Deflection formula for beam with distributed load over center portion 2

[electricpete]

Lets say I have a massless beam of length L simply supported on both ends. (x coordinates 0..L)

Apply a uniform distributed load over the center distance d of the beam
(i.e. from L/2 – d/2 to L/2 + d/2). The total load is F (distributed load is F/d force per unit legnth).

What is the deflection at the center of the beam?

(I am going to try to calculate it from beam theory but I’d like a way to double-check that calculation).

The reason for the calculation is to quantify the effect of spider construction upon susceptibility to magnetic pull for induction motor. The less flexible the rotor is, the more it can deflect under influence of unbalance or magnetic force. The more it deflects the higher the magnetic force pulling it further off-center. Rotor operates far below first critical.

Most motors have fairly long spider similar to length d which distributes load from rotor core along the shaft. We have one motor with only a single center set of spokes which will act similar to point load. It has shown itself to be susceptible to vibration from small increases in bearing clearance or shaft runout. My theory is that it is due to the low static stiffness of this rotor configuration (point load).... others don’t quite see the connection so I want to quantify it a little bit more with a calculation of deflection under point and distributed load.


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[rb1957]

isn't it just that you have two unknowns (the force reactions at the ends) and two equations of equilibrium (sum Forces and sum Moments) ?

as you know the moment distribution M(x), the slope is the integral of M(x)/EI and the deflection is the integral of the slope. you actually have a whole bunch of constants of integration, depending on how many segments you have to integrate over. In your case it looks like you've got three segments (either side of the distributed load and the distributed load) as you've got three different moment equations. so you've got six constants of integration, and six boundary conditions defelection = 0 at x=0 and x=L, and slope and deflection at the junctions (where the moment equation changes) ... if that makes sense [at the junctions you've got two ways to define the slope and deflection of the beam (from either moment distribution), obviously they have to be the same].

 

[electricpete]

“isn't it just that you have two unknowns (the force reactions at the ends) and two equations of equilibrium (sum Forces and sum Moments) ?”
Yes, solving those two static equilibrium equations led directly to reaction forces RR and RL.

Each additional segment beyond one provides an addition integration constant for each parameter (M,Theta, X) and one additional continuity boundary condition for each parameter (M, Theta X). So the number of segments doesn’t seem to change the fundamental picture since each new segment adds one equation per boundary condtiion.

I have to think a little bit about the rest of your comments.


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