RE: [CAD_CAM_EDM_DRO] Shear Pin Size

Roughly, the equation is:

(Radius of the pin) times (the radius of the pin) times PI (3.14159) times
(the shear strength of the material of the pin) times (two for the two side
in shear) times (three inches radius of the shear point) will give the
shearing of the pin.

Site for shear strengths �

HYPERLINK
"http://www.accuratescrew.com/info/material.htm"http://www.accuratescrew.com
/info/material.htm



I do not know how to determine the cutting forces involved but there are
several calculators available on the web for determining that.

Site for calculating - HYPERLINK
"http://books.google.com/books?id=uiM56pDFdqsC&pg=PT506&lpg=PT506&dq=cutting
+forces+in+turning+calculation&source=web&ots=hB-y2ogYsI&sig=JmscI9bkUN8lKcd
bRt1S0lAtJdI"http://books.google.com/books?id=uiM56pDFdqsC&pg=PT506&lpg=PT50
6&dq=cutting+forces+in+turning+calculation&source=web&ots=hB-y2ogYsI&sig=Jms
cI9bkUN8lKcdbRt1S0lAtJdI



R. Wink









From: CAD_CAM_EDM_DRO@yahoogroups.com
[mailto:CAD_CAM_EDM_DRO@yahoogroups.com] On Behalf Of roboticscnc
Sent: Sunday, December 30, 2007 5:06 AM
To: CAD_CAM_EDM_DRO@yahoogroups.com
Subject: [CAD_CAM_EDM_DRO] Shear Pin Size



Hi Group,
I've upgraded my Lathe Motor and want to include a shear pin on the
spindle pulley. The reason for this is that a threading wizard
recently decided to rapid traverse the tool into the chuck.
The lathe needed a rebuild after this. I had a 3 phase 4Kw motor
lying around so I decided to replace the 750w spindle motor with it as
the lathe was always underpowered.
I need to work out what diameter shear pin to use on the pulley and
what material. The shear pin will be at a radius of 75mm ( 3" ) from
the centre of the spindle and I want to be able to machine steel (
chip breaking ) at a similar radius with carbide tools.
I would prefer to use aluminium or brass for the pin as the pulley is
cast iron.
The shear pin would be in double shear ( 1 driven plate with a
driving plate on each side ) parallel to the axis of the spindle.
Obviously the loads mentioned are ambiguous and I'm only after a ball
park answer.
Thanks,
Glen.




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