Question

The maximum stress of a shaft of 100 mm diameter and 2.7 m length subjected to a torque of 125 kNm with shear modulus of 75 GPa is
1. \(\frac{{2000}}{{\rm{\pi }}}{\rm{MPa}}\)
2. \(\frac{{200}}{{\rm{\pi }}}{\rm{MPa}}\)
3. \(\frac{{100}}{{\rm{\pi }}}{\rm{MPa}}\)
4. \(\frac{{125}}{{\rm{\pi }}}{\rm{MPa\;}}\)

Answer 1

Correct Answer - Option 1 : \(\frac{{2000}}{{\rm{\pi }}}{\rm{MPa}}\)

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:

Given: Diameter, D = 100 mm, Length, L = 2700 mm, Torque, T_R = 125 kNm = 125 × 10⁶ Nmm, and Shear modulus, G = 75 GPa = 75 × 10³ MPa

By taking,

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

\(\therefore {\rm{\theta }} = \frac{{125 \times {{10}^6}}}{{\frac{{\rm{\pi }}}{{32}}{{100}^4}}} \times \frac{{2700}}{{75 \times {{10}^3}}} = \frac{{1.44}}{{\rm{\pi }}}\)

Now taking,

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

\(\therefore {{\rm{\tau }}_{{\rm{max}}}} = \frac{{75 \times {{10}^3}}}{{2700}} \times \frac{{1.44}}{{\rm{\pi }}} \times 50 = \frac{{2000}}{{\rm{\pi }}}\)