Coining/axially-symmetric compression

[1350-F]

Homework Statement

You are asked to figure out the force required to coin a 25-cent piece and are given the final dimensions and an average flow stress. Sticking friction is "reasonable"

Hosford and Caddell 2nd Ed. Q 7-3

Homework Equations

P_a = Y + 2kR₀/3h₀

The Attempt at a Solution

I have Y. I can figure out k since k=0.577Y. I have R and h. Not R₀ and h₀

Can you do this without the initial workpiece dimensions? As long as your workpiece has the same volume as the coin, you could start with any height to radius ratio you like, however different ratios would require different pressures to flow. Are we supposed to guess at the original dimensions? I suppose to avoid barrelling we'd want the ratio of h to r to be small. In addition, without the original dimensions, I can't find an area to calculate force from. I tried to constrain my geometry in terms of strain and I get

F_a =AYexp(ε) + 2kAR exp(2.5ε)/3h.

Where ε is ε-bar or ln(A₀/A).

I could get the effective strain from a flow law but I'm only given the average yield stress.

Perhaps I'm missing a useful approximation here...

 

[Bystander]

different ratios would require different pressures to flow.

How do those pressures change from beginning to end of "strike?"

 

[1350-F]

How do those pressures change from beginning to end of "strike?"

Good Point!

The pressure to overcome friction would increase up until the end of the "strike," when R/h is greatest. I suppose in that case I can just use the final geometry. Seems intuitive I guess, but every problem I've encountered so far uses the initial geometry.

 

[Bystander]

Seems intuitive I guess

I kept looking for pitfalls, and couldn't find any. No guarantee there aren't.

 

[1350-F]

I kept looking for pitfalls, and couldn't find any. No guarantee there aren't.

If we do it that way

P = 25ksi + 1.154*25ksi(0.95/3*0.060) ~180 ksi
F = 180ksi * pi * 0.95^2 = 500 000 lb = 250 tons

Seems like a lot for a little coin

Thanks for your help!