Beyond Rigidity: The Science of Elasticity and Plasticity

In previous chapters, we treated solids as “rigid bodies”—perfectly stiff objects that never change shape. In the real world, every solid is a bit like a spring. Whether it’s a steel bridge, a rubber band, or a bone in your body, every material stretches, compresses, or twists when a force is applied.

Understanding the mechanical properties of solids is the difference between a skyscraper that stands for a century and one that collapses under its own weight.

The Core Pillars of Solid Mechanics

1. Stress and Strain

These are the two fundamental metrics of deformation:

• Stress (σ): The internal restoring force per unit area. It’s what the material “feels” internally.
• Strain (ε): The fractional change in dimension (length, volume, or shape). It’s how the material “reacts.”

2. Hooke’s Law and Moduli

For small deformations, Stress is proportional to Strain. The constant of proportionality is the Modulus of Elasticity.

• Young’s Modulus (Y): Resistance to change in length (stretching/compression).
• Bulk Modulus (B): Resistance to change in volume (squeezing from all sides).
• Shear Modulus (G): Resistance to change in shape (sliding layers).

3. The Stress-Strain Curve

This graph is the “personality profile” of a material. It shows exactly when a material will bounce back (Elastic region), when it will stay permanently bent (Plastic region), and when it will snap (Fracture point).

The Gauntlet: 10 Challenging Aptitude Questions

Question 1: The Dual-Wire Hanging

Two wires, one of steel and one of brass, are joined end-to-end. The steel wire has a length of 1.5m and the brass wire 1.0m. Both have a diameter of 0.25 cm. When a load is hung from the bottom, which wire will stretch more, and what is the ratio of their elongations? (Y_steel = 2.0 × 10¹¹ Pa, Y_brass = 0.91 × 10¹¹ Pa).

Question 2: Thermal Stress Trap

A steel rod is clamped at both ends so that it cannot expand. If the temperature is lowered, the rod experiences “Thermal Stress.” If the rod’s length is L, its area is A, and the temperature change is ΔT, find the tension developed in the rod. (Use coefficient of linear expansion α).

Question 3: The Squeezed Sphere

A solid copper sphere is taken to the bottom of the ocean where the pressure is 10⁷ Pa. If the Bulk Modulus of copper is 1.4 × 10¹¹ Pa, what is the percentage change in the volume of the sphere?

Question 4: Shear in a Tall Building

A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force of 9 × 10⁴ N on its narrow face. The lower edge is riveted to the floor. How much will the upper edge be displaced? (Shear Modulus G = 5.6 × 10⁹ Pa).

Question 5: Potential Energy of a Stretched Wire

Prove that the work done in stretching a wire (Elastic Potential Energy) is equal to ½ × Stress × Strain × Volume. Why is there a factor of ½?

Question 6: The Breaking Stress Limit

A heavy chandelier of mass M is suspended by a steel wire. If the breaking stress of steel is S, what is the minimum radius the wire must have to support the chandelier with a safety factor of 5?

Question 7: The “Rubber vs. Steel” Paradox

In physics, we say steel is “more elastic” than rubber. Explain why this is true conceptually using the definition of Young’s Modulus.

Question 8: Poisson’s Ratio and Density

When a wire is stretched, its length increases but its diameter decreases. If Poisson’s ratio (σ) is 0.5, what happens to the total volume of the wire? (Hint: Calculate the fractional change in volume).

Question 9: The Energy Density Challenge

Two wires of the same material have lengths in the ratio 1:2 and radii in the ratio 2:1. If they are stretched by the same force, find the ratio of the elastic potential energy stored in them.

Question 10: The Composite Rod Equilibrium

A rod consists of two halves—one aluminum and one steel—joined together. If the rod is subjected to a compressive force, do both halves experience the same stress? Do they experience the same strain?

Detailed Explanations & Solutions

1. Elongation Ratio

Since the wires are in series, the Force (Tension) is the same for both.

ΔL = (F × L) / (A × Y). Since F and A are constant, ΔL ∝ L/Y.

Ratio (Steel/Brass) = (Lₛ/Yₛ) / (L_b/Y_b) = (1.5 / 2.0) / (1.0 / 0.91) = 0.75 / 1.09 ≈ 0.68.

Result: Brass stretches more.

2. Thermal Stress

ΔL (thermal) = LαΔT. But the clamps prevent this change, so the “elastic” strain is αΔT.

Stress = Y × Strain = YαΔT.

Tension (Force) = Stress × Area.

Result: F = YAαΔT.

3. Bulk Compression

Bulk Modulus B = ΔP / (ΔV/V).

(ΔV/V) = ΔP / B = 10⁷ / (1.4 × 10¹¹).

Fractional change = 0.71 × 10⁻⁴.

Result: Percentage change = 0.0071%.

4. Shearing Displacement

Δx = (F × h) / (A × G).

Here, height h = 0.5m, Area A (where force is applied) = 0.5m × 0.1m = 0.05 m².

Δx = (9 × 10⁴ × 0.5) / (0.05 × 5.6 × 10⁹).

Result: Δx ≈ 0.16 mm.

5. Elastic Potential Energy

The force is not constant; it starts at 0 and increases linearly to F as the wire stretches. The average force is F/2.

Work = Average Force × Extension = (F/2) × ΔL.

Multiply and divide by (A × L) to get the Stress/Strain form.

Result: Energy = ½ × Stress × Strain × Volume.

6. Safety Factor

Required Stress = Breaking Stress / Safety Factor = S/5.

Force = Mg. Area = πr².

Mg / πr² = S/5 → r² = 5Mg / πS.

Result: r = √(5Mg / πS).

7. The Elasticity Paradox

Elasticity is defined by how much force is required to produce a certain strain. Because steel requires a massive force to stretch even a tiny bit, its Young’s Modulus is much higher than rubber’s.

Result: Higher Modulus = More Elastic.

8. Poisson’s Ratio (σ)

Fractional change in volume dV/V = (dL/L)(1 – 2σ).

If σ = 0.5, then (1 – 2(0.5)) = 0.

Result: The volume remains constant.

9. Energy Ratio

Energy U = F²L / (2AY). Since F and Y are constant, U ∝ L/A ∝ L/r².

Ratio = (L₁/r₁²) / (L₂/r₂²) = (1 / 2²) / (2 / 1²) = (1/4) / 2 = 1/8.

Result: 1 : 8.

10. Composite Rod

Since they are in series, the Force is the same, so the Stress (F/A) is the same.

However, because they have different Young’s Moduli, the Strain will be different.

Result: Same Stress, Different Strain.

Key Summary: The “Stiffness” Hierarchy