Mechanical Properties of Solids

This chapter answers a single practical question — how much does a solid deform under a given force, and when does that deformation become permanent or fatal — by building a toolkit in a fixed order. It opens with stress and strain, the three kinds of loading a solid can experience: pulling or pushing along a length, sliding one face past another, and squeezing from every side. It then states Hooke's law and shows how a real material's stress-strain curve departs from that line as load grows, marking out the elastic limit, yield point, ultimate tensile strength and fracture. The heart of the chapter is the three elastic moduli — Young's, shear and bulk — each a ratio of a stress to its matching strain, together with Poisson's ratio and elastic potential energy. The chapter closes with the applications that shape real engineering: how thick a crane rope must be, why deep beams sag less, why beams are I-shaped, and why mountains cannot grow indefinitely tall.

Elasticity, plasticity and the restoring force

A rigid body, in the sense used for rotational motion, has a perfectly fixed shape and size. Real solids are not like this: apply a force and every solid deforms by some amount, however small. If a body returns to its original size and shape the instant the deforming force is removed, that behaviour is elasticity, and the deformation is elastic deformation — a stretched helical spring that snaps back is the everyday example. If the body stays deformed with no tendency to recover, that is plasticity; putty and wet mud are close to ideal plastic materials. Almost every engineered structure — buildings, bridges, aeroplanes, artificial limbs — is designed to stay firmly in the elastic regime under normal use, which is why elastic behaviour matters so much to engineering design. When a deforming force acts on a body held in static equilibrium, the body develops an internal restoring force equal and opposite to the applied force. This restoring force per unit area is what the chapter calls stress.

Stress and the three ways a solid can be loaded

where: $F$ is the magnitude of the applied (equally, the internal restoring) force acting normal or tangential to a surface, and $A$ is the cross-sectional area it acts over. The SI unit is $\text{N m}^{-2}$, the pascal (Pa), with dimensional formula $[\text{ML}^{-1}\text{T}^{-2}]$. There are exactly three geometrically distinct ways a solid can be stressed, each with its own name and matching strain. A cylinder pulled by two equal, opposite forces normal to its cross-section develops tensile stress; pushed together, the same geometry gives compressive stress — together, longitudinal stress, since the change is along the body's length. If two equal, opposite forces instead act tangentially, parallel to a face, the restoring force per unit area is a shearing stress, sliding one face relative to the opposite face rather than stretching the body. If a solid is immersed in a fluid under pressure acting perpendicular to the surface at every point equally, the body is under hydraulic stress, numerically equal to the fluid pressure, shrinking in volume with no change of shape.

Strain: how we measure the deformation

Each stress in the figure above produces its own matching strain, and all three strains are pure numbers with no units, since each is a ratio of two lengths, areas or volumes of the same kind. Longitudinal strain is the fractional change in length,

Shearing strain is the ratio of the sideways displacement $\Delta x$ of one face to the perpendicular distance $L$ between the faces, which equals $\tan\theta$ for the angular displacement $\theta$ of the body from its original shape,

where: the small-angle approximation is excellent for the tiny deformations solids actually undergo — at $\theta = 10^{\circ}$ the difference between $\theta$ (in radians) and $\tan\theta$ is only about $1\%$, and real elastic shear angles are far smaller than that. Volume strain, produced by hydraulic stress, is the fractional change in volume,

Hooke's law and the stress-strain curve

For small deformations, stress and strain are directly proportional to each other in almost every material — this empirical statement is Hooke's law:

where: $k$, the constant of proportionality, is called the modulus of elasticity, and its value and its physical meaning depend on which stress-strain pair is in play. Hooke's law is empirical, not a fundamental law, and some materials — notably elastomers such as rubber and the elastic tissue of the aorta — never obey it over any appreciable range even though they are highly elastic. The relationship between stress and strain for a real material, found by gradually loading a test wire or cylinder and recording the stress it develops against the strain produced, traces out a stress-strain curve. For a typical metal this curve has a well-defined shape and vocabulary that every subsequent calculation in the chapter depends on.

Between $O$ and $A$ the curve is a straight line, the region where Hooke's law holds. Between $A$ and $B$, stress and strain are no longer proportional, but the body still returns fully to its original dimensions when unloaded; $B$ is the yield point (elastic limit), and the stress there is the yield strength $\sigma_y$. Beyond $B$, a small stress increase produces a rapid strain increase, and unloading anywhere between $B$ and $D$ leaves a permanent set — plastic deformation. Point $D$ is the ultimate tensile strength $\sigma_u$, the maximum stress the material can bear; beyond it the material finally snaps at the fracture point $E$. A material is ductile if $D$ and $E$ are far apart (structural steel), and brittle if they are close together (glass, cast iron).

The three elastic moduli: Young's, shear and bulk

Within the proportional region $OA$, the ratio of stress to strain is a constant characteristic of the material, called its modulus of elasticity. Three different moduli correspond to the three kinds of stress described above.

Young's modulus $Y$ is the ratio of tensile or compressive stress to the resulting longitudinal strain, and it applies only to solids, since only a solid has a definite length to stretch or compress:

where: $\sigma = F/A$ is the longitudinal stress, $\varepsilon = \Delta L/L$ the longitudinal strain, and since strain is dimensionless, $Y$ carries stress's unit, $\text{N m}^{-2}$. Metals have large Young's moduli — steel's is about $2.0\times10^{11}\ \text{N m}^{-2}$ — needing a large force for even a small strain; wood, bone and glass have smaller values, stretching more readily for the same stress. The everyday intuition that "stretchier means more elastic" is backwards: resisting stretching (larger $Y$) is what makes a material more elastic technically. The shear modulus $G$, also called the modulus of rigidity, is the ratio of shearing stress to shearing strain:

where: $\sigma_s = F/A$ is the tangential stress and $\theta \approx \Delta x/L$ the shearing strain in radians. $G$ shares $Y$'s unit, and for most solids $G \approx Y/3$ — shape distorts more readily than length changes. Like $Y$, shear modulus applies only to solids, since liquids and gases have no fixed shape to shear. The bulk modulus $B$ is the ratio of hydraulic stress to volume strain:

where: $p$ is the hydraulic pressure and $\Delta V/V$ the volume strain. The minus sign exists because increasing pressure always decreases volume, so $B$ itself comes out positive. Its reciprocal, $k = 1/B$, is compressibility. Unlike $Y$ and $G$, bulk modulus applies to solids, liquids and gases alike; solids have the largest bulk moduli, liquids intermediate values, gases the smallest by far — air is roughly a million times more compressible than a solid, since tight interatomic bonding resists volume change far more than loose liquid or gas coupling.

Poisson's ratio: the sideways shrink

When a wire is stretched, it does not merely elongate — it also grows very slightly thinner. The strain perpendicular to the applied force is lateral strain, and Siméon Poisson showed that within the elastic limit it is proportional to the longitudinal strain that caused it. The ratio of the two,

where: $d$ is the original diameter of the wire and $\Delta d$ its (negative) change as the wire stretches, $L$ is the original length and $\Delta L$ the elongation. Being a ratio of two strains, Poisson's ratio is a dimensionless pure number depending only on the material — about $0.28$ to $0.30$ for steels, about $0.33$ for aluminium alloys — and is why a single modulus never fully describes a general deformation: a force in one direction always strains the perpendicular directions too.

Elastic potential energy in a stretched wire

Stretching a wire does work against the interatomic forces holding it together, and that work does not vanish — it is stored as elastic potential energy, exactly as work done compressing a spring is stored in the spring. This energy, and the energy density $u = \tfrac{1}{2}\sigma\varepsilon$ obtained from it, is derived in full below and used throughout the applications that follow, from the crane rope to the loaded beam.

Engineering applications: ropes, beams, pillars and mountains

Every application here is the moduli above put to work with a safety margin. A steel crane rope must stay comfortably below the yield strength $\sigma_y$, so its minimum cross-section is $A \geq W/\sigma_y$ (with a generous safety factor, typically about ten, on top); it is then braided from many thin wires rather than cast as one rigid rod, for flexibility as much as strength. A beam of length $l$, breadth $b$ and depth $d$, loaded at its centre and supported near its ends, sags by an amount

where: $W$ is the central load and $Y$ the Young's modulus of the beam's material. Because $\delta$ falls off as $d^{-3}$ but only as $b^{-1}$, increasing depth reduces sagging far more effectively than increasing breadth — why joists and rails stand tall rather than lie flat. Pushed too far, though, a very deep, thin bar can buckle sideways instead of bending cleanly, so real beams compromise with an I-shaped cross-section: material concentrated far from the centre, where it resists bending most, with just enough web to hold the flanges apart and prevent buckling.

The same reasoning, applied vertically, explains why a heap of sand settles into a pyramidal shape, why load-bearing pillars distribute their end loads rather than concentrating them at a point, and — pushed to its geological limit in the third derivation below — why no mountain on Earth stands taller than about ten kilometres.

Board level - stress, strain and elongation of a loaded steel rod

A structural steel rod has a radius of $10\ \text{mm}$ and a length of $1.0\ \text{m}$. A $100\ \text{kN}$ force stretches it along its length. Take Young's modulus of structural steel as $Y = 2.0\times10^{11}\ \text{N m}^{-2}$. Find (a) the stress, (b) the elongation, and (c) the strain in the rod.

Step 1. Find the area and stress. $A = \pi r^2 = \pi (0.01\ \text{m})^2 = 3.14\times10^{-4}\ \text{m}^2$, so $\sigma = F/A = (1.0\times10^{5}\ \text{N})/(3.14\times10^{-4}\ \text{m}^2) = 3.18\times10^{8}\ \text{N m}^{-2}$.

Step 2. Find the elongation using Eq. (6) rearranged: $\Delta L = \dfrac{(F/A)\,L}{Y} = \dfrac{(3.18\times10^{8})(1.0)}{2.0\times10^{11}} = 1.59\times10^{-3}\ \text{m}$.

Step 3. Convert to strain: $\text{Strain} = \Delta L/L = 1.59\times10^{-3}$. Answer: stress $= 3.18\times10^{8}\ \text{N m}^{-2}$, elongation $\Delta L = 1.59\ \text{mm}$, strain $\approx 0.16\%$.

Board level - fractional compression of ocean water from its bulk modulus

The average depth of the Indian Ocean is about $3000\ \text{m}$. The bulk modulus of water is $B = 2.2\times10^{9}\ \text{N m}^{-2}$. Taking the density of water as $1000\ \text{kg m}^{-3}$ and $g = 10\ \text{m s}^{-2}$, find the fractional compression $\Delta V/V$ of water at the bottom of the ocean.

Step 1. Find the hydraulic pressure at the bottom, treating the ocean as a column of height $h = 3000\ \text{m}$: $p = h\rho g = (3000)(1000)(10) = 3\times10^{7}\ \text{N m}^{-2}$.

Step 2. Apply Eq. (8): $\dfrac{\Delta V}{V} = \dfrac{p}{B} = \dfrac{3\times10^{7}}{2.2\times10^{9}} = 1.36\times10^{-2}$. Answer: the water is compressed by about $1.36\%$ even at the ocean floor — a striking demonstration of how large water's bulk modulus really is.

Competitive level - composite copper-steel wire (JEE and NEET pattern)

A copper wire of length $2.2\ \text{m}$ and a steel wire of length $1.6\ \text{m}$, both of diameter $3.0\ \text{mm}$, are joined end to end and stretched by a load. The net elongation is found to be $0.70\ \text{mm}$. Take $Y_{\text{copper}} = 1.1\times10^{11}\ \text{N m}^{-2}$ and $Y_{\text{steel}} = 2.0\times10^{11}\ \text{N m}^{-2}$. Find (a) the applied load, and (b) the total elastic potential energy stored in the two wires together.

Step 1. Use the series-wire result from the second derivation above: $\dfrac{\Delta L_c}{\Delta L_s} = \dfrac{Y_s}{Y_c}\times\dfrac{L_c}{L_s} = \dfrac{2.0\times10^{11}}{1.1\times10^{11}}\times\dfrac{2.2}{1.6} = 2.5$. With $\Delta L_c + \Delta L_s = 7.0\times10^{-4}\ \text{m}$, solving gives $\Delta L_s = 2.0\times10^{-4}\ \text{m}$ and $\Delta L_c = 5.0\times10^{-4}\ \text{m}$.

Step 2. Recover the load from copper's Hooke's-law relation. $A = \pi(1.5\times10^{-3})^2 = 7.07\times10^{-6}\ \text{m}^2$, so $F = \dfrac{A\,Y_c\,\Delta L_c}{L_c} = \dfrac{(7.07\times10^{-6})(1.1\times10^{11})(5.0\times10^{-4})}{2.2} \approx 1.8\times10^{2}\ \text{N}$, the tension every cross-section of the composite wire carries.

Step 3. Find the elastic potential energy using the shortcut $U = \tfrac{1}{2}F\,\Delta L$ from the first derivation, since $F \approx 176.7\ \text{N}$ acts throughout. $U_c \approx 0.044\ \text{J}$ and $U_s \approx 0.018\ \text{J}$. Answer: load $F \approx 1.8\times10^{2}\ \text{N}$; total stored energy $U_c + U_s \approx 0.062\ \text{J}$. Timing note: about three minutes. The problem collapses once you notice "joined end to end, same diameter" forces identical stress in both wires — write that ratio first. Recognising that $\tfrac{1}{2}F\,\Delta L$ gives the stored energy, rather than a fresh integral, saves time under exam pressure.

The single most common slip is applying Hooke's law, or a modulus value, beyond the elastic limit — $\sigma = Y\varepsilon$ is only true up to point $A$; using it between $B$ and $D$ silently gives a wrong, too-large strain. A related error is confusing the yield point ($B$, where permanent deformation begins) with the ultimate tensile strength ($D$, the maximum stress the material can bear) — only $\sigma_y$, not $\sigma_u$, is the safe design limit. Students often treat "stretches more easily" as "more elastic," which is backwards: a large Young's modulus means a material resists stretching, and that resistance is what makes it more elastic technically. Sign errors are common with bulk modulus: forgetting the minus sign in $B = -p/(\Delta V/V)$ misses that $B$ is always positive even though $\Delta V$ is negative under compression. Unit slips too — strain and Poisson's ratio have no units, while stress and every modulus share the unit $\text{N m}^{-2}$. In series-wire problems, students often assume the wires share the same strain rather than the same stress — it is the force that is identical along an unbranched series connection.

• Identify tensile, compressive, shearing and hydraulic stress from how forces act on a body, and write $F/A$ for each.
• Compute longitudinal, shearing and volume strain from given deformation data, remembering all three are dimensionless.
• State Hooke's law and identify the proportional region on a stress-strain curve where it holds.
• Read a stress-strain graph, correctly label the elastic limit, ultimate tensile strength and fracture point, and classify a material as ductile or brittle.
• Calculate Young's, shear and bulk modulus from experimental data, and use $G \approx Y/3$ as a sanity check for solids.
• Define Poisson's ratio and recall it is a dimensionless number, roughly $0.3$ for common metals.
• Derive and apply the elastic potential energy per unit volume of a stretched wire, $u = \tfrac{1}{2}\sigma\varepsilon$, and $U = \tfrac{1}{2}F\,\Delta L$.
• Solve series (composite wire) problems by recognising equal tension, not equal strain, as the shared quantity.
• Apply elastic-modulus reasoning to engineering: sizing a crane rope, computing beam sag from Eq. (10), the I-beam cross-section, and the maximum height of a mountain.

Ask the tutor: if you would like, I can walk through a mixed problem combining all three moduli on one composite structure, or work backwards from a given stress-strain graph to identify a material.