Hello everyone and welcome to this video!
In finite element analysis, boundary conditions are a critical part that can ‘make or break’ our results!
What makes boundary conditions so special?
We know that there are different sources of variability in our analysis - for example, we have geometry, material property, loads, etc.
Let’s say that, in a simulation, the load we apply on a system deviates by 5%, then we know that the resulting stress or deformation too will be off by 5% in a linear analysis.
However, for boundary conditions, it’s another story.
If an inappropriate boundary condition is used, the simulation may be far from correct.
In many cases, the choice of boundary conditions can entirely change the nature of the problem we are solving.
Hence, it is important to give proper thought to the choice of the supports for the model.
Given a simulation problem to solve, determining which supports to use can be the very first step in our thinking process.
This video will help you to understand the different support types so that you can determine which support to use for different situations.
Now, let’s get started.
So, let us see what is a Support?
Supports help to truncate the domain
Suppose we want to model a Vehicle’s engine block – we do not want to have the roads, tires and other parts attached to the engine to be modeled.
Truncating the model helps in efficiently obtaining numerically accurate results without modeling other parts of the geometry that are not of primary interest.
Thus, supports help to model the parts that are not present in the model but are interacting with it.
For this engine block example, we can model only the engine block that is highlighted, but we also want to include the effect of the rest of the parts through the use of supports, or boundary conditions.
Let us now investigate the different support types available in Ansys Mechanical and their meaning.
The first one is a Fixed Support.
All degrees of freedoms are fixed or constrained for the scoped locations when this support type is used.
For a surface or line body, the rotations will be constrained as well.
But one should be aware of the possibility of overconstraint because the nodes on fixed face cannot move at all.
Let’s look at a simple block with a Fixed Support on one end and a Pressure load on the opposite side.
Since the Fixed Support prevents any motion, we will end up with a bulging of the material at the sides, as shown here, which may or may not be what we want.
Now let us move to the next support type, the displacement support.
A Displacement support allows us to specify a zero or non-zero value to any of three orthogonal directions.
This support type can be used to model very common scenarios like displacement-controlled testing of materials, such as tensile or compression test.
Here, we can see tensile testing of a metallic specimen where one rigid jaw is fixed, holding the specimen and the other rigid jaw holding the specimen is pulling it.
Using supports, we can model this scenario of tensile test without modeling the jaws.
In the simulation, we can apply a fixed support at one end, which models the fixed rigid jaw and a displacement support at other end, which models the rigid jaw pulling the specimen.
Let’s move on to the next support– the Frictionless support.
This support only constraints translational movement in the direction normal to the surface.
For example, let’s say that we have a similar block as shown earlier, this time with a Frictionless Support on one side and a pressure on the opposite face.
Looking at this animation, we see that when the block is compressed, the face grows, which indicates that Frictionless support does not restrict the in-plane motion.
If using fixed support, the shape of the face would have remained the same.
Same phenomenon can be seen when we pull the block; the face where the Frictionless Support will shrink in size since it is free to contract in-plane.
Next, we discuss the cylindrical support.
This support is specific to cylindrical surfaces and constrains axial, radial, and/or tangential directions.
This support can be thought of as a rigid pin fitting snugly inside of a cylindrical surface of our part of interest.
We specify whether or not the nodes of the cylindrical surface can move freely in these three directions.
A simple practical application of this support is for machinery with a rigid shaft.
You can apply a cylindrical support on the hole of the machinery.
If the shaft is free only to rotate, then the tangential motion of the cylindrical support can be set to free and the radial and axial motion can be set to be fixed.
However, we should be careful to only use a Cylindrical Support in small-deflection analyses - never use them in large-deflection analyses!
The reason for this is that the nodes of an element have Cartesian behavior.
Suppose we have a cylinder, and the nodes can move only in X and Y directions as indicated.
It might happen that we constrain only X and not Y and when large deflection happens, Y moves in "tangential" direction, not "circumferential" direction,
so motion will continue to move tangent to the initial cylinder, and we will see that the cylindrical surface grows in diameter, which is not realistic.
Now let us move on to next support – the Compression Only support.
This support can be thought of as “Frictionless Contact” with a rigid object not modeled.
The body is free to slide in tangential direction, free to ‘lift off’ in negative normal direction but cannot move in the positive normal direction.
This support is useful to model a rigid pin in a hole in large-deflection analyses or something ‘resting’ on a stiffer part, for example.
A point to note here is that this is a nonlinear support, so it takes extra computations.
After the Compression Only support, let us talk about the Elastic Support. .
This support is useful to model the parts that are not rigid, but you can define a ‘stiffness’ in the direction normal to the surface, such as a structure connected to a ‘ground’ or ‘foundation".
When this support is applied to a surface then the surface is free to slide in the tangential direction.
A practical example where this support can be used is a building’s foundation.
The foundation of the building rests on soil, and the 'stiffness' of the soil may be represented by an Elastic Support, although the tangential stiffness is neglected.
We don't want to model all of the soil, but its stiffness may be important since it may not be rigid compared with the building materials.
Next let us discuss the Remote Displacement support.
Consider a rigid object bonded to the part’s surface, but its motion can be ‘controlled’ at a specific location.
Using the Remote Displacement support, we can transfer this displacement or rotation from rigid body to our part of interest without modeling the rigid body.
This is useful, for example, if you want to apply a rotation from a location in space to a surface.
For example, here we have a part connected to the hinge and the part can rotate about that hinge.
However, the hinge is not part of the model.
In this case, we can apply a remote displacement to represent the effect of the hinge part.
The center of the remote displacement should correspond to the center of the hinge, and we can then apply rotation to rotate the part without modeling the hinge.
Now we know that supports help to truncate the domain, but where we are truncating the domain is also important.
We should always think about the meaning of the support we want to apply and consider if the situation is appropriate.
If the boundary condition is not rigid (non-zero displacements prevented at that location), think about whether the ‘rigid’ condition will lead to conservative or unconservative results.
If needed, you may want to model the additional flexible parts and connect the parts via contact.
St. Venant principle says that difference between the effect of two statistically equivalent loads become smaller at larger distance.
The same can be said for boundary conditions.
After truncating our domain by applying the relevant supports, we solve the model.
We can drag and drop the support boundary conditions on the Solution Branch to obtain the reaction force at that support.
This will help us to check the balance between the applied and reaction forces.
But we should be careful while evaluating the support’s reaction forces.
When one support shares vertices or edges with another support, then the sum of the reaction forces may not balance
This can be explained by the fact that the shared edge only has one set of nodes, even though it is common to two or more Supports.
In such cases, the reported reaction force on nodes of the common edge will appear in reaction forces for both Supports.
Thus, if we need to obtain accurate reaction forces for each Support, we should ensure that Supports do not share edges or vertices with other items.
Now let’s use a simple demo to illustrate the effect of supports in a simulation.
Here, we have a cylindrical geometry sitting on a recessed base.
Sliding is allowed between the surfaces of the cylindrical part and the base part.
You can see that the cylinder can rotate along the axis and can be lifted from the base.
To simulate such a situation, we can use frictionless contact between the two parts.s
On the top surface of the cylinder, a force is applied to a section of the area in the negative y-direction.
For materials in this model, the base is made of steel and the cylinder is made of aluminum.
Because of this, we know that the base is much stiffer than the cylinder.
Now in this model, we will use an appropriate boundary condition to replace the base,
so that we don’t need to worry about frictionless contact,
and doing so simplifies the calculations.
First of all, we notice that the behavior of the model is expected to be symmetric about this plane.
So, we just need to have half of the geometry.
And also, the base can be removed, as we will use a boundary condition to replace it.
Now, let’s go to Ansys Workbench.
We can drag a static structural analysis from Toolbox.
The geometry file is prepared, so we can just right mouse click on geometry, import geometry,
then select the geometry file for this problem.
Now, we can just double click on the model to open Ansys Mechanical.
First of all, let’s check the unit system, here we are using millimeter and kilogram unit system.
For the material, we assign aluminum alloy for the part.
Next, let’s define the symmetry region for this half cylinder.
We can right-mouse-click on Model from tree outline and insert “Symmetry”.
Then, right-mouse-click on Symmetry, insert “Symmetry Region”.
Here, we select all exposed surfaces on the symmetry plane and also make sure that the symmetry normal is the x axis.
For the mesh, we apply a sizing control on the part and set the size to be 10 millimeters.
Generate the mesh to visualize the elements.
Now we can apply the load to the cylinder.
Let’s right-mouse-click on Static Structural on the Tree outline, then insert a Force.
Then scope the small area on the top surface of the cylinder.
We can switch the mesh view on and off by clicking on “Show mesh”.
The force is defined by “component”, and the value is 50,000 Newton in the negative y-direction.
Now, about the boundary conditions, since sliding is allowed between the base and the cylinder,
it definitely shouldn’t be a fixed boundary condition used here.
As we mentioned before, if the base is modeled, frictionless contact should be defined between the two parts.
So, off the top of our head, frictionless support could be a good option here as the boundary condition.
Let’s have a try.
We can right-mouse-click on Static Structural on the Tree outline and insert a “frictionless support”.
We can see from the geometry that the outer surface of the half-cylinder is already sliced to have an area corresponding to the area in contact with the base.
So, for the frictionless support, we should scope both this area and the bottom surface.
The model is ready to solve. Since everything is linear, this model solves really fast.
Let’s insert total deformation and equivalent stress results under Solution and evaluate the results.
Here, we can go to the Result Tab and go to Edges, turn on “show undeformed wireframe” to have a better idea of how the model deforms.
Note that the deformation is exaggerated to visualize the deformation easily.
If we change the scale to be 1, we can see that the real deformation is fairly small.
So the question is, does this result reflect reality?
Intuitively, if we have an eccentric load on a cylinder, we expect that the opposite side would lift.
But if we probe results on the bottom surface, we can see that the results are all almost 0.
Also, since the cylinder is allowed to lift up, the stress on the bottom surface can be compressive or zero, but it definitely cannot be tensile, because nothing should be preventing the bottom of the cylindrical part from lifting off.
To check this, we can insert a normal stress result over the bottom surface of the cylinder.
We can right-mouse-click on Solution, insert Stress, choose Normal stress. Then, scope the bottom surface and set the direction to be y.
From the color scale, we can see the range of normal stress in y-direction goes from negative values to positive values.
In this case, a negative value indicates compression in the y-direction, and a positive value indicates tensile in the y-direction.
If we probe values on the surface, we can see that near the loading area, the stress is compressive and gradually it becomes tensile if we leave the area directly under force.
Apparently, this does not reflect reality, as the material on the bottom should not be in tension for this problem.
We should recall that although frictionless support allows the surface to slide freely, it does not allow movement in the normal direction, including separation between the part and the ground.
Therefore, frictionless support is not the right boundary condition to use for this problem.
This makes us realize that we should have a boundary condition that allows not only for free sliding but also separation.
Luckily, we do have such an option, which is the compression-only support.
As indicated by the name, only compression is constrained by this support, with no resistance for separation and sliding.
Now let’s go back to the workbench schematic, copy the model we just worked on, and open the copied model in Mechanical.
Let’s suppress the frictionless support first.
Then, we can right-mouse-click on Static Structural on the tree outline and insert a compression-only support.
We scope this to the sliced bottom area on the cylinder surface and the bottom surface.
Solve the problem again.
This time, you will notice that the model takes more time to solve.
And if you click on Solution Information, and set the Solution output to be Force Convergence, you will notice that the solver is trying to solve the problem iteratively.
Why is this?
This is because compression only support is a nonlinear support since we don’t know beforehand what portion of the supported surface may separate, or lift off.
For nonlinear analyses, an iterative solution is needed.
Now the problem is solved successfully.
Let’s have a look of the results.
First of all, for deformation, you can see that, at the bottom, the area far from the force actually has some deformation value
If we scale the deformation, now we can see clearly that the cylinder is lifting up on the opposite side, which is expected.
Also, for the normal stress in the y-direction on the bottom surface, if we probe the results for the area far from the force, the values are almost 0, indicating that for these areas, the support is not providing any tensile or compressive force.
Here’s a side-by-side comparison of total deformation and equivalent stress from the two models with different boundary conditions.
Same legend scales are used.
It’s quite obvious that the results are different.
In conclusion, we can see that for this problem, frictionless support is not the appropriate boundary condition to use and leads to incorrect simulation, while compression support provides a more realistic simulation of the boundary condition.
Let’s summarize the key takeaways from this video.
Supports are used to represent components that are not a part of the numerical solution but are interacting with the mesh,
of the numerical solution but are interacting with the mesh,
Using the appropriate support is essential for two reasons
First, choosing the right support helps assure that
the simulation model will properly represent the boundary condition.
using wrong boundary conditions can entirely change the nature of a problem,
thereby significantly skewing the result.
Secondly, selecting the proper support can help to truncate the model.
Truncating the model can represent the right approximations of
of interacting bodies and the working environment without actually
modeling them. This can reduce computational effort
while still obtaining numerically accurate results. Also
the reaction forces are summed when reporting the total reaction force for
each Support. However, if there is a common edge shared between
two Supports, be careful, as the reaction forces of the nodes
the common edge will appear in the total reaction force for each support.
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