Question 3 of 7: Geometric isotropy of a bilinear square element [20 marks]

Question 3: Geometric isotropy of a bilinear square 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 nodes numbered counter-clockwise from the bottom-left, two coordinate frames, and a bilinear interpolation of the field variable:

Find. The meaning of geometric isotropy and the two polynomial properties that guarantee it, and a demonstration that the given bilinear interpolation retains exactly the same functional form when it is re-expressed in the $\xi,\eta$ frame centred at node 4.

(a) Meaning of geometric isotropy [4 marks]

Geometric isotropy — equivalently geometric invariance or spatial isotropy — is the property that an element’s interpolated field is independent of the position and orientation of the coordinate frame used to describe it. If the same element is re-described in a frame that is translated and/or rotated relative to the original, the interpolation must retain the identical functional form (the same set of polynomial terms), with only the values of the undetermined coefficients changing. The physical consequence is that the element behaves identically no matter how it is oriented in the mesh: two identical elements rotated relative to one another give identical stiffness in their own local directions, and the assembled answer does not depend on how the analyst happened to number or orient the mesh.

(b) The two required properties of the polynomial [6 marks]

Property 1 — completeness. The polynomial must be complete to the highest order it retains: if a term of order $m$ is present, then every term of order $0,1,\dots,m$ from Pascal’s triangle must also be present. This guarantees that constant and linear fields — rigid-body motion and constant strain — can be reproduced exactly, which is the requirement for convergence.

Property 2 — symmetry (balance) of any incomplete higher-order terms. Terms of an order beyond the complete order may be retained only in symmetric pairs about the vertical axis of Pascal’s triangle, so that no coordinate direction is favoured. If $x^{2}y$ is retained then $xy^{2}$ must be retained as well; if $x^{2}$ is retained then $y^{2}$ must be too. Retaining $x^{2}$ without $y^{2}$ would make the element stiffer along one axis than the other, and rotating that element would change its behaviour.

The bilinear form of this question satisfies both: it is complete to first order (it contains $1$, $x$ and $y$), and the single higher-order term $xy$ is symmetric in $x$ and $y$, so it is its own symmetric pair.

(c) Demonstration in the $\xi,\eta$ frame centred at node 4 [10 marks]

Approach. Write the coordinate transformation implied by the figure — a translation of the origin from node 1 to node 4 combined with a reversal of the second axis — substitute it into the given interpolation, and collect terms to show that the result has the identical bilinear form with redefined constants.

1. Write the transformation between the two frames. In the $x,y$ frame the nodes are at $1(0,0)$, $2(L,0)$, $3(L,L)$ and $4(0,L)$. The $\xi$ axis is centred at node 4 and points in the same direction as $x$, while $\eta$ is centred at node 4 and points opposite to $y$. Therefore $$\xi = x - 0 = x ,\qquad \eta = -\,(y-L) = L - y$$ and the inverse relations needed for substitution are $$x = \xi ,\qquad y = L - \eta$$ As a check, node 4 maps to $(\xi,\eta)=(0,0)$, node 3 to $(L,0)$, node 1 to $(0,L)$ and node 2 to $(L,L)$, which is exactly the numbering shown in the figure.
2. Substitute into the given 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 the products, $$u(\xi,\eta)=C_1+C_2\xi+C_3L-C_3\eta+C_4L\,\xi-C_4\,\xi\eta$$ and grouping the constant, the $\xi$ term, the $\eta$ term and the $\xi\eta$ term, $$u(\xi,\eta)=\big(C_1+C_3L\big)+\big(C_2+C_4L\big)\xi+\big(-C_3\big)\eta+\big(-C_4\big)\xi\eta$$
4. Define the transformed constants and compare the forms. Setting $$D_1=C_1+C_3L ,\qquad D_2=C_2+C_4L ,\qquad D_3=-C_3 ,\qquad D_4=-C_4$$ the interpolation in the new frame becomes $$\boxed{\,u(\xi,\eta)=D_1+D_2\xi+D_3\eta+D_4\,\xi\eta\,}$$ which is term for term the same bilinear polynomial as $u(x,y)=C_1+C_2x+C_3y+C_4xy$. No term has appeared, none has disappeared, and only the numerical values of the four undetermined constants have changed. The element therefore possesses geometric isotropy.
5. Confirm that the transformation is invertible and reproduces nodal values. The relation $D_i \leftrightarrow C_i$ is a one-to-one linear map (its determinant is $(1)(1)(-1)(-1)=1$), so the two descriptions carry exactly the same information. As a numerical spot check, the field at node 4 is $u(0,L)=C_1+C_3L$ in the original frame and $u(\xi,\eta)=(0,0)\Rightarrow D_1=C_1+C_3L$ in the new frame — the same value, as it must be. The demonstration is unchanged if the axes are rotated by $90^\circ$ instead of reflected, which is the general statement of the property.