# External moments when assembling internal force functions

1. Dec 11, 2013

### Bill Nye Tho

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

I have a 6 meter long beam that is supported by a roller at 0m and a pin at 6m.

There is a triangular distributed load that begins at 0m to 3m which is 2kN/m. (Max load is at 0m)

There is an additional triangular distributed load that begin at 3m which mirrors the first triangular load (2kN/m, max load is at 6m)

And there is a counter-clockwise moment acting at 0m of 18 kNm

2. Relevant equations

ƩFx = 0
ƩFy = 0
ƩMr = 0

Vertical Reaction @ A (0m) = 6kN
Vertical Reaction @ C (6m) = 0kN

3. The attempt at a solution

So my shear functions are fine but my bending functions always run into a similar problem.

From 0m to 3m my bending function is f(x) = -(2/3)^2(x)^3(.5) + 6(x)
From 3m to 6m my bending function is f(x) = -(.5)(2)(3)(x-1) + 6(x) - ((.5)(2/3)(x-3)^3)/3

My method for drawing internal force diagrams involves checking shared values of each function. So in this case, I like to use 3m in each function to see if I get the same thing, since I know that the bending function is continuous.

When I plug 3 into my first equation I get: -12 kNm
When I plug 3 into my second equation I get: -12 kNm

Everything checks out!

Now, if I calculate my bending forces at 0 m, I get 0 kNm; If I calculate my bending forces at 6 m I get -18kNm.

So basically I now have a bending force diagram that begins at 0, hits -12 kNm at 3m, finally hits -18 kNm at 6m.

I'm looking at the solution and for some reason it's reversed. So I decided to see what I did wrong by including the external force into my functions and while I get the correct answer at 0m and 6m, my 3m answer is completely off.

Is my solution correct? Should it be reversed? Why/Why not? (I solve left to right)

2. Dec 12, 2013

### SteamKing

Staff Emeritus
You've got to include the applied couple of 18 kN-m @ x = 0 m in your bending moment function, just like you include the reaction forces at x = 0 m and at x = 6 m in your shear force functions.

3. Dec 12, 2013

### SteamKing

Staff Emeritus
With these problems, make a sketch of the shear force and bending moment curves. It's much easier to construct your shear force and bending moment functions if you have a sketch.

4. Dec 12, 2013

### Bill Nye Tho

Thanks SteamKing. When I include the couple at x = 0 m, it checks out but at x = 6 m i still get -18.

5. Dec 12, 2013

### SteamKing

Staff Emeritus
Why don't you post your work? That way, everything you've done should be clear.

6. Dec 12, 2013

### Bill Nye Tho

What part should I post? My functions are the same on paper as my original post.

7. Dec 13, 2013

### SteamKing

Staff Emeritus
That's your problem. They shouldn't be.

When x = 0, M = 18 kN-m

8. Dec 13, 2013

### Bill Nye Tho

This is where my confusion lies.

I include this couple in calculation only at x=0? That I can accept but then at 6m I have to also include it to get a bending moment of 0.

9. Dec 13, 2013

### SteamKing

Staff Emeritus
Look, it will be easier to construct the shear force and bending moment functions if you construct the shear force and bending moment diagrams to check yourself. The applied couple at x = 0 affects ALL of the subsequent bending moment values, just like the reaction force RA affects ALL subsequent shear force values.