Question

A thin-walled, closed cylindrical vessel of inside diameter $d$ and wall thickness $t$ contains a fluid under pressure $p$. The figure shows a part of the cylindrical vessel; end caps are not shown. Consider the small element with sides parallel and perpendicular to the axis of the cylinder. The stresses $\sigma_1$ and $\sigma_2$ are:

• A. $\sigma_1 = \dfrac{pd}{2t}, \; \sigma_2 = \dfrac{pd}{4t}$
• B. $\sigma_1 = \dfrac{pd}{t}, \; \sigma_2 = \dfrac{pd}{2t}$
• C. $\sigma_1 = \dfrac{pd}{4t}, \; \sigma_2 = \dfrac{pd}{2t}$
• D. $\sigma_1 = \dfrac{pd}{2t}, \; \sigma_2 = 0$

Hint:

In thin-walled cylinders under pressure: - Hoop stress = $\dfrac{pd}{2t}$ (largest). - Longitudinal stress = $\dfrac{pd}{4t}$. Always ignore radial stress since $t \ll d$.

Answer 1

The Correct Option is A

Step 1: Recall thin-walled pressure vessel assumptions.
For a thin-walled cylinder under internal pressure: - **Hoop stress (circumferential stress)** is the largest. - **Longitudinal stress (axial stress)** is half of hoop stress. - Radial stress is neglected (very small compared to $p$).
Step 2: Formula for hoop stress.
\[ \sigma_{hoop} = \frac{p d}{2t} \] where $d$ = internal diameter, $t$ = wall thickness.
Step 3: Formula for longitudinal stress.
\[ \sigma_{longitudinal} = \frac{p d}{4t} \]
Step 4: Identify stresses.
From the figure: - $\sigma_1$ is along circumferential (hoop) direction. - $\sigma_2$ is along longitudinal axis. Thus, \[ \sigma_1 = \frac{pd}{2t}, \sigma_2 = \frac{pd}{4t} \] Final Answer: \[ \boxed{\sigma_1 = \dfrac{pd}{2t}, \sigma_2 = \dfrac{pd}{4t}} \]