joints issue in dymanic simulation

[Anonymous]

I do not have much knowledge about joints in simuation. I am still learning.
Could someone please help me in this assembly?

I want to test a RHS which is coming down the slope and friction between the contact is 0.25. I can do the hand calculation but i want to see this happening in inventor

Question___why the block down the slope keep moving in a straight line in the assembly although I have specified the gravity? It should fall down as soon as it leaves the slope contact.

where am I wrong, is this joint or constrain or anything else that I have not specified?

Could someone please help me out?

[henderh]

Hi skumar26,

Please see the attached assembly.

1) The planar joint in the original model will restrict the block from dropping off the inclined plane.

2) The coefficient of static friction at 0.25 seems too high. The threshold for movement seems to be between 0.17 and 0.18.

3) Set up the Planar joint perpendicular to the inclined plane. Then use a 2D contact using the inclined plane and block's side profiles.

4) Set the friction coefficient in the 2D Contact Joint.

Hope this helps!

Thanks and Best regards, -Hugh (Autodesk)

[Anonymous]

Thanks Huge,

I can not understand one thing, why have you set friction in the 2D contact joint? When the RHS is sliding down, the friction should be between rubbing face that is fixed plane and RHS bottom face. It means there will be a joint to fulfill that requirement

No doubt the assembly is working, but this friction concept I could not understand.

Look at the other asseembly I have attached .In this assembly why should we not give Rolling Cylinder on plane Joint. when I try to do that it shows me some error. With Planar joint same like other assembly is working.

thanks
s.kumar

[henderh]

Hi S. Kumar,

There is a part missing from your second assembly (AS 1163-C250L0 - 168.3x4.8 CHS - 50.ipt) but I'll try to answer your questions without it.

To step back, DS behavior is a major paradigm shift in comparison to assembly modeling using Constraints. Most notably, in assembly-land we create constraints to lock-down some or all six DOFs. Conversely, in DS, all DOFs are locked down and Standard Joints are used to open up DOFs. This is because the more open DOF's there are, the larger the matrix is to solve. Also, if gravity was enabled and all DOF's were open, all components would "fall to infinity".

There are four major types of Joints in DS (open the Joints Table to view them in categories):

1) Standard joints: These are used to define the mechanism's DOFs

2) Rolling and Sliding Joints: These are to define motion relationships between the components and their DOFs

3) Contact Joints: 2D Contacts are "precise" meaning the Inventor geometry is exactly what is used in the solver. 3D Contacts are "imprecise" since we take advantage of facetizing the geometry. 2D contacts are preferred over 3D when possible, especially if the contacts nature is 2D. Also, 3D takes much more computation compared to 2D.

4) Force Joints: Spring, Damper and Jack (or a combination of these three) can define an actuator, spring and damper, etc type of forces

First, one must define the Standard Joints so that the mechanism DOFs are well defined, and the other (non-Standard) Joint types can be defined. This is needed so that (for instance) a 2D contact joint has only "planar" DOFs open, not all six. For Rolling Cylinder on Cylinder, the cylinder's axes must be parallel, such as in "real-life" etc. This is so that equations can be solved, where extraneous or conflicting DOFs can be "locked out" in the DOF matrix.

We use geometry to define the Standard Joints. After that, DS uses the mass properties and the joint relationships to calculate the physics. (e.g. DS doesn't know how large the top of the table is in a Planar joint. It treats both faces as infinite, much like a workplane). For a 2D contact joint, if we use the profile (side) of the table, we know where the table ends.

"I can not understand one thing, why have you set friction in the 2D contact joint? When the RHS is sliding down, the friction should be between rubbing face that is fixed plane and RHS bottom face. It means there will be a joint to fulfill that requirement"

There are two types of friction. One you can define in the Standard, or 3D Joint Properties, and one you can define in the 2D Contact Joint. They are both Coulombic friction, but only 2D is real-non linear and the one defined in Joints and 3D Contacts are regularized (see attached image of the help). Therefore, using a 2D contact we can compute the real tangential force at zero velocity.

The main reason I used a 2D contact was that your requirement is that the block fall off the inclined plane. If using a planar joint on the mating faces, this will not be possible, since DS treats the plane as infinite (this is consistent with Assembly mate constraint). Alternately, we could have used a spatial joint and 3D contact (as in real-life) but this would be much more computation intensive to solve and be less precise than using a 2D contact. We always try to use a 2D over a 3D if possible for this reason.

"Look at the other assembly I have attached .In this assembly why should we not give Rolling Cylinder on plane Joint. when I try to do that it shows me some error. With Planar joint same like other assembly is working."

I'm assuming that you didn't create the planar joint for the cylinder before attempting to create the Rolling Cylinder on Plane. If this is the case, you'll see an error message like "Impossible to assemble the mechanism. Please check joints' nature and geometry" This is because there are too many open DOF's between the components. For a Rolling Cylinder on plane, there can only be rotation about the cylinder's axis, and translation tangential to the plane, or along an axis. This DOFs need to be created in order for the rolling joint to be created.

In addition, for Rolling Cylinder on Plane there is an option to define either a rolling constraint, or a rolling constraint with tangency (2 constraints). In the first case, the cylinder needs to maintain tangency via another Joint. In the second, there is tangency "built in" to the Rolling joint. If you did create the planar joint, you much also check the tangency option in the creation of the Rolling Cylinder on Plane joint If you don't, you'll see a different error message such as "...distance between plane and rotation center is not equal with the radius".

Please see the attached Simulation to see how a Rolling Cylinder on plane can be used instead of 2D Contact. But then again, since we are defining the translation on a plane, the cylinder won't fall down below planar face because DS treats the plane as having infinite bounds.

I hope this cleared things up a bit instead of causing more confusion. Please let us know if you have any additional questions, comments or suggestions.

Thanks! -Hugh (Autodesk)

[Anonymous]

Huuuuuuuuuuuuuuuuuuuuuuu..................................

Now i Got it.

thanks Huge.

[henderh]

Hi Sushil,

Excellent! I'm glad we could be of service.

Cheers, -Hugh (Autodesk)

[Anonymous]

Can we expect such type of elaborated information in the help of inventor. I have seen that Inventor help are very basic. It should be a bit elaborated especially for FEA and DS. It is very limited info there.

1. Like what is remote force?_________only definition,supposed to be a bit elaborated with picture.
difference between remore force and normal force

2. body load?_____________________"only definition,supposed to be a bit elaborated with picture.

Other as well.

Once agian thanks for the help you provided.

[henderh]

I can agree with you Sushil.

I'm planning on beefing up the help wiki with these kinds of elaborations. There is a new Wiki available for customers (as well as Autodesk Employees) that can contribute or edit content. I just need to find the time to add the information. I greatly appreciate you asking these questions, since this will help me to find good starting points of what needs more in-depth explaination. I'd really like to create examples of how to use our simulation tools in video format, as well as providing the datasets for the user's own practice and to follow along.

For a brief overview of Remote Force:

What is the difference between Remote Force and "conventional" Force? It is the same as force, except the application of the force can be off the entity picked for the application of the load (face, edge or vertex) or it can be on the entity, but away from the centroid of the entity. e.g. if a centriod of the face is at (0,0,0) you can define the location of the remote force at say (0,10,0). We evenly distrubute the force on the entity, if multiple entities are picked, we divide the force by the number of entities and apply the portional load accordingly.

How is Remote Force different than "conventional" Force? Because the remote force location can be defined a distance away from the centroid, Subsequently, there is an implicit moment load associated with the force (since Moment = Force * perpendicular distance) if located away from the centroid of the entity picked for the application of the Force load. In vector terms is (Moment = distance_vector by the cross_product of the Force_vector).

Why would you want to use a remote force? In some situations, it may be easier to apply only the remote force, instead of creating the CAD geometry. For example, say that you have a weight hanging off a crane boom. You want to neglect modeling the boom initially, but you still want the effect of the weight, a distance away from the turret. You would apply the remote force to the turret faces, but make the Remote location where the weight hangs from the boom. Then, you could also create a remote force to represent the weight of the boom with it's location at the centriod of the boom.

Brief overview of Body Loads

A Body load is used when a dynamic load is applied to the entire body (geometry). Wait a minute, how can Stress Analysis (which is static) compute this type of dynamic loading? The reason is that since body loads can be defined as steady-state, it is possible to do in SA.

A specialized type of Body Load is Gravity. It is simply a linear acceleration of the mass with Earth's gravitational constant of -32.2 ft/s^2 or -9.81 m/s^2 (of course).

The other types of body loads are Rotational Velocity and Rotational Acceleration. Say that you have a centrifuge (like a test tube in a laboratory) or a mass at the end of a cable, rotating with a constant angular velocity. We can find the Centripetal Force in the cable (or plate) using Rotational velocity and the Reaction force of the constraint. The same would apply for a linearly increasing Rotational Velocity, in this case we'd use Rotational Acceleration.

The Angular Velocity and Acceleration also allows the use of a remote location, as we do for Remote Force. This is used to define the point of rotation. For instance, with the mass rotating via the cable example, you would want to define the point of rotation of the mass / cable, not at the center of the mass.

As always, it is very helpful that you're bringing up these examples of what we need to do to increase our communication (via documentation) of how our Simulation Application tools can be used, and what needs more explaination in the Online Help, Tutorials, White Papers, AOTC, etc. If you have any more suggestions, please let us know!

Thanks and Best regards, -Hugh (Autodesk)

[Edit: Changed implied to implicit]

[Anonymous]

Hello Hugh.

As you know i am learning DS and last time you had elaborated some of answers and those answers really helped me to clear out doubts i had in my minds.

This time i want to know" what is this "enable joint force" values. how do we choose value for joint force? It may be a foolish question to ask.

Coefficient of friction is easy to put as we can check values.
spring parameters are easy to put as we can find values

But what about these damping and joint force? The fact is that I do not know what is joint force?

For example lets see there are two plane surface one over another.one surface is fixed and another is sliding. I insert a planar joint between them. now if we enable joint force, we can friction value but joint force how will iget this?

[henderh]

Hi Sushil,

I will try to answer the best I can.

This time i want to know" what is this "enable joint force" values. how do we choose value for joint force? It may be a foolish question to ask.

Not a foolish question at all. There are basically two types of Forces / Torque in DS. Ones that you can apply externally such as force or torque on say a crank handle or lever. This would be similar to an operator running a machine. This will take into account the moment (lever) arm of an externally applied force to a crank handle or lever (for example).

The second type (that you're referring to) is one in which we can apply a force or torque in a translational or rotational DOF, respectively. The joint force or torque is applied internally in the joint itself. An example would be a torque in an electric motor, or a force in a linear actuator. We can also add stiffness, damping and friction as well. It is simulating that there is actually a spring and damper between the two components, which will try to return them to the initial position(s)

Coefficient of friction is easy to put as we can check values.
spring parameters are easy to put as we can find values

But what about these damping and joint force? The fact is that I do not know what is joint force?

To obtain the damping value, you would want to match it to spring stiffness and forces the joint will see. We provide an example in the online help with the theoretical cases for validation (Displacement, the Mass-Spring case). The wiki link: http://wikihelp.autodesk.com/Product_Help/Autodesk_Inventor/Autodesk_Inventor_2011/116DynamicSimulation/3061DynamicSimulationHe/3072Theoreticalcasesfor

Basically, you may have to play around with the damping value. Too much damping will restrict the mechanism's movement, and too little will cause oscillations that could continue too long, or the positions will change too quickly. Also see http://en.wikipedia.org/wiki/Damping_ratio

For example lets see there are two plane surface one over another.one surface is fixed and another is sliding. I insert a planar joint between them. now if we enable joint force, we can friction value but joint force how will iget this?

We can add friction without adding a joint force. The frictional force will depend on the normal force (weight) of the mobile body.

Cheers, -Hugh

[Anonymous]

Huuuuuuuuuuuuuuuuuuuu......................good one.
Nicely elaborated.............good anough to clear out my doubts.

Thanks Huge