A shaft of 50 mm diameter and 0.7 m long is subjected to a torque of 1200 Nm. Calculate the shear stress.

Concept:

Equation of Pure torsion

\(\frac{T}{{{I_P}}} = \frac{τ }{r} = \frac{{Gθ }}{L}\)  where,

T = Maximum torque, IP = Polar moment of inertia, τ = shear stress at any point at a distance r from center, G = Modulus of rigidity, θ = angle of twist in radians, R = radius of shaft

Considering the rigidity of shaft, The maximum shear stress induced in the solid shaft of diameter d, to transmit twisting moment T is given by,

\(\tau_{max}\;=\frac{{16T}}{{\pi d^3 }}\)

The maximum shear stress induced in the hollow shaft of outer diameter do and inner diameter di to transmit the twisting moment T.

\(\tau_{max}\;=\frac{{16T}}{{\pi d_o^3 (1-K^4)}}\), where K is the ratio of outer and inner diameter, \(K =\frac{{d_i}}{{d_o}}\)

Calculation:

Given:

d = 50 mm, L = 0.7 M, T = 1200 N-m

Maximum shear stress-induced, \(\tau_{max}\;=\frac{{16T}}{{\pi d^3 }}\)

⇒ \(\tau_{max}\;=\frac{{16~\times~1200~\times~1000}}{{\pi ~\times~50^3 }}~=~48.9~MPa\)