2
$\begingroup$

I'm used to calculate the minimum radius required for bending operations on both aluminum and steel sheets, given the following parameters: https://imgur.com/a/7akHC

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?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – wwarriner Nov 6 '17 at 18:04
  • $\begingroup$ @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 Nov 6 '17 at 21:03
  • $\begingroup$ @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 Nov 6 '17 at 21:28
  • $\begingroup$ @starrise I updated the question in order to provide more information about the problem, please check it out. $\endgroup$ – user6039980 Nov 6 '17 at 21:42
  • 1
    $\begingroup$ 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. $\endgroup$ – wwarriner Nov 6 '17 at 22:36
2
$\begingroup$

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


enter image description here

enter image description here

enter image description here

enter image description here

enter image description here

enter image description here


Minimum Bend Radius

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

enter image description here


| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I highlighted the min bend radius equation $\endgroup$ – Derkooh Nov 6 '17 at 23:23
  • 1
    $\begingroup$ 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. $\endgroup$ – Derkooh Nov 7 '17 at 1:07
  • 1
    $\begingroup$ R/T is the y-axis, and r (tensile reduction of area) is the x-axis. $\endgroup$ – Derkooh Nov 7 '17 at 1:55
  • 1
    $\begingroup$ 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) $\endgroup$ – Derkooh Nov 7 '17 at 2:02
  • 2
    $\begingroup$ Still available on Amazon : a good investment? amazon.ca/s/… $\endgroup$ – Solar Mike Nov 7 '17 at 6:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.