Flatting Round Stock 1

[Turn488]

How do you determine the geometry of a round bar if it has been flatten in a punch press? The area of the resulting part should be equal to the area of the original round bar stock. If you know the height of the flatten portion, can you determine the width of the flatten area and the resulting radii on the sides? I figure there has to be a formula that relates the two areas.

 

[FOETS]

Unless the flattening has also lengthened the rod thereby reducing the cross sectional area

 

[Turn488]

I would assume that bar would push out significantly more than it would extend. I only need an approximation, so I would assume that amount it extends is negligible.

 

[rb1957]

theoretically, as you say area1 = area2
area1 = pi*d*t
area2 = W*2t+pi*t^2
so W = (pi*d*t-pi*t^2)/2t
W = pi*(d-t)/2

practically you need tobe careful squeezing the tube, to avoid cracking on the bend radius; there are several ways of doing this.

 

[MintJulep]

The area is NOT constant.

Get out your materials text book and lookup "Poison's ratio".

 

[rb1957]

i'm sure you're right mintjulep,
but i think it's a second (or third) order effect ...

i've got a terminal lug on my desk, and to the accuracy of a scale in 1/64" the equation above works.

 

[deanc]

http://mathworld.wolfram.com/Flattening.html

Handy site.

 

[Turn488]

I believe the tip given by "rb1957" is for a tube, not a solid round bar. How do you relate this formula to a solid bar?

Thanks rb1957 for your input!

 

[rb1957]

yes, i was thinking of a tube.

if it's a rod then
area1 = pi*d^2/4
area2 = Wt+pi*t^2/4 (t now is the thickness of your flattened bar)
W = pi/4*(d^2-t^2)/t

neglecting 2nd order effects

 

[rb1957]

btw, doesn't poisson effect only impose a constant volume relationship ... if something strains in one direction, constant volume means that it has to shrink in the other ?

what will happen to the rod, is that that as it is being flattened, the end will bulge, (an effect not particularly significant with a tube) so the post above is not quite right ... instead of considering cross sectional area you need a constant volume approach. unfortunately this is alot more complex

 

[MintJulep]

It's not constant volume. This is a sophmore engineering problem. Block of metal a x b x c gets squashed, what are the new dimensions in the other axis?

All the students go for constant volume, except for the few that read the book and apply poison.

 

[CoryPad]

Poisson's Ratio applies to elastic deformation. For plastic deformation, you do use volume constancy.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[GregLocock]

Otherwise you could make dense lead by hitting normal lead with a hammer. It doesn't work to any useful extent.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[MintJulep]

Ok, maybe I need to dig up the old books.

 

[GregLocock]

You are right for elastic deformation , if nu is not 0.5 then there is a volume change in response to forces. One way of thinking about it is that the elastic energy that is stored is by deforming all the little springs in the crystals. That implies a change in dimensions, and so density. When we plastically deform something the crystals are all relaxed again, so there are no strains, and no stored energy, and no change in volume.

Is that an awful analogy?

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[CoryPad]

The spring analogy is spot on. The metallic bonds are stretched, interatomic spacing increases, density decreases.

Upon release of the elastic strain (either during elastic-only deformation or after elastic + plastic strain), the bonds relax, interatomic spacing decreases, density increases to its original value. Hence, volume constancy is an acceptable assumption.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.