Hi, I'm a mechanical drawer, I'll use Elmer for test tensions on some piece.

I studied and done Tutorial 2 Linear elasticity equation – Loaded elastic beam

Questions:

- Why the volume isn't considered in the formula? Model\BodyForce\Name = Gravity\Linear Elasticity\Force 2 = $ -9.81 * 550

- How to interpret results?

with Gravity I get Displacement_abs = 1.5e+002 instead of 6.36 cm of Tutorial or 6.4 cm by a manual calculation

removing gravity (27 N over 2000 N of load) I get Displacement_abs = -91 and vonmises = max 1.4e+08 (I'm using SI units, so what is it? N/m^2 ?)

Notes: I can't find Directory: ElasticBeam3D\beam3d.grd (installer: Elmer7.0-rev5817-2012-08-22.exe),

so on my CAD program I recreate and export a .step (in mm). If I get some mistake mm <-> m the result could be similar with only a difference of the decimal position, I suppose.

Thanks for explanations.

## Displacement and tensions value in a elastic beam

### Re: Displacement and tensions value in a elastic beam

Hi,

I think it is indeed the problem of using mm instead of m. You can either scale your mesh with "ElmerGrid" or draw it again.

By the way, you can download stuff at https://elmerfem.svn.sourceforge.net/sv ... fem/trunk/

For example in your case the geometry is here https://elmerfem.svn.sourceforge.net/sv ... mples/grd/

or https://elmerfem.svn.sourceforge.net/sv ... beam3d.grd

Your question:

- The body force is actually a "force density" and not "force". So it is the density of the material that determines the body force on each point of the body. Volume is thus not needed.

- please specify what you mean by "interpreting". In the tutorial picture you can clearly see the corresponding beam bending.

Cheers,

Eelis

I think it is indeed the problem of using mm instead of m. You can either scale your mesh with "ElmerGrid" or draw it again.

By the way, you can download stuff at https://elmerfem.svn.sourceforge.net/sv ... fem/trunk/

For example in your case the geometry is here https://elmerfem.svn.sourceforge.net/sv ... mples/grd/

or https://elmerfem.svn.sourceforge.net/sv ... beam3d.grd

Your question:

- The body force is actually a "force density" and not "force". So it is the density of the material that determines the body force on each point of the body. Volume is thus not needed.

- please specify what you mean by "interpreting". In the tutorial picture you can clearly see the corresponding beam bending.

Cheers,

Eelis

### Re: Displacement and tensions value in a elastic beam

Hi,

to use a mesh with mm as length unit write

Coordinate Scaling = 0.001

into the simulation section of your sif (ElmerGUI: free text field in Model->Setup->Simulation).

HTH,

Matthias

to use a mesh with mm as length unit write

Coordinate Scaling = 0.001

into the simulation section of your sif (ElmerGUI: free text field in Model->Setup->Simulation).

HTH,

Matthias

### Re: Displacement and tensions value in a elastic beam

Thanks for reply.

my second question was about units, but as I show below my troubles caused by trying to make ends meet keeping for good some wrong data.

I note theese mistakes in the tutorial:

Now I use beam3d.grd:
so Boundary area are:
Tutorial 2 says:

In fact if I set corrected data into ElmerGUI, I keep these results from Post:

Thanks.

my second question was about units, but as I show below my troubles caused by trying to make ends meet keeping for good some wrong data.

I note theese mistakes in the tutorial:

Now I use beam3d.grd:

Code: Select all

```
BOUNDING BOX
X-coordinate: [ 0 , 0.1 ]
Y-coordinate: [ 0 , 0.05 ]
Z-coordinate: [ 0 , 1 ]
```

Code: Select all

```
Z*X = 0.1 m^2
Z*Y = 0.05 m^2
```

From my knowledge, the approximated formulas are:The second boundary condition distributes

the load of 2000 N uniformly on the area of 5.0e-3 m2 (wrong).<--0.1 m^2Z*X

...

Linear elasticity

Force 2 = -4.0e5 (wrong)<---20000N/m^2

...

The maximum displacement is 6.36 cm (wrong)<--0.024 m or 2.4 cmsee below

...

Extra task: Gravity in x direction

...

Linear elasticity

Force 1 = -4.0e5 (wrong)<---4.0e4Z*X

Code: Select all

```
Displacement_y = Force * length^3 / (8 * Youngs_Modulus * I)
I = bh^3/12 = 0.1*0.05^3 /12 = 1.041667e-6 m^4
Displacement_y = (-2027 N * 1 m^3) / (8 * 10*10^9 N/m^2,* 1.0416e-6 m^4) = -0.0243 m
max stress = 3 * length * Force / (b*h^2) = 3* 1 * 2027 / (0.1 * 0.05^2 ) = 24324000 N/m^2
```

In fact if I set corrected data into ElmerGUI, I keep these results from Post:

Code: Select all

```
Displacement_y = -0.0239 ("m") OK
Vonmises = 2.3006e+007 ("N/m^2") OK
```

**Is it right?**Thanks.

### Re: Displacement and tensions value in a elastic beam

No.

The beam tutorial is correct. The force works on the smallest area of 0.1 * 0.05 = 0.005 m^2. The confusing label "Force" in the Linear Elastic boundary condition is actually a pressure. So you enter something with for example the units N/mm^2. To get 2000 N on this surface (one end of the beam) you calculate:

p = F / A = 2000 / 0.005 = 400000 N/mm^2

These inputs lead to a result of a maximum absolute displacement of 0.06359 m. To see the result actually deformed in Elmer Post, enter the following two lines in the Elmer Post command line (bottom of main window):

math n0=nodes

math nodes=n0+Displacement

display

To get an exaggerated deformation (for example 5 times):

math n0=nodes

math nodes=n0+5*Displacement

display

Instead of entering the last command "display" you can also left-mouseclick the psychedelic looking button on the right.

The beam tutorial is correct. The force works on the smallest area of 0.1 * 0.05 = 0.005 m^2. The confusing label "Force" in the Linear Elastic boundary condition is actually a pressure. So you enter something with for example the units N/mm^2. To get 2000 N on this surface (one end of the beam) you calculate:

p = F / A = 2000 / 0.005 = 400000 N/mm^2

These inputs lead to a result of a maximum absolute displacement of 0.06359 m. To see the result actually deformed in Elmer Post, enter the following two lines in the Elmer Post command line (bottom of main window):

math n0=nodes

math nodes=n0+Displacement

display

To get an exaggerated deformation (for example 5 times):

math n0=nodes

math nodes=n0+5*Displacement

display

Instead of entering the last command "display" you can also left-mouseclick the psychedelic looking button on the right.

### Re: Displacement and tensions value in a elastic beam

p = F / A = 2000N / (0.05m x 0.1m) = 400,000 N/m^2

i think is in meter square and not millimeter square

i think is in meter square and not millimeter square

### Re: Displacement and tensions value in a elastic beam

Vespa

Your Formula is faulty :

<Displacement_y = Force * length^3 / (8 * Youngs_Modulus * I)>

You divide with 8 it should be 3

Displacement_y = Force * length^3 / (3 * Youngs_Modulus * I)

Multiplying your result : 8/3 gives the 0.064m.

Your Formula is faulty :

<Displacement_y = Force * length^3 / (8 * Youngs_Modulus * I)>

You divide with 8 it should be 3

Displacement_y = Force * length^3 / (3 * Youngs_Modulus * I)

Multiplying your result : 8/3 gives the 0.064m.