Contents

- 1 What is the Hooke’s Law? What are the Different Types of Moduli of Elasticity?
- 1.1 Units and dimension of the modulus of elasticity:
- 1.2 Different Kinds of Strain And Moduli of Elasticity
- 1.3 Sagging of an elastic body due to its own weight
- 1.4 Bulk or Volume Strain : Bulk or Volume Stress : Bulk Modulus
- 1.5 Units of bulk modulus:
- 1.6 Units of compressibility:
- 1.7 Shearing Strain : Shearing Stress : Modulus of Rigidity or Shear Modulus
- 1.8 Units of the modulus of rigidity:

Some of the most important Physics Topics include energy, motion, and force.

## What is the Hooke’s Law? What are the Different Types of Moduli of Elasticity?

The fundamental law of elasticity was propounded by Robert Hooke in 1676. Later Thomas Young expressed this law in the following way.

Statement: Within the elastic limit of a substance, stress is directly proportional to strain.

Hence, stress ∝ strain or, \(\frac{\text { stress }}{\text { strain }}\) = constant

This constant is called the modulus of elasticity for the material of the body. Therefore, the stress developed for unit strain is defined as the modulus of elasticity. This modulus depends on the material of the body. The modulus of elasticity changes with temperature, in general. Usually its value decreases with an increase in temperature.

### Units and dimension of the modulus of elasticity:

Since strain has no unit, the modulus of elasticity has the unit of stress.

Relation: 1 N ᐧ m^{-2}(1 Pa) = \(\frac{1 \mathrm{~N}}{1 \mathrm{~m}^2}\) = \(\frac{10^5 \mathrm{dyn}}{10^4 \mathrm{~cm}^2}\)

= 10dyn ᐧ cm^{-2}

Similarly, the dimensions of modulus of elasticity and of stress are identical, which is ML^{-1}T^{-2}.

### Different Kinds of Strain And Moduli of Elasticity

Longitudinal Strain : Longitudinal Stress : Young’s Modulus

If force is applied along the length of a body usually whose length is much greater than all of its other dimensions (like breadth, height, etc., as in the case of a long thin wire or other rod-like bodies), then the body undergoes longitudinal strain. In other words, if, under the influence of an external force, a body undergoes an increase or decrease primarily in length, then this body is associated with a longitudinal strain.

The ratio of the change in length (Increase or decrease) to the original length of the body is the measure of its longitudinal strain. Longitudinal strain is possible only in the case of a solid substance. The stress developed inside the body, i.e., the reaction force developed per unit cross sectional area while it undergoes longitudinal strain is called longitudinal stress.

Young’s modulus: Within the elastic limit, longitudinal stress divided by longitudinal strain is called Young’s modulus.

Youngs modulus (Y) \(=\frac{\text { longitudinal stress }}{\text { longitudinal strain }}\)

If longitudinal stress tends to infinity and longitudinal strain tends to zero, Young’s modulus tends to infinity. This is the case for a perfectly rigid body, for which Young’s modulus Y is infinite. On the other hand, if any external force is applied on a plastic body, its longitudinal stress is zero and Young’s modulus (Y) too becomes zero.

Liquids and gases cannot produce longitudinal stress at all, i.e., on application of even a very small longitudinal force, they begin to flow. Therefore, Young’s modulus (Y) is a characteristic only of solids and not of liquids and gases.

Let us consider a wire of length L and cross sectional area A suspended from a rigid support [Fig.]. Now a force of magnitude F is applied perpendicularly to the cross sectional area A on the wire i.e., the force will act downwards along the length of the wire. As a result, the wire will show a slight increase in length.

If the wire is of circular cross section of radius r and a mass m is hung from its lower end, then, A = πr^{2}, F = mg

So, Y = \(\frac{m g L}{\pi r^2 l}\) ……… (2)

Units of Young’s modulus:

Young’s modulus for copper is 1.26 × 10^{12} dyn ᐧ cm^{-2} — means that a force of 1.26 × 10^{12} dyn must be applied per cm^{2} area of cross section of a copper wire to produce unit longitudinal strain in it.

### Sagging of an elastic body due to its own weight

In our daily life, we observe many bodies sag due to their own weights when they are suspended from a rigid support. Examples: a wet towel suspended across a horizontal rope, a cloth bag containing a few kilograms of rice suspended from a hook etc. However, these are not perfectly elastic bodies, and no theoretical expression for their sagging can be obtained.

Now, we turn our attention to elastic bodies. Suppose a cylindrical rod made of an elastic metal (say, steel) is hung vertically from a rigid support. It definitely sags due to its own weight. although the sagging is relatively small. But using elastic properties of the material, a clear expression for the sagging can be found out.

Let m be the mass of body B made of an elastic material. One of its end is hung from a rigid support [Fig.].

Then mg weight of B, L = its length and A = area of its cross section.

If B is a body of uniform density and its cross section is also uniform throughout its height, the centre of gravity G is situated at the mid-point at a depth \(\frac{L}{2}\) from the rigid support.

A force due to the weight mg acts downwards at the centre of mass (G). It may be assumed that this force stretches the body above the point G. So the effective initial length that is strained is \(\frac{L}{2}\). If the body B sags downwards through a length l, then,

longitudinal strain \(=\frac{\text { elongation }}{\text { initial length }}\) = \(\frac{l}{\frac{L}{2}}\) = \(\frac{2 l}{L}\)

Also, logitudinal stress = \(=\frac{\text { applied force }}{\text { area of cross section }}\) = \(\frac{2 l}{L}\)

Again the mass of the body B is,

m = Volume × density = LAρ

[ρ = density of the material]

So Youngs modulus,

Y = \(\frac{\text { stress }}{\text { strain }}\) = \(\frac{\frac{m g}{A}}{\frac{2 l}{L}}\) = \(\frac{L A \rho g L}{2 A l}\) = \(\) = \(\frac{\rho g L^2}{2 l}\)

Hence the sagging of the body is,

l = \(\frac{\rho g L^2}{2 Y}\) ………. (3)

It is important to note that this expression is independent of A. So the sagging of a thin wire is the same as that of a thick rod, if their initial lengths are equal and they are made of the same elastic material.

### Bulk or Volume Strain : Bulk or Volume Stress : Bulk Modulus

When a body is subjected to uniform pressure acting normally at every point on its surface, then it undergoes a change in volume, without any change in its shape. For example, when a cube or a sphere is subjected to a uniform normal pressure, their shape remains unchanged, but they undergo a volume strain [Fig.].

Under the influence of external forces, when a body undergoes an Increase or decrease in volume without any change in its shape, then the strain of the body Is called bulk or volume strain. The ratio of the change in volume (increase or decrease) of the body to its initial volume gives the measure of the volume strain of the body. The stress, developed in the body, i.e., the reaction force developed per unit surface area due to its volume strain is called bulk or volume stress.

Bulk modulus: Within the elastic limit, volume stress divided by volume strain is called the bulk modulus of elasticity.

Bulk modulus (K) = \(=\frac{\text { volume stress }}{\text { volume strain }}\)

Every substance—solid, liquid or gas—has some volume and hence the bulk modulus is meaningful for all substances.

The bulk modull of a perfectly rigid body and a perfectly plastic body are infinite and zero respectively.

Since, liquid and gaseous substances undergo only volume strain, the bulk modulus is the only elastic modulus for them. Among the different bulk moduli of a gas, very useful are its isothermal and adiabatic bulk moduli [for details see the chapter First and Second Law of Thermodynamics]. The isothermal bulk modulus of air 1.01 × 10^{5} Pa and the adiabatic bulk modulus of air = 1.42 × 10^{5} Pa.

Let V = initial volume of a body,

p = applied force per unit surface area of the body

= applied pressure,

v = corresponding decrease in volume,

so, -v = change in volume.

Therefore, the bulk modulus of elasticity of the material of the body is,

K \(=\frac{\text { volume stress }}{\text { volume strain }}\) = \(\frac{p}{-\frac{v}{V}}\) = \(-\frac{p V}{v}\) ……. (4)

On application of pressure, the change in volume of all gases, and of a few materials like rubber, cotton etc., is fairly large. This means that t’ is large even for a relatively small applied pressure p. So, the bulk moduli of all gases, and of the said materials, are fairly low. On the other hand, almost all solids and liquids show very small changes in volume on application of external forces. As a result, their bulk moduli are much higher.

For more mathematical rigour, the applied pressure is denoted by Δp (instead of p) and the corresponding decrease in volume by v (instead of v). Then, equation (4) becomes,

K = -V\(\frac{\Delta p}{\Delta v}\) ……….. (5)

This equation (5) is treated as the defining equation of the bulk modulus of elasticity.

For magnitudes only, the negative sign on the right-hand side of equations (4) and (5) is ignored.

### Units of bulk modulus:

Compressibility: It is defined as the change in volume due to a unit change in pressure, if a unit volume of a substance is taken initially. Like bulk modulus, compressibility is also a characteristic property of all substances—solids, liquids and gases.

Again, let V = initial volume Δp = applied pressure, and -Δ V = corresponding change in volume.

Then, from definition,

Comparing equations (5) and (6), we observe that

compressibility \(=\frac{1}{\text { bulk modulus }}\)

So, bulk modulus and compressibility are not independent properties. If one is known, the other can be calculated. The compressibility of a solid or a liquid is very small, compared to that of a gas due to lack of space between molecules and larger intermolecular forces. The compressibility of a perfectly rigid body is zero.

The compressibility of water is 44 × 10^{-6} atm^{-1}, meaning that the volume strain of water becomes 44 × 10^{-6} on the application of 1 standard atmosphere pressure.

### Units of compressibility:

Another useful unit of compressibility is atmosphere^{-1}

### Shearing Strain : Shearing Stress : Modulus of Rigidity or Shear Modulus

If a match-box is held firmly on a table and a tangential force is applied on the upper surface of the box with the help of a finger, the shape of the box changes [Fig.]. This is known as shearing strain. In this condition, however, the change in the volume of the box is negligible as well as irrelevant.

Under the influence of an external force, if a body under goes a change in its shape, then the strain of the body is called shearing strain or shear. The corresponding stress developed inside the body is called shearing stress.

Modulus of rigidity: Within the elastic limit, shearing stress divided by shearing strain is called the modulus of rigidity or shear modulus.

Modulus of rigidity (n) \(=\frac{\text { shearing stress }}{\text { shearing strain }}\)

The modulus of rigidity ¡s a characteristic property of solids because only solids have definite shapes.

Let us consider a rectangular parallelepiped ABCD [Fig.]. ABCD is only the front surface of the parallelepiped; for convenience, the depth has not been shown in the figure. AD and BC denote the vertical side faces, whereas AB and CD are the top and the bottom faces respectively. The lower face CD is kept fixed to the horizontal surface.

A tangential force F is applied on its upper face AB. An equal but opposite reaction force F will act tangentially on the lower surface DC of the block. These two forces constitute a couple and due to its action, each layer parallel to the surface DC will be displaced in the direction of the applied force. The layers that are farther from the surface DC will have larger relative displacements during deformation. As a result, the block will undergo a change in its shape.

It will be seen that each vertical surface parallel to F takes the form of a parallelogram from the original rectangular shape, i.e., the surface ABCD will become a parallelogram A’B’CD. This kind of strain is called a shearing strain. It should be noted that, if the displacement AA’ is large, the surface A’B’ will come down a bit, and its position will be lower relative to the original position AB. As a result, the area of the parallelogram and the volume of the parallelepiped will decrease. However, in most cases, particularly for metals, AA’ is so small that the change in volume may safely be ignored.

The angle formed between the initial and the final positions of any vertical line drawn perpendicular to the direction of the applied force is called shearing strain or the angle of shear. Usually this angle is small, as the displacement AA’ is small.

Let ∠ADA’ = ∠BCB’ = θ; AD = L; AA’ = l;

Since θ is very small, tan θ ≈ θ.

[where θ is expressed in radian]

∴ Shearing strain = θ tanθ = \(\frac{l}{L}\)

Now, if L = 1, then θ = 1, which means that the relative displacement between two layers situated at a unit distance apart is the measure of shearing strain.

Shearing stress is measured by the applied tangential force per unit area. If the area of the surface AB is a, then shearing stress = \(\frac{F}{a}\).

∴ Modulus of rigidity,

n \(=\frac{\text { shearing stress }}{\text { shearing strain }}\) = \(\begin{aligned}

& \frac{F}{a} \\

& \frac{a}{\theta}

\end{aligned}\) = \(\frac{F}{a \theta}\) = \(\frac{F}{a} \cdot \frac{L}{l}\)