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

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

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

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

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