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The displacement equation of the linear elastic branch: rho_0 partial_t^2 xi = rho_0 b + mu_s Laplacian xi + (lambda + mu_s) grad div xi

Physical scope

Conditional on the adopted reference-coordinate balance and small-strain isotropic linear constitutive law: their stress divergence is exactly mu_s Laplacian xi plus (lambda+mu_s) grad div xi. This yields the displacement equation on the reference region; no exact finite-deformation identification, approximation error bound, measured modulus or solution existence is asserted.

Statement

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and the reference-coordinate branch of Adopted isotropic linear elastic constitutive model with its small-strain regime and strain energy on $I \times O$ : $ξ$ is $C^{2}$ , $ρ_{0} > 0$ is constant, $b : = b_{r e f}$ is continuous, and $λ, μ_{s}$ are constant with $μ_{s} > 0$ and $λ + 2 μ_{s} / 3 > 0$ . Its stress is $σ = λ (tr e) I_{3} + 2 μ_{s} e$ , $e = \frac{1}{2} (D_{X} ξ + D_{X} ξ^{T})$ .

Then the adopted reference momentum balance

$\begin{matrix} ρ_{0} \partial_{t}^{2} ξ = ρ_{0} b + {div}_{X} σ & (M) \end{matrix}$

is equivalent to the displacement equation

$\begin{matrix} ρ_{0} \partial_{t}^{2} ξ = ρ_{0} b + μ_{s} Δ ξ + (λ + μ_{s}) \nabla (div ξ), & (E) \end{matrix}$

equivalently componentwise $ρ_{0} \partial_{t}^{2} ξ_{i} = ρ_{0} b_{i} + μ_{s} Δ ξ_{i} + (λ + μ_{s}) \partial_{i} (div ξ)$ for $i = 1, 2, 3$ , where $Δ = \sum_{j} \partial_{j}^{2}$ and $div ξ = \sum_{j} \partial_{j} ξ_{j}$ (Euclidean spatial calculus: gradient, divergence and material derivative). In particular, for $b \equiv 0$ equation (E) is linear with constant coefficients. The statement asserts no solution existence, no boundary conditions and no measured modulus.

Facts & Assumptions

Given: ${A C}_{ω}$ ; the reference-coordinate fields and adopted linear balance (M).

[F1]

Reference balance (Adopted isotropic linear elastic constitutive model with its small-strain regime and strain energy clause (iv)): $ρ_{0} \partial_{t}^{2} ξ = ρ_{0} b + g$ , with $g_{i} = \sum_{j} \partial_{X_{j}} σ_{i j}$ on $I \times O$ .

[F2]

Reference constitutive law (Adopted isotropic linear elastic constitutive model with its small-strain regime and strain energy clauses (i)-(iii), The small-strain isotropic linear elastic constitutive law and its strain energy): $σ = λ (tr e) I_{3} + 2 μ_{s} e$ , $e = \frac{1}{2} (D_{X} ξ + D_{X} ξ^{T})$ , with the stated constant coefficients and $ξ \in C^{2}$ .

[F3]

Symmetry of mixed partials (Clairaut--Schwarz theorem for continuous second partial derivatives): for a function of class $C^{2}$ on an open set of $R^{3}$ , the second partial derivatives commute, $\partial_{j} \partial_{i} ξ_{k} = \partial_{i} \partial_{j} ξ_{k}$ .

[F4]

Algebra of derivatives and matrix identities (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The trace $tr (A)$ as the sum of the diagonal entries, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, Euclidean spatial calculus: gradient, divergence and material derivative): $tr e = div ξ$ , $\sum_{j} \partial_{j}^{2} ξ_{i} = Δ ξ_{i}$ , and the product rule for the derivatives of the matrix-valued field $σ$ .

Proof

technique · direct; write the adopted reference momentum equation and compute the divergence of the linear elastic stress with the commuting mixed partials

1.1givenF1F2

By [F1], the adopted reference equation (M) reads $ρ_{0} \partial_{t}^{2} ξ = ρ_{0} b + g$ with $g_{i} = \sum_{j} \partial_{j} σ_{i j}$ , with all derivatives taken in $X$ .

1.2givenF2F3F4algebra

Compute the divergence of the stress: from $σ_{i j} = λ (tr e) δ_{i j} + 2 μ_{s} e_{i j}$ of [F2], the product rule of [F4] gives $\sum_{j} \partial_{j} σ_{i j} = λ \partial_{i} (tr e) + 2 μ_{s} \sum_{j} \partial_{j} e_{i j}$ . Since $tr e = \sum_{j} \partial_{j} ξ_{j} = div ξ$ by [F4], the first term is $λ \partial_{i} (div ξ)$ . For the second, $2 \sum_{j} \partial_{j} e_{i j} = \sum_{j} \partial_{j} (\partial_{j} ξ_{i} + \partial_{i} ξ_{j}) = Δ ξ_{i} + \partial_{i} (div ξ)$ , where the interchange of $\partial_{j}$ and $\partial_{i}$ on $\partial_{i} ξ_{j}$ uses the commuting mixed partials of [F3]. Hence $g_{i} = μ_{s} Δ ξ_{i} + (λ + μ_{s}) \partial_{i} (div ξ)$ .

2.1step 1.1step 1.2algebra∎

Substituting step 1.2 into step 1.1 gives the componentwise form of (E), equivalently the vector identity $ρ_{0} \partial_{t}^{2} ξ = ρ_{0} b + μ_{s} Δ ξ + (λ + μ_{s}) \nabla (div ξ)$ ; the equivalence with (M) follows by substituting the same stress divergence in either direction, and for $b \equiv 0$ the displayed operator is linear with constant coefficients, while no solution, boundary condition or measured modulus is asserted.

Remarks

Approximation status. The reference balance is adopted in Adopted isotropic linear elastic constitutive model with its small-strain regime and strain energy clause (iv). A rigorous passage from a finite-strain model with an error bound is a separate approximation problem.
Divergence of a symmetric matrix. The row divergence used in (M) coincides with the column divergence because $σ$ is symmetric by direct construction in this branch, so writing $div σ$ in (E) is unambiguous.
Dimensions. With $[ξ] = ℓ$ and $[ρ_{0}] = μ ℓ^{- 3}$ , both sides of (E) have dimension $μ ℓ^{- 2} τ^{- 2}$ (force per volume); the moduli carry $[λ] = [μ_{s}] = μ ℓ^{- 1} τ^{- 2}$ (Continuum body data, the fixed inertial frame and the dimensions of the fields).

The displacement equation of the linear elastic branch: rho_0 partial_t^2 xi = rho_0 b + mu_s Laplacian xi + (lambda + mu_s) grad div xi

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources