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Mechanical Properties of Solids

NCERT Class 11 · Physics Based on NCERT Class 11 Physics textbook · Free CBSE study kit

Chapter Notes

MECHANICAL PROPERTIES OF SOLIDS

Introduction and Definitions

A **solid body** has a definite shape and size. When external forces are applied to a solid, it deforms (changes shape or size). Two key properties determine how materials respond to deformation:

**Elasticity** — The property by which a body tends to regain its original shape and size when the applied external force is removed. The deformation produced is called **elastic deformation**. Example: A stretched steel spring returns to its original length when released.

**Plasticity** — The property by which a body does NOT regain its original shape and size after the deforming force is removed. The deformation is permanent (**plastic deformation**). Example: Putty, mud, or wet clay undergo permanent deformation.

Most real materials are partially elastic and partially plastic. Understanding elastic behaviour is essential for engineering design of buildings, bridges, aeroplanes, and artificial limbs.

Stress and Strain

When a body is subjected to external deforming forces while remaining in static equilibrium, internal **restoring forces** develop inside the body. These restoring forces are equal in magnitude but opposite in direction to the applied forces.

**Stress** is defined as the restoring force per unit area:

**Stress = F/A**

where:

  • F = magnitude of applied force (in N)
  • A = cross-sectional area perpendicular to force (in m²)
  • **SI unit of stress:** N m⁻² or Pascal (Pa)

    **Dimensional formula:** [ML⁻¹T⁻²]

    Types of Stress and Strain

    There are three types of stress-strain pairs that solids can experience:

    **(1) Tensile/Compressive Stress and Longitudinal Strain**

    When two equal and opposite forces are applied **perpendicular to the cross-sectional area** of a body (pulling apart or pushing together along the length):

  • **Tensile stress** occurs when the body is stretched/pulled
  • **Compressive stress** occurs when the body is compressed/pushed
  • Both are forms of **longitudinal stress**.

    The resulting deformation is **longitudinal strain**, defined as the fractional change in length:

    **Longitudinal strain = ΔL/L**

    where:

  • ΔL = change in length (m)
  • L = original length (m)
  • Strain has **no units** and **no dimensional formula** (it is dimensionless).

    Example: A steel rod of length 1 m is stretched to 1.002 m. Strain = 0.002/1 = 0.002 or 0.2%.

    **(2) Shearing Stress and Shearing Strain**

    When two equal and opposite forces are applied **parallel to the cross-sectional area** (tangential forces that cause one face to slide relative to the opposite face):

  • **Shearing stress (τ or σₛ)** = F/A (tangential force per unit area)
  • The resulting deformation is **shearing strain**, defined as:

    **Shearing strain = Δx/L = tan θ**

    where:

  • Δx = relative displacement between opposite faces (m)
  • L = length of body in the direction perpendicular to the displacement (m)
  • θ = angular displacement from the vertical (in radians)
  • For small deformations (typical in elastic region), θ is very small, so:

    **Shearing strain ≈ θ** (since tan θ ≈ θ for small angles)

    Example: When you push a book horizontally while holding it from above, the pages shear relative to each other.

    **(3) Hydraulic (Volume) Stress and Volume Strain**

    When a body is **submerged in a fluid under high pressure** or subjected to forces perpendicular to the surface **at every point uniformly**:

  • **Hydraulic stress** = pressure (p) = F/A (force per unit area acting perpendicular at all surfaces)
  • The body undergoes **volumetric compression** (no change in shape, only size):

    **Volume strain = ΔV/V**

    where:

  • ΔV = change in volume (m³)
  • V = original volume (m³)
  • Example: A solid sphere placed deep in the ocean experiences uniform compression from all sides without any change in shape.

    Hooke's Law

    **Statement:** For small deformations within the elastic limit of a material, the applied **stress is directly proportional to the produced strain**.

    **Mathematically:**

    **stress ∝ strain**

    **stress = k × strain**

    where k is a constant called the **modulus of elasticity** (or elastic modulus).

    This can be rewritten as:

    **Modulus of elasticity = stress/strain**

    **Validity:** Hooke's law is an **empirical law** — it is valid for most materials when deformations are small. Beyond a certain limit (elastic limit), many materials deviate from this linear relationship.

    **Importance:** This proportional relationship allows engineers to design structures that remain elastic and return to original shape after loading.

    Stress-Strain Curve and Material Behavior

    A **stress-strain curve** is obtained experimentally by gradually increasing tensile force on a material sample and measuring the corresponding strain at each force level. The curve reveals important information about material behavior.

    Regions of the Stress-Strain Curve

    **Region OA (Linear/Elastic Region):**

  • Stress and strain are **directly proportional** (Hooke's law obeyed)
  • The curve is **linear with slope = modulus of elasticity**
  • When the load is removed, the body **completely recovers** its original dimensions
  • The material behaves as a **perfect elastic body** in this region
  • **Point A: Proportional Limit**

  • The stress at point A is called the **proportional limit**
  • Beyond this point, stress and strain are no longer proportional
  • **Region AB (Elastic Non-linear Region):**

  • Stress and strain are **not proportional**
  • The relationship becomes **curved**
  • Despite non-proportionality, the body **still returns to original dimensions** when the load is removed
  • The material is still elastic but no longer obeys Hooke's law
  • **Point B: Yield Point (Elastic Limit)**

  • Marked by the symbol **σᵧ** (yield strength)
  • **Critical point** — beyond this, permanent deformation begins
  • If load is removed before point B, no permanent deformation occurs
  • If load is applied beyond point B, irreversible (plastic) deformation begins
  • **Region BC (Plastic Region — Early Stage):**

  • Strain increases **rapidly** for small increases in stress
  • Large **plastic (permanent) deformation** occurs
  • If the load is removed at any point between B and D (say at point C), the material **does not return to original dimensions**
  • A **permanent set** remains — the material shows residual strain even when stress becomes zero
  • **Point D: Ultimate Tensile Strength**

  • Marked by symbol **σᵤ**
  • Maximum stress the material can withstand
  • Beyond this point, additional strain occurs even with **reduced applied force** (strain increases while stress decreases)
  • **Point E: Fracture/Breaking Point**

  • The material **breaks and fails**
  • Stress drops to zero as the sample separates
  • Classification of Materials Based on Stress-Strain Curve

    **Ductile Materials:**

  • Points D and E are **far apart**
  • Material can undergo **large plastic deformation** before fracture
  • Shows a long plastic region
  • Examples: Steel, copper, aluminum, brass
  • Advantage: Can be drawn into wires and bent without breaking
  • **Brittle Materials:**

  • Points D and E are **very close or coincide**
  • Little or **no plastic deformation** before breaking
  • Fracture occurs suddenly with little warning
  • Examples: Glass, cast iron, ceramics
  • Disadvantage: Can suddenly break under load
  • **Elastomers (Elastic Substances):**

  • Materials like rubber and elastic tissues (e.g., aorta in heart)
  • Can be **stretched to several times original length** and still return to original shape
  • **Elastic region is very large**
  • **Do NOT obey Hooke's law** over most of the region
  • **No well-defined plastic region** — transition from elastic to breaking is gradual
  • Stress-strain curve is non-linear throughout
  • Elastic Moduli

    The **modulus of elasticity** (or elastic modulus) is the ratio of stress to strain in the proportional (linear) region of the stress-strain curve. It is a **material property** — different materials have different elastic moduli. Three types of elastic moduli correspond to three types of stress-strain:

    Young's Modulus (Y)

    **Definition:** Young's modulus is the ratio of **tensile (or compressive) stress** to the **longitudinal strain** produced.

    **Formula:**

    **Y = (tensile/compressive stress)/(longitudinal strain)**

    **Y = σ/ε = (F/A)/(ΔL/L)**

    Rearranging:

    **Y = (F × L)/(A × ΔL)**

    where:

  • F = applied force (N)
  • A = cross-sectional area (m²)
  • L = original length (m)
  • ΔL = change in length (m)
  • **SI unit:** N m⁻² or Pascal (Pa)

    **Dimensional formula:** [ML⁻¹T⁻²]

    **Physical Meaning:** Young's modulus represents the **stiffness** of a material in tension or compression. A **larger Y means the material is harder to deform** (more elastic). A **smaller Y means the material is easier to deform** (less elastic).

    **Interpretation from Table 8.1:**

  • **Metals (Steel, Iron, Copper)** have **very large Y values** (order of 10¹¹ N m⁻²) — require large forces to produce small changes in length — preferred for structural applications
  • **Wood, bone, concrete, glass** have **relatively small Y values** (order of 10¹⁰ N m⁻² or less)
  • Steel (Y ≈ 2 × 10¹¹ N m⁻²) is more elastic than copper (1.1 × 10¹¹), brass (0.91 × 10¹¹), and aluminum (0.7 × 10¹¹) — that's why steel is preferred for heavy-duty machinery
  • **Worked Example:** A structural steel rod of radius 10 mm and length 1 m is stretched by a force of 100 kN. Young's modulus Y = 2.0 × 10¹¹ N m⁻².

    **Find:** (a) stress, (b) elongation, (c) strain

    **Solution:**

    (a) **Stress:**

    Stress = F/A = 100 × 10³ N / [π(10 × 10⁻³)²] m²

    = 100 × 10³ / [3.14 × 10⁻⁴] = **3.18 × 10⁸ N m⁻²**

    (b) **Elongation (ΔL):**

    From Y = (F × L)/(A × ΔL), we get:

    ΔL = (F × L)/(A × Y) = (100 × 10³ × 1) / [3.14 × 10⁻⁴ × 2.0 × 10¹¹]

    = 10⁵ / [6.28 × 10⁷] = **1.59 × 10⁻³ m = 1.59 mm**

    (c) **Strain:**

    Strain = ΔL/L = (1.59 × 10⁻³)/1 = **1.59 × 10⁻³ or 0.159%**

    Shear Modulus (G) or Modulus of Rigidity

    **Definition:** Shear modulus is the ratio of **shearing stress** to the **shearing strain** produced.

    **Formula:**

    **G = (shearing stress)/(shearing strain)**

    **G = (F/A)/(Δx/L) = (F × L)/(A × Δx)**

    For small angles:

    **G = (F/A)/θ**

    where:

  • F = applied tangential force (N)
  • A = area of face parallel to force (m²)
  • Δx = relative displacement of opposite faces (m)
  • L = distance between opposite faces (m)
  • θ = angular displacement (radians)
  • **SI unit:** N m⁻² or Pascal (Pa)

    **Physical Meaning:** Shear modulus represents the **resistance of a material to shape change** without change in volume. A **larger G means the material resists shearing** (more rigid).

    **Relationship with Young's Modulus:** For most materials:

    **G ≈ Y/3** (Shear modulus is typically one-third of Young's modulus)

    This indicates that materials are generally **more resistant to longitudinal stress** than to shearing stress.

    From Table 8.2, typical values:

  • Steel: G = 84 × 10⁹ N m⁻²
  • Copper: G = 42 × 10⁹ N m⁻²
  • Lead: G = 5.6 × 10⁹ N m⁻² (smallest among common materials)
  • **Worked Example:** A square lead slab (side 50 cm, thickness 10 cm) is fixed at the bottom. A shearing force of 9 × 10⁴ N is applied horizontally on the top narrow face. How much will the top edge be displaced?

    **Given:** G (lead) = 5.6 × 10⁹ N m⁻²

    **Solution:**

    Area of narrow face: A = 0.5 m × 0.1 m = 0.05 m²

    Shearing stress: σₛ = F/A = 9 × 10⁴ / 0.05 = 1.8 × 10⁶ N m⁻²

    Distance between fixed and moving faces: L = 0.5 m

    From G = (Shearing stress × L)/(Displacement):

    Displacement (Δx) = (σₛ × L)/G = (1.8 × 10⁶ × 0.5) / (5.6 × 10⁹)

    = **1.6 × 10⁻⁴ m = 0.16 mm**

    Bulk Modulus (B)

    **Definition:** Bulk modulus is the ratio of **hydraulic stress (pressure)** to the **volumetric strain (fractional volume change)** produced.

    **Formula:**

    **B = -p/(ΔV/V)**

    where:

  • p = hydraulic pressure (N m⁻²)
  • ΔV = change in volume (m³)
  • V = original volume (m³)
  • **Negative sign explanation:** When pressure increases (p is positive), volume decreases (ΔV is negative). The negative sign ensures B is always **positive** for a stable material in equilibrium.

    **SI unit:** N m⁻² or Pascal (Pa)

    **Physical Meaning:** Bulk modulus represents the **resistance of a material to volume change**. A **larger B means the material is harder to compress** (incompressible). A **smaller B means the material is easier to compress** (compressible).

    **Compressibility (k):**

    **Compressibility k = 1/B = -(1/p) × (ΔV/V)**

    Compressibility is the **fractional change in volume per unit increase in pressure**. It is the **reciprocal of bulk modulus**.

    **Comparison of Bulk Moduli** (from Table 8.3):

  • **Solids:** Very large B (10¹⁰ to 10¹¹ N m⁻²) — highly incompressible
  • Copper: 140 × 10⁹ N m⁻²
  • Steel: 160 × 10⁹ N m⁻²
  • Nickel: 260 × 10⁹ N m⁻²
  • **Liquids:** Much smaller B (10⁹ N m⁻²) — more compressible than solids
  • Water: 2.2 × 10⁹ N m⁻²
  • Mercury: 25 × 10⁹ N m⁻² (largest among liquids)
  • **Gases:** Extremely small B (10⁻⁴ N m⁻²) — highly compressible
  • Air (at STP): 1.0 × 10⁻⁴ N m⁻² (approximately)
  • **Gases are ~10⁶ times more compressible than solids!**
  • **Physical Reason:** In solids, atoms/molecules are tightly coupled to neighbors, requiring large pressure to change volume. In liquids, coupling is weaker. In gases, molecules are very poorly coupled, so even small pressure changes cause significant volume changes.

    **Worked Example:** The average depth of the Indian Ocean is 3000 m. Calculate the fractional compression (ΔV/V) of water at the bottom, given B (water) = 2.2 × 10⁹ N m⁻², g = 10 m s⁻².

    **Solution:**

    Hydraulic pressure from 3000 m column of water:

    p = hρg = 3000 × 1000 × 10 = 3 × 10⁷ N m⁻²

    Fractional compression:

    ΔV/V = p/B = (3 × 10⁷) / (2.2 × 10⁹) = **1.36 × 10⁻² or 1.36%**

    This shows that even at extreme ocean depths, water compresses by only ~1.4%, demonstrating its incompressibility.

    Summary Table of Elastic Moduli

    See Table 8.4 in the chapter for complete classification. Key distinctions:

    | Property | Young's Modulus | Shear Modulus | Bulk Modulus |

    |---|---|---|---|

    | **Type of Stress** | Tensile/Compressive (perpendicular to area) | Shearing (parallel to area) | Hydraulic (uniform pressure) |

    | **Type of Strain** | Longitudinal (ΔL/L) | Shearing (Δx/L or θ) | Volumetric (ΔV/V) |

    | **Change in Shape** | Yes | Yes | No |

    | **Change in Volume** | No (approximately) | No | Yes |

    | **Applicable to** | Solids only | Solids only | Solids, liquids, gases |

    | **Formula** | Y = (F×L)/(A×ΔL) | G = (F×L)/(A×Δx) = F/(A×θ) | B = -p/(ΔV/V) |

    | **Typical Range** | 10¹⁰-10¹¹ N m⁻² | 10¹⁰ N m⁻² | 10⁹-10¹¹ N m⁻² (solids) |

    Applications of Elastic Behavior

    **Engineering Design:**

  • **Building structures:** Steel and concrete chosen based on their elastic properties; knowledge of Young's modulus ensures buildings can withstand loads without excessive deformation
  • **Bridges:** Design must account for elastic deformation under traffic loads; cable materials chosen for high Y and tensile strength
  • **Aeroplanes:** Materials must be lightweight (low density) but strong (high Y); aluminum alloys used extensively
  • **Machinery:** Steel preferred in heavy-duty machines because of its high elastic modulus
  • **Design of Materials:**

  • **I-shaped railway tracks:** The wide flanges and narrow web distribute stress efficiently, reducing deformation for given load
  • **Artificial limbs:** Must be strong (high Y) yet lightweight and flexible
  • **Why glass is brittle:** Low plastic deformation region; fractures suddenly with little warning
  • **Why brass is ductile:** Large plastic region; can be drawn into wires and bent before breaking
  • **Elastic Tissue Engineering:**

  • Heart tissues (like aorta) are elastomers that can stretch significantly without following Hooke's law; design of artificial heart valves requires matching natural tissue properties
  • **Real-world Example (Human Pyramid):** In the circus stunt described in Example 8.3, the thighbone (femur) of a performer supporting 220 kg compresses by only 0.046 mm despite enormous load. This demonstrates:

  • High incompressibility of bone (Y = 9.4 × 10⁹ N m⁻²)
  • Large safety margin before permanent deformation
  • Evolution has designed bones to be light yet strong
  • Key Points for Board Exam

    1. **Stress = F/A** (always); **Strain = ΔL/L** (or other dimensional ratios, always dimensionless)

    2. **Three types of deformation:** (i) Tensile/compressive with longitudinal strain, (ii) Shearing with angular strain, (iii) Hydraulic with volumetric strain

    3. **Hooke's law:** stress ∝ strain (valid only in elastic/linear region); **proportionality constant is elastic modulus**

    4. **Stress-strain curve regions:** OA (elastic, linear), AB (elastic, non-linear), B (yield point), BD (plastic), D (ultimate strength), E (fracture)

    5. **Ductile vs. Brittle:** separation between D and E determines whether material can deform plastically before breaking

    6. **Three elastic moduli:**

  • **Y** (Young's) for tension/compression
  • **G** (Shear) for shearing, typically G ≈ Y/3
  • **B** (Bulk) for volumetric change; solids have B >> liquids >> gases
  • 7. **SI units:** All elastic moduli in N m⁻² or Pa; dimensionless strains; stresses in N m⁻²

    8. **Numerical problems:** Always use Y = (F×L)/(A×ΔL) systematically; identify which modulus to use from type of deformation described

    MCQs — 10 Questions with Answers

    Q1. A rod of original length 1 m is stretched to 1.005 m. What is the longitudinal strain?

    • A. 0.005 ✓
    • B. 0.05
    • C. 5
    • D. 0.5

    Answer: A — Longitudinal strain = ΔL/L = (1.005 - 1)/1 = 0.005/1 = 0.005 (dimensionless).

    Q2. Which statement about elasticity and plasticity is correct?

    • A. Elasticity and plasticity are the same property with different names
    • B. An elastic material returns to original shape after force removal; a plastic material does not ✓
    • C. Plastic materials are always stronger than elastic materials
    • D. Steel is purely plastic and mud is purely elastic

    Answer: B — Elasticity is recovery of shape after force removal; plasticity is permanent deformation—they are opposite properties.

    Q3. A copper wire of length L and cross-sectional area A is stretched by a force F. The stress on the wire is:

    • A. F × A
    • B. F/A ✓
    • C. L/A
    • D. F × L

    Answer: B — Stress is defined as force per unit area: σ = F/A, measured in Pa or N m⁻².

    Q4. A cylinder is compressed uniformly from all sides by hydraulic pressure. Which quantity remains unchanged?

    • A. Volume
    • B. Geometric shape (spherical remains spherical) ✓
    • C. Density
    • D. Length

    Answer: B — Under hydraulic compression, the body is squeezed uniformly; volume decreases but shape is preserved (sphere stays sphere, cube stays cube).

    Q5. In a stress-strain graph for a metal, Hooke's law is valid in the region marked as:

    • A. O to A (linear region only) ✓
    • B. A to B (curved region)
    • C. Beyond B (plastic region)
    • D. Entire curve from O onwards

    Answer: A — Hooke's law (stress ∝ strain) holds only in the linear elastic region O to A; beyond A, the relationship becomes nonlinear.

    Q6. A wire experiences shearing strain of 0.02 rad when subjected to a shear force. If the original length is 2 m, the relative displacement between opposite faces is:

    • A. 0.01 m
    • B. 0.02 m
    • C. 0.04 m ✓
    • D. 0.1 m

    Answer: C — Shearing strain = Δx/L; therefore Δx = strain × L = 0.02 × 2 = 0.04 m.

    Q7. A material obeys Hooke's law. If stress increases from 2 × 10⁸ Pa to 4 × 10⁸ Pa, the strain will:

    • A. Decrease by half
    • B. Double (increase by factor of 2) ✓
    • C. Remain constant
    • D. Become zero

    Answer: B — Under Hooke's law, stress ∝ strain; doubling stress directly doubles strain within the elastic limit.

    Q8. Which of the following is NOT a correct statement? Assertion (A): Strain is a dimensionless quantity. Reason (R): Strain is the ratio of two quantities with the same dimension.

    • A. Both A and R are true, and R is the correct reason for A ✓
    • B. Both A and R are true, but R is not the correct reason for A
    • C. A is true, R is false
    • D. A is false, R is true

    Answer: A — Strain (ΔL/L or ΔV/V) is indeed dimensionless because it's a ratio of identical quantities, making both assertion and reason correct.

    Q9. A steel rod of diameter 1 cm and length 1 m experiences a tensile stress of 10⁸ Pa. Calculate the tensile force applied (use π = 3.14). Which force is closest?

    • A. 785 N
    • B. 7850 N ✓
    • C. 78500 N
    • D. 785000 N

    Answer: B — Stress = F/A; F = stress × A = 10⁸ × π(0.005)² = 10⁸ × 7.85 × 10⁻⁵ ≈ 7850 N.

    Q10. A solid sphere of bulk modulus K is subjected to an external pressure P. Derive an expression for the fractional change in volume (ΔV/V). Which relationship is correct?

    • A. ΔV/V = K/P
    • B. ΔV/V = P/K ✓
    • C. ΔV/V = K × P
    • D. ΔV/V = K + P

    Answer: B — Bulk modulus K = P/(ΔV/V); rearranging gives ΔV/V = P/K, showing volume strain inversely proportional to bulk modulus.

    Flashcards

    What is elasticity?

    The property of a body to regain its original size and shape when the applied deforming force is removed.

    Define stress and state its SI unit.

    Stress is the restoring force per unit area (F/A); SI unit is pascal (Pa) or N m⁻².

    What is longitudinal strain?

    The ratio of change in length (ΔL) to the original length (L) of the body: strain = ΔL/L.

    Distinguish tensile stress from shearing stress.

    Tensile stress acts perpendicular to the cross-section and changes length; shearing stress acts parallel and causes relative displacement between opposite faces.

    State Hooke's law.

    For small deformations, stress is directly proportional to strain: stress = k × strain, where k is the modulus of elasticity.

    What is volume strain?

    The ratio of change in volume (ΔV) to the original volume (V) produced by hydraulic pressure: volume strain = ΔV/V.

    What does the linear region of a stress-strain curve represent?

    The elastic region (O to A) where Hooke's law is obeyed and the body regains its original dimensions when the force is removed.

    Why does strain have no units?

    Strain is a dimensionless ratio of two lengths (or volumes), so units cancel out.

    What is plasticity?

    The property of a material to undergo permanent deformation without regaining its original shape when the applied force is removed.

    Why is the I-shaped cross-section used in railway tracks?

    The I-shape maximizes the moment of inertia and bending resistance while minimizing material use, providing strength with low weight.

    Important Board Questions

    Define stress and strain. Give one example each for tensile stress and shearing stress in everyday life. [2 marks]

    State stress = F/A (with units Pa) and strain = ΔL/L (dimensionless). Example: tension in a rope pulling a load (tensile); pushing a book horizontally on a table (shearing).

    A brass wire of length 2 m and cross-sectional area 2 × 10⁻⁶ m² is stretched by a force of 50 N. If the extension produced is 0.01 m, calculate the (a) tensile stress, (b) longitudinal strain, and (c) Young's modulus of brass. [5 marks]

    Use stress = F/A = 50/(2×10⁻⁶); strain = ΔL/L = 0.01/2; Young's modulus E = stress/strain. Show working for all three parts separately.

    Explain with a stress-strain diagram why the I-shaped cross-section is preferred in railway tracks over a square or circular cross-section. Derive the relationship between Young's modulus, stress, and strain, and explain why engineers must ensure materials operate within the elastic region. [6 marks]

    Describe the linear elastic region (O to A) where Hooke's law applies and material recovers. Derive E = (F/A)/(ΔL/L). Explain that beyond elastic limit, permanent deformation occurs; I-shape maximizes moment of inertia (resistance to bending) while minimizing material—connect this to elastic properties and safety limits.

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