Shaft deflection problem

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
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
wrote:

For a first estimate, why not do this:
for a beam of the length of the thick portion, find the loaded deflection.
For a beam of the full length but the thin section the whole length, and point-loaded at the ends of the imaginary thick center and supported at the actual end positions, check the deflection from the end to the load near the end.
Add the two deflections - that from the thick center, and that from the thin ends and that approximates the actual deflection value...
Brian Whatcott Altus OK
Brian Whatcott
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

The correct solution (I think) is to cut the thick bit in half, and treat it as a cantilver. Work out the deflection DT of the end of the thick bit, and the rotation R (standard result) with half the load on it. Work out the tip deflection dt of the thin bit considered as a cantilver length L, Then the total deflection will be DT+dt+L*R
Cheers
Greg Locock
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
wrote:

Here's a possible objection to Greg's approach. Let's make the thin shaft length, infinitesimal, so the shaft is essentially all thick. We can work the central deflection on this shaft, no problem.
But following Greg's lead, we cut the shaft in two, and measure the cantilever deflection at the tip for an encased center of half length, and half the distributed load length. That shorter cantilever gives 30% less deflection on my beam program....
Brian Whatcott
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Hi everyone,
Thanks for your replies I appreciate your input.
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.
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 roller length ?
If I try to follow Gregs advice and treat the problem as a cantilever, do I point load the small shaft at the center of the 3mm bushing length ? This gives a cantilever overhang of 0.1" from the edge of the .1875" OD section. With a load of 100 pounds on the end of a 2mm OD shaft, cantilever mounted & having 0.1" of overhang, I calculate 0.001" deflection at the dowel end at 0.1", a bending stress of 209 KSI, and a slope of .506 degrees. This is about 1/2 of the .909 slope, so I gues the ends would seem to deflect about .004" with this method, plus whatever the thick portion deflect under cantilever load.
I am not sure I am completely clear on Brian's method.
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. 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
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Greg Locock wrote:

If the stepped beam is symmetrical along its length, then cutting it in half would give a guided-simply supported beam, where the guided end it at the symmetry plane.
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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/inē
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 inē
#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 inē
#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 inē
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/inē No Limit specified Compressive = 27034.99 lb/inē No Limit specified Shear (Avg) = 20484.67 lb/inē 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/inē Compressive = 0.0 lb/inē Shear stress = 0.0 lb/inē
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/inē Compressive = 0.0 lb/inē Shear stress = 0.0 lb/inē
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/inē Compressive = 6285.415 lb/inē Shear stress = 3636.475 lb/inē
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/inē Compressive = 20719.99 lb/inē Shear stress = 3606.844 lb/inē
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/inē Compressive = 6292.608 lb/inē Shear stress = 3606.844 lb/inē
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/inē Compressive = 0.0 lb/inē Shear stress = 0.0 lb/inē
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/inē Compressive = 0.0 lb/inē Shear stress = 0.0 lb/inē
OLD DESIGN WITH .1875" LONG ROLLER
BEAM LENGTH = 0.5047204 in
MATERIAL PROPERTIES steel: Modulus of elasticity = 29000000.0 lb/inē
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 inē
#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 inē
#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 inē
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/inē No Limit specified Compressive = 84130.45 lb/inē No Limit specified Shear (Avg) = 20484.67 lb/inē 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/inē Compressive = 0.0 lb/inē Shear stress = 0.0 lb/inē
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/inē Compressive = 0.0 lb/inē Shear stress = 0.0 lb/inē
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/inē Compressive = 84034.27 lb/inē Shear stress = 20484.67 lb/inē
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/inē Compressive = 20719.99 lb/inē Shear stress = 3606.844 lb/inē
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/inē Compressive = 84130.45 lb/inē Shear stress = 20317.75 lb/inē
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/inē Compressive = 0.0 lb/inē Shear stress = 0.0 lb/inē
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/inē Compressive = 0.0 lb/inē Shear stress = 0.0 lb/inē
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
wrote:

///
///
I think it is praiseworthy when a person wants to take a device analysis as far as possible with his own resources.
Having monitored the thread, I think the very best advice I can offer is to consult a mech engineer in a similar field.
If you lack the background to realise when extreme pressures or undesirable wear properties are proposed, you are going to be disappointed, either sooner, or worse, later.
Brian W.
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
John: Looks good!
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

What is the intended application? Are people envolved? Did you consider tear-out of the yoke? These are some of the other aspects to the design. As for bushings, take a look at oil-impregnated bronze bushings (sometimes called Oilite bushings). They are self lubricating for a period of time determined by use and load. After that, it is simple to just replace them. If human safety is envolved, you will have to be very careful when applying a safety factor and do not forget the effects of stress concentrations in the stepped shaft and yoke.
-Matt
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

I doubt very much that there is such a formula. Not that it can't be done, but that it's difficult to find it.
You better solve the integral (twice):
d2y/dx2 = M(x)/EI
taking care of boundary conditions in each region of the shaft i.e.: each time the equation of M(x) or I (section) varies.
And considering that at each supported end y = 0.
I, myself, would do it numerically in an Excel sheet, for instance. That way you don't have to worry about boundary conditions, but you have to have care defining the equations of V(x) (shear force) and M(x) bending moment.
HTH.
JV
d2y/dx2 = second derivative of deflection vs x
M(x) = equation of bending moment along x
E = Young modulus
I = Moment of inertia of section (variable in this case)
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.