Question 2.7: A hollow shaft transmits 300 kW at 80 rpm. If the shear stre...

The power transmitted is

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

by putting P = 300 × 10³ W and N = 80 rpm, we get

T=\frac{300 \times 10^3 \times 60}{2 \pi \times 80}=35.82 \times 10^3  Nm

This is the mean torque, so

\begin{aligned} \text { Maximum torque } & =1.4 \times T_{\text {mean }} \\ & =1.4 \times 35.82 \times 10^3 \\ & =50.15 \times 10^3  Nm \end{aligned}

Now \tau_{\max } or allowable shear stress is given by

\tau_{\max }=\frac{T_{\max } r}{J}

or        60=\frac{50.15 \times 10^6 \times \frac{d}{2}}{\left\lgroup \frac{d_1^4-d_2^4}{32} \right\rgroup \pi}

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

60=\frac{50.15 \times 10^6 \times \frac{d_1}{2}}{\frac{d_1^4-\left(0.6 d_1\right)^4}{32} \times \pi}

or        d_1=169  mm

Therefore,

d_2=0.6 \times 169=101.4  mm

Therefore, the external and internal diameters are 169 mm and 101.4 mm, respectively. From practical point view, we can consider them to be 170 mm and 102 mm, respectively.