A thin cylindrical shell provided with hemispherical end-caps is subjected to the same internal pressure and material allowables as the spherical ends. Compare the required wall thickness of the cylindrical shell with that of the hemispherical ends.

Introduction:Pressure vessels commonly use cylindrical shells with hemispherical ends. For the same internal pressure and allowable stress, the required thickness differs for cylinder versus sphere due to different membrane stress distributions.

Given Data / Assumptions:

• Same internal pressure p and radius r.
• Same allowable tensile stress sigma_t.
• Thin-shell (membrane) theory; negligible bending.

Concept / Approach:For a thin cylinder: hoop stress sigma_h = p * r / t ⇒ t_cyl = p * r / sigma_t. For a thin sphere: membrane stress sigma_sph = p * r / (2 * t) ⇒ t_sph = p * r / (2 * sigma_t). Thus, for identical p, r, and sigma_t, t_cyl = 2 * t_sph. Hence, the cylindrical shell requires more thickness than the hemispherical ends.

Step-by-Step Solution:1) Write cylinder thickness: t_cyl = p * r / sigma_t.2) Write sphere thickness: t_sph = p * r / (2 * sigma_t).3) Compare: t_cyl / t_sph = 2 ⇒ t_cyl is double.4) Conclude: cylinder requires more thickness.

Verification / Alternative check:Design codes and handbooks consistently show spherical shells as the most material-efficient shape for internal pressure, validating the relation.

Why Other Options Are Wrong:Equal to / less than: contradict membrane theory.

Indeterminate / depends only on joint efficiency: primary dependence is on geometry and membrane equations; joint efficiency modifies required t but does not invert the basic relation.

Common Pitfalls:Using longitudinal stress instead of hoop stress for cylinders; overlooking the factor of 2 between cylinder and sphere equations.

Final Answer:more than