Modifying Operators

*e*

Following the above assignments, one defines

$$
\stackrel{\mathcal{\mathcal{C}}}{\dot{\mathcal{R}}} = \begin{bmatrix} \mathcal{R}^1 & \mathbf{0} \\ \mathbf{0} & \mathcal{R}^2 \end{bmatrix}. \tag{69}
$$

Within the block *R*1, the following overlapping sub-blocks can be distinguished

.

$$\mathbf{R}^1 = \begin{bmatrix} \vdots & \vdots \\ \cdots & \mathbf{R}\_{lk} & \cdots & \mathbf{R}\_{kl} & \cdots \\ & \vdots & \vdots \\ \cdots & \mathbf{R}\_{lk} & \cdots & \mathbf{R}\_{ll} & \cdots \\ & \vdots & & \vdots \end{bmatrix}. \tag{70}$$

.

Sub-blocks *Rkk*, *Rkl* are responsible for summation of the top and bottom displacement dofs, while *Rlk* and *Rll* for subtraction of these dofs. In turn, the block *R*<sup>2</sup> possesses the following structure based on the identity sub-blocks *I*

$$\mathcal{R}^2 = \text{diag}\left[ \dots , \mathcal{R}\_{\mathfrak{m}} , \dots \right], \quad \mathcal{R}\_{\mathfrak{m}} \equiv I = \text{diag}\left[ 1, 1, 1 \right]. \tag{71}$$

This means that neither the summation nor the subtraction is performed for the dofs *m*.

One more remark concerns the sums and differences of displacements present in the constraints (27). In the case of the symmetric-thickness geometry, these sums define mid-surface displacements. The first two differences, after division by the thickness *t*, give rotations of the normal to the mid-surface, while the third difference determines a transverse elongation of the corresponding normal. Note that only the latter difference needs to be constrained (equal to zero) in accordance with the last Equation (27). Because of this, multiplication of the penalty stiffness by the diagonal matrix *Z* is necessary. This matrix possesses the following structure:

$$\stackrel{\circ}{Z} = \left[ \begin{array}{cc} \mathbf{Z}^1 & \mathbf{0} \\ \mathbf{0} & \mathbf{Z}^2 \end{array} \right]. \tag{72}$$

The component block *Z*<sup>1</sup> is composed of the diagonal blocks *Zk* and *Zl* corresponding to the degrees of freedom *k* and *l*:

$$\mathbf{Z}^1 = \text{diag}\left[ \dots, \mathbf{Z}\_k, \dots, \mathbf{Z}\_l, \dots \right], \quad \mathbf{Z}\_k = \text{diag}\left[ 0, 0, 0 \right], \quad \mathbf{Z}\_l = \text{diag}\left[ 0, 0, 1 \right], \tag{73}$$

where the degrees of freedom *k* and *l* correspond to the sums and differences of the top and bottom displacement dofs. The second component block has the simpler form

$$\mathbf{Z}^2 = \text{diag}\left[\ldots, \mathbf{Z}\_m, \ldots\right], \quad \mathbf{Z}\_m = \text{diag}\left[0, 0, 0\right], \tag{74}$$

which results from the fact that, for the degrees of freedom *m*, other than the sums and differences of the displacement dofs, the penalty terms are not necessary.
