Chapter 9 — Mechanical Properties of Solids

Solids resist deformation. The microscopic spring-like bonds between atoms give rise to macroscopic quantities — stress, strain, and elastic moduli — that govern how a beam bends, how a wire stretches, and how high a mountain can rise before its base crushes. This chapter develops the linear theory (Hooke's law) and walks to the brink of fracture.

Concept Map

• 9.1 Elasticity vs plasticity; intermolecular forces
• 9.2 Stress — longitudinal, shear, hydraulic; units & dimensions
• 9.3 Strain — longitudinal, shear, volumetric
• 9.4 Hooke's law
• 9.5 Stress–strain curve — elastic limit, yield, ultimate, fracture; brittle vs ductile
• 9.6 Elastic moduli — Young's $Y$, bulk $B$, shear $G$, Poisson's ratio $\sigma$
• 9.7 Applications — beam depression, I-girders, maximum height of a mountain
• 9.8 Elastic potential energy
• 9.9 Thermal stress

9.1 Elasticity vs Plasticity; Intermolecular Forces

Definition

A body is elastic if it regains its original shape and size completely after the deforming force is removed. It is plastic if it retains the deformed shape. A perfectly elastic body (e.g. quartz fibre, very close) follows Hooke's law all the way to fracture; a perfectly plastic body (e.g. wet clay, putty) shows no recovery.

Real materials lie between the two limits. The behaviour depends on the magnitude of stress: at small stress most metals are elastic; beyond the elastic limit they become plastic.

Derivation — origin of the restoring force

Treat two neighbouring atoms as connected by a potential $U(r)$ with equilibrium spacing $r_0$. Expand $U$ around $r_0$:

$U(r) = U(r_0) + \tfrac{1}{2} k (r - r_0)^2 + \mathcal{O}\big((r-r_0)^3\big),\qquad k \equiv \left.\frac{d^2 U}{dr^2}\right|_{r_0}.$

For small displacements the restoring force is linear:

$F = -\frac{dU}{dr} = -k(r - r_0).$

Summed over an Avogadro-scale population of bonds this gives the macroscopic Hooke's law. Beyond the inflexion point of $U(r)$ the curvature softens and the response becomes non-linear — this is the onset of plasticity, and well past it, fracture.

Worked Example

A copper rod has interatomic spacing $r_0 \approx 2.5\times 10^{-10}$ m and a bond stiffness $k \approx 40$ N/m. Estimate Young's modulus.

$Y \approx \frac{k}{r_0} = \frac{40}{2.5\times 10^{-10}} = 1.6 \times 10^{11}\ \text{N/m}^2,$

which matches the measured $Y_\text{Cu} \approx 1.1\times 10^{11}$ Pa within a factor of order one.

Pitfalls

• "Elastic" in physics means recovers shape, not stretchy. Steel is more elastic than rubber because steel resists deformation more.
• Plasticity is not a failure of Hooke's law alone — it is permanent rearrangement of dislocations in the crystal lattice.

9.2 Stress

Definition

Stress is the restoring force per unit area developed inside a body when an external force deforms it.

$\boxed{\ \sigma = \frac{F}{A}\ }$

SI unit: $\text{N}/\text{m}^2 = \text{Pa}$. Dimensions: $[\text{ML}^{-1}\text{T}^{-2}]$.

Derivation — stress is a tensor in general

Consider an internal plane with unit normal $\hat n$. The traction (force per area) on it, $\vec t$, generally has three components: one along $\hat n$ (normal stress) and two perpendicular to it (shear). For a uniform tensile rod the only non-zero component of the stress tensor is $\sigma_{xx} = F/A$.

Worked Example

A steel wire of cross-section $2\ \text{mm}^2$ supports a $20$ kg load.

$\sigma = \frac{(20)(9.8)}{2\times 10^{-6}} = 9.8 \times 10^{7}\ \text{Pa} = 98\ \text{MPa}.$

Comparison: yield stress of mild steel is $\sim 250$ MPa, so the wire is safely elastic.

Pitfalls

• Stress is defined on the deformed cross-section in the true stress convention; NCERT uses engineering stress (original area) — both agree for small strains.
• Hydraulic stress in a fluid equals pressure but is taken positive when compressive, whereas tensile stress in a solid is positive.

9.3 Strain

Definition

Strain is the fractional deformation — dimensionless.

• Longitudinal strain: $\ \varepsilon = \dfrac{\Delta L}{L}$
• Shear strain: $\ \gamma = \dfrac{\Delta x}{L} = \tan\theta \approx \theta$ for small angles
• Volumetric strain: $\ \varepsilon_V = \dfrac{\Delta V}{V}$

Derivation — relating volumetric strain to linear strain (small-strain limit)

For an isotropic block stretched by factors $1+\varepsilon_x,1+\varepsilon_y,1+\varepsilon_z$:

$\frac{V+\Delta V}{V} = (1+\varepsilon_x)(1+\varepsilon_y)(1+\varepsilon_z) \approx 1 + \varepsilon_x + \varepsilon_y + \varepsilon_z,$

$\boxed{\ \frac{\Delta V}{V} \approx \varepsilon_x + \varepsilon_y + \varepsilon_z\ }$

For uniform pressure all three are equal, giving $\Delta V/V = 3\varepsilon$.

Worked Example

A cube of side $10$ cm shows a side increase of $0.05$ mm when heated. Linear strain: $\varepsilon = 5\times 10^{-4}$, so volumetric strain $\approx 1.5 \times 10^{-3}$.

Pitfalls

• Strain is unitless but often quoted in microstrain ($1\ \mu\varepsilon = 10^{-6}$).
• Shear strain $\gamma$ is the total angle, not half of it (some textbooks use the engineering convention).

9.4 Hooke's Law

Definition

Within the elastic limit, stress is directly proportional to strain:

$\boxed{\ \sigma = E\,\varepsilon\ }$

where $E$ is the appropriate modulus ($Y, B, G$).

Derivation — from the linearised bond potential

We already showed in 9.1 that for $\vert r-r_0\vert \ll r_0$, $F = -k(r-r_0)$. Summing over $N$ bonds per unit area and using $\varepsilon = (r-r_0)/r_0$:

$\sigma = \frac{NF}{1} = -Nk r_0 \,\varepsilon \quad\Longrightarrow\quad Y = N k r_0.$

Robert Hooke (1676) published this as the anagram ceiiinosssttuv — Ut tensio, sic vis ("as the extension, so the force").

Worked Example

A wire of $Y = 2\times 10^{11}$ Pa is stretched by $0.1\%$ strain. Stress?

$\sigma = (2\times 10^{11})(10^{-3}) = 2\times 10^{8}\ \text{Pa}.$

Pitfalls

• Hooke's law is not universal — only the initial portion of the stress-strain curve is linear.
• Springs obey $F = -kx$, an analogous force–extension law. The spring constant $k$ already wraps geometry into the material modulus.

9.5 Stress–Strain Curve

A typical ductile metal (mild steel) exhibits:

       stress
         ^
       U |          ___
         |         /   \
       Y |       _/     \
         |     _/        \_ B (fracture)
       P |   _/
         | _/
         |/__________________> strain
         O

Brittle vs Ductile

• Brittle (glass, cast iron, ceramics): fracture point very close to elastic limit; little plastic deformation; sharp break.
• Ductile (copper, mild steel, aluminium): large plastic region between yield and fracture; can be drawn into wires; absorbs energy.
• Elastomers (rubber): no linear region; large strain recovered fully; hysteresis loop.

Worked Example

A steel cable shows ultimate strength $4\times 10^8$ Pa. Maximum load on a $1\ \text{cm}^2$ cable?

$F_\text{max} = \sigma_U A = (4\times 10^8)(10^{-4}) = 4\times 10^4\ \text{N} \approx 4\ \text{tonnes}.$

Standard practice uses a factor of safety $\sim 4$, so the rated load $\approx 1$ tonne.

Pitfalls

• "Elastic limit" and "yield point" are close in mild steel but not identical in general.
• For ductile materials necking starts past $U$, so the curve dips before fracture even though the cable is closer to failure.

9.6 Elastic Moduli

Young's modulus $Y$

For a wire of length $L$, area $A$, elongation $\Delta L$ under axial force $F$:

$\boxed{\ Y = \frac{F/A}{\Delta L / L} = \frac{FL}{A\,\Delta L}\ }$

Units: Pa. Typical values:

Bulk modulus $B$

Volumetric stress is just pressure $-\Delta P$ (compressive); volumetric strain $\Delta V/V$:

$\boxed{\ B = -\frac{\Delta P}{\Delta V/V}\ }$

The minus sign keeps $B>0$. Compressibility $\kappa = 1/B$.

Shear / Rigidity modulus $G$ (or $\eta$)

For tangential force $F$ on area $A$ producing shear angle $\theta$:

$\boxed{\ G = \frac{F/A}{\theta} = \frac{F\,L}{A\,\Delta x}\ }$

Fluids cannot sustain shear, so $G=0$ for liquids and gases.

Poisson's ratio $\sigma$

A stretched wire becomes thinner. The ratio:

$\boxed{\ \sigma = -\frac{\varepsilon_\text{lateral}}{\varepsilon_\text{longitudinal}} = -\frac{\Delta d/d}{\Delta L/L}\ }$

Theoretical bounds for isotropic materials: $-1 < \sigma < 0.5$. Typical metals: $\sigma \approx 0.28 - 0.33$. Rubber: $\sigma \to 0.5$ (nearly incompressible).

Derivation — relations among $Y$, $B$, $G$, $\sigma$

For an isotropic solid only two of the four are independent. Standard relations:

$Y = 2G(1+\sigma) = 3B(1-2\sigma),\qquad \frac{1}{Y}=\frac{1}{9B}+\frac{1}{3G}.$

(Derivation via the strain matrix for uniaxial loading; outside NCERT scope.)

Worked Example

A 4-m steel wire, area $2.0\ \text{mm}^2$, stretches $1.0$ mm under $100$ N.

$Y = \frac{(100)(4)}{(2.0\times 10^{-6})(10^{-3})} = 2.0\times 10^{11}\ \text{Pa}.\ \checkmark$

Pitfalls

• $Y$ is a property of the material, not the wire — geometry cancels.
• For composite wires in series, extensions add; in parallel, forces add. The equivalent $Y$ is not the simple sum.

9.7 Applications

Beam bending — depression of a cantilever

A beam of length $\ell$, breadth $b$, depth $d$, fixed at one end and loaded with $W$ at the free end. Standard result (derived from $Y \cdot I$ analysis):

$\boxed{\ \delta = \frac{W \ell^3}{3 Y\,I}\ },\qquad I = \frac{b d^3}{12}\ \text{(rectangle)}.$

So $\delta \propto 1/d^3$ — doubling the depth makes the beam 8× stiffer. This is why girders are tall, not wide.

I-shaped girders

Maximum stress in bending occurs at the outer fibres (top and bottom). The neutral axis carries no longitudinal stress. By concentrating material at the top and bottom flanges (I-section) we maximise $I$ for a given mass, achieving high bending stiffness with minimum weight.

Maximum height of a mountain (yield-stress argument)

The base of a mountain of density $\rho$ and height $h$ is under hydrostatic-like stress $\rho g h$. The mountain stays standing only if this is below the rock's elastic limit $\sigma_Y$:

$h_\text{max} = \frac{\sigma_Y}{\rho g}.$

With $\sigma_Y \approx 3\times 10^8$ Pa for granite, $\rho \approx 3\times 10^3\ \text{kg/m}^3$:

$h_\text{max} \approx \frac{3\times 10^8}{(3\times 10^3)(10)} = 10^4\ \text{m} = 10\ \text{km}.$

Mt Everest at $8.85$ km is within this bound — barely.

Worked Example

A cantilever steel ruler $30$ cm long, $2$ cm wide, $2$ mm thick, loaded by $0.2$ kg at the tip. Find depression.

$I = \frac{(0.02)(0.002)^3}{12} = 1.33\times 10^{-11}\ \text{m}^4,$

$\delta = \frac{(0.2)(9.8)(0.30)^3}{3(2\times 10^{11})(1.33\times 10^{-11})} = \frac{0.0529}{7.98} \approx 6.6\ \text{mm}.$

Pitfalls

• The $\ell^3$ dependence is dramatic — a beam twice as long sags 8× more under the same load.
• Engineers add ribs or webs to channel sections; same idea as I-girders.

9.8 Elastic Potential Energy

Derivation

Stretching a wire by $dx$ against tension $F(x) = (YA/L)x$:

$dW = F(x)\,dx = \frac{YA}{L} x\,dx.$

Integrate from $0$ to $\Delta L$:

$W = \frac{YA}{L}\cdot \frac{(\Delta L)^2}{2} = \tfrac{1}{2} F\,\Delta L.$

Per unit volume ($V = AL$):

$\boxed{\ u = \frac{W}{V} = \tfrac{1}{2}\,\text{stress} \times \text{strain} = \tfrac{1}{2}\,Y\varepsilon^2 = \tfrac{\sigma^2}{2Y}\ }$

Worked Example

Energy stored in a $1$-m steel wire of area $1\ \text{mm}^2$ stretched by $0.5$ mm.

$\varepsilon = 5\times 10^{-4}, \quad \sigma = Y\varepsilon = (2\times 10^{11})(5\times 10^{-4}) = 10^{8}\ \text{Pa},$

$U = \tfrac{1}{2}\sigma\varepsilon V = \tfrac{1}{2}(10^8)(5\times 10^{-4})(10^{-6})(1) = 0.025\ \text{J}.$

Pitfalls

• The factor of $\tfrac{1}{2}$ — same origin as the $\tfrac{1}{2}kx^2$ of a spring.
• Beyond the elastic limit the area under the curve is the total work done, but it is not all recoverable energy; the hysteresis area is lost to heat.

9.9 Thermal Stress

Definition

If a rod of length $L$ is prevented from expanding when heated by $\Delta T$, an internal thermal stress develops. Free expansion would have been $\Delta L = L\alpha\Delta T$. The clamped rod is effectively compressed by this amount.

Derivation

$\varepsilon = \frac{\Delta L}{L} = \alpha\,\Delta T,\qquad \sigma = Y\varepsilon = Y\alpha\,\Delta T.$

$\boxed{\ F = \sigma A = Y A \alpha \Delta T\ }$

Worked Example

A steel rail of cross-section $40\ \text{cm}^2$ is rigidly clamped at $20^\circ$C. Force on the supports when temperature rises to $50^\circ$C? ($\alpha = 1.2\times 10^{-5}\ \text{K}^{-1}$, $Y = 2\times 10^{11}$ Pa).

$F = (2\times 10^{11})(40\times 10^{-4})(1.2\times 10^{-5})(30) = 2.88\times 10^{5}\ \text{N}.$

Hence the small gaps between railway tracks.

Pitfalls

• Thermal stress depends on $Y$, $\alpha$, $\Delta T$ — not on length or area (stress is intensive, force is not).
• If the rod is free to expand there is no stress, even with huge $\Delta T$.

Solved Problems

1. A steel wire $1.5$ m long, radius $0.25$ mm, carries a $5$ kg mass. Find elongation. ($Y = 2\times 10^{11}$ Pa.)

$A = \pi r^2 = \pi (0.25\times 10^{-3})^2 = 1.96\times 10^{-7}\ \text{m}^2,$ $\Delta L = \frac{F L}{A Y} = \frac{(49)(1.5)}{(1.96\times 10^{-7})(2\times 10^{11})} = 1.87\times 10^{-3}\ \text{m} \approx 1.87\ \text{mm}.$

2. Two wires of same length and same load, one of steel ($Y_s$), one of copper ($Y_c$), have radii in ratio $1:2$. Ratio of elongations?

$\frac{\Delta L_s}{\Delta L_c} = \frac{Y_c}{Y_s}\cdot \frac{A_c}{A_s} = \frac{Y_c}{Y_s}\cdot 4.$

With $Y_c/Y_s = 110/200 = 0.55$, ratio $= 2.2$.

3. A cube of rubber ($B = 10^9$ Pa) is dropped to a depth of $1$ km in sea water ($\rho = 1.03\times 10^3\ \text{kg/m}^3$). Fractional volume change?

$\Delta P = \rho g h = (1030)(9.8)(1000) \approx 1.01\times 10^{7}\ \text{Pa},$ $\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{10^7}{10^9} = 10^{-2} = 1\%.$

4. A copper wire of length $2$ m and area $1\ \text{mm}^2$ is stretched by $1$ mm. Energy stored? ($Y_\text{Cu} = 1.1\times 10^{11}$ Pa.)

$U = \frac{1}{2}\frac{YA(\Delta L)^2}{L} = \frac{(1.1\times 10^{11})(10^{-6})(10^{-3})^2}{2(2)} = 2.75\times 10^{-2}\ \text{J}.$

5. Two identical wires of steel and copper ($Y_s = 2\times 10^{11}, Y_c = 1.1\times 10^{11}$) hang side by side, each loaded so they have the same strain $10^{-3}$. Stress ratio?

$\frac{\sigma_s}{\sigma_c} = \frac{Y_s}{Y_c} = \frac{2.0}{1.1} \approx 1.82.$

6. A wire elongates by $\ell$ under load $W$. If half its length is cut off and the same load applied, the new elongation is:

$\Delta L = \frac{WL}{AY} \implies L \to L/2 \implies \Delta L \to \ell/2.$

7. A brass rod is heated through $30^\circ$C while clamped. Compressive stress? ($\alpha = 2\times 10^{-5}/$K, $Y = 9\times 10^{10}$ Pa.)

$\sigma = Y\alpha\Delta T = (9\times 10^{10})(2\times 10^{-5})(30) = 5.4\times 10^{7}\ \text{Pa}.$

JEE/NEET Edge Cases

• Wire under its own weight: Tension varies linearly with height $\Rightarrow$ elongation $= \tfrac{1}{2}\rho g L^2/Y$, not $\rho g L^2/Y$. The half-factor is the trap.
• Composite wires (in series): $\Delta L_\text{tot}=\sum \Delta L_i$; equal force on each. (Same as resistors in series with current.)
• Composite wires (in parallel, same elongation): forces share $\propto YA/L$; equal strain on each.
• Two wires identical except for material, stretched by the same load: equal stress, strains in inverse ratio of $Y$.
• Two wires identical except for material, stretched to the same length: equal strain, stresses in ratio of $Y$.
• Bulk modulus of an ideal gas: isothermal $B_T = P$; adiabatic $B_S = \gamma P$.
• Negative Poisson's ratio (auxetics): rare materials that expand sideways when stretched — used in body armour. Excluded from NCERT but tested in advanced problems.
• Energy stored as $\tfrac{1}{2}F\Delta L$, not $F\Delta L$ — the work is averaged because $F$ rises from $0$ to its final value.

Quick Recap

• Stress is force per unit area; strain is fractional deformation.
• Hooke's law: stress $\propto$ strain (within elastic limit), slope = modulus.
• Young's, Bulk, Shear moduli for tensile, volumetric, shear deformations.
• Poisson's ratio: lateral contraction over longitudinal extension; $0 < \sigma < 0.5$ in practice.
• Stress–strain curve: proportional limit → elastic limit → yield → ultimate → fracture.
• Brittle = small plastic region; ductile = large plastic region.
• Elastic PE per unit volume = $\tfrac{1}{2}\sigma\varepsilon$.
• Thermal stress = $Y\alpha\Delta T$; depends only on material properties and temperature change.
• Beam depression $\delta \propto W\ell^3 / (Y b d^3)$ — depth dominates.
• Mountain height bounded by $\sigma_Y/(\rho g)$.

Formula Sheet