ynn1980 发表于 2008-7-24 23:35:50

求复合材料非线性屈曲例子??

如题,谢谢

[ 本帖最后由 ynn1980 于 2008-7-28 22:14 编辑 ]

hdm106 发表于 2008-8-14 17:03:06

我也想要,能给我一份吗?
hdm106@hotmail.com

pengbo_buaa 发表于 2008-9-17 09:27:56

我也比较关心,请高人指点!

wb21c2005 发表于 2009-2-10 19:21:52

my qq:15834362

kaochao 发表于 2009-7-3 19:27:35

我也想要 kaochao@tom.com

rhp2008 发表于 2009-8-6 09:09:14

我也想要,谢谢renhongpeng2006@126.com] 5# kaochao

iamperisher 发表于 2009-8-7 17:25:41

/TITLE,Orthotropic plate with edge load, Bifurcation Analysis, SHELL91
/UNITS,MPA            ! Units are in mm, MPa, and Newtons
/PREP7                  ! Pre-processor module
! Material properties for lamina
uimp,1,ex,ey,ez,145880,13312,13312
uimp,1,gxy,gyz,gxz,4386,4529,4386
uimp,1,prxy,pryz,prxz,0.263,0.470,0.263
ET,1,SHELL91,,1         ! Chooses Shell91 element for analysis
                  ! Set KEYOPT(2)=1, then supply 12+(6*NL) const.
KEYOPT,1,1,12       ! Set KEYOPT(1)=12, Max number of layers = 12
KEYOPT,1,5,1      ! Set KEYOPT(5)=1, Element output: Middle layer
! Real constant set #1, [(0/90)3]s, NL=12, lamina thick=0.85 mm
R,1,12,1,               ! 12 layers symmetrical
RMORE,,,,,,,
RMORE,1,0,0.85          ! 1st layer: mat. #1,   0 deg, Th=0.85 mm
RMORE,1,90,0.85         ! 2nd layer: mat. #1,90 deg, Th=0.85 mm
RMORE,1,0,0.85          ! 3nd layer: mat. #1,   0 deg, Th=0.85 mm
RMORE,1,90,0.85         ! 4th layer: mat. #1,90 deg, Th=0.85 mm
RMORE,1,0,0.85          ! 5th layer: mat. #1,   0 deg, Th=0.85 mm
RMORE,1,90,0.85         ! 6th layer: mat. #1,90 deg, Th=0.85 mm

! Geometry and mesh
RECTNG,0,500,0,250      ! Creates a rectangle with x=1 m and y=1 m
ESIZE,,25               ! 25 divisions for edge
AMESH,all               ! Mesh the area

FINISH                  ! Exit pre-processor module

/SOLU                   ! Solution module, (i) STATIC ANALYSIS
ANTYPE,STATIC         ! Set static analysis
PSTRESS,ON            ! Calculate the stress stiffness matrix
DL,2,1,uz,0             ! Impose Simple Supported BC
DL,3,1,uz,0
DL,1,1,symm             ! Impose Symmetry BC
DL,4,1,symm
!d,all,rotz             ! Constraint rotations about z axes (optional)
!Load application
SFL,2,PRES,1            ! Apply uniform pressure in x=500 mm
SFL,3,PRES,1            ! Apply uniform pressure in y=250 mm
SOLVE                   ! Solve current load state
FINISH                  ! Exit solution module

/SOLU                   ! Solution module, (ii) find the BIFURCATION LOADS
ANTYPE,BUCK
BUCOPT,SUBSP,10         ! Find the first 10 bifurcations loads
SOLVE                   ! Solve
FINISH                  ! Exit solution module

/SOLU                   ! Solution module, (ii) find the BUCKLING MODES
EXPASS
MXPAND
SOLVE                   ! Solve
FINISH                  ! Exit solution module

/POST1                  ! Post-processor module
SET,LIST                ! List the critical loads
SET,1,2               ! Set mode number 2 shape
PLDISP,1                ! Display the mode 2 shape displacements
FINISH                  ! Exit post-processor module

iamperisher 发表于 2009-8-7 17:26:40

/TITLE,Orthotropic plate with edge load, Bifurcation Analysis, SHELL91
/UNITS,MPA            ! Units are in mm, MPa, and Newtons

/PREP7                  ! Pre-processor module
! Material properties for lamina
uimp,1,ex,ey,ez,145880,13312,13312
uimp,1,gxy,gyz,gxz,4386,4529,4386
uimp,1,prxy,pryz,prxz,0.263,0.470,0.263

ET,1,SHELL91,,1         ! Chooses Shell91 element for analysis
                  ! Set KEYOPT(2)=1, then supply 12+(6*NL) const.
KEYOPT,1,1,12       ! Set KEYOPT(1)=12, Max number of layers = 12
KEYOPT,1,5,1      ! Set KEYOPT(5)=1, Element output: Middle layer

! Real constant set #1, [(0/90)3]s, NL=12, lamina thick=0.85 mm
R,1,12,1,               ! 12 layers symmetrical
RMORE,,,,,,,
RMORE,1,0,0.85          ! 1st layer: mat. #1,   0 deg, Th=0.85 mm
RMORE,1,90,0.85         ! 2nd layer: mat. #1,90 deg, Th=0.85 mm
RMORE,1,0,0.85          ! 3nd layer: mat. #1,   0 deg, Th=0.85 mm
RMORE,1,90,0.85         ! 4th layer: mat. #1,90 deg, Th=0.85 mm
RMORE,1,0,0.85          ! 5th layer: mat. #1,   0 deg, Th=0.85 mm
RMORE,1,90,0.85         ! 6th layer: mat. #1,90 deg, Th=0.85 mm

! Geometry and mesh
RECTNG,0,500,0,250      ! Creates a rectangle with x=1 m and y=1 m
ESIZE,50                       ! 50 size element edge
AMESH,all               ! Mesh the area

FINISH                  ! Exit pre-processor module

/SOLU                   ! Solution module, (i) STATIC ANALYSIS
ANTYPE,STATIC         ! Set static analysis
PSTRESS,ON            ! Calculate the stress stiffness matrix
DL,2,1,uz,0             ! Impose Simple Supported BC
DL,3,1,uz,0
DL,1,1,symm             ! Impose Symmetry BC
DL,4,1,symm
!d,all,rotz             ! Constraint rotations about z axes (optional)
!Load application
SFL,2,PRES,1            ! Apply uniform pressure in x=500 mm
SFL,3,PRES,1            ! Apply uniform pressure in y=250 mm
SOLVE                   ! Solve current load state
FINISH                  ! Exit solution module

/SOLU                   ! Solution module, (ii) find the BIFURCATION LOADS
ANTYPE,BUCK
BUCOPT,SUBSP,10         ! Find the first 10 bifurcations loads
SOLVE                   ! Solve
FINISH                  ! Exit solution module

/SOLU                   ! Solution module, (ii) find the BUCKLING MODES
EXPASS
MXPAND
SOLVE                   ! Solve
FINISH                  ! Exit solution module

/PREP7                  ! Pre-processor module
ftr=(10.8/10)         ! Multiplicator shape factor (Th/10)
UPGEOM,ftr,1,1,file,rst
    ! ftr: Multiplier for displacements added to coordinates
    ! 1,1 : Load step 1, substep=1, equivalent to mode =1
    ! file,rst: results file to obtain displacements
FINISH                  ! Exit pre-processor module

/SOLU                   ! Solution module, Continuation loads
ANTYPE,STATIC         ! Set static analysis
NLGEOM,1                ! Use large displacements analysis
OUTRES,ALL,ALL          ! Keep results of each substep

mult=225                ! Apply loads until N = 225 N/mm
SFL,2,PRES,1*mult       ! Apply uniform pressure in x=500 mm
SFL,3,PRES,1*mult       ! Apply uniform pressure in y=250 mm

ARCLEN,1,10,0.1         ! Use ARCLENG method to obtain solution
NSUBST,50,0,0         ! #Substeps
SOLVE                   ! Solve current load state
FINISH                  ! Exit solution module

/POST26               ! Post-processor module
LINES,1000            ! List without breaks between pages
NSOL,2,1,U,Z,UZ_node1   ! Load deflexion in central plate node
PLVAR,2               ! DISPLAY VARIABLES evolution
PRVAR,2               ! PRINT VARIABLES evolution
FINISH

rongwei0912 发表于 2009-8-11 20:37:47

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kuning19901 发表于 2015-11-17 18:37:10

同求 329063105@qq.com, 谢谢各位,谢谢楼主
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