## Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

## Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

(OP)

(See image)

**P = Q**(They are the applied loads)

**Unknowns**are the reactions

**A, B, and C**.

**A and C are known to be equal however**, so there are really

**only 2 unknowns**.

A, B, and C are known to all be pointing

**downward**.

At first glance this looks like a pretty easy/standard statics problem. But this has actually been giving me a lot of trouble, and I suspect it might be because I'm forgetting some sort of basic principle that I've perhaps been taking for granted up til now, which maybe happens to uniquely cause issues with this problem.

I've actually been a practicing engineer for a few years now. I make free-body diagrams all the time and solve statics problems on a regular basis. I feel that my statics foundation is pretty solid. But this problem has kind of humbled me--maybe it's just a brain fart, or maybe I've constructed a problem that isn't actually possible to solve for reasons not currently apparent to me.

Every time I try to solve this problem, attempting to solve the moment-sum equation combined with the force-sum equation

**eventually results in a trivial 0 = 0 identity**, which means (by my understanding) that the equations were not truly independent. Normally, these equations absolutely should be independent however, so I suspect there is some other kind of issue going on here--OR maybe they really are not independent, in which case I would like to know what makes them non-independent, and how I can make sure to avoid constructing such problems in the future.

**Or maybe it's statically indeterminate, despite there only being 2 unknowns?**

(There was also a time I tried to solve a less-simplified version of this problem, and the solution told me that B was actually pointing upward. I definitely insist it should be pointing downward with the others, though)

Or maybe one of you guys will just solve this in five minutes and I'll realize I've just missed something obvious.

Curious to hear your thoughts/commentary.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

Solving your system of equations should result in solving for two unknowns that will both be the same value. The net downward applied force is A + C, regardless, so there are two forces.

TTFN (ta ta for now)

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## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

does the black dot at B symbolise anything ? (like a break in the beam (discontinuous moment) ? ... then the structure is 2 SS beams (and not a beam on 3 supports).

I would solve for a beam, on 3 supports, with one load at a variable point, say x = a, and probably use the 3 moment equation.

Then use superposition for the two loads.

"Hoffen wir mal, dass alles gut geht !"

General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

The black dot at B is just there to show that I'm approximating B as a pinned support (just as a formality so that the system isn't technically under- or over-constrained).

After posting, I found this definition for "statically indeterminate" that I hadn't seen before and seems more helpful/specific than other definitions I had seen before and gotten used to.

"

A system is externally statically indeterminate, if the"number of support reactionsexceeds the number of possible movement directions.So it seems to me that since the number of support reactions is 3, then the number "3" is the number relevant for determining statical-determinancy (if that's a word), and not the number 2, even though I know that there are truly only 2 variables I don't know the value of. The fact that the two equal reactions (A and C) occur at separate supports is more important than the fact that they are equal to one another (for the purpose of determining if statically determinate, that is).

So in conclusion

I think the problem is statically indeterminateand that's why I've had so much trouble solving it the normal/basic way using sum of forces in Y and sum of moments in Z.Please definitely let me know if you disagree though.P.S. After realizing the above, I went ahead and just plugged the system into a structural analysis program. I ran it for multiple different values of P (which is equal to Q, remember) and multiple different values of S. The answer is that B = 4.4*A. That's the result I consistently end up with for all values of P and S that I've tried.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

Still indeterminate if fully symmetrical, Ra = Rc = x*P, Rb = (1-x)*2P ... but what is "x" ? You've used your 2 available equations to get this far ...

1) symmetry is effectively sum M (as Pa = Pc)

2) sum F gave us Rb

3) need something else to get "x".

Use the 3 moment equation. or unit force method (at Rb)

"Hoffen wir mal, dass alles gut geht !"

General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

My structures professors were so bored with people solving every beam problem with MacAulay's method that they often included problems using superposition in exams to sort the wheat from the chaff.

Cheers

Greg Locock

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## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

B/2=2.2A, whereBis the reaction in the original beam, andB/2 is of course the reaction in the cut beam So your result is correct!prex

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## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

If the beam is not rigid and is of uniform section properties, then it acts like a beam with one end fixed in rotation, by symmetry, at B and free to rotate at A or C. See Beam Fixed at One End, Supported at Other, diagram 13 on http://faculty-legacy.arch.tamu.edu/anichols/index...

If the beam is not rigid and is not uniform, particularly not symmetrical, then the problem has no general solution.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

"Hoffen wir mal, dass alles gut geht !"

General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

~~If P and Q were not equal, the structure would be indeterminate.~~Actually, the structure is indeterminate whether or not they are equal, but if they are not equal, the slope at B is not zero.The reaction at the fixed end is 11P/16, which means that your reaction at B is 22P/16. Reactions at A and C are 5P/16.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

A propped cantilever is a redundant structure ... since it has more reactions than the statically determinate simply supported beam.

The fact that solutions exist in texts is beside the point ... the problem is solved by bringing another equation into play ... the fact that the slope at the fixed end is zero.

This is an additional equation that solves the indeterminacy. I challenge anyone to solve the problem with only the equations of equilibrium (in this case there are two sum Forces and sum Moments).

What you're doing is using the unit force method, with the end moment as the redundancy.

Solve the simply supported beam, calculate the slope at end B

Now, apply a unit moment (opposing this deflection) at end B, and calculate the slope at B,

Then determine the moment at B for zero slope,

Then load doesn't Have to be at the mid-span, well it does for the solution provided. But it can be anywhere on the beam, and of course the solution will change, but it can be solved by the same methods.

"Hoffen wir mal, dass alles gut geht !"

General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

The structure is indeterminate because it cannot be solved by statics alone, whether or not P = Q.## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

About reinventing the wheel: when I was a student, long long ago, there was a story that was told to engineering students who took the monster exam on Theory of elasticity.

The story read: Do you know who was the student who obtained the highest mark? The one who answered in the written assignment: "This problem is solved on page xxx of the engineer's handbook."

prex

http://www.xcalcs.com : Online engineering calculations

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## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

This would serve as a check...

x=L/2, a=L/4

From Fig. 9, x1 = (Pa/6EI)(3Lx-3x^2-a^2) = 11PL^3/384EI

From Fig. 7, x2 = BL^3/48EI

If x1 = x2, then B = 1.375P, which is equal to 22P/16 as found above. okay

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

equally easy to solve with unit force ... deflection a B is an easy simplification ... the two spans become one long span, with two point loads, superposition and the deflection curve for a SS beam with a point load ... easy !

General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

## RE: Statics Problem with 3 supports but only 2 Unknowns (because 2 reactions are known to be equal)

General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.