If p is the internal pressure, d the diameter and t the wall thickness of a thin cylindrical shell, the hoop stress will be:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 23/03/2021 (Morning); SSC JE 29/10/2020 (Morning); SSC JE 2005
Detailed Explanation
Imagine cutting a unit length of the cylinder in half along its axis. The pressure pushes the two halves apart, and the two cut walls hold them together.
Bursting force on unit length: \( p \times d \times 1 \)
Resisting force: \( \sigma_h \times 2t \times 1 \)
\[ \sigma_h \cdot 2t = p\,d \;\Rightarrow\; \sigma_h = \frac{pd}{2t} \]
Why “pd/2t” Is Correct
Equating the bursting force to the resistance of the two wall sections gives the hoop (circumferential) stress pd/2t.
| 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. pt/4d: pt/4d has t and d inverted and gives far too small a value.
- B. pd/4t: pd/4t is the longitudinal stress.
- D. td/4p: td/4p does not have units of stress.
Key Point
Thin cylinder: σh = pd/2t, σl = pd/4t.
Key Concepts for Students
- Hoop stress governs the design of thin cylinders.
- Required thickness t = pd/(2σ_allow) (× joint efficiency).
- For a thin sphere, stress = pd/4t in every direction.
Practice more: this is question 2 in Thin Cylindrical and Spherical Shells MCQs – Part 1. Try the whole set and check every answer instantly.