Back to Basics FBD
Back to Basics FBD
(OP)
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
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





RE: Back to Basics FBD
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
RE: Back to Basics FBD
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 ?.
RE: Back to Basics FBD
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
RE: Back to Basics FBD
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.
RE: Back to Basics FBD
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 ?.
RE: Back to Basics FBD
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
RE: Back to Basics FBD
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.
RE: Back to Basics FBD
if the directions at 1 and 2 aren't predetermined, then a couple will react the couple.
Quando Omni Flunkus Moritati
RE: Back to Basics FBD
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.
RE: Back to Basics FBD
RE: Back to Basics FBD
Quando Omni Flunkus Moritati
RE: Back to Basics FBD
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.
RE: Back to Basics FBD
RE: Back to Basics FBD
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??
RE: Back to Basics FBD
RE: Back to Basics FBD
Quando Omni Flunkus Moritati
RE: Back to Basics FBD
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.
RE: Back to Basics FBD
"
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.
RE: Back to Basics FBD
Quando Omni Flunkus Moritati
RE: Back to Basics FBD
I'll give one example similar to OQ' slot idea.
One bracket hole is circular and has a very small clearance with one pipe while the other bracket "hole" is a parallelogram, loosely fitted around the other pipe. Now rotate the bracket around the circular hole and see what you get.
RE: Back to Basics FBD
RE: Back to Basics FBD
@zeke, the point is that there is a redundant load (along the line 1-2) that can't be determined by equilibrium, but any load in this direction (applied at 1, reacted at 2) effectively reduces the arm for the couple, ie increases the resultant force and so increases the strain energy in the plate. minimum strain energy is a solution for redundant structures, and i suspect/propose that the minimum strain energy solution has the longest effective arm, the smallest couple.
Quando Omni Flunkus Moritati
RE: Back to Basics FBD
Rb,
Read my post.
Not for the scenario I propose where there are pipes in two loosely fitted holes. There is a real answer for this.
RE: Back to Basics FBD
RE: Back to Basics FBD
Quando Omni Flunkus Moritati
RE: Back to Basics FBD