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Ironworker shear apparatus

Ironworker shear apparatus

(OP)
Hello,

I am currently designing an Ironworker machine and I need help designing the shear. See my blog post for more info on the puropse of the machine: http://blog.opensourceecology.org/2011/12/ironworker-blade-design-help/

The shear will need to shear 1"x12" of A36 steel. I need to know what force is necessary when you are shearing with a rotating blade, or what moment would be necessary at the shearing point furthest from the pin. Even just how to calculate this would be valuble. Since the angle is changing therefore the thickness, I'm assuming you'd need some ridiculously complicated equation to find the necessary force.

I would like to use a rotating blade vs. a blade which moves vertically because the rotating blade Ironworkers usually have much simpler designs and need less force.

If it helps, I was told by an ironworker blade manufacturer that with a vertically moving blade with a 5 degree rake, you can shear it with 120T.

Any info helps, even just the right terminology.

Thanks,
Brianna

RE: Ironworker shear apparatus

30 tons per square inch is a rule of thumb for mild steel when constructing punch press type of tooling.  An angled blade will help to reduce tonnage, as will the rotary motion.  Actually, if I understand it correctly, your 5 deg angle will become a moot point because the blade angle of entry into the work piece will far exceed 5 deg on a portion of the work being sheared.

I think 120 T might be a little optimistic, however, especially on a directly perpendicular shear, guillotine blade or not.

Good luck on the project.

It is better to have enough ideas for some of them to be wrong, than to be always right by having no ideas at all.

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