Mechanical Properties of Solids — Class 11 Physics

1. Elasticity and Plasticity

When an external force changes the shape or size of a body, the body develops internal restoring forces. Elasticity is the property by which a material tends to regain its original shape and size when the deforming force is removed, provided the elastic limit is not exceeded. Plastic deformation is permanent deformation that remains after the load is removed.

Elasticity should not be confused with “ability to stretch a lot.” A material can undergo a small deformation and still have a high elastic modulus.

2. Stress

Stress is the restoring force developed per unit area of cross-section. For normal tensile or compressive loading, stress = F/A. Its SI unit is pascal (Pa), or N/m².

Tensile stress tends to increase length; compressive stress tends to decrease length. Shearing stress acts tangentially to a surface.

3. Strain

Strain measures fractional deformation and has no unit. Longitudinal strain is ΔL/L. Volumetric strain is ΔV/V. Shear strain is the angular deformation, commonly written as Δx/L for small deformations.

4. Hooke’s Law

Within the elastic limit, stress is proportional to strain: stress ∝ strain. For a material in the linear elastic region, the ratio of stress to strain is constant and defines the appropriate elastic modulus.

5. Young’s Modulus

Young’s modulus measures resistance to longitudinal deformation:

Y = longitudinal stress / longitudinal strain = (F/A)/(ΔL/L) = FL/(AΔL).

A large Young’s modulus means a relatively small longitudinal strain is produced by a given stress.

6. Bulk Modulus

Bulk modulus measures resistance to change in volume:

K = −ΔP/(ΔV/V).

The negative sign reflects that an increase in pressure generally produces a decrease in volume. The magnitude of K is positive for ordinary stable materials.

7. Shear Modulus

Shear modulus, also called modulus of rigidity, measures resistance to change in shape:

G = shear stress / shear strain.

8. Poisson’s Ratio

When a wire is stretched, its length increases while its lateral dimensions generally decrease. Poisson’s ratio is defined as the magnitude of lateral strain divided by longitudinal strain. With the conventional sign definition, ν = − lateral strain / longitudinal strain.

9. Stress–Strain Curve

A typical stress–strain curve for a ductile material contains important regions and points:

• Proportional region: stress is directly proportional to strain.
• Elastic region: deformation is recoverable when the load is removed.
• Yielding: significant plastic deformation can occur with relatively small changes in stress.
• Ultimate tensile stress: maximum engineering stress reached on the curve.
• Fracture: the specimen finally breaks.

The exact shape and positions of these points depend on the material and loading conditions. Brittle materials show comparatively little plastic deformation before fracture.

10. Elastic Potential Energy

Work done in deforming an elastic body is stored as elastic potential energy, provided the deformation remains within the elastic regime. For a Hookean spring, U = ½kx². The energy density is the elastic energy stored per unit volume.

11. Practical Importance of Elasticity

Engineers select materials according to required stiffness, strength, toughness and allowable deformation. Bridges, buildings, cranes, cables and machine parts must be designed so that working stresses remain within safe limits. Factors such as temperature, repeated loading and material defects can affect mechanical behaviour.

Worked Numerical 1 — Stress

Question: A force of 200 N acts normally on a wire having cross-sectional area 2×10⁻⁶ m². Find the stress.

Solution: Stress = F/A = 200/(2×10⁻⁶) = 1×10⁸ Pa.

Worked Numerical 2 — Young’s Modulus

Question: A wire of length 2 m and cross-sectional area 1×10⁻⁶ m² extends by 1 mm when a force of 100 N is applied. Find Young’s modulus.

Y=FL/(AΔL).

Y=(100×2)/[(1×10⁻⁶)(1×10⁻³)] = 2×10¹¹ Pa.

Worked Numerical 3 — Extension of a Wire

Question: A wire has Y=2×10¹¹ Pa, length 1 m, area 2×10⁻⁶ m² and is subjected to 400 N. Find its extension.

ΔL=FL/(AY) = (400×1)/[(2×10⁻⁶)(2×10¹¹)] = 1×10⁻³ m = 1 mm.

Worked Numerical 4 — Volumetric Strain

Question: A material of initial volume 0.020 m³ changes in volume by −2×10⁻⁵ m³ under pressure. Find the volumetric strain.

Volumetric strain=ΔV/V=(−2×10⁻⁵)/0.020=−1×10⁻³. The negative sign indicates a decrease in volume.

Worked Numerical 5 — Elastic Energy in a Spring

Question: A spring with force constant 400 N/m is stretched by 0.10 m within its elastic limit. Find the energy stored.

U=½kx²=½(400)(0.10)²=2 J.

Worked Numerical 6 — Comparing Two Wires

Question: Two wires of the same material and length carry the same load. Wire A has twice the cross-sectional area of wire B. Compare their extensions.

From ΔL=FL/(AY), extension is inversely proportional to area. Therefore ΔL_A:ΔL_B=1:2. Wire A extends half as much.

Conceptual Derivation — Young’s Modulus

For a wire, longitudinal stress is F/A and longitudinal strain is ΔL/L. Therefore their ratio is (F/A)/(ΔL/L), giving Y=FL/(AΔL). Rearranging gives ΔL=FL/(AY), which shows directly how extension depends on force, length, area and material stiffness.

Explained MCQs

1. What is the SI unit of stress?
   Answer: Pascal (Pa). Stress is force per unit area, so 1 Pa = 1 N/m².
2. Which quantity is dimensionless?
   Answer: Strain. It is a ratio of two quantities having the same dimensions.
3. A material with a larger Young’s modulus is, in the usual engineering sense:
   Answer: Stiffer. A larger stress is required to produce the same small longitudinal strain.
4. If the cross-sectional area of a wire is doubled while force, length and material remain unchanged, its extension becomes:
   Answer: Half. ΔL=FL/(AY), so extension is inversely proportional to area.
5. What does the slope of the linear portion of a stress–strain graph represent?
   Answer: Young’s modulus. In that region, Y=stress/strain.
6. Can a body be elastic without undergoing any deformation?
   Answer: Yes. Elasticity is a material property describing recovery from deformation; zero applied load simply means there is no current deformation to recover.

Competency Question

Two suspension cables are made from different materials but have the same length and cross-sectional area. The same load is applied to each. One cable stretches much less. Explain which material has the larger Young’s modulus and why.

HOTS

A designer wants a cable that can safely support a large load while keeping its extension small. Explain why choosing a material with a high Young’s modulus alone is not enough; discuss the role of cross-sectional area and allowable stress.

Common Mistakes

• Confusing stress with force; stress is force per unit area.
• Assigning a unit to strain.
• Using Young’s modulus without converting area and extension into SI units.
• Assuming a material with a high Young’s modulus is necessarily the strongest or toughest in every sense.
• Forgetting the negative sign in the conventional definition of bulk modulus and Poisson’s ratio.
• Assuming every material has exactly the same stress–strain curve.

Quick Revision

• Stress=F/A
• Longitudinal strain=ΔL/L
• Volumetric strain=ΔV/V
• Y=FL/(AΔL)
• K=−ΔP/(ΔV/V)
• G=shear stress/shear strain
• ν=−lateral strain/longitudinal strain
• Spring energy U=½kx²
• Stress unit: Pa=N/m²
• Strain: dimensionless