Finite Element Analysis is widespread nowadays with all types of users having a go at a massive range of problems. Experience levels vary from very little all the way up to long-in-the-tooth users such as myself and Laurence Marks. The perception nowadays is that all the “issues” have been ironed out and that we can plug in our best data, press the button, and out comes the right result. Yes, many of the intricacies of FEA have been ironed out but there is one area, that of constraint, that still causes great concern and difficulty. Note that the so-called CAD-embedded systems (which in some cases can ONLY fully-constrain a face) “encourage” the much-simplified method of reducing all structures to that of a cantilever beam – load it this end, hold it at this end and run it. You can’t fully-constrain a face in practice so I say that the remaining “artistry” of FEA is designing a realistic constraint which enable sensible stresses comparable with reality.
The starting point for me then is to determine whether the component (or assembly) that you’re considering can be classified as a “static” problem or not. In other words, the action of the applied loads is balanced by a set of reactive loads such that the structure deforms but doesn’t fly off into space. If your structure does fly off then refer to Laurence and the other clever types at SSA and they will help you solve your dynamic problem (‘cause I can’t help you). The vast majority of structures are static problems as long as we can convince ourselves that traveling round this lovely planet at a fixed angular velocity doesn’t invalidate the idea. OK there’s some complexity in that some structures are said to be “determinate” (mostly simple structures where we can deduce the load path) and some are said to be “indeterminate” (where the relative stiffness of the structure governs where the loads go) but for now we will proceed as follows: –
If you have a component with no symmetry then must build a full model and you should refer to the first (leftmost) column of Figure 1. You have no symmetry and without constraint the part is free to translate in the X, Y and Z and free to rotate about the X, Y and Z axes. There are therefore 6 (*) rigid motions which you must suppress. Failing to constrain all six will give you zero pivot problems (stress-free movement) and constraining more than six will give you dangerous over-constraint. The desired 3-2-1 constraint pattern will therefore constrain X,Y and Z at the first point; YZ at the second point (where the first point and the second point are parallel to the X-axis) and a single constraint AWAY from the axis (between the first and second points) in order to stop the final rotation. See Figure 2 for an example.
[(*) MAGIC NUMBERS: our mathematical friends see the number 6 as the first “magic number” because it’s factors (1,2 and 3) add up to 6. The second magic number is 28 so check for yourself now].
If you’re lucky enough to have one plane of symmetry then you move to the second column of Figure 1. The effect of one plane of symmetry is to remove the “3” and you are left with the requirement to apply a 2-1 constraint pattern. If you consider the action of one plane of symmetry then this removes one translation and two rotations (three in total). Let’s consider an example where the part is symmetric about the Y-Z plane: – All the nodes on the symmetry plane must be constrained such that there is no movement in the X-direction (the X-direction being normal to the symmetry plane). The action of this group of constraints is to stop translation in the X and rotation about the Y and Z axes. The remaining freedoms (Y and Z translation plus X-rotation) must be suppressed with a 2-1 constraint (say Y and Z translation at one node and Y-translation at a second node). Less than a 2-1 is under-constraint and more is over-constraint. Plane stress and plane strain 2D runs come into this category of constraint except that you are usually forced to make your model in the X-Y plane.
If your luck is running even better and have a model with two planes of symmetry then you should refer to the third column of Figure 1. The first plane of symmetry removes the “3” and the second plane of symmetry removes the “2” leaving you to apply a constraint at a single node. Two nodes at this final point is over-constraint and no constraint means that your quarter model will honour the two symmetry planes but will fly off in a direction parallel to the two planes of symmetry. Consider a quarter model of a pressure vessel – apply two planes of symmetry in the normal way and then constraint one node in order to suppress the final rigid body translation. Axi-symmetric analyses also fall into this category because the elements have hoop stiffness such that they naturally resist movement in the radial direction (X in Abaqus) and the overall model only has a tendency to translate as a free body along the axial (Y) direction.
If you’re super-lucky then you have three planes of symmetry. The first plane removes the “3”, the second plane removes the “2” and the third plane removes the “1”. Adding further support to your model is over-constraint. The main problem with over-constraint is that the supports help to carry the loads and mask otherwise high stresses that occur in practice.
There is therefore a definite level of “artistry” in the application of meaningful and realistic constraints. Under-constraint will allow rigid body movements (and if you are considering a contact analysis then Abaqus will struggle to converge without falsely damping the movement down) and over-constraint will suppress the stresses and you will sign-off on a job which is potentially NOT fit for purpose.
Bob Johnson – Realistic Engineering Analysis Limited