cosine distribution

[ABV97]

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

 

[BAretired]

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

 

[ABV97]

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

 

[BAretired]

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

 

[BAretired]

meant to say continuous function which may be integrated.

BA

 

[ABV97]

Thank you BA for your thoughts!

 

[rb1957]

and don't multi-post !

Quando Omni Flunkus Moritati

 

[JStephen]

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.

 

[ABV97]

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

 

[rb1957]

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

 

[BAretired]

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

 

[JStephen]

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.

 

[ABV97]

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!!!

 

[BAretired]

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

 

[rb1957]

so the beam is being loaded by torque ?

Quando Omni Flunkus Moritati

 

[BAretired]

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

 

[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"

 

[rb1957]

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

 

[BAretired]

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

 

[ABV97]

...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...