Understanding the Need to Fully Constrain the Model Using Ansys Mechanical — Lesson 2 - バイリンガル字幕

Hello everyone.
When performing a static analysis, we need to always keep in mind the need to fully constrain the model.
That does not imply that we must over constrain the model, but rather we must have optimal boundary conditions.
to complete analysis successfully as well as accurately.
There are two common mistakes that people make while performing a static analysis.
The first one is encountering the DOF limit exceeded and similar errors that arise while performing.
while the other is over-constraining the model while performing free thermal expansion or inertial relief resulting in unrealistic stresses.
In this video, we will discuss the importance of properly constraining the model and the best strategies that users may follow to achieve so.
So let's get started.
Let's start with the explanation of what is the static
The key assumption in static analysis are that the loads and boundary conditions do not change over time,
and that the bodies do not have rigid body motion.
The analysis is solved using the final element equation f equals ku, where k is the stiffness matrix, u is the nodal displacement vector.
Consider two-bar element with three nodes i, j, and k.
Each node is limited to only one degree of freedom.
These are the finite element equations for the current model where ui,
uj, and uk denote the nodal displacements, and k is the stiffness of the bar.
For the first case, let us assume there are no boundary conditions to find.
mind.
If we solve the equations,
we obtain ui equals uj equals uk, and we can see there are an infinite number of solutions that satisfy this condition.
This case represents a model with not enough boundary conditions to restrict rigid bodies.
Rigid body motion is the situation where the displacement solution can be any non-zero value representing unconstrained movement.
In the second case, let us add a boundary condition ui equals 0 and a force F acting at node k.
When we solve the equation, we get ui equals 0, uj equals uk over 2 and uk equals 2f over k.
We also have a reaction force of F resulting at node I as a of the restricted motion.
This reaction force is minus K times UJ and it's equal and opposite to our applied force F at node K.
This is the case where the model, namely the bar elements, are adequately constrained resulting in a unique solution for the null displacements.
Additionally, to avoid rigid body motion, we must also ensure that all directions are properly
constrained, even if no forces or loads are applied in that direction.
For example, if we imagine a block sitting on another block, only vertical forces will be at work.
Even though no lateral forces are present,
we still must prevent rigid body motion in the lateral directions, as well in order to obtain a unique solution.
This requirement is specific to static analyses and does not appear in dynamic analyses.
If we do not properly prevent rigid body motion, we will receive a solver error such as DWEF limit.
We now understand how critical it is to appropriately constrain a model in order to prevent all six rigid body motions,
three translational and three rotational.
However, what should we do if we have a system that's in air or underwater and thus not attached to ground?
With the help of inertial relief, the body motions of such systems can be balanced by inertial forces with minimal constraints.
It's a technique in which the applied forces and torques are precisely balanced by inertial forces induced by an acceleration field computed by the solver.
If we have inertial relief, we will still want to define properly our constraints to motion, even in directions that are not excited.
Recall that even if the model is not excited in a given direction,
the possibility of a rigid body motion is there, hence the matrix is still singular.
The inertial relief method will be discussed later in the workshop for Now,
if all else fails, there is one other option, the weak springs, that can be activated in mechanical.
These are automatically created springs, with just enough stiffness to constrain the rigid body motion.
The weak springs feature will create an imaginary bounding box for every part,
where springs at each of the eight corners are added for each part.
So, if we imagine a box covering each part, each node nearest to these eight corners gets
weak springs attached in the three orthogonal directions, x, y, and z.
The free ends of the springs are constrained to ground,
and the spring stiffness is automatically calculated defined by a to be of very low value.
We should always check the weak spring reaction forces to verify they're low and not doing any work or influencing the results.
As a note, weak springs are available in nearly all analyses, while inertial relief is only available for static and buckling analyses.
Now, let's get a better understanding of these concepts with the help of a walkthrough example in Ansis-McCanada.
For the walkthrough, we will use a simple foam glider model, which taking forward flight.
Our geometry consists of the complete model with the material expanded polystyrene foam assigned to every part.
The glider is subjected to a simplified pressure load to represent
One could use computational fluid dynamics to compute the pressure loading and map those on our model,
but since the focus of this video is constrained in the model,
we will use a simplified pressure loading just to impart some lift onto the foam glider.
A point mass is used to account for the portions of the foam glider that are not modeled with
the geometry and is precisely located in the cavity of the glider such that our glider has lift and weight balance both translational and rotational.
This implies that our glider will not rise up or down nor pitch a roll with these balance forces and accelerations.
Therefore, our goal is to achieve near 1G translation acceleration, which is essentially our 1G gravitational acceleration necessary to balance.
and near-zero rotation acceleration, as reported by the inertial relief summary.
We will use the 3-2-1 rule to provide minimal yet sufficient constraints to prevent rigid body motion.
The results of stress and deformation will be analyzed,
and the force reaction values will be verified to confirm that our model is not over-constrained.
Without further ado, let's get into the workshop.
Go to the project page and pick file open.
Select the workbench project archive file foamglider.ubpz.
Save the project to the desired location.
We will keep the file name as specified.
Double-click on the model cell to open mechanical.
The model has been partially set up.
So Expand the geometry foam glider to see the parts that make up the glider.
By clicking on each part, we can observe that the material assignment is foam expanded polystyrene.
Right click on Fuselash Canopy and pick Hide Body option to hide the selected body.
Click on Point Mask to reveal the detail.
We're using a 0.02 kilogram point mass to account for the portions of the foam glider
that are not modeled with the geometry and to balance the glider weight distribution.
Rotate and zoom model is needed to view the location of the point mass and the scope faces, which is highlighted in red.
The point mass is located such that our glider has lift and weight balance.
To again view the complete geometry, click the right mouse button and pick Show All Bodies option.
Now, click on Geometry and expand the Properties tab.
We can see the total mass of the model is approximately 0.021 kilograms.
The next step is to mesh the model.
Our model has shared topology so all the parts are connected via conformal mesh.
Click Mesh and change the element size to 2mm.
Now right-click on Mesh and pick Generate Mesh.
Let's now apply the Loads and Boundaries.
We're considering a situation where the foam glider is taking forward flight.
To represent the lift, click on static structural.
Pick all four bottom surfaces of the wings and horizontal stabilizers.
Now right-click on Static Structural, Insert and Pick Pressure.
Change the magnitude to 30 Pascal.
Click on Analysis Settings.
Turn on inertia relief.
Let's first understand the significance of inertia relief.
Inertia relief applies an acceleration field to all the elements in our model, in a way to balance out the applied loads.
It works to cancel out all rigid body accelerations, leaving a state of static equilibrium on it.
In our case,
when the foam glider is taking forward flight,
it has a net lift force acting on it and an equal and opposite force due to its mass times acceleration.
This equal and opposite force is the computed inertial acceleration loading on the glider.
Now to properly constrain the model, we will use the 3-2-1 rule.
Use of the 3.2.1 rule can help us provide a minimal,
yet sufficient constraints to prevent six rigid body modes while preventing over-constraint.
It means you apply three transitional degree of freedom constraints to one.
node.
Then two transitional degree of freedom constraints to another node that is aligned with one of the three axes of the nodal coordinate system of the nodes.
By the way, the nodal coordinate system is by default aligned with the global Cartesian system.
The degrees of freedoms that should be constrained at this second node should be orthogonal to the line between the first two nodes.
So in other words, do not constrain the degree of freedom in the direction of the two.
This now prevents rigid body rotation about two axes, yet still leaves the body free to expand.
Finally, pick a node that is not collinear with the first two nodes and constrain it in a degree of freedom direction
that is most aligned with the normal that is formed by the three nodes.
This will constrain the third rigid body rotation.
but still allow the body to freely expand without resistance, such as to accommodate unconstrained growth of the body, for example, free thermal expansion.
Let's do that now.
Zoom in on the tail region.
Select top vertex on the horizontal stabilizer as shown.
Right-click on Static Structural, Insert, Pick Displacement.
Specify zero value for all three translations in X, Y, and components.
This constraint will hold the glider from translation, but it can still rotate about the three axes.
Repeat the same procedure for the other horizontal stabilizer.
Specify zero value for X and Z components.
These constraints will hold the glider for rotations about the Z and X axes respectively.
Finally, select a vertex on the tail of the glider and repeat the same procedure.
Specify zero value for the Z component.
This constraint will hold the glider from rotation about the Y axis.
Now the model has been fully constrained, it's ready to be solved.
Right-click on Static Structural and Solve.
As is now solved, right-click on Solution, Insert Deformation, and Total Deformation.
Right-click on Solution again and pick Evaluate All Results.
Click on the Z-axis of the triad to analyze the deformation results from the front of the glider.
Also, verify the animations created to get a better understanding of the
We can observe that the maximum deformation value is 9.96 mm and has occurred on the extreme parts of the glider wings indicating that the wings are bending upward due to the lift.
Now let's check the stresses generated in the glider.
Right click on solution insert stress and pick maximum principal stress.
us.
Right click on solution again and pick evaluate all results.
Click on the x-axis of the triad to analyze the stress results from the bottom of the glider.
We can observe that the maximum principal stress value is 0.0.
which is tensile and has occurred on the bottom of the wings at the few slash junction.
Next go to solution information, click on the text output, type control F for find and type in inertia relief summary.
Scroll down to view the summary.
We can observe that the inertia release summary shows near zero values for translational and rotational accelerations required to balance the glider,
except for the minus 9.86 meters squared per second acceleration value,
which is close to 1G, and it's needed to balance the glider in the vertical direction.
Since we did not apply any gravity during the simulation, the inertial relief has computed the precise value to balance the applied lift pressure force.
Now let's check the force reactions at the three displacement boundary conditions.
Click on static Drag and drop the three displacement boundary conditions onto solution.
Right click on solution again and pick evaluate all results.
Click each force reaction to view the details.
We can observe that the reaction force values at the three displacement boundary conditions is near zero.
confirming proper specification to fully but not over constrain the model.
I hope this simulation example has helped you understand the importance of properly constraining the model
and has also provided you with the best strategies to achieve the same.
With that said, the workshop demo is completed.
Let's discuss some important points.
Performing a successful analysis requires us to properly constrain the model.
If our model is under constrained, we'll have rigid body motion, while an over constrained model may behave too stiff and may develop unrealistic stresses.
Users performing free thermal expansion or inertial relief need to define constraints properly to prevent rigid body motion,
even in the directions that are not excited.
The use of the 3-2-1 rule or a weak springs options can be utilized.
Use of symmetry regions when applicable naturally prevent rigid body motion in directions normal to symmetry and rotation through the plane of symmetry.
With the understanding of these key concepts, we have higher confidence that our simulation model has been properly constrained.
I hope you found this video informative, thank you for watching and do check out our other courses to discover more useful learning resources.

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