Request for correct way to model beam / shaft forces (special case)

[John2004]

Hi everyone,

I would like to ask if anyone could please help me with the following problem.

If you go to this link

http://www.ice9.zoomshare.com

I have uploaded two dimensioned JPEG drawings of two different options I have for implementing a cantilevered mounted bearing housing. Just click on photo's, then "housing drawings", then click on the drawings and maximize the window.

As shown in the two drawings, I can either use a 3/8" OD shaft with the bearings spaced further apart, or a 7/16" OD shaft with the bearings spaced closer together. The shaft is stationary, and the needle roller bearings are pressed into the housing. The housing oscillates on the shaft.

I can determine how much force is on the bearings, due to the load on the end of the housing, using the formula...

Where (LA) = the housing load

Load on bearing #1 = (LA) * B / A

Load on bearing #2 = (LA) * C / A


However, I am not sure exactly "how" the force will be applied to the shaft. Will the force be more of a torque moment, having a rotation axis in the middle of the two bearings ? Will there also be a vertical load and if so, how do I divide the forces between the two types of loading ?

The shaft is a hardened steel dowel pin, meeting ASME B18.8.2 standards.

The main two things I am trying to determine are, how far will the end of the dowel or housing deflect, due to the housing load, and will the deflection and stresses be within the elastic limit and/or yield strength of the hardened dowel pin ? I don't want to overstress the shaft.

The maximum clearance between the bearing ID and shaft OD will be .002". With AutoCAD, it looks like the end of housing #1 will move .005" from it's longitudinal axis (Just due to bearing play & not considering shaft deflection) and the end of housing # 2 will move .002" from it's longitudinal axis just due to bearing play and not considering shaft deflection. This would be a worse case scenario, using maximum bearing / shaft clearance.

My main goal is to minimize the amount of deflection of the end of the housing from it's longitudinal axis, and insure that the dowel / shaft is not going to be overstressed. The maximum load on the end of the housing will be about 130 pounds, but it would be nice if it could handle 200 pounds for a little safety factor.

I have a beam design program that can model vertical and torque moment forces, but I don't have FEA. Perhaps the beam program can do the job, if I know how much force to apply as a moment, and how much force to apply vertically, to each bearing, in each case.

I would really appreciate any advice or suggestions on the correct way to model this.

Thanks for your help.
John

 

[JStephen]

Sum moments about each bearing to find the load on the other. The two loads you get should total up to the applied load- there isn't any additional load. The bearing loads you calculated are vertical loads, up or down, at that point. For simplicity, treat them as point loads, though they would be distributed over some area in reality. You could actually develop a moment in one bearing (essentially a load up at one end, down at the other end) but don't have enough information to calculate that.

For modeling, treat it as a beam with simple supports at the bearing points, applied load at the end, no moments or torques. You don't need structural software but can use it if that's easier. Add elastic deflection to the deflection due to slack in the bearings that you're calculating. There will be some additional deflection due to the bearings themselves, the support plate, etc.

Maximum stress should be at or near the connection to the plate in either case; the moment is just the load times the distance. Check on fatique stress as well as yielding.

It's possible you may yield the support piece rather than the pin, or may work the pin out of the hole under cyclic loading.

 

[John2004]

Hi JStephen,

Thanks for your reply.

>JStephen wrote:
>The bearing loads you calculated are vertical loads, up or >down, at that point. For simplicity, treat them as point >loads, though they would be distributed over some area in >reality. You could actually develop a moment in one >bearing (essentially a load up at one end, down at the >other end) but don't have enough information to calculate >that.

John2004:
Using the formula I gave, could the load on bearing #2 be considered downward and the load on bearing # 1 upward ? Would this be more accurate than modeling two downward loads ?

The housing / bearing loading seems the same as it would be if you would slide a hollow pipe over the shaft, with two contact points on the shaft, and then try to bend the shaft with the pipe. It seems the housing load is trying to bend the end of the shaft, between the two bearing contact points, in the same manner. With this in mind, I was considering modeling this in my beam design program as a torque moment, with the rotation axis in-between the two bearings, and the moment arm length running from the rotation axis to the housing loading point . Is this wrong ?

I have been using point loads in the center of the bearings, to try to keep it as simple as possible.

>JStephen wrote:
>For modeling, treat it as a beam with simple supports at >the bearing points, applied load at the end, no moments or >torques. You don't need structural software but can use >it if that's easier.

John2004:
I'm not sure I follow you here. The shaft is cantilever mounted, and pressed into a steel support at one end. If I put simple supports at each bearing, in addition to the cantilever support at one end, it seems the beam will be to well supported to give an accurate model.

>JStephen wrote:
>Sum moments about each bearing to find the load on the >other. The two loads you get should total up to the >applied load- there isn't any additional load.

John2004:
I'm not sure I completely follow you here. Can I use the formula I gave, and/or the torque moment method I suggested, to get the bearing loads ? How do I calculate the moment about each bearing individually ?

Thanks again for your help.
John