When a thin cylindrical shell is subjected to an internal pressure, there will be:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 03/03/2017 (Morning)
Detailed Explanation
A closed thin cylinder under internal pressure has two tensile stresses in its wall: hoop stress pd/2t and longitudinal stress pd/4t. Each strain depends on both, through Poisson's ratio μ.
\[ \varepsilon_h = \frac{pd}{4tE}(2 - \mu) > 0 \qquad \varepsilon_l = \frac{pd}{4tE}(1 - 2\mu) > 0 \ \ (\mu < 0.5) \]
Why “None of these” Is Correct
Both strains are positive for any real material (μ < 0.5), so the diameter and the length both increase. None of options (a)–(c) says that, so the answer is (d) None of these.
Disputed answer: the book's key gives (b), an increase in diameter and a decrease in length. That happens only in an open-ended cylinder (no longitudinal stress), where the Poisson effect of the hoop stress shortens the cylinder. For the usual closed thin shell, (d) is correct.
| Quantity | Thin cylinder (closed ends) | Thin sphere |
|---|---|---|
| Hoop stress | σh = pd/2t | — |
| Longitudinal stress | σl = pd/4t | — |
| Stress in any direction | — | pd/4t |
| Hoop strain | (pd/4tE)(2 − μ) | (pd/4tE)(1 − μ) |
| Longitudinal strain | (pd/4tE)(1 − 2μ) | — |
| Volumetric strain | (pd/4tE)(5 − 4μ) | (3pd/4tE)(1 − μ) |
Thin shell: wall thickness less than about 1/20 of the diameter, so stress is taken as uniform through the wall.
Why the Other Options Are Wrong
- A. A decrease in diameter and length of the shell: Internal pressure makes the hoop strain positive, so the diameter increases.
- B. An increase in diameter and decrease in length of the shell: Printed in the book's key; true only for an open-ended cylinder. A closed cylinder also gets longer.
- C. A decrease in diameter and increase in length of the shell: The diameter increases, not decreases.
Key Point
Closed thin cylinder under internal pressure: diameter ↑ and length ↑ (εl = pd(1 − 2μ)/4tE).
Key Concepts for Students
- Volume of a closed cylinder increases by (pd/4tE)(5 − 4μ) × V.
- If μ = 0.5, the longitudinal strain becomes zero.
- An open-ended cylinder (σl = 0) gets slightly shorter by the Poisson effect.
Practice more: this is question 4 in Thin Cylindrical and Spherical Shells MCQs – Part 1. Try the whole set and check every answer instantly.