A flexible coupling, is used to transmit 15 $kW$ power at100 $rpm$. There are six pins and their pitch circle diameter is 200 $mm$. The effective length of the bush $(lb)$, the gap between two flanges and the length of the pin in contact with the right hand flange are 35, 5 and 23 $mm$ respectively. The permissible shear and bending stresses for the pin are 35 and 152 $N/mm^2$ respectively.

Calculate:

(i) Pin diameter by shear consideration; and

(ii) Pin diameter by bending consideration.

Subject: Machine Design -I

Topic: Shaft, Keys and Coupling

Difficulty: Medium

Step 1: Pin diameter by shear consideration:

$M_t=\frac{60*10^6*kW}{2 \pi n}=\frac{60*10^6*15}{2 \pi 100}=1432394.49 N-mm$

$\tau=\frac{8M_T}{\pi d_1^2DN} or 35=\frac{8(1432349.49}{\pi d_1^2*200*6}$

$d_1=9.32mm (i)$

Step 2: Pin diameter by bending consideration.

The force acting on each bush P and torque M_t are related by the following expression,

$M_t=P*\frac{D}{2}*N$

$P=\frac{2M_t}{DN}=\frac{2(1432394.49)}{200*6}=2387.32N$

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

$M_b=P[5+\frac{35}{2}]=2387.32*22.5 N-mm$

$\sigma_b=\frac{32M_b}{\pi d_1^3} or 152=\frac{32(2387.32*22.5)}{\pi d_1^3}$

$d_1=15.33mm$