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13 November, 22:40

A solid circular steel shaft having an outside diameter of 2.00 in. is subjected to a pure torque of T = 8380 lb-in. The shear modulus of the steel is G = 12,000 ksi. Determine the magnitude of the angle of twist in a 5.00-ft length of shaft.

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  1. 13 November, 22:52
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    0.02667 rad (counterclockwise)

    Explanation:

    Parameters given:

    Diameter of shaft = 2 in

    Torque = 8380 lbin

    Shear modulus = 12000 ksi = 12000 * 1000 pounds force per inch

    Length of shaft = 5 ft = 60 in

    The angle of twist is given as

    A = (T*L) / (P*G)

    Where T = Torque

    L = length of shaft

    P = polar moment of the shaft

    G = Shear modulus

    Polar moment of a cylindrical shaft is given as

    P = (pi * R⁴) / 2

    R = radius

    P = (3.142 * 1⁴) / 2

    P = 1.571 m⁴

    A = (8380 * 60) / (12000 * 1000 * 1.571)

    A = 0.02667 rad (counterclockwise)
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