Thin cylinder, open ends – longitudinal strain due to internal pressure A thin-walled cylinder of radius r and thickness t is open-ended (no end caps). It is subjected to internal pressure p. The Young’s modulus is E and Poisson’s ratio is μ. What is the longitudinal (axial) strain?

Introduction / Context:
Thin pressure vessels develop hoop and longitudinal stresses under internal pressure. Whether the ends are closed or open changes the longitudinal stress state and hence the axial strain. Understanding this distinction is essential in vessel design.

Given Data / Assumptions:

• Thin cylinder (t ≪ r), linear elasticity.
• Open-ended ⇒ no net end force ⇒ longitudinal stress σ_l = 0.
• Hoop (circumferential) stress σ_h = p r / t for thin cylinders.
• Material constants: E (Young’s modulus), μ (Poisson’s ratio).

Concept / Approach:
Generalized Hooke’s law for axial strain in a biaxial stress state is epsilon_l = (1/E) * (σ_l − μ σ_h) (neglecting radial stress). With σ_l = 0 for open ends, the axial strain arises only from Poisson’s effect of the hoop stress and is negative (axial contraction) if μ > 0.

Step-by-Step Solution:
Compute hoop stress: σ_h = p r / t.Set σ_l = 0 (no end caps).Axial strain: epsilon_l = (1/E) * (0 − μ σ_h) = − μ (p r) / (E t).

Verification / Alternative check:
For closed ends, σ_l = p r / (2 t), giving epsilon_l = (1/E) * (σ_l − μ σ_h) = (p r / (2 t E)) − μ (p r / (E t)), different from the open-ended result, confirming the role of end restraint.

Why Other Options Are Wrong:

• (a) would be true only for μ = 0, which is not general.
• (b), (d), (e) ignore the sign and Poisson coupling; they predict axial extension instead of contraction.

Common Pitfalls:
Assuming the same longitudinal stress for open and closed cylinders; always check end conditions before computing strains.

Final Answer:
− μ (p r) / (E t)