Question 4 of 7: Geometric isotropy of a bilinear element

Question 4: Geometric isotropy of a bilinear element (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A square element of side $L$ with the bilinear interpolation $u = C_1 + C_2x + C_3y + C_4xy$ written in Cartesian axes centred at node 1 ($x$ towards node 2, $y$ towards node 4), and a second frame $\xi,\eta$ centred at node 4 with $\xi$ directed towards node 3 and $\eta$ directed downwards towards node 1.

Find. A statement of what geometric isotropy means, the two polynomial properties that guarantee it, and a demonstration that the given interpolation retains its form when re-expressed in the node-4 frame.

Parts (a) and (b) are short-answer theory and are best answered in prose before the algebra of part (c).

(a) The meaning of geometric isotropy. An element is geometrically isotropic — equivalently, it possesses spatial isotropy or geometric invariance — when the field it interpolates does not depend on the position or orientation of the coordinate system used to describe it. Rotate the axes, translate the origin to a different node, or renumber the nodes consistently, and the assumed polynomial must retain exactly the same form, so that the computed displacements, temperatures or stresses at a physical point are unchanged. The practical consequence is that an element behaves identically no matter how the analyst chooses to orient it in the mesh: a bar of material does not become stiffer because the modeller rotated it by thirty degrees.

(b) The two required properties of the polynomial. First, the polynomial must be complete: it must contain every term of every order up to the highest order it uses, so that no lower-order term is missing. Completeness up to first order — the constant and both linear terms — is what allows the element to represent rigid-body motion and a state of constant strain, and it is a necessary condition for convergence as well as for isotropy. Second, any terms of an order that is only partially represented must be included in symmetric (balanced) pairs drawn from the corresponding row of Pascal's triangle. If $x^{2}$ appears then $y^{2}$ must appear; if the incomplete quadratic term $xy$ is used, it is itself symmetric in $x$ and $y$ and may stand alone. Taking a lopsided selection such as $\{1,x,y,x^{2}\}$ destroys isotropy, because rotating the axes generates a $y^{2}$ contribution the element cannot represent.

(c) Demonstration for the given bilinear field. The polynomial $\{1, x, y, xy\}$ satisfies both tests by inspection: it is complete through first order, and its single second-order term $xy$ is symmetric under interchange of $x$ and $y$. The formal demonstration is a change of frame.

1. Relate the two coordinate systems. Node 4 sits directly above node 1 at a distance $L$. The $\xi$ axis at node 4 points towards node 3, i.e. in the same sense as $x$; the $\eta$ axis points downwards, i.e. against $y$. Reading the geometry off the figure, $$\xi = x,\qquad \eta = L - y \quad\Longleftrightarrow\quad x = \xi,\qquad y = L - \eta$$ Every node maps consistently: node 1 at $(x,y) = (0,0)$ becomes $(\xi,\eta) = (0,L)$, node 2 at $(L,0)$ becomes $(L,L)$, node 3 at $(L,L)$ becomes $(L,0)$, and node 4 at $(0,L)$ becomes the new origin $(0,0)$.
2. Substitute into the interpolation. Replacing $x$ and $y$ in $u = C_1 + C_2x + C_3y + C_4xy$, $$u(\xi,\eta) = C_1 + C_2\xi + C_3(L-\eta) + C_4\,\xi\,(L-\eta)$$
3. Expand and collect like terms. Multiplying out, $$u(\xi,\eta) = \big(C_1 + C_3L\big) + \big(C_2 + C_4L\big)\xi + \big(-C_3\big)\eta + \big(-C_4\big)\xi\eta$$ which is of the form $$\boxed{\,u(\xi,\eta) = D_1 + D_2\xi + D_3\eta + D_4\,\xi\eta\,}$$ with $$D_1 = C_1 + C_3L,\qquad D_2 = C_2 + C_4L,\qquad D_3 = -C_3,\qquad D_4 = -C_4$$
4. Confirm the transformation is information-preserving. The coefficient map is linear, and in matrix form $$\begin{Bmatrix}D_1\\ D_2\\ D_3\\ D_4\end{Bmatrix} = \begin{bmatrix}1&0&L&0\\ 0&1&0&L\\ 0&0&-1&0\\ 0&0&0&-1\end{bmatrix}\begin{Bmatrix}C_1\\ C_2\\ C_3\\ C_4\end{Bmatrix}$$ The matrix is upper triangular with determinant $(1)(1)(-1)(-1) = 1 \ne 0$, so it is invertible: the two descriptions carry exactly the same information and either set of coefficients can be recovered from the other.
5. Check that the physical field is unchanged. Evaluating both expressions at each of the four nodes gives identical values — for example at node 2, $u = C_1 + C_2L$ from the first form and $D_1 + D_2L + D_3L + D_4L^{2} = (C_1+C_3L) + (C_2+C_4L)L - C_3L - C_4L^{2} = C_1 + C_2L$ from the second. The element therefore interpolates the same physical field regardless of which node hosts the origin and which way the second axis points.

The demonstration is worth reading as a statement about what the algebra proves and what it does not. Nothing above required the two frames to be related by a pure rotation: the transformation used here combines a translation of the origin from node 1 to node 4 with a reflection of the second axis, which is the more demanding test of the two, since a reflection changes the sign of $\eta$ and would expose any unbalanced odd-order term immediately. Because the four monomials $\{1,\xi,\eta,\xi\eta\}$ close under this substitution, the element passes. Had the assumed field instead been $u = C_1 + C_2x + C_3y + C_4x^{2}$, the same substitution would have produced a $\xi^{2}$ term with no $\eta^{2}$ partner available to absorb the corresponding term generated by a rotation, and the element would have been orientation-dependent — the classic symptom of a violated Pascal-triangle pairing.