用ANSYS建立钢筋混凝土梁模型.docx
《用ANSYS建立钢筋混凝土梁模型.docx》由会员分享，可在线阅读，更多相关《用ANSYS建立钢筋混凝土梁模型.docx（24页珍藏版）》请在冰豆网上搜索。
用ANSYS建立钢筋混凝土梁模型
用ANSYS1立钢筋混凝土梁模型
问题描述：
钢筋混凝土梁在受到中间位移荷载的条件下的变形以及个组成部分的应力情况。
P=5mm位移
L=2000mm二
图1钢筋混凝土结构尺寸图
一、用合并节点的方法模拟钢筋混凝土梁
1.用solid65号单元以及beam188单元时
材料特性
钢材的应力应变关系
混凝土的弹性模量采用线弹性
建立钢筋线
TEMP
对钢筋线划分网格后形成钢筋单元
合并单元节点后施加约束以及位移载荷
进入求解器进行求解
钢筋单元的受力云图
.4fZE；-jJ
75.26?
15C-.53E275.SOt
112*S03186.172
301.075
263.44330*709
37.635
混凝土的应力云图
3.1119,E2214,13319.64425-155
混凝土开裂
2使用单元solid45号单元与beam188
钢筋的应力应变关系不变，而混凝土应力应变关系为：
混凝土单元
钢筋单元
力与位移曲线
11.IT7733
4.QQ314tGJ20P21725.704
p（kn）
uy
合并节点时的命令流
fini
/clear,nostart/config,nres,5000/prep7/title,rc-beamb=150h=300a=301=2000fcu=40ec=2.85e4displacement=10!
定义单元类型et,1,solid45et,2,beam188et,3,plane42!
定义截面类型sectype,1,beam,csolid,,0secoffset,centsecdata,8,0,0,0,0,0,0,0,0,0sectype,2,beam,csolid,,0secoffset,centsecdata,4,0,0,0,0,0,0,0,0,0!
定义材料属性，混凝土材料属性mp,ex,1,ecmp,prxy,1,0.2tb,kinh,1,,16tbpt,,0.000179067,5.10tbpt,,0.000358133,9.67tbpt,,0.0005372,1.37e1tbpt,,0.000716267,1.72e1tbpt,,0.000895333,2.01e1tbpt,,0.0010744,2.26e1tbpt,,0.001253467,2.44e1tbpt,,0.001432533,2.58e1tbpt,,0.0016116,2.66e1tbpt,,0.001790667,2.69e1tbpt,,0.0019916,2.65e1tbpt,,0.002393467,2.57e1tbpt,,0.002795333,2.48e1tbpt,,0.0031972,2.40e1tbpt,,0.003599067,2.32e1tbpt,,0.0038,2.28e1
tb,conc,1,1,9tbdata,,0.4,1,3,-1
!
纵向受拉钢筋mp,ex,2,2e5
mp,prxy,2,0.3
tb,bkin,2,1,2,1tbdata,,350
!
横向箍筋，受压钢筋材料属性
mp,ex,3,2e5mp,prxy,3,0.25tb,bkin,3,1,2,1tbdata,,200
!
生成钢筋线
k,,
k,,b
kgen,2,1,2,,,h
k,,a,a
k,,b-a,a
kgen,2,5,6,,,h-2*a
kgen,21,5,8,,,,-100*do,i,5,84,1
l,i,i+4
*enddo
*do,i,5,85,4
l,i,i+1
l,i,i+2
*enddo
*do,i,8,88,4
1.1.1-1
1.1.1-2
*enddo
!
受拉钢筋
lsel,s,loc,y,alsel,r,loc,x,alsel,a,loc,x,b-alsel,r,loc,y,acm,longitudinal,linetype,2
mat,2
secnum,1lesize,all,50lmesh,allallscmsel,u,longitudinalcm,hoopingreinforcement,line
!
箍筋，受压钢筋
type,2
mat,2
secnum,2lesize,all,50lmesh,all/eshape,1
!
将钢筋节点建为一个集合cm,steel,node
!
生成面单元，以便拉伸成体单元a,1,2,4,3
lsel,s,loc,y,0lsel,a,loc,y,hlesize,all,,,8lsel,alllsel,s,loc,x,0lsel,a,loc,x,blesize,all,,,10type,3amesh,all
!
拉伸成混凝土单元
type,1
real,3
mat,1extopt,esize,20extopt,aclear,1vext,all,,,,,-lalls
!
合并节点
nummrg,all
d,all,uy
d,all,ux
!
施加外部荷载
/solunsel,allnsel,s,loc,y,hnsel,r,loc,z,-1000d,all,uy,-displacementalls
!
求解
nlgeom,on
nsubst,50outres,all,allneqit,50pred,on
cnvtol,f,,0.05,2,0.5allsel
solvefinish
/post1allsel
/device,vector,1
!
时间历程后处理
/post26
nsel,s,loc,z,-l/2
*get,Nmin,node,0,num,minnsol,2,nmin,u,yprod,3,2,,,,,,-1
nsel,s,loc,y,0
nsel,r,loc,z,0
*get,Nnum,node,0,count*get,Nmin,node,0,num,minn0=Nmin
rforce,5,Nmin,f,y
*do,i,2,ndinqr（1,13）ni=ndnext（nO）rforce,6,ni,f,yadd,5,5,6
n0=ni
*enddo
prod,7,5,,,,,,1/1000/axlab,x,uy
/axlab,y,p（kn）xvar,3
numcmp,all!
边界条件约束nsel,s,loc,y,0nsel,r,loc,z,0d,all,uyd,all,uxnsel,s,loc,y,0nsel,r,loc,z,-l
plvar,7
二、用约束方程法模拟钢筋混凝土梁
1.用solid65号单元以及beam188单元时
混凝土以及钢筋采用线弹性关系：
建立钢筋线
对钢筋线划分网格后形成钢筋单元
建立混凝土单元
对钢筋线节点以及混凝土节点之间建立约束方程
后施加约束以及位移载荷
进入求解器进行求解；钢筋单元的受力云图
K2,
+0ll9a517.787155233.337311.11；
318汕丄11苛5194.45272.225350
混凝土的应力云图
.147216t.96913.1920.£1227.433
3.55BL0*37&17*20124.02230.644
混凝土开裂
2使用单元solid45号单元与beam188
使用混凝土的本构关系曲线
钢材的本构关系曲线
pfkn）
EPS
钢筋的vonmises应力
11^49177.656155,2233.372311.124
38.992116.7441列”牝272.246350
混凝土的应力
1287995.R951I17-42923.195
3.0126.77914.54520.31226.079
用在solid45号单元下，用合并节点法、约束方程法建立模中钢筋与混凝土之间的关系
的时候的一个力与位移全程曲线的比较。
—合并节点
T-约束方程
约束方程法命令流：
fini
/clear,nostart
/config,nres,5000
/prep7
/title,rc-beam
b=150
h=300
a=30
1=2000
fcu=40
ec=2.85e4
displacement=5
!
定义单元类型
et,1,solid65
et,2,beam188
et,3,plane42
!
定义截面类型
sectype,1,beam,csolid,,0
secoffset,cent
secdata,8,0,0,0,0,0,0,0,0,0sectype,2,beam,csolid,,0secoffset,cent
secdata,4,0,0,0,0,0,0,0,0,0
!
定义材料属性，混凝土材料属性
mp,ex,1,ec
mp,prxy,1,0.2
tb,kinh,1,,16
tbpt,,0.000179067,5.10
tbpt,,0.000358133,9.67
tbpt,,0.0005372,1.37e1
tbpt,,0.000716267,1.72e1
tbpt,,0.000895333,2.01e1tbpt,,0.0010744,2.26e1tbpt,,0.001253467,2.44e1
tbpt,,0.001432533,2.58e1tbpt,,0.0016116,2.66e1tbpt,,0.001790667,2.69e1tbpt,,0.0019916,2.65e1tbpt,,0.002393467,2.57e1tbpt,,0.002795333,2.48e1tbpt,,0.0031972,2.40e1tbpt,,0.003599067,2.32e1tbpt,,0.0038,2.28e1
!
tb,conc,1,1,9
!
tbdata,,0.4,1,3,-1
!
纵向受拉钢筋mp,ex,2,2e5
mp,prxy,2,0.3
tb,bkin,2,1,2,1tbdata,,350
!
横向箍筋，受压钢筋材料属性
mp,ex,3,2e5mp,prxy,3,0.25tb,bkin,3,1,2,1tbdata,,200
!
生成钢筋线
k,,
k,,b
kgen,2,1,2,,,h
k,,a,a
k,,b-a,a
kgen,2,5,6,,,h-2*a
kgen,21,5,8,,,,-100*do,i,5,84,1
l,i,i+4
*enddo
*do,i,9,81,4
l,i,i+1
l,i,i+2
*enddo
*do,i,12,84,4
1.1.1-1
1.1.1-2
*enddo
!
受拉钢筋
lsel,s,loc,y,alsel,r,loc,x,alsel,a,loc,x,b-alsel,r,loc,y,acm,longitudinal,linetype,2
mat,2
d,all,ux
nsel,s,loc,y,0
nsel,r,loc,z,-l
d,all,uy
d,all,ux
!
施加外部荷载
/solu
nsel,all
nsel,s,loc,y,hnsel,r,loc,z,-1000d,all,uy,-displacementalls
!
求解
nlgeom,onnsubst,200outres,all,allneqit,1OOpred,on
cnvtol,f,,0.05,2,0.5
allsel
solve
finish
/post1allselplcrack,0,1plcrack,0,2
!
时间历程后处理
/post26
nsel,s,loc,z,-l/2
*get,Nmin,node,0,num,minnsol,2,nmin,u,yprod,3,2,,,,,,-1
nsel,s,loc,y,0
nsel,r,loc,z,0
*get,Nnum,node,0,count*get,Nmin,node,0,num,minn0=Nmin
rforce,5,Nmin,f,y
*do,i,2,ndinqr（1,13）ni=ndnext（nO）rforce,6,ni,f,yadd,5,5,6n0=ni
*enddo
secnum,1lesize,all,50lmesh,allallscmsel,u,longitudinalcm,hoopingreinforcement,line
!
箍筋，受压钢筋
type,2
mat,2
secnum,2lesize,all,50lmesh,all/eshape,1
!
将钢筋节点建为一个集合cm,steel,node
!
生成面单元，以便拉伸成体单元a,1,2,4,3
lsel,s,loc,y,0lsel,a,loc,y,hlesize,all,,,10lsel,alllsel,s,loc,x,0lsel,a,loc,x,blesize,all,,,20type,3amesh,all
!
拉伸成混凝土单元
type,1
real,3
mat,1extopt,esize,30extopt,aclear,1vext,all,,,,,-lalls
!
建立约束方程
cmsel,s,hoopingreinforcementcmsel,a,longitudinalnsll,s,1ceintf,,ux,uy,uzallsel,all
!
边界条件约束
nsel,s,loc,y,0
nsel,r,loc,z,0
d,all,uy
xvar,3
plvar,7
prod,7,5,,,,,,1/1000
/axlab,x,uy
/axlab,y,p（kn）
升级会员 