How can I define material other than acrylic,steel, etc... in motion analysis contact? For example I want to have contact between 2 hdpe material, how can I define this contact?

How can I define material other than acrylic,steel, etc... in motion analysis contact? For example I want to have contact between 2 hdpe material, how can I define this contact?

I know the coefficients of friction between the material I'm using, but not the stiffness,exponent, damping etc... how can I run this? I do not know how to obtain all these parameters

1. Search on Google.

2. Check with the manufacturer of the material. They should be able to provide this number for you

3. You can also try Matereality.com(Need subscription)

If you have SW Simulation Premium/Professional, Add Simulation Add-in, Open any part file > RMB on material > Edit MaterialIn Material dialog, left bottom corner, Click "here". This will launch Matereality.com

**Stiffness**

It is the parameter used to

define the compression that occurs when parts contact each other. This is the

equivalent of spring stiffness where the force generated is equal to the

stiffness times the penetration between the parts. The other aspect of this

value is that the stiffness is a function of geometry.

For example,

theoretically, for a steel rod, the stiffness k = A*E / LA is the cross-sectional area

E is the Young's

Modulus

L is the effective length which is free to deform**E****xponent**It is used for the proportionality of Force with penetration. F=k*x^e, where

e is the exponent. You can therefore have non-linear relationships which is

especially important for incompressible materials. For something like Rubber,

use an exponent of 2, maybe even 3. For metals, a value between 1.3 and 1.5 is

normally used.

- When you choose k and e to describe a curve of force vs.

displacement, remember that the units you are using affect curve stiffness.

Because one millimeter is much smaller than one inch, the same k and e create a

stiffer spring in a Newton-millimeter system than they create in a pound-inch

system. - Evaluate k to determine if it is too high for the defined

problem. - Avoid using values of e that are less than or equal to one. These

curves have a slope discontinuity at x=0. For e < 1, the slope is infinity at

x=0.

**Max Damping**This term (c) indicates the

loss of energy that occurs between parts when they collide. This is the maximum

damping applied. Normally, damping is between 0.1 and 1% of the stiffness value.

**Penetration Distance**This is actually the depth at which maximum damping occurs. When the parts first touch, there is no

damping force, but then at they deform or penetrate, the damping force increases

until at a certain depth it will be applying the maximum damping force

Make sure you don't have d so small that is causing discontinuities or so large

that the damping is never turning on

- When you choose k and e to describe a curve of force vs.
The stiffness value above is based on the blocks of my design? and what E is taken? for what material?

For plastic, e =2 is logical?

Thank you!

Is it possible if I apply a force with several values on a block, and find the maximum displacement with each force, and then curve fit the force vs displacement to find the stiffness and the exponent?

Yes, you can setup a No Penetration contact problem of 2 blocks with the appropriate materials. Measure the contact force

and the displacement at the interacting faces. Then, take an average stiffness = force/displacement and use that in the Impact model.

I tried finding stiffness myself, as well as other properties, but the first 2.25 seconds gave the force attached in this message, which doesn't seem very logical. It is the same problem of dropping the wall. I didnt understand the drop in the force. I stopped the simulation before finishing, and before breaking the wall.

You can try increasing Frame Per second.

Also you can share files with your setup. I will see.

Thanks,

Amit

Hello, attached is the file I am testing now. I applied a force when blocks are in contact and found the contact force then divided it the displacement at the contact, for several forces and displacements i got a linear result with the stiffness given and the exponent to be 1. I tried to do the same thing with the steel rail and the plastic block, but applying a force gave a very small contact displacement, which gives a high stiffness. I took the stiffness for steel. I tried to run the study, it took 3 days to reach 2 seconds then after i stopped it it reached about 20000N. These are plastic blocks mounted on top of each other and on a steel rail bolted to the ground, how can 20000 newtons not even tilt the wall? I find something not logical and I cannot figure out why.

Regards and hope you can help.

Hi,

To define contact material other that standard materials, you have to uncheck "Materials" check box and type all properties manually.

see below image:

Regards,

Amit