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Application of Elasticity

Application of Elasticity :- Elasticity has important applications in the fields of mechanical, civil, aeronautical and materials engineering, for example determining the strength and load carrying ability of engineering structures including buildings, bridges, cars, planes, and thousands of machine parts. Some of the application of elasticity are :-

(1) Metallic ropes used in cranes :-

A thick metallic rope is used in cranes to lift heavy objects from one place to another. Determination of the maximum load that can be lifted by the crane is important. The load to be lifted must be within the elastic limit of the wire.

Let us suppose that the wire is made of steel and the maximum load to be lifted is 10⁵ kg. The elastic limit of steel is 3 × 10⁸ N/m². So the stress developed in the wire must be less than elastic limit.

$\displaystyle Stress=\frac{Weight(mg)}{Area(A)}\le Elastic\text{ }Limit$

$\displaystyle \Rightarrow \text{Area of cross section of wire(}A\text{) }\ge \frac{mg}{Elastic\text{ }Limit}$

$\displaystyle {{A}_{\min }}=\frac{{{10}^{5}}\times 9.8}{3\times {{10}^{8}}}\approx 3.3\times {{10}^{-3}}{{m}^{2}}$

If r is the radius of circular procession of the wire then,

$\displaystyle \pi r_{\min }^{2}=3.3\times {{10}^{-3}}$

$\displaystyle \Rightarrow {{r}_{\min }}=\sqrt{\frac{3.3\times {{10}^{-3}}}{\pi }}\approx 0.032m=3.2cm$

So to lift an object of mass 10⁵ kg, the minimum radius of the wire used in the crane should be 3.2cm.

For safety point of view, the area of cross section of the wire is taken approximately 10 times larger than the calculated value. In that case the radius of the rope comes out to be approximately 10 cm.

Keeping in mind the flexibility, strength and convenience of construction, the rope is made by wrapping many thin wires together.

(2) In designing beam to support load (in construction of roofs and bridges) :-

When a horizontal rod is supported between two rigid supports and a weight W is applied at its midpoint, then the depression(δ) in the rod is given by :-

$\displaystyle \delta =\frac{W{{l}^{3}}}{4b{{d}^{3}}Y}\text{ }.....(1)$

Here l, b and d are the length, breath and depth of the girder respectively.

It is clear from equation (1) that for a given load and length of the beam, the depression(δ) will be low if width(b), Young’s modulus(Y) and the depth of the beam(d) is large. From equation (1),

$\displaystyle \delta \propto \frac{1}{{{d}^{3}}}$

So increasing the depth(d) of the beam will be more economical because with decrease in depression, the amount of steel for making of beam will be low.

Note :- There is one problem with increasing depth of the beam. As we increase the depth of the beam, it may bend as shown in the figure below. This is called buckling.

To prevent buckling, the girder is given the shape of letter I.

(3) Estimating the maximum height of a mountain :-

In a mountain of height h having rock density ρ the pressure at the base of it is given by :-

P = hρg = stress

The elastic limit of a typical rock is 3 × 10⁸ N m^–2. The stress must be less than the elastic limit, otherwise the rock begins to flow. Hence

$\displaystyle \Rightarrow h<\frac{P}{\rho g}$

$\displaystyle \therefore {{h}_{\max }}=\frac{3\times {{10}^{8}}}{3\times {{10}^{3}}\times 10}=10km$

This height is higher than the height of Mount Everest(8849m), the highest mountain in the world.

(4) Preference of hollow shaft than the solid shaft :-

A shaft is a cylindrical road which is used to provide rotational motion to machines. A hollow shaft is found to be stronger than a solid shaft made of same size and material. If one end of a shaft is kept fixed and other end is a twisted, then a shearing strain is produced in the shaft. The torque τ required for angle of twist θ is given by :-

$\displaystyle \tau =\frac{\eta \pi {{r}^{4}}\theta }{2l}$

here,

r = radius of shaft, l = length of shaft and η = shearing strain.

To produce unit twist(θ = 1 rad.) in a solid shaft, the torque required,

$\displaystyle {{\tau }_{solid}}=\frac{\eta \pi {{r}^{4}}}{2l}\text{ }.....(2)$

Similarly to produce unit twist in a hollow shaft, the torque required,

$\displaystyle {{\tau }_{hollow}}=\frac{\eta \pi (r_{2}^{4}-r_{1}^{4})}{2l}\text{ }.....(3)$

From equation (2) and (3),

$\displaystyle \frac{{{\tau }_{hollow}}}{{{\tau }_{solid}}}=\frac{(r_{2}^{4}-r_{1}^{4})}{{{r}^{4}}}=\frac{(r_{2}^{2}+r_{1}^{2})(r_{2}^{2}-r_{1}^{2})}{{{r}^{4}}}\text{ }.....(4)$

As volume of both shaft is same

$\displaystyle \pi {{r}^{2}}l=\pi (r_{2}^{2}-r_{1}^{2})l$

$\displaystyle \Rightarrow {{r}^{2}}=r_{2}^{2}-r_{1}^{2}$

Using this in equation (4), we get,

$\displaystyle \frac{{{\tau }_{hollow}}}{{{\tau }_{solid}}}=\frac{r_{2}^{2}+r_{1}^{2}}{{{r}^{2}}}>1$

Hence,

$\displaystyle {{\tau }_{hollow}}>{{\tau }_{solid}}$

That is, in the solid and hollow shaft of the same material and mass, more work will have to be done to twist a hollow shaft in comparison to a solid shaft. So a hollow shaft is stronger than a solid shaft.

Note :- Torque required for unit twist(θ = 1 rad.) in a rod is called it’s Torsion constant.

$\displaystyle \text{Torsion constant(C) =}\frac{\eta \pi {{r}^{4}}}{2l}\text{ }$

(a) Toque required for twisting by angle θ, $\displaystyle \tau =C\theta$

(b) Work done in twisting by angle θ, $\displaystyle W=\frac{1}{2}C{{\theta }^{2}}$

(5) Bicycle and rickshaw frames are made of hollow pipes :-

If a cylindrical bar is loaded with weight W then depression(δ) is given by

$\displaystyle \delta =\frac{W{{l}^{3}}}{12Y\pi {{R}^{4}}}$

here R is the radius of cross section of the bar.

A hollow cylindrical rod of the same mass is stronger than a solid rod because the radius of the hollow rod is greater. This is why bicycle and rickshaw frames are made of hollow pipes.

Application of Elasticity

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