Please scroll down to get to the study materials.

- Note
- Things to remember
- Videos

In order to stretch a wire, certain work must be done against restoring force. This work done is stored in the form of potential energy (U) of the stretched wire.

Suppose a wire having length 'L' and cross-section area (A) is being deformed by a deforming force 'F' . Then the small work done 'dw' to elongated in by 'dl' is given by

$$dw = F dl \dots (i)$$

Hence, the total work done 'w' to elongated the wire by length 'l' is obtained by integrating dw between the limits o and 'l'.

Therefore,

$$W = \int dW $$

$$W = \int _o^l F dl \dots {ii}$$

Here 'F' depends upon the elongation. So when the elongation is 'l' the definition of young's Modulus (Y) gives

$$Y =\frac {stress}{strain} = \frac {F/A}{l/L} = frac {FL}{Al}$$

$$\therefore F = \frac {YAl}{L} \dots (iii)$$

Using this in equation (ii) and integrating, we get

$$dW = \int_0^l \frac {YAl}{L} dl $$

$$= \frac {YA}{L} \left [\frac l2 \right ]_0^l$$

$$= \frac {YA}{L} \left [\frac {l^2 - 0^2}{2} \right ]$$

$$= \frac 12 \frac {YA}{L} l^2 $$

$$= \frac 12 \left (\frac {YAl}{L} \right ) l $$

$$= \frac 12 F \times l $$

$$\therefore W= \frac 12F \times l $$

$$\boxed {= \frac 12 \text {force} \times \text {elongation}}$$

This work done is equal to the potential energy stored (U).

Energy stored (U) = work done (w)

$$ U = \frac 12\text {force} \times \text {elongation} $$

**Energy Density (U)**

The energy stored per unit volume of the stretched wire is called its energy density.

$$ \text {Energy density} (U) =\frac {\text {enrgy stored}}{\text {volume}} $$

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

$$=\frac 12 \frac FA \times \frac lL $$

$$= \frac 12 \text {stress} \times \text {strain} $$

$$\therefore Energy Density = \frac 12 \text {stress} \times \text {strain} $$

When a deforming force is applied at the free end of a suspended wire, the length of the wire increases and its dimension decreases. Thus there occurs longitudinal strain as well as lateral strain. The change in dimensions per unit original dimension along the force is applied is called longitudinal strain. It is denoted by α, while the change in dimension in the per unit dimension in perpendicular direction is called lateral strain and is denoted by β. Experimentally, it has been found that within the elastic limit the lateral strain β is directly proportional to the longitudinal strainα, i.e

$$ \beta \propto \alpha $$

$$\text {or} \beta = \sigma \alpha $$

$$\text {or} \sigma = \frac {\beta}{\alpha} $$

whereσ is proportionality constant which is known as Poisson's ratio and is defined as the ratio of lateral strain to the longitudinal strain within elastic limit.

$$\text {poisson's ratio,} \:\sigma = \frac {\beta}{\alpha} = \frac {\text {llateral strain}}{\text {longitudinal strain}}$$

Let L, D be the original length and diameter respectively of the wire, and dL and d be the slight increase in length and a corresponding slight decrease in diameter of the wire as shown in the figure.

$$\text {Longitudional strain,} \:\alpha = \frac {dL}{L} $$

$$\text {Lateral strain,} \:\beta = \frac {d}{D} $$

Therefore, Poisson 's ratio for the material of the wire is given by

$$ \sigma = \frac {\beta}{\alpha} \frac {d/D}{dL/L}$$

$$\boxed {\therefore\sigma = \frac dD \times \frac {L}{dL}}$$

Poisson's ratio is a unitless and dimensionless quantity.

We know that on removing the deforming forces from the elastic bodies, they regain their original configuration. Some elastic bodies regain their original configuration immediately, but a few others take some tome to recover their original configurations. The temporary delay in regaining the original configuration by an elastic body after removal of a deforming force is called elastic after effect.

When a body is repeating deformed and released, its elasticity is lost. This is known as elastic fatigue. Due to elastic fatigue it is possible to break an iron nail after deforming repeatedly.

Due to elastic after-effect the strain produced in an elastic lags behind the stress to which it is subjected and this effect is known as elastic hysteresis.

When a body is subjected to the road cycle of increasing and decreasing stress then the strain produced is larger when the body is unloaded then when it is loaded.

In order to stretch a wire, certain work must be done against restoring force. This work done is stored in the form of potential energy (U) of the stretched wire.

The energy stored per unit volume of the stretched wire is called its energy density.

The change in dimensions per unit original dimension along the force is applied is called longitudinal strain.

The change in dimension in the per unit dimension in perpendicular direction is called lateral strain and is denoted by β.

The temporary delay in regaining the original configuration by an elastic body after removal of a deforming force is called elastic after effect.

When a body is repeating deformed and released, its elasticity is lost. This is known as elastic fatigue.

Due to elastic after effect the strain produced in an elastic lags behind the stress to which it is subjected and this effect is known as elastic hysteresis.

.-
## You scored /0

You must login to reply

## Anup Tiwari

Why is rubber used as vibration absorber?

Mar 02, 2017

0 Replies

Successfully Posted ...

## Please Wait...