Back to Basics FBD 3

[galaxy212]

Hi All,

I have a really simple FBD problem though its been a while since I have had to do them so I was wondering if somebody would be so kind as to check my logic. FBD attached. I have a component that is subjected to a couple Tc and is restrained in two places P1 & P2. I want to work out what the reactions are on P1 & P2 so forces F1 & F2 respectively. Note that I have neglected self weight of the part as it is insignificant compared to the coupling forces. My method is to take the sum about to places firstly at P1 to give me F2 than about O to give me F1. Is this the correct approach to the problem ?. Any help welcome.

A

 

[rb1957]

nope ...
you sum moments about any one point ... 0 would be reasonably obvious (since all the forces are normal).

then you sum forces in the x- and y- directions.
which way is x- and y- ? i'd take x- in the direction 0 to pt4 (where force 4 acts) and y- thru 0 in the same direction as your force 4
with these directions the sum of your applied forces = 0 (in both directions), so the sum of the two reaction forces will also some to 0.

Quando Omni Flunkus Moritati

 

[galaxy212]

Thanks for the response rb1957.

That was my initial pass to take moments about a single point then resolve in the vertical and horizontal directions but I found that as F3 & F4 are equal an opposite that there x & y components cancel each other out so I get F2x+F1x=0 and F2y+F1y=0 which if you rearrange either one gives you a negative ratio which cant be correct i.e F1=-0.37F2. This is because two points are resisting a couple so its a moment problem not a vert-horz equilibrium problem was my thinking ?.

 

[rb1957]

actually, the problem has four unknowns (F1x,F1y,F2x,F2y) and only three equations.

if you take sumM at pt2, you'll get F1 ... and there is -F1 acting at pt2 to balance things.

i'd suggest that the solution is a couple based on the 1-2 arm.

Quando Omni Flunkus Moritati

 

[zekeman]

"I want to work out what the reactions are on P1 & P2 so forces F1 & F2 respectively. "

What does this mean?
What are F1 and F2 and F3 and

http://www.eng-tips.com

Are they real forces?

If you have a couple and only 2 pin restraints, you don't need the F's.

 

[galaxy212]

Thanks for the response zekeman.

To give you some more details on the problem. All the forces are real; F3 and F4 are resultant pulley bearing forces which diverts a vertical rope horizontally then exits vertically again, so the two 90 degree turns produce a couple between the two pulley resultant bearing forces. F1 and F2 are where this pulley bracket is supported by two pieces of scaffolding tubing which I want to determine the stress in. I have just idealized with pin restraints to simplify. Can you please explain why if I have 2 pin restraints and a couple you believe I do not need the forces ?.

 

[rb1957]

if your two force directions are pre-determined (it looked like they were selected for easy of analysis), then you have an under-defined system (the opposite of redundant); i'm pretty darned sure there's a valid solution (though there might be if the directions were carefully chosen).

you have two unknowns (your two force magnitudes) and three equations (sumFx, sumFy and sumM) to satisfy.

something you can do is to solve the couple reacting the moment (a couple at 3 and 4, perpendicular to 3-4, yes?) then add forces at 3 and 4 along the line 3-4 that are equal and opposite. with vector addition you can get the resultant at one point to be in your desired direction, and see where the force resultant points ...



Quando Omni Flunkus Moritati

 

[zekeman]

So, what you are saying ( correct me if I am wrong), you basically have 2 pin restraint points on a plate to which you impose a pure couple,(F3x(r3+r4) and you want to know the reactions at the pins.

If so, then draw a line between p1 and p2 and call the distance D.
The forces in each pin would be
(F3x(r3+r4)/D. and they would be perpendicular to line between p1 and p2.

You would get this result by summing the moments of the system about either pin and setting the result to zero.

 

[rb1957]

i don't get the directions of F3 and F4 ... if there's a rope vertical at one point and horizontal at the other. they look close for a rope vertical at both, but they act along a line bisecting the angle the rope turns through, and not perdendicular to the line 3-4 ... yes? no?

if the directions at 1 and 2 aren't predetermined, then a couple will react the couple.

Quando Omni Flunkus Moritati

 

[zekeman]

Well, looks like more confusion.
Why don't you show a sketch of the pulley and bracket arrangement
so we can better assist you.
Looks like there is no pure couple as Rb indicates.

 

[galaxy212]

Good points. Iv included a more detailed sketch below. P1 and P2 are actually scaffolding tubes. F3 & F4 are the pulley resultants on the bracket which are equal and opposite in direction hence the couple. I think I may have made a wrong assumption in my initial sketch where by I assumed the scaffolding reactions would be perpendicular to the center of the couple and so I don't know the the angles of F2 and F1. Taking into account these angles call them THETA1 & THETA2, I have four unknowns and there for the problem is statically indeterminate ??.

 

[rb1957]

i think the plate is smart enough to react your applied couple with an opposite one. the plate will bear up on one side of one hole, and on the other side of the other hole; the forces would be normal to the line 1-2.

Quando Omni Flunkus Moritati

 

[zekeman]

Apparently, you do have a pure couple.
The easiest way to find that couple is to consider the pulleys as part of the bracket assembly and then the external forces to the assembly are the 2 vertical pulley rope forces,I'll call T, the tension in the rope.
These 2 forces cause the couple T*x, where x is the distance between the entering and leaving rope. Your method of finding the individual components will surely yield the same result ( but why complicate a trivially simple problem). Now the forces on the tubes must yield a couple equal in value. To get these forces...

Paraphrasing my earlier post
.. then draw a line between p1 and p2 and call the distance S.
The forces on each tube would be T*x/S
and they would be perpendicular to that line between p1 and p2.

 

[galaxy212]

Thanks for your rb1957 and zekeman. You've both pretty much come to the same conclusion which I also agree upon reflection so thanks again.

 

[oq172]

I took a look at this for a day or so before replying and I am not so sure about the simplicity of the problem. I do a agree about the pure couple of magnitude T*x (using Zekemems nomincalature). However, I am not sure what we can say about the reactions. To assume the reaction at each pin is all in the y direction or all in the x direction would be wrong. so what can we say with certainty:

1. The two forces acting on the pins are equal and opposite.
2. There is an equivalent couple at the pins equal to the applied couple T*x.
3. We can say that the magnitude of the resulting force at the pins is AT LEAST equal to T*(x/s). But because we don;t know direction, this reaction force could be higher.

This is deceptively complex probelm. I don't think there is a solution for the reaction forces.

Tell me, is there no reaction in the x-direction??

 

[galaxy212]

Thanks og172. If by x-you mean the horizontal axis on my most recent sketch then no (see above). The two scaffolding tubes illustrated as pins for simplicity counter the full torsion of the couple. I agree that there is no accurate fbd solution due to static indeterminancy for this problem though zeks and rbs solution should make a good approximation for now.

 

[rb1957]

"I don't think there is a solution for the reaction forces." ... why wouldn't the reaction be a couple based on 1-2 ?


Quando Omni Flunkus Moritati

 

[oq172]

Rb1957 - "why wouldn't the reaction be a couple based on 1-2 ?"

Well, look I am not suggesting that I have all the answers by anymeans. I just wanted to join the discussion with my persepective.

But to answer your quesiton, if you attempt to get the reaction forces at 1 and 2 based on the agreed and known couple, then you have to assume a direction for the resultant reaction forces. Remember there is a reaction in X (horizontal) and in Y (vertical) at each pin. I think you might be assuming that the there is not an x reaction. If this were true, then it would be easy to solve.

This my friend appears to be statically indeterminate.

Look, make one of the holes a slot and it will be deterministic.

 

[zekeman]

"This my friend appears to be statically indeterminate.
"
You are right!
Depends on the size and shape of the clearance holes in the plate.
If the holes were perfectly concentric then Rb and my solution would work.
Otherwise, pick your clearance geometry and hole shapes and you will get a variety of answers to the force vectors, but in all cases their component normal to the line between the two points, P1 and p2 are the one that we gave.
Accordingly, I withdraw my solution and agree with the indeterminacy of the problem.

 

[rb1957]

i'm assuming that the plate will react a couple with a couple. i agree that generally this problem is redundant (see one of my posts above). however in this special case there is only one load being applied (a torque) and there is one simple reaction to this (being a torque). of course there could be equal and opposite forces at 1 and 2, along the line 1-2; but i suggest that the minimum strain energy solution would be for these to be zero.

Quando Omni Flunkus Moritati