# I Mechanics, statically indeterminate reactions

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1. Sep 12, 2016

### anton_kiziloff

Suppose I have a beam and I would like to constrain it (two dimensional structure). At one end I've chosen to use support which constrains motion only in "x" and "y" direction (horizontally and vertically), but it doesn't prevent a body from rotation. Because of that, another beam's end have to be constrained as well. If I use the same type of support, the system would have four unknown and three equations. Is it still possible to solve this kind of system and how should I proceed with it? Would it be correct to split the problem in two cases and solve them separately, choosing the unknown resultants depending on the loading conditions?

2. Sep 12, 2016

### Stephen Tashi

You need to state the forces acting on the beam. For example, why does the beam need horizontal support? As far a vertical support goes, is there a downward force on the beam in addition to the force of the beams own weight ?

3. Sep 12, 2016

### Staff: Mentor

This isn't correct, you should have the same number of equations and variables. You should only have three variables, two for position and one for rotation.

4. Sep 12, 2016

### anton_kiziloff

You are right, I forgot to mention what kind of load beam has. Okay, in my case there are two forces to consider. The first is a weight and the second is a force directed downwards at some angle "alpha" (basically it has horizontal and vertical component). And there is another issue here: the resultants of the support reacts only in one direction, like with the wall or the floor.

5. Sep 12, 2016

### anton_kiziloff

Exactly. But such cases do exist and they are even mentioned in several books on vector mechanics. I've red somewhere that one additional equation could be added by considering material you are using in a structure. Basically using the material property like elasticity limit or yield strength. Though I'm not sure of it.

6. Sep 12, 2016

### Staff: Mentor

Yes, that is correct. For systems that are statically indeterminate the additional equations come from Hooke's law.

7. Sep 12, 2016

### Stephen Tashi

The two supports preventing horizontal motion make it an indeterminate problem. When the beam is held motionless, you can always have the support on the left push toward the right with X more pounds and have the support on the right push toward the left with X more pounds to cancel things out.

This type of indeterminate problem can't be solved just by knowing the elastic properties of the materials involved. Solving it involves making a decision about how much "internal stress" you want to build into your design.

8. Sep 12, 2016

### Staff: Mentor

Hmm, I think @Stephen Tashi is right. It has been too long since my statics class. Please ignore my comments.

9. Sep 12, 2016

### Staff: Mentor

Can you please provide a diagram. I'd like to chime in on this, but I want to be sure that I understand the problem before doing so.

10. Sep 13, 2016

### A.T.

The simplest example is some mass fixed to the wall with two nails arranged vertically. You know the total vertical force on both nails together, but not how it's distributed among them.

11. Sep 13, 2016

### Staff: Mentor

OK. But, if the wall is perfectly rigid, then you can determine the forces if you know the elastic modulii and Poisson's ratios of the nails and the mass. It's just a typical solid mechanics problem that can be solved using finite element. So, even though the problem is statically indeterminate for rigid mass and nails, it can still be solved if the deformational mechanics are included.

12. Sep 13, 2016

### Nidum

This is a possible interpretation of what @anton_kiziloff describes .

Not sure what was meant by 'a weight' .

As for : ' the resultants of the support reacts only in one direction, like with the wall or the floor '. I don't think that requirement can be met entirely without modifying the loads and/or the constraints .

Why not tell us what you are actually trying to do ?

Last edited: Sep 13, 2016
13. Sep 13, 2016

### Nidum

There are standard methods for analysing statically indeterminate beam problems .

14. Sep 13, 2016

### Stephen Tashi

You may also need to know the assembly procedure. For example, one may hang a plaque to the wall with a nail on top of the plaque and then drive another nail through the bottom of the plaque "just to be safe" or one may hang the plaque by driving the nails in the reverse order.

15. Sep 13, 2016

### Staff: Mentor

Sure. The initial geometry and state of stress and strain are important in solving a solid deformation mechanics problem.

16. Sep 13, 2016

### anton_kiziloff

Thank you @Nidum for the picture you've provided. This is an almost perfect graphical description of the system I've had in mind. As for the thing I was trying to achieve, it is quit simple. I was trying to apply the knowledge from the statical mechanics to the real life situations. For example, if one had some body under the load and he would like to restrain its motion, how would he do that? What needs to be considered? Where to start and how to move next? I mean, in the textbooks we have already prepared examples with the described forces and connections. But what if only body is provided with the forces acting on it, and nothing more? How to choose the right connection/support?!

Returning to the main topic here, there is something else worth trying... How about using the third Newton law? If there is no action, then there is no reaction. So in the case of reaction acting only in one direction, it should work. Like in the @Nidum figure. Each of beam's end has an upwards directed vertical component of a resultant and a horizontal component directed opposite to support. But there are no components in other directions. So what matters here is the angle of the force acting on beam. Just thinking logically, when the angle is less than 90 degrees, one of the support will compensate the horizontal component of the force, while the other will do nothing because it doesn't have the ability to prevent the motion. The same situation changes when angle will change from "less than 90" to "more than 90". When you think it that way, there are no problems with the system, because it will always have only 3 unknowns.

17. Sep 13, 2016

### Staff: Mentor

I don't think your analysis of the horizontal force situation is correct, particularly if the right end is pinned so that the horizontal displacement at this end is zero.

18. Sep 13, 2016

### Stephen Tashi

That's only "logical" if you assume the configuration is assembled so that there is initially no horizontal load on the beam, including no horizontal force from the two supports. That's a safe assumption when you solve problems about "academic reality".

19. Sep 14, 2016

### Staff: Mentor

Here is the deformable solid mechanics analysis of the problem where the right support prevents the right end of the bar from moving:

This refers to Nidum's figure in post #12. Let E be the Young's modulus of the bar and A be the cross sectional area of the bar. Let $\delta$ be the displacement to the left for the bar cross section directly underneath the applied concentrated load F, and let x represent the distance of the concentrated load from the left end of the bar. The total length of the bar is L.

With the displacement $\delta$ being directed to the left, the section of the bar to the left of the applied load will be in compression, and the section of the bar to the right of the applied load will be in tension. The tension in the portion of the bar to the left of the applied load will be $-EA\frac{\delta}{x}$, and the tension in the portion of the bar to the right of the applied load will be $+EA\frac{\delta}{(L-x)}$. Therefore the reaction force exerted by the left support will be directed to the right, and will be equal to $EA\frac{\delta}{x}$; the reaction force exerted by the right support will also be directed to the right, and will be equal to $+EA\frac{\delta}{(L-x)}$. From a force balance on the bar in the x direction, we then have:
$$EA\frac{\delta}{x}+EA\frac{\delta}{(L-x)}-F\cos{\theta}=0$$ Solving for $\delta$, we obtain:
$$\delta=\frac{x(L-x)}{L}F\cos{\theta}$$Substituting this into the relationships for the reaction forces gives:$$R_{left}=\frac{(L-x)}{L}F\cos{\theta}$$
$$R_{right}=\frac{x}{L}F\cos{\theta}$$
Note that these final result do not involve the elastic modulus E or cross sectional area A. Both reaction forces are to the right. If the applied load is very close to the left end, the reaction force on the left end carries most of the load. If the applied load is very close to the right end, the reaction force on the right end carries most of the load.

20. Sep 14, 2016

### anton_kiziloff

Thank you friends for all your replies. And thank you @Chestermiller for your explanation. I think I get your point. I have to learn Mechanics of Materials...