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coining formula
- Thread starter bachelier
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As a "mechanical", I'm right there with you, Nick, but I think you over-simplified. You basically defined elastic shear failure, but the OP's question, I think, deals more with viscoplasticity. I don't know the answer, and would be willing to start with AxFc, but certainly wouldn't design a roller or press based on it.
The problem here is that OP has not provided any sort of information. I wrote my answer the way I did for that reason. Two of the parts we make in my plant have a coined feature. They are both .120" SAE J403: 1008/1010 material. There woudl be a huge difference in the forces from that part to one I designed last year that was SAE J404:4130.
I based my answer off of the basic need for greater than yield strength stresses to casue material to permanently deform.
I based my answer off of the basic need for greater than yield strength stresses to casue material to permanently deform.
No, those specs I quoted are materials specifications, I could have just as easily said ASTM A1010:1008/1010 or ASTM A507-00:4130.....
Or 1008/1010 and 4130...
I've been trying to get out of the habit of expecting people to understand things like "ten oh eight ten ten"
Or 1008/1010 and 4130...
I've been trying to get out of the habit of expecting people to understand things like "ten oh eight ten ten"
Chapter 17, Indentation Processes, of Handbook of Metal Forming (Kurt Lange editor, published by SME in 1985) has some good information on this topic. One estimate of the forming force is given by the following equation:
F = 4.5[σ]fAp
where [σ]f is the flow stress and Ap is the cross-sectional area of the punch
Flow stress obviously depends on the material & thermomechanical processing history. The stated accuracy of this method is ~ + 10%.
F = 4.5[σ]fAp
where [σ]f is the flow stress and Ap is the cross-sectional area of the punch
Flow stress obviously depends on the material & thermomechanical processing history. The stated accuracy of this method is ~ + 10%.
Why not regard this as a shear? In that case, shear is simply force over area as previously mentioned, but shear stress is 0.577 X material yield. (i.e. Energy Distortion Theory).
In practice, I found more ductile metals such as aluminum, bronze, brass, copper, etc to behave like 0.52 X yield. This idea is used extensively in the oil industry for shear load computations on knock out plugs, etc.
Works very well, well within 5% error theory verses practice.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
In practice, I found more ductile metals such as aluminum, bronze, brass, copper, etc to behave like 0.52 X yield. This idea is used extensively in the oil industry for shear load computations on knock out plugs, etc.
Works very well, well within 5% error theory verses practice.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
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