20}^3}\left( {1 - {{0.8}^4}} \right)}} = 43.13\;N/m{m^2}\)

The shear stress outside of the shaft is 43.13 N/mm2.