Strain $=\frac{\text{ Change in size of the body }}{\text{ Original size of the body }}$

Longitudinal strain $=\frac{\text{ change in length of the body }}{\text{ initial length of the body }}=\frac{\Delta L}{L}$

Volume strain $=\frac{\text{ change in volume of the body }}{\text{ original volume of the body }}=\frac{\Delta \mathrm{V}}{\mathrm{V}}$

Shear strain, $\tan \varphi =\frac{\ell }{\mathrm{L}}\text{ or }$

$\varphi =\frac{\ell }{L}=\frac{\text{ displacement of upper face }}{\text{ distance between two faces }}$

$\text{ Stress }=\frac{\text{ Internal restoring force }}{\text{ Area of cross }-\text{ section }}=\frac{F_{\text{in }}}{A}$

$\Rightarrow$ $\sigma =\frac{F_{in}}{A}$

Compressibility,

$C=\frac{1}{\text{ Bulk modulus }}=\frac{1}{\mathrm{K}}$

Work done in stretching wire

$W=\frac{1}{2}\times \text{ stress }\times \text{ strain }\times \text{ volume: }$

$\mathrm{W}=\frac{1}{2}\times \frac{\mathrm{F}}{\mathrm{A}}\times \frac{\Delta \ell }{\ell }\times \mathrm{A}\times \ell =\frac{1}{2}\mathrm{F}\times \Delta \ell$

$(\sigma )=\frac{\text{ lateral strain }}{\text{ Longitudinal strain }}$

Let a bar be of length l and diameter d. When a tensile force is applied along the length of the bar, the length increases l + $\Delta$l and the diameter decreases d $-$ $\Delta$d.

$\therefore$ lateral strain = $\frac{\Delta d}{d}$

Longitudinal strain = $\frac{\Delta l}{l}$

$\therefore$ Poisson's ratio of bar components $\sigma =\frac{\frac{\Delta d}{d}}{\frac{\Delta l}{l}}$

$\therefore$ $\sigma =\frac{l\Delta d}{d\Delta l}$

$\Delta \ell=\frac{\mathrm{MgL}}{2 \mathrm{AY}}=\frac{\rho g \mathrm{~L}^2}{2 \mathrm{Y}}$

Let an object of mass m have volume V and density $\rho$ under pressure P.

If pressure increases P + dP, then density becomes $\rho$ + d$\rho$ and volume becomes V + dV.

$\therefore$ Bulk Modulus, $K=$ $\rho \frac{dP}{d\rho }$

Within elastic limit,

Stress $\propto$ strain

stress = k $\times$ strain

where k is the proportionality constant and is known as modulus of elasticity.

$$ \mathrm{K}=\frac{\text { Volume stress }}{\text { Volume strain }}=\frac{\mathrm{F} / \mathrm{A}}{\left(\frac{-\Delta \mathrm{V}}{\mathrm{V}}\right)}=\frac{\mathrm{P}}{\left(\frac{-\Delta \mathrm{V}}{\mathrm{V}}\right)} $$

Bulk modulus of an ideal gas is process dependent.

- For isothermal process, $\mathrm{PV}=$ constant

$$ \Rightarrow \mathrm{PdV}+\mathrm{VdP}=0 \Rightarrow P=\frac{-\mathrm{dP}}{\mathrm{dV} / \mathrm{V}} $$

So bulk modulus $=P$

- For adiabatic process $PV^\gamma=$ constant

$$ \begin{aligned} &\Rightarrow \gamma \mathrm{PV}^{\gamma-1} \mathrm{dV}+\mathrm{V}^\gamma \mathrm{dP}=0 \\\\ &\Rightarrow \gamma \mathrm{PdV}+\mathrm{VdP}=0 \Rightarrow \gamma \mathrm{P}=\frac{-\mathrm{dP}}{\mathrm{dV} / \mathrm{V}} \end{aligned} $$

So bulk modulus $=\gamma \mathrm{P}$

- For any polytropic process $\mathrm{PV}^{\mathrm{n}}=$ constant

$$ \begin{aligned} &\Rightarrow \mathrm{nPV}^{\mathrm{n}-1} \mathrm{dV}+\mathrm{V}^{\mathrm{n}} \mathrm{dP}=0 \Rightarrow \mathrm{PdV}+\mathrm{VdP}=0 \\\\ &\Rightarrow \mathrm{nP}=\frac{-\mathrm{dP}}{\mathrm{dV} / \mathrm{V}} \end{aligned} $$

So bulk modulus $=n P$

$\mathrm{C}=\frac{\pi \eta r^4}{2\ell }$ Where $\eta$ is modulus of rigidity $r$ and $\ell$ is radius and length of wire respectively.

(a) Toque required for twisting by an angle $\theta ,\tau =\mathrm{C}\theta$.

(b) Work done in twisting by an angle $\theta ,\mathrm{W}=\frac{1}{2}\mathrm{C}{\theta }^2$.

$\mathrm{Y}=\frac{\text{ Longitudinal stress }}{\text{ Longitudinal strain }}=\frac{\mathrm{F}\ell }{\mathrm{A}\Delta \ell }$

If $L$ is the length of wire, $r$ is radius and $\ell$ is the increase in length of the wire by suspending a weight Mg at its one end then Young's modulus of elasticity of the material of wire

$\mathrm{Y}=\frac{\left(\mathrm{Mg}/\pi {\mathrm{r}}^2\right)}{(\ell /\mathrm{L})}=\frac{\mathrm{MgL}}{\pi {\mathrm{r}}^2\ell }$

Young's coefficient (Y), Bulk Modulus (K), Modulus of rigidity (n) of a solid, these three are called elastic coefficients and Poisson's ratio ($\sigma$) are the elastic constants of the material.

Among them the following two relations can be proved,

$\frac{Y}{3K}=1-2\sigma$ ...... (1)

$\frac{Y}{2n}=1+\sigma$ ...... (2)

Doing (1) + (2), $\frac{Y}{3K}+\frac{Y}{n}=3$

or, $Y.\frac{n+3K}{3Kn}=3$

or, $Y=\frac{9Kn}{n+3K}$

Doing (1) $÷$ (2) , $\frac{Y/3K}{Y/2n}=\frac{1-2\sigma }{1+\sigma }$

or, $\frac{2n}{3K}=\frac{1-2\sigma }{1+\sigma }$

or, $2n+2n\sigma =3K-6K\sigma$

or, $(6K+2n)\sigma =3K-2n$

$\therefore$ $\sigma =\frac{3K-2n}{6K+2n}$

Let a bar of circular cross-section be of length l and radius r.

$\therefore$ Bar volume $v=\pi r^{2}l$

by differentiating we get, $dv=\pi r^{2}dl+\pi l.2rdr$

or, $\frac{dv}{v}=\frac{1}{v}\left(\pi r^{2}dl+2\pi lrdr\right)$

$=\frac{\pi r^{2}l}{\pi r^{2}l}\left(\frac{dl}{l}+2\frac{dr}{r}\right)$ [$\because$ $v=\pi r^{2}l$]

$\therefore$ $\frac{dv}{v}=\frac{dl}{l}+2\frac{dr}{r}$

If Poisson's ratio of bar components is $\sigma$, then

$\sigma =-\frac{dr/r}{dl/l}$

or, $\frac{dr}{r}=-\sigma \frac{dl}{l}$

$\therefore$ $\frac{dv}{v}=\frac{dl}{l}-2\sigma \frac{dl}{l}$

or, $\frac{dv}{v}=\frac{dl}{l}\left(1-2\sigma \right)$

$\therefore$ Volume Strain = Longitudinal Strain $\times$ (1 $-$ 2 $\times$ Poisson's ratio)

Note : Poisson's ratio under linear deformation of an object where volume deformation is negligible i.e. volume remains almost unchanged is $\sigma$ = 0.5

If volume deformation is negligible compared to linear deformation, then $\frac{dv/v}{dl/l}\to 0$

$\therefore$ $1-2\sigma =0$

or, $\sigma =0.5$.