Correct Answer - Option 1 : 80/81
Concept:
The equation for Shaft Subjected to Torsion T
\(\frac{τ }{R}\) = \(\frac{T}{J}\) = \(\frac{{Gθ }}{L}\)
Solid shaft J = \(\frac{{\pi D_s^4}}{{32}}\)
Hollow Shaft J = \(\frac{{\pi \left( {D_0^4 - \;D_i^4} \right)}}{{32}}\)
τ = Shear stress induced due to torsion T, G = Modulus of Rigidity, θ = Angular deflection of the shaft, R, L = Shaft radius and length respectively, Ds = Diameter of the solid shaft, D0 = Outer Diameter of Hollow shaft, Di = Inner Diameter of Hollow shaft
Calculation;
Given
D0 = 3 × Di
Ds = D0 (Solid shaft diameter is equal to an outer diameter of hollow shaft)
Material is the same for the hollow and solid shaft (i.e. G is the same for both because G depends upon the material)
\(\frac{T}{\tau }\) = \(\frac{{G\theta }}{L}\)
∴ T \( \propto \;\) J (θ, G, L same for both or constant)
\(\frac{{{T_H}}}{{{T_s}}}\) = \(\frac{{{J_H}}}{{{J_s}}}\) = \(\frac{{\frac{\pi }{{32}}\left( {D_0^4 - D_i^4} \right)}}{{\frac{\pi }{{32}}{{\left( {{D_s}} \right)}^4}}}\)
Put, D0 = 3 × Di
Ds = D0
\(J_H\over J_s\) = \(80\over 81\)
Polar moment of inertia
J = \(\mathop \smallint \limits_0^R 2\pi \)r3dr
For Solid Shaft, J = \(\pi D^4 \over 32\)
For Hollow Shaft, J = \(\mathop \smallint \limits_r^R 2\pi \)r3dr = \(\frac{{\pi \left( {D_0^4 - \;D_i^4} \right)}}{{32}}\)
The assumption for Torsion equation:
\(\tau\over R\) = \(T\over J\) = \(G\theta\over L\)
- The bar is acted upon by pure torque.
- The section under consideration is remote from the point of application of the load and from a change in diameter.
- The material must obey Hooke's Law
- Cross-sections rotate as if rigid, i.e. every diameter rotates through the same angle.