I don't have a video explaining how to handle more than one indeterminacy with a force based method. Have a search for "flexibility matrix".
@patrikengas6479Ай бұрын
when you at 5:35 make that statement, how was that derived? i tried using the known curvature - moment relation to find a slope at B due to the moment "Mab" at A where i got that slope to be: θ(B) = Mab*L/2EI and using the virtual method i got the fbb' = L/3EI, and finally plugging that into the compatibility equation i got the Mba = (3/2)*Mab, so obviously this doesn't match. But then my question would be, is this a correct approach or just wrong?, and if the latter, whats the correct way?
@ProgrammedMechanicsАй бұрын
follow the method at 5:07, I get \theta_B = M_{ab}*L/6EI and f_{bb} = L/3EI
@patrikengas6479Ай бұрын
Okay, do you get this: "theta_B = M_{ab}*L/6EI" using the virtual moment method? does that mean that you place the unit moment in B and from there gain the moment equation m(x) and use that together with M(x) caused by the real moment Mab ?
@patrikengas6479Ай бұрын
update: actually that seemed to work out trying it, i get the angle at B to be Mab*L/6EI, using the virtual moment method. only weird thing is that the relation: EI*θ = ∫M(x)dx provides a different angle. thought these should be equivalent. the M(x) i get = -(Mab/L)x + Mab, if that checks out
@CapsuleParrots2 ай бұрын
At 2:58, can you explain further? Are joints at any supports not zfm's because the support is a reaction?
@ProgrammedMechanics2 ай бұрын
There is no way to declare that they are definitely zero force members. However, you can't rule out members connecting to a support being a zero force member, but unlikely.
@marcosrangel77842 ай бұрын
You are incorrect, in the first problem, CE is not a ZFM, because you have a 5KN external load
@ProgrammedMechanics2 ай бұрын
The external load is applied at E. If you were to draw a free body diagram of joint C then F_CE would be the only force with a vertical component.
@eliasvusi34513 ай бұрын
Thank you, why is the deflection due to real load uses moment from virtual unit load?
Saludos estoy analizando una cercha fink con angulo doble pero al correr el modelo los resultados en las reacciones son diferentes en el peso propio de la estructura ?
@ProgrammedMechanics4 ай бұрын
yes, be careful to check what loads are applied, I find the load table is the best place to check
@krstev297 ай бұрын
I hate these problems with two blocks, especially when there are a few scenarios.Thanks for explaining.
@ProgrammedMechanics7 ай бұрын
They can look more terrifying than they are. Once you work out the possible scenarios (the problem solving aspect of question), draw good clear free body diagrams and apply the equations of equilibrium.
@JesusMartinez-zu3xl11 ай бұрын
Thank you so much!! I'm preparing to take the FE exam and you made zero force members very clear.
@ProgrammedMechanics11 ай бұрын
Glad it helped!
@sohailahmed5039 Жыл бұрын
Sir you have stopped making videos. There is almost 95% work left undone by you both in structure analysis I and II. Any reasone for not making any video.
@ProgrammedMechanics Жыл бұрын
Never say never, but I only record videos for topics that I teach. For each topic there is only a limited number of examples as I follow up with many hours of on-campus problem solving workshops.
@sohailahmed5039 Жыл бұрын
@@ProgrammedMechanics i completely agree sir and i knew it might be the answer too. But do you really teach these few topics, as structure analysis surely does not contain only these topics....anyways your explaination really helps
@ProgrammedMechanics Жыл бұрын
@sohailahmed5039 the team of us teach a full stream from statics -> strength of materials -> structural mechanics -> computer methods -> structural dynamics -> finite element theory (and practice). We have reduced the number of hand calculation methods we teach, as they are never(/ very rarely) used in practice, but we do make sure that students are comfortable with both the theory of underlying computer packages they will use and the safe practice of these packages.
@hermangaviria690 Жыл бұрын
great video, thank you very much. I'm surprised more people haven't commented or liked.
@ProgrammedMechanics Жыл бұрын
I just use YT to host my videos, not concerned with likes/subscribes etc. but glad it was helpful for you.
@sohailahmed5039 Жыл бұрын
This guy is wonderful
@kashifanihaan4032 Жыл бұрын
Isn't this a determinate truss? How do we approach a problem with indeterminate truss??
@ProgrammedMechanics Жыл бұрын
yes the truss in this example is determinate, for indeterminate see kzbin.info/www/bejne/oWfbiItpa8qFmdk
@gotemnausiah6495 Жыл бұрын
Thank you for this video, just wanted to ask whether the horizontal unit load applied is always to the right or could it be applied either side?
@ProgrammedMechanics Жыл бұрын
You can assume any direction, a positive final answer will confirm your assumption.
@kakabeesh8091 Жыл бұрын
Really useful. Clear explanation and good and easy example. Thank you
@ProgrammedMechanics Жыл бұрын
thank you for taking the time to leave feedback
@ProgrammedMechanics Жыл бұрын
correction - 9:29 off- diagonal terms should be zero, NOT -12/L^2.
@mohamedsherif4096 Жыл бұрын
nice tutorial, but how is that different from the force method?
@ProgrammedMechanics Жыл бұрын
virtual work is a force method
@kvasios2 жыл бұрын
Richard Ayoade is this you?
@ProgrammedMechanics2 жыл бұрын
No my name is Maurice Moss ;-)
@kvasios2 жыл бұрын
@@ProgrammedMechanics 😆
@ProgrammedMechanics2 жыл бұрын
34:45 other solutions include the unit load in AC to calculate the flexibility of AC. This gives a more accurate solution, but the compatibility equation is less clear.
@pedrocalorio16552 жыл бұрын
this is assuming that your dynamical system is linear and your M, K and D matrices does not change with time, correct?
@ProgrammedMechanics2 жыл бұрын
The time integration doesn't change for a nonlinear system, but you would need to incorporate an iteration loop to achieve convergence in an implicit scheme.
@thabangmakhayi89622 жыл бұрын
Damn love the intro
@ProgrammedMechanics2 жыл бұрын
not sure if you're being serious or sarcastic, but I'll take it! Hope the rest of the vide was useful for your studies.
@Mohamed-ef5zw2 жыл бұрын
How can I check if this deflection is acceptable or not?
@ProgrammedMechanics2 жыл бұрын
deflection (serviceability) limits will be set in design codes / standards
@Mohamed-ef5zw2 жыл бұрын
@@ProgrammedMechanics What is the deflection limits for trusses in Euro code? I couldn’t find it.
@RuinedCityGamer2 жыл бұрын
how do we get thoses constants to calculate a1 a2 a3 ... ?
@ProgrammedMechanics2 жыл бұрын
set \alpha = 0.25 and \delta = 0.5 (this will give constant average acceleration), \delta T is problem dependent - aim for 25 timesteps over the shortest period of interest
@hoyinmatthewcheung82863 жыл бұрын
I want to ask a question , to determine whether a structure is stable , do we need to make sure both internally and externally do determine its unstable or not ? Or we just need to consider it externally?
@ProgrammedMechanics3 жыл бұрын
check both stability criteria
@wyleong43263 жыл бұрын
Thank you Program Mechanics. I’m unsure if it’s my Bluetooth headphones, but I’m getting a slight humming static (sounds like an airplane’s engine before taking off) as I’m listening to the lesson. This video has been helpful.
@ProgrammedMechanics3 жыл бұрын
it's the CPU of my surface pro getting picked up by the mic, didn't use an external mic when I recorded.
@mohammedlasmi6493 жыл бұрын
Thanks too much
@ProgrammedMechanics3 жыл бұрын
No problem
@reivila49813 жыл бұрын
Hello, your force for the middle support you found to be -171 kN therefore you should draw R_By going downwards at 35:30 Did I misinterpret that? EDIT: I realize my mistake, we assumed the unit load going down so a negative value means it goes upwards
@ProgrammedMechanics3 жыл бұрын
glad the meaning of the minus sign clicked.
@marshallmather26383 жыл бұрын
Please could you do another video on multiple loadings as some of the steps are a bit ambiguous.
@ProgrammedMechanics3 жыл бұрын
worked example here, if this is what you're after: kzbin.info/www/bejne/j4aThmqsj6x9ptE
@ProgrammedMechanics3 жыл бұрын
14:40 We eventually look at horizontal equilibrium of the entire structure to generate an extra equation, but need to get the reactions first from FBDs of the columns
@bobamericanu21933 жыл бұрын
Poggers video
@ProgrammedMechanics3 жыл бұрын
I'm far too old to know if that is good/bad feedback, but thanks for taking time to comment either way.
@gracemuthoni953 жыл бұрын
The N at BD, should be - 11.67 Thanks for the tutorial!
@ProgrammedMechanics3 жыл бұрын
@32:36? yes, typo
@ProgrammedMechanics3 жыл бұрын
3:47 error P = F/A
@ProgrammedMechanics3 жыл бұрын
4:22 is correct
@sushankm77293 жыл бұрын
Could you please tell, where do we perform Newton-Raphson iterations in this implicit algorithm.
@ProgrammedMechanics3 жыл бұрын
This is just Newmark for time stepping linear problems only. You would need to have N-R loop inside of each time step for non-linear problems.
@mahdy26823 жыл бұрын
Hard to follow when constantly scrolling down, thus being unable to simultaneously see the diagram and working out. Solution could be to zoom out when working out is completed or zoom out as doing working out.
@ProgrammedMechanics3 жыл бұрын
Thank you for taking the time to give feedback. My on campus students have a copy of my notes with the solution typed. You could take your own notes whilst following along?
@nehachauhan40503 жыл бұрын
Hey! Thanks for this, this is helpful Can you provide the code?
@ProgrammedMechanics3 жыл бұрын
thanks for the comment, it's probably a good exercise to type along with the video to understand each line of the code
@tanmoysahoo23813 жыл бұрын
Can you please explain why FEMf=0 but FEMn is not equal to 0 though both are not fixed ( I understood about FEMf=0)?
@ProgrammedMechanics3 жыл бұрын
there is a moment at n, but not at f.
@tanmoysahoo23813 жыл бұрын
I have seen in a book. There its considered that both end is fixed then they find fem in both end. I can't find similarity with hibbler's book.
@tanmoysahoo23813 жыл бұрын
If we consider fem=0 then how can I solve for theta f?
@AnythingETC3 жыл бұрын
Thanks for your lecture! But there is a typo,,, In 8:26 14:00, you have to time delta_t at the end. Becuase V(+dt) = V(t) + [(1-beta)... + (beta)...] * "delta_t". It was a helpful lecture. Thanks Again!!!
@ProgrammedMechanics3 жыл бұрын
oops, yes 2nd eqn at these timestamps should have \Delta T after the square brackets. Hope the procedure was useful anyway.
@hutheifa48803 жыл бұрын
Any video about the determinacy of compound structure and specifically king-post truss beam.
@ProgrammedMechanics3 жыл бұрын
the same principles apply, but truss members don't have moments at the member ends.
@hutheifa48804 жыл бұрын
Great video. Any formula for frame structure to check the stability?
@ProgrammedMechanics4 жыл бұрын
try this: kzbin.info/www/bejne/gX6yhIyqi813j9k
@hutheifa48804 жыл бұрын
@@ProgrammedMechanics I meant a formula for stability of beams and frames. it is not addressed.
@ProgrammedMechanics3 жыл бұрын
there's no simple formula, you need to look at individual elements
@TheSterlingArcher164 жыл бұрын
Good detail of the process. Very unintuitive however that summing the axial displacements would result in a total deflection in the direction specified.
@ProgrammedMechanics4 жыл бұрын
there are 2 prior videos in the playlist to help with the concept - kzbin.info/aero/PLG7h9Zm4rTYxo36g3Rl6C7_8gLBUGKv16
@lukeamper58314 жыл бұрын
Sir is it possible to solve this using a matrix?
@ProgrammedMechanics4 жыл бұрын
There is only 1 degree of indeterminacy, so no need to solve using matrices if using unit-load method. However, the problem could always be solved using the direct stiffness method (kzbin.info/aero/PLG7h9Zm4rTYy55h3KRZoQw6VKnxZlwMD8).
@lukeamper58314 жыл бұрын
Ohhh okay sir, thanks!! 💯
@tajsay4 жыл бұрын
Thank you so much it's perfectly explained... Appreciate your time and efforts
@ProgrammedMechanics4 жыл бұрын
Glad it was helpful!
@RICHARD-nd1sn4 жыл бұрын
On the BD beam, there is a positive moment or negatif moment?
@ProgrammedMechanics4 жыл бұрын
I always draw bending moments on the tension side of members.
@jsntndzdlvllrsls4 жыл бұрын
Hello. When you explained how to calculate the flexibility coefficient, is it possible that there is a small mistake in the notation in the integral? You wrote m(x)*m(m) in the numerator, shouldn't it be M(x)*m(x)? Or is it m(x)*m(x) maybe? What you're saying and what you're writing is different. Thanks.
@ProgrammedMechanics4 жыл бұрын
definitely x in the brackets
@manallambarki61754 жыл бұрын
Wow that was very helpful , specially going through all the details. Thank you so much!
@ProgrammedMechanics4 жыл бұрын
I use this pre-recorded material for my own classes mainly, but happy to hear if it is useful for others
@krokodilvomnil53274 жыл бұрын
Great video greetings from germany
@ProgrammedMechanics4 жыл бұрын
Glad it was useful. I just use YT for video hosting content for my own students, but always glad to hear if they're of use to others.
@pradipguhathakurta4994 жыл бұрын
DETERMINE ALL THE MEMBER FORCES IN THE LOADED TRUSS BY METHODS OF JOINTS . POSSIBLE TO SOLVE THE PROBLEM
@ProgrammedMechanics4 жыл бұрын
Identifying the zero force members speed up the method of joints. My on-campus students do this problem in method of joints tutorials before we introduce this, so they already know that the members are zero force members.
@kratobalaco40314 жыл бұрын
great tutorial! although it kinda too slow for me
@ProgrammedMechanics4 жыл бұрын
Slow is good, it is always possible to skip or speed-up the play back. I've tried to not gloss over any bits of the calculation which many students can find frustrating.
@kratobalaco40314 жыл бұрын
@@ProgrammedMechanics 👍👍👍
@thefedefede984 жыл бұрын
Do you know a software for qualitative analysis? I only find software which require specific quantities but I need parameters
@ProgrammedMechanics4 жыл бұрын
use any frame analysis package, try to keep the EI values for all members constant. You could use unit values for loads (pt or udl), as analysis is elastic.
@thefedefede984 жыл бұрын
@@ProgrammedMechanics Aldready tried using unit values, but I'd need the relation e.g. For momentum like ql^2 dependent on load and length.. Can't find anything parametrical
@ProgrammedMechanics4 жыл бұрын
ql^2/12? use q=1, L=length of beam / column. This is the Fixed end moment, quite often hidden in the calculation engine.
@pawanacharya29154 жыл бұрын
can you help me to find the moments ??
@ProgrammedMechanics4 жыл бұрын
This video is just the derivation of stiffness coefficients for use in displacement based methods. You will need to use the direct stiffness or slope-deflection equations to calculate moments.