The Finite Element Method | Part 13: SFD and BMD of a Frame

[@donaldkhanye9011]

Hello Dr. CEE, Thank you so much for adding this example and its associated “extras = SFD, BMD, and the AFD” to the series. Such packages do really help a lot to cement many of the previously learned concepts and ideas. Fun fact: I “knew it” the moment I saw the video thumbnail and the title or the description of the video that we are still in the 2D domain and that we are here because we will be doing the extras ^_^. I then immediately decided to try this problem out myself before I watched the video. Lo and behold, this time I got all of the solutions to this exercise problem correctly, all the way through including the global reactions, and the extras [except that I did not verify if the BMD for element #2 does stay in the negative, a point that was honestly made and was also well explained by Dr. CEE at 16:15, thank you Doc for this, you never cease to amaze me]. Boy oh boy, having successfully completed the exercise upfront did feel so good and it still does ^_^ as I write this comment. [I am aware that this particular problem is a “cute and convenient” problem but to me it is the principles and the seeing of the applications of those principles coming together that is priceless beyond the “cuteness and convenience” of the particular problem itself]. Thank you for doing this exercise, your explanation after have been a real joy to watch and have been informative in many other regards. There are always gems of information to be found throughout your explanations. 03:13 😊 indeed nobody knows. 06:04 [(GSM = Global Stiffness Matrices) Assemble!], Yes, the TA is on fire here, well done TA, the colour coding and the animation boxes ideas are sleek. 09:40 those “seemingly little” diagrams intended just to understand are gems, know that those are really appreciated, always. 11:50 I 1000% agree. 15:16 Nice, this is another one of those gems like at 09:40, thank you Dr., and these (alternative worlds) are very much appreciated. Here are some of the classic lines from the CEE Channel that gets me every time I hear them (LOL): ….rinse and repeat…..; ….imagine, imagine in another world....; …. if you are getting bored, then you are perfectly fine…… Now, imagine in another world where Element #2 was inclined at some angle, and you had two loading scenarios as follows: Scenario 1 = vertically applied UDL like DL or LL, and Scenario 2 = normally applied UDL like wind load acting normal to the surface. Would it be correct to consider that in scenario 2, one would have had to first “sine and cosine” the equivalent global nodal forces resulting from the wind load UDL on this (now inclined) element #2 into their global sense coordinates equivalents? And that such “sining and cosining” would not be required for the loads under scenario 1 even if the element #2 was inclined? Does it then further mean that the incline would again have to be “sine and cosine” when considering the equivalent forces on the LOCAL element during post-processing? [in that case (local element post-processing), the wind UDL would not require any further “sining and cosining” because it is already in the LOCAL element coordinates sense? But the DL and LL UDL loading scenario 1 would now need to be “sine and cosine” in the calculations of the LOCAL element equivalent forces?] In chapter three, we dealt with the issue of “inclined or skewed supports”. When I look at support #3 in our current problem, I can’t help to think that this support is “inclined” at an angle of 90-degrees counter-clock-wise from the global x-axis? I can “see” that at this angle [also at 180, and 270 degrees] the “support” reactions will ends up as being coinciding with the global X and Y axes (although it seems that the x-and-the-y are being swapped), Yes as long as we look at “Vertical” to be representing the Y-direction sense and the “Horizontal” to be representing the X-direction sense, the above “supposed swopping of the principal directions that I am talking about” is immaterial and the support can be considered as “not skewed, not inclined, and not rotated” even though strictly speaking such a support is in fact rotated relative to the global x-axis. It seems that the only times when the support rotation makes a difference and therefore when it starts to matter is for any values of the support rotation angle that is not equal to 0, 90, 180, 270, or 360 degrees is this line of thinking correct? Lastly, when one does post-processing on the LOCAL element; the NODAL forces (global nodal forces, like the global reactions at node #1 refer to 09:21 are NOT considered separately as it is being done (separately) for the case with the “Local equivalent forces” but rather these other (global nodal forces) are taken care of by the usual rotation or transformation matrix inclusion in the local element post-processing right? I mean, the only additional loads that one need to consider on the LOCAL element during post-processing are the LOCAL equivalent forces caused on that element by the actions of the externally applied global loads that were applied along the length of that particular element, correct? I have enjoyed this video and yes, it was very beneficial for me in many regards. I am concerned though that I was not getting bored ^_^. I am looking forward to the next CEE videos on the series. Thank you Dr. CEE and keep up the excellent work! Regards, DK

[@CivilEngineeringEssentials]

Hi there Engr DK. Wow, I am really happy you tried out that example prior to me explaining it. You even tried drawing the SFD BMD and AFD, so I guess great minds think a like. It is a thrilling experience when things come full circle for a learner. I am happy that your solution coincided with mine. Yep, the problem is cute, as I still have to keep it in the realm of "human possible" ^_^ besides, we are still in the basics. TA had a really good time, I had to give him props for his ideas. It is kind of cool to see the mildness and tenderness of the TA when compared to the wittyness and savageness of the editor. I am happy that you liked those little nuggets and diagrams. I try to provide alternative ideas to kind of expand my examples to try cover more than what the respected author: Daryl L. Logan intended. LOL ^_^ yeah, I guess they come exactly at those points were I kinda read your mind, especiall the "bored" part. If I get your proposed scenario correctly, element 2 will have a transformation matrix coz it is now not horizontal. Having a vertical applied UDL "like what you do in Robot", you would calculate the equivalent loads on the nodes. Let me here differentiate between two scenarios based on what you could do in Robot: 1) Global Vertical and Not Projected Here you would wL/2 and wL^2/12 for your nodal forces, apply them on the nodes, and follow similar steps in assembling the Global Structural K, taking into account that Element #2 is inclined. 2) Global Vertical and Projected. Here there is a pitfall, having a projected load means that it is acting on the projection of the element, kind of like what you see in Live loads on staircases. (depends on code). or snow (depends on code). Here, you still use the load value BUT!! The (L) is now the projection L. So if your element is 5meters long and inclined at 45degrees, your projected L is 5cos45. This projected L is then used in wL/2 and wL^2/12 and the rest follows naturally. 3) A load that is arbitrary in space (more general than just perpendicular on element). Note, we are assuming non-projected load. In this case, cosing and sine the load into the global directions and rinse repeat. L in this case is your element length. 4) A load that is arbitrary in space. BUT projected. Here it is cosined and sined first, then multiplied by the projections. A load in y would be multiplied by 5cos45. A load in x would be multiplied by 5sin45 before finding the equivalent nodal forces. I tried to cover more generalized cases than your question, so I hope it helped. If you still need further clarification, feel free to ask a follow-up question. About the skewed support, you are kind of right and not at the same time. It sounds crazy so hear me out: The idea of skewed support is only applicable when you have a surface specific support. Sounds odd and philosophical, so let me simplify this. A pin is a non-surface specific support. In 2D, it generates 2 forces perpendicular to each other. And you can always cosine and sine those 2 initial forces to generate 2 new perpendicular forces on any other orientation. Similarly, a fixed support generates a set of 2 perpendicular forces and a moment. The 2 original perpendicular forces are the same as the pin, and the generated moment does not need to be cosined or sined as it is perpendicular to the 2D space (NOTE it becomes crazy if it is in 3D). The pin equivalent in 3D is a ball and socket connection. But I will keep it 2D. Now why is the roller so amazing and needs to be considered if skewed? it is because it generates only 1 reactions and it is direction specific. it is perpendicular to the surface the roller is rolling on. Hope it made sense. Your question is really cool and shows deep understanding, so I tried to address it as clear as I could. Please feel free to ask follow-up questions in case you still need further clarification or have anything on your mind. If you do post-processing on the local element, you get the forces at the ends of those elements. Those are not equal to the external forces. If I understood your comment correctly (I think I did not 100% get it, so sorry bout that, feel free to clarify), I want to mention the following: By Global nodal forces, I mean now the applied nodal forces, not those equivalent forces we calculate. No, real nodal forces. So in Robot: Nodal Force. Not bar loads. The local forces calculated on elements at shared nodes, need to be flipped for node equilibrium. For example (you need a pen and paper for this), let's say element 1 is horizontal. And has a force on its node 1 (left side) of f1x = 400 to the right. When you perform nodal equilbrium, you would have to first transform the nodal forces on element 1 to global coordinates (in our case, it remains 400 because the element is horizontal) Then, you have to flip the force (remember how the method of joints in trusses work). And only then, you can do your node equilibrium calculations. There was a video where I did node calculations (in springs or something, back when we were still Level 1 FEM specialists).. Of course, the difference between that level and this level is huge, not only because of transformation. Anways, Sorry for the larger comment. If you have any further clarifications or questions, feel free. Stay tuned for more content, great to know you enjoyed it. Regards CEE

[@donaldkhanye9011]

@@CivilEngineeringEssentials Hello Dr. CEE, Apologies it took me sooooo loooong, to reply to your feedback (life happened, but all is well now). Thank you so much for the response, it certainly has addressed my comments and then some. Also, it is always a joy to read through your feedback, and I really do not mind a large comment from you as your feedback bring to light so many key ideas that does a LOT of good. Regards, DK

[@CivilEngineeringEssentials]

Hi there Engr. DK, no worries, I hope all is well and good. thnx a lot, take care, CEE

[@francescoindolfo]

When you calculated the local nodal force on each element should you have subtracted the equivalent nodal force of the distribuited load??

[@CivilEngineeringEssentials]

Hi there and thnx for the comment. (If possible, please write the timestamp related to your question) But, to answer the question: if you find the local forces by multiplying the local stiffness and displacement, you get the local forces immediately, without the need to remove Eq. Nodal Forces. if possible, please provide the time stamp. I would help you as good as I can. Regards, CEE

[@francescoindolfo]

Hi, my doubt is at about 12:38 when you calculate the local nodal force for the element number 2 I think you should have subtracted the equivalent nodal force due to the udl at least in the book it was done in this way

[@CivilEngineeringEssentials]

Sorry for the late reply. For some reason, youtube did not allow your comment in. Now I understood your question. F + F_equivalent = kx you are right. Here I missed mentioning that. I will add it in the description of the video.