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@chidubemoleka8729
@chidubemoleka8729 7 сағат бұрын
I have some problems, I don't know if you could help me
@chidubemoleka8729
@chidubemoleka8729 7 сағат бұрын
Some structural engineering problems that I need the solution
@QuickQuestionEngineering
@QuickQuestionEngineering 5 күн бұрын
Great explanation! Thank you for the video!
@kianzedkay
@kianzedkay 6 күн бұрын
thanks a lot man u saved me with this video!!
@user-sn9lw2sz2g
@user-sn9lw2sz2g 6 күн бұрын
❤❤❤
@danvining2186
@danvining2186 7 күн бұрын
Would an RCC wall such as an ICF wall have any load, deflection or shear effect on an RCC beam if the wall was equal to the effectiive length of the beam once the wall was fully cured?
@jackyeh7512
@jackyeh7512 13 күн бұрын
There is a design usable life of the structure, maybe 70 years or so. There should slso be inspection and maintenance requirements on the building, just like cars. The roof has been functional for 50 years. I would say the design was proper.
@bunnuakhil2920
@bunnuakhil2920 14 күн бұрын
well demonstrated. Enjoyed it.
@RK-tx5xb
@RK-tx5xb 17 күн бұрын
Your videos are extremely exceptional sir. Please keep on posting. It’s hard to find such a good ref these days. Thanks once again.
@davidtettleton2732
@davidtettleton2732 19 күн бұрын
How do your beams compare to wood, the second beam for instance?
@HermelaMerishet
@HermelaMerishet 21 күн бұрын
Thank you
@WG-ft6tz
@WG-ft6tz 25 күн бұрын
Hi Professor, Thanks for the video as always. Was wondering in practice where the cracked moment comes into play? It seems that engineer's are almost always focused on the reinforced design moment strength.
@StructuresProfH
@StructuresProfH 23 күн бұрын
The cracking moment is typically only used for serviceability concerns (i.e., deflections). When estimating deflections of concrete beams, the moment of inertia will change depending on whether the beam is cracked or uncracked. I have another video on that topic: kzfaq.info/get/bejne/oK9mqtqbx9-dfac.html
@sothattheng8624
@sothattheng8624 27 күн бұрын
Thanks for your sharing Prof.H!
@chichha420
@chichha420 27 күн бұрын
Thanks sir from germany. I would be grateful to you if fürther videos of approximate analysis of frames are uploaded.
@bekehbenjamin775
@bekehbenjamin775 Ай бұрын
Does it mean that when ever your calculations gives you zero as an answer, it shows that the trusses is determinate?
@StructuresProfH
@StructuresProfH Ай бұрын
Yep, zero means determinate (unless there is some other issue that causes instability).
@user-ui3jp4pq6g
@user-ui3jp4pq6g Ай бұрын
Please can you make a video how to calculate the bending moments at the end of a fixed concrete beam on both sides when one side moves down (settlement) this condition usually is imposed by soil engineers
@ferenczbabinszki7699
@ferenczbabinszki7699 Ай бұрын
i thought you're supposed to waith at least 28 days for proper cure?
@StructuresProfH
@StructuresProfH Ай бұрын
Sorry, I wasn’t clear in the video. We did a wet cure (with burlap) for 7 days. We tested roughly one month after casting. That said, there are many applications in the real world where concrete is loaded well before 28 days, and concrete will continue to gain strength even after 28 days. It’s more accurate to say that 28 days is the laboratory standard.
@marcoromanelli6000
@marcoromanelli6000 Ай бұрын
And now, a video about real buildings where concrete beams are loaded that way...
@cleisonarmandomanriqueagui9176
@cleisonarmandomanriqueagui9176 Ай бұрын
best youtube channel for civil indsutry ever . experience is better than theory
@cminor1477
@cminor1477 Ай бұрын
This is the best video I have ever seen
@MohammedMinhajUlHaq
@MohammedMinhajUlHaq Ай бұрын
Excellent understanding after watching this lecture
@PremKumar-ky7gl
@PremKumar-ky7gl Ай бұрын
Thank you very much sir, this is a great work. Thanks alot
@RK-tx5xb
@RK-tx5xb Ай бұрын
Sir, absolutely mind blowing. Your hinge method introduction is phenomenal. Please keep on posting great videos. I do have some few FEA uncleared questions. Is there any possible way to reach you?
@zhilyaismael8435
@zhilyaismael8435 Ай бұрын
thank you dear prof . could you please cane you give me some idea how it's come in in exam?
@zhilyaismael8435
@zhilyaismael8435 Ай бұрын
🥰🥰🥰
@tsegazeabdejene6685
@tsegazeabdejene6685 Ай бұрын
Great explanation
@patrikengas6479
@patrikengas6479 Ай бұрын
hey, (i'm just taking a peek into the next semester curriculum), but one thing i dont get is why you use this method to find support reactions, are they initially unknown?
@StructuresProfH
@StructuresProfH Ай бұрын
Yes, the reactions are unknown. For an indeterminate structure, equilibrium alone is not enough to solve for the reactions, so that's why we need the force method (or other techniques for indeterminate structural analysis).
@joealex2463
@joealex2463 Ай бұрын
how do you find r?
@StructuresProfH
@StructuresProfH Ай бұрын
"r" is the number of releases, where a release counts as a single known internal force. So for example, a hinge in a beam forces the bending moment to be zero at that location - that counts as one release, because we know that M = 0 at that point. You can also have releases for shear and axial forces as well. These can even be combined. For example, a roller-like expansion joint in a bridge might count as two releases if both the axial force N = 0 and the bending moment M = 0 at that location.
@joealex2463
@joealex2463 Ай бұрын
@@StructuresProfH thank you!
@Sarissa99
@Sarissa99 Ай бұрын
What if I have a curved (semi-circular in my instance) frame? How do I implement NQM there?
@StructuresProfH
@StructuresProfH Ай бұрын
The full answer might be a full video, but in the meantime, I'd probably take slices of the free body diagram at some angle theta (sort of like polar coordinates) rather than at some position x. For a semi-circular frame, the axial force will be tangential, the shear force will be radial, and the bending moment will still just be a moment. Even for a concentrated force at the peak of frame, the axial and shear force values will not be constants - they will vary by the angle theta. From there, you can apply equilibrium (sum of forces and sum of moments) to find the internal forces.
@Havoner
@Havoner Ай бұрын
Dear Professor! In the "Model type" option, if I select "Cable", will it be a compressed structure? 12:51
@StructuresProfH
@StructuresProfH Ай бұрын
No, the "Cable" option will not automatically prestress the structure, though you can do that with a little work. If you have a linear mesh, Cable creates the LINK180 element designation, which is the same as you would use for a simple truss. If you have a quadratic mesh, Cable creates the CABLE280 element. This element requires tensile stress to provide adequate transverse stiffness, and may be unstable under compression - it works essentially like a hanging string or cable. For most applications, either Link/Truss or Cable would work for applying prestressing, but I think Link/Truss is a bit easier to deal with. Either way, to apply prestressing you need to define an initial stress in the link or cable. The INISTATE command using Mechanical APDL is the way to go.
@Havoner
@Havoner Ай бұрын
@@StructuresProfH Thanks Dear Professor Could you create such a 3D video on KZfaq? What material do I need to define for the cables? I must set Link/Truss option?
@fadi747
@fadi747 Ай бұрын
I encourage you to add more tutorial videos related to the new ACI 318-19 code in the future. Your videos are very helpful and easy to understand.
@fadi747
@fadi747 Ай бұрын
How did you compute Av min (at 8:35 of this video).
@ymsh2
@ymsh2 Ай бұрын
Thank you for uproading this video.
@benonp3622
@benonp3622 2 ай бұрын
in mesh, there's no squares visible. how?
@Havoner
@Havoner 2 ай бұрын
Hello i have other questions Why can't I see the "Edit Sketch" button in SpaceClaim. I want create construction frame model in 3D. I have version - Ansys 2024R1 Student
@nisarahmadriaz3117
@nisarahmadriaz3117 2 ай бұрын
Brilliant dear sir!
@burhanuddinmotiwala1428
@burhanuddinmotiwala1428 2 ай бұрын
for the last example if there were no hinges in the frame then would it be indeterminate to the fourth degree?
@StructuresProfH
@StructuresProfH 2 ай бұрын
That is correct!
@Ans_aka_Inzi1073
@Ans_aka_Inzi1073 2 ай бұрын
That's an amazing lecture! Everything was explained with great proficiency.
@skablerpro
@skablerpro 2 ай бұрын
It seems to me like beam 5 needs to be longer. The weight can't break it at the center so it breaks it at the end since thats where the opposing force is
@BumBleBee-em9tp
@BumBleBee-em9tp 2 ай бұрын
is the BC's BMD correct
@areskei
@areskei 2 ай бұрын
hi, may I ask if how did you get the -7.5 and the displacement 0.625 ?
@Snowbird4164
@Snowbird4164 2 ай бұрын
So the Ax and Cx reactions would dictate the required thrust force to be resisted. Would you do an example where a collar tie is installed and no ceiling tie as in many wood framed homes?
@StructuresProfH
@StructuresProfH 2 ай бұрын
Yes, the Ax and Cx reactions are the thrust forces. Normally these could be resisted by a ceiling or rafter tie. A collar tie near the peak of roof does not effectively reduce these thrust forces. I believe the primary purpose for collar ties is to better secure the rafters to the ridge beam. Anyway, if you don't have a ceiling tie, that thrust force needs to be resisted by the walls that support the roof. The other option is to have a structural ridge beam. In structural analysis terms, that means adding a third vertical support at the roof peak - this effectively eliminates the horizontal thrust forces because the truss is no longer being "squashed" out by the vertical load.
@darcyschmidt2239
@darcyschmidt2239 2 ай бұрын
Does the negative/positive component of the M diagram become relevant if some parts are negative and some are positive? In the second example, even though it was negative you didn't include it?
@StructuresProfH
@StructuresProfH 2 ай бұрын
Yes, the moment diagram sign is relevant if some is negative and some is positive. The sign gives you the direction of curvature (positive opens up or "smiles" while negative opens down or "frowns"). It is important to keep this consistent in general when taking the moment of the area. That said, if the sign is all positive or all negative, I usually handle the signs intuitively (some might say lazily!) for the moment area method just by considering the deflected shape geometry. The negative sign in the second example is reflected by the fact that the deflected shape is below the tangent, whereas the positive sign in the first example is reflected by the fact that the deflected shape is above the tangent.
@darcyschmidt2239
@darcyschmidt2239 2 ай бұрын
@@StructuresProfH Awesome, did some examples and its making much more sense.
@GitauKimani
@GitauKimani 2 ай бұрын
Excellent Video!!
@israelcrawford8734
@israelcrawford8734 2 ай бұрын
Hello Sir, can you use the reinforcement connect a post-tension tendon in the concrete beam?
@ioannaathineos5269
@ioannaathineos5269 2 ай бұрын
THANKS!!!!YOU saved my staying in collage
@StructuresProfH
@StructuresProfH 2 ай бұрын
Happy to help!
@WG-ft6tz
@WG-ft6tz 2 ай бұрын
Thanks for the video. Is it correct to state that the only reason the reactions have an X component is due to the hinge at the peak of the roof?
@StructuresProfH
@StructuresProfH 2 ай бұрын
The hinge certainly doesn't help, but you will still get horizontal (X-component) reactions even without the hinge. As the load presses down on the structure, it tends to "flatten out" which will be resisted as horizontal reactions. The hinge means it tends to "flatten" more under vertical load than an equivalent system without a hinge, but the phenomena is present for either case.
@WG-ft6tz
@WG-ft6tz 2 ай бұрын
Why is it then with truss analysis vertical loads never transfer lateral force to the roller/pin? Due to a 3 force member in equilibrium? And in the roof case, the bottom chord would resist the tension?
@StructuresProfH
@StructuresProfH 2 ай бұрын
@@WG-ft6tz For truss analysis, we typically assume that one support is a pin and the other is a roller (meaning, it allows for horizontal movement). In this configuration, any horizontal force is resisted by tension in the bottom chord. However, if both supports are pins (meaning both resist horizontal motion), then you will once again get horizontal reactions even if you have a bottom chord. This is because the deformation of the truss will push these two supports outward, which is resisted by inward horizontal reactions.
@WG-ft6tz
@WG-ft6tz 2 ай бұрын
Ah thank you! So used to analyzing pin/roller trusses in examples. Subscribed!
@Mujungu
@Mujungu 2 ай бұрын
hello professor, is it possible to carryout numerical modelling of crack propagation using ANSYS
@user-ug2ec5cp3b
@user-ug2ec5cp3b 2 ай бұрын
Thank you very much. I like the video and I understand your accent very well.
@alfredjrtovar1499
@alfredjrtovar1499 2 ай бұрын
This video should be shown to every prospective engineering student, guarentee people will go a different route.
@alfredjrtovar1499
@alfredjrtovar1499 2 ай бұрын
Im officallly changing my major, this shit is extremely boring!!!!! Math is fun, but this is just plain cruel