--- title: "Shape functions, ten-node triangular element" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:shape-function-ten-node-triangle" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/shape-function-ten-node-triangle site: BCT Engineering --- # Shape functions, ten-node triangular element Computational Techniques in Civil Engineering Β· Chapter 4: Finite element method - Asked in: 2078 Chaitra, 6 marks; 2070 Bhadra, 8 marks (asked 2 times) - 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/shape-function-ten-node-triangle - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Derive the shape function for the triangular element shown in the figure. [Figure: triangular element with corner nodes 1, 2, 3, two nodes on each side (4 and 5 on side 1-2, 6 and 7 on side 2-3, 8 and 9 on side 3-1) and a central node 10; area coordinates run $L_1 = 1, 2/3, 1/3, 0$ along the sides, i.e. a cubic triangle] ## Short answer Using area coordinates L1,L2,L3, the ten-node cubic triangle has corner functions Ni=0.5Li(3Li-1)(3Li-2); side-node functions N4=4.5L1L2(3L1-1), N5=4.5L1L2(3L2-1), and cyclic forms for other sides; central node N10=27L1L2L3. These satisfy N_i=1 at node i, zero at other nodes, sum to one, and are a complete cubic. ## Answer #### Approach A cubic triangle has 10 nodes (3 corners, 6 side nodes at the third-points, 1 centroid) and a complete cubic polynomial has 10 terms ($1,x,y,x^2,xy,y^2,x^3,x^2y,xy^2,y^3$). It is easiest to use the **area (triangular) coordinates** $L_1,L_2,L_3$ with $L_1+L_2+L_3=1$, where $L_i=\dfrac{a_i+b_ix+c_iy}{2A}$. The displacement is $$ u=\sum_{i=1}^{10}N_i\,u_i $$ and each $N_i$ is a cubic that equals 1 at node $i$ and 0 at the other nine nodes. Node coordinates in area coordinates ($L_1,L_2,L_3$): nodes 1, 2, 3 = $(1,0,0),(0,1,0),(0,0,1)$; node 4 = $(\tfrac23,\tfrac13,0)$, 5 = $(\tfrac13,\tfrac23,0)$; 6 = $(0,\tfrac23,\tfrac13)$, 7 = $(0,\tfrac13,\tfrac23)$; 8 = $(\tfrac13,0,\tfrac23)$, 9 = $(\tfrac23,0,\tfrac13)$; 10 = $(\tfrac13,\tfrac13,\tfrac13)$. (Side nodes are numbered anticlockwise round the triangle: 4 and 5 go from node 1 to node 2, 6 and 7 from node 2 to node 3, and 8 and 9 from node 3 to node 1.) ```text 3 / \ 8 / \ 7 / \ 9 / 10 \ 6 / \ 1---4---5---2 ``` #### Corner node 1 $N_1$ must vanish on the lines $L_1=0$ (nodes 2, 3, 6, 7), $L_1=\tfrac13$ (nodes 5, 8, and the centre 10) and $L_1=\tfrac23$ (nodes 4, 9). So $$ N_1=C\,L_1\left(L_1-\tfrac13\right)\left(L_1-\tfrac23\right) $$ At node 1, $L_1=1$: $C\times1\times\tfrac23\times\tfrac13=1\Rightarrow C=\tfrac92$: $$ N_1=\tfrac12L_1(3L_1-1)(3L_1-2) $$ Similarly $N_2=\tfrac12L_2(3L_2-1)(3L_2-2)$ and $N_3=\tfrac12L_3(3L_3-1)(3L_3-2)$. #### Side node 4 (on side 1-2, $L_1=\tfrac23$, $L_2=\tfrac13$) $N_4$ vanishes on $L_1=0$, $L_2=0$ and $L_1=\tfrac13$, so $N_4=C\,L_1L_2(L_1-\tfrac13)$. At node 4: $C\times\tfrac23\times\tfrac13\times\tfrac13=1\Rightarrow C=\tfrac{27}{2}$: $$ N_4=\tfrac92L_1L_2(3L_1-1) $$ By the same reasoning (node 5 is the mirror image, vanishing at $L_2=\tfrac13$): $$ N_5=\tfrac92L_1L_2(3L_2-1) $$ and for the other sides, cyclically: $$ \begin{aligned} N_6&=\tfrac92L_2L_3(3L_2-1),& N_7&=\tfrac92L_2L_3(3L_3-1)\\ N_8&=\tfrac92L_3L_1(3L_3-1),& N_9&=\tfrac92L_3L_1(3L_1-1) \end{aligned} $$ #### Central node 10 $N_{10}$ must vanish on all three sides: $N_{10}=C\,L_1L_2L_3$; at the centre $C/27=1$: $$ N_{10}=27\,L_1L_2L_3 $$ #### Checks - $N_i=1$ at node $i$ and $0$ at all other nodes (checked for all ten nodes). - $\sum_{i=1}^{10}N_i=1$ everywhere, so rigid-body motion is represented. - Along a side the displacement is a cubic defined by the four nodes on that side, so adjacent elements are compatible. The element is a complete cubic, so it gives a quadratic variation of strain across the element. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the cubic triangle have 10 nodes? A. A complete cubic polynomial in x and y has 10 terms, needing 10 nodal values B. Each side needs 3 nodes and there are 3 sides plus one C. A quadratic has 10 terms D. The area coordinates L1, L2, L3 give 10 combinations Answer: A. The terms 1, x, y, x^2, xy, y^2, x^3, x^2y, xy^2, y^3 are 10, so 10 nodes (3 corners, 6 side nodes, 1 centroid) are needed. (Key point: Why 10 nodes) 2. N1 is built as C L1 (L1 - 1/3)(L1 - 2/3). Why these factors? A. They make N1 equal 1 at nodes 2 and 3 B. N1 must vanish on the lines L1 = 0, 1/3 and 2/3 that contain all the other nine nodes C. They make the sum of all Ni equal zero D. They come from the area of the triangle Answer: B. All other nodes lie on L1 = 0, 1/3 or 2/3, so N1 is zero there; C is then fixed by N1 = 1 at L1 = 1. (Key point: Corner shape function form) 3. What is the shape function of the central node 10 and why? A. N10 = 27 L1 L2 L3, zero on all three sides and equal to 1 at the centroid B. N10 = 9 L1 L2 L3 because there are 9 other nodes C. N10 = L1 L2 L3, since area coordinates already sum to 1 D. N10 = 27 L1 (L1-1/3)(L1-2/3) Answer: A. L1L2L3 vanishes on all sides; at the centre it is 1/27, so C = 27. (Key point: Centre node function) 4. What does sum of Ni = 1 everywhere guarantee? A. Compatibility between adjacent elements B. Constant strain in the element C. Rigid-body motion is represented D. The stiffness matrix is symmetric Answer: C. If all nodal displacements are equal, the interpolated displacement equals that same value, so a rigid translation is reproduced. 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