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Rigid motions in a free-body stiffness nullspace

Statement

Let D⊂Rd, d∈{2,3}, be a connected polygonal or polyhedral domain with a finite conforming affine simplicial mesh Th and no essential boundary constraints. Define the continuous P1 vector space Sh={v∈C0(D‾;Rd):v∣K is affine for every K∈Th} and its global nodal basis (ϕi)i=1N. For constants μ>0 and λ≥0, set ε(v)=(∇v+∇vT)/2, k(v,w)=2μ∫Dε(v):ε(w)+λ∫Ddiv⁡v div⁡w, and Kij=k(ϕj,ϕi). For every infinitesimal rigid motion r(x)=a+Ωx, with a∈Rd and ΩT=−Ω, its nodal coefficient vector lies in ker⁡K. Thus the range of any complete nullspace basis supplied to Algebraic gauge for a rigid-body stiffness nullspace must contain these translation and rotation vectors. This example does not claim they exhaust the nullspace on every mesh.

Facts & Assumptions

Given: A conforming affine simplicial mesh, its global continuous P1 vector space without essential boundary constraints, and the displayed isotropic stiffness bilinear form.

[F1]

The algebraic gauge procedure requires a complete basis R of the stiffness kernel and a compatible load before it is applied. (Algebraic gauge for a rigid-body stiffness nullspace)

Verification

1.1givenconstruct

On each nondegenerate simplex, an affine map is uniquely determined by its values at the d+1 affinely independent vertices, so its nodal P1 interpolant equals it. On each shared face the affine traces are determined by the same face-vertex values, so they agree and define a global continuous member of Sh.

2.1F1step 1.1algebra∎

Since Ω is skew-symmetric, ε(r)=(Ω+ΩT)/2=0 and div⁡r=tr⁡Ω=0. Thus the stiffness bilinear form pairs r to zero with every test basis function, which is exactly Kr=0 in coefficient coordinates. By [F1], every complete gauge basis must span these null vectors, although other null modes are not excluded here.

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