# Shear and moment diagram help

1. May 5, 2012

### editheraven

1. The problem statement, all variables and given/known data
Given the simple beam in the picture, draw the moment and shear diagram.
distributed load : q on a lenght of 2a

where M - is the moment, T - is the shear

2. Relevant equations

3. The attempt at a solution

I draw the shear diagram, found all the ractions but i'm stuck at the moment diagram.
On section AB, is the qa load giving any moment value? I know that the moment diagram value is the area of the shear diagram, but for example, on AB, the area would be 2qa^2, but the moment value seems to be qa^2. I'm a little confused here.

What i did so far is in the attachament.

#### Attached Files:

• ###### DSC00085.jpg
File size:
13.4 KB
Views:
109
Last edited: May 5, 2012
2. May 5, 2012

### pongo38

There are two definitions of bending moment, each one of which can be derived from the other. At the moment, you seem to be using the definition that M is the integral of shear force. That definition has the disadvantage that you need to find the arbitrary constant of integration by looking at boundary conditions. The alternative definition is more helpful: "The bending moment at a section is the algebraic sum of the moments on one side (or the other) of that section". Your diagram was not clear enough to read, but at B the M is RA * distance AB. You can check this by taking moments to the right of B. And you can use your first definition to check that the moment obtained is indeed equal to the area of the shear force diagram (or, rather more easily, you can check that the gradient of the M diagram in AB is equal to the shear force in AB - that is what most engineers would do).

3. May 5, 2012

### editheraven

And the vertical reaction from A is not giving any moment in B, right? Its arm is nule. This is how i started also : qa * distance AB equals a moment of "qa (point load in B) * a (length of AB)). But the diagram seems to get kinga high (low, if i keep in mind the sign) but at the right end the moment should be zero. But i don't see any way to get there. Because the moment at the end is 12qa^4 (vertical reaction from the end and total lenght of the bend). And i already get to a value of moment of 42, so that value of 12 can't get it to zero.

4. May 5, 2012

### pongo38

To apply the definition correctly, put a blank sheet of paper on top of the diagram so that the edge goes vertically through B and covers up everything to the right of B. What you are left with is AB. The bending moment at B is then RA * distance AB. Move the paper along till its edge goes through C. What then is the moment at C? MC = RA*length AC - qa*length BC.
You should do this and then check that you get precisely the same result by moving the paper across to cover up AB, and take moments to the right of B. Then again, a similar process for MC. If you are applying the sign conventions consistently, you should get the same values, whichever side of the section you take. To answer your specific point, the qa at B has no moment effect at B itself. Your first statement/question is incorrect. The reaction at A does impose a moment at B because it is some distance from it. (But maybe you didn't say exactly what you meant?). The thing you started to work out "This is how i started also : qa * distance AB equals a moment of "qa (point load in B) * a (length of AB))" was the beginning of working out the bending moment at A but missed out all the other loads. This is probably how you worked out the reactions. Anyway, do try the cover up technique. With a bit of practice you will learn to manage without the cover up paper.

5. May 5, 2012

### editheraven

Oh, now i got it. Thanks for the sheet tip. If i'll have any new questions after i redo my work, i'll ask :)