📌 Snapshot
- Establishes that real solids are not perfectly rigid: under a deforming force they develop an internal restoring force, and depending on the material they either recover (elastic) or stay deformed (plastic).
- Introduces the three forms of stress (tensile/compressive, shearing, hydraulic) and their matched strains, then ties them via Hooke's law and the three elastic moduli (Y, G, B).
- Builds the full stress-strain curve — proportional limit, yield point, plastic region, ultimate tensile strength, fracture — and contrasts metallic curves with elastomers (e.g., aorta tissue).
- Closes with quantitative applications: design of crane ropes, bending of beams (the I-section), pillar shapes, and the upper bound on mountain height (~10 km) from rock shear strength.
- CUET regularly tests these as numerical plug-ins (Y from F, L, A, ΔL), tabular comparison of moduli, and statement-based items on the stress-strain curve.
📖 Detailed Notes
2.1 Core concepts
The rigid body of mechanics is an idealisation. In reality every solid changes shape and size, however slightly, when forces act on it; steel beams sag, springs stretch, and even concrete pillars compress. The macroscopic origin is microscopic — the inter-atomic spacing rearranges when external forces disturb it, and the inter-atomic restoring forces try to bring the atoms back. The property of a body to regain its original size and shape on removal of the deforming force is called elasticity, and the residual permanent deformation seen in materials like putty or mud is called plasticity (NCERT §8.1, p. 167). Engineering choices — what to build a bridge cable, a clock spring, a bullet-proof vest or an artificial joint out of — are essentially choices among these mechanical responses.
When a deforming force acts on a body in equilibrium, an equal and opposite internal restoring force develops. The intensity of this restoring force is captured by stress, defined as restoring force per unit area: stress = F/A (NCERT §8.2, p. 168). The SI unit is the newton per square metre, N m⁻², also called the pascal (Pa); its dimensional formula is [M L⁻¹ T⁻²]. A force of a few newtons spread over a square metre is a tiny stress; the same force concentrated on a needle tip is enormous — which is why a tailor's needle pierces cloth easily.
There are three independent modes in which a body can be deformed, each with its own stress and strain (NCERT §8.2, Fig. 8.1, p. 168–169).
(a) Tensile/compressive stress (longitudinal). Two equal and opposite forces normal to the cross-section pull a rod apart (tensile) or push it together (compressive). The rod's length changes by ΔL and the longitudinal strain is ε = ΔL/L. Both stress (F/A) and strain are along the length.
(b) Shearing (tangential) stress. Equal and opposite forces act parallel to opposite faces, so the cross-section slides relative to itself. A vertical face initially perpendicular to the base now tilts by an angle θ; the shearing strain is Δx/L = tan θ, which for small deformations is ≈ θ. A book pushed sideways on a table is the everyday picture.
(c) Hydraulic stress. A body is immersed in a fluid (or surrounded by gas) so that the fluid exerts the same normal pressure at every point on its surface. The body's shape stays the same but its volume decreases by ΔV; the volume strain is ΔV/V.
Because strain is a ratio of like quantities, it is dimensionless and unit-less — an answer of "1.2 × 10⁻⁴ m" for a strain is automatically wrong (NCERT §8.2, p. 169).
The key empirical observation, due to Robert Hooke (1676), is that for small enough deformations stress is proportional to strain: stress = k × strain (NCERT §8.3, p. 169). The proportionality constant k is the modulus of elasticity of the material, with the same units as stress. Hooke's law is empirical, holds only for "small" deformations whose magnitude depends on the material, and is the foundation of nearly every engineering calculation in this chapter.
The full behaviour of a metal is captured in the stress-strain curve (NCERT Fig. 8.2, §8.4, p. 169). Plotting stress on the y-axis and strain on the x-axis, the curve has five distinct regions:
- O → A: A straight line where Hooke's law strictly holds; the slope is Young's modulus. The material is purely elastic.
- A → B: The curve bends but the deformation is still elastic — unloading returns the body to O. The point B is the yield point or elastic limit, and the stress at B is the yield strength σ_y.
- B → D: Beyond B the body enters the plastic region. If unloaded at, say, point C, it does not return to O but to a non-zero permanent set; the body has "yielded".
- D: The maximum stress the material can take, called the ultimate tensile strength σ_u.
- D → E: The strain keeps increasing even though the applied force begins to drop, until fracture occurs at E. If the interval D-to-E is small, the material fractures soon after reaching σ_u — it is brittle (cast iron, glass). If D and E are far apart, the material is ductile: it can be drawn into wires (copper, steel). The same metal can be made more or less ductile by heat treatment. Elastomers like rubber bands and the elastic tissue of the aorta show a very different curve (NCERT Fig. 8.3, p. 170). They can be stretched to several times their original length and still recover their shape — i.e., the elastic region is enormous — but the curve is almost entirely non-linear, so Hooke's law fails over most of the region; and there is no well-defined plastic region preceding fracture. The three elastic moduli — one for each kind of stress-strain pair — quantify the slope of the linear region. Young's modulus Y (NCERT §8.5.1, p. 170). Y = tensile (or compressive) stress / longitudinal strain = (F·L)/(A·ΔL). Units Pa. Y is a property of the material, not the specimen — a thicker wire of the same steel has the same Y. Metals have large Y (steel ≈ 2 × 10¹¹ Pa, copper ≈ 1.2 × 10¹¹ Pa), so they need huge stresses for small strains; wood, bone, glass and concrete have much smaller Y. As a useful benchmark (NCERT p. 171), to stretch a thin steel wire of cross-section 0.1 cm² by 0.1 % needs 2000 N — far more than the 690 N needed for aluminium of identical dimensions, so steel is "more elastic" than aluminium (the material that returns more for a given load is the one that hardly let itself be deformed in the first place). Shear modulus G (modulus of rigidity) (NCERT §8.5.2, p. 172). G = shearing stress / shearing strain = (F·L)/(A·Δx) = F/(A·θ). Equivalent statement: σ_s = G·θ. For most materials G ≈ Y/3. Soft metals like lead have small G (5.6 × 10⁹ Pa) and tungsten has very high G (150 × 10⁹ Pa). Bulk modulus B (NCERT §8.5.3, p. 173). B = −p/(ΔV/V). The minus sign is essential because increasing pressure produces a decrease in volume — the ratio p/(ΔV/V) is itself negative, and the minus sign restores B to a positive number, which is what we need for a stable solid or liquid. Solids have the largest B (steel ≈ 160 × 10⁹ Pa), liquids smaller (water 2.2 × 10⁹ Pa), and gases by far the smallest (air at STP ≈ 10⁵ Pa). Hence gases are about 10⁶ times more compressible than solids. The reciprocal compressibility k = 1/B captures the same information. Poisson's ratio σ (NCERT §8.5.4, p. 174). When a wire is stretched longitudinally, its lateral dimensions contract. The ratio (lateral strain)/(longitudinal strain) is Poisson's ratio σ. It is dimensionless and a property of the material alone — for steel about 0.28–0.30, for aluminium alloys about 0.33. Elastic potential energy (NCERT §8.5.5, Eq. 8.14, p. 174). Work done against the restoring force during elastic stretching is stored as potential energy. The energy density (per unit volume) of a stretched specimen is u = ½ × stress × strain = ½ × Y × ε². Total stored energy U = ½ × stress × strain × volume. The closing applications (NCERT §8.6, p. 174–176) take these moduli into engineering practice. A steel rope for a crane lifting 10 tonnes needs only ~1 cm radius if pushed to the yield stress, but real designs use ~3 cm radius so the working stress is well below σ_y and safety is built in. A horizontal beam of length l, breadth b, depth d, supported at both ends and loaded at the centre with weight W, sags by δ = W·l³/(4·b·d³·Y). The depth appears as d³ so doubling d cuts sag by a factor of 8 — a far better strategy than doubling b — but a deep thin beam tends to buckle sideways. The compromise that engineers settled on is the I-section: most of the material at the top and bottom, a thin web in between, balancing strength against buckling. Similarly, pillars with distributed (flared) ends carry larger loads than rounded-end pillars. Finally, the maximum height of a mountain on Earth (~10 km, comparable to Everest) is set by the shear elastic limit of rock (~3 × 10⁸ N m⁻²) via the condition hρg ≤ σ_critical; rocks at greater depth would flow plastically.
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Rigid body | Idealised body whose shape and size do not change under any force | 167 |
| Elasticity | Property by which a body regains its original size and shape when the applied force is removed | 167 |
| Plasticity | Property by which a body undergoes permanent deformation (no tendency to regain shape) | 167 |
| Deforming force | External force that tries to change the shape or size of a body | 167 |
| Restoring force | Equal and opposite internal force that develops in a deformed body | 168 |
| Stress | Restoring force per unit area; SI unit N m⁻² (Pa); dimensions [M L⁻¹ T⁻²] | 168 |
| Tensile/compressive (longitudinal) stress | Normal forces stretching/compressing a body; F/A | 168 |
| Shearing (tangential) stress | Tangential forces on opposite faces; F/A parallel to surface | 168 |
| Hydraulic stress | Equal normal pressure at every point of body surface (fluid) | 169 |
| Longitudinal strain | ΔL/L — change in length / original length | 168 |
| Shearing strain | Δx/L = tan θ ≈ θ for small θ | 168 |
| Volume strain | ΔV/V — fractional change in volume under hydraulic stress | 169 |
| Hooke's law | For small deformations, stress = k × strain; k is the modulus of elasticity | 169 |
| Proportional (linear) limit | Stress up to which stress-strain curve is straight (point A) | 169 |
| Yield point (elastic limit) | Point B on stress-strain curve beyond which deformation becomes plastic; stress at B is σ_y | 169 |
| Plastic region | B → D portion of stress-strain curve in which permanent deformation occurs | 169 |
| Ultimate tensile strength | Maximum stress σ_u the material can withstand (point D) | 169 |
| Fracture point | Point E at which the body breaks | 169 |
| Brittle / ductile | Materials with small / large D-to-E gap on stress-strain curve | 170 |
| Elastomer | Material like rubber or aorta tissue with very large elastic strains, not obeying Hooke's law, no well-defined plastic region | 170 |
| Young's modulus (Y) | Tensile (or compressive) stress / longitudinal strain; unit Pa | 170 |
| Shear modulus (G) | Shearing stress / shearing strain; also called modulus of rigidity; typically G ≈ Y/3 | 172 |
| Bulk modulus (B) | B = −p/(ΔV/V); always positive in equilibrium; unit Pa | 173 |
| Compressibility (k) | Reciprocal of bulk modulus: k = 1/B; fractional change in volume per unit increase in pressure | 173 |
| Poisson's ratio (σ) | Lateral strain / longitudinal strain; dimensionless; steel 0.28–0.30, aluminium alloys ≈ 0.33 | 174 |
| Elastic potential energy density | u = ½ × stress × strain = ½ Y ε² | 174 |
2.3 Diagrams / processes to remember
- Fig. 8.1 (p. 168) — Four cases of deformation: (a) cylinder under tensile stress, elongated by ΔL; (b) cylinder under shearing stress, deformed by angle θ; (c) book pushed horizontally, illustrating shear; (d) sphere under hydraulic stress with ΔV/V but no shape change. A standard CUET match-the-following template.
- Fig. 8.2 (p. 169) — Typical stress-strain curve for a metal showing OA (linear/Hookean), A–B (elastic, non-linear, ending at yield point B), B–D (plastic region; permanent set if unloaded at C), D (ultimate tensile strength), E (fracture). Identifying the labelled regions correctly is a recurring exam item.
- Fig. 8.3 (p. 170) — Stress-strain curve for the elastic tissue of aorta (an elastomer): very large elastic region, mostly non-linear, no clear plastic region.
- Table 8.1 (p. 170) — Young's moduli and yield strengths of common materials (steel highest among the listed metals, hence preferred in heavy-duty structures).
- Table 8.2 (p. 172) — Shear moduli G: lead 5.6, aluminium 25, brass 36, copper 42, iron 70, nickel 77, steel 84, tungsten 150 (×10⁹ N m⁻²).
- Table 8.3 (p. 173) — Bulk moduli B: glass 37, brass 61, aluminium 72, iron 100, copper 140, steel 160, nickel 260 (solids, ×10⁹ Pa); water 2.2, glycerine 4.76, mercury 25 (liquids); air at STP 1.0 × 10⁻⁴.
- Table 8.4 (p. 173) — Consolidated chart of stress type, strain, change in shape/volume, modulus formula, and state of matter.
- Fig. 8.6 (p. 175) — Beam loaded at the centre and supported at the ends (used in deriving δ = W·l³/(4·b·d³·Y)).
- Fig. 8.7 (p. 175) — Cross-sections of a beam: (a) rectangular; (b) deep thin bar buckling; (c) I-section, the engineering compromise.
- Fig. 8.8 (p. 176) — Pillars: rounded ends (less load) vs distributed ends (more load).
2.4 Common confusions / NTA trap points
- "More elastic = stretches more" is wrong. A material which stretches less for a given load is more elastic (NCERT Points to Ponder 6, p. 177). Steel is more elastic than rubber.
- The wire-suspended-from-ceiling trap: ceiling pulls up with F, weight pulls down with F, but tension at any cross-section is just F (not 2F), so tensile stress is F/A (Points to Ponder 1, p. 177).
- Young's modulus and shear modulus apply only to solids (they need a definite length/shape). Bulk modulus is defined for solids, liquids and gases (Points to Ponder 3 & 4, p. 177).
- Yield point B is also called the elastic limit — same point on the curve, two names.
- Stress is not a vector: it cannot be assigned a unique direction (Points to Ponder 8, p. 177). Force is the vector; stress is force per area.
- Strain is dimensionless — beware MCQs offering "m" or "m²" as the unit of strain.
- The negative sign in B = −p/(ΔV/V) is required so that B comes out positive (Δp positive ⇒ ΔV negative).
- For an elastomer like aorta tissue or rubber, the curve is mostly elastic but non-linear, so Hooke's law does not hold over most of it (NCERT §8.4, p. 170).
- Compressibility k = 1/B has units of Pa⁻¹ — a frequent unit trap.
- Don't confuse Poisson's ratio σ with shearing stress σ_s — both use σ in NCERT but mean different things.
- Confusing ultimate tensile strength σ_u (max stress) with yield strength σ_y (onset of plastic region) — they are distinct points on the curve.
- Forgetting that beam depth d enters as d³ in the sag formula δ = W·l³/(4·b·d³·Y) — depth matters far more than breadth.
2.5 Key formulas table
| Symbol | Formula | Meaning | NCERT page |
|---|---|---|---|
| Stress | σ = F/A | Restoring force per unit area | 168 |
| Longitudinal strain | ε = ΔL/L | Fractional change in length | 168 |
| Shearing strain | γ = Δx/L = tan θ ≈ θ | Tangential strain angle | 168 |
| Volume strain | ε_V = ΔV/V | Fractional change in volume | 169 |
| Hooke's law | σ = k × ε | Stress proportional to strain | 169 |
| Young's modulus | Y = (F·L)/(A·ΔL) | Tensile stress / longitudinal strain | 170, Eq. 8.7–8.8 |
| Shear modulus | G = (F·L)/(A·Δx) = F/(A·θ) | Shearing stress / shearing strain | 172, Eq. 8.10–8.11 |
| Bulk modulus | B = −p/(ΔV/V) | Pressure / volume strain | 173, Eq. 8.12 |
| Compressibility | k = 1/B = −(1/Δp)(ΔV/V) | Inverse of bulk modulus | 173, Eq. 8.13 |
| Poisson's ratio | σ = (Δd/d)/(ΔL/L) | Lateral over longitudinal strain | 174 |
| Strain energy density | u = ½ σ ε = ½ Y ε² | PE per unit volume | 174, Eq. 8.14 |
| Total elastic PE | U = ½ × stress × strain × V | Energy stored in stretched body | 174 |
| Shear modulus rule-of-thumb | G ≈ Y/3 | Empirical metal ratio | 172 |
| Yield stress σ_y | σ at point B of stress-strain curve | Onset of plastic flow | 169 |
| Ultimate tensile strength σ_u | σ at point D of stress-strain curve | Max stress before failure | 169 |
| Beam sag | δ = W·l³ / (4·b·d³·Y) | Mid-span sag of beam supported at ends | 175 |
| Mountain-height limit | h ρ g ≤ σ_critical (≈ 3 × 10⁸ Pa) | Upper bound from rock shear strength | 176 |
| Rope minimum radius | A_min = F/σ_y, r = √(A/π) | Crane-rope design | 174 |
| F for length change ΔL | F = Y·A·(ΔL/L) | Inverse of Y formula | 170 |
| Pressure for ΔV/V | Δp = −B·(ΔV/V) | Inverse of B formula | 173 |
🎯 Practice MCQs
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Q1. The SI unit of stress and its dimensional formula are respectively
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Answer: B
Stress = F/A, so its unit is N m⁻² (pascal) and dimensions are [M L⁻¹ T⁻²]. Option (C) gives the dimensions of force, not stress.
Q2. Which of the following statements about strain is correct?
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Answer: C
Strain is a ratio of like quantities; it has no units or dimensions. Shearing strain is Δx/L (not Δx × L), and volume strain is ΔV/V which obviously depends on V.
Q3. On a typical stress-strain curve for a metal, the point at which the material loses its ability to regain its original shape on unloading and just begins to deform permanently is called
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Answer: B
Up to B (the yield point), the body returns to its original dimensions on unloading; beyond B it shows plastic deformation. Ultimate tensile strength (D) and fracture (E) lie further along the curve.
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Q4. A steel rod of cross-sectional area 1.0 × 10⁻⁴ m² is stretched along its length by a force of 2.0 × 10⁴ N. Given that Young's modulus of steel is 2.0 × 10¹¹ N m⁻², the longitudinal strain produced in the rod is
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Answer: C
Stress = F/A = (2.0 × 10⁴)/(1.0 × 10⁻⁴) = 2.0 × 10⁸ N m⁻². Strain = stress/Y = (2.0 × 10⁸)/(2.0 × 10¹¹) = 1.0 × 10⁻³.
Q5. The wires of aluminium, brass, copper and steel each have the same cross-sectional area (0.1 cm²) and the same length. To increase the length of each by 0.1%, the forces required are 690 N, 900 N, 1100 N and 2000 N respectively. Which of the following is the correct ordering of these materials in increasing order of Young's modulus?
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Answer: B
For the same strain and same A, Y is proportional to the required force. Forces are 690 (Al) < 900 (Br) < 1100 (Cu) < 2000 (St), so Y orders the same way — steel is the most elastic of the four.
Q6. Match List I (elastic modulus) with List II (defining ratio) and choose the correct option. | List I | List II | |---|---| | (P) Young's modulus | (i) −p/(ΔV/V) | | (Q) Shear modulus | (ii) F/(A·θ) | | (R) Bulk modulus | (iii) (F·L)/(A·ΔL) | | (S) Compressibility | (iv) 1/B |
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Answer: A
Y is tensile stress over longitudinal strain ⇒ (F·L)/(A·ΔL); G can be written F/(A·θ); B = −p/(ΔV/V); compressibility k = 1/B.
Q7. Assertion (A): In the relation B = −p/(ΔV/V), the negative sign is essential. Reason (R): For a system in equilibrium, the bulk modulus B is always taken as a positive quantity, while an increase in pressure produces a decrease in volume.
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Answer: A
Because Δp and ΔV have opposite signs, the ratio p/(ΔV/V) is negative. Inserting the minus sign keeps B positive — exactly the reason given in R.
Q8. Which of the following statements is/are correct about elastomers (such as the elastic tissue of the aorta or rubber)? I. They can be stretched to several times their original length and still return to the original shape. II. They obey Hooke's law over most of their elastic region. III. They do not have a well-defined plastic region.
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Answer: B
Elastomers can be stretched to very large strains and recover (I correct), but they do *not* obey Hooke's law over most of the region (II wrong) and have no well-defined plastic region (III correct).
Q9. The elastic potential energy stored per unit volume of a stretched wire is
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Answer: B
u = ½ × stress × strain = ½ Y ε². Option (C) is twice the correct value; option (A) is dimensionally wrong.
Q10. Among the following, the material with the greatest bulk modulus is
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Answer: C
Steel B = 160 × 10⁹ Pa; glass 37 × 10⁹ Pa; water 2.2 × 10⁹ Pa; air ≈ 10⁵ Pa. Solids dominate over liquids, which in turn dominate over gases.
Q11. A horizontal beam of length l, breadth b and depth d is supported at both ends and loaded at the centre with weight W. The mid-span sag δ is
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Answer: C
δ = W·l³/(4·b·d³·Y). Sag grows as the cube of the length but falls as the cube of the depth — that is why beams are made deep, and (with the buckling caveat) why I-sections are used.
Q12. The maximum height of a mountain on Earth (~10 km) is essentially limited by
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Answer: C
The weight of a rock column hρg must not exceed the rock's shear elastic limit (~3 × 10⁸ N m⁻²); otherwise the material flows plastically. This caps h at about 10 km.
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