Question

The angular twist of a shaft of radius R mm and length l mm subjected to a torque of 70 kN mm (G = 70 GPa) is
1. \(\frac{{10{\rm{\;l}}}}{{{\rm{\pi }}{{\rm{R}}^4}}}\)
2. \(\frac{{2{\rm{l}}}}{{{\rm{\pi }}{{\rm{R}}^4}}}\)
3. \(\frac{{4{\rm{l}}}}{{{\rm{\pi }}{{\rm{R}}^4}}}\)
4. \(\frac{{15{\rm{l}}}}{{{\rm{\pi }}{{\rm{R}}^4}}}\)

Answer 1

Correct Answer - Option 2 : \(\frac{{2{\rm{l}}}}{{{\rm{\pi }}{{\rm{R}}^4}}}\)

Concept:

Torsional equation of a shaft is given by,

\(\frac{{{{\rm{\tau }}_{{\rm{max}}}}}}{{\rm{R}}} = \frac{{{\rm{G\theta }}}}{{\rm{L}}} = \frac{{{{\rm{T}}_{\rm{R}}}}}{{\rm{J}}}\)

Where,

τ_max = Maximum shear stress, R = radius of shaft, G = Shear modulus, θ = Angle of twist, L = Length of shaft, T_R = Torque and J = Polar moment of inertia

Calculation:

\(\frac{{{\rm{G\theta }}}}{{\rm{L}}} = \frac{{{{\rm{T}}_{\rm{R}}}}}{{\rm{J}}}\)

\(70 \times {10^3} \times {\rm{\theta }} = \frac{{70 \times {{10}^3}}}{{\frac{{\rm{\pi }}}{{32}} \times {{\left( {2{\rm{R}}} \right)}^4}}}\)

\(\therefore {\rm{\theta }} = \frac{{2{\rm{l}}}}{{{\rm{\pi }}{{\rm{R}}^4}}}\)