5.0 Types of Modulus of Elasticity

  Elasticity

1. Young’s Modulus of Elasticity

It is defined as the ratio of the longitudinal stress to the longitudinal strain.

$$Young's\ modulus\left( Y \right) = \frac{{Longitudinal\ Stress}}{{Longitudinal\ Strain}}$$

Consider a metal wire $AB$ of negligible mass of length $l$, radius $r$ and of uniform area of cross-section '$a$'. Let it be suspended from a rigid support at $A$, as shown in Fig. E. 9.

Let a longitudinal force $F$ be applied at its free end $B$ and let its length increases by $\Delta L$. Then, $$LongitudinalStrain = \frac{{\Delta l}}{l}$$ and $$Longitudinal\ Stress = \frac{F}{a} = \frac{F}{{\pi {r^2}}}$$

Therefore, $$Y = \frac{{\frac{F}{{\pi {r^2}}}}}{{\frac{{\Delta l}}{l}}} = \frac{{Fl}}{{\pi {r^2}\Delta l}}$$

Example 2. One end of a steel wire $1\ m$ long an $0.1\ m^2$ in cross section is fixed in a ceiling and a load of $5\ kg$ is attached to the free end. Find the extension in the wire.

Young’s modulus of steel = $2 \times {10^{11}}N{m^{ - 2}}$. Take $g = 10m{s^{ - 2}}$.

Solution: $$l = 1m,{\text{ }}A = 0.1{m^2},{\text{ }}m = 5kg$$ With reference to the Fig. E. 9.

The tension ($T$) produced in the wire due to the load is $mg$.

$$\begin{equation} \begin{aligned} T = mg \\ T = 5 \times 10 = 50N \\\end{aligned} \end{equation} $$

So, the stress developed in the wire is,

$$Stress = \frac{{Tension}}{{Area}} = \frac{{50}}{{0.1}} = 500N{m^{ - 2}}$$

We know that,

$$\begin{equation} \begin{aligned} Y = \frac{{Stress}}{{Strain}} \\ Stain = \frac{{Stress}}{Y} \\ \frac{{\Delta l}}{l} = \frac{{Stress}}{Y} \\ \Delta l = \frac{{l \times Stress}}{Y} = \frac{{1 \times 500}}{{2 \times {{10}^{11}}}} \\ \Delta l = 2.5 \times {10^{ - 9}}m = 2.5nm \\\end{aligned} \end{equation} $$