Maximum shear stress in a thin cylindrical shell subjected to internal pressure $ p $ is

Question

A rectangular section has width $b$ and depth $d$ and a symmetrical triangular section has base $b$ and depth $d$. The ratio of moment of inertia of rectangular section to that of triangular section with respect to their respective centroidal axis will be:

Answer 1

Step 1: The three principal stresses on the cylinder wall are hoop $ \sigma_h = \dfrac{pd}{2t} $, longitudinal $ \sigma_l = \dfrac{pd}{4t} $, and radial $ \sigma_r \approx 0 $ (thin-wall approximation).
Step 2: The absolute maximum shear stress is half the difference between the algebraically largest and smallest principal stresses, here $ \sigma_h $ and $ \sigma_r $.
Step 3: Substituting these values:
\[ \tau_{\max} = \dfrac{\sigma_h - \sigma_r}{2} = \dfrac{\dfrac{pd}{2t} - 0}{2} = \dfrac{pd}{4t} \]

Answer 2

A simpler route uses the relation that, whenever the radial stress on a thin shell is negligible, the absolute maximum shear stress works out to exactly half of the largest in-plane principal stress, the hoop stress. Checking this ratio against each option:

$ \dfrac{pd}{t} $: Dividing this by the hoop stress $ \dfrac{pd}{2t} $ gives a ratio of 2, not the expected $ \tfrac{1}{2} $, so this value is too large to be the maximum shear stress.
$ \dfrac{pd}{2t} $: This is equal to the hoop stress itself (a ratio of 1), but the maximum shear stress must be half the hoop stress, not equal to it.
$ \dfrac{pd}{4t} $: This is exactly half of the hoop stress $ \dfrac{pd}{2t} $, matching the expected ratio of $ \tfrac{1}{2} $ between maximum shear stress and the largest principal stress when the third principal stress is negligible.
$ \dfrac{pd}{8t} $: This is one-quarter of the hoop stress, smaller than the expected half-hoop-stress ratio, corresponding instead to the in-plane shear between hoop and longitudinal stress alone, not the true maximum.

The half-of-hoop-stress relation confirms the maximum shear stress is $ \dfrac{pd}{4t} $.

Therefore, the correct answer is $ \dfrac{pd}{4t} $.

Answer 3

A simpler route uses the relation that, whenever the radial stress on a thin shell is negligible, the absolute maximum shear stress works out to exactly half of the largest in-plane principal stress, the hoop stress. Checking this ratio against each option:

$ \dfrac{pd}{t} $: Dividing this by the hoop stress $ \dfrac{pd}{2t} $ gives a ratio of 2, not the expected $ \tfrac{1}{2} $, so this value is too large to be the maximum shear stress.
$ \dfrac{pd}{2t} $: This is equal to the hoop stress itself (a ratio of 1), but the maximum shear stress must be half the hoop stress, not equal to it.
$ \dfrac{pd}{4t} $: This is exactly half of the hoop stress $ \dfrac{pd}{2t} $, matching the expected ratio of $ \tfrac{1}{2} $ between maximum shear stress and the largest principal stress when the third principal stress is negligible.
$ \dfrac{pd}{8t} $: This is one-quarter of the hoop stress, smaller than the expected half-hoop-stress ratio, corresponding instead to the in-plane shear between hoop and longitudinal stress alone, not the true maximum.

The half-of-hoop-stress relation confirms the maximum shear stress is $ \dfrac{pd}{4t} $.

Therefore, the correct answer is $ \dfrac{pd}{4t} $.

Maximum shear stress in a thin cylindrical shell subjected to internal pressure \( p \) is

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The Correct Option is C

Approach Solution - 1

Approach Solution -2

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