Question

The A-36 solid steel shaft is 2 m long and has a diameter of 60 mm. It is required to transmit 60 kW of power from the motor M to the pump P. Determine the smallest angular velocity the shaft if the allowable shear stress is ?

Answer 1

The smallest angular velocity is 17.7rad/s

Step-by-step explanation:

We know,

`T = (Tr)/(J)`

For solid shaft,
`J = (\pi )/(2) r^4`

For hollow shaft,
`J = (\pi )/(2) (r_o^4 - r_i^4)`

Therefore,

`T_m_a_x = (Tr)/(J) \\\\`

`80 X 10^6 = (T (0.03))/((\pi )/(2)(0.03)^4 ) \\\\T = 3392.92 Nm\\`

We know,

P = Tω

`w = (P)/(T) \\\\w = (60 X 10^3)/(3392.92) \\\\w = 17.7rad/s`

Thus, the smallest angular velocity is 17.7rad/s