The variational formulation is a(u,v)=(f,v) , or ∬𝐴 ∇𝑢∇𝑣𝑑𝐴 = ∬𝐴 𝑓𝑣𝑑𝐴, where 𝑑𝐴 = 𝑑𝑥𝑑𝑦
First the 1D problem was solved using four elements and linear Lagrange FE basis functions. Figure 1-1
displays the exact and FEM solutions plotted together. The solutions for parts 1b and 1c were integrated
for the sake of comparison. The values of the integrated solutions are shown below in Table 1b-1.
Table 1b-1. Integration of various solutions to Problem 1
1D
2D
COMSOL
Linear
4 Elem
78.125
79.668*
COMSOL
Quadratic
4 Elem
83.333
83.333*
Maple
Linear
4 Elem
78.125
---
Maple
Maple
Linear Quadratic
8 Elem
4 Elem
82.03
83.333
-----
*In 2D, the solution was integrated over the entire solution
Maple
Exact
Sln
83.333
---
Figure 1b-1. Exact and FEM solutions, 4 elements, linear
Next the 1D problem was solved using four elements and quadratic Lagrange FE basis functions. Figure
1-2 displays the exact and FEM solutions plotted together. See Table 1b-1 for the integration of the
solution over the interval.
Figure 1b-2. Exact and FEM solutions, 4 elements, quadratic
As expected, quadratic elements significantly increase the accuracy of the solution. The FEM solution is
in fact the exact solution, and so there is no error.
Next, the COMSOL solutions were plotted. Both solutions utilize 4 elements. The solutions are shown
below. See Table 1b-1 for the integration of the solutions over the interval.
Figure 1b-3. COMSOL solution, linear elements
Figure 1b-4. COMSOL solution, quadratic elements
The 2D problem was then solved in COMSOL. The solution across the plate is shown below in Figure
1c-1 and the solution along the upper boundary is shown below in Figure 1c-2. See Table 1b-1 for the
integration of the solutions over the entire surface.
Figure 1c-1. 2D problem, COMSOL solution
Figure 1c-2. 2D problem, COMSOL solution, upper boundary
Finally, the 2D problem was solved in Abaqus. The surface solution is shown below in Figure 1d-1, and
the solution across the upper boundary is shown in Figure 1d-2.
Figure 1d-1. Abaqus solution across surface
Figure 1d-2. Abaqus solution across upper edge
The problem was first solved in COMSOL with quadratic, triangular elements and an approximate
element size of 0.25. The solution is shown below in Figure 2a-1. The mesh was then refined, using an
approximate element size of 0.1. The solution is shown below in Figure 2a-2. The solution was
integrated over the surface in order to evaluate the validity of the solutions. The values of the integrals
are displayed in Table 2a-1 below.
Table 2a-1. Integration of various solutions to Problem 2
COMSOL
Quadratic
0.25 Elem Size
COMSOL
Quadratic
0.1 Elem Size
0.33063
0.33059
Maple
Exact
Sln
0.3306
Figure 2a-1. COMSOL solution, quadratic triangular elements, 0.25 element size
Figure 2a-2. COMSOL solution, quadratic triangular elements, 0.1 element size
The problem was first solved in Abaqus with linear elements and a mesh size of 0.25. The results are
shown in Figure 2b-1. The problem was next solved with quadratic elements and a mesh size of 0.25, and
finally with quadratic elements and a mesh size of 0.08. The results for the quadratic and quadratic
refined meshes are shown in Figures 2b-2 and 2b-3.
Figure 2b-1. Abaqus, linear elements, 0.25 mesh size
Figure 2b-2. Abaqus, quadratic elements, 0.25 mesh size.
Figure 2b-3. Abaqus, quadratic elements, 0.08 mesh size
This is a 2-D Cartesian problem, that can be treated as a heat transfer problem where the given boundary
conditions are applied to a slab. The problem was first solved in COMSOL using quadratic elements with
a maximum element size of 0.25. The solution across the upper boundary is plotted below in Figure 3a-1.
The temperature was also solved for at the point (0.5,0.5) in order to compare with the Maple and Abaqus
solutions. The point evaluation is given in Table 3b-1.
Figure 3a-1. Temperature vs x, upper boundary, COMSOL
The problem was next solved in Abaqus with quadratic elements and a maximum element size of 0.25.
The temperature contour plot is shown below in Figure 3b-1. The temperature across the upper boundary
is shown in Figure 3b-2. It is evident that the Abaqus and COMSOL solutions are very similar.
However, the point evaluation (see Table 3b-1) shows that the solutions are not exactly the same.
Figure 3b-1. Temperature contour plot, Abaqus
Figure 3b-2. Temperature vs x, upper boundary, Abaqus
The exact solution was plotted in Maple (Figure 3b-3) for comparison and validation of the COMSOL
and Abaqus solutions. It is evident that all three solutions are similar.
Figure 3b-3. Exact solution, Maple
In order to validate the solution, the temperature was evaluated at (x,y) = (0.5,0.5). This allows for
comparison between the Maple, COMSOL, and Abaqus results. The point values are displayed in Table
3b-1. It is evident that the Abaqus solution is more accurate than the COMSOL solution. This is likely
due to the fact that the COMSOL solution utilizes triangular elements. The COMSOL mesh was
significantly refined and the problem was solved a second time in COMSOL, and the solution did not
significantly change.
Table 3b-1. Point evaluations
(x,y)
(0.5,0.5)
Maple (Exact)
0.5
COMSOL
0.574
Abaqus
0.5
The cantilever beam problem was first solved in COMSOL using triangular elements and an approx mesh
size of 0.25. The results are shown in Figure 4a-1.
Figure 4a-1. COMSOL solution, Z-direction displacement field
Need to verify validity – could do by comparing at points… or plotting along a curve – need to write out
the steps for doing this…
The problem was solved in Abaqus first with a coarse mesh (0.25 approx element size) and second with a
finer mesh (0.1 approx element size). The solution for the finer mesh is shown below in Figure 4b-1.
Quadratic elements were used.
Figure 4b-1. Solution with fine mesh, quadratic elements, Abaqus