Basic statics question on shear moment equation

[brinethery]

Sorry, I meant to title it "Basic statics question on bending moment equation." Oh well.

Homework Statement

Problem F7-7
http://www.scribd.com/doc/94480961/Chapter-7-Fundamental-Problems

Homework Equations

Summation of the forces in y, summation of the moments

The Attempt at a Solution

I calculated the reaction at A, then found the shear force equation.

ƩF_y = 0 = A_y - 6kN , so A_y = 6 kN

The answer to the moment equation is supposed to be:
ƩM = 0 = M + 18 -6x
So M = 6x -18

I am particularly interested in why the M is positive in the beginning part of the moment equation. If someone could provide an explanation of the part along with a picture of the free-body diagram that's needed to make the moment equation, that would help out a lot. I am trying to understand chapter 7 but the more I read, the more confused I become.

 

[jimep]

Hello,

I will try to explain the problem. What you are dealing with is an encastre' support. This is actually called a cantilever.

To understand this please try to imagine a beam stuck into a wall. When you apply the force, the wall will provide reactions in all directions, that is X,Y and will also provide a Moment reaction to ensure that the point in the wall does not try to rotate.

I have tried my best to sketch the diagrams so you can understand. Refer to the 1st picture and you will see that since there are no forces in the X direction, reaction from the wall will only give an Ay. You have already solved this in your problem.

Now, logically the wall will also give you a moment reaction as I said before to ensure that the constrained point does not rotate.

Now please refer to the 2nd picture and you will see that I have derived the equations of equilibrium for this problem. I have taken M as a positive moment because it is an arbitrary selection. You could take M as a negative moment and then you would get a positive scalar, that is M=18.

Now, the reason we want to do the cross-section cut is to have a mathematical equation which gives us a moment for every point on the beam. We continue with the standard as you can read on Hibbelers, Chapter 7. Refer to picture Fig. 7-1 if you have Twelfth Eidition. The cross section allows us to derive an equation for the bendind moment and that defines for any distance 'x' in the beam.

Finally please refer to the 3rd picture in which you can see that the static diagrams are drawn. Since the shear forces are constant - that is T(x) = c; we can conclude that the M(x) will be a linear equation. That is from calculus:

M = ∫Tdx

Since your equation T(x)=6

We apply the integral, M(x) = ∫6dx = 6x+c1

Since we already have c1 = -18 from the equilibrium equations we can conclude that the moment equation is satisfied M(x)=6x-18.

This is again proven by the fact that a x=0 means the distance from point A is zero therefore the moment is equal to -18. M(0)=-18 which is in agreement with the derived M from the equations.

 

[jimep]

I am particularly interested in why the M is positive in the beginning part of the moment equation. If someone could provide an explanation of the part along with a picture of the free-body diagram that's needed to make the moment equation, that would help out a lot. I am trying to understand chapter 7 but the more I read, the more confused I become.

Please also refer to the two pictures below which I have created to further explain your problem. As I have said there must be no movement of on the wall side so in picture 1 you have what is the correct solution.

In picture 2 you have the opposite of the correct - and that is exactly what the wall prevents the beam from doing. Imagine a beam being able to lift the whole wall like that - not good engineering there.

I hope my explanations have been at least a bit of help to you. Please feel free to ask any questions you might have regarding the problem.

 

[brinethery]

Thank you thank you thank you. You really helped me to understand this problem and I now feel comfortable if I write a moment wrong. For some reason, I felt comfortable with drawing forces in the wrong direction but not moments when we got to chapter 7. I have the 12th edition of Hibbeler and maybe I'm just slow... it takes me a long time to understand these fundamental concepts.

If I have any questions about the other fundamental problems on this page, may I send you a private message with the link?

Seriously, your explanation helped out A LOT. This problem is crystal clear to me now. Thank you! :-)

 

[jimep]

Glad to have been able to help. Of course, you may contact me with the link.

I personally prefer to integrate over the shear forces. But if you are not comfortable with calculus then you might prefer the cross-section method. Either way you'll come to the same conclusion.

Just make sure to keep the directions clear, once you write down the equilibrium equations, keep the same directions on your further analysis.

 

[brinethery]

Glad to have been able to help. Of course, you may contact me with the link.

I personally prefer to integrate over the shear forces. But if you are not comfortable with calculus then you might prefer the cross-section method. Either way you'll come to the same conclusion.

Just make sure to keep the directions clear, once you write down the equilibrium equations, keep the same directions on your further analysis.

I am fine with integration, but I'd like to really understand how to do these problems before using calculus. Thank you again. I really appreciate your help :-)

 

[jimep]

I am fine with integration, but I'd like to really understand how to do these problems before using calculus. Thank you again. I really appreciate your help :-)

Yeah that's a good idea. It would be a really good idea if you check out the problems with loads. They give a better insight on the nature of the problem.

F7-8 gives a really good example on that. Try solving that and we can discuss it afterwards.