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bcockrell1-Visitor
1-Visitor
May 29, 2019
Question

shrink

  • May 29, 2019
  • 3 replies
  • 1785 views

I add .0208 shrink to the model and the mass properties are the same as with no shrink?

there should be a difference between the two?

because the .0208 shink on the model is bigger?

I doing this because I work in a steel foundry so the pattern must be bigger to allowed the steel to shrink to a common dimension.

thank you 

3 replies

Patriot_1776
Patriot_177622-Sapphire II
22-Sapphire II
May 30, 2019

Never played with that before, but it MIGHT have to do with your accuracy.  Try setting it to absolute, and a value of, say, .0001.  It may be a rounding error.

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pausob19-Tanzanite
19-Tanzanite
June 2, 2019

Well how does it work in real world?  Does the material that is injected into the mold stay at the same density after it's cooled down? Mass before and after "shrinking" should surely be the same?

 

But what if the shrinkage feature is used to account for dimensional changes due to some evaporative process?

 

Patriot_1776
Patriot_177622-Sapphire II
22-Sapphire II
June 4, 2019

In plastic injection molding, there is shrinkage and theoretically you lose some negligable mass as the plastic outgasses.  I'd imagine that would be so small as to be almost immeasurable though.  Also, if the mass stayed the same and the volume got smaller, the density would INCREASE, right?  Although, with plastics, you tend to get slight density (and strength) variations due to the material flow into the mold.

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pimm15-Moonstone
15-Moonstone
June 4, 2019

Bcockrell:

 

I don't understand why your weight would not update to the larger value.

 

I work in the Forging industry so we also apply shrink value to our die models.

 

I just did a rough test of applying shrink to a part model and it does update the weight value.

 

I am assuming that you have the mold package and are applying shrink from this by going to File/Prepare/Model Properties and shrinkage.

 

The only possibility that comes to mind is that somehow the shrink did not truly apply.

 

It is an interesting idea in regards to the fact that in reality the weight of the part would be the same with the larger shrink when the temperature is greatly elevated, but I haven't messed around with material properties and temperature to see how that works.

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