EX03: Solved Problem: Conjugate Beam Method (beam with internal hinge)

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.

[@sushantbhusal935]

Best explanation in such a vivid way. Thanks dr.structure

[@OGTennyson]

Thanks so much for this... tried using moment area method to solve for deflection at B and still got the same answer 👍🏾

[@user-bt5hg2bh7u]

thank you

[@rahulmalpekar1]

Great explanation. Thank you.

[@biniambelayneh4774]

no word ! thank you very much

[@dthanna2720]

Thanks for sharing

[@HemantKumar-qu8yg]

thanks..

[@seblewongelminlargh5060]

mark saver in the final exam
top leve leve explanation

[@dthanna2720]

Thanks for sharing.👏🏽👏🏽

[@shantanuaggarwal7777]

Why there is no need to find moment at B due to whole system of reactions and loadings?

[@DrStructure]

Keep in mind that when a structural system is in equilibrium, the equilibrium equations are satisfied not only for the system as a whole but also for any segment/part of it. So, if we can find what we are looking for (the unknowns) using only a part of the beam, then that is what we shall do. In this problem, there is no reason to consider the entire system when examining only a part of it produces the result we are after.

[@glennn.micarandayo8657]

I wonder What application you used in making this Audio-visual presentation?

[@DrStructure]

Camtasia Studio, Adobe Illustrator, Reallusion software apps.

[@inzimamurrehman7997]

Thank you , need more vedios of deflection methods

[@kissflute662]

Hello ..i just wonder why you multiply 6/EI with (3/2)[(3×2)/3]. That one i dont get it.

[@DrStructure]

6/EI is the height of the right load triangle; The base of the triangle is 3 m. So, (6/EI)(3/2) is the area of the triangle, which we treat as the equivalent concentrated load for the distributed load.
The location of the concentrated load (the center of the load triangle) is 1/3 of the base from point A; or 2/3 of the base from point B. Since the base is 3 m, the location of the equivalent concentrated load from B becomes: (2/3)(3).
The moment of the equivalent concentrated load about point B, then is: (The Load Magnitude) ( Moment Arm). Or,
(6/EI)(3/2) (2/3)(3).

[@raisaferdousi5638]

We got By as -2kn . So when we are calculating left segment reactions, why is it Ay+2=0? Shouldn’t that be Ay-2=0?

[@DrStructure]

Based on the free-body diagram we have drawn for the left segment of the beam, the equilibrium equation for AB is: Ay - By = 0. So if By = -2, then the equation becomes: Ay - (-2) = Ay + 2 = 0.

[@kittebanty2171]

plz calculate the slope at B

[@structuralanalysis2790]

64/15,however ......

[@vitrovogan9635]

Why there is no need to calculate reaction when we are calculating moment at B i.e why are we only considering one side of support B to calculate moment at B?

[@DrStructure]

A similar question was asked today. The response is pasted below. Feel free to elabortate further if your question is not answered.
“Keep in mind that when a structural system is in equilibrium, the equilibrium equations are satisfied not only for the system as a whole but also for any segment/part of it. So, if we can find what we are looking for (the unknowns) using only a part of the beam, then that is what we shall do. In this problem, there is no reason to consider the entire system when examining only a part of it produces the result we are after.”

[@vitrovogan9635]

@@DrStructure thank you😎

[@vitrovogan9635]

@@DrStructure may I know your profession❓

[@DrStructure]

Structural engineering educator :)

[@vitrovogan9635]

@@DrStructure Do you teach some where?

[@rahuldey9220]

u took clockwise negative then again anticlock wise positive why?

[@DrStructure]

Clockwise negative and anticlockwise positive are not contradictory, if one is negative the other has to be positive. Regardless, when writing the equilibrium equations, the positive/negative direction is quite arbitrary. We can pick either direction to be positive which make the opposite direction negative.

[@akocgateway]

IS THIS WRONG???

[@DrStructure]

What is wrong?

[@structuralanalysis2790]

Dr structure you are WRONG on this buddy.what's your email?

[@DrStructure]

Please elaborate. It would be best to have a constructive dialogue here on the comment section. Such an exchange could serve as a learning tool for students that come across it later on.

[@structuralanalysis2790]

Come to think of it, you are right;
Any contact email?
Thx.

[@DrStructure]

Technical questions that involve diagrams and such, which cannot be easily posted on the comment section can be emailed to [e-mail]

[@structuralanalysis2790]

Much better, thx

[@sayandey1048]

Feels sleepy after hearing the voice...please change it