Simple static problem

[autothesis]

Hello All,

I am having problem in solving a very simple static problem. I need to find the resultant load and its location. See the attached snapshot of the problem description.

Thank you

 

[rb1957]

that's confusing ... the sketch shows the reactions and is asking to determine the lines of action of these forces. research "force polygons" and "three force members" and "equilibrium of forces"

Quando Omni Flunkus Moritati

 

[autothesis]

Sorry for drawing the ABC forces are reactions. I am trying to find the resultant location based on three forces. I can calculate Y location by moment calculation. Having trouble in X location.
I am reading information on force polygons based on your comments.

 

[rb1957]

is it not "P" ?

the question is asking what are the co-ordinates of the load, so that the three force directions (AP, BP, CP) resolve the applied load. this is going to be easier to solve through "equations of equibilirium" ... summing the force components so that sum Fx = P and sum Fy = 0. force components depend on their angles determined by x and y. the "lazy" way to solve is to up a spreadsheet with x and y as input, ie guess these, that'll determine the lines of action of the three reactions, and so their components in the x and y directions and so check against the required summs, and change your guesses.

Quando Omni Flunkus Moritati

 

[rb1957]

i took your "R" values as reactions ... but they are only the x-component of the reactions (sum R = P).

that said there should be a line of solutions (since there are two variables and one equation)





Quando Omni Flunkus Moritati

 

[rb1957]

looking at your reactions element PA is in compression; not a problem, just an observation.

so for i get sumFy = 0,
-1006.1*(y/x)+3416.1*(y/(21-x)-1150*((16.3-y)/(21-x)) = 0
(sum Fx already balances, sumM balances as all forces go through a common point)

to solve the redundancy assume all three are the same area, so sum forces is a measure of strain energy; and pick (x,y) that minimise this sum.

my answer, x=8.7325

you can check by drawing a force polygon of the three reactions, and see that the applied load closes the polygon.

Quando Omni Flunkus Moritati

 

[hokie66]

Y=(1150 / 5572.2) x 16.3 = 3.36
X can be anything.

 

[rb1957]

yes, that solves for moments about A ... i put more meaning into the lines joining A, B, and C to P, and assumed that Ry at the reaction points was allowed.

Quando Omni Flunkus Moritati

 

[hokie66]

There will be vertical reactions, but you can't determine them without knowing the X dimensions.

 

[autothesis]

Thanks Hokie66 and Rb.

Even i came up with the same answer. But what threw me off here is, X can be anywhere.
Check the picture attached.
I started thinking that 5572.1 lbs applied at one end (D) at a 3.36 inch distance will load differently than at D' or at D".

 

[autothesis]

See this picture for D' and D".

 

[hokie66]

Yes, I was incorrect. If B, C, and D are collinear, and the joints are all pins, there will be no horizontal reaction at B or C. So without knowing the geometry, you could not derive the reactions. You have given the only the X component of the reactions, so the geometry cannot be derived uniquely, as rb1957 suggested.

 

[rb1957]

actually i disagree.

if the reactions at A, B and C are horizontal, and the load is horizontal then your solution is correct. A, B and C are all roller reactions, with no y-component.

put a big solid rectangle between A, B and C (instead of three lines) and apply to force. C reacts the overturning moment.

i guess "x" is a trick part of the problem.

btw, review rules on student posts.

Quando Omni Flunkus Moritati

 

[hokie66]

I suppose I originally thought rollers, then changed to thinking he meant pins. Or something.

 

[autothesis]

What I am trying to find here is a one resultant 'point' load which will induce X load at three ABC points. Even if it induces little bit of moment Y loads, I am OK with that.
A resultant load dimension (X Y) is what i need to find and apply. The fixture can be a co-linear square tube transferring load to three end points.

(By the way, I am a professional who needs to go back to school for basic statics )

 

[hokie66]

Not sure why I am persisting with this, but...

They can't all be rollers, as that would make the assembly unstable. With axial inclined members, if there is a horizontal component, there has to be a vertical component. Even with your "big solid" block, which makes the assembly a moment resisting frame, at least one of the supports has to be a pin to assure stability.

 

[autothesis]

There are no rollers. Each end point ABC will be bolted down.

 

[hokie66]

Then you have vertical reactions, as well as the horizontal ones you showed. Without these reactions, the geometry cannot be determined.

 

[rb1957]

well, if you have a solid box structure, then yes.

but if your structure is more linear (like the three dotted lines), then you have a simpler problem (you know the line of action of the reactions) and can solve the single redundancy that can be solved along the lines i've posted (but i suspect that you haven't understood them).

this has become an extremely frustrating thread ... a badly phrased question, and now with missing key info ...

Quando Omni Flunkus Moritati

 

[autothesis]

Guys I am sorry for not giving all information on the first place. Lets end this here. Thanks for your valuable time. I really appreciate your help here.