--- title: "Shape functions of four-node rectangular element" subject: "Finite Element Method" chapter: 5 key: "semester_8/finite_element_method:rectangular-element-shape-functions" source: https://www.bctengineer.com/q/semester_8/finite_element_method/rectangular-element-shape-functions site: BCT Engineering --- # Shape functions of four-node rectangular element Finite Element Method · Chapter 5: Interpolation Functions - Asked in: year not given, 6 marks - Video: [FEA 24: 2-D Rectangular Elements](https://www.youtube.com/watch?v=jq_L2jnZR30) — Schuster Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/rectangular-element-shape-functions - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-5 ## Question Derive the shape functions of a four-node rectangular element of sides $a$ (along x) and $b$ (along y) with nodes 1 (0, 0), 2 ($a$, 0), 3 ($a$, $b$) and 4 (0, $b$). Verify two properties and comment on the strain variation. ## Short answer Shape functions are N1=(1-x/a)(1-y/b), N2=(x/a)(1-y/b), N3=xy/(ab), N4=(1-x/a)y/b. They satisfy Kronecker delta and partition of unity. Strains vary linearly within the element, giving better bending response than a CST. ## Answer ```text 4 (0,b) o-----------o 3 (a,b) | | | | b | | 1 (0,0) o-----------o 2 (a,0) a ``` #### Displacement function Four nodes per scalar field need four terms. Taking terms symmetrically from Pascal's triangle gives the bilinear polynomial $$ u(x, y) = \alpha_1 + \alpha_2x + \alpha_3y + \alpha_4xy $$ (The $xy$ term is chosen; $x^2$ or $y^2$ alone would make the function unsymmetrical.) Apply the four nodal conditions $u(0,0) = u_1$, $u(a,0) = u_2$, $u(a,b) = u_3$, $u(0,b) = u_4$: $$ \begin{aligned} \alpha_1 &= u_1 \\ \alpha_2 &= \frac{u_2 - u_1}{a} \\ \alpha_3 &= \frac{u_4 - u_1}{b} \\ \alpha_4 &= \frac{u_1 - u_2 + u_3 - u_4}{ab} \end{aligned} $$ Substituting and collecting the coefficients of $u_i$ gives $u = \sum N_iu_i$ with $$ \begin{aligned} N_1 &= \left(1 - \frac{x}{a}\right)\left(1 - \frac{y}{b}\right), & N_2 &= \frac{x}{a}\left(1 - \frac{y}{b}\right) \\ N_3 &= \frac{xy}{ab}, & N_4 &= \left(1 - \frac{x}{a}\right)\frac{y}{b} \end{aligned} $$ The same functions interpolate $v$, i.e. $v = \beta_1 + \beta_2x + \beta_3y + \beta_4xy = \sum N_iv_i$. #### Properties verified - At node 3 ($x = a$, $y = b$): $N_3 = 1$, $N_1 = N_2 = N_4 = 0$. The same holds at the other nodes. - Sum: $N_1 + N_2 + N_3 + N_4 = \left(1 - \frac{x}{a}\right)\left(1 - \frac{y}{b}\right) + \frac{x}{a}\left(1 - \frac{y}{b}\right) + \frac{xy}{ab} + \left(1 - \frac{x}{a}\right)\frac{y}{b} = 1$. - Along an edge, e.g. $y = 0$, only $N_1$ and $N_2$ are non-zero and they are linear in $x$, so the displacement is continuous between neighbouring elements. #### Strain variation $$ \varepsilon_x = \frac{\partial u}{\partial x} = \alpha_2 + \alpha_4y,\qquad \varepsilon_y = \frac{\partial v}{\partial y} = \beta_3 + \beta_4x $$ $\varepsilon_x$ varies linearly with $y$ and $\varepsilon_y$ varies linearly with $x$, so the element represents bending better than the CST. A drawback is that it can only be used for rectangles with sides parallel to the axes; the (isoparametric) quadrilateral element removes this limitation. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which polynomial is used for the four-node rectangular element? A. u = a1 + a2 x + a3 y + a4 xy B. u = a1 + a2 x^2 + a3 y^2 + a4 xy C. u = a1 + a2 x + a3 y + a4 x^2 D. u = a1 + a2 x + a3 y Answer: A. Four nodes need four terms, and the xy term keeps the function symmetric, unlike x^2 or y^2. (Key point: Displacement function) 2. Which is the correct N3 for node 3 at (a, b)? A. (x/a)(1 - y/b) B. (1 - x/a)(y/b) C. (1 - x/a)(1 - y/b) D. xy/(ab) Answer: D. N3 must be 1 at (a, b) and zero at the other nodes, which gives xy/(ab). (Key point: Shape function) 3. How is continuity of displacement between neighbouring elements verified? A. The sum of shape functions is 1 B. Strains are constant along the edge C. Each shape function is quadratic along the edge D. Along an edge only the two end-node functions are non-zero and linear, so the edge displacement depends only on those nodes Answer: D. On y = 0 only N1 and N2 survive and are linear in x, so adjacent elements sharing the edge nodes give the same edge displacement. (Key point: Verification) 4. How does strain vary in this element compared with the CST? A. Strain is zero at the centre B. Strain is constant, same as CST C. Strain varies quadratically in both directions D. ex varies linearly with y and ey linearly with x, so it represents bending better Answer: D. ex = a2 + a4 y and ey = b3 + b4 x, so strain varies linearly, which captures bending better than the CST. 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