Statics question
Statics question
(OP)
Hi,
This is probably a very basic question, but it's been a long time since I took statics.
I have a body with 1 known force and three unknown, all acting in the Y direction. Can I figure this out by summing the Y forces and the moments at 2 different points, or is this ambigious? Based on what I am getting, and from what I think I remember from statics, you can't solve this problem. Am I correct on this?
Thanks
This is probably a very basic question, but it's been a long time since I took statics.
I have a body with 1 known force and three unknown, all acting in the Y direction. Can I figure this out by summing the Y forces and the moments at 2 different points, or is this ambigious? Based on what I am getting, and from what I think I remember from statics, you can't solve this problem. Am I correct on this?
Thanks





RE: Statics question
RE: Statics question
The proof is simply
x+y+z=f
ax+by+cz=df
(a+u)x+(b+u)y+(c+u)z=(d+u)f
where a,b and c and d are the distances from the moment axis 1 to the force vectors and u is the distance between moment axis 2 to axis 1 and f is the known force.
Since the last equation is a linear combination of the first two it proves that you only have two independent equations.This is manifestly clear by seeing that the determinant of the linear set is zero.
RE: Statics question
Given your problem, I'd consider the middle unknown Y to be redundant. Solve for the other two unknowns (statically determinate). Now determine the deflection that this hypotheoretical beam at the redundant unknown. Now apply a unit force at the unknown position, in a direction opposite to the static deflection. solve this beam (determine the reactions at the other two unknowns), then calculate the deflection at the unit load. Finally the value of the redunant reaction is the static deflection (at the redundant unknow)/unit deflection (ie calculated for the unit load) ... this method is called (predictably enough) the unit load method; i prefer it 'cause it is very intuitive. Remember to adjust the other reactions for the full value of the redundant unknown. Check your beam is in equilibrium at the end of this !
good luck
RE: Statics question
THE PROBLEM AS STATED IS NOT A BEAM PROBLEM.It is not necessarily an elastic system and none of the unknowns is a moment force so, as stated, the problem has no solution.
RE: Statics question
RE: Statics question
and JStephen ... as a planar problem, it can be solved. For example IF it is a classic beam on three supports.
RE: Statics question
V-------------------^-------V----------^
120 lb
V=downward load
^ = upward load
all unknown except the 120 lb load
RE: Statics question
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RE: Statics question
RE: Statics question
And TheTick- summing moments at each point won't give you the solutions as the equations you get are not independent- see the post up above. Or, to put it another way, you could assume any value whatever for one of those forces, and still solve for the other two (as a statics problem.) (And you'd get different answers for those other two depending on what you assumed for the first one.)
Even solving as an elastic beam problem assumes you have information that you may or may not have. Suppose the beam is supported at each end. What are those other three forces then? You don't know and can't find them. You can solve it if those points are support points, but that's not a given.
RE: Statics question
RE: Statics question
Which are the supports? OP lists everything as "forces".
Is this a real-world problem? If so, who says that the supports are rigid?
IF the two upward forces were actually rigid supports, the one unknown downward force could be anything from the minimum required to balance the known force up to infinity and still satisfy statics. There are still too many ambiguities, even with the diagram posted, to determine a solution method, much less a solution.
RE: Statics question
Is the stiffness of each reaction point known?
As drawn it is indeterminate.
Cheers
Greg Locock
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RE: Statics question
RE: Statics question
Cheers
Greg Locock
Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Statics question
RE: Statics question
loads/forces, the OP said there is one known and 3 unknowns ... this is a classic single redundant beam, can easily be solved by my post yesterday.
RE: Statics question
If you are confident that this can be done, and that it can be done easily and intuitively, why not make up some lengths, check the OP's rudimentary FBD, and post a solution? After all, the distances between the loads are the only givens that aren't actually posted. If you do, please tell us which "loads" or "forces" you transform into reactions at supports.
RE: Statics question
and i'll take you up on your offer. From the sketch, with 120 lbs load on the left end, we have three line segments, a=3",b=4",c=5" for a total length of 12". We've three reactions A, B, and C (C is at the extreme right end of the beam). I prefer to to have th eload acting down, and all the reactions are assumed to react up (I suspect that B will actually act down).
Assume reaction B is redundant. Solve the beam for reactions A and C ... 120*12 = A*9 >> A = 160, and C = -40
At B this statically determinate beam deflects upwards (indicating that the reaction here is downwards). From my handy-dandy spreadsheet (calculating the deflections on a statically supported beam, the deflection at B is 1867 (to be accurate, this is deflection*EI).
Now consider the beam, supported at A and C loaded at B with -1 lbs. The reactions are A = 5/9, C = 4/9. The deflection at B is 14.815. Therefore the redundant reaction B is -1867/14815 = -126 (ie acts down)
and A = 160+126*5/9 = 230
and C = -40+126*4/9 = 16
RE: Statics question
What if the "reactions" are not reactions at all?
I accept your solution, but with some pretty big reservations.
Here are just a few of the assumptions you've made:
1. All unknowns are reactions at supports. This actually conflicts with the wording of every statement made by the OP. All unknowns are referred to as "forces" or "loads".
2. All "loads" that you've assumed to be supports are rigid.
3. The "loads" that you've assumed are supports are capable of acting in the opposite direction that the OP stated that they act.
4. Without the applied 120lb load on the left end there is no preload at the supports. I.E. the beam is not already clamped.
5. Without the applied 120lb load, the beam is actually in contact with each support. I.E. there is no deformation required to make the beam contact one of the rigid supports.
Unless the OP is trying to solve a problem from a textbook, I don't think any one of these four assumptions has any support from the information provided. If any one of them is false your solution is invalidated, but you didn't state a single one until your "solution" post, and even then you didn't really recognize that you had made them.
RE: Statics question
1) i think that's semantics
2) yeap, supports are assumed rigid, as opposed to elastic foundations.
3) yeap, supports are assumed to be able to react load in either direction.
4) yeap, assumed no initial preload
5) teap, assumed no initial deformation.
whilst we're getting picky, i assumed no weight either.
RE: Statics question
RE: Statics question
ok zekeman, ASSUMING the problem is elastic ... it is solvable ... the body with the forces acting on it may not be a classical beam but the method outlined applies ...
and JStephen ... as a planar problem, it can be solved. For example IF it is a classic beam on three supports."
I still submit it is NOT solvable by the socalled classical methods, since they imply some type of restraint on the system. In your analysis you are asuming that all of the loads points do not move which is NOT in the statement of the problem.
Just consider the FBD THE OP posted and assume the right end force is zero;. You still get a valid solution. Now give it any other value and you still get a valid solution. The result of two "valid" solutions suggests that there are NO valid solutions. Yor problem is that you solved the problem assuming zero for one of the forces, get a deflection at that point and then by superposition you push at thaty point to reverse the deflection which implies that you are restraining the system to have no deflection under the loads, which is NOT a condition given by the OP.
RE: Statics question
i have assumed that the three reaction points are restrained, since the OP doesn't say they are elastic. If they are elastic the problem is just as solvable (tho' quite a bit more complicated).
my solution to the problem, given my assumptions as both you and handleman are want to point out, is a correct solution to a beam on three rigid supports. this may not be hte OP's problem, but let him say so, eh ?