UNSW - Aerospace Structures - Aeroelasticity

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  • เผยแพร่เมื่อ 17 มิ.ย. 2014
  • • Definition of Aeroelasticity
    • Range of Aeroelastic effects
    • Static Aeroelasticity
    ○ Load redistribution
    ○ Divergence
    ○ Control Reversal
    • Dynamic Aeroelasticity
    ○ Flutter
    § 1D flutter (control surface example)
    § 2D flutter (wing example)

ความคิดเห็น • 27

  • @SuryaSarav_
    @SuryaSarav_ 3 ปีที่แล้ว +1

    This is truly amazing. Learning as a beginner, this is great. Thanks for the lecture.

  • @eriklan4014
    @eriklan4014 3 ปีที่แล้ว +5

    I felt like I was out of my body as the bridge music continued to be played. 1:10:59

  • @rodrigom7686
    @rodrigom7686 6 ปีที่แล้ว

    I learn alot from this video. Ty for upload it.

  • @frederickmannings8700
    @frederickmannings8700 7 ปีที่แล้ว +1

    Awesome lecture

  • @jeremygunetilleke7672
    @jeremygunetilleke7672 9 ปีที่แล้ว +6

    Hi, these are brilliant videos! I am currently teaching myself about aircraft stresses and loads. These are great explanations and help when used alongside textbooks. Is there any way that we can access the rest of the lecture videos? Many thanks

  • @d.joseph4356
    @d.joseph4356 9 ปีที่แล้ว

    Good information. Thanks.

  • @arvindyadav-jt8jq
    @arvindyadav-jt8jq 4 ปีที่แล้ว

    Very helpful lectures

  • @haidem5179
    @haidem5179 5 ปีที่แล้ว +2

    at 10:32, I'd strongly suggest to also take into account the increase in induced drag if the lift distribution is shifted outward due to aeroelastic effects. Since drag is (usually) the most important thing you want to avoid, this could be a more intuitive answer for students. The root bending moment comes much later.

    • @AerospaceStructures
      @AerospaceStructures  5 ปีที่แล้ว +6

      Great point Haidem. I'll absolutely include that when I teach the course again next year.

    • @haidem5179
      @haidem5179 5 ปีที่แล้ว +4

      Aeroelasticity is certainly one of the toughest lectures to hold....you are doing a great Job! Keep it up

  • @lucasmathijssen1116
    @lucasmathijssen1116 3 ปีที่แล้ว +3

    I'm trying to understand flutter a bit better, and I'm having some doubts about this statement in the video: can the failure of Tacoma Narrow's Bridge really be attributed to flutter (as discussed at 1:08:38)? Of course the crosswind on the bridge is not oscillating from left to right at the structures eigenfrequency. However, as far as I was aware, the original design of the bridge didn't allow air to pass through the bridge, forcing the flow to stall and move around the body. Therefore, the flow around the bridge started behaving as that around a bluff body, leading to vortex shedding and a von Kármán street. This vortex shedding is periodic, such that the pressure distribution around the bridge is also periodically changed. If this shedding happens at the structures eigenfrequency, the loads on the bridge can definately lead to resonance. This periodic loading is independent of the bridge's own movement, as the crowsswind on the bridge is still present, and up and down movement of the bridge won't change that. This means the aerodynamic forcing of the bridge is not coupled to the displacement of the bridge, as would be required for it to be called aeroelastic flutter: it is not the up and down movement of this bridge interacting with the flow that creates the periodic forcing, its the vortex shedding that creates it. Hence, due to the periodic forcing of the bridge by the vortex shedding, I think this should be classified as resonance, and not as flutter. Correct?
    Edit: Just investigated a bit more and it seems the failure can be attributed to flutter! It should be noted that the bridge did experience resonance in a bending mode due to vortex shedding. However, as can be seen in footage of the bridge's failure, the bridge failed in a combined bending and torsion mode. The torsional oscillations were created due to vortices as well, but these vortices could only be created due to the I-beam shape of the bridge and motion of the bridge itself, making the interaction a coupled interaction, and thus flutter! See this footage for a nice (and short) explanation: th-cam.com/video/mXTSnZgrfxM/w-d-xo.html.

    • @vallurusailakshminrusimhap2873
      @vallurusailakshminrusimhap2873 3 ปีที่แล้ว

      @@lucasmathijssen1116 So are you stating that any coupled mode vibration can only be classfied as flutter?

    • @vallurusailakshminrusimhap2873
      @vallurusailakshminrusimhap2873 3 ปีที่แล้ว

      Just trying to understand the difference between flutter and vortex induced vibrations. Vortex induced vibrations are caused due to flow separation and ther by creating a periodic shedding, where as the flutter happens even for an attached flow. Am I right?

    • @lucasmathijssen1116
      @lucasmathijssen1116 3 ปีที่แล้ว +1

      ​@@vallurusailakshminrusimhap2873 Hi Valluru, to answer your question, I think first of all we have to be careful when talking about 'coupled modes'. I assume you are talking about a coupling between the airstream and the structure, not a coupling between for example a bending and a torsion mode. This can be confusing, because classic wing flutter is basically caused by a 'coupling'/combination/coalescence of a bending and torsion mode interacting with the flow, which reduced the damping coefficient of the combined mode considerably such that it becomes unstable.
      So, if we are talking about coupling in the sense of an airflow and structure interacting, I think that indeed the flutter instability is always a coupling between the flow and the structure, but only in such a way that the vibration is self-excited. In the bridge example, it was the oscillation of the bridge itself which allowed the creation of vortices on the deck (not the shedded vortices as explained in that video I linked), hence it was self-excited. If the bridge was not rotating in a torsion mode, these vortices could not have existed, and the system would not be excited any further. (Increasing the torsional stiffness of the bridge would have solved that problem, the bridge was too flexible).
      However, I think your answer can be found by looking at Callar's diagram. According to Collar's triangle, a dynamic aeroelastic problem like flutter should always be the result of a 'coupling' between aerodynamic, elastic and inertial forces. However, flutter is not the only dynamic aeroelastic problem. Others are for example 'buffeting' and 'dynamic response'. Therefore, not every coupled mode vibration can be classified as flutter. Flutter has that coupling, but not all interactions with that coupling can be called flutter.
      So, then the question remains, what is this vortex induced vibration then? If I may quote Wikipedia: "vortex shedding is an oscillating flow that takes place when a fluid such as air or water flows past a bluff (as opposed to streamlined) body at certain velocities". In the example of the bridge, first the flow separation made the cross-section of the bridge behave like a bluff body, and then secondly, the flow around the bluff body created the vortex shedding. For it to happen, it doesn't matter if the flow is (locally) separated or attached, as long as the object experienced by the flow is bluff. In addition, vortex shedding itself doesn't need the movement of a structure for it to occur. It does not involve any elastic or inertial forces and is therefore, at least according to Collar's triangle, only an Aerodynamic effect/vibration, and not an aeroelastic effect. It is not self-induced, and therefore not flutter.
      If you were to solve for the flutter problem mathematically, you would be forced to solve an equation where this coupling takes place. Basically such that the aerodynamic forcing is dependent on(/a function of) the shape of the structure. For the vortex shedding that is not the case. Instead, for the vortex induced vibrations, you could do so simply by moddelling them as a periodic forcing term at the frequency of vortex shedding in your equations of motion/ lagrange equation, which would allow you to solve a (much simpler!) vibrations problem.

  • @lucao240
    @lucao240 5 ปีที่แล้ว +1

    Amazing lecture!
    I have a doubt: How can I find the location of the twist center and the stiffness of the airfoil?

    • @FlynnCz
      @FlynnCz 4 ปีที่แล้ว +1

      Any wing is deformable so if you apply a force (not in the shear center) or a twisting moment in some point, the structure will twist and bend. If you calculate the static displacement relative to the wing root as function of an arbitrary twisting moment you will get something like S = M/K where S is the (relative) twist angle, M the moment and K the equivalent stiffness. To do so you need to study the theory of beam.
      The shear center is the point where if you apply a force the section will not twist or, better, the conjugated internal work between shear strains and twist stresses and viceversa is 0. Again, you need to study the beam's theory in order to calculate it.

  • @patrickamstad5091
    @patrickamstad5091 5 ปีที่แล้ว +1

    @ 16:50 : In the lecture it is said, that we picked our moments about the aerodynamic center, and therefore M_0 is not a function of the angel of attack. Is that true? If we actually picked the aerodynamic center for our moment balance, then the lift force wouldn't have a contribution? Or where is my fault? I think that the moment M_0 isn't a function of the angle of attack, because we are considering the static case....
    Thanks for the great lecture

    • @AerospaceStructures
      @AerospaceStructures  5 ปีที่แล้ว

      I will have to re-watch the video to know exactly what I said, but in principle I think I can describe the problem. We define the aerodynamic centre (AC) as the location at which we apply lift, drag and moment, precisely because it has the property that moments do not change with angle of attack (AoA). Therefore the AC is independent of AoA.
      The AC isn't the same as the centre of pressure (Cp), which is the location of the resultant lift and drag of the aerodynamic surface pressures. The Cp varies with AoA, so isn't a useful point to do structural calculations.
      Once we know that the moment about the AC (M_0) is constant, we can then take our moment centre to be anywhere (in this case the elastic axis). M_0 will have the same contribution to our moment equation, regardless of the location that we take moments about.

  • @TennisGvy
    @TennisGvy 8 ปีที่แล้ว

    at 1:44:51 why do the restorative spring force and the lift force have the same moment arm? Isn't the spring force b*x_alpha away and the lift force even farther away?

  • @NASHThoughts
    @NASHThoughts 3 ปีที่แล้ว

    In this video...there is an calculation is missing at the end..its not covered in camera.....its in 2:04:11

  • @Desrayyy_
    @Desrayyy_ 9 หลายเดือนก่อน

    Could you make a pdf of the equations needed for aerospace structures and materials in the most simplified way please ❤

  • @mechtrotip5034
    @mechtrotip5034 4 ปีที่แล้ว

    Which book are you referring to?

  • @velos26b
    @velos26b 7 ปีที่แล้ว +1

    Please could you tell me about the title of the book you are using? Thank you so much!.

    • @adu3422
      @adu3422 7 ปีที่แล้ว

      Aircraft Structures for engineering students by Megson

    • @velos26b
      @velos26b 7 ปีที่แล้ว +1

      thanks a lot!

  • @DGG.85
    @DGG.85 ปีที่แล้ว

    Forgive me for being a naive MechE but why are you using h for displacement?