Chapter 6: Mechanical Properties of Solids
In 1638, an old and nearly blind Galileo Galilei published his last and, in his own view, most important book — Two New Sciences. Tucked into it was a question that had puzzled builders for centuries. Why is it that a small wooden beam can carry its own weight easily, yet a giant beam of exactly the same shape, scaled up many times, cracks and collapses under nothing but its own weight?
A carpenter might have shrugged and said “big things are just weaker.” Galileo refused to. He reasoned that when you double the size of a beam, its weight grows much faster than the area of the surface that has to hold it together. The load on each square of that surface — what we now call stress — climbs until the material can bear no more, and the beam fractures.
With that single insight, Galileo founded the science of the strength of materials — the study of how solids stretch, bend, twist and break. This chapter follows the trail he opened: we will learn how to measure exactly how hard a solid is being pushed or pulled, how much it deforms in response, why some materials spring back while others stay bent, and why solids grip, slide and roll the way they do.
Diagram
Fig. 6.1 – A small wooden beam resting easily across two supports beside a much larger geometrically similar beam of the same shape cracking under its own weight, illustrating how stress rises with scale.
By the end of this chapter, you will be able to answer each of these questions for yourself.
6.1 Introduction
How Atoms Are Arranged Inside a Solid
A solid is made up of atoms, or of groups of atoms, arranged in a definite geometric pattern. This arrangement is not accidental. Nature settles into an arrangement in which the resultant force acting on each constituent (each atom or group of atoms) due to all the others is zero.
A “resultant force of zero” means all the pulls and pushes from the neighbouring atoms cancel out exactly. This balanced arrangement is called the equilibrium state of a solid at room temperature, as shown in Fig. 6.2.
Diagram
Fig. 6.2 – Regular geometric arrangement of atoms in a solid, each atom shown with balanced interatomic forces from its neighbours.
Once formed, this equilibrium arrangement does not change with time. It can change only when an external stimulus acts on the solid — for example, a compressive force applied from all sides.
The atoms are not frozen still, however. Even at very low temperatures the constituents keep vibrating about their equilibrium positions, but they cannot leave those fixed positions. This is exactly why a solid has a definite shape and a definite size, unlike a liquid or a gas, whose particles are free to wander.
Restoring Forces: The Solid Fights Back
When an external force is applied to a solid, its constituents get slightly displaced from their equilibrium positions. The moment this happens, restoring forces develop inside the solid.
These restoring forces are simply the interatomic forces trying to pull (or push) the displaced atoms back to where they belong. They act so that the solid can regain its shape.
When the deforming forces are removed, these interatomic forces restore the original positions of the molecules, and so the body regains its original shape and size. As we will see later in the chapter, this recovery is possible only within certain limits — beyond a certain amount of stretching or squeezing, the atoms do not go back fully.
Shape, Size and the Deforming Force
The “form” of a body is decided by two things — its size and its shape. For example, a tennis ball and a football are both spherical, so they have the same shape. But the tennis ball is smaller in size than the football. Shape and size are therefore two independent descriptions of a body.
Now suppose a force is applied to a solid that is not free to move away (say, a wire clamped at the top, or a block held on a table). The force cannot simply accelerate the body, so it does something else: it changes the relative positions of the molecules. As a result the size, or the shape, or both, change. Such a force is called a deforming force.
The change produced by this force has its own name.
The larger the deforming force acting on a body, the larger is its deformation. Deformation can show up in three different ways, as illustrated in Fig. 6.3:
- a change in the length of a wire,
- a change in the volume of an object,
- a change in the shape of a body.
Diagram
Fig. 6.3 – Three panels showing the three kinds of deformation: a wire elongating, a cube shrinking in volume under all-round pressure, and a block tilting sideways with change of shape.
These three ways of deforming a body are worth remembering, because the whole of the next few sections is built on them — each type of deformation will get its own kind of stress and its own kind of strain.
Rubber and Clay Behave Differently
Consider a familiar experiment. Apply a deforming force to a rubber band by stretching it. The rubber band gets deformed — it elongates. Now remove the force. The rubber band regains its original length.
Next, apply a similar force to a lump of dough or clay. It also gets deformed. But when the force is removed, it does not regain its original shape and size. It stays in its new, altered form.
These two simple observations tell us something important: rubber and clay are different in nature. The property that decides this nature — whether a body springs back or stays deformed — is called elasticity/plasticity. We will study these properties of solids in detail in this chapter.
What This Chapter Will Build
In the sections that follow, we will make these ideas precise. We will define quantities that measure how hard a body is being deformed (stress) and how much it has actually deformed (strain), connect the two through Hooke’s law, and then use them to characterise materials by their moduli of elasticity. Towards the end, we will look at what happens when a solid is pushed past its limits, at the energy stored inside a stretched wire, and at friction, which is another mechanical property arising at the surfaces of solids in contact.
6.2 Elastic Behavior of Solids
Bodies That Come Back
We ended the previous section with two ordinary objects behaving in two very different ways under the same kind of deforming force. That difference is now given a precise name.
Some bodies recover completely once the deforming force is taken away. Stretch a metal wire gently, release it, and it shortens back to its original length. The reason lies inside the material: here the restoring forces between the molecules are strong enough to bring the displaced molecules back to their original positions.
Examples of elastic materials are metals, rubber and quartz.
Notice what the definition does and does not demand. It says the body must return — but it does not say how quickly or how completely. That is why a stricter category is needed.
The two words in that definition are exam-critical. “Completely” means nothing at all is left behind — not the smallest permanent change. “Instantaneously” means there is no time lag, no slow creeping back to the original size.
Diagram
Fig. 6.4 – Two-panel comparison: a stretched rubber band returning to its original length after the load is removed (elastic), and a lump of clay keeping its squashed shape after the load is removed (plastic).
Bodies That Stay Deformed
Now the opposite case, shown in the second panel of Fig. 6.4. Press a lump of clay and let go; it keeps the new shape you have given it.
Why does the clay not recover? The molecular reason is the mirror image of the elastic case: here the restoring forces are not strong enough to bring the molecules back to their original positions. The molecules have been displaced so far that they simply settle down in their new positions and stay there.
Examples of plastic materials are clay, putty, plasticine and thick mud.
The Two Nearest-Ideal Materials
Since no material is a perfect example of either behaviour, physicists point instead to the closest available approximations. The best example of a near ideal elastic solid is quartz fibre, and the best example of a plastic body is putty.
Why the Distinction Matters
Elasticity is not “the ability to stretch.” A rubber band stretches a great deal and is elastic; a steel wire stretches barely at all and is also elastic. What makes both elastic is that they return. Similarly, plasticity is not “weakness” — it is the tendency to keep the change. Students who confuse “stretches easily” with “is elastic” go wrong in reasoning questions, so it is worth fixing the correct meaning now: elasticity is about recovery, not about how much a body deforms.
6.3 Stress and Strain
Two Quantities to Describe a Deformed Body
The elastic properties of a body are described in terms of two quantities: stress and strain. One of them measures how hard the body is being deformed; the other measures how much it has actually deformed.
When a body gets deformed under an applied force, restoring forces are set up inside it. These internal forces oppose the change in shape or size of the body — they try to undo whatever the external force is doing.
Now suppose the body has settled down: it is deformed, but it is no longer changing. It is in equilibrium in its altered shape or size. In this state the deforming force and the restoring force are equal and opposite. This is an important bridge, because it means the internal restoring force has exactly the same magnitude as the external force we applied — and that lets us calculate an internal quantity from the external force we can measure.
Stress
\[\text{stress} = \frac{\text{deforming force}}{\text{area}} = \frac{|\vec{F}|}{A} \qquad \text{--- (6.1)}\]
Here \(\vec{F}\) is the internal restoring force, which (as argued above) is equal in magnitude to the externally applied deforming force, and \(A\) is the area over which the force acts, in \(\text{m}^2\).
The SI unit of stress is \(\text{N m}^{-2}\), also called the pascal (Pa). The dimensions of stress are \([L^{-1} M^{1} T^{-2}]\).
Strain
Strain is a measure of the deformation of a body. When two equal and opposite forces are applied to an elastic body, there is a change in the dimensions of the body, and strain records how large that change is compared with the size the body started with.
\[\text{Strain} = \frac{\text{change in dimensions}}{\text{original dimensions}} \qquad \text{--- (6.2)}\]
Because it is the ratio of two similar quantities (a length divided by a length, or a volume divided by a volume), strain is a dimensionless physical quantity. It has no units.
This is worth pausing on. A strain of \(0.002\) does not mean “0.002 metre.” It means the body has changed by 0.002 of itself — two parts in a thousand. Writing a unit next to a strain value is a common and costly slip in exams.
There are three types of stress and three corresponding strains, and we now take them one at a time.
Type 1: Stress Along the Length of a Body
Stress produced by a deforming force acting along the length of a body or a rod is called tensile stress or longitudinal stress. The strain produced is called tensile strain.
Tensile Stress or Compressive Stress
Suppose a force \(\vec{F}\) is applied along the length of a wire, that is, perpendicular to its cross section of area \(A\). This produces an elongation in the wire, and the length of the wire increases accordingly, as shown in Fig. 6.5 (a).
Diagram
Fig. 6.5 (a) – A cylindrical rod of length L pulled outward at both ends by forces F, elongating by Δl; tensile stress.
\[\text{Tensile stress} = \frac{|\vec{F}|}{A} \qquad \text{--- (6.3)}\]
Now reverse the direction of the two forces. When a rod is pushed at two ends with equal and opposite forces, its length decreases. The restoring force per unit area in this situation is called compressive stress, as shown in Fig. 6.5 (b).
Diagram
Fig. 6.5 (b) – The same rod pushed inward at both ends by forces F, shortening by Δl; compressive stress.
\[\text{Compressive stress} = \frac{|\vec{F}|}{A} \qquad \text{--- (6.4)}\]
Notice that Equations (6.3) and (6.4) look identical. The formula does not change; only the direction of the force does. In a tensile situation the forces pull the body apart and it gets longer; in a compressive situation they push it together and it gets shorter. The same wire can experience either.
Tensile Strain
The strain produced by a tensile deforming force is called tensile strain, or longitudinal strain, or linear strain — the three names mean the same thing.
If \(L\) is the original length of the body and \(\Delta l\) is the change in length produced by the deforming force, then
\[\text{Tensile strain} = \frac{\Delta l}{L} \qquad \text{--- (6.5)}\]
Both \(\Delta l\) and \(L\) are lengths, measured in metres, so their ratio is a pure number, as expected from Equation (6.2).
Type 2: Stress That Changes the Volume
When a deforming force acting on a body produces a change in its volume, the stress is called volume stress and the strain produced is called volume strain.
Volume Stress or Hydraulic Stress
Let \(\vec{F}\) be a force acting perpendicular to the entire surface of the body. It acts normally and uniformly all over the surface area \(A\) of the body — that is, it presses inward everywhere at right angles to the surface, with the same intensity at every point. Such a stress, which produces a change in size but no change in shape, is called volume stress.
\[\text{Volume stress} = \frac{|\vec{F}|}{A} \qquad \text{--- (6.6)}\]
Because volume stress produces a change in size without a change in shape of the body, it is also called hydraulic or hydrostatic volume stress, as shown in Fig. 6.6. A sphere squeezed uniformly from all sides stays a sphere; it just becomes a smaller sphere.
Diagram
Fig. 6.6 – A sphere of volume V and a cube, each with inward arrows pressing normally on the whole surface, shrinking to a smaller volume V′; volume stress.
Volume Strain
A deforming force acting perpendicular to the entire surface of a body produces a volume strain. Let \(V\) be the original volume of the body and \(\Delta V\) be the change in volume produced by the deforming force. Then
\[\text{Volume stain} = \frac{\Delta V}{V} \qquad \text{--- (6.7)}\]
Once again this is a ratio of two volumes, so volume strain is dimensionless and unitless.
Type 3: Stress That Changes the Shape
When a deforming force acting on a body produces a change in the shape of a body, shearing stress and shearing strain are produced. Here the size of the body stays the same; only its shape is distorted.
Shearing Stress
Let \(\vec{F}\) be a tangential force acting on a surface of area \(A\). This force produces a change in the shape of the body without changing its size, as shown in Fig. 6.7.
\[\text{Shearing stress} = \frac{\text{Tangential force}}{\text{Area}} \qquad \text{--- (6.8)}\]
Diagram
Fig. 6.7 – Square front face ABCD of a cube with the bottom edge AB fixed and the top edge DC pushed sideways by a tangential force F to D′C′, the side edges tilting through angle θ.
Suppose ABCD is the front face of a cube, as in Fig. 6.7. A force \(\vec{F}\) is applied to the cube so that the bottom of the cube is fixed and only the top surface is slightly displaced. Such a force is called a tangential force. Tangential force is parallel to the top and the bottom surface of the block. This is the crucial difference from the earlier cases, where the force acted perpendicular to the surface.
Shearing Strain
Look again at Fig. 6.7. There is a relative displacement, \(\Delta l\), of the bottom face and the top face of the cube — the top has shifted sideways while the bottom has stayed put. Such a relative displacement of two surfaces is called shear strain. It can be calculated as follows.
If \(l\) is the height of the block (the perpendicular distance between the two faces) and \(\theta\) is the small angle through which the vertical edges have tilted, then
\[\text{Shearing strain} = \frac{\Delta l}{l} = \tan \theta = \theta \qquad \text{--- (6.9)}\]
when the relative displacement \(\Delta l\) is very small.
The last step, \(\tan\theta = \theta\), deserves a word. For a small angle measured in radians, the tangent of the angle is very nearly equal to the angle itself. So for small deformations we may simply use \(\theta\) (in radian) as the shear strain. Being an angle in radian, it is again a pure number with no unit — consistent with all the strains we have met.
6.4 Hooke’s Law
We now have a way of measuring how hard a body is being deformed (stress) and how much it has deformed (strain). The obvious next question is how the two are related. For small deformations, the relation is remarkably simple.
Robert Hooke (1635–1703), an English physicist, studied the tension in a wire and the strain produced in it. His study led to a law now known as Hooke’s law.
The condition “within the elastic limit” is not decoration — it is part of the law. Stretch the wire too far and the proportionality fails, as we shall see in Section 6.6.
Since stress is directly proportional to strain, their ratio must be a fixed number:
\[\frac{\text{Stress}}{\text{Strain}} = \text{constant}\]
This constant is called the modulus of elasticity.
The word “corresponding” matters: a tensile stress must be divided by the tensile strain it produced, a volume stress by the volume strain, and so on. Mixing them gives a meaningless number.
The Modulus as a Slope
There is a second, more graphical way to read the same constant. If stress is plotted on the y-axis and strain on the x-axis, Hooke’s law says the graph is a straight line through the origin, and the ratio stress/strain is simply its slope. Hence the modulus of elasticity is defined as the slope of the stress-strain curve in the elastic deforming region, as shown in Fig. 6.8.
Diagram
Fig. 6.8 – Stress (y-axis) versus strain (x-axis) for an elastic body within the elastic limit: a straight line through the origin, its slope marked as the modulus of elasticity.
The value of this slope depends on the nature of the material. A steep line means a large modulus: a big stress is needed to produce even a small strain, so the material is hard to deform. A gentle line means a small modulus and an easily deformed material. The modulus is therefore a fingerprint of the material itself, not of the particular wire or block being tested.
The Elastic Limit
Hooke’s law does not hold for ever. As the load on a wire is increased, a point is reached beyond which stress and strain are no longer proportional.
The straight-line graph of Fig. 6.8 is therefore the stress-strain curve within the elastic limit. What happens past that limit — the bending of the curve, the permanent set, and finally fracture — is taken up in Section 6.6.
6.5 Elastic Modulus
In Section 6.4 we saw that the ratio of stress to the corresponding strain is a constant called the modulus of elasticity. But we have already met three separate types of stress and strain — those related to a change in length, those related to a change in volume, and those related to a change in shape.
Since each type of stress must be divided by the strain that corresponds to it, one single constant cannot serve all three. Hence we have three moduli of elasticity, one corresponding to each type of stress and strain, as summarised in Fig. 6.9.
Diagram
Fig. 6.9 – Three-branch chart: change in length → Young’s modulus (Y); change in volume → Bulk modulus (K); change in shape → Modulus of rigidity (η).
Each of these three is taken up in turn in the subsections that follow, together with Poisson’s ratio, which describes a related effect: the sideways thinning that accompanies stretching.
6.5.1 Young’s Modulus (Y)
What Young’s Modulus Describes
Because it deals entirely with changes in length, it is also called the elasticity of length. It is named after the British physicist Thomas Young (1773–1829).
Setting Up the Standard Experiment
Consider a metal wire of length \(L\) and radius \(r\), suspended from a rigid support, as shown in Fig. 6.10. A load \(Mg\) is attached to the free end of the wire. Because of this, a deforming force is applied at the free end of the wire in the downward direction.
Diagram
Fig. 6.10 – A wire of length L and radius r hanging from a rigid support, with a mass M attached at the lower end, showing the elongation l produced.
The wire quickly settles into equilibrium: the downward pull of the load and the upward restoring force in the wire balance each other. In this equilibrium position, the stress in the wire is the applied force divided by the area of cross section.
The wire is a cylinder of radius \(r\), so its area of cross section is \(\pi r^2\). This gives
\[\text{Longitudinal stress} = \frac{\text{Applied force}}{\text{Area}} = \frac{F}{A} = \frac{Mg}{\pi r^2} \qquad \text{--- (6.10)}\]
Here \(M\) is the mass of the load in kg, \(g\) is the acceleration due to gravity in \(\text{m s}^{-2}\), so \(Mg\) is the force in newton, and \(\pi r^2\) is the area in \(\text{m}^2\). The stress is therefore in \(\text{N/m}^2\).
This stress produces a change in the length of the wire. If \((L + l)\) is the new length of the wire, then \(l\) is the extension or elongation in the wire — the increase in length, not the total length. Confusing \(l\) with the full new length is one of the commonest errors in numericals of this type.
\[\text{Longitudinal strain} = \frac{\text{change in length}}{\text{original length}} = \frac{l}{L} \qquad \text{--- (6.11)}\]
The Formula
\[Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}} \qquad \text{--- (6.12)}\]
Substituting the two expressions we have just obtained,
\[Y = \frac{\dfrac{Mg}{\pi r^2}}{\dfrac{l}{L}}\]
\[Y = \frac{MgL}{\pi r^2 l} \qquad \text{--- (6.13)}\]
Read Equation (6.13) physically before you use it. The elongation \(l\) sits in the denominator, so for a given load a material with a large \(Y\) stretches only a little. The length \(L\) is in the numerator, meaning a longer wire of the same material stretches more under the same load — which is why \(l\) alone tells you nothing about the material until you divide by \(L\).
The SI unit of Young’s modulus is \(\text{N/m}^2\). Its dimensions are \([L^{-1} M^{1} T^{-2}]\) — the same as those of stress, because strain is dimensionless.
Young’s modulus indicates the resistance of an elastic solid to elongation or compression. Young’s modulus of a material is useful for the characterization of an object subjected to compression or tension. Young’s modulus is the property of solids only.
Values for Familiar Materials
Table 6.1 lists Young’s modulus for some familiar materials. The larger the value, the stiffer the material in tension.
Table 6.1: Young’s modulus of some familiar materials
| Material | Young’s modulus Y ×10¹⁰ Pa (N/m²) |
|---|---|
| Lead | 1.5 |
| Glass (crown) | 6.0 |
| Aluminium | 7.0 |
| Silver | 7.6 |
| Gold | 8.1 |
| Brass | 9.0 |
| Copper | 11.0 |
| Steel | 21.0 |
Steel stands at the top of this table, with a value about fourteen times that of lead. That single fact explains why steel wires and steel beams carry the loads in cranes, lifts and buildings: for the same stress, steel gives away the least in strain.
6.5.2 Bulk Modulus (K)
What Bulk Modulus Describes
Since it deals with changes of volume, it is also called the elasticity of volume. Unlike Young’s modulus, the bulk modulus of elasticity is a property of solids, liquids and gases — because a volume can be squeezed even in a substance that has no fixed shape of its own.
Setting Up the Situation
If a sphere made from rubber is completely immersed in a liquid, it will be uniformly compressed from all sides, as shown in Fig. 6.11. The liquid presses inward everywhere, at right angles to the surface, with the same intensity at every point.
Diagram
Fig. 6.11 – A rubber sphere immersed in a liquid, with inward arrows pressing normally on its entire surface, shrinking from volume V to a smaller volume.
Suppose this compressive force is \(F\). Let the change in pressure on the sphere be \(dP\) and let the change in its volume be \(dV\). If the original volume of the sphere is \(V\), then the volume strain is defined as
\[\text{Volume strain} = \frac{\text{change in volume}}{\text{original volume}} = -\frac{dV}{V} \qquad \text{--- (6.14)}\]
The negative sign indicates that there is a decrease in volume: an increase in pressure squeezes the body, so \(dV\) is itself negative and the strain, written with the minus sign in front, comes out positive. The magnitude of the volume strain is \(\dfrac{dV}{V}\).
The Formula
The volume stress here is simply the change in pressure \(dP\) applied to the surface, since pressure is already a force per unit area acting normally on the surface.
\[\text{Bulk modulus} = \frac{\text{volume stress}}{\text{volume strain}} = K\]
\[K = \frac{dP}{\left(\dfrac{dV}{V}\right)} = V \frac{dP}{dV} \qquad \text{--- (6.17)}\]
Here \(dP\) is the change in pressure in \(\text{N/m}^2\), \(V\) is the original volume in \(\text{m}^3\), and \(dV\) is the change in volume in \(\text{m}^3\). Because the volume strain \(dV/V\) is dimensionless, \(K\) carries the unit of pressure.
The SI unit of bulk modulus is \(\text{N/m}^2\). The dimensions of \(K\) are \([L^{-1} M^{1} T^{-2}]\) — the same as those of Young’s modulus, as they must be, since every modulus is a stress divided by a pure number.
Table 6.2 gives the bulk moduli of some familiar materials.
Table 6.2: Bulk modulus of some familiar materials
| Material | Bulk modulus K ×10¹⁰ Pa (N/m²) |
|---|---|
| Lead | 4.1 |
| Brass | 6.0 |
| Glass (crown) | 6.0 |
| Aluminium | 7.5 |
| Silver | 10.0 |
| Copper | 14.0 |
| Steel | 16.0 |
| Gold | 18.0 |
Bulk modulus measures the resistance offered by gases, liquids or solids while an attempt is made to change their volume. A large \(K\), as for gold, means the substance stubbornly holds on to its volume; a small \(K\) means it gives way more readily.
Compressibility
It is often more convenient to speak of how easily a material can be squeezed, rather than how strongly it resists. For this we simply invert the bulk modulus.
\[\text{Compressibility} = \frac{1}{\text{Bulk modulus}} \qquad \text{--- (6.18)}\]
Compressibility is the fractional decrease in volume, \(-\Delta V / V\), per unit increase in pressure. In words: it tells you what fraction of its volume a body loses for each pascal of extra pressure applied to it.
The SI unit of compressibility is \(\text{m}^2/\text{N}\) or \(\text{Pa}^{-1}\), and its dimensions are \([L^{1} M^{-1} T^{2}]\) — exactly the reciprocal of those of bulk modulus, as expected.
6.5.3 Modulus of Rigidity (η)
What Rigidity Modulus Describes
It is the property of solids only, as they alone possess a definite shape. A liquid or a gas has no shape of its own to defend, so the question of resisting a change of shape does not arise for them.
The Block Under a Couple
The block shown in Fig. 6.12 is made of a uniform isotropic material — “isotropic” meaning its properties are the same in every direction, so it responds identically no matter which way it is sheared. It has a uniform cross-section area \(A\) and height \(l\). A cross section of the block is defined as any plane parallel to the top and the bottom surface that cuts the block.
Two forces of magnitude \(F\) are applied along the top and bottom surfaces, as shown in Fig. 6.12. They are equal in magnitude, opposite in direction, and act along different lines — so together they constitute a couple. The upper surface is displaced relative to the lower surface by a small distance \(\Delta l\), and the corresponding angles change by a small amount \(\theta = \Delta l / l\).
Diagram
Fig. 6.12 – A rectangular block of height l with force F applied along its top surface and −F along its bottom surface, the top face shifting by Δl and the vertical edges tilting through angle θ = Δl/l.
A couple is applied by pushing the top and the bottom surfaces as shown in Fig. 6.12. A similar couple would be applied if the bottom of the block were fixed and only the top pushed. Both arrangements produce the same distortion, which is why the two pictures of shear you have seen — one with the base clamped, one with equal opposite forces on top and bottom — describe the same physics.
The forces \(\vec{F}\) and \(-\vec{F}\) are parallel to the cross section. This is different from the tensile stress, where the force is normal to the cross section. That single difference in direction is what changes the shape instead of the length.
Shear Stress and Shear Strain
As a result of the way in which the forces are applied, the block is subjected to a shear stress defined by
\[\text{shear stress} = \frac{F}{A}\]
The SI unit of shear stress is \(\text{N/m}^2\) or Pa. Here \(F\) is the tangential force in newton and \(A\) is the area of the surface on which it acts, in \(\text{m}^2\). Note that \(A\) is the area of the face being pushed, not the area of some side face — using the wrong face is a frequent source of error in numericals.
The block is distorted as a result of the shear stress. The top and bottom surfaces are relatively displaced by a small distance \(\Delta l\). The corner angle changes by a small amount \(\theta\), which is called the shear strain and is expressed in radian. For small \(\Delta l\),
\[\theta = \frac{\Delta l}{l}\]
The Formula
\[\eta = \frac{\text{shear stress}}{\text{shear strain}} = \frac{F/A}{\theta} = \frac{F}{A\theta} \qquad \text{--- (6.17)}\]
In this expression \(F\) is the tangential force in newton, \(A\) is the area over which it acts in \(\text{m}^2\), and \(\theta\) is the shear strain in radian, a pure number. The SI unit of \(\eta\) is therefore \(\text{N/m}^2\), and its dimensions are the same as those of the other two moduli.
Table 6.3 gives values of the rigidity modulus \(\eta\) of some familiar materials.
Table 6.3: Rigidity modulus η of some familiar materials
| Material | Rigidity modulus η ×10¹⁰ Pa (N/m²) |
|---|---|
| Lead | 0.6 |
| Aluminium | 2.5 |
| Glass (crown) | 2.5 |
| Silver | 2.7 |
| Gold | 2.9 |
| Brass | 3.5 |
| Copper | 4.4 |
| Steel | 8.3 |
Rigidity modulus indicates the resistance offered by a solid to a change in its shape.
6.5.4 Poisson’s Ratio
Stretching in One Direction, Shrinking in Another
So far we have watched a wire get longer when it is pulled. But something else happens at the same time, in a direction at right angles to the pull.
Suppose a wire is fixed at one end and a force is applied at its free end so that the wire gets stretched. The length of the wire increases and, at the same time, its diameter decreases. In other words, the wire becomes longer and thinner, as shown in Fig. 6.13 (a).
Diagram
Fig. 6.13 (a) – A wire of radius r and length l pulled at both ends: length increases by Δl while the radius shrinks by Δr, so the wire becomes longer and thinner.
The reverse happens under compression. If equal and opposite forces are applied to an object along its length inwards, the object gets compressed, as shown in Fig. 6.13 (b). There is a decrease in dimensions along its length and, at the same time, there is an increase in its dimensions perpendicular to its length. When the length of the wire decreases, its diameter increases.
Diagram
Fig. 6.13 (b) – The same wire pushed inward at both ends: its length decreases while its diameter increases, so the wire becomes shorter and fatter.
This makes physical sense once you remember that the material is not vanishing or being created. Pull the atoms further apart along one direction and they draw a little closer together in the perpendicular directions; push them together along the length and they spread out sideways.
Linear Strain and Lateral Strain
To describe the two changes we need two separate strains.
The linear strain is the one we have been calling tensile strain all along; the lateral strain is the new one, describing the sideways thinning or fattening. Both are ratios of two lengths, so both are pure numbers.
The Formula
If \(l\) is the original length of the wire, \(\Delta l\) is the increase or decrease in the length of the wire, \(D\) is the original diameter and \(d\) is the corresponding change in the diameter of the wire, then Poisson’s ratio is given by
\[\sigma = \frac{\text{Lateral strain}}{\text{Linear strain}}\]
\[= \frac{d/D}{l/L}\]
\[= \frac{d \cdot L}{D \cdot l} \qquad \text{--- (6.18)}\]
Here \(d/D\) is the fractional change in the diameter and \(l/L\) is the fractional change in the length. Poisson’s ratio has no unit. It is dimensionless — a ratio of one dimensionless strain to another.
Values for Familiar Materials
Table 6.4 gives values of Poisson’s ratio, \(\sigma\), of some familiar materials.
Table 6.4: Poisson ratio, σ, of some familiar materials
| Material | Poisson ratio σ |
|---|---|
| Glass (crown) | 0.2 |
| Steel | 0.28 |
| Aluminium | 0.36 |
| Brass | 0.37 |
| Copper | 0.37 |
| Silver | 0.38 |
| Gold | 0.42 |
Every entry is well below 1, which tells you that the sideways thinning is always a smaller fraction than the lengthwise stretching. Steel, for example, thins sideways by only about 0.28 of the fraction by which it lengthens.
6.6 Stress-Strain Curve
The Experiment Behind the Graph
Hooke’s law told us what happens for small deformations. To find out what happens when a wire is pushed to its limit, we must go beyond that region — and the way to do it is to keep loading a wire until it breaks.
Suppose a metal wire is suspended vertically from a rigid support and stretched by applying a load to its lower end. The load is gradually increased in small steps until the wire breaks. The elongation produced in the wire is measured during each step.
The stress and the strain are noted for each load, and a graph is drawn by taking tensile strain along the x-axis and tensile stress along the y-axis. This is the stress-strain curve, as shown in Fig. 6.14.
Diagram
Fig. 6.14 – Stress (y-axis) versus strain (x-axis) for a metal wire loaded to breaking: straight line OA to the proportional limit, curve to yield point B, plastic region B to fracture point D, with unloading line CE showing the permanent set OE.
Note the choice of axes: strain, the effect, is on the x-axis, and stress, the cause, on the y-axis. This is the reverse of what students often expect, and it matters when you are asked to identify the slope.
From O to A: Hooke’s Region
The initial part of the graph is a straight line OA. This is the region in which Hooke’s law is obeyed and stress is directly proportional to strain. The straight line portion ends at A.
From A to B: Still Elastic, but No Longer Proportional
If the load is further increased till point B is reached, stress and strain are no longer proportional and Hooke’s law is not valid. The graph has begun to bend over.
But something important still holds in this stretch. If the load is gradually removed starting at any point between O and B, the curve is retraced until the wire regains its original length. The change is reversible. The material of the wire shows elastic behaviour in the region OB.
Point B is called the yield point. The corresponding point is called the elastic limit.
Beyond B: Permanent Set
When the stress is increased beyond point B, the strain continues to increase. Now unload the wire: if the load is removed at any point beyond B — at C, for example — the material does not regain its original length. It follows the line CE.
Look carefully at where CE lands in Fig. 6.14. The length of the wire when there is no stress is now greater than the original length. The deformation is irreversible, and the material has acquired a permanent set. The distance OE on the strain axis is precisely this leftover strain.
Further increase in load causes a large increase in strain for a relatively small increase in stress, until a point D is reached at which fracture takes place. The wire snaps.
The material shows plastic flow or plastic deformation from point B to point D. The material does not regain its original state when the stress is removed.
Not Every Material Shows the Whole Curve
The curve described above shows all the possibilities for an elastic substance. In particular, many metallic wires (copper, aluminium, silver, etc.) exhibit this type of behaviour. However, the majority of materials in everyday life exhibit only some part of it. That is why materials are classified by which portion of the curve they actually show.
Materials such as glass, ceramics, etc., break within the elastic limit. They are called brittle. On their graph, fracture arrives before there is any plastic region at all — which is why a glass rod snaps cleanly rather than bending first.
Metals such as copper, aluminium, wrought iron, etc., have a large plastic range of extension. They lengthen considerably and undergo plastic deformation till they break. They are called ductile.
Metals such as gold and silver, which can be hammered into thin sheets, are called malleable.
Rubber has a large elastic region. It can be stretched so that its length becomes many times its original length, and after removal of the stress it returns to its original state — but the stress-strain curve is not a straight line.
Note the pairing of ideas here: rubber has an enormous elastic region but does not obey Hooke’s law over most of it. Being elastic and obeying Hooke’s law are not the same thing.
Elastic Hysteresis
In the case of some materials like vulcanized rubber, when the stress applied on a body decreases to zero, the strain does not return to zero immediately. The strain lags behind the stress.
Figure 6.15 shows the stress-strain curve for increasing and decreasing load. Because the unloading path does not coincide with the loading path, the two curves enclose a loop. The area of the loop gives the energy dissipated during deformation of a material.
Diagram
Fig. 6.15 – Two stress-strain curves for the same rubber specimen, one for increasing load and one for decreasing load, not coinciding, enclosing a hysteresis loop.
Think about what an enclosed area on this graph must mean. Stress is force per unit area and strain is a length per unit length; their product has the dimensions of energy per unit volume. So the area trapped inside the loop is the energy per unit volume that goes into the material during loading but does not come back out during unloading — it is lost, mostly as heat.
6.7 Strain Energy
Where the Work Goes
Stretching a wire takes effort. A force is applied and the end of the wire moves, so work is done. That work does not vanish — it is stored inside the stretched wire, ready to be given back when the wire is released. This stored energy has a name.
The Force Is Not Constant
Consider a wire of original length \(L\) and cross-sectional area \(A\), stretched by a force \(F\) acting along its length. The wire gets stretched and an elongation \(l\) is produced in it. As the load is added, the stress and the strain increase proportionately.
We already have the two ingredients:
\[\text{Longitudinal stress} = \frac{F}{A}\]
\[\text{Longitudinal strain} = \frac{l}{L}\]
\[Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}}\]
\[Y = \frac{\left(\dfrac{F}{A}\right)}{\left(\dfrac{l}{L}\right)} = \frac{FL}{Al}\]
\[\therefore F = \frac{YAl}{L} \qquad \text{--- (6.19)}\]
Equation (6.19) is the key to the whole derivation. It says the force needed is proportional to the elongation already produced. So the stretching force is not constant during the process — it starts at zero and grows.
The magnitude of the stretching force increases from zero to \(F\) during the elongation of the wire. This is why we cannot simply write work = \(F \times l\); that would assume the full force \(F\) acted through the entire extension, which it did not. Instead, we must add up the work in tiny steps.
Adding Up the Work
At a certain stage of the stretching, let \(f\) be the force applied and \(x\) be the corresponding extension. The force at this stage is given by Equation (6.19) as
\[f = \frac{YAx}{L}\]
For a further extension \(dx\) in the wire — so small that \(f\) does not change appreciably while it happens — the work done is given by
Work = (force) \(\cdot\) (displacement)
\[dW = f \, dx\]
\[\therefore dW = \frac{YAx}{L} \, dx\]
When the wire gets stretched from \(x = 0\) to \(x = l\), the total work done is obtained by integrating:
\[W = \int_{0}^{l} dW\]
\[\therefore W = \int_{0}^{l} \frac{YAx}{L} \, dx\]
\(Y\), \(A\) and \(L\) are all constants for a given wire, so they come outside the integral:
\[\therefore W = \frac{YA}{L} \int_{0}^{l} x \, dx\]
\[\therefore W = \frac{YA}{L} \left[ \frac{x^2}{2} \right]_{0}^{l}\]
\[\therefore W = \frac{YA}{L} \left[ \frac{l^2}{2} - \frac{0^2}{2} \right]\]
\[W = \frac{YAl^2}{2L}\]
Now we simply regroup the terms so that Equation (6.19) can be recognised inside the expression:
\[W = \frac{1}{2} \frac{YAl}{L} \, l\]
\[W = \frac{1}{2} F l\]
\[\text{Work done} = \frac{1}{2} (\text{load}) \cdot (\text{extension}) \qquad \text{--- (6.20)}\]
The factor \(\tfrac{1}{2}\) is the whole point of the derivation, and it is exactly what a graph of force against extension shows: the graph is a straight line from the origin up to \(F\), and the work done is the area under it — the area of a triangle, \(\tfrac{1}{2} \times F \times l\), as in Fig. 6.16. Forgetting this \(\tfrac{1}{2}\) is the single most common error in numericals on strain energy.
Diagram
Fig. 6.16 – Stretching force f (y-axis) against extension x (x-axis): a straight line from the origin to the point (l, F), with the shaded triangular area under it labelled as the work done, ½Fl.
This work done by the stretching force is equal to the energy gained by the wire. This energy is the strain energy.
\[\text{Strain energy} = \frac{1}{2} (\text{load}) \cdot (\text{extension}) \qquad \text{--- (6.21)}\]
Here the load is in newton and the extension in metre, so the strain energy comes out in joule.
Strain Energy per Unit Volume
Equation (6.21) depends on the particular wire — a longer or thicker wire stores more energy for the same material. To get a quantity that describes the material itself, we divide by the volume of the wire, which is \(A \times L\).
Strain energy per unit volume can be obtained by using Equation (6.20) together with the formulae for stress, strain and Young’s modulus.
\[\text{Work done per unit volume} = \frac{\text{work done in stretching wire}}{\text{volume of wire}}\]
\[= \frac{1}{2} \frac{F \cdot l}{A \cdot L}\]
Splitting this product deliberately into the two familiar factors:
\[= \frac{1}{2} \left( \frac{F}{A} \right) \left( \frac{l}{L} \right)\]
The first bracket is the stress and the second is the strain. Hence
\[\text{Work done per unit volume} = \frac{1}{2} (\text{stress}) \cdot (\text{strain})\]
\[\text{Strain energy per unit volume} = \frac{1}{2} (\text{stress}) \cdot (\text{strain}) \qquad \text{--- (6.22)}\]
The SI unit of this quantity is \(\text{J/m}^3\).
Two More Forms of the Same Result
Since \(Y = \dfrac{\text{stress}}{\text{strain}}\), we may write
Stress \(= Y \cdot\) (strain), and strain \(= \dfrac{\text{stress}}{Y}\)
Substituting the first of these into Equation (6.22) removes the stress and leaves only the strain:
\[\therefore \text{Strain energy per unit volume} = \frac{1}{2} Y \cdot (\text{strain})^2 \qquad \text{--- (6.23)}\]
Substituting the second instead removes the strain and leaves only the stress:
\[\therefore \text{Strain energy per unit volume} = \frac{1}{2} \frac{(\text{stress})^2}{Y} \qquad \text{--- (6.24)}\]
Thus Equations (6.22), (6.23) and (6.24) give the strain energy per unit volume in various forms. All three describe the same physical quantity; which one you use depends only on what a problem gives you — stress and strain together, strain and \(Y\), or stress and \(Y\).
6.8 Hardness
What Hardness Means
Hard materials have little ductility, and they are brittle to some extent. This follows from the definition itself: a material that refuses to deform permanently cannot be drawn out into a wire, so it has almost no plastic range — and when it finally gives way, it breaks rather than stretches.
The term hardness also refers to stiffness or resistance to bending, scratching, abrasion or cutting. It is the property of a material which gives it the ability to resist permanent deformation when a load is applied to it. The greater the hardness, the greater is the resistance to deformation.
The most well-known example of the hard materials is diamond. It is incredibly difficult to scratch a diamond. A metal with very low hardness is aluminium.
Hardness Is Not Strength, and Not Toughness
Hardness of a material is different from its strength and toughness. Students routinely treat these three words as loose synonyms for “tough stuff,” and lose marks for it. Each has a precise and separate meaning.
If a force is applied to a body, it produces deformation in it. The higher the force required for deformation, the stronger is the material — that is, the material has more strength.
Steel has high strength, whereas plasticine clay is not strong, because it gets easily deformed even by a small force.
Plasticine clay is relatively tough, as it can be stretched and deformed due to an applied force without breaking. Note what this reveals: plasticine is weak (a small force deforms it) but tough (it does not snap). The two properties are entirely independent of each other.
Diagram
Fig. 6.17 – Three-column comparison chart: Hardness (resists scratching / permanent dent), Strength (resists deformation under large force), Toughness (resists fracture); with an example material listed under each.
The Three Properties Together
A single material may be hard, strong and tough. For example:
- Bulletproof glass is hard and tough but not strong.
- Drill bits must be hard, strong and tough for their work.
- Anvils are very tough and strong but they are not hard.
Read the third example carefully. An anvil takes hammer blow after hammer blow without cracking (tough) and without being pushed out of shape (strong), yet its surface can be dented and scratched over the years (not hard). The three properties genuinely come apart in real objects.
6.9 Friction in Solids
Surfaces Push Back Sideways
Everything so far in this chapter has concerned what happens inside a solid when it is deformed. We now turn to what happens at the surface where two solids touch — another mechanical property of solids, and one that governs almost every movement you make.
Whenever the surface of one body slides over another, each body exerts a certain amount of force on the other body. These forces are tangential to the surfaces — that is, they act along the surfaces, not into them.
The force on each body is opposite to the direction of motion between the two bodies. It prevents or opposes the relative motion between the two bodies, as shown in Fig. 6.18.
Diagram
Fig. 6.18 – A block on a horizontal surface with an applied force F acting to the right and the frictional force acting tangentially to the left, along the surface of contact.
Note the careful wording: friction opposes relative motion between the two bodies, not motion in general. Which way friction points therefore depends on which way one surface is trying to slide over the other — a point we shall return to when we discuss self-adjusting static friction.
Everyday Evidence
It is a common experience that an object placed on any surface does not move easily when a small force is applied to it. This is because of a certain force of opposition acting between the surface of the object and the surface on which it is placed.
Even a rolling ball comes to rest after covering a finite distance on a playground, because of such an opposing force. Nothing pushed it backwards, yet it stopped — so something at the contact must have been quietly taking away its motion.
Our footwear is provided with designs at the bottom of its sole so as to produce a force of opposition and avoid slipping. It is difficult to walk without such an opposing force. You know what happens when you try to walk fast on polished flooring at home with soap water spread on it: there is a possibility of slipping, due to the lack of the force of opposition.
Friction, Defined
To initiate any motion between a pair of surfaces, we need a certain minimum force. Also, after the motion begins, it is constantly opposed by some natural force. This mechanical force between two solid surfaces in contact with each other is called frictional force.
When Friction Is Unwanted
In some cases it is necessary to avoid friction, because friction causes dissipation of energy in machines, due to which the efficiency of machines decreases. Energy that should have driven the machine’s output is instead lost as heat and sound at every rubbing contact.
In such cases friction should be reduced, by using polished surfaces, lubricants, and so on. The detailed methods of doing this are taken up later in this section.
Friction Is Not Confined to Solids
Relative motion between solids and fluids (i.e. liquids and gases) is also naturally opposed by friction. For example, a boat on the surface of water experiences opposition to its motion — the water resists being pushed aside, and drags on the hull.
In this section, however, we are going to study friction in solids only.
6.9.1 Origin of Friction
The Old Explanation: Interlocking
If smooth surfaces are observed under a powerful microscope, many irregularities and projections are observed. What looks like a flat, polished face to the eye turns out to be a landscape of tiny hills and valleys.
The oldest explanation of friction rests on this picture. Friction arises due to the interlocking of these irregularities between two surfaces in contact: the projections of one surface sit in the hollows of the other, and sliding one surface over the other means dragging those projections out of place.
Why That Explanation Fails
The interlocking picture makes a clear prediction, and the prediction can be tested. If interlocking were the whole story, then removing the irregularities should remove the friction.
The surfaces can be made extremely smooth by polishing to avoid irregularities — but it is noticed that in this case also, friction does not decrease. It may even increase. Hence the interlocking of irregularities is not the real cause of friction.
This is a good example of how physics advances: a reasonable idea is tested against experiment, the experiment refuses to agree, and a better idea must be found.
The Modern Explanation: Molecular Attraction
According to the modern theory, the cause of friction is the force of attraction between the molecules of the two surfaces in actual contact, in addition to the force due to the interlocking between the two surfaces. Interlocking is not discarded altogether; it is demoted to being only a part of the story.
When one body is in contact with another body, the real microscopic area in contact is very small, due to the irregularities in contact. The two surfaces touch only at the tips of their projections, so the true contact area is far smaller than the apparent area you would measure with a ruler. Figure 6.19 shows the microscopic view of two polished surfaces in contact.
Diagram
Fig. 6.19 – Microscopic view of two polished surfaces in contact, showing jagged projections meeting only at a few points, with the true area of contact far smaller than the apparent area.
Due to this small area, the pressure at the points of contact is very high. Recall that pressure is force divided by area: the whole weight of the upper body is being carried on a handful of tiny high spots, so the force per unit area there is enormous. Hence there is a strong force of attraction between the surfaces in contact — the molecules at those crushed-together points are pressed close enough for their mutual attraction to take hold.
The Surprising Consequence
Now the modern theory explains the very observation that defeated the old one. When the surfaces in contact become more and more smooth, the actual area of contact goes on increasing — polishing flattens the projections, so the two faces meet over a wider real area, not a narrower one. Due to this, the force of attraction between the molecules increases, and hence the friction also increases.
How Lubricants Fit In
Putting some grease or other lubricant — a different material — between the two surfaces reduces the friction. The lubricant separates the two faces so that the molecules of one surface are no longer close enough to those of the other to attract them strongly. In effect, the strong solid-to-solid attraction is replaced by a much weaker interaction across a thin film.
6.9.2 Types of Friction
Friction shows up in three distinct situations: when a body is at rest and refuses to start moving, when it is already sliding, and when it is rolling. Each is given its own name and its own laws.
1. Static Friction
Suppose a wooden block is placed on a horizontal surface, as shown in Fig. 6.20. A small horizontal force \(F\) is applied to it. The block does not move with this force, as it cannot overcome the frictional force between the block and the horizontal surface. In this case, the force of static friction is equal to \(F\) and balances it.
Diagram
Fig. 6.20 – A block resting on a horizontal surface with a small applied force F acting horizontally, balanced by the force of static friction acting in the opposite direction along the surface.
If we keep increasing \(F\), a stage will come when, for \(F = F_{max}\), the object will start moving. For \(F < F_{max}\), the force of static friction is equal to \(F\). For \(F \geq F_{max}\), the kinetic friction comes into play.
Static friction opposes impending motion — that is, the motion that would take place in the absence of the frictional force under the applied force. The word “impending” is worth holding on to: the block is not moving, so friction is opposing a motion that has not happened, and would have happened but for friction itself.
Static Friction Is Self-Adjusting
The force of static friction is a self-adjusting force. When the applied force \(F\) is very small, the block remains at rest, and here the force of friction is also small. When \(F\) is increased by a small value, the block still remains at rest, as the force of friction is increased to balance the applied force. If the applied force is increased further, the friction also increases, and reaches a maximum value.
Just before the body starts sliding over another body, the value of the frictional force is maximum.
Static friction adjusts its direction as well as its magnitude. If the direction of the applied force is reversed, the direction of the static friction is also reversed.
Laws of Static Friction
1] The limiting force of static friction is directly proportional to the normal reaction (N) between the two surfaces in contact.
\[F_L \propto N\]
\[\therefore F_L = \mu_s N \qquad \text{--- (6.25)}\]
Here \(F_L\) is the limiting force of friction in newton, \(N\) is the normal reaction in newton — the perpendicular push of the surface on the body, which for a block resting on a horizontal surface is simply \(mg\) — and \(\mu_s\) is the constant of proportionality. It is called the coefficient of static friction.
\[\therefore \mu_s = \frac{F_L}{N} \qquad \text{--- (6.26)}\]
Being a ratio of two forces, \(\mu_s\) has no unit and no dimensions. Table 6.4 gives the coefficient of static friction for some materials.
2] The limiting force of friction is independent of the apparent area between the surfaces in contact, so long as the normal reaction remains the same.
This law surprises most students. Lay the same block on its broad face or on its narrow edge, and the friction is the same. Section 6.9.1 explains why: what matters is the real microscopic area of contact, and spreading the same weight over a larger apparent area simply reduces the pressure at each contact point, leaving the real contact area — and hence the friction — unchanged.
3] The limiting force of friction depends upon the materials in contact and the nature of their surfaces.
Table 6.4: Coefficient of static friction
| Material | Coefficient of static friction μ_s |
|---|---|
| Teflon on Teflon | 0.4 |
| Brass on steel | 0.51 |
| Copper on steel | 0.53 |
| Aluminium on steel | 0.61 |
| Steel on steel | 0.74 |
| Glass on glass | 0.94 |
| Rubber on concrete (dry) | 1.0 |
Notice that every entry names a pair of materials, never a single one. Friction is a property of two surfaces together, not of one surface by itself — which is exactly what the third law states.
2. Kinetic Friction
Once the sliding of the block on the surface starts, the force of friction decreases. The force required to keep the body sliding steadily is thus less than the force required to just start its sliding. This is a fact you can feel: a heavy carton is hardest to get moving, and easier to keep moving once it is under way.
The force of friction that comes into play when a body is in a steady state of motion over another surface is called the force of kinetic friction.
Laws of Kinetic Friction
1. The force of kinetic friction (\(F_k\)) is directly proportional to the normal reaction between the two surfaces in contact.
\[\therefore F_k \propto N\]
\[\therefore F_k = \mu_k N \qquad \text{--- (6.27)}\]
Here \(\mu_k\) is the constant of proportionality. It is called the coefficient of kinetic friction.
\[\therefore \mu_k = \frac{F_k}{N} \qquad \text{--- (6.28)}\]
Table 6.5 gives the coefficient of kinetic friction for some materials.
2. The force of kinetic friction is independent of the shape and apparent area of the surfaces in contact.
3. The force of kinetic friction depends upon the nature and material of the surfaces in contact.
4. The magnitude of the force of kinetic friction is independent of the relative velocity between the object and the surface, provided that the relative velocity is neither too large nor too small.
Table 6.5: Coefficient of kinetic friction
| Material | Coefficient of kinetic friction μ_k |
|---|---|
| Rubber on concrete (dry) | 0.25 |
| Glass on glass | 0.40 |
| Brass on steel | 0.40 |
| Copper on steel | 0.44 |
| Aluminium on steel | 0.47 |
| Steel on steel | 0.57 |
| Teflon on Teflon | 0.80 |
Compare Table 6.5 with Table 6.4, pair by pair. For steel on steel, \(\mu_s = 0.74\) while \(\mu_k = 0.57\); for rubber on concrete, \(\mu_s = 1.0\) while \(\mu_k = 0.25\). In every case the kinetic coefficient is the smaller of the two — the quantitative statement of what we just said in words, that a sliding body is easier to keep moving than a resting body is to start.
3. Rolling Friction
Motion of a body over a surface is said to be rolling motion if the point of contact of the body with the surface keeps changing continuously. A wheel does not drag one patch of itself along the road; it sets down a fresh patch at every instant.
For the same pair of surfaces, the force of static friction is greater than the force of kinetic friction, while the force of kinetic friction is greater than the force of rolling friction. In symbols, for the same two surfaces:
\[F_{\text{static}} > F_{\text{kinetic}} > F_{\text{rolling}}\]
As rolling friction is the minimum, ball bearings are used to reduce friction in the parts of machines, and thereby to increase their efficiency. A ball bearing replaces a shaft that would otherwise slide inside its housing with a set of balls that roll — converting a large kinetic friction into a small rolling friction.
Advantages of Friction
Friction is necessary in our daily life.
- We can walk due to friction between the ground and our feet.
- We can hold an object in the hand due to static friction.
- Brakes of vehicles work due to friction; hence we can reduce speed or stop vehicles.
- Climbing on a tree is possible due to friction.
Disadvantages of Friction
- Friction opposes motion.
- Friction produces heat in different parts of machines. It also produces noise.
- Automobile engines consume more fuel due to friction.
Methods of Reducing Friction
- Use of lubricants, oil and grease in different parts of a machine.
- Use of ball bearings, which converts kinetic friction into rolling friction.
6.10 Summary
Deformation and Elastic Behaviour
- The constituents of a solid sit in a definite geometric arrangement in which the resultant force on each constituent due to the others is zero. They vibrate about these equilibrium positions but cannot leave them, which gives a solid its definite shape and size.
- A force that changes the size or shape or both of a body, by changing the relative positions of its molecules, is a deforming force. The resulting change in shape or size or both is called deformation. The larger the deforming force, the larger the deformation.
- Elasticity: if a body regains its original shape and size after removal of the deforming force, it is an elastic body. A body that regains its original shape and size completely and instantaneously is perfectly elastic.
- Plasticity: if a body does not regain its original shape and size and retains the altered form, it is a plastic body.
- No solid is perfectly elastic or perfectly plastic. The nearest ideal elastic solid is quartz fibre; the nearest plastic body is putty.
Stress and Strain
- Stress is the internal restoring force per unit area of a body: \[\text{stress} = \frac{\text{deforming force}}{\text{area}} = \frac{|\vec{F}|}{A} \qquad \text{(6.1)}\] SI unit: \(\text{N m}^{-2}\) or pascal (Pa). Dimensions: \([L^{-1} M^{1} T^{-2}]\).
- Strain is the ratio of the change in dimensions of the body to its original dimensions: \[\text{Strain} = \frac{\text{change in dimensions}}{\text{original dimensions}} \qquad \text{(6.2)}\] Strain is a dimensionless quantity and has no unit.
- Three types of stress and the corresponding strains:
| Type | Stress | Strain | Effect |
|---|---|---|---|
| Longitudinal | Tensile / compressive stress \(= \dfrac{\lvert \vec{F} \rvert}{A}\) (6.3), (6.4) | Tensile strain \(= \dfrac{\Delta l}{L}\) (6.5) | Change in length |
| Volume | Volume (hydraulic) stress \(= \dfrac{\lvert \vec{F} \rvert}{A}\) (6.6) | Volume strain \(= \dfrac{\Delta V}{V}\) (6.7) | Change in size, not shape |
| Shear | Shearing stress \(= \dfrac{\text{Tangential force}}{\text{Area}}\) (6.8) | Shearing strain \(= \dfrac{\Delta l}{l} = \tan\theta = \theta\) (6.9) | Change in shape, not size |
- In tensile and volume stress the force is perpendicular to the surface; in shearing stress the tangential force is parallel to the top and bottom surfaces of the block. Shearing strain \(\theta\) is in radian, and \(\tan\theta = \theta\) only when \(\Delta l\) is very small.
Hooke’s Law
- Hooke’s law: Within the elastic limit, stress is directly proportional to strain. \[\frac{\text{Stress}}{\text{Strain}} = \text{constant}\]
- The constant is the modulus of elasticity — the ratio of stress to the corresponding strain. It is the slope of the stress-strain curve in the elastic deforming region and depends on the nature of the material.
- The elastic limit is the maximum value of stress up to which stress is directly proportional to strain.
The Three Moduli and Poisson’s Ratio
- Young’s modulus (Y) — related to change in length; property of solids only. \[Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}} \qquad \text{(6.12)}, \qquad Y = \frac{MgL}{\pi r^2 l} \qquad \text{(6.13)}\] SI unit \(\text{N/m}^2\); dimensions \([L^{-1} M^{1} T^{-2}]\). It indicates the resistance of an elastic solid to elongation or compression.
- Bulk modulus (K) — related to change in volume; a property of solids, liquids and gases. \[\text{Volume strain} = -\frac{dV}{V} \qquad \text{(6.14)}, \qquad K = \frac{dP}{\left(\dfrac{dV}{V}\right)} = V\frac{dP}{dV} \qquad \text{(6.17)}\] The negative sign in the volume strain shows a decrease in volume. SI unit \(\text{N/m}^2\); dimensions \([L^{-1} M^{1} T^{-2}]\).
- Compressibility is the reciprocal of bulk modulus: \[\text{Compressibility} = \frac{1}{\text{Bulk modulus}} \qquad \text{(6.18)}\] It is the fractional decrease in volume, \(-\Delta V/V\), per unit increase in pressure. SI unit \(\text{m}^2/\text{N}\) or \(\text{Pa}^{-1}\); dimensions \([L^{1} M^{-1} T^{2}]\).
- Modulus of rigidity (η) — related to change in shape; property of solids only. \[\eta = \frac{\text{shear stress}}{\text{shear strain}} = \frac{F/A}{\theta} = \frac{F}{A\theta} \qquad \text{(6.17)}\] It indicates the resistance offered by a solid to a change in its shape.
- Poisson’s ratio (σ) — within the elastic limit, the ratio of lateral strain to linear strain: \[\sigma = \frac{\text{Lateral strain}}{\text{Linear strain}} = \frac{d/D}{l/L} = \frac{d \cdot L}{D \cdot l} \qquad \text{(6.18)}\] It has no unit and is dimensionless. For most commonly used metals \(\sigma\) lies between 0.25 and 0.35.
- When a wire is stretched it becomes longer and thinner; when compressed it becomes shorter and its diameter increases.
The Stress-Strain Curve
- Plotting tensile stress (y-axis) against tensile strain (x-axis) for a wire loaded to breaking gives the stress-strain curve.
- OA: straight line — Hooke’s law obeyed; stress at A is the proportional limit.
- A to B: stress and strain no longer proportional, but the change is still reversible; the material is elastic in the region OB. Point B is the yield point; the corresponding point is the elastic limit.
- Beyond B: unloading (e.g. from C) does not retrace the curve; it follows CE, and the wire is left longer than it began — a permanent set. The deformation is irreversible.
- B to D: plastic flow / plastic deformation. At D, fracture takes place.
- Brittle materials (glass, ceramics) break within the elastic limit. Ductile metals (copper, aluminium, wrought iron) have a large plastic range. Malleable metals (gold, silver) can be hammered into thin sheets. An elastomer (rubber) can be elastically stretched to a large strain, though its stress-strain curve is not a straight line.
- Elastic hysteresis: in materials such as vulcanized rubber, the strain lags behind the stress on unloading. The loading and unloading curves enclose a loop, and the area of the loop gives the energy dissipated during deformation.
Strain Energy
- Strain energy is the elastic potential energy gained by a wire during elongation by a stretching force.
- From \(Y\): \(\quad F = \dfrac{YAl}{L} \qquad \text{(6.19)}\)
- The stretching force grows from zero to \(F\), so the work is found by integration, giving \[\text{Work done} = \frac{1}{2}(\text{load}) \cdot (\text{extension}) \qquad \text{(6.20)}\] \[\text{Strain energy} = \frac{1}{2}(\text{load}) \cdot (\text{extension}) \qquad \text{(6.21)}\]
- Strain energy per unit volume, in three equivalent forms: \[= \frac{1}{2}(\text{stress}) \cdot (\text{strain}) \qquad \text{(6.22)}\] \[= \frac{1}{2} Y \cdot (\text{strain})^2 \qquad \text{(6.23)}\] \[= \frac{1}{2} \frac{(\text{stress})^2}{Y} \qquad \text{(6.24)}\]
Hardness, Strength and Toughness
- Hardness is the property of a material which enables it to resist plastic deformation; it also refers to stiffness or resistance to bending, scratching, abrasion or cutting. The greater the hardness, the greater the resistance to deformation. Diamond is the best-known hard material; aluminium has very low hardness.
- Strength: the higher the force required to deform a material, the stronger it is. Steel has high strength; plasticine clay does not.
- Toughness: the ability to resist fracturing when a force is applied. Plasticine clay is relatively tough.
- These three properties are independent: bulletproof glass is hard and tough but not strong; drill bits must be hard, strong and tough; anvils are tough and strong but not hard.
Friction
- Friction is the property which resists the relative motion between two surfaces in contact. The frictional forces are tangential to the surfaces.
- Origin: interlocking of surface irregularities is not the real cause, since polishing does not reduce friction and may increase it. According to modern theory, friction is caused by the force of attraction between the molecules of the two surfaces in actual contact, in addition to interlocking. The real area of contact is very small, so the pressure at the contact points is very high. The attraction is called cohesive force between like materials and adhesive force between unlike materials. Lubricants reduce friction by separating the surfaces.
- Static friction balances the applied force while the body is static, and is self-adjusting in both magnitude and direction. Its maximum value, just before sliding begins, is the limiting force of friction \(F_L\). \[F_L \propto N, \qquad F_L = \mu_s N \qquad \text{(6.25)}, \qquad \mu_s = \frac{F_L}{N} \qquad \text{(6.26)}\] Laws: (1) \(F_L \propto N\); (2) \(F_L\) is independent of the apparent area of contact, so long as \(N\) is unchanged; (3) \(F_L\) depends on the materials in contact and the nature of their surfaces.
- Kinetic (dynamic) friction acts when one body is actually sliding over another. \[F_k \propto N, \qquad F_k = \mu_k N \qquad \text{(6.27)}, \qquad \mu_k = \frac{F_k}{N} \qquad \text{(6.28)}\] Laws: (1) \(F_k \propto N\); (2) independent of shape and apparent area of contact; (3) depends on the nature and material of the surfaces; (4) independent of the relative velocity, provided that velocity is neither too large nor too small.
- \(\mu_s\) and \(\mu_k\) are ratios of two forces, and so are dimensionless and unitless.
- Rolling friction acts when the point of contact of the body with the surface keeps changing continuously.
- For the same pair of surfaces: \(F_{\text{static}} > F_{\text{kinetic}} > F_{\text{rolling}}\). Since rolling friction is the least, ball bearings are used to reduce friction in machines and increase their efficiency.
- Friction is useful (walking, holding objects, brakes, climbing) and also harmful (opposes motion, produces heat and noise, wastes fuel). It is reduced by lubricants, oil and grease, and by ball bearings, which convert kinetic friction into rolling friction.
6.11 Check Your Concepts
A rubber band can be stretched to several times its original length, while a steel wire of the same length can be stretched by only a fraction of a millimetre. Yet steel is said to be more elastic than rubber. Explain why this statement is correct.
Stress is defined as the internal restoring force per unit area, but in every numerical problem we divide the externally applied load by the area. Justify this practice.
Why does strain have no unit, even though both the change in dimension and the original dimension are measured in metre?
Tensile stress, volume stress and shearing stress are all written as force divided by area. What then distinguishes them from one another? Answer in terms of the direction of the force and the change it produces.
Young’s modulus and modulus of rigidity are properties of solids only, but bulk modulus applies to solids, liquids and gases. Explain the reason for this difference.
Two wires of the same material but different lengths and different areas of cross section are stretched by different loads within the elastic limit. Will they have the same Young’s modulus? Will they have the same elongation? Justify both answers.
Why is a negative sign included in the definition of volume strain, when we anyway use only its magnitude while solving numericals?
The proportional limit and the elastic limit are two different points on the stress-strain curve. What exactly happens to the wire between these two points?
A wire is loaded beyond its yield point and then the load is completely removed. Sketch what happens on the stress-strain curve and explain what a permanent set means physically.
Glass is very hard, yet a glass rod snaps without bending. Copper is far softer, yet a copper wire can be drawn out to great lengths without breaking. Use the terms brittle and ductile to explain these observations.
In the derivation of strain energy, why is the work done taken as \(\tfrac{1}{2} F l\) and not \(F l\)? What would go wrong physically if we used \(F l\)?
The area enclosed by the hysteresis loop of a rubber specimen represents energy dissipated per unit volume. What happens to this energy, and where in everyday life is this effect noticeable?
An anvil is described as tough and strong but not hard, while bulletproof glass is described as hard and tough but not strong. Explain, in terms of the definitions of hardness, strength and toughness, how such combinations are possible.
Interlocking of surface irregularities was once believed to be the cause of friction. Which experimental observation shows that this cannot be the real cause, and what does the modern theory propose instead?
Two surfaces are polished until they are extremely smooth. Would you expect the friction between them to increase or decrease? Give reasons based on the real area of contact.
A block of wood is placed first on its broad face and then on its narrow edge on the same table. Will the limiting force of friction change? Which law of static friction does your answer illustrate?
Why is the force of static friction called self-adjusting? Describe how both its magnitude and its direction adjust.
A heavy carton is hard to get moving but comparatively easy to keep moving once it has started sliding. Explain this using the coefficients of static and kinetic friction, and support your answer with values from the tables in the chapter.
Ball bearings are used in the moving parts of machines. Which type of friction do they replace, and which type do they introduce? Why does this increase the efficiency of the machine?
Friction is described as a “necessary evil.” Give two situations in which friction is deliberately increased and two in which it is deliberately reduced, and explain the purpose in each case.
Define elasticity.
What do you mean by deformation?
State the SI unit and dimensions of stress.
Why are bridges unsafe after a very long use?
How should a force be applied on a body to produce shearing stress?
State the conditions under which Hooke’s law holds good.
Define Poisson’s ratio.
What is an elastomer?
What do you mean by elastic hysteresis?
State the names of the hardest material and the softest material.
Name two factors on which the coefficient of friction depends.
Distinguish between elasticity and plasticity.
State any four methods to reduce friction.
What is rolling friction? How does it arise?
Explain how lubricants help in reducing friction.
State the laws of static friction.
State the laws of kinetic friction.
State the advantages of friction.
State the disadvantages of friction.
What do you mean by a brittle substance? Give any two examples.
Distinguish between Young’s modulus, bulk modulus and modulus of rigidity.
Define stress and strain. What are their different types?
What is Young’s modulus? Describe an experiment to find out Young’s modulus of a material in the form of a long straight wire.
Derive an expression for strain energy per unit volume of the material of a wire.
What is friction? Define coefficient of static friction and coefficient of kinetic friction. Give the necessary formula for each.
State Hooke’s law. Draw a labelled graph of tensile stress against tensile strain for a metal wire up to the breaking point. In this graph show the region in which Hooke’s law is obeyed.
6.12 Practice with Numericals
A steel wire of length 3.0 m and cross-sectional area \(2.0 \times 10^{-6} \ \text{m}^2\) carries a load of 400 N. Calculate the tensile stress in the wire and the tensile strain produced. [Take Young’s modulus of steel from Table 6.1.]
A copper wire 2.0 m long and 1.0 mm in diameter is stretched by a load of 50 N. Using the value of Young’s modulus of copper from Table 6.1, calculate the elongation produced in the wire.
An aluminium wire and a brass wire have the same length and the same area of cross section, and are stretched by the same load within the elastic limit. Using Table 6.1, find the ratio of the elongation of the aluminium wire to that of the brass wire.
A load of 20 kg is suspended from a steel wire of radius 0.5 mm and length 4.0 m. Calculate the stress in the wire and the extension produced. [Take \(g = 9.8 \ \text{m s}^{-2}\) and Young’s modulus of steel from Table 6.1.]
A metal cube of side 0.50 m is subjected to a uniform pressure increase of \(4.0 \times 10^{6} \ \text{N/m}^2\) acting normally on its entire surface. If the bulk modulus of the metal is \(8.0 \times 10^{10} \ \text{N/m}^2\), calculate the change in its volume.
A solid silver sphere of volume \(1.5 \times 10^{-3} \ \text{m}^3\) is taken to a depth in the sea where the pressure on it is greater by \(1.0 \times 10^{7} \ \text{N/m}^2\) than at the surface. Using Table 6.2, calculate the change in the volume of the sphere and its volume strain.
The bulk modulus of a certain liquid is \(2.0 \times 10^{9} \ \text{N/m}^2\). Calculate its compressibility, and state the SI unit of your answer.
A cube of copper of side 0.20 m is fixed at its bottom face. A tangential force of \(8.0 \times 10^{5} \ \text{N}\) is applied to its top face. Using the rigidity modulus of copper from Table 6.3, calculate the shearing strain produced and the sideways displacement of the top face.
A metal block of square face of side 0.10 m and height 0.10 m has its base fixed. A tangential force of 500 N applied to the top face displaces it through 0.05 mm. Calculate the shearing stress, the shearing strain, and the modulus of rigidity of the metal.
A brass wire of original length 1.5 m and original diameter 0.80 mm is stretched so that its length increases by 0.60 mm. Using the value of Poisson’s ratio for brass from Table 6.4, calculate the change in the diameter of the wire.
A wire of length 2.0 m, cross-sectional area \(1.0 \times 10^{-6} \ \text{m}^2\) and Young’s modulus \(1.2 \times 10^{11} \ \text{N/m}^2\) is stretched by a slowly applied load of 60 N. Calculate (a) the elongation produced, (b) the strain energy stored in the wire, and (c) the strain energy per unit volume.
The strain produced in a stretched wire is \(5.0 \times 10^{-4}\), and the Young’s modulus of its material is \(2.0 \times 10^{11} \ \text{N/m}^2\). Calculate the strain energy per unit volume of the wire, using two different formulae, and check that both give the same answer.
A wire of volume \(2.0 \times 10^{-6} \ \text{m}^3\) stores a strain energy of \(1.6 \times 10^{-2} \ \text{J}\) when stretched. Calculate the strain energy per unit volume of the wire.
A block of mass 8.0 kg rests on a horizontal glass surface. The two surfaces are glass on glass. Using Table 6.4, calculate the least horizontal force required to just start the block moving. [Take \(g = 9.8 \ \text{m s}^{-2}\).]
A brass block of mass 12 kg is sliding steadily on a horizontal steel surface. Using the coefficient of kinetic friction for brass on steel from Table 6.5, calculate the force of kinetic friction acting on the block. [Take \(g = 9.8 \ \text{m s}^{-2}\).]
A block of mass 15 kg rests on a rough horizontal surface. A horizontal force of 60 N is just able to start it moving, and a force of 45 N keeps it sliding with constant velocity. Calculate the coefficient of static friction and the coefficient of kinetic friction between the block and the surface. [Take \(g = 9.8 \ \text{m s}^{-2}\).]
An aluminium block is placed on a steel surface. If the limiting force of friction between them is 47.0 N, calculate the mass of the block, using the coefficient of static friction for aluminium on steel from Table 6.4. [Take \(g = 9.8 \ \text{m s}^{-2}\).]
A rubber tyre of mass 10 kg is dragged over dry concrete. Using Tables 6.4 and 6.5, calculate the horizontal force needed to just start it sliding and the horizontal force needed to keep it sliding steadily. Find the difference between the two forces. [Take \(g = 9.8 \ \text{m s}^{-2}\).]
Calculate the coefficient of static friction for an object of mass 50 kg placed on a horizontal table and pulled by attaching a spring balance. The force is increased gradually, and it is observed that the object just moves when the spring balance shows 50 N.
A block of mass 37 kg rests on a rough horizontal plane having coefficient of static friction 0.3. Find out the least force required to just move the block horizontally.
A body of mass 37 kg rests on a rough horizontal surface. The minimum horizontal force required to just start the motion is 68.5 N. In order to keep the body moving with constant velocity, a force of 43 N is needed. What is the value of (a) the coefficient of static friction, and (b) the coefficient of kinetic friction?
A wire gets stretched by 4 mm due to a certain load. If the same load is applied to a wire of the same material with half the length and double the diameter of the first wire, what will be the change in its length?
Calculate the work done in stretching a steel wire of length 2 m and cross-sectional area 0.0225 mm² when a load of 100 N is slowly applied to its free end. [Young’s modulus of steel \(= 2 \times 10^{11} \ \text{N/m}^2\)]
A solid metal sphere of volume 0.31 m³ is dropped in an ocean where the water pressure is \(2 \times 10^{7} \ \text{N/m}^2\). Calculate the change in volume of the sphere if the bulk modulus of the metal is \(6.1 \times 10^{10} \ \text{N/m}^2\).
A wire of mild steel has initial length 1.5 m and diameter 0.60 mm, and is extended by 6.3 mm when a certain force is applied to it. If Young’s modulus of mild steel is \(2.1 \times 10^{11} \ \text{N/m}^2\), calculate the force applied.
A composite wire is prepared by joining a tungsten wire and a steel wire end to end. Both the wires are of the same length and the same area of cross section. If this composite wire is suspended from a rigid support and a force is applied to its free end, it gets extended by 3.25 mm. Calculate the increase in length of the tungsten wire and the steel wire separately. [Given: \(Y_{steel} = 2 \times 10^{11} \ \text{N/m}^2\), \(Y_{Tungsten} = 3.40 \times 10^{8} \ \text{N/m}^2\)]
A steel wire having cross-sectional area 1.2 mm² is stretched by a force of 120 N. If a lateral strain of 1.455 mm is produced in the wire, calculate the Poisson’s ratio. [Given: \(Y_{steel} = 2 \times 10^{11} \ \text{N/m}^2\)]
A telephone wire 125 m long and 1 mm in radius is stretched to a length of 125.25 m when a force of 800 N is applied. What is the value of Young’s modulus for the material of the wire?
A rubber band originally 30 cm long is stretched to a length of 32 cm by a certain load. What is the strain produced?
What is the stress in a wire which is 50 m long and 0.01 cm² in cross section, if the wire bears a load of 100 kg?
What is the strain in a cable of original length 50 m whose length increases by 2.5 cm when a load is lifted?
6.13 Objective Questions
- Change in dimensions is known as…..
- deformation
- formation
- contraction
- strain
- The point on the stress-strain curve at which strain begins to increase even without increase in stress is called….
- elastic point
- yield point
- breaking point
- neck point
- Strain energy of a stretched wire is \(18 \times 10^{-3}\) J and strain energy per unit volume of the same wire and same cross section is \(6 \times 10^{-3} \ \text{J/m}^3\). Its volume will be….
- 3 cm³
- 3 m³
- 6 m³
- 6 cm³
- —– is the property of a material which enables it to resist plastic deformation.
- elasticity
- plasticity
- hardness
- ductility
- The ability of a material to resist fracturing when a force is applied to it, is called……
- toughness
- hardness
- elasticity
- plasticity