Good evening
So I understand that you are a student given the grammar used and the SMS style (not very well seen on the forum)
I am attaching an image to agree at least on the vocabulary used.
Kind regards
Thank you
You're right, sorry. As far as gravity is concerned, how can I apply it to the center. I redid the middle part, and I drew the tube rings and shafts in one piece. And I created a rectangular area along the top of the tube.
Kind regards
Good evening
It seems to me – if I may say so – that you are making an error in your reasoning. Gravity applies everywhere to any object, from top to bottom. It is the load and the direction of load given by the direction of gravity that counts.
What should be indicated, but distributed over the entire flat part of the top of the tube (or preferably on an area made with a separation line) is the distributed load (especially not a force). If the total mass of the rubber is, for example, 82 Kg, it is 820 N for the distributed mass.
This allows you to have an approximate value but it will not be totally accurate. Indeed, the load is offset because it is tangent to the tube with a second offset of: either the half thickness of the rubber; or a thickness and a half. These rubbers are quite heavy, especially depending on the height and width.
Kind regards
Thank you
For gravity, it's settled, you're right, a big mistake. For the distributed load it's also adjust, thank you. I created a delimited area on the upper surface of the tube.
On the other hand, I remain completely mixed up with connections and movements. In addition I want to create a delimited area in the tree at the level of the bearing/shaft contacts, I don't know how to do it. I tried I get like two surfaces on each side
Kind regards
Ps: I want to consider the bearings to simulate bending
Hello@Avatar
When you say "" I remain completely mixed up with connections and movements. "We have to separate the two problems.
In your case AMHA you have to make do with global contacts for connections. For travel, since you don't have any imposed travel, I don't see why you should worry about it.
For the second point and the creation of a bounded area, it is normal that you get two surfaces on each side. But you just have to select only the one that suits you and not both (the one above in this case)
You also say ""I want to consider the bearings to simulate bending""
I would point out to you, if I may , that if you want to simulate what is really happening, you have to use the function provided for it. Namely the load function on the bearing. However, in your case you will not see much because the bearings purchased on the shelves are sized to have no significant deformation (except in the case of an extreme overload as on the roads of the impossible) so if you take as in your image square flange bearings: just refer to the charts of any brand. We don't do a simulation on this, we stick to the manufacturer's calculations (except the in china). In any case, if you have an oscillating bearing, you will not be able to easily simulate the displacement due to a possible bending of the tube.
At worst, even if you do a simulation considering that the shaft and the bearing are one and the same part, you can get a result that shows the stresses inside the material. The Von Mises and Flexion results show only the external results, but the other possibility shows the cross-sectional stresses inside the material (all other things being equal).
To make your life easier, start with the total weight (tube + shaft + rubber) and as you will have half a load per step, you can immediately compare it to the permissible load given by the supplier. If your tube does not flex more than 2 centimeters in the center, which would already be a lot and may likely, you will not have any deformation at the level of the bearing which is oscillating.
Kind regards
Hello
Thank you for your explanation
Kind regards
Hello
Let's move on to the second part of your request, namely the wind.
For the pre-calculations for the wind tests , a lot of information about the curtain will have to be given. Thickness, but above all is reinforced rubber in polyester, aramid or metal weft. Depending on the width, it is necessary to indicate which are the anti-spinnaker reinforcements. In general, there are horizontal bars more or less spaced according to a height-width ratio. But today with reinforced rubbers there is no longer a bar but with "sliding curtain track" (horizontal tensor). The weight, nature and method of attachment to the rubber and in any case the method of attachment in the side guides of the rubber, must be indicated precisely.
You will also have to be precise about the side rails, as well as how they are fixed to the support (concrete wall or metal beam on metal frame, etc.), but also about the "self-locking sliding curtain track" you have chosen. All these elements are involved in the calculation of the resistance of the whole
Can you give us the reference of the rubber curtain (at least the supplier).
Kind regards
Hello
For the second part, indeed the rubber (EPDM60) is reinforced with polyester and the curtain has a thickness of 6 mm and it has no horizontal bars, it has an over-thickness on the sides (end cap system), in order to hold it between the two rails. These are fixed on an angle iron, a welded U-shaped rail and the other Z-shaped rail is buttoned on the angle iron. The entire door structure is welded to a steel frame that is also anchored to the building.
Kind regards.
Hello @AVATAR100
For the wind what is important is to have more information on the "self-locking sliding curtain track" because if I read you correctly you say """ it has an over-thickness on the sides (end cap system), """
Please show a picture of this part
Certainly enough for an interior curtain but not for a curtain subject to the wind.
Kind regards
@ All
You will notice that the symptoms of covid-19 show a loss of smell and taste, but it appears that the need to say thank you also disappears.
As we see on this forum, covid-19 causes an inhibition of the likes (0), scientists are looking for if it is an alteration of vision, combined with an aphasic neurological syndrome of the limbic cortex, in this case caseous tonsillitis.
I hope I don't get an IG-Nobel for this structuring discovery for the industry and let's not hesitate to say it for humanity. ;-)
Kind regards
Hello
You know, you have to snatch information from you because you don't always respond to the requests made to you.
You invite us to go boating, but we are the ones who have to row in your place.
How wide and high is the curtain???? I don't think I read the news.
In any case, the guidance is insufficient for a curtain exposed to the wind. The curtain will inevitably come out of the U-shaped rails as soon as the wind picks up a little.
Remember the spinnaker effect already mentioned!
Kind regards
Funny better answer
don't understand well
I must be infected myself
+1 @ Zozo
@+ ;-)
Hello
For the simulation on SW it's ok, thank you all. Currently I need help to theoretically calculate the resistance due to the wind load of the industrial garage door, Dimensions of the door 8900 mm wide and 6500 mm high, wind of 140 km/h.
Kind regards.
Hello
Indeed thank you @tous ;-)
After a long wait with my sister Anne - who didn't see anything coming either - we are finally informed of the size of the rubber door.
But alas, as the requester indicates that "the simulation on SW is ok" implied to be over, we are stuck since the calculation or more precisely the simulation of the deformation of the rubber can only be done with SW Simulation or by calculations with no cockchafers.
I have asked several times for detailed details, but in vain on the "sliding curtain track". The weight, the nature and the method of attachment to the rubber. We were just told that there was an extra thickness of rubber at the points of friction in the side guides.
As well as saying that the side guides are fixed, it makes us -as my grandmother used to say- a beautiful leg, as long as we don't know the exact shape of these guides.
In other words, if the guide is a simple open U, nothing prevents movements in X and -X, which is sure to make the curtains come out of the side guides.
So without a cross-sectional view of the side guides (which should look like a square or rectangular tube with an opening of a few centimeters in radial) and an image of the horizontal tensors, no simulation, no calculation at the mano not possible for me but that some Xs would know how to do.
In addition, without having the characteristics of the rubber, namely the name of the supplier and the exact reference of the rubber , we will not have access to the information of the three friends Hooke, Poisson and the young Young.
But what I can tell you without having read in the coffee grounds is that it will only take 20 knots of wind for the curtain to more or less come off the rails. From 50 to 60 knots of wind everything will be off the rails and will hit and destroy everything within 18m on either side of the door including the roof until the winding tube gives way and turns into destructive UFOs that will destroy, nearby, the rabbit hutch of Père Jules
Also destroying the coffee machine, the canteen, the toilets and the owner (ranked in order of importance).
@tous I inform you that an indefinite strike notice is being given on the grounds of "clear withholding of information and obstructions to the manifestation of the clear light that was supposed to tear apart the darkness of ignorance.
Kind regards
Hello
Thank you for answering me.
Need help, if you allow, I am writing to you again, I would like to calculate or determine, the performance (theoretically) of a garage door, such as the resistance due to the load due to the wind which is the most important to calculate (WIND SPEED 140 km/h) (exp, door 8900 mm wide and 4270 mm high, SBR elastomer synthetic rubber curtain.
Note that the curtain (the tarpaulin) is held between two rails (containing a distance of 6.35 mm), at the level of each side of the door
The rails are fixed on angled irons (angles), by welding and bolts
The tarpaulin or curtain has a system of end caps like an over-thickness (19.05 mm thick) to help the hold.
Unfortunately, there is no sliding curtain track, the only way to hold the tarpaulin between the two rails, and that' s enough. However, perhaps for large doors a wind bar could be installed. I looked to find out how to do the calculations, a model for example, a method, I can't find anything (I thought about the strength of the materials, assuming a plate simply supported under pressure, but how to do ???) Thank you in advance.
Kind regards.
Show a cross-sectional drawing or a picture of the guide rails, I don't understand why you don't make this minimum effort. It's not complicated.
What you describe is of no use.
How likely this is to be a never-ending discussion!!!
You can make a simple calculation with a rubber of the right dimensions resting on a flat surface and applying pressure to the entire surface in solidworks simulation. This means perpendicular to the guide rails.
The curtain must be vertical and you must apply a vertical load on the bottom of the curtain (load equal to the total weight of the rubber with dimensions of 8900 mm wide and 4270 mm high.
After that, we have to see if there is a material in solidworks that corresponds more or less to the SBR elastomer that I told you about a long time ago.
You do not say whether it is an academic exercise or what you were asked to do as part of an internship.
For pure calculation you must remember
calculation is defined by the following formula:
This formula is valid for any area where the strength of the wind must be determined. (Source : Guide de la construction bois)
This gives for 140 km/h = 38.89 knots which gives 975.52 N per M² reduced to the 38 m² of your curtain it gives 37069.76 N or 3.7 tons. How can you imagine for a moment that without support in the radial direction your curtain, which is very flexible, does not fly away. The average weight of a car is 1.5 tons, so put two cars in a hammock, what will happen
Kind regards
PS: I maintain that your curtain is not at all held on the sides in the radial direction it will not resist the wind on such a span ( ¤/#~&!,w¤/#~&!,w¤/#~&!,w¤/#~&!,w)
Hello
thank you for the answer, here is an image that shows the rails (right side)
Kind regards.
