# Slope and deflection diagrams using conjugate beam method

1. Dec 2, 2011

### musicmar

1. The problem statement, all variables and given/known data
Construct the slope and deflection diagrams. I've attached the problem with the original diagram (problem #1).

3. The attempt at a solution
Considering the number of diagrams required, I thought it would be best to attach a photo of my work.
I drew the shear and moment diagrams for the original beam, with equations for V and M as follows:
V = {150 (0<x<3), -75x (3<x<5)
M = {150x (0<x<3), -75/2*x2 + 375x (3<x<5)

I then drew the conjugate beam, which is essentially a mirror image of the original beam. I then loaded it with M/EI, where M is the moment from the original beam. Now, I know I need to draw the shear and moment diagrams for the conjugate beams, and these will be the slope and deflection diagrams. I'm just not sure how to go about this, considering the complicated loading.
Also, for another problem, I am told to find the max deflection. Is there a way to do this without constructing the shear and moment diagrams(for the conjugate beam)?

Thank you.

#### Attached Files:

File size:
25.4 KB
Views:
803
• ###### IMG00185-20111202-2343.jpg
File size:
26.5 KB
Views:
533
Last edited: Dec 2, 2011
2. Dec 4, 2011

### afreiden

Use an "integration" method, rather than a "cutting" method.

1) Get your reactions in you conjugate beam.

2) Integrate to get the slope (makes no difference if you start at the left end or the right end). You should find that you get zero at the left end of your conjugate beam, and something nonzero (equal and opposite to your reaction) at the right end of your conjugate beam. So, in the actual beam, the slope is zero at the fixed end and nonzero at the free end (as expected).

3) Integrate (2) to get the deflection.
You will find that you again get zero at the left end of your conjugate beam, and something nonzero at the right end of your conjugate beam. So, in the actual beam, the deflection is zero at the fixed end and nonzero at the free end (as expected).

Hope that helps,

3. Dec 4, 2011

### musicmar

I got RC'y=1425/EI and MC'=4387.5/EI .
I also found the equations for V and M of the conjugate beam. Do I need to consider end conditions to find integration constants? As it stands right now, my shear diagram goes down to -900/EI at x=3, and with the equation I have for 3<x<5, it would go towards the axis. However, the reaction I found at C' would need the diagram to end at -1425/EI at x=5.

Attached is more of my work.
Thank you.

File size:
29.5 KB
Views:
246
4. Dec 4, 2011

### afreiden

To answer your question regarding "integration constants" -- yes, I would expect to see some constant values in your equations. My advice would be to consult a statics textbook if you aren't familiar with the "integration" method for finding shear and moment diagrams.