How do you simulate an earth push on a face?

Simulation Generality
1

Hello

I have sheets to make to make a soil reservoir, I know the quantity of earth that will be put in front of it, I started with an approximate value of 1.2 tons per m^3.

Here on my example I blocked the bottom of my bent sheet metal in fixed geometry and then I exerted a force distributed on the face where the earth is located...

but I'm not sure of the value used (put pressure? a force?)

 

Thank you


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1 Like
2

Hello

In my opinion, this is rather difficult to simulate because earth is not a homogeneous material and it seems difficult to predict its behaviour. If you take clay the earth will hold on its own because of its low "viscuosity" if you take sand then we are a little closer to the behavior of a fluid.

Let's assume that you have the density of the earth and let's assume that it can be assimilated to a "fluid", the force applied to your folded sheet metal is not constant as you defined it here. It's equal to zero at the top of your sheet. It is proportional to the height of the column of earth. The effort increases with depth.

 

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3

Hello

I would proceed as a fluid as well.

Edit: https://www.my-cad.fr/actualites/tutoriels/communication-technique/be-sous-traitance/pression-hydrostatique-modeliser-solidworks-simulation/

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4

Hello

You have to define a pressure afterwards according to the height of the ground to be remembered, your value will change. Your model is quite close to a hydraulic dam if we consider the fluid to be earth. Of course, the earth is not a homogeneous fluid but I invite you to consider it as such. You will also need values such as the density of the earth.

If you have problems calculating the pressure, don't hesitate otherwise you can find this kind of calculation on the internet.

Dimitri

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5

I considered a density of earth at 1250kg/m^3 with a sheet height of 0.5m so I have my pressure which is equal to 6131N/m²

I can't really apply the tutorial that DBZ advised me to fill in solidworks...

 

6

What's your width?

7

j'ai une longueur de retenu de terre de 14.7m

I have a length of earth retention of 14.7m

8

6131 Pa is your pressure at the base of your structure. If you consider that your overpressure you have to integrate it on your surface. 

Personally I find an  effort according to your width of 1532*width. 

Dimitri

9

Here is my Solid Works 2016 file


partie_courbe_terrasse_03_cde002683.sldprt
10

So 22,531 N =)

 

But remember that this effort was obtained by considering the earth as a homogeneous fluid. It is likely that this effort is very over-enorsing.

Dimitri

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11

how do you find 1532 Dimitri?

12

To calculate that your pressure is 6131 N/m², you did P(z) = density * 9.81 * z

and for z = 0.5 => P(z) = 6131 

Now if you integrate P(z) = density * 9.81 * z on the height (from 0 to 0.5) and then on the width (from 0 to 14.7) you have 

int (from 0 to width) [ int (from 0 to height) [ density * 9.81 * z] dz ] dl

 = width * density*9.81 *(height²/2)

= width * 1532 

= 22,531 N

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13

See this link

https://books.google.fr/books?id=O7eWDgAAQBAJ&pg=PA27&lpg=PA27&dq=calculer+une+retenue+de+terre&source=bl&ots=UXnZrJi5Ek&sig=rvCnqHxEdO5vSXE0xBRJh97ZFLc&hl=fr&sa=X&ved=0ahUKEwiz0vW31MDWAhUhCMAKHenWAD44ChDoAQguMAI#v=onepage&q=calculer%20une%20retenue%20de%20terre&f=false

@+

14

Hello 

Have you solved your problem?

Dimitri

15

yes perfect thank you Dimitri