Posted:
5 years ago
Feb 25, 2013, 2:43 PM EST

Hi

are you integrating the volume to see it change or the length of the deformed edges/bundaries ? In that caase I believe you need to turn on the Solver - "non linear geometries"

Bit the deformation displacements remain u,v,w

Do not forget the difference of the frames in "solid" between lower and upper case x,y,z and X,Y,Z

--

Good luck

Ivar

Hi
are you integrating the volume to see it change or the length of the deformed edges/bundaries ? In that caase I believe you need to turn on the Solver - "non linear geometries"
Bit the deformation displacements remain u,v,w
Do not forget the difference of the frames in "solid" between lower and upper case x,y,z and X,Y,Z
--
Good luck
Ivar

Posted:
5 years ago
Feb 25, 2013, 3:25 PM EST

Dear Ivar

Thanks for your quik reply. Very grateful to you.

But still my problem is unsolved. Let it put in a different way:

I want either of following things done:

1. Want the co ordinates of the profile , so that i can calculate the area based on that.

or

2. Want the area of the deformed body.

Can you help me in this.

Dear Ivar
Thanks for your quik reply. Very grateful to you.
But still my problem is unsolved. Let it put in a different way:
I want either of following things done:
1. Want the co ordinates of the profile , so that i can calculate the area based on that.
or
2. Want the area of the deformed body.
Can you help me in this.

Posted:
5 years ago
Feb 26, 2013, 2:00 AM EST

Hi

but you can integrate the volume or the surfaces in COMSOL by a Derived Value with an operand of "1"

But you must be sure you define the correct frame material or spatial, and that you use the correct settings of the geometrical non linearity.

try it out on a simple model:

i.e. make a 3D Solid Stationary model with a geometry of a unit cube, add a material and use some dummy value E=1E6, nu=0.3, rho = 1 so long they are physical it doesn't matter

Fix the bottom surface, apply a prescribed displacement of Z=-1[cm] to the top surface and solve

Go to the Derived Valus enter a integration Volume, apply to all domain (there is only 1 you can also select it), replace the Expression by "1" and solve, you get 1 the initial volume, as by default your results are on the material frame

Then Go to the Data Set , select Solution 1 and Duplicate and select the Spatial Frame for Solution 2 (it's just a copy of "1"), then go back to the Derived Values and Duplicate your integration Have it pointing to Data-Set Solution 2 and solve, you get again "1" as the results, as for the linear geometry case there is NO change in volume.

Now add a new stationary study to your model, Select this time in the Study 2 - Step 1 Stationary node Include Geometry Non-linearities, and Solve (this takes a bit longer to solve ;)

Duplicate your first "Volume integration 1" and point it to the new Solution 3 case and calculate, you get again "1" in the Material frame, this is normal,

Now Duplicate Data Set Solution 3 and make a "4" that you point to the Spatial Frame of the results, duplicate your Integration, have it pointing to Solution 4 and calculate: NOW you get 0.99539.

but perhaps you expected 0.99 since we compressed it by 1 cm for a 1 m length. So what is this ?

Try to set the material Poisson coefficient nu = 0 and solve Study 2 again Then recalculate (Evaluate icon) the last Volume integration, now you get 0.99.

Then what if nu =0.5 => this is not physical, so you could try 0.495 now you get 0.99971 which is almost "1"

Do not forget to compare the stress levels for the 3 cases too

If you still want to have fun:

a) what about the boundary areas ? try the same ...

b) what about the fixed side, it forces the area to 1 and does not allow for any natural expansion, this has some nasty effects on the stress at this fixed interface. Often your true model is not like that, try to set up some boundary conditions such that you get less residual stress but still have a fixed surface (hint you can play with symmetry or with roller conditions and some fixed points or ...)

--

Have fun COMSOLing

Ivar

Hi
but you can integrate the volume or the surfaces in COMSOL by a Derived Value with an operand of "1"
But you must be sure you define the correct frame material or spatial, and that you use the correct settings of the geometrical non linearity.
try it out on a simple model:
i.e. make a 3D Solid Stationary model with a geometry of a unit cube, add a material and use some dummy value E=1E6, nu=0.3, rho = 1 so long they are physical it doesn't matter
Fix the bottom surface, apply a prescribed displacement of Z=-1[cm] to the top surface and solve
Go to the Derived Valus enter a integration Volume, apply to all domain (there is only 1 you can also select it), replace the Expression by "1" and solve, you get 1 the initial volume, as by default your results are on the material frame
Then Go to the Data Set , select Solution 1 and Duplicate and select the Spatial Frame for Solution 2 (it's just a copy of "1"), then go back to the Derived Values and Duplicate your integration Have it pointing to Data-Set Solution 2 and solve, you get again "1" as the results, as for the linear geometry case there is NO change in volume.
Now add a new stationary study to your model, Select this time in the Study 2 - Step 1 Stationary node Include Geometry Non-linearities, and Solve (this takes a bit longer to solve ;)
Duplicate your first "Volume integration 1" and point it to the new Solution 3 case and calculate, you get again "1" in the Material frame, this is normal,
Now Duplicate Data Set Solution 3 and make a "4" that you point to the Spatial Frame of the results, duplicate your Integration, have it pointing to Solution 4 and calculate: NOW you get 0.99539.
but perhaps you expected 0.99 since we compressed it by 1 cm for a 1 m length. So what is this ?
Try to set the material Poisson coefficient nu = 0 and solve Study 2 again Then recalculate (Evaluate icon) the last Volume integration, now you get 0.99.
Then what if nu =0.5 => this is not physical, so you could try 0.495 now you get 0.99971 which is almost "1"
Do not forget to compare the stress levels for the 3 cases too
If you still want to have fun:
a) what about the boundary areas ? try the same ...
b) what about the fixed side, it forces the area to 1 and does not allow for any natural expansion, this has some nasty effects on the stress at this fixed interface. Often your true model is not like that, try to set up some boundary conditions such that you get less residual stress but still have a fixed surface (hint you can play with symmetry or with roller conditions and some fixed points or ...)
--
Have fun COMSOLing
Ivar

Posted:
5 years ago
Feb 27, 2013, 1:58 PM EST

Thanks Ivan for your great help.

Thanks Ivan for your great help.