幾個(gè)耦合的例子

1、一般說來，ANSYS的流固耦合主要有 4種方式：1, sequential這需要用戶進(jìn)行 APDL編程進(jìn)行流固耦合seque ntia指的是順序耦合以采用MpCCI為例，你可以利用ANSYS和一個(gè)第三方CFD產(chǎn)品執(zhí)行流固耦合分析。在這個(gè) 方法中，基于網(wǎng)格的平行代碼耦合界面（MpCCI）將ANSYS和CFD程序耦合起來。即使網(wǎng)格上存在差別，MpCCI也能夠?qū)崿F(xiàn)流固界面的數(shù)據(jù)轉(zhuǎn)換。ANSYS CD中包含有MpCCI庫和一個(gè)相關(guān)實(shí)例。關(guān)于該方法的詳細(xì)信息，參見ANSYS Coupled-Field Analysis Guide中的Seque ntial Coupli n2， FSI solver流固

2、耦合的設(shè)置過程非常簡單，推薦你使用這種方式3， multi-field solver這是FSI solver的擴(kuò)展，你可以使用它實(shí)現(xiàn)流體，結(jié)構(gòu)，熱，電磁等的耦合4,直接采用特殊的單元進(jìn)行直接耦合，耦合計(jì)算直接發(fā)生在單元?jiǎng)偠染仃囈粋€(gè)流固耦合的例子len gth=2width=3height=2/prep7et,1,63et,2,30 !選用FLUID30單元，用于流固耦合問題r,1,0.01mp,ex,1,2e11mp, nu xy,1,0.3mp,de ns,1,7800mp,dens,2,1000 !定義Acoustics材料來描述流體材料-水mp,s on c,2,1400mp,mu,0,

3、!block,le ngth,width,heightesize,0.5mshkey,1!type,1mat,1real,1asel,u,loc,y,widthamesh,allalls!type,2mat,2vmesh,allfini/soluan type,2modopt,unsym,10 !非對稱模態(tài)提取方法處理流固耦合問題 eqslv,fro ntmxpa nd,10,1n sel,s,loc,x,n sel,a,loc,x,le ngthn sel,r,loc,yd,all,ux,uy,u z, n sel,s,loc,y,width, d,all,pres,0 alls asel,

4、u,loc,y,width,sfa,all,fsi !定義流固耦合界面allssolv fini/postl set,first pl nsol,u,sum,2,1 fini再給大家一個(gè)實(shí)例！考慮結(jié)構(gòu)在水中的自振頻率：例子是一加筋板在水中的模態(tài)分析。 命令流如下：FINISH /CLEAR /FILENAME,pla ne /UNITS,SI /TITLE,pla ne/PREP7*!*ELEMENT DEFINE*ET,63,63ET,4,beam4et,30,fluid30!*MATERIAL DEFINE*MP,EX,1,2.10E11MP,DENS,1,7850MP,NUXY,1,0.

5、3mp,de ns,30,1025mp,so nc,30,1500mp,mu,30,0.5!*REAL CONSTANT* *r,30,1e-06r,50,0.05r,75,0.375e-02,0.78125e-06,0.000016406 k,1k,4,1kfill,1,4,2,1kgen ,4,1,4,1,1/3,10 a,1,2,12,11*do,i,0,2*do,j,0,2*10,10 a,1+i+j,2+i+j,12+i+j,11+i+j *enddo*enddo!*fluideleme nt*k,100,-14.5,-14.5k,101,-14.5,15.5k,102,15.5,1

6、5.5k,103,15.5,-14.5k,140,-14.5,-14.5,30k,141,-14.5,15.5,30k,142,15.5,15.5,30k,143,15.5,-14.5,30a,100,101,102,103,4,14,24,34,33,32,31,21,11,1a,1,2,3,4,103,100a,140,141,142,143a,100,101,141,140a,101,102,142,141a,142,143,103,102a,140,143,103,100a,14,24,34,33,32,31,21,11,1,2,3,4asel,u,1,FLST,2,8,5,ORDE,

7、2FITEM,2,10FITEM,2,-17VA,P51Xnummrg,allallsMSHKEY,0MSHAPE,0 esize,1 lsel,s,loc,y,1/3 lsel,r,loc,x,0,1 lsel,r,loc, z,0 latt,1,75,4 lmesh,all lsel,s,loc,y,2/3 lsel,r,loc,x,0,1 lsel,r,loc, z,0 latt,1,75,4 lmesh,all lsel,s,loc,x,1/3 lsel,r,loc,y,0,1 lsel,r,loc, z,0 latt,1

8、,75,4 lmesh,all lsel,s,loc,x,2/3 lsel,r,loc,y,0,1 lsel,r,loc, z,0 latt,1,75,4 lmesh,all asel,s,1,9 aatt,1,50,63 amesh,all allsMSHAPE,1,3d esize,3 vsel,s,1 type,30 $mat,30 $real,30 vmesh,all allsFINISH/solu alls*求解*I*ANTYPE,MODALMODOPT,la nb,25,0SOLVEFINISH 總是出現(xiàn)error說矩陣不對稱，不可以用lanb計(jì)算?？偨Y(jié)：流體單元不能用對稱的解法

9、應(yīng)該采用非對稱解法。例子是一圓環(huán)在水中的模態(tài)分析。命令流如下：finish/clear/PREP7!定義單元類型ET,1,PLANE42 ! structural eleme ntET,2,FLUID29 ! acoustic fluid eleme nt with ux & uyET,3,129 ! acoustic infin ite line eleme ntr,3,0.31242,0,0ET,4,FLUID29,1,0 ! acoustic fluid eleme nt without ux & uy !材料屬性MP,EX,1,2.068e11MP,DENS,1,7929MP,NUXY

10、,1,0MP,DENS,2,1030MP,SONC,2,1460!創(chuàng)建四分之一模型CYL4,0,0,0.254,0,0.26035,90CYL4,0,0,0.26035,0,0.31242,90!選擇屬性，網(wǎng)格劃分ASEL,S,AREA,1AATT,1,1,1,0LESIZE,1,16,1LESIZE,3,16,1LESIZE,2,1,1LESIZE,4,1,1MSHKEY,1MSHAPE,0,2D ! map ped quad meshAMESH,1ASEL,S,AREA,2AATT,2,1,2,0LESIZE,5,16,1LESIZE,7,16,1LESIZE,6,5LESIZE,8,5M

11、SHKEY,0MSHAPE,0,2D ! map ped quad meshAMESH,2!關(guān)于Y軸鏡像nsym,x,1000,all ! offset n ode number by 1000esym,1000,all!關(guān)于y軸鏡像n sym,y,2000,all ! offset n ode number by 2000esym,2000,allNUMMRG,ALL ! merge all qua ntitiesesel,s,type,1nsle,ses In ,s,0nsle,sesel,i nvensle,semodif,all,type,4esel,allnsel,all!指定無限吸

12、收邊界csys,1nsel,s,loc,x,0.31242type,3real,3mat,2esurfesel,allnsel,all!標(biāo)識流固交接面nsel,s,loc,x,0.26035esel,s,type,2sf,all,fsi,1nsel,allesel,allFINISH/soluan type,modalmodopt,damp,10mxpa nd,10,yessolvefinish為了便于對比，也對圓環(huán)在空氣中做了模態(tài)分析finish/clear/PREP7!定義單元類型ET,1,PLANE42 ! structural eleme nt!材料屬性MP,EX,1,2.068e11

13、MP,DENS,1,7929MP,NUXY,1,0!創(chuàng)建四分之一模型CYL4,0,0,0.254,0,0.26035,90!選擇屬性，網(wǎng)格劃分ASEL,S,AREA,1AATT,1,1,1,0LESIZE,1,16,1LESIZE,3,16,1LESIZE,2,1,1LESIZE,4,1,1MSHKEY ,1MSHAPE,0,2D ! map ped quad meshAMESH,1!關(guān)于Y軸鏡像nsym,x,1000,all ! offset n ode number by 1000 esym,1000,all!關(guān)于y軸鏡像n sym,y,2000,all ! offset n ode nu

14、mber by 2000 esym,2000,allNUMMRG,ALL/soluan type,modalmodopt,la nb,10mxpa nd,10,yessolvefinish在水中的自振頻率為SET TIME/FREQ LOAD STEP SUBSTEP CUMULATIVE1-0.19544E-10 1 1 12 0.29640E-03 1 1 13- 0.21663E-10 1 2 24- 0.29640E-03 1 2 25 0.30870E-03 1 3 36 0.0000 1 3 37-0.30870E-03 1 4 48 0.0000 1 4 49-0.53726E-

15、03 1 5 510 0.57522E-11 1 5 511 0.53726E-03 1 6 612-0.89057E-11 1 6 613 0.98059E-01 1 7 714 35.232 1 7 715 0.98059E-01 1 8 816 -35.232 1 8 817 0.98061E-01 1 9 918 35.233 1 9 919 0.98061E-01 1 10 1020 -35.233 1 10 10在空氣中的自振頻率為SET TIME/FREQ LOAD STEP SUBSTEP CUMULATIVE1 0.0000 1 1 12 0.0000 1 2 23 0.73

16、609E-03 1 3 34 60.805 1 4 45 60.805 1 5 56 172.97 1 6 67 172.97 1 7 78 334.40 1 8 89 334.40 1 9 910 546.59 1 10 10主要有以下疑問：1） 考慮流固耦合，做模態(tài)分析時(shí)流體單元是否只能用fluid29（ 2d）和fluid30 （3d）,對于fluid129 和fluid130在耦合中具體起到什么作用，能不能不設(shè)，而用邊界約束條件代替？2） 流體范圍怎樣確定，如本例中（CYL4,0,0,0.26035,0,0.31242,90），外半徑為 0.31242。如 果不是環(huán)形的，如一塊當(dāng)水板，

17、該怎樣考慮？3） 如果不考慮流體的壓縮性，把聲速設(shè)的很大，MP,SONC,2,1e20,就可以了。4）從自振頻率可以看出，在水中和在空氣中，圓環(huán)的自振頻率差別特別大，且振型也大相徑庭，為什么？在水中時(shí)，模態(tài)提取方法用damp （為什么不能用unsym），特征值的虛部代表角頻率，為什么第一階為正，第二階為負(fù)，而第三階和第四階都為0,第六階、八階、十階都為負(fù)。應(yīng)該是從小到大才對？5） 在空氣中時(shí)，模態(tài)提取方法用Ianb，為什么第一階第二階的頻率都為0。請高手指點(diǎn)迷津，急盼中對以上問題的解答：頻率為零，一般是發(fā)生了剛體位移，估計(jì)你是把水抽走，而沒有限制圓環(huán)。1。 圓環(huán)在水中振動(dòng)必然導(dǎo)致波動(dòng) （其實(shí)就

18、是聲波）在水中傳播，當(dāng)聲波到達(dá)水的另一個(gè)界面 時(shí)就會(huì)發(fā)生反射（除非水和另一個(gè)相鄰體的聲阻抗是匹配的）。水和金屬中的聲速相差不大，即可壓縮性相差不大。 兩種可壓縮性相差不大的物質(zhì)的相互作用對兩者影響都很大。圓環(huán)在水中振動(dòng)，水對圓環(huán)的反作用是由于反射波引起的，流固耦合中采用fluid129和130就是最大程度的減弱反射波。2。聲波從圓環(huán)開始傳播，隨著傳播距離的增加，波陣面不斷增大，振幅不斷減小。同時(shí)由于水的衰減，聲波也不斷減弱。如果水的空間越大，則反射波返回圓環(huán)的路徑越長，衰減也就越多，影響也就越小。fluid129和130對反射波的衰減（通過很小的反射實(shí)現(xiàn)）有限，因此還需要水要有足夠的空間。 f

19、luid129和130離結(jié)構(gòu)應(yīng)該大于 0.2入（入=c/f為水中聲速）。以上 的做法在誤差允許的情況下等效于水在無限大水空間中的情況。如果是擋水板，水就是有限空間了，情況也不一樣。3。聲速加大情況也不一樣，就是不知是不是你所要的情況？4?？諝庾鳛榻橘|(zhì)，由于其聲速比金屬小很多，可壓縮性大很多，影響可以忽略不計(jì)。而水的影響就不同了。這可能就是頻率和振型不同的原因吧？我試了你的例子，各種提取方法都可以。5?？諝獾挠绊懞雎圆挥?jì)，因此需要對圓環(huán)進(jìn)行約束。你沒有約束，那么就會(huì)發(fā)生靜態(tài)位移 即頻率為零。圓環(huán)有兩個(gè)對稱軸，因此會(huì)發(fā)生頻率成對出現(xiàn)的情況。也就是說，兩個(gè)方向上 有同樣的振型。接觸分析實(shí)例-包含初始

20、間隙fini/clear, no start/prep7et,1,82KEYOPT,1,3,3r,1,0.5 mp,ex,1,1e9 mp,prxy,1,0.3k,1,0,0k,2,10,0k,3,10,5k,4,6.2,5k,5,7.5,3.4k,6,2.5,3.4k,7,3.8,5k,8,0,5a,1,2,3,4,5,6,7,8LFILLT,6,5,0.18,LFILLT,5,4,0.18,FLST,2,3,4FITEM,2,9FITEM,2,11FITEM,2,10AL,P51XFLST,2,3,4FITEM,2,13FITEM,2,14FITEM,2,12AL,P51XFLST,2,3

21、,5,ORDE,2FITEM,2,1FITEM,2,-3AADD,P51Xrect,0,10,4.8,5ASBA,4,1gap=0.02 k,24,6.2-gap,5 k,25,7.5-gap,3.4 k,26,2.5+gap,3.4 k,27,3.8+gap,5 a,24,25,26,27LFILLT,4,3,0.2,LFILLT,3,2,0.2,FLST,2,3,4FITEM,2,7FITEM,2,10FITEM,2,8AL,P51XFLST,2,3,4FITEM,2,13FITEM,2,14FITEM,2,11AL,P51XFLST,3,2,5,ORDE,2FITEM,3,3FITEM,

22、3,-4ASBA,1,P51Xrect,3.8+g ap,6.2-gap,5,10rect,3.8+g ap,3.8+gap+8,10,12FLST,2,3,5,ORDE,3FITEM,2,1FITEM,2,3FITEM,2,5AADD,P51Xrect,3.8+gap+8,3.8+gap+8+2,10,12FLST,2,2,5,ORDE,2FITEM,2,1FITEM,2,4AGLUE,P51XCYL4,2.0,1.8,0.6CYL4,7.0,1.8,0.6FLST,2,3,5,ORDE,3FITEM,2,2FITEM,2,4FITEM,2,-5AOVLAP,P51Xesize,0.2ame

23、sh,allFLST,5,135,2,ORDE,32FITEM,5,485FITEM,5,576FITEM,5,-577FITEM,5,621FITEM,5,-625FITEM,5,707FITEM,5,-711FITEM,5,716FITEM,5,741FITEM,5,-745FITEM,5,750FITEM,5,-751FITEM,5,766FITEM,5,797FITEM,5,-798FITEM,5,854FITEM,5,888FITEM,5,-938FITEM,5,1101FITEM,5,1103FITEM,5,1420FITEM,5,1628FITEM,5,1653FITEM,5,1

24、696FITEM,5,1699FITEM,5,-1702FITEM,5,1726FITEM,5,-1728FITEM,5,1852FITEM,5,-1874 FITEM,5,2044 FITEM,5,-2066 CM,_Y,ELEM ESEL, , , ,P51X CM,_Y1,ELEM CMSEL,S,_Y CMDELE,_Y EREF,_Y1, , ,1,0,1,1 CMDELE,_Y1ET,2,TARGE169 ET,3,CONTA172R,3,R,3,0,0,0.1, 10,0,0R,4,R,4,0,0,0.1, 10,-0.02,0 lsel,s,9lsel,a,5 lsel,a,1

25、2 nsll,s,1 type,3 real,3 esurf,all alls,lsel,s,19 lsel,a,20 nsll,s,1 type,3 real,4 esurf,all alls,lsel,s,7 lsel,a,3 lsel,a,11 nsll,s,1 type,2 real,3 esurf,allalls, lsel,s,25 lsel,a,26 nsll,s,1 type,2 real,4 esurf,all alls,FLST,2,2,5,ORDE,2 FITEM,2,4 FITEM,2,-5 DA,P51X,ALL, FLST,2,1,4,ORDE,1 FITEM,2,

26、6SFL,P51X,PRES,500,/soluan type,0 nl geom ,on outres,all,all nsubst,200,200,2 neqit,1OOO solve耦合小程序最近用到耦合，寫了一段小程序，奉獻(xiàn)出來,與大家共享。如果有很多節(jié)點(diǎn)，每兩個(gè)節(jié)點(diǎn)位置相同，如果將這些雜亂無章的節(jié)點(diǎn)耦合，是件很麻煩的事，可用這段程序，輕松解決。cpnum=0cmsel,s ,n-zho ng !需要耦合的節(jié)點(diǎn) *GET,n_num,NODE,COUNT, , , , !節(jié)點(diǎn)總數(shù) cmsel,s ,n-zho ng*GET,no west,NODE,NUM,MIN, , , , ! 號

27、碼最小的節(jié)點(diǎn)*GET,n_x,NODE,no west,LOC,X !該節(jié)點(diǎn)坐標(biāo)*GET,n_y,NODE,no west,LOC,Y*GET,n_z,NODE,no west,LOC,ZNSEL,s,LOC,X,n_x-O.3,n_x+O.3 !尋找與該節(jié)點(diǎn)位置相同的節(jié)點(diǎn)NSEL,R ,L OC,Y,n_y-O.3,n_y+O.3NSEL,R, LOC,z,n_z-O.3,n_z+O.3cm,n_cp_cp, node !位置相同的節(jié)點(diǎn)形成一個(gè)組cmsel,s ,n-zho ngcmsel,u ,n _cp_cpcm,n-zho ng,node !取消這些點(diǎn)后剩余的點(diǎn)形成組*GET,n_ n

28、u m_1,NODE,COUNT, , , , !節(jié)點(diǎn)總數(shù)*if,n_num_1,lt,2,exit !如果節(jié)點(diǎn)數(shù)小于二則退出cmsel,s ,n _cp_cp*GET,n_ nu m,NODE,COUNT,*if, n_nu m,gt,1,the nCP,cp num+1,ux,allCP,cp nu m+2,uy,allCP,cp nu m+3,u z, allcpnum=cp nu m+3 *else*en dif*enddo該段程序可用CPINTF，UX，0.001CPINTF，UY，0.001CPINTF，UZ，0.001代替*DO,I,2,296,3CP,l,UX,l,l+2*EN

29、DDO*DO,I,2,296,3CP,I,UY,I,I+2*ENDDO*DO,I,2,296,3CP,I,UZ,I,I+2*ENDDODK,1, , , ,0,UX,UY ,UZ,以上幾句改為：*DO,I,2,296,3CP,NEXT,ALL,l,l+2*ENDDODK,1, , , ,0,ALL或 CPINTF,ALL,0.001因?yàn)槟氵x用的單元有六個(gè)自由度，如果只約束三個(gè)，程序是不會(huì)運(yùn)行的另:三次循環(huán)語句的I相等約束UY時(shí),UX的耦合就被刪掉了 ，最后只剩UZ 了這樣修改：c*耦合練習(xí)/PREP7K,1,0,0K,2,0.1,0L, 1,2K,300,0,-10000LGEN,100,1,

30、0.1,2ET,1,BEAM188MP,EX,1,2.1e11MP,PRXY,1,0.3MP,DENS,1,0.783e4SECTYPE, 1, BEAM, T, , 0SECOFFSET, CENTSECDATA,0.06,0.03,0.003,0.006,0,0,0,0,0,0 LSEL,ALLLATT,1,1,1,300LESIZE,ALL,1,1,1LMESH,ALLcpi ntf,allDK,1,ux,0, , ,UY ,UZDK,200, , , , ,UY ,UZACEL,0,9.8,0,FINISH一個(gè)流固耦合的例子這個(gè)例子關(guān)于裝有水的水杯旋轉(zhuǎn)，是軸對稱問題，為了簡化，所以選擇

31、了平面模型。*SET,RAD,0.8*SET,h,1*SET,g,9.8*SET,OMEGAR,2*SET,ROU,1000/PREP7ET,1,FLUID79KEYOPT,1,3,1MP,EX,1,2E9MP,DENS,1,ROUK,1K,2,RADK,3,RAD,HK,4,HK,4,HA,1,2,3,4LESIZE,ALL,10AMESH,ALLFINISH/SOLDL,2,UXDL,1,UYNSEL,S ,L OC,XDSYM,SYMM,XD,ALL,UXD,ALL,UXNSEL,ALLACEL,GOMEGA,OMEGARSOLVEFINISH/POST1SET,LASTPLNSOL,U

32、,X,0,1*SET,UCENT,UY(22)*SET,UEDGE,UY(12) *SET,UELEV,UEDGE-UCENTansys從9.0發(fā)展到10.0，一個(gè)最大的進(jìn)步就是流固耦合計(jì)算更加規(guī)范，這一點(diǎn)已遠(yuǎn)領(lǐng)先于 其他同類軟件，實(shí)現(xiàn)了單向耦合到即時(shí)雙向耦合的飛躍，使用戶對于解決流固耦合問題又多了一種選擇，希望大家對多種方法物理環(huán)境轉(zhuǎn)換，fsi，mfx等進(jìn)行討論，提供一下案例本人拋磚引玉：使用物理環(huán)境法進(jìn)行流固耦合的實(shí)例及講解流道中有一橡膠墊阻礙水的流動(dòng)，入口速度為2m/s，其他參數(shù)將在命令流中詳細(xì)給出。求解水通過此流道的壓力降，以及穩(wěn)態(tài)條件下橡膠墊的變形。/prep7/sho,gasket

33、,grphshpp,offET,1,141! Fluid - static meshET,2,56,! Hyperelastic eleme nt! Fluid Structure In teraction - Multiphysics!Deformatio n of a gasket in a flow field.! Eleme nt plots are writte n to the file gasket.grph.!- Water flows in a vertical pipe through a con structi on! formed by a rubber gasket.

34、!- Determ ine the equilibrium positi on of the gasket and!the result ing flow field!|!|!|1 Boun dary of morphi ng fluid! | |!|gasket!|!|1 Boun dary of morphi ng fluid (sf)!|! 1. Build the model of the en tire doma in:! Fluid regi on - static mesh! Gasket leaves a hole in the cen ter of the duct! Mor

35、ph ing Fluid regi on is a user defi ned regi on around! the gasket. The fluid mesh here will deform and be! updated as the gasket deforms.! Parameterize Dime nsions in the flow direct ion!yent = 0.0! Y coord in ate of the entrance to the pipedye n = 1.0! Un deformed geometry flow entrance len gthysf

36、l = yen t+dye n ! Y coord in ate of entrance to the morph ing fluid regi ondsfl = 0.5! Thick ness of upstreamygas = ysf1+dsf1 ! Y coord in ate of the bottom of the gasketdg = 0.02! Thick ness of the gasketdg2=dg/2.ytg = ygas+dg ! Y coord in ate of the in itial top of the gasketdsf2 = 0.5! Thick ness

37、 of dow nstream regi onysf2 = ytg + dsf2! Y of Top of the dow nstream morphi ng fluids regi ondyex = 6.0! Exit fluid len gthx = 0.! Locati on of the axisymmetric Cen terl inedgasr =.20! In itial spa n of gasketpiper = 0.3! Radius of the pipexrgap = piper-dgasr! radius of completely un obtructed flow

38、 passage! Create geometry!rect,xrgap,piper,ygas,ytg ! A1:Gasket (keypo ints 1-4)rect,x,piper,ysf1,ysf2 ! A2: Morph ing fluid regi on rect,x,piper,ye nt,ysf1 ! A3: Fluid regi on with static mesh rect,x,piper,ysf2,ysf2+dyex ! A4: Fluid regi on with static mesh aovlap,allk,22,xrgap+dg2,ygas+dg2rarc = d

39、g2*1.1larc,1,4,22,rarcal,6,4adelete,7al,6,3,22,7,8,5,21,1!Mesh Divisio n in formatio nn gap = 10! Number eleme nts across the gapn gas = 10! Number of eleme nts along the gasketrgas = -2! Spaci ng ratio along gasketnflu = n gap+ngas ! Number of eleme nts across the fluid regi on raflu = -3! Space fl

40、uid eleme nts n ear the walls and cen ternenty =8! Eleme nts along flow - entranceraent =5! Size ratio in the inlet regi onnfl1 = 20! Eleme nts along flow - first morph.fluid.n thgas = 4! Eleme nts in the gasketnfl2 = 3! Eleme nts along flow - sec ond morph.fluid.n ext = 30! Eleme nts along flow - e

41、xit regi onrext = 6! Size ratio in flow directio n of outletrafls = 12! In itial eleme nt spac ing ratio - morph.fluidlesize,1, ”n gas,rgas lesize,3, ”n gas,rgas nfl11= n fl1*2+9lsel,s,2,4,2 ! (Modify lesize of line 8 if cha nging gasket mesh) lesize,all, ”n thgas allslesize,5, ”n flu,raflulesize,7,

42、 ”n flu,raflulesize,9, ”n flu,raflulesize,15, ”n flu,raflulesize,18, ”nen ty,1./rae ntlesize,17, ”nen ty,1./rae nt lesize,21, nfl1,raflslesize,8, nfl11,-1./(rafls+3)lesize,22, nfl1,rafls lesize,19, ”n ext,rext lesize,20，”n ext,rext! AATT,MA T,REAL,TYPE asel,s,1,2-Set the attributes for the areasaatt

43、,2,2,2! Gasket (material 2)asel,s,3cm,area2,areaalist ! List area selected for further morph ingasel,a,5,6aatt,1,1,1! Fluid area (material 1)allseshape,2asel,u,2,3amesh,all eshape,0 asel,s,2,3 amesh,all! Create eleme nt plot and write to the file gasket.grph asel,s,1,3esla,s/Title, In itial mesh for

44、 gasket and n eighborhoodeplot/ZOOM,1,RECT,0.3,-0.6,0.4,-0.5alls! 2. Create Physics Environment for the Fluidet,1,141et,2,0! Gasket becomes the Null Eleme ntvin=3.5e-1! Inlet water velocity (meters/sec ond)!CFD Solution Con trolflda,solu,flow,1flda,solu,turb,1 flda,iter,exec,400 flda,outp,sumf,10!CF

45、D Property In formatio nflda,prot,de ns,c on sta nt flda,prot,visc,c on sta ntflda ,n omi,de ns,1000.! 1000 kg/m3 for den sity - waterflda ,no mi,visc,4.6E-4 ! 4.6E-4 kg-s/m (viscosity of water) flda,c on v,pres,1.E-8! Tighte n pressure equati on con verge nee! CFD Bou ndary Co nditio ns (Applied to

46、 Solid Model) lsel,s,8,17,9lsel,a,20dl,all,vx,0.,1! Cen terl ine symmetryIsel,s,9dl,all,vx,0.,1 dl,all,vy,vi n,1 lsel,s,2 lsel,a,18,19 lsel,a,21,22 dl,all,vx,0.,1 dl,all,vy,0.,1 lsel,s,1,3,2 lsel,a,6 dl,all,vx,0.,1 dl,all,vy,0.,1 lsel,s,15 dl,15,pres,0.,1! In let Con dition! Outer Wall! Gasket! Outl

47、et pressure con diti on! create n amed comp onent of no des at the bottom of gasket lsel,s,1 nsll,1cm,gasket, noden list ! List in itial no dal positi ons of the bottom of the gasket/com, + STARTING gasket coord in ates alls/title,Fluid An alysisphysics,write,fluid,fluid! 3. Create Physics Environme

48、nt for the Structure !physics,clearet,1,0! The Null eleme nt for the fluid regio net,2,56! Gasket eleme nt - material 2mp,ex,2,2.82E+6! Youn gs modulus for rubbermp, nu xy,2,0.49967 ! Poisso ns ratio for the rubber tb,m oon ey,2tbdata,1,0.293E+6 ! Moo ney-Rivlin Con sta nts tbdata,2,0.177E+6 ! Fix t

49、he end of the gasketlsel,s,2 nsll,1 d,all,ux,0. d,all,uy,0. alls/title,structural an alysis finish/soluan type,staticnl geom ,oncnvtol,f,-1physics,write,struc,strucphysics,clearsave! 4. Fluid-Structure In teraction Loop!loop=25! Maximum allowed nu mber of loopstoler=0.005! Con verge nee tolera nee f

50、or maximum displaceme nt*dim,dismax,array,loop ! Define array of maximum displaceme nt values *dim,strcri,array,loop ! Define array of con verge nee values *dim,i ndex,array,loop*do,i,1,loop/solu physics,read,fluid *if,i, ne,1,the n flda,iter,exec,100 *en difsolvefini! Execute fluid - structure solu

51、ti ons! Read in fluid environment! Execute 100 global iterati ons for! each new geometry! FLOTRAN solution! end of fluid portion physics,read,struc! Read in structures environment/assig n, esave,struc,esav ! Files for restart ing non li near structure/assig n, emat,struc,emat*if,i,gt,1,then parsave,

52、all resume! Structural restart loop! Save parameters for con verge nee check! Resume DB - to retur n origi nal node positi onsparresume/prep7! Resume parameters n eeded for con verge nee checkan type,stat,rest fini*en dif/solusolc,offlsel,s,1,3,2 lsel,a,6 nsll,1! Select proper lines to apply fluid p

53、ressures! to the en tire gasket surfaceesel,s,type,2ldread,pres,last,rfl ! Apply pressure surface load from Flotra nallsrescon trol” none! Do not use multiframe restart for non li nearsolve*if,i,eq,1,the nsave! save origi nal node locatio ns at the first run*en diffini/post1cmsel,s,gasketn sort,u,sum,1,1*get,dismax(i),sort,0,max ! Get the maximum displaceme nt value strcri(i)=toler*dismax(i)allsfini/pre