Deformation and the restoring force
A solid has a definite shape and size. To change either, a force is required. When such a deforming force acts on a body held in static equilibrium, the body is deformed — stretched, sheared or compressed — to an extent that depends on the material and on the magnitude of the force. The deformation may be invisible to the eye, but it is always present. The moment the force is removed, a body that returns to its original shape and size is said to be elastic; one that retains the deformed shape is plastic. Putty and mud are close to ideal plastics.
The key idea is the restoring force. When a body is subjected to a deforming force, internal forces develop inside it that oppose the deformation. At equilibrium this restoring force is equal in magnitude and opposite in direction to the applied force. Stress and strain are simply the two normalised numbers that describe this situation — one for the force side, one for the deformation side.
What stress is — and its unit
Stress is the restoring force per unit area developed inside a deformed body. If $F$ is the deforming force and $A$ is the area over which it acts, the magnitude of the stress is
$$\text{Stress} = \frac{F}{A}$$
Because the restoring force is equal and opposite to the applied force at equilibrium, the magnitude of the stress also equals the applied force per unit area. The SI unit of stress is the newton per square metre ($\text{N m}^{-2}$), also called the pascal (Pa). Its dimensional formula is $[\mathrm{M\,L^{-1}\,T^{-2}}]$ — identical to that of pressure, because both are force divided by area.
| Quantity | Defining ratio | SI unit | Dimensions |
|---|---|---|---|
| Stress | Restoring force / area, \(F/A\) | N m⁻² = Pa | \([\mathrm{M\,L^{-1}\,T^{-2}}]\) |
| Strain | Change in dimension / original dimension | none (pure number) | \([\mathrm{M^{0}\,L^{0}\,T^{0}}]\) |
| Modulus of elasticity | Stress / strain | N m⁻² = Pa | \([\mathrm{M\,L^{-1}\,T^{-2}}]\) |
A subtle but examinable point: stress is not a vector. Unlike a force, which acts on a specified side of a section with a definite direction, stress cannot be assigned a single direction — it depends on the orientation of the surface and on whether the deformation is along, across or all around the body. NCERT states this in its "Points to ponder": stress is treated as a tensor, not a vector.
The three types of stress
There are exactly three ways a solid can be stressed, set by the direction of the deforming force relative to the surface it acts on. NCERT Fig. 8.1 illustrates all three.
| Type of stress | How the force acts | Sub-types | Effect on body |
|---|---|---|---|
| Longitudinal (tensile / compressive) | Normal to the cross-section, along the length | Tensile (stretch) or compressive (squeeze) | Change in length; shape preserved |
| Shearing (tangential) | Parallel / tangential to the cross-section | Pure shear | Change in shape; volume preserved |
| Hydraulic | Normal to the surface, equal at every point (a fluid all around) | Uniform compression | Change in volume; shape preserved |
Longitudinal (tensile and compressive) stress
When two equal and opposite forces are applied normal to the cross-section of a body, pulling it apart, the restoring force per unit area is tensile stress. If the forces push the body inward instead, the restoring force per unit area is compressive stress. Both are bracketed under the term longitudinal stress, because in each case the length of the body changes. A wire carrying a hanging weight is under tensile stress; the thighbone of a circus performer at the base of a human pyramid is under compressive stress.
Shearing (tangential) stress
When two equal and opposite forces are applied parallel to the cross-sectional faces — tangential rather than normal — opposite faces of the body slide relative to one another. The restoring force per unit area developed against this tangential force is the shearing or tangential stress. Pressing a thick book with the hand and pushing it sideways, with the bottom face held by friction, is the standard picture: the shape of the stack distorts while its volume does not change. Shearing stress is possible only in solids, because only solids resist a change of shape.
Hydraulic stress
When a body is immersed in a fluid under high pressure, the fluid pushes inward, perpendicular to the surface, at every point and by the same amount. The body is under hydraulic compression: its volume decreases while its geometrical shape is preserved. The internal restoring force per unit area that develops is the hydraulic stress, equal in magnitude to the applied hydraulic pressure. This is the only stress type that also applies to liquids and gases, since it requires no resistance to change of shape — only resistance to change of volume.
What strain is — and why it is unitless
Strain is the response side of the bargain: the fractional change in dimension produced by the stress. In general, strain is the change in a dimension (length, shape or volume) divided by the original value of that dimension. Because it is the ratio of two quantities of the same kind, the units cancel and strain is a pure number with no unit and no dimensional formula. This single fact explains why every modulus of elasticity — being stress divided by strain — carries the same unit as stress, the pascal.
The three types of strain
Each type of stress produces its own kind of strain. The three pair up one-to-one.
| Strain | Definition | Formula | Produced by |
|---|---|---|---|
| Longitudinal strain | Change in length per unit length | \(\dfrac{\Delta L}{L}\) | Tensile / compressive stress |
| Shearing strain | Relative displacement of faces per unit height; equals the shear angle | \(\dfrac{\Delta x}{L} = \tan\theta \approx \theta\) | Shearing stress |
| Volume strain | Change in volume per unit volume | \(\dfrac{\Delta V}{V}\) | Hydraulic stress |
Longitudinal strain is the change in length $\Delta L$ divided by the original length $L$, written $\Delta L / L$. It is positive for stretching and negative for compression, but the magnitude of strain for a given material is the same whether the stress is tensile or compressive — which is why a single Young's modulus describes both.
Shearing strain is the relative displacement $\Delta x$ between opposite faces divided by the separation $L$ between them, which equals $\tan\theta$, where $\theta$ is the angle through which a line perpendicular to the fixed face is turned. For the small deformations of NEET problems $\theta$ is tiny, so $\tan\theta \approx \theta$; at $\theta = 10^\circ$ the two differ by only about one per cent. Shearing strain is therefore reported directly as the angle $\theta$ in radians.
Volume strain is the change in volume $\Delta V$ divided by the original volume $V$, written $\Delta V / V$. It accompanies hydraulic stress, where the body is squeezed uniformly from all sides and shrinks without changing shape.
The three stress–strain pairs and their moduli
Within the elastic limit, stress is directly proportional to strain — this is Hooke's law, and the constant of proportionality is the modulus of elasticity. There are exactly three such pairings, one per stress type, and each defines a different modulus. The full proportionality and the moduli themselves are developed on the dedicated pages; here is the map that links them.
| Stress | Strain | Change in shape? | Change in volume? | Modulus |
|---|---|---|---|---|
| Tensile / compressive (longitudinal) | Longitudinal, \(\Delta L/L\) | Yes | No | Young's modulus \(Y\) |
| Shearing (tangential) | Shearing, \(\theta\) | Yes | No | Shear modulus \(G\) |
| Hydraulic | Volume, \(\Delta V/V\) | No | Yes | Bulk modulus \(B\) |
Young's modulus and the shear modulus are defined only for solids, because only solids possess a definite length and shape to begin with. The bulk modulus, by contrast, applies to solids, liquids and gases alike, since every state of matter has a definite volume that a uniform pressure can compress.
The numerical values, formulae and physical meaning of $Y$, $G$ and $B$ are worked out in Elastic Moduli. The proportionality itself is set up in Hooke's Law.
The molecular origin of stiffness
The reason a solid resists deformation at all lies in the forces between its atoms. Matter is built from atoms and molecules held together by intermolecular forces. When two molecules are far apart the force between them is weakly attractive; as they approach, the attraction grows to a maximum and then, at very small separations, turns strongly repulsive. At one particular separation $R_0$ — the equilibrium separation, about $10^{-10}\,\text{m}$ in a solid — the net force is zero.
In a solid the atoms sit almost exactly at $R_0$, locked in place by these strong forces. When a deforming force pulls the atoms apart so that their separation exceeds $R_0$, attractive forces revive and try to draw them back. When the force pushes them closer than $R_0$, repulsive forces build and push them apart. Either way an internal restoring force appears — and the restoring force per unit area is precisely the stress. The atoms behave like a lattice of balls connected by tiny springs: displace them and the springs pull them home.
This same picture explains the hierarchy of compressibility. Solids resist volume change strongly because neighbouring atoms are tightly coupled; liquids are bound but more loosely; gases are barely coupled at all and so are about a million times more compressible than solids. It also explains why only solids sustain a static shearing stress: their fixed atomic arrangement resists a change of shape, whereas the freely sliding molecules of a fluid cannot.
Worked numericals
A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate (a) the stress and (b) the longitudinal strain. Take Young's modulus of structural steel as $Y = 2.0\times 10^{11}\,\text{N m}^{-2}$.
Stress. The cross-sectional area is $A = \pi r^2 = 3.14\times(10\times 10^{-3})^2 = 3.14\times 10^{-4}\,\text{m}^2$. The tensile stress is $\sigma = F/A = (100\times 10^{3})/(3.14\times 10^{-4}) = 3.18\times 10^{8}\,\text{N m}^{-2}$.
Strain. Longitudinal strain $= \sigma/Y = (3.18\times 10^{8})/(2.0\times 10^{11}) = 1.59\times 10^{-3}$, i.e. about $0.16\%$. The number is a pure ratio — no unit — consistent with strain being dimensionless.
A load of 100 kg is suspended by a wire of length 1.0 m and cross-sectional area $0.10\,\text{cm}^2$. The wire stretches by 0.20 cm. Find (i) the tensile stress and (ii) the longitudinal strain. Take $g = 9.80\,\text{m s}^{-2}$.
Tensile stress. $A = 0.10\,\text{cm}^2 = 0.10\times 10^{-4}\,\text{m}^2$. The force is the weight, $F = Mg = 100\times 9.80 = 980\,\text{N}$. Stress $= F/A = 980/(0.10\times 10^{-4}) = 9.8\times 10^{7}\,\text{N m}^{-2}$.
Longitudinal strain. $\Delta L / L = (0.20\times 10^{-2})/(1.0) = 0.20\times 10^{-2} = 2.0\times 10^{-3}$, again a dimensionless ratio.
Stress and strain in one breath
- Stress is the internal restoring force per unit area, $F/A$; unit pascal ($\text{N m}^{-2}$); dimensions $[\mathrm{M\,L^{-1}\,T^{-2}}]$; not a vector.
- Three stresses: longitudinal (tensile/compressive, normal to face), shearing (tangential to face), hydraulic (uniform pressure all around).
- Strain is the fractional change in dimension; it is a pure number with no unit.
- Three strains: longitudinal $\Delta L/L$, shearing $\Delta x/L = \tan\theta \approx \theta$, volume $\Delta V/V$.
- The three pairings give the three moduli: longitudinal → Young's $Y$; shear → rigidity $G$; hydraulic → bulk $B$.
- $Y$ and $G$ exist only for solids; $B$ applies to solids, liquids and gases.
- Stress and pressure share the unit pascal but differ in physics; "more elastic" means smaller strain for a given stress.