Deflection of beam in "X" and "Y"

[karthur]

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?

 

[Lutfi]

You can compute it in same manner as for the y-direction. However, use the proper moment in inertia (I). To get a final resultant displacement:

R = ((Dx)^2+(Dy)^2)^0.5

In other words, square both deflections, add them and then take the square root.

Good luck.

 

[Lutfi]

Follow up,

I am assuming that you would have a load applied in the other direction. Otherwise, your deflection would be zero!

 

[karthur]

Lutfi,

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

 

[corus]

I think in general beam theory is based upon small deflections so the deflection into the support is negigible. The forumla for deflection is y = (F * x^2)/(6 * E * I) * (x - 3 * L) which gives the maximum at x=L and not at a point x<L. You'd have to look at large deformation theory to get the correct answer.
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.

 

[steve1]

The deflection equations noted above are based on a &quot;model&quot; of the beam. The model has the load applied perfectly concentric with the cg of the beam, the beam is perfectly prismatic, the compression flange is braced against rotation, etc. Thus there is no lateral displacement or rotation.

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 &quot;exactly&quot; 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.

 

[karthur]

Steve,
&quot;If it has to do with needing to know &quot;exactly&quot; what the lateral displacement is you may have a bigger problem on your hands.&quot;

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

 

[steve1]

Karthur,

You had us all confused. We thought you were looking for displacement &quot;into&quot; 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.

 

[Denial]

Now we can see your problem in its context, it is apparent that the formula you need is not the one for the X displacement of the cantilever's tip, but the one for its rotation:
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.

 

[dbuzz]

karthur, how did you paste that picture in your post?

 

[Lutfi]

karthur

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 &quot;x&quot; movement.

How did you paste the image, I never tried before, but I think this could be helpful for the future posts.

 

[flamby]

I think corus had it right even before looking at your pic.

His equation will give you quite an accurate answer to your problem.

 

[haynewp]

Here is a discussion on posting images. thread507-16013

 

[Binary]

If you assume that the line from the centroid of the root to the centroid of the tip becomes a curve of the same length under loading then perhaps you could use the equations for deflection and angular displacement as functions of x to plot in CAD a spline of the same length as the undeflected centroid-centroid line and see where the tip of the curve ends up.

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 &quot;W&quot; applied distance &quot;a&quot; from the tip as:

th = th_A + M_Ax/EI +R_Ax²/2EI-W/2EI<x-a>²

y = y_A + th_Ax + M_Ax²/2EI + R_Ax³/6EI - W/6EI<x-a>³

where:

th_A = W(l-a)²/2EI
M_A = 0
R_A = 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.

 

[corus]

Denial gave the correct method, basically using the simple expression S = Theta * R for arcs. This is ok for small angles but to be exact use X = L * sin (theta) where theta is the rotation given in Denial's formula and L is the length of the vertical member. (sin theta is approximately equal to theta for small angles)