# Having some issues deriving internal force functions for a section.

1. Dec 8, 2013

### Bill Nye Tho

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

I have a 13 foot long beam supported by a pin at x = 0 feet and a roller at x = 9 feet.

There is a triangular distributed load of 50 lb/ft from 0 ft to 9 ft. (Increasing as it approaches 9 ft)

At the end of the beam there is a moment of 200 lb-ft counter-clockwise.

2. Relevant equations

ƩFx = 0
ƩFy=0
ƩMr = 200 lb-ft

Vertical Force at Pin: 52.78 lbs upward
Vertical Force at Roller: 172.22 lbs upward
Normal Force = 0

3. The attempt at a solution

For the first section, it seems my equations are correct;

0 < x < 9:

Shear: -.5(50/9)x^2 + 52.78
Bending: -.5(50/9)x^3/3 + 52.78x

----

This is where I'm having an issue with my Bending equation;

9 < x < 13:

Shear: -225 + 52.78 + 172.22
Bending: 52.78x + 172.22(x-9) - 225(x-3)

-----

At exactly 9 feet, both bending functions should give a bending moment of -200 lb-ft but for some reason, I can't seem to get that answer with the second one. I tried to rework this function a few times but it's not happening.

2. Dec 8, 2013

### SteamKing

Staff Emeritus
The applied couple (200 ft-lb) at the end of the beam (x = 13 ft) means there is a jump in the bending moment value at this location. However, for x > 9, there is zero shear. In your equations above, it's not clear what the factor (x-3) means, since the beam is 13 feet long and the right support occurs at x = 9 ft.

3. Dec 8, 2013

### Bill Nye Tho

(x-3) is the distance of the resultant force (225 lbs) from the distributed load at any given point between 9 and 13 feet. <---- I think this is my problem. It makes no sense.

Since there is 0 shear after 9 feet, I'm just getting that the bending moment should be -200 lb-ft from 9 feet to 13 feet. This equation is not reflecting that though and I'm trying to figure out where I made my mistake. :(

Last edited: Dec 8, 2013
4. Dec 8, 2013

### Bill Nye Tho

I think I got it... I need some sleep

Bending Function: 52.78x + 172.22(x-9) - 225(x/3)

5. Dec 8, 2013

### SteamKing

Staff Emeritus
No ... I don't know what you mean by 225(x/3).

Also, what is 13 - 9 = ?

6. Dec 8, 2013

### Bill Nye Tho

4, I can't believe I botched that up.

7. Dec 9, 2013

### pongo38

When you first calculated the reaction forces, did you check them by (say) taking moments about some axis not already used? You should show us all your working.