Evenly distributed beam with overhang and multiple supports

[gmreit]

Homework Statement

I have an evenly distributed load with overhang and symmetry. I want to express the reaction loads as a function of support distances. I did the the 3 support problem with integrating a load structure and redundant and solving for the center reaction to find the outter supports.

Homework Equations

As the number of supports goes up so does the unknowns. I need to do this so my customers know how to place the supports in order to not overload the structures. We do have symmetry to work in our favor which should simplify it alittle.

The Attempt at a Solution

Go about it like I did for the 3 support system or is there a different approach to take. I included a picture of the next system I want to write an equation for.

 

[haruspex]

You don't explain how you solved it for three supports.
The obvious approach is to treat each L/n section as effectively being borne by one centrally placed support. That's fine as long as the load really is uniform. But if it is only that there is a uniform upper limit on the load per unit length, there might be a better arrangement.

 

[gmreit]

For 3 supports I solved it by superposition. I solved for deflections with double integration and set the supports equal to 0. I want to do this for up to 6 supports and was thinking unit force would be the way to.

 

[haruspex]

For 3 supports I solved it by superposition. I solved for deflections with double integration and set the supports equal to 0. I want to do this for up to 6 supports and was thinking unit force would be the way to.

So is the principle to minimise the maximum deflection (as opposed to, say, minimising the maximum moment)?
And you did not answer my question: is it really a uniform load, or is it a uniform limit to the load density? In many real world situations it is the second.

 

[gmreit]

It is an actual uniform load. Packs of commercial glass weighing about 21,000 lbs

 

[haruspex]

It is an actual uniform load. Packs of commercial glass weighing about 21,000 lbs

Ok, what about my other question in the last post?

 

[gmreit]

I think I will try and solve it with force analysis and generate a matrix that I can solve on the computer. I haven't had to do a beam with this many redundants in awhile. I am even complicate the problem further once I get this nailed down.

 

[gmreit]

The supports that the packs sit on can move. I was really interested in the reaction loads that develops at each support. I do like your question though because it is very important to minimize deflection between supports because glass does not tolerate that well. They are 11' tall and very stiff. I guess Thr real question now is what is the placement that minimizes deflection and then what would the support reactions then be

 

[haruspex]

I believe there is an essential difficulty with what you are trying to calculate in a practical setting. As soon as you go past two supports, the stiffness of the beam becomes important. If the beam is very stiff then the slightest discrepancy in the heights of the supports will alter the load distribution. The closer the supports, the less tolerance in the heights.
This suggests the use of a spongy layer between the supports and the beam.

Leaving that aside, I had been unable to read your detailed calculation for the three support case, so I attacked it myself. First, I considered what goes wrong with the obvious solution of putting the outside supports at L/6 from the ends. If you imagine cutting the beam at L/3 and 2L/3 then this would give equal loading. But with the uncut beam there must be a pair of opposite torques at each of those points to twist them into forming a straight line. At one end support, that torque will tend to tip that third of the beam up at L/2 and down at the beam end. That will reduce the load on the central support. To counter this, the end supports must be moved out a little. So we can see that they will have to be a little less than L/6 from the ends.
Ploughing through the algebra, I found they should be set at (√142-11)L/6 from the ends, which is just the tiniest bit less than L/6 (about 11L/72), so it looks reasonable.
With more supports, the ones near the middle will get even closer to the obvious spacing, with the greatest departure being near the ends. But it looks to me that the discrepancy between the obvious spacing and the ideal one is never going to be significant.

 

[gmreit]

Thanks for the explanation. What is the method you used to analyze this? I would like to do more reading up on it. I take it you are separating the beams and connecting them together with moments. Seems like it should be easier this way especially as the number of supports increase. I haven't had to mess with continuous beam problems in a long time.

 

[gmreit]

Also the supports have about 1" thick rubber on top of them.

 

[gmreit]

Is this how you would approach this problem?

 

[haruspex]

Number the supports 1 to N, left to right. Each exerts force F upwards and is at height 0. So total weight is NF.
Beam has length L, weight per unit length λ=NF/L.
Beam segment lengths are d_i, i=0 to N, where d₁ is from 1st to second support, etc.
At support i, the beam has zero deflection, some gradient g_i and some curvature. The segment to the left exerts a torque T_i on the segment to the right (anticlockwise positive).
Between supports i and i+1, the vertical deflection y at distance x from support i satisfies
ky"=-T_i+Fx-x²λ for i>0 (1)
since the torques must balance at support i+1:
-T_i+Fd_i-d_i²λ/2=-T_i+1 (1a)
For the segment left of the 1st support,
-d₀²λ/2=-T₁ (1b)
Integrating (1):
ky'=-xT_i+Fx²/2-λx³/6+g_i for i>0 (2)
Since the gradient must be smooth across i+1:
-d_iT_i+Fd_i²/2-λd_i³/6+g_i=g_i+1 (2a)
Integrating 2:
ky=g_ix-T_ix²/2+Fx³/6-λx⁴/24 for i>0 (3)
(No constant of integration since y=0 at x=0.)

Can you make some progress with that?

 

[gmreit]

I do follow what you are doing by sectioning the beam at each support and apply a bending moment instead. Am I understanding this correctly? Can this still be done if the distance between the supports is not the same resulting in a varying reaction forces. It will be symmetric though. Again thanks for the help. Doing this through virtual force is becoming very cumbersome to do by hand

 

[haruspex]

Can this still be done if the distance between the supports is not the same resulting in a varying reaction forces.

I am assuming the forces from the supports are all the same, but the distances will vary, so the moments vary.

Edit: I found a mistake in my equations...

 

[gmreit]

I believe I am not accounting for the torque values the same way you are referring to. You are indexing them. When you integrate do you only go between sections l, 0-di of just that section or do you always have the same start. I believe with your method it doesn't matter how many supports there are then which I like

 

[gmreit]

I am getting stuck here.

 

[gmreit]

Took another shot at this

 

[haruspex]

This appears a rather tougher problem than I first thought. The equations get painful.
Anyway, here's my revised attempt...

Let the i^th segment, i=0 to N, have length d_i.
Each support provides force F. Weight per unit length of load is $\lambda$.
The deflection at x from the start of a segment is y_i. (Omitting the stiffness constant.)
y_i" = the torque = $-F(ix+\Sigma_1^{i-1}jd_j)+\frac{\lambda}2(x+\Sigma_0^{i-1}d_j)^2$, for i > 0. [1]
Integrating:
$y_i'=-\frac F2(ix+\Sigma_1^{i-1}jd_j)^2+\frac{\lambda}6(x+\Sigma_0^{i-1}d_j)^3+g_i$, for i > 0. [2]
By continuity with start of following segment:
$-\frac F2(id_i+\Sigma_1^{i-1}jd_j)^2+\frac{\lambda}6(d_i+\Sigma_0^{i-1}d_j)^3+g_i=-\frac F2(\Sigma_1^{i}jd_j)^2+\frac{\lambda}6(\Sigma_0^{i}d_j)^3+g_{i+1}$, for i > 0. [2a]
Integrating [2] and using the fact that the deflection is 0 at each support:
$y_i=-\frac F6(ix+\Sigma_1^{i-1}jd_j)^3+\frac{\lambda}{24}(x+\Sigma_0^{i-1}d_j)^4+xg_i+\frac F6(\Sigma_1^{i-1}jd_j)^3-\frac{\lambda}{24}(\Sigma_0^{i-1}d_j)^4$, for i > 0. [3]
By continuity with start of following segment:
$0=-\frac F6(\Sigma_1^{i}jd_j)^3+\frac{\lambda}{24}(\Sigma_0^{i}d_j)^4+d_ig_i+\frac F6(\Sigma_1^{i-1}jd_j)^3-\frac{\lambda}{24}(\Sigma_0^{i-1}d_j)^4$, for i > 0.
$d_ig_i=\frac F6((\Sigma_1^{i}jd_j)^3-(\Sigma_1^{i-1}jd_j)^3)-\frac{\lambda}{24}((\Sigma_0^{i}d_j)^4-(\Sigma_0^{i-1}d_j)^4)$ [3a]
In principle, we can use 3a to eliminate the g constants from 2a, but it gets very messy.
If we normalise the length of the beam to 1 and F to 1, we can set $\lambda=N$. That leaves us with a set of relationships between the d_i.
Maybe something an iterative bit of software can handle.

 

[gmreit]

When you wrote the initial torque equation, why do you have i and j in the reaction part of the equation, F(i*x + "SUMMATION"j*dj).

 

[haruspex]

When you wrote the initial torque equation, why do you have i and j in the reaction part of the equation, F(i*x + "SUMMATION"j*dj).

There are i instances of an upward force F to the left of the point under consideration. The nearest has moment Fx, the next nearest has moment F(x+d_i-1), the third nearest has moment F(x+d_i-1+d_i-2), and so on. Summing these gives the expression I wrote.

 

[pongo38]

I think you seem intent on using elastic analysis, which is fine at working loads, if can do it. However, plastic analysis is easier, and arguably safer, whilst admittedly not giving you deflections.