Model deformed rubber parts using the volume of the original part as constraint

[frank.gavilar]

Current work-flow:

My company is in the Gas Regulator business and we use a lot of NBR rubber parts. Think of O-Rings, Gaskets and Membranes. These are often modeled to represent the shape they have during their production (i.e. injection molding). The though thing is these parts often have a different shape inside the final assembly. It is up to the engineer to imagine/calculate how this part will deform in the assembly. The engineer will create a second .ipt file where the part is deformed with the exact same iProperties but a different file name. It is very hard to keep the volume (mm3) of the part the same whilst deforming it. 

The Problem:

Though it is possible to work it would be of great help if we where able to deform the original using the original's volume as a constraint. Currently volume is only a end-state iProperty, calculated after modeling the part. 

[frank.gavilar]

Recently had a colleague draw up a more detailed description of this issue and why is hard to solve. I hope this clarifies this issue a bit more. Should be a nice challenge for the Autodesk team at Cambridge.

Lianne Mostert Wrote:

Changing dimensions at a constant volume for curved objects

For dimensional purposes it maybe needed to calculate the dimensions of a curved object in different circumstances.

Take for instance an o-ring seal. The o-ring is made of a string with radius r₁, the radius of the center of the o-ring itself to the middle of the string is R₁. The internal radius of the o-ring, P₁, thus is P₁=R₁-r₁.

When the o-ring is at rest, the volume I₁ which is 2π²r₁²R₁.

When however the o-ring is placed around a metal axis which is just a bit wider than the inner circle of the o-ring, it will stretch. However, the volume will remain constant.

Now, we do want to know the new dimensions of the stretched o-ring, say r₂ and R₂.

We do know I₂ because I₂ = I₁ and we do know the new internal radius of the o-ring P₂, which is equal to the radius of the used metal axis. The internal radius of the o-ring is therefore stretched by δP = P₂ -P_1.

Assuming the o-ring is stretched out equally, so its form remains an o-ring, you might think R₂ is R₁ + δP. This however cannot be true as r₂ has to change, which has the effect that the position of the middle of the string changes and therefore R also. In other words: r and R are not independent of each other. Therefore we have one equation ( I₂ = 2π²r₂²R₂) with three variables of which only one (I₂) is known. This is not solvable.

Another way to handle this is to relate the volume to r and P, which are independent variables. As δP=R-r, the volume equation can be rewritten as I = 2π²r²(P+r) or I = 2π²r²P+ 2π²r³.

This equation cannot be rewritten to an equation of r as a function of P and I, so the values have to be found by iteration. This is possible of course, but it is quite an annoying job. What’s more, this is the most simple example as rubber parts often have much more complex dimensions.