Subtopic: Bending of Unsymmetrical Thin-Walled Sections
Question Statement:
A thin-walled cylindrical pressure vessel with closed ends, having mean radius $r$ and wall thickness $t$, is subjected to an internal gauge pressure $p$. Under the plane stress assumption, the ratio of the circumferential (hoop) stress $\sigma_h$ to the longitudinal stress $\sigma_l$ is:
Options:
(A) 0.5
(B) 1.0
(C) 2.0
(D) 4.0
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2021 Q.22 / AE Q.12)**
**1. Intuitive Physical Setup & Given Parameters**:
- Thin-walled closed cylinder under internal pressure $p$ ($t \ll r$).
**2. Step-by-Step Structural Mechanics Derivation**:
1. **Circumferential (Hoop) Stress ($\sigma_h$)**:
From equilibrium of a diametral cut:
2(\sigma_h t L) = p (2r L) \implies \sigma_h = \frac{p r}{t}
2. **Longitudinal (Axial) Stress ($\sigma_l$)**:
From equilibrium of an end cap cut:
\sigma_l (2\pi r t) = p (\pi r^2) \implies \sigma_l = \frac{p r}{2t}
3. **Ratio of Stresses**:
\frac{\sigma_h}{\sigma_l} = \frac{\frac{p r}{t}}{\frac{p r}{2t}} = 2.0
The hoop stress is exactly **twice** the longitudinal stress.
**3. Comprehensive Options & Distractor Analysis**:
- **(A)** 0.5: ❌ Incorrect. Ratio of longitudinal to hoop stress.
- **(B)** 1.0: ❌ Incorrect. Ratio for spherical pressure vessels.
- **(C)** 2.0: ✅ **CORRECT**. $\sigma_h = 2\sigma_l$.
- **(D)** 4.0: ❌ Incorrect.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **C**
- **Exam Rule**: In thin cylinders: $\sigma_{hoop} = \frac{pr}{t}$, $\sigma_{longitudinal} = \frac{pr}{2t}$, so $\sigma_h / \sigma_l = 2$.