Question

A thin-walled circular tube is made of a material whose magnitude of the ultimate strength, both in tension and compression, is \(200\) MPa. The mean radius of the tube is \(0.2\) m and the wall thickness is \(0.004\) m. Based on the maximum stress criteria, the maximum torque that the tube can sustain is _______ kN-m (round off to the nearest integer).

Hint:

Pure torsion on a thin tube gives pure shear, whose principal stresses equal the shear stress. Set that shear stress equal to the ultimate strength and use the thin-wall torsion formula \(\tau = T/(2 A_m t)\).

Answer 1

Correct Answer: 201

Step 1: Set up the failure condition.
Pure torsion produces pure shear on the tube wall, and pure shear gives principal stresses of $+\tau$ and $-\tau$. Under the maximum (normal) stress theory, the tube fails when $\tau$ reaches the ultimate strength, so $\tau_{max} = 200$ MPa.

Step 2: Use the torsion constant of a thin circular tube.
Treat the tube like a thin ring and write its torsion (polar) constant directly,
\[ J = 2\pi r_m^3 t \]
This comes from integrating $r^2\,dA$ around a thin ring of radius $r_m$ and thickness $t$: $dA = r_m\,t\,d\phi$, so $J = \int_0^{2\pi} r_m^2 (r_m t)\,d\phi = 2\pi r_m^3 t$.
The torsion shear stress is then $\tau = T r_m / J$.

Step 3: Solve for T and substitute.
\[ T = \frac{\tau_{max} J}{r_m} \]
$J = 2\pi (0.2)^3 (0.004) = 2\pi(0.008)(0.004) = 2.0106\times 10^{-4}$ m$^4$.
\[ T = \frac{(200\times10^6)(2.0106\times10^{-4})}{0.2} = \frac{40212}{0.2}= 201060 \text{ N-m} \]

Final Answer:
\[ \boxed{T_{max} \approx 201 \text{ kN-m}} \]