Question

Poisson's ratio is 0.4, longitudinal strain is 2*10^3, so what will be the volume percentage?

Answer 1

Answer on Question #51730, Physics, Molecular Physics | Thermodynamics

Poisson's ratio is 0.4, longitudinal strain is $2*10^{-3}$, so what will be the volume percentage?

Solution:

Poisson's Ratio can be expressed as

$$\nu = - \frac {\varepsilon_ {t}}{\varepsilon_ {l}}$$

where

$\nu =$ Poisson's ratio

$\varepsilon_{t} =$ transverse strain

$\varepsilon_{l} =$ longitudinal or axial strain

Strain can be expressed as

$$\varepsilon = \frac {\Delta L}{L}$$

where

$\Delta L =$ change in length (m, ft)

$L =$ initial length (m, ft)

For a cube stretched in the x-direction with a length increase of $\Delta L$ in the x direction, and a length decrease of $\Delta L'$ in the y and z directions

$$\nu \approx \frac {\Delta L ^ {\prime}}{\Delta L}$$

The relative change of volume $\Delta V / V$ of a cube due to the stretch of the material can now be calculated. Using $V = L^3$ and

$$V + \Delta V = (L + \Delta L) (L - \Delta L ^ {\prime}) ^ {2}$$ $$\frac {\Delta V}{V} = \left(1 + \frac {\Delta L}{L}\right) \left(1 - \frac {\Delta L ^ {\prime}}{L}\right) ^ {2} - 1$$

Using the above derived relationship between $\Delta L$ and $\Delta L'$:

$$\frac {\Delta V}{V} = \left(1 + \frac {\Delta L}{L}\right) ^ {1 - 2 v} - 1$$

and for very small values of $\Delta L$ and $\Delta L'$, the first-order approximation yields:

$$\frac {\Delta V}{V} \approx (1 - 2 \nu) \frac {\Delta L}{L}$$

Hence,

$$\frac {\Delta V}{V} \approx (1 - 2 * 0.4) * 2 * 10 ^ {- 3} = 0.0004 \text{ or } 0.04 \%$$

Answer: $\frac{\Delta V}{V} = 0.04\%$

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