Question

A shaft required to transmit 50 kW of power will be running at 100 rev/min. The maximum twist in 3m length of the shaft is limited to 1.5 degree. If the maximum torque is 30 percent more than mean or average torque. Assume entirely elastic action and take modulus of rigidity for the shaft material G = 135 Gn/m².

Find the following:

Diameter of the shaft.

Answer 1

The diameter of the shaft is approximately 92.5 mm.

Step-by-step explanation:

To find the diameter of the shaft, we can use the formula for power transmitted by a shaft: P = (2Tnπ)/60, where P is the power, T is the torque, n is the rotational speed, and π is the constant pi.

Given: P = 50 kW, n = 100 rev/min (which is equivalent to 100/60 = 5/3 rad/s).

Since the maximum torque is 30% more than the average torque, we can write T = (1 + 0.30) * (P * 60) / (2 * n * π) = 0.00162144 GNm.

Now, we can use the formula for torque to find the diameter of the shaft: T = (π/32) * (Gd^4)/L, where G is the modulus of rigidity, d is the diameter, and L is the length.

Given: G = 135 GN/m², L = 3m, T = 0.00162144 GNm.

Now, we can rearrange the formula to solve for d: d = ((32Tl)/(Gπ))^0.25.

Plugging in the values, we find that the diameter of the shaft is approximately 0.0925 meters or 92.5 mm.