Shear and Bending Moment Diagrams

1. Jan 25, 2008

Shear and Bending Moment Diagrams!!

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

2. Relevant equations
Newton's 2nd and 3rd

3. The attempt at a solution

So my main confusion is this:

I found the reaction at A to be 19 kN

If I make a vertical cut through some arbitrary point along the beam and and do some analysis I get:

From the left portion:
$$\downarrow +\sum F_y=0\Rightarrow V-19+3x=0\Rightarrow V=19-3x$$

Now if I take the right portion I get:
$$\uparrow +\sumFy=0\Rightarrow V-10-[3(6-x)]=0\Rightarrow V=28-3x$$

Now which do I use? Or How do I go from what I have to a shear diagram?

Last edited: Jan 25, 2008
2. Jan 25, 2008

Or am I making this too difficult? Should I just look at the left portion?

3. Jan 25, 2008

Who am I kidding. . . no one looks in this forum.

4. Jan 25, 2008

Staff: Mentor

5. Jan 25, 2008

Thanks Astronuc. I screwed up my reaction force at A. It should be 28. Thus, the equation is the same.

6. Jan 25, 2008

WHat the hell am I doing now?! Why can't I get this. My shearing function should be the derivative of my moment function, which is not the case. Here's my work.

7. Jan 25, 2008

Pyrrhus

When you cut (section) a distributed force i see you use the incorrect lever arm, you know the force if it was concentraded it'll be acting at the centroid of the figure formed by the distribution force, therefore the lever arm for 3x should be x/2.

8. Jan 26, 2008

How is there no shear at B. If V=28-3x than at B V(6)=28-3(6)=10kn

9. Jan 26, 2008

Pyrrhus

Yes, there is shear at B because of the point load at the end. Astronuc probably didn't notice the point load, because he is correct for the case without it.

10. Jan 26, 2008

Thanks!!

11. Jul 26, 2011

the_red_jeste

Re: Shear and Bending Moment Diagrams!!

given that the problem is one of a cantilever there will be no shear force at B since it is a free end and there is nothing for the beam end to bear upon or shear across.