Explain with illustration cranes regarding the applications of elastic behaviour of materials.

For in all engineering designs, elastic behaviour of materials play important role. Let us consider the illustration of cranes for this.

Cranes used for lifting and moving heavy loads from one place to another have a thick metal rope to which the load is attached and so rope (Cable) is under stress.

Suppose we want to make a crane, which has a lifting capacity of $10$ tonnes or metric tons ( $1$ metric ton $=1000 \mathrm{~kg}$ ). How thick should the steel rope be ?

For that load does not deform the rope permanently. Therefore, the extension should not exceed the elastic limit.

Means the value of yield strength $\mathrm{S}_{y}$ produced in rope is less than the value of elastic limit. Suppose the least cross sectional area of rope of mild steel is A and yield strength of mild steel $\left(\mathrm{S}_{y}\right)$ is $300 \times 10^{6} \mathrm{~N} \mathrm{~m}^{-2}$

$\therefore \mathrm{A} \geq \frac{\mathrm{W}}{\mathrm{S}_{y}}$

$=\frac{\mathrm{Mg}}{\mathrm{S}_{y}}$

$\quad=\frac{10^{4} \mathrm{~kg} \times 10 \mathrm{~ms}^{-2}}{300 \times 10^{6} \mathrm{Nm}^{-2}}$

$\quad=3.3 \times 10^{-4} \mathrm{~m}^{2}$

$\therefore \mathrm{A} \geq 3.3 \times 10^{-4} \mathrm{~m}^{2}$

If $g=3.1 \pi \mathrm{ms}^{-2}$ and $\mathrm{A}=\pi r^{2}$ then

From $\mathrm{A}=\frac{\mathrm{Mg}}{\mathrm{S}_{y}} \quad\left[\because g=9.8=3.1 \times \pi \mathrm{ms}^{-2}\right]$