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Hi all. I am building a machine to deform sheet metal into corrugations. I have attached an image that I got off a presentation from MIT's website. In the example, they provide a formula for the creation of a single indentation.

I edited the image to give you an idea of the shape I am going for. It is essentially a "W" shape after pressing.

I do not know how to apply the formula to my design. Is it the same amount of force that I would require? Double the force or even triple the force. My gut feel is that it should be somewhere between the same and double the force.

Any help would be appreciated.

Kind regards Stefan

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  • $\begingroup$ Try turning your lower diagram over and using two separate punches for your analysis... $\endgroup$
    – Solar Mike
    Jan 28 '19 at 11:14
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The plastic section modulus for the plate is

$S = \frac {bh^2}{6} = \frac {LT^2}{6} \ $

W defines the narrowness of the grove, the narrower W, the bigger F is required.

I your case because you have two groves next to each other with no flat band in between there is tension in the two groves by the lateral dragging of the sheet into the jig.

If we were to estimate force needed to bend only the two indentations it would be just 2F. But because there is some extra force spent on dragging the sheet up the inclined walls of the center ridge, it's sate to allow 2.5 F.

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The answer is yes, the total force required equals to the number of corrugations times the force required to produce a single corrugation, with one condition - the support condition must agree with the setup, for which the equation was derived.

The MIT equation, if my understanding is correct, was based on a beam with both ends fixed, then the punch causes plastic hinges to form at the ends first, then forms in the midspan, through continues effort/operation. However, in your sketch of the two corrugations, 2F is correct only if the punch force is applied from bottom up, so the support arrangement agrees with the original study.

Support Conditions:

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