以下两段命令流除了加载初始温度的部分不同外,其余相同,请大家帮我分析一下,这样我自己的那个复杂的铸造分析估计就有思路了。
谢谢!
先加高温后加低温(照抄ANSYS里的例子):
/TITLE,CASTING SOLIDIFICATION !Give the analysis a title
/PREP7
K,1,0,0,0
K,2,22,0,0
K,3,10,12,0
K,4,0,12,0
/TRIAD,OFF
/REPLOT
!
A,1,2,3,4
SAVE
RECTNG,4,22,4,8
APLOT
AOVLAP,1,2
ADELE,3,,,1
SAVE
!
MP,DENS,1,0.054
MP,KXX,1,0.025
MP,C,1,0.28
!
MPTEMP,1,0,2643,2750,2875,,,
MPDATA,KXX,2,1,1.44,1.54,1.22,1.22,,,
MPDATA,ENTH,2,1,0,128.1,163.8,174.2
MPPLOT,KXX,2,,,,,
MPPLOT,ENTH,2,,,,,
SAVE
!
ET,1,PLANE55
!
SAVE
SMRT,5
MSHAPE,0,2D
MSHKEY,0
AMESH,5
!
TYPE,1
MAT,2
REAL
ESYS,0
AMESH,4
!
SAVE
SFL,1,CONV,0.014,,80,,
SFL,3,CONV,0.014,,80,,
SFL,4,CONV,0.014,,80,,
SAVE
FINISH
/SOLU
!
ANTYPE,4
SOLCONTROL,ON,0
!
APLOT
ASEL,S,,,4
NSLA,S,1
NPLOT
IC,ALL,TEMP,2875
NSEL,INVE
/REPLOT
IC,ALL,TEMP,80
ALLSEL,ALL
SAVE
!
TIME,3
AUTOTS,-1
DELTIM,0.01,0.001,0.25,1
KBC,1
!
OUTRES,ALL,ALL
SAVE
/STAT,SOLU
/REPLOT
APLOT
SOLVE
FINISH
先加低温后加高温(改了一下ANSYS里例子的初始温度加载顺序):
/TITLE,CASTING SOLIDIFICATION !Give the analysis a title
/PREP7
K,1,0,0,0
K,2,22,0,0
K,3,10,12,0
K,4,0,12,0
/TRIAD,OFF
/REPLOT
!
A,1,2,3,4
SAVE
RECTNG,4,22,4,8
APLOT
AOVLAP,1,2
ADELE,3,,,1
SAVE
!
MP,DENS,1,0.054
MP,KXX,1,0.025
MP,C,1,0.28
!
MPTEMP,1,0,2643,2750,2875,,,
MPDATA,KXX,2,1,1.44,1.54,1.22,1.22,,,
MPDATA,ENTH,2,1,0,128.1,163.8,174.2
MPPLOT,KXX,2,,,,,
MPPLOT,ENTH,2,,,,,
SAVE
!
ET,1,PLANE55
!
SAVE
SMRT,5
MSHAPE,0,2D
MSHKEY,0
AMESH,5
!
TYPE,1
MAT,2
REAL
ESYS,0
AMESH,4
!
SAVE
SFL,1,CONV,0.014,,80,,
SFL,3,CONV,0.014,,80,,
SFL,4,CONV,0.014,,80,,
SAVE
FINISH
/SOLU
!
ANTYPE,4
SOLCONTROL,ON,0
!
APLOT
ASEL,S,,,5
NSLA,S,1
NPLOT
IC,ALL,TEMP,80
NSEL,INVE
/REPLOT
IC,ALL,TEMP,2875ALLSEL,ALL
SAVE
!
TIME,3
AUTOTS,-1
DELTIM,0.01,0.001,0.25,1
KBC,1
!
OUTRES,ALL,ALL
SAVE
/STAT,SOLU
/REPLOT
APLOT
SOLVE
FINISH
不好意思,为了把问题说清楚,有点啰嗦,彩色部分就是改变的内容。望大家不吝赐教! |