# how to apply a constant velocity to a rigid wall in a crashing simulation

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Hi Thomas,

there is no other way to get a good estimation.

Execute this line in a command window, copy the batch file into the calculation directory and call it. It generates a d3kil command, which LSDyna detects and look for the switch (sw2).

You can also create the "d3kil" (only "sw2" is written into it) file manualy in the calculation directory.

You can read this in the LSDyna manual.

Regards,

Mario

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Hi Mario,

I have run a simulation with MAT20 applied to a plane in a tube crashing event. I have constrainted y and z translational and x, y and z rotational dof in the MAT20 card editor for CON1 and CON2 respectively.

But result shows me there is displacement in the plane. (see the attachment, color changed stands for large deformation as reference by the bar chart on the left) As I understand, if the wall is rigid, there should not be any displacement in the impact plane (like a concrete wall), but it is not. Why?

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Hello,

if i remember it is a rigid which is moving in x !? Then there are still displacements of course (x-direction).

Switch to Y or Z (instead of mag) and look at the displacements, they have to be zero.

Regards,

Mario

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Hi Mario,

I did checked the y and z directions and also the mat20 to make sure the same parameters applied as the tube.

These two directions are zero. If the displacement is near zero in x direction, i still can accept. What I don't understand is as follows:

1, If the impact plane has a displacement which means the impact plane deforms instead of rigid, and this deformation will reduce the impact load in contact surface between the tube and plane. As a result, the result will not be accurate.

2, The impact plane has a displacement more than 100mm which you can see from the pic. Instead, the displacement on the tube is normal (I mean accurate). Why it act like this?

Regards,

Thomas

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Hi,

i can't follow, what you mean.

The rigid (shell plane) can't deform with MAT20. So only a "rigid" displacement in x is possible (if the bc's are correct, CMO=1, CON1=5, CON2=7). This displacememt is for all nodes the same.

If you don't want a displacement, then you have to set both CON to 7, but then no movement is possible and you have to move the tube against the wall.

If the tube have bc's at the end (e.g. all dofs), then the tube deforms. It is simple...

Regards,

Mario

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Hi Mario,

I think I have mixed up the displacement and deformation. ^-^

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hi good day mario.

mario do you have an example 3 point bending simulation that using LS-Dyna.?

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