Answer to Question #193694 in Mechanical Engineering for Nimesh Chathuranga

Question

The roofing design consists of a retractable roof where the roof is opened using a shaft that is

connected to a motor. The motor of 1hp transmits power to a rotor at 1200rev/min. Neglecting any

transmission losses, determine the maximum shear stress in the shaft due to the same applied

torque, if the maximum shear stress in the shaft is limited to 2MPa. The external diameter of the

shaft is 0.05m and a wall thickness of 0.005m What will be the resulting angle of twist of the shaft,

due to the applied torque, over a length of 2.5m, given that the rigidity modulus, G=70GPa.

(1hp=745.7W)

Answer 1

Given,

power of the motor = 1 hp

$=745.7 W$

Rotor speed (f) = 1200 rev/ min = 20 rev/sec

Maximum shear stress = ?

Modulus of rigidity (G)=70GPa.

$=70\times 10^3 MPa$

Length of the shaft $(l)=2.5m$

Let the torque is represented as T

$P=2\pi f T$

$T=\frac{P}{2\pi f}$

Now, substituting the values,

$T=\frac{745.7}{2\times 3.14\times 20}$

$=5.93$ N-m

Applying the torsion equation,

$\Rightarrow \frac{\tau_{max}}{r}=\frac{T}{I}=\frac{G\theta}{l}$

$\Rightarrow \frac{\tau_{max}}{\frac{D}{2}}=\frac{T}{\frac{\pi D^4}{32}}$

$\Rightarrow \tau_{max}=\frac{16T}{\pi D^3}$

Now, substituting the values,

$\tau_{max}=\frac{16T}{\pi D^3}$

$=\frac{16 \times 5.93}{3.14\times 0.05^3}$

$=2.41\times 10^5Pa$

Now again, $\frac{\tau_{max}}{r}=\frac{G\theta}{l}$

Substituting the values,

$\theta = \frac{\tau_{max} l}{r G}$

Now, substituting the values,

$\Rightarrow \theta = \frac{2.41\times 10^5\times 2.5\times 40}{70\times 10^9}=0.0.00344 \ rad$