Shape functions, ten-node triangular element

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.)

              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.