Approximate Solutions - The Galerkin Method 

I realize that I wasn't clear enough about this, but I made the assumption that f(x) was constant (at the time when I took it out of the integral) - so this is the result for a constantly distributed load. I should have made a point of mentioning that! Thanks for catching it.

@@Freeball99 oh ok Thanks!

@@Freeball99 I had the same question; now I can quit scritching my head.

Good explanation, but I was having the same question that initially he took f as a function of x and finally converted it into constant without explanation. But he answered you so.....

I have a feeling distributed load in real life also depends on displacement: f = f(x, w)

I hope so too!

From my experience, almost nobody seems to explain this.

For more on weak forms and strong forms, watch this: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4f1601w4syI.html

You are most welcome!

Thank you very much for your comprehensive and complete explanation.

Glad you enjoyed it

Glad you liked it

You are welcome!

We're going to get to finite element theory soon in the context of variational principles. However, in the meantime, I do have a playlist with some introductory FEM videos and Python code to accompany it. ru-vid.com/group/PL2ym2L69yzkaue8Ly2Oz51LALRzUV8LZ0

Thank you for working out the example, but this is only the solution for a constant load f correct? The discussion on orthogonality was helpful.

Yes. The final result is for a uniformly distributed load. I make this assumption at 29:35 where I take f outside of the integral because it is constant. Up to that point, it is valid for any distributed load.

@@Freeball99 Thank you!

Yes, you're exactly right! I misspoke. I should have said that the integral of the function is 1. The math, however, is correct.

Nope. I found Dym [sic] and Shames to be somewhat lacking in their explanation of the Galerkin Method. For this video I used a combination of my class notes and some articles that I found online. I didn't follow a single source, but rather just gathered a bunch of information from various places and tried to include what I thought were the best parts.

Try Dym & Shames, "Solid Mechanics. A Variational Approach"

For classical FEM, anything by Klaus-Jurgen Bathe and for more modern approaches, meshless methods, etc. I'd recommend the books by Satya Atluri.

I will certainly be making additional video showing examples of the Galerkin Method. Was impossible to cram all of it into one video. So I began with something simple for which we knew an exact result and will extend it from there. This will include (likely next) showing how to solve this same problem using a weak formulation (including 1st order shape functions) and also the effect of increasing the number of terms in the approximation of the displacement. Will also include a video showing example(s) for which exact results are not possible. Just trying to break this all down into bite-sized chunks.

Yes, your comment makes sense. I am unsure too how Galerkin arrived at his choice of shape function. Since his paper was written in Russian (which I don't understand), and since I have not found any suitable translations or explanations for why he chose this weight function, I too don't have the answer to this. Interestingly, I haven't been able to find any articles or papers that explain it. All of them seem to just accept it rather than offer an explanation as to why it is. I tried working through it mathematically where I assumed that the weight could be treated as the variation of the displacement (even though all the texts tell us that the Method of Weighted Residuals is not based on a variational principle). I seemed to get close to proving that using Galerkin's weighting method was the equivalent to converting the Method of Weighted Residuals into a variational method (like the Ritz Method), but I couldn't quite get the math to work out. This is why, the best answer I could arrive at is that the shape functions provide a basis to the space and that the orthogonal error projection means there is no error in the directions of each basis vector and thus no error within that space spanned by the basis vectors. BTW - I don't believe this has to do with assumptions regarding the nature of the beam since this method can be used for solving differential equations in general for all sorts of problems - not just beams. Thanks for your feedback. If you manage to figure it out, I would love to hear about it.

The app is called "Paper" by WeTransfer. It is running on an iPad Pro 13 inch and I am using an Apple Pencil. Parts of the slide can be deleted and then recovered by using the "undo" function. By selectively deleting pieces of the slide (in reverse order) and then undoing this, the pieces of the slide can be displayed as I present it. I am using Quicktime to capture the screen recordings and audio and then I edit the videos using iMovie to allow them to flow. Pro Tip: Make sure to duplicate your slides before attempting this. The app can be a little buggy at times and quit midway - in which case the deleted parts of the slide cannot be undone and remain deleted and you can end up losing your slide.

We are attempting to solve the equations of motion, which is why we apply the orthogonality condition there. As a result, we get a constraint on the derivatives of the approximate solution (4th derivative in this case). This is actually a more (much much more) stringent constraint than any constraint involving simply w itself. This is because differentiating an approximate solution increases the error and integrating it smooths out the error.

With each derivative, you are effectively dividing by distance. The unit of displacement is the unit of distance (like m or ft). So w, x is the slope and has units of radians (which are dimensionless). The 2nd derivative w, xx has units of "per unit length", but really it's radians/length and is a measure of curvature. When multiplying by EI (which has units like Force x distance^2), then this becomes a moment (force x distance). Taking the 3rd derivative of the displacement, w,xxx gives us units of radians/(length^2) and multiplying this by EI give us the shear force. So we get from the approximate displacement to the shear force by taking the 3rd derivative and multiplying by EI.

It's hard to know exactly why Galerkin came up with this method because I haven't yet found a translation of his paper nor an explanation in English. My understanding, however, is that since we are finding an approximate solution, which exists in the function space defined by the shape functions used in the response, it seems that by ensuring that R(x) is orthogonal to the shape functions, it means that R(x) does not appear in the solution/response space. By adding more shape functions to the response, we thereby drive the residual error to zero.

Not sure I follow this. How does this check for co-linearity?

No, x, x^2, x^3, and x^4 are not our shape functions because our shape functions need to satisfy ALL the boundary conditions. Instead I am showing here how one can easily find a shape function using a polynomial approach and then applying the boundary conditions to determine (some of) the constants. Since we have 4 boundary conditions, we need at least 5 constants; 4 of them in order to satisfy the BC's and then at least one more which will be determined using the Galerkin Method. After substituting the boundary conditions to eliminate 4 of the constants, I arrived at the form in equation 25 (27:35). This now in the form of a constant multiplied by a shape function (which satisfies all BCs).

Thank you for the explanations and for taking the time to share your insights! Much appreciated! 🙏 @@Andy-hy8px

Yes, it is a linear algebra topic. Try to find something on function spaces.

This might be of interest... math.stackexchange.com/questions/1209408/why-is-a-function-space-considered-to-be-a-vector-space-when-its-elements-are

The φ_i's are the shape functions. The process begins by assuming an approximate displacement field consisting of constants and shape functions - i.e. w_approx = a_1 φ_1 + a_2 φ_2 + a_3 φ_3 +... with the φ_i's satisfing the boundary conditions.

For Variational Calculus, you can try Dym & Shames, "Solid Mechanics: A Variational Approach" I don't really have any good ones for Galerkin and weighted residuals on the top of my head. Many textbooks explain this stuff rather poorly, which is why I wanted to make a video - although it's going to require several more to really cover the topic well. Perhaps your could try KJ Bathe's "Finite Element Procedures" if you can find a copy of it somewhere. Anything by Bathe on the subject should be good.

I misspoke. I should have said that the integral of the function is 1.

Yes. I was trying to demonstrate that the integral did, in fact, go to zero as it's supposed to.

I will need to make a video on this. It's going to be impossible to explain it in the comments I'm afraid.

Using a shape function is a technique to distribute the nodal displacements of the beam along its length. In general, these shape functions need to satisfy exactly the boundary conditions, but beyond that they simply approximate the displacement at all other points along the beam - ie it provides a spatial representation of the displacement and other properties at every point along the beam. The mode shape can be thought of as a special case of the shape function which represents the EXACT distribution of the displacement over the length of the beam. While using the mode shape will always give more accurate (exact) results, it is often hard to find. For this reason, we use shape functions that simply satisfy the boundary conditions in order to provide approximate solutions.

Ja. From Durbs originally.

@@Freeball99 Lekker by die see ! ... on a more serious note ...do you know of any nice and easy reference books etc that could help me with recasting PDE's into variational form

@@barrysilver2075 My goto text on Calculus of Variations is "Solid Mechanics: A Variational Approach" by Dym & Shames. You can probably find a PDF online. Not sure that it specifically shows how to cast PDEs into variational form, but there's likely enough background on variational principles in there to allow you to figure it out.

Sorry for the delayed response, but I somehow missed this until here... The problem with this is that we do not know what w is. This is why we are trying to find an approximate solution.

Yes. I will be showing several additional examples using the Galerkin Method.

The a's and c's are just dummy variables which is why they can be readily interchanged.

Try to find a book or articles on function spaces (which is a topic of linear algebra). Here is an example math.stackexchange.com/questions/1209408/why-is-a-function-space-considered-to-be-a-vector-space-when-its-elements-are