# Homework Help: Statics, beam attached to a wall, reaction forces

1. Feb 27, 2013

### jonjacson

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

I´ll show it with a picture:

2. Relevant equations

Static equilibrium, Sum Forces=0 ; Sum Moments= 0

3. The attempt at a solution

With the Fx, and Fy equations I can calculate the reactions Rx and Ry but I don´t understand the moments equation:

Sum M around point O =0= 1,4*1,2 + 15 - 3* cos(30)*4.8 = 4,2 KN m = 0

I didn´t use the moments of the reactions at O because I thought that their moment arms were zero, but in that case I get an equation that doesn´t have any sense.

I can calculate the moments of the forces to the point O, and I can add the 15 Knm, but the point is in order to get equilibrium Should I consider the moments of the reaction forces at O? In that case why the problem is not giving me any data about the lenghts x,y,z that I showed near O?

2. Feb 27, 2013

### voko

Compute all the moments about B. Don't forget the weight of the beam.

3. Feb 27, 2013

### jonjacson

Yes I forgot the weight.

I calculate Rx and Ry with the forces equations:

Rx = 1,5 kN
Ry=6,1 KN

Then I have the equation for the moments:

Moment around B= -1,4 * 1,8 -6,1 * 3 -3*cos(30)*1,8 + 0,5*9.81*0.6 + 15=-7.6 KN m = 0

The point is I have only two unkown quantites Rx, and Ry but three equations and I don´t know what can I get from the third equation.

4. Feb 27, 2013

### SteamKing

Staff Emeritus
First, you did not construct a free body diagram for this beam..

Second, you did not show the sum of the forces equations for both the x and y directions. (It helps to show the force at C in terms of its components)

Third, writing a moment equation about B is unnecessary. A moment equation about point O is sufficient.

The reason your moment equation doesn't make sense is that you neglected to put the unknown reaction moment at O in your equation. If you had started with a FBD, this would have been clear.

5. Feb 27, 2013

### voko

The fixed end at O, in addition to being able to exert a reaction force, may exert a reaction couple (because it prevents rotation about itself). So you should include that couple in the moments equation.

6. Feb 27, 2013

### jonjacson

You are right I was missing the moment reaction.

Well I went directly to the equation I didn´t understand, but I´m goint to do what you said.

1.- Hopefully this body diagram is correct:

2.- Ok, these are the equations:

Sum Fx= -3KN *sin (30) + Rx= 0 -----> Rx= 1,5KN

Sum Fy=-3KN*cos(30) -0,5KN*9.81 +1,4KN +Ry= 0 ----> Ry=6,103

3.- Well I didn´t understand what was going to change but I tried to do that to see if the reactions at O entered the equation and it made sense.

4.- I see, well I thought that the reactions on the wall were only forces. It´s difficult to imagine to me that there is a moment reaction, since a moment could be created through the use of two forces.

Now the equation for the moments looks like this one:

1,4*1,2 +15 - 0,5*2.4*9.81 - 3*cos(30)*4.8 + Mro=0

Mro= 7.57 KN m, ccw

Which is the answer in the book.

I think that the main problem to me was that I thought that moments could me "counteracted" through forces, it was difficult to realize that even if the reaction forces at O are equal to the forces applied I didn´t solve the problem because not only is important the force, in the equation for the moments the moment arms are equally important and there must be a reaction moment.

I´ll show what I mean with an example to check if I´m right about this, lets take the same situation but without a wall, now there are three points as it´s shown in the image.

And we know the distances a=b=c=0,3 meters.

It is not necessary to calculate forces and moments, from a general point of view, you have three points acting normally to the beam and that means that their line of action passes through the point O, as a consequence their moment arms are zero around O.

If an external moment is applied to the beam around point O, it doesn´t matter if it´s gravity or any other force, there must be a moment reaction from the three points to "counteract" that moment. But how is that possible? If the moment arms of the points are zero around O how are they able to create a moment around O of equal magnitude but opposite sense to the applied moment?

I can´t imagine moments without forces and moment arms.

Last edited: Feb 27, 2013
7. Feb 27, 2013

### voko

A fixed end modeled as a single point is an idealization. Any real fixed end will have a non-zero length, the reaction forces will be acting along this entire length (not unlike the distributed force of gravity in the example), so there will be a moment. In the idealization, you lose those distributed forces, but you still have their moment - without it, the joint can't prevent rotation about itself.

8. Feb 27, 2013

### jonjacson

Well I agree with your first part of the post, but I think that it looks like impossible to create any moment in the idealization, so I think that the joint in that case could not prevent rotation, I´m not sure of course.

Let´s create another model of the wall, but now with points able to create moment around O:

Now there are two points able to create a moment arm, so the equilibrium equations are:

Sum Fx= -3sin(30)+Rc=0 ---> Rc=1.5KN

Sum Fy= 1,4 -mg -3 +Rd+Re-Rb-Ra=0

For the moments these are the distances:

ba=de=0.3m

O to f1=1,2m

f1 to f2=1,8m

f2 to f3=1,8m

The equation for the moments is:

Sum Mo= Mb - Md +1,2*1,4 -2.25*0.5*9.81 - 3cos(30)*4.8 +15=0

Now the problem is that I have three equations and 5 unknown quantities, unless I start making assumtions it´s impossible to resolve the system :(

At least do you agree with the approach now?

9. Feb 27, 2013

### voko

If you model joints not as single points, you end up having more unknowns than equations. For the record you could end up in the same situation when you have additional joints. These problems are known as statically indeterminate, and they cannot be resolved assuming that bodies are rigid.

In the rigid body limit, you have to idealize things. That's how you end up having a single point joint that can have a couple.

Here is one way to deidealize a fixed end joint. Assume that the joint is made of two "fittings". One is at point O. It can exert force in the Y direction only. The second fitting is at the deeper end of the joint, and it can exert force in X and Y directions.

10. Feb 27, 2013

### SteamKing

Staff Emeritus
You are getting hung up on the details of how the reaction couple may be produced.

The purpose of the FBD is to show that the reaction couple must be present in order for the beam to be in equilibrium.

This is what is called a cantilever beam. There are other types of beams with different support conditions which are more difficult to analyze. This problem helps to show you the proper procedure for solving all types of beam problems.

11. Feb 27, 2013

### jonjacson

I see, I read yesterday about statically indeterminate and I´m not used to think about them. Well I understand that the solid rigid model has its limitations, but it was the first problem with moment reactions and I was just curious about how the couple was created.

Yes it´s a mess trying to guess how the wall reacts exactly to the applied forces and moments, now it´s much more clear what can I expect from the solid rigid model.

If I want more precision about how exactly forces are distributed on the wall I suppose that you will need to use another model more complex.

Sorry if I´ve made stupid questions, it´s just that it has been quite surprising to see that without the reaction couple anything made sense, and I didn´t expect that without talking about forces and moment arms, I´ll remember this problem next time I´ll solve a problem.

Thank you very much Voko and SteamKing.