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T6 · Mechanics / StrengthRubber & Elastomer TechnologyGATE 2027 Syllabus (XE5)
Mooney-Rivlin Strain Energy Density Equation (Hyperelasticity)
Phenomenological hyperelastic model expressing strain energy density function W for vulcanized elastomers undergoing moderate deformations up to 100–200% strain.
Mathematical Canonical Relation
Plain Formula: W = C10 * (I1 - 3) + C01 * (I2 - 3)
Live Engineering Parameter Solver
Input your production or test values below for instant, verified output.
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lambda_val
Range: 1.05 – 2.5 extension ratio
c10_val
Range: 0.05 – 2 MPa
c01_val
Range: 0.01 – 1 MPa
Calculated Engineering Result:
W = 0.278 MJ/m³
Engineering Verification:Initial shear modulus G0 = 1.00 MPa; Young's Modulus E0 ≈ 3.00 MPa.
When & Where to Apply
Use in FEA nonlinear simulation of rubber O-rings, tyres, engine mounts, and elastomeric seals undergoing large elastic deformation.
Common Engineering Pitfalls
- Extrapolating beyond 250% elongation where finite chain extensibility causes severe upward stress deviation
- Using linear Hooke’s law Young’s modulus E for large deformation rubber problems
Physical Assumptions & Validity Limits
- Isotropic, incompressible material behavior (I3 = 1)
- Valid for moderate elongation ratios (lambda < 2.5) before non-Gaussian strain hardening
- Reversible hyperelastic deformation without Mullins stress-softening effect
Variable Symbols, SI Units & Dimensional Specifications
| Symbol | Variable Meaning | SI Unit | Typical Industrial Range |
|---|---|---|---|
| W | Strain energy density per unit undeformed volume | MJ/m³ (or MPa) | — |
| C_{10} | First Mooney-Rivlin material constant (crosslink network term) | MPa | 0.40 |
| C_{01} | Second Mooney-Rivlin material constant (chain entanglement term) | MPa | 0.10 |
| I_1 | First invariant of Cauchy-Green deformation tensor | dimensionless | 3.25 |
| I_2 | Second invariant of Cauchy-Green deformation tensor | dimensionless | 3.25 |
Worked Industrial Numerical Example: Natural Rubber Tyre Tread Strain Energy at 50% Elongation
Problem Statement:
For vulcanized NR compound with C10 = 0.40 MPa and C01 = 0.10 MPa under uniaxial stretch lambda = 1.5 (50% tensile strain). Calculate invariants I1, I2 and strain energy density W.
Step-by-Step Numerical Solution:
- For uniaxial incompressible stretch: I1 = lambda^2 + 2/lambda = 1.5^2 + 2/1.5 = 2.25 + 1.333 = 3.583
- Calculate I2 = 2*lambda + 1/lambda^2 = 2*1.5 + 1/2.25 = 3.0 + 0.444 = 3.444
- Evaluate W: 0.40 * (3.583 - 3) + 0.10 * (3.444 - 3)
- W = 0.40 * 0.583 + 0.10 * 0.444 = 0.2332 + 0.0444 = 0.278 MPa
Final Answer: W = 0.278 MJ/m³ (0.278 MPa)
💡 Engineering Interpretation: The compound stores 0.278 MJ/m³ elastic strain energy at 50% extension. Initial shear modulus G = 2*(C10 + C01) = 2*(0.40 + 0.10) = 1.00 MPa, with Young’s modulus E approx 3.0 MPa.
Authoritative Technical Reference & Source Provenance
The Physics of Rubber Elasticity by L.R.G. Treloar (Oxford University Press, 2005), 3rd Edition, Page 211-224.
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