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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[JStephen]

Refer to beam loading formulas in Roark's Formulas For Stress and Strain or Strength of Material textbooks.

If the formulas show loading from one end to an arbitrary point, you can superimpose two cases to remove the loading at the other end.

If the loaded area is fairly small, you can treat it as a point load. If it is fairly larger, consider treating the beam as a uniformly loaded beam.

I recall the subject of "whirling" coming up in one of my classes long long ago- that being the case where an initial deflection in a spinning shaft throws it off center, which then gives a centrifical force which deflects if further- a stability problem, in other words. You might want to check into that- sounds sort of like what you're getting into.

 

[electricpete]

Thanks. I will try to get hold of a Roark's.

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[GregLocock]

Failing that you can solve it by superposition of teh deflections due to a moving point laod, to suffcient accuracy.

y=a^2*b^2/(3*L)*F/(E*I)

a and b being the distances from the ends of the beam to the point load.


But yes, you do need to buy Roark.




Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[prex]

electricpete,
you can go to the site below instead of using the Roark.
Anyway the formula you are looking for is as follows:
f=P(d³-4Ld²+8L³)/384EJ
This correctly reduces to
f=5PL³/384EJ for d=L (the formula for uniformly distributed load over L)
and to
f=PL³/48EJ for d=0 (the formula for a concentrated center load)


prex

http://www.xcalcs.com

Online tools for structural design

 

[electricpete]

Thanks Prex. That is exactly what I was looking for.

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[electricpete]

A question about xcalcs input:

I am used to calculating beam response using E, I, lengths and loading.

Here instead of I they are asking for
J = Moment of inertia in load plane:
Z = Section modulus in load plane:

What are these two parameters and how are they related to I?


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[rb1957]

"Most motors have fairly long spider similar to length d which distributes load from rotor core along the shaft."
i think you mean that most motors have several rotors, distributing the load onto the shaft (the length of the shaft per se doesn't help to distribute the load.

"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."

i think it is more like a dynamic instability, like JStephen mentioned, "whirling", but then the shaft would need to have mass (like most real things). It could be theat the single rotor motor might be badly designed in the first place, so that the shaft has a marginal dynamic stability.

 

[mrMikee]

From Blodgett "Design of Welded Structures" Case 3c for beam supported at both ends with uniform load partially distributed over span:

when a=c the deflection at center is... wb(8L³-4b²L+b³)/(384EI)

When a=c uniform load is length 'b' centered on span.

w = unit load
L = length
E and I as you would expect

Regards,
-Mike

 

[rb1957]

sort of like prex's post ...

 

[electricpete]

ThanksMike - now I have a reference for the formula.

rb1957 - No, we have one rotor. It includes a shaft, a spider and a core. The core contains the large majority of the mass (and also has magnetic forces applied to it). The spider mounts the core to the shaft. Most spiders run almost the length of the core and will transmit force from the core to the shaft along their length. This particular spider is very short axially, and transmits force from core to the shaft approxiatmely at one point.

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[sreid]

I like rb1957's whril idea. I visulize a "Rocking Mode" caused by the spider transmitting a moment to the center of the shaft. The natural frequency of this mode might be much lower than the usual shaft first critical frequency.

 

[electricpete]

I realized this particular motor rotor is attached for off-center toward the left side of the shaft (actually toward the bottom,it's a vertical, but I work my beam problems left to right).

I solved the general case for distributed load extending from distance a to distance b to the left end. Solution method was integrating force 4 times (with factor of 1/EI between moment and slope).

This is 71" long shaft between bearings, 10" diameter in the middle (less at the ends but I treated it as uniform for simplicity), 19,000 pounds rotor weight.

I plotted V, M, Theta (slope), and Y (deflection) for loading corresponding to rotor weight in horizontal orientation for two different cases:
1 - a=25, b=30
2 - a=15, b=40
To my surprise there was hardly any difference in the max deflection (can't see any by eyeballing the graph). I guess that part of the reason is that the parts of the shaft far away from the load can't tell the difference between the two load patterns.

One thing I didn't model was the stiffening effect of the spider (increase in I at that portion of the shaft). Seems like that could be significant for two reasons:
1 - the center of the shaft is the most important for preventing bending... that is apparently why shafts are designed bigger in the center.
2 - The spider extends out to a 3-foot diamater (4 spokes in this design, 6 in many others)... that could be a significantly higher I than the shaft alone over that length.
I'm not sure if I have the interest or energy to add that to my model.

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[electricpete]

Here is link to the calculation I mentioend above comparing case 1 and 2

http://home.houston.rr.com/electricpete/BeamLoad3.pdf

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[electricpete]

By the way, in the middle of page 3 you can see I checked my model out by substituting
a=L/2-d/2
b=L/2+d/2
x=L/2
The results agreed with the expression given by Prex and Mikey which gives me confidence that I didn't make an unintended math error in my approach.

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[electricpete]

One weird think I noticed.

Given applied force and reaction forces we know V(x).

Then we determine M(x), Theta(x), and Y(x) by three integrations which involve three integration constants.

There are 4 available boundary conditions for the simply supported case M(0)=0, M(L)=0, Y(0)=0, Y(L)=0.

I only used three of them:M(0)=0, Y(0)=0, Y(L)=0.
(I didn't write out M(0)=0 was it done in my head by leaving out the integration constant that would have to be 0).

Then sure enough in the plot it shows M(L)=0.
Could I have chosen any 3 of the 4 and expected it to work?
Are beam problems usually that way>?
It seems like I should be able to change one of the 4 boundary conditions without affecting the other 3 but if that were true I would need all 4 to solve the problem.

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[electricpete]

Maybe somehow one of those boundary conditions are rolled into solving V(x)? It doesn't seem like it because I just used forces but maybe there is an assumption required for calculating V(x) related to boundary conditions that I'm overlooking ?

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[prex]

You got it: one boundary condition is implicit in the derivation of V(x), as you obtain it by a first integration from q(x) (the load per unit length). And yes you can use any 3 of the 4 conditions, once you have the correct V(x).
However at this point can't see where are you exactly heading with your calculations.
Concerning some of your questions:
-J=I
-Z is sometimes called W, it is necessary to calculate stresses
-no surprise on the small difference in deflection, as the 2 load distributions have the same resultant (in amplitude and position)
-and yes your spider could significantly add inertia to the shaft, but this depends critically on how and where it is connected; as you mentioned somewhere else a short spider, this one should give a limited impact on inertia and consequently on deflections.


prex

http://www.xcalcs.com

Online tools for structural design

 

[Beggar]

J = Moment of inertia in load plane:
Z = Section modulus in load plane:

What are these two parameters and how are they related to I?

The section modulus, Z, is simply the moment of inertia, I, divided by the distance from the neutral axis, c. That is, Z=I/c. Therefore, the flexure equation for bending stress becomes [σ]=M/Z.

I've typically seen "J" used as the symbol for the polar moment of inertia which is useful in torsional and rotational calculations.

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[electricpete]

Thanks guys.

Thinking about Prex' comments, I guess can see that the combination of forces F, RLeft, and RRight which are in static equilibrium from both force and moment considerations will only remain in static equilibrium only if any moment applied on the left side is also applied on the right side. In other words M(L)=M(0). So when taken together with M(0)=0 BC applied later, this form of V does ensure M(L)=0.

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