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Question 6.109: A torque of T = 100 kip-in. will be applied to the hollow, t...

A torque of T = 100 kip-in. will be applied to the hollow, thin-walled aluminum alloy section shown in Figure P6.109. If the section has a uniform thickness of 0.100 in., determine the magnitude of the maximum shear stress developed in the section. (Note: The dimensions shown are measured to the wall centerline.)

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The maximum shear stress for a thin-walled section is given by Eq. (6.25)

τmax⁡=T2Amt\tau_{\max }=\frac{T}{2 A_{m} t}

For the aluminum alloy section,

Am=π8(10 in.)2=39.2699 in.2A_{m}=\frac{\pi}{8}(10  in .)^{2}=39.2699  in .^{2}

The maximum shear stress developed in the section is:

τmax⁡=T2Amt=100 kip–in.2(39.2699 in.2)(0.100 in.)=12.73 ksi\tau_{\max }=\frac{T}{2 A_{m} t}=\frac{100  kip – in .}{2\left(39.2699  in .^{2}\right)(0.100  in .)}=12.73  ksi

 

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