Reduced Basis Approximation and A Posteriori Error Estimation: Applications to Elasticity Problems in Several Parametric Settings

Abstract

In this work we consider (hierarchical, Lagrange) reduced basis approximation and a posteriori error estimation for elasticity problems in affinely parametrized geometries. The essential ingredients of the methodology are: a Galerkin projection onto a low-dimensional space associated with a smooth “parametric manifold”—dimension reduction; an efficient and effective greedy sampling methods for identification of optimal and numerically stable approximations—rapid convergence; an a posteriori error estimation procedures—rigorous and sharp bounds for the functional outputs related with the underlying solution or related quantities of interest, like stress intensity factor; and Offline-Online computational decomposition strategies—minimum marginal cost for high performance in the real-time and many-query (e.g., design and optimization) contexts. We present several illustrative results for linear elasticity problem in parametrized geometries representing 2D Cartesian or 3D axisymmetric configurations like an arc-cantilever beam, a center crack problem, a composite unit cell or a woven composite beam, a multi-material plate, and a closed vessel. We consider different parametrization for the systems: either physical quantities—to model the materials and loads—and geometrical parameters—to model different geometrical configurations—with isotropic and orthotropic materials working in plane stress and plane strain approximation. We would like to underline the versatility of the methodology in very different problems. As last example we provide a nonlinear setting with increased complexity.

Appendix

8.1.1 Stress-Strain Matrices

In this section, we denote E _i, i = 1, 3 as the Young’s moduli, ν _ij; i, j = 1, 2, 3 as the Poisson ratios; and G ₁₂ as the shear modulus of the material.

8.1.1.1 Isotropic Cases

For both of the following cases, E = E ₁ = E ₂, and ν = ν ₁₂ = ν ₂₁.

Isotropic plane stress:

$$\displaystyle \begin{aligned}{}[{\mathbf{E}}] = \frac{E}{(1-\nu^2)}\left[ \begin{array}{ccc} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & 2(1+\nu) \\ \end{array} \right]. \end{aligned}$$

Isotropic plane strain:

$$\displaystyle \begin{aligned}{}[{\mathbf{E}}] = \frac{E}{(1+\nu)(1-2\nu)}\left[ \begin{array}{ccc} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & 2(1+\nu) \\ \end{array} \right]. \end{aligned}$$

8.1.1.2 Orthotropic Cases

Here we assume that the orthotropic material axes are aligned with the axes used for the analysis of the structure. If the structural axes are not aligned with the orthotropic material axes, orthotropic material rotation must be rotated by with respect to the structural axes. Assuming the angle between the orthogonal material axes and the structural axes is θ, the stress-strain matrix is given by $[{\mathbf {E}}] = [{{\boldsymbol {T}}}(\theta )][\hat {{\mathbf {E}}}][{{\boldsymbol {T}}}(\theta )]^T$, where

$$\displaystyle \begin{aligned}{}[{{\boldsymbol{T}}}(\theta)] = \left[ \begin{array}{ccc} \cos^2\theta & \sin^2\theta & -2\sin\theta\cos\theta \\ \sin^2\theta & \cos^2\theta & 2\sin\theta\cos\theta \\ \sin\theta\cos\theta & -sin\theta\cos\theta & \cos^2\theta-\sin^2\theta\\ \end{array} \right]. \end{aligned}$$

Orthotropic plane stress:

$$\displaystyle \begin{aligned}{}[\hat{{\mathbf{E}}}] = \frac{1}{(1-\nu_{12}\nu_{21})}\left[ \begin{array}{ccc} E_1 & \nu_{12}E_1 & 0 \\ \nu_{21}E_2 & E_2 & 0 \\ 0 & 0 & (1-\nu_{12}\nu_{21})G_{12} \\ \end{array} \right]. \end{aligned}$$

Note here that the condition

$$\displaystyle \begin{aligned} \nu_{12}E_1 = \nu_{21}E_2 \end{aligned} $$

(8.44)

must be required in order to yield a symmetric [E].

Orthotropic plane strain:

$$\displaystyle \begin{aligned}{}[\hat{{\mathbf{E}}}] = \frac{1}{\varLambda}\left[ \begin{array}{ccc} (1-\nu_{23}\nu_{32})E_1 & (\nu_{12}+\nu_{13}\nu_{32})E_1 & 0 \\ (\nu_{21}+\nu_{23}\nu_{31})E_2 &(1-\nu_{13}\nu_{31})E_2 & 0 \\ 0 & 0 & \varLambda G_{12} \\ \end{array} \right]. \end{aligned}$$

Here Λ = (1 − ν ₁₃ ν ₃₁)(1 − ν ₂₃ ν ₃₂) − (ν ₁₂ + ν ₁₃ ν ₃₂)(ν ₂₁ + ν ₂₃ ν ₃₁). Furthermore, the following conditions,

$$\displaystyle \begin{aligned}\nu_{12}E_1 = \nu_{21}E_2, \quad \nu_{13}E_1 = \nu_{31}E_3, \quad \nu_{23}E_2 = \nu_{32}E_3,\end{aligned}$$

must be satisfied, which leads to a symmetric [E].

An reasonable good approximation for the shear modulus G ₁₂ in orthotropic case is given by [10] as

$$\displaystyle \begin{aligned} \frac{1}{G_{12}} \approx \frac{(1+\nu_{21})}{E_1} + \frac{(1+\nu_{12})}{E_2}. \end{aligned} $$

(8.45)