Question regarding shaft deflection problem

[John2004]

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.,

http://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 and/or 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

 

[Cockroach]

Interesting problem.

You probably would need to use superpositioning, i.e. decompose the system into the various elements and then transfer loads across to those members in contact. This would be similar to analysing a truss for example, the load in one member becomes a reaction in the adjoining member, equal in magnitude but of the opposite sense.

Obviously your situation is as complex, but of no more difficulty. You should try to use Roarks's or Marks as a reference and consider a stepped shaft of equal geometry. The assumption is that your composite shaft "acts like one" so you should get very similar answers. Note that there will be some differences, losses I imagine due to connections not being so rigid. None-the-less, you should be able to obtain a second, independent answer that would closely match the first analysis.

I will read your posting once more, perhaps try it myself over the long weekend. Maybe I can post a FEA for you.

In the meantime, good luck with it.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[GregLocock]

You could consider it as being a 4 point loaded beam, with the loading points being at the middle of each bush and the outer end of the pressed on shaft, as a way in.

However many things about this approach make me nervous, I'd certainly be more confident after an FEA confirmed the estimate.

On the other hand, can a 2mm by half inch long rod even deflect by 0.6 thou? And what sort of plastic bush will react a 300N/mm^2 (WAG) stress, reliably? I don't think you have an elastic problem at all.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Philrock]

At your relatively high load level, I don't think the center section will behave as a solid piece - intutition tells me that ignoring the stiffness contribution of the roller would be more accurate, though I agree with GregLocock that FEA is called for. I think modelling the roller/shaft assembly accurately will be very difficult. For accurate numbers - do experiments.

While the shaft will certainly bend in the right direction to make the ends go up, they may not actually end up higher than their starting position - depends on the load distribution you end up with inside the bearings. For a first order approximation, assume a linear pressure distribution. My guess is that the bearings will be loaded along their full length and the entire shaft will deflect downward.

 

[John2004]

Hi everyone,

Thanks for your replies,

Cockroach, you mentioned FEA, if you or anyone else could run a quick FEA, I would be grateful.

I don't have FEA software and that's not my area of expertise. Most of the FEA software I have seen seems to have a steep learning curve, but perhaps someone knows of a freeware FEA or online FEA I could try that would do the job.

I don't have the formula for a stepped shaft or beam, where the loaded center portion of the shaft has a larger diameter than the shaft ends. If anyone could please give me this formula, it may shed some light on the issue, but as Philrock & cockroach pointed out, the pressed on roller may not really act just like one solid piece with the shaft. I think it will be close though.

Thanks again for your replies.
John

 

[John2004]

Hi everyone,

Thanks for your replies I appreciate your input.

I wanted to try to fuurther clarify the problem in case it can help:

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.

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 or deforming it past it's elastic limit, at least that's what my instincts say.

If a 2mm OD X .269" long steel beam is simply supported at the very ends, and loaded in the middle with a point load of 200 pounds, I calculate that the middle of the dowel will deflect by .0007". The bending stress will be 280 KSI, the bending moment will be 13.4 lb-in, and the slope is .909 degrees.

However, when you press the steel tube onto the center of the dowel, everything stiffens up, and all of this changes. Plus, as I understand it, a small beam like this, with this aspect ratio, might not conform to standard deflection formulas.

The double shear stress for the 2mm OD dowel is 741 pounds.

If a .505" long dowel is supported with two simple supports, spaced .269" apart, with the center of the .269" wide span in line with the center of the .505" length, then loaded with a 200 pound point load at the center of the .505" length, the dowel center will still deflect by .0007", but I think the very ends of the .505" long dowel will curl up or deflect by .008".

So, the question becomes how much will the .1875" OD X .1875" long tube decrease center deflection, slope, and thus the .008" curl up at the dowel ends ? Also, will the fact that the load is distributed over the 3/16" length help, since the cam plate is 3/16" thick which matches the cam-follower roller length ?

I found a copy of a free demo program called "DTbeam"

http://www.dtware.com/

that appears to let you run beam deflections for beams that have different OD sections and properties along their length. You can then select, copy and paste the results. However, it seems to use a "pinned" joint for the supports and I am not sure this would act like a simple support.

Nevertheless, if it can calculate the deflection and slope at the center, then I may be able to figure out the deflection at the ends. The program looks easy to use, perhaps someone could take a look and we could compare results ?

The demo is fully functional, but you cannot save the results, and it only runs for 45 minutes at a time, then you must restart the program. My first attempt at modeling my problem with this program seemed to produce extremely small deflections, and I am not sure they are valid.

Thanks again guys, I appreciate your input & would appreciate any further feedback.

John

 

[John2004]

Hi everyone,

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

The only question is whether the roller being pressed onto the center of the dowel, will basically act like a stepped shaft made from one solid piece ?

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