A torque of T = 2.5 kN-m will be applied to the hollow, thin-walled aluminum alloy section shown in Figure P6.108. If the maximum shear stress must be limited to 50 MPa, determine the minimum thickness required for the section. (Note: The dimensions shown are measured to the wall centerline.)
Question 6.108: A torque of T = 2.5 kN-m will be applied to the hollow, thin...
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}=\frac{1}{4} \pi(100 mm )^{2}=7,853.982 mm ^{2}The minimum thickness required for the section if the maximum shear stress must be limited to 50 MPa is thus:
t_{\min }=\frac{T}{2 \tau_{\max } A_{m}}=\frac{(2.5 kN – m )(1,000 N / kN )(1,000 mm / m )}{2\left(50 N / mm ^{2}\right)\left(7,853.982 mm ^{2}\right)}=3.18 mm
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