Elastic Energy and Material Stability
Separate volumetric and distortional energy to determine whether an isotropic elastic law resists every small strain.
On this page
The question for today
What you’ll understand
- Derive stress from a quadratic small-strain energy.
- Test isotropic material stability using shear and bulk moduli.
- Distinguish a stable material law from a stable loaded structure.
A positive tensile modulus is encouraging, but a three-dimensional material can deform in more ways than a tensile test reveals. It can change volume, distort without changing volume, or combine both motions. An energy description asks a stronger question: does every nonzero, sufficiently small strain require positive stored energy? That question leads to precise restrictions on elastic constants and also explains why one negative coefficient need not make a material unstable.
Work accumulated into a state function
Imagine loading a small material element slowly, without heat exchange changing its elastic constants and without plastic flow. The incremental mechanical work per unit volume is the double contraction of stress with the strain increment. For a linear elastic material loaded from its unstressed reference state, stress grows linearly along a proportional loading path. Integrating therefore produces the familiar factor of one half. Energy density has units of pascals, equivalently joules per cubic metre; it is not the total energy of the whole specimen.
For small symmetric strain, an isotropic constitutive law uses two Lamé constants. Differentiating its quadratic energy with respect to strain recovers stress. Here repeated indices are summed, so off-diagonal tensor shear entries contribute twice in the contraction. Engineering shear strain is twice the corresponding tensor component. Mixing these conventions is a common source of an incorrect factor of two. The energy formula assumes a stress-free reference and linear response; an already highly stressed material requires a more careful incremental analysis.
Volume and shape expose the stability test
Split the strain into its trace, which describes the first-order fractional volume change, and a traceless part that describes distortion. These parts are orthogonal under the tensor contraction. Substituting the split into the energy removes the coupling between them. The result says that a pure volume change is charged according to the bulk modulus, while a pure shape change is charged according to the shear modulus. Testing these two independent deformation families tests all symmetric small strains in this isotropic model.
Strict positive energy requires a positive shear modulus and a positive bulk modulus. It does not require the first Lamé constant to be positive. For example, a moderately negative first Lamé constant can coexist with both required positive moduli; such a material has an allowed auxetic elastic response. If either required modulus is zero, some nonzero strain costs no quadratic energy, so strict stability is lost. If either is negative, a strain in the corresponding family lowers the energy of the reference state.
A stable material can belong to an unstable structure
Material stability is a local statement about the constitutive energy. A slender compressed column can buckle even when its material has positive shear and bulk moduli. The column gains lateral deflection and releases work from the compressive load; the total potential energy includes that external loading contribution as well as stored bending energy. Boundary supports and geometry then determine whether the complete structure is stable. A check on elastic constants cannot replace a structural buckling calculation.
Use the quadratic-function lab as a one-coordinate picture of energy curvature. Before changing anything, predict what happens to the restoring slope at a fixed positive position when the coefficient doubles. Check the tangent readout: for the displayed quadratic, the slope doubles. This illustrates positive curvature in one chosen direction. The lab plots a dimensionless scalar quadratic and its area, not the six-dimensional strain energy, a material specimen, or a buckling test. Our actual stability conclusion comes from the volumetric and deviatoric decomposition, not from that picture alone.
Work it through
Stable constants with a negative first Lamé constant
An illustrative isotropic material has μ = 30 GPa and λ = −10 GPa. A small strain has trace θ = 0.002 and traceless diagonal entries 0.001, −0.001, and 0. Find the energy density and test strict material stability.
Compute K = −10 + 2(30)/3 = 10 GPa. Both K and μ are positive, despite the negative value of λ.
The contraction e:e is 0.001² + (−0.001)² = 0.000002. The distortional energy is 30 × 10⁹ × 0.000002 = 60,000 J/m³.
The volumetric contribution is (10 × 10⁹/2) × 0.002² = 20,000 J/m³. Adding the independent contributions gives 80,000 J/m³.
The energy density is 80 kJ/m³, and these constants satisfy the stated isotropic material-stability conditions. This does not certify a particular loaded structure against buckling.
The living notebook
Explore the model
Interactive lab · function
Inspect an energy curvature
At a fixed positive position, predict the change in tangent slope when the coefficient a doubles; then compare the readout for y = ax². This is a dimensionless one-coordinate analogy for a positive quadratic energy.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forPositive curvature gives a restoring slope in this scalar analogy; actual isotropic elastic stability requires separate positive bulk and shear moduli.
Pause & reason
Check your understanding
Your turn
Try it yourself
A proposed law has μ = 12 GPa and λ = −9 GPa. Test it using a purely volumetric strain with θ = 0.003. Is a positive shear modulus enough?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
First calculate K. A pure volume change has e = 0, so only the bulk-energy term remains.
Compare with the solution
K = −9 + 2(12)/3 = −1 GPa. The volume-change energy is (−10⁹/2) × 0.003² = −4,500 J/m³. The reference state therefore fails the positive-energy test even though its shear modulus is positive. The two kinds of deformation must be tested independently.
Keep the idea
Take it with you
- Stored elastic energy is work per volume, measured in J/m³ or Pa.
- For this isotropic linear model, strict material stability requires μ > 0 and K > 0; λ may be negative.
- Geometry, supports, and load potential can destabilize a structure made from stable material.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.