Hooke’s Law

Hooke’s Law & Elastic Constants

Young’s Modulus, Shear Modulus, Bulk Modulus, Poisson’s Ratio & Their Relationships

Last Updated: March 2026

📌 Key Takeaways

  • Hooke’s Law: Within the elastic limit, stress is proportional to strain. σ = Eε (normal), τ = Gγ (shear).
  • Young’s modulus (E): Resistance to axial deformation. σ = Eε. Units: GPa.
  • Shear modulus (G): Resistance to shear deformation. τ = Gγ. Also called modulus of rigidity.
  • Bulk modulus (K): Resistance to volumetric compression. p = −KΔV/V.
  • Poisson’s ratio (ν): Ratio of lateral strain to axial strain. Range: 0 to 0.5.
  • Key relationships: E = 2G(1+ν) = 3K(1−2ν). Only two independent constants needed for isotropic materials.

1. Hooke’s Law — Statement & Limits

Hooke’s Law

Within the elastic limit, stress is directly proportional to strain.

Normal: σ = Eε

Shear: τ = Gγ

Volumetric: p = Kεv

Hooke’s law was first stated by Robert Hooke in 1678 (as the Latin anagram “ceiiinosssttuv” — “ut tensio, sic vis” meaning “as the extension, so the force”). It is the starting point for virtually all structural analysis in engineering.

Limits of validity: Hooke’s law is valid only within the proportional limit of the material — the region where the stress-strain curve is linear. Beyond this point, the material enters the non-linear elastic and then plastic region, and Hooke’s law no longer applies. For design purposes, engineers use the yield strength as the practical limit.

2. Young’s Modulus (E)

Young’s Modulus (Modulus of Elasticity)

E = σ / ε = (F/A) / (ΔL/L) = FL / (AΔL)

SI unit: Pa (typically GPa for engineering materials)

Measures stiffness — resistance to axial deformation.

Young’s modulus is the slope of the linear portion of the stress-strain diagram. A material with high E deforms very little under load (steel, diamond); a material with low E deforms significantly (rubber, polymers).

E is a material property — it does not depend on the dimensions of the specimen. A thin wire and a thick bar of the same steel have the same Young’s modulus.

3. Shear Modulus (G)

Shear Modulus (Modulus of Rigidity)

G = τ / γ

SI unit: Pa (typically GPa)

Measures resistance to shear deformation — angular distortion.

The shear modulus is always less than Young’s modulus for the same material. For most metals, G ≈ 0.35E to 0.4E. This makes physical sense — materials resist shape change (shear) less than they resist volume change (axial).

4. Bulk Modulus (K)

Bulk Modulus

K = −ΔP / (ΔV/V) = −P / εv

SI unit: Pa (typically GPa)

Measures resistance to uniform volumetric compression.

Compressibility = 1/K

The negative sign ensures K is positive (pressure increase causes volume decrease). Bulk modulus is relevant for hydraulic systems, underwater structures, and any situation involving uniform pressure on all sides.

5. Poisson’s Ratio (ν)

Poisson’s Ratio

ν = −εlateral / εaxial

Dimensionless. Theoretical range: 0 ≤ ν ≤ 0.5

For most metals: ν ≈ 0.25 to 0.35

When you pull a rubber band, it gets longer but also thinner. The thinning (lateral contraction) is quantified by Poisson’s ratio. A material with ν = 0.5 is perfectly incompressible (volume doesn’t change during deformation — only shape changes). A material with ν = 0 (like cork) has no lateral contraction when compressed, which is why corks work well as bottle stoppers.

Why ν ≤ 0.5: If ν > 0.5, the material would expand in volume when compressed — violating thermodynamic stability. The limit ν = 0.5 corresponds to an incompressible material (K → ∞).

6. Relationships Between Elastic Constants

For an isotropic, homogeneous material (same properties in all directions), only two independent elastic constants are needed. The others can be derived:

The Three Key Relationships

E = 2G(1 + ν)

E = 3K(1 − 2ν)

E = 9KG / (3K + G)

From these, you can derive:

G = E / [2(1 + ν)]

K = E / [3(1 − 2ν)]

ν = (E − 2G) / (2G) = (3K − E) / (6K)

ν = (3K − 2G) / (2(3K + G))

GATE favourite: “Given E and ν, find G and K” is one of the most common 1-mark questions. Memorise E = 2G(1+ν) and E = 3K(1−2ν) — they cover almost every exam scenario.

GivenFind GFind K
E, νG = E/[2(1+ν)]K = E/[3(1−2ν)]
E, G—ν = E/(2G) − 1, then K from E, ν
E, Kν = (3K−E)/(6K), then G from E, ν—
G, K—E = 9KG/(3K+G), ν = (3K−2G)/[2(3K+G)]

7. Worked Numerical Examples

Example 1: Find G and K from E and ν

Problem: A material has E = 200 GPa and ν = 0.3. Find G and K.

Solution

G = E/[2(1+ν)] = 200/[2(1.3)] = 200/2.6 = 76.92 GPa

K = E/[3(1−2ν)] = 200/[3(0.4)] = 200/1.2 = 166.67 GPa

Example 2: Find ν from E and G

Problem: E = 70 GPa, G = 26 GPa. Find Poisson’s ratio.

Solution

E = 2G(1+ν) → 70 = 2(26)(1+ν) → 70 = 52(1+ν)

1+ν = 70/52 = 1.346

ν = 0.346

This is consistent with aluminium (ν ≈ 0.33).

Example 3: Composite Bar — Two Materials

Problem: A steel bar (E₁ = 200 GPa, A₁ = 500 mm²) and an aluminium bar (E₂ = 70 GPa, A₂ = 1000 mm²) are rigidly joined end-to-end. Total length = 1.5 m (steel = 0.8 m, Al = 0.7 m). A tensile load of 100 kN is applied. Find total elongation.

Solution

ΔLsteel = FL/(AE) = 100,000 × 0.8 / (500 × 10⁻⁶ × 200 × 10⁹) = 80,000 / 100,000 = 0.8 mm

ΔLAl = 100,000 × 0.7 / (1000 × 10⁻⁶ × 70 × 10⁹) = 70,000 / 70,000 = 1.0 mm

ΔLtotal = 0.8 + 1.0 = 1.8 mm

8. Common Mistakes Students Make

  • Confusing E and G: E is for normal stress/strain (axial). G is for shear stress/strain. Using E in τ = Gγ or G in σ = Eε gives wrong results.
  • Forgetting the factor of 2 in E = 2G(1+ν): Students sometimes write E = G(1+ν), missing the factor of 2. Always double-check this formula.
  • Thinking elastic constants depend on dimensions: E, G, K, and ν are material properties. They do not change with specimen size, shape, or loading. A wire and a beam of the same steel have identical elastic constants.
  • Using ν > 0.5: Poisson’s ratio cannot exceed 0.5 for a stable material. If your calculation gives ν > 0.5, you have an error. Most metals are 0.25–0.35.
  • Applying these relationships to anisotropic materials: The relations E = 2G(1+ν) and E = 3K(1−2ν) are valid only for isotropic materials (same properties in all directions). Wood, composites, and single crystals are anisotropic and have different elastic constants in different directions.

9. Frequently Asked Questions

What is the relationship between E, G, and ν?

E = 2G(1 + ν). This means Young’s modulus is always greater than twice the shear modulus (since ν > 0 for all real materials). For steel: E = 200 GPa, ν = 0.3, so G = 200/(2 × 1.3) = 76.9 GPa. If you know any two constants, the third is determined.

What is the relationship between E, K, and ν?

E = 3K(1 − 2ν). Since ν is between 0 and 0.5, the factor (1 − 2ν) is between 0 and 1, meaning K is always greater than or equal to E/3. For an incompressible material (ν = 0.5), K → ∞ (cannot be compressed).

Next Steps