ANSYS热应力分析经典例题.doc

ANSYS热应力分析例题实例1圆简内部热应力分折： 有一无限长圆筒，其核截面结构如图131所示，简内壁温度为200，外壁温度为20，圆筒材料参数如表131所示，求圆筒内的温度场、应力场分布。 该问题属于轴对称问题。由于圆筒无限长，忽略圆筒端部的热损失。沿圆筒纵截面取宽度为10M的如图132所示的矩形截面作为几何模型。在求解过程中采用间接求解法和直接求解法两种方法进行求解。间接法是先选择热分析单元，对圆筒进行热分析，然后将热分析单元转化为相应的结构单元，对圆筒进行结构分析；直接法是采用热应力藕合单元，对圆筒进行热力藕合分析。/filname,exercise1-jianjie /title,thermal stresses in a long/prep7$Et,1,plane55 Keyopt,1,3,1 $Mp,kxx,1,70 Rectng,0.1,0.15,0,0.01 $Lsel,s,1,3,2Lesize, all,20$Lsel,s,2,4,2Lesize,all,5 $Amesh,1 $Finish/solu $Antype,static Lsel,s,4 $Nsll,s,1$d,all,temp,200lsel,s,2$nsll,s,1$d,all,temp,20allsel $outpr,basic,allsolve $finish/post1$Set,last/plopts,info,onPlnsol,temp $Finish/prep7 $Etchg,ttsKeyopt,1,3,1$Keyopt,1,6,1Mp,ex,1,220e9$Mp,alpx,1,3e-6$Mp,prxy,1,0.28Lsel,s,4 $Nsll,s,1 $Cp,8,ux,allLsel,s,2 $Nsll,s,1 $Cp,9,ux,allAllsel $Finish/solu$Antype,staticD,all,uy,0$Ldread,temp,rthAllsel $Solve $Finish/post1/title,radial stress contoursPlnsol,s,x/title,axial stress contoursPlnsol,s,y/title,circular stress contoursPlnsol,s,z/title,equvialent stress contoursPlnsol,s,eqv$finish/filname,exercise1-zhijie /title,thermal stresses in a long/prep7$Et,1,plane13 Keyopt,1,1,4 $Keyopt,1,3,1Mp,ex,1,220e9 $Mp,alpx,1,3e-6$Mp,prxy,1,0.28MP,KXX,1,70 Rectng,0.1,0.15,0,0.01 $Lsel,s,1,3,2Lesize, all,20$Lsel,s,2,4,2Lesize,all,5 $Amesh,1Lsel,s,4$Nsll,s,1 $Cp,8,ux,allLsel,s,2$Nsll,s,1 $Cp,9,ux,allALLSEL$Finish/solu $Antype,static Lsel,s,4 $Nsll,s,1$d,all,temp,200lsel,s,2$nsll,s,1$d,all,temp,20allsel $outpr,basic,allsolve $finish/post1$Set,last/plopts,info,onPlnsol,temp/title,radial stress contoursPlnsol,s,x/title,axial stress contoursPlnsol,s,y/title,circular stress contoursPlnsol,s,z/title,equvialent stress contoursPlnsol,s,eqv$finish318页实例2冷却栅管的热应力分析图中为一冷却栅管的轴对称结构示意图，其中管内为热流体，温度为200，压力为10Mp,对流系数为110W(m2)；管外为空气，温度为25，对流系数为30w(mz)。栅管材料参数如表132所示，求栅管内的温度场和应力场分布。根据对称性，并在图示边界线段上施加对称边界约束，进行热应力分析求解。FINISH$/CLEA/filname,exercise2/title,thermal stresses in an axisymmetrical pipe/prep7$et,1,plane13$Keyopt,1,1,4mp,ex,1,200e9 $Mp,alpx,1,1.5e-5mp,prxy,1,0.3$mp,kxx,1,30rectang,0.12,0.16,0,0.07$ rectang,0.16,0.4,0.025,0.045 $ rectang,0.38,0.4,0.015,0.055k,100,0.15,0.055$k,101,0.15,0.015aadd,all $numcmp,linelfillt,8,12,0.01$lfillt,7,9,0.01ldiv,9,0.8$ldiv,12,0.8L2tan,12,-6 $L2tan,9,5al,15,16,17 $al,18,19,20al,14,22,23 $al,13,21,24aadd,1,2,4$aadd,5,6aadd,1,3$numcmp,lineesize,0.0025 $wpstyle,1csys,4$kwpave,100wprot,0,0,90 $asbw,2wprot,0,90$asbw,3kwpave,101 $asbw,4kwpave,16$wprot,0,-90 $asbw,5kwpave,19$asbw,6kwpave,12$asbw,7amesh,1,3$amesh,5,8,3amap,4,15,16,18,17$amap,6,19,20,9,12allsel$wpstyle,0csys,0 $nsel,s,loc,y,0cp,1,uy,all$allsel$finish/solu$antype,staticSfl,3,pres,10e6$Sfl,3,conv,110,200Lsel,s,4,18$Lsel,a,20,21Sfl,all,conv,30,25 $Lsel,s,17,20,3Dl,all,symm $lsel,s,18,21,3Dl,all,symm $allselOutpr,basic,all$Solve$Finish/post1 $Set,last/title, temperature contoursPlnsol,temp/title,sum displamentl contoursPlnsol,u,sum/title,radial stress contoursPlnsol,s,x/title,axial stress contoursPlnsol,s,y/title,circular stress contoursPlnsol,s,z/title,equvialent stress contoursPlnsol,s,eqv/expand,9,axis,10/view,1,1,1,1/title,temperature contoursPlnsol,temp/title,equvialent stress contoursPlnsol,s,eqv$finish332页实例3两无限长平扳热膨胀分析：有两块厚度均为0.02mm的无限长平板1和2，受如图1352所示约束。平板初始温度为20，求将其加热到800时平板内部的应力场分布(平板材料参数见表)。根据题意，忽略平板沿长度方向的端面效应，将问题简化为平面应变问题。在分析过程中取两平板的横截面建立几何模型，并选取plane13热一结构锅台单元进行求解。/filname,exercise3/title,thermal expansion between two infinite flat/prep7 $et,1,plane13 $Keyopt,1,1,4 mp,alpx,1,1.5e-5 $mp,ex,1,1.0e11mp,prxy,1,0.25 $mp,kxx,1,65mp,prxy,2,0.3$mp,ex,2,2.0e11mp,kxx,2,30 $mp,alpx,2,2.5e-5rectng, 0,0.1,0,0.02$ rectng, 0.1,0.3,0,0.02esize,0.02$mat,1$amesh,1$mat,2$amesh,2nummrg,all $numcmp,all/solu$antype,staticAutots,on$lsel,s,4,6,2Nsll,s,1$d,all,uxTref,20$bfunif,temp,800Allsel $solve/post1 $Set,last/plopts,info,on/title, temperature contoursPlnsol,temp/title,sum displamentl contoursPlnsol,u,sum/title,x direction displament contoursPlnsol,u,x/title,y direction displament contoursPlnsol,u,y/title,sum direction displament contoursPlnsol,u,sum/title,x direction stress contoursPlnsol,s,x/title, y direction stress contoursPlnsol,s,y/title,equvialent stress contoursPlnsol,s,eqv$finish340页实例4包含焊缝的金属板热膨胀分析某一平板由钢板和铁板焊接而成，焊接材料为铜，平板尺寸为110.2，横截面结构如图1368所示。平板初始温度为800，将平板放置于空气中进行冷却，周围空气温度为30，对流系数为110W(m2)。求10分钟后平板内部的温度场及应力场分布(材料参数见表134)。属于瞬态热应力问题，选择整体平板建立几何模型，选取solid5热一结构耦合单元进行求解。/filname,exercise4/title,thermal stresses in secti* including welding seam/prep7$et,1,plane13Keyopt,1,1,4 $et,2,solid5Mp,alpx,1,1.06e-5$mp,kxx,1,66.6Mp,dens,1,7800$mp,c,1,460Mptemp,30,200,400,600,800Mpdata,ex,1,2.06e11,1.92e11,1.75e11,1.53e11,1.25e11Mpdata,prxy,1,0.3,0.3,0.3,0.3,0.3Tb,bkin,1,5 !指定材料模型Tbtemp,30$Tbdata,1,1.40e9,2.06e10Tbtemp,200 $Tbdata,1,1.330e9,1.98e10Tbtemp,400 $Tbdata,1,1.15e9,1.83e10Tbtemp,600 $Tbdata,1,0.92e9,1.56e10Tbtemp,800 $Tbdata,1,0.68e9,1.12e10MP,ALPX,2,1.75E-5$MP,KXX,2,383MP,DENS,2,8900 $MP,C,2,390MPDATA,EX,2,1.03E11,0.99E11,0.90E11,0.79E11,0.58E11MPDATA,PRXY,2,0.3,0.3,0.3,0.3,0.3TB,BKIN,2,5TBTEMP,30$TBDATA,1,0.9E9,1.03E10 TBTEMP,200 $TBDATA,1,0.85E9,0.98E10TBTEMP,400 $TBDATA,1,0.75E9,0.89E10TBTEMP,600 $TBDATA,1,0.62E9,0.75E10TBTEMP,800 $TBDATA,1,0.45E9,0.52E10MP,ALPX,3,5.87E-6 $MP,KXX,3,46.5MP,DENS,3,7000$MP,C,3,450MPDATA,EX,3,1.18E11,1.09E11,0.93E11,0.75E11,0.52e11MPDATA,PRXY,3,0.3,0.3,0.3,0.3,0.3TB,BKIN,3,5TBTEMP,30$TBDATA,1,1.04E9,1.18E10TBTEMP,200 $TBDATA,1,1.01E9,1.02E10TBTEMP,400 $TBDATA,1,0.91E9,0.86E10TBTEMP,600 $TBDATA,1,0.76E9,0.69E10TBTEMP,800 $TBDATA,1,0.56E9,0.51E10K,1 $K,2,0.5K,3,1$K,4,0,0.2K,5,0.4,0.2 $K,6,0.6,0.2 $K,7,1,0.2A,1,2,5,4 $A,2,3,7,6CSYS,1$L,6,5$AL,2,8,9MSHKEY,1 $ESIZE,0.05AMESH,1,3,1 $ESIZE,10TYPE,2 $MAT,1$VOFFST,1,1MAT,3 $VOFFST,2,1MAT,2 $VOFFST,3,-1$ACLEAR,1,3,1NUMMRG,ALL $NUMCMP,ALL$ALLSEL/SOLU $ANTYPE,4TRNOPT,FULL$TIMINT,1,STRUCTTIMINT,1,THERM $TIMINT,0,MAG TIMINT,0,ELECT$TINTP,0.005,-1,0.5,0.2TIME,600 $DELTIM,30,10,100AUTOTS,ON$KBC,1OUTRES,ALL $BFUNIF,TEMP,800ASEL,U,6,13,7$NSLA,S,1SF,ALL,CONV,110,30 $ALLSELSOLVE $FINISH/POST1SET,LAST/TITLE,TEMPERATURE CONTOURSPLNSOL,TEMP/TITLE,X DIRECTION DISPLAMENT CONTOURSPLNSOL,U,X/TITLE,Y DIRECTION DISPLAMENT CONTOURSPLNSOL,U,Y/TITLE,Z DIRECTION DISPLAMENT CONTOURSPLNSOL,U,Z/TITLE,SUMDIRECTION DISPLAMENT CONTOURSPLNSOL,U,SUM/TITLE,X DIRECTION STRESS CONTOURSPLNSOL,S,X/TITLE,Y DIRECTION STRESS CONTOURSPLNSOL,S,Y/TITLE,Z DIRECTION STRESS CONTOURSPLNSOL,S,Z/TITLE, EQUVIALENT DIRECTION STRESS CONTOURSPLNSOL,S,EQV353页:实例5连杆热应力分析图示为一连杆，其初始温度为100，现将连杆放置于空气中进行冷却，周围空气温度为20，对流系数为95W(m2)。求：1)5分钟后平板内部的温度场及应力场分布； 2)A、B两点温度随时间的变化关系曲线(材料参数见表135)。该问题属于瞬态热应力问题，选取连杆建立几何模型，并选强solid98十节点四面体热结构耦合单元进行求解。/filname,exercise5/title,thermal stress analysis to a connecting rod/prep7 $Et,1,solid98Mp,alpx,1,2.05e-5$Mp,prxy,1,0.3Mp,kxx,1,16.3 $Mp,ex,1,220e9Mp,c,1,300$Mp,dens,1,7800K,1,-0.18,0,0$K,2,-0.18,0.035,0K,3,-0.14,0.035,0 $L,1,2L,2,3 $Csys,1K,4,0.14,135,0 $K,5,0.1,180,0K,6,0.1,135,0$L,3,4L,5,6 $LFILLT,2,3,0.02Wprota,45 $Csys,4Allsel $Lsymm,x,1,5Wprota,-45 $Wpoffs,0.7Cswpla,11,1K,16,0.07,0,0 $K,17,0.07,135,0K,18,0.04,0,0 $K,19,0.04,180,0L,16,17$L,18,19Csys,0Lsymm,x,6,7 $Lsymm,x,4,9,5K,28,0.25,0.055,0 $K,29,0.52,0.045,0Bsplin,17,29,28,22Lfillt,14,17,0.03 $Lfillt,11,17,0.15Lsymm,y,1,19 $Nummrg,allLsel,s,12,31,19 $Al,allLsel,s,4,9,5 $Lsel,a,15,16Lsel,a,23,28,5 $Lsel,a,34,35$Al,allLsel,s,1,3 $Lsel,a,5,8Lsel,a,10,11 $Lsel,a,13,14Lsel,a,17,22 $Lsel,a,24,27Lsel,a,29,30 $Lsel,a,32,33Lsel,a,36,38 $Al,all $AllselAsba,3,1 $Asba,4,2Numcmp,all$Voffst,1,0.02$Smrt,6Mshape,1,3d$Mshkey,0Vmesh,1 $Wpstyle,0 $Save/SOLUANTYPE,4$TRNOPT,FULLTIMINT,1,STRUCT $TIMINT,1,THERMTIMINT,0,ELECT$TINTP,0.005,-1,0.5,-1TIME,300 $DELTIM,4,4,20AUTOTS,ON$KBC,1OUTRES,ALL$BFUNIF,TEMP,100ASEL,s,all $NSLA,S,1SF,ALL,CONV,95,20 $ALLSELSOLVE$FINISH/POST1SET,LAST/TITLE,TEMPERATURE CONTOURSPLNSOL,TEMP/TITLE,SUM DISPLAMENT CONTOURSPLNSOL,U,SUM/TITLE,X DIRECTION STRESS CONTOURSPLNSOL,S,X/TITLE,Y DIRECTION STRESS CONTOURSPLNSOL,S,Y/TITLE,Z DIRECTION STRESS CONTOURSPLNSOL,S,Z/TITLE, EQUVIALENT DIRECTION STRESS CONTOURSPLNSOL,S,EQV/POST26/AXLAB,A,TIME,(SEC)/AXLAB,Y,TEMPERATURE/GTHK,AXIS,3/GTHK,CURVE,3/COLOR,CURVE,MRED,1WPCSYS,-1NSEL,S,LOC,X,-0.18NSEL,R,LOC,Y,0NSEL,R,LOC,Z,0*GET,NODE1,NODE,NUM,MAXNSEL,S,LOC,X,0.77NSEL,R,LOC,Y,0NSEL,R,LOC,Z,0*GET,NODE2,NODE,NUM,MAXNSOL,2,NODE1,TEMPNSOL,3,NODE2,TEMP/TITLE,CURVE DECRIBED THE RELATION BETWEEN TEPERATURE AND TIME AT POINT APLVAR,2/TITLE,CURVE DECRIBED THE RELATION BETWEEN TEPERATURE AND TIME AT POINT BPLVAR,3FINISH371页：实例6热喷涂过程中熔滴夜基体表面沉积凝固后的残余应力分析热喷徐过程中，一金属N6熔滴以一定速度在无限大碳钢基体表面沉积后，发生散流变形的同时与基休有热变换作用，最终凝固成圆片状固体颗粒。其纵剖面形状及几何尺寸如图13l16所示。金属Ni与碳钢的基本物性参数如表136所示。假定颗粒由最初的熔点温度1454冷却到室温25，忽略对流带来的影响，月假定平均制备温度625为参考温度。若定义路径1为线段AB，定义路径2为线段BC，求：1)熔滴经过Iooy5时的温度场分布。2)点A、点B的温度随时间的变化曲线。3)100隅时，沿路径1、路径2颗粒的轴向应力变化曲线与径向应力变化曲线。4)100y5时，熔滴的轴向应力场与径向应力场。5)100隅时，在歹维方向熔滴(取I4份)的轴向应力场与径向应力场。热喷徐过程中，一金属N6熔滴以一定速度在无限大碳钢基体表面沉积后，发生散流变形的同时与基休有热变换作用，最终凝固成圆片状固体颗粒。其纵剖面形状及几何尺寸如图13l16所示。金属Ni与碳钢的基本物性参数如表136所示。假定颗粒由最初的熔点温度1454冷却到室温25，忽略对流带来的影响，月假定平均制备温度625为参考温度。若定义路径1为线段AB，定义路径2为线段BC，求：1)熔滴经过Iooy5时的温度场分布。2)点A、点B的温度随时间的变化曲线。3)100隅时，沿路径1、路径2颗粒的轴向应力变化曲线与径向应力变化曲线。4)100y5时，熔滴的轴向应力场与径向应力场。5)100隅时，在歹维方向熔滴(取I4份)的轴向应力场与径向应力场。根据题意，可将问题简化为二维轴对称问题，分析过程中取纵剖团的一半进行求解，如图13117所示，对称向上取对称约束。对于无限大碳钢基体可在建模过程中取远远大于颗粒的尺寸近似代替(本例按矩形取2000四1肋opn)。求解热应力采用间接辊合法，即先用P1抓e55单元完成温度场计算后，再转换成PlMe42单元，计算热应力。为建模及分析问题方便，统一取微米毫克微秒单位制。经单位制转换后，各材料属性的数值如表13。7所示。注意，取不同的单位制，属性数值虽然会有所不同，且将导致最终计算结果数值也相应不同，但换算成相同单位制后，结果是一致的。/filname,exercise6/title,thermal stress analysis of molten in particle in thermal spraying/prep7 $Et,1,55,1Mp,ex,1,0.204$Mp,nuxy,1,0.28Mp,alpx,1,13.36e-6$Mp,kxx,1,87.86e-6Mptemp,1,0,25,355,360,365,1454Mpdata,enth,1,1,0,8.48e-5,1.45e-3,1.5e-3,1.55e-3,6.6.5e-3Mp,ex,2,0.215$Mp,nuxy,2,0.26Mp,alpx,2,11.3e-6 $Mp,dens,2,7.9e-9Mp,kxx,2,7.1e-5$Mp,c,2,460.24K,1,0,22.5,0 $K,2,10,22K,3,20,19.3$K,4,30,16.5K,5,40,13.8$K,6,50,12K,7,60,9.5 $K,8,70,7.5K,9,80,5.5 $K,10,90,4.5K,11,100,2.8 $K,12,110,2.8K,13,200,0$K,14,0,0*do,i,1,13Lstr,i,i+1*enddoLstr,14,1$Al,allRectng,0,2000,-1000,0 $Aglue,all $Numcmp,areaAsel,s,area,2$Aatt,2Asel,all $Amesh,allNummrg,node $Save,exercise61,db/solu$antype,4Solcontrol,on $Esel,s,ma