Contents

Physics Topics are often described using mathematical equations, making them precise and quantifiable.

## What are the Applications of Elasticity?

Suppose a wire of uniform cross section is clamped at its upper end with a rigid support and a load is applied at its lower end which is then gradually increased. As a result, the length of the wire goes on increasing. In Fig., the stress- strain graph of a ductile metallic wire has been shown. The different parts of this graph are described below.

i) Straight line OA: In this section, the stress on the wire is proportional to the strain, which means that the metal follows Hooke’s law. The wire behaves like a completely elastic body up to the point A.

ii) Point A: This point indicates the proportionality limit.

iii) Line segment AB: In this section, the ratio of stress and strain is comparatively less, which means that the metal does not follow Hooke’s law. However, after reaching this section of strain, if the stress is removed, then the wire will regain its original length, which means that the strain will again be zero.

iv) Point B : This point indicates the elastic limit. In the case of most metals, the two points A and B are found to be quite close to each other. In the case of glass, A and B are identical points and in the case of rubber, the distance between A and B is quite high.

v) Line segment BC: In this section, stress divided by strain becomes even less and the metal gradually loses its elastic property and becomes plastic. After reaching this section of strain, if the stress is removed, then the wire is unable to regain its original length. So, the wire undergoes a permanent deformation.

vi) Point C : This point is called the yield point or the upper yield point. At this point, the stress level is known as yield stress. The yield point for most substances cannot be determined accurately.

vii) Line segment CD : In this section, stress divided by strain is negative. It implies that even if stress is decreased, strain will increase.

viii) Point D: This point is called the lower yield point. If the stress is gradually decreased after the strain reaches this point, the return graph is not along DO, but along DO’. In this case OO’ indicates the permanent deformation. For bodies with nearly perfect elasticity, the points A, B, C and D are situated so close to one another that practically the four points can be assumed to be identical.

ix) Line segment DE : In this section, stress divided by strain is the least and the metal becomes plastic. The cross section of certain parts of a wire becomes comparatively lower than that of the remaining parts.

x) Point E: At point E the stress reaches its maximum value. At this point, the material of the wire flows like a viscous liquid and the wire becomes thin. Now, even if the load is decreased, the wire goes on thinning down.

xi) Line segment EF: In this section, the area of the cross section at different parts of the wire starts decreasing fast.

xii) Point F: At this point, the wire snaps from its weakest part. The stress corresponding to the point F is called breaking stress or the ultimate stress. This point F is called fracture or breaking point.

Ductile material: The materials which have large plastic range of extension are called ductile materials. As shown in the stress-strain curve in Fig., their fracture or breaking point is widely separated from the elastic limit. Such materials undergo an irreversible increase in length before snapping. So they can be drawn into thin wires. Copper, silver, iron, aluminium etc., are examples of ductile materials.

Brittle material: The materials which have very small range of plastic limit of extension are called brittle materials.

Such materials break as soon as the stress is increased beyond the elastic limit. Their breaking point lies just close to their elastic limit as shown in Fig.

Cast iron, glass, ceramic etc., are examples of brittle materials.

Necessity of the stress-strain graph: For practical purposes, knowledge of the load-extension graph of a metal is absolutely essential. From this graph, the elastic limit of the material can be known. For example, during the use of a machine, the stress developed on the axle or the other parts of the machine should be kept below the elastic limit of its material. For this, the stress-strain or the load-extension graphs of different materials are highly useful.

Elastic fatigue: If the force (or load) applied on an elastic body rises and falls rapidly and this periodic fluctuation continues for a long time, then the elastic property of the body gets degraded, even if the elastic limit for the material is not exceeded. It means that the body remains permanently deformed in some respect, i.e., some part of,the body becomes thinner and weaker, even after the deforming force is withdrawn. The body may then break or snap at a load less than the normal breaking load. This kind of degradation of the elastic property of material due to rapid changes in stress is called elastic fatigue.

### Experimental Verification of Hooke’s Law : Determination of Young’s Modulus

Searle’s Experiment: A uniform metal wire, about 2 to 3 m long, is hung from the roof of the laboratory. Initially, a small weight, usually called zero-load, is attached to the lower end of the wire. With the help of a screw gauge, the average diameter of the experimental wire is measured. Its length (L) is determined using a metre scale.

The load at the lower end of the wire is then gradually increased. Measuring the corresponding extensions with suitably fitted verniers, a graph can be plotted with the load along the X-axis and the elongation along the Y-axis. The graph passes through the origin and is a straight line [Fig.],

As the graph is a straight line, we can say that, load oc elongation, i.e., stress oc strain (within the elastic limit). This proves the validity of Hooke’s law.

Calculation: From any point P on the graph, two per-pendiculars are drawn on the axes.

Here, OQ = load (mg); OR = elongation (l).

Therefore, the longitudinal stress = \(\frac{m g}{\pi r^2}\) and the longitudinal strain = \(\frac{l}{L}\)

Since the quantities on the right-hand side of the above expression are known, the value of Young’s modulus (Y) can be determined.

### Applications of Elasticity in Daily Life

Most materials used in our daily life undergo some kind of stress. It is therefore important to design things in a way that they continue working under the stress suffered by them. The following examples will illustrate this point.

i) In cranes: The thickness of metallic ropes used in cranes to lift and move heavy weights is decided on the basis of the elastic limit of the material of the rope and the factor of safety.

ii) In designing abeam: When a transverse load is put across a horizontal beam, it causes a depression leading to bending in the beam. In various applications like bridges, buildings, etc., it is important that the beam can withstand the load or the weight and it should not bend too much or break. When the bending is not accompanied by any torsion or shear, it is said to be simple bending.

iii) A bridge is declared unsafe after long use.

During its long use, a bridge undergoes quick alternating strains repeatedly. It results in the loss of elastic strength of the bridge. After a long period such a bridge starts developing large strains corresponding to the same usual value of stress and ultimately it may lead to collapse of the bridge. To avoid such situation a bridge is declared unsafe after its long use.

### Numerical Examples

**Example 1.**

To increase the length of an elastic string of radius 3.5 mm by \(\frac{1}{20}\) th of its initial length, within its elastic limit, a 10 N force is required. Calculate the Young’s modulus for the material of the string.

**Solution:**

We know that Y = \(\frac{F L}{\pi r^2 l}\)

Here, F = 10 N, l = \(\frac{1}{20} L\) and r = 3.5 mm = 0.0035m.

∴ Y = \(\frac{10 \times L}{3.14 \times(0.0035)^2 \times \frac{L}{20}}\) = 5.2 × 10^{6} N ᐧ m^{-2}

**Example 2.**

Two wires of the same length but of different materials have diameters of 1 mm and 3 mm respectively. If both of them are stretched by the same force, then the elongation of the first wire becomes thrice that of the second. Compare their Young’s moduli.

**Solution:**

Let the initial length of each wire be L and the force applied on each be F.

**Example 3.**

If the elastic limit of a typical rock is 3 × 10^{8} N ᐧ m^{-2} and its mean density is 3 × 10^{3} kg ᐧ m^{-3}, estimate the maximum height of a mountain on the earth, (g = 10 m ᐧ s^{-2})

**Solution:**

Let us assume that the maximum height of the mountain is h and g is nearly uniform along this height. Then the maximum pressure at its bottom = hρg.

According to the problem,

hρg = breaking stress = elastic limit.

∴ h = \(\frac{\text { elastic limit }}{\rho g}\)

Here, ρ = 3 × 10^{3} kg ᐧ m^{-3} = 3 × 10^{8} N ᐧ m^{-2}

∴ h = \(\frac{3 \times 10^8}{3 \times 10^3 \times 10}\) = 10^{4} m

**Example 4.**

Two equal and opposite forces are applied tangentially to two mutually opposite faces of an aluminium cube of side 3 cm to produce a shear of 0.01°. If the modulus of rigidity for aluminium is 7 × 10^{10} N ᐧ m^{-2}, then calculate the force applied.

**Solution:**

Let the magnitude of applied force be F.

**Example 5.**

A rubber cord of length 20 m is suspended from a rigid support by one of its ends and it hangs vertically. What will be the elongation of the cord due to its own weight? The density of rubber = 1.5 g ᐧ cm^{-3} and Young’s modulus = 49 × 10^{7} N ᐧ m^{-2}.

**Solution:**

Here the downward force,

F = weight of the cord

= volume of the cord × density × acceleration due to gravity

= 20 × A × 1.5 × 1000 × 9.8 N

[A = area of cross section of the cord]

The centre of gravity of the rubber cord is at a vertical distance of 10 m from the fixed end. The weight of the cord acts through its centre of gravity. Hence the length of the upper half, above the centre of gravity, is taken as the initial length to estimate the elongation of the cord. So, L = 10 m.

∴ Y = \(\frac{F L}{A l}\)

or, l = \(\frac{F L}{A Y}\) = \(\frac{20 \times A \times 1.5 \times 1000 \times 9.8 \times 10}{A \times 49 \times 10^7}\) = 0.006 m

**Example 6.**

A steel wire of diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart in the same horizontal line. A body is hung from the middle point of the wire such that the middle point sags 1 cm from the original position. Calculate the mass of the body. [F = 2 × 10^{12} dyn ᐧ cm^{-2}] [HS ’02]

**Solution:**

In Fig., C is the mid-point of the wire AB. When a mass m is hung from C, it sags 1 cm and comes down to the point O.

Here, tension in the part OA or OB is T. The two horizontal components of T balance each other and the vertical com-ponents balance the weight mg.

∴ 2Tcosθ = mg or, T = \(\frac{m g}{2 \cos \theta}\)

The length AC of the wire changes into AO.

∴ Elongation = l = AO – AC = 50.01 – 50 = 0.01 cm

Young’s modulus, Y = \(\frac{T L}{A l}\) = \(\frac{m g L}{2 \cos \theta \cdot A l}\)

or, m = \(\frac{2 Y A l \cos \theta}{g L}\)

= \(\frac{2 \times\left(2 \times 10^{12}\right) \times 3.14 \times(0.04)^2 \times 0.01}{980 \times 50} \times \frac{1}{50.01}\)

= 82.01 g

**Example 7.**

**I**f the work done in stretching a uniform wire, of cross section 1 mm^{2} and length 2 m, by 1 mm is 0.05 joule, find the Young’s modulus for the material of the wire. [HS 03]

**Solution:**

Work done, W = \(\frac{1}{2} \frac{Y \alpha l^2}{L}\)

So, Y = \(\frac{2 W L}{\alpha l^2}\)

W = 0.05 J; α = 1 mm^{2} = 10^{-6} m^{2} ; L = 2 m ;

l = 1 mm = 10^{-3} m

∴ Y = \(\frac{2 \times 0.05 \times 2}{10^{-6} \times\left(10^{-3}\right)^2}\) N ᐧ m^{-2}

**Example 8.**

A cylindrical pipe of uniform cross section and of length 120 cm is closed at one end and is completely filled with water. Kept upright, the cylinder is stretched to increase its length. It elongates by 1 cm, but the length of the water-column increases by 0.7 cm. Calculate the Poisson’s ratio of the material of the pipe.

**Solution:**

Poisson’s ratio, σ = \(=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

Here, longitudinal strain = \(\frac{l}{L}\) = \(\frac{1}{120}\)

Let the initial diameter of the pipe be D, the decrease in its diameter due to elongation be d.

Initial volume of water = \(\frac{\pi}{4} D^2\) × 120

and final volume of water = \(\frac{\pi}{4}\)(D – d)^{2} × ( 120 + 0.7)

= \(\frac{\pi}{4}\)(D – d)^{2} × 120.7

Since the volume of water inside the cylinder remains unchanged,

**Example 9.**

A light bar of length 2 m is suspended horizontally by means of two wires of equal lengths connected to its two ends. One wire is of steel having a cross sectional area of 0.1 cm^{2}, the other is of brass with a cross sectional area of 0.2 cm^{2}. Find the point on the bar from where a weight must be suspended so that both the wires experience

(i) the same stress,

(ii) the same strain. Given, the Young’s modulus for steel = 2 × 10^{11} N ᐧ m^{-2} and that for brass = 10 × 10^{10} N ᐧ m^{-2}.

**Solution:**

Let the horizontal bar be AB [Fig.]. A load W is hung from the point C. Let AC be x. In this condition, let the tension in the steel wire be T_{1} and that in the brass wire be T_{2}

Stress on the steel wire = \(\frac{T_1}{\alpha_1}\) and

stress on the brass wire = \(\frac{T_2}{\alpha_2}\).

i) If the stresses in both the wires are the same, then

\(\frac{T_1}{0.1 \times 10^{-4}}\) = \(\frac{T_2}{0.2 \times 10^{-4}}\) or, \(\frac{T_1}{T_2}\) = \(\frac{1}{2}\) ……. (1)

Since the system is in equilibrium, taking moments about the point C, we get,

T_{1}x = T_{2}(2 – x) or, \(\)

So the weight should be suspended at a distance of 1.33 m from the steel wire.

ii) longitudinal stress \(\frac{\text { longitudinal stress }}{Y}\)

= \(\frac{T}{\frac{\alpha}{Y}}\) = \(\frac{T}{\alpha Y}\)

For the same strain in the two wires,

Since the system is in equilibrium, taking moments about the point C, we get,

T_{1}x = T_{2}(2 – x) or, x = 2 – x or, x = 1 m

So, the weight should be suspended from the midpoint of the horizontal bar.

**Example 10.**

A sphere of mass 25 kg and radius 0.1 m is hung from the ceiling of a room with the help of a steel wire. The height of the ceiling from the floor is 5.21 m. When the sphere is hung just like a pendulum, its lower surface touches the floor of the room. What will be the velocity of the sphere at the lowest point of its oscillation? The Young’s modulus for steel = 2 × 10^{11} N ᐧ m^{-2}; the initial length of the wire = 5 m and the radius of the wire = 5 × 10^{-4} m.

**Solution:**

According to Fig., the elon-gation of the wire at the lowest position of the sphere (diameter =0.2m),

l = 5.21 – (5 + 0.2) = 0.01 m

If the velocity of the sphere at the lowest point of its oscillation is v, then the tension in the wire,

T – mg = \(\frac{m v^2}{r}\)

or, T = mg + \(\frac{m v^2}{r}\) ……. (1)

Here, m = mass of the sphere; r = distance of the centre of gravity of the sphere from the point of suspension = 5.21 – 0.1 = 5.11m. Suppose x = radius of the wire.