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

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/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

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以下是计算的结果

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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

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以下是问题的讨论

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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)?

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