Question 5 of 7: Six-node transition element — shape functions and field recovery

Question 5: Six-node transition element — shape functions and field recovery (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 six-node transition element on the parent square: corners 1$(-1,-1)$, 2$(1,-1)$, 3$(1,1)$, 4$(-1,1)$, a mid-side node 5 at $(0,-1)$ on the bottom edge and a mid-side node 6 at $(-1,0)$ on the left edge. The bottom and left edges are therefore quadratic; the top and right edges remain linear.

Find. All six shape functions in closed form, three specific values of $N_6$, and the interpolated displacement field for a prescribed set of nodal values.

Approach. Start from the four bilinear corner functions, add a mid-side function for each enriched edge built to vanish at every other node, then correct each corner function by subtracting its own value at the added nodes multiplied by those nodes' functions. Verify with the Kronecker-delta property, the partition of unity, and degeneration of each edge to its own one-dimensional Lagrange set.

Part (a) — the six shape functions. The construction runs in five steps, the first four building the functions and the last verifying them.

1. Write the uncorrected bilinear corner functions. For corner $i$ at $(\xi_i,\eta_i)$, $$N_i^{0}(\xi,\eta) = \tfrac{1}{4}\big(1+\xi\xi_i\big)\big(1+\eta\eta_i\big)$$ giving $N_1^{0} = \tfrac14(1-\xi)(1-\eta)$, $N_2^{0} = \tfrac14(1+\xi)(1-\eta)$, $N_3^{0} = \tfrac14(1+\xi)(1+\eta)$ and $N_4^{0} = \tfrac14(1-\xi)(1+\eta)$. These already form a valid bilinear set but they do not vanish at nodes 5 and 6, so they cannot stand as they are.
2. Construct the two mid-side functions. Each must equal unity at its own node, vanish at all four corners and at the other mid-side node, and reduce to the quadratic Lagrange bubble along its own edge. The standard parabolic forms do exactly this: $$N_5 = \tfrac{1}{2}\big(1-\xi^{2}\big)\big(1-\eta\big),\qquad N_6 = \tfrac{1}{2}\big(1-\xi\big)\big(1-\eta^{2}\big)$$ $N_5$ vanishes on $\xi = \pm1$ and on $\eta = 1$; $N_6$ vanishes on $\eta = \pm1$ and on $\xi = 1$. Cross-checking, $N_5(-1,0) = 0$ and $N_6(0,-1) = 0$, so the two do not interfere with one another.
3. Correct the corner functions. The general recipe removes from each corner function whatever it contributes at the added nodes: $$N_i = N_i^{0} - \sum_{m\,\in\,\{5,6\}} N_i^{0}(\xi_m,\eta_m)\,N_m$$ The required values are $N_1^{0}(0,-1) = \tfrac12$, $N_2^{0}(0,-1) = \tfrac12$, $N_1^{0}(-1,0) = \tfrac12$, $N_4^{0}(-1,0) = \tfrac12$, while $N_3^{0}$ vanishes at both added nodes and $N_2^{0}$, $N_4^{0}$ each vanish at the node they do not share an edge with. Node 1 therefore receives two corrections, nodes 2 and 4 receive one each, and node 3 receives none.
4. Simplify to closed form. Carrying out the subtractions and factoring, $$\boxed{\begin{aligned} N_1 &= -\tfrac{1}{4}(1-\xi)(1-\eta)(\xi+\eta+1), & N_2 &= \tfrac{1}{4}\,\xi\,(1+\xi)(1-\eta),\\ N_3 &= \tfrac{1}{4}(1+\xi)(1+\eta), & N_4 &= \tfrac{1}{4}\,\eta\,(1-\xi)(1+\eta),\\ N_5 &= \tfrac{1}{2}(1-\xi^{2})(1-\eta), & N_6 &= \tfrac{1}{2}(1-\xi)(1-\eta^{2}). \end{aligned}}$$ Note that $N_3$ emerges unchanged from its bilinear parent, because node 3 touches neither enriched edge.
5. Verify the set. Three independent checks all pass. The Kronecker-delta property $N_i(\xi_j,\eta_j) = \delta_{ij}$ holds at all six nodes. The partition of unity $\sum_{i=1}^{6} N_i = 1$ holds identically, which guarantees the element can represent a rigid-body translation. And each edge degenerates to its own one-dimensional interpolation: on $\eta = -1$ the surviving functions are $\tfrac12\xi(\xi-1)$, $\tfrac12\xi(\xi+1)$ and $1-\xi^{2}$, the quadratic Lagrange trio; on $\xi = -1$ they are the same trio in $\eta$; while on $\eta = +1$ only $\tfrac12(1-\xi)$ and $\tfrac12(1+\xi)$ survive, and on $\xi = +1$ only $\tfrac12(1-\eta)$ and $\tfrac12(1+\eta)$. The linear top and right edges are exactly what allows this element to sit against ordinary four-node quadrilaterals without a gap.

Part (b). Reading the values of $N_6 = \tfrac12(1-\xi)(1-\eta^{2})$ directly:

The first two are required by the Kronecker-delta property. The third is worth pausing on: a mid-side function reaching one-half at the element centre is entirely normal, and by contrast the corner functions of this transition element take the values $N_1 = -\tfrac14$, $N_2 = 0$, $N_3 = \tfrac14$ and $N_4 = 0$ there. A negative shape function inside the element is not an error — it is the price of enforcing zero at the mid-side nodes — and the four corner values plus $N_5 = \tfrac12$ and $N_6 = \tfrac12$ still sum to unity, as they must.

Part (c). All $u_i$ are zero, so $u(\xi,\eta) = \sum N_iu_i = 0$ identically: the element carries no displacement in the $\xi$ direction anywhere. For $v$, the three loaded nodes 1, 4 and 6 are exactly the three nodes of the left edge, so

It is tempting to expect the three left-edge functions to collapse to the bare linear ramp $-0.025\cdot\tfrac12(1-\xi)$, which is what happens when the loaded nodes form a complete edge and nothing else. That is not the case here, and the reason is worth stating: node 1 lies on the left edge and on the bottom edge, so its correction against $N_5$ survives the sum and leaves the extra $(1+\xi)(1-\eta)$ term. The surviving term is precisely what forces $v = 0$ at node 5, which sits on the loaded bottom-left corner's other edge and must remain undisplaced. Checking the closed form at all six nodes confirms it: $v = -0.025$ at nodes 1, 4 and 6, and $v = 0$ at nodes 2, 3 and 5. At the centroid the field takes $v(0,0) = -0.00625$ mm, one quarter of the imposed edge value, and along the entire right edge $\xi = +1$ it vanishes, as the shape-function support requires.