EX02: Solved Problem: Conjugate Beam Method (beam with overhang)

This video shows the steps for calculating slope in a beam using the Conjugate Beam Method.
The problem and its solution were contributed by Vian Abu-Bakir. You can reach her at: [e-mail]
Or at: / @engineervian3013
Numerous contributions of Galina Jorgic in designing and creating the video lectures and related content are gratefully acknowledged.

[@johnpaine7191]

Thank you Dr. Structure, very informative especially those beginners like me... More power to you!

[@travelvlogger8509]

Its amazing to see... Such a good explanation... Its understood in first attempt only... Without any doubt

[@jenniecave2533]

Clear Concept 👍👍 👍

[@alanbecker4814]

Superb work. Keep it up.. make more examples ☺☺😊😊

[@Dipesh155]

i like this video

[@stupidfilip23]

Can you make a variable rigidity question?

[@notetaker1141]

does this method work for indeterminate beams?

[@DrStructure]

The method works for any beam. In the case of an indeterminate beam, to draw its moment diagram, we need to use a technique such as slope-deflection to determine the beam's reactions. Once the reactions are determined, then we can draw the moment diagram, turn it into the M/EI diagram, and proceed as usual.

[@Laris2725]

how did you get 60 kN.m in the moment diagram? I kept multiplying 25x2... I feel like I'm missing something hehehe great video by the way

[@DrStructure]

Not quite sure what you are referring to, please elaborate and point to the specific timeframe where your question is based.

[@Laris2725]

At 2:10

[@DrStructure]

@@Laris2725 Okay, now I see what you are referring to. We can get to that peak moment value of -60 kN.m either from the left or from the right side of B.
Going from left to right: We have a negative moment of 25 kN.m just to the right of the concentrated moment at the midpoint of Segment AB. The area under the shear diagram from that point to point B is 2(-17.5) = -35. So, the moment at the right end of that segment (at B) equals to -25 + (-35) = -60.
Going from right to left: The moment at the right end of the beam (at point C) is zero. The area under the shear diagram from B to C is (3 m)(20 kN) = 60 kN.m, and since the moment at C must be equal to the moment at B plus the area under the shear diagram. That is,
Mc = Mb + 60, then, Mb = Mc - 60. Or, Mb = 0 - 60 = -60 kN.m