8.1 Introduction
Chapter 8 — Mechanical Properties of Solids
In the 1660s, an English scientist named Robert Hooke became fascinated by a very ordinary object — a coiled spring. He noticed something that seems obvious today but was not obvious then: if he hung a small weight on a spring it stretched a little, and if he hung double the weight it stretched twice as much. The stretch, he realised, was neatly proportional to the pull.
Hooke was certain he had found a genuine law of nature, but he was also protective of his discovery. Rather than announce it openly while he worked out its uses, he published it in 1676 as a scrambled Latin anagram — a jumble of letters that meant nothing to anyone but himself. Two years later he revealed the solution: ut tensio, sic vis — “as the extension, so the force.”
That short phrase became the foundation of the entire study of how solids deform. It let engineers predict how much a cable would stretch, how far a beam would bend, and how much load a bridge could carry — all before a single piece of metal was cut. In this chapter we study exactly these ideas: how real solids stretch, compress, twist, and bend, and why understanding this is essential to building almost everything around us.
Figure to come
Fig. 8.0 – An illustration of a helical spring hanging vertically from a support, stretching by increasing amounts as heavier weights are hung on it, with a small portrait-style figure of a 17th-century scientist observing.
By the end of this chapter, you will be able to answer each of these questions using the physics of stress, strain, and elasticity.
In our earlier study of rotational motion (Chapter 6), we saw that how a body moves and rotates depends on how its mass is spread out inside it. To keep those problems manageable, we treated objects as rigid bodies. A rigid body is an idealised solid whose shape and size never change, no matter how hard you push, pull, or twist it — the distance between any two points inside it always stays fixed.
But this is only an approximation. In the real world, no solid is perfectly rigid. If you press hard enough, even a thick steel bar will bend slightly. Every real solid can be stretched, squeezed, or bent to some extent when a large enough force acts on it. The perfectly rigid body is a useful fiction, not a real object.
A solid has a definite shape and a definite size. To change (deform) that shape or size, a force must act on it. Consider a helical (coiled) spring. If you gently pull its two ends, its length increases a little, as suggested in Fig. 8.1a. The moment you let go, the spring pulls itself back to exactly its original length and shape.
This “spring-back” behaviour has a name. The property of a body by which it returns to its original shape and size once the deforming force is removed is called elasticity, and the temporary change in shape during the process is called elastic deformation.
Not every material behaves this way. If instead you press a lump of putty or wet mud, it changes shape and simply stays that way — it shows almost no tendency to spring back. Materials like this are called plastic, and the property is called plasticity. Putty and mud come close to being ideal plastic substances.
Figure to come
Fig. 8.1a – A helical spring being stretched by a pull at both ends and returning to its original length when released (elastic behaviour), shown beside a lump of putty that stays flattened after being pressed (plastic behaviour).
So the difference between the spring and the putty is not the force you apply — it is what the material does afterwards. The spring stores the effect of the force temporarily and gives it back; the putty keeps the deformation permanently. This single distinction — whether a body recovers or not — is the starting point of the whole chapter.
Why does any of this matter? Because the elastic behaviour of materials sits at the heart of almost all engineering design. When engineers design a building, they must know the elastic properties of steel, concrete, and other materials so that the structure carries its load without collapsing or sagging permanently. The same knowledge governs the design of bridges, automobiles, and ropeways.
The chapter also raises questions you may not yet be able to answer: Why is a railway track shaped like the letter I? Why does glass shatter while brass simply bends? The study of how relatively simple loads and forces deform different solid bodies gives us the tools to answer all of these. In the sections that follow, we build these tools step by step — beginning with a precise way to measure how much force a solid feels and how much it actually deforms.
8.2 Stress and Strain
Suppose you apply forces to a body but arrange them so that the body does not move off as a whole — it stays in static equilibrium. Under such balanced forces the body does not fly away; instead it is deformed, stretched or squeezed by a small or large amount. How much it deforms depends on two things: the nature of the material and how large the deforming force is.
This deformation is always present, even when your eyes cannot see it. Press on a thick metal table and it sags by an amount far too tiny to notice — but the sag is real.
When a deforming force acts on a body, the body fights back. Its internal structure develops a restoring force that tries to return the body to its original shape. In equilibrium, this restoring force is equal in magnitude and opposite in direction to the applied deforming force. It is this internal restoring force, spread over the area of the body, that we measure as stress.
If \(F\) is the force applied normal (perpendicular) to a cross-section of area \(A\), then the magnitude of the stress is
\[\text{Magnitude of the stress} = \frac{F}{A} \tag{8.1}\]
where \(F\) is the applied force in newtons (N) and \(A\) is the cross-sectional area in square metres (m²).
The SI unit of stress is newton per square metre (N m⁻²), also called the pascal (Pa). Its dimensional formula is \([\text{ML}^{-1}\text{T}^{-2}]\) — the same as that of pressure.
There are three distinct ways in which a solid can change its dimensions when an external force acts on it. All three are shown together in Fig. 8.1.
Figure to come
Fig. 8.1 – Four bodies: (a) a cylinder pulled by equal outward forces, elongating by ΔL (tensile); (b) a cylinder with equal opposite forces parallel to its faces, tilting by angle θ with the top face shifted sideways by Δx (shearing); (c) a book pressed by a hand and pushed sideways so its pages slide (shearing); (d) a solid sphere squeezed inward from all sides by a surrounding fluid (hydraulic).
(1) Longitudinal stress and strain. In Fig. 8.1(a), a cylinder is pulled by two equal forces acting outward, normal to its end faces. The restoring force per unit area in this case is called tensile stress. If instead the two forces push inward and compress the cylinder, the restoring force per unit area is called compressive stress. Because both act along the length of the body, tensile and compressive stress are together called longitudinal stress.
In both cases the length of the cylinder changes — it grows longer under tension and shorter under compression. If the original length is \(L\) and the change in length is \(\Delta L\), the resulting strain is the longitudinal strain:
\[\text{Longitudinal strain} = \frac{\Delta L}{L} \tag{8.2}\]
where \(\Delta L\) and \(L\) are both lengths (in metres), so their ratio is a pure number.
(2) Shearing stress and strain. Instead of pulling along the length, suppose two equal and opposite forces are applied parallel to the cross-sectional area of the cylinder, as in Fig. 8.1(b). The two faces of the body then slide relative to each other. The restoring force per unit area developed against this tangential (sideways) force is called tangential stress or shearing stress.
Under this force, one face is displaced by a small distance \(\Delta x\) relative to the opposite face, which lies a distance \(L\) away. The strain produced is called shearing strain, defined as the ratio of this relative displacement to the length of the cylinder:
\[\text{Shearing strain} = \frac{\Delta x}{L} = \tan\theta \tag{8.3}\]
Here \(\theta\) is the angle by which the cylinder is tilted from its original vertical position, and \(\Delta x\) is the sideways shift of the top face relative to the bottom.
In most practical situations \(\theta\) is very small. For a small angle measured in radians, \(\tan\theta\) is very nearly equal to \(\theta\) itself — for instance, even at \(\theta = 10^\circ\) the two differ by only about 1%. So for small shear we may write
\[\text{Shearing strain} = \tan\theta \approx \theta \tag{8.4}\]
You can picture the same kind of deformation by pressing down on a thick book with your hand and pushing it horizontally, as in Fig. 8.1(c): the pages slide over one another and the stack tilts, exactly like the sheared cylinder.
(3) Hydraulic stress and volume strain. In Fig. 8.1(d), a solid sphere is placed in a fluid under high pressure. The fluid presses inward, perpendicular to the surface, at every single point. The body is squeezed uniformly from all sides — this is called hydraulic compression — and its volume decreases while its shape stays the same.
Just as before, the body resists. It develops internal restoring forces equal and opposite to the inward push of the fluid, and it springs back to its original shape and size once removed from the fluid. This internal restoring force per unit area is called hydraulic stress, and in magnitude it equals the hydraulic pressure (the applied force per unit area) of the fluid.
The strain produced by this uniform pressure is a change in volume, not in length or shape. It is called volume strain, defined as the ratio of the change in volume \(\Delta V\) to the original volume \(V\):
\[\text{Volume strain} = \frac{\Delta V}{V} \tag{8.5}\]
Notice a feature common to all three kinds of strain: each is a ratio of a change in some dimension to the original dimension. Length divided by length, or volume divided by volume, always cancels its units. Therefore strain — of every type — is a pure number with no unit and no dimensional formula.
8.3 Hooke’s Law
All three cases in Fig. 8.1 — tensile, shearing, and hydraulic — produce stress and strain in different forms. Yet they share one simple relationship whenever the deformation is small.
For small deformations, the stress produced in a body is directly proportional to the strain. This experimental result is called Hooke’s law, named after the scientist who first stated it.
Written as a proportionality, \[\text{stress} \propto \text{strain}\] and, removing the proportionality sign by introducing a constant, \[\text{stress} = k \times \text{strain} \tag{8.6}\]
Here \(k\) is the constant of proportionality. It is a property of the material and is called the modulus of elasticity.
The words “for small deformations” matter: the straight-line relationship holds only while the strain stays small, and it breaks down under larger loads, as the next section will show.
Figure to come
Fig. 8.2a – A straight-line graph of stress (y-axis) versus strain (x-axis) passing through the origin, its slope marked as the modulus of elasticity k, valid for small deformations.
Hooke’s law is an empirical law — obtained from experiment and observation rather than derived from deeper theory — and it holds for most materials over a useful range. It is not universal, though: some materials, such as rubber, do not follow this simple linear relationship at all.
8.4 Stress-Strain Curve
For a given material, the exact relationship between stress and strain is not guessed — it is measured. In a standard tensile test, a sample in the form of a wire or a test cylinder is gripped at its ends and slowly stretched by a steadily increasing force.
At each step, two things are recorded: the applied force, and the small increase in length it causes. From the force we calculate the stress (force per unit area), and from the length change we calculate the strain (fractional change in length). Plotting stress on the vertical axis against strain on the horizontal axis gives the material’s stress-strain curve.
A typical curve for a metal is shown in Fig. 8.2. Similar curves can be obtained for compression and for shear, and every material has its own characteristic shape. These curves are extremely useful: they tell an engineer, at a glance, how a material will behave as the load on it grows.
Figure to come
Fig. 8.2 – A typical stress-strain curve for a metal, showing origin O, a straight portion O–A ending at the proportional limit A, point B the yield point (elastic limit) at stress σy, a curved plastic region B–D peaking at the ultimate tensile strength σu at D, fracture at point E, and a dashed unloading line from a point C showing the permanent set on the strain axis; strain axis marked from <1% up to 30%.
Let us walk along the curve from the origin.
The elastic region (O to A). In the first part of the curve, from O to A, the graph is a straight line. Here stress is proportional to strain — this is exactly the region where Hooke’s law is obeyed. If the force is removed anywhere in this range, the body springs back completely to its original dimensions. So in region OA the solid behaves as a perfectly elastic body. Point A, where the straight line ends, is called the proportional limit.
Still elastic, but not proportional (A to B). Just past A, from A to B, stress and strain are no longer proportional — the curve begins to bend. Even so, the material is still elastic: remove the load anywhere up to B and the body still returns to its original size. Point B marks the end of this elastic behaviour. It is called the yield point, or elastic limit, and the stress at this point is the yield strength, written \(\sigma_y\).
The plastic region (B to D). If the load is pushed past the yield point, the behaviour changes sharply. Now the strain shoots up rapidly even for a small increase in stress — the material begins to “give way.” This is the portion from B to D.
Suppose we stop at some point C in this region and remove the load. The body does not return to its original length. Even when the stress has dropped to zero, some strain remains. This left-over, permanent deformation is called a permanent set, and deformation of this kind is called plastic deformation.
Continuing to D, the curve reaches its highest point. The stress here is the ultimate tensile strength, \(\sigma_u\) — the greatest stress the material can withstand.
Fracture (point E). Beyond D, something surprising happens: the material keeps stretching and finally breaks even though the applied force is now reducing. The point where it snaps is the fracture point, E.
The spacing between D and E tells us what kind of material we have. If D and E lie close together, the material breaks soon after reaching its peak stress, with little further stretching — such a material is called brittle. If D and E are far apart, the material stretches a great deal before breaking — such a material is called ductile.
Materials that behave differently: elastomers. The shape of the stress-strain curve varies enormously from one material to another. Rubber is a striking example: it can be pulled to several times its original length and still snap back to its original shape when released.
Fig. 8.3 shows the stress-strain curve for the elastic tissue of the aorta — the large blood vessel that carries blood away from the heart. Two features stand out. First, the elastic region is very large — the tissue can stretch a lot and still recover. Second, over most of this region the material does not obey Hooke’s law; the curve is not a straight line.
Figure to come
Fig. 8.3 – Stress-strain curve for the elastic tissue of the aorta, showing a large, upward-curving elastic region that is non-linear over most of its length, with no well-defined plastic region.
Notice also that there is no well-defined plastic region for such tissue. Materials like the tissue of the aorta and rubber, which can be stretched to produce very large strains while still returning to their original shape, are called elastomers.
8.5 Elastic Moduli
Recall the straight-line region OA of the stress-strain curve in Fig. 8.2 — the range within the elastic limit where stress and strain stay proportional. This region matters greatly in practice, because almost all structural and manufacturing designs are meant to work within it, where the material stays elastic and returns to its original shape.
Within this region, the ratio of stress to strain is a fixed number for a given material. This ratio is the modulus of elasticity (introduced in Section 8.3 as the constant \(k\)), and its value is a characteristic of the material — a measure of how stiff or how yielding it is.
Since stress and strain each come in three forms — longitudinal, shearing, and hydraulic — there are three corresponding elastic moduli: Young’s modulus (for longitudinal deformation), shear modulus (for shearing), and bulk modulus (for hydraulic compression). The next subsections take these up in turn, followed by two closely related ideas — Poisson’s ratio and the elastic potential energy stored in a stretched body.
8.5.1 Young’s Modulus
Experiments show a useful fact: for a given material, the size of the strain produced is the same whether the applied stress stretches the body (tensile) or squeezes it along its length (compressive). This lets us describe both cases with a single number.
The ratio of the tensile (or compressive) stress \(\sigma\) to the longitudinal strain \(\varepsilon\) is called Young’s modulus, denoted by the symbol \(Y\):
\[Y = \frac{\sigma}{\varepsilon} \tag{8.7}\]
Here \(\sigma\) is the longitudinal stress (in N m⁻²) and \(\varepsilon\) is the longitudinal strain (a pure number).
We can write this in terms of the directly measured quantities. Using stress \(= F/A\) (Eq. 8.1) and longitudinal strain \(= \Delta L / L\) (Eq. 8.2),
\[Y = \frac{F/A}{\Delta L / L} = \frac{F \times L}{A \times \Delta L} \tag{8.8}\]
where \(F\) is the applied force (N), \(A\) is the cross-sectional area (m²), \(L\) is the original length (m), and \(\Delta L\) is the change in length (m).
Rearranging Eq. (8.8) gives \(\Delta L = \dfrac{F L}{A Y}\). This is exactly the tool we needed: once the material’s Young’s modulus is known, we can predict in advance precisely how much a wire of given length and thickness will stretch under a given load — no experiment required.
Because strain is a dimensionless quantity, dividing stress by strain does not change the units. So Young’s modulus has the same unit as stress: N m⁻² or pascal (Pa). Table 8.1 lists Young’s moduli, along with ultimate and yield strengths, for some common materials.
[Table 8.1 – Young’s moduli and yield strengths of some materials]
| Substance | Density \(\rho\) (kg m⁻³) | Young’s modulus \(Y\) (\(10^{9}\) N m⁻²) | Ultimate strength \(\sigma_u\) (\(10^{6}\) N m⁻²) | Yield strength \(\sigma_y\) (\(10^{6}\) N m⁻²) |
|---|---|---|---|---|
| Aluminium | 2710 | 70 | 110 | 95 |
| Copper | 8890 | 110 | 400 | 200 |
| Iron (wrought) | 7800–7900 | 190 | 330 | 170 |
| Steel | 7860 | 200 | 400 | 250 |
| Glass# | 2190 | 65 | 50 | — |
| Concrete | 2320 | 30 | 40 | — |
| Wood# | 525 | 13 | 50 | — |
| Bone# | 1900 | 9.4 | 170 | — |
| Polystyrene | 1050 | 3 | 48 | — |
# Substance tested under compression.
Looking at Table 8.1, the Young’s moduli of metals are large. A large \(Y\) means that a large force is needed to produce even a small change in length. For instance, to stretch a thin steel wire of cross-sectional area 0.1 cm² by just 0.1%, a force of about 2000 N is needed. To produce the same strain in wires of the same area made of aluminium, brass, and copper requires only about 690 N, 900 N, and 1100 N respectively.
Steel needs the largest force of the four, and this is precisely why we say steel is more elastic than copper, brass, and aluminium. Here lies a common surprise: in physics, “more elastic” does not mean “stretches more” — it means the material resists stretching more strongly and restores its shape more forcefully, which shows up as a larger Young’s modulus. Rubber stretches far more than steel, yet its Young’s modulus is tiny, so steel is by far the more elastic of the two. It is for this reason that steel is preferred in heavy-duty machines and structural designs, while wood, bone, concrete, and glass — with much smaller Young’s moduli — are not.
8.5.2 Shear Modulus
When a body is subjected to a shearing (tangential) stress, it responds with a shearing strain — the sideways tilt we met in Fig. 8.1(b). The ratio of the shearing stress to the shearing strain it produces is called the shear modulus of the material, written \(G\). It is also called the modulus of rigidity, because it measures how strongly a material resists a change in shape.
Writing the two quantities out, \[G = \frac{\text{shearing stress } (\sigma_s)}{\text{shearing strain}}\]
Using shearing stress \(= F/A\) and shearing strain \(= \Delta x / L\) (from Eq. 8.3), \[G = \frac{F/A}{\Delta x / L} = \frac{F \times L}{A \times \Delta x} \tag{8.10}\] where \(F\) is the tangential force (N), \(A\) the area of the face on which it acts (m²), \(\Delta x\) the sideways displacement of that face (m), and \(L\) the distance between the two faces (m).
Since for small angles the shearing strain equals the angle of shear \(\theta\) (Eq. 8.4), we can also write \[G = \frac{F/A}{\theta} = \frac{F}{A \times \theta} \tag{8.11}\]
Rearranging gives a handy expression for the shearing stress in terms of the shear angle: \[\sigma_s = G \times \theta \tag{8.12}\]
The SI unit of shear modulus is the same as that of stress: N m⁻² or pascal (Pa). Table 8.2 lists the shear moduli of some common materials.
[Table 8.2 – Shear moduli (G) of some common materials]
| Material | \(G\) (\(10^{9}\) N m⁻² or GPa) |
|---|---|
| Aluminium | 25 |
| Brass | 36 |
| Copper | 42 |
| Glass | 23 |
| Iron | 70 |
| Lead | 5.6 |
| Nickel | 77 |
| Steel | 84 |
| Tungsten | 150 |
| Wood | 10 |
Comparing Table 8.2 with Table 8.1, notice that the shear modulus of a material is generally smaller than its Young’s modulus. In fact, for most materials \(G \approx Y/3\). Physically, this means it is usually easier to change a body’s shape (shear it) than to stretch its length.
8.5.3 Bulk Modulus
In Section 8.2 we saw that when a body is submerged in a fluid under pressure, it experiences a hydraulic stress equal in magnitude to the hydraulic pressure. This uniform squeezing decreases the body’s volume, producing the volume strain of Eq. (8.5).
The ratio of the hydraulic stress to the resulting volume strain is called the bulk modulus of the material, denoted by \(B\):
\[B = -\,\frac{p}{\Delta V / V} \tag{8.12}\]
Here \(p\) is the hydraulic stress (equal to the applied pressure, in N m⁻²) and \(\Delta V / V\) is the volume strain (a pure number).
Why the negative sign? When the pressure on a body increases, its volume decreases — so \(\Delta V\) is negative. The minus sign in front cancels this, ensuring that the bulk modulus \(B\) comes out positive for a body in equilibrium. In other words, if \(p\) is positive, \(\Delta V\) is negative, and the two minus signs together make \(B > 0\).
The SI unit of bulk modulus is the same as that of pressure — N m⁻² or pascal (Pa). Table 8.3 lists the bulk moduli of some common materials.
[Table 8.3 – Bulk moduli (B) of some common materials]
| Material | \(B\) (\(10^{9}\) N m⁻² or GPa) |
|---|---|
| Solids | |
| Aluminium | 72 |
| Brass | 61 |
| Copper | 140 |
| Glass | 37 |
| Iron | 100 |
| Nickel | 260 |
| Steel | 160 |
| Liquids | |
| Water | 2.2 |
| Ethanol | 0.9 |
| Carbon disulphide | 1.56 |
| Glycerine | 4.76 |
| Mercury | 25 |
| Gases | |
| Air (at STP) | \(1.0 \times 10^{-4}\) |
Notice a clear pattern in Table 8.3: the bulk moduli of solids are far larger than those of liquids, which in turn are far larger than that of a gas such as air.
The reciprocal of the bulk modulus is called the compressibility, denoted by \(k\). It is defined as the fractional change in volume per unit increase in pressure:
\[k = \frac{1}{B} = -\frac{1}{\Delta p} \times \frac{\Delta V}{V} \tag{8.13}\]
Since \(B\) is largest for solids, their compressibility is the smallest. Thus solids are the least compressible, while gases are the most compressible — gases are about a million times more compressible than solids. A gas’s compressibility also varies with its pressure and temperature.
Why this huge difference? It comes down to how tightly the particles are bound. In a solid, neighbouring atoms are locked together by strong forces, so the material strongly resists any change in volume. In a liquid, molecules are still bound to their neighbours, but less firmly. In a gas, the molecules are so far apart and so weakly coupled that the volume can be squeezed easily.
Table 8.4 collects the three types of stress, their strains, the corresponding elastic moduli, and the states of matter to which each applies — a useful summary of everything in this section.
[Table 8.4 – Stress, strain and various elastic moduli]
| Type of stress | Stress | Strain | Change in shape | Change in volume | Elastic modulus | Name of modulus | State of matter |
|---|---|---|---|---|---|---|---|
| Tensile or compressive (\(\sigma = F/A\)) | Two equal and opposite forces perpendicular to opposite faces | Elongation or compression parallel to force direction, \(\Delta L/L\) (longitudinal strain) | Yes | No | \(Y = (F \times L)/(A \times \Delta L)\) | Young’s modulus | Solid |
| Shearing (\(\sigma_s = F/A\)) | Two equal and opposite forces parallel to opposite surfaces, in each case such that total force and total torque on the body vanish | Pure shear, \(\theta\) | Yes | No | \(G = F/(A \times \theta)\) | Shear modulus (modulus of rigidity) | Solid |
| Hydraulic | Forces perpendicular everywhere to the surface; force per unit area (pressure) same everywhere | Volume change (compression or elongation), \(\Delta V/V\) | No | Yes | \(B = -p/(\Delta V/V)\) | Bulk modulus | Solid, liquid and gas |
8.5.4 Poisson’s Ratio
When you stretch a wire, it does two things at once. It gets longer along the direction of the pull — the change we have studied so far — but it also gets slightly thinner across its width. This sideways change is a strain too.
The strain measured perpendicular to the applied force is called lateral strain, to distinguish it from the longitudinal strain measured along the force.
The ratio of the lateral strain to the longitudinal strain in a stretched wire is called Poisson’s ratio.
To write it in symbols, suppose the wire has original diameter \(d\), and under stress its diameter contracts by \(\Delta d\). Then the lateral strain is \(\Delta d / d\). If the original length is \(L\) and it elongates by \(\Delta L\), the longitudinal strain is \(\Delta L / L\). Poisson’s ratio is therefore
\[\text{Poisson's ratio} = \frac{\Delta d / d}{\Delta L / L} = \frac{\Delta d}{\Delta L} \times \frac{L}{d}\]
Figure to come
Fig. 8.5a – A wire of original length L and diameter d stretched by a force, shown becoming longer by ΔL while its diameter contracts by Δd.
Because it is a ratio of two strains — each already a pure number — Poisson’s ratio has no dimensions and no units. Its value does not depend on the size or shape of the object, but only on the nature of the material. For steels it lies between 0.28 and 0.30, and for aluminium alloys it is about 0.33. In practice, most common solids have Poisson’s ratios between roughly 0.2 and 0.5, with a larger value indicating a stronger sideways contraction for a given stretch.
8.5.5 Elastic Potential Energy in a Stretched Wire
When you stretch a wire, you must pull against the forces that hold its atoms together — the inter-atomic forces. Doing work against these forces does not waste the energy; it is stored inside the wire as elastic potential energy, ready to be released when the wire is let go and springs back.
Let us calculate how much energy is stored. Take a wire of original length \(L\) and cross-sectional area \(A\), stretched along its length by a deforming force. Suppose that at some stage the wire has already been elongated by an amount \(l\).
At this stage the stretching force needed, from Eq. (8.8), is \[F = Y A \times \frac{l}{L}\] where \(Y\) is the Young’s modulus of the wire’s material and \(l/L\) is the strain at that instant.
Now stretch the wire by a further tiny amount \(dl\). The work done in this small step is force times distance: \[dW = F\, dl = \frac{Y A l}{L}\, dl\]
To find the total work done in stretching the wire from its natural length (\(l = 0\)) all the way to a final extension \(l\), we add up all these tiny bits of work by integrating: \[W = \int_{0}^{l} \frac{Y A l}{L}\, dl = \frac{Y A}{2} \times \frac{l^2}{L}\]
This result can be rewritten in a more meaningful form. Pulling out the strain \(l/L\), \[W = \frac{1}{2} \times Y \times \left(\frac{l}{L}\right)^2 \times A L\]
Since \(AL\) is just the volume of the wire, and since \(Y \times (l/L)\) is the stress while \((l/L)\) is the strain, this becomes \[W = \frac{1}{2} \times \text{Young's modulus} \times \text{strain}^2 \times \text{volume}\] \[= \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}\]
Figure to come
Fig. 8.5b – A straight-line graph of stretching force F (y-axis) versus elongation l (x-axis) through the origin; the shaded triangular area under the line represents the work done, equal to the stored elastic potential energy.
All of this work is stored in the wire as elastic potential energy \(U\) (so \(U = W\)). Dividing by the volume of the wire gives the elastic potential energy stored per unit volume, written \(u\): \[u = \frac{1}{2}\,\sigma \varepsilon \tag{8.14}\] where \(\sigma\) is the stress and \(\varepsilon\) is the strain in the wire.
8.6 Applications of Elastic Behaviour of Materials
Everything we have built up in this chapter — stress, strain, and the elastic moduli — comes together in real engineering. The elastic behaviour of materials plays an important role in everyday life, and every engineering design depends on knowing it precisely.
When engineers design a building, the columns, beams, and supports must all be sized using the strength of the materials they are made from. The same is true for bridges. This is why certain questions have definite physical answers: Why do the beams used in bridges and supports so often have a cross-section shaped like the letter I? Why does a heap of sand, or a hill, settle into a roughly pyramidal shape? Such questions belong to structural engineering, which rests on exactly the ideas developed in this chapter. Let us look at four concrete examples.
1. How thick should a crane’s rope be?
Cranes lift and move very heavy loads using a thick metal rope, pulled by pulleys and motors. Suppose we want a crane that can lift 10 tonnes (10 metric tons; 1 metric ton = 1000 kg, so the load is \(10^4\) kg). How thick must its steel rope be?
We must make sure the load never stretches the rope past its elastic limit — otherwise the rope would deform permanently and eventually fail. So the stress in the rope must stay below the yield strength of the steel. From Table 8.1, mild steel has a yield strength \(\sigma_y \approx 300 \times 10^6\ \text{N m}^{-2}\).
The stress is the weight \(W = Mg\) divided by the rope’s cross-sectional area \(A\). Requiring this to stay at or below \(\sigma_y\) gives a minimum area:
\[A \geq \frac{W}{\sigma_y} = \frac{Mg}{\sigma_y} \tag{8.15}\]
Putting in the numbers,
\[A \geq \frac{10^4\ \text{kg} \times 9.8\ \text{m s}^{-2}}{300 \times 10^6\ \text{N m}^{-2}} = 3.3 \times 10^{-4}\ \text{m}^2\]
For a circular rope this corresponds to a radius of about 1 cm. In practice, engineers build in a large safety margin — typically a factor of about ten in the load — so a thicker rope of radius about 3 cm is recommended.
But a single solid steel wire of 3 cm radius would be almost as stiff as a rigid rod, impossible to wind over a pulley. That is why crane ropes are never single rods: they are made of many thin wires braided together, like the strands of a pigtail, giving the same strength while remaining flexible and easy to manufacture.
2. Why beams bend — and the shape of the I-beam
A bridge must carry flowing traffic, resist the wind, and hold up its own weight without bending too much or breaking. Buildings face the same challenge with their beams and columns. In every case, the key problem is the bending (sagging) of a beam under load.
Consider a beam of length \(l\), breadth \(b\), and depth \(d\), supported near its two ends and loaded at the centre by a weight \(W\), as shown in Fig. 8.6. The beam sags in the middle by an amount \(\delta\) (called the deflection) given by
\[\delta = \frac{W l^3}{4 b d^3 Y} \tag{8.16}\]
Figure to come
Fig. 8.6 – A horizontal beam supported near both ends, loaded at its centre by a weight W, sagging downward by an amount δ; length l and depth d marked.
where \(W\) is the load (N), \(l\) the length of the span (m), \(b\) the breadth (m), \(d\) the depth (m), and \(Y\) the Young’s modulus of the beam’s material (N m⁻²). This relation can be derived from what you have already learned, together with a little calculus.
Equation (8.16) tells us how to build a beam that bends as little as possible. First, use a material with a large Young’s modulus \(Y\) — since \(\delta \propto 1/Y\), a stiffer material sags less. Second, and more interestingly, look at how \(\delta\) depends on the beam’s shape: it is proportional to \(d^{-3}\) but only to \(b^{-1}\).
This means increasing the depth \(d\) is far more effective at reducing sag than increasing the breadth \(b\). Doubling the depth cuts the sag to one-eighth; doubling the breadth only halves it. (The span length \(l\) should also be kept as short as possible, since \(\delta \propto l^3\).)
So why not simply make every beam very deep and thin? Because a tall, thin beam has a new weakness. Unless the load sits exactly in the right place — nearly impossible with moving traffic on a bridge — such a deep bar tends to bend sideways and collapse, as shown in Fig. 8.7(b). This sideways collapse is called buckling.
Figure to come
Fig. 8.7 – Three beam cross-sections: (a) a solid rectangular bar; (b) a tall thin bar buckling sideways under load; (c) the I-shaped section used for load-bearing beams.
The practical solution is a compromise between depth and stability: the I-shaped cross-section shown in Fig. 8.7(c). The two wide horizontal flanges provide a large depth to resist sagging, while the central vertical web keeps them apart and resists buckling. Crucially, the material is concentrated where it does the most work — at the top and bottom — so the beam achieves great strength using far less material. This lowers both the weight and the cost of the beam without sacrificing strength.
3. Pillars and columns
Pillars and columns are just as common as beams in buildings and bridges, and their shape matters too. A pillar with rounded ends, as in Fig. 8.8(a), can support less load than one with a broadened, distributed shape at its ends, as in Fig. 8.8(b).
Figure to come
Fig. 8.8 – Two pillars: (a) a pillar with rounded ends; (b) a pillar with broadened, distributed ends that can bear a larger load.
The distributed ends spread the load over a wider area and give the column more stability, letting it bear a greater weight before it buckles. In real design, engineers must weigh all of this against the conditions the structure will face, its cost, and how long the material must remain reliable.
4. Why mountains cannot grow taller than about 10 km
Elasticity even sets a limit on the height of mountains. Why can’t a mountain on Earth be much taller than about 10 km? The answer lies in the elastic properties of rock.
A mountain’s base is not squeezed uniformly from all sides — the sides of the mountain are free, while the enormous weight of rock above pushes straight down. This uneven loading sets up a shearing stress in the rock at the base, and rock can take only so much shear before it begins to flow (deform plastically) and spread out. For the mountain to stay standing, the stress from the weight above must stay below this critical shearing stress.
At the bottom of a mountain of height \(h\), the force per unit area due to the weight of the material above is \(h\rho g\), where \(\rho\) is the density of the rock and \(g\) is the acceleration due to gravity. Because the material feels this force vertically while its sides are free, this is not a case of uniform pressure or bulk compression — instead there is a shear component of roughly \(h\rho g\) itself.
The elastic limit for a typical rock is about \(30 \times 10^7\ \text{N m}^{-2}\). Setting the base stress equal to this limit, and taking the rock density \(\rho = 3 \times 10^3\ \text{kg m}^{-3}\):
\[h\rho g = 30 \times 10^7\ \text{N m}^{-2}\] \[h = \frac{30 \times 10^7\ \text{N m}^{-2}}{3 \times 10^3\ \text{kg m}^{-3} \times 10\ \text{m s}^{-2}} = 10\ \text{km}\]
So the greatest height a mountain can reach before its own base starts to give way is about 10 km — which is more than the height of Mt. Everest. Nature, it turns out, obeys the same elasticity we have studied all chapter.
8.7 Summary
Stress is the restoring force per unit area, and strain is the fractional change in dimension. In general there are three types of stress: (a) tensile stress — a longitudinal stress associated with stretching — or compressive stress, associated with compression; (b) shearing stress; and (c) hydraulic stress.
For small deformations, stress is directly proportional to strain for many materials. This is known as Hooke’s law. The constant of proportionality is called the modulus of elasticity. Three elastic moduli — Young’s modulus, shear modulus, and bulk modulus — are used to describe the elastic behaviour of objects as they respond to deforming forces acting on them. A class of solids called elastomers does not obey Hooke’s law.
When an object is under tension or compression, Hooke’s law takes the form \[\frac{F}{A} = Y\,\frac{\Delta L}{L}\] where \(\Delta L / L\) is the tensile or compressive strain of the object, \(F\) is the magnitude of the applied force causing the strain, \(A\) is the cross-sectional area over which \(F\) is applied (perpendicular to \(A\)), and \(Y\) is the Young’s modulus of the object. The stress is \(F/A\).
When a pair of forces is applied parallel to the upper and lower faces of a solid, the solid deforms so that the upper face moves sideways with respect to the lower one. The horizontal displacement \(\Delta L\) of the upper face is perpendicular to the vertical height \(L\). This type of deformation is called shear, and the corresponding stress is the shearing stress. This type of stress is possible only in solids. In this kind of deformation, Hooke’s law takes the form \[\frac{F}{A} = G \times \frac{\Delta L}{L}\] where \(\Delta L\) is the displacement of one end of the object in the direction of the applied force \(F\), and \(G\) is the shear modulus.
When an object undergoes hydraulic compression due to a stress exerted by a surrounding fluid, Hooke’s law takes the form \[p = B\left(\frac{\Delta V}{V}\right)\] where \(p\) is the pressure (hydraulic stress) on the object due to the fluid, \(\Delta V / V\) (the volume strain) is the absolute fractional change in the object’s volume due to that pressure, and \(B\) is the bulk modulus of the object.
8.8 Points to Ponder
In the case of a wire suspended from a ceiling and stretched by a weight (\(F\)) hung from its other end, the force exerted by the ceiling on the wire is equal and opposite to the weight. However, the tension at any cross-section \(A\) of the wire is just \(F\), and not \(2F\). Hence the tensile stress, which equals the tension per unit area, is \(F/A\).
Hooke’s law is valid only in the linear part of the stress-strain curve.
Young’s modulus and shear modulus are relevant only for solids, since only solids have definite lengths and shapes.
Bulk modulus is relevant for solids, liquids, and gases. It refers to the change in volume when every part of the body is under uniform stress, so that the shape of the body remains unchanged.
Metals have larger values of Young’s modulus than alloys and elastomers. A material with a large value of Young’s modulus requires a large force to produce small changes in its length.
In daily life, we feel that a material which stretches more is more elastic, but this is a misnomer. In fact, a material which stretches to a lesser extent for a given load is considered to be more elastic.
In general, a deforming force in one direction can produce strains in other directions as well. The proportionality between stress and strain in such situations cannot be described by just one elastic constant. For example, for a wire under longitudinal strain, the lateral dimensions (the radius of the cross-section) undergo a small change, which is described by another elastic constant of the material — the Poisson ratio.
Stress is not a vector quantity, since — unlike a force — a stress cannot be assigned a specific direction. Force acting on the portion of a body on a specified side of a section does have a definite direction.
8.9 Physical Quantities in this Chapter
| Physical quantity | Symbol | SI unit | Dimensional formula |
|---|---|---|---|
| Stress | \(\sigma\) | N m⁻² (Pa) | \([\text{ML}^{-1}\text{T}^{-2}]\) |
| Strain | \(\varepsilon\) | — (dimensionless) | \([\text{M}^{0}\text{L}^{0}\text{T}^{0}]\) |
| Young’s modulus | \(Y\) | N m⁻² (Pa) | \([\text{ML}^{-1}\text{T}^{-2}]\) |
| Shear modulus (modulus of rigidity) | \(G\) | N m⁻² (Pa) | \([\text{ML}^{-1}\text{T}^{-2}]\) |
| Bulk modulus | \(B\) | N m⁻² (Pa) | \([\text{ML}^{-1}\text{T}^{-2}]\) |
| Compressibility | \(k\) | N⁻¹ m² (Pa⁻¹) | \([\text{M}^{-1}\text{L}\,\text{T}^{2}]\) |
| Poisson’s ratio | — | — (dimensionless) | \([\text{M}^{0}\text{L}^{0}\text{T}^{0}]\) |
| Elastic potential energy density | \(u\) | J m⁻³ | \([\text{ML}^{-1}\text{T}^{-2}]\) |
8.10 NCERT Questions
A steel wire of length 4.7 m and cross-sectional area \(3.0 \times 10^{-5}\ \text{m}^2\) stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area \(4.0 \times 10^{-5}\ \text{m}^2\) under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Figure 8.9 shows the strain–stress curve for a given material. What are (a) the Young’s modulus and (b) the approximate yield strength for this material?
Figure to come
Fig. 8.9 – A strain–stress curve for a material; stress axis (in \(10^{6}\) N m⁻²) marked from 0 to 300, strain axis marked from 0 to 0.004, rising almost linearly to about 300 and then curving over near the top.
- The stress–strain graphs for materials A and B are shown in Fig. 8.10. The graphs are drawn to the same scale.
- Which of the materials has the greater Young’s modulus?
- Which of the two is the stronger material?
Figure to come
Fig. 8.10 – Two stress–strain graphs side by side, both to the same scale: material A with a gentler-sloped curve, material B with a steeper initial slope.
- Read the following two statements carefully and state, with reasons, whether each is true or false.
- The Young’s modulus of rubber is greater than that of steel.
- The stretching of a coil is determined by its shear modulus.
- Two wires of diameter 0.25 cm, one made of steel and the other made of brass, are loaded as shown in Fig. 8.11. The unloaded length of the steel wire is 1.5 m and that of the brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Figure to come
Fig. 8.11 – A steel wire of length 1.5 m hanging from a rigid support, carrying a 4.0 kg mass, from which a brass wire of length 1.0 m hangs carrying a 6.0 kg mass.
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
A piece of copper having a rectangular cross-section of 15.2 mm \(\times\) 19.1 mm is pulled in tension with a 44,500 N force, producing only elastic deformation. Calculate the resulting strain.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed \(10^{8}\ \text{N m}^{-2}\), what is the maximum load the cable can support?
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is \(0.065\ \text{cm}^2\). Calculate the elongation of the wire when the mass is at the lowest point of its path.
Compute the bulk modulus of water from the following data: initial volume = 100.0 litre, pressure increase = 100.0 atm (\(1\ \text{atm} = 1.013 \times 10^{5}\ \text{Pa}\)), final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature), and explain in simple terms why the ratio is so large.
What is the density of water at a depth where the pressure is 80.0 atm, given that its density at the surface is \(1.03 \times 10^{3}\ \text{kg m}^{-3}\)?
Compute the fractional change in volume of a glass slab when it is subjected to a hydraulic pressure of 10 atm.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of \(7.0 \times 10^{6}\ \text{Pa}\).
How much should the pressure on a litre of water be changed to compress it by 0.10%?
8.11 Check Your Concepts
Stress is defined as force per unit area, yet it is not treated as a vector quantity. Explain why, and how stress differs from pressure.
On a typical stress–strain curve for a metal, distinguish clearly between the proportional limit and the elastic limit. Why can a material be beyond its proportional limit but still be elastic?
“A material that stretches more under a given load is more elastic.” State whether this is true or false, and justify your answer using the meaning of Young’s modulus. Which is more elastic — steel or rubber?
Explain why Young’s modulus and shear modulus are defined only for solids, whereas bulk modulus is defined for solids, liquids, and gases.
Why does the formula for bulk modulus, \(B = -p/(\Delta V/V)\), contain a negative sign? What would be physically wrong if the sign were omitted?
Define buckling. Explain, using the beam-sag relation, why an I-shaped cross-section is preferred for load-bearing beams and railway tracks over a solid rectangular bar of the same material.
Why are the ropes of a crane made from many thin wires braided together rather than from a single thick steel rod of the same total cross-sectional area?
Using the idea of critical shearing stress in rock, explain why there is an upper limit of about 10 km to the height of a mountain on Earth.
The elastic potential energy stored in a stretched wire is \(\tfrac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}\). Explain the physical origin of the factor \(\tfrac{1}{2}\).
Define Poisson’s ratio and lateral strain. Explain why Poisson’s ratio has no units, and describe what a value close to zero (as in cork) tells you about a material’s behaviour.
A steel wire hangs from a ceiling with a weight \(F\) attached to its lower end. Some students argue that the tension in the wire is \(2F\) because the ceiling also pulls up with force \(F\). Explain why the tensile stress is \(F/A\) and not \(2F/A\).
Two beams of the same material and the same length carry the same central load. Beam X is twice as deep as beam Y but has the same breadth. Which beam sags less, and by what factor? Justify using the dependence of sag on depth.
8.12 Practice with Numericals
A wire of length 2.0 m and cross-sectional area \(1.0\ \text{mm}^2\) is stretched by a force of 100 N, producing an elongation of 1.0 mm. Calculate the Young’s modulus of the material of the wire.
A load of 4.0 kg hangs from a vertical wire of diameter 1.0 mm. Taking \(g = 9.8\ \text{m s}^{-2}\), calculate the tensile stress in the wire.
A metal cube of side 8.0 cm has its lower face fixed. A tangential force of \(0.20\ \text{kN}\) is applied to its upper face. If the shear modulus of the metal is \(8.0 \times 10^{10}\ \text{N m}^{-2}\), find the horizontal displacement of the upper face.
A solid sphere of volume \(0.50\ \text{m}^3\) is subjected to a uniform pressure of \(2.0 \times 10^{7}\ \text{Pa}\). If the bulk modulus of its material is \(1.0 \times 10^{11}\ \text{N m}^{-2}\), find the decrease in its volume.
The bulk modulus of a certain liquid is \(2.0 \times 10^{9}\ \text{N m}^{-2}\). Calculate its compressibility.
In a stretched wire the stress is \(1.0 \times 10^{8}\ \text{N m}^{-2}\) and the strain is \(5.0 \times 10^{-4}\). If the volume of the wire is \(2.0 \times 10^{-6}\ \text{m}^3\), calculate the elastic potential energy stored in it.
A wire of diameter 2.0 mm undergoes a longitudinal strain of \(1.0 \times 10^{-3}\) when stretched. If the Poisson’s ratio of the material is 0.25, find the change in the diameter of the wire.
A steel beam of length 2.0 m, breadth 5.0 cm, and depth 10 cm is supported at its ends and loaded at the centre by a weight of 1000 N. Taking the Young’s modulus of steel as \(2.0 \times 10^{11}\ \text{N m}^{-2}\), calculate the sag at the centre of the beam.
Two wires P and Q are made of the same material and have the same length. The diameter of wire Q is twice that of wire P. If both are stretched by the same load, find the ratio of the elongation of P to that of Q.
A steel cable is to support a maximum load of 2000 kg without the stress in it exceeding \(1.0 \times 10^{8}\ \text{N m}^{-2}\). Taking \(g = 9.8\ \text{m s}^{-2}\), find the minimum cross-sectional area required for the cable.