--- title: "Shape functions for quadrilateral element" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:shape-function-quadrilateral" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/shape-function-quadrilateral site: BCT Engineering --- # Shape functions for quadrilateral element Computational Techniques in Civil Engineering Β· Chapter 4: Finite element method - Asked in: 2079 Shrawan, 6 marks - Video: [Shape function using Serendipity Function](https://www.youtube.com/watch?v=hErpbZchdf0) β€” Basant BhattπŸ‡³πŸ‡΅ - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/shape-function-quadrilateral - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Derive shape functions for a quadrilateral element. ## Short answer For a four-node quadrilateral in natural coordinates (xi,eta), the bilinear shape functions are Ni=0.25(1+xi*xi_i)(1+eta*eta_i). They satisfy Ni=1 at node i and 0 at others, and sum to 1. ## Answer #### Element and coordinates A four-node quadrilateral has nodes 1 to 4 anticlockwise and 8 DOFs $(u_i,v_i)$. Shape functions are derived for the parent square in natural coordinates $(\xi,\eta)$, $-1\le\xi,\eta\le1$, with nodes $$ 1:(-1,-1),\quad 2:(1,-1),\quad 3:(1,1),\quad 4:(-1,1) $$ ```text 4-----------3 eta | | ^ | + | | | | +--> xi 1-----------2 ``` #### Assumed displacement A bilinear polynomial (4 terms, one for each node) is used for each displacement component: $$ u=\alpha_1+\alpha_2\xi+\alpha_3\eta+\alpha_4\xi\eta $$ Applying the nodal values: $$ \begin{Bmatrix}u_1\\u_2\\u_3\\u_4\end{Bmatrix}=\begin{bmatrix}1&-1&-1&1\\1&1&-1&-1\\1&1&1&1\\1&-1&1&-1\end{bmatrix}\begin{Bmatrix}\alpha_1\\\alpha_2\\\alpha_3\\\alpha_4\end{Bmatrix} $$ Inverting: $$ \begin{Bmatrix}\alpha_1\\\alpha_2\\\alpha_3\\\alpha_4\end{Bmatrix}=\frac14\begin{bmatrix}1&1&1&1\\-1&1&1&-1\\-1&-1&1&1\\1&-1&1&-1\end{bmatrix}\begin{Bmatrix}u_1\\u_2\\u_3\\u_4\end{Bmatrix} $$ Substituting into $u=[1\ \xi\ \eta\ \xi\eta]\{\alpha\}$ and collecting coefficients of the nodal values: $$ \begin{aligned} N_1&=\tfrac14(1-\xi-\eta+\xi\eta)=\tfrac14(1-\xi)(1-\eta)\\ N_2&=\tfrac14(1+\xi-\eta-\xi\eta)=\tfrac14(1+\xi)(1-\eta)\\ N_3&=\tfrac14(1+\xi+\eta+\xi\eta)=\tfrac14(1+\xi)(1+\eta)\\ N_4&=\tfrac14(1-\xi+\eta-\xi\eta)=\tfrac14(1-\xi)(1+\eta) \end{aligned} $$ **General form:** $$ \boxed{N_i=\tfrac14(1+\xi\xi_i)(1+\eta\eta_i)},\quad i=1,\dots,4 $$ #### Properties - $N_i=1$ at node $i$ and $0$ at the other three nodes. - $\sum N_i=1$ (rigid-body motion), and $\sum N_ix_i=x$, so a linear field is reproduced exactly. - Along each side the variation is linear, so neighbouring elements are compatible. #### Use for a general quadrilateral For an arbitrary quadrilateral the same $N_i$ are used as isoparametric functions: $x=\sum N_ix_i$, $y=\sum N_iy_i$, $u=\sum N_iu_i$. The derivatives with respect to $x,y$ are obtained with the Jacobian: $$ [J]=\begin{bmatrix}\partial x/\partial\xi&\partial y/\partial\xi\\\partial x/\partial\eta&\partial y/\partial\eta\end{bmatrix},\qquad \begin{Bmatrix}\partial N/\partial x\\\partial N/\partial y\end{Bmatrix}=[J]^{-1}\begin{Bmatrix}\partial N/\partial\xi\\\partial N/\partial\eta\end{Bmatrix} $$ For a rectangle of sides $2a\times2b$ centred at the origin, $\xi=x/a$ and $\eta=y/b$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which displacement polynomial is assumed for the 4-node quadrilateral? A. u = a1 + a2 xi + a3 eta B. u = a1 + a2 xi + a3 eta + a4 xi eta C. u = a1 + a2 xi^2 + a3 eta^2 + a4 xi eta D. u = a1 + a2 xi + a3 xi^2 + a4 xi^3 Answer: B. A bilinear polynomial has four terms, one per node. (Key point: Assumed polynomial) 2. The general form of the quadrilateral shape functions is A. Ni = (1/4)(1 + xi xi_i)(1 + eta eta_i) B. Ni = (1/2)(1 + xi xi_i) C. Ni = (1 - xi^2)(1 - eta^2) D. Ni = (1/4)(1 - xi xi_i)(1 - eta eta_i) Answer: A. At node i this equals (1/4)(2)(2) = 1 and it vanishes at the other nodes. (Key point: General shape function) 3. Why are neighbouring elements compatible? A. Variation along each side is linear and fixed by the two end nodes B. Because the sum of Ni is 1 C. Because the Jacobian is constant D. Because the mesh is square Answer: A. A linear variation along a shared side is defined by the same two nodal values for both elements. (Key point: Compatibility) 4. Why is the Jacobian needed for a general quadrilateral? A. To compute N in natural coordinates B. To convert derivatives of N with respect to xi, eta into derivatives with respect to x, y C. To find the number of nodes D. To impose boundary conditions Answer: B. {dN/dx, dN/dy} = [J]^-1 {dN/dxi, dN/deta}. 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