The study of a material’s elastic behavior is very important. Almost every engineering design necessitates an understanding of material elastic behavior. In the building of different constructions such as bridges, columns, pillars, and beams, among others. It is critical to understand the strength of the materials used in building. The elastic behavior of materials is considered and learned mostly in three main cases:

• The thickness of the Steel Rope used in Cranes.
• Design of the bridges,
• Maximum Height of the mountains.

Thickness of the Steel Rope used in Cranes

Cranes used to lift loads use ropes that are designed so that the stress due to the maximum load does not exceed the breaking stress. It is also found that a collection of thinner wire strands when compacted together makes the rope stronger than a solid rope of the same cross-section. That is why crane ropes are made of several strands instead of one.

We can find how thick rope is required for a crane; The load should not deform the rope permanently. Therefore, the extension should not exceed the elastic limit. Mild steel has a yield strength \(\left(\sigma_y\right)\) of about 300 \(\times 10^6 \mathrm{Nm}^{-2}\).

Thus, the area of cross-section (A) of the rope should at least be:

Elastic limit, \(\sigma_y=\) Load to be lifted \(/\) Cross-sectional Area \(=W / A\)

The load to be lifted by the crane is equal to the weight (W) that is to lift which must be equal to the product of the mass \((M)\) of the weight lifted and gravitational acceleration \((g)\) that is, \(M g\).

And since, the rope is circular-shaped of radius (r) so, its area of cross-section (A) is \(\pi r^2\).
Therefore,
\(
\sigma_y=W / A=M g / \pi r^2
\)

Thus, the area of cross-section \((A)\) of the rope should at least be
\(A \geq W / \sigma_y=M g / \sigma_y \dots(9.16)\)
i.e.
\(
\begin{aligned}
A & \geq\left(10^4 \mathrm{~kg} \times 10 \mathrm{~ms}^2\right) /\left(300 \times 10^6 \mathrm{Nm}^{-2}\right) \\
& =3.3 \times 10^{-4} \mathrm{~m}^2
\end{aligned}
\)
This corresponds to the radius of about \(1 \mathrm{~cm}\) for a rope of circular crosssection. Generally, a large margin of safety (of about a factor of ten in the (load) is provided. Thus, a thicker rope of radius about \(3 \mathrm{~cm}\) is recommended.

Design of the Bridges

A bridge has to be designed in such a way that it should have the capacity to withstand the load of the flowing traffic, the force of winds, and even its own weight. In the design of buildings, the use of beams and columns is very common. In both cases, the overcoming of the problem of bending of the beam under a load is of prime importance. The beam should not bend too much or break.

Let us consider the case of a beam loaded at the centre and supported near its ends as shown in the Figure below. A bar of length \(l\), breadth \(b\), and depth \(d\) when loaded at the centre by a load \(W\) sags by an amount given by
\(
\delta=W l^3 /\left(4 b d^3 Y\right) \dots(9.17)
\)

Maximum Height of the Mountains

Let us assume, the height of the mountains at the bottom is \(h\), the force per unit area due to the weight of the mountain is \(h \rho g\) where \(\rho\) is the density of the material of the mountain and \(g\) is the acceleration due to gravity.

The material at the bottom experiences this force in the vertical direction, and the sides of the mountain are free. Therefore, this is not a case of pressure or bulk compression. There is a shear component, approximately \(h \rho g\) itself. Now the elastic limit for a typical rock is \(30 \times 10^7 \mathrm{~N} \mathrm{~m}^{-2}\).

Equating this to \(h \rho g\), with \(\rho=3 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\) gives
\(
\begin{aligned}
h \rho g & =30 \times 10^7 \mathrm{~N} \mathrm{~m}^{-2} . \\
h & =30 \times 10^7 \mathrm{~N} \mathrm{~m}^{-2} /\left(3 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} \times 10 \mathrm{~ms}^{-2}\right) \\
& =10 \mathrm{~km}
\end{aligned}
\)
which is more than the height of Mt. Everest!