Question

What is the expression for the determination of Poisson's ratio for rubber?

Answer 1

Step 1: Poisson's ratio for rubber:

1. Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of the stretching force. The definition of Poisson's ratio contains a minus sign so that normal materials have a positive ratio.
2. The formula for calculating the volume of rubber is given as,

$V=AL...\left(1\right)$

Where, $V$ is the volume of the rubber, $A$ is the area of the rubber, and $L$ is the length of the rubber.

Step 2: Differentiating the expression:

Derivate equation (1).

$$\begin{gathered} V=AL \\ \frac{dV}{dL}=-L\frac{dA}{dL}+A...\left(2\right) \end{gathered}$$

Where, $dV$ is the small change in the volume, and $dL$ is the corresponding increase in length, and $dA$ is the change in the area.

As the duration rises, the negative sign reduces.

Step 3: Calculate the area of the rubber given as,

$$\begin{gathered} A=\pi R^2 \\ \Rightarrow \frac{dA}{A}=2\frac{dR}{R} \end{gathered}$$

Apply the chain rule on the left-hand side,

$$\begin{gathered} \frac{dA}{dL}\times \frac{dL}{A}=2\frac{dR}{R} \\ \frac{dA}{dL}=\frac{2A\frac{dR}{R}}{dL}...\left(3\right) \end{gathered}$$

Substitute equation (3) in equation (1).

$$\begin{gathered} \Rightarrow \frac{dV}{dL}=\frac{-2A\frac{dR}{R}}{\frac{dL}{L}}+A \\ \Rightarrow \frac{dV}{dL}=(-2\sigma +1)A \end{gathered}$$

Where, $\sigma$ is the Poisson's ratio for rubber.

Hence, $\sigma =\frac{1}{2}(1-\frac{dV}{AdL})$ is the expression for the determination of Poisson's ratio for rubber.