A steel tube (G 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 1.5 in. When twisted by a torque T, the tube develop

Question

A steel tube (G 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 1.5 in. When twisted by a torque T, the tube develops a maximum normal strain of 170 106 . What is the magnitude of the applied torque T?

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Latifah 5 months 2021-08-22T05:37:45+00:00 1 Answers 3 views 0

Answers ( )

    0
    2021-08-22T05:38:51+00:00

    Explanation:

    Given data:

    G = 11.5×10⁶psi

    d₂ = 2.0 inch

    d₁ = 1.5 inch

    ε_{max} = 170 × 10⁻⁶

    Y_{max} = 2ε

    T/J = τ_{max} /R

    \frac{Td_{2} }{2J} = τ_{max}         (1)

    τ_{max} = G Y

    from 1 and 2

    \frac{T d_{2} }{2J} = G Y_{max}

    T = \frac{2 G Y_{max}J }{d_{2} }

    \frac{2* 11.5*10^{6}*0.006895*10^{6}*340*10*^{-6} *\frac{\pi }{32}[2^{4}-1.5^{4}]*(0.0254)^{4}       }{2* 0.0254}

      = 474.14 Nm

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