Simple Mechanics problem - stepped shaft uniform load

[silkyslim]

I need to know the general stress equation for a stepped shaft with a uniform load on section 2. The reactions R represent roller bearings, and are located at their centerlines. I don't know which equations from the charts to apply to each section. Thanks.

 

[rb1957]

are the reactions truely on the edge of the shaft ? or is the arrow showing the CL of the bearing ?

section 2 is not very clearly shown ...

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[silkyslim]

The arrows are at the centerlines of the bearing. the shaft continues on one side coupled to a drive and on the other it ends near the end of the bearing. What I mean by section 2 is the part of the shaft subjected to the uniformly distributed load W. My end goal is to see how the stress at the shoulders changes as L2 changes, while the total length and the total load W stay the same.

 

[rb1957]

so you have a beam with a partial-span uniform load ? bearings are reasonably simple supports ? does that help point to a table ?

btw, why in the FE forum ? checking a model result ?

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[silkyslim]

I can't find any partial span beams in the tables. I only see total span. I also don't know how to reconcile the different diameters in each section.

As far as the forum, is there a better subforum for these questions? This is basically "homework help" but yes I am trying to validate a model.

I am trying to develop a standard 3D setup for briquetting rolls for FEA. The problem is I don't have access to 2D software (I have ProE mechanica basic), the engineer that usually does this analysis is located off site and uses 2D models, which I am essentially copying and trying to do by hand.

 

[rb1957]

oh no ... you've mentioned the h- word .... sorry, student posting is NOT allowed.

please take your question to your TA or peer group.

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[silkyslim]

I was referring to the subject matter. I am not a student.

 

[rb1957]

ok ... that was confusing ... maybe you're helping someone else with their homework ?

the moment in the beam is easy to calculate ...
assuming simply supported, length "L", then end reactions are (w*a)/2 ... where "a" = the length of the distributed load "w"

then define "b" = (L-a)/2 the unloaded length at each end of the load, yes?

then for x<b, M = Rx = (wa/2)*x

and for b<x<a/2, M = Rx-w(x-b)*(x-b)/2 ... Mmax(x=L/2) = RL/2-w(L/2-b)(L/2-b)/2 = waL/4-w*a/2*a/4

and the stress = My/I = M(d/2)/(pi*d^4/64) = 32M/(pi*d^3)

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[silkyslim]

Thanks for your time rb. Following your steps, it leads me to believe that reducing w*a to a point load W will not change the stress in the smaller diameter sections, but will double the stress in the roll journal, keeping the max stress (at x=b) the same for my diameters.

 

[rb1957]

"will double the stress in the roll journal" ? ... i see the loading/stresses in the smaller diameter shaft the same for both loadings, ie either a distributed load on the larger diameter shaft or an equivalent (=wa) point load in the middle.

the reaction is the same in both cases, no?

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[silkyslim]

Yes the reactions are the same, and the stresses in the smaller portions will be the same. The max stress @ x=L/2 will be double with a point load vs distributed.

 

[rb1957]

oh, that's what you meant ... depends on the geometry a/L ... if a = L the point load moment (P/2*L/2) is double the distributed moment (P/2*L/2-P/2*L/4). for a partially loaded span the ratio is lower (the moment due to the distributed load is higher)

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