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.
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:
**SI unit of stress:** N m⁻² or Pascal (Pa)
**Dimensional formula:** [ML⁻¹T⁻²]
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):
Both are forms of **longitudinal stress**.
The resulting deformation is **longitudinal strain**, defined as the fractional change in length:
**Longitudinal strain = ΔL/L**
where:
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):
The resulting deformation is **shearing strain**, defined as:
**Shearing strain = Δx/L = tan θ**
where:
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**:
The body undergoes **volumetric compression** (no change in shape, only size):
**Volume strain = ΔV/V**
where:
Example: A solid sphere placed deep in the ocean experiences uniform compression from all sides without any change in shape.
**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.
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.
**Region OA (Linear/Elastic Region):**
**Point A: Proportional Limit**
**Region AB (Elastic Non-linear Region):**
**Point B: Yield Point (Elastic Limit)**
**Region BC (Plastic Region — Early Stage):**
**Point D: Ultimate Tensile Strength**
**Point E: Fracture/Breaking Point**
**Ductile Materials:**
**Brittle Materials:**
**Elastomers (Elastic Substances):**
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:
**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:
**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:**
**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%**
**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:
**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:
**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**
**Definition:** Bulk modulus is the ratio of **hydraulic stress (pressure)** to the **volumetric strain (fractional volume change)** produced.
**Formula:**
**B = -p/(ΔV/V)**
where:
**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):
**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.
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) |
**Engineering Design:**
**Design of Materials:**
**Elastic Tissue Engineering:**
**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:
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:**
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
Q1. A rod of original length 1 m is stretched to 1.005 m. What is the longitudinal strain?
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?
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:
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?
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:
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:
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:
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.
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?
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?
Answer: B — Bulk modulus K = P/(ΔV/V); rearranging gives ΔV/V = P/K, showing volume strain inversely proportional to bulk modulus.
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.
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.
Practice with interactive flashcards, mind maps, upload your own chapters and get AI study kits instantly
Try StudyOS Free →