Shape functions of a seven-node transition element: 22-Mec-B10 Finite Element Analysis, December 2014, Q3

Question 3: Shape functions of a seven-node transition 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 parent element carrying seven nodes: the four corners, one mid-side node on the right-hand edge, and two interior-edge nodes that divide the top edge into three equal parts — so the element makes a transition from a linear edge, through a quadratic edge, to a cubic edge.

Nodal positions in the parent domain
Node	(ξ, η)	Role
1	(−1, −1)	corner
2	(+1, −1)	corner
3	(+1, +1)	corner
4	(−1, +1)	corner
5	(+1, 0)	mid-side, right edge (quadratic)
6	(+1/3, +1)	top edge (cubic)
7	(−1/3, +1)	top edge (cubic)

Find. All seven shape functions in the natural coordinates, the value of N₃ at node 6 and at the element centroid, and the interpolated displacement field for the stated set of nodal values.

Question 3 — the seven-node transition element in the parent domain: linear bottom and left edges, a quadratic right edge through node 5, and a cubic top edge through nodes 7 and 6 at ξ = ∓1/3.

Approach. Start from the four bilinear corner functions, construct a function for each added node that vanishes at every other node, then correct the corner functions by subtracting their (non-zero) values at the added nodes multiplied by those new functions — the standard hierarchical node-addition construction.

Write the bilinear starting set. The four-node parent element has $N_i^{0}=\tfrac14(1+\xi\xi_i)(1+\eta\eta_i)$, i.e.$$N_1^{0}=\tfrac14(1-\xi)(1-\eta),\quad N_2^{0}=\tfrac14(1+\xi)(1-\eta),\quad N_3^{0}=\tfrac14(1+\xi)(1+\eta),\quad N_4^{0}=\tfrac14(1-\xi)(1+\eta)$$These already satisfy the Kronecker-delta property at the four corners; the task is to keep that property while adding three more nodes.
Add the mid-side node 5 on the right edge. The function must equal one at $(1,0)$ and vanish on $\xi=-1$ (nodes 1 and 4) and on $\eta=\pm1$ (nodes 2, 3, 6, 7). The serendipity mid-side function does exactly this:$$N_5=\tfrac12(1+\xi)(1-\eta^{2})$$Check: $N_5(1,0)=\tfrac12(2)(1)=1$, and it is zero at every other node.
Add the two cubic-edge nodes. Nodes 7 and 6 lie at $\xi=-1/3$ and $\xi=+1/3$ on $\eta=1$. Each function must vanish on $\xi=\pm1$ (killing the corners and node 5), on $\eta=-1$, and at the other cubic node:$$N_7=\tfrac{9}{32}(1-\xi^{2})(1-3\xi)(1+\eta),\qquad N_6=\tfrac{9}{32}(1-\xi^{2})(1+3\xi)(1+\eta)$$Verifying node 7: $N_7(-\tfrac13,1)=\tfrac{9}{32}\left(\tfrac89\right)(2)(2)=1$, and $N_7(\tfrac13,1)=0$ because $(1-3\xi)$ vanishes there.
(a) Correct the corner functions and collect the seven shape functions. A newly added function spoils the delta property of any old function that is non-zero at the new node, so subtract that value times the new function: $N_i\leftarrow N_i^{0}-\sum_m N_i^{0}(\xi_m,\eta_m)\,N_m$. Evaluating the bilinear functions at the three new nodes gives $N_2^{0}(1,0)=N_3^{0}(1,0)=\tfrac12$; $N_4^{0}(-\tfrac13,1)=\tfrac23$, $N_3^{0}(-\tfrac13,1)=\tfrac13$; $N_4^{0}(\tfrac13,1)=\tfrac13$, $N_3^{0}(\tfrac13,1)=\tfrac23$. Hence the complete set is$$\boxed{\begin{aligned}N_1&=\tfrac14(1-\xi)(1-\eta)\\N_2&=\tfrac14(1+\xi)(1-\eta)-\tfrac12 N_5\\N_3&=\tfrac14(1+\xi)(1+\eta)-\tfrac12 N_5-\tfrac13 N_7-\tfrac23 N_6\\N_4&=\tfrac14(1-\xi)(1+\eta)-\tfrac23 N_7-\tfrac13 N_6\\N_5&=\tfrac12(1+\xi)(1-\eta^{2})\\N_6&=\tfrac{9}{32}(1-\xi^{2})(1+3\xi)(1+\eta)\\N_7&=\tfrac{9}{32}(1-\xi^{2})(1-3\xi)(1+\eta)\end{aligned}}$$Node 1 needs no correction because the bilinear $N_1^{0}$ already vanishes at nodes 5, 6 and 7.
Confirm the two mandatory properties. Summing the seven functions, every correction cancels — $-\tfrac12-\tfrac12+1=0$ for node 5, $-\tfrac23-\tfrac13+1=0$ for node 7, $-\tfrac13-\tfrac23+1=0$ for node 6 — so $\sum N_i=1$ everywhere, which guarantees that a rigid-body displacement is reproduced exactly. Restricting the set to an edge shows the promised transition: on $\eta=-1$ only the linear $\tfrac12(1\mp\xi)$ survive; on $\xi=1$ the three functions $N_2,N_5,N_3$ reduce to the quadratic Lagrange set; and on $\eta=1$ the four functions $N_4,N_7,N_6,N_3$ reduce to the cubic set, e.g. $N_4\big|_{\eta=1}=\tfrac1{16}(1-\xi)(9\xi^{2}-1)$.
(b) Evaluate $N_3$ at node 6 and at the centroid. At node 6, $(\xi,\eta)=(\tfrac13,1)$, the Kronecker-delta property must hold, and indeed $\tfrac14\left(\tfrac43\right)(2)-0-0-\tfrac23(1)=\tfrac23-\tfrac23=0$. At the centroid $(0,0)$, $N_3^{0}=\tfrac14$, $N_5=\tfrac12$ and $N_6=N_7=\tfrac{9}{32}$, so$$N_3(0,0)=\tfrac14-\tfrac12\left(\tfrac12\right)-\tfrac13\left(\tfrac{9}{32}\right)-\tfrac23\left(\tfrac{9}{32}\right)=\tfrac14-\tfrac14-\tfrac{3}{32}-\tfrac{3}{16}$$$$\boxed{N_3(\text{node }6)=0,\qquad N_3(0,0)=-\tfrac{9}{32}=-0.28125}$$A negative shape function is perfectly legitimate for higher-order elements; only the sum is constrained.
(c) Interpolate the given nodal displacements. With $u_3=u_4=u_6=u_7=0.025$ mm and all other nodal components zero, $u=\sum N_iu_i=0.025\,(N_3+N_4+N_6+N_7)$. The corrections cancel in that sum (each of $N_6$ and $N_7$ appears with total coefficient $1-\tfrac13-\tfrac23=0$), leaving$$N_3+N_4+N_6+N_7=\tfrac12(1+\eta)-\tfrac14(1+\xi)(1-\eta^{2})=\tfrac14(1+\eta)\big[2-(1+\xi)(1-\eta)\big]$$$$\boxed{u(\xi,\eta)=\tfrac{0.025}{4}(1+\eta)\big[2-(1+\xi)(1-\eta)\big]\ \text{mm},\qquad v(\xi,\eta)=0}$$The expression returns 0.025 mm all along the top edge (where $\eta=1$), zero along the bottom edge and at node 5, and 0.00625 mm at the centroid — the correct behaviour for a uniform displacement imposed on one edge only.

Final results — Question 3
Quantity	Value
N₁	¼(1−ξ)(1−η)
N₂	¼(1+ξ)(1−η) − ½N₅
N₃	¼(1+ξ)(1+η) − ½N₅ − ⅓N₇ − ⅔N₆
N₄	¼(1−ξ)(1+η) − ⅔N₇ − ⅓N₆
N₅	½(1+ξ)(1−η²)
N₆ / N₇	(9/32)(1−ξ²)(1±3ξ)(1+η)
N₃ at node 6 / at centroid	0 / −9/32 = −0.28125
Field variables	u = 0.00625(1+η)[2 − (1+ξ)(1−η)] mm, v = 0

Topic. Transition elements and the hierarchical addition of nodes.

Key relations. Bilinear base $N_i^{0}=\tfrac14(1+\xi\xi_i)(1+\eta\eta_i)$; correction rule $N_i\leftarrow N_i^{0}-\sum_m N_i^{0}(\xi_m,\eta_m)N_m$; delta property $N_i(\xi_j,\eta_j)=\delta_{ij}$; partition of unity $\sum N_i=1$.

Why this works. A transition element exists so that a coarse (linear) mesh region can be joined to a refined (quadratic or cubic) region without hanging nodes. Compatibility across an inter-element boundary depends only on the restriction of the shape functions to that edge, so an element whose edges carry different numbers of nodes is legitimate provided each edge degenerates to the standard one-dimensional Lagrange set of the matching order. The node-addition procedure guarantees precisely that: subtracting the old function value at each new node restores the delta property without disturbing the sum, so completeness and inter-element continuity both survive.

Common pitfall. Building the added-node functions correctly and then forgetting to correct the corners, which leaves N₃ and N₄ non-zero at nodes 6 and 7 and destroys interpolation. A second trap is assuming a mid-side node at ξ = 0 on the top edge; the figure places two nodes there, so the spacing is one third and the edge is cubic, not quadratic.

Source. Cook, Malkus, Plesha & Witt, Concepts and Applications of Finite Element Analysis (4th ed.), §6.3–6.5 (isoparametric families and transition elements); Zienkiewicz, Taylor & Zhu (7th ed.), Ch. 6.

Question 3 of 7: Shape functions of a seven-node transition element

Question 3: Shape functions of a seven-node transition element (20 marks)