Outside diameter of a hollow shaft is twice its inside diameter. Ratio of torque carrying capacity to that of a solid shaft of same outside diameter and same material is

Concept:

Stress in the hollow shaft is given by:

\(τ_h= \frac{{16{d_0}{T_1}}}{{\pi \left( {d_0^4 - d_i^4} \right)}}\)

where do = outside diameter, di = inside diameter

Stress in the solid shaft is given by:

\(τ_s= \frac{{16{T_2}}}{{\pi d_0^3}}\)

Calculation:

Given:

For hollow shaft = do = 2di

For solid shaft d = do

Stress in the hollow shaft:

\(τ_h= \frac{{16{d_0}{T_1}}}{{\pi \left( {d_0^4 - d_1^4} \right)}} = \frac{{16{d_0}{T_1}}}{{\pi \left( {d_0^4 - {{\left( {\frac{{{d_0}}}{2}} \right)}^4}} \right)}} = \frac{{16{}T_{1}}}{{\frac{{15}}{{16}}\pi d_0^3}}\)

Stress in solid shaft:

\(τ_s= \frac{{16{T_2}}}{{\pi d_0^3}}\)

Now, As τh = τs

 \(\frac{{16 \times 16{T_1}}}{{15\pi d_0^3}} = \frac{{16{T_2}}}{{\pi d_0^3}}\)