strain hardening coefficient? 1

[salmon2]

A year or so ago, I remember I saw one practical way to interpret or get strain hardening coefficient. It is like dividing tensile strength by yield strength, then minus 1, you will get the cofficient. I am give you example here to tell you what I am asking. It is something like this, but I can't remember.

Does this ring any bell to anyone? Thanks a lot. How Christmas eve dinner treat everyone? I have been super sick, didn't have anything.

 

[arunmrao]

http://en.wikipedia.org/wiki/Strain_hardening_exponent

Please follow the link and you might get some useful values for "n". Hope,it helps.

 

[salmon2]

That was informative, at one point, I thought it is exactly what I want, then it is not. Thanks a lot Arun.

 

[TERIO]

Perhaps you are thinking of Considere's criterion.

F=[σ]A
dF=d[σ]A+[σ]dA

At maximum load (ultimate strength [σ]_u) dF=0 so

d[σ]A=-[σ]dA or d[σ]/[σ]=-dA/A

For constant volume process

V=AL; dV=AdL+LdA=0
-dA/A=dL/L=d[ε]

So d[σ]/[σ]=d[ε] at ultimate load or

d[σ]/d[ε]=[σ]

If [σ]=K[ε]ⁿ
d[σ]/d[ε]=Kn[ε]^n-1

K[ε]ⁿ=Kn[ε]^n-1 -> n=[ε]

So the strain hardening exponent is equal to the true strain at ultimate load.

 

[salmon2]

Thanks a lot TERIO, well done. Very possible, but I thought it is something else. Right now I still need one more equation or relationship to close the loop.