SA15: The Conjugate Beam Method (Part 1)
ฝัง
- เผยแพร่เมื่อ 7 ก.พ. 2025
- This lecture is a part of our online course on introductory structural analysis. Sign up using the following URL: courses.struct...
In addition to updated, expanded, and better organized video lectures, the course contains quizzes and other learning content.
Part 2 of this lecture can be found here: • SA15-2: The Conjugate ...
Best Explanation So Farrr
Nice to upload new videos, thank you :)
I don't get it @ 4:35 where's the 5/2 & 5/3 coming from and did you calculate the area of the triangle to get 12.5 and lastly the 122/3EI. I'm new to this topic i hope you don't mind answering my questions
Yes, 12.5/EI is the reaction force at A. It equals half of the total triangular load placed on the beam. The total load due to the triangle is half of its area. So, the total triangular load is (5/EI)(10)/2 = 25/EI. Then each support reaction is half of this load, or 12.5/EI.
Note the free-body diagram drawn below the conjugate beam. There is a triangular load with a base 5 m and height 5/EI acting on the diagram. In order to determine the area under the load diagram, we need to compute the triangle's area, which equals (5/EI)(5/2). That is where 5/2 come from. The moment of this force about the cut point equals the load magnitude (5/EI)(5/2) times the distance from the center of the triangle to the cut point, which is (5/3).
Thank you, robot friends.
Thanks for clarification. It’s clear for me now
Amazing work!!
Very useful for us. Thank u
Appreciated SIR
jus discovered tis channel!!
realli helpful...!!
thnx a lot:D
amazing video!!
So am I correct to assume that you can not use this method for a simple beam with fixed ends? And please explain why
This method works for statically determinate beams only. A beam fixed at both ends is indeterminate.
Thanks. Very informative
I don't understand how you got 125/3EI,
the numbers you used in Mc don't add up to 125/3EI,
and why is the 12.5/EI placed on the end of the Free body diagram rather than 2/3 up this is at time- 4:38
The conjugate beam is simply supported having a vertical support reaction at the left end and one at the right end.
The load applied to the conjugate beam is triangular. The area of the triangle is (5/EI)(10)/2 = 25/EI. Since the load is symmetrical, each support reaction is going to be half of the total load magnitude, or 12.5/EI. Also, since the M/EI diagram is positive, the reactions at A and B are going to be downward.
So, the 12.5/EI is the support the reaction at A for the entire beam shown above the free-body diagram.
Now, considering the free-body diagram, if we take sum of the moments about point C, we get the equation shown in the video. It is:
Mc = (5/EI)(5/2)(5/3) - (12.5/EI)(5).
The first term in the equation is the moment of the equivalent concentrated load for the triangular load placed on the left half of the conjugate beam.
The area of the triangle is (5/EI)(5/2), the moment arm is the distance from the center of the triangle to point C; this distance is 5/3.
The above equation can be written as:
Mc = 125/6EI - 375/6EI
Or,
Mc = -250/6EI = -125/3EI
thank you so much, understand it better now
Deep thanks 🙏🏻
I want to use Conjugate Beam Method to find the moments at the ends of fixed beam subjected to central concentrated loading. Please help me out!
Conjugate Beam Method is for calculating displacements not moments. The moment you are looking for can be easily calculated by multiplying the load magnitude by the moment arm.
Why is the load on the conjugate beam pointed in upward direction...?
How can this method be used by pointing the load on conjugate beam downwards as usually done...?
I tried to do that but the deflection came out to be positive instead of negative as in your case...
Also, why is v(x) = integral of w(x)...? Shouldn't it be v(x) = integral of -w(x) ....? , since dv/dx = - w and not = w....
please explain.... Confused !!!
+VIVEK SINGH
The sign convention here could be a source of confusion. We call shear in a beam segment positive when the shear force tends to rotate the beam segment in the clockwise direction. So, when the shear force at the right end of the segment is acting downward and shear at the left end of the segment is upward, shear is positive.
The load function, w(x), is considered positive if it is pointing up, if it tends to pull the beam upward along the positive y direction.
With this sign convention in mind, to see why v(x) = integral w(x), consider a cantilever beam free at the left end and fixed at the other end. The beam is subjected to an upward (positive) distributed load.
If we cut the beam at some distance x from the left end, then the free-body diagram of the left segment entails the upward distributed load w(x) and a shear force at the right end of the segment. Let’s place a positive shear force at the right end of the segment. This would be a downward force labeled v(x).
Since the segment has to be in equilibrium, we can write: integral w(x) - v(x) = 0. Or, v(x) = integral w(x)
The load on the conjugate beam is the M/EI on the real beam. So, if M is positive, we treat it as a positive load on the conjugate beam. This means, according to our sign convention, an upward distributed load on the beam. But, that should not make any difference when applying the equilibrium equations to the conjugate beam. If the load is downward, the reactions are upward, if the load is upward, the reactions would be downward.
;Let me know if additional explanation is necessary.
+Dr. Structure
Thanks for the explanation.... Will try few problems with this.
well done
Nice video!
But where does the 12 come from for the reaction in A? I thought it would be 25/(6*EI)
Edit: And the arrow would face upwards..
The downward reaction at A, written at 4:30 mark, equals to half of the triangular load applied to the conjugate beam. The area of the triangle is 25/EI. This load distributed equally between the two support reactions. Therefore, the reaction at A is 12.5/EI.
Dr. Structure Oh I understand where the 12.5 comes from. We write a point whenever we multiply and a comma for decimals. But shouldn't the load be distributed by 1/3 left and 2/3 right instead of equally?
Now I see. The distributed load comes from the whole conjugated beam! Thx
Ben Verbergt Exactly.
Ben Verbergt Yes, this difference in notation could become confusing sometimes.
Thanks a lot
please do a Moment-Area method with advanced examples
Hello!, what software was used for presentations?
The main ones are Camtasia Studio, VideoScribe and CrazyTalk.
Thanks you!!
Thnks, it realy helpd. But plse upload a vid of momnt area methd. Please!
👍🏻
thank u
what software is being used to generate these videos
For this particular video, we traced the writing manually using a stylus on an Ipad, and saved it in SVG format. We then used VideoScribe to turn the SVG files into video segments. Finally, Camtasia Studio was used to put the video segment together, and to synchronize video and audio.
Go check my first video ever, explaining conjugate beam method
Great explanation. If you would like to contribute content to our channel and have a broader impact, feel free to send me a note.
Dr. Structure I would love that
Sir...can you explain a little bit the INTERNAL HINGE. I mean can you give me a PRACTICAL EXAMPLE OF AN INTERNAL HINGE in a structure ... say in a building?
Imagine a long beam supporting the roof of a building. But, the beam is actually made of two shorter beams connected together using what is called a beam splice, basically a plate that connects to both beams creating a single beam. If the connection is made using bolts only, that is a bolt or two connect the plate to each of the two beams, then the connection is considered to be an internal hinge since the bolts allow relative rotation between the two beam segments.
thanks...good explanation
and again where's part 2 ? :D
Part 2: th-cam.com/video/MR1DmMnLTvw/w-d-xo.html