Question

It is desired to transmit 90 kW by means of a solid circular shaft roating at 3.5 rad/s. the allowable shearing stress it 45 MPa. Find the required shaft diameter.

Answer 1

Question #80243

It is desired to transmit 90 kW by means of a solid circular shaft rotating at 3.5 rad/s. the allowable shearing stress it 45 MPa. Find the required shaft diameter.

Answer:

The torque $T$ applied to the shaft is the ratio of transmitted power $P$ to the rotating velocity $\omega t$

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

Substitute into (1):

$$T = \frac {90,000}{3.5} = 25,714 \mathrm{Nm}.$$

The shear stress in the shaft is given by:

$$\tau = \frac {Td}{2j},$$

where $d$ is diameter of the shaft,

$J$ is the second are moment, which is given by:

$$J = \frac {\pi d ^ {4}}{32}$$

for a circular shaft.

Let us substitute (3) in (2) and derive equation for $d$:

$$\tau = \frac {Td}{2 \cdot \frac {\pi d ^ {4}}{32}} = \frac {16T}{\pi d ^ {3}},$$ $$d = \sqrt[3]{\frac{16T}{\pi\tau}}$$

Now, substitute into (4):

$$d = \sqrt[3]{\frac{16 \cdot 25,714}{\pi \cdot 45 \cdot 10^6}} = 0.1428 \mathrm{m} = 143.8 \mathrm{mm}.$$

Thus, the diameter of the shaft should be more than $143.8 \mathrm{mm}$