Elastic Constants Calculator

Related Calculators

Known constants pair: The two elastic constants you already know. For an isotropic linear elastic material, any valid two independent constants are enough to solve the others.

Young's modulus, E: Stretching stiffness. Higher E means the material changes length less under the same normal stress.

Poisson's ratio, nu: A unitless number that shows sideways strain compared with lengthwise strain. In classical isotropic elasticity, stable values must be greater than -1 and less than 0.5 [4].

Shear modulus, G: Resistance to shape change when layers slide sideways.

Bulk modulus, K: Resistance to uniform squeezing. A larger K means the material is harder to compress in volume.

First Lame constant, lambda: A standard isotropic elasticity constant used together with G in Hooke's law forms [1]. It can be negative and still be physically valid.

Physical feasibility note: A short message that tells whether the solved set is stable isotropic, auxetic, near-incompressible, or invalid.

Equation set used: The closed-form relation used for your selected input pair.

Mistake: Entering 30 instead of 0.30 when Known constants pair includes Poisson's ratio, nu .
Fix: In First known value (selected unit or ratio) or Second known value (selected unit or ratio) , enter nu as a unitless decimal such as 0.30, not a percent.

Mistake: Changing Modulus unit to MPa or psi but typing values that are really in GPa.
Fix: Make sure both First known value (selected unit or ratio) and Second known value (selected unit or ratio) use the exact Modulus unit you selected for modulus-type constants.

Mistake: Choosing the wrong Known constants pair , such as selecting Young's modulus and shear modulus when your second value is actually bulk modulus.
Fix: Recheck the pair first, because the calculator maps First known value (selected unit or ratio) and Second known value (selected unit or ratio) based on that selection.

Mistake: Thinking a negative First Lame constant, lambda must be an error.
Fix: Do not force lambda positive. Read the Physical feasibility note ; negative lambda can still be valid if the full isotropic stability conditions are satisfied.

Mistake: Ignoring a warning when Poisson's ratio, nu is very close to 0.5 or a selected pair is near a singular relation.
Fix: Check the Physical feasibility note and Equation set used . Near these limits, Bulk modulus, K and First Lame constant, lambda can become extremely sensitive to tiny input changes.

Mistake: Comparing raw numbers without noticing the display choice.
Fix: If values look unusually large or small, review Number display format and Result precision (significant figures) so you read the outputs consistently.

• This calculator assumes an isotropic, linear elastic material, so one set of constants is used in every direction and stress is proportional to strain.
• Results apply only when the chosen input pair is mathematically independent and physically feasible for isotropic stability.
• Poisson's ratio, nu must stay between -1 and 0.5 for the stable isotropic check used here; values outside that range are rejected.
• Some input pairs become singular or extremely sensitive near special values such as nu near 0.5, nu near 0, E near 3G, 9K near E, lambda plus G near 0, or 3K near lambda.
• Negative First Lame constant, lambda is allowed when the full solved set still gives positive E, G, and K with valid nu.
• Unit conversion changes only the modulus-type quantities. Poisson's ratio, nu is always unitless.
• Displayed values may differ slightly from hand calculations because of rounding, chosen significant figures, and display format.

How the solver works

The calculator treats the material as isotropic linear elastic, where any two independent elastic constants determine the others [1][3]. It reads the selected pair, converts modulus-type inputs into a consistent internal unit, applies the matching closed-form relation, then converts results back to the chosen display unit.

Equation sets

From E and ν: G = E / (2(1 + ν))

From E and ν: K = E / (3(1 - 2ν))

From E and ν: λ = Eν / ((1 + ν)(1 - 2ν))

From E and G: ν = E / (2G) - 1

From E and G: K = EG / (3(3G - E))

From E and G: λ = G(E - 2G) / (3G - E)

From E and K: ν = (3K - E) / (6K)

From E and K: G = 3EK / (9K - E)

From E and K: λ = 3K(3K - E) / (9K - E)

From E and λ: R = sqrt(E^2 + 9λ^2 + 2Eλ)

From E and λ: ν = 2λ / (E + λ + R)

From E and λ: G = (E - 3λ + R) / 4

From E and λ: K = (E + 3λ + R) / 6

From ν and G: E = 2G(1 + ν)

From ν and G: K = 2G(1 + ν) / (3(1 - 2ν))

From ν and G: λ = 2Gν / (1 - 2ν)

From ν and K: E = 3K(1 - 2ν)

From ν and K: G = 3K(1 - 2ν) / (2(1 + ν))

From ν and K: λ = 3Kν / (1 + ν)

From ν and λ: E = λ(1 + ν)(1 - 2ν) / ν

From ν and λ: G = λ(1 - 2ν) / (2ν)

From ν and λ: K = λ(1 + ν) / (3ν)

From G and K: E = 9KG / (3K + G)

From G and K: ν = (3K - 2G) / (2(3K + G))

From G and K: λ = K - 2G / 3

From G and λ: E = G(3λ + 2G) / (λ + G)

From G and λ: ν = λ / (2(λ + G))

From G and λ: K = λ + 2G / 3

From K and λ: E = 9K(K - λ) / (3K - λ)

From K and λ: ν = λ / (3K - λ)

From K and λ: G = 3(K - λ) / 2

Stability and feasibility checks

After solving, the calculator checks the standard isotropic stability conditions:

E > 0

G > 0

K > 0

-1 < ν < 0.5

If these fail, the pair is marked physically invalid for isotropic linear elasticity. It also blocks singular or nearly singular cases where a denominator is zero or too close to zero, because tiny input changes can cause huge output jumps.

Unit conversion

Modulus-type values are converted with standard pressure-unit relations. The calculator uses 1 kPa = 1000 Pa, 1 MPa = 1000000 Pa, 1 GPa = 1000000000 Pa, 1 psi = 6894.757293168 Pa, and 1 ksi = 1000 psi. Poisson's ratio, nu is unitless and is never converted.

Mini example

Suppose you choose Young's modulus and Poisson's ratio with E = 200 GPa and ν = 0.30. Then:

G = 200 / (2(1 + 0.30)) = 76.92 GPa

K = 200 / (3(1 - 2(0.30))) = 166.67 GPa

λ = 200(0.30) / ((1 + 0.30)(1 - 2(0.30))) = 115.38 GPa

Because E, G, and K are positive and ν = 0.30 stays between -1 and 0.5, the result is classified as stable isotropic.

Interpretation notes

A negative Poisson's ratio, nu means auxetic behavior, where the material expands sideways when stretched. Values of ν very close to 0.5 indicate near-incompressible behavior, so Bulk modulus, K and First Lame constant, lambda can become very large for a finite E. Negative First Lame constant, lambda can still be valid if the full solved set passes the stability checks.

Assumptions behind the result

This method assumes small-strain, linear elastic, isotropic behavior and does not model anisotropy, plastic deformation, viscoelastic effects, temperature dependence, damage, porosity changes, or nonlinear large-strain response. Real measured values can differ from the ideal relations because actual materials may not follow these assumptions exactly.