Coefficients of the stiffness matrix - Derivation - Beam element

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TM'sChannel

TM'sChannel

Күн бұрын

Пікірлер: 43
@dkscidmore
@dkscidmore 4 жыл бұрын
It took forever to find this video, but this was the concept I was stuck on. Thank you so much for making this video.
@tariq-salzengineering9650
@tariq-salzengineering9650 2 жыл бұрын
Hi, @TM'sChannel: I found a confirm mistake for the value of a2 which should be: a2=-3/L3(v1-v2) - 1/L(2 Θ1+Θ2) [Not "...-1/L( 1/L(Θ1+Θ2)" as per your slides]. I checked the after derivation part with matrix and also the algebra so I have confirmed this by all means. You also check and confirm. [P.S: Just checked you already mentioned that in description and also uploaded corrected video. Thanks for that. ] updated.
@evazhang3232
@evazhang3232 2 жыл бұрын
a2=-3/L2(v1-v2) - 1/L(2 Θ1+Θ2)
@nicologrilli9931
@nicologrilli9931 Жыл бұрын
I agree, a factor 2 is missing Thank you for pointing this out
@Amhara3K
@Amhara3K 4 жыл бұрын
Thanks. FYI, I think thet the General displacement equation(4:42) needs some revision on the fourth term.
@pratikdutta5136
@pratikdutta5136 6 жыл бұрын
a2 = -3/L^2(v1-v2) - 1/L(2phi1+ phi2)
@MaltaLumpie
@MaltaLumpie 5 жыл бұрын
Haha I can relate
@igorbarcelos9531
@igorbarcelos9531 4 жыл бұрын
@@MaltaLumpie Yes, there is this mistake in your calculation.
@ashrafultamjeed8746
@ashrafultamjeed8746 4 жыл бұрын
@@igorbarcelos9531 this mistake destroyed my day
@joelstains4366
@joelstains4366 7 жыл бұрын
Thank you so very much. That was very helpful.
@koushikr7039
@koushikr7039 3 жыл бұрын
TM's channel, can you show the derivation of stiffness matrix of 2 noded beam element
@TMsChannels
@TMsChannels 9 жыл бұрын
@Yong-Min Jeong I have noticed this and posted the corrected video at kzbin.info/www/bejne/iH-si2uAqpiSeas Thank you for noticing... :)
@gowthamthotapalli8867
@gowthamthotapalli8867 5 жыл бұрын
can you tell me how to find out the missing element in the matrix using shape function
@rajithodlme
@rajithodlme 9 жыл бұрын
hi, one doubt, at 6:47, i am not getting the expansion of m1 as mentioned in the video, i get it as EI/l3 * (6LV1+fi1L^2-6LV2+fi2L^2) instead of EI/l3 * (6LV1+4fi1L^2-6LV2+2fi2L^2)...could you please upload a screenshot of the derivation .. thanks
@TMsChannels
@TMsChannels 9 жыл бұрын
Hi, thanks for pointing it out :). I went back to my derivation and I discovered a typing error on the slide when calculating the displacement function v(x) at 4:44. The expression for the second term i.e. a2 should be: a2 = -3 (v1-v2) /L^2-( 2fi1 + fi2 )/L . Using this adjustment you can obtain the correct expression for v(x) and from there calculate the correct coefficients as in the video. Sorry for the inconvenience, I will upload a revision of this video with the typing error corrected. Please let me know if you encounter any more errors/doubts.
@mrajsma01
@mrajsma01 9 жыл бұрын
a beam element has 3 dofs at each node
@TMsChannels
@TMsChannels 9 жыл бұрын
+Chris10B Thanks for your comment. Are you referring to the degree of freedom associated with axial force and displacement as the one I left out?
@mrajsma01
@mrajsma01 9 жыл бұрын
+TM'sChannel I realise you left it out since there is no load in the axial direction :) I am trying to solve a problem with three degrees of freedom at each node and looking for a way to deal with the stifness matrices.
@TMsChannels
@TMsChannels 9 жыл бұрын
+Chris10B I see :). It sounds like you are describing a frame element. It is the combination of a beam and a bar element and thus have three degrees of freedom at each note. You can have a look at my derivation video of this element at kzbin.info/www/bejne/raKagKZmncmMqs0 . The stiffness matrix is shown there. I also have a video showing an example on how to approach a problem with frame elements. I hope this comment was helpful :)
@thomasdakin9108
@thomasdakin9108 7 жыл бұрын
You teach me how to obtain stiffness coefficients, I teach you how to record on something other than a potato
@TMsChannels
@TMsChannels 7 жыл бұрын
Hahaha, As long as you supply the computer :)
@dexbuild1206
@dexbuild1206 6 жыл бұрын
😂
@abdofaisel4968
@abdofaisel4968 5 жыл бұрын
Can you prove it with cojugate method or virutual work
@DANNY40379
@DANNY40379 7 жыл бұрын
Can you elaborate more on how the geometric stiffness matrix for a beam element is derived? thx
@TMsChannels
@TMsChannels 7 жыл бұрын
Hi, I came across this document: people.duke.edu/~hpgavin/cee421/frame-finite-def.pdf Which has a version of the derivation. I will add it to my list for future videos. :)
@enginkeskin4829
@enginkeskin4829 10 жыл бұрын
How v(x) changes when beam is parabolic haunch beam ? Or what changes?
@TMsChannels
@TMsChannels 10 жыл бұрын
Hi, thanks for the comment. Honestly I don't really know. But the theory used to derive the matrix models the center line displacement of the (straight) beam. If that line is curved, I presume the theory also changes. Unfortunately I cannot say how. Another thing that also changes is the inertia (I), this becomes a function of position due to the change in height (h). Sorry if this is not very helpful, I have not yet dealt with the analysis of haunch beams.
@enginkeskin4829
@enginkeskin4829 10 жыл бұрын
Hi, thanks for reply. Video is a very helpful. Haunch beam is not a easy topic. Thanks again.
@madalitsonjobvucristoclear
@madalitsonjobvucristoclear 7 жыл бұрын
hi,how can i derive the stiffness matrix for authotropic beam?
@TMsChannels
@TMsChannels 7 жыл бұрын
Hi, I actually have no idea. I haven't worked with such beams yet. Maybe you can find a paper or a book which deals with this topic and post it here? I would love to read it.
@zackbristol1
@zackbristol1 7 жыл бұрын
If you have a cantilever does the stiffness matrix change ?
@TMsChannels
@TMsChannels 7 жыл бұрын
Hi, no it does not. Only the boundary conditions are different i.e. the end degrees of freedom are not prescribed.
@zackbristol1
@zackbristol1 7 жыл бұрын
TM'sChannel can you add a video of 6 degree of freedom derivation ?
@TMsChannels
@TMsChannels 7 жыл бұрын
Hi, I have a video on the derivation of a frame element (which has 6 degrees of freedom, 3 at each node). You are welcome to view it here: kzbin.info/www/bejne/raKagKZmncmMqs0
@zackbristol1
@zackbristol1 7 жыл бұрын
TM'sChannel sorry I meant 12, 6 at each node. I am interested in twist
@TMsChannels
@TMsChannels 7 жыл бұрын
I see, I shall try to make a video for such an element, although it may take me a while. For torsion the basis is similar to that of the truss element, only with AE/L replaced by GJ/L. For the time being you may look at the book "a first course in the finite element method" by D.L. Logan which does provide a derivation of a 3D frame element. Hope this helps
@azadyadav5080
@azadyadav5080 2 жыл бұрын
thanks
@salmankhi
@salmankhi 8 жыл бұрын
I can't find a1 and a2 at 04:15. Has anyone done this?
@TMsChannels
@TMsChannels 8 жыл бұрын
+Salman Khan Hi, I have made a mistake in the video. Please see kzbin.info/www/bejne/iH-si2uAqpiSeas for the corrected version.
@salmankhi
@salmankhi 8 жыл бұрын
+TM'sChannel thanks for the fast response.
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