第1节  挠度计算例子

1、用opensees计算杆端内力

TCL：

logFile log.txt

wipe

 

set m 1.

set KN 1.

set sec 1.

set LunitTXT "m"

set FunitTXT "kN"

set TunitTXT "sec"

set mm [expr $m/1000.]

set m2 1.

set mm2 [expr $m2/1e6]

set KPa [expr $KN/pow($m,2)];

set MPa [expr $KPa*1000.]

set sg [expr $KN/pow($m,3)]

set PI [expr 2*asin(1.0)]

set Ubig 1.e10

set Usmall [expr 1/$Ubig]

 

model basic -ndm 3 -ndf 6

node 1 0 0 -0.2

node 2 10 0 -0.2

 

 

fix 1 1 1 1 1 1 1

fix 2 1 1 1 1 1 1

geomTransf Linear 1 0 0 1

set E 206000000 

set G 79000000

set A1 0.020496

set A2 0.013216

set J1 2812267e-12

set J2 891733e-12

set Iy1 99155648e-12

set Iz1 880954688e-12

set Iy2 21415125e-12

set Iz2 729076565e-12

element elasticBeamColumn  1  1  2  $A2 $E $G $J2 $Iy2 $Iz2 1;

timeSeries Constant 1

system BandGeneral

numberer RCM

constraints Penalty 1e160 1e160

integrator LoadControl 1.0

test NormDispIncr 1.0e-10 6

algorithm Newton

 

pattern Plain 1 1 {

eleLoad -ele 1 -type -beamUniform -1.037 0

eleLoad -ele 1 -type -beamUniform -100 0

}

 

recorder Node -file NodeReaction.txt -time -nodeRange 1 1 -dof 1 2 3 4 5 6 reaction

recorder Element -file ElementForce.txt -time -eleRange 1 2 localForces

 

analysis Static

 

set res1 [analyze 1]

if { $res1 != 0 } {

puts "失败"

wipe

return $res1

}

remove loadPattern 1

remove recorders

wipe

#end

2、openSEES计算结果

ElementForce.txt内容：

1 0 505.185 0 0 0 841.975 0 505.185 0 0 0 -841.975

其中 Ma=841.975,Mb=-841.975;

3、挠度计算

double E=206000000.0;

         double I=729076565e-12;

         double 挠度[11];

         ::ZeroMemory(挠度,11*sizeof(double));

1)     计算荷载引起的挠度   

         double q,L;

         q=-(100+1.037);

         L=10;

         int 分段数=10;

         简支梁_均布荷载_挠度计算(q,L,E,I,分段数,挠度);

2)     计算端部内力引起的挠度

         简支梁_两端弯矩荷载_挠度计算(L,E,I,841.975,-841.975,分段数,挠度);

3)     结果比较

以上计算结果：

http://s3/middle/900f3a81gb0e41e853602&690
SS2000结果：

http://s7/middle/900f3a81gb0e420185e96&690

 

正确!