Hello everyone,

I have a 2mm OD hardened steel dowel pin used as a bushing shaft, that
is 0.505" inches long. Each end of the dowel is supported by a small
Igus plastic bushing, i.e., www.igus.com Part # GSM-0203-03, which are
each 3mm long. The bushings are located flush with the ends of the
dowel. The clearance between the shaft OD and the bushing ID will be
from .0006" Minimum to 0.003" Maximum.

The distance between the inside edges of the two bushings will be
0.269" inch. In the center of this .269" span, a 3/16" OD X
3/16" long steel tube is pressed onto the 2mm OD dowel pin. The tube
acts as a small roller or cam-follower. When the roller rotates, the
2mm OD dowel "rotates with" the roller, since the roller is pressed
onto the dowel and is basically like one piece.

I have a .031" thick thrust washer located on each side of the
roller, within the 0.269" wide span.

If a shaft is simply supported at each end, and you put a load in the
middle of the supports, not only does the shaft deflect down at the
center, but the ends of the shaft will tend to deflect & curl up as
well.

I can calculate the shaft deflections if the 3/16" OD X 3/16" long
tube were not pressed onto the shaft, but after the tube is pressed
onto the shaft, it's as if the center of the shaft has a 3/16" OD
and the two ends have a 2mm OD. The 3/16" OD tubing stiffens
everything up.

Can anyone please tell me how to calculate the deflection of the shaft
after the 3/16" OD X 3/16" long steel tubing is pressed onto the
dowel ?

I will have a 200 pound load on the roller, the cam is 3/16" wide
just like the roller, so I suppose you would consider this to be a
distributed load. I am not concerned with deflections inside of the
..269" span because I think they will be very small, probably less
than .0005". I am concerned with how far the very ends of the dowel
will curl up or deflect, since this could produce misalignment and /or
binding of the dowel / shaft in the ID of the bushing.

I would appreciate any feedback anyone can offer.

Thanks for your help.
John

Hi everyone,

Does anyone know the formula to calculate the deflection for a
"stepped" shaft or beam, where the loaded center portion of the beam
has a larger diameter than the simply supported ends ?

I think this would at least give me a very close approximation to the
information I need. I have not been able to find the formula on the
internet.

Thanks
John

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

Does anyone know the formula to calculate the deflection for a
"stepped" shaft or beam, where the loaded center portion of the beam
has a larger diameter than the simply supported ends ?

I think this would at least give me a very close approximation to
the
information I need. I have not been able to find the formula on the
internet.

Thanks
John


You probably won't find a general formula. Problems like these are
sometimes solved graphically or by the moment-area method. Any good
mechanics of materials book will cover the subject...

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

Does anyone know the formula to calculate the deflection for a
"stepped" shaft or beam, where the loaded center portion of the beam
has a larger diameter than the simply supported ends ?

I think this would at least give me a very close approximation to
the
information I need. I have not been able to find the formula on the
internet.

Thanks
John


The problem can also be solved analytically, of course. I would
imagine there are computer programs these days to do the same thing.
Might make for an interesting exercise.....

In article 1128657645.296912.19850
@f14g2000cwb.googlegroups.com,
says...
Hi everyone,

Does anyone know the formula to calculate the deflection for a
"stepped" shaft or beam, where the loaded center portion of the beam
has a larger diameter than the simply supported ends ?

I think this would at least give me a very close approximation to the
information I need. I have not been able to find the formula on the
internet.


You can treat the parts of the shaft independently and sum
the effects of bending (superposition). Note that you
*cannot* simply sum the deflections, but will probably have
to calculate the slope and/or radius of curvature at the
various points on the shaft. I haven't thought this thru
completely, but a good strength of materials text
(Timoshenko) or handbook (Roark's) should cover this.

Ned Simmons

"John2005" wrote in message
ups.com...
Hello everyone,

I have a 2mm OD hardened steel dowel pin used as a bushing shaft,
that
is 0.505" inches long. Each end of the dowel is supported by a small
Igus plastic bushing, i.e., www.igus.com Part # GSM-0203-03, which
are
each 3mm long. The bushings are located flush with the ends of the
dowel. The clearance between the shaft OD and the bushing ID will be
from .0006" Minimum to 0.003" Maximum.

The distance between the inside edges of the two bushings will be
0.269" inch. In the center of this .269" span, a 3/16" OD X
3/16" long steel tube is pressed onto the 2mm OD dowel pin. The tube
acts as a small roller or cam-follower. When the roller rotates, the
2mm OD dowel "rotates with" the roller, since the roller is pressed
onto the dowel and is basically like one piece.

I have a .031" thick thrust washer located on each side of the
roller, within the 0.269" wide span.

If a shaft is simply supported at each end, and you put a load in
the
middle of the supports, not only does the shaft deflect down at the
center, but the ends of the shaft will tend to deflect & curl up as
well.

I can calculate the shaft deflections if the 3/16" OD X 3/16" long
tube were not pressed onto the shaft, but after the tube is pressed
onto the shaft, it's as if the center of the shaft has a 3/16" OD
and the two ends have a 2mm OD. The 3/16" OD tubing stiffens
everything up.

Can anyone please tell me how to calculate the deflection of the
shaft
after the 3/16" OD X 3/16" long steel tubing is pressed onto the
dowel ?

I will have a 200 pound load on the roller, the cam is 3/16" wide
just like the roller, so I suppose you would consider this to be a
distributed load. I am not concerned with deflections inside of the
.269" span because I think they will be very small, probably less
than .0005". I am concerned with how far the very ends of the dowel
will curl up or deflect, since this could produce misalignment and
/or
binding of the dowel / shaft in the ID of the bushing.

I would appreciate any feedback anyone can offer.

Thanks for your help.
John


I don't know what the allowable shear stress is of the pin, but that
may be the limiting factor-or the bearing load on the plastic....

"Rick" wrote in message
link.net...

"John2005" wrote in message
ups.com...
Hello everyone,

I have a 2mm OD hardened steel dowel pin used as a bushing shaft,
that
is 0.505" inches long. Each end of the dowel is supported by a small
Igus plastic bushing, i.e., www.igus.com Part # GSM-0203-03, which
are
each 3mm long. The bushings are located flush with the ends of the
dowel. The clearance between the shaft OD and the bushing ID will be
from .0006" Minimum to 0.003" Maximum.

A quick sanity check says the 2mm pin is too small, and indicates little
need for further analysis. The 200 lb load on the projected cross section is
right at the sleeve's rated compressive strength of 78 MPa (11.3 ksi vs.
10.8 ksi). Motion, temperature, and shock will derate this further. Flexural
stress on the pin (simplifying and treating both ends fixed with a centered
point load) is 49 ksi. Static double shear is 21 ksi. I would investigate
enlarging the pin to 3mm before analysing further. Service life for the 2mm
pin and bearing will be quite short. Their GFM-0304-05 flange bearing would
be, I think, a better match for your application.

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


The distance between the inside edges of the two bushings will be
0.269" inch. In the center of this .269" span, a 3/16" OD X
3/16" long steel tube is pressed onto the 2mm OD dowel pin. The tube
acts as a small roller or cam-follower. When the roller rotates, the
2mm OD dowel "rotates with" the roller, since the roller is pressed
onto the dowel and is basically like one piece.

I have a .031" thick thrust washer located on each side of the
roller, within the 0.269" wide span.

If a shaft is simply supported at each end, and you put a load in
the
middle of the supports, not only does the shaft deflect down at the
center, but the ends of the shaft will tend to deflect & curl up as
well.

I can calculate the shaft deflections if the 3/16" OD X 3/16" long
tube were not pressed onto the shaft, but after the tube is pressed
onto the shaft, it's as if the center of the shaft has a 3/16" OD
and the two ends have a 2mm OD. The 3/16" OD tubing stiffens
everything up.

Can anyone please tell me how to calculate the deflection of the
shaft
after the 3/16" OD X 3/16" long steel tubing is pressed onto the
dowel ?

I will have a 200 pound load on the roller, the cam is 3/16" wide
just like the roller, so I suppose you would consider this to be a
distributed load. I am not concerned with deflections inside of the
.269" span because I think they will be very small, probably less
than .0005". I am concerned with how far the very ends of the dowel
will curl up or deflect, since this could produce misalignment and
/or
binding of the dowel / shaft in the ID of the bushing.

I would appreciate any feedback anyone can offer.

Thanks for your help.
John


I don't know what the allowable shear stress is of the pin, but that
may be the limiting factor-or the bearing load on the plastic....

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

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

Hi everyone,

Hi Mike,

Thanks for your reply and for plugging this into Solidworks. I
appreciate it.

Your results were much different than the one's I got from the DT beam
program.

I have pasted the results of the Igus Expert system below, so you can
see how the bushings are being used in my application, and why there is
such a difference in the our life estimates. Under the operating
conditions for my application, the service life is 2,400 hours with a
100 pound bushing load. With my application, 1,600 hours will give
about 10 years of product life. This thing is just oscillated manually
& intermittently by hand, and always used at the ambient temperature of
the standard atmosphere, it never gets hot & the environment is very
clean. It's just activated about 3 seconds at a time, once or twice per
minute, over the course of about one or two hours per day. Plus, the
100 pound load is is probably only about 50 pounds for 80% of the time.


Your flanged bushing suggestion is a good idea, I have been thinking
about the shaft so much, I forgot about the possibility of eliminating
the thrust washers with a flanged bushing.

Someone suggested to make the roller-shaft combination from a ground
down 3/16" OD dowel, but I don't know if I can assemble it as one
piece.

Would reducing the .031" thick thrust washers to .005" thick, or
smaller, and extending the length of the 3/16" OD tube so that it is
just about even with the .269" wide notch help significantly ? This
would mean the tube is almost exactly even with the start of the
bushing support at the .269" wide span.

What about going with a different shaft material like Tungsten,
carbide, or Cobalt ? They may have drill or reamer blanks like this, I
know Mcmaster Carr has a Cobalt rod in this size PN# 8813A22 grade M42
Cobalt tool steel, but I don't know if the yield is higher than regular
steel.

I am very surprised the dowel is overstressed at a 200 pound load with
a span of only .269", especially when it has the 3/16" OD tube for
added support. It seems my only choice is to go with a stiffer material
with a higher yield, and/or go with thinner thrust washers and increase
the length of the 3/16" OD tube so it fits as snug as possible inside
the .269" notch. If the roller fits as snug as possible in the .269"
wide notch, it seems you almost have more of a shear situation than
deflection.

I would love to see the report. Can I email you at the email given on
this site ? My email address on this site has been deactivated, and I
was not able to reactivate it for some reason. I prefer not to give out
my good email on newsgroups.

Thanks again for your help, and to everyone for their replies.
John

Sliding bearing analysis iglide® G300 (Igus Expert system)

Measurement:
Type: Type S
Shaft diameter d1 = 0.079 inch
Bearing length b1 = 0.118 inch

Load:
Maximum bearing load F = 100 lb
Average surface pressure P = 10752.71 PSI
Exposure to shocks occurs
Edge loading occurs

Type of motion:
Pivoting

Speed:
Pivoting angle ß = 21.12 °
Frequency waving = 180 1/min
Pivoting speed = 0.87 fpm

Temperatu
Maximum ambient temperature = 68 °F
Max. Temperature = 68 °F
Minimum temperature = 68 °F

Mating surface:
Shaft material: Cold Rolled 1018 Carbon Steel
Shaft roughness = 0.2 µm
Thermal conductivity = 48 W/m*K

Housing:
Housing material: Steel
Thermal conductivity = 48 W/m*K

Intermittent service:
Turn-on time t1 = 3 Seconds
Standstill period t2 = 17 Seconds

Overall result
The material is suitable for use under the specified conditions.
84 points were allocated for the evaluation of costs per sliding
bearing.
The product is not available in the desired dimensions.

Test of actual sliding speeds
The applicable type of motion is Pivoting.
The permissible oscillating rate is 0.87 fpm.
The maximum permissible oscillating rate is 196.85 fpm.

The material is suitable for use at the specified speed.

Test of the maximum permissible surface pressure
The applicable surface pressure is 10752.71 PSI.
The surface pressure permissible for the material at the specified
temperature is 11602.557 PSI.

The material is suitable for use at the specified surface pressure.

Test of the maximum permissible PxV value
The PxV value of the described application is 8466.901 PSI * fpm.
The maximum permissible PxV value is 11991.16 PSI * fpm.

The P*V value of the arrangement lies within the permissible limits.

Test of the temperatures occurring under real operating conditions
The ambient temperature during operation is 68 °F.
The highest permissible temperature for this application is 68 °F.
The minimum temperature for this application is 68 °F.
The maximum temperature permissible for the material is 266 °F.
The minimum temperature permissible for the material is -40 °F.
The material can withstand a temperature of 428 °Ffor brief periods.

The material is suitable for use at the specified temperatures.

It was nothing. Just a couple of sleeves and cylinders. Thinking back on it,
the default mesh size is likely too coarse for good results. I'll go back
and have another go with it this evening.

Re. Igus's calculator: I figured for 100 rpm, continuous duty. I likely also
picked the worst possible conditions from those listed. That would explain
why the results differed so.

I'd love to see pictures when you're ready to share. It's tough (for me) to
imagine a design so tight that another 1/2 millimeter of radius would break
it.

I don't fully know the assumptions made for "shrink fit" in the software, in
regards the sleeve and the pin. It's clear that it thinks the deflections
are large enough that contact is localized to three points: at the ends on
top, toward the applied force, and at midspan along the bottom. The results
should be more favorable if it were to treat it as "bonded". Silver solder
would seem reasonable if you couldn't turn down shoulders from one piece.
Shear stress is low(er) toward midspan and centroid of the section. That's
not enough for your use, though; it still failed when modeled as a solid
piece.

The flange bushing is available for shaft sizes 3mm and up.

One would think a 2mm shaft can span 1/4" and still carry a man's weight. I
would just silver solder something up and see, but I only have 3mm shafts on
hand. (No joking there, and only by coincidence. They're for small BLDC
motors for model planes.) On the other hand, thinking about it as 3/32"
piano wire brings to mind multiple failure modes. I'm not used to thinking
so small.

Mike.

---------------

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

Hi Mike,

Thanks for your reply and for plugging this into Solidworks. I
appreciate it.

Your results were much different than the one's I got from the DT beam
program.

I have pasted the results of the Igus Expert system below, so you can
....

Hi everyone,

I think I have found a way to greatly reduce the stresses on the 2mm OD
shaft.

Since the stresses were so high, I decided to take a closer look at
this problem, rather than just relying on a physical test. With
variances in steel, I could have a few that would test OK, but others
that would not. Also, if I tested one and it seemed OK, I am afraid it
could yield a little more with each use, and then cause problems down
the road.

I found a beam deflection program called "beam 2d"

http://www.orandsystems.com/Bm2DShow/show0.html

This program lets you model stepped shafts. You get 30 unrestricted
uses with the demo. I found that increasing the roller from .1875"
long to .243" long so that it fits more snug inside of the .269"
support span, causes a drastic reduction in the bending stress of the
beam.

I have pasted the program printout for both the .1875" long roller and
a .243" long roller below. I will just use .010" thick delrin thrust
washers on each side of the roller instead of .03 to .04" thick thrust
washers.

The highest bending stress seems to occur right where the 2mm shaft
comes out of the 3/16" OD portion. With the .1875" long roller, the
maximum bending stress is 84,130.45 PSI, but with the .243" long roller
the maximum bending stress is reduced to 27,034.99 which surprised me.

With the .1875" long roller, the center of the roller deflected by
..0001" and the very ends of the 2mm OD end portions curled up by
..0002". With the new .243" long roller, the center of the roller
deflected by only 0.00005", and the very ends of the 2mm OD end
portions deflected up by .0001".

With the new longer roller, the first portion of the shaft is 2mm OD X
..131" long, then the second portion is .1875" OD X .243" long, and the
last portion is .2mm OD X .131" long.

The question now becomes, will the pressed on 3/16" OD center portion
act fairly close to a stepped shaft made from one solid piece as
modeled by the program ?

I do have one way to use a 1/8" OD shaft, but I must sacrifice the Igus
plastic bushings. The roller and shaft is held in a yoke, I could make
the yoke itself out of a bushing material, so the shaft turns right in
the yoke instead of the plastic bushings. This gives me room for a 1/8"
OD shaft.

However, this is a high load oscillating application, and I can only
lube the shaft once at assembly then never again. I am a Little
concerned about wear. I hear 0-6 tool steel makes good bushings, and
has a self lubricating graphitic property. The walls are so thin on the
yoke I don't think I can harden it without cracking, so I would just
have to lube the shaft at assembly, and hope for the best as far as
wear is concerned. This thing is just intermittently oscillated by
hand, so perhaps it would wear well.

Here are the printouts from the beam design program. I would appreciate
any other feedback anyone may have. If the new longer pressed on roller
acts close to a one piece stepped shaft, I think I should be OK.

NEW DESIGN WITH .243" LONG ROLLER

BEAM LENGTH = 0.5047204 in

MATERIAL PROPERTIES
steel:
Modulus of elasticity = 29000000.0 lb/in²

CROSS-SECTION PROPERTIES
#1: from 0.0 in to 0.1308602 in
Moment of inertia = 0.000001911958 in^4
Top height = 0.0395 in
Bottom height = 0.0395 in
Area = 0.00490167 in²

#2: from 0.1308602 in to 0.3738602 in
Moment of inertia = 0.00006067014 in^4
Top height = 0.09375 in
Bottom height = 0.09375 in
Area = 0.02761165 in²

#3: from 0.3738602 in to 0.5047204 in
Moment of inertia = 0.000001911958 in^4
Top height = 0.0395 in
Bottom height = 0.0395 in
Area = 0.00490167 in²

EXTERNAL CONCENTRATED FORCES
200.0 lb at 0.252 in

SUPPORT REACTIONS ***
Simple at 0.1181 in
Reaction Force =-100.4091 lb

Simple at 0.387 in
Reaction Force =-99.59093 lb

MAXIMUM DEFLECTION ***
-0.0000780247 in at 0.5047204 in
No Limit specified

MAXIMUM BENDING MOMENT ***
13.44477 lb-in at 0.252 in

MAXIMUM SHEAR FORCE ***
100.4091 lb from 0.1181 in to 0.252 in

MAXIMUM STRESS ***
Tensile = 27034.99 lb/in² No Limit specified
Compressive = 27034.99 lb/in² No Limit specified
Shear (Avg) = 20484.67 lb/in² No Limit specified

ANALYSIS AT SPECIFIED LOCATIONS ***
Location = 0.0 in
Deflection = -0.00007765793 in
Slope = 0.03767546 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.07930511 in
Deflection = -0.00002550999 in
Slope = 0.03767546 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.1586102 in
Deflection = 0.00002143606 in
Slope = 0.02681157 deg
Moment = 4.067595 lb-in
Shear force = 100.4091 lb
Tensile = 6285.415 lb/in²
Compressive = 6285.415 lb/in²
Shear stress = 3636.475 lb/in²

Location = 0.2523602 in
Deflection = 0.00004730955 in
Slope = 0.00002452662 deg
Moment = 13.4089 lb-in
Shear force = -99.59093 lb
Tensile = 20719.99 lb/in²
Compressive = 20719.99 lb/in²
Shear stress = 3606.844 lb/in²

Location = 0.3461102 in
Deflection = 0.00002163168 in
Slope = -0.0266601 deg
Moment = 4.07225 lb-in
Shear force = -99.59093 lb
Tensile = 6292.608 lb/in²
Compressive = 6292.608 lb/in²
Shear stress = 3606.844 lb/in²

Location = 0.4254153 in
Deflection = -0.00002546155 in
Slope = -0.03797543 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.5047204 in
Deflection = -0.0000780247 in
Slope = -0.03797543 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²



OLD DESIGN WITH .1875" LONG ROLLER

BEAM LENGTH = 0.5047204 in

MATERIAL PROPERTIES
steel:
Modulus of elasticity = 29000000.0 lb/in²

CROSS-SECTION PROPERTIES
#1: from 0.0 in to 0.1586102 in
Moment of inertia = 0.000001911958 in^4
Top height = 0.0395 in
Bottom height = 0.0395 in
Area = 0.00490167 in²

#2: from 0.1586102 in to 0.3461102 in
Moment of inertia = 0.00006067014 in^4
Top height = 0.09375 in
Bottom height = 0.09375 in
Area = 0.02761165 in²

#3: from 0.3461102 in to 0.5047204 in
Moment of inertia = 0.000001911958 in^4
Top height = 0.0395 in
Bottom height = 0.0395 in
Area = 0.00490167 in²

EXTERNAL CONCENTRATED FORCES
200.0 lb at 0.252 in

SUPPORT REACTIONS ***
Simple at 0.1181 in
Reaction Force =-100.4091 lb

Simple at 0.387 in
Reaction Force =-99.59093 lb

MAXIMUM DEFLECTION ***
-0.0002312445 in at 0.5047204 in
No Limit specified

MAXIMUM BENDING MOMENT ***
13.44477 lb-in at 0.252 in

MAXIMUM SHEAR FORCE ***
100.4091 lb from 0.1181 in to 0.252 in

MAXIMUM STRESS ***
Tensile = 84130.45 lb/in² No Limit specified
Compressive = 84130.45 lb/in² No Limit specified
Shear (Avg) = 20484.67 lb/in² No Limit specified

ANALYSIS AT SPECIFIED LOCATIONS ***
Location = 0.0 in
Deflection = -0.0002310493 in
Slope = 0.1120927 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.07930511 in
Deflection = -0.00007589781 in
Slope = 0.1120927 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.1586102 in
Deflection = 0.00005918867 in
Slope = 0.02695565 deg
Moment = 4.067595 lb-in
Shear force = 100.4091 lb
Tensile = 84034.27 lb/in²
Compressive = 84034.27 lb/in²
Shear stress = 20484.67 lb/in²

Location = 0.2523602 in
Deflection = 0.0000852979 in
Slope = 0.0001685984 deg
Moment = 13.4089 lb-in
Shear force = -99.59093 lb
Tensile = 20719.99 lb/in²
Compressive = 20719.99 lb/in²
Shear stress = 3606.844 lb/in²

Location = 0.3461102 in
Deflection = 0.00005985576 in
Slope = -0.02651603 deg
Moment = 4.07225 lb-in
Shear force = -99.59093 lb
Tensile = 84130.45 lb/in²
Compressive = 84130.45 lb/in²
Shear stress = 20317.75 lb/in²

Location = 0.4254153 in
Deflection = -0.00007546126 in
Slope = -0.1125491 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.5047204 in
Deflection = -0.0002312445 in
Slope = -0.1125491 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

What's the allowable shear stress of your material? You're at 20,000
PSI...


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

I think I have found a way to greatly reduce the stresses on the 2mm
OD
shaft.

Since the stresses were so high, I decided to take a closer look at
this problem, rather than just relying on a physical test. With
variances in steel, I could have a few that would test OK, but others
that would not. Also, if I tested one and it seemed OK, I am afraid it
could yield a little more with each use, and then cause problems down
the road.

I found a beam deflection program called "beam 2d"

http://www.orandsystems.com/Bm2DShow/show0.html

This program lets you model stepped shafts. You get 30 unrestricted
uses with the demo. I found that increasing the roller from .1875"
long to .243" long so that it fits more snug inside of the .269"
support span, causes a drastic reduction in the bending stress of the
beam.

I have pasted the program printout for both the .1875" long roller and
a .243" long roller below. I will just use .010" thick delrin thrust
washers on each side of the roller instead of .03 to .04" thick thrust
washers.

The highest bending stress seems to occur right where the 2mm shaft
comes out of the 3/16" OD portion. With the .1875" long roller, the
maximum bending stress is 84,130.45 PSI, but with the .243" long
roller
the maximum bending stress is reduced to 27,034.99 which surprised me.

With the .1875" long roller, the center of the roller deflected by
..0001" and the very ends of the 2mm OD end portions curled up by
..0002". With the new .243" long roller, the center of the roller
deflected by only 0.00005", and the very ends of the 2mm OD end
portions deflected up by .0001".

With the new longer roller, the first portion of the shaft is 2mm OD X
..131" long, then the second portion is .1875" OD X .243" long, and the
last portion is .2mm OD X .131" long.

The question now becomes, will the pressed on 3/16" OD center portion
act fairly close to a stepped shaft made from one solid piece as
modeled by the program ?

I do have one way to use a 1/8" OD shaft, but I must sacrifice the
Igus
plastic bushings. The roller and shaft is held in a yoke, I could make
the yoke itself out of a bushing material, so the shaft turns right in
the yoke instead of the plastic bushings. This gives me room for a
1/8"
OD shaft.

However, this is a high load oscillating application, and I can only
lube the shaft once at assembly then never again. I am a Little
concerned about wear. I hear 0-6 tool steel makes good bushings, and
has a self lubricating graphitic property. The walls are so thin on
the
yoke I don't think I can harden it without cracking, so I would just
have to lube the shaft at assembly, and hope for the best as far as
wear is concerned. This thing is just intermittently oscillated by
hand, so perhaps it would wear well.

Here are the printouts from the beam design program. I would
appreciate
any other feedback anyone may have. If the new longer pressed on
roller
acts close to a one piece stepped shaft, I think I should be OK.

NEW DESIGN WITH .243" LONG ROLLER

BEAM LENGTH = 0.5047204 in

MATERIAL PROPERTIES
steel:
Modulus of elasticity = 29000000.0 lb/in²

CROSS-SECTION PROPERTIES
#1: from 0.0 in to 0.1308602 in
Moment of inertia = 0.000001911958 in^4
Top height = 0.0395 in
Bottom height = 0.0395 in
Area = 0.00490167 in²

#2: from 0.1308602 in to 0.3738602 in
Moment of inertia = 0.00006067014 in^4
Top height = 0.09375 in
Bottom height = 0.09375 in
Area = 0.02761165 in²

#3: from 0.3738602 in to 0.5047204 in
Moment of inertia = 0.000001911958 in^4
Top height = 0.0395 in
Bottom height = 0.0395 in
Area = 0.00490167 in²

EXTERNAL CONCENTRATED FORCES
200.0 lb at 0.252 in

SUPPORT REACTIONS ***
Simple at 0.1181 in
Reaction Force =-100.4091 lb

Simple at 0.387 in
Reaction Force =-99.59093 lb

MAXIMUM DEFLECTION ***
-0.0000780247 in at 0.5047204 in
No Limit specified

MAXIMUM BENDING MOMENT ***
13.44477 lb-in at 0.252 in

MAXIMUM SHEAR FORCE ***
100.4091 lb from 0.1181 in to 0.252 in

MAXIMUM STRESS ***
Tensile = 27034.99 lb/in² No Limit specified
Compressive = 27034.99 lb/in² No Limit specified
Shear (Avg) = 20484.67 lb/in² No Limit specified

ANALYSIS AT SPECIFIED LOCATIONS ***
Location = 0.0 in
Deflection = -0.00007765793 in
Slope = 0.03767546 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.07930511 in
Deflection = -0.00002550999 in
Slope = 0.03767546 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.1586102 in
Deflection = 0.00002143606 in
Slope = 0.02681157 deg
Moment = 4.067595 lb-in
Shear force = 100.4091 lb
Tensile = 6285.415 lb/in²
Compressive = 6285.415 lb/in²
Shear stress = 3636.475 lb/in²

Location = 0.2523602 in
Deflection = 0.00004730955 in
Slope = 0.00002452662 deg
Moment = 13.4089 lb-in
Shear force = -99.59093 lb
Tensile = 20719.99 lb/in²
Compressive = 20719.99 lb/in²
Shear stress = 3606.844 lb/in²

Location = 0.3461102 in
Deflection = 0.00002163168 in
Slope = -0.0266601 deg
Moment = 4.07225 lb-in
Shear force = -99.59093 lb
Tensile = 6292.608 lb/in²
Compressive = 6292.608 lb/in²
Shear stress = 3606.844 lb/in²

Location = 0.4254153 in
Deflection = -0.00002546155 in
Slope = -0.03797543 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.5047204 in
Deflection = -0.0000780247 in
Slope = -0.03797543 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²



OLD DESIGN WITH .1875" LONG ROLLER

BEAM LENGTH = 0.5047204 in

MATERIAL PROPERTIES
steel:
Modulus of elasticity = 29000000.0 lb/in²

CROSS-SECTION PROPERTIES
#1: from 0.0 in to 0.1586102 in
Moment of inertia = 0.000001911958 in^4
Top height = 0.0395 in
Bottom height = 0.0395 in
Area = 0.00490167 in²

#2: from 0.1586102 in to 0.3461102 in
Moment of inertia = 0.00006067014 in^4
Top height = 0.09375 in
Bottom height = 0.09375 in
Area = 0.02761165 in²

#3: from 0.3461102 in to 0.5047204 in
Moment of inertia = 0.000001911958 in^4
Top height = 0.0395 in
Bottom height = 0.0395 in
Area = 0.00490167 in²

EXTERNAL CONCENTRATED FORCES
200.0 lb at 0.252 in

SUPPORT REACTIONS ***
Simple at 0.1181 in
Reaction Force =-100.4091 lb

Simple at 0.387 in
Reaction Force =-99.59093 lb

MAXIMUM DEFLECTION ***
-0.0002312445 in at 0.5047204 in
No Limit specified

MAXIMUM BENDING MOMENT ***
13.44477 lb-in at 0.252 in

MAXIMUM SHEAR FORCE ***
100.4091 lb from 0.1181 in to 0.252 in

MAXIMUM STRESS ***
Tensile = 84130.45 lb/in² No Limit specified
Compressive = 84130.45 lb/in² No Limit specified
Shear (Avg) = 20484.67 lb/in² No Limit specified

ANALYSIS AT SPECIFIED LOCATIONS ***
Location = 0.0 in
Deflection = -0.0002310493 in
Slope = 0.1120927 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.07930511 in
Deflection = -0.00007589781 in
Slope = 0.1120927 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.1586102 in
Deflection = 0.00005918867 in
Slope = 0.02695565 deg
Moment = 4.067595 lb-in
Shear force = 100.4091 lb
Tensile = 84034.27 lb/in²
Compressive = 84034.27 lb/in²
Shear stress = 20484.67 lb/in²

Location = 0.2523602 in
Deflection = 0.0000852979 in
Slope = 0.0001685984 deg
Moment = 13.4089 lb-in
Shear force = -99.59093 lb
Tensile = 20719.99 lb/in²
Compressive = 20719.99 lb/in²
Shear stress = 3606.844 lb/in²

Location = 0.3461102 in
Deflection = 0.00005985576 in
Slope = -0.02651603 deg
Moment = 4.07225 lb-in
Shear force = -99.59093 lb
Tensile = 84130.45 lb/in²
Compressive = 84130.45 lb/in²
Shear stress = 20317.75 lb/in²

Location = 0.4254153 in
Deflection = -0.00007546126 in
Slope = -0.1125491 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

Location = 0.5047204 in
Deflection = -0.0002312445 in
Slope = -0.1125491 deg
Moment = 0.0 lb-in
Shear force = 0.0 lb
Tensile = 0.0 lb/in²
Compressive = 0.0 lb/in²
Shear stress = 0.0 lb/in²

(Hi, John. Your email bounced. I tried to send you the FEM report.)

"John2005" wrote in message
ups.com...
The highest bending stress seems to occur right where the 2mm shaft
comes out of the 3/16" OD portion. With the .1875" long roller, the
maximum bending stress is 84,130.45 PSI, but with the .243" long roller
the maximum bending stress is reduced to 27,034.99 which surprised me.
-----
The magnitude is a bit surprising, but that it reduces stress is quite
reasonable. Stress distribution along the beam span is parabolic for uniform
loads, highest at mid-span and 0 at the supports.
******

The question now becomes, will the pressed on 3/16" OD center portion
act fairly close to a stepped shaft made from one solid piece as
modeled by the program ?
-----

My gut feel is to just go with it, but I don't have the investment and
responsibility you do. Anything you can do to smooth the transition between
the two should help. A fillet of silver solder, for example, will help
minimize the localized stress where they come together.

Hi Mike & Rick,

Thanks for your replies,

My gut feel is to just go with it, but I don't have the investment and
responsibility you do. Anything you can do to smooth the transition
between
the two should help. A fillet of silver solder, for example, will help
minimize the localized stress where they come together.

I'm just going to give it a try with the longer roller, I was just very
surprised at how much a .06" increase in the roller length made towards
stress reduction. The space is so tight it may be very hard to assemble
with solder, I may be able to wipe a thin film on the shaft first, then
heat the outside of the roller to seal after it is assembled (like
plumbers do with pipes sometimes). Another option may be loc-tite (if I
can keep it off the bushing ID during assembly). I think they make a
loc-tite that even works in conjunction with a press fit.

What's the allowable shear stress of your material? You're at 20,000 PSI...

The shear stress for most hardened dowel pins I have seen is 130,000
PSI. Mcmaster Carr lists 130,000 PSI for the inch size pins meeting
ASME B18.8.2 standards. The metric pins meet DIN 6325. I am assuming
the hardened metric pins would have very similar maximum shear stress
ratings. Mcmaster Carr lists the single shear load rating for the 2mm
OD pin at 710 pounds, but does not give a PSI rating for metric. I have
a copy of ASME B18.8.2 standards but not the DIN 6325.

As long as the pressed together roller and dowel act fairly close to a
solid piece, I think everything will be fine, I will know for sure
soon.

Anyway, thanks again for all your help guys, especially Mike for
running the FEA for me. It sure helps to get some different opinions
and viewpoints sometimes.

Take care,
John