Thin-walled cylindrical vessel under internal pressure: what is the expression for the longitudinal (axial) stress?

Difficulty: Medium

Correct Answer: σ_l = p D / (4 t)

Explanation:


Introduction / Context:
Design of thin-walled pressure vessels distinguishes between circumferential (hoop) stress and longitudinal (axial) stress. Correct formulas are critical for sizing wall thickness, selecting materials, and verifying code compliance. For thin cylinders (t ≪ D), membrane theory provides simple expressions used in preliminary design and quick checks.


Given Data / Assumptions:

  • Internal pressure p (uniform).
  • Cylinder mean diameter D and wall thickness t, with t ≪ D.
  • Closed ends carrying internal pressure load.


Concept / Approach:
The longitudinal stress arises from the pressure force on the end caps, balanced by the axial stress in the cylindrical wall across the cross-sectional metal area. Force balance yields σ_l = p D / (4 t), whereas the hoop stress from tangential force balance is σ_h = p D / (2 t). The longitudinal stress is thus half the hoop stress in a thin cylinder.


Step-by-Step Solution:

End force = pressure * end area = p * (π D^2 / 4).Metal resisting area (axial) = π D t.Set equilibrium: σ_l * (π D t) = p * (π D^2 / 4).Solve: σ_l = p D / (4 t).


Verification / Alternative check:
Compare with hoop stress formula σ_h = p D /(2 t); longitudinal is half of hoop for thin cylinders, matching classic results.


Why Other Options Are Wrong:

  • p D /(2 t) is the hoop stress, not longitudinal.
  • 2 p D / t and p t / D do not follow membrane equilibrium for thin shells.
  • p / 2 lacks dimensional consistency.


Common Pitfalls:
Using internal radius rather than mean diameter inconsistently; applying thin-wall formulas when t is not small relative to D.


Final Answer:
σ_l = p D / (4 t)

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