"Are you tired of being called a scrub because you can only solve a 2nd order PDE, well worry not this video will teach you how to solve a 4th order PDE like a chad" Lmfao
If the wave propagation is in the x-direction and wave amplitude is vertical (u), Shouldn't the moment of inertia be w.r.t the axis orthogonal to the plane of the drawing?
I'm not terribly familiar with beam mechanics, so I can't speak much to how alpha was derived, but based on this link: en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory#Static_beam_equation The moment of inertia is 'calculated with respect to the axis which passes through the centroid of the cross-section and which is perpendicular to the applied loading.' In this case, the cross section is in the uz-place (i.e. the circular cross-section of the cylindrical beam with its length along the x-axis), and the applied loading is in the u-direction. The axis which satisfies both of these stipulations is the x-axis (passes through the centroid of the cross-section in the uz-plane and is perpendicular to the applied loading in the u-direction). Hope that helps!
I've typically heard of Neumann in the context of a boundary condition on the first derivative. In this problem, we have boundary conditions on the function itself (i.e. Dirichlet) and on the second derivative (not sure that would classify as Neumann though).
Thanks to Skillshare for sponsoring this video! The first 1000 people to use the link will get a free trial of Skillshare Premium Membership: skl.sh/facultyofkhan02211
I've finally uploaded part 2 of this video! ruclips.net/video/bEOih8cRWag/видео.html
Thanks for the video. Would also appreciate a part 2... I am not seeing one on your page...
Thank you for this wonderful video but I don't know if it is me, part 2 of the video is missing. Thank you!!!
"Are you tired of being called a scrub because you can only solve a 2nd order PDE, well worry not this video will teach you how to solve a 4th order PDE like a chad"
Lmfao
This is a great video Sir. May I know where I can watch the part 2 of it??? Thank you
Here you go (2 years late to the party lol): ruclips.net/video/bEOih8cRWag/видео.html
@@FacultyofKhan Ahahhahahha
Thanks a lot such a good video, but where's the next part of this?
Is there a part 2?
I did it :)
ruclips.net/video/BrXcb2-_rZc/видео.html
Please continue the real analysis series
Thank you so much for these videos! Want the part 2 :)
I did the second part for you :) Please watch and give me your feedback if you wish
ruclips.net/video/BrXcb2-_rZc/видео.html
Amazing video!
Thanks for this! Can I ask what software you use for the anumation?
it's just some kind of manual drawing program combined with screen recording, but i would also like to know what program that is specifically
which drawing tablet you are using???
Thanks!!!
Cool. Waiting for more PDEs connected with physics. Best regards
great video very helpful
next level
and the best is the derivation of the equation by using the Euler Lagrange equation on a beam
Raos continuous vibrations book does this for beams, plates, shells, etc. Real fancy stuff !
wheres the part 2
If the wave propagation is in the x-direction and wave amplitude is vertical (u), Shouldn't the moment of inertia be w.r.t the axis orthogonal to the plane of the drawing?
I'm not terribly familiar with beam mechanics, so I can't speak much to how alpha was derived, but based on this link: en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory#Static_beam_equation
The moment of inertia is 'calculated with respect to the axis which passes through the centroid of the cross-section and which is perpendicular to the applied loading.' In this case, the cross section is in the uz-place (i.e. the circular cross-section of the cylindrical beam with its length along the x-axis), and the applied loading is in the u-direction. The axis which satisfies both of these stipulations is the x-axis (passes through the centroid of the cross-section in the uz-plane and is perpendicular to the applied loading in the u-direction). Hope that helps!
I missed you!!!
where is the second part!?!?!?!?
Thank you.
where is part 2
GOD BLESS YOU
We are using neumann boundary conditions correct
I've typically heard of Neumann in the context of a boundary condition on the first derivative. In this problem, we have boundary conditions on the function itself (i.e. Dirichlet) and on the second derivative (not sure that would classify as Neumann though).
Thanks to Skillshare for sponsoring this video! The first 1000 people to use the link will get a free trial of Skillshare Premium Membership: skl.sh/facultyofkhan02211
Very bad because it's quick, the writing is fast and I couldn't follow up with what you say
FAIL.