Question

Design a solid steel shaft to transmit 14 hp at a speed of 2400rpm f the allowable shearing stress is given as 3.5 ksi.

Answer 1

Is required a 0.8 inches diameter steel shaft.

Step-by-step explanation:

With the power P and the rotating speed n (RPM), we can find the torque applied:

T = P/N

Before calculating the torque, we convert the power and rotating speed units:

`P = 14\ HP * 550\ ((lb.ft)/(s))/(HP) *(12\ in)/(ft) = 92400\ (lb.in)/(s)`

`n=2400\ RPM .(2\pi/60(rad)/(s))/(RPM)= 251(rad)/(s)`

Replacing the values, the torque obtained is:

`T = (92400\ lb.in/s)/(251\ rad/s) = 368\ \ lb.in`

Then the maximum shearing stress will be located at the edge of the shat at the Maximus radius:

`Smax =\ (T.R)/(J)`

Here J is the moment of inertia and R a radius. For a solid shaft, it is calculated by:

`J =(\pi.D^4)/(32)`

Where D is shaft's diameter. Replacing the expression of J in

`Smax =(T.R)/((\pi.D^4)/(32))`

As the radius is half of the diameter:

`Smax =(T.D)/((2*\pi.D^4)/(32\\) ) = (16T)/(\pi.D^3)`

For the maximum stress of 3.5 ksi (3500 psi = 3500\ lb/in^2) and the calculated torque:

`Smax = (16.368\ lb.in)/(3500\ lb/in^2*\pi.D^3)`

Solving for D:

`D =\sqrt[3]{16.368\ lb.in / (3500\pi\ lb/in^2)}} = 0.8\ in`