Hello'
That can help me, how to find the resistance to the wind load at the curtain of an industrial garage door. To find out if the rubber curtain can be removed from the rails that are welded and buttoned on the door mounts (see pictures). The curtain has a system of end caps and a rigid part at the bottom of the door, these systems are supposed to hold them in the tracks.
The whole door is welded to the H-shaped frame, and the frame is anchored to the building.
I don't know how really, I'm going to go about it, I think, apply a pressure on the face of the curtain of about 70 Mpa and see what happens, but nothing works. Beginner on solidworks
What to put as boundary conditions, fixed geometry and connections...
Your help will be truly appreciated.
Kind regards.
Ps: Thank you for the answer, Michel, for the first question
capture_avatar600.png
EDIT: message deleted and transferred to the duplicate of May 11, 2020.
Let's first sell the previous message before replying to this one or we don't have the slightest information to answer correctly.
Kind regards
Thank you for your answer,
Well yes, my goal is to find the load due to the wind, and pk its not working with solidworks simulation ??, I will have unusable results??
I'm not going to put the load at the tube, but rather, find the resistance to the wind load on the curtain. So that it does not detach from the door rails.
I guess, I have to apply normal pressure to the face of the curtain.
Kind regards
Hello
I only answered the first part of the question and depending on your image attached in the other post, this one becomes a duplicate.
You said then """"" but first I wanted to check the resistance of the main unit (see attached image ), in order to model the bending of this sub-assembly composed of: pipe , shafts, discs and bearing (bearing), ""
Kind regards
PS: there was at least one big outage on the forum for two days! Itis possible to answer the questions asked.
Ok, so for the second question?
Good evening
Have you done the simulation successfully for the tube and bearings?
Kind regards
Not really, I look at the level of the fixed geometry to select the face of the shaft to put it on option: "On spherical surfaces", it does not accept solidworks, gives me a message saying only for spherical face, and yet I know that normally function but not
Good evening AVATAR100
I suggest you drop the first message that is corrupted (impossible to answer it).
So if we answer the first part of the question correctly (tube and turntable), only the outer faces of the turntables should be declared fixed.
The tube and the other parts (shaft and rings) if it is declared solid, there will be no rotation. On the other hand, to apply the vertical load on the tube, you have to make a delimited area with the "separation line" function.
See you
Thank you
What do you mean the turntables, the shafts by ? And for the bearings I declared it in the part of making or fixed travel?.
Kind regards
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
2020-05-14_18_49_01-capture_avatar100.png_image_png_794_x_486_pixels.jpg
2 Likes
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
1 Like
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
1 Like
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
2 Likes
Hello
Thank you for your explanation
Kind regards
1 Like
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
1 Like
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
2 Likes
@ 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
2 Likes
Hello
Thanks for the follow-up, attached is the photo of the door curtain.
Kind regards
avatar100.png