Hydraulic tonnage pressure to punch a hole through a hot rolled plate 2

[arunmrao]

Typically not a question from me. But my client wanted an answer for this, I hope I have given all the details necessary.



How much hydraulic tonnage of pressure is required to punch a 2-1/2” whole through .375” hot rolled steel plate.

Learn the rules,so you know how to break them properly.
Dalai Lama

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[desertfox]

Hi arunmrao

The punching force can be obtained from the following formula:-

F = ultimate shear stress of material * shearing area

the shearing area in the case you have posted would be:-

2.5" * pi * .375"

desertfox

 

[arunmrao]

Thanks,shear stress can be assumed to be 50% of UTS safely. Now all that I need is UTS of a hot rolled plate.

Learn the rules,so you know how to break them properly.
Dalai Lama

_____________________________________

 

[desertfox]

hi arunmrao

I would use 70% of the UTS it would give you a larger tonnage than using 50% and give you some margin, you don't want the press to be to weak to punch the disc.
If your working with hot rolled plate to european standards here is a site that gives strength values:-

http://www.roymech.co.uk/Useful_Tables/Matter/Steel_Data/Steel-10025.html

desertfox

 

[arunmrao]

Desertfox Thanks! and a star for your quick and valuable guidance.

Learn the rules,so you know how to break them properly.
Dalai Lama

_____________________________________

 

[desertfox]

your welcome

desertfox

 

[racookpe1978]

In general terms, remember it is not the area of the round hole, rounded slot, or notch that is being sheared that matters.

It is the circumference of the hole or slot that you must use, times the thickness of the plate being punched.

(If you ever have to specify eqpt to use in the field to punch plate, the throat depth of the punch is critical also: The deeper the throat, the more rigid (more expensive!!) and heavier the machine must become.)

 

[desertfox]

hi racookpe1978

I covered the shear area in my first post ie:-

"The punching force can be obtained from the following formula:-F = ultimate shear stress of material * shearing areathe shearing area in the case you have posted would be:-2.5" * pi * .375""

 

[racookpe1978]

I agree: The shear area calculated is correct as you described in the formula. I wanted to point out that other shaped slots or openings can also be punched - most often faster than trying to drill two holes, or sawcut-and-file a notch.