Question

A cylinder made of rubber (length $= L$ and diameter $= d$) is inserted in a rigid container as shown in the figure. The rubber cylinder fits snugly in the rigid container. There is no wall friction. The modulus of elasticity of the rubber is $E$ and its Poisson’s ratio is $\nu$. The cylinder is subjected to a small uniform pressure $p$ as shown in the figure. The resulting axial strain ($\varepsilon_{zz}$) is:

• A. $-\dfrac{p}{E}$
• B. $-\dfrac{p}{E}(1 - 2\nu)$
• C. $-\dfrac{p}{E}\dfrac{(1 + \nu)(1 - 2\nu)}{(1 - \nu)}$
• D. $-\dfrac{p}{E}\dfrac{(1 - \nu)(1 - 2\nu)}{(1 + \nu)}$

Hint:

When lateral strain is constrained ($\varepsilon_x = \varepsilon_y = 0$), convert the problem into a triaxial stress state with $\sigma_x = \sigma_y = \dfrac{\nu}{1-\nu}\sigma_z$. This is a standard plane strain formulation in elasticity.

Answer 1

The Correct Option is C

Step 1: Identify the stress state.
- The rubber cylinder is compressed axially by uniform pressure $p$ along $z$-axis. - Because the cylinder fits snugly in the rigid container, there is **no lateral strain**: \[ \varepsilon_x = \varepsilon_y = 0 \] This is a **uniaxial strain** problem under triaxial stress conditions.
Step 2: Generalized Hooke’s law (3D).
For isotropic materials: \[ \varepsilon_x = \frac{1}{E} \big( \sigma_x - \nu (\sigma_y + \sigma_z) \big) \] \[ \varepsilon_y = \frac{1}{E} \big( \sigma_y - \nu (\sigma_x + \sigma_z) \big) \] \[ \varepsilon_z = \frac{1}{E} \big( \sigma_z - \nu (\sigma_x + \sigma_y) \big) \]
Step 3: Apply boundary conditions.
- Since lateral strains vanish: \[ \varepsilon_x = 0, \varepsilon_y = 0 \] By symmetry: $\sigma_x = \sigma_y$. Let $\sigma_x = \sigma_y = \sigma_r$ (radial stress). \[ 0 = \frac{1}{E} (\sigma_r - \nu(\sigma_r + \sigma_z)) \] \[ \sigma_r - \nu \sigma_r - \nu \sigma_z = 0 \] \[ \sigma_r (1 - \nu) = \nu \sigma_z \] \[ \sigma_r = \frac{\nu}{1 - \nu}\sigma_z \]
Step 4: Substitute into axial strain expression.
Axial strain: \[ \varepsilon_z = \frac{1}{E} \big(\sigma_z - \nu(\sigma_x + \sigma_y)\big) \] \[ \varepsilon_z = \frac{1}{E} \big(\sigma_z - \nu(2\sigma_r)\big) \] Substitute $\sigma_r$: \[ \varepsilon_z = \frac{1}{E} \Big(\sigma_z - 2\nu . \frac{\nu}{1 - \nu}\sigma_z \Big) \] \[ \varepsilon_z = \frac{\sigma_z}{E} \Big(1 - \frac{2\nu^2}{1 - \nu}\Big) \] \[ \varepsilon_z = \frac{\sigma_z}{E} . \frac{1 - \nu - 2\nu^2}{1 - \nu} \]

Step 5: Simplify.
Factor numerator: \[ 1 - \nu - 2\nu^2 = (1 + \nu)(1 - 2\nu) \] So: \[ \varepsilon_z = \frac{\sigma_z}{E} . \frac{(1 + \nu)(1 - 2\nu)}{1 - \nu} \] Since $\sigma_z = -p$ (compression): \[ \varepsilon_{zz} = -\frac{p}{E} . \frac{(1 + \nu)(1 - 2\nu)}{1 - \nu} \] Final Answer: \[ \boxed{-\dfrac{p}{E}\dfrac{(1 + \nu)(1 - 2\nu)}{(1 - \nu)}} \]