Thin-walled cylinder under internal pressure If p is the internal pressure in a thin cylinder of diameter d and wall thickness t, what is the circumferential (hoop) stress σ_h developed in the shell?
-
Aσ_h = p * d / t
-
Bσ_h = p * d / (2 * t)
-
Cσ_h = p * d / (4 * t)
-
Dσ_h = 2 * p * d / t
-
Eσ_h = p * t / d
Answer
Correct Answer: σ_h = p * d / (2 * t)
Explanation
Introduction / Context:Thin-cylinder formulas are used widely for boilers, pipes, and pressure vessels where wall thickness is small relative to diameter (t ≪ d). Two principal membrane stresses develop: hoop and longitudinal.
Given Data / Assumptions:
- Internal pressure p, cylinder diameter d, wall thickness t.
- Thin-wall assumption: stress is uniform through thickness; t/d is small.
- No end caps or fittings altering membrane stress locally.
Concept / Approach:Equilibrium of half the cylinder under pressure gives the hoop stress. Cutting along a diametral plane and balancing pressure force with resisting stresses yields the standard expression.
Step-by-Step Solution:
Pressure force on projected area = p * d * L (per unit length L).Resisting hoop force = 2 * σ_h * t * L.Equate: p * d * L = 2 * σ_h * t * L ⇒ σ_h = p * d / (2 * t).Verification / Alternative check:Longitudinal stress σ_l from end-cap equilibrium: σ_l = p * d / (4 * t), exactly half the hoop stress—consistent with thin-cylinder theory.
Why Other Options Are Wrong:
- p d / t and 2 p d / t overestimate; they ignore the factor 2 from two walls resisting.
- p d / (4 t) is the longitudinal stress, not hoop stress.
- p t / d is dimensionally inconsistent for stress.
Common Pitfalls:Confusing hoop and longitudinal formulas or missing the factor of 2 in the hoop-stress derivation.
Final Answer:σ_h = p * d / (2 * t)