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

A wire of metal is stretched with a load. Prove that in my unit volume of the wire, The stored potential energy is

\frac{1}{2} Y {(∆ X)}^{2}

we're

Y

is the Young's modulus the material of the wire and is the longitudinal on the strain.

Accepted Answer

When a wire is stretched, some work is done against the internal restoring 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 of the wire is fixed and the other end is being staged by an external force $F . Y$

Is Young's modulus of the material of the wire. Due to the application of force $F$ the increase in length of occurs by $l$ .

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

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

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

The work done by external force to extend by an amount $d l$ is

$d w = F d l$

From equation (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.

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

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

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

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

The 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 $(i i)$ is written as

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

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

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

Volume $U = \frac{1}{2} \times$ Stress $\times$ starin .................... $(i i i)$

Now $Y = \frac{Stress}{strain}$

stress $=$ $Y \times$ strain ...................... $(i v)$

From equation $(i i i)$ and $(i v)$

$U = \frac{1}{2} Y \times s t r a i n \times s t r a i n$

$U = \frac{1}{2} \times Y \times {(s t r a i n)}^{2}$

A wire of metal is stretched with a load. Prove that in my unit volume of the wire, The stored potential energy is $\frac{1}{2} Y {(∆ X)}^{2}$ we're $Y$ is the Young's modulus the material of the wire and is the longitudinal on the strain.

Important Questions on Elasticity

Learn and Prepare for any exam you want !

A wire of metal is stretched with a load. Prove that in my unit volume of the wire, The stored potential energy is 12Y∆X2 we're Y is the Young's modulus the material of the wire and is the longitudinal on the strain.

Important Questions on Elasticity

Learn and Prepare for any exam you want !

A wire of metal is stretched with a load. Prove that in my unit volume of the wire, The stored potential energy is $\frac{1}{2} Y {(∆ X)}^{2}$ we're $Y$ is the Young's modulus the material of the wire and is the longitudinal on the strain.