1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Coining/axially-symmetric compression

  1. Dec 8, 2014 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

    Pa = Y + 2kR0/3h0

    3. 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

    Fa =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...
  2. jcsd
  3. Dec 9, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    How do those pressures change from beginning to end of "strike?"
  4. Dec 10, 2014 #3
    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.
  5. Dec 10, 2014 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I kept looking for pitfalls, and couldn't find any. No guarantee there aren't.
  6. Dec 10, 2014 #5
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted