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- /filname,earthquake,1
- keyw,pr_struc,1
- /prep7
- !定义单元类型
- et,1,beam3
- !定义实常数
- r,1,0.0256,0.546e-4,0.16
- !定义材料属性
- mp,ex,1,2.1e11
- mp,prxy,1,0.3
- mp,dens,1,7800
- !建立几何模型
- k,1
- k,2,100
- k,3,50,25
- l,1,2
- larc,1,2,3
- ldiv,1,,,10
- ksel,s,kp,,4,12,1
- kgen,2,all,,,,30
- kdele,3
- *do,i,4,12,1
- l,i,i+9
- *enddo
- lsel,s,line,,12,20,1
- lsel,a,line,,2
- lsbl,2,all
- l,4,22
- *do,i,5,11,1
- l,i,i+19
- *enddo
- l,12,23
- allsel
- numcmp,all
- !网格划分、施加边界条件
- lesize,all,,,1
- type,1
- real,1
- mat,1
- lmesh,all
- !边界条件
- nsel,s,loc,x,0
- nsel,a,loc,x,100
- d,all,uy
- nsel,s,loc,x,0
- d,all,ux
- allsel
- !自重载荷
- acel,,9.81
- allsel
- !地震谱响应分析
- /solu
- !模态分析
- antype,2
- MODOPT,SUBSP,28 !子空间法,模态提取数为28
- solve
- finish
- !谱分析
- /solu
- antype,8 !spectrum
- SPOPT,SPRS,28,1 !单点谱分析,参与计算的模态数为28,打开盈利响应开关
- SVTYP,2,1, !设置反应谱类型为加速度谱,
- SED,1,1,0, !激励方向为x和y
- FREQ,0.333,0.357,0.416,0.5,0.625,0.833,1.25,2.5,2.778
- FREQ,3.333,3.571,4.167,5,6.25,8.333,10,12.5,16.667
- FREQ,25,50
- SV,0,0.013,0.013884,0.01595,0.018794,0.022974,0.029763,0.042871,0.08,0.08,
- SV,0,0.08,0.08,0.08,0.08,0.08,0.08,0.08,0.0712,0.0624,
- SV,0,0.0536,0.0448,
- solve
- finish
- !模态扩展
- /solu
- antype,2
- MODOPT,SUBSP,28
- mxpand,28,0,0,1,0.005 !扩展模态数设为28,计算单元的计算结果设置为yes,临界点设为0.005
- solve
- finish
- !模态合并
- /solu
- antype,8
- SRSS,0.02,DISP !组合方式为srss,significant threshold为0.02
- solve
- finish
- !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
- !地震波瞬态分析
- *DIM,tjx,ARRAY,2,50,1
- *DIM,tjy,ARRAY,2,50,1
-
- *CREATE,ansuitmp
- *VREAD,tjx(1,1),'tjx','txt',' ',50
- (e9.3,e11.3)
- *END
- /INPUT,ansuitmp
-
- *CREATE,ansuitmp
- *VREAD,tjy(1,1),'tjy','txt',' ',50
- (e9.3,e11.3)
- *END
- /INPUT,ansuitmp
- !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!以下是一个矩阵形成的例子
- !*dim,aa,,2,3
- !*CREATE,ansuitmp
- !*vread,aa(1,1),'data','dat',,jik,3,2
- !!!!!!!!!!!!!!!!!!!!!!!!!!!!!
- antype,4
- TRNOPT,FULL
- LUMPM,0
- *do,t,1,50,1 !地震波时间5秒,时间间隔0.1
- time,0.1*t !每一个加载步的求解时间
- kbc,0 !每个加载步的加载形式为rump
- nsub,10 !每一个加载步分为10个子步
- acel,tjx(2,t),tjy(2,t) !施加横向、竖向加速度
- nsel,all
- solve !求解
- *enddo
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