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A longitudinal plane wave in the isotropic elastic model: dispersion check and strain energy at explicit parameters

Example

Take $λ = 2$ , $μ = 1$ , $ρ_{0} = 1$ (The small-strain isotropic linear elastic constitutive law and its strain energy: here $λ + \frac{2 μ}{3} = \frac{8}{3} > 0$ ). Let $k = (π, 0, 0)$ , $a = (1, 0, 0)$ and

$ξ (t, x) = a \cos (π x_{1} - ω t), ω : = 2 π .$

Then $ξ$ is a longitudinal plane wave with speed $\sqrt{(λ + 2 μ) / ρ_{0}} = 2$ , and the associated strain and stress satisfy the elastic momentum identity; the strain energy density at time $t$ equals both $W (e) = 2 π^{2} \sin^{2} (π x_{1} - ω t)$ and the decomposition $μ ∣ e_{0} ∣_{F}^{2} + (λ / 2 + μ / 3) (tr e)^{2}$ with the same value.

Facts & Assumptions

Given: The parameters $λ = 2, μ = 1, ρ_{0} = 1$ ; the vectors $a = (1, 0, 0)$ , $k = (π, 0, 0)$ , $ω = 2 π$ ; the field $ξ (t, x) = a \cos (π x_{1} - ω t)$ .

[L1]

Plane-wave dispersion: the momentum identity holds for this ansatz exactly when $ρ_{0} ω^{2} a = μ ∣ k ∣_{2}^{2} a + (λ + μ) ⟨ k, a ⟩_{2} k$ ; for $a ∥ k$ this is $ρ_{0} ω^{2} = (λ + 2 μ) ∣ k ∣_{2}^{2}$ (Exact plane-wave solutions of the isotropic linear elastic model, with longitudinal and transverse speeds).

[L2]

Strain and stress: $e = \frac{1}{2} (D_{x} ξ + (D_{x} ξ)^{T})$ and $σ = λ (tr e) I_{3} + 2 μ e$ ; here $D_{x} ξ$ has the single entry $\partial_{1} ξ_{1} = - π \sin θ$ with $θ : = π x_{1} - ω t$ , so $e = diag (- π \sin θ, 0, 0)$ , $tr e = - π \sin θ$ (Exact plane-wave solutions of the isotropic linear elastic model, with longitudinal and transverse speeds, The small-strain isotropic linear elastic constitutive law and its strain energy, A total derivative computes every directional derivative, and its matrix is the Jacobian, Euclidean spatial calculus: gradient, divergence and material derivative, The trace $tr (A)$ as the sum of the diagonal entries, The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations).

[L3]

Strain-energy decomposition: $W (e) = μ ∣ e_{0} ∣_{F}^{2} + (λ / 2 + μ / 3) (tr e)^{2}$ with $e_{0} = e - \frac{tr e}{3} I_{3}$ and $∣ A ∣_{F}^{2} = tr (A^{T} A)$ (The isotropic elastic strain energy is $μ ∣ e_{0} ∣_{F}^{2} + (λ / 2 + μ / 3) (tr e)^{2}$ , hence positive definite under $μ > 0$ , $λ + 2 μ / 3 > 0$ ); trigonometric values and derivatives as usual (The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, Parity and the Pythagorean identity for sine and cosine).

Verification

technique · direct; substitute the numbers into the dispersion relation and the two energy expressions

1.1givenL1algebra

Check the dispersion relation: $μ ∣ k ∣_{2}^{2} = 1 \cdot π^{2} = π^{2}$ , $(λ + 2 μ) ∣ k ∣_{2}^{2} = (2 + 2) π^{2} = 4 π^{2}$ , and $ρ_{0} ω^{2} = 1 \cdot (2 π)^{2} = 4 π^{2}$ ; the relation of [L1] holds, so the ansatz solves the elastic momentum equation at these parameters.

2.1step 1.1L2L3algebra

Compute the strain energy directly: from [L2], $tr (e^{2}) = e_{11}^{2} = π^{2} \sin^{2} θ$ , so $W = \frac{λ}{2} (tr e)^{2} + μ tr (e^{2}) = π^{2} \sin^{2} θ + π^{2} \sin^{2} θ = 2 π^{2} \sin^{2} θ$ .

3.1step 2.1L2L3algebra∎

Compute the decomposition: $e_{0} = e - \frac{tr e}{3} I_{3} = diag (- \frac{2 π}{3} \sin θ, \frac{π}{3} \sin θ, \frac{π}{3} \sin θ)$ , so $∣ e_{0} ∣_{F}^{2} = \frac{4 + 1 + 1}{9} π^{2} \sin^{2} θ = \frac{2}{3} π^{2} \sin^{2} θ$ ; also $(λ / 2 + μ / 3) (tr e)^{2} = (1 + \frac{1}{3}) π^{2} \sin^{2} θ = \frac{4}{3} π^{2} \sin^{2} θ$ . Then $μ ∣ e_{0} ∣_{F}^{2} + (λ / 2 + μ / 3) (tr e)^{2} = \frac{2}{3} π^{2} \sin^{2} θ + \frac{4}{3} π^{2} \sin^{2} θ = 2 π^{2} \sin^{2} θ$ , in agreement with step 2.1.

Remarks

The example is a finite parametric check of the theorem and of the energy identity; it does not show that every elastic solution is such a wave, and no material is identified with these Lamé parameters.

A longitudinal plane wave in the isotropic elastic model: dispersion check and strain energy at explicit parameters

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources