Welcome to the forums. Please post in English or French.

You are not logged in.

#1 2020-04-15 19:32:00

konyaro
Member
From: Genève
Registered: 2016-04-02
Posts: 255

Friction force depending on the timestep

Hello,
if the COEF_PENA_FROT coefficient is close to the convergence limit, the friction force depends on the timestep (CONTINUE, PENALISATION). Dividing the timestep by a factor 2 leads to the same results than dividing the COEF_PENA_FROT by a factor 2.

Is that a normal behaviour? It is quite dangerous because if the timestep is reduced automatically the friction force becomes wrong.

Below is depicted the friction force of a part sliding in the x direction on a fixed part. The timestep is divided twice and the sliding force depicted on the right gets also reduced twice.
COEF_PENA_FROT

Konyaro

Last edited by konyaro (2020-04-15 19:32:19)


Attachments:
Friction_force.png, Size: 44.29 KiB, Downloads: 548

失敗は成功のもと (L'échec est la base de la réussite)

Offline

#2 2020-04-16 07:21:51

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

if we look at standard HP formula in friction

HP = Load x Velocity x Coefficient of friction / 4500

Load - kg
Velocity - mtr/minute
COF -  as define by yourself depending on application

Now, Load in inversely proportional to speed & coef. of friction from above formula

&

Speed is inversely proportional to coef. of friction.

To understanding, Code Aster Says What is Load for 1 second time step, Now Divided TimeStep to 1/2 Sec, Then Code Aster Divides the Load by 2, thus 1 kg reduces to 1/2kg. Means Reading for 1sec travel, load is 1kg & for 1/2sec travel, load read is 1/2kg. keeping coefficient of friction constant.

So, In Physical Calculation the Reading by Code-Aster is True.
Velocity is mtr./sec or mtr./minute taken in HP formula.
So Load Reading is per sec or per minute
If you are taking 1/2 sec time step, your reading will be 1/2kg.
But When Defining HP or Total Load per second or per minute, you have to total all timestep reading.

Hope, above interpretation is acceptable. Any argument is welcome.

Offline

#3 2020-04-16 17:21:06

konyaro
Member
From: Genève
Registered: 2016-04-02
Posts: 255

Re: Friction force depending on the timestep

Hello Sameer,
Thank you for thinking about my question. I don’t think it is the right explanation for the following reasons:

- This phenomena exists for both dynamic and static analysis, for which time has no real meaning.

- If the COEF_PENA_FROT is high enough, the sliding force is constant for different timesteps. Unfortunately it is often not possible to set a high coefficient as convergence fails.

- Coulomb friction is used, which means the tangential force should equal the normal force multiplied by the friction coefficient when sliding. Power is not supposed to be constant.

Konyaro

Last edited by konyaro (2020-04-16 17:22:01)


失敗は成功のもと (L'échec est la base de la réussite)

Offline

#4 2020-04-16 21:39:31

mib
Member
Registered: 2019-11-10
Posts: 85

Re: Friction force depending on the timestep

Try using both bodies like rigid body.

for the slave body impose only Dx D.O.F. like free

for the fix plate try to impose DX=0, DY=0 DZ=0 (like fixed) to contact surface.

This for understand how reaction force and friction coefficient goes.

do you think which is possible which the solver compute the friction cone locally in deformed zone?

(i usually , for my application, use, coulomb low = 0.15, DISCRETE formulation, and manual time step, sets)

link (copy and past on your browser , add the third "w")

ww.youtube.com/watch?v=ps5-vfUoggw&list=UU3-HKuZqNwHmmTpe4JF9_qw&index=39

Last edited by mib (2020-04-16 22:02:44)

Offline

#5 2020-04-17 23:10:13

AsterO'dactyle
Administrator
Registered: 2007-11-29
Posts: 343

Re: Friction force depending on the timestep

Hello,

Did you deactivate the automatic adjustment of the coefficients? What is the value of the keyword ADAPTATION?


Code_Asterの開発者

Offline

#6 2020-04-18 06:00:05

konyaro
Member
From: Genève
Registered: 2016-04-02
Posts: 255

Re: Friction force depending on the timestep

Hello,
Thank you for looking at my message.

It has the default value:

ADAPTATION='CYCLAGE'

Attached comm, med and mess.

Konyaro


Attachments:
ctc_et.zip, Size: 76.75 KiB, Downloads: 80

失敗は成功のもと (L'échec est la base de la réussite)

Offline

#7 2020-04-19 09:01:37

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

paraview - point over time has output results as attached in zip.
you can check.
using ubuntu 18.04 with astk 14.4

the graphs seems different wave in windows & ubuntu


Attachments:
zip.zip, Size: 144.69 KiB, Downloads: 75

Offline

#8 2020-04-19 18:50:57

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

Mr. Mib,

Saw your youtube video's, 1 question in contact analysis.

How you decide on values of,

COEF_PENA_CONT=
    COEF_PENA_FROT=

Any specific theory formula explanation.

Offline

#9 2020-04-19 22:45:22

mib
Member
Registered: 2019-11-10
Posts: 85

Re: Friction force depending on the timestep

Sameer,

see attachment.

:-)


Attachments:
values.png, Size: 123.74 KiB, Downloads: 88

Offline

#10 2020-04-20 07:16:53

konyaro
Member
From: Genève
Registered: 2016-04-02
Posts: 255

Re: Friction force depending on the timestep

@mib : thank you for your advices. I made a few tests, normal force is always ok, only the friction force gets reduced, even for small deformations.

With ALGO_FROT='STANDARD' and ALGO_CONT='PENALISATION', the friction force is constant for any timesteps. I will try it on complex models to see if convergence is ok.

With ADAPTATION='ADAPT_COEF' convergence is hard to reach, so difficult to use with complex models I guess.

Nevertheless I remain surprised of the dependence of friction to timestep with ALGO_FROT='PENALISATION'.

Konyaro

Last edited by konyaro (2020-04-20 07:21:03)


失敗は成功のもと (L'échec est la base de la réussite)

Offline

#11 2020-04-20 07:59:43

mib
Member
Registered: 2019-11-10
Posts: 85

Re: Friction force depending on the timestep

@Konyaro

in the version which i use
ALGO_FROT='PENALISATION
is the only choice available

usually i use paravis for understand if we are out transitory state

:-)

see attachment


Attachments:
time.png, Size: 305.38 KiB, Downloads: 58

Offline

#12 2020-04-20 08:05:52

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

Mr. Mib,

Your Below line ,

"for the fix plate try to impose DX=0, DY=0 DZ=0 (like fixed) to contact surface."

fixed the contact surface,,,,why?. you can fix the bottom of plate or side of plate.

Offline

#13 2020-04-20 08:21:35

mib
Member
Registered: 2019-11-10
Posts: 85

Re: Friction force depending on the timestep

@Sameer

this face (group of element and node after mesh) see attachment

so you can exclude (only for test)  the local deformation of the plate.

:-)


Attachments:
fixed.png, Size: 144.72 KiB, Downloads: 65

Offline

#14 2020-04-20 08:39:37

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

Got your point, so no vibration induced due to contact and smooth reaction curve. Appreciable.

Offline

#15 2020-04-20 09:08:40

mib
Member
Registered: 2019-11-10
Posts: 85

Re: Friction force depending on the timestep

Exactly Sameer

so you can go near to ideal state, excluding local acceleration and local inertia phenomenon from surface deformation which changes everytime the friction cone (i think).

See attachment ( in Italian , translate with google :-))


Attachments:
train wheel.png, Size: 307.92 KiB, Downloads: 60

Offline

#16 2020-04-20 09:25:50

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

@mib

COEF_PENA_CONT=
    COEF_PENA_FROT=

Once, we keep on increasing above parameters , the convergence is diffcult.

What above 2 parameters define in terms of Output Results, or say at lower values--it is easy to converge and got x result....but if running at high values taking more time to converge...how output result will change.

and as we face more difficulty in convergence, increase timestep is only solution. e.g 0.01 to 0.001 ...

Last edited by sameer21101970 (2020-04-20 10:08:53)

Offline

#17 2020-04-20 11:50:35

mib
Member
Registered: 2019-11-10
Posts: 85

Re: Friction force depending on the timestep

@Sameer

i do the same (on manual mode)

for example :

initial time step set : 0.01;0.02;0.03;0.04;0.05
run and at 0.03 the solver goes in time cut and after don't converge

i modify the time step set like this : 0.01;0.02;0.022;0.024;0.026;0.028;0.03;0.04;0.05
run the solver and i obtain the convergence

during temporal window from 0.02 to 0.03 i observe in paravis which the cursor scale of DPL goes to left and after 0.03 to 0.05 goes totally to right.

P.S. in my application i'm interested to final deformed geometry.

:-)

Offline

#18 2020-04-21 08:08:31

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

@mib,

You are right, by fixing the Contact Face ...The REaction curve smoothen...attached photo of difference


Attachments:
Screenshot from 2020-04-21 12-28-05.png, Size: 26.67 KiB, Downloads: 66

Offline

#19 2020-04-21 08:11:08

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

photo.
contact face boundary not fixed


Attachments:
Screenshot from 2020-04-19 13-18-22.png, Size: 45.9 KiB, Downloads: 64

Offline

#20 2020-05-12 06:39:53

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

@mib @konyaro

"
"i modify the time step set like this : 0.01;0.02;0.022;0.024;0.026;0.028;0.03;0.04;0.05
run the solver and i obtain the convergence

during temporal window from 0.02 to 0.03 i observe in paravis which the cursor scale of DPL goes to left and after 0.03 to 0.05 goes totally to right."

On above message of MIB.

When i am testing 2 Gears in contact, The Centre Distance Between Gears Shifts. As we fix Single Axis Say DZ,,,but DY,DX are free for movement. if i fix DY & / or DX Gear stops rotation.

To be Realistic Simulation, I have to Apply Torque in Reverse to Rotation to Driven Gear, to Obtain Reaction Load Realistic.
How this can be done. Any Hints.

Offline

#21 2020-05-12 07:52:42

mib
Member
Registered: 2019-11-10
Posts: 85

Re: Friction force depending on the timestep

there will be a tab on the driven wheel
on the slot of the tongue on the wheel, apply pressure that produces the resistant torque

:-)

Offline

#22 2020-05-12 08:06:09

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

I did not get very clear,

There is a extra say rectangle object on Driven Gear Which will be given pressure against direction of rotation of driving gear.

a small screen shot can you send

Offline

#23 2020-05-12 10:48:57

mib
Member
Registered: 2019-11-10
Posts: 85

Re: Friction force depending on the timestep

see attachment  :-)


Attachments:
screen shot.jpg, Size: 237.38 KiB, Downloads: 55

Offline

#24 2020-05-12 12:42:21

sameer21101970
Member
Registered: 2019-09-06
Posts: 263

Re: Friction force depending on the timestep

@mib


Attachments:
Compound_Mesh_1.med, Size: 635.58 KiB, Downloads: 66

Offline

#25 2020-05-12 14:39:00

mib
Member
Registered: 2019-11-10
Posts: 85

Re: Friction force depending on the timestep

@ sameer

to test the tooth, the direction is that shown in the last figure

Salome don't open your file give me back an error message

probably you have free DOF

when i tried Worm without results i gone for the way in pictures (see attachment) :-)

also for me to apply a torque is a mystery,

normally i use this concept : M=F*a

M moment
F force
a arm

Last edited by mib (2020-05-12 14:49:37)


Attachments:
worm_gear.zip, Size: 445.71 KiB, Downloads: 58

Offline