A torque of T = 150 kip-in. will be applied to the hollow, thin-walled aluminum alloy section shown in Figure P6.107. If the maximum shear stress must be limited to 10 ksi, determine the minimum thickness required for the section. (Note: The dimensions shown are measured to the wall centerline.)
Question 6.107: A torque of T = 150 kip-in. will be applied to the hollow, t...
The maximum shear stress for a thin-walled section is given by Eq. (6.25)
\tau_{\max }=\frac{T}{2 A_{m} t}For the aluminum alloy section,
A_{m}=(6 in .)(8 in .)+\frac{\pi}{4}(6 in. )^{2}=76.274 in .{ }^{2}The minimum thickness required for the section if the maximum shear stress must be limited to 10 ksi is thus:
t_{\min }=\frac{T}{2 \tau_{\max } A_{m}}=\frac{150 kip – in .}{2(10 ksi )\left(76.274 in .^{2}\right)}=0.0983 in.
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