Question

The relationship between the hoop stress \(\sigma_1\) and the longitudinal stress \(\sigma_2\) of a closed cylindrical thin-walled pressure vessel is

• A. \(\sigma_1 = 2\sigma_2\)
• B. \(\sigma_1 = \frac{1}{3}\sigma_2\)
• C. \(\sigma_1 = \sigma_2\)
• D. \(\sigma_1 = \frac{\sigma_2}{2}\)

Hint:

A simple way to remember the relationship is to think about how a cylinder would fail. It's more likely to split open along its length (due to hoop stress) than to be pulled apart at its ends (due to longitudinal stress). This implies that the hoop stress is the larger of the two. Specifically, Hoop Stress = 2 \(\times\) Longitudinal Stress. For a thin-walled sphere, the stress is uniform in all directions and equals the longitudinal stress of a cylinder (\(pD/4t\)).

Answer 1

The Correct Option is A

Step 1: Understanding Thin-Walled Stresses:
In a thin-walled cylinder (where thickness $t<D/20$), internal pressure $P$ creates two primary tensile stresses. 1. Hoop Stress ($\sigma_1$): Also called circumferential stress. It acts along the circumference of the cylinder. It is derived by considering a longitudinal section of the cylinder. \[ \sigma_1 = \frac{PD}{2t} \]
Step 2: Calculating Longitudinal Stress:
2. Longitudinal Stress ($\sigma_2$): Also called axial stress. It acts along the length of the cylinder, trying to push the ends off. It is derived by considering a transverse (cross-sectional) cut of the cylinder. \[ \sigma_2 = \frac{PD}{4t} \]
Step 3: Comparing the Stresses:
Comparing the two expressions: \[ \sigma_1 = \frac{PD}{2t} = 2 \times \left( \frac{PD}{4t} \right) = 2\sigma_2 \] This shows that the stress trying to burst the cylinder open along its side is twice as high as the stress trying to pull it apart axially.