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, it is as if you have 1/2 W reversed in between, so we estimate the force needed to be 2.5 F.

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.