Question 9.21: A rigid coupling is used to transmit 50 kW power at 300 rpm....

Given kW = 50 n = 300 rpm μ = 0.15.           (i).

\text { For bolts, } \quad S_{y t}=380 N / mm ^{2} \quad(f s)=3 \quad N=6 .

\text { For flanges, } \quad D_{o}=200 mm \quad D_{i}=150 mm .

Step I Permissible tensile stress

\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{380}{3}=126.67 N / mm ^{2}     (ii).

Step II Preload in bolts
The torque transmitted by the shaft is given by,

M_{t}=\frac{60 \times 10^{6}( kW )}{2 \pi n}=\frac{60 \times 10^{6}(50)}{2 \pi(300)} .

= 1 591 549.4 N-mm              (iii).

R_{f}=\frac{2}{3} \frac{\left(R_{o}^{3}-R_{i}^{3}\right)}{\left(R_{o}^{2}-R_{i}^{2}\right)}=\frac{2}{3} \frac{\left(100^{3}-75^{3}\right)}{\left(100^{2}-75^{2}\right)}=88.1 mm .

From Eq. (9.43),

M_{t}=\mu P_{i} N R_{f}                (9.43).

P_{i}=\frac{M_{t}}{\mu N R_{f}}=\frac{1591549.4}{0.15(6)(88.1)}=20072.51 N .

Step III Diameter of bolts
Due to pre-load of 20 072.51 N, the bolts are subjected to tensile stresses.

P_{i}=\left(\frac{\pi}{4}\right) d_{1}^{2} \sigma_{t} .

or            d_{1}^{2}=\frac{4 P_{i}}{\pi \sigma_{t}}=\frac{4(20072.51)}{\pi(126.67)} .

∴            d_{1}=14.2 mm .