Question

The hydrofoil boat has an A-36 steel propeller shaft that is 100ft long. It is connected to an inline diesel engine that delivers a maximum power of 2500 hp and causes the shaft to rotate at 1700 rpm. If the outer diamater of the shaft is 8 in. and the wall thickness is 3/8 in., determine the maximum sheer stress developed in the shaft. Also, what is the "wind up," or angle of twist in the shaft at full power?

Answer 1

To solve this problem it is necessary to apply the concepts related to internal torque, shear stress and the angle generated from the modulo of elasticity, torque, length and polar moment of inertia.

We will start by finding the angular velocity in units that allow us to find the total torque:

$\omega = 1700rev/min (\frac{2\pi rad}{1rev})(\frac{1min}{60s})$

$\omega = 56.67\pi rad/s$

From the indicated variable of power, in English system this would be

$P = 2590HP (\frac{550ft \cdot lb/s}{1hp})$

$P = 1424500ft \cdot lb/s$

With the power and angular velocity found, the torque would then be:

$T = \frac{P}{\omega}$

$T = \frac{1424500}{56.67\pi}$

$T = 8001.27lb\cdot ft$

The torsion formula will allow us to find maximum stress due to the shear force acting on the body:

$\tau_{max} = \frac{Tc}{J}$

$\tau_{max} = \frac{800.27(12)(4)}{\pi/4(4^4-3.625^4)}$

$\tau_{max} = 2.93ksi$

And finally the angle of twist

$\phi = \frac{TL}{JG}$

$\phi =\frac{(8001.27)(12)(100)(12)}{\pi/4(4^4-3.625^4)(4^4-3.625^4)(11)(10^6)}$

$\phi = 0.08002rad$

$\phi = 4.58\°$

Therefore the anle of twist in the shaft at full power is 4.58°