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Question: The role of the shear center when calculate the blade torsion? #2100

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winder365 opened this issue Mar 16, 2024 · 5 comments
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@winder365
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winder365 commented Mar 16, 2024

Dear @jjonkman,

I have read a post on the forum about "Aerodynamic center offset effects on blade torsion" (https://forums.nrel.gov/t/aerodynamic-center-offset-effects-on-blade-torsion/4841), and I have known that the offset between the aerodynamic center and pitch axis center will affect the moment arm and the torsional deformation of blade.

However, some questions still confuse me:
(1) The pitch axis is actually a virtual axis, why it will have effect on the torsional deformation of the blade? For some special case, such as backward swept blade, the pitch axis is actually far away from the aerodynamic center.
(2) What is the role of the shear center (of the blade section) when calculate the blade deflection (especially the torsion deformation). Will the pitch moment (from the aerodynamic force crossed with the moment arm as well as the pitching moment (Cm) term) eventually act on the shear center when calculate the blade torsional deformation in BeamDyn?
(3) Does the offset between the aerodynamic center (or pitch axis center) and the shear center affect the calculation of the torsional deformation in BeamDyn?

Thank for your attention

@jjonkman
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Dear @winder365,

Here are my responses:

  1. I'm not sure what you mean be "virtual axis", but the pitch axis is the axis about which the blade pitches about. At the cylindrical root of a blade that pitches, all of the centers (shear, mass, tension, aerodynamic) lie along the pitch axis. And the aerodynamic center in AeroDyn is simply the reference point for the aerodynamic load calculation, i.e., the point where the aerodynamic forces (lift, drag) are calculated and applied and the reference point about which the aerodynamic pitching moment is calculated and applied about.
  2. The shear center is the point in the cross section where a shear force will not induce torsion. So, if the reaction forces are applied at the shear center, the cross section will not twist. But it is rare for the shear center to be at the same location along the blade. So, even if the aerodynamic center is at the shear center at given cross section, that is likely not the case along the entire blade, and so, both aerodynamic forces and the pitching moment can induce blade torsion.
  3. Yes, it is the offset between the aerodynamic center and shear center that can induce torsion, as well as the aerodynamic pitching moment.

Best regards,

@winder365
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Dear @jjonkman ,

Thank for your reply.

(1) As your mentioned, the shear center is the point in the cross section where a shear force will not induce torsion. So, if the reaction forces are applied at the shear center, the cross section will not twist. The force you mentioned is in narrow sense or in broad sense? Namely, is it the force like the lift and drag, or it also include the moment? My understand is that, if the force such as lift and drag are applied on the shear center, it will not result in torsion. However, if the moment such as the pitch moment is applied on the shear center, it will result in torsion. Right?

(2) As your mentioned, the offset between the aerodynamic center and shear center can induce torsion. Besides in the post I mentioned above, the offset between the aerodynamic center and pitch axis can also induce torsion. This confused me a lot. As mentioned in the post, the extra moment caused by the aerodynamic force cross with the offset is accounted for within the internal spatial mesh-to-mesh mapping algorithm used by OpenFAST, such as indicate in the following code:

DO j = 1,2 ! number of nodes on dest 

   jNode = Dest%ElemTable(ELEMENT_LINE2)%Elements(jElem)%ElemNodes(j)   
   DisplacedPosition =   Src%Position(:,i)     +  SrcDisp%TranslationDisp(:,i)     &
                      - Dest%Position(:,jNode) - DestDisp%TranslationDisp(:,jNode)  
                  
   torque = Src%Force(:,i)
   torque = CROSS_PRODUCT( DisplacedPosition, torque )       
        
   Dest%Moment(:,jNode) = Dest%Moment(:,jNode) + torque*MeshMap%MapLoads(i)%shape_fn(j)

END DO

What the var DisplacedPosition includes? As I understand, it includes the offset between the aerodynamic center and pitch axis (namely the BlSwpAC ), and includes the component due to the blade deformation (especially the edgewise deflection). Right? Besides, what the var torque includes, does it only include the the aerodynamic force crossed with the moment arm between the AC and pitch axis, as well as the pitching moment (Cm) term.

(3) I am not very clear about the effect of so many offsets on torsion (such as the offset between the aerodynamic center and shear center, the offset between the aerodynamic center and pitch center, and the offset between the shear center). As your mentioned, the offset between the aerodynamic center and shear center can induce torsion. What about the offset between the aerodynamic center and pitch center, and the offset between the shear center? Do these offsets affect torsion and how these offsets affect torsion?

Look forward to your direction.

@jjonkman
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Dear @winder365,

Here are my responses:

  1. I agree.
  2. The code snippet you show indicates how offsets between nodes in mesh-mapping mean that forces in a source mesh will generate moments in a destination mesh associated with the offset. The offset includes contributions from the reference positions and displacement of the nodes. A moment applied at the source mesh will transfer directly to a destination mesh. See our papers that discuss mesh mapping for more information: https://www.nrel.gov/docs/fy14osti/60742.pdf and
    https://dx.doi.org/10.2514/6.2015-1461.
  3. As I mentioned before, all centers generally lie along the pitch axis at the root of a blade with active pitching. However, it is not likely that all centers (including shear and aerodynamic center) lie along the pitch axis across the entire blade. Thus, even if the aerodynamic and shear centers are aligned at each cross section (not likely), there is typically an offset relative between stations such that lift and drag forces will generate moments regardless.

Best regards,

@andrew-platt andrew-platt changed the title Feature request -- The role of the shear center when calculate the blade torsion? Question: The role of the shear center when calculate the blade torsion? Mar 26, 2024
@winder365
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Dear @jjonkman ,

Thank you for your reply. I still have two question:

(1) As we discussed above, shear center is important in blade deformation calculation, but how can I get the position of the shear center along the blade through the 6*6 stiffness matrix? just like this:
image

(2)How the BeamDyn code consider the effect of the shear center position or shear center offset? Will the BeamDyn code itself calculate the position of the shear center when make simulation?

@jjonkman
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jjonkman commented Apr 1, 2024

Dear @winder365,

Regarding (1), the shear center can be inferred from the shear and torsion elements of the 6x6 cross-sectional stiffness matrix. My forum post dated July 20, 2020 contains an attachment, BeamDynInput_05_2020.pdf, that explains how the 6x6 stiffness matrix would change for different reference axes. To see how the shear center effects the shear and torsion elements of the stiffness matrix, see Section 2.3.

Regarding (2), the BeamDyn code makes use of the full 6x6 cross-sectional mass and stiffness matrices, and so, does not need to calculate and make use of the shear center directly.

Best regards,

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