Beam Deflections - Double Integration Method Example (Part 3/3) - Structural Analysis
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- เผยแพร่เมื่อ 7 ก.พ. 2025
- Finishing up the example problem for the beam deflection calculation using the double integration. Solving for the remaining constants.
I love your videos!!! Those silly jokes light up the mood and i learn more in 20 mins watching your videos than 1hour of a university lecture.
Thank you. You sound really enthusiastic on your explanations, it makes me feel more positive about my homework :-)
thanks for watching.
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such a great lecture , started a new course which needed us to revise this method. One of the best videos to refresh your memory thank you :)
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This was incredibly helpful. I'm about to write an exam on this stuff and it didn't make sense till now. GOD BLESS!!!!!!
I'm glad it was useful.
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You are really good... Got some stuff in my head now :) Thanks
Great example! as always.... Thank you very much
+Ryan Griffin you are very welcome. thank you for the kind words.
2020 Thanks... U made this sound easy :)
I can't understand a word my professor says so thanks for the videos.
Excellent explanation
James Gesell Thank you!
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Very helpful THANKS!
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so what exactly is the answer?
If they ask for deflection and angle at right end of beam, you would plug 0 in for x2 in equations 3 and 4 and solve for theta/phi. Because the end of the beam from right end is a value of x2 = 0m based on his specified coordinate system. But if you wanted to solve for 2m from start of beam at the left, you would plug in x1 = 2m into equations 1 and 2 and then solve for unknown variable.
This video is kind of confusing because he doesn't ask for anything specific in the question, he just solves the 4 equations and leaves it open. You could solve for deflection/angle at any point on the beam as long as you plug in the correct value of x1 or x2 that corresponds with the locations you are trying to solve for. So it depends on what location(s) your prof asks for in your question.
Great !! It Helped a lot
It helped a lot . Thank you same so so much .
These videos are dope
Thanks for the vids, nice work!! BC's finally make sense :)
great you explained so well in slope1 = -slope2, but that is a very headache stuff for me to find x value with power greater than 3, the 4, 5, 6+++++++
you're the man
more tutorials please, frames.
could you please do a part 4 to calculate max deflection coz i gets tricky you wouldn't know which of the equations you use
Thank you so much, incredibly helpful
Many Many Thanks!!!!!!!!
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if we wanted to find the elastic curve for this beam, do we just add all the equations together.?
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great vids!
Thank you
at the end (5:00) you say that units work out because all the units of the deflection equation are kn.M^3 but what happens to the kN/m and kn.m^2?
+shahrukh Nadeem For the kN/m , when you divide by EI you will get kN/m^3 , but you will multiply by x^4(take another look at the equation. x is in m) and its a similar situation with kN.m^2 where dividing by EI cancels the kN.m^2 but there is an x that multiplies with that unit.
Sorry for explaining it poorly.
So if the question asks to calculate the maximum deflection, we'd just solve equation 4 @x=10? or would we have to add that with equation 2@ x=5 ?
Great!
great !
Those boundary conditions are confusing me a bit, so basically for every change in the moment diagram we get 2 more constants? Hope u reply soon got my exam tomorrow haha really helpful video thanks!
pretty much. we need two boundary conditions per moment function.
Oh and for each major change in moment it gives us a moment function? Not sure what im saying missing lectures is horrible...
Rabih Hawwa I'd recommend reviewing how to write moment functions from statics, then jumping into double integration method.
Took statics 2 years ago im in my 3rd year now so i dont really remember much i quickly jumped to structure made an issue for me :/ but thanks for ur help and ur videos are awesome keep posting them haha!
nice, i like when you say " baaaaam"
anyway, EI is in units of kN mm^2 now, i think then need to be changed or just change the meter to be consistent. am i right ?
So the E is in GPa and the I is in mm^4, do we keep them in those units?
what if the beam isnt a cantilever beam and is instead a simply supported beam which a force of say 20KN on both supports?
thanks bro. yo da man!
i know this is an old video, but could you not have done (from the x2 reference) V(10) = 0 and theta(10)=0? i mean using the coordinate system it makes sense...
I was thinking the same way but I am not sure ?
Thanks man :D
awesome!!!
How did you derive the boundary condition -theta1(5m)=theta2(5m) ? Looking at the beam it doesn't make sense why one should be negative of the other since it's technically the same angle...
+tullao1979 Check out part two of this series. He clearly states his coordinate system starts on two different ends of the beam resulting in opposite slopes hence the negative sign.
hi mr. check out your si units is not correct i.e you write moment of x2 is KN/m instead of KNm and the si unit value of C4 is KNm power 4 which can not be added with other constant value
hi how do you calculate the maximum deflection?
bannaghbeg finding location of maximum deflection is like finding the critical point...in most cases, where the slope is zero is a location of max. deflection. The ends of cantilever beams are one exception.
3:26 I think you made a mistake when finding c4... You put a negative sign on the right side of the equation.
i think you're 2 years too late
So the answer for both slope and deflection have two equations? Or should I plus them together?
+Wang Zirui I am confused because there's the same problem in the book but without numbers. If I use the number in this example, the slope and deflection at free end in the book equals to the answer from equations in M2(x2) in this example. Why the answer is that but not add two slope equations together and use x1=10m, x2=0m?
Please don't put them together. The functions only work between their respective discontinuities.
wow.....awesome :D thnx a lot :)
For the cut where the dist load is i chose the LHS so that included vertical reaction and moment at A. Then u get a different moment function M2(x2). Does this matter/affect the final equations ?
So is the final answer only theta 2 and V2 and not theta 1 and V1? Thanks.
for others that may wonder about this.....Because those were found in step 4. They both equal 0 because they are at a fixed end.
Are you a professor, structurefree?
I for theta 2, I got 5.2e^-9m. for v2, I got -1.519e^-8 KN. Anyone else get that? My prof never has EI's that large so the numbers seem weird to me.
clutch af
_/\_ just awesome
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