Homework Help: ME statics - Shear Forces and moments inside beam

1. Nov 2, 2014

Feodalherren

1. The problem statement, all variables and given/known data
W= 3kip/ft.
Find the shear force and moment at C.

2. Relevant equations

3. The attempt at a solution
Skipping some initial steps, I found the reactions to be Ax=0 and Ay=9

These are known to be correct.

Now, slicing the beam from a to c.

I get Ay=9.

Now I take 3(1/2)6= 9 from the W equation and I know that this acts at 4ft from A.

So then the sum of the forces in the Y direction.

V+9-9=0 therefore V=0. This is incorrect. V=3 by the solutions manual but I have no idea how they got that.

2. Nov 2, 2014

OldEngr63

Draw an FBD of the segment of the beam from a to c only. The shear force acts on the right end, the previously determined reaction Ay acts on the left end, and the distributed load over a-c acts on the top.

3. Nov 2, 2014

SteamKing

Staff Emeritus
Don't you have this already? It's the support reaction at A. What's the change in loading between A and C?

Instead of just writing out numbers that you think you know, why don't you write the equations of statics and show how you calculate the reactions at A and B?

You can construct the shear force diagram for the beam after finding the reactions at both A and B. There's no need to worry about any reactions in the x-direction since the beam is loaded only vertically.

Once you have constructed the shear force diagram for the beam, the bending moments at a particular point are just the area of the shear force diagram from A up to the point of interest.

4. Nov 2, 2014

Dr.D

SteamKing, I think you are mistaken when you said, "Once you have constructed the shear force diagram for the beam, the bending moments at a particular point are just the area of the shear force diagram from A up to the point of interest."

If we look at an FBD for the left end, it will show the left end reaction, the distributed load, and the internal shear force, all combined in equilibrium. I think you left out the left end reaction in your statement.

5. Nov 2, 2014

SteamKing

Staff Emeritus
The shear force diagram implicitly includes the reactions of the beam, which is why one writes and solves the equations of static equilibrium before constructing the shear force diagram. This is all standard technique for working these types of problems.

6. Nov 2, 2014

Dr.D

The problem asks for the shear force and bending moment at C. If we take your approach, SteamKing, then we have to integrate a dirac delta to account for the reaction load on the left end. (That may be a bit heavy for the OP, I suspect.) I think it is much simply to simply draw the FBD with a cut at C and apply the equations of statics to this FBD. We don't need the shear diagram at all points; we are only interested in pt C.

7. Nov 3, 2014

SteamKing

Staff Emeritus
I'm afraid I don't follow this explanation at all. You can't apply the equations of statics to the cut free body unless you know the reaction force acting at point A, which even the OP has found, but his mistake was in not properly computing the loading on the beam when trying to find the shear force at point C.

The shear force value at point A is the reaction at that point. The value of w at the right end of the beam is 3 kip/ft, so the value of the loading at any intermediate point is simply proportional to the distance from point A, or w(x) = 3*x/18 kip/ft. The total load on the beam is w(L)*L/2 = 3*18/2 = 27 kip. As stated by the OP, RA = 9 kip, so the shear force at point C, which is located at x = 6 ft. from A is then RA-w(6)*6/2 kip, using the previous expression derived to calculate w(x). Evaluating this expression, we get 9 - (3*6/18)*6/2 = 9 - 3 = 6 kip as the shear force at x = 6 ft. from A.

8. Nov 3, 2014

OldEngr63

SteamKing, I don't think you are hearing me.

My only point was that we do not have to find V(x) in order to find V(6). We can deal with the FBD to the left of C using the equations of statics without the need to find the more general expression for V(x). I am not saying that your approach is incorrect, but only that it is more general than it needs to be.

9. Nov 3, 2014

SteamKing

Staff Emeritus
I'm not saying you need to find V(x) in general, but you do need to find the reaction at A in order to find the shear at x = 6 feet.

I advocate using standard methods of solution (of which there may be more than one) as opposed to just writing numbers down on the fly. I also think for beam problems of this type, it is helpful to know the relationship between the loading of the beam, the shear force diagram, and the bending moment diagram, as opposed to just calculating the values of these quantities at a given location, especially for the student who is inexperienced in solving these types of problems. IMO, this approach saves a lot of time later when you want to calculate the slopes and deflections of a beam under load.

10. Nov 3, 2014

Feodalherren

I skipped a couple of steps because I have the solutions manual - I know that my Ay and By are correct. That's why I ignored showing you how I got to them.

But here's the FBD that I used.

The question is how exactly they got a force of 3kip at 4 ft.

If I take the W=3x equation, then on 6 ft of the beam I get W=.5*6*3=9kip load. By centroids I know that it will act 2/3 from the origin so at 4 ft.

11. Nov 3, 2014

SteamKing

Staff Emeritus
First things first. On the diagram in the OP, point C is located at 6 feet from point A, so you have to calculate the load on the beam between x = 0 ft. and x = 6 ft. You know that the distributed load has a magnitude of 3 kip/ft when x = 18 ft. and zero at point A. You can find out what w is at x = 6 feet by using a simple ratio. The shear force at point C = RA - w[@x = 6] * 6 ft./2 kip. You (or somebody) have already calculated RA = 9 kip, so the shear force VC at x = 6 ft. is going to be something less than 9 kip. In other words, your calculations have omitted the loading on the beam between points A and C.

12. Nov 3, 2014

Feodalherren

I didn't omit the loading... I said W=.5*6*3=9kip between A and C. I don't understand where the book gets V=3 from. That is what I need help with. The rest of this problem is easy. I'm confused on how the calculated the load of the 6ft section.

13. Nov 3, 2014

OldEngr63

14. Nov 3, 2014

SteamKing

Staff Emeritus
But the 3 kip/ft is the magnitude of the distributed load only at x = 18 ft. That's why you have to use a ratio to find out what the value of the distributed load is at x = 6 ft., and then calculate the total applied load between x = 0 and x = 6 feet.

15. Nov 5, 2014

pongo38

If I may chip in here, the problem the OP has relates to his diagram in #10 which gives a point load of 9 instead of a triangularly distributed load. If he calculates the proportion of w at x=6, then the value of load from 0 to 6 will give a total load in that region, but it is misleading to describe that as a concentrated load (and particularly not the value of 9) when considering shear force and bending moment. Also, to be pedantic, one cannot have a shear force at a section. But you can have a shear force just to the right of A, and just to the right (or left) at C. This is one of those rare cases where it is helpful to recognise that the loading function is linearly varying. Therefore the shear force equation is second order, and the bending moment a third order expression.

16. Nov 5, 2014

Dr.D

Pongo38, you have some amazing statements here, things that are news to me. Let me take the one by one:

"... one cannot have a shear force at a section." Why not? If we can have a shear force just to the left of C and just to the right of C, as you later say, why can we not have one at C?

:... the shear force equation is second order, and the bending moment a third order expression." Where does this statement come from, and what value is it in this problem?

17. Nov 6, 2014

pongo38

Thank yo Dr D. It comes from the definition of shear force. Where you have a point load on a beam, the shear force is not well-defined at the point load, because it is a discontinuous function, as the graph so clearly shows. What I would like to encourage the OP to do in this case is to consider a section distance x from A and obtain an expression for w(x) at that point. Check that expression by substituting x = 0 and x = 18. Then add up all the shear forces V(x) on one side of that section (say, to the left); then check that by summing all the shear forces to the right of that section; then obtain an expression for the Moment M(x) to the left of x. Check it by taking moments to the right. Then apply x = 6, or whatever is needed to solve the problem All that would in the end be more valuable as a learning experienece. When students go through this process, they frequently make mistakes. The sequence of loading f(x^1), shear force f(x^2) and moment f(x^3) acts as a check on making trivial errors.

18. Nov 6, 2014

SteamKing

Staff Emeritus
I'm not sure what is going on here, but it looks like the OP has skedaddled back into his burrow.

The original beam had only a single distributed triangular load applied over its full length. The 9 kip concentrated load located at 4 feet from point A (as shown in Post #10) was the result of an incorrect analysis of the problem by the OP, and there is no discontinuity in the shear force curve for this particular beam.

19. Nov 6, 2014

Dr.D

pongo38, what do you mean by the notations f(x^1), f(x^2), etc? I have never seen this notation before.

I agree that the shear in the beam is discontinuous at a point where there is an appied load, but this does not apply at point C at all. Still, your statement "Also, to be pedantic, one cannot have a shear force at a section." seems far too broad. The only places where shear is discontinuous in the problem at hand is at the two ends where the support reactions are applied; at every other section, the shear is continuous and well defined. Do you not agree?