Calculation of resulting force from out of axis Torque

[Jony S]

Homework Statement

[/B]
Hello, I would like to request some help in solving this mechanics problem.
Consider the attached drawing of a solid structure which is pinned to the ground on points A and B.
Distances AC=Xa, CB=Xb, CD=H1 and DE=H2
A torque with magnitude M is applied on point E. How do I determine the resulting forces from this torque Fx and Fy on point A and point B ?

Homework Equations

This would be the equation for torque M
M=r*F

The Attempt at a Solution

My guess is that we can bypass the "middle" geometry of the structure, and simply use the distance from point A to point E (let's call it R_AE), and its angle with the horizontal (θ_AE).
The resulting force would thus be F = M/R_AE, with Fx = Fcos(θ_AE) and Fy = Fsin(θ_AE). A similar approach would be applied for point B.
Is this correct ?

Thanks in advance !

 

[BvU]

Hello Jony, $\qquad$ $\qquad$ !

When I try to look at the picture I get something almost black on black. Can you fix that ?

How do I determine the resulting forces from this torque Fx and Fy on point A and point B ?

The principle is simple: make force and torque balances. If you pick your axis conveniently, the work you need to do can be minimum.

 

[haruspex]

How do I determine the resulting forces from this torque Fx and Fy on point A and point B ?

You can't, at least not in that general sense.
Suppose you have a solution. You could add a force to the right at A and an equal and opposite force at B and it would still be a solution.
Other than that, follow @BvU's advice.

 

[Jony S]

Hello Jony, $\qquad$ $\qquad$ !

When I try to look at the picture I get something almost black on black. Can you fix that ?
The principle is simple: make force and torque balances. If you pick your axis conveniently, the work you need to do can be minimum.

Sorry about the picture, it's a png with transparency that's why it appears weird, something to do with the website's interface. If you click on the name of the file instead of the picture, a new tab opens with the correct display.

What do you mean pick my axis conveniently ? is my proposed solution the convenient axis ? i.e., the line AE ? where F = M * R_AE ?, with the force vector having a 90deg angle with line AE ?

 

[BvU]

You have shown an Fx and an Fy. Are those the ones with the same names in post#1 ?
What other forces are there to complete the balances mentioned ?
As @haruspex says, it llooks as if you need to make an assumption (or find another given in the problem statement) since otherwise you end up with one equation less than the number of unknowns ...

 

[haruspex]

we can bypass the "middle" geometry of the structure

Yes, even more so than you have done.

The resulting force would thus be F = M/RAE, with Fx = Fcos(θAE) and Fy = Fsin(θAE).

I see no basis for this reasoning.
Always start these statics problems with a collection of torque and force balance equations:
Σtorques about a chosen axis = 0,
Σforces in a chosen direction = 0.
In 2D problems such as this, you generally have three such equations available. Commonly, two linear and one rotational are used, but usually one linear and two rotational also works.
Careful choice of axis can avoid involving one or more forces that you do not know and do not need to find.
In the present case, as I indicated, there is insufficient info to find the x direction forces, so that is a guide to where to put your axis/axes.
Hint: with a good choice of axis, you can write down an equation that gives you the Fy at A immediately.

 

[Jony S]

View attachment 238611
You have shown an Fx and an Fy. Are those the ones with the same names in post#1 ?
What other forces are there to complete the balances mentioned ?
As @haruspex says, it llooks as if you need to make an assumption (or find another given in the problem statement) since otherwise you end up with one equation less than the number of unknowns ...

Fx and Fy that I want are simply the forces from applied momentum M. If we approach it as a traditional "static" problem then I guess we can simply imagine that A and B are "screwed" to the ground, and whatever forces result from the momentum are the normal reaction forces on this screw, i.e., the screw reacts on the structure with opposite forces to the resulting ones from the torque.

I have drawn what I meant initially in this updated picture. From here we get that Fx = -Fcos(theta2) and Fy = Fsin(theta2), with F = M / R_AE

 

[Jony S]

Yes, even more so than you have done.

I see no basis for this reasoning.
Always start these statics problems with a collection of torque and force balance equations:
Σtorques about a chosen axis = 0,
Σforces in a chosen direction = 0.
In 2D problems such as this, you generally have three such equations available. Commonly, two linear and one rotational are used, but usually one linear and two rotational also works.
Careful choice of axis can avoid involving one or more forces that you do not know and do not need to find.
In the present case, as I indicated, there is insufficient info to find the x direction forces, so that is a guide to where to put your axis/axes.
Hint: with a good choice of axis, you can write down an equation that gives you the Fy at A immediately.

I see your approach but I think this is a bit more complicated than I wanted, since there was no need to make this a "static" problem (last comment for more details :P) I simply need the forces on A or B that result from momentum M.

edit: ok I mentioned the structure was pinned to the ground on the initial comment so that might have sparked some confusion. If we assume it is pinned to the ground than we just make the resulting force from momentum be the reaction forces on points A and B, and we have our sums on X and Y be zero that way

 

[haruspex]

I have drawn what I meant initially in this updated picture.

As I posted, you need to specify your axis. From your diagram, you appear to be taking point E as your axis, but your equation must then include all forces that have moment about that axis. This will include the forces at B.

I see your approach but I think this is a bit more complicated

On the contrary, it is far simpler. You do not need to work with angles at all. Better still, it is a valid method, whereas yours does not appear to be.
As I wrote, with a well chosen axis, you can instantly write down an equation that gives the y force at A. For what axis is that the only unknown force that has a moment?

 

[Jony S]

As I posted, you need to specify your axis. From your diagram, you appear to be taking point E as your axis, but your equation must then include all forces that have moment about that axis. This will include the forces at B.

On the contrary, it is far simpler. You do not need to work with angles at all. Better still, it is a valid method, whereas yours does not appear to be.
As I wrote, with a well chosen axis, you can instantly write down an equation that gives the y force at A. For what axis is that the only unknown force that has a moment?

of course i see what you mean now, yes the axis of rotation is on point E, but of course the forces from the moment applied there will distributed on points A and B, so it's the sum of the momentum from both those forces will equal M. Can't do the math now but I'll come back here tomorrow :)

 

[Jony S]

I drew a new picture to explain the new approach.

We have points A, B and C of a rigid structure. The force "F" is such that it's momentum on point C is M, so that M = F * L4.
In order for the structure to be static, the reaction forces on points A, B and C will have to obey the system of equations:

F=M/L4

ΣFy = 0 -> F_Ay + F_By + Fsin(θ) + F_Cy = 0

ΣTorque(A) = 0 -> F*L3 + F_By*L1 + F_Cy(L1+L2) = 0

ΣTorque(B) = 0 -> F_Ay*L1-F_Cy*L2 = 0

Is this the right approach ? anything wrong/missing on my equation system ?

 

[haruspex]

The force "F" i

But what is this force F? And what is F_Cy?
There are only the forces at A and B and the moment M.

If you won't follow my hints in post #9, start by considering the X components of the forces at A and B. Write down that force balance.

 

[PhanthomJay]

For clarity here, you are making several incorrect assumptions. Firstly, Forget about the Fx reaction forces at A and B. They are essentially zero in this case, whether both supports are pinned or one is pinned and the other is roller or simply supported. And even if there were Fx forces, they would not be related to the Fy forces in the manner you have shown.

Secondly, Point E is on the axis of the applied moment M (which you incorrectly call momentum, leave off the -um). But you can choose ANY point to be the axis about which to sum moments, as has been hinted at in prior posts.

Thirdly, did you know that the moment of an applied moment about ANY point is the applied moment itself??

Focus on solving your Fy forces at A and B by conveniently choosing the point about which to sum moments (as has been noted by Haruspex).

 

[haruspex]

Forget about the Fx reaction forces at A and B. They are essentially zero in this case,

I had been carefully avoiding stating that, preferring to lead Jony to discovering it.

 

[PhanthomJay]

I had been carefully avoiding stating that, preferring to lead Jony to discovering it.

That is not readily apparent to even the best young ‘computer schooled’ engineers, never-mind a beginner, and especially since there are no applied vert or horizontal forces. You can trust me on that.

 

[haruspex]

That is not readily apparent to even the best young ‘computer schooled’ engineers, never-mind a beginner, and especially since there are no applied vert or horizontal forces. You can trust me on that.

It should have followed swiftly from post #12.

 

[PhanthomJay]

That tells me that the x forces at A and B are equal and opposite.

 

[haruspex]

That tells me that the x forces at A and B are equal and opposite.

And in the same line, so whatever they are they cancel completely and can be taken as zero.

 

[Jony S]

Hey guys, sorry about the delay,

yes forget about Fcy, that's silly, and yes Fxs will cancel, I got that :P

Now, if I exclude my force "F" (the one that makes moment M) and just assume there's an applied moment M on point C /(with no "translational" force), I will ultimately get something which doesn't sound physical to me.

ΣFy = FAy + FBy = 0
ΣFx = FAx + FBx = 0

ΣTorque_C = M + FAy(L1+L2) + FBy*L1 - FAx*h - FBx*h = 0

Since FAx*h + FBx*h = (FAx+FBx)*h = 0, the last equation is simplified.

From here i get that FAy = -Fby = -M/L1

Now, this means that if L1 tends to zero, i.e., the two pinned points are very close together, FAy tends to infinity, which isn't intuitive at all, so I still think there's something missing here...

 

[Jony S]

If I add forces F1 and F2 that produce moment M, but are symmetric so that it doesn't introduce translation to the system, I get the same solution (in the picture, sorry for bad handwriting :) )

 

[haruspex]

FAy = -Fby = -M/L1

Correct, depending on your sign convention for the forces and moments.
The easier route is to take moments A (and likewise B). This has the merit that M and F_By are the only terms.

if L1 tends to zero, i.e., the two pinned points are very close together, FAy tends to infinity,

True. Try turning a long bar only holding it in the middle when someone else is turning it the other way while holding it near the ends.

 

[Jony S]

The easier route is to take moments A (and likewise B). This has the merit that M and F_By are the only terms.

True, the end solution is the same, but I wasn't sure if the moments at A/B were simply M+F_B/Ay*L1, because it's wasn't clear to me if moment M is somehow distributed between A and B or if it actually is M in both points (plus the Fy*L1 of the other point)

Thanks for the help :)