Name: Poisson's Ratio
Uploaded: 2019-07-19
Duration: 9 min 46 s
Description: In this video, let us discuss Poisson's ratio, the most interesting and important measure of a property of a material. Watch this to explore.

Question

If a wire is the stretched by applying an external force, then prove that the elastic potential energy per unit volume of the wire is equal to

\frac{1}{2}

x

(stress

x

strain).

Accepted Answer

When a wire is stretched, some work is done against the internal re-storing forces acting between the particles of the wire. This work done appears as elastic potential energy in the wire. Now we sell derive an expression for the increase in elastic potential energy of the wire when it is stretched from its natural length.

Let the initial length of the wire L and area of cross-section is $A$ . One and off the wire is fixed and the other end is being a staged by an external force $F . Y$
Is the Young's modulus of the material of the wire. Due to the application of force F the increase in length of occurs
by $l$ .

But $Y = \frac{Longitudinal stress}{Longitudinal strain}$

$Y = \frac{\frac{F}{A}}{\frac{l}{L}} = \frac{F L}{A l}$

$F = \frac{Y A l}{L} . . . . . . . . . . . . . . . . . . (i)$

The work done by external force to extend by an amount dl is

$d w = F d l$

From (i)

$d w = \frac{Y A l}{L} d l$

Total work done in producing an extension from $0$ to $1$ is obtained by integrating
both sides under proper limits.

$\int_{0}^{w} d w = \int_{0}^{l} \frac{Y A l}{L} d l$

${[w]}_{0}^{w} = \frac{Y A}{L} \int_{0}^{l} l d l = \frac{Y A}{L} {[\frac{l^{2}}{2}]}_{0}^{l}$

$w - 0 = \frac{Y A}{L} [l^{2} - 0^{2}]$

$w = \frac{Y A l^{2}}{2 L}$

This work done is stored in the form of potential
energy called elastic potential energy denoted by $U$ .

$U = \frac{A Y}{2 L} l^{2} . . . . . . . . . . . . . . . . . . . . . (i i)$

Equation (ii) is written as

$U = \frac{1}{2} [Y \frac{l}{L}] \frac{l}{L} \cdot A L$ [ $∴$ [Multiply and divide by $L$ ]

$U = \frac{1}{2}$ stress $\times$ strain $\times$ volume

if volume $= 1$ means elastic P.E. per unit

Volume $U = \frac{1}{2}$ stress $\times$ strain H.P.

If a wire is the stretched by applying an external force, then prove that the elastic potential energy per unit volume of the wire is equal to $\frac{1}{2}$ $x$ (stress $x$ strain).

Important Questions on Elasticity

Learn and Prepare for any exam you want !

If a wire is the stretched by applying an external force, then prove that the elastic potential energy per unit volume of the wire is equal to 12 x (stress x strain).

Important Questions on Elasticity

Learn and Prepare for any exam you want !

If a wire is the stretched by applying an external force, then prove that the elastic potential energy per unit volume of the wire is equal to $\frac{1}{2}$ $x$ (stress $x$ strain).