# Statics problem: finding forces on symmetric supports due to beam

## Homework Statement

I only just started thinking about this, so I apologize if I can't frame it correctly... But say I have a completely uniform beam sitting on, say, 8 supports all distributed evenly about the beam's center of mass (which is also its geometric center). That is, for every one of the 4 supports on the left side of the center of mass, there is an equivalent support the same distance from the center of mass on the right side.

Let's say I know the mass of the beam and the distances from the center of mass of the 8 supports. It seems like the symmetry of this problem makes finding the force exerted by the beam on any one support impossible. That is, we can only find information about the distribution of forces... eg, all the supports on the left add up to be taking only 1/2 the total weight of the beam, etc.

I'd like to ask if this is true. Even though I know the mass of the beam, that the beam is uniform, and the distances of each support from the center of mass of the beam, is ultimately the force on each support from the beam arbitrary - IE, the only constraint is the way the forces add up? If I (in the "real world") take a more or less uniform beam, arrange supports more or less symmetrically around its center of mass, and find the force on each support, how will the distribution look? Will all the forces on the left side add up to the same thing as all the forces on the right?

More specifically. Say my 8 supports are symmetric about the CoM. Say the beam is 50 ft long. The first support on the left and the first support on the right are 5 ft away from the CoM. The second support on the left and the second support on the right are 15 ft away from the CoM. The third support on the left and the third support on the right are 18 ft away from the CoM. The fourth support on the left and the fourth support on the right are 25 ft away from the CoM, IE at the ends of the beam.

In this scenario, the supports are still symmetric about the CoM. But the distribution on each side is still uneven, IE the first support is 5 ft away, the second 15 ft, the third 18 ft, the fourth 25 ft, corresponding to spacings between supports of 10, 3 and 7 ft.

Is finding the forces on the supports still arbitrary?

Sigma F = 0
Sigma M = 0

## The Attempt at a Solution

I wrote out the equations of equilibrium for the beam and support system. I saw no way of pulling out of just these equations and the information I know (mass of beam, distances from CoM of supports) the forces on each beam.

Is this how it is? Or am I overlooking something? If the supports were not symmetric about the CoM, could I find the force on each?

How would an engineer do this problem? If I have an actual beam that will rest on actual supports distributed in some way under the beam, how can I predict the force on each support (not taking into consideration bending or flexing of the beam, lack of uniformity, etc.)?

If the supports were on a circular track, the beam attached to the supports (no sliding), and the support/beam system were rotating at some rate, would the distribution of forces on the supports change? (I think it should!) In this case could I calculate the forces on individual supports?

Thank you so much... this is driving me crazy.

## Answers and Replies

cepheid
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Isn't this just an underspecified system? You have two equations and eight unknowns. As a result, you certainly can't arrive at a unique solution. It seems possible that there are infinitely many solutions.

Last edited:
radou
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You need to look into a displacement method, and you could consider the weight of the beam acting as a uniformly distrubuted load along the beam.

It definitely is an underspecified system, heh.

radou: What do you mean, "displacement method"?

PhanthomJay
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The only thing "arbitrary" about this system is that is "underspecified". Given the proper specified parameters, the reactions are definitely solveable. For example, if all supports are near rigid, and the beam is flexible, Radou's method will work. And assuming the supports were equally spaced, I believe the 6 interior supports would more or less be equal in value, and the edge supports would be about 1/2 the value of any interior support. On the other hand, if the beam is considered rigid in comparison to its supports, which might be spring supports of equal stiffness, then all supports would equally support the load regardless of the support spacing. So it all depends on what is given or assumed in the problem.

I assume (as is the case) that both the supports and the beam are near-rigid. But the beam is "more flexible" than the supports, so perhaps I'd assume that the beam is flexible (somewhat) and the supports are not, if this would help me work the problem and obtain a somewhat meaningful answer.

So, if I assume that the beam is flexible, the supports are not, and that the load is uniformly distributed (instead of being concentrated at the center of mass), the problem would become soluble?

PhanthomJay
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I assume (as is the case) that both the supports and the beam are near-rigid. But the beam is "more flexible" than the supports, so perhaps I'd assume that the beam is flexible (somewhat) and the supports are not, if this would help me work the problem and obtain a somewhat meaningful answer.

So, if I assume that the beam is flexible, the supports are not, and that the load is uniformly distributed (instead of being concentrated at the center of mass), the problem would become soluble?
Yes. If you try to assume, for a statically indeterminate beam, that the uniform load is concentrated at the center of mass (mid point of beam length), you'll be way off in getting the support reactions. For your particular example, you have 8 unknown reactions, but only 2 equations of statics (sum of forces in y direction =0, and sum of moments about any point =0), so you need more equations to solve the problem. One way is to use the knowledge that the deflection (displacement) of the beam at the support points is 0 (and non zero in between). It's a bit of a tedious process by hand calcs. I haven't done one in ages. A computer would do nicely.

Now, fyi, I'm looking at my old steel manual for a homogeneous 4-span beam on 5 equally spaced supports (a distance of 'L' apart) subject to a uniform load w (force/length units), and note that the reactions at the supports are, respectively, from left to right, .393 wL, 1.143 wL, 0.928 wL, 1.143 wL, and 0.393 wL. It's always nice to have tables.

What parameters of the metal would be enough to help me generate that sort of support reaction distribution (IE, 0.393wL, 1.143wL, etc.)? In addition to the mass of the beam what else would we need...? Elasticity modulus?

Is there a name for this process that I'd be doing to find the reactions? It seems like this is a problem that would occur frequently in statics. I believe it's a tedious process by hand, I don't even know where to begin!

Those numbers you gave for the reactions of the "4-span beam" on equally space supports seem like they came from nowhere... but those are exactly the kind of numbers I'm looking for, but for my own distribution (not equally spaced, etc.)... What is the name of the book you got them from?

Thank you so much, I am really interested in this problem now.

Does this amount to solving the beam equation with a weight distribution of w(x) = pg (p density) and boundary conditions (IE, the displacement at the ends is zero, and also at all the supports)?

Or do I need to do "more"?

PhanthomJay
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The deflection of a beam on rigid supports is a function of its loading, length, support points, boundary conditions (simple or fixed supports), elasticity modulus (E), and moment of inertia (I). For a homogeneous beam with constant E and I, the support reactions will be the same regardless of E and I (that is, you get the same reactions whether the beam is steel or wood, large or small, as long as the suport conditions are the same for each case (simple supports vs, fixed supports, etc.). There are a number of ways to solve statically indetermimate problems. The displacement method is one of them. It involves calculating the displacement of the beam as if the interior supports were not there (using tables or virtual work or other methods), then one by one placing a concentrated upward load "R1", "R2", etc., at each support point and calculate the deflection at those points under those loads, noting that the sum total of the deflections at those points from the applied and unknown loads must equal zero, and solve the simultaneous equations. A tedious process for 8 supports, without a computer. Let's look at a simpler case of a beam of length L on three simple supports (one at each end and one in the middle) and with a uniform load distribution (w) on the beam. First, you remove the middle support and calculate the beam deflection at the middle of the beam under the uniform load. You will find that the deflection is
5wL^4/384EI using calculus, tables, virtual work, or other methods. Then you place an unknown R force at midpoint, and calculate the deflection at midpoint under that load, and you will get deflection =
RL^3/48EI. So set both deflections equal (since they sum to 0) and solve for
R = 5wL/8. The end reactions are then each 3wL/16.

I am starting to see how this work. To get that deflection (5WL^4 / 384EI), did you use the Euler-Bernoulli beam equation, u''''(x) = w(x) / EI? (Is this what yo mean by "calculus"?) For w(x) = W/L (uniform load), I got a deflection of WL^3 / 384EI... I probably messed up, I just wanted to confirm.

I can see how much of a pain this process would be! Thank you for all your help!

PhanthomJay
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I am starting to see how this work. To get that deflection (5WL^4 / 384EI), did you use the Euler-Bernoulli beam equation, u''''(x) = w(x) / EI? (Is this what yo mean by "calculus"?) For w(x) = W/L (uniform load), I got a deflection of WL^3 / 384EI... I probably messed up, I just wanted to confirm.

I can see how much of a pain this process would be! Thank you for all your help!
Actually, I looked up the deflection in a table in my worn but trusty steel handbook , but, yes, your equation is quite correct for the distributed load case; there are other methods as well. You have the correct deflection; I used 'little w' in weight per unit length, whereas you used 'big W' (W =wL), same result, wL^4/384EI = WL^3/384EI, nice job.