Thanks for your sharing. It's useful for civil engineering. Could you also introduce "Qualitative Structural Analysis of Beams and Frames"? It seems hard to understand.
Thank you, I'm building an Excel spreadsheet for these calcs . A client is asking for load charts for lift beams, where the position of single and dual point loads vary. I've been looking for a more efficient method than dozens of FEA models. Much appreciated.
@@StructuresProfH super helpful - I'd watched my Uni's lectures a bunch of times over and it just didn't click. Your explanation was super clear and it made sense straight away - thanks heaps for putting this video out!
Does the negative/positive component of the M diagram become relevant if some parts are negative and some are positive? In the second example, even though it was negative you didn't include it?
Yes, the moment diagram sign is relevant if some is negative and some is positive. The sign gives you the direction of curvature (positive opens up or "smiles" while negative opens down or "frowns"). It is important to keep this consistent in general when taking the moment of the area. That said, if the sign is all positive or all negative, I usually handle the signs intuitively (some might say lazily!) for the moment area method just by considering the deflected shape geometry. The negative sign in the second example is reflected by the fact that the deflected shape is below the tangent, whereas the positive sign in the first example is reflected by the fact that the deflected shape is above the tangent.
Are you struggling specifically with locating the point of maximum deflection? Maximum deflection will occur at the point of zero slope. I'll illustrate the idea with the second problem presented in this video, but I'm afraid it will be tough for me to extrapolate here in the comments section to whatever specific problem you are trying to tackle. At any rate, I hope this helps... Let's say I want to find the maximum upward displacement in Span AB in the second example problem in the video (starting at 10:00). I can find the slope at point A using the deviation t_B|A. Slope at A is equal to rise/run, where the rise (vertical distance) is t_B|A and the run (horizontal distance) is 8 meters. The area under the curvature diagram (which is just the moment diagram divided by EI) is the change in slope. So my goal is to find distance X from point A where the slope becomes zero. Here are the numbers: - t_B/A = 2048/EI (from video 13:20) - Slope Theta_A = (2048/EI)/(8 meters) = 256/EI - Area under moment diagram from A to X = 1/2*X*(24X) = 12 * X^2 ... [note that at X = 8m, this is 768, which is the area under the triangle at 10:45] - By extension, the area under the curvature plot is 12 * X^2/EI, because curvature = M/EI - Therefore, the value of X with zero slope is at 12 * X^2/EI = 256/EI, which gives X = 4.6188 meters. Now that I know the location of where I need to compute the deflection, the actual deflection amount can be computed using the similar triangle methods illustrated in both examples. Here I would want to find the deviation t_X|A, where X is that point of maximum displacement. From similar triangles, I can see that... (Delta + t_X|A)/4.6188 meters = (t_B|A)/8 meters where Delta is the displacement I want to compute. Deviation t_B|A = 2048/EI, and calculating t_X|A gives... t_X|A = (256/EI)*(4.6188/3) = 394.14/EI Now solving for Delta gives... Delta = 4.6188/8 * t_B|A - t_X|A = 4.6188/8*2048/EI - 394.14/EI = 788.28/EI Plugging in the values of E = 200 GPa and I = 2.5e8 mm^4 gives a displacement Delta = 15.77 mm Again, I hope this helps, but I apologize if this was not exactly what you were looking for. Sometimes finding the areas and moment arms of the shapes under the moment diagram can be tricky. For problems with really complicated moment diagrams, I'd probably choose to use some method other than moment-area.
Yep. Curvature is given by moment M divided by EI. Therefore, the deviation from the tangent tb|a equals the first moment of the area under the moment diagram divided by EI. Rather than divide by EI, I just moved it to the other side of the equation as a multiplication.
Already have one, but it’s for frames instead of beams. Same idea though. Deflection of Frames using Principle of Virtual Work - Intro to Structural Analysis ruclips.net/video/JAD-8ATyu6k/видео.html
you have done for me in 16 minutes what my teacher couldn't achieve for WEEKS. thank you so much, professor.
I just couldn't go on without thanking you, you literally made it a piece of cake. thanks alot
Clear explanations with excellent graphics. Your progressive solution path is the best. Thank you.
You're a very good lecturer.
Excellent understanding after watching this lecture
very good examples and smart way of solving
Thanks for your sharing. It's useful for civil engineering. Could you also introduce "Qualitative Structural Analysis of Beams and Frames"? It seems hard to understand.
Thank you, I'm building an Excel spreadsheet for these calcs . A client is asking for load charts for lift beams, where the position of single and dual point loads vary. I've been looking for a more efficient method than dozens of FEA models. Much appreciated.
thanks a lot man u saved me with this video!!
thanks for this video it is very helpful for civil engineering
You are welcome!
Very helpful for me who studies mechanical engineering as well!
Beautiful and clear
Thank you!
Thank you sir you made it so easy💎
This is great, thanks!
Glad it was helpful!
@@StructuresProfH super helpful - I'd watched my Uni's lectures a bunch of times over and it just didn't click. Your explanation was super clear and it made sense straight away - thanks heaps for putting this video out!
THANKS!
Solid video
thank you sir
How about a cantilever beam with a span of 8m half of it has a uniform distribited load of 12kn/m at wnd span😊
How can I find the maximun deflection?
Does the negative/positive component of the M diagram become relevant if some parts are negative and some are positive? In the second example, even though it was negative you didn't include it?
Yes, the moment diagram sign is relevant if some is negative and some is positive. The sign gives you the direction of curvature (positive opens up or "smiles" while negative opens down or "frowns"). It is important to keep this consistent in general when taking the moment of the area.
That said, if the sign is all positive or all negative, I usually handle the signs intuitively (some might say lazily!) for the moment area method just by considering the deflected shape geometry. The negative sign in the second example is reflected by the fact that the deflected shape is below the tangent, whereas the positive sign in the first example is reflected by the fact that the deflected shape is above the tangent.
@@StructuresProfH Awesome, did some examples and its making much more sense.
Perfect!
Thanks for watching!
is the BC's BMD correct
How do you solve this kinda of problem on a simple beam for max deflection with UDLs and point loads Im struggling immensely with area moment.
Are you struggling specifically with locating the point of maximum deflection? Maximum deflection will occur at the point of zero slope. I'll illustrate the idea with the second problem presented in this video, but I'm afraid it will be tough for me to extrapolate here in the comments section to whatever specific problem you are trying to tackle. At any rate, I hope this helps...
Let's say I want to find the maximum upward displacement in Span AB in the second example problem in the video (starting at 10:00). I can find the slope at point A using the deviation t_B|A. Slope at A is equal to rise/run, where the rise (vertical distance) is t_B|A and the run (horizontal distance) is 8 meters. The area under the curvature diagram (which is just the moment diagram divided by EI) is the change in slope. So my goal is to find distance X from point A where the slope becomes zero. Here are the numbers:
- t_B/A = 2048/EI (from video 13:20)
- Slope Theta_A = (2048/EI)/(8 meters) = 256/EI
- Area under moment diagram from A to X = 1/2*X*(24X) = 12 * X^2 ... [note that at X = 8m, this is 768, which is the area under the triangle at 10:45]
- By extension, the area under the curvature plot is 12 * X^2/EI, because curvature = M/EI
- Therefore, the value of X with zero slope is at 12 * X^2/EI = 256/EI, which gives X = 4.6188 meters.
Now that I know the location of where I need to compute the deflection, the actual deflection amount can be computed using the similar triangle methods illustrated in both examples. Here I would want to find the deviation t_X|A, where X is that point of maximum displacement. From similar triangles, I can see that...
(Delta + t_X|A)/4.6188 meters = (t_B|A)/8 meters
where Delta is the displacement I want to compute. Deviation t_B|A = 2048/EI, and calculating t_X|A gives...
t_X|A = (256/EI)*(4.6188/3) = 394.14/EI
Now solving for Delta gives...
Delta = 4.6188/8 * t_B|A - t_X|A = 4.6188/8*2048/EI - 394.14/EI = 788.28/EI
Plugging in the values of E = 200 GPa and I = 2.5e8 mm^4 gives a displacement Delta = 15.77 mm
Again, I hope this helps, but I apologize if this was not exactly what you were looking for. Sometimes finding the areas and moment arms of the shapes under the moment diagram can be tricky. For problems with really complicated moment diagrams, I'd probably choose to use some method other than moment-area.
at 6:47 can you explain why you multiplied the EI by the tb/a
Yep. Curvature is given by moment M divided by EI. Therefore, the deviation from the tangent tb|a equals the first moment of the area under the moment diagram divided by EI. Rather than divide by EI, I just moved it to the other side of the equation as a multiplication.
any chance of making a video about deflection by virtual work?
Already have one, but it’s for frames instead of beams. Same idea though.
Deflection of Frames using Principle of Virtual Work - Intro to Structural Analysis
ruclips.net/video/JAD-8ATyu6k/видео.html
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