Hooke's Law, Stress-Strain & Spring Elastic Energy Calculator

Materials Science & Solid Mechanics: Hooke's Law

First stated in anagram form by Robert Hooke in 1676 as "Ut tensio, sic vis" ("As the extension, so the force"), Hooke's Law describes the linear relationship between applied force/stress and resulting deformation in elastic bodies:

• Spring Mechanics vs. Continuum Mechanics:

• Spring Formulation): Discrete macroscopic model where spring stiffness k depends on wire diameter, coil diameter, and active coils.

• Continuum Formulation): Microscopic material model independent of geometry, where stress \ is normalized by cross-sectional area and strain \ is normalized by length.

• Young's Modulus (E): The intrinsic atomic stiffness of a material. Adding heat treatment changes a steel's yield strength, but its Young's modulus (E ≈ 200 GPa) remains virtually unchanged because it is dictated by atomic bond stiffness.

• Proportional Limit & Plasticity: Hooke's law is valid strictly within the proportional elastic limit. Beyond this point, materials undergo plastic yielding, where atomic planes slip permanently and deformation is non-reversible.

How to Use This Calculator

Enter your Hooke's Law Application Domain — choose from Solid Material Mechanics — Stress & Strain (σ = E · ε | ΔL = σL₀/E), Spring Mechanics — Force & Displacement (F = k · Δx | U = ½kΔx²), Applied Tensile/Compressive Stress (σ in MPa) (e.g. 150 (1 MPa = 10⁶ N/m²)), Young's Modulus of Elasticity (E in GPa) (e.g. 200 (Steel: 200, Titanium: 116, Aluminum: 69)), Original Unstretched Length (L₀ in mm) (e.g. 500), Spring Stiffness Constant (k in N/m [for spring mode]) (e.g. 2500 (N/m = 2.5 N/mm)), and Spring Elongation / Compression (Δx in mm [for spring mode]) (e.g. 35). Results update instantly as you type — adjust any value to compare scenarios.

The Formula

Hooke's Law (Robert Hooke, 1676) \& Elasticity equations:

Material Solid Mechanics: σ = E · , = (σ) / (E) = (Δ L) / (L_0)

Elastic Elongation: Δ L = (σ · L_0) / (E) = (F · L_0) / (A · E)

Modulus of Resilience (Energy Density): u_r = (1) / (2) σ = (σ^2) / (2E)

Helical Spring Restoring Force: F = -k · Δ x

Stored Spring Potential Energy: U = (1) / (2) k (Δ x)^2

Worked Example

For example, applying tensile stress to a structural steel rod of initial length \ with Young's modulus \ (200 × 10^9 Pa): Engineering strain is . Total elastic elongation is . Modulus of resilience is 0.000750) = 56.25 kJ/m^

• For a helical spring with \ compressed by 35 mm (0.035 m), restoring force is \ storing 0.035^2) = 1.531 Joules.

Understanding Your Result

Evaluate your result in the context of your specific use case and relevant industry standards or reference values. A single calculated result is a starting point — test two to three scenarios to understand how sensitive the outcome is to your key input assumptions. For high-stakes decisions, verify your result against a secondary source or subject-matter expert before acting.