Chapter 9: Mechanical Properties of Solids

Introduction

Key Points

• Solid bodies have definite shape and size but are not perfectly rigid.
• Elasticity: Property of a body to regain original shape and size after removal of deforming force.
• Elastic substances: Materials that regain original shape and size after deformation.
• Plasticity: Inability to regain original shape; plastic substances like plastic, bakelite exhibit this.

Solved Examples

Practice Set

• Q1: What property allows a solid to regain its shape after deformation?
  A) Plasticity
  B) Elasticity
  C) Rigidity
  D) Viscosity
• Q2: Which of the following is a plastic substance?
  A) Steel
  B) Rubber
  C) Bakelite
  D) Glass
• Q3: A solid that does not regain its shape after force removal is called:
  A) Elastic
  B) Plastic
  C) Rigid
  D) Fluid

Answer Key

• Q1: B - Elasticity is the property to regain original shape after force removal.
• Q2: C - Bakelite is a plastic substance showing plasticity.
• Q3: B - Plastic solids do not regain original shape after deformation.

Memory Aids

• Elasticity = "Elastic" = returns to original shape.
• Plasticity = "Plastic" = permanent deformation.

Elastic Behaviour of Solids

Key Points

• Particles in solids are closely packed with stable interatomic distances.
• Deforming force changes interatomic distances; removal causes restoring forces to return to original state.
• Spring-ball model: balls = atoms, springs = interatomic forces.

Solved Examples

Practice Set

• Q1: What represents interatomic forces in the spring-ball model?
  A) Balls
  B) Springs
  C) Bonds
  D) Atoms
• Q2: When deforming force is removed, what causes the solid to regain shape?
  A) Gravity
  B) Interatomic restoring forces
  C) External force
  D) Friction
• Q3: Which arrangement explains solid's shape and properties?
  A) Random particle distribution
  B) Crystal lattice
  C) Liquid arrangement
  D) Gas particles

Answer Key

• Q1: B - Springs represent interatomic forces.
• Q2: B - Interatomic restoring forces return solid to original shape.
• Q3: B - Crystal lattice arrangement defines solid's shape and properties.

Memory Aids

• Spring-ball model: Springs = forces, balls = atoms.
• Crystal lattice = regular 3D atomic pattern.

Stress and Strain

Stress

• Stress (σ) = Restoring force per unit area = \( \sigma = \frac{F}{A} \)
• SI unit: N/m² (Pascal), Dimensions: [ML⁻¹T⁻²]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Shearing (tangential) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force
• Types of strain:
  • Longitudinal strain: \( \varepsilon_l = \frac{\Delta L}{L} \)
  • Shearing strain: \( \varepsilon_T = \frac{\Delta x}{L} = \tan \theta \)
  • Volume strain: \( \varepsilon_v = \frac{\Delta V}{V} \)

Solved Examples

Practice Set

• Q1: Stress is defined as:
  A) Force per unit volume
  B) Force per unit area
  C) Change in length per original length
  D) Change in volume per original volume
• Q2: Longitudinal strain for a rod stretched from 2 m to 2.002 m is:
  A) 0.001
  B) 0.0001
  C) 0.01
  D) 0.1
• Q3: Shearing strain is given by:
  A) \( \frac{\Delta V}{V} \)
  B) \( \frac{\Delta L}{L} \)
  C) \( \frac{\Delta x}{L} = \tan \theta \)
  D) \( \frac{F}{A} \)

Answer Key

• Q1: B - Stress is force per unit area.
• Q2: B - \( \varepsilon_l = \frac{0.002}{2} = 0.001 \)
• Q3: C - Shearing strain = relative displacement over length = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL/L
• Shearing strain = tan θ
• Volume strain = ΔV/V

Hooke's Law

Key Concepts

• Discovery: Robert Hooke, English physicist, 1676.
• Law Statement: For small deformations, stress is directly proportional to strain.
• Mathematical Form: Stress \( \propto \) Strain; Stress = K × Strain.
• Modulus of Elasticity (K): \( K = \frac{\text{Stress}}{\text{Strain}} \), a constant for a material.
• Units: SI unit of K is N/m² (Pascal), dimension \([ML^{-1}T^{-2}]\).

Solved Examples

Ex 1: A wire experiences a stress of 200 N/m² and strain of 0.001. Find modulus of elasticity.

• Given: Stress = 200 N/m², Strain = 0.001
• Use: \( K = \frac{\text{Stress}}{\text{Strain}} = \frac{200}{0.001} = 2 \times 10^{5} \text{ N/m}^2 \)

Practice Set

Q1: Hooke's Law states that stress is proportional to:

• A) Force
• B) Strain
• C) Displacement
• D) Time

Q2: If a material has a modulus of elasticity \( 5 \times 10^{7} \) N/m² and strain is 0.002, what is the stress?

• A) 1 \times 10^{5} N/m²
• B) 2.5 \times 10^{5} N/m²
• C) 1 \times 10^{8} N/m²
• D) 2.5 \times 10^{4} N/m²

Q3: A wire stretches by 0.5% under a stress of 1 \times 10^{7} N/m². What is its modulus of elasticity?

• A) 2 \times 10^{9} N/m²
• B) 5 \times 10^{8} N/m²
• C) 2 \times 10^{6} N/m²
• D) 5 \times 10^{7} N/m²

Answer Key

• Q1: B — Stress is directly proportional to strain as per Hooke's Law.
• Q2: B — Stress = K × Strain = \(5 \times 10^{7} \times 0.002 = 1 \times 10^{5} \) N/m² (Option A is correct; correction: calculation shows 1e5, so option A correct)
• Q3: A — Modulus \( K = \frac{\text{Stress}}{\text{Strain}} = \frac{1 \times 10^{7}}{0.005} = 2 \times 10^{9} \) N/m².

Memory Aids

• Stress \( \propto \) Strain: "Stress follows strain's line."
• Modulus of Elasticity \( K = \frac{\text{Stress}}{\text{Strain}} \) — ratio constant.
• SI unit of K: Pascal (N/m²), remember as pressure unit.

Stress-Strain Curve

Key Points

• Stress-strain curve shows material behavior under tension.
• From O to A: Linear region obeying Hooke's Law; stress \( \propto \) strain; material returns to original shape.
• From A to B: Non-linear but elastic; material still regains shape.
• Point B: Yield point (elastic limit); beyond this, permanent deformation occurs.
• From B to D: Plastic deformation; strain increases rapidly with small stress increments.
• Point D: Ultimate tensile strength; maximum stress material can withstand.
• Point E: Fracture point; material breaks; stress here is breaking stress.
• Large D-E gap: Material is ductile; small D-E gap: Material is brittle.

Solved Examples

• Ex 1: Calculate stress at point D if force is 1000 N and cross-sectional area is 0.5 mm².
  • Stress \( \sigma = \frac{F}{A} = \frac{1000}{0.5 \times 10^{-6}} = 2 \times 10^{9} \) Pa

Practice Set

• Q1: Which point on the stress-strain curve marks the elastic limit?
  • A) A
  • B) B
  • C) D
  • D) E
• Q2: What does a large gap between points D and E indicate?
  • A) Brittle material
  • B) Ductile material
  • C) Elastic material
  • D) Plastic material
• Q3: If a wire has a cross-sectional area of 1 mm² and breaks under a force of 500 N, what is its breaking stress?
  • A) 5 \( \times 10^{8} \) Pa
  • B) 5 \( \times 10^{7} \) Pa
  • C) 5 \( \times 10^{6} \) Pa
  • D) 5 \( \times 10^{5} \) Pa

Answer Key

• Q1: B) B is the yield point or elastic limit where permanent deformation starts.
• Q2: B) Large D-E gap means the material is ductile and can undergo large plastic deformation before breaking.
• Q3: A) Breaking stress \( = \frac{F}{A} = \frac{500}{1 \times 10^{-6}} = 5 \times 10^{8} \) Pa.

Memory Aids

• O-A: Hooke's Law (Elastic, linear)
• B: Yield point = Elastic limit
• D: Ultimate tensile strength (max stress)
• E: Breaking point (fracture)
• D-E gap: Ductility indicator

Elastic Properties

Key Points

• Elastic after effect: Delay in regaining original shape after removing force.
• Elastic fatigue: Reduced elasticity due to repeated deformation cycles.
• Elastomer: Materials like rubber and aorta tissue that can stretch to large strains.

Solved Examples

• Ex 1: Identify an elastomer from given options: rubber, steel, glass, copper.
  • Answer: Rubber (large strain capability)

Practice Set

• Q1: What is elastic fatigue?
  • A) Permanent deformation
  • B) Temporary delay in shape recovery
  • C) Loss of elasticity due to repeated stress
  • D) Increase in elasticity
• Q2: Which material is an elastomer?
  • A) Steel
  • B) Rubber
  • C) Glass
  • D) Copper
• Q3: Elastic after effect means:
  • A) Instant recovery
  • B) Permanent set
  • C) Delayed recovery after force removal
  • D) No recovery

Answer Key

• Q1: C) Elastic fatigue is loss of elasticity due to repeated deformation.
• Q2: B) Rubber is an elastomer with large strain capacity.
• Q3: C) Elastic after effect is the delayed recovery after removing force.

Memory Aids

• Elastic fatigue = repeated stress reduces elasticity
• Elastomers = large strain materials (rubber, aorta)
• Elastic after effect = delayed shape recovery

Variation in Elasticity

Key Points

• Effect of temperature: Elasticity decreases as temperature increases.
• Annealing (heating and cooling) reduces elasticity.
• Effect of impurities:
  • More elastic impurity increases material elasticity.
  • More plastic impurity decreases material elasticity.
• Elastic relaxation time: Time delay in regaining original shape after force removal.
• Relaxation time negligible for phosphor, bronze, silver, gold.
• Minimum for quartz; high for glass.

Solved Examples

• Ex 1: Which material has the highest elastic relaxation time?
  • Answer: Glass

Practice Set

• Q1: How does temperature affect elasticity?
  • A) Increases elasticity
  • B) Decreases elasticity
  • C) No effect
  • D) First increases then decreases
• Q2: What happens to elasticity when more plastic impurities are added?
  • A) Increases
  • B) Decreases
  • C) No change
  • D) Becomes infinite
• Q3: Which material has negligible elastic relaxation time?
  • A) Glass
  • B) Quartz
  • C) Silver
  • D) Rubber

Answer Key

• Q1: B) Elasticity decreases with increasing temperature.
• Q2: B) Plastic impurities reduce elasticity.
• Q3: C) Silver has negligible elastic relaxation time.

Memory Aids

• Temperature ↑ → Elasticity ↓
• Plastic impurities ↓ elasticity; elastic impurities ↑ elasticity
• Relaxation time: Glass > Quartz > Metals (negligible)

Elastic Moduli

Definition and Types

• Modulus of elasticity = Ratio of stress to strain; characteristic of material.
• Young's modulus (Y): Ratio of longitudinal stress to longitudinal strain.
  Formula: \( Y = \frac{\text{Longitudinal stress}}{\text{Longitudinal strain}} = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta L / L} = \frac{F \times L}{A \times \Delta L} \)
• Ideal rigid body has infinite Young's modulus.
• Higher Young's modulus means more elasticity (e.g., steel > rubber).
• Shear modulus (G): Ratio of shearing stress to shearing strain.
  Formula: \( G = \frac{F/A}{\Delta x / L} = \frac{F \times L}{A \times \Delta x} = \frac{F}{A \tan \theta} \)
• For small angles, \( \tan \theta \approx \theta \), so shearing stress \( \sigma_s = G \times \theta \).
• Typically, \( G \approx \frac{Y}{3} \).
• Bulk modulus (B): Ratio of normal stress to volumetric strain.
  Formula: \( B = -\frac{F/A}{\Delta V / V} = -\frac{P V}{\Delta V} \)
• Bulk modulus order: solids > liquids > gases.
• For gases:
  Isothermal bulk modulus \( B_{iso} = P \)
  Adiabatic bulk modulus \( B_{ad} = \gamma P \), where \( \gamma = \frac{C_p}{C_v} > 1 \).
• Adiabatic bulk modulus is greater than isothermal bulk modulus.
• Compressibility (K): Reciprocal of bulk modulus.
  Formula: \( K = \frac{1}{B} = \frac{\Delta V}{P V} \)
• Poisson's ratio (\( \nu \)): Ratio of lateral strain to longitudinal strain within elastic limit.
  Formula: \( \nu = \frac{\text{Lateral strain}}{\text{Longitudinal strain}} = -\frac{\Delta D / D}{\Delta l / l} = -\frac{1}{D} \times \frac{\Delta D}{\Delta l} \)

Solved Examples

• Ex 1: Calculate Young's modulus if a wire of length 2 m and area 1 mm² stretches by 1 mm under a force of 10 N.
  Solution:
  \( Y = \frac{F \times L}{A \times \Delta L} = \frac{10 \times 2}{1 \times 10^{-6} \times 0.001} = 2 \times 10^{10} \text{ Pa} \)

Practice Set

• Q1: Young's modulus is defined as:
  A) Stress/Strain
  B) Strain/Stress
  C) Force/Area
  D) Force/Length
• Q2: Shear modulus is approximately:
  A) Equal to Young's modulus
  B) One-third of Young's modulus
  C) Twice Young's modulus
  D) Half of Young's modulus
• Q3: Bulk modulus of a gas under adiabatic process is:
  A) \( P \)
  B) \( \gamma P \)
  C) \( \frac{P}{\gamma} \)
  D) Zero

Answer Key

• Q1: A) Stress/Strain — Definition of Young's modulus.
• Q2: B) One-third of Young's modulus — Typically, \( G \approx Y/3 \).
• Q3: B) \( \gamma P \) — Adiabatic bulk modulus is \( \gamma \) times pressure.

Memory Aids

• Young's modulus: \( Y = \frac{F L}{A \Delta L} \)
• Shear modulus: \( G \approx \frac{Y}{3} \)
• Bulk modulus: \( B = -\frac{P V}{\Delta V} \)
• Compressibility: \( K = \frac{1}{B} \)
• Poisson's ratio: Negative ratio of lateral to longitudinal strain.

Elastic Potential Energy

Definition and Formula

• Work done in stretching wire is stored as elastic potential energy (U).
• Force applied: \( F \), extension: \( \Delta L \).
• Work done: \( W = \text{Average force} \times \text{extension} = \frac{F}{2} \times \Delta L \).
• Using stress and strain:
  \( U = W = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume} = \frac{1}{2} \times \frac{F}{A} \times \frac{\Delta L}{L} \times A L \).
• Elastic potential energy per unit volume:
  \( u = \frac{U}{V} = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2} Y (\text{strain})^2 \).

Solved Examples

• Ex 1: Calculate elastic potential energy stored in a wire of volume 0.001 m³, Young's modulus \( 2 \times 10^{11} \) Pa, and strain 0.001.
  Solution:
  \( u = \frac{1}{2} Y (\text{strain})^2 = \frac{1}{2} \times 2 \times 10^{11} \times (0.001)^2 = 1 \times 10^{5} \text{ J/m}^3 \)
  Total energy \( U = u \times V = 1 \times 10^{5} \times 0.001 = 100 \text{ J} \)

Practice Set

• Q1: Elastic potential energy per unit volume is:
  A) \( \frac{1}{2} \times \text{stress} \times \text{strain} \)
  B) \( \text{stress} \times \text{strain} \)
  C) \( \text{stress} / \text{strain} \)
  D) \( \text{strain}^2 \)
• Q2: If strain doubles, elastic potential energy per unit volume becomes:
  A) Same
  B) Double
  C) Four times
  D) Half
• Q3: Work done in stretching a wire is:
  A) \( F \times \Delta L \)
  B) \( \frac{F}{2} \times \Delta L \)
  C) \( 2 F \times \Delta L \)
  D) Zero

Answer Key

• Q1: A) \( \frac{1}{2} \times \text{stress} \times \text{strain} \) — Formula for elastic potential energy per unit volume.
• Q2: C) Four times — Energy depends on square of strain.
• Q3: B) \( \frac{F}{2} \times \Delta L \) — Average force times extension.

Memory Aids

• Elastic energy per volume: \( u = \frac{1}{2} Y (\text{strain})^2 \)
• Work done = average force × extension = \( \frac{F}{2} \times \Delta L \)

Material Properties Table

Key Properties

• Density: Mass per unit volume (kg/m³).
• Young's modulus: Stiffness of material (Pa).
• Ultimate strength: Maximum stress before breaking (Pa).
• Yield strength: Stress at which permanent deformation starts (Pa).

Solved Examples

• Ex 1: Which material is stiffer: steel with Young's modulus \( 2 \times 10^{11} \) Pa or aluminum with \( 7 \times 10^{10} \) Pa?
  Answer: Steel is stiffer due to higher Young's modulus.

Practice Set

• Q1: Which material has highest Young's modulus?
  A) Wood
  B) Steel
  C) Glass
  D) Aluminum
• Q2: Ultimate strength indicates:
  A) Elastic limit
  B) Maximum stress before breaking
  C) Density
  D) Yield point
• Q3: Yield strength is:
  A) Stress causing permanent deformation
  B) Stress causing fracture
  C) Stress at zero strain
  D) Stress at maximum strain

Answer Key

• Q1: B) Steel — Highest Young's modulus among options.
• Q2: B) Maximum stress before breaking — Definition of ultimate strength.
• Q3: A) Stress causing permanent deformation — Definition of yield strength.

Memory Aids

• Steel: High Young's modulus and strength.
• Ultimate strength = breaking point stress.
• Yield strength = start of permanent deformation.

Applications of Elastic Behaviour of Materials

Key Applications

• Bridges rely on elastic strength; declared unsafe after long use due to loss of elasticity.
• Maximum mountain height estimated using elastic limit of rocks.
• Pressure at mountain base: \( \text{stress} = h \rho g \).
• Elastic limit of typical rock: \(3 \times 10^{8} \text{ N/m}^2\).
• Stress must be less than elastic limit to prevent rock flow.
• Maximum height \( h < \frac{3 \times 10^{8}}{\rho g} = 10^{4} \text{ m} = 10 \text{ km} \).
• Height of Mount Everest ~9 km, close to calculated limit.

Solved Examples

Ex 1: Calculate max height of mountain if \( \rho = 3 \times 10^{3} \text{ kg/m}^3 \), \( g = 10 \text{ m/s}^2 \), elastic limit = \(3 \times 10^{8} \text{ N/m}^2\).

• Use \( h < \frac{3 \times 10^{8}}{3 \times 10^{3} \times 10} = 10^{4} \text{ m} = 10 \text{ km} \).

Practice Set

• Q1: What is the elastic limit of a typical rock?
  A) \(3 \times 10^{6} \text{ N/m}^2\) B) \(3 \times 10^{7} \text{ N/m}^2\) C) \(3 \times 10^{8} \text{ N/m}^2\) D) \(3 \times 10^{9} \text{ N/m}^2\)
• Q2: Maximum height of a mountain with rock density \(3 \times 10^{3} \text{ kg/m}^3\) and \(g=10 \text{ m/s}^2\) is?
  A) 1 km B) 5 km C) 10 km D) 15 km
• Q3: If stress at base exceeds elastic limit, rock will:
  A) Remain solid B) Flow C) Compress D) Expand

Answer Key

• Q1: C - Given elastic limit is \(3 \times 10^{8} \text{ N/m}^2\).
• Q2: C - Calculated max height \(=10 \text{ km}\) using \(h < \frac{elastic\ limit}{\rho g}\).
• Q3: B - Stress beyond elastic limit causes rock to flow (plastic deformation).

Memory Aids

• Max mountain height \(h = \frac{elastic\ limit}{\rho g}\).
• Elastic limit rock ~\(3 \times 10^{8} \text{ N/m}^2\).
• Stress = pressure = \(h \rho g\).

Beam Deflection

Key Concepts

• Beam supported at ends, load \(W\) at center causes deflection \(\delta\).
• Beam dimensions: length \(l\), breadth \(b\), depth \(d\).
• Deflection formula: \(\delta = \frac{W l^{3}}{4 b d^{3} Y}\), where \(Y\) is Young's modulus.


Solved Examples

Ex 1: Calculate deflection \(\delta\) for \(W=100\,N\), \(l=2\,m\), \(b=0.05\,m\), \(d=0.01\,m\), \(Y=2 \times 10^{11} \text{ Pa}\).

• Use \(\delta = \frac{100 \times (2)^3}{4 \times 0.05 \times (0.01)^3 \times 2 \times 10^{11}}\).
• Calculate \(\delta\) accordingly.

Practice Set

• Q1: Deflection \(\delta\) is directly proportional to:
  A) \(W\) B) \(b\) C) \(d^3\) D) \(Y\)
• Q2: If depth \(d\) doubles, deflection changes by factor:
  A) 2 B) 4 C) 8 D) 16
• Q3: For fixed \(W, l, b\), increasing Young's modulus \(Y\) decreases deflection by:
  A) Linear B) Quadratic C) Cubic D) No effect

Answer Key

• Q1: A - \(\delta \propto W\).
• Q2: C - \(\delta \propto \frac{1}{d^3}\), doubling \(d\) reduces \(\delta\) by 8.
• Q3: A - \(\delta \propto \frac{1}{Y}\), linear inverse relation.

Memory Aids

• Deflection \(\delta = \frac{W l^{3}}{4 b d^{3} Y}\).
• Deflection decreases sharply with depth \(d\) (cube power).
• Higher Young's modulus means less deflection.

Hollow vs Solid Shaft Strength

Key Concepts

• Hollow shaft stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: \(\tau_{solid} = \frac{\pi \eta r^{4}}{2 L}\).
• Torque for hollow shaft: \(\tau_{hollow} = \frac{\pi \eta (r_2^{4} - r_1^{4})}{2 L}\).
• Ratio: \(\frac{\tau_{solid}}{\tau_{hollow}} = \frac{r^{4}}{r_2^{4} - r_1^{4}}\).
• Also, \(\frac{\tau_{hollow}}{\tau_{solid}} = \frac{(r_2^{2} + r_1^{2})(r_2^{2} - r_1^{2})}{r^{4}} = \frac{r_2^{2} + r_1^{2}}{r^{2}} > 1\).

Solved Examples

Ex 1: Compare torque for solid shaft radius \(r=5\,cm\) and hollow shaft with \(r_2=6\,cm\), \(r_1=4\,cm\).

• Calculate \(\tau_{solid}/\tau_{hollow} = \frac{5^{4}}{6^{4} - 4^{4}}\).
• Evaluate ratio to confirm hollow shaft is stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to:
  A) \(r^{2}\) B) \(r^{3}\) C) \(r^{4}\) D) \(r\)
• Q2: Hollow shaft torque formula includes:
  A) \(r_2^{2} - r_1^{2}\) B) \(r_2^{4} - r_1^{4}\) C) \(r_2^{3} - r_1^{3}\) D) \(r_2 - r_1\)
• Q3: Ratio \(\frac{\tau_{hollow}}{\tau_{solid}}\) is always:
  A) <1 B) =1 C) >1 D) =0

Answer Key

• Q1: C - Torque \(\propto r^{4}\) for solid shaft.
• Q2: B - Hollow shaft torque depends on \(r_2^{4} - r_1^{4}\).
• Q3: C - Hollow shaft torque is greater than solid shaft (ratio >1).

Memory Aids

• Torque \(\propto r^{4}\) solid shaft.
• Hollow shaft torque uses difference of fourth powers.
• Hollow shaft stronger due to larger polar moment of inertia.