A multiflex flexible coupling, transmitting 15 kW power at 720 rpm, is shown in Fig. 9.46. It consists of two flanges keyed to the driving and driven shafts and connected together by means of a number of coils. The coils are placed in the slots on the periphery of the flanges at a distance of 30 mm from the axes of the shafts. The coils have a rectangular cross-section 4 × 2 mm and are made of steel FeE 220 \left(S_{y t}=220 N / mm ^{2}\right) .
The factor of safety is 2.5. Calculate the required number of coils.
Question 9.25: A multiflex flexible coupling, transmitting 15 kW power at 7...
Given kW = 15 n = 720 rpm
\text { For coils, } \quad S_{y t}=220 N / mm ^{2} \quad(f s)=2.5 .
R = 30 mm
cross-section = 4 × 2 mm
Step I Permissible shear stress
S_{s y}=0.577 S_{y t}=0.577(220)=126.94 N / mm ^{2} .
\tau=\frac{S_{s y}}{(f s)}=\frac{126.94}{2.5}=50.78 N / mm ^{2} .
Step II Number of coils
The torque transmitted by the coupling is given by,
M_{t}=\frac{60 \times 10^{6}( kW )}{2 \pi n}=\frac{60 \times 10^{6}(15)}{2 \pi(720)} .
= 198 943.68 N-mm
Each coil is subjected to double shear and the expression for torque is derived in the following way.
\text { Shear resistance of one coil }=2 A \tau .
\text { Shear resistance of } N \text { coils }=2 A \tau N .
\text { Resisting torque }=M_{t}=2 A \tau N R (a).
where,
A = cross-sectional area of wire for the coil
N = number of coils
R = radial distance of coil from the axis of the shaft
Rearranging the terms of Eq. (a),
N=\frac{M_{t}}{2 A \tau R}=\frac{198943.68}{2(4 \times 2)(50.78)(30)} .
= 8.19 or 9 coils.
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