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Hydraulic Cutting Forces

Hydraulic Cutting Forces

Hydraulic Cutting Forces

(OP)
Can anyone tell me how to calculate the hydraulic cylinder diameter requred to cut steel plate using a shear tool operated from a hydraulic power pack.

RE: Hydraulic Cutting Forces

Hi Sparky,

  First you'll have to find the shear strength of the thickness and type of steel you're trying to cut (material engineering).  Then we need to know the pressure of your particular power pack.

RE: Hydraulic Cutting Forces

(OP)
Hi, vr7brz

Thanks for your reply but I am looking for a calculation method or general formula so I can calculate for general tool design.
Generaly using Mild steel at 1000psi to 1500psi.

RE: Hydraulic Cutting Forces

First, the shear strength of the work material is of no use here.  To cut in lay language means to SEPARATE not deform to a certain strain.  Second, there is such a thing as strain hardening.


To start to solve this relatively simple problem, you should draw a model of your process showing the tool and the workpiece in their initial and final positions.  Then, you calculate the energy required to reach the final position, then, knowing the process velocity, you can easily calculate the force needed to deliver this energy at the required speed.  This is as simple as that.

Viktor
http://viktorastakhov.tripod.com

RE: Hydraulic Cutting Forces

I think what you're asking for is the cylinder diameter to punch out a part using a punch and die. Look in the Metals Handbook or Machinery's Handbook for more information. If you're designing tooling you are going to have a lot to learn.

Multiply the perimeter, P, of the cut piece by the material thickness,t, to get the area to be blanked. Multiply this by 0.8 (to get the shear strength for mild steel only) x ultimate tensile strength for mild steel (approximately 60,000 psi) to get the force required.

So, P x t x 0.8 x 60,000 = force

Since area = force/pressure and area = (pi x diameter^2)/4, then

Diameter = sqrt[(4 x force)/(pi x pressure)]

 Also allow yourself a factor of safety of 1.5 or more in your force calculations.

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