Post Buckling of Thin Shells

Submitted by Himayat Ullah on Fri, 2008-10-10 03:56.

education

It is difficult to conduct post buckling analysis of thin shell under axial compression using Arc Length method in Ansys.
The Minimum & Maximum Arc Length radii are chosen by hit & trial.When imperfections are incorporated in the FE model, some times the solution diverges at a higher critical load that the eigen value buckling, which should be vice versa in actual. Some times , negative eigen values are also shown.

Is there any easy way to solve this nonlinear stability problem?

Himayat

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Here a good example as a reference

Submitted by Pengfei Liu on Fri, 2008-10-17 14:52.

It is indeed difficult to solve the convergence problem for post-buckling FEA solution. The following is a good example and wish a good luck for you!

/PREP7
smrt,off
/TITLE, VM17, SNAP-THROUGH BUCKLING OF A HINGED SHELL
:COM    CHANG, C.C.,"PERIODICALLY RESTARTED QUASI-NEWTON UPDATES IN
:COM    IN CONSTANT ARC-LENGTH METHOD", COMPUTERS AND STRUCTURES,
:COM    VOL. 41, NO. 5, PP. 963-972, 1991.
ANTYPE,STATIC            ! STATIC ANALYSIS
ET,1,SHELL63,,1
R,1,6.350                ! SHELL THICKNESS
MP,EX,1,3102.75
MP,NUXY,1,0.3
:COM CREATE FINITE ELEMENT MODEL
R1 = 2540                ! SHELL MID-SURFACE RADIUS
L = 254                 ! HALF THE LENGTH
PI = 4*ATAN(1)           ! VALUE OF PI COMPUTED
THETA = 0.1*180/PI       ! 0.1 RADIANS CONVERTED TO DEGREES
CSYS,1                   ! CYLINDRICAL CO-ORDINATE SYSTEM
N,1,R1,90                ! NODES 1 AND 2 ARE CREATED AT POINTS
N,2,R1,90,L              ! A AND B RESPECTIVELY.
K,1,R1,90
K,2,R1,(90-THETA)
K,3,R1,90,L
K,4,R1,(90-THETA),L
ESIZE,,2                 ! TWO DIVISION ALONG THE REGION BOUNDARY
A,1,3,4,2
AMESH,1
NUMMRG,NODE
:COM APPLY BOUNDARY CONDITIONS
NSEL,S,LOC,Z,0
DSYM,SYMM,Z
NSEL,S,LOC,Y,90
DSYM,SYMM,X
NSEL,S,LOC,Y,(90-THETA)
D,ALL,UX,,,,,UY,UZ
NSEL,ALL
FINISH
:COM   SOLUTION PHASE
:COM SINCE THE SOLUTION OUTPUT IS SUBSTANTIAL IT IS DIVERTED TO A
:COM SCRATCH FILE
/OUTPUT,SCRATCH
/SOLUTION
NLGEOM,ON                ! LARGE DEFLECTION TURNED ON
OUTRES,,1                ! WRITE SOLUTION ON RESULTS FILE FOR EVERY SUBSTEP
F,1,FY,-250              ! 1/4 TH OF THE TOTAL LOAD APPLIED DUE TO SYMMETRY
NSUBST,30                ! BEGIN WITH 30 SUBSTEPS
ARCLEN,ON,4              ! ARC-LENGTH SOLUTION TECHNIQUE TURNED ON WITH
                         ! MAX. ARC-LENGTH KEPT AT 4 TO COMPUTE AND STORE
    ! SUFFICIENT INTERMEDIATE SOLUTION INFORMATION
SOLVE
FINISH
/OUTPUT
:COM   POSTPROCESSING PHASE
/POST26
NSOL,2,1,U,Y             ! STORE UY DISPLACEMENT OF NODE 1
NSOL,3,2,U,Y             ! STORE UY DISPLACEMENT OF NODE 2
PROD,4,1,,,LOAD,,,4*250 ! TOTAL LOAD IS 4*250 DUE TO QUARTER SYMMETRY
PROD,5,2,,,,,,-1         ! CHANGE SIGNS OF THE DISPLACEMENT VALUES
PROD,6,3,,,,,,-1
*GET,UY1,VARI,2,EXTREM,VMIN
*GET,UY2,VARI,3,EXTREM,VMIN
PRVAR,2,3,4              ! PRINT STORED INFORMATION
/AXLAB,X, DEFLECTION (MM)
/AXLAB,Y, TOTAL LOAD (N)
/GRID,1
/XRANGE,0,35
/YRANGE,-500,1050
XVAR,5
PLVAR,4                  ! PLOT LOAD WITH RESPECT TO -UY OF NODE 1
/NOERASE
XVAR,6
PLVAR,4