The maximum torque that can be safely applied to a shaft of 100 mm diameter if the permissible angle of twist is 1 degree in a length of 3 m and the permissible shear stress is 30 N/mm². Take G = 0.8 × 10⁵ N/mm².

Concept:

Using, Torsion formula

\(\frac{{\rm{T}}}{{{{\rm{I}}_{\rm{P}}}}} = \frac{{\rm{τ }}}{{\rm{R}}} = \frac{{{\rm{G\theta }}}}{{\rm{L}}}\)

Where, T = Applied Torsion (kN-m), IP = Polar moment of inertia

R = Radius of the shaft, G = Shear modulus

θ = permissible angle of twist, L = Length of the shaft

τ = permissible shear stress in the shaft

Calculation:

Given,

Diameter = 100 mm, θ = 1 ° , L = 3 m, τ = 30 N/mm2

Torque T = ?, G = 0.8 × 105 N/mm2

R = 100/2 = 50 mm

θ = 1.0° = π /180  radian = 0.0175 radian

\({{\rm{I}}_{\rm{P}}} = \frac{{\rm{π }}}{{32}}\left[ {{{\rm{D}}^4}} \right] \)

\({{\rm{I}}_{\rm{P}}} = \frac{{\rm{π }}}{{32}}{\left( {0.10} \right)^4} = 9.81 × {10^{ - 6}}{\rm{\;}}{{\rm{m}}^4}\) = 9.81 × 10^-6 ×  10¹² mm⁴

Stress Developed = R × (Gθ/L) 

\({{\rm{τ }}_{{\rm{developed}}}} = \frac{{0.8 × {{10}^5} × 0.0175 × 50}}{3000}\)

\({τ _{developed}}\) = 23.33 N/mm2 < 30 N/mm2

Hence, limit the τ _max = 30 N/mm2

So,

\(T~=~\frac{G~\times~J~\times~\theta}{L}~=~\frac{0.8~\times~10^5~\times~9.81~\times~10^6~\times~\pi}{3000~\times~180~\times~1000~\times~1000}~=~4.57~KN.m\)