Open die forging, von Mises related questions

[Twigg]

TL;DR Summary

Hung up on two steps in a derivation of mechanics in open die forging. First question is a supposedly von Mises stress that looks to me like a Tresca stress. Second question is a confusing boundary condition.

Hey folks!

I had a hard time finding accessible resources on open die forging. If you have a better source, that'd be just as good as an answer to these questions. I've been following this slideshow, and my questions are about steps in this derivation I don't follow.

First up, one page 6 "Force balance in the x-direction", I don't follow the last step. It says it's the von Mises yielding criterion, but to me $\sigma_x + p$ looks like the Tresca criterion. I would've expected $\frac{1}{\sqrt{3}}[\sigma_{x}^2 + \sigma_x p + p^2]^{1/2}$ for von Mises. Am I missing something? I have a funny feeling it has something to do with how the author is working in 2D but I can't put a finger on it.

Second question is about the boundary condition on page 8 "Sliding region". The author explicitly says at $x=0$, that $\sigma_x = 2k$. But in the lower limit of integration on the left hand side integral, he says at $x=0$, that $p = 2k$. They can't both be true because per the yield criterion on page 6, $\sigma_x + p = 2k$. What gives? My understanding is that the stress won't exceed the yield criterion because if the material temporarily exceeds the yield point it will squish and flow around until the stress returns to the yield point. Is that right? Additionally I don't feel like I understand where this boundary condition $\sigma_x = 2k$ at $x=0$ comes from. Can anyone fill me in?

Thanks all!

 

[Twigg]

Ok, I answered the first of my two questions.

I was confused by the equation given for the yield condition: $$\sigma_x + p = 2k = \frac{2\sigma_y}{\sqrt{3}}$$ The text claimed it was a von Mises criterion, I thought it looked like a Tresca criterion, but that was just me not knowing the difference between $k$ and $\sigma_y$. From what I've learned, $\sigma_y$ is the actual yield strength, whereas $k = \sigma_y / \sqrt{3}$ is the von Mises shear strength. My understanding is that when the maximum shear stress in the material equals $k$, then the material will yield per the von Mises criterion. That's why it looks like a Tresca criterion, with the important caveat that the right hand side of the equation is $k$ not $\sigma_y$. Can anyone verify?

For now, still stumped on my second question.

 

[Twigg]

I think I've made a little more progress. At least I think I understand why the boundary condition $\sigma_x = 2k$ @ $x=0$ exists, but it doesn't explain why the author uses $p=2k$ @ $x=0$ in the lower limit of integration.

I threw in a little quick diagram to help illustrate what I think is going on. As the dies close in on the workpiece, the edges of the workpiece deform by bulging outwards. Where this bulge starts, the dies no longer touch the workpiece, so there is no pressure by the dies on the workpiece at this point, so p=0 at this point. I exagerrated the bulge in my little sketch, but my gut tells me there has to be some bulge. I can do this experiment with playdough and the playdough would bulge, so it shouldn't be qualitatively different for steel?

If my "bulging" hypothesis is correct, then since the material is yielding, $\sigma_x + 0 = 2k$ gives $\sigma_x = 2k$ at x=0 where the bulge starts. However, that conclusion contradicts the author's statement that $p=2k$ at $x=0$. What gives?? I don't think $\int \frac{dp}{p}$ is even integrable from p=0, so I feel like I'm missing something.

 

[Twigg]

Nope, nevermind my last reply, it's totally wrong. I skimmed through the slideshow to check the author's numerical results and see if p=0 at the edges, and he explicitly says $p_{edge} = 2k$ and the numerics have non-zero edge pressure.

So apparently there were typos in the slideshow, and what the author meant to say was $\sigma_x = 0$ and $p = 2k$ @ $x=0$. Only remaining question for me is where that comes from. Why is there no stress in the x-axis on the edge of the material? If it's plastically deforming in that direction, doesn't that mean there has to be stress along the x axis??

 

[Twigg]

Ok, last post I think. I think I understand why $\sigma_x = 0$ at the edge of the workpiece. Internal stresses can only push or pull on the edge layer from the inside. There's nothing outside the edge to apply any balancing forces. If there was an unbalanced force on the edge layer by the inside material, it would flow/yield until the force vanished. As a consequence, the edge layer is in uniaxial compression under the forging dies.