Пікірлер

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  @proticchobi960215 сағат бұрын

  Well explained. Just a small error, the solve doesn't occur at nodes it occurs at a point inside the element which is called integration point and then its extrapolated to the node.

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  @StructuresProfH11 сағат бұрын

  First, thanks for watching! It is a bit more nuanced than just saying everything happens at a node or integration point (I was trying to keep it simple). In the context of elasticity, the typical finite element formulation constructs the solution from nodal displacements that are interpolated over an element using shape functions. Integration points are used with these shape functions for the construction of the stiffness matrix, essentially turning a kind of energy minimization problem over a volume (requiring integration over that volume) into a set of algebraic equations. Once we reduce the problem to a system of equations in the form of Stiffness*Displacement = Force, we solve for the displacements, and these quantities reflect nodal displacements. However, the strains and stresses that we derive from these displacements (because they come from derivatives of displacement which depends on the particular choice of shape function) are NOT nodal values. This means it is correct to say, as you have done, that the stress and strain solutions are evaluated at the integration points and then extrapolated to the nodes, but at the same time the displacements themselves are nodal quantities that are interpolated throughout the element.

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  @abuibrahim69122 күн бұрын

  I hope if it's in SI unites 😢

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  @StructuresProfHКүн бұрын

  I can help with that. The equations are all the same for SI units except for the following exceptions. Here are the conversions from psi to MPa and inches to millimeters, with the video timestamps included... Converting equations for Vc in Section 22.5.5 from psi to MPa (2:44) - change the 2 coefficient to 0.17 and change the 8 coefficient to 0.66 Converting Av,min in Section 9.6.3.4 from psi to MPa (2:44) - change the 0.75 coefficient to a 0.062 and change the 50 coefficient to 0.35 Converting size factor lambda,s from Section 22.5.5.1.3 from inches to millimeters (5:20) - equation becomes sqrt(2/(1+0.004d)) All the steel strength equations are the same - obviously you'll have the stirrup Av in square millimeters and the steel strength fy in MPa. It's a bit more work to convert the examples, but the process is identical except with the small equations changes above. If you want to covert my numbers... 1 MPa = 145 psi (so the 5000 psi concrete is 35 MPa), 25.4 mm = 1 inch (so the 24-in deep beam is 610-mm deep), and 4.45 Newtons = 1 pound or 4.45 kN = 1 kip (so the Example 1 design capacity of 24 kips is 107 kN). Those are the only US to SI conversions you'll basically ever need for any strength calculations. Rebar are specified differently, but a #3 rebar is very close to a 10M (10-mm diameter) bar and a #8 bar is very close to a 25M (25-mm diameter) bar.

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  @abuibrahim691212 сағат бұрын

  @StructuresProfH thank you

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  @siddharthtyagi21232 күн бұрын

  If your rebar’s are not. Extending till the end you are stupid

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  @Aishu-p4r4 күн бұрын

  Informative demonstration!!

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  @ytrew97174 күн бұрын

  Asked Claude 3.5 : Let me explain why rebars are often placed at the bottom of beams: Tension zone: When a beam is loaded from the top, the bottom of the beam experiences tension. Concrete is strong in compression but weak in tension. Steel rebars are added to the bottom to handle these tensile forces. Bending moment: The maximum bending moment in a simply supported beam typically occurs at the middle of the span. This creates the greatest tension at the bottom of the beam in this area. Compression zone: The top of the beam is typically in compression when loaded from above. Concrete is already strong in compression, so it often doesn't need as much reinforcement there. Testing to failure: In tests, placing the rebar at the bottom allows researchers to observe how the beam behaves under typical loading conditions. They can see how the concrete cracks and how the steel reinforcement performs. However, it's important to note that in real-world applications, beams often do have some reinforcement at the top as well, especially for: Negative bending moments over supports in continuous beams Shrinkage and temperature control Torsional resistance Compression reinforcement in some designs

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  @ishniff33474 күн бұрын

  THANKS!

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  @chidubemoleka87296 күн бұрын

  I have some problems, I don't know if you could help me

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  @chidubemoleka87296 күн бұрын

  Some structural engineering problems that I need the solution

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  @QuickQuestionEngineering11 күн бұрын

  Great explanation! Thank you for the video!

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  @kianzedkay12 күн бұрын

  thanks a lot man u saved me with this video!!

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  @user-sn9lw2sz2g12 күн бұрын

  ❤❤❤

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  @danvining218613 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?

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  @jackyeh751219 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.

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  @bunnuakhil292020 күн бұрын

  well demonstrated. Enjoyed it.

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  @RK-tx5xb24 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.

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  @davidtettleton273225 күн бұрын

  How do your beams compare to wood, the second beam for instance?

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  @HermelaMerishet27 күн бұрын

  Thank you

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  @WG-ft6tzАй 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.

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  @StructuresProfH29 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: kzread.info/dash/bejne/nH1q2dmjktyffdY.html

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  @sothattheng8624Ай бұрын

  Thanks for your sharing Prof.H!

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  @chichha420Ай бұрын

  Thanks sir from germany. I would be grateful to you if fürther videos of approximate analysis of frames are uploaded.

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  @bekehbenjamin775Ай бұрын

  Does it mean that when ever your calculations gives you zero as an answer, it shows that the trusses is determinate?

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  @StructuresProfHАй бұрын

  Yep, zero means determinate (unless there is some other issue that causes instability).

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  @user-ui3jp4pq6gАй 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

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  @ferenczbabinszki7699Ай бұрын

  i thought you're supposed to waith at least 28 days for proper cure?

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  @StructuresProfHАй 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.

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  @marcoromanelli6000Ай бұрын

  And now, a video about real buildings where concrete beams are loaded that way...

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  @cleisonarmandomanriqueagui9176Ай бұрын

  best youtube channel for civil indsutry ever . experience is better than theory

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  @cminor1477Ай бұрын

  This is the best video I have ever seen

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  @MohammedMinhajUlHaqАй бұрын

  Excellent understanding after watching this lecture

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  @PremKumar-ky7glАй бұрын

  Thank you very much sir, this is a great work. Thanks alot

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  @RK-tx5xbАй 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?

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  @zhilyaismael8435Ай бұрын

  thank you dear prof . could you please cane you give me some idea how it's come in in exam?

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  @zhilyaismael8435Ай бұрын

  🥰🥰🥰

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  @tsegazeabdejene6685Ай бұрын

  Great explanation

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  @patrikengas6479Ай 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?

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  @StructuresProfHАй 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).

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  @joealex24632 ай бұрын

  how do you find r?

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  @StructuresProfHАй 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.

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  @joealex2463Ай бұрын

  @@StructuresProfH thank you!

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  @Sarissa992 ай бұрын

  What if I have a curved (semi-circular in my instance) frame? How do I implement NQM there?

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  @StructuresProfHАй 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.

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  @Havoner2 ай бұрын

  Dear Professor! In the "Model type" option, if I select "Cable", will it be a compressed structure? 12:51

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  @StructuresProfHАй 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.

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  @HavonerАй бұрын

  @@StructuresProfH Thanks Dear Professor Could you create such a 3D video on KZread? What material do I need to define for the cables? I must set Link/Truss option?

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  @fadi7472 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.

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  @fadi7472 ай бұрын

  How did you compute Av min (at 8:35 of this video).

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  @ymsh22 ай бұрын

  Thank you for uproading this video.

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  @benonp36222 ай бұрын

  in mesh, there's no squares visible. how?

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  @Havoner2 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

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  @nisarahmadriaz31172 ай бұрын

  Brilliant dear sir!

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  @burhanuddinmotiwala14282 ай бұрын

  for the last example if there were no hinges in the frame then would it be indeterminate to the fourth degree?

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  @StructuresProfH2 ай бұрын

  That is correct!

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  @Ans_aka_Inzi10732 ай бұрын

  That's an amazing lecture! Everything was explained with great proficiency.

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  @skablerpro2 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

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  @BumBleBee-em9tp2 ай бұрын

  is the BC's BMD correct

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  @areskei2 ай бұрын

  hi, may I ask if how did you get the -7.5 and the displacement 0.625 ?

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  @Snowbird41642 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?

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  @StructuresProfH2 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.

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  @darcyschmidt22392 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?

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  @StructuresProfH2 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.

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  @darcyschmidt22392 ай бұрын

  @@StructuresProfH Awesome, did some examples and its making much more sense.

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  @GitauKimani2 ай бұрын

  Excellent Video!!

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  @israelcrawford87342 ай бұрын

  Hello Sir, can you use the reinforcement connect a post-tension tendon in the concrete beam?