Question

A thin-walled spherical gas balloon of radius \(R\) and wall thickness \(t\) (\(t \ll R\)) is subjected to an internal (gauge) pressure \(p\). The maximum tensile and shear stresses in the balloon wall are, respectively:

• A. Zero and \(pR/2t\)
• B. \(pR/2t\) and Zero
• C. \(pR/2t\) and \(pR/4t\)
• D. \(pR/4t\) and Zero

Hint:

For a sphere, hoop stress and meridional stress are equal, so the in-plane shear stress collapses to zero.

Answer 1

The Correct Option is B

Step 1: Recall the thin-wall (membrane) stress formula for a spherical shell.
For a thin-walled sphere of radius \(R\) and wall thickness \(t\) (\(t \ll R\)), under internal gauge pressure \(p\), the wall carries load by membrane action alone. By symmetry, the hoop (circumferential) stress and the meridional (longitudinal) stress are equal at every point on the sphere, and each equals
\[ \sigma_1 = \sigma_2 = \frac{pR}{2t} \]
This comes from cutting the sphere along a diametral plane and balancing the internal pressure force, \(p \cdot \pi R^2\), against the wall's resisting force, \(\sigma \cdot 2\pi R t\), which gives \(\sigma = pR/2t\).

Step 2: Identify the maximum tensile stress.
Since both in-plane principal stresses equal \(pR/2t\) and this is the largest stress in the thin membrane wall, the maximum tensile stress in the balloon wall is
\[ \sigma_{max} = \frac{pR}{2t} \]

Step 3: Find the maximum shear stress on the thin-walled element.
For the plane-stress element used in thin-shell membrane theory, which carries only the two in-plane stresses \(\sigma_1\) and \(\sigma_2\), the maximum in-plane shear stress is
\[ \tau_{max} = \frac{\sigma_1 - \sigma_2}{2} \]
Since \(\sigma_1=\sigma_2=pR/2t\) exactly, because the sphere is symmetric in every direction (unlike a cylinder, where hoop and longitudinal stress differ), this difference is zero:
\[ \tau_{max} = \frac{(pR/2t)-(pR/2t)}{2} = 0 \]
This equal biaxial stress state is exactly why a sphere is the most efficient shape for a pressure vessel: there is no shear distortion tendency in the wall's own plane, unlike a cylindrical vessel where the hoop stress is twice the longitudinal stress and a nonzero shear stress appears.

Step 4: Match to the options.
Maximum tensile stress \(=pR/2t\), maximum shear stress \(=0\), which is option (B). Option (A) reverses the two values. Options (C) and (D) bring in \(pR/4t\), which would only enter if a third principal stress (the through-thickness radial stress) were also being compared, but the standard membrane thin-wall treatment used here works with the two equal in-plane stresses only.

Final Answer:
Maximum tensile stress \(= pR/2t\) and maximum shear stress \(=\) Zero, so option (B) is correct. \[ \boxed{pR/2t \text{ and } 0} \]