地震反应谱分析实例
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结构地震反应谱分析实例
在多位朋友的大力帮助下,经过半个多月的努力,鄙人终于对结构地震反应谱分析有了一定的了解,现将其求解步骤整理出来,以便各位参阅,同时,尚有一些问题,欢迎各位讨论!
为叙述方便,举一简单实例:
在侧水压与顶部集中力作用下的柱子的地震反应谱分析,谱值为加速度反应谱,考虑X与Y向地震效应作用。已知地震影响系数a与周期T的关系:
a(T)= 0.4853*(0.4444+2.2222*T) 0 0.4853*(0.10/T)^(-0.686) 0.04 0.4853 0.1 0.4853*(1.2/T)^1.5 1.2 以下是命令流程序 ---------------------------------------------------------------------------------------------------- /filname,SPEC,1 /PREP7 !定义单元类型及材料特性 ET,1,45 MP,EX,1,2.8E10 MP,DENS,1,2.4E3 MP,NUXY,1,0.18 !建立模型 BLOCK,0,1,0,1,0,5 !网格剖分 ESIZE,0.5 VMESH,all /VIEW,,-0.3,-1,1 EPLOT FINISH /SOLU !施加底部约束 ASEL,,LOC,Z,0 DA,ALL,ALL ALLSEL !施加自重荷载 ACEL,0,0,10 !进行模态求解 ANTYPE,MODAL MODOPT,LANB,30 SOLVE FINISH !进行谱分析 /SOLU ANTYPE,SPECTR SPOPT,SPRS,30,YES SVTYP,2 !加速度反应谱 SED,1,1 !X与Y向 FREQ,0.2500,0.2632,0.2778,0.2941,0.3125,0.3333,0.3571,0.3846,0.4167 FREQ,0.4545,0.5000,0.5556,0.6250,0.7143,0.8333,1.1111,2.0000,10.0000 FREQ,25.0000,1000.0000 SV,0.05,0.0797,0.0861,0.0934,0.1018,0.1114,0.1228,0.1362,0.1522,0.1716 SV,0.05,0.1955,0.2255,0.2642,0.3152,0.3851,0.4853,0.4853,0.4853,0.4853 SV,0.05,0.2588,0.2167 SOLVE FINISH !进行模态求解(模态扩展) /SOLU ANTYPE,MODAL EXPASS,ON MXPAND,30,,,YES,0.005 SOLVE FINISH !进行谱分析(合并模态) /SOLU ANTYPE,SPECTR SRSS,0.15,disp SOLVE FINISH /POST1 SET,LIST !结果1 /INP,,mcom LCASE,11 PRRSOL, !结果2 SET,FIRST PRRSOL, !结果3 SET,NEXT PRRSOL, !结果4 SET,NEXT PRRSOL, !结果5 SET,NEXT PRRSOL, !结果6 FINISH !静力分析 /SOLU ANTYPE,STATIC !施加水压荷载 NSEL,,LOC,Y,0 NSEL,R,LOC,Z,0,5 SFGRAD,PRES,0,Z,0,-10000 SF,ALL,PRES,50000 !施加集中荷载 NSEL,,LOC,Y,0 NSEL,R,LOC,z,5 F,ALL,FY,10000 ALLSEL EPLOT SOLVE FINISH /POST1 set,last lcwrite,12 Lcase,11 Lcoper,add,12 Lcwrite,13 LCASE,12 PRRSOL, !结果7 PRRSOL, !结果8 FINISH --------------------------------------------------------------------------------------------------- 以下是计算的结果 --------------------------------------------------------------------------------------------------- 结果1:(Results Summary) 1 21.647 2 21.647 3 121.51 4 121.51 结果2:(单独谱分析反力LCASE,11) VALUE 2467.9 2290.1 18384. 结果3:(单独谱分析反力SET,FIRST) VALUE 0.13334E+06-0.15785E+07-0.18819E-06 结果4:(单独谱分析反力SET,NEXT) VALUE -0.15785E+07-0.13334E+06-0.48918E-06 结果5:(单独谱分析反力SET,NEXT) VALUE -0.87805E+07 0.27008E+08 0.86846E-07 结果6:(单独谱分析反力SET,NEXT) VALUE 0.27008E+08 0.87805E+07 0.79325E-06 结果7:(单独静力分析反力LCASE,12) VALUE 0.22901E-08-0.15500E+06 0.12000E+06 结果8:(谱分析与静力分析叠加反力LCASE,13) VALUE 2467.9 -0.15271E+06 0.13838E+06 --------------------------------------------------------------------------------------------------- 以下是问题的讨论 --------------------------------------------------------------------------------------------------- 1、模态提取数为30,即取前30阶振型数,但在谱分析时得到的是4阶,这4阶是什么意思? 2、在单独谱分析时,为何结果2、 3、 4、 5、6会相差如此之大?(其应力和位移也是如此。) 3、在进行谱分析的合并模态步骤中,模态合并方法应该选用SRSS,0.15,DISP还是SRSS,0.15,ACEL(其对结果影响很大)?