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I'm used to calculate the minimum radius required for bending operations on both aluminum and steel sheets, given the following parameters: https://i.sstatic.net/yxEBc.jpg

Problem statement:

For a sheet metal stock with inch thickness, determine the minimum tool radius for both the steel and aluminum alloys that will not tear the material. Assume the sheet of material is in pure bending (i.e., no additional tension is applied during forming).

I have already learned about the following formulas:

  • R = 1/k
  • k(y) = e(y) / (h/2)
  • e(y) = k.y

Given that R = 1/k, the next step is to find the maximum curvature k for each sheet.

The problem is that I don't know how to find the maximum curvature, and I'm still stuck.

Any thoughts?

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    $\begingroup$ Is there an error in your equations? Does R depend on little k, or big K? What are y and h? My guess is that y is distance from neutral plane and h is total wall thickness. If I have this all correct in my head, your equations are monotonic in all the variables, so there is no maximum. Are you perhaps missing a constraining equation that relates radius to stress in the material? $\endgroup$
    – do-the-thing-please
    Commented Nov 6, 2017 at 18:04
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    $\begingroup$ @starrise There is no big K, sorry but I meant 'k' the curvature. y = distance from the neutral axis to the top of the sheet cross-section. h = thickness of the sheet. $\endgroup$
    – user6039980
    Commented Nov 6, 2017 at 21:03
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    $\begingroup$ @starrise "Are you perhaps missing a constraining equation that relates radius to stress in the material?" - There is no constraining equation specific to the problem. But I'm pretty sure that there is a solution for finding the minimalist radius. $\endgroup$
    – user6039980
    Commented Nov 6, 2017 at 21:28
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    $\begingroup$ @starrise I updated the question in order to provide more information about the problem, please check it out. $\endgroup$
    – user6039980
    Commented Nov 6, 2017 at 21:42
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    $\begingroup$ Thanks for the update. I think that there is a constraint, but based on the answer from Derkooh it is empirically derived. I suppose what I was getting at is that the tighter the radius, the greater the plastic strain. If the strain exceeds the failure strain the material ruptures. So stress/strain must be related to this in some way. $\endgroup$
    – do-the-thing-please
    Commented Nov 6, 2017 at 22:36

1 Answer 1

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I hope the following helps. This is from a book I used in college. Manufacturing Engineering and Technology, 5th Ed. by Kalpakjian and Schmid

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Minimum Bend Radius

Where **r** is the tensile reduction of area.

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    $\begingroup$ I highlighted the min bend radius equation $\endgroup$
    – user13416
    Commented Nov 6, 2017 at 23:23
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    $\begingroup$ r is tensile reduction of area, and is approximated through the chart in Fig. 16.18.. I also added Fig 16.16 to the derivation. $\endgroup$
    – user13416
    Commented Nov 7, 2017 at 1:07
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    $\begingroup$ R/T is the y-axis, and r (tensile reduction of area) is the x-axis. $\endgroup$
    – user13416
    Commented Nov 7, 2017 at 1:55
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    $\begingroup$ R/T is the y-axis, and r (tensile reduction of area) is the x-axis. Minimum bend radius R is a function of material thickness and material, and is empirically defined from table 16.3. Eq. 16.5 can be solved for r as follows: r = 50 / (R/T +1) $\endgroup$
    – user13416
    Commented Nov 7, 2017 at 2:02
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    $\begingroup$ Still available on Amazon : a good investment? amazon.ca/s/… $\endgroup$
    – Solar Mike
    Commented Nov 7, 2017 at 6:42

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