这是Curtain兄给我的答复:
这么说压力应该是550E6 Pa
修改后的apdl如下:
fini
/cle
/PREP7
ET,1,95
ET,2,63
MP,EX,1,193E9
MP,NUXY,1,.3
TB,BKIN,1,1 ! BILINEAR KINEMATIC HARDENING
TBTEMP,70
TBDATA,1,476e6,0 ! YIELD STRESS AND ZERO TANGENT MODULUS
arc_rad=0.05
k,1,0.009+arc_rad !define the arc origin
k,2,0.009
k,3,0.025,0.04
WPCSYS,,1
kwpave,1
csys,4
l,2,3
csys,0
k,4,0,0
k,5,0,0.08
k,6,0.025,0.08
l,2,4
l,4,5
l,5,6
l,6,3
a,3,6,5,4,2
aplo
wpcsys,,0
cyl4,,0.06,0.006
asba,1,2
aplo
lesize,1,,,20,3
lesize,2,,,6
lesize,4,,,6
lesize,5,,,8
lesize,6,,,6
lesize,9,,,6
lesize,10,,,20
lesize,11,,,6,1/3
amesh,all
extopt,esize,4
VOFFST, 3,0.006 !thickness=0.006
!vsweep,all
aclear,all
da,6,all
da,7,all
da,5,symm
da,8,symm
asel,s,,,9
nsla
cp,1,uy,all
sfa,9,,press,-550e6
fini
/solu
alls
solve
我原来做的是整体的模型,后来也按curtain的方法改为了对称的模型。curtain是按我的大概描述做的。以下是我按真实结构及材料性能参数修改后的程序:
fini
/cle
/PREP7
ET,1,95
ET,2,63
MP,EX,1,193E9
MP,NUXY,1,.3
tbdele,miso,all
tb,miso,1
tbtemp,20,
tbpt,defi,0.0014,270082700
tbpt,defi,0.0022,378784700
tbpt,defi,0.003,435644200
tbpt,defi,0.0038,460729300
tbpt,defi,0.0058,475780400
tbplot,miso,1
arc_rad=0.05
k,200,0.009
k,1,0.009+arc_rad,0.00334 !define the arc origin
k,2,0.009,0.00334
k,3,0.025,0.04
WPCSYS,,1
kwpave,1
csys,4
l,2,3
csys,0
k,4,0,0
k,5,0,0.08
k,6,0.025,0.08
l,200,4
l,4,5
l,5,6
l,6,3
l,2,200
a,3,6,5,4,200,2
aplo
wpcsys,,0
cyl4,,0.06,0.006
asba,1,2
aplo
lesize,1,,,20,3
lesize,2,,,6
lesize,4,,,10
lesize,5,,,14
lesize,6,,,2
lesize,7,,,6
lesize,10,,,6
lesize,11,,,24,3
lesize,12,,,6
amesh,all
extopt,esize,3
VOFFST, 3,0.006 !thickness=0.006
!vsweep,all
aclear,all
da,6,all
da,4,all
da,5,symm
da,8,symm
asel,s,,,9
nsla
cp,1,uy,all
sfa,9,,press,-550e6
allsel,all
fini
/solu
alls
NLGEOM,on
pred,on
nsubst,800,1000,200
time,1
auto,on
solve
这样做的结果就不再收敛了,是否是由于大变形选项的原因?但必须得用大变形选项,因为应变值已远超过0.05。 |