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For linear uniaxial loading, elastic energy per unit volume is the area under the stress–strain curve.
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Formula
u in J/m3 = Pa; σ in Pa; ε dimensionless; E in Pa
| Symbol | Meaning | Unit |
|---|---|---|
| Stored elastic energy per unit volume | J/m3 | |
| Uniaxial normal stress | Pa | |
| Uniaxial strain | dimensionless | |
| Young’s modulus in the linear range | Pa |
u=\frac{1}{2}\sigma\varepsilon=\frac{\sigma^2}{2E}=\frac{1}{2}E\varepsilon^2For linear-elastic behavior, work per volume is the integral ∫σdε; a straight line from zero gives triangular area ½σε. The equivalent forms use σ=Eε. Nonlinear materials require integration under the actual curve. Energy density has pressure dimensions because one joule per cubic metre equals one pascal.
For linear-elastic behavior, work per volume is the integral ∫σdε; a straight line from zero gives triangular area ½σε. The equivalent forms use σ=Eε. Nonlinear materials require integration under the actual curve. Energy density has pressure dimensions because one joule per cubic metre equals one pascal.
For linear uniaxial loading, elastic energy per unit volume is the area under the stress–strain curve.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Elastic strain-energy density under uniaxial stress” and check its unit and magnitude.
For σ=100 MPa and E=200 GPa, ε=5×10−4 and u=σ2/(2E)=25,000 J/m3.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: u in J/m3 = Pa; σ in Pa; ε dimensionless; E in Pa.
Formula cluster
For linear uniaxial loading, elastic energy per unit volume is the area under the stress–strain curve.
For σ=100 MPa and E=200 GPa, ε=5×10−4 and u=σ2/(2E)=25,000 J/m3.
Estimate elastic energy storage in materials, springs and components.
Here is how to work through a typical Elastic strain-energy density under uniaxial stress (u = ½σε = σ²/(2E) = ½Eε²) task step by step:
Task
Work through the numerical example for “Elastic strain-energy density under uniaxial stress” and check its unit and magnitude.
Solution path
For σ=100 MPa and E=200 GPa, ε=5×10−4 and u=σ2/(2E)=25,000 J/m3.
In a linear-elastic plot, σ rises proportionally from zero to its final value, so the area under the curve is a triangle. Its area is ½σε. For a nonlinear material, energy is instead the integral ∫σdε; the factor one-half cannot be applied automatically. This energy density applies to uniaxial linear-elastic loading from zero load to the stated stress. It is the area under the stress–strain curve, not an additional force. Use E and σ from the same elastic material state. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
u is energy per volume. A joule is a newton-metre; divided by m3 it becomes N/m2, or pascals. This is dimensionally correct even though u is energy density rather than ordinary mechanical pressure. The unit check helps distinguish total energy U from density u. This energy density applies to uniaxial linear-elastic loading from zero load to the stated stress. It is the area under the stress–strain curve, not an additional force. Use E and σ from the same elastic material state. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
For uniform stress and strain over volume V, total energy is U=uV. If the stress field varies through a component, integrate u spatially. For a linear tensile specimen, U=½FΔL also applies when force and extension rise proportionally and remain elastic. This energy density applies to uniaxial linear-elastic loading from zero load to the stated stress. It is the area under the stress–strain curve, not an additional force. Use E and σ from the same elastic material state. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
Write the quantity and unit from the variable list next to every value: u in J/m3 = Pa; σ in Pa; ε dimensionless; E in Pa. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example For σ=100 MPa and E=200 GPa, ε=5×10−4 and u=σ2/(2E)=25,000 J/m3. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. This energy density applies to uniaxial linear-elastic loading from zero load to the stated stress. It is the area under the stress–strain curve, not an additional force. Use E and σ from the same elastic material state. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
The background states this starting condition: For linear-elastic behavior, work per volume is the integral ∫σdε; a straight line from zero gives triangular area. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. This energy density applies to uniaxial linear-elastic loading from zero load to the stated stress. It is the area under the stress–strain curve, not an additional force. Use E and σ from the same elastic material state. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
Quanta exam set
Curated exam set for "Elastic strain-energy density under uniaxial stress":
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Which formula describes Elastic strain-energy density under uniaxial stress?
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