Question

Two coaxial solid shafts of diameter 120mm each are to be connected by a flanged coupling using M12 X 1.5 steel bolts equally spaced around a 320mm pitch circle diameter of the flange. The flanges have thickness of 15mm each and must be welded to the shafts. The coupling must deliver 500kW of power at 400rpm.If the allowable shear stress and bearing pressure are 90N/mm2 and 65N/mmw respectively,on the basis of power being delivered,design the coupling by; Determine (a) Number of bolts required (b) size of the weld (c) efficiency of the coupling

Answer 1

Given :

Load on shaft: $P = \sigma_b \times \frac{\pi}{4} \times d^2$

$P = 65 \times \frac{\pi}{4} \times 120 ^2 = 735.13 \times 10^3 N$

Torque Transmitted : $T = \frac { P \times 3600} { 2\times \pi N } =\frac { 500 \times 10 ^3 \times 3600} { 2\times \pi 400} =716.19 \times 10^3 N.m$

1) Size of bolt :

Load on each bolt $= \tau_b \times \frac{\pi}{4} \times {d_1}^2$

$d_1 = \frac {735.13 \times 10 ^3 \times 4} {90\times \pi} =\sqrt {10399.96}=101.98 mm$

2) Number of bolts :

Torque transmitted $T = \frac{\pi}{4}\times {d_1}^{2} \times \tau_b \times n \times \times \frac{D_1}{2}$

$n = \frac{716.19 \times 10^6\times 2\times 4}{101.98^2\times 90\times \pi \times 320} = 6$

Output Torque

$T_{out} = \frac{\pi}{4}\times {d_1}^{2} \times \tau_b \times n \times \times \frac{D_1}{2} = \frac{\pi}{4}\times {101.98}^{2} \times 90 \times 6 \times \times \frac{320}{2} = 705.72 \times 10^6 Nmm$

3) efficiency of coupling $= \frac{T_{out}}{T_{in}} = \frac{705.72 \times 10 ^6 }{716.19\times 10^ 6} =0.9853 = 98.53$