ANSYS热应力分析经典例题

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13—1

所示，简内壁温度为200℃，外壁温度为20℃，圆筒材料参数如表13．1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13—2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。 fini clear

/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,,,1,3,2 Lesize, all,,,20 $Lsel,s,,,2,4,2 Lesize,all,,,5 $Amesh,1 $Finish /solu $Antype,static Lsel,s,,,4 $Nsll,s,1 $d,all,temp,200 lsel,s,,,2 $nsll,s,1 $d,all,temp,20 allsel $outpr,basic,all solve $finish /post1 $Set,last /plopts,info,on

Plnsol,temp $Finish /prep7 $Etchg,tts Keyopt,1,3,1 $Keyopt,1,6,1

Mp,ex,1,220e9 $Mp,alpx,,1,3e-6 $Mp,prxy,1,0.28 Lsel,s,,,4 $Nsll,s,1 $Cp,8,ux,all Lsel,s,,,2 $Nsll,s,1 $Cp,9,ux,all Allsel $Finish /solu $Antype,static D,all,uy,0 $Ldread,temp,,,,,,rth Allsel $Solve $Finish /post1

/title,radial stress contours Plnsol,s,x

/title,axial stress contours Plnsol,s,y

/title,circular stress contours Plnsol,s,z

/title,equvialent stress contours

Plnsol,s,eqv $finish

/filname,exercise1-zhijie /title,thermal stresses in a long

/prep7 $Et,1,plane13 Keyopt,1,1,4 $Keyopt,1,3,1

Mp,ex,1,220e9 $Mp,alpx,,1,3e-6 $Mp,prxy,1,0.28 MP,KXX,1,70

Rectng,0.1,0.15,0,0.01 $Lsel,s,,,1,3,2 Lesize, all,,,20 $Lsel,s,,,2,4,2 Lesize,all,,,5 $Amesh,1

Lsel,s,,,4 $Nsll,s,1 $Cp,8,ux,all Lsel,s,,,2 $Nsll,s,1 $Cp,9,ux,all ALLSEL $Finish

/solu $Antype,static Lsel,s,,,4 $Nsll,s,1 $d,all,temp,200 lsel,s,,,2 $nsll,s,1 $d,all,temp,20 allsel $outpr,basic,all solve $finish /post1 $Set,last /plopts,info,on Plnsol,temp /title,radial stress contours Plnsol,s,x

/title,axial stress contours Plnsol,s,y

/title,circular stress contours Plnsol,s,z

/title,equvialent stress contours Plnsol,s,eqv $finish

318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13—2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,1,1,4 mp,ex,1,200e9 $Mp,alpx,1,1.5e-5 mp,prxy,1,0.3 $mp,kxx,1,30

rectang,0.12,0.16,0,0.07 $ rectang,0.16,0.4,0.025,0.045 $ rectang,0.38,0.4,0.015,0.055 k,100,0.15,0.055 $k,101,0.15,0.015 aadd,all $numcmp,line lfillt,8,12,0.01 $lfillt,7,9,0.01 ldiv,9,0.8 $ldiv,12,0.8 L2tan,12,-6 $L2tan,9,5 al,15,16,17 $al,18,19,20 al,14,22,23 $al,13,21,24 aadd,1,2,4 $aadd,5,6 aadd,1,3 $numcmp,line esize,0.0025 $wpstyle,,,,,,,,1 csys,4 $kwpave,100 wprot,0,0,90 $asbw,2 wprot,0,90 $asbw,3 kwpave,101 $asbw,4

kwpave,16 $wprot,0,-90 $asbw,5 kwpave,19 $asbw,6 kwpave,12 $asbw,7 amesh,1,3 $amesh,5,8,3

amap,4,15,16,18,17 $amap,6,19,20,9,12 allsel $wpstyle,,,,,,,,0 csys,0 $nsel,s,loc,y,0 cp,1,uy,all $allsel $finish /solu $antype,static

Sfl,3,pres,10e6 $Sfl,3,conv,110,,200 Lsel,s,,,4,18 $Lsel,a,,,20,21 Sfl,all,conv,30,,25 $Lsel,s,,,17,20,3 Dl,all,,symm $lsel,s,,,18,21,3 Dl,all,,symm $allsel

Outpr,basic,all $Solve $Finish /post1 $Set,last /title, temperature contours Plnsol,temp

/title,sum displamentl contours Plnsol,u,sum

/title,radial stress contours Plnsol,s,x

/title,axial stress contours Plnsol,s,y

/title,circular stress contours Plnsol,s,z

/title,equvialent stress contours Plnsol,s,eqv /expand,9,axis,,,10

/view,1,1,1,1

/title,temperature contours Plnsol,temp

/title,equvialent stress contours Plnsol,s,eqv $finish

332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13—52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,1,1.0e11 mp,prxy,1,0.25 $mp,kxx,1,65 mp,prxy,2,0.3 $mp,ex,2,2.0e11 mp,kxx,2,30 $mp,alpx,2,2.5e-5

rectng, 0,0.1,0,0.02 $ rectng, 0.1,0.3,0,0.02 esize,0.02 $mat,1 $amesh,1 $mat,2 $amesh,2 nummrg,all $numcmp,all /solu $antype,static Autots,on $lsel,s,,,4,6,2 Nsll,s,1 $d,all,ux Tref,20 $bfunif,temp,800 Allsel $solve /post1 $Set,last /plopts,info,on

/title, temperature contours Plnsol,temp

/title,sum displamentl contours Plnsol,u,sum

/title,x direction displament contours Plnsol,u,x

/title,y direction displament contours Plnsol,u,y

/title,sum direction displament contours Plnsol,u,sum

/title,x direction stress contours Plnsol,s,x

/title, y direction stress contours Plnsol,s,y

/title,equvialent stress contours Plnsol,s,eqv $finish

340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1×1×0.2，横截面结构如图13—68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,plane13 Keyopt,1,1,4 $et,2,solid5 Mp,alpx,1,1.06e-5 $mp,kxx,1,66.6 Mp,dens,1,7800 $mp,c,1,460 Mptemp,,30,200,400,600,800

Mpdata,ex,1,,2.06e11,1.92e11,1.75e11,1.53e11,1.25e11 Mpdata,prxy,1,,0.3,0.3,0.3,0.3,0.3 Tb,bkin,1,5 !指定材料模型

Tbtemp,30 $Tbdata,1,1.40e9,2.06e10 Tbtemp,200 $Tbdata,1,1.330e9,1.98e10 Tbtemp,400 $Tbdata,1,1.15e9,1.83e10 Tbtemp,600 $Tbdata,1,0.92e9,1.56e10 Tbtemp,800 $Tbdata,1,0.68e9,1.12e10 MP,ALPX,2,1.75E-5 $MP,KXX,2,383 MP,DENS,2,8900 $MP,C,2,390

MPDATA,EX,2,,1.03E11,0.99E11,0.90E11,0.79E11,0.58E11 MPDATA,PRXY,2,,0.3,0.3,0.3,0.3,0.3 TB,BKIN,2,5

TBTEMP,30 $TBDATA,1,0.9E9,1.03E10 TBTEMP,200 $TBDATA,1,0.85E9,0.98E10 TBTEMP,400 $TBDATA,1,0.75E9,0.89E10 TBTEMP,600 $TBDATA,1,0.62E9,0.75E10 TBTEMP,800 $TBDATA,1,0.45E9,0.52E10 MP,ALPX,3,5.87E-6 $MP,KXX,3,46.5 MP,DENS,3,7000 $MP,C,3,450

MPDATA,EX,3,,1.18E11,1.09E11,0.93E11,0.75E11,0.52e11 MPDATA,PRXY,3,,0.3,0.3,0.3,0.3,0.3 TB,BKIN,3,5

TBTEMP,30 $TBDATA,1,1.04E9,1.18E10 TBTEMP,200 $TBDATA,1,1.01E9,1.02E10 TBTEMP,400 $TBDATA,1,0.91E9,0.86E10 TBTEMP,600 $TBDATA,1,0.76E9,0.69E10 TBTEMP,800 $TBDATA,1,0.56E9,0.51E10 K,1 $K,2,0.5 K,3,1 $K,4,0,0.2

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