Thin-walled cylindrical pressure vessel: what is the circumferential (hoop) stress under internal pressure p for diameter d and wall thickness t?

Difficulty: Easy

Correct Answer: pd/2t

Explanation:


Introduction / Context:
Mechanical design of thin-walled cylindrical vessels (e.g., columns, drums, and pipes) requires calculating membrane stresses due to internal pressure. The largest stress is the circumferential or hoop stress, which dictates wall thickness for a given allowable stress and corrosion allowance.


Given Data / Assumptions:

  • Thin-walled assumption: t << d (typically t ≤ d/10).
  • Uniform internal pressure p.
  • Neglecting stress concentrations near nozzles and supports.


Concept / Approach:
The thin-cylinder model treats the wall as a membrane carrying tension. Force balance on a longitudinally cut half-cylinder equates internal pressure force to resisting hoop stress resultant. This yields the standard hoop stress relation used in ASME-based preliminary sizing.


Step-by-Step Solution:
Pressure force on projected area: F_p = p * d * L (for unit length L).Resisting force: 2 * (σ_h * t * L) = hoop stress times two wall strips.Set equilibrium: p * d * L = 2 * σ_h * t * L → σ_h = p d / (2 t).


Verification / Alternative check:
Longitudinal stress from end caps is σ_L = p d / (4 t), exactly half of hoop stress, matching classical thin-wall results and validating the derivation.


Why Other Options Are Wrong:
pd/4t is the longitudinal stress, not hoop; pd/t and 2pd/t overestimate; pd/3t has no basis in thin-wall equilibrium.


Common Pitfalls:
Using outside diameter vs. mean diameter inconsistently; applying thin-wall formula to thick shells; ignoring joint efficiency factors during final code calculations.


Final Answer:
pd/2t

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