Deflection of beam in "X" and "Y"
Deflection of beam in "X" and "Y"
(OP)
I am trying to figure out how much the end of a cantilevered beam will displace if loaded at the end. I know the equation that will give me the displacement in the "Y" direction, but how do I figure the displacement in the "X" direction?






RE: Deflection of beam in "X" and "Y"
R = ((Dx)^2+(Dy)^2)^0.5
In other words, square both deflections, add them and then take the square root.
Good luck.
RE: Deflection of beam in "X" and "Y"
I am assuming that you would have a load applied in the other direction. Otherwise, your deflection would be zero!
RE: Deflection of beam in "X" and "Y"
The beam is horizontal and the (point) load is acting vertical. Every equation that I have only solves for the deflection in the "Y" direction. In reality as the cantilevered beam deflects down, the end of the beam will move toward the wall support.
I am trying to determine how much is moved toward the support.
Thanks
RE: Deflection of beam in "X" and "Y"
Alternatively you could integrate along the length of the curve to equate the original length of the beam to the curve length to evaluate x, ie. Integrate from 0 to x the curve length [root(1 + f'(x)^2)]=L to find x.
RE: Deflection of beam in "X" and "Y"
In reality, of course, the conditions are not perfect, and there is a lateral displacement and/or rotation.
Depending on why you need to know there are a couple of different approaches. If it's a matter of allowable stress and/or displacement you could assume a horizontal load of say 0.1 of gravity load and/or apply the load eccentric.
If it has to do with needing to know "exactly" what the lateral displacement is you may have a bigger problem on your hands. Load testing is a possibility. Another choice would be to brace the tip of the cantilever back to the support, thus greatly reducing any lateral motion.
Hope this helps.
RE: Deflection of beam in "X" and "Y"
"If it has to do with needing to know "exactly" what the lateral displacement is you may have a bigger problem on your hands."
That is exactly what I need.
This is what I am going to have to figure. I am interested in the amount the center of the circle shown here moves.
Thanks
RE: Deflection of beam in "X" and "Y"
You had us all confused. We thought you were looking for displacement "into" the page.
Your problem is really quite simple.
If you have access to an analysis program, simply model it.
If you're doing hand calcs, and I haven't done a problem like this by hand in a while, you can use the slope deflection equations to solve. Unfortunately I'm in transit and my books are in temporary storage or I'd give you a better reference.
You can approximate the displacement by noting that it's due to the rotation at the end of the long cantilever. There are published equations for calculating this. From the end rotation (and dispacement y) you use trig to get the displacement at the point you want.
Sorry I don't have the equations you need at hand. I'm sure someone else will provide them.
RE: Deflection of beam in "X" and "Y"
FL^2/(2EI)
where L is the length of your horizontal member.
Your required X displacement at the top of your vertical member is then the product of this rotation and the length of the vertical member.
RE: Deflection of beam in "X" and "Y"
RE: Deflection of beam in "X" and "Y"
I was confused as well. I think you solve it by simple trig as well, once you determined the vertical deflection. You may have to exaggerate the deflection and by simple triangulations, you should be able to be to compute the "x" movement.
How did you paste the image, I never tried before, but I think this could be helpful for the future posts.
RE: Deflection of beam in "X" and "Y"
His equation will give you quite an accurate answer to your problem.
RE: Deflection of beam in "X" and "Y"
RE: Deflection of beam in "X" and "Y"
I'm not sure how accurate this would be but should at least give you an idea of what's happening.
Roark (6e) gives equations for the slope and deflection of a cantilevered beam of length l, left end (point A) free, right end (point B) fixed, with point load "W" applied distance "a" from the tip as:
th = thA + MAx/EI +RAx2/2EI-W/2EI<x-a>2
y = yA + thAx + MAx2/2EI + RAx3/6EI - W/6EI<x-a>3
where:
thA = W(l-a)2/2EI
MA = 0
RA = 0
<x-a>2 = (x-a)2*<x-a>0
<x-a>3 = (x-a)3
<x-a>0 = 0 when x<a and <x-a>0 = 1 if x>a (undefined for x=a)
I haven't tried it and don't know if it would work but this is how I would start to attack the problem.
RE: Deflection of beam in "X" and "Y"