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Apply load 2

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mojtaba pourgholami

Civil/Environmental
Mar 28, 2019
35
how can I apply the load look like picture in abaqus
 
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I can’t see any picture here. Please try to attach it again.
 
bar_jp3lqo.png

here you are
 
This will require direct cyclic analysis step. In the step definition you specify the time of a single loading cycle. If fatigue is included you also give the maximum number of cycles in this step. Cyclic load is defined using amplitude. Remember that start value and end value of amplitude must be equal. It is of course possible to use multiple steps with different amplitudes.
 
Loading frequency of 1 Hz means that there is 1 cycle per second. Now if you look at the top part of the picture you will notice that it shows the variation of load (loading, holding and unloading) in terms of time but we can easily convert it to cycles. It shows 2 seconds so these are two cycles. Now look at the bottom part of the picture. It shows the full magnitude of load (peak value on the previous plot) for each range of cycles. As you can see in cycles 1 to 1000 full load is 10 kN, then for the next thousand cycles it’s 20 kN and so on.

Knowing this you can create an amplitude that defines load magnitude with respect to time. You can do it fast in spreadsheet and then import to Abaqus. Your amplitude will look like this:

t — F[kN]
0 — 0
0.4 — 0
0.65 — 10
0.75 — 10
1 — 0 _________(1st cycle completed)
1.4 — 0
1.65 — 10
1.75 — 10
2 — 0 _________ (2nd cycle completed)
.
.
.
1000 — 0
1000.4 — 0
1000.65 — 20
1000.75 — 20
1001 — 0 _________(1001st cycle completed)
1001.4 — 0
1001.65 — 20
1001.75 — 20
1002 — 0 _________(1002nd cycle completed)
 
Yes, import the data created in spreassheet to tabular amplitude type in Abaqus.
 
And what can do when frequeny is 16 hz ?
The cycle must be 16times repeat ?
 
In case of 16 Hz frequency there are 16 cycles per second. This means that a single cycle (no load, load, hold, unload) will have to be completed in 1/16 of a second. It can be represented in the amplitude using the same approach as in previous case.
 
0 0
0.025 0
0.040625 10
0.046875 10
0.0625 0
0.0875 0
0.103125 10
0.109375 10
0.125 0
0.15 0
0.165625 10
0.171875 10
0.1875 0
0.2125 0
0.228125 10
0.234375 10
0.25 0
0.275 0
0.290625 10
0.296875 10
0.3125 0
0.3375 0
0.353125 10
0.359375 10
0.375 0
0.4 0
0.415625 10
0.421875 10
0.4375 0
0.4625 0
0.478125 10
0.484375 10
0.5 0
0.525 0
0.540625 10
0.546875 10
0.5625 0
0.5875 0
0.603125 10
0.609375 10
0.625 0
0.65 0
0.665625 10
0.671875 10
0.6875 0
0.7125 0
0.728125 10
0.734375 10
0.75 0
0.775 0
0.790625 10
0.796875 10
0.8125 0
0.8375 0
0.853125 10
0.859375 10
0.875 0
0.9 0
0.915625 10
0.921875 10
0.9375 0
0.9625 0
0.978125 10
0.984375 10
1 0

,,,,,,,,
is that true for one cycle?
There is no easier way to do this?
 
Yes, this is correct amplitude for 1 second (16 cycles). It can be prepared relatively fast in spreadsheet. In this case it will be easier because in the article for the study at 1 Hz they use constant full load magnitude of 200 kN.

To do it another way you would have to switch from explicit to direct cyclic (implicit) step or use fatigue software such as fe-safe. But if you want to follow the article you have to use explicit in Abaqus.
 
Analysis time must be set to overall time of the whole test (all cycles). So for the first case of 1 Hz it should be 6000 seconds (there are 6000 cycles) and for the second case of 16 Hz it should be 6250 seconds (there are 100000 cycles). However you don’t have to request output for each second. In the article for the second case they request output every 10 seconds (160 cycles).
 
another question witch category and load I must be used to applied?
mechanical/pressure?
 
In the article they say that they used equivalent pressure load. So you have to divide the value in newtons by the area of the surface where you apply load to get appropriate pressure value. In Abaqus select Mechanical —> Pressure load. Be careful about the units. And don’t forget that in the article they used additional 100 s long step with only gravity load applied. This step was preceding the actual analysis step with cyclic load.
 
my model was finish and i analyze that, i have an error and i don't know how to fix that
The elements contained in element set ErrElemExcessDistortion have distorted excessively.
and with this warning :
87 elements are distorted. Either the isoparametric angles are out of the suggested limits or the triangular or tetrahedral quality measure is bad. The elements have been identified in element set WarnElemDistorted.
can you help me to fix it.
the abaqus file is in attach
 
 https://files.engineering.com/getfile.aspx?folder=c4ff07a3-b303-4b33-a1ec-0cbfac131be9&file=geocell_railway-98.5.28.zip
I checked the model, fixed some errors and it’s running now. Here’s what you should change to make it work:
- those back faces should have XSYMM instead of YSYMM (BC-4)
- one instance of a part called „tie” was misaligned with embankment (overclosure on one side and gap on the other side), you can use „Translate To” tool to fix it
- geocell was slightly sticking out of embankment, I translated it a bit in the positive X direction (towards the front face of the model)
- there was a problem with contact between geotextile and embankment (Int-12), I changed it to tie constraint

It solves with these changes but I think that you should also refine the mesh. Especially rail’s mesh.
 
What distribution of loading should I use?
load_vyytpo.png

when I use uniform my answers are far from reality.
 
Either use uniform with magnitude 1 and amplitude values specified as load in Newtons divided by surface area in mm^2 or use total force with magnitude 1 and amplitude values specified in newtons.

This is all assuming that SI(mm) units are used (material properties seem to be defined with this unit system). But now I see that the dimensions are probably in SI(m). This is the reason - you have to fix unit consistency in your model.
 
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