Class 11 Physics · NCERT chapter 9 · 6 topics · free to read
Stress & Strain
Stress ( F/A ) measures the internal restoring force per unit area when a body is deformed. Strain ( L / L ) measures how much it deforms relative to its original size. Both are defined for three modes: tensile/compressive (length change), shearing (shape change), and volumetric (volume change). NEET loves unit-based elimination — stress has units of Pa, strain is dimensionless.
Stress = F A | Strain = L L
Stress
When you apply an external force to a solid, the body develops an internal restoring force. Stress is this restoring force divided by the cross-sectional area over which it acts.
When you pull, push, or twist a solid, its atoms get displaced from equilibrium. The inter-atomic forces resist this displacement, creating an internal restoring force equal and opposite to the applied force (in static equilibrium). Stress is this restoring force per unit area of the cross-section.
Stress = F A (SI unit: N/m^2 = Pa)
Stress is NOT the applied force — it is the restoring force per unit area developed inside the body
SI unit: pascal (Pa) = N/m ^2 . Also expressed in MPa, GPa for engineering materials
Dimensions: [ML^ -1 T^ -2 ] — same as pressure
Stress has the same units as pressure, but pressure acts on a surface from outside while stress acts within the material
Types of Stress
The type of stress depends on how the deforming force is applied: pulling apart (tensile), pushing together (compressive), or sliding layers (shearing).
There are three fundamental types of stress, each linked to a different way of applying force:
Tensile stress: Force pulls the body along its length (stretching a wire). The restoring force resists elongation
Compressive stress: Force pushes the body inward (compressing a pillar). The restoring force resists compression
Shearing stress (tangential stress): Force is applied tangentially to a surface, causing layers to slide. Restoring force resists the sliding
Tensile and compressive stresses are normal stresses (force area). Shearing stress is a tangential stress (force area)
There's also hydraulic stress (or volume stress): when a body is submerged in a fluid, the fluid exerts uniform pressure on all surfaces, compressing the body equally in all directions. This is just pressure P , and it acts inward on the entire surface.
Hydraulic stress = P = F A (uniform on all surfaces)
Strain
Strain quantifies deformation as a ratio: change in dimension divided by original dimension. It is dimensionless — no units.
Strain measures how much a body deforms relative to its original size. It is always a ratio of two lengths (or volumes, or angles), so it is a pure number — dimensionless, no units.
Longitudinal strain = L L — change in length / original length (for tensile or compressive deformation)
Shearing strain = (in radians) — the angle through which a face is displaced, where = x L
Volumetric strain = V V — change in volume / original volume (for hydraulic compression)
Strain is dimensionless — no units. It is often expressed as a percentage
Longitudinal strain = L L | Shearing strain = = x L | Volume strain = V V
Hooke's Law & Elastic Moduli
Hooke's Law states stress strain within the elastic limit. The proportionality constant is the elastic modulus — Young's modulus ( Y ) for length, bulk modulus ( B ) for volume, and shear modulus ( G ) for shape. Higher modulus = stiffer material. NEET frequently tests Y = FL/A L numerical problems.
Y = FL A L | B = PV V | G = F/A
Hooke's Law
For small deformations, stress and strain are linearly proportional. This is the foundation of all elasticity — and its limit defines where 'elastic' ends and 'plastic' begins.
Hooke's Law: For small deformations, stress is directly proportional to strain. Stress = E Strain , where E is the elastic modulus (modulus of elasticity). This linear relationship holds only up to the proportional limit — beyond that, the material may still be elastic but the graph curves.
Stress = E Strain E = Stress Strain
Hooke's Law is valid only for small deformations within the proportional limit
The elastic modulus E has the same units and dimensions as stress ( [ML^ -1 T^ -2 ] , Pa) because strain is dimensionless
A higher modulus means the material is stiffer — it deforms less for the same stress
Hooke's Law applies to all three types of deformation — tensile, shearing, and volumetric — with different moduli
Young's Modulus ( Y )
Young's modulus relates longitudinal stress to longitudinal strain. It is the most tested elastic modulus in NEET — especially in wire-stretching numerical problems.
Young's modulus ( Y ) is defined as the ratio of longitudinal (tensile or compressive) stress to longitudinal strain. It tells you how much a rod or wire resists being stretched or compressed along its length.
Y = Longitudinal stress Longitudinal strain = F/A L / L = FL A L
Y for steel 2 10^ 11 Pa, for copper 1.3 10^ 11 Pa, for rubber 5 10^ 8 Pa
Y depends on the material, NOT on dimensions (length, area) — it's a material property
A wire's extension: L = FL AY — inversely proportional to both area and Y
Doubling the length doubles the extension; doubling the diameter (4x area) reduces extension to 1/4
Quick recall
A steel wire of length 2 m and cross-sectional area 1 10^ -6 m ^2 is stretched by a force of 100 N. Find the extension. ( Y_ steel = 2 10^ 11 Pa)
L = FL AY = 100 2 10^ -6 2 10^ 11 = 200 2 10^ 5 = 10^ -3 m = 1 mm
Bulk Modulus ( B ) & Shear Modulus ( G )
Bulk modulus measures resistance to uniform compression (volume change). Shear modulus measures resistance to shape distortion. Together with Young's modulus, these three cover all types of elastic deformation.
Bulk modulus ( B ): When a body is compressed uniformly from all sides (hydraulic stress), B measures how much it resists volume change. Shear modulus ( G or ): When a tangential force distorts the shape without changing volume, G measures how much it resists.
B = Hydraulic stress Volume strain = - P V / V = - PV V
G = Shearing stress Shearing strain = F/A
The negative sign in B ensures B is positive — when pressure increases ( P > 0 ), volume decreases ( V < 0 )
Compressibility = 1/B — how easily a material can be compressed. Gases are highly compressible (low B ), solids are nearly incompressible (high B )
B for water 2.2 10^9 Pa, for steel 1.6 10^ 11 Pa, for air 1.01 10^5 Pa
Solids have all three moduli ( Y , B , G ). Fluids have only B (they can't sustain tensile or shearing stress)
Resists volume change
Resists shape change
Stress = hydraulic pressure P
Stress = tangential force / area
Strain = V / V
Strain = angle
Defined for solids, liquids, gases
Defined for solids ONLY
Deep ocean: water compressed by high P
Book pages sliding when pushed sideways
Stress-Strain Curve
The stress-strain curve is the complete biography of a material under load — from elastic obedience (Hooke's Law region) through yielding, plastic flow, and finally fracture. NEET asks you to identify proportional limit, elastic limit, yield point, ultimate strength, and fracture point on the curve, and to distinguish ductile from brittle behavior.
Curve regions: O A (proportional) B (elastic limit) C (yield) D (ultimate) E (fracture)
Regions of the Curve
The stress-strain curve for a ductile material (like mild steel) has distinct regions that reveal the material's behavior at every stage of loading.
When you gradually increase the load on a wire and plot stress vs. strain, you get a characteristic curve. For a ductile metal like mild steel, the curve has five key landmarks:
O A (Proportional limit): Stress strain — Hooke's Law holds perfectly. The curve is a straight line. Slope = Young's modulus Y
A B (Elastic limit): Stress-strain relationship is no longer linear, but the body still returns to original shape on unloading. No permanent deformation
B C (Yield point): The material 'gives in.' Strain increases rapidly with little or no increase in stress. Permanent (plastic) deformation begins
C D (Plastic region / strain hardening): Material deforms permanently. Stress increases again as the material hardens. Point D = ultimate tensile strength — maximum stress the material can withstand
D E (Necking and fracture): Cross-section narrows at one point (necking). Stress appears to decrease (calculated on original area). At E , the wire snaps — fracture point
Elastic vs Plastic Deformation
Below the elastic limit, deformation is reversible. Beyond it, the material retains a permanent set — it never goes back to its original dimensions.
Elastic deformation: The body returns to its original shape when the deforming force is removed. All the energy stored during deformation is recovered. Plastic deformation: The body does NOT return to its original shape. Some energy is lost as heat during the permanent rearrangement of atoms.
Reversible — body returns to original shape
Irreversible — permanent change in shape
Occurs below elastic limit
Occurs above elastic limit
Stress-strain: follows same path on loading and unloading
Stress-strain: different paths (hysteresis)
Atoms return to equilibrium positions
Atoms settle in new positions
Example: rubber band stretched gently
Example: clay molded into a shape
Permanent set: If you stretch a wire beyond the elastic limit and then remove the force, it doesn't return to its original length. The remaining deformation is called the permanent set. The amount of permanent set depends on how far beyond the elastic limit you went.
Ductile vs Brittle Materials
Ductile materials (steel, copper) have large plastic regions and undergo significant deformation before breaking. Brittle materials (glass, cast iron) snap suddenly with almost no plastic deformation.
The shape of the stress-strain curve reveals whether a material is ductile or brittle. Ductile materials have a large plastic region — they warn you before breaking. Brittle materials have almost no plastic region — they shatter suddenly.
Large plastic deformation before fracture
Little or no plastic deformation
Fracture point far from elastic limit
Fracture point close to elastic limit
Gives warning before failure (necking, bending)
Fails suddenly without warning
Can be drawn into wires (ductility)
Cannot be drawn — they snap
Examples: mild steel, copper, aluminum, gold
Examples: glass, cast iron, concrete, ceramics
Elastomers (rubber, tissue) have very large elastic regions but can stretch to many times their original length. Their stress-strain curve has no well-defined plastic region — they either snap back or tear
The area under the stress-strain curve up to fracture represents toughness — total energy absorbed before breaking
Ductile materials are tough (large area under curve). Brittle materials are typically strong but not tough
Quick recall
A material has a stress-strain curve that is nearly a straight line up to the fracture point with very little strain. Is it ductile or brittle?
Brittle. Almost no plastic deformation before fracture — the material breaks while still nearly in the elastic region. Examples: glass, ceramics.
Poisson's Ratio
When you stretch a wire, it gets longer but also thinner. Poisson's ratio quantifies this coupling: = -(lateral strain)/(longitudinal strain) . For most materials, 0 < < 0.5 . Theoretically -1 0.5 . The three elastic moduli ( Y , B , G ) are related through . NEET tests the range of and the inter-relations.
= - Lateral strain Longitudinal strain = - d / d L / L
Poisson's Ratio ( )
When a rod is stretched, its length increases but its diameter decreases. Poisson's ratio measures this lateral contraction relative to the longitudinal extension.
Stretch a rubber band — it gets longer AND thinner. Compress a rubber ball — it gets shorter AND fatter. The length change and the width change are coupled. Poisson's ratio ( ) is the ratio of lateral strain to longitudinal strain, with a negative sign to make positive for normal materials.
= - Lateral strain Longitudinal strain = - d / d L / L
Poisson's ratio is a dimensionless scalar — it's a ratio of two strains
The negative sign ensures > 0 for normal materials (lateral strain is opposite in sign to longitudinal strain)
For most engineering materials: 0.2 < < 0.4 . Steel: 0.28 , rubber: 0.5 , cork: 0
Theoretical range: -1 0.5 (from thermodynamic stability). For NEET, remember: 0 0.5 for real materials
Relations Between Y , B , G , and
Young's modulus, bulk modulus, and shear modulus are not independent — they are related through Poisson's ratio. These relations are important for NEET.
For an isotropic material (same properties in all directions), the three elastic moduli and Poisson's ratio are connected by two fundamental relations. Knowing any two of Y , B , G , lets you find the other two.
Y = 3B(1 - 2 )
Y = 2G(1 + )
From these: 9 Y = 3 G + 1 B or equivalently Y = 9BG 3B + G
From Y = 3B(1 - 2 ) : if = 0.5 , then Y = 0 — material is perfectly incompressible (like an ideal liquid)
From Y = 2G(1 + ) : if = -1 , then Y = 0 — not physically realizable for normal materials
For = 0.5 : B (incompressible) and Y = 3G
For most metals ( 0.3 ): Y 2.6G and Y 1.2B , so roughly G < B < Y
Quick recall
If Poisson's ratio = 0.25 , find the ratio Y : B : G .
Y = 3B(1 - 0.5) = 1.5B B = Y/1.5 . Y = 2G(1.25) = 2.5G G = Y/2.5 = 0.4Y . So Y : B : G = 1 : 0.667 : 0.4 = 5 : 3.33 : 2 .
Elastic Potential Energy
A stretched or compressed elastic body stores energy — elastic potential energy. For a wire: U = 1 2 stress strain volume . This equals 1 2 F L (analogous to spring PE). The energy density (energy per unit volume) is u = 1 2 stress strain . NEET tests the formula and its rearranged forms.
U = 1 2 stress strain V = 1 2 F^2 L AY
Energy Stored in a Stretched Wire
When you stretch a wire within the elastic limit, you do work against the internal restoring force. This work is stored as elastic potential energy in the wire.
Stretching a wire is like stretching a spring — the force increases linearly with extension (Hooke's Law region). The work done equals the area under the force-extension graph, which is a triangle: W = 1 2 F L . This work is stored as elastic potential energy.
U = 1 2 F L = 1 2 stress strain V
The factor 1 2 appears because the force builds up from 0 to F during stretching — the average force is F/2 . This is exactly like the spring energy formula U = 1 2 kx^2 .
U = 1 2 F L where F is the final stretching force and L is the total extension
U = 1 2 stress strain V where V = A L is the volume of the wire
Using stress = Y strain : U = 1 2 Y (strain)^2 V = 1 2 (stress)^2 Y V
Using F = YA L L : U = 1 2 YA( L)^2 L = 1 2 F^2 L AY
Elastic Energy Density
Energy density is the elastic potential energy stored per unit volume of the material. It equals half the product of stress and strain, and is represented by the area under the stress-strain curve.
Energy density ( u ) is the elastic PE stored per unit volume. It's useful because it's a material property — independent of the wire's dimensions.
u = U V = 1 2 stress strain = 1 2 Y ^2 = _ stress ^2 2Y
Energy density u = area under the stress-strain curve (up to the strain considered)
SI unit of u : J/m ^3 = Pa (same dimensions as stress and pressure)
u = 1 2 Y ^2 — proportional to Y and to the SQUARE of strain
A stiffer material (higher Y ) stores MORE energy for the same strain, but LESS energy for the same stress (since u = ^2 / 2Y )
Quick recall
A wire of length 1 m and area 10^ -6 m ^2 is stretched by 0.5 mm under a force of 100 N. Find the elastic PE stored.
U = 1 2 F L = 1 2 100 0.5 10^ -3 = 0.025 J = 25 mJ . Alternatively: stress = 10^8 Pa, strain = 5 10^ -4 , V = 10^ -6 m ^3 , so U = 1 2 10^8 5 10^ -4 10^ -6 = 0.025 J.
Applications of Elasticity
Elasticity concepts are used everywhere in engineering: designing crane ropes, setting bridge load limits, choosing materials for beams, understanding why metallic ropes outperform rubber for lifting, and calculating thermal stress when materials expand against constraints. NEET tests crane/rope problems and thermal stress.
Thermal stress = Y T | L_ thermal = L T
Crane Design & Metallic Ropes
Designing a crane rope means ensuring the stress never exceeds the elastic limit. Steel's high Young's modulus means minimal stretching under load — safer and more predictable than rubber.
When a crane lifts a load of mass m , the rope must support a weight mg . The stress in the rope is = mg/A , where A is the cross-sectional area. For safe design, this stress must stay well below the elastic limit — typically by a factor of safety of 10 .
A_ = mg factor of safety _ elastic limit
Factor of safety = elastic limit / working stress. Typical value: 8-10 for structural applications
Steel rope: small L (high Y ), predictable, doesn't bounce. Rubber: huge L (low Y ), dangerous oscillations
Rope design question: find minimum radius/diameter from _ working = mg / r^2 _ elastic limit / safety factor
Real crane ropes use multiple thin steel wires twisted together — for flexibility without sacrificing strength
Quick recall
A crane must lift 1000 kg. The elastic limit of steel is 3 10^8 Pa. Using a safety factor of 10, find the minimum cross-sectional area of the steel rope.
A bridge beam bends (depresses) under load. The depression depends on the material, dimensions, and cross-sectional shape. I-shaped cross-sections maximize strength while minimizing weight.
A horizontal beam supported at both ends sags (depresses) in the middle when loaded. NCERT gives the depression formula for a beam of rectangular cross-section loaded at the centre:
= Wl^3 4Ybd^3
Here W = mg is the load, l is the span (distance between supports), b is the breadth, d is the depth (vertical thickness), and Y is Young's modulus. Notice: depression is inversely proportional to d^3 — increasing depth is far more effective than increasing breadth.
To reduce depression: use a material with high Y (steel > wood), increase depth d (most effective, goes as d^3 ), increase breadth b , or reduce span l
I-shaped girders concentrate material at the top and bottom flanges (far from the neutral axis) — maximizing effective d while saving material in the middle
Bridges have expansion joints — gaps between sections that allow thermal expansion without generating thermal stress that could crack the bridge
Railway tracks are laid with small gaps between sections for the same reason — thermal expansion in summer
Thermal Stress
If a rod is clamped at both ends and heated, it cannot expand freely. The blocked expansion generates a compressive thermal stress that can be large enough to buckle or crack the material.
When a rod of length L is heated by T , it would normally expand by L = L T (where is the coefficient of linear expansion). If the rod is rigidly clamped at both ends and cannot expand, the blocked expansion creates a compressive stress inside the rod.
Thermal stress = F A = Y strain = Y L L = Y T
Thermal force = F = YA T
Thermal stress depends on Y , , and T — NOT on the length or area of the rod (they cancel out)
Heating a clamped rod → compressive stress (rod tries to expand but can't). Cooling → tensile stress (rod tries to contract but can't)
Thermal force F = YA T — this DOES depend on area
Materials with low and low Y develop less thermal stress — useful for applications with large temperature variations
Quick recall
A steel rod ( Y = 2 10^ 11 Pa, = 1.2 10^ -5 /°C) is clamped at both ends and heated by 50°C. Find the thermal stress.
= Y T = 2 10^ 11 1.2 10^ -5 50 = 1.2 10^8 Pa = 120 MPa . This is significant — comparable to the yield stress of mild steel!