The ratio of strengths of solid to hollow shafts, both having outside diameter D and hollow shaft having inside diameter D / 2, in torsion, is

Explanation:

Let the strength of the solid and hollow shaft is T_S and T_H 

\({{\bf{T}}_{\bf{S}}} = \frac{\pi }{{16}}{{\bf{d}}^3}{τ _{per}}\), 

\({{\bf{T}}_{\bf{H}}} = \frac{\pi }{{16}}{{\bf{D}}^3}\left( {1 - {{\bf{k}}^4}} \right){τ _{per}}\)

where τ_per = permissible shear strength, k = Ratio of inner diameter (din) to outer diameter (D_o) i.e. \(\left( \frac{d_{in}}{D_o} \right)\)

The ratio of the strength of the hollow shaft to that of the solid shaft when the outer diameter of both the shafts are the same provided RPM, Material and length are the same is given by

\(\frac{{{{\bf{T}}_{\bf{H}}}}}{{{{\bf{T}}_{\bf{S}}}}} = 1 - {{\bf{k}}^4}\)

Here the outer diameter of both the shaft is the same 

D_o = D, d_in = \(\frac{{\rm{D}}}{2}\)

∴ k = \(\frac{{\rm{1}}}{2}\)

\(\frac{{{{\bf{T}}_{\bf{H}}}}}{{{{\bf{T}}_{\bf{S}}}}} = 1 - {\left( {\frac{1}{2}} \right)^4} = \frac{{15}}{{16}}\)

\(\frac{{{{\bf{T}}_{\bf{S}}}}}{{{{\bf{T}}_{\bf{H}}}}}\; = \;\frac{{16}}{{15}}\)

Alternate Method 

This problem can also be solved by using \({M \over I} = {σ \over y}\) 

Where strength can be correlated with M (MOMENT RESISTED BY THE SECTION)\(I_{Solid} = {\pi × D^4 \over 64} \ and\ I_{Hollow} = {\pi [D^4 - ({D/2})]^2\over 64}\)

\(Y_{Solid} = D/2\ and\ Y_{Hollow} = D/2\)

Putting the above values in M/I = σ/y

We get M = (I/Y) × σ (Assuming both are having same stress), we can say that strength depends on I/Y only

Putting the value of I and Y from the above equation, we get

\({{Strength \ of \ Solid} \over {Strength \ of \ hollow\ shaft} } = {I\over Y}\)

\({{{\pi \times D^4\over 64 } \times {D \over 2} } \over {{\pi [D^4 - {(D/2})^2]\over 64 } \times {D \over 2} }} = {1 \over (15/16)} = {16 \over 15}\)