Question 9.23: A flexible coupling, illustrated in Fig. 9.39, is used to tr...

Given kW = 15 n = 100 rpm.

\text { For pins, } \quad \tau=35 N / mm ^{2} \quad \sigma_{b}=152 N / mm ^{2} .

N = 6 D = 200 mm
Step I Pin diameter by shear consideration

M_{t}=\frac{60 \times 10^{6}( kW )}{2 \pi n}=\frac{60 \times 10^{6}(15)}{2 \pi(100)} .

= 1 432 394.49 N-mm
From Eq. (9.51),

\tau=\frac{8 M_{t}}{\pi d_{1}^{2} D N}                   (9.51).

\tau=\frac{8 M_{t}}{\pi d_{1}^{2} D N} \quad \text { or } \quad 35=\frac{8(1432394.49)}{\pi d_{1}^{2}(200)(6)} .

\therefore \quad d_{1}=9.32 mm               (i).

Step II Pin diameter by bending consideration
The force acting on each bush P and torque M_{t} are related by the following expression,

M_{t}=P \times \frac{D}{2} \times N .

\therefore \quad P=\frac{2 M_{t}}{D N}=\frac{2(1432394.49)}{(200)(6)}=2387.32 N .

It is assumed that the force P is uniformly distributed over the bush length of 35 mm as shown in Fig. 9.42. At the section-XX,

M_{b}=P\left[5+\frac{35}{2}\right]=(2387.32 \times 22.5) N – mm .

\sigma_{b}=\frac{32 M_{b}}{\pi d_{1}^{3}} \text { or } 152=\frac{32(2387.32 \times 22.5)}{\pi d_{1}^{3}} .

\therefore \quad d_{1}=15.33 mm                 (ii).