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The outside diameter of a hollow shaft is thrice to its inside diameter. The ratio of its torque carrying capacity to that of a solid shaft of the same material and the same outside diameter is: 
1. 80/81
2. 1/81
3. 8/9
4. 1/9

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Correct Answer - Option 1 : 80/81


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



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.

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