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Young's modulus describes the ratio of tensile stress to tensile strain for a material in the linear-elastic range. Tensile stress is force per cross-sectional area; tensile strain is relative length change.
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Formula
E and σ in Pa (N/m2); ε dimensionless; F in N; L0 and ΔL in m; A0 in m2
| Symbol | Meaning | Unit |
|---|---|---|
| Young’s modulus, a measure of axial material stiffness | Pa | |
| Tensile stress, here F/A0 | Pa | |
| Tensile strain, here ΔL/L0 | dimensionless | |
| Axial tensile force normal to the cross-section | N | |
| Initial cross-sectional area of the specimen | m2 | |
| Initial length before loading | m | |
| Length change under tensile load | m |
E = \frac{\sigma}{\varepsilon} = \frac{F L_0}{A_0 \Delta L}The definitions are σ=F/A0 and ε=ΔL/L0. In the proportional-elastic range, Hooke’s material law gives σ=Eε; division yields E=σ/ε. E is a material property under specified conditions, not a direct geometric property of a component. The law does not remain unchanged beyond yielding, plastic deformation, strong nonlinearity or viscoelastic effects. Tensile tests often use the original area; engineering and true stress differ at larger deformations.
Applies in the uniaxial linear-elastic range; strain is small and area and material response follow the stated convention.
Young’s modulus is the slope of the linear relation between tensile stress and tensile strain.
Change in length
Only in the linear-elastic range under axial loading.
Tensile strain and stress
Strain is dimensionless; stress is expressed in Pa.
A bar with E=200 GPa, L0=1 m and A0=1 cm2 is pulled with 10 kN. Find ΔL.
ΔL=FL0/(EA0)=104·1/(2·1011·10−4)=5·10−4 m=0.5 mm.
A specimen lengthens by 0.2 mm from an initial 0.50 m. What is ε?
ε=0.0002/0.50=4·10−4.
Strain has no unit.
Using the extension ΔL as strain without dividing by initial length.
ε=ΔL/L0 is a relative, dimensionless change in length.
Assuming the calculated E remains constant beyond the yield region.
Here E describes only the linear-elastic material range.
Formula cluster
Stress per strain is a material response; component geometry also determines total deformation.
A bar with L0=2.0 m and A0=3.0·10−4 m2 is extended by ΔL=0.60 mm under F=18 kN. Then σ=60 MPa, ε=3.0·10−4 and E=60 MPa/(3.0·10−4)=200 GPa.
Estimate elastic extension, compare materials and size axially loaded components in the linear range.
Here is how to work through a typical Young's modulus (E = σ/ε = FL₀/(A₀ΔL)) task step by step:
Task
A bar with E=200 GPa, L0=1 m and A0=1 cm2 is pulled with 10 kN. Find ΔL.
Solution path
ΔL=FL0/(EA0)=104·1/(2·1011·10−4)=5·10−4 m=0.5 mm.
Task
A specimen lengthens by 0.2 mm from an initial 0.50 m. What is ε?
Solution path
ε=0.0002/0.50=4·10−4.
Strain has no unit.
Young’s modulus describes how a material responds to a given strain under axial tension or compression in the linear-elastic range. A larger E means more stress is needed for the same strain; the material is stiffer in this sense. E is neither force nor strain, but their ratio. Its value depends on material and test conditions. Real components also depend on length and cross-section, so a stiff material alone does not automatically make every component stiff. Component design must therefore account for geometry, not just the material property.
In tension, engineering stress is σ=F/A0, force divided by original cross-sectional area. Engineering strain is ε=ΔL/L0, the extension relative to initial length. Within the linear-elastic range, σ=Eε, so E=σ/ε. Stress has units of pascals and strain is dimensionless, so E is also measured in pascals. The proportionality ends when the material no longer responds proportionally and elastically. For larger deformations, true stress and strain may use different definitions. Area enters stress, while length enters relative strain. Under compression, use a consistent sign convention for both stress and strain.
Strain compares two lengths: the change ΔL and original length L0. Both use the same unit, so metres cancel. A strain of 0.001, for example, means a relative length change of 0.1 percent. This dimensionless quantity can be compared across specimens of different lengths. Tables may express it as a fraction, percent or microstrain. Convert these representations consistently before calculating: 0.1 percent equals 0.001, not 0.1. In a tensile test, L0 is the initial gauge length between the marks. Gauge marks should span the region being evaluated.
Rearrange E=FL0/(A0ΔL) to get ΔL=FL0/(EA0). Use force in newtons, length in metres, area in square metres and E in pascals. For example, 10 kN, 1 m, 1 cm2 and 200 GPa give an extension of 0.5 mm. Then check that the assumed strain is small and stress remains in the linear-elastic range. If cross-section or material varies along the bar, the simple constant formula is insufficient. Units and orders of magnitude make a quick plausibility check especially useful here. This example corresponds to strain 0.0005, or 0.05 percent.
No. In this formula, E is the slope in the proportional-elastic part of a stress–strain diagram. Beyond the proportional or elastic limit, the curve may become nonlinear; after yielding, permanent deformation often remains. A tangent or secant modulus can still be defined over a chosen interval, but that is a different evaluation. Polymers and biological materials can also depend on time and temperature. State E together with material condition, temperature and the strain range in which it applies. Within the elastic range, unloading should bring the specimen approximately back to its initial length.
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Which formula describes Young's modulus?
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How do you rearrange Young's modulus for Change in length?
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Which common mistake happens with Young's modulus?
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