TAGGED: apdl, cross-section
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April 25, 2023 at 12:39 pm
User196534
SubscriberHello to all,
I am trying to define an ellipsoidal section of a beam (BEAM 188) via the ASEC function.
Ansys documentation shows the following data to be entered in SECDATA:
A, Iyy, Iyz, Izz, Iw, J, CGy, CGz, SHy, SHz, TKz, TKy
where
A = Area of the section
Iyy = Moment of inertia about the y axis
Iyz = Product of inertia
Izz = Moment of inertia about the z axis
Iw = Deformation constant
J = Torsion constant
CGy = y coordinate of the centroid
CGz = z coordinate of the centroid
SHy = y coordinate of the shear center
SHz = z coordinate of the shear center
TKz = Thickness along the Z axis (maximum height)
TKy = Thickness along the Y axis (maximum width)
My ellipse has a dimension of 60 cm along the y axis and 30 cm along the x axis. Starting from the centroid, I have TKy = 30 cm and TKz = 15 cm.
Here is the data I get by entering (A, Iyy, Izz, J, TKz, TKy):
Area = 1360.5
Iyy = 77939.
Iyz = 0.0000
Izz = 0.29275E+06
Warping Constant = 0.0000
Torsion Constant = 0.58550E+06
Centroid Y = 0.0000
Centroid Z = 0.0000
Shear Center Y = 0.0000
Shear Center Z = 0.0000
Shear Correction-xy = 1.0000
Shear Correction-yz = 0.0000
Shear Correction-xz = 1.0000
Maximum height = 15
Maximum width = 30
My question is the following.
No matter what value I enter for TKz and TKy, the displacement of my beam is always the same. I tried with zero values (TKz = 0 cm, TKy = 0 cm) and extreme values (TKz = 100,000 cm, TKy = 100,000 cm).
Do you know what is the impact of TKz and TKy and if they correspond to the values I entered?
Here is a piece of my Apdl code if it helps:
SECTYPE,1, BEAM, ASEC,di, 0
SECOFFSET, CENT
SECDATA, A_di, Iy_di, , Iz_di, , J_di, , , ,TKz ,TKy, ,
Thanks to those who will take the time to answer, I hope I have been clear enough in my explanations.
Do not hesitate to ask me questions if this is not clear enough.
Have a good day ! -
April 26, 2023 at 12:14 pm
Ashish Khemka
Ansys EmployeeHi,
How are you applying the load? Are you applying a force load or a displacement load?
Regards,
Ashish Khemka
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April 26, 2023 at 12:16 pm
User196534
SubscriberHi,
I am applying a force load.
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