Question 2.10: A solid shaft is to transmit 300 kW at 120 rpm. If the shear...

We know that the power transmitted is given by

P=\frac{2 \pi N T}{60}

Here power, P = 300 kW and N = 120 rpm. Substituting these values, we get

T=\frac{300 \times 10^3 \times 60}{2 \pi(120)}=23873.24  Nm

For the solid shaft, we know that

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

or        d=\sqrt[3]{\frac{16 T}{\pi \tau_{\max }}}

Now putting T = 23873.24 N m and \tau_{\max }=100  MPa , we get d = 106.73 mm as the shaft diameter.
For the hollow shaft, similarly, we get

\tau_{\max }=\frac{16 T d^2}{\pi\left(d_2^4-d_1^4\right)}

where d_1 is the internal diameter and d_2 is the external diameter. Putting T = 23873.24 Nm and d_1=0.6 d_2 , we get

\tau_{\max }=100=\frac{16 \times 23873.24 \times d_2}{\pi\left[d_2^4-(0.6)^4 d_2^4\right]}

or        d_2=\sqrt[3]{\frac{16 \times 23873.24 \times 10^3}{\pi\left(1-0.6^4\right) \times 100}}=111.79  mm

The internal diameter d_1   is 0.6 × 111.79 = 67.07 mm. Thus, clearly for same material and length of the two shafts, we have percent saving in weight as

\frac{A_{\text {solid }}-A_{\text {hollow }}}{A_{\text {solid }}} \times 100 \%

where A is the area of shaft. Substituting the values, we get

\frac{(106.73)^2-\left(111.79^2-67.07^2\right)}{(106.73)^2} \times 100 \%=29.78 \%

Thus, the percentage saving in weight is 29.78%.