Statics question 1

[frankyg]

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

 

[JStephen]

If the three unknown forces are not all in a line, you should be able to sum forces and sum moments about two different axes and find the forces. If you sum moments about two different points, but in the same direction, you'll just get more or less the same equation twice.

 

[zekeman]

As you surmised, you can't solve this since at most you have two equations ( sum of the vert forces=0 and sum of moments about a point=0).As pointed out summing moments about another point does not add an independent equation.
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.

 

[rb1957]

you can solve a statically indeterminate problem like you're describing. it involves applying other equations (boundary displacements or slopes, displacement of an intermideiate point).

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

 

[zekeman]

quote"you can solve a statically indeterminate problem like you're describing. it involves applying other equations (boundary displacements or slopes, displacement of an intermideiate point)".
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.

 

[JStephen]

As I understand the problem statement, a stool with 3 legs with a vertical force on the seat would fit the description. And that is solvable. If it's a plane problem, then it's not.

 

[rb1957]

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.

 

[frankyg]

Just to clarify the problem a bit, I will try to draw the FBD.



V-------------------^-------V----------^
120 lb



V=downward load
^ = upward load
all unknown except the 120 lb load

 

[TheTick]

Taking moment at point that each force is applied will yield 3 equations and 3 unknown forces. This can be solved.

Honesty may be the best policy, but insanity is a better defense.

http://www.EsoxRepublic.com

-SolidWorks API VB programming help

 

[frankyg]

That was what I originally tried, and running my equations in an equation solver gave me different results every time I ran it.

 

[JStephen]

I should say it is unsolvable as a statics problem to be more accurate; it was posed as a 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.

 

[JsTyLz]

Use either moment equations and integrate, or use distributed load as the first scalar input. At the supports, the deflection is equal to zero. If the cross section is continuous between supports/loads, then it can be solved.

 

[handleman]

At the supports

Which are the supports? OP lists everything as "forces".

the deflection is equal to zero

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.

 

[GregLocock]

Are the beam properties known, along its length?

Is the stiffness of each reaction point known?

As drawn it is indeterminate.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[TheTick]

I did the math. zekeman is correct.

 

[GregLocock]

How do we know it is indeterminate? Because the right hand support can be removed entirely and you'll still have (unstable) equilibrium.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[handleman]

Assumptions!! Assumptions!! In no post did the OP say that the upward forces were reactions at supports! In his first post they are all "forces". In his second post they are "loads". Until some of those "forces" or "loads" are specified as supports you can't begin to think about what steps are needed to start figuring out how to solve the problem.

 

[rb1957]

i jumpoed several posts ... since "the tick" suggested summing moments a different points gave you different equations ... it doesn't, there's only one equation of equilbrium for moments (summing at different points gives you the same equation, albeit written differently).

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.

 

[handleman]

Sorry, rb, but I really think you're off base on this one. You're making too many assumptions. As I said previously, you cannot simply assume that the two forces acting upward are supports. The OP defines them as "forces" and "loads". All of them. Perhaps all three unknowns are really reactions at supports. Perhaps none of them are. All of these elastic beam, boundary condition, slope, etc methods depend on actually knowing some boundary conditions. No boundary conditions are known from the information given.

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.

 

[rb1957]

handleman, i don't care what assumptions are made in the direction of the reactions (it only changes the sign of the reaction).

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