Thin cylindrical pressure vessel: For a cylinder of diameter d and thickness t under internal pressure p, what is the longitudinal (axial) membrane stress in the shell?
-
Ap d / t
-
Bp d / (2 t)
-
Cp d / (4 t)
-
Dp d / (6 t)
-
E2 p d / t
Answer
Correct Answer: p d / (4 t)
Explanation
Introduction / Context:Thin-walled pressure vessels develop two principal membrane stresses: circumferential (hoop) and longitudinal (axial). Correct formulas are vital for sizing thickness and checking allowable stress compliance in boilers, tanks, and pipelines.
Given Data / Assumptions:
- Thin cylinder, t ≪ d.
- Internal pressure p; ends closed.
- Uniform membrane stresses; neglect radial stress.
Concept / Approach:Equilibrium on a free-body of half the vessel gives hoop stress sigma_h = p d / (2 t). Equilibrium on the end cap gives longitudinal stress sigma_L = p d / (4 t). The longitudinal stress is half the hoop stress in a thin cylinder with closed ends.
Step-by-Step Solution:
Hoop: sigma_h = p d / (2 t).End equilibrium: p * (π d^2 / 4) balanced by 2 * sigma_L * t * (d / 2) * π → sigma_L = p d / (4 t).Hence the longitudinal stress is p d / (4 t).Verification / Alternative check:Check ratio: sigma_h / sigma_L = 2, consistent with thin cylinder theory.
Why Other Options Are Wrong:
- p d / t and 2 p d / t are excessive (do not match equilibrium).
- p d / (2 t) is the hoop, not longitudinal, stress.
- p d / (6 t) has no basis in thin-cylinder derivation.
Common Pitfalls:Using hoop formula for both directions; forgetting that end closures create axial stress; applying thick-cylinder relations to thin shells.
Final Answer:p d / (4 t)