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
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
BA
RE: cosine distribution
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
BA
RE: cosine distribution
BA
RE: cosine distribution
RE: cosine distribution
Quando Omni Flunkus Moritati
RE: cosine distribution
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
RE: cosine distribution
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
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
RE: cosine distribution
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
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
Quando Omni Flunkus Moritati
RE: cosine distribution
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
RE: cosine distribution
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
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
RE: cosine distribution
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
RE: cosine distribution
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