BEAM188 is a linear 2-noded beam element whereas BEAM189 is a quadratic 3-noded beam element. To obtain BEAM189 element, you can insert mesh method and keep the Element midside nodes turned on. Alternatively you can right click on geometry and add the command 'et,matid,beam189' to apply BEAM189 element.

Thank you so much sir Shlok K Laddha for your answer. After adding the command ''et,matid,beam189'' below the geometry, the Solution Output showed that Beam189 was the element type being used.

Sir, I need to know if i should also change the element order in the mesh settings from program controlled into quadratic or not?

I don't think there is a need to change the element order to quadratic from program controlled after having changed the element type to BEAM189 as it is a 3-noded quadratic element. But in case you do run into some problem, try changing it to quadratic and keep element midside nodes.

sir Shlok K Laddha , i tried to add only the command ''et,matid,beam189'' and keeping the element order at the program controlled but the result does not agree with the analytical ones. When i changed also the element order to quadratic, the results were better.

For beam problems when element order is set to program controlled it creates linear elements of BEAM188, maybe that is why the results didn't agree with analytical result. Maybe the command only cannot force the BEAM189 element on the geometry and that is why you have to change to quadratic order. I am glad you are getting better results now.

Best wishes to you & Good luck !

Thank you so much sir Shlok K Laddha, really your answers were so usefull. Can i ask you sir about more things related to changing the beam elements types in ANSYS?

Yes, sure.

Thank you so much Sir.

This time i need to change the beam theory: Usually in my simulation i use the beam188 or beam189 and i've found in the web that this element type is based on the Timoshenko beam theory. I would like to know how to change Timoshenko to Euler-Bernoulli since in the most of cases i need to compare my result with analytical one which are derived using Euler Bernoulli's theory.

I've read in some articles that this is possible by using beam4 instead of beam188 but when I add "et,matid,beam4" under the geometry , the solution can't be achieved.

Thank you in advance sir.

Dear Hajar Rhylane

BEAM188 gives section control via the SECCONTROL command for defining the transverse shear stiffness and added mass and it doesn't take into account any real constant data. In case of BEAM4 along with 'et,matid,beam4' you also need to add the real constant command 'r, set number, Area, Izz, Iyy, thickness in z, thickness in y, theta'. You can get the whole real constant list for BEAM4 from ANSYS files.

Really thank you so much sir Shlok K Laddha for your answers and for your help. but i have a query regarding the assignation of the real constant and specially the Torsional constant for a non circular cross-section (rectangular cross section in my case).

In the Ansys help, i found that for the special case of a circular cross-sections, the torsionnal constant is equal to the polar moment of inertia and calculated by using this formula: Ixx=J=Iyy+Izz. Could you please if it is possible to clarify for me as to what formulation Ansys follows for the calculation of the Torsional constant of a non circular cross section.

Thank you so much sir!

Dear Hajar Rhylane ,

The torsional stiffness for circular cross section is indeed equal to the polar moment of inertia as the cross section is axisymmetric. But as the non-circular cross sections are not axisymmetric their cross section would bulge or warp when the shaft is twisted. If no value is entered for Ixx then ANSYS will compute the torsional stiffness (Ixx) as 'Iyy+Izz' which is correct for circular cross section but not for non-circular cross sections.

The expressions for allowable twist angle, allowable shear stress and torsional stiffness are available in literature. You can input this value in ANSYS as Ixx.

Best Regards,

Shlok

Thank you so much sir Shlok K Laddha for all the help and for your important answers.