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30 November, 22:01

A solid titanium alloy [G 114 GPa] shaft that is 720 mm long will be subjected to a pure torque of T 155 N m. Determine the minimum diameter required if the shear stress must not exceed 150 MPa and the angle of twist must not exceed 7 degree. Report both the maximum shear stress and the angle of twist at this minimum diameter.

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  1. 1 December, 01:44
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    a) minimum diameter = 16.90 mm

    b) minimum diameter = 17.39 mm

    Explanation:

    The formula to apply here is;

    Angular deformation of a solid shaft is

    α=32LT/GπD⁴

    where

    α=angular shaft deformation in radians

    L=length of shaft in meters

    T=twisting moment in Nm

    G=shear modulus of rigidity in Pa

    D=diameter of shaft in m

    Given

    α=7° = 0.1222 rad

    L=720 mm = 0.72 m

    T = 155 N

    G=114*10⁹ Pa

    D=diameter of shaft in m

    For the angle of twist

    α=32LT/GπD⁴

    0.1222 = 32*0.72*155/114*10⁹*3.142 * D⁴

    0.1222*114*10⁹*3.142 * D⁴=32*0.72*155

    D⁴=32*0.72*155 / 0.1222*114*10⁹*3.142

    D = (32*0.72*155 / 0.1222*114*10⁹*3.142) ^1/4

    D=0.01690 m = 16.90 mm

    For maximum shear stress apply

    D=1.72 * (Tmax/tmax) ^1/3

    where

    D=diameter of solid shaft in m

    Tmax = maximum twisting moment in Nm

    tmax = maximum shear stress in Pa

    Given

    tmax=150*10⁶ Pa

    Tmax=155 Nm

    D = 1.72 * (155/150*10⁶) ^1/3

    D = 1.72 * 0.0101098989

    D=0.01739 m = 17.39 mm
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