For a permissible shear stress fs, the torque transmitted by a thin tube of mean diameter D and wall thickness t is:
Try it first: pick an option, then read the explanation.
Correct answer: A. (πD²/2)·t·fs
Asked in SSC JE 22/01/2018 (Evening)
Detailed Explanation
In a thin tube the shear stress is nearly uniform through the wall. The shear force on the wall is fs × (πD × t), acting at radius D/2.
\[ T = (f_s\,\pi D t)\times\frac{D}{2} = \frac{\pi D^2}{2}\,t\,f_s \qquad\text{(Bredt: } T = 2A_m t f_s,\ A_m = \pi D^2/4) \]
Why “(πD²/2)·t·fs” Is Correct
The torque is (πD²/2)·t·fs.
Why the Other Options Are Wrong
- B. (πD/2)·t·fs: Missing one D — the lever arm D/2 has been dropped.
- C. πD²·t·fs: Twice the correct value (lever arm taken as D).
- D. (πD²t²/4)·fs: t² is dimensionally wrong for a thin tube (stress × length³ needs D²t).
Key Point
Thin tube: T = 2Am·t·τ = (πD²/2)·t·τ.
Key Concepts for Students
- Bredt–Batho applies to any thin closed section.
- Shear flow q = τt is constant round the wall.
- Same result as π(D⁴ − d⁴)τ/16D when t ≪ D.
Practice more: this is question 18 in Torsion MCQs – Part 2. Try the whole set and check every answer instantly.