ansys连续钢构桥变截面底板纵向为二次抛物梁单元模拟

bjtchenjing

ansys连续钢构桥变截面底板纵向为二次抛物梁单元这段命令流怎么写呀？谢谢！

lscookie

信息量不足，给出图片说明

bjtchenjing

我举书上个例子，书上是用壳单元的。如果用梁单元怎么写。关键点都是一样的。抛物线的那段变截面梁单元怎么写？
/prep7
!上顶板关键点位置0m
k,1,
k,2,-2.1
k,3,-2.6,-0.125
k,4,-2.8,-0.125
k,5,-3,-0.125
k,6,-3.4857,-0.1036
k,7,-3.9714,-0.0821
k,8,-4.4571,-0.0607
k,9,-4.9429,-0.0393
k,10,-5.4286,-0.0179
k,11,-5.9143,0.0036
k,12,-6.4,0.025
!下底板关键点位置
k,13,-2.8,-1.85
k,14,,-1.85
!上顶板关键点位置49m
kgen,9,1,12,1,0,0,49/8,100
!上顶板关键点位置50m
kgen,2,1,12,1,,,50,900
!上顶板关键点位置50+34.5m
kgen,9,901,912,1,,,34.5/8,100


c1=0.000843399 !边跨二次抛物线系数
c2=0.001701323 !中跨二次抛物线系数
!yb＝1.85＋c1*x1^2 !边跨梁高变化曲线
!ym＝1.85＋c2*x2^2 !中跨梁高变化曲线
*dim,x1,array,8
x1(1)=49/8
x1(2)=2*49/8
x1(3)=3*49/8
x1(4)=4*49/8
x1(5)=5*49/8
x1(6)=6*49/8
x1(7)=7*49/8
x1(8)=8*49/8
*dim,x2,array,8
x2(1)=34.5/8
x2(2)=2*34.5/8
x2(3)=3*34.5/8
x2(4)=4*34.5/8
x2(5)=5*34.5/8
x2(6)=6*34.5/8
x2(7)=7*34.5/8
x2(8)=8*34.5/8
*dim,yb,array,8
yb(1)=c1*x1(1)*x1(1)
yb(2)=c1*x1(2)*x1(2)
yb(3)=c1*x1(3)*x1(3)
yb(4)=c1*x1(4)*x1(4)
yb(5)=c1*x1(5)*x1(5)
yb(6)=c1*x1(6)*x1(6)
yb(7)=c1*x1(7)*x1(7)
yb(8)=c1*x1(8)*x1(8)
*dim,ym,array,8
ym(1)=-c2*(x2(1)-34.5)**2+2.2025
ym(2)=-c2*(x2(2)-34.5)**2+2.2025
ym(3)=-c2*(x2(3)-34.5)**2+2.2025
ym(4)=-c2*(x2(4)-34.5)**2+2.2025
ym(5)=-c2*(x2(5)-34.5)**2+2.2025
ym(6)=-c2*(x2(6)-34.5)**2+2.2025
ym(7)=-c2*(x2(7)-34.5)**2+2.2025
ym(8)=-c2*(x2(8)-34.5)**2+2.2025
kgen,2,13,14,1,0,-yb(1),x1(1),100
kgen,2,13,14,1,0,-yb(2),x1(2),200
kgen,2,13,14,1,0,-yb(3),x1(3),300
kgen,2,13,14,1,0,-yb(4),x1(4),400
kgen,2,13,14,1,0,-yb(5),x1(5),500
kgen,2,13,14,1,0,-yb(6),x1(6),600
kgen,2,13,14,1,0,-yb(7),x1(7),700
kgen,2,13,14,1,0,-yb(8),x1(8),800

kgen,2,813,814,1,0,0,1,100

kgen,2,913,914,1,0,ym(1),x2(1),100
kgen,2,913,914,1,0,ym(2),x2(2),200
kgen,2,913,914,1,0,ym(3),x2(3),300
kgen,2,913,914,1,0,ym(4),x2(4),400
kgen,2,913,914,1,0,ym(5),x2(5),500
kgen,2,913,914,1,0,ym(6),x2(6),600
kgen,2,913,914,1,0,ym(7),x2(7),700
kgen,2,913,914,1,0,ym(8),x2(8),800

kgen,2,1,14,1,,,-0.5,50