§ 29. Hooke's Law in Differential Form & Young's Modulus (Grade 7–10)
Elastic uniaxial tension of a rod, mechanical stress σ, strain ε, and Poisson necking. Within the proportional limit, mechanical normal stress in a rod is directly proportional to its relative longitudinal elongation.
Theory Summary & Scientific Mechanism
While the elementary spring formula F = kΔ L depends on specimen dimensions, continuum mechanics expresses elasticity entirely in terms of intrinsic material quantities: normal mechanical stress σ = (F)/(S) (in Pascals, Pa = N/m²) and dimensionless longitudinal strain ε = (Δ L)/(L₀).
Within the proportional elastic limit, stress varies linearly with strain (Hooke’s Law in differential form): σ = Eε, yielding total elongation Δ L = (F L₀)/(E S) and effective axial stiffness k = (ES)/(L₀). The material constant E is Young’s modulus, measuring interatomic bond stiffness (ranging from sim 0.01 GPa in elastomers to 70 GPa in aluminum and 200 GPa in structural steel).
Elastic strain energy stored per unit volume equals the area under the stress–strain curve: w = (1)/(2)σε = (1)/(2)Eε² = (σ²)/(2E), with total potential energy Eₚ = (1)/(2)FΔ L. Axial stretching is accompanied by lateral Poisson contraction (ε_⊥ = -νε_∥), visible on the 3D tensile specimen.
Everyday Analogy
When you pull a spring scale, stretch a rubber band, or load a suspension-bridge steel cable, the material stretches slightly. Within the elastic limit, elongation Δ L grows in exact proportion to the pulling force!
Interactive Question & Derivation
Two cylindrical rods of the same material (Young's modulus E) and equal initial length L₀ are stretched by the same force F. The diameter of the first rod is twice as small as the second (d₁ = d₂ / 2). How many times greater is the elongation Δ L₁ of the thinner rod?
- A: 2 times greater, because elongation is inversely proportional to diameter d
- B: 4 times greater, because cross-sectional area A = (π d²)/(4) is 4 times smaller and Δ L = (FL₀)/(EA)
- C: Identical, because both rods share the same Young's modulus E
Study Hint: From Hooke’s law Δ L = (F L₀)/(E A): if cross-sectional area A or Young’s modulus E increases, does elongation Δ L increase or decrease?
Proof & Explanation: By Hooke's Law for uniaxial tension, normal stress σ = (F)/(A) is related to relative strain ε = (Δ L)/(L₀) via Young's modulus: σ = Eε, giving Δ L = (F L₀)/(E A). Since cross-sectional area scales with the square of diameter (A = (π d²)/(4)), halving diameter reduces A by a factor of 4, quadrupling normal stress σ and absolute elongation Δ L!
Key Equations & 3D Simulation Parameters
σ = (F)/(A), ε = (σ)/(E) = (F)/(E A), Δ L = (F L₀)/(E A)
- Tensile load force F: 2 … 16 kN (default: 8 kN)
- Young's modulus E: 2 … 20 ×10¹⁰ Pa (default: 7 ×10¹⁰ Pa)
- Specimen diameter d: 4 … 10 mm (default: 6 mm)
- Initial gauge length l₀: 1.2 … 2.8 m (default: 2 m)