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FEA Meshing Guidelines

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Introduction

This page and the next are a bit of a detour from the theoretical development presented thus far. Instead of classical theory, these pages address the very practical tasks of (i) generating finite element meshes for linear elastic fracture mechanics (LEFM) analyses, and (ii) determining stress intensity factors, \(K\), from displacement data. This page will focus on meshing guidelines at crack tips.

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Meshing Guidelines for LEFM Analyses

The sketch here shows the ideal finite element mesh at a crack tip for performing LEFM analyses. The key feature of the mesh is the placement of the midside nodes of the inner ring of elements at quarter-point locations. The reason for this is related to the element shape functions and will be explained in the next section. But first, here is a more complete list of meshing guidelines.

1. As stated already, the inner ring of elements are quadratic and their midside nodes are at 25% of the edge length. This is the one guideline here that borders on being an absolute requirement.
   
   
2. Note that all element edges are straight in the figure. It may be tempting to move the midside nodes so the outer edges of the elements form a perfect circle. But this would be wrong. The reason is buried deep in quadratic element shape function theory.
   
   
3. The midside nodes on the outer ring of elements are at the midpoints of the edges, not at the quarter points. This should in fact be true of all elements that are not part of the inner ring.
   
   
4. Finally, an anti-guideline. While the sketch here shows eight elements forming the inner ring around the crack tip, this number is not cast in stone. Six elements are also popular. Any number greater than five should be acceptable.

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Quadratic Element Shape Functions

The reason for the quarter point node locations is... finite element shape functions. Unfortunately, the subject of shape functions is far too expansive to thoroughly review here. Nevertheless, it is absolutely necessary to be familiar with them in order to follow the discussion below. This pdf file is a pretty good reference for shape functions.

For LEFM analyses, elements at the crack tip should be quadratic with quarter-point nodes, regardless of whether they are 2-D or 3-D. Elements with triangular cross-sections, 2-D triangles and 3-D wedges, are by far the most popular.

An example element is highlighted in the lower chart to the right. It includes the element's 6 nodes and local \(s\) and \(t\) coordinates.

The mapping equations for position, \({\bf x}\), and displacement, \({\bf u}\), are

\[ {\bf x} = \sum_{i=1}^6 {\bf x}_i \, \phi_i \qquad \qquad \text{and} \qquad \qquad {\bf u} = \sum_{i=1}^6 {\bf u}_i \, \phi_i \]
and the 6 shape functions are

\[ \begin{eqnarray} \phi_1 & = & (1 - s - t) (1 - 2s - 2t) \\ \\ \phi_2 & = & s ( 2s - 1 ) \\ \\ \phi_3 & = & t ( 2t - 1 ) \\ \\ \phi_4 & = & 4 s (1 - s - t) \\ \\ \phi_5 & = & 4 s t \\ \\ \phi_6 & = & 4 t (1 - s - t) \end{eqnarray} \]
where both \(s\) and \(t\) vary between 0 and 1.

The crack edge corresponds to nodes 1, 4, and 2, and \(t = 0\). So setting \(t = 0\) and focusing on the three key nodes gives

\[ \begin{eqnarray} \phi_1 & = & (1 - s) (1 - 2s) \\ \\ \phi_2 & = & s ( 2s - 1 ) \\ \\ \phi_4 & = & 4 s (1 - s) \end{eqnarray} \]
Rename the three nodes to make the following discussions easier to follow. Rename "1" to "0" because it is at the crack tip. Rename "4" to "mid" because it is the midside node. And rename "2" to "L" because it is at the full length of the finite element. The shape functions for the position vector, \({\bf x}\), and the displacement vector, \({\bf u}\), can now be written as

\[ \begin{eqnarray} {\bf x} = {\bf x}_0 \, (1 - s) (1 - 2s) + {\bf x}_\text{mid} \, 4 s (1 - s) + {\bf x}_L \, s (2s - 1) \\ \\ {\bf u} = {\bf u}_0 \, (1 - s) (1 - 2s) + {\bf u}_\text{mid} \, 4 s (1 - s) + {\bf u}_L \, s (2s - 1) \end{eqnarray} \]
The components of the position vector are \({\bf x} = (x, y, z)\) and we only care about the \(x\) position. Also, the components of the displacement vector are \({\bf u} = (u_x, u_y, u_z)\) and we only care about the vertical displacement, \(u_y\). This leads to

\[ \begin{eqnarray} x & = & x_0 \, (1 - s) (1 - 2s) & + & x_\text{mid} \, 4 s (1 - s) & + & x_L \, s (2s - 1) \\ \\ u_y & = & u_{y,0} \, (1 - s) (1 - 2s) & + & u_{y,\text{mid}} \, 4 s (1 - s) & + & v_{y,L} \, s (2s - 1) \end{eqnarray} \]
Now set the following variables. For the \(x\) position, set \(x_0 = 0\), \(x_L = L\), and finally set \(x_\text{mid}\) equal to a fraction, \(f\), of the element length, \(L\), or \(x_\text{mid} = f*L\) where \((0 \le f \le 1)\). So the position function can finally be written as

\[ \begin{eqnarray} x & = & f \, L \, 4 s (1 - s) + L \, s (2s - 1) \\ \\ & = & \Big[ f \, 4 s (1 - s) + s (2s - 1) \Big] \, L \end{eqnarray} \]
Also set \(u_{y,0} = 0\) and \(u_{y,L} = \delta_L\) where \(\delta_L\) is the vertical displacement of Node 2 at \(x_L\). These are the easy displacements. The remaining displacement at \(x_\text{mid}\) is the tricky one. Its vertical displacement is \(u_{y,\text{mid}} = \sqrt{f \,} \, \delta_L\) because the crack profile follows a square root function. All this gives

\[ \begin{eqnarray} u_y & = & \sqrt{f\,} \, \delta_L \, 4 s (1 - s) + \delta_L \, s (2s - 1) \\ \\ & = & \Big[ \sqrt{f\,} \, 4 s (1 - s) + s (2s - 1) \Big] \, \delta_L \end{eqnarray} \]
So in summary, we have

\[ \begin{eqnarray} x & = & \Big[ f \, 4 s (1 - s) + s (2s - 1) \Big] \, L \\ \\ u_y & = & \Big[ \sqrt{f\,} \, 4 s (1 - s) + s (2s - 1) \Big] \, \delta_L \end{eqnarray} \]
Now we can finally start choosing different values for \(f\) and analyzing the results. We'll start by showing how and why \(f = 1/4\) is the proper value for LEFM analyses. Insert \(f = 1/4\) and \(\sqrt{f\,} = 1/2\) into the equations. (They lead to remarkable simplifications.)

\[ \begin{eqnarray} x & = & \Big[ f \, 4 s (1 - s) + s (2s - 1) \Big] \, L \\ \\ & = & \Big[ (1/4) \, 4 s (1 - s) + s (2s - 1) \Big] \, L \\ \\ & = & s^2 \, L \end{eqnarray} \]

\[ \begin{eqnarray} u_y & = & \Big[ \sqrt{f\,} \, 4 s (1 - s) + s (2s - 1) \Big] \, \delta_L \\ \\ & = & \Big[ \sqrt{1/4\,} \, 4 s (1 - s) + s (2s - 1) \Big] \, \delta_L \\ \\ & = & s \, \delta_L \end{eqnarray} \]
In summary, \(x = s^2 L\) and \(u_y = s \, \delta_L\). The final step is to combine the two equations by eliminating \(s\). Do this by solving the \(x\) equation for \(s\) and substituting the expression into the other for \(u_y\). Solving the \(x\) equation for \(s\) gives

\[ s = \sqrt{ x \over L} \]
and inserting this into the \(u_y\) equation results in

\[ u_y = \sqrt{ x } \, \left( \delta_L \over \sqrt{L} \right) \]
This shows the vertical displacement in a quadratic element is indeed dependent on \(\sqrt{x}\), just as LEFM theory requires. It cannot be overstated how critical this is. The quadratic element with the quarter-point node properly captures the square-root dependence, and just as important, the infinite slope of \(d u_y/dx\) at \(x = 0\) that arises from the square-root function. This infinite slope at \(x = 0\) is known as the singularity.

Any other meshing strategy, either linear elements or quadratic elements with midside nodes at any location other than the quarter-point, will not capture this singularity regardless of how fine the mesh is because the slope at \(x = 0\) will not be infinite. It may be very large, but even very large is still finite, not infinite. In fact, one could say that very large is still infinitely far from infinite.