--- title: "Derive shape functions of CST element" subject: "Finite Element Method" chapter: 5 key: "semester_8/finite_element_method:cst-shape-functions-derivation" source: https://www.bctengineer.com/q/semester_8/finite_element_method/cst-shape-functions-derivation site: BCT Engineering --- # Derive shape functions of CST element Finite Element Method · Chapter 5: Interpolation Functions - Asked in: year not given, 8 marks - Video: [CST (2D) Triangular Element Shape Function - Derivation](https://www.youtube.com/watch?v=P3LdjdpmID8) — FEA Simulation Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/cst-shape-functions-derivation - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-5 ## Question Derive the shape functions of a three-node linear triangular (constant strain triangle) element in terms of nodal coordinates. State their properties and explain why the element is called a constant strain triangle. ## Short answer For a linear triangle, shape functions are N_i=(a_i+b_i x+c_i y)/(2A), derived by solving nodal conditions. They satisfy Kronecker delta, sum to unity, and equal area coordinates. Strains are constant because b_i and c_i do not depend on x or y. ## Answer Consider a triangle with nodes 1, 2, 3 at $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ numbered anticlockwise. Each node has two displacements $(u_i, v_i)$, so the element has 6 degrees of freedom. ```text 3 (x3,y3) / \ / \ / \ 1-------2 ``` #### Displacement field Choose a complete linear polynomial (from Pascal's triangle): $$ u = \alpha_1 + \alpha_2x + \alpha_3y,\qquad v = \alpha_4 + \alpha_5x + \alpha_6y $$ Apply the nodal conditions for $u$: $$ \begin{Bmatrix} u_1 \\ u_2 \\ u_3 \end{Bmatrix} = \begin{bmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{bmatrix}\begin{Bmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{Bmatrix} $$ Solving by Cramer's rule, where the determinant of the matrix equals $2A$ ($A$ = area of triangle): $$ u = N_1u_1 + N_2u_2 + N_3u_3,\qquad N_i = \frac{1}{2A}\left(a_i + b_ix + c_iy\right) $$ with $$ \begin{aligned} a_1 &= x_2y_3 - x_3y_2, & b_1 &= y_2 - y_3, & c_1 &= x_3 - x_2 \\ a_2 &= x_3y_1 - x_1y_3, & b_2 &= y_3 - y_1, & c_2 &= x_1 - x_3 \\ a_3 &= x_1y_2 - x_2y_1, & b_3 &= y_1 - y_2, & c_3 &= x_2 - x_1 \end{aligned} $$ $$ 2A = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) $$ The same $N_i$ apply to $v$, so $$ \begin{Bmatrix} u \\ v \end{Bmatrix} = \begin{bmatrix} N_1 & 0 & N_2 & 0 & N_3 & 0 \\ 0 & N_1 & 0 & N_2 & 0 & N_3 \end{bmatrix}\{d\} $$ #### Properties - $N_i = 1$ at node $i$ and $0$ at the other two nodes. - $N_1 + N_2 + N_3 = 1$ at every point (and $\sum N_ix_i = x$, $\sum N_iy_i = y$). - $N_i$ varies linearly along each edge and depends only on the two nodes of that edge, so displacements are continuous between adjacent elements. - $N_i$ equal the area coordinates $L_i = A_i/A$. #### Why constant strain The strains are $$ \varepsilon_x = \frac{\partial u}{\partial x} = \frac{1}{2A}\sum b_iu_i,\quad \varepsilon_y = \frac{1}{2A}\sum c_iv_i,\quad \gamma_{xy} = \frac{1}{2A}\sum (c_iu_i + b_iv_i) $$ $b_i$ and $c_i$ depend only on the nodal coordinates, not on $x$ or $y$. So the strains (and stresses) are the same everywhere inside the element. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which displacement field is assumed for the three-node triangle? A. u = a1 + a2 x + a3 y (and likewise v), a complete linear polynomial B. u = a1 + a2 x + a3 y + a4 xy C. u = a1 + a2 x only D. u = a1 + a2 x^2 + a3 y^2 Answer: A. Three nodes per field need three terms; the complete linear polynomial 1, x, y is used for u and v separately. (Key point: Displacement field) 2. What is the form of the shape function Ni obtained by Cramer's rule? A. Ni = (bi x + ci y)/(ai) B. Ni = (ai + bi x + ci y)/(2A) C. Ni = 2A(ai + bi x + ci y) D. Ni = (ai + bi x + ci y)/A^2 Answer: B. The determinant of the coordinate matrix is 2A, so each Ni is a linear function divided by 2A. (Key point: Shape function form) 3. Which set of properties is correct for the CST shape functions? A. Ni = 1 at node i and 0 at the other nodes; N1 + N2 + N3 = 1 B. Ni = 0 at node i and 1 at the others; sum = 0 C. Ni = 1 everywhere; sum = 3 D. Ni is quadratic along edges; sum = 2 Answer: A. Ni is unity at its own node, zero at the others and the three sum to 1 everywhere (they are the area coordinates). (Key point: Properties) 4. Why is the element called a constant strain triangle? A. bi and ci depend only on nodal coordinates, so the strains do not vary with x or y B. The shape functions are constant C. The element area is constant D. The nodal displacements are all equal Answer: A. Strains are derivatives of linear functions, giving sums of bi ui and ci vi over 2A with no x or y dependence, so strain and stress are uniform inside the element. 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