I studied planar 10-bar Cantilever truss ( buckling analysis), but I didn't understand the results of the ANSYS apdl software well:
/TITLE, Planar 10-Bar Cantilever Truss
/FILENAME,tenbar/UNITS,SI ! unites en systeme international
/PREP7 ! entree en mode pre-traitement!*
!**************scalar parameters*******************!
E=68.95E9 ! module d'Young en N/m2
Poisson=0.334 ! Coefficient de Poisson du materiau
A1=0.0129032 ! Section du materiau en m2
A2=0.0129032
A3=0.0129032
ET,1,LINK1 ! element fini de type LINK1
R,1,A1 ! constantes reelles du materiau 1:
R,2,A2
R,3,A3
MP,EX,1,E ! module d'elasticite
MP,PRXY,1,Poisson ! coefficient de Poisson
MP,DENS,1,2768 ! Aluminum density, kg/m3
! Geometrie du 10-Bars: numero et coordonnees des noeuds
N,1,18.28,9.14,0
N,2,18.28,0,0
N,3,9.14,9.14,0
N,4,9.14,0,0
N,5,0,9.14,0
N,6,0,0,0
! Creation des elements constitutifs du structure
real ,1
EN,2,1,3
EN,5,3,4
EN,6,1,2
EN,10,1,4
real ,2
EN,4,2,4
EN,8,3,6
EN,9,2,3
real ,3
EN,1,3,5
EN,7,4,5
EN,3,4,6
! Conditions aux limites en deplacement
D,5,ALL,0
D,6,ALL,0! Forces appliquees au treillis
F,2,FY,-444800 ! application de la composante de F suivant y sur le noeud 2
F,4,FY,-444800 ! application de la composante de F suivant y sur le noeud 4
!acel,0,9.8,0
FINISH ! fin du mode pre-traitement
!************DEFINING ARRAYS FOR OUTPUT*****************!
*GET,BL,ACTIVE,0,SET,FREQ
*DIM,NDISPX ,ARRAY,NCOUNT,1
*GET,NCOUNT,NODE,,COUNT
*DIM,NDISPX ,ARRAY,NCOUNT,1
*DIM,NDISPY ,ARRAY,NCOUNT,1
*cfopen,results10bars1b,txt
!***************SOLUTION PHASE**************************!
/SOLU ! mode solution
ANTYPE,STATIC ! analyse statique
PSTRES,ON
SOLVE ! demarrage de la resolution
SAVE ! sauvegarde de la base de donnees
FINISH
/SOLU
ANTYPE,BUCKLE
BUCOPT,SUBSP
OUTPR,ALL,ALL
SAVESOLVE
FINISH
/SOLU
EXPASS,ON
MXPAND, 1
OUTRES,ALL,ALL
SOLVE
FINISH
!*************POST PROCESSING PHASE*********************!
/POST1 ! entree en mode post-traitement
*GET,BL,ACTIVE,0,SET,FREQ
/OUTPUT,tenbarBuckling,txt,C:\Users\pc\ANSYS
*VWRITE,BL(F24.5)
/OUTPUT,,,
FINISH
What exactly represents the value of the LOAD MULTIPLIER = -572.72670
and VECTOR DOFS in this analysis?
Is Ansys code correct?
First of all, You are doing linear buckling analysis for the attached truss.For the first sight, your code seems true. The value -572.72670 represents the minimum load required to buckle the truss. Regarding DOF in your analysis is 2 (translation in both X&Y) and that because you chose link1 as a meshing element.
Dear Mr Kareem N Salloomi,
Thanks for your answer
My experience in static analysis
What do I do to prove that the system is stable?
Displacement limits +- 0.0508 m
ALL VECTOR DOFS ARE IN NODAL COORDINATE SYSTEMS.
for buckling analysis (mode1)
NODE UX UY
1 -0.261024 0.994484
2 0.622163 1.00000
3 -0.204640 0.117440
4 0.354446 0.292473
5 0.00000 0.00000
6 0.00000 0.00000
I do not think the results mean that the displacement have exceeded the limit?
Dear Claudio Pedrazzi
Thanks for your answer
I have seen the Chapter 7: Buckling Analysis
https://www.sharcnet.ca/Software/Ansys/16.2.3/en-us/help/ans_str/Hlp_G_STR7.html
Your explanation was useful and important to me
I now understand the meaning of load multiplier in results
Are there other important outputs to study this problem?
I want to study the optimization of structures with discrete variables
I am now looking at the constraint functions (stress, displacement,buckling...)
Hi Ayoub,
In order to be more realistic with your study, you have to consider nonlinear buckling analysis and Mr. Claudio Pedrazzi explained to you the reasons behind this fact. Concerning optimization of structures, ANSYS can solve design optimization problems. I recommend you to use APDL feature so that you can change any parameter easily. Good luck.
Dear sirs,
Thanks for your answer
I hope you will answer the two questions I posed before
1/ Are there other important outputs to study this problem?
2/ I do not think the results mean that the displacement have exceeded the limit?
VECTOR DOFS for buckling analysis (mode1)
NODE UX UY
1 -0.261024 0.994484
2 0.622163 1.00000
3 -0.204640 0.117440
4 0.354446 0.292473
5 0.00000 0.00000
6 0.00000 0.00000
Displacement limits +- 0.0508 m
I appreciate your cooperation.
First I'm gonna talk analytically, in order to investigate buckling possibility of your truss you should apply Euler buckling formula to calculate the critical load for each member or you can apply critical buckling stress formula for each element and that would be the analytical solution of the problem. Then you have to calculate the reaction forces in each member due to the applied external loads. If the reaction force is larger than the calculated buckling load then the member will buckle.
By the way, Euler's formula has limitations. It can be used only if the slenderness ratio (i.e. ratio of effective length of a column to the least radius of gyration) is less than 80. Otherwise, Rankine's relation is used
Now, let's talk numerically. When you used ANSYS to predict buckling load, your simulation was linear and without imperfections and the results was 572 (i.e. buckling will not occur until you apply a value of 572). We advised you to carry out a non-linear buckling analysis ( i.e. the behavior of the structure for large displacements, nonlinear materials (plastic behavior of ductile materials or cracking for brittle materials), imperfections in the geometry of the structure ( effect of manufacturing tolerances)), here the value of the buckling should be decreased (less than 572), it will be more realistic.
Considering displacement, theory says the displacement increases without limit in buckled columns.
I do not understand what you mean with "Displacement limits +- 0.0508 m". A structure in buckling just collapses! if breaks down! it is not stable anymore.
It is constraints for optimization problem
See the attached picture
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