$Y = /\left(\frac{(F / A)}{(\Delta l/l_0)}\right)=\frac{\text { Stress }}{\text { Strain }}$

$[Y]=\left[\frac{M L T^{-2}}{L^2} \times 1\right]$

Unit of Y $=\frac{N}{m^2}$.

Load causes an elongation.

Reading is noted of the difference between the two vernier scales when there are equal weights as compared to when there are unequal weights.

Difference between readings is taken as the elongation.

Examples

Which material has larger $Y$ ? $\rightarrow$ (b)

Which of the two is a stronger material? $\rightarrow$ (b)

Complete Bulk modulus of water from the data.

Initial volume $V_i=100.0 \mathrm{L}$

Pressure increase, $\Delta P=100.0 \mathrm{atm}$

$1 atm = 1.013 \times 10^5 \mathrm{Pa}$

Final volume, $V_f=100.5 \mathrm{L}$

$B=\Delta P / \frac{\Delta V}{V_i} \quad \Delta V=V_f-V_i=0.5 \mathrm{L}$

=$\frac{100 \times 1.013 \times 10^5 \times 100 \mathrm{L}}{0.5 \mathrm{L}}=2.026 \times 10^9 \mathrm{Pa} =2.026 \times 10^9 \mathrm{N} /\mathrm{m}^2$

Determine the volume contraction of solid copper cube 10cm on each edge when subjected to a hydraulic pressure of $7.0\times {10^6} Pa$.

Given: $B_{cu} =140 \times 10^9 \mathrm{N} / \mathrm{m}^2$

$B = \frac{\Delta P}{( \Delta V / \mathrm{V_i})}$

$V_i =(10 \mathrm{cm})^3$=$(0.1 \mathrm{m})^3=0.001 \mathrm{m}^3$

$\Delta V=\frac{\Delta P}{B} \times V_i =\frac{7 \times 10^6}{140 \times 10^9} \times 0.001 \mathrm{m}^3$

$=0.5 \times 10^{-6} \mathrm{m}^3$

Two wires each of diameter 0.25 cm, one made of steel and the other of brass as shown below. The unloaded length of the steel wire is 1.5m and that of the brass wire is 1.0m. Compute the elongations of the steel and brass wires.

Given: $Y_{\text {steel }}=20 \times 10^{10} \mathrm{N} / \mathrm{m}^2 \text { and } Y_{\text {brass}}=9.0 \times 10^{10} \mathrm{N} / \mathrm{m}^2$

$d=0.25 \mathrm{cm}=25 \times 10^{-4} \mathrm{m}$

$Y=\frac{F}{A} \times \frac{l_0}{\Delta l}$

$\Delta l=\frac{F \times l_0}{A \times Y}$

Steel

$\Delta l_{\text {steel}} =\frac{100 \times 1.5 \times 4}{\pi(25)^2 \times 10^{-8} \times 20 \times 10^{10}}m$

$=1 \times 10^{-4} \mathrm{m}$

Brass

$\Delta l_{\text {brass}} =\frac{60 \times 1 \times 4}{\pi(25)^2 \times 10^{-8} \times 9 \times 10^{10}}=1.35 \times 10^{-4} \mathrm{m}$

The initial area of the cylinder is $A_0=\pi\left(\frac{l_0}{2}\right)^2 \quad d_0=3.9 \mathrm{mm}$

Original length, $L_0 = \frac{\Delta l \times Y}{F/A_0}$

$= \frac {(0.4 \times 10^{-3}) \times (108 \times 10^6 N/m^2) \pi (3.9 \times 10^{-3})^2}{4 \times 2000 N}$

$=0.257 m=257 \mathrm{mm}$

1) Toughness

2) Brittleness

3) Hardness

4) Resilience

5) Stiffness

Toughness

Ability of a material to absorb energy and plastically deform without rupturing.

Ceramics- low toughness

Ruller- high toughness

Brittleness

Breaks being subjected to stress without undergoing significant deformation under strain.

Hardness

Measure of how resistant a material in to a permanent shape change when subjected to an applied force.

Resilience

Resilience is the capacity of a material to absorb energy when it is deformed elastically and then the energy in released upon unloading.

It is obtained from the area of the stress v/s strain graph.

$U_{\text {res }}=\frac{1}{2} \int_0^{(\Delta x) d} \sigma d x=\frac{1}{2} \frac{F}{A} \int_0^{(\Delta x) d} x$

$U_{\text {res }}=\frac{1}{2} \frac{F}{A}(\Delta x) d$

Stiffness

Stiffness is the ratio of the steady force acting on an elastic body to the resulting displacement. As such, a stiffer material has a higher elastic modulus.