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I am struggling with the shape function of finite element method. For sake of computation, most often we work with local coordinate. For 1d things are bit clear to me. but for 2d things getting hard to understand. Like what will be the best way to get $N_{1}^{[e]},N_{2}^{[e]}$ and $N_{3}^{[e]}$ for three node 2d triangular element?

$$ \begin{align} N^{[e]}_1&=\frac12 (\xi+\eta)\\ N^{[e]}_2 &= \frac12 (1+\xi)\tag1\\ N^{[e]}_3 &= \frac12 (1+\eta) \end{align} $$

For four node 2d rectangle it's easy to understand the $\xi-\eta$ system,

image

But I didn't get how they manage to get $N^{[e]}_i$ for $(1)$. Because I feel confused how to pick the $\xi-\eta$ system for triangle.

image2

Didn't give me same $N^{[e]}_i$.

And for the integration we also need the variable transformation and Jacobian,

$$ \begin{align} x&=\sum x_i N^{[e]}_i\\ y&=\sum y_i N^{[e]}_i\\ J&=\left|\matrix{\frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi}\\\frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta}}\right| \end{align} $$

Which seems also painful if the nodes are increasing.

I didn't find any resource where shape function was described details for three and four node 2d triangle and rectangle element respectively. It will be a great help if anyone help me to figure out this. TIA.

1 Answers1

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The shape functions defined on the element domain $\Omega_E$ with edgepoints $x_j$ must satisfy the properties $$ N_i(x_j) = \delta_{ij} \\ \sum^{n}_{i=1} N_i(x) = 1 \quad \forall x \in \Omega_E $$ with $\delta_{ij}$ denoting the Kronecker-delta

For linear shape functions on a triangle in reference coordinates we can write down the general equation for all shape functions: $$ N_1(\xi,\eta) = c_{11} + c_{12} \xi + c_{13}\eta \\ N_2(\xi,\eta) = c_{21} + c_{22} \xi + c_{23}\eta \\ N_3(\xi,\eta) = c_{31} + c_{32} \xi + c_{33}\eta $$ which can be written as $$ N(\xi,\eta) = \begin{bmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \\ \end{bmatrix} \cdot \begin{bmatrix}1 \\ \xi \\ \eta\end{bmatrix} $$ Using the reference triangle with edgepoints $(0,0),(1,0),(0,1)$ we can derive 9 equations for the unknowns $c_{ij}$.

Another Approach

The given shape functions are Lagrangian shape functions in disguise: If we use barycentric coordinates as a base we arrive at the following form: $$ N_i(\xi_1,\xi_2,\xi_3) = \xi_i \\ \xi_1 + \xi_2 + \xi_3 = 1 $$ But as we have a 2-dimensional domain, we need two coordinates to have a quadratic jacobian. Therefore we map the baricentric coordinates to the $(\xi,\eta)$-coordinates via the mapping: $$ \xi_1 = \xi \\ \xi_2 = \eta \\ \xi_3 = 1 - \xi - \eta \\ $$ which also lead us to shape functions in the $\xi, \eta$-coordinate system.

Noiv
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  • Thanks for your response @Noiv. But I was wondering how they got $$N^{[e]}_1=\frac12 (\xi+\eta), N^{[e]}_2 = \frac12 (1+\xi), N^{[e]}_3 = \frac12 (1+\eta)$$ – N00BMaster Nov 01 '23 at 03:52
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    @NOOBMaster The $\frac{1}{2}$ hints at the reference triangle being different, than the one in the figure. I arrived at the shape function via using $(1,-1),(-1,-1),(-1,1)$ as coordinates and allowing for the second property to not be fulfilled. Using the second property i arrive at $N_1(\xi,\eta) = -\frac{1}{2} (\xi + \eta)$ – Noiv Nov 01 '23 at 10:08