Hello Dr. CEE, Thank you for the video, great explanations as always (and # assembled 😊). I have some questions. 06:13 [and later 08:24 and 13:33] so Doctor, are you saying that all those nicely “boxed and coloured ^_^” matrices can change (move around) depending on what “forces” are released or correspond to the released direction(s)? Would one then have to “crunch the math and condense” a special element local stiffness matrix each time depending on the nature of the release(s) on that element of interest? 22:23 In the first instance, and just like in this video example, I used (applied on element #2) the condensed stiffness matrix for the beam element with a moment hinge at its RIGHT end (Eq. 4.6.13), all worked out fine and the results matched the video and the referenced text. I then changed and I used (applied on element #3) the condensed stiffness matrix for the beam element with a moment hinge at its LEFT end (Eq. 4.6.15). And a “regular” stiffness matrix for element #2. I had expected the results to be the same (except for the slope value of phi-3, as the use of the latter condensed stiffness matrix would have resulted in a phi-3 value associated with element two… the element that got the “regular treatment this time around ^_^”). However, all the results are now different from those I had obtained from the first instant of using the RIGHT end condensed element stiffness matrix (what gives? did I miss something?). Aren’t the two approaches (either using the RHS or the LHS condensed stiffness matrix) supposed to produce the exact same overall structural forces? results in this case? I have enjoyed this video and I am looking forward to the next CEE videos. Keep up the good work. Kind regards, DK
Hi there Engr. DK, Yep, the matrix not only can change and move around, the resulting result will differ. Each release will have its own matrix, but softwares "as mentioned at the end of the video" do not do this, they have their own trick up their sleeve. Also, from a practical point of view, internal hinges are the most common releases so I guess the Stiffness Matrix that was presented is good. Even if you have a double released beam (release on both ends), you could circumvent this by splitting your beam into two beams, each having a hinge. The results should be the same. I am surprised that there are different. I think I will just quickly check it from my side. So your expectation is valid, you are not missing anything, or do you? (dun dun duuuuuun). Nah really, you are correct. There might be a slight miscalc or smthn. But I will check it out myself. Regards, CEE
I found out the error. In the stiffness matrix of element #1: it is written 0.67 (2/3) although it was 3/2 in my notes. You can see it "magically" become 3/2 in the global stiffness matrix. I am pretty sure that when you tried the second approach, you used the k1 stiffness matrix with the 2/3 values. which is of course my fault. ^_^ sry bout that.
@@CivilEngineeringEssentials Hi Doc, I have a little set-up in Mathcad. When I try these things out, I ussually just input raw question parameters into my little set-up and let it munch the various matrices from there. In this case, I have used the equations for both cases of the moment hinge at its RIGHT end (Eq. 4.6.13) and the case of the moment hinge at its LEFT end (Eq. 4.6.15) directly from the referenced text. I did not use the values from your notes [I did however used and got all the IDEAS from your wonderful notes, and without your notes, none of this would have happened, so thank you for those notes] Perhaps you could run the calculations from your side assuming the moment hinge to be on the left side and compare your results? or have you done and made this comparison alreday? do the results match? If so, did you use Eq. 4.6.15 from the referenced text with the assumption of the moment hinge on the left hand side? It is a curious thing for sure. Thank you for your reply and kind regards DK.
