# Classical mechanics problem

1. Jul 27, 2008

### Dafe

1. The problem statement, all variables and given/known data

Find the expression for the forces on support A and B. Fax, Fay, Fbx and Fby.

2. Relevant equations

I'm thinking;

Moment , M=F*l
Fa^2=Fax^2+Fay^2, same for b
Fax=Fbx

3. The attempt at a solution

(1) M = F*L1
Since I know that the sum of moments around the "joint" is zero, and that Fax=Fbx I can find both Fax and Fbx.

What I don't understand is how to find the forces in the y direction. Could someone please give me a hint on how to proceed? Thanks alot!

2. Jul 27, 2008

### PhanthomJay

You know from Statics that the forces at A and B in the Y direction must sum to F. Since the problem is statically indeterminate, you need another equation to find out how that force F splits between the 2 supports. Think about axial deformations in the member AB under the Load F applied at L3 up from A (Hint: Think about proportioning the load).

3. Jul 27, 2008

### Dafe

I just can't see what other equation is needed. About the axial deformations you mentioned; do you mean hos the rod tries to "pull" support A upwards?
Thank you

4. Jul 27, 2008

### PhanthomJay

No, the force F, applied at the point where the horizontal and vertical members meet, puts the 'squeeze' on part L3, but 'extends' part L2. From compatability, these 2 deformations must be equal. So most of the load goes to the shorter (stiffer) part of AB. For example, if L3 were very short in comparison to the total length of AB, just about all of the load F would go to A.

5. Jul 29, 2008

### Dafe

I have a little question about something you wrote earlier.

Doesn't statically indeterminate mean that the problem has more supports then needed for it to be in equilibrium?
To me it looks like this particular problem is not statically indeterminate, since it has just enough supports for equilibrium.

I thank you for your patience, I'm a slow learner.

6. Jul 29, 2008

### moemoney

Let the sumnation of all X, and Y forces = 0.
You could then take the moment about the intersection of L1, L2 , L3 and let it = 0.
You would then have 3 equations for 3 variables which is enough to solve for the solution.

7. Jul 29, 2008

### PhanthomJay

A statically indeterminate structure is one in which there are more unknown reaction forces (or internal forces) than there are the static equilibrium equations. In this problem, there are 3 static equilibrium equations: Sum of forces in X direction =0, Sum of forces in Y direction =0, and sum of torques (moments) about any point =0. But there are 4 unknown reaction forces, namely, Fax, Fay, Fbx, and Fby. Using the static equilibrium equations, you can solve for Fax and Fbx, but try as you might, you can't solve for Fby or Fay without having to consider compatability and deformations. All that the static equilibrium equation will tell you in the Y direction is that Fay + Fby = F (the applied force); it cannot determine how that force F splits amongst each of those terms. That is why you need another equation beyond those 3 equilibrium equations.

If one of those supports were a roller support instead of a pinned support, that is, it was free to slide vertically, then the problem would be statically determinate, since all of F would have to be taken out at the pinned support. But such is not the case here.

Let's assume in this problem that L2 and L3 are equal. Can you visualize that for this case, The reaction upward at A would be F/2 (placing segment L3 in compression), and the reaction upward at B would be F/2 (placing segment L2 in tension)? Now you have to develop the general relationship when L2 and L3 are not equal, using the deformation approach. Are you familiar with the deformation formula? It is a linear equation, such that you can proportion the vertical load reactions as a function of the segment lengths. The shorter member carries the greater share of the vertical load.

8. Jul 29, 2008

### PhanthomJay

Might seem that way, but not true. Put in some simple numbers like F =1, and L1 = L2 =L3 =1, and try it. You won't be able to get Fay and Fby no matter how hard you try.

9. Jul 29, 2008

### moemoney

Ahh yes very true. Fay and Fby in this case cant be represented in any moment equations other then having Fay = Fby (moment about point F1) just because they are in-line with each other right?

10. Jul 29, 2008

### PhanthomJay

Be careful, if you take moments about F1, you have to consider the moments due to support reactions Ax and Bx as well. Ay does not generally = By. You have to use the compatability/deformation approach to find them, in conjunction with Ay + By = F. Statics alone and taking moments won't cut it.

11. Jul 30, 2008

### Dafe

I used Hooks law and found that Fby/Fay = L3/L2. I then use F=Fay+Fby with it and get that
Fay=F/(1+L3/L2)

Is this correct? Thank you

12. Jul 30, 2008

### PhanthomJay

Yes, excellent. You might want to rearrange terms a bit, to get $$F_{ay} = FL2/(L2 + L3)$$, but it's just an algebraic manipulation to put it in a somewhat better form. You'll also find that $$F_{by} = FL3/(L2 + L3)$$.

I assume you've calculated Fax and Fbx correctly?

13. Jul 30, 2008

### Dafe

That does look better.

For Fax and Fbx I just used the fact that the moment about the joint is zero and that Fax=Fbx.
That should be correct.
Should I post a proper solution to this problem for others to see? I'm a bit slow with Latex formatting..

Thank you very much!

By the way. You wrote earlier :

"No, the force F, applied at the point where the horizontal and vertical members meet, puts the 'squeeze' on part L3, but 'extends' part L2."

Did you just imagine the force F moved to the point where the members meet?

14. Jul 30, 2008

### PhanthomJay

Fax and Fbx are equal, but they point in opposite directions, as you show on your diagram. Their values are functions of F, L1, L2 and L3. If you are comfortable with your result, fine. I notice, however, that you have not shown the proper direction for the support reactions Fay and Fby. Be sure to catch it.
your call. Latex is nice, but you've done well so far without it.

My pleasure.

I think I may have unfortunately left my imagination in Disney World, but regardless, I was envisioning a free body diagram of member AB, where at that joint you have a downward internal force F and an internal counterclockwise couple of (F)(L1). The actual applied force F is where it is. Nice job!

15. Jul 31, 2008

### Dafe

If I envision the member AB with a internal force F which "squeezes" the bottom part and extends the part above the joint I would say the supporting forces Fay and Fby should face in the opposite direction.

The way I saw it was that the force F "twisted" the construction around the joint. That's why I drew the supporting forces as I did.. Hm, I think I'm a little bit confused. Care to explain?
Thanks again.

16. Jul 31, 2008

### PhanthomJay

That is correct. Fay points upward (toward the member, compression forces in segment L3), and Fby also points upward (away from the member, tension forces in segment L2). It takes a little imagination to determine this.

You are confusing the axial forces in AB, which causes no twisting, with the moment couple, (F)*(L1), which does the twisting. It is this couple which produces the horizontal reactions Fax and Fbx in the directions you have shown, which, for some reason, you have been very secretive of the values you have determined for same. Care to disclose them? I'd very much like to see what you got.

17. Jul 31, 2008

### Dafe

I think i get it now. I thought you didn't have any imagination :) Thank you very much.
Here is how I find Fax and Fbx:

Moments around the "joint":
F*L1=Fbx*L2+Fax*L3

since Fax=Fbx

Fax=F*L1/(L2+L3)

18. Jul 31, 2008

### PhanthomJay

Thanks to you and John Lennon, I'm a dreamer again. Imagine!
"You may say that I'm a dreamer
But I'm not the only one
And the world will live as one"

Yup, you got it! I was just testing you, because sometimes as we focus on the difficult, we mess up the simpler stuff. Also, way back in your original post you noted that Fa^2 = Fax^2 + Fay^2, which is quite correct. But it is perfectly acceptable (and preferred) to leave the support reactions in their x and y component forms. Peace.

19. Aug 1, 2008

### Dafe

Great stuff! You've been of great help.
Until next time!