Question 2 of 7: Six-node transition element — shape functions and field interpolation

Question 2: Six-node transition element — shape functions and field interpolation (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. The parent-domain node positions read directly off the figure — four corners in counter-clockwise order plus two mid-side nodes on adjacent edges:

Find. All six shape functions, the two requested values of $N_4$, and the interpolated displacement fields $u(\xi,\eta)$ and $v(\xi,\eta)$ for the given nodal data.

Approach. Build the two mid-side functions first so each vanishes at every other node, then correct the four bilinear corner functions by subtracting their spurious values at the added nodes, and finally verify the Kronecker-delta and partition-of-unity properties before using the set.

1. Recognise the element as a transition element. The bottom edge $(\eta=-1)$ carries three nodes 1–5–2, so it interpolates quadratically; the left edge $(\xi=-1)$ carries three nodes 1–6–4 and is likewise quadratic; the top and right edges carry only two nodes each and stay linear. Such an element is used to join a quadratic mesh to a bilinear mesh without leaving a gap along the shared edge, and its shape functions cannot simply be lifted from either standard family.
2. Construct the mid-side functions. Each added function must equal unity at its own node and vanish at every other node, including along the two linear edges. Taking the product of the lines that pass through all the other nodes, $$N_5 = \tfrac{1}{2}\big(1-\xi^{2}\big)(1-\eta),\qquad N_6 = \tfrac{1}{2}(1-\xi)\big(1-\eta^{2}\big)$$ Checking $N_5$ at node 5, $\tfrac12(1-0)(1+1)=1$; at nodes 3 and 4 the factor $(1-\eta)$ vanishes, and at nodes 1 and 2 the factor $(1-\xi^{2})$ vanishes. $N_6$ verifies identically with the roles of $\xi$ and $\eta$ exchanged.
3. Write the uncorrected bilinear corner functions. The starting point is the standard four-node set $$N_i^{0} = \tfrac{1}{4}\big(1+\xi\xi_i\big)\big(1+\eta\eta_i\big),\qquad i = 1\ldots4$$ These already satisfy $N_i^{0}(\text{node }j)=\delta_{ij}$ for the corners, but they do not vanish at the two mid-side nodes, so they must be corrected.
4. Correct each corner function. The general correction removes the value that each bilinear function leaves at every added node: $$N_i = N_i^{0} - \sum_{m=5,6} N_i^{0}\big(\xi_m,\eta_m\big)\, N_m$$ Evaluating the required corner values at the two mid-side positions gives $N_1^{0}(0,-1)=\tfrac12$, $N_1^{0}(-1,0)=\tfrac12$, $N_2^{0}(0,-1)=\tfrac12$, $N_4^{0}(-1,0)=\tfrac12$, while $N_3^{0}$ vanishes at both added nodes and needs no correction at all.
5. Part (a): assemble the six shape functions. Substituting those corner values, $$N_1 = \tfrac14(1-\xi)(1-\eta) - \tfrac12 N_5 - \tfrac12 N_6,\qquad N_2 = \tfrac14(1+\xi)(1-\eta) - \tfrac12 N_5$$ $$N_3 = \tfrac14(1+\xi)(1+\eta),\qquad N_4 = \tfrac14(1-\xi)(1+\eta) - \tfrac12 N_6$$ Expanding and factoring, the set takes the compact closed form $$\boxed{\;\begin{aligned} N_1 &= -\tfrac14(1-\xi)(1-\eta)\,(\xi+\eta+1), & N_2 &= \tfrac14\,\xi(1+\xi)(1-\eta),\\ N_3 &= \tfrac14(1+\xi)(1+\eta), & N_4 &= \tfrac14\,\eta(1-\xi)(1+\eta),\\ N_5 &= \tfrac12(1-\xi^{2})(1-\eta), & N_6 &= \tfrac12(1-\xi)(1-\eta^{2}). \end{aligned}\;}$$
6. Verify the set before using it. Two checks are mandatory. The cardinal (Kronecker-delta) property $N_i(\xi_j,\eta_j)=\delta_{ij}$ holds at all six nodes — for instance $N_1$ at node 1 gives $-\tfrac14(2)(2)(-1+(-1)+1)= +1$, and $N_1$ at node 5 gives $-\tfrac14(1)(2)(0-1+1)=0$. Partition of unity also holds, $$\sum_{i=1}^{6} N_i = 1 \quad \text{for all } (\xi,\eta)$$ which is what guarantees that a rigid-body translation is reproduced exactly. As a third check, setting $\eta=-1$ collapses the set to the one-dimensional quadratic Lagrange trio on nodes 1, 5, 2, and setting $\eta=+1$ collapses it to the linear pair on nodes 4 and 3 — exactly the transition behaviour intended.
7. Part (b): evaluate $N_4$ at the sixth node and at the centroid. Using the closed form $N_4=\tfrac14\eta(1-\xi)(1+\eta)$, at node 6 $(\xi,\eta)=(-1,0)$, $$N_4(-1,0)=\tfrac14(0)(2)(1)=0$$ and at the centroid $(0,0)$, $$N_4(0,0)=\tfrac14(0)(1)(1)=0$$ so $$\boxed{\, N_4\big|_{\text{node }6}=0,\qquad N_4\big|_{\text{centroid}}=0\,}$$ The first zero is required by the cardinal property. The second is not automatic and is the point of the sub-question: the bilinear part contributes $\tfrac14$ at the centroid but the correction $-\tfrac12 N_6(0,0)=-\tfrac14$ cancels it exactly, because the factor $\eta$ appearing after simplification makes $N_4$ antisymmetric about the line $\eta=0$. For comparison, the six values at the centroid are $N_1=-\tfrac14$, $N_2=0$, $N_3=\tfrac14$, $N_4=0$, $N_5=\tfrac12$, $N_6=\tfrac12$, which duly sum to unity — note that a corner function of a transition element can legitimately be negative inside the element.
8. Part (c): interpolate the given nodal displacements. Every $u_i$ is zero, and the shape-function expansion is linear in the nodal values, so $$u(\xi,\eta)=\sum_{i=1}^{6} N_i u_i = 0 \quad\text{identically}$$ For the transverse component only nodes 1, 2 and 5 are displaced, all by $v_0=-0.025$ mm, hence $$v(\xi,\eta)=v_0\big(N_1+N_2+N_5\big)$$ The three loaded nodes are exactly the nodes of the quadratic bottom edge, so the $N_5$ corrections in $N_1$ and $N_2$ cancel against $N_5$ itself: the $(1-\xi^{2})(1-\eta)$ terms carry coefficients $-\tfrac14-\tfrac14+\tfrac12 = 0$. What survives is the bilinear pair plus node 1's left-edge correction, which does not cancel because node 1 is shared by both enriched edges: $$N_1+N_2+N_5 = \tfrac12(1-\eta)-\tfrac14(1-\xi)(1-\eta^{2})$$ Therefore $$\boxed{\;u(\xi,\eta)=0,\qquad v(\xi,\eta) = -0.025\,\tfrac14(1-\eta)\Big[\,2-(1-\xi)(1+\eta)\,\Big]\ \text{mm}\;}$$ Spot-checking, at nodes 1, 2 and 5 the bracket and prefactor combine to give $-0.025$ mm, while at nodes 3, 4 and 6 the expression returns zero — at node 6, for example, $\tfrac12(1)-\tfrac14(2)(1)=0$. At the element centroid $v=-0.025(\tfrac12-\tfrac14)=-0.00625$ mm.