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Force imparted on pins form shrinkage

Force imparted on pins form shrinkage

Force imparted on pins form shrinkage

(OP)
Hi, I am a little stumped one trying to calculate the force acting on three pins which are in an aluminium die casting mold.

basicly I want to work out what force is acting on these pins due to the shrinkage of the molded component material IE aluminium 1.3% shrinkage about pin PCD cenrer point, is there a reasnobly simple way to calculate the force imparted from the shrinkage onto the pins?

Any help is much appriciated.

RE: Force imparted on pins form shrinkage

That depends on way that the aluminum freezes and cools in the mold. You can easily form shrinkage voids in an aluminum casting. if the pin is in that location there will be no pressure on the pin. If you mean load between the pin that will vary with depth from the surface.

RE: Force imparted on pins form shrinkage

(OP)
the pins are part of the die itself, so I assume the aluminium would form around the die and cool and therefore shrink which as the part is central about the pins PCD then the shrinkage would act to force the pins inwards to the center of the pcd causing shear and bending on the pins, I just want to try come up with a rough idea of the shear force on the pins ideally.

RE: Force imparted on pins form shrinkage

The below link looks much scarier than it really is, but investigate "stress in thick walled cylinders"

http://www.roymech.co.uk/Useful_Tables/Mechanics/Cylinders.html

This link is less scary and has some calculators:
http://www.engineeringtoolbox.com/stress-thick-walled-tube-d_949.html

-Dustin
Professional Engineer
Certified SolidWorks Professional
Certified COSMOSWorks Designer Specialist
Certified SolidWorks Advanced Sheet Metal Specialist
 

RE: Force imparted on pins form shrinkage

The forces on the pins will be a function of the stresses/strains induced by thermal expansion.  But the stresses are dependent on highly-temperature-dependent material properties.  Young's modulus and yield strength are very low at elevated temperatures.  

Your solution will only be as good as the material properties data that you use.  You might find some interesting titles at this website:

http://www.aluminum.org/Content/NavigationMenu/BookStore/BestSellers/default.htm




 

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