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:

Question

When a thin cylindrical shell is subjected to internal fluid pressure, the stresses developed in its wall are

Answer 1

Use Mohr's circle for the biaxial stress state on the sphere's wall, instead of the direct membrane-formula argument.

Plot the stress state: at any point on the thin spherical wall, the two in-plane principal stresses (hoop and meridional) are $\sigma_1=\sigma_2=pR/2t$, both tensile and equal in every direction around that point, since a sphere has no preferred direction the way a cylinder does.
Draw Mohr's circle: Mohr's circle for principal stresses $\sigma_1$ and $\sigma_2$ has its centre at $(\sigma_1+\sigma_2)/2$ and radius $(\sigma_1-\sigma_2)/2$. Since $\sigma_1=\sigma_2$ here, the radius is $0$, so the circle collapses to a single point on the normal-stress axis.
Read off the maximum values: a Mohr's circle of radius $0$ means the normal stress is the same, $pR/2t$, on every plane through that point, and the shear stress is $0$ on every plane. So the maximum tensile (normal) stress is $pR/2t$ and the maximum shear stress is $0$.

This matches the physical picture: a sphere spreads pressure load equally in all directions on its surface, so there is no tendency for one direction to shear against another within the wall, which is exactly why spherical pressure vessels are structurally efficient compared to cylindrical ones, where hoop stress is twice the longitudinal stress and a nonzero shear stress appears.

The maximum tensile and shear stresses in the balloon wall are $pR/2t$ and Zero, respectively, matching option (B).

Answer 2

Use Mohr's circle for the biaxial stress state on the sphere's wall, instead of the direct membrane-formula argument.

Plot the stress state: at any point on the thin spherical wall, the two in-plane principal stresses (hoop and meridional) are $\sigma_1=\sigma_2=pR/2t$, both tensile and equal in every direction around that point, since a sphere has no preferred direction the way a cylinder does.
Draw Mohr's circle: Mohr's circle for principal stresses $\sigma_1$ and $\sigma_2$ has its centre at $(\sigma_1+\sigma_2)/2$ and radius $(\sigma_1-\sigma_2)/2$. Since $\sigma_1=\sigma_2$ here, the radius is $0$, so the circle collapses to a single point on the normal-stress axis.
Read off the maximum values: a Mohr's circle of radius $0$ means the normal stress is the same, $pR/2t$, on every plane through that point, and the shear stress is $0$ on every plane. So the maximum tensile (normal) stress is $pR/2t$ and the maximum shear stress is $0$.

This matches the physical picture: a sphere spreads pressure load equally in all directions on its surface, so there is no tendency for one direction to shear against another within the wall, which is exactly why spherical pressure vessels are structurally efficient compared to cylindrical ones, where hoop stress is twice the longitudinal stress and a nonzero shear stress appears.

The maximum tensile and shear stresses in the balloon wall are $pR/2t$ and Zero, respectively, matching option (B).

Air Pollutants		Treatment techniques
P	NO₂	i	Flaring
Q	SO₂	ii	Cyclonic separator
R	CO	iii	Lime scrubbing
S	Particles	iv	NH₃ injection

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:

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The Correct Option is B

Solution and Explanation

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