John,

I meant nothing by my remark; just didn't want to do some kid's homework for
him. The static forces are so close to material limits I suspected a 3rd
year ME asking for help. You obviously put quite a bit of thought and effort
into this.

Igus calculated 70 hours for the 100 lb load on the 3mm x 2mm contact face.
I don't know where we might have differed.

Here's what Solidworks has to say (static analysis):

The entire span of the pin is overstressed, peaking at 120 ksi (failed) in
flexural tension near midspan. Stress is also localized at the ends of the
sleeve, where it pinches into the flexed pin, peaking at 216 ksi
compression. For the bushing, highest displacement and stress occur where
the pin enters. Stress at the free ends are quite a bit lower, which makes
sense in hindsight.

I modeled the bushing and pin in contact. COSMOSworks requires a bit more
mesh setup if they start out of contact. Node-to-node contact was specified
for the bushing-pin contact; shrink-fit for the sleeve-pin contact.

Maximum displacement occurs at midspan: .0020". The maximum "crush" on the
bushing, at the entrance, is .0010". These values are likely inaccurate
because of the large displacements; all components exceeded yield stress at
multiple locations. I could analyze it again with a different rule set, but
the stresses are such that this is not warranted. Reducing the load to 50 lb
(from 200 lb) shows a reasonable factor of safety everywhere, again with a
strong stress concentration where the flexed pin pinches the sleeve. You
might consider turning this from one solid piece to avoid the localized
stress.

Materials:
Pin: AISI 4130 normalized (106 ksi ultimate tensile, 66 ksi yield,
2.97(10^6) psi elastic modulus).
Follower sleeve: AISI 1020 cold rolled. (61 ksi, 57 ksi, 2.97e6)
Bushings: Igus G material (from their data sheet: 30.5 ksi tensile, 11.3
compressive, 11.6 ksi yield, 1.13(10^6) psi elastic modulus).

I'll send you the report if you like.

I suggested the flanged bearing because the flange can act as thrust
washers. It seemed a reasonable improvement.

Mike.

"John2005" wrote in message
oups.com...
Hi everyone,

Thanks for your replies, I appreciate it.

Hi Mike,

The www.igus.com GSM-0203-03 bushing has a very long life with only a
100 pound load on it, remember the 200 pound load on the .1875" OD X
.1875" long roller, is in-between and in the middle of the two plastic
bushings that support the ends of the 2mm OD shaft. Therefore, the 200
pound load is cut in half for each bushing. The loaded roller or steel
tube, is just pressed onto the 2mm shaft with an interference fit,
in-between and in the middle of the two plastic bushings. The online
"expert system" at igus calculated excellent service life for the
GSM-0203-03 bushings with a 100 pound load. This force is just the
maximum load, and the bushings don't normally have that much force on
them, but even if the 100 pound load is constant, the service life is
still excellent, even with shock loads and edge loading selected within
the Igus expert system computerized bushing life estimator.

I don't have the space for a 3mm OD pin, I am already using the largest
possible parts and I am working in a very confined space.

If a 2mm OD X .269" long beam is simply supported at each end, and you
put a point load of 200 pounds at the center of the beam, I calculate
that the center of the beam will deflect by 0.0007". The bending moment
will be 13.4 lb-in. and the bending stress is 280 KSI. The slope will
be .909 degrees. However, when you press the .1875" OD X .1875" long
tube onto the 2mm OD dowel at the center of the .269" long span, this
all changes and the system stiffens up. Also, the load is not really a
point load, but is more like a distributed load, since the cam is 3/16"
thick, just like the cam-follower roller. The cam fits in the middle of
the .269" wide span, and makes contact with the roller.

You mention the sleeves compressive strength, with the sleeve being
pressed onto the 2mm OD dowel, it's just about like one piece. I see no
problem with a 3/16" OD rod supporting a 200 pound load across a .269"
long span. I don't see how a 200 pound load would cause any significant
deformation to a 3/16" OD rod that short, if I understood your comment
correctly.

The double shear strength of the 2mm OD pin is 741 pounds. I am curious
why you would suggest a flange bearing when I have no need for a flange
?

Let me know if you still want a quick FEA check. It smells of a homework
problem, though.

This is not homework & I am not a student ( I am always trying to learn
more, but no longer in school).

If you could do a quick FEA check, I would be grateful. However,
something about your reply made me think that perhaps you were not
completely clear as to how the parts are arranged.

To Clarify:

The .1875" OD X .1875" long roller or steel tube is pressed onto the
center of a 2mm OD X .505" long steel dowel with an interference fit,
so when the roller or tube turns, the shaft or dowel turns with it. The
center of the .1875" roller length is in line with the center of the
.505" length of the dowel pin. The roller is located in the center of a
.269" wide notch or span. Imagine a .269" wide notch or groove cut out
of a solid block of steel. The two ends of the dowel are supported by
plastic bushings. If you measured the distance between the inside edges
of each of the two spaced apart plastic bushings, the distance would be
.269". The bushings are pressed into the steel walls on each side of
the .269" wide notch. From that point, each bushing extends to a length
of 3mm. If you measured the distance between the outer edges of the
two spaced apart bushings, it would be .504". The entire 3mm length of
the bushings is surrounded and supported by steel, since they are
pressed into a steel wall that is thicker than the length of the
bushings.

If you are pretty clear on the arrangement, a quick FEA would be just
what I need to find out how much the ends of the dowel will curl up or
deflect, and find out the stresses and deflections at the center. I
could send a 2D or 3D AutoCAD DWG or JPEG image file to your email, to
avoid any confusion.

I weigh about 185 pounds, and it seems to me a 2mm OD dowel across a
short .269" wide span could support me with no problem at all. I think
a 200 pound person could tie a rope around the center of the dowel and
swing on it without overloading the dowel, at least that's what my
instincts say.

Anyway, I appreciate your reply, even with the homework comment.

Thanks again,
John