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12 April, 07:35

The maximum stress in a section of a circular tube subject to a torque is τmax = 22 MPa. If the inner diameter is Di = 3.75 cm and the outer diameter is Do = 5.5 cm, what is the torque on the section? Express your answer with appropriate units to three significant figures. View Available Hint (s)

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  1. 12 April, 10:00
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    The torque on the section is 563.373 Nm

    Explanation:

    The torque in the section can be calculated through torsion formula. The torsion formula is given as follows:

    Torque = T = τJ / r

    where,

    r = maximum radius = 5.5 cm/2 = 2.75 cm = 2.75 x 10^-2 m

    τ = maximum shear stress = 22 MPa = 22 x 10^6 Pa

    J = Polar moment of inertial = π/2 (ro^4 - ri^4)

    J = π/2 (0.0275^4 - 0.01875^4) m^4 = 7.042 x 10^-7 m^4

    Therefore:

    T = (7.042 x 10^-7 m^4) (22 x 10^6 Pa) / (2.75 x 10^-2 m)

    T = 563.373 Nm
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