Catia Composites Course
Introduction
In order to learn more about composites and FEA in CAD and CAE environment, I'll be following a course created by Nader G. Zamani, professor of mechanical engineering at University of Windsor, Canada.
I would like to thank him for the time and effort spent to produce all his content.
Exercises
In the following sections, only notes and results about the tutorials are going to be shown. For further information, please follow the tutorials linked on each exercise.
Single ply composite
This case is neither symmetric, since it only has one ply, nor balanced, since the fiber direction is not aligned with the applied load (the part will experience shear shear deformation apart from the stretching deformation).
A key option of the Imported Composite Property dialog box is the symmetry. If symmetry is selected, the applied loads are going to be executed in the middle plane of the carbon sheet. If the symmetrical tick box is not selected, the loads are going to be applied as if all the material was above or below the action plane, producing a curvature on the sheet. All these results are with the displacement function after computing the case.
Symmetrical case
Non-symmetrical case
After that, we can take a look at the different stress tensor components. The interesting visualizations are obtained by selecting the global and local axis systems or rosettes and then selecting the component.
[Static case solution>(Generate Image)>Stress full tensor component (nodal values)>Ok and click again>More>Axis System ...>Type: User>Name: select axis system>Ok>Component:C11/C22/C12]
If the global axis system is selected, the C11, C22 and C12 correspond to the global X, Y and XY components.
X stress component
Y stress component
XY stress component
If the local axis system (aligned with the fiber) is selected, the C11, C22 and C12 correspond to the component on the fiber direction, the component on the transverse fiber direction and the shear component, which, all in all, are the principal orthotropy axis.
Fiber direction stress component
Transverse fiber direction stress component
Shear stress component
All three failure criteria is met since all the values are below 1.
Tsai-Wu criterion
Hoffman criterion
Maximum failure criterion
Two ply composite
Now, the same case is repeated but with two plies instead of one. However, the thickness of the whole laminate will remain the same, since each ply is half the thickness. This exercise will demonstrate that the same result is obtained with a certain thickness no matter the number of plies if all of them are exactly the same material and orientation.
This time, the laminate is symmetric, therefore, the composite terminology matrix B is null, in-plane deformation and bending are decoupled and there should not be any out-of-plane deformation.
Exactly the same deformation is obtained, however, the stress representation is not showing up as in the previous exercise when the global or user reference is selected. We might pass on that detail. The results are indeed the same as with the thicker single ply laminate.
In conclusion, if several plies of the exact same fiber lamina are going to be used, a single ply with the total thickness is going to give the same result and it will be less complex to configure.
Laminate with drop-off section
In this exercise, three zones are defined. There is zone A, with 4 plies, zone B with two plies and zone AB, also called "drop-off" zone, being a discrete transition from A to B. This is this way to avoid severe discontinuities that could wildly affect the performance of the part (quite prone to delamination).
Once the whole laminate is done, all plies are checked with the Ply Explorer function. Here we have the plies correctly oriented (90º in blue and 0º in grey).
After taht, the meshing process is done, always making sure the different zones nodes match each other.
There is something to take into account when performing the structural analysis. This exercise is just one half of the model. This approach has been done due to the symmetry of the problem. There is no need to compute the whole part if this is done correctly, therefore, reducing computational costs. The key is to constrain the part appropriately, in this case, clamping the back edge and making restrains to the left edge (mid line of the part), so that it doesn't have translation on hte Y axis and no rotations on the X and Z axis (the mid line must be on the XZ plane and the immediate surface to that plane must maintain the same angle on both sides).
The analysis is computed and the expected results are achieved.
Drive shaft
This tutorial covers the definition of a cylinder by using two U shapes, top and bottom, which tends to present less incongruity along the mesh cells. The approach is to fix one end of the drive shaft and twist the other end by applying the specified torque.
For the definition of the shape in the Generative Shape Design environment, two sketches are perfomed with two semicircles and then they are extruded.
Each of them are joined and their limits are defined with the Boundary Definition function.
Once the drive shaft shape is done, instead of applying the torque on the edge of the cylinder and clamping the opposite edge (which could be done), we are going to do it by using a rigid virtual part.
We are going to create a point and with the Rigid Virtual Part function, we are going to fix the whole edge to that point. Therefore, The clamping and torque action can be applied directly to those points.
After that, the mesh is defined (one side at a time). Once done, we take a closer look around the cylinder surface to check if all the cell directions. That is achieved by computing only the mesh, then generating an image from Properties (tree) and finally selecting Composite angle symbol. There are no discontinuities.
After that, we can compute the whole analysis and see the results. The deformation is as expected and the Von Mises stress match with those from the original reference. Notice that the outer layer has a tad higher values of stress, this is due to the cured thickness of the material is set to be 2.54 mm (0.1 in) which is oddly thick. The thinner the plies, the less diference in stress will be between them.
