Question 4 of 7: Shape functions of a seven-node transition element

Question 4: 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 seven-node parent element whose node positions are read directly off the figure: four corners, one mid-side node on the bottom edge, and two third-point nodes on the left edge.

Find. (a) all seven shape functions; (b) $N_{4}$ at node 6 and at the centroid; (c) the interpolated displacement field for the given nodal values.

Approach. Read the node spacing from the figure to establish the order of each edge, write a function for every added node that vanishes at all other nodes, then correct the four bilinear corner functions by subtracting their own values at those added nodes.

1. Classify the edges. The bottom edge carries three nodes (1, 5, 2) with node 5 at the mid-point, so it interpolates quadratically in $\xi$. The left edge carries four nodes (1, 7, 6, 4) spaced at the third points $\eta = -1, -\tfrac13, +\tfrac13, +1$, so it interpolates cubically. The top and right edges carry two nodes each and remain linear. This is exactly the role of a transition element: it joins a coarse linear mesh to a finer higher-order mesh without hanging nodes.
2. Build the mid-side function for node 5. A function that is unity at (0, −1) and vanishes on the other three edges is the standard serendipity mid-side form:$$N_{5} = \tfrac{1}{2}\bigl(1-\xi^{2}\bigr)\bigl(1-\eta\bigr)$$ It vanishes at $\xi = \pm 1$ (so at all corners on the left and right) and at $\eta = +1$, and at the left-edge nodes 6 and 7 because $1-\xi^{2} = 0$ there.
3. Build the cubic-edge functions for nodes 6 and 7. On the left edge take the form $(1-\xi)(1-\eta^{2})(1 \pm 3\eta)$, which already vanishes at $\xi = +1$ and at $\eta = \pm1$. Normalising each at its own node (where the product equals $2 \cdot \tfrac89 \cdot 2 = \tfrac{32}{9}$) gives$$N_{6} = \tfrac{9}{32}\bigl(1-\xi\bigr)\bigl(1-\eta^{2}\bigr)\bigl(1+3\eta\bigr),\qquad N_{7} = \tfrac{9}{32}\bigl(1-\xi\bigr)\bigl(1-\eta^{2}\bigr)\bigl(1-3\eta\bigr)$$ Each vanishes at the other third point because $1 \mp 3\eta = 0$ there.
4. Evaluate the bilinear corner functions at the added nodes. Starting from $N_{i}^{0} = \tfrac14(1+\xi\xi_{i})(1+\eta\eta_{i})$, the values that must be corrected away are$$N_{1}^{0}(5) = \tfrac12,\; N_{1}^{0}(6) = \tfrac13,\; N_{1}^{0}(7) = \tfrac23;\qquad N_{2}^{0}(5) = \tfrac12;\qquad N_{4}^{0}(6) = \tfrac23,\; N_{4}^{0}(7) = \tfrac13$$ with $N_{3}^{0}$ already zero at all three added nodes (it sits on the two linear edges).
5. Correct the corner functions. Applying $N_{i} = N_{i}^{0} - \sum_{m} N_{i}^{0}(\text{node } m)\,N_{m}$ over the added nodes gives the four corrected corner functions:$$N_{1} = \tfrac14(1-\xi)(1-\eta) - \tfrac12 N_{5} - \tfrac13 N_{6} - \tfrac23 N_{7}$$$$N_{2} = \tfrac14(1+\xi)(1-\eta) - \tfrac12 N_{5}$$$$N_{3} = \tfrac14(1+\xi)(1+\eta)$$$$N_{4} = \tfrac14(1-\xi)(1+\eta) - \tfrac23 N_{6} - \tfrac13 N_{7}$$
6. Verify the set. Substituting all seven node positions confirms the Kronecker property $N_{i}(\xi_{j},\eta_{j}) = \delta_{ij}$, and summing the seven functions gives$$\boxed{\;\sum_{i=1}^{7} N_{i}(\xi,\eta) = 1 \quad \text{for all } (\xi,\eta)\;}$$ Each edge also degenerates correctly: on $\eta = +1$ only $N_{3} = \tfrac12(1+\xi)$ and $N_{4} = \tfrac12(1-\xi)$ survive (linear), and on the left edge the four surviving functions are the one-dimensional cubic Lagrange set.

(b) Evaluating $N_{4}$. At the sixth node $(\xi,\eta) = (-1, +\tfrac13)$ the Kronecker property must hold, and the arithmetic confirms it:$$N_{4}(-1, \tfrac13) = \tfrac23 - \tfrac23(1) - \tfrac13(0) = 0$$ At the centroid (0, 0) the bilinear term contributes $\tfrac14$ while $N_{6} = N_{7} = \tfrac{9}{32}$, so$$\boxed{\;N_{4}(0,0) = \tfrac14 - \tfrac23\!\left(\tfrac{9}{32}\right) - \tfrac13\!\left(\tfrac{9}{32}\right) = \tfrac14 - \tfrac{9}{32} = -\tfrac{1}{32} = -0.03125\;}$$ The negative value is not an error. Corrected serendipity corner functions routinely go slightly negative in the interior, which is why higher-order elements do not satisfy a maximum principle even though they still sum to unity and reproduce constant and linear fields exactly.

(c) The interpolated field. Every $u_{i}$ is zero, so the interpolation $u = \sum N_{i}u_{i}$ gives $u \equiv 0$ throughout the element. For the transverse component only nodes 1, 2 and 5 are displaced, all by the same amount, so$$v(\xi,\eta) = -0.025\,\bigl(N_{1} + N_{2} + N_{5}\bigr)\ \text{mm}$$ Because nodes 1, 2 and 5 are exactly the nodes of the bottom edge, the corrections to $N_{1}$ collapse: $N_{1} + N_{2} + N_{5}$ reduces to $\tfrac12(1-\eta) - \left[\tfrac13 N_{6} + \tfrac23 N_{7}\right]$, and the bracket factorises, leaving the compact closed form

$$\boxed{\;u(\xi,\eta) = 0, \qquad v(\xi,\eta) = -0.025\,(1-\eta)\left[\tfrac12 - \tfrac{9}{32}(1-\xi)\bigl(1-\eta^{2}\bigr)\right]\ \text{mm}\;}$$

The expression can be checked at every node without expanding it. At nodes 1, 2 and 5 the factor $(1-\eta^{2})$ vanishes and $(1-\eta) = 2$, giving $-0.025$ mm; at nodes 3 and 4 the factor $(1-\eta) = 0$; and at nodes 6 and 7 the bracket itself vanishes because $\tfrac{9}{32}(2)\left(\tfrac89\right) = \tfrac12$. At the centroid the field is $v(0,0) = -0.025\left(\tfrac12 - \tfrac{9}{32}\right) = -0.00547$ mm, which is smaller in magnitude than a bilinear element would predict — the cubic edge pulls the interior back towards the undisplaced side.