This question asks for the ratio between the longitudinal strain and the hoop strain experienced by a thin cylindrical shell when subjected to an internal pressure.

Understanding Stresses in a Thin Cylindrical Shell

When a thin cylindrical shell is subjected to an internal pressure ($p$), it experiences stresses along its circumference (hoop stress) and along its length (longitudinal stress).

• Hoop Stress ($\sigma_h$): This stress acts tangentially to the circumference. The formula is: $$ \sigma_h = \frac{pd}{2t} $$ where $p$ is the internal pressure, $d$ is the diameter of the shell, and $t$ is the thickness of the shell wall.
• Longitudinal Stress ($\sigma_l$): This stress acts parallel to the axis of the cylinder. The formula is: $$ \sigma_l = \frac{pd}{4t} $$ Notice that the hoop stress is twice the longitudinal stress.

Calculating Strains Based on Stresses and Poisson's Ratio

The strains in the shell are related to these stresses through the material's Young's modulus ($E$) and Poisson's ratio ($\nu$).

• Hoop Strain ($\epsilon_h$): This is the strain in the circumferential direction. It is caused by the hoop stress ($\sigma_h$) and is affected by the longitudinal stress ($\sigma_l$) due to Poisson's effect. The formula is: $$ \epsilon_h = \frac{\sigma_h}{E} - \nu \frac{\sigma_l}{E} $$ Substituting the stress formulas: $$ \epsilon_h = \frac{1}{E} \left( \frac{pd}{2t} - \nu \frac{pd}{4t} \right) $$ Simplifying this expression gives: $$ \epsilon_h = \frac{pd}{2Et} \left( 1 - \frac{\nu}{2} \right) $$
• Longitudinal Strain ($\epsilon_l$): This is the strain in the axial direction. It is caused by the longitudinal stress ($\sigma_l$) and is affected by the hoop stress ($\sigma_h$) due to Poisson's effect. The formula is: $$ \epsilon_l = \frac{\sigma_l}{E} - \nu \frac{\sigma_h}{E} $$ Substituting the stress formulas: $$ \epsilon_l = \frac{1}{E} \left( \frac{pd}{4t} - \nu \frac{pd}{2t} \right) $$ Simplifying this expression gives: $$ \epsilon_l = \frac{pd}{4Et} \left( 1 - 2\nu \right) $$

Determining the Ratio of Longitudinal Strain to Hoop Strain

To find the required ratio, we divide the expression for longitudinal strain by the expression for hoop strain:

$$ \frac{\epsilon_l}{\epsilon_h} = \frac{\frac{pd}{4Et} \left( 1 - 2\nu \right)}{\frac{pd}{2Et} \left( 1 - \frac{\nu}{2} \right)} $$

We can cancel out the common terms $\frac{pd}{Et}$:

$$ \frac{\epsilon_l}{\epsilon_h} = \frac{\frac{1}{4} \left( 1 - 2\nu \right)}{\frac{1}{2} \left( 1 - \frac{\nu}{2} \right)} $$

Multiply the numerator and denominator by 4 to simplify:

$$ \frac{\epsilon_l}{\epsilon_h} = \frac{1 \cdot \left( 1 - 2\nu \right)}{2 \cdot \left( 1 - \frac{\nu}{2} \right)} = \frac{1 - 2\nu}{2 - \nu} $$

The question provides the Poisson's ratio as $\nu = \frac{1}{m}$. Substituting this value into the ratio:

$$ \frac{\epsilon_l}{\epsilon_h} = \frac{1 - 2\left(\frac{1}{m}\right)}{2 - \left(\frac{1}{m}\right)} $$

To simplify further, multiply the numerator and denominator by $m$:

$$ \frac{\epsilon_l}{\epsilon_h} = \frac{m \left( 1 - \frac{2}{m} \right)}{m \left( 2 - \frac{1}{m} \right)} = \frac{m - 2}{2m - 1} $$

This result matches one of the given options.