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: Understand what "maximum stress criteria" means for this tube.
When a thin-walled tube carries only torque, every point on its wall is in a state of pure shear, stress \(\tau\). For pure shear, the two principal stresses are \(\sigma_1 = \tau\) and \(\sigma_2 = -\tau\), acting at 45 degrees to the tube axis. The maximum stress (maximum normal stress) theory says the material fails when the larger principal stress magnitude reaches the ultimate strength. So the allowable shear stress equals the given ultimate strength directly: \(\tau_{max} = 200\) MPa.

Step 2: Write the thin-wall torsion formula.
For a thin-walled closed tube, the shear stress from a torque \(T\) is given by Bredt's formula,
\[ \tau = \frac{T}{2 A_m t} \]
where \(A_m\) is the area enclosed by the mean wall line and \(t\) is the wall thickness. For a circular tube of mean radius \(r_m\), \(A_m = \pi r_m^2\).

Step 3: Substitute the numbers.
\(r_m = 0.2\) m, so \(A_m = \pi (0.2)^2 = 0.12566\) m\(^2\).
\[ T_{max} = \tau_{max} \times 2 A_m t = (200 \times 10^6)(2)(0.12566)(0.004) \]
\[ T_{max} = (200 \times 10^6)(0.0010053) = 2.0106 \times 10^5 \text{ N-m} \]

Final Answer:
\(T_{max} \approx 201.06\) kN-m, which rounds to
\[ \boxed{T_{max} \approx 201 \text{ kN-m}} \]