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

Answer:

 Is required a 0.8 inches diameter steel shaft.

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:

$P = 14\ HP * 550\ \frac{\frac{lb.ft}{s}}{HP} *\frac{12\ in}{ft} = 92400\ \frac{lb.in}{s}$

$n=2400\ RPM .\frac{2\pi/60\frac{rad}{s}}{RPM}= 251\frac{rad}{s}$

Replacing the values, the torque obtained is:

$T = \frac{92400\ lb.in/s}{251\ rad/s} = 368\ \ lb.in$

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

$Smax =\ \frac{T.R}{J}$

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

$J =\frac{\pi.D^4}{32}$

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

$Smax =\frac{T.R}{\frac{\pi.D^4}{32}}$

As the radius is half of the diameter:

$Smax =\frac{T.D}{\frac{2*\pi.D^4}{32\\} } = \frac{16T}{\pi.D^3}$

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

$Smax = \frac{16.368\ lb.in}{3500\ lb/in^2*\pi.D^3}$

Solving for D:

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