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

[caridea]

(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.

 

[IRstuff]

The fact that they are equal through symmetry does not mean you lost an unknown. If P <> Q, then A <> C

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.

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[rb1957]

yes, symmetry requires symmetry of structure and symmetry of loads.

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.

 

[caridea]

The structure is symmetrical and the beam is continuous. (it actually represents a shaft freely rotating inside 3 plates, which are the 3 "supports")
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 reactions exceeds 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 indeterminate and 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.

 

[rb1957]

yes, the problem is statically indeterminate, a beam on three supports.

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.

 

[GregLocock]

You could use superposition. Consider the case where B is zero, the deflection due to P and Q is x1. Now consider the case where P and Q are 0, what is the deflection due to a force at B, call it x2. Obviously for the combined case x1 =x2, hence you know the force at B.

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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[prex]

Owing to symmetry, if you cut the beam at B, you end up with a beam on two supports, one fixed and one simple (pinned). It's a statically determinate problem that's solved in many engineering books (remember: an engineer should never reinvent the wheel ). With P midway of A and B the solution gives B/2=2.2 A, where B is the reaction in the original beam, and B/2 is of course the reaction in the cut beam So your result is correct!

prex
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[3DDave]

If the beam is infinitely rigid then no deflection can allow load to transfer from B to either A or C. If either A or C moves the resulting loads at A, B, and C are infinite.

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_files/courses/arch331/NS8-2beamdiagrams.pdf

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

 

[rb1957]

"a beam on two supports, one fixed and one simple (pinned). It's a statically determinate problem" ... no, a propped cantilever is still indeterminate, but yes it is a common problem in most texts. But solving this using 3 moment equation or by unit force method is not "reinventing the wheel" ... it's applying tools in our toolbox.



"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[BAretired]

Since P and Q are equal, the structure is symmetrical, which means that the slope is zero at point B as mentioned by prex. 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.

 

[rb1957]

we'll probably get into an argument over how statically indeterminate is defined.

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.

 

[BAretired]

You are quite right, rb1957, and I apologize. The structure is indeterminate because it cannot be solved by statics alone, whether or not P = Q.

 

[prex]

Yes, it is indeterminate, according to the accepted definition.
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
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[rb1957]

I would have failed that smartar$e. It is important (IMHO) to be able to derive common results.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[BAretired]

You could use superposition. Consider the case where B is zero, the deflection due to P and Q is x1. Now consider the case where P and Q are 0, what is the deflection due to a force at B, call it x2. Obviously for the combined case x1 =x2, hence you know the force at B.

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

 

[BridgeSmith]

For the case where P does not equal Q, the spans are not equal lengths, and/or the beam properties are different in two spans, I believe there is no general solution, thus the need for methods to approximate a result. Our go-to for this would be moment distribution, but I've also used the direct stiffness matrix method for problems of this type.

 

[rb1957]

"I believe there is no general solution" ... nah, it is easy to throw this into 3 moment equation, with any P, Q, L1, L2, a, b, c, d ...
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 !

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[BridgeSmith]

If it's so easy, why did they teach me all the hard ways to do it in my structural analysis classes?

 

[rb1957]

I really don't know. The only thing that'd make this really hard to solve would be a varying I.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.