## Displacement and tensions value in a elastic beam

Post processing utility for Elmer
vespa_83
Posts: 2
Joined: 12 Dec 2012, 19:15
Antispam: Yes

### Displacement and tensions value in a elastic beam

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.

Takala
Posts: 183
Joined: 23 Aug 2009, 23:59

### 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.
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

- 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

mzenker
Posts: 1969
Joined: 07 Dec 2009, 11:49
Location: Germany

### 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

vespa_83
Posts: 2
Joined: 12 Dec 2012, 19:15
Antispam: Yes

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

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 ]``````
so Boundary area are:

Code: Select all

`````` Z*X = 0.1 m^2
Z*Y = 0.05 m^2``````
Tutorial 2 says:
The second boundary condition distributes
the load of 2000 N uniformly on the area of 5.0e-3 m2 (wrong). <-- 0.1 m^2 Z*X
...
Linear elasticity
Force 2 = -4.0e5 (wrong) <-- -20000 N/m^2
...

The maximum displacement is 6.36 cm (wrong) <-- 0.024 m or 2.4 cm see below
...
Extra task: Gravity in x direction
...
Linear elasticity
Force 1 = -4.0e5 (wrong) <-- -4.0e4 Z*X
From my knowledge, the approximated formulas are:

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.

jeroenr
Posts: 20
Joined: 16 Oct 2013, 21:41
Antispam: Yes

### 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.

laileng
Posts: 23
Joined: 15 Mar 2016, 08:01
Antispam: Yes

### 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

Oldbelt
Posts: 11
Joined: 24 Dec 2018, 14:15
Antispam: Yes

Vespa