Dynamic Analysis to calculate reaction force

[Anonymous]

Hi everyone, 

I would like to send my concern about making dynamic simulation for a 180 degree opening door.

I got some struggles and cannot solve it. If possible, please take a look and send me some advice.

Now I want to calculate reaction force at O, B and A. 

Cylinder has force 75kN to operate this door.

When opening, there is no external force, when closing, there are an external force 30 kN at the center of gravity of the door

  

In this figure, OC and AC are two short link, B is hinge of the door. CD is cylinder, D is at the door (let see the attachment file )

I did dynamic simulation, but I don't know how to put cylinder force and get reaction force at these point and also how to make sure this door can fully open and close (180 degree)

Thanks in advance, 

Looking forward to hearing from you soon.

[JDMather]

Can you attach *.zip rather than *.rar?

I do not have a RAR extractor, but I do have Windows OS on my Inventor machine.

[Anonymous]

Dear @JDMather,

Please take a look an attached zip file and give me some suggestion

Thanks so much

[Anonymous]

Dear @JDMather,

Is it possible to use Spring/Damper joint to simulate cylinder force?

I still wonder because theoretically, spring force is changed based on sin-function during the time whereas, cylinder force remain constantly its force and seem to be reached its peak immediately 

[JDMather]

Unfortunately it would take me all day to solve this problem.

Perhaps one of the Autodesk employees will take on the challenge.  Kelly? Johnson?

If I were to spend time on this I would prefer to go one of two ways.

1. Simplify the linkages as single line elements and nodes (I like to use a line with circle at each end).

2. Fully modeled parts.

Clearly Option 1 would be the easier path if the parts are not already modeled.

[Anonymous]

Dear @JDMather,

I've simplified this model already by some single line elements and circle nodes at its end. 

Let see and send me any suggestion.

Thanks & Best Regards

[kelly.young]

Hello @Anonymous interesting design. Thanks for the additional assembly, will take a look.

Will also tap our simulation guy and see if he has anything fun to say. 

[JDMather]

---

@Anonymous wrote:

I've simplified this model already by some single line elements and circle nodes at its end. 

---

This is an example that we do in class.

I have reduced the mechanism to single line elements.

The circles are not really needed - but I like to use them for visual reference of where I have Revolution (or similar Joints, like Spherical or Cylindrical).

[Anonymous]

Dear @JDMather,

This is the simplified mechanism assembly, please take a look and tell me whether it meets your point or not

Thanks & Best Regards, 

[Anonymous]

Hello @kelly.young,

Really nice if I have a chance to discuss with you, thanks in advance for your kind support 

Best Regards, 

[kelly.young]

@Anonymous looks like you've been progressing on setting it up.

I see that you used a point-line and cylindrical instead of a revolution with the short link to shortlink on door constraint and cylinder to door_Blade.

Also setup a revolution for piston to short link.

I believe setting them up all with revolution will get you joints on all of the circles, can use the cylindrical where you have the piston, and load your forces to the front of the door.

I think you're on the right track but the constraints need to be re-evaluated.

Please select the Accept as Solution button if a post solves your issue or answers your question.

[KubliJ]

Hi @Anonymous,

I've looked at the model and do not see anything wrong with the setup.

From your original post it looks like you are looking for reactions at O, B, and A.

Here is a video of how to do that.  Remember though that FORCE is a magnitude and values will always be positive and without direction.  You'll have to look at the component forces to calculate the direction yourself.

I have validated the final reaction forces by running a static stress analysis in Simulation Mechanical.

Inventor Dynamic Sim:

FO = 288490 N

FA = 364623 N

FB = 317296 N

Sim Mech:

FO = 288469 N

FB = 317276 N

(I don't get a FA in Sim Mech as there is no boundary condition)

I am not sure I understand your question about applying a force on the cylinder.  Currently you have defined a forced displacement, if you want a applied force instead of an applied displacement, then:

1. Right click on Prismatic 1
2. Select the middle icon 
3. Select Enable Joint Force
4. Define as necessary.
5. Disable the imposed motion

Let me know if this helps.

Thanks,

James

[Anonymous]

Hi @KubliJ,

Very happy when I received your explanation, and knew that my first set up is correct. However, let me ask you that

1. If I use 2D sketch like that, do I can get the same value with using 3D model? Because, if the answer is yes, so I can use 2D sketch for decrease time consuming and can get reasonable results.

2. As you can see, in order to close this door under the external force 30 kN and self-weight, this cylinder's force is more than 244 kN (driving force) -> it means that with 75kN, this cylinder cannot drive the door under this external force --> this statement is correct or not?

and these reaction forces at each point (A, B, D and O) are shown in this picture as well.

However, in fact, this door is operated by cylinder with its force is only 75kN. If I apply 75kN in joint force (and ofcourse disable imposed motion)

so at this case, do I need to set up angle between door blade from 180 degree to 0 degree? If not, this door is moved very fast and it seems uncontrol, so I have to set the operation angle for the door blade

I got the results (I don't set up operation angle of the door blade, just set the value of cylinder force 75kN)

so now, I can get new reaction forces caused by the effect of 75kN cylinder force.

It is a bit confusing, because, at the first step, this cylinder force is at least 244kN to close the door, but now, with 75kN this cylinder still operate this door? How to make this point more clear?

Please take a look and help me to solve this problem

Thanks in advance for your kind help

[Anonymous]

Hi @kelly.young,

As you can see that, the first attached file I use these joints as you mentioned. But I think that, it should be use cylindrical joint between cylinder house and cylinder rod, because there are two dof (translation and rotation), besides, there is a bit tricky when I joined two short link and cylinder rod. If I used point line joint, is it correct or not?

In addition, if I use 3D model, do I have to used different joints in comparision with using 2D sketch? and do I can use 2D sketch when doing analysis dynamic simulation instead of using 3D model?

Thanks for your support

Best Regards, 

[KubliJ]

Hi @Anonymous,

To answer your questions:

1) Yes, 2D can be used but I believe the mass of the system and therefor its inertia are not accounted for.  If the mass is small compared to the load, the direction of motion is perpendicular to gravity and the speed/acceleration of the system is small, then it can be ignored for reasonable results.  If not, then you may want to consider solid bodies.

2) Well, you see, the issue with the applied motion and then the force required is that the load initially is in the direction of rotation.  So the load is assisting the motion and the applied displacement is actually countering it (the results output is a magnitude and not directional).  That is why when you use the force instead it moves very quickly.

At this point it may be easier to understand the intent of the analysis to figure out the best means to setup the model instead of troubleshooting the results.

Some questions I have are:

Which way is up/down/gravity?

What is the mass of the system?

How fast does the door need to close?

Besides its own weight are there any external loads?

I am assuming that the load is the weight of the door, and it's applied direction is in the direction of gravity.  And the door needs to rotate 180 degrees to close over a period of 1 second.  Only the piston is the control for the door, then the results seem good.  Well at least the last step does.  I could look into the initial results, get reaction forces and check them.

Thanks,

James

[Anonymous]

Hi @KubliJ,

I would like to attach this figure to more clear about this door

So  by using this picture, the questions 1, 2 and 4 can be answered

Regarding to the third question, the speed of closing is quite small, the closing time is about 30 - 40 seconds 

My concern is about which way can be used to get feasible result: using imposed motion or add joint force. 

Please take a look and give me any suggestions. 

Thanks & Best Regards, 

[KubliJ]

Hi @Anonymous,

So it looks like gravity is not going to affect the motion of the door.

Another question, what is the source of the load on the door?  Does the load follow the rotation or does it always apply in the same direction?

So if the load follows as the door rotates 180 degrees the direction of the load will also change direction and always stay perpendicular to the door face.

If it always applies to the same direction, then they way it is currently defined is correct.

30-40 seconds seems slow enough to me to ignore the mass and inertia of the door.

The results seems right to me.  A static verification of the initial and final position looks correct.

This is what a Static Stress analysis looks like based on the initial position of the door:

I get an axial stress of 3452 N/mm^2

Multiply that by the cross sectional area of the truss (314 mm^2) and I get an axial force of 1084 kN.

Which matches closely to Inventor Dynamic value of 1091kN:

Results are correct for the setup.

Thanks,

James

[Anonymous]

Hi @KubliJ,

It can assume that this force always apply same direction, however, in fact, this force will against this door blade when the door is nearly closed (example wind pressure will act on the door blade when this door prepares to closed). I don't know how to apply this force like that so I give force at the beginning of the stage.

I still have a bit confusing because in fact the cylinder force is 75kN but based on this result the required driving force is much more than

Thanks for your kind support, it is very useful for me.

Kind Regards, 

[KubliJ · Accepted solution]

Hi @Anonymous,

We have validated through other analysis that the static results are correct.  At this point I would say that the design or the requirements must either be incorrect, or we are missing information.

Looking at the model, the line of action from the hinge to the joint where the piston connects to and the other end of the hinge put both of them fairly close to parallel.  And as the two lines become parallel, the required force goes to infinity.

If the load isn't applied until the door partially rotates, you would then just ignore the results until they are appropriate.  Unfortunately it does not appear to be possible to apply a time varying load.

Thanks,

James

[Anonymous]

Hi @KubliJ,

Thanks for your support, 

I would say that your explanation is really useful and I got your point that using static results to check the analysis, thanks for that. 

I think that this problem is solved and this is a lesson learned for me.

Anyway, thank you for your kind help,

Cheer,