Glad it helped

This is such a broad topic that I could not fit it all into one video. This one was already on the longer side. So I started with a simple static problem. There will be additional videos for sure which will deal with Rayleigh's Quotient and the dynamic case.

Glad it was helpful!

Will add it to my list.

Thanks for this and so noted. I always appreciate getting feedback from my viewers... There was a lot of material that was crammed into this one video (really requires additional videos to better describe the topic). As a result of trying to fit all the material into the slides, it was useful to derive W,xx just before I used it on the next slide.

Sorry, but it can't be edited at this stage. Will pay closer attention to this is future videos. Thanks for the feedback.

Much of it comes from my class notes that I took back in the day, however, my "go-to" textbook for the more introductory material is "Mechanical Vibrations" by S.S. Rao and for the more advanced material tends to be Dym & Shames "Solid Mechanics: A Variational Approach". In addition, I look at various online articles and papers and use these for insights. For some topics, Wikipedia is also a useful reference.

Yes, grew up in Durban...a long time ago.

I believe you are confusing two different concepts. Work is defined as force x displacement. 1. In the case of a tip load, the force P is CONSTANT throughout the displacement. When the tip displacement is 0, the force is still P and when the displacement reaches w(L), the force is still P. So no 1/2 needed here. The work is P x w(L) 2. On the other hand if we consider the work done by internal, linear elastic forces we are, in effect, saying that the elastic load is 0 initially when the displacement is 0 AND LINEARLY APPROACHES P when the displacement reaches d, so we can think of the average load as 1/2 P. The work is then 1/2 P x d. This is when get get the factor of 1/2 . Typically, we will see the 1/2 when dealing with energy (strain energy or kinetic energy), but not when dealing with external work.

I didn't just pluck it out of thin air. I needed to satisfy 4 boundary conditions and I knew that for this I could use a polynomial with 4 constants (ie a 3rd degree polynomial) and apply the 4 BC's to solve for the four constants. Because I wanted to use the Ritz Method, I added a degree to the polynomial (4th degree) so that I have a 5th constant to solve for. Solving for the 5th constant required application of the Ritz method. That said, there are literally entire books that have been written on shape functions. You can generally pick these from a book. And, yes, some shape functions work better than others. The closer the shape function is to approximating the actual mode shape, the more accurate the approximate results that are produced. Two things to take away from this: 1) you can get shape functions from a book 2) you can always use a polynomial of higher-enough degree to satisfy the required boundary conditions and then just add a degree (or more) to that.

It's a variational method because in invokes the Principle of Minimum Potential Energy which tells us that the potential energy for a system in equilibrium in an extremum. This implies that the variation of the potential energy (the functional) must be zero which, in turn, leads to equation 5. In applying this condition, we are able to find the values for the Ritz Coefficient(s) which minimize the potential energy.

Yes, you certainly can. I will make a future video showing that.

Correct. If you substitute the actual mode shape for the shape function, then you'll get the exact result. In general a "better" choice of shape function will cause a more rapid convergence of the approximate result to the exact result.

Not sure which equation you are referring to, but eqn 24 shows the tip deflection to be negative.

The positive direction of displacement is upwards.

@@SourabhBhat I was wondering the same thing as Umangkumar. But this makes sense he's talking about the coordinate system

At 15:17 video says, that when displacement and force are in opposite direction then the work is negative. That is correct. But it still seems to me that they are both in negative direction, hence work should be positive. However, at 15:10 you say that displacement is positive upward, but eq.(15) kinda gives that displacement is downwards, not upwards. Sorry for my confusion.