--- title: "Shape functions, six-node triangular element" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:six-node-triangle-shape-function" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/six-node-triangle-shape-function site: BCT Engineering --- # Shape functions, six-node triangular element Computational Techniques in Civil Engineering Β· Chapter 4: Finite element method - Asked in: 2072 Asoj, 6 marks - Video: [Shape function of noded Traingle || CT](https://www.youtube.com/watch?v=PZR2e6emqRc) 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/six-node-triangle-shape-function - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Derive the shape function for the element as shown in the figure. [Figure: six-noded triangle with corner nodes 1, 2, 3 and mid-side nodes 4 (side 1-2), 5 (side 2-3), 6 (side 3-1); area coordinates $L_1 = 1$ at node 1, $L_2 = 1$ at node 2, $L_3 = 1$ at node 3, and $L = 1/2$ values at mid-side nodes] ## Short answer The six-node quadratic triangle uses area coordinates L1,L2,L3. Corner shape functions are N1=L1(2L1-1), N2=L2(2L2-1), N3=L3(2L3-1), and midside functions are N4=4L1L2, N5=4L2L3, N6=4L3L1. ## Answer The six-node (quadratic) triangle, also called the linear strain triangle (LST), has three corner nodes and three mid-side nodes. Its shape functions are written in **area (triangular) coordinates** $L_1, L_2, L_3$. ```text 3 / \ / \ 6 * * 5 / \ / \ 1-----*-----2 4 ``` #### Area coordinates For a point $P$ inside the triangle, $L_i = A_i/A$, where $A_i$ is the area of the sub-triangle opposite node $i$. - $L_1 + L_2 + L_3 = 1$. - $L_i=1$ at node $i$ and $L_i=0$ on the opposite side. - Mid-side nodes: 4 $(L_1,L_2,L_3)=(\tfrac12,\tfrac12,0)$, 5 $(0,\tfrac12,\tfrac12)$, 6 $(\tfrac12,0,\tfrac12)$. - $L_i=\dfrac{a_i+b_ix+c_iy}{2A}$, so a quadratic in $x,y$ is a quadratic in $L_i$. #### Conditions on the shape functions The displacement is $u=\sum_{i=1}^{6}N_iu_i$ with $N_i=1$ at node $i$ and $N_i=0$ at the other five nodes. Hence $\sum N_i=1$ (rigid-body motion is represented). #### Corner node 1 $N_1$ must be zero at nodes 2, 3, 5 (all on the line $L_1=0$) and at nodes 4, 6 (on the line $L_1=\tfrac12$). The quadratic that does this is $$ N_1=c\,L_1\left(L_1-\tfrac12\right) $$ At node 1, $L_1=1$: $c\cdot1\cdot\tfrac12=1\Rightarrow c=2$. $$ N_1=L_1(2L_1-1) $$ By the same argument for nodes 2 and 3: $$ N_2=L_2(2L_2-1),\qquad N_3=L_3(2L_3-1) $$ #### Mid-side node 4 (between 1 and 2) $N_4$ must vanish at nodes 1, 6 ($L_2=0$), at nodes 2, 5 ($L_1=0$) and at node 3 (both zero). The product $L_1L_2$ does this: $$ N_4=c\,L_1L_2,\qquad \text{at node 4: } c\cdot\tfrac12\cdot\tfrac12=1\Rightarrow c=4 $$ $$ N_4=4L_1L_2,\qquad N_5=4L_2L_3,\qquad N_6=4L_3L_1 $$ #### Result | Node | Shape function | |---|---| | 1 | $N_1=L_1(2L_1-1)$ | | 2 | $N_2=L_2(2L_2-1)$ | | 3 | $N_3=L_3(2L_3-1)$ | | 4 | $N_4=4L_1L_2$ | | 5 | $N_5=4L_2L_3$ | | 6 | $N_6=4L_3L_1$ | #### Check - At node 1 ($L_1=1$): $N_1=1$; all others contain $L_2$ or $L_3$ and are 0. - At node 4 ($L_1=L_2=\tfrac12$): $N_4=1$, $N_1=N_2=\tfrac12(0)=0$, $N_3=N_5=N_6=0$. - Sum: $\sum N_i = (2L_1^2+2L_2^2+2L_3^2-L_1-L_2-L_3)+4(L_1L_2+L_2L_3+L_3L_1) = 2(L_1+L_2+L_3)^2-(L_1+L_2+L_3)=2-1=1$. Each $N_i$ is quadratic, so strains vary linearly over the element. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What properties do area coordinates L1, L2, L3 have? A. L1 + L2 + L3 = 1; L_i = 1 at node i and 0 on the opposite side B. L1 + L2 + L3 = 0 C. L_i = 1 on the opposite side D. Each L_i is always 1/3 Answer: A. L_i is the ratio of the opposite sub-triangle area to the total, so L_i = 1 at node i and 0 on the opposite side. (Key point: Area coordinates) 2. What is the shape function of corner node 1? A. N1 = L1 B. N1 = L1(2 L1 - 1) C. N1 = 4 L1 L2 D. N1 = L1(L1 - 1) Answer: B. N1 = c L1 (L1 - 1/2) vanishes at the other five nodes; N1 = 1 at node 1 gives c = 2. (Key point: Corner shape function) 3. What is the shape function of mid-side node 4 (between 1 and 2)? A. N4 = 2 L1 L2 B. N4 = 4 L1 L2 C. N4 = L1 L2 D. N4 = 4 L1 L3 Answer: B. c L1 L2 equals c/4 at node 4, so c = 4. 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