A torque of T = 2.75 kN-m will be applied to the hollow, thin-walled aluminum alloy section shown in Figure P6.110. If the section has a uniform thickness of 4 mm, determine the magnitude of the maximum shear stress developed in the section. (Note: The dimensions shown are measured to the wall centerline.)
Question 6.110: A torque of T = 2.75 kN-m will be applied to the hollow, thi...
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}=(150 mm )(50 mm )+\frac{\pi}{2}(25 mm )^{2}=8,481.748 mm ^{2}The maximum shear stress developed in the section is:
\tau_{\max }=\frac{T}{2 A_{m} t}=\frac{(2.75 kN – m )(1,000 N / kN )(1,000 mm / m )}{2\left(8,481.748 mm ^{2}\right)(4 mm )}=40.5 MPa
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