Efficient uncertainty quantification in stochastic finite element analysis based on functional principal components

Abstract

The great influence of uncertainties on the behavior of physical systems has always drawn attention to the importance of a stochastic approach to engineering problems. Accordingly, in this paper, we address the problem of solving a Finite Element analysis in the presence of uncertain parameters. We consider an approach in which several solutions of the problem are obtained in correspondence of parameters samples, and propose a novel non-intrusive method, which exploits the functional principal component analysis, to get acceptable computational efforts. Indeed, the proposed approach allows constructing an optimal basis of the solutions space and projecting the full Finite Element problem into a smaller space spanned by this basis. Even if solving the problem in this reduced space is computationally convenient, very good approximations are obtained by upper bounding the error between the full Finite Element solution and the reduced one. Finally, we assess the applicability of the proposed approach through different test cases, obtaining satisfactory results.

Appendices

Appendix 1

This Appendix describes the general procedure to derive matrix $\mathbb {W}$. As suggested in Sect. 3, the bivariate function that represents the horizontal and vertical displacement can be written as

$$\begin{aligned} \varvec{v}_i(\varvec{x})=\sum _{f=1}^{2N_{f}}c_{if}\phi _f(\varvec{x}), \quad i=1, \dots , N , \end{aligned}$$

where $N_{f}$ is the number of nodes in the grid, equal to $(n+1)^2$, and

$$\begin{aligned} \varvec{c}_i = (c_{i1}^u, c_{i1}^v, c_{i2}^u,c_{i2}^v,\dots ,c_{iN_{f}}^u,c_{iN_{f}}^v). \end{aligned}$$

where $c_{ij}^u$ and $c_{ij}^v$ are the horizontal and vertical nodal displacements, respectively. Similarly,

$$\begin{aligned} \varvec{\phi } = (\tilde{\phi }_1, \tilde{\phi }_1, \tilde{\phi }_2,\tilde{\phi }_2, \dots ,\tilde{\phi }_{N_{f}},\tilde{\phi }_{N_{f}}) \end{aligned}$$

where $\tilde{\phi }_f$ is the finite element basis function centered in node f, $f=1,\dots ,N_{f}$. Hence,

$$\begin{aligned} \mathbb {W}= & {} \int \varvec{\phi } \ \varvec{\phi }'\\= & {} \begin{pmatrix} \int \phi _1\phi _1 &{} \int \phi _1\phi _2 &{} \dots &{} \int \phi _1\phi _{2N_{f}} \\ \int \phi _2\phi _1 &{} \int \phi _2\phi _2 &{} \dots &{} \int \phi _2\phi _{2N_{f}} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ \int \phi _{2N_{f}}\phi _1 &{} \int \phi _ {2N_{f}}\phi _2 &{} \dots &{} \int \phi _{2N_{f}}\phi _{2N_{f}} \\ \end{pmatrix}\\= & {} \begin{pmatrix} \int \tilde{\phi }_1\tilde{\phi }_1 &{} \int \tilde{\phi }_1\tilde{\phi }_1 &{} \dots &{} \int \tilde{\phi }_1\tilde{\phi }_{2N_{f}} &{}\int \tilde{\phi }_1\tilde{\phi }_{2N_{f}} \\ \int \tilde{\phi }_1\tilde{\phi }_1 &{} \int \tilde{\phi }_1\tilde{\phi }_1 &{} \dots &{} \int \tilde{\phi }_1\tilde{\phi }_{2N_{f}} &{}\int \tilde{\phi }_1\tilde{\phi }_{2N_{f}} \\ \int \tilde{\phi }_2\tilde{\phi }_1 &{} \int \tilde{\phi }_2\tilde{\phi }_1 &{} \dots &{} \int \tilde{\phi }_2\tilde{\phi }_{2N_{f}} &{} \int \tilde{\phi }_2\tilde{\phi }_{2N_{f}} \\ \int \tilde{\phi }_2\tilde{\phi }_1 &{} \int \tilde{\phi }_2\tilde{\phi }_1 &{} \dots &{} \int \tilde{\phi }_2\tilde{\phi }_{2N_{f}} &{} \int \tilde{\phi }_2\tilde{\phi }_{2N_{f}} \\ \vdots &{}\vdots &{} \ddots &{} \vdots &{} \vdots \\ \int \tilde{\phi }_{2N_{f}}\tilde{\phi }_1 &{} \int \tilde{\phi }_{2N_{f}}\tilde{\phi }_1 &{} \dots &{} \int \tilde{\phi }_{2N_{f}}\tilde{\phi }_{2N_{f}} &{} \int \tilde{\phi }_{2N_{f}}\tilde{\phi }_{2N_{f}} \\ \end{pmatrix} \end{aligned}$$

The procedure outlined till now is general and does not depend on the specific case, i.e., the shape of the domain and the mesh. In Appendix 2, we apply the procedure to the specific test case considered in this paper.

Appendix 2

In this Appendix, we compute the matrix $\mathbb {W}$ specific for our test case, with rectangular domain and structured triangular mesh.

Thanks to an appropriate change of variables, we obtain that the number of elements of the FE basis with a peak in node i, $i=1, \dots , F$, is $\int \tilde{\phi }_i\tilde{\phi }_i = \frac{h_1 h_2}{12}$ and that the number of FE basis with a peak in node i and the other in node k is $\int \tilde{\phi }_i\tilde{\phi }_k = \frac{h_1 h_2}{24}.$ Then, due to the structured mesh, $\mathbb {W}$ is as follows.

Diagonal entries:

• $w_{kk}=\frac{h_1h_2}{6}$ if $k=1, n^2$;

• $w_{kk}=\frac{h_1h_2}{12}$ if $k=n, n(n-1)+1$;

• $w_{kk}=\frac{h_1h_2}{4}$ if k belongs to a border that it is not an angle;

• $w_{kk}=\frac{h_1h_2}{2}$ otherwise.

Non-diagonal entries:

• If i corresponds to an internal node

  $w_{ij} = \frac{h_1h_2}{12}\,j = i+1, i-1, i+n, i-n, i+n+1, i-n-1$;

  $w_{ij} = 0$ otherwise.

• If i corresponds to a node belonging to the bottom edge

  $w_{ij} = \frac{h_1h_2}{12}\,j = i+n, i+n+1$;

  $w_{ij} = \frac{h_1h_2}{24}\,j = i+1, i-1$;

  $w_{ij} = 0$ otherwise.

• If i corresponds to a node belonging to the top edge

  $w_{ij} = \frac{h_1h_2}{12}\,j = i-n, i-n-1$;

  $w_{ij} = \frac{h_1h_2}{24}\,j = i+1, i-1$;

  $w_{ij} = 0$ otherwise.

• If i corresponds to a node belonging to the right edge

  $w_{ij} = \frac{h_1h_2}{12}\,j = i-1, i-n-1$;

  $w_{ij} = \frac{h_1h_2}{24}\,j = i+n, i-n$;

  $w_{ij} = 0$ otherwise.

• If i corresponds to a node belonging to the left edge

  $w_{ij} = \frac{h_1h_2}{12}\,j = i+1, i+n+1$;

  $w_{ij} = \frac{h_1h_2}{24}\,j = i+n, i-n$;

  $w_{ij} = 0$ otherwise.

• If $i=1$

  $w_{ij} = \frac{h_1h_2}{12}\,j = i+n+1$;

  $w_{ij} = \frac{h_1h_2}{24}\,j = i+1, i+n$;

  $w_{ij} = 0$ otherwise.

• If $i=n$

  $w_{ij} = \frac{h_1h_2}{24}\,j = i-1, i+n$;

  $w_{ij} = 0$ otherwise.

• If $i=n(n-1)+1$

  $w_{ij} = \frac{h_1h_2}{24}\,j = i+1, i-n$;

  $w_{ij} = 0$ otherwise.

• If $i=n^2$

  $w_{ij} = \frac{h_1h_2}{12}\,j = i-n-1$;

  $w_{ij} = \frac{h_1h_2}{24}\,j = i-1, i-n$;

  $w_{ij} = 0$ otherwise.

The resulting sparsity pattern of this matrix $\mathbb {W}$ is shown in Fig. 4.