Question

A bushed pin type flexible coupling is used to connect two shafts and transmit 15 kW power at 600 rpm Shafts, keys and pins are made of commercial steel (σyt = σyc = 240 N/mm2) and the factor of safety is 3. The flanges are made of grey cast iron FG 200 (σut = 200 N/mm2) and the factor of safety is 6. Assume, τy = 0.5σyt and τu = 0.5σut. The permissible shear stress for the pins is 35 N/mm2. The permissible bearing pressure for the rubber bushes is 1 N/mm2. The keys have a square cross-section. Design the coupling using manufacturer’s catalogue.

Answer 1

Permissible stresses;

Shaft, $\tau=\frac{S_{Sy}}{FS}$

$=\frac{0.55*240}{3}=44N/mm^2$

This is same as permissible stress in key

Pins, $\tau=35N/mm^2$

$\sigma_t=\frac{240}{3}=80N/mm^2$

Flanges, $\tau=\frac{S_{Sy}}{FS}$

= $\frac{0.5*200}{6}=16.67N/mm^2$

Diameter of shaft

$P=\frac{2πnM_t}{60}$

$5*10^3=\frac{2π*720M_t}{60}$

$M_t=66314.6Nmm$

$\tau=\frac{16M_t}{πd^3}$

$d^3=\frac{16*66314.6}{π*44}$

d=20mm

Diameter of hub $d_h=2d=40mm$

Length of hub $l_h=1.5d=30mm$

Pitch circle diameter $D=4d=80mm$

Thickness of flange $t=0.5d=10mm$

Thickness of protective rim $t_1=0.25d=5mm$