Unit 8: Elasticity
1. Hooke's Law and Force Constant
When a material is stretched or compressed within its elastic limit, the deformation is directly proportional to the applied force. This linear relationship is known as Hooke's Law.
Mathematical Formulation
Hooke's Law is expressed as:
F = kx
where:
F= applied force (in newtons, N)k= force constant or stiffness of the material (in N/m)x= extension or compression from the natural length (in meters, m)
The force constant k depends on the material's intrinsic properties as well as its shape and size. For example, a longer wire of the same material has a smaller k than a shorter one because it extends more under the same force.
Units
The SI unit of the force constant is newton per meter (N/m).
Example
A steel spring requires a force of 10 N to produce an extension of 0.02 m. Its force constant is:
k = F/x = 10 N / 0.02 m = 500 N/m
If the same spring is stretched by 0.05 m, the required force is:
F = kx = 500 N/m × 0.05 m = 25 N
2. Stress, Strain, Elasticity and Plasticity
When a deforming force acts on a body, internal forces develop that resist the deformation. The intensity of these internal forces is called stress, while the resulting deformation relative to the original dimensions is called strain.
Stress
Stress is defined as the restoring force per unit area:
σ = F/A
where:
σ= stress (pascal, Pa)F= restoring force (N)A= cross‑sectional area perpendicular to the force (m²)
Types of stress:
- Tensile stress: forces act to elongate the body.
- Compressive stress: forces act to shorten the body.
- Shear stress: forces act parallel to the surface, causing layers to slide.
Strain
Strain is a dimensionless ratio representing the fractional change in dimension:
Strain = ΔL / L₀ (for linear dimensions)
Depending on the type of deformation, we have:
- Longitudinal strain:
εₗ = ΔL / L - Volumetric strain:
εᵥ = ΔV / V - Shear strain:
γ = θ(angle in radians through which a line originally perpendicular to the fixed face rotates)
Elasticity
Elasticity is the property of a material to regain its original shape and size after the deforming force is removed, provided the force does not exceed the elastic limit.
Plasticity
Plasticity is the property of a material to undergo a permanent deformation when the applied force exceeds the elastic limit; the material does not return to its original shape upon removal of the force.
Example: Stress‑Strain Curve
Consider a metallic rod subjected to increasing tensile force. Initially, stress is proportional to strain (linear region) – this obeys Hooke's Law. The slope of this linear region is Young's modulus. Beyond the yield point, the curve deviates, indicating plastic deformation. If the force is removed after entering the plastic region, the rod retains a permanent elongation.
3. Elastic Modulus
Elastic moduli quantify the stiffness of a material relating stress to strain in the linear elastic region. There are three primary moduli corresponding to the three basic types of strain.
Young's Modulus (Y)
Young's modulus measures resistance to longitudinal deformation:
Y = (Tensile Stress) / (Longitudinal Strain) = (F/A) / (ΔL/L)
Units: pascal (Pa) = N/m².
Example: A copper wire of length 2 m and cross‑sectional area 1 mm² (1×10⁻⁶ m²) stretches by 1 mm under a load of 100 N. Compute Y:
Stress = F/A = 100 N / 1×10⁻⁶ m² = 1×10⁸ Pa
Strain = ΔL/L = 0.001 m / 2 m = 5×10⁻⁴
Y = Stress / Strain = (1×10⁸) / (5×10⁻⁴) = 2×10¹¹ Pa ≈ 200 GPa
Bulk Modulus (B)
Bulk modulus measures resistance to uniform compression (volumetric stress):
B = - (Volumetric Stress) / (Volumetric Strain) = -P / (ΔV/V)
V ensures B is positive because an increase in pressure (positive P) produces a decrease in volume (negative ΔV).
Compressibility is the reciprocal of bulk modulus:
Compressibility = 1/B
Example: If a pressure of 5×10⁵ Pa reduces the volume of a fluid by 0.1 %, compute B:
ΔV/V = -0.001 (negative because volume decreases)
B = -P / (ΔV/V) = -(5×10⁵ Pa) / (-0.001) = 5×10⁸ Pa
Compressibility = 1 / (5×10⁸) = 2×10⁻⁹ Pa⁻¹
Shear Modulus (G) – Modulus of Rigidity
Shear modulus relates shear stress to shear strain:
G = (Shear Stress) / (Shear Strain) = (F/A) / θ
Units: pascal (Pa).
Example: A rectangular block of gel (area 0.01 m², height 0.05 m) is subjected to a tangential force of 200 N, causing the top surface to shift by 0.002 m relative to the bottom. The shear strain θ ≈ Δx / height = 0.002 / 0.05 = 0.04 rad. Shear stress = F/A = 200 / 0.01 = 2×10⁴ Pa. Hence:
G = (2×10⁴ Pa) / 0.04 = 5×10⁵ Pa = 0.5 MPa
4. Poisson's Ratio (ν)
When a material is stretched in one direction, it tends to contract in the perpendicular directions. Poisson's ratio quantifies this lateral contraction relative to longitudinal extension.
ν = - (Lateral Strain) / (Longitudinal Strain) = - (Δd/d) / (ΔL/L)
The negative sign ensures ν is positive for ordinary materials that become thinner when stretched.
Theoretical limits: -1 ≤ ν ≤ 0.5 (derived from stability considerations).
Practical range for most solids: 0 ≤ ν ≤ 0.5.
Typical values: metals ≈ 0.25‑0.35, rubber ≈ 0.49‑0.50 (nearly incompressible), cork ≈ 0 (almost no lateral expansion).
Example: A steel rod of original diameter 10 mm and length 1 m is pulled, increasing its length by 0.5 mm. Its diameter decreases by 0.01 mm. Compute ν:
Longitudinal strain = ΔL/L = 0.5 mm / 1000 mm = 5×10⁻⁴
Lateral strain = Δd/d = (-0.01 mm) / 10 mm = -1×10⁻³
ν = - (Lateral strain) / (Longitudinal strain) = -(-1×10⁻³) / (5×10⁻⁴) = 2
This result exceeds the theoretical limit, indicating the assumed deformation is beyond the elastic limit; within the elastic regime, ν for steel is about 0.30.
5. Elastic Potential Energy
When a body is deformed elastically, work is done against the internal restoring forces. This work is stored as elastic potential energy, which can be recovered when the force is removed.
Energy in a Spring (Hooke's Law System)
For a spring obeying F = kx, the incremental work dW for a small extension dx is:
dW = F·dx = kx·dx
Integrating from 0 to x gives the total stored energy:
U = ∫₀ˣ kx·dx = ½ kx²
Using F = kx, alternative forms are:
U = ½ Fx = F² / (2k)
Energy Density in a Continuum
For a material stressed uniformly, the elastic potential energy per unit volume (energy density) is:
u = ½ × stress × strain
Substituting stress = Y·strain (for longitudinal deformation):
u = ½ Y (strain)²
Area Under Force‑Extension Graph
The work done (and thus the elastic potential energy) equals the area under the force‑extension (F‑x) curve. For a linear spring, this area is a triangle with base x and height F, giving ½ Fx, consistent with the formula above.
Example: Energy Stored in a Stretched Wire
A steel wire (Y = 2×10¹¹ Pa) of length 2 m and cross‑sectional area 1 mm² is stretched by 2 mm. Compute the elastic potential energy stored.
- Stress = Y × strain = 2×10¹¹ Pa × (ΔL/L) = 2×10¹¹ Pa × (0.002 m / 2 m) = 2×10¹¹ Pa × 0.001 = 2×10⁸ Pa.
- Force = Stress × A = 2×10⁸ Pa × 1×10⁻⁶ m² = 200 N.
- Extension x = 0.002 m.
- Using U = ½ Fx = 0.5 × 200 N × 0.002 m = 0.2 J.
- Alternatively, using U = ½ kx²: first find k = YA/L = (2×10¹¹ Pa × 1×10⁻⁶ m²) / 2 m = 1×10⁵ N/m. Then U = 0.5 × 1×10⁵ N/m × (0.002 m)² = 0.5 × 1×10⁵ × 4×10⁻⁶ = 0.2 J.
The wire stores 0.2 joules of elastic potential energy.
Example: Energy Density in a Compressed Cube
A cube of rubber (bulk modulus B = 2×10⁶ Pa) experiences a uniform pressure increase in length of side? Let's compute using volumetric strain.
Suppose the cube's volume decreases by 0.5 % under pressure P = 1×10⁴ Pa.
Volumetric strain = ΔV/V = -0.005
Energy density u = ½ × P × |ΔV/V| = 0.5 × 1×10⁴ Pa × 0.005 = 25 J/m³
Thus each cubic meter of the rubber stores 25 joules of elastic energy under this pressure.