请教:ansys如何模拟混凝土水化热

谢谢楼上的兄弟，真是及时啊
目前导师让我用ANSYS分析浇注大体积混凝土水化热产生的应力
我在《ANSYS土木工程应用实例》找了一个例子，不过是ANSYSY7.0的，不是很好，
我用的9.0。
哪位兄弟有没有关于这方面的GUI模式的例子？谢谢

让我来给你一个算例，看看对不对你有帮助。
!100*80*10基岩上浇筑混凝土50*40*10，每三天浇筑1米高（第一天浇完，剩下两天窝工），
!共用一个月时间,分为十层（layer)浇筑，模型、条件均为假设
!建模
/filename,test
/PREP7
BLOCK,,50,,10,,40,
BLOCK,-25,75,-10,0,-20,60
ET,1,70
LSEL,S,,,9,12,1
LESIZE,ALL,,,10
LSEL,S,,,21,24,1
LESIZE,ALL,,,20
LSEL,S,,,2,4,2
LSEL,A,,,5,7,2
LESIZE,ALL,,,10
LSEL,S,,,14,16,2
LSEL,A,,,17,19,2
LESIZE,ALL,,,20
LSEL,S,,,1,3,2
LSEL,A,,,6,8,2
LESIZE,ALL,,,10
LSEL,S,,,13,15,2
LSEL,A,,,18,20,2
LESIZE,ALL,,,5
allsel
MP,DENS,1,2400.0 !混凝土材料
MP,KXX ,1,300.89
MP,C,1,0.94500
MAT,1
VMESH,1
MP,DENS,2,2600.0 !基岩材料
MP,KXX ,2,300.89
MP,C,2,0.94500
MAT,2
VMESH,2
nummrg,all

!热分析
/SOLU
NROPT,FULL !Specifies the Newton-Raphson options in full transient analysis
ESEL,S,MAT,,2 !MAT2 IS 基岩
NSLE,S
IC,all,TEMP,16.4, !加基岩初温
ASEL,S,,,7,9,1
ASEL,A,,,11,12,1
NSLA,S,1
SF,all,HFLUX,0 !加基岩绝热边界条件
NSEL,S,LOC,Y,-0.1,0.1
ASEL,S,,,3
NSLA,U,1
SF,ALL,CONV,2016,20 !加基岩第三边界条件（和空气接触面）
!加初始条件
ALLSEL,ALL
ESEL,U,MAT,,2
NSLE,S
EKILL,ALL !KILL 混凝土
IC,all,TEMP,35, !混凝土浇筑时温度为35度
ANTYPE,TRANSIENT,new !分析类型
!日子的流逝
*DO,LAYER,1,10,1
!激活第N层单元
ALLSEL,ALL
NSEL,S,LOC,Y,(LAYER-1),LAYER !第一层从Y=0~1米，第二层从Y=1~2米,...
ESLN,S ,1
EALIVE,ALL
*DO,DAY,LAYER*3-2,LAYER*3,1 !每三天浇筑一层
ALLSEL,ALL
ASEL,S,,,1,2,1
ASEL,A,,,5,6,1
NSLA,S,1
NSEL,R,LOC,Y,0,LAYER
NSEL,A,LOC,Y,LAYER-0.1,LAYER+0.1
SF,ALL,CONV,2016,20 !加混凝土第三类对流边界条件
*DO,LOOP,1,LAYER,1
ALLSEL,ALL
NSEL,R,LOC,Y,LOOP-1,LOOP
ESLN,S,1
TDAY=DAY-(LOOP-1)*3 !已经完工的天数
HE00=47880*(TDAY/(0.862+TDAY)-(TDAY-1)/(0.862+TDAY-1))
BFE,ALL,HGEN, ,HE00 !加水化热
*ENDDO
!计算设置
ALLSE
TRNOPT,FULL
LUMPM,0
TIME,DAY
AUTOTS,0
DELTIM,1, , ,1
KBC,0
TSRES,ERASE
OUTRES,ALL,ALL,
SOLVE
*ENDDO
ESEL,S,MAT,,1
NSLE,S,
SFEDELE,ALL,ALL,CONV !删对流边界条件
SFDELE,ALL,CONV
*ENDDO
FINISH

!结构分析
/PREP7
ET,1,SOLID65 !变热单元为结构单元
*DIM,EXX,ARRAY,30 !考虑变化的EX
EXX(1)=0.325E10,0.57778E10,0.78E10,0.94545E10,0.10833E11,0.12E11,0.13E11,0.13867E11,0.14625E11,0.15294E11, 0.15889E11, 0.1642E11 ,0.169E11, 0.17333E11, 0.17727E11, 0.18087E11,0.18417E11,0.1872E11
EXX(19)=0.190E11, 0.19259E11, 0.195E11, 0.19724E11, 0.19933E11, 0.20129E11,0.20312E11, 0.20485E11, 0.20647E11, 0.208E11, 0.20944E11, 0.21081E11
*DO,I,1,30,1 !定义材料，混凝土浇筑完每过一天变一种材料
MP,DENS,I,2400
MP,EX,I,EXX(I)
MP,PRXY,I,0.167
MP,ALPX,I,0.9E-5
MP,REFT,I,15
*ENDDO
MP,DENS,31,2600 !定义基岩材料
MP,EX,31,0.20800E+11
MP,PRXY,31,0.167
MP,ALPX,31,0.9E-5
ESEL,S,MAT,,,2
MPCHG,31,ALL, !!变热材料为结构材料
ALLSEL
/SOLU
NROPT,FULL !Specifies the Newton-Raphson options in a full transient analysis
NLGEOM,ON
NSLA,S,,,7,9,1
NSLA,A,,,11,12,1
D,all,ALL !加基岩约束
!KILL THE CONCRETE ELEMENT
ESEL,U,MAT,,1
NSLE,S
EKILL,ALL !KILL 混凝土
*DO,DAY,1,30,1 !随时间改变，变换材料
*DO,LAYER,1,10
*IF,DAY,GT,LAYER*3-3,THEN
NSEL,S,LOC,Y,LAYER-1,LAYER
ESLN,S,1
EALIVE,ALL !ACTIVE NEXT LAYER
MPCHG,DAY-(LAYER-1)*3,ALL, !CHANG MAT
ALLSEL,ALL
*ENDIF
*ENDDO
ALLSEL,ALL
ACEL,0,0,-9.8, !GRAVITY LOAD
LDREAD,TEMP,,,DAY, ,’TEST’,’rth’,’ ’ !TEMP LOAD Reads results from the results file and applies them as loads.
CNVTOL,F, ,0.2,2,200, !Sets convergence values for nonlinear analyses.
TIME,DAY
AUTOTS,0
DELTIM,1, , ,1
KBC,0
TSRES,ERASE !Defines an array of keytimes at which the time-stepping strategy changes
OUTRES,ALL,ALL,
SOLVE
*ENDDO

!1、定义重力集散地是应为正值：ACEL,0,0,9.8
!2、砼单元边长（1*4*5）相差5倍，对计算精度的影响会不会较大？
!3、热单元SOLID70对应的结构单元为SOLID45，改为SOLID65是否合适？
!4、原分析得到的结果与实际情况比较如何？
!5、如何体现气温的变化？

不懂，
期待高手来解答
。。。。。。。。。。。。。。

有知道的兄弟,请替楼主帮帮忙了!
我来悬赏10分给提供解决方案的高手!