How to model an industrial garage door ON SOLIDWORKS, in order to find the resistance to wind load at the garage door curtain (the curtain is made of rubber)

Simulation
21

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

1 Like
22

Funny better answer 

don't understand well 

I must be infected myself

+1 @ Zozo

@+ ;-)

23

 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.

24

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

25

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.

 

26

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:

T= {1 \over 2} . Cx . \rho . v^{2} . S

  • T : Wind force [N] in Newton So be careful to respect the units.
  • Cd: Air Penetration Coefficient
  • P  : Wind density or density [kg/m3]
  • v : Wind speed [m/s]
  • S  : Wind-exposed area [m²]

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)

 

27

Hello 

thank you for the answer, here is an image that shows the rails (right side)

Kind regards.

 


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1 Like
28

Hello 

Here is another image,

Kind regards.


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1 Like
29

Hello.


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