What press force is needed ?
On Sat, 6 Feb 2010 05:54:14 +0000 (UTC), Andre Majorel
wrote: Thanks for the hint, Ned. How do you know what press force is needed for a given job ? Specifically, punching holes about 1 cm² in 1 mm steel or 3 mm aluminium. Go to a site such as http://www.matweb.com/, look up the shear strength of your material of choice noting the units psi..N/m^2 etc and converting if required. The force you need will be the area you're cutting through multiplied by the shear strength, plus a bit for luck. HTH Mark Rand RTFM |
What press force is needed ?
In article ,
Mark Rand wrote: On Sat, 6 Feb 2010 05:54:14 +0000 (UTC), Andre Majorel wrote: Thanks for the hint, Ned. How do you know what press force is needed for a given job ? Specifically, punching holes about 1 cm² in 1 mm steel or 3 mm aluminium. Go to a site such as http://www.matweb.com/, look up the shear strength of your material of choice noting the units psi..N/m^2 etc and converting if required. The force you need will be the area you're cutting through multiplied by the shear strength, plus a bit for luck. Area, or is it perimeter? Joe Gwinn |
What press force is needed ?
Joseph Gwinn wrote:
The force you need will be the area you're cutting through multiplied by the shear strength, plus a bit for luck. Area, or is it perimeter? Area I think. Which would be permimeter x material depth x shear Wes -- "Additionally as a security officer, I carry a gun to protect government officials but my life isn't worth protecting at home in their eyes." Dick Anthony Heller |
What press force is needed ?
In article ,
Wes wrote: Joseph Gwinn wrote: The force you need will be the area you're cutting through multiplied by the shear strength, plus a bit for luck. Area, or is it perimeter? Area I think. Which would be permimeter x material depth x shear Area of the sheared material, but not of the piece being punched out. That's the distinction I was making. Joe Gwinn |
What press force is needed ?
On Sat, 06 Feb 2010 10:46:43 -0500, Joseph Gwinn wrote:
In article , Wes wrote: Joseph Gwinn wrote: The force you need will be the area you're cutting through multiplied by the shear strength, plus a bit for luck. Area, or is it perimeter? Area I think. Which would be permimeter x material depth x shear Area of the sheared material, but not of the piece being punched out. That's the distinction I was making. Joe Gwinn Yes, I guess I skipped that part of the explanation. My bad. Mark Rand RTFM |
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