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Request for help with shaft deflection problem Question
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Posted by: john2003 Ū

10/03/2005, 19:00:20

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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, 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






Modified by Administrator at Mon, Oct 03, 2005, 19:04:01

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Re: Request for help with shaft deflection problem
Re: Request for help with shaft deflection problem -- john2003 Post Reply Top of thread Forum
Posted by: john2003 Ū

10/10/2005, 13:42:09

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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 3/16" OD roller being pressed onto the center of the 2mm OD 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"

https://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







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