Question 4 of 7: Basis functions, shape functions and geometric isotropy

Question 4: Basis functions, shape functions and geometric isotropy (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.

(a) Basis function versus shape function. A basis function is one of the independent terms of the polynomial space chosen to represent the field inside an element — the monomials 1, x, y, xy in the present case — and it is multiplied by a generalised coefficient C_i that has no direct physical meaning. A shape function is what remains after those generalised coefficients have been eliminated in favour of the nodal values of the field: each shape function is tied to one specific node, equals unity there and zero at every other node, and the two descriptions span exactly the same polynomial space. In short, the basis spans the space, while the shape functions are the particular basis of that same space that is dual to the nodal degrees of freedom.

(b) Geometric isotropy. An element possesses geometric isotropy (also called spatial isotropy or geometric invariance) when the interpolated field keeps the same polynomial form, and the element the same behaviour, no matter where the reference axes are placed or how they are oriented — the element has no preferred direction. Physically it means that two analysts who set up the same mesh with differently oriented global axes must obtain the same answer.

(c) The two required properties of the polynomial. Completeness: the polynomial must contain a complete lower-order set, i.e. the constant term and every term of each order up to the highest complete order retained (this also guarantees that rigid-body motion and constant strain states can be represented). Symmetry in the Pascal triangle: any incomplete higher order must be filled symmetrically, so that whenever a term x^myⁿ is included the mirror term xⁿy^m is included as well. The bilinear set {1, x, y, xy} satisfies both: it is complete to first order, and the second-order terms are represented only by the symmetric pair xy (x² and y² are both absent, so no direction is favoured).

Given. A square element of side L with a bilinear interpolation written about node 1, and a second, rotated-and-reflected frame (ξ, η) whose origin sits at node 3, with ξ measured towards node 4 and η measured towards node 2.

Find. The same field expressed in the (ξ, η) frame, so as to demonstrate that its polynomial form is unchanged.

Approach. Write the coordinate transformation implied by the schematic, substitute it into the assumed polynomial, and collect terms to show that the result is a bilinear polynomial of exactly the same form with relabelled coefficients.

1. State the coordinate transformation. Node 1 sits at $(x,y)=(0,0)$ and node 3 at $(L,L)$. The $\xi$ axis starts at node 3 and points towards node 4, i.e. in the $-x$ direction, and the $\eta$ axis points from node 3 towards node 2, i.e. in the $-y$ direction. Therefore$$\xi=L-x,\qquad \eta=L-y\qquad\Longleftrightarrow\qquad x=L-\xi,\quad y=L-\eta$$The new frame is the old one translated to the opposite corner and reflected in both axes — the most demanding of the simple axis changes.
2. Substitute into the assumed field. Replacing $x$ and $y$ in $u=C_1+C_2x+C_3y+C_4xy$ gives$$u=C_1+C_2(L-\xi)+C_3(L-\eta)+C_4(L-\xi)(L-\eta)$$and expanding the product $(L-\xi)(L-\eta)=L^{2}-L\eta-L\xi+\xi\eta$.
3. Collect the terms. Grouping by powers of $\xi$ and $\eta$,$$u(\xi,\eta)=\underbrace{\big(C_1+C_2L+C_3L+C_4L^{2}\big)}_{D_1}+\underbrace{\big(-C_2-C_4L\big)}_{D_2}\xi+\underbrace{\big(-C_3-C_4L\big)}_{D_3}\eta+\underbrace{C_4}_{D_4}\xi\eta$$so that$$\boxed{u(\xi,\eta)=D_1+D_2\xi+D_3\eta+D_4\xi\eta}$$which is the same bilinear polynomial as before, only with new constants. No new monomial has appeared and none has been lost.
4. Draw the conclusion. Because the transformation is invertible ($C_4=D_4$, $C_2=-D_2-D_4L$, $C_3=-D_3-D_4L$, $C_1=D_1+D_2L+D_3L+D_4L^{2}$), the two descriptions carry identical information: the four nodal values determine the same surface in either frame. The element therefore exhibits geometric isotropy. The structural reason is visible in the monomial set $\{1,\xi,\eta,\xi\eta\}$: it is complete to first order and its only second-order member is the symmetric product, so translating, rotating through a multiple of 90° or reflecting the axes can only permute and recombine terms already present.