Which one is more work-hardened than the other?

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TL;DR
The effect on number of cold drawing to work-hardening.
Hi.

I wonder which one is more work-hardened in cold drawing process.

① φ7 -> φ6.5
② φ7 -> φ6.8 -> φ6.5

Final diameter is same, but the number of cold drawing is different.
In my guess, ② is more work-hardened even thought I can't explain the reason exactly.

If you know the answer and the mechanism, I hope you tell me about that.
 
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When a strain hardening material is loaded beyond yield, the stress reduced to zero, then again loaded beyond yield, the result is as shown below. The first loading takes it to Point A, then unloading takes to Point B. The material has yielded. On further loading, the material follows the dashed line back to Point A, after which it yields some more.
Plastic Strain.jpg

That is for simple tensile loading. A cold drawing process involves a more complex stress-strain path that complicates matters. Answering your question requires finding the exact strain path as the part goes through the cold drawing process. The image below shows a part that was cold drawn in two halves. Each half had lines scribed before drawing. After drawing, the two halves were separated. The lines are no longer straight. I believe this means that the actual strain path is more complex than that of a simple tensile test, which would imply (but does not prove) that drawing in two stages might cause more work hardening than drawing in one stage.
Cold Draw.jpg


I cannot explain it exactly either, but I think you are right. Proving it would require studying the strain path through the cold drawing process.
 
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