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:
[tex]P = 14\ HP * 550\ \frac{\frac{lb.ft}{s}}{HP} *\frac{12\ in}{ft} = 92400\ \frac{lb.in}{s} [/tex]
[tex]n=2400\ RPM .\frac{2\pi/60\frac{rad}{s}}{RPM}= 251\frac{rad}{s}[/tex]
Replacing the values, the torque obtained is:
[tex]T = \frac{92400\ lb.in/s}{251\ rad/s} = 368\ \ lb.in[/tex]
Then the maximum shearing stress will be located at the edge of the shat at the Maximus radius:
[tex]Smax =\ \frac{T.R}{J}[/tex]
Here J is the moment of inertia and R a radius. For a solid shaft, it is calculated by:
[tex]J =\frac{\pi.D^4}{32}[/tex]
Where D is shaft's diameter. Replacing the expression of J in
[tex]Smax =\frac{T.R}{\frac{\pi.D^4}{32}}[/tex]
As the radius is half of the diameter:
[tex]Smax =\frac{T.D}{\frac{2*\pi.D^4}{32\\} } = \frac{16T}{\pi.D^3}[/tex]
For the maximum stress of 3.5 ksi (3500 psi = 3500\ lb/in^2) and the calculated torque:
[tex]Smax = \frac{16.368\ lb.in}{3500\ lb/in^2*\pi.D^3}[/tex]
Solving for D:
[tex]D =\sqrt[3]{16.368\ lb.in / (3500\pi\ lb/in^2)}} = 0.8\ in[/tex]
