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cosine distribution

cosine distribution

cosine distribution

(OP)
Hi,
I am dealing with a beam having a reaction moment in the middle of the beam. Beam is simply supported. Assumption is that the load have a cosine distribution with a running load w = (pi*2 x M)/L*2 x 2 [lb/in].
Does anybody know where can I find more information about cosine load distribution, confirming that formula?

Thank you in advance all

RE: cosine distribution

I don't understand the question. The formula doesn't mention cosine. Also, if the beam is simply supported, how can it have a reaction moment at midspan?

BA

RE: cosine distribution

(OP)
BA,
yes, you are right, consider that the moment is applied, not reacted. I am at the case where the author presents a beam with a beam having a moment midspan and presents the abovementioned formula for running load in which formula the moment is included and he said the cosine distribution is considered. that's why I am looking for more information about cosine distribution (it is similar to normal distribution from statistics, but I couldn't find that formula anywhere...)
ABV97

RE: cosine distribution

It sounds a bit like a Fourier Series but I haven't used that in many years, so would have to review it. If the load is represented as a cosine function, a variety of effects can be represented including concentrated loads and possibly concentrated moments. The advantage is that it provides a continuous which may be readily integrated.

BA

RE: cosine distribution

meant to say continuous function which may be integrated.

BA

RE: cosine distribution

(OP)
Thank you BA for your thoughts!

RE: cosine distribution

and don't multi-post !

Quando Omni Flunkus Moritati

RE: cosine distribution

Take the standard beam bending equations.
Set deflection = y(x) = Yocos(x*pi/L), where x=0 is the center of the beam and Yo is the deflection at the center. This is a cosine distribution with maximum load at the center, tapering to zero at either end.
The distributed load is q=EI*d4y/dx4
The moment distribution is M=-EI*d2y/dx2.
Differentiate the deflection equation 4 times and plug it into the q equation, differentiate twice and plug it into the M equation, then you can calculate q/M = (pi/L)^2 or q = M (pi/L)^2

I think this is on the right track. You have a 2 factor in there that I don't get. Maybe I goobered it up somewhere, or maybe your source did. Maybe something is defined differently somewhere.

I'm using "*" for multiplication. I think you used it for "raised to the power of". It looks like pi*2 is meant as pi-squared, etc.
I'm using "^" for "raised to the power of".
I'm using d2y/dx2 to mean second derivative of y with respect to x, and using "Yo" for Y-nought, that being a constant.

I had to look the beam equations up on the internet, and they use "w" for deflection, not load, so that's a potential source of confusion.

Where this might come up: In some vibration applications, you assume a deflected shape of the vibrating beam, and that is a reasonable selection for a simple beam.

RE: cosine distribution

(OP)
JStphen, thank you so much for your thorough answer, highly appreciated!

RE: cosine distribution

personally i'd challenge a cosine deflection, centered on the mid-span, as being appropriate for a moment loading.

i'd've thought it was more a sine dist'n ? but personally i'd use a linear distributed load to represent an applied moment ... assuming that such a load can be applied to the structure ... else it'd propably be a couple.

Quando Omni Flunkus Moritati

RE: cosine distribution

Using JStephen's approach, the deflection must be zero at the center and at each end. A sine curve works better with y = C*sin(2πx/L) where x varies from 0 to L.

Then, when x = 0, L/2 or L, y = 0 which it should
and when x = L/4, 3L/4, y = C, -C respectively.

The constant C would depend on the magnitude of moment and the stiffness of the beam.


BA

RE: cosine distribution

My response is based on my understanding of the question, which was not that clear to me, and changing how the question is interpreted will obviously invalidate any conclusions.

RE: cosine distribution

(OP)

First, thank you again to JStephen, BAretired and rb1957 for your thoughts and comments, I think your thoughts are quire handy and helpful. I'll try to clear the questions, since some confusions raised. All numbers are made up for ease of explination.
I have a box laying over a beam. For 9g forward load, a 10,000lb load located into the box centroid, 15" above the beam, will create a moment, which moment will be reacted at the beam supports. Beam is simply supported at the beginning and at the end with L = 20". So the author said that, assuming a cosine distribution, we will have : W = (pi^2 x M)/L^2 x 2 [lb/in]. I am trying to confirm and validate that this running load is true and correct. I haven't been in Calculus for years, but what I remember is that we are making first derivative equal to zero to find max/min of the function. I hope this clarification of the problem helps, and thank you again all for your thoughts!!!

RE: cosine distribution

Okay. Correct me if I'm wrong. You have a beam AB spanning 20" with hinge at A, roller at B and a horizontal force of 10,000# applied to a rigid arm 15" above the beam at midspan. This force produces a moment at the neutral axis of the beam of 10,000#x15" = 150,000#".

The vertical reactions at A and B are M/L or 150,000/20 = 7,500#, downward at A and upward at B. The horizontal reactions are 15,000# at A and 0 at B.

Shear is -7,500# from A to midspan and +7,500# from midspan to B. Moment varies linearly from 0 at A to -75,000#" just left of midspan and from 0 at B to +75,000#" just right of midspan. The 150,000#" discontinuity at midspan is the applied moment.

Where does the cosine distribution enter the picture?

BA

RE: cosine distribution

so the beam is being loaded by torque ?

Quando Omni Flunkus Moritati

RE: cosine distribution

rb1957,
That was not my assumption, but you may be correct. I assumed the 10,000# load is applied in the positive x direction at a 15" eccentricity to the beam centroid.

BA

RE: cosine distribution

(OP)
BAretired - that is correct, only that horizontal reactions at A and B are by 5000#, not 15,000#, and shear is a constant +7500 all the way from A to B. And moment, as you said, has a jump in midspan, I think. That's why I was confused that the author is stating that "assuming a cosine distribution, has ...etc"

RE: cosine distribution

so we're a beam running port/stbd, with a load of 10,000 lbs fwd, 15" above the beam ... that'd be a torque on the beam of 150,000 in.lbs ... no?

of course, the beam could be running fwd/aft ... in which case it would indeed be a moment on the beam ...
so there'd be a fwd reaction of 5,000 lbs at each end, a vertical reaction +-7500 lbs (balancing the moment of 150,000 in.lbs)
then the question is how does this moment get into the beam ? and the answer depends on how the box is attached to the beam ...

1) a couple (at obvious fasteners),
2) a couple at the compression bearing edge and the obvious tension fastener,
3) a linear varying load,
4) a full wave cosine, peaking at the ends, zero in the middle ...

Quando Omni Flunkus Moritati

RE: cosine distribution

Quote (ABV97)

BAretired - that is correct, only that horizontal reactions at A and B are by 5000#, not 15,000#, and shear is a constant +7500 all the way from A to B. And moment, as you said, has a jump in midspan, I think. That's why I was confused that the author is stating that "assuming a cosine distribution, has ...etc"

Actually, the horizontal reactions at A and B are 10,000# and 0# respectively since B is a roller but you are correct about shear being constant from A to B, namely -7,500# by the usual convention. My bad in my earlier post.

I still don't know what the guy meant by "cosine distribution".

BA

RE: cosine distribution

(OP)
...cross section of the beam is I beam extrusion. bottom of the box has welded "G" extrusion - I mean square box with a slot on the bottom and this "G"extrusion runs over entire length L over the top section of the I beam extrusion. That means thru the entire lengt L both extrusion bear each other and that's how the load is being transferred...

RE: cosine distribution

so the beam is fwd/aft ?

is the beam an I section with a G extrusion welded onto it (so the extrusion is supporting the I beam) ?

if the box has a slot in it (for the G extrusion) how is the box attached to the extrusion ?

Quando Omni Flunkus Moritati

RE: cosine distribution

(OP)
...the beam is fwd/aft oriented, G extrusion is welded on the bottom of the box and the box and the G extrusion with his slot facing down slides over the I extrusion....than having stops in the front and aft and the box is completely fixed...

RE: cosine distribution

"having stops in the front and aft and the box is completely fixed" this sounds like there are discrete (two?) attachments between the box and the beam ... so i think you can represent the beam loading as either ...

1) a simple couple at the discrete attmts (interpolate, linear varying, if more than 3)
2) discrete tension loads and linear varying compression (contact)

a full wave cosine distribution (peaks at ends) sounds overly and unnecessarily complicated; but you could apply this loading ... just make the toatl moment of the applied load equal to the applied ... should be easy enough.

Quando Omni Flunkus Moritati

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