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1、流固耦合模态分析-水箱模态盛水水箱的模态分析这是一个流固耦合模态分析的典型爭•例,采fflANSYS/MECHANICAL可以完成。处理过程中需要注意以下几个方面的问题:1、单元的选择;2、流体材料模式;3、流固耦合关系的定义;4、模态提取方法。水箱采用SHELL63单元,水箱中的水采用FLUTD30单元,以卜•即为整个流固耦合模态计算的命令流文件:length=lwidth=O.6height=O.8/prep7et,1,63et,2,30!选用FLUID30单元,用于流固耦合问题r,1,0.
2、01mp,ex,1,2ellmp,nuxy,1,0.3mp,dens,1,7800mp,dens,2,1000!定义Acoustics材料来描述流体材料-水mp,sone,2,1400mp,mu,2,block,,length,,width,,heightesize,0・1mshkey,1i■type,1mat,1real,1asel,u,loc,y,widthamesh,allalls!type,2mat,2vmesh,al1fini/soluantype,2modopt,unsym,10!非
3、对称模态提取方法处理流固耦合问题eqslv,frontmxpand,10,,,1nsel,s,loc,x,nsel,a,loc,x,lengthnsel,r,loc,yd,all,…,ux,uy,uznsel,s,loc,y,widthd,all,pres,0allsasel,u,loc,y,widthsfa,all,,fsi!定义流固耦合界面fluid-structureinteractionallssolvfini/postlset,firstplnsol.u,sum,2,1finiSTEP
4、=1SUB=1RFRQ=OIFRQ=.931E-06MODERealpartUSUM(AVG)RSYS=ODMX=.193E-04SMX=.193E-04.430E-05.860E-05.129E-04.172E-04.215E-05.645E-05.107E-04.150E-04.193E-04.150E-04STEP=1SUB=9RFRQ=101.178IFRQ=OMODERealpartUSUM(AVG)RSYS=ODMX=.150E-04SMX=.150E-040.334E-05.669
5、E-05.100E-04.134E-04.167E-05.502E-05.836E-05.117E-04STEP=10.222E-05.444E-05.667E-05.889E-05.111E-05.333E-05.556E-05.778E-05.100E-04SUB=8RFRQ=97.801IFRQ=OMODERealpartUSUM(AVG)RSYS=ODMX=.100E-04SMX=.1OOE-O4STEP=1SUB=7MX0.266E-05.531E-05.797E-05.106E-04
6、.133E-05.398E-05.664E-05•929E-05.119E-04对应部分振动方向相同。最大位移处出现在上侧振动处ORFRQ=77.658IFRQ=OMODERealpartUSUM(AVG)RSYS=ODMX=.119E-04SMX=.119E-04STEP=1SUB=6.126E-05.377E-05.629E-05.881E-05.113E-04另外左右两边也有小幅振动。最大位移出现在下两边。RFRQ=73.913IFRQ=OMODERealpartUSUM(AVG)RSYS
7、=ODMX=.113E-04SMX=.113E-04STEP=1SUB=5RFRQ=43.178IFRQ=OMODERealpartUSUM(AVG)RSYS=ODMX=.172E-040.382E-05.765E-05.115E-04.153E-04.191E-05.574E-05SMX=.172E-04STEP=1SUB=4RFRQ=33.791IFRQ=OMODERealpartUSUM(AVG)RSYS=ODMX=.236E-040.525E-05.105E-04.158E-04.210
8、E-04.263E-05.788E-05.131E-04SMX=.236E-04STEP=1SUB=3RFRQ=28.186IFRQ=OMODERealpartUSUM(AVG)RSYS=ODMX=.239E-04SMX=.239E-04.531E-05.106E-04.159E-04.212E-04.266E-05.797E-05.133E-04STEP=1SUB=2RFRQ=25.416IFRQ=OMODERealpartUSUM(AVG)RSYS=ODMX=.297E-04SMX=.297