Shape functions, six-node triangular element

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 $L_1, L_2, L_3$.

            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

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.