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What is Young’s modulus ? Explain. and Give its unit and dimensional formula.
Solution
Experimental observation show that for a given material the magnitude of the strain produced is same whether the stress is tensile or compressive.
The ratio of tensile (or compressive) stress $(\sigma)$ to the longitudinal strain $(\varepsilon)$ is defined as Young's modulus and is denoted by the symbol Y.
$\text { Young modulus }=\frac{\text { Tensile stress }(\sigma)}{\text { Longitudinal } \operatorname{strain}(\varepsilon)}$
$\quad \mathrm{Y}=\frac{\sigma}{\varepsilon}$
$\therefore \mathrm{Y}=\frac{(\mathrm{F} / \mathrm{A})}{(\Delta \mathrm{L} / \mathrm{L})}=\frac{(\mathrm{F} \times \mathrm{L})}{(\mathrm{A} \times \Delta \mathrm{L})}$
Here, strain is a dimensionless quantity, the unit of Young's modulus is the same as that of stress. Mean $\mathrm{N} \mathrm{m}^{-2}$ or Pascal $(\mathrm{P} a)$.
Dimensional formula : $\left[\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]$.
Young's moduli, elastic limit and tensile strength of some materials are given below.
Substance |
Young's Modulus $10^{9} \mathrm{~N} / \mathrm{m}^{2}$ $\sigma_{y}$ |
Elastic limit $10^{7} \mathrm{~N} / \mathrm{m}^{2}$ $%$ |
Tensile strength $10^{7} \mathrm{~N} / \mathrm{m}^{2}$ $\sigma_{u}$ |
Aluminium | $70$ | $18$ | $20$ |
Copper | $120$ | $20$ | $40$ |
Iron (Wrought) | $190$ | $17$ | $33$ |
Steel | $200$ | $30$ | $50$ |
Bone (Tensile) (Compressive) |
$16$ $9$ |
$12$ $12$ |
For metals Young's moduli are large therefore these materials require a large force produce small change in length.
Steel is more elastic than copper, brass and aluminium. It is for this reason that steel is preferred in heavy-duty machines and in structural designs.
Wood, bone, concrete and glass have rather small Young's moduli.