A 7.5-ft-long solid steel shaft (G = 11.2 x 106 psi) is to transmit 15 hp at a speed of 1500 rpm. The shaft is limited to an allowable shearing stress of 4.5 ksi and an allowable angle of twist of 4 degrees.

Determine:

a. The power transmitted by the steel shaft in units of in-lb/sec.

b. The corresponding torque for the steel shaft in units of in-lb.

c. The required diameter of the shaft (Show all units in your calculations).

P = 99000 lb. in / s

T = 630.25 lb-in

dr = 0.9272 in

Explanation:

Given:

- The length of the shaft L = 7.5 ft

- The power transmitted P = 15 hp

- The rotational speed f = 1500 rpm

- Allowable shearing stress of 4.5 ksi

- Allowable angle of twist θ = 4°

Find:

a. The power transmitted by the steel shaft in units of in-lb/sec.

b. The corresponding torque for the steel shaft in units of in-lb.

c. The required diameter of the shaft

Solution:

- The power transmitted can be calculated as:

P = 15 hp

P = 15*6600 lb. in / s

P = 99000 lb. in / s

- The Torque T for the steel shaft can be related to P and w as follows:

T = P / 2*pi*f

T = 99000 / 2*pi * (1500 / 60)

T = 630.25 lb-in

- Use allowable shearing stress (τ) for design of shaft:

τ = 16*T / pi*d^3

Where, d is the diameter of the shaft:

d^3 = 16*T / pi*τ

d = cbrt (16*T / pi*τ)

d = cbrt (16*630.25 / pi*4.5*10^3)

d = 0.89345 in

- Use allowable angle of twist (θ) for design of shaft:

θ = 32*T*L / pi*d^4*G

d^4 = 32*T*L / pi*θ*G

d = (32*T*L / pi*θ*G) ^0.25

d = (32*630.25*7.5*12 / pi^2 * (4/180) * 11.2*10^6) ^0.25

d = 0.9272 in

- The required diameter should be kept in lieu to both allowable angle of twist and allowable shear stress so dr

dr = max (0.89345, 0.9272)

dr = 0.9272 in