Question

. A flexible coupling is to be used to connect two shafts, carrying 24 kW power @ 1200 rpm. If the connecting strip is placed 2.5” away from the center of the shafts, how many folds of 0.015” by 0.15” steel made from SAE 1045, quenched in water and drawn at 600^oF) are required? (SAE 1045, quenched in water and drawn at 600^oF has s_u = 1 GPa and s_y = 775 MPa)

NOTE: Solve this problem (in SI Units) by using a Factor of Safety of 2.5 on s_u and knowing that the Maximum Shear Stress is the half of the Ultimate Tensile Stress.

Answer 1

First of all let us find the diameter of the shaft that is d.

We know, Mean Torque is given by, T(mean)=
60p/2an
=$60p/2\pi n=24*10^3*60/2\pi *1200=600Nm$

where p is the power given and n is the rpm.

Further, there is term called service factor which help us to find tthe maximum torque such that

service factor=1.35

Next , $T_m_a_x=\pi /16*\tau *d^3$

$810*10^3=\pi /16*1*10^6*d^3$

$d=40mm$

$D=2d=2*40=80mm$

$L=1.5d=1.5*40=60mm$

Approximately , 28 folds are required.