Class 11 Physics · Part IINCERT 2026–27

Chapter 8: Mechanical Properties of Solids

Study elasticity, stress, strain, and elastic moduli with worked examples and exercise solutions.

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Learn the chapter

Study elasticity, stress, strain and elastic moduli alongside the examples that use them.

8.1

Elasticity, plasticity and rigid-body limits

  • A rigid body is an idealisation: real solids can stretch, compress or bend under a sufficient force. A solid has a definite shape and size, but is not perfectly rigid.
  • Elasticity: tendency to recover the original shape and size after the deforming force is removed. A gently stretched spring exhibits elastic deformation.
  • Plasticity: permanent deformation after unloading. Putty and mud are close to ideal plastics.
  • Elastic properties guide buildings, bridges, automobiles, ropeways, aircraft and artificial limbs: stiffness, strength and weight must be balanced.
8.2

Stress and three types of strain

  • In static equilibrium, an applied deforming load is balanced by internal restoring forces. Stress is restoring force per unit area: $\sigma=F/A$. Its SI unit is pascal ($1\,\mathrm{Pa}=1\,\mathrm{N\,m^{-2}}$); dimensions $[ML^{-1}T^{-2}]$.
  • Longitudinal: forces normal to the cross-section produce tensile or compressive stress. Longitudinal strain $\varepsilon=\Delta L/L$ (positive for extension, negative for compression).
  • Shear: tangential forces produce relative face displacement $\Delta x$. Shearing strain $\gamma=\Delta x/L=\tan\theta\approx\theta$ for small angles in radians; at $10^\circ$, the approximation differs by about 1%. Balance both total force and torque in a static shear arrangement.
  • Hydraulic: equal pressure normal to every surface compresses volume without changing geometrical shape. Hydraulic stress equals pressure; volume strain $\varepsilon_V=\Delta V/V$.
  • All strains are ratios: they have no units and are dimensionless. Stress is not an ordinary force vector; its components depend on the chosen internal surface.
LongitudinalShear · ΔxHydraulic
Fig. 8.1 · Three idealised modes of deformation. A pushed book illustrates shear.
Do not double the tension: two opposite end forces each of magnitude $F$ give internal tension $F$, so stress is $F/A$, not $2F/A$.
8.3

Hooke’s law and its domain

  • For small deformations in the proportional region, $\text{stress}=k\,\text{strain}$; $k$ is the appropriate modulus of elasticity.
  • Hooke’s law is empirical, established experimentally; it is not a universal theorem to be proved. Some elastic materials have a nonlinear response.
  • The proportional limit and elastic limit need not coincide: elasticity means recovery, while Hooke’s law means a linear stress–strain relationship.
  • A modulus has stress units (Pa) because strain has no units. It characterises the material under the stated conditions, not the size of the specimen.
8.4

Stress–strain curve, strength and elastomers

Strain εStress σOABCDEPermanent set
Fig. 8.2 · Schematic metal curve; the dashed unloading path leaves permanent strain. Not to scale.
  • Experimental method: increase a wire/cylinder’s tensile load in steps, measure extension, calculate stress and strain, then plot stress vertically against strain horizontally. Compression and shear curves are analogous.
  • OA: straight line; Hooke’s law holds and unloading restores the original dimensions. Its slope gives Young’s modulus.
  • AB: nonlinear but still recoverable. B is the textbook’s yield point/elastic limit; the corresponding stress is yield strength.
  • Beyond B: large strain develops for small stress changes. Unloading from C leaves a permanent set; this is plastic deformation.
  • D: ultimate tensile strength (largest engineering stress). Beyond D, further strain can occur with a smaller force; E is fracture.
  • D and E close: brittle material. Far apart: ductile material. Stiffness (initial slope), yield strength and ultimate strength are different properties.
Strain εStress σOLarge elastic regionVariable slope: not Hookean
Fig. 8.3 · Aorta tissue and rubber can be elastic over large strains without a linear stress–strain relationship. Schematic.
  • Rubber and aorta tissue are elastomers: large recoverable strain, a nonlinear curve over most of the elastic region, and no well-defined plastic region.
  • An elastic material need not obey Hooke’s law; “stretches easily” is not the physics definition of “more elastic”.
8.5.1

Young’s modulus: axial extension and compression

$$Y=\frac{\sigma}{\varepsilon}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L},\qquad |\Delta L|=\frac{FL}{AY}$$
  • Derivation: divide longitudinal stress by longitudinal strain. Use magnitudes when calculating a compression; its signed strain is negative.
  • Within the linear range, the same magnitude of tensile or compressive stress produces the same magnitude of longitudinal strain for a given material.
  • Larger Y means smaller strain for the same stress: steel is stiffer and, in this sense, more elastic than copper, brass or aluminium. A larger Y does not by itself imply a larger breaking stress.
  • For a wire, $\Delta L\propto F L/A$; with $A=\pi r^2$, extension varies as $1/r^2$. A wire acts like a spring with axial stiffness $k_{\rm wire}=YA/L$.
  • Reference comparison: to produce $0.1\%$ strain in a $0.1\,\mathrm{cm^2}$ wire, the stated forces are steel 2000 N, aluminium 690 N, brass 900 N and copper 1100 N. Values are approximate and depend on composition.
Table 8.1 · NCERT material data
MaterialDensity (kg m⁻³)Y (GPa)Ultimate strength (MPa)Yield strength (MPa)
Aluminium27107011095
Copper8890110400200
Iron (wrought)7800–7900190330170
Steel7860200400250
Glass #21906550—
Concrete23203040—
Wood #5251350—
Bone #19009.4170—
Polystyrene1050348—
“#” denotes specimens tested under compression. These are material-specific reference values, not universal strength rankings. For the brass-wire exercises, use Y = 90 GPa; the text’s crane calculation separately uses mild-steel yield strength 300 MPa.
NCERT Example 8.1

Stress, elongation and strain

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) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel is $2.0\times10^{11}\,\mathrm{N\,m^{-2}}$.

Step 1. Convert: $r=0.010\,\mathrm m$, $F=1.00\times10^5\,\mathrm N$; $A=\pi r^2=3.14\times10^{-4}\,\mathrm{m^2}$.
Step 2. Stress: $\sigma=F/A=3.18\times10^8\,\mathrm{Pa}$.
Step 3. Extension: $\Delta L=\sigma L/Y=1.59\times10^{-3}\,\mathrm m=1.59\,\mathrm{mm}$.
Step 4. Strain: $\varepsilon=\Delta L/L=1.59\times10^{-3}\approx0.16\%$.
$$\Delta L_{\rm series}=F\sum_i\frac{L_i}{A_iY_i}$$
  • Wires connected end to end carry the same tension (neglect their weights); total extension is the sum of the individual extensions.
  • Parallel wires attached to a rigid level bar undergo the same extension if their original lengths are equal; their tensions divide in proportion to AY/L.
NCERT Example 8.2

Two wires in series

A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.

Step 1. The same tension $W$ acts in both wires; $A=\pi(1.5\times10^{-3})^2=7.07\times10^{-6}\,\mathrm{m^2}$. Use $Y_c=1.1\times10^{11}$ and $Y_s=2.0\times10^{11}\,\mathrm{Pa}$.
Step 2. $\Delta L_c/\Delta L_s=(L_c/Y_c)/(L_s/Y_s)=2.5$. With $\Delta L_c+\Delta L_s=7.0\times10^{-4}\,\mathrm m$, obtain $\Delta L_c=5.0\times10^{-4}\,\mathrm m$ and $\Delta L_s=2.0\times10^{-4}\,\mathrm m$.
Step 3. $W=AY_c\Delta L_c/L_c=1.77\times10^2\,\mathrm N\approx1.8\times10^2\,\mathrm N$.
NCERT Example 8.3

Compression of thighbones

In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig. 8.4). The combined mass of all the persons performing the act, and the tables, plaques etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.

Extra load: (280 − 60) kg1078 N1078 NTwo femurs · L = 0.50 m
Fig. 8.4 · Load schematic: the bottom performer’s own mass is excluded from the extra load.
Step 1. Supported weight: $(280-60)g=220(9.8)=2156\,\mathrm N$. Each bone carries $F=1078\,\mathrm N$.
Step 2. $L=0.50\,\mathrm m$, $r=0.020\,\mathrm m$, $A=\pi r^2=1.26\times10^{-3}\,\mathrm{m^2}$; $Y_{\rm bone}=9.4\times10^9\,\mathrm{Pa}$.
Step 3. Compression magnitude: $|\Delta L|=FL/(AY)=4.56\times10^{-5}\,\mathrm m\approx4.55\times10^{-3}\,\mathrm{cm}$.
Step 4. Fractional compression: $|\Delta L|/L\approx9.1\times10^{-5}=0.0091\%$.

See Young’s modulus: stretch two wires

The same tensile load extends copper more than steel because its Young’s modulus is lower.

Interactive 3D model
Steel · less extensionCopper · more extension

Calculated extensions appear here.

Both model wires have length 2 m and diameter 10 mm. Shape extension is exaggerated 80× so the change is visible; read the calculated millimetres for the actual extension.

8.5.2

Shear modulus (modulus of rigidity)

$$G=\frac{F/A}{\Delta x/L}=\frac{FL}{A\Delta x}\approx\frac{F}{A\theta},\qquad \tau\approx G\theta$$
  • Derivation: divide tangential stress by shear strain; replace tan θ by θ only at small angles. The area is the face parallel to which the shear force acts; L is the separation of the sliding faces.
  • Units: Pa. Only solids can support a sustained static shear stress; fluids flow under it. A helical coil stretches chiefly because its wire twists, so coil extension depends on shear modulus.
  • Typically $G$ is smaller than $Y$. The textbook estimate $G\approx Y/3$ is approximate, not an exact universal identity.
Table 8.2 · Shear moduli
MaterialG (GPa)
Aluminium25
Brass36
Copper42
Glass23
Iron70
Lead5.6
Nickel77
Steel84
Tungsten150
Wood10
NCERT Example 8.4

Shear displacement of lead

A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of $9.0\times10^4\,\mathrm N$. The lower edge is riveted to the floor. How much will the upper edge be displaced?

Step 1. The sheared narrow face has $A=0.50(0.10)=0.050\,\mathrm{m^2}$; the separation of top and bottom is $L=0.50\,\mathrm m$.
Step 2. $\tau=F/A=9.0\times10^4/0.050=1.80\times10^6\,\mathrm{Pa}$; use $G_{\rm lead}=5.6\times10^9\,\mathrm{Pa}$.
Step 3. $\Delta x=\tau L/G=1.61\times10^{-4}\,\mathrm m\approx0.16\,\mathrm{mm}$.
Source correction: the printed working shows 9.4 × 10⁴ N, but the question gives 9.0 × 10⁴ N. The stated stress and displacement agree with 9.0 × 10⁴ N.

See shear strain: slide the top face

The bottom stays fixed while a tangential force shifts the top face.

Interactive 3D model
Shear: top displacement Δx, height L

Shear strain and face shift appear here.

For small angles, γ = Δx/L ≈ θ when θ is measured in radians. The model exaggerates displacement to show the changing shape.

8.5.3

Bulk modulus and compressibility

$$B=-\frac{\Delta p}{\Delta V/V},\qquad \frac{\Delta V}{V}=-\frac{\Delta p}{B},\qquad \kappa=\frac1B=-\frac{1}{\Delta p}\frac{\Delta V}{V}$$
  • An increase in pressure produces a decrease in volume: Δp > 0 and signed ΔV < 0. The minus sign makes B positive. Fractional compression is the positive magnitude |ΔV|/V.
  • B applies to solids, liquids and gases under uniform hydraulic stress. Compressibility κ has unit Pa⁻¹; smaller κ means harder to compress.
  • Neighbouring atoms are tightly coupled in solids, less strongly coupled in liquids and poorly coupled in gases. Typical solids are least compressible, liquids intermediate and gases most compressible.
  • Gas compressibility depends strongly on pressure and temperature; typical gases can be about a million times more compressible than solids.
Table 8.3 · Bulk moduli
MaterialStateB (GPa)
AluminiumSolid72
BrassSolid61
CopperSolid140
GlassSolid37
IronSolid100
NickelSolid260
SteelSolid160
WaterLiquid2.2
EthanolLiquid0.9
Carbon disulphideLiquid1.56
GlycerineLiquid4.76
MercuryLiquid25
Air (at STP)Gas$1.0\times10^{-4}$
NCERT Example 8.5

Compression of ocean water

The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, $\Delta V/V$, of water at the bottom of the ocean, given that the bulk modulus of water is $2.2\times10^9\,\mathrm{N\,m^{-2}}$. (Take $g=10\,\mathrm{m\,s^{-2}}$)

Step 1. Pressure increase relative to the surface: $\Delta p=\rho gh=1000(10)(3000)=3.0\times10^7\,\mathrm{Pa}$.
Step 2. Signed volume strain: $\Delta V/V=-\Delta p/B=-1.36\times10^{-2}$.
Step 3. Fractional compression magnitude is $|\Delta V|/V=1.36\times10^{-2}=1.36\%$. The negative signed strain means volume decreases.
Table 8.4 · Idealised deformation modes
ModeApplied forceStrainShape / volumeModulusMatter
LongitudinalNormal to opposite faces$\Delta L/L$Shape changes; volume change neglected in this simplified axial modelYSolid
ShearParallel to opposite faces; force and torque balanced$\Delta x/L\approx\theta$Shape changes; volume unchanged to first orderGSolid
HydraulicUniform pressure normal to every surface$\Delta V/V$Shape retained; volume changesBSolid / liquid / gas
Real axial deformation can also produce lateral strain and a volume change. The simplified mode table does not override Poisson’s effect.
8.5.4

Poisson’s ratio and lateral contraction

$$\nu=\frac{\Delta d/d}{\Delta L/L}=\frac{\Delta d}{\Delta L}\frac{L}{d}$$
  • Here Δd is the positive magnitude of diameter contraction and ΔL is extension. Lateral contraction accompanies longitudinal stretching.
  • With signed strains, the equivalent definition is $\nu=-\varepsilon_{\perp}/\varepsilon_{\parallel}$. Do not omit this minus sign if the lateral strain is negative.
  • Poisson’s ratio is a pure number, independent of specimen dimensions in the elastic regime. Typical values: steel 0.28–0.30; aluminium alloys about 0.33.
  • A force in one direction can produce strains in other directions, so one elastic constant cannot describe every deformation.
8.5.5

Proof: elastic potential energy in a wire

  • Work is done against interatomic restoring forces and stored as elastic potential energy. Stretch slowly within the linear elastic region. The force grows from zero with the instantaneous extension x; the final force must not be used as a constant throughout the work integral.
Step 1. From Young’s modulus, $F(x)=YAx/L$.
Step 2. Integrate: $U=W=\int_0^{\Delta L}F(x)\,dx=\frac{YA}{L}\int_0^{\Delta L}x\,dx=\frac{YA(\Delta L)^2}{2L}=\frac12F\Delta L$.
Step 3. Since wire volume is $AL$ and $\varepsilon=\Delta L/L$, $U=\frac12Y\varepsilon^2(AL)=\frac12\sigma\varepsilon(AL)$.
$$u=\frac{U}{AL}=\frac12\sigma\varepsilon=\frac12Y\varepsilon^2=\frac{\sigma^2}{2Y}$$
Energy density u has unit J m⁻³, numerically equivalent to Pa. It is the area under the linear stress–strain graph. For nonlinear elastic loading, integrate the actual curve rather than assuming a triangular area.
8.6

Applications: crane ropes and safe loads

  • Use yield strength to avoid permanent deformation; use ultimate strength only when considering breaking. Structural design also accounts for traffic, wind, self-weight, reliability, service life and cost.
Step 1. For a load $W=Mg$, require $W/A\le\sigma_y$, hence $A\ge Mg/\sigma_y$.
Step 2. For a 10-tonne load and mild steel with $\sigma_y=300\times10^6\,\mathrm{Pa}$, $A_{\min}=10^4(9.8)/(300\times10^6)=3.27\times10^{-4}\,\mathrm{m^2}$.
Step 3. Circular radius $r_{\min}=\sqrt{A/\pi}\approx1.02\,\mathrm{cm}$. A load safety factor 10 multiplies area by 10 and radius by $\sqrt{10}$: $r\approx3.2\,\mathrm{cm}$ (about 3 cm).
  • A thick single wire behaves like a rigid rod. Braided thin wires provide practical flexibility, ease of manufacture and strength.
8.6

Proof: beam bending and I-section design

$$\delta=\frac{Wl^3}{4Ybd^3}$$
  • For a rectangular beam: span l, breadth b, vertical depth d; simply supported near its ends and carrying a central load W. The formula assumes small elastic bending, a uniform beam and negligible shear deflection.
  • Increase Y or depth d; reduce span l. Doubling depth reduces sag by 8, doubling breadth reduces it by 2, and doubling span increases it by 8.
Step 1. For a rectangular section, the second moment of area about the neutral axis is $I=\int y^2\,dA=b\int_{-d/2}^{d/2}y^2\,dy=bd^3/12$.
Curvature relation. A fibre at distance y from the neutral axis has bending strain $\varepsilon=-y/R$ and stress $\sigma=-Yy/R$. The bending-moment magnitude is $M=\left|\int\sigma y\,dA\right|=(Y/R)\int y^2\,dA=YI/R$. For small slopes, $1/R\approx|v''(x)|$, so $|M|=YI|v''|$.
Step 2. Each support reaction is $W/2$. For $0\le x\le l/2$, bending moment $M(x)=Wx/2$. In the small-slope limit, the curvature magnitude follows $YI\,v''(x)=Wx/2$ (choose a consistent deflection sign).
Step 3. Integrate: $v'(x)=Wx^2/(4YI)+C_1$. Symmetry gives $v'(l/2)=0$, hence $C_1=-Wl^2/(16YI)$.
Step 4. Integrate again and use $v(0)=0$: $v(x)=Wx^3/(12YI)-Wl^2x/(16YI)$. Therefore $\delta=|v(l/2)|=Wl^3/(48YI)=Wl^3/(4Ybd^3)$.
RectangularThin deep barI-section
Fig. 8.7 · A deep section resists bending; broad flanges help resist lateral instability while saving material.
  • A very thin deep beam can buckle sideways, especially under an off-centre moving load. An I-section combines useful depth and broad flanges with lower weight and cost.
  • Columns with well-distributed end support carry larger stable loads than comparable rounded/pinned-end arrangements; end restraint and load alignment affect buckling.
  • Pyramidal hills/heaps and broad structural bases spread loads. A heap of sand also depends on granular friction; it is not a uniform elastic beam.

See beam bending under a centre load

Raise the load and watch the centre sag between the supports.

Interactive 3D model
Supports · central load W · centre sag δ

Calculated centre sag appears here.

The displayed bend is exaggerated 60×. The readout uses δ = Wl³/(48YI), for a simply supported beam with a central load.

8.6

Proof: maximum mountain height

  • The sides of a mountain are free: the base is not under uniform hydraulic pressure. Its self-weight produces a shear component of roughly the same order as the vertical stress.
Step 1. For a column of rock of height h and base area A: mass $m=\rho Ah$, so weight per unit area is $mg/A=\rho gh$.
Step 2. Require the characteristic shear stress to stay below the critical rock strength: $\rho gh\lesssim\tau_{\rm critical}$. Thus $h_{\max}\sim\tau_{\rm critical}/(\rho g)$.
Step 3. With $\tau_{\rm critical}=30\times10^7\,\mathrm{Pa}$, $\rho=3\times10^3\,\mathrm{kg\,m^{-3}}$ and $g\approx10\,\mathrm{m\,s^{-2}}$, $h\sim10^4\,\mathrm m=10\,\mathrm{km}$.
This is an order-of-magnitude estimate, not an exact maximum for every rock or planet; it explains the terrestrial mountain-height scale.
Quiz

Concept quiz

Choose an answer to see the reasoning and key idea.

0 / 29 Answered0 (0%)

8.1 Elasticity, plasticity and rigid-body limits

1. Recovery after removal of a deforming force is called:
Elasticity means recovery of the original shape and size. Putty illustrates permanent plastic deformation.
Review: Elasticity means recovery of the original shape and size. Putty illustrates permanent plastic deformation.
19. Why is a rigid body described as an idealisation in this chapter?
Real solids stretch, compress, shear and bend. Treating one as rigid is a useful approximation when those deformations can be neglected.
Remember: Elastic properties quantify how real solids respond to load.

8.2 Stress and three types of strain

2. Two opposite end forces, each F, stretch a wire of area A. Its tensile stress is:
Internal tension equals F, not the sum of the two opposite external forces.
Review: Internal tension equals F, not the sum of the two opposite external forces.
3. Which quantity is dimensionless?
Strain is a change in dimension divided by the original dimension. Stress and all moduli have units Pa.
Review: Strain is a change in dimension divided by the original dimension. Stress and all moduli have units Pa.
4. For a small shear angle measured in radians, the shear strain is:
Geometry gives γ = Δx/L = tan θ. The small-angle approximation requires θ in radians.
Review: Geometry gives γ = Δx/L = tan θ. The small-angle approximation requires θ in radians.

8.3 Hooke’s law and its domain

20. Hooke’s law relates stress and strain in which region?
Hooke’s law is an experimentally observed linear relation within the proportional limit; it is not a universal law for every deformation.
Remember: Do not extend the initial straight-line relation beyond its domain.
21. Which statement correctly distinguishes elastic behaviour from Hooke’s law?
Elasticity means recovery; Hooke’s law means stress is proportional to strain. The AB region of a stress–strain curve can be nonlinear yet recoverable.
Remember: Recoverable does not always mean linear.

8.4 Stress–strain curve, strength and elastomers

5. An elastic but nonlinear region of a metal stress–strain curve is:
AB is recoverable but stress is not proportional to strain. Elasticity and Hooke’s law are different conditions.
Review: AB is recoverable but stress is not proportional to strain. Elasticity and Hooke’s law are different conditions.
6. Which determines Young’s modulus from a tensile stress–strain graph?
Y = σ/ε in the linear range. Maximum stress measures ultimate strength, not stiffness.
Review: Y = σ/ε in the linear range. Maximum stress measures ultimate strength, not stiffness.
7. A material with nearby ultimate-strength and fracture points is:
A small additional strain between ultimate strength and fracture indicates brittle behaviour.
Review: A small additional strain between ultimate strength and fracture indicates brittle behaviour.
8. Rubber and aorta tissue illustrate:
Elastomers can recover after very large strains without obeying Hooke’s law over most of that range.
Review: Elastomers can recover after very large strains without obeying Hooke’s law over most of that range.

8.5.1 Young’s modulus: axial extension and compression

9. If a wire’s radius doubles at the same load, length and Y, its extension becomes:
ΔL = FL/(πr²Y); doubling radius multiplies area by four.
Review: ΔL = FL/(πr²Y); doubling radius multiplies area by four.
22. Two wire segments connected in series carry the same axial load. Their total extension is:
The tension is common to both series segments, while their extensions add. For segment i, ΔLᵢ = FLᵢ/(AᵢYᵢ).
Remember: Series: same force and additive extensions. Parallel equal-length wires: same extension.

8.5.2 Shear modulus (modulus of rigidity)

10. A helical coil spring’s extension primarily depends on its wire’s:
The wire twists as the helical spring stretches, so shear rigidity governs this deformation.
Review: The wire twists as the helical spring stretches, so shear rigidity governs this deformation.
23. For shear modulus G = FL/(AΔx), what are A and L?
Shear stress is tangential force divided by the sheared face area. Shear strain is displacement divided by the separation of the faces.
Remember: Match the area and length to the geometry before substituting.

8.5.3 Bulk modulus and compressibility

11. For pressure increase Δp > 0, signed volume strain is:
Pressure increase contracts volume. The minus sign gives a positive bulk modulus.
Review: Pressure increase contracts volume. The minus sign gives a positive bulk modulus.
12. Which statement about elastic moduli is correct?
Static axial/shear stiffness is a solid property; bulk compression applies to all three states.
Review: Static axial/shear stiffness is a solid property; bulk compression applies to all three states.
24. Bulk compressibility κ is related to bulk modulus B by:
B = −Δp/(ΔV/V), so κ = 1/B. A pressure increase gives negative signed volume strain and positive compressibility.
Remember: A larger B means a smaller compressibility.

8.5.4 Poisson’s ratio and lateral contraction

13. With signed strains, Poisson’s ratio is:
Axial extension is positive while lateral contraction is negative; the signed definition includes a minus sign.
Review: Axial extension is positive while lateral contraction is negative; the signed definition includes a minus sign.
25. A wire is stretched so its diameter contracts. If positive Δd denotes the contraction magnitude, Poisson’s ratio is:
When transverse strain is written with its signed negative value, ν = −ε⊥/ε∥. If Δd is defined as a positive contraction magnitude, the ratio is positive without an extra minus.
Remember: State whether lateral change is signed or a positive magnitude.

8.5.5 Proof: elastic potential energy in a wire

14. In linear elastic stretching, energy stored in a wire is:
Force grows linearly from zero; integrating the force–extension triangle gives U = ½FΔL.
Review: Force grows linearly from zero; integrating the force–extension triangle gives U = ½FΔL.
26. For a linearly elastic wire, the elastic energy density is:
The force rises linearly from zero, so U = ∫F dx = ½FΔL. Dividing by the wire volume gives u = ½σε = ½Yε² = σ²/(2Y).
Remember: The triangular area under the loading graph gives the factor ½.

8.6 Applications: crane ropes and safe loads

16. For crane-rope area chosen to avoid permanent deformation, use:
Require σ ≤ σ_y and include a safety margin. Stiffness and allowable strength are distinct.
Review: Require σ ≤ σ_y and include a safety margin. Stiffness and allowable strength are distinct.
17. A factor of ten in allowable load multiplies a circular rope’s required radius by:
Area scales with load; radius scales as the square root of area.
Review: Area scales with load; radius scales as the square root of area.
27. For design load W and safety factor n, an ideal rope area based on yield strength σᵧ should satisfy:
Limit working stress to σᵧ/n: W/A ≤ σᵧ/n. Rearranging gives A ≥ nW/σᵧ; a separate practical design also checks other loads and service conditions.
Remember: Use yield strength to prevent permanent deformation, with a safety margin.

8.6 Proof: beam bending and I-section design

15. If a rectangular beam’s depth doubles, its sag under the same central load becomes:
δ ∝ d⁻³ for the stated rectangular simply supported beam.
Review: δ ∝ d⁻³ for the stated rectangular simply supported beam.
28. Why does an I-section beam place much of its material far from the neutral axis?
I = ∫y²dA weights area by distance squared from the neutral axis. Flanges far from that axis add bending rigidity efficiently; the web joins them and carries shear.
Remember: Material farther from the neutral axis contributes more strongly to I.

8.6 Proof: maximum mountain height

18. A mountain base is approximately limited by:
Free sides make the loading non-hydraulic. Comparing ρgh with critical rock shear strength gives the height scale.
Review: Free sides make the loading non-hydraulic. Comparing ρgh with critical rock shear strength gives the height scale.
29. The order-of-magnitude mountain-height estimate follows from which balance?
Free mountain sides make the loading non-hydrostatic. Compare the self-weight stress scale ρgh with critical rock strength; typical values give a scale near 10 km, not an exact universal limit.
Remember: This is an order-of-magnitude estimate for typical terrestrial rock.
Exercises

NCERT Exercises

Try each question before opening its worked solution.

Ex 8.1young

A steel wire of length 4.7 m and cross-sectional area $3.0\times10^{-5}\,\mathrm{m^2}$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0\times10^{-5}\,\mathrm{m^2}$ under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Prerequisite:Review the matching concept
Step 1. Same force and extension: $FL_s/(A_sY_s)=FL_c/(A_cY_c)$.
Step 2. $Y_s/Y_c=L_sA_c/(L_cA_s)=4.7(4.0)/(3.5(3.0))=1.79$.
Ex 8.2curve

Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?

00.0010.0020.0030.004050100150200250300StrainStress (10⁶ N m⁻²)
Fig. 8.9 · Redrawn from the NCERT graph. Use the straight-line slope for Y; the yield stress is approximately 3 × 10⁸ Pa.
Prerequisite:Review the matching concept
Step 1. In the linear region, $(\varepsilon,\sigma)=(0.002,150\times10^6\,\mathrm{Pa})$. Thus $Y=\sigma/\varepsilon=7.5\times10^{10}\,\mathrm{Pa}$.
Step 2. The departure from the linear region is near $\sigma\approx300\times10^6\,\mathrm{Pa}$; approximate yield strength $\approx3\times10^8\,\mathrm{Pa}$. Read this as a graph estimate, not an exact material constant.
Ex 8.3curve

The stress-strain graphs for materials A and B are shown in Fig. 8.10. The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?

ABStrainStrainStressStress
Fig. 8.10 · Same axis scales. Compare initial slopes for stiffness and maximum stresses for strength.
Prerequisite:Review the matching concept
Step 1. (a) A has the steeper initial straight-line slope, so A has the larger Young’s modulus.
Step 2. (b) B reaches the greater maximum stress on the same scale, so B is the stronger material. Do not compare slopes to decide ultimate strength.
Ex 8.4curve · shear

Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.

Prerequisite:Review the matching concept
Step 1. (a) False. Steel has a much larger Young’s modulus; rubber stretches more for the same stress.
Step 2. (b) True for a helical coil spring: axial extension twists the spring wire, so its shear modulus governs the response.
Ex 8.5young

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 steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Steel · 1.5 m4.0 kgBrass · 1.0 m6.0 kg
Fig. 8.11 · Steel supports both masses (10 kg); brass supports only the lower mass (6 kg).
Prerequisite:Review the matching concept
Step 1. $A=\pi(0.0025)^2/4=4.91\times10^{-6}\,\mathrm{m^2}$. Steel carries $(4+6)g=98\,\mathrm N$; brass carries $6g=58.8\,\mathrm N$.
Step 2. Use $Y_s=2.0\times10^{11}\,\mathrm{Pa}$ and $Y_b=9.0\times10^{10}\,\mathrm{Pa}$.
Step 3. $\Delta L_s=98(1.5)/(AY_s)=1.50\times10^{-4}\,\mathrm m=0.150\,\mathrm{mm}$.
Step 4. $\Delta L_b=58.8(1.0)/(AY_b)=1.33\times10^{-4}\,\mathrm m=0.133\,\mathrm{mm}$.
Ex 8.6shear

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?

Prerequisite:Review the matching concept
Step 1. $F=mg=980\,\mathrm N$, $A=(0.10)^2=0.010\,\mathrm{m^2}$, $L=0.10\,\mathrm m$, $G=25\times10^9\,\mathrm{Pa}$.
Step 2. $\Delta x=FL/(AG)=3.92\times10^{-7}\,\mathrm m=0.392\,\mathrm{\mu m}$.
Ex 8.7young

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 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Prerequisite:Review the matching concept
Step 1. Each column carries $F=Mg/4=50000(9.8)/4=1.225\times10^5\,\mathrm N$.
Step 2. $A=\pi(R^2-r^2)=\pi(0.60^2-0.30^2)=0.848\,\mathrm{m^2}$.
Step 3. With $Y=2.0\times10^{11}\,\mathrm{Pa}$, compression magnitude $|\varepsilon|=F/(AY)=7.22\times10^{-7}$. Signed strain is negative.
Ex 8.8young

A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Prerequisite:Review the matching concept
Step 1. $A=(15.2\times10^{-3})(19.1\times10^{-3})=2.9032\times10^{-4}\,\mathrm{m^2}$.
Step 2. Use $Y_c=1.1\times10^{11}\,\mathrm{Pa}$. $\varepsilon=F/(AY)=44500/[2.9032\times10^{-4}(1.1\times10^{11})]=1.39\times10^{-3}$.
Ex 8.9applications

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 $10^8\,\mathrm{N\,m^{-2}}$, what is the maximum load the cable can support ?

Prerequisite:Review the matching concept
Step 1. $A=\pi(0.015)^2=7.07\times10^{-4}\,\mathrm{m^2}$.
Step 2. Maximum load (force) $W_{\max}=\sigma_{\max}A=7.07\times10^4\,\mathrm N$. Corresponding supported mass $M=W/g\approx7.21\times10^3\,\mathrm{kg}$.
Ex 8.10young

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 ratios of their diameters if each is to have the same tension.

Prerequisite:Review the matching concept
Step 1. Symmetry and rigidity keep the bar level: equal original lengths therefore have equal extension. Equal tensions require $A_cY_c=A_iY_i$, hence $d_c^2Y_c=d_i^2Y_i$.
Step 2. Using Table 8.1, $Y_i=190\,\mathrm{GPa}$ and $Y_c=110\,\mathrm{GPa}$, $d_c/d_i=\sqrt{190/110}=1.31$. The diameter ratio is copper : iron : copper $=1.31:1:1.31$.
Step 3. Each tension is $Mg/3=49\,\mathrm N$, but the mass cancels from the diameter ratio.
Ex 8.11young

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 $0.065\,\mathrm{cm^2}$. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Prerequisite:Review the matching concept
Step 1. At the bottom, $T-mg=m\omega^2R$. Convert $\omega=4\pi\,\mathrm{rad\,s^{-1}}$, $A=6.5\times10^{-6}\,\mathrm{m^2}$; for a small extension use $R\approx L=1.0\,\mathrm m$.
Step 2. $T=m(g+\omega^2L)=14.5[9.8+(4\pi)^2]=2.432\times10^3\,\mathrm N$.
Step 3. $\Delta L=TL/(AY)=1.87\times10^{-3}\,\mathrm m=1.87\,\mathrm{mm}$ with $Y=2.0\times10^{11}\,\mathrm{Pa}$.
Step 4. If the radius increase is included, substitute $R=L+\Delta L$: $\Delta L=\frac{mL(g+\omega^2L)}{AY-m\omega^2L}\approx1.874\,\mathrm{mm}$. The school approximation is consistent.
Ex 8.12bulk

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = $1.013\times10^5\,\mathrm{Pa}$), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Prerequisite:Review the matching concept
The question is preserved as supplied. “Final volume = 100.5 litre” is inconsistent with a pressure increase: it would give a negative bulk modulus. For the intended compression calculation, use a volume decrease of 0.5 litre (final volume 99.5 litre).
Step 1. $\Delta p=100(1.013\times10^5)=1.013\times10^7\,\mathrm{Pa}$; $\Delta V/V=-0.5/100=-0.005$.
Step 2. $B=-\Delta p/(\Delta V/V)=2.026\times10^9\,\mathrm{Pa}$.
Step 3. For air at constant temperature, $pV=\text{constant}$ gives $dp/dV=-p/V$ and $B_{\rm air}=-V(dp/dV)=p$. At 1 atm, $B_{\rm air}=1.013\times10^5\,\mathrm{Pa}$. Thus $B_{\rm water}/B_{\rm air}=2.0\times10^4$.
Step 4. Liquid molecules are close together and more strongly coupled; gases contain much more empty space, so their volume decreases far more readily.
Ex 8.13bulk

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03\times10^3\,\mathrm{kg\,m^{-3}}$?

Prerequisite:Review the matching concept
Step 1. Use water’s $B=2.2\times10^9\,\mathrm{Pa}$ from Table 8.3 and a constant mass. Since $V=V_0(1-\Delta p/B)$, $\rho=\rho_0/(1-\Delta p/B)$.
Step 2. If the surface is at 1 atm, $\Delta p=(80-1)(1.013\times10^5)=8.0027\times10^6\,\mathrm{Pa}$. Then $\rho=1030/(1-8.0027\times10^6/2.2\times10^9)\approx1.034\times10^3\,\mathrm{kg\,m^{-3}}$.
Step 3. Taking 80 atm as the pressure increase instead also gives $1.034\times10^3\,\mathrm{kg\,m^{-3}}$ at the stated precision. This small-strain calculation treats B as constant.
Ex 8.14bulk

Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Prerequisite:Review the matching concept
Step 1. $B_{\rm glass}=37\times10^9\,\mathrm{Pa}$ and $\Delta p=10(1.013\times10^5)=1.013\times10^6\,\mathrm{Pa}$.
Step 2. $\Delta V/V=-\Delta p/B=-2.74\times10^{-5}$. The fractional contraction magnitude is $2.74\times10^{-5}$.
Ex 8.15bulk

Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of $7.0\times10^6\,\mathrm{Pa}$.

Prerequisite:Review the matching concept
Step 1. $V=(0.10)^3=1.0\times10^{-3}\,\mathrm{m^3}$; $B_c=140\times10^9\,\mathrm{Pa}$.
Step 2. $|\Delta V|=V\Delta p/B=5.0\times10^{-8}\,\mathrm{m^3}=0.050\,\mathrm{cm^3}$. Signed $\Delta V$ is negative.
Ex 8.16bulk

How much should the pressure on a litre of water be changed to compress it by 0.10%?

Prerequisite:Review the matching concept
Step 1. $|\Delta V|/V=0.10/100=1.0\times10^{-3}$. Use $B=2.2\times10^9\,\mathrm{Pa}$.
Step 2. Increase pressure by $\Delta p=B|\Delta V|/V=2.2\times10^6\,\mathrm{Pa}\approx21.7\,\mathrm{atm}$. One litre decreases by 1 mL.
Revision

Revision

Review the definitions, equations and conditions for each topic.

8.1 · Elasticity, plasticity and rigid-body limits

Review concept and reasoning
  • Real solids deform; “rigid body” is a useful approximation when deformation is negligible.
  • Elastic deformation recovers after unloading; plastic deformation leaves permanent set.
  • Applications balance stiffness, strength, weight, reliability, service life and cost.

8.2 · Stress and three types of strain

Review concept and reasoning
  • Stress is restoring force per area, σ = F/A (Pa; [ML⁻¹T⁻²]); internal tension in a wire is F, not the sum 2F of opposite end forces.
  • Longitudinal strain ε = ΔL/L is positive in extension and negative in compression; strain has no unit.
  • Shear strain γ = Δx/L = tanθ ≈ θ for small θ in radians; hydrostatic volume strain is ΔV/V. Balance total force and torque in shear arrangements.

8.3 · Hooke’s law and its domain

Review concept and reasoning
  • Hooke’s law is empirical: stress is proportional to strain only in the proportional region.
  • Elasticity means recovery after unloading; it does not require a linear stress–strain curve.
  • A modulus is stress divided by strain, so it has units Pa; it characterises material response under specified conditions.

8.4 · Stress–strain curve, strength and elastomers

Review concept and reasoning
  • Measure load and extension in steps, calculate stress and strain, and plot stress vertically against strain horizontally.
  • OA is linear Hookean; AB may remain recoverable but nonlinear; B marks the yield/elastic limit; beyond B, unloading leaves permanent set.
  • D is ultimate tensile strength and E is fracture. Close D/E indicates brittle behaviour; separation indicates ductility. Elastomers can show large recoverable nonlinear strain.

8.5.1 · Young’s modulus: axial extension and compression

Review concept and reasoning
  • Y = σ/ε = FL/(AΔL); hence |ΔL| = FL/(AY), k_wire = YA/L. Use magnitudes for compressive extension calculations.
  • At equal stress, larger Y gives smaller strain. For a wire, ΔL ∝ FL/(AY) ∝ 1/r²; stiffness does not by itself specify breaking strength.
  • Series segments carry the same tension and their extensions add; equal-length parallel wires share extension.

8.5.2 · Shear modulus (modulus of rigidity)

Review concept and reasoning
  • G = (F/A)/(Δx/L) = FL/(AΔx) ≈ F/(Aθ) for small θ; shear stress τ ≈ Gθ.
  • A is the face parallel to shear and L is the separation of sliding faces. Solids resist static shear; a stretched helical spring twists its wire.
  • G is generally below Y; G ≈ Y/3 is a rough estimate, not a universal identity.

8.5.3 · Bulk modulus and compressibility

Review concept and reasoning
  • B = −Δp/(ΔV/V); therefore ΔV/V = −Δp/B and κ = 1/B = −(ΔV/V)/Δp.
  • For increased pressure, Δp > 0 and signed ΔV < 0; compressive strain magnitude is |ΔV|/V. B is positive; κ has units Pa⁻¹.
  • Bulk compression applies to solids, liquids and gases. Typically gas > liquid > solid in compressibility; gas values depend strongly on pressure and temperature.

8.5.4 · Poisson’s ratio and lateral contraction

Review concept and reasoning
  • With positive contraction magnitude Δd, ν = (Δd/d)/(ΔL/L); with signed strains, ν = −ε⊥/ε∥.
  • Poisson’s ratio is dimensionless and describes lateral contraction accompanying axial extension; typical steel values are 0.28–0.30 and aluminium alloys about 0.33.
  • Axial force can produce lateral strain too; one elastic constant alone cannot describe every deformation state.

8.5.5 · Proof: elastic potential energy in a wire

Review concept and reasoning
  • Within the linear elastic region, force grows with extension: F(x) = YAx/L.
  • Integrate rather than treating final force as constant: U = ∫₀^ΔL F(x)dx = ½FΔL = YA(ΔL)²/(2L).
  • For wire volume AL, energy density u = U/(AL) = ½σε = ½Yε² = σ²/(2Y), the area under the linear stress–strain curve.

8.6 · Applications: crane ropes and safe loads

Review concept and reasoning
  • To avoid yield, W/A ≤ σᵧ; with safety factor n, choose A ≥ nW/σᵧ. Yield prevents permanent deformation; ultimate strength concerns breaking.
  • For a circular rope r = √(A/π); multiplying allowable load factor by n multiplies area by n and radius by √n.
  • Braided thin wires improve flexibility; design also accounts for dynamic loads, reliability and service conditions.

8.6 · Proof: beam bending and I-section design

Review concept and reasoning
  • For a rectangular section, I = ∫y²dA = bd³/12 about its neutral axis. A fibre has strain ε = −y/R and stress σ = −Yy/R, giving moment magnitude M = YI/R.
  • For a simply supported beam with central load W, δ = Wl³/(48YI) = Wl³/(4Ybd³), under small elastic bending with negligible shear deflection.
  • Sag varies as l³/(Ybd³); I-sections place material far from the neutral axis to increase I efficiently. Deep narrow beams can buckle.

8.6 · Proof: maximum mountain height

Review concept and reasoning
  • A rock column of height h and base area A has weight per area mg/A = ρgh.
  • Because mountain sides are free, loading is not uniform hydraulic compression; characteristic shear stress is of order ρgh.
  • Require ρgh ≲ τ_critical: h_max ~ τ_critical/(ρg), about 10 km for typical terrestrial values. It is an order-of-magnitude estimate, not a universal limit.
Tests

Practice tests

Choose a level, answer the questions and submit to check your score.

1. Which property corresponds to permanent deformation?
2. The SI unit of Young’s modulus is:
3. A larger Y at the same stress means:
4. Compressibility is:
5. Which material response need not be Hookean?