Question 2 of 7: Seven-node transition element and mesh compatibility

Question 2: Seven-node transition element and mesh compatibility (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 parent element in natural coordinates $-1\le\xi,\eta\le 1$ with corner nodes 1 $(-1,-1)$, 2 $(1,-1)$, 3 $(1,1)$, 4 $(-1,1)$. Reading the figure, the three additional nodes lie at the mid-points of three sides only: node 5 at $(1,0)$ on the right side 2–3, node 6 at $(0,1)$ on the top side 3–4, and node 7 at $(-1,0)$ on the left side 4–1. The bottom side 1–2 carries no mid-side node and is therefore a two-node (linear) edge; the other three sides are three-node (quadratic) edges. Part (c) supplies a mesh of three eight-node Serendipity elements, in which element a sits above elements b and c, offset by half an element width.

Find. The seven shape functions $N_1 \ldots N_7$ satisfying $N_i(\text{node }j)=\delta_{ij}$ and $\sum N_i = 1$; a reasoned verdict on the $C^0$ continuity of $N_2$ on the two sides through node 2; and the defect in the 2-3-4 interface of the mesh.

Approach. Build each mid-side function first as a product that vanishes on every side not containing its node, then correct each bilinear corner function by subtracting the value it takes at every mid-side node multiplied by that node’s function — the standard Serendipity construction. Verify with the Kronecker-delta and partition-of-unity conditions, then restrict $N_2$ to each of its two sides and inspect the degree.

1. Write the mid-side functions. Each mid-side function must equal 1 at its own node and vanish at all six other nodes. Node 5 sits at $(1,0)$: the factor $(1+\xi)$ kills the whole left side, and $(1-\eta^2)$ kills both the top and bottom sides, leaving only node 5 non-zero; normalising by the value $2\cdot 1=2$ at $(1,0)$ gives the coefficient $\tfrac12$. Applying the same reasoning at nodes 6 and 7, $$N_5 = \tfrac12(1+\xi)\left(1-\eta^{2}\right),\qquad N_6 = \tfrac12(1+\eta)\left(1-\xi^{2}\right),\qquad N_7 = \tfrac12(1-\xi)\left(1-\eta^{2}\right).$$ Each is quadratic along its own side and linear across it, exactly as a three-node edge requires.
2. Start from the bilinear corner functions. The uncorrected corner functions are the ordinary four-node bilinear set $$N_i^{0}=\tfrac14\left(1+\xi\xi_i\right)\left(1+\eta\eta_i\right),\qquad i=1,\ldots,4,$$ which satisfy $N_i^{0}(\text{corner }j)=\delta_{ij}$ but take non-zero values at the mid-side nodes. Since a shape function must vanish at every node other than its own, each $N_i^{0}$ must be corrected.
3. Apply the standard correction. Subtract from each corner function its own value at every added node, weighted by that node’s shape function: $$N_i = N_i^{0}-\sum_{m=5}^{7} N_i^{0}\!\left(\xi_m,\eta_m\right) N_m .$$ Evaluating the bilinear functions at the three mid-side nodes gives the correction table below; a value of $\tfrac12$ occurs whenever the corner lies on the same side as the mid-side node, and zero otherwise.

   Values of the bilinear corner functions at the three added nodes

   	node 5 $(1,0)$	node 6 $(0,1)$	node 7 $(-1,0)$
   $N_1^{0}=\tfrac14(1-\xi)(1-\eta)$	0	0	$\tfrac12$
   $N_2^{0}=\tfrac14(1+\xi)(1-\eta)$	$\tfrac12$	0	0
   $N_3^{0}=\tfrac14(1+\xi)(1+\eta)$	$\tfrac12$	$\tfrac12$	0
   $N_4^{0}=\tfrac14(1-\xi)(1+\eta)$	0	$\tfrac12$	$\tfrac12$

4. Correct nodes 1 and 2 (one enriched side each). Node 1 lies on the left side only, so a single correction applies: $$N_1 = \tfrac14(1-\xi)(1-\eta)-\tfrac12 N_7 = \tfrac14(1-\xi)\left[(1-\eta)-\left(1-\eta^{2}\right)\right] = -\tfrac14(1-\xi)\,\eta\,(1-\eta).$$ Node 2 lies on the right side only, so by the mirror argument $$N_2 = \tfrac14(1+\xi)(1-\eta)-\tfrac12 N_5 = -\tfrac14(1+\xi)\,\eta\,(1-\eta).$$ Both corners also touch the linear bottom side, which contributes no correction because that side has no mid-side node.
5. Correct nodes 3 and 4 (two enriched sides each). Node 3 sits at the junction of the right and top sides, so two corrections survive and the result factors neatly: $$N_3 = \tfrac14(1+\xi)(1+\eta)-\tfrac12 N_5-\tfrac12 N_6 = \tfrac14(1+\xi)(1+\eta)\Big[1-(1-\eta)-(1-\xi)\Big] = \tfrac14(1+\xi)(1+\eta)\left(\xi+\eta-1\right).$$ Node 4 sits at the junction of the top and left sides, giving the mirror form $$N_4 = \tfrac14(1-\xi)(1+\eta)-\tfrac12 N_6-\tfrac12 N_7 = \tfrac14(1-\xi)(1+\eta)\left(\eta-\xi-1\right).$$ Notice that these two are the familiar eight-node Serendipity corner functions, whereas $N_1$ and $N_2$ are not — the element is genuinely a hybrid, which is the whole point of a transition element.
6. Collect the complete set. $$\boxed{\begin{aligned} N_1 &= -\tfrac14(1-\xi)\,\eta\,(1-\eta), &\qquad N_2 &= -\tfrac14(1+\xi)\,\eta\,(1-\eta),\\[2pt] N_3 &= \tfrac14(1+\xi)(1+\eta)(\xi+\eta-1), &\qquad N_4 &= \tfrac14(1-\xi)(1+\eta)(\eta-\xi-1),\\[2pt] N_5 &= \tfrac12(1+\xi)\left(1-\eta^{2}\right), &\qquad N_6 &= \tfrac12(1+\eta)\left(1-\xi^{2}\right),\\[2pt] N_7 &= \tfrac12(1-\xi)\left(1-\eta^{2}\right). & & \end{aligned}}$$
7. Verify the two admissibility conditions. Substituting each of the seven nodal coordinate pairs into each function reproduces the identity matrix, $N_i(\text{node }j)= \delta_{ij}$; for instance $N_3(0,0)=\tfrac14(1)(1)(-1)=-\tfrac14$ is legitimately negative inside the element, while $N_3(1,0)=\tfrac14(2)(1)(0)=0$ at node 5 as required. Summing the seven functions and expanding gives $$\sum_{i=1}^{7}N_i = 1 \quad\text{identically in }\xi,\eta,$$ so a uniform field is reproduced exactly and rigid-body motion is admissible. Each side also degenerates to the correct one-dimensional set: on $\eta=-1$ only $N_1$ and $N_2$ survive as the linear pair, and on $\xi=1$ only $N_2$, $N_5$, $N_3$ survive as the quadratic Lagrange triple.
8. Part (b) — restrict $N_2$ to the bottom side $\eta=-1$. Setting $\eta=-1$, $$N_2\big|_{\eta=-1} = -\tfrac14(1+\xi)(-1)\left(1-(-1)\right)=\tfrac12(1+\xi),$$ a linear function of $\xi$ that equals 1 at node 2 and 0 at node 1. On this side the field variable depends only on the two nodal values belonging to that side.
9. Restrict $N_2$ to the right side $\xi=1$. Setting $\xi=1$, $$N_2\big|_{\xi=1} = -\tfrac14(2)\,\eta\,(1-\eta)=\tfrac12\,\eta(\eta-1),$$ a quadratic function of $\eta$ taking the values $1,\,0,\,0$ at nodes 2 $(\eta=-1)$, 5 $(\eta=0)$ and 3 $(\eta=1)$ — precisely the end-node member of the three-node one-dimensional Lagrange family. Again the trace depends only on nodes lying on that side. $$\boxed{\text{Yes: } N_2 \text{ satisfies } C^{0}\text{ continuity on both sides through node 2}}$$ Because the trace of $N_2$ on each side is a one-dimensional interpolation of the nodes on that side alone, two elements sharing that side and using the same edge order will predict identical field values along it — no gap or overlap can open. For completeness, $N_2$ vanishes identically on the two sides that do not contain node 2 ($\xi=-1$ and $\eta=1$), which is the complementary half of the same requirement.
10. Part (c) — diagnose the 2-3-4 interface. In the mesh, node 3 is the mid-side node of element a’s lower edge but is a corner node shared by elements b and c; conversely nodes 2 and 4 are corner nodes of element a but mid-side nodes of b and c respectively. Element a therefore interpolates the field along the whole segment 2–3–4 with a single quadratic in its own edge coordinate, whereas element b interpolates only the half-segment 2–3 with a different quadratic — the restriction of the quadratic defined over b’s full top edge. A quadratic and half of another quadratic agree at their two shared nodes but not in between.
11. State the consequence. The interface is therefore incompatible (non-conforming): inter-element $C^{0}$ continuity is violated along 2–3 and along 3–4, so gaps and overlaps open between element a and the pair b, c as soon as the field varies. $$\boxed{\text{Major problem: loss of }C^{0}\text{ compatibility — the mesh does not conform across 2-3-4}}$$ The practical symptoms are failure of the patch test, spurious stress discontinuities that do not diminish with refinement, and loss of guaranteed monotonic convergence. The cure is either to align the elements corner-to-corner and mid-side-to-mid-side, or to insert a transition element of the kind derived in part (a), whose linear side mates with a linear neighbour while its quadratic sides mate with the quadratic region.