Question

The figure shows a thin-walled open-top cylindrical vessel of radius \(r\) and wall thickness \(t\). The vessel is held along the brim and contains a constant-density liquid to height \(h\) from the base. Neglect atmospheric pressure, the weight of the vessel and bending stresses in the vessel walls. Which one of the plots depicts qualitatively CORRECT dependence of the magnitudes of axial wall stress (\(\sigma_1\)) and circumferential wall stress (\(\sigma_2\)) on \(y\)? 

• A. A
• B. B
• C. C
• D. D

Hint:

For thin-walled pressure vessels: - {Hoop Stress (\(\sigma_{hoop}\)):} Always related to the internal pressure at that point. For a liquid, pressure increases with depth. - {Axial Stress (\(\sigma_{axial}\)):} Related to pressure on the end caps (for a closed vessel) or the weight of contents/structure below the point of interest (for an open, suspended vessel). Always think about what force is causing the stress to determine how it varies.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to determine the axial stress (\(\sigma_1\), also known as longitudinal or meridional stress) and the circumferential stress (\(\sigma_2\), also known as hoop stress) in the wall of a cylindrical vessel containing a liquid. The stresses will depend on the forces acting on the cylinder walls.
Step 2: Key Formula or Approach:
- Circumferential (Hoop) Stress (\(\sigma_2\)): This stress is caused by the internal pressure (\(p\)) of the fluid trying to burst the cylinder. At any depth \(y\) from the free surface of the liquid, the gauge pressure is \(p(y) = \rho g y\). The hoop stress at this location is given by: \[ \sigma_2 = \frac{pr}{t} = \frac{(\rho g y) r}{t} \] This shows that the hoop stress is zero at the free surface (\(y=0\)) and increases linearly with depth, reaching a maximum at the bottom of the liquid (\(y=h\)). The question defines \(y\) from the base, so let's use a coordinate \(y'\) from the top, where \(y' = h-y\). The pressure is \(p = \rho g y' = \rho g (h-y)\). \[ \sigma_2 = \frac{\rho g (h-y) r}{t} \] This means \(\sigma_2\) is maximum at the base (\(y=0\)) and decreases linearly to zero at the free surface (\(y=h\)).
- Axial (Longitudinal) Stress (\(\sigma_1\)): This stress is caused by the forces acting along the axis of the cylinder. Since the vessel is open at the top and held at the brim, the axial stress at any cross-section is caused by the weight of the liquid column below that cross-section. Let's find the axial stress at a height \(y\) from the base. The force causing this stress is the weight of the liquid in the volume below \(y\). - Weight of liquid below height \(y\): \(W(y) = (\text{Volume}) \times \rho g = (\pi r^2 y) \rho g\).
- This weight is supported by the cross-sectional area of the cylinder wall at height \(y\), which is \(A_{wall} = 2\pi r t\).
- The axial stress is then: \[ \sigma_1(y) = \frac{\text{Force}}{\text{Area}} = \frac{W(y)}{A_{wall}} = \frac{\pi r^2 y \rho g}{2\pi r t} = \frac{\rho g r y}{2t} \] This shows that the axial stress is zero at the base (\(y=0\)) and increases linearly to a maximum at the top of the liquid (\(y=h\)).
Step 3: Detailed Explanation of Plots:
- We found \(\sigma_2 \propto (h-y)\): A line with negative slope, maximum at \(y=0\) and zero at \(y=h\).
- We found \(\sigma_1 \propto y\): A line with positive slope, zero at \(y=0\) and maximum at \(y=h\).
Let's check the options:
- (A): Shows \(\sigma_2\) (circumferential) starting at a maximum at \(y=0\) and decreasing linearly to zero at \(y=h\). It shows \(\sigma_1\) (axial) starting at zero at \(y=0\) and increasing linearly to a maximum at \(y=h\). This matches our derivation perfectly.
- (B): Shows both stresses increasing linearly from non-zero values. Incorrect.
- (C): Shows \(\sigma_1\) decreasing and \(\sigma_2\) increasing. This is the opposite of our result. Incorrect.
- (D): Shows both stresses increasing from zero. Incorrect because hoop stress is maximum at the bottom.
Step 4: Final Answer:
Plot (A) correctly depicts the dependence of the axial and circumferential stresses on the height \(y\).
Step 5: Why This is Correct:
The derived formulas for hoop stress (proportional to local pressure) and axial stress (proportional to the weight of the fluid below) correctly predict the linear variations shown in plot (A). The hoop stress is highest where the pressure is highest (at the bottom), and the axial stress is highest where the suspended weight is greatest (at the top).