4.4 Shape Functions — Solved Exam Questions

We use area (natural) coordinates $L_1, L_2, L_3$ with the constraint $L_1 + L_2 + L_3 = 1$. The element is a complete cubic triangle requiring $m = \frac{(3+1)(3+2)}{2} = 10$ terms.

Corner Nodes ($L_i = 1$ at vertex $i$, cubic in $L_i$):

For $N_1$: must be zero at $L_1 = 0$, $L_1 = \frac{1}{3}$, and $L_1 = \frac{2}{3}$, and equal to 1 at $L_1 = 1$.

At $L_1 = 1$: $1 = C(1)\left(\frac{2}{3}\right)\left(\frac{1}{3}\right) = \frac{2C}{9} \implies C = \frac{9}{2}$

Side Nodes (between corners, at $\frac{1}{3}$ and $\frac{2}{3}$ positions):

For $N_4$ (between nodes 1 and 2, at $L_1 = \frac{2}{3}$, $L_2 = \frac{1}{3}$):

At node 4: $1 = C\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right) \implies C = \frac{27}{2}$

Interior Node (centroid at $L_1 = L_2 = L_3 = \frac{1}{3}$):

At node 10: $1 = C\left(\frac{1}{3}\right)^3 = \frac{C}{27} \implies C = 27$

We use the Serendipity concept (Ref: Section 4.4 — Serendipity Family). Each $N_i$ must equal 1 at node $i$ and 0 at all other nodes.

Corner Nodes:

For $N_1$ at $(\xi_1, \eta_1) = (-1, -1)$:

• $N_1 = 0$ when $\xi = 1 \implies (1-\xi)$ is a factor
• $N_1 = 0$ when $\eta = 1 \implies (1-\eta)$ is a factor
• $N_1 = 0$ at node 5 ($\xi=0,\eta=-1$) and node 8 ($\xi=-1,\eta=0$): the line through these mid-side nodes is $\xi + \eta + 1 = 0$, so $(1+\xi+\eta)$ is a factor

Assume $N_1 = C(1-\xi)(1-\eta)(1+\xi+\eta)$.

At node 1: $1 = C(1+1)(1+1)(1-1-1) = C(2)(2)(-1) = -4C \implies C = -\frac{1}{4}$

By similar reasoning for the other corners:

General corner formula: $N_i = \frac{1}{4}(1+\xi\xi_i)(1+\eta\eta_i)(\xi\xi_i + \eta\eta_i - 1)$

Mid-side Nodes:

For $N_5$ at $(0, -1)$:

• $N_5 = 0$ when $\xi = \pm 1 \implies (1-\xi^2)$ is a factor
• $N_5 = 0$ when $\eta = 1 \implies (1-\eta)$ is a factor

Assume $N_5 = C(1-\xi^2)(1-\eta)$. At node 5: $1 = C(1)(2) \implies C = \frac{1}{2}$

Method 1: Polynomial (Cartesian Coordinate) Approach

6 nodal displacements ($u_1, u_2, u_3, v_1, v_2, v_3$) $\implies$ select polynomial with 3 coefficients each:

Applying boundary conditions for $u$:

Inverting:

where $2A = \det \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{vmatrix}$ and:

Substituting back: $u = [1 \;\; x \;\; y] \cdot \frac{1}{2A}[C]\{u_i\}$

Method 2: Natural (Area) Coordinate Approach

Using area coordinates $L_1, L_2, L_3$ with $L_1 + L_2 + L_3 = 1$:

At node 1 $(L_1=1, L_2=0, L_3=0)$: $u_1 = \alpha_1$. Similarly $u_2 = \alpha_2$, $u_3 = \alpha_3$.

This is identical to Q4 above. Using the natural coordinate mapping $\xi = x/a$, $\eta = y/b$ (so that $\xi, \eta \in [-1,1]$), the derivation follows the Serendipity approach from Section 4.4.

The shape functions are derived exactly as in Q4. For completeness, restating the results:

Corner Nodes ($\xi_i = \pm 1$, $\eta_i = \pm 1$):

Mid-side Nodes:

Verification:

This is again the 8-noded serendipity element as derived in Q4 and Q6. The Serendipity approach yields:

Expanding individually:

The parent element has vertices at $(0,0)$, $(1,0)$, $(0,1)$ with constraint $\xi + \eta \leq 1$.

For $N_1$ at $(0,0)$: Must be zero on the hypotenuse $\xi + \eta - 1 = 0$.

For $N_2$ at $(1,0)$: Must be zero at $\xi = 0$.

For $N_3$ at $(0,1)$: Must be zero at $\eta = 0$.

Step 1: Node Coordinates

From the figure, with origin at the midpoint of the base:

Step 2: Area Coordinates for this Triangle

Vertices: $(x_1, y_1) = (-a, 0)$, $(x_2, y_2) = (a, 0)$, $(x_3, y_3) = (0, 2b)$.

So the total area $A = 2ab$.

The area coordinates $L_i = \frac{1}{2A}(a_i + b_i x + c_i y)$ where:

Step 3: Shape Functions in Area Coordinates

For a 6-noded quadratic triangle, using the standard area coordinate formulas (Ref: Sec. 4.4):

Corner nodes ($N_i$ must vanish at $L_i = 0$ and $L_i = \frac{1}{2}$):

Mid-side nodes ($N_k$ at midpoint of side between $i$ and $j$):

Step 4: Shape Functions in Cartesian Coordinates $(x, y)$

Substituting the expressions for $L_1, L_2, L_3$:

Verification at nodes:

Using area coordinates $L_1, L_2, L_3$ with $L_1 + L_2 + L_3 = 1$ (Ref: Section 4.4 — Shape Functions for Isoparametric Triangular Elements).

Corner Nodes:

$N_1$ at node 1 ($L_1=1, L_2=0, L_3=0$): Must vanish when $L_1 = 0$ and when $L_1 = \frac{1}{2}$.

Mid-side Nodes:

$N_4$ at node 4 ($L_1 = \frac{1}{2}, L_2 = \frac{1}{2}, L_3 = 0$): Must vanish when $L_1 = 0$ and $L_2 = 0$.

Similarly for nodes 5 and 6:

Verification: