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23 May, 22:21

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).

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  1. 23 May, 23:26
    0
    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
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