Thin Cylindrical Pressure Vessel Stress Analysis

When a thin cylindrical pressure vessel with closed ends is subjected to internal pressure, its walls experience various stresses. The two primary stresses relevant to this question are the circumferential (hoop) stress and the longitudinal stress. Understanding these stresses and their relationship is crucial for the design and safety of such vessels.

Circumferential (Hoop) Stress Derivation

The circumferential stress, also known as hoop stress ($\sigma_h$), acts around the circumference of the cylinder. It is caused by the internal pressure pushing outwards, which tends to split the vessel along its length, parallel to the axis.

To derive the formula for hoop stress, imagine cutting the cylinder longitudinally across its diameter for a small length \(L\). The total force exerted by the internal pressure on this cut section is resisted by the tensile forces in the two wall thicknesses.

• Force due to internal pressure (\(F_P\)): The pressure \(P\) acts on the projected area \(D \times L\), where \(D\) is the internal diameter and \(L\) is the length. $$F_P = P \times (D \times L)$$
• Resisting force from the walls (\(F_R\)): This force is developed in the two wall sections, each with thickness \(t\) and length \(L\). The area resisting the force is \(2 \times t \times L\). $$F_R = \sigma_h \times (2 \times t \times L)$$

For equilibrium, the pressure force must balance the resisting force: $$F_P = F_R$$ $$P D L = \sigma_h (2 t L)$$

Solving for the circumferential stress ($\sigma_h$): $$\sigma_h = \frac{P D L}{2 t L}$$ $$\sigma_h = \frac{P D}{2 t}$$

Longitudinal Stress Derivation

The longitudinal stress, also called axial stress ($\sigma_l$), acts along the length or axis of the cylinder. It is caused by the internal pressure acting on the closed ends of the vessel, which tends to pull the ends apart.

To derive the formula for longitudinal stress, consider a cross-section of the cylinder. The internal pressure acts on the circular area of the end cap, and this force is resisted by the tensile stress developed in the cylindrical wall around its circumference.

• Force due to internal pressure (\(F_P\)): The pressure \(P\) acts on the internal cross-sectional area of the cylinder, which is \(\frac{\pi D^2}{4}\). $$F_P = P \times \left(\frac{\pi D^2}{4}\right)$$
• Resisting force from the wall (\(F_R\)): This force is developed in the wall material around the entire circumference. The approximate area resisting this force is \(\pi D t\) (circumference multiplied by thickness, for thin vessels). $$F_R = \sigma_l \times (\pi D t)$$

For equilibrium, the pressure force must balance the resisting force: $$F_P = F_R$$ $$P \left(\frac{\pi D^2}{4}\right) = \sigma_l (\pi D t)$$

Solving for the longitudinal stress ($\sigma_l$): $$\sigma_l = \frac{P \pi D^2}{4 \pi D t}$$ $$\sigma_l = \frac{P D}{4 t}$$

Stress Ratio Calculation

The question asks for the ratio of circumferential (hoop) stress to the longitudinal stress. We have derived the formulas for both:

• Circumferential (hoop) stress: $$\sigma_h = \frac{P D}{2 t}$$
• Longitudinal stress: $$\sigma_l = \frac{P D}{4 t}$$

Now, let's find the ratio \(\frac{\sigma_h}{\sigma_l}\): $$\frac{\sigma_h}{\sigma_l} = \frac{\frac{P D}{2 t}}{\frac{P D}{4 t}}$$

To simplify this expression, we can multiply the numerator by the reciprocal of the denominator: $$\frac{\sigma_h}{\sigma_l} = \frac{P D}{2 t} \times \frac{4 t}{P D}$$ The terms \(P\), \(D\), and \(t\) cancel out, leaving: $$\frac{\sigma_h}{\sigma_l} = \frac{4}{2}$$ $$\frac{\sigma_h}{\sigma_l} = 2$$

Therefore, the ratio of circumferential (hoop) stress to longitudinal stress in a thin cylindrical pressure vessel is 2.0. This means the hoop stress is always twice the longitudinal stress.

Key Takeaways for Pressure Vessel Design

This fundamental relationship is vital in mechanical engineering and materials science for ensuring the structural integrity of pressure vessels, pipelines, and tanks.