--- title: "Shape functions of eight-node serendipity element" subject: "Finite Element Method" chapter: 7 key: "semester_8/finite_element_method:eight-node-serendipity-shape-functions" source: https://www.bctengineer.com/q/semester_8/finite_element_method/eight-node-serendipity-shape-functions site: BCT Engineering --- # Shape functions of eight-node serendipity element Finite Element Method · Chapter 7: Higher order Elements - Asked in: year not given, 6 marks - Video: [8 Nodded | Quadrilateral element | Shape function | Derivation | Serendipity | FEA | L-11](https://www.youtube.com/watch?v=kBDCRMpKA7w) — Mech Proverse - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/eight-node-serendipity-shape-functions - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-7 ## Question For the eight-node serendipity quadrilateral in natural coordinates $(\xi, \eta)$, with corner nodes 1 (-1, -1), 2 (1, -1), 3 (1, 1), 4 (-1, 1) and mid-side nodes 5 (0, -1), 6 (1, 0), 7 (0, 1), 8 (-1, 0): derive the shape functions $N_5$ and $N_1$ and write the general formulae. Verify that $N_1$ is 1 at node 1. ## Short answer N5=½(1–ξ²)(1–η). N1=¼(1–ξ)(1–η)(–ξ–η–1). General corner: N_i=¼(1+ξξ_i)(1+ηη_i)(ξξ_i+ηη_i–1); mid-side ξ_i=0: N_i=½(1–ξ²)(1+ηη_i); mid-side η_i=0: N_i=½(1+ξξ_i)(1–η²). At node 1, N1=1. ## Answer Each shape function must equal 1 at its own node and 0 at the other seven nodes, and the sum must be 1. #### Mid-side node 5 (0, -1) $N_5$ must vanish at the other two nodes of the edge $\eta = -1$ ($\xi = \pm1$) and on the edge $\eta = +1$, and at nodes 6 and 8 ($\xi = \pm1$). Use a quadratic in $\xi$ times a linear function in $\eta$: $$ N_5 = c\,(1 - \xi^2)(1 - \eta) $$ At node 5, $N_5 = c(1)(2) = 1$, so $c = \dfrac{1}{2}$: $$ N_5 = \tfrac{1}{2}(1 - \xi^2)(1 - \eta) $$ #### Corner node 1 (-1, -1) Start with the bilinear function $\tfrac{1}{4}(1 - \xi)(1 - \eta)$. It is 1 at node 1 and 0 at nodes 2, 3, 4, but it is non-zero at mid-side nodes 5 and 8 (value $\tfrac{1}{2}$ each). Subtract $\tfrac{1}{2}N_5$ and $\tfrac{1}{2}N_8$: $$ N_1 = \tfrac{1}{4}(1 - \xi)(1 - \eta) - \tfrac{1}{2}N_5 - \tfrac{1}{2}N_8 $$ With $N_8 = \tfrac{1}{2}(1 - \xi)(1 - \eta^2)$, simplifying gives $$ N_1 = \tfrac{1}{4}(1 - \xi)(1 - \eta)(-\xi - \eta - 1) $$ **Check:** at node 1 ($\xi = \eta = -1$): $\tfrac{1}{4}(2)(2)(1 + 1 - 1) = 1$. At node 2 ($\xi = 1$): factor $(1 - \xi) = 0$. At node 5 ($\xi = 0$, $\eta = -1$): factor $(1 - \eta)(-\xi - \eta - 1) = 2(0) = 0$. At node 8 similarly zero. #### General formulae ($\xi_i$, $\eta_i$ = nodal values) - Corner nodes ($i = 1$ to 4): $$ N_i = \tfrac{1}{4}(1 + \xi\xi_i)(1 + \eta\eta_i)(\xi\xi_i + \eta\eta_i - 1) $$ - Mid-side nodes with $\xi_i = 0$ (nodes 5, 7): $$ N_i = \tfrac{1}{2}(1 - \xi^2)(1 + \eta\eta_i) $$ - Mid-side nodes with $\eta_i = 0$ (nodes 6, 8): $$ N_i = \tfrac{1}{2}(1 + \xi\xi_i)(1 - \eta^2) $$ The edge shape functions are quadratic along each edge, so adjacent elements share three nodes per edge and compatibility is maintained. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is N5 for the mid-side node 5 at (0, -1)? A. (1/2)(1 - xi^2)(1 - eta) B. (1/2)(1 + xi)(1 - eta^2) C. (1/4)(1 - xi)(1 - eta) D. (1 - xi^2)(1 - eta^2) Answer: A. Quadratic in xi (zero at xi = +-1) times linear in eta (zero at eta = 1); c = 1/2 makes N5 = 1 at node 5. (Key point: Mid-side function) 2. How is N1 constructed from the bilinear function? A. Multiply by N5 B. Subtract N5 and N8 fully C. Subtract (1/2)N5 and (1/2)N8 from (1/4)(1 - xi)(1 - eta) D. Add (1/2)N5 and (1/2)N8 Answer: C. The bilinear function has value 1/2 at nodes 5 and 8, so half of N5 and N8 are subtracted to make N1 zero there. (Key point: Corner function method) 3. Which is the general corner-node formula? A. (1/4)(1 + xi xi_i)(1 + eta eta_i)(xi xi_i + eta eta_i + 1) B. (1/4)(1 + xi xi_i)(1 + eta eta_i)(xi xi_i + eta eta_i - 1) C. (1/2)(1 - xi^2)(1 + eta eta_i) D. (1/4)(1 + xi xi_i)(1 + eta eta_i) Answer: B. At node 1 the last factor is -xi - eta - 1 which equals 1 at (-1,-1), as the check confirms. (Key point: Corner formula) 4. Why is compatibility maintained between adjacent elements? A. Functions are linear along edges B. Functions are constant on edges C. Elements share only corner nodes D. Functions are quadratic along each edge and adjacent elements share three nodes per edge Answer: D. A quadratic along the edge is fixed by three shared nodes, so the displacement along the edge is the same for both elements. (Key point: Compatibility) ## Related in Finite Element Method - [Shape functions of four-node rectangular element](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Arectangular-element-shape-functions) — Finite Element Method, chapter 5 - [Derive shape functions of CST element](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Acst-shape-functions-derivation) — Finite Element Method, chapter 5 - [Jacobian of a four-node isoparametric element](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Aisoparametric-jacobian-quadrilateral) — Finite Element Method, chapter 7 - [Linear shape functions for one-dimensional element](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Alinear-bar-shape-functions) — Finite Element Method, chapter 5 - [Area coordinates and temperature at a point](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Acst-area-coordinates-interpolation) — Finite Element Method, chapter 5 - [Derive bar element stiffness matrix](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Abar-element-stiffness-derivation) — Finite Element Method, chapter 3 ## Related in other subjects - [Shape functions, eight-noded rectangular element](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Ashape-function-eight-noded-rectangle) — Computational Techniques in Civil Engineering, chapter 4, asked 2× - [Shape functions, six-node triangular element](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Asix-node-triangle-shape-function) — Computational Techniques in Civil Engineering, chapter 4 - [Shape functions, ten-node triangular element](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Ashape-function-ten-node-triangle) — Computational Techniques in Civil Engineering, chapter 4, asked 2× - [Isoparametric formulation; eight-noded shape functions](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Aisoparametric-eight-noded) — Computational Techniques in Civil Engineering, chapter 4