Solidworks simulation problem

Simulation
1

Hi all 

I have tried many times to get the right deformation on the tubular structure in the picture, but without success.

This structure (3D sketch) is composed of 4 120x80 tubes and a 50x30 tube (crossbar) (all constructions with the functions of "welded constructions"). I used the fit function to adjust the crossbar length relative to the frame of the structure.

To apply the targeted load, I created 2 extrusions (volumetric bodies) to be able to apply the forces in the desired places (2000N on the surfaces of 2 extruded bodies).

So I run the study and there, surprise, only the crossbar deforms (see attached image).

I don't know how to solve this problem at all... probably the management of the connections between mechanically welded elements that I don't master at all...

I'm on SDW 2017.

Thank you in advance for your help


simu.png
2

Hello

You have an inertia of a ratio of 10 between 120x80x3 tube and 50x30x3 tube. Displaying the Maxi and Mini constraints in the constraints and option of the graph will validate the weak part that will deform.

I had asked the question to locate a force on a shear axis, you have to go through separation lines, and a much older exchange of another person and the same procedure.

https://forums.futura-sciences.com/technologies/813215-declarer-une-simulation-xpress-solidworks.html

Kind regards;

 

 

3

Hello spectrum,

Are you talking about the 2 boxes "show annotations?"

I checked them, but nothing I change, or I didn't understand how to validate the part that is deforming, as you say. :/


simu2.png
4

Hello

In fact, this shows that the deformation is indeed on this element more fragile because it has the smallest cross-section.

Your 120x80 exterior frame should undergo a slight deformation but it should appear, you can load the beam by choosing the beam element in the external applied load.

You have two loads concentrated on each center of your two elements, the choice of the 50x30 element will be the point that will undergo the greatest displacement but it remains small 1mm no factor E+000 means 10^0= factor of 1.

and 9.8 E 10^-1 = 0.98 mm displacement;

You can put in general in the results instead of scientific.

Will avoid powers of 10.

Kind regards.

Spectrum.

 

5

I think I understood.

I would like to load the beam at a specific point and not with a distributed load, is this possible on SDW? 

 

6

Thank you very much spectrum,

With your link ("separation line" tool) + the "probe" tool, I managed to understand everything, thank you^^


simu3.png
7

Yes, for charging at a specific point, you have to go through remote actions.

In addition, here is the tutorial on the use of dividing lines

https://www.lynkoa.com/contenu/les-lignes-de-s%C3%A9paration-1

You have to familiarize yourself with the tool, I have obtained help very often by using this forum, I am not a specialist but it gives an idea by simulating the behavior of an element.

We must pay attention to the concentration of stress which masks and distorts the problem posed.

Happy for you, have a good week.

 

8

Hello @spectrum  ;-)   :-)

Thank you for the links to futura and to the dividing lines tutorial.

I wonder who is the author of these things   ;-)

Kind regards

 

9

Hello Zozo,

Yes it tells me something, thank you again for all the help provided, it's the spirit of the beginning of the internet............

Have a good week to the team!.

 

10

Hello @simonpbmeca

Have you solved your problem?

In the meantime, it is quite normal that you try different deformations between the side tube and the central tube.

Explanation (without the need for simulation
1°) The side tube can only work in bending like a beam placed on two non-deformable plates under normal loads

2°) the central tube is attached to two opposite beams but which obviously undergo a twist and a bending. That's why you don't have the same result on both beams.

A few remarks on simulation.

A) You cannot see the deformation described in point 2, because you are using the beam mode instead of doing your entire simulation in volume.
I repeat over and over again on this forum that for sections less than 300 X 300 mm it is better to use volume to see better (if you are not used to predictable deformations without simulation). This does not change much in terms of simulation time, even for larger structures.
B) For "external loading"  it is often preferable to use "a pressure" rather than a force (even if our eminent colleague @obi wan tells us that the force is with you ;-)    )  indeed a force will always remain normal to the plane which distorts the result in case of significant bending and especially in case of spillage. While the pressure will remain normal on the surface defined by the dividing lines, which means that it will follow the deformation of the beam or the part. Be careful, however, because "pressure because it is a force divided by a surface"" and in M² the pressure must therefore be reduced to M² at the surface defined by the separation lines.
C) for fixed movements on four legs, this is only valid if your feet are all anchored in the ground. Otherwise, you also distort the results.

Kind regards

1 Like
11

 

Hello @Zozo_mp 

Thank you for all your informative answers^^

Without the need for simulation, I have enough knowledge of mecha to estimate the most used elements. It was mainly the displacement values due to bending that interested me.

My main problem (on which I am working more and more) is to know how to "correctly" simulate such simple solicitations and understand the operation, the graphics, the functions, the results displayed by solidworks...

A) So I see that I have a lot of shortcomings on this subject, I would like to know what do you mean by volume simulation? Is this a parameter to select when creating a finite element study? Or should we convert a mechanically welded part (which would therefore not be volumetric) before a simulation by FE? Or maybe nothing to do with it...

How do you choose between volume or surface when you have designed a structure (part or asm) entirely composed of mechanically welded profiles (i.e. via mechanically welded functions)? 

B) Thanks for the info, I didn't know that in SDW simulation, a pressure followed the deformation of the element on which it is applied^^  

I had still tried to simulate with pressure, but I still had the error message "not possible to apply force on a beam element". So what to do? The only solution I had found thanks to @spectrum was to go through "stress reference points" on which I directly applied point forces.

C) For the four legs, fixed to the ground?

So what do I do if my 4 legs are fixed on 2 horizontal beams at the same height (2 feet per beam, recessed beams at the ends)? 

If the feet are fixed, how do I get the deflection due to the weight that these feet have to support??

Thank you in advance for your help

12

Good evening @ simonpbmeca

Answers:

a) When you in your simulation, look into the feature manager and do the following. Be careful, sometimes you have to develop all the pieces to see if there is a mix of volume and beam. You have to pass everything without exception in volume ;-)

B) if I understood my trainer at the time correctly (and also for the reasons I developed before on the beams) it is not possible to use pressure. Except maybe in mixed modes (to be checked)

C) Not clear what you are saying;-)   If your feet are fixed on beams, the entire load supported by your feet will be transmitted to the horizontal recessed beams.   You have to be wary because the word embedded has several meanings. In other words, do the horizontal  beams have one end fixed and the other with a plane support or are both ends totally fixed (bolted or welded to something reputed to be indefinitely solid). If it is totally fixed , only the elongation of the beam under the load can give a bending. If the beam has one side on a flat support, then you will have a deflection that will be greater than if the beam is totally fixed.
Deflection (or strain under load) can be measured by placing probes when you are in URES mode.

But for the explanation to be valid, you have to know if what is above your feet. Because if the vertical beams do not suffer any buckling or spillage, this will contribute to the stiffening of your horizontal beams and therefore reduce the deformation of the said horizontal beams.
We would have to see the complete model to be more precise.

Kind regards

 

13

Good evening @Zozo_mp 

A) Great, thank you, I hadn't paid attention to this option at all, I will be able to consider the beam in volume and be able to apply the appropriate pressures on the dividing lines

B) Ok, I would remember that beam of dim less than 300x300 = mandatory volume simulation.

C) I actually wanted to explain very simply what I wanted to simulate, here is an image that will be more telling

Above is a tray on which a container will be fixed that will weigh 500kg

4 legs supported by the 2 rectangular tubes

The chassis that will be mounted on legs (not shown in the photo)

 

---> And so, the equivalent bond between straight tubes . and the frame of the chassis is indeed of the embedded type, but the frame of the chassis being fixed on ball feet, the load applied on the tubes will make the whole thing flex (if I'm not mistaken^^ )

I don't have the buckling analysis module, so I went through the theory (Euler, Rankine) to determine the maximum load before buckling the legs of the platen. The feet largely support the applied load.

If I understand correctly, despite the weight, it stiffens the straight beams? The flexion will therefore be reduced, but we agree that the forces transmitted through the feet remain to be monitored? (i.e., not focusing only on travel and looking at the efforts at the connections?)


chariott.png
14

Hello

you say this """ And so, the equivalent connection between the straight tubes and the frame of the chassis is indeed of the embedded type, but the frame of the chassis being fixed on ball legs, the load applied on the tubes will make the whole thing flex (if I'm not mistaken^^)"""

You're not wrong but your conclusion deserves discussion ;-)

Because you put casters you are not in the case of a recess effectively,  but in the case of a contact between a cylinder and a plane support.
In your case, the top table is quite close to the casters (at the top in your image) so your effort will be 5 times more on its busiest casters, which means that your two bottom beams (lengthwise) will not flex (except for a huge load) but also that your casters must be resized because the load is not centered between the casters. The load to be supported by the castors is not equal to 1/4 of the total load (with the margin which is fine for a minimum of 3 for potentially rolling stock).

A small friendly remark (since you accept the dialogue) For the load you have to take into account the CG of it, because the load is not on the table plane but much higher. I always take into account the CG of the load and I put a remote load facing up to do the simulation.
This also affects the possible bending of the horizontal beams of the first and second level of your tiered structure.
This makes it possible to better control a possible spill (warping, etc.) and above all to put in the user or validation notebook such as Veritas that this frame and its load should not be used on a slope greater than so many degrees. You must have a popey plate fixed on the frame and which must be explicit with the pictograms that go well. But that's probably what you're already doing in this company (sorry to drive nails into an open door)

Kind regards

15

Hello Zozo_mp, first of all thank you for any explanations, I am of course open to any positive or negative remarks, we always have to learn from others, it would be a shame to get angry when we have things to learn^^

 

If we put ourselves in the case where the legs are centered on either side of the tubular structure (frame), the casters will each have to support 150kg/4 = 32kg.

Even if we admit that the load is off-center in relation to the CDG of the chassis, I don't see at all how you deduce that casters would be 5 times more loaded... How do you deduce that?

The linear connection would allow me to deduce the pressure exerted by the ground on the wheels (32kg/wheel thickness, oversized calculation because real contact = surface/surface). So I don't see at all how you deduce that there would be 5 times more force...

 

For the spill, are you talking about the tipping resistance?

If so, why do you put a remote charge up? Wouldn't it be more of a downward charge? And I don't see how solidworks can deduce a tilt if the load is only vertical? Shouldn't we have a component oriented more horizontally?

I've never used remote charging on SDW before, I'll probably still have things to understand about it^^

 

What is a "popée"?

Otherwise, for the maximum slope of use, no, it's not a contractual one, in general we go to the field to analyze the location and if ok, we don't bother to specify the conditions of proscribed positioning ^^

16

Hello @simonpbmeca

I expressed myself badly, sorry.
You know very well the difference between a point load and a distributed load. If you look closely at your assembly if the CG of your load is perfectly centered on the top table, then the mass is distributed over the four legs. On the other hand, on the lower chassis the load is directly above the CG of the upper load. This means that the load is not evenly distributed on the four wheels. I always think center of gravity as for the loading of a truck (see  attachment) and this regardless of the number of pallets in height or in your case the number of stacked structures.
I formulated the numbers wrong, I should have said that on the 500 kilos of the load the feet will not  receive 125 kg each but that the wheels on one side will support more. Easy to calculate (L1 x P1 = L2 x P2) knowing that L2 is larger than L1 (L is the length and P is the weight.)

Why I put the load up!!! Here too, it's a question of vocabulary. The CG of the load is, let's say, 50 cm from the table, so the mass at a distance is well above the frame, but in the parameterization, you have to put the direction of the load which will obviously be downwards, and in addition, you have to indicate the gravity and its direction downwards too. (look at how the remote ground works, it will seem obvious to you ;-)

I am talking about a spill but I could have said what the tipping point is. SW doesn't calculate it but SW gives you the CG permanently so you just have to draw with a sketch a straight line between the CG and the ground. Then a line between the CG and a few centimeters beyond the busiest wheels. The angle recorded gives the slope not to be exceeded before everything goes into the gutter;-)   If at your customers' place everything is static then no PB on the other hand if they move the whole then if the wheels are blocked by an extra thickness in the ground then the risk of tipping is obvious depending on the speed if your angle calculate higher is low.

Popée is not a Greek, Roman or Celtic deity but it is a pseudo verb of 'fix with pop rivets'. It is a barbarism of the trades that transforms the act of riveting pop rivets into a verb of action. Poper verb and in the feminine ""the plate is popée"" which in this case becomes a linguistic shorthand. Like Stabilobosser ;-)           (end of the cultural minute)

My experience in industrial expertise is that if the precise conditions are not indicated, then the insurers, or even the judges in the event of a bodily accident, will  hold your company liable, etc. And this according to the principle of it's not me, it's the other who is naughty. It's the cat in the microwave syndrome.

Kind regards

 


chargement_a_lessieu.png
1 Like
17

Hello @Zozo_mp 

No worries, so I understand better, so we're in agreement. For the distribution of the load on the wheels, the famous PFS which says that, qqch is static implies that the sum of the moments of the forces ext. to this qqch is zero at any point of this qqch, I know well ;)

 

Ok ok, I'd have to take a closer look at this function, in the old days I would draw a cylindrical or parallelepiped container, edit the mass properties by forcing the mass I wanted, and activate gravity in the simulation to simulate the actual weight. So the procedure is a bit cumbersome...  

Remote charging could save me a lot of time if I understood the usefulness

 

Here too, we are generally connected, to save time and if possible, I prefer to place the CDG of a set in the middle of its support. 

There are also cases where I use the calculation of the tipping resistance to estimate the minimum horizontal force that would tip the carriage (static initially). (the famous formula of the adhesion coef = Ftangential or horizontal./Fnormal)

Then I make sure that this horizontal force is high enough that it doesn't reach it while handling the truck.

 

Ha ha, thank you for this cultural mn, there are so many terms lol, it would be good to make a lexicon of them to avoid it being lost in time, who knows^^

 

You're absolutely right, maybe one day, if I have a status with the responsibilities that go with it, I'll think strongly about this aspect of legal protection via the lines that are needed in the contracts and lots of POPEES everywhere lol

 

Thank you very much for your help;)