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1、 用ANSYS建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。 B=150mm H=300mm L=2000mm P=5mm位移载荷 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
2、凝土的弹性模量采用线弹性 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 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 混凝土的应力云图 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 精品资料 合并节点时的命令流: fini /clear,nostart /config,nres,5000 /prep7 /title,rc-b
3、eam b=150 h=300 a=30 l=2000 fcu=40 ec=2.85e4 displacement=10 !定义单元类型 et,1,solid45 et,2,beam188 et,3,plane42 !定义截面类型 sectype,1,beam,csolid,,0 secoffset,cent secdata,8,0,0,0,0,0,0,0,0,0 sectype,2,beam,csolid,,0 secoffset,cent secdata,4,0,0,0,0,0,0,0,0,0 !定义材料属性
4、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,,0.000895333,2.01e1 tbpt,,0.0010744,2.26e1 tbpt,,0.001253467,2.44e1 tbpt,,0.001432533,2.58e1 tbpt,,0.0016116,2.66e1 tbpt,,0.001790667,2.
5、69e1 tbpt,,0.0019916,2.65e1 tbpt,,0.002393467,2.57e1 tbpt,,0.002795333,2.48e1 tbpt,,0.0031972,2.40e1 tbpt,,0.003599067,2.32e1 tbpt,,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,2,1,2,1 tbdata,,350 !横向箍筋，受压钢筋材料属性 mp,ex,3,2e5 mp,prxy,3,0.
6、25 tb,bkin,3,1,2,1 tbdata,,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 l,i,i-1 l,i,i-2 *enddo !受拉钢筋 lsel,s,loc,y,a lsel,r,loc,x,a lsel,a,loc,x,b-a lsel
7、r,loc,y,a cm,longitudinal,line type,2 mat,2 secnum,1 lesize,all,50 lmesh,all alls cmsel,u,longitudinal cm,hooping reinforcement,line !箍筋，受压钢筋 type,2 mat,2 secnum,2 lesize,all,50 lmesh,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,0 lsel,a,loc,
8、y,h lesize,all,,,8 lsel,all lsel,s,loc,x,0 lsel,a,loc,x,b lesize,all,,,10 type,3 amesh,all !拉伸成混凝土单元 type,1 real,3 mat,1 extopt,esize,20 extopt,aclear,1 vext,all,,,,,-l alls !合并节点 nummrg,all numcmp,all !边界条件约束 nsel,s,loc,y,0 nsel,r,loc,z,0 d,all,uy d,all,ux nsel,s,loc,y,
9、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,h nsel,r,loc,z,-1000 d,all,uy,-displacement alls !求解 nlgeom,on nsubst,50 outres,all,all neqit,50 pred,on cnvtol,f,,0.05,2,0.5 allsel solve finish /post1 allsel /device,vector,1 !时间历程后处理 /p
10、ost26 nsel,s,loc,z,-l/2 *get,Nmin,node,0,num,min nsol,2,nmin,u,y prod,3,2,,,,,,-1 nsel,s,loc,y,0 nsel,r,loc,z,0 *get,Nnum,node,0,count *get,Nmin,node,0,num,min n0=Nmin rforce,5,Nmin,f,y *do,i,2,ndinqr(1,13) ni=ndnext(n0) rforce,6,ni,f,y add,5,5,6 n0=ni *enddo prod,7,5,,,,,
11、1/1000 /axlab,x,uy /axlab,y,p(kn) xvar,3 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 混凝土的应力云图 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
12、曲线 钢筋的von mises 应力 混凝土的应力 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。 约束方程法命令流: fini /clear,nostart /config,nres,5000 /prep7 /title,rc-beam b=150 h=300 a=30 l=2000 fcu=40 ec=2.85e4 displacement=5 !定义单元类型 et,1,solid65 et,2,beam188 et
13、3,plane42 !定义截面类型 sectype,1,beam,csolid,,0 secoffset,cent secdata,8,0,0,0,0,0,0,0,0,0 sectype,2,beam,csolid,,0 secoffset,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.
14、37e1 tbpt,,0.000716267,1.72e1 tbpt,,0.000895333,2.01e1 tbpt,,0.0010744,2.26e1 tbpt,,0.001253467,2.44e1 tbpt,,0.001432533,2.58e1 tbpt,,0.0016116,2.66e1 tbpt,,0.001790667,2.69e1 tbpt,,0.0019916,2.65e1 tbpt,,0.002393467,2.57e1 tbpt,,0.002795333,2.48e1 tbpt,,0.0031972,2.40e1 tbpt,,0.00359906
15、7,2.32e1 tbpt,,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,2,1,2,1 tbdata,,350 !横向箍筋，受压钢筋材料属性 mp,ex,3,2e5 mp,prxy,3,0.25 tb,bkin,3,1,2,1 tbdata,,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,,
16、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 l,i,i-1 l,i,i-2 *enddo !受拉钢筋 lsel,s,loc,y,a lsel,r,loc,x,a lsel,a,loc,x,b-a lsel,r,loc,y,a cm,longitudinal,line type,2 mat,2 secnum,1 lesize,all,50 lmesh,all alls cmsel,u,longitudinal cm,hoo
17、ping reinforcement,line !箍筋，受压钢筋 type,2 mat,2 secnum,2 lesize,all,50 lmesh,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,0 lsel,a,loc,y,h lesize,all,,,10 lsel,all lsel,s,loc,x,0 lsel,a,loc,x,b lesize,all,,,20 type,3 amesh,all !拉伸成混凝土单元 type,1
18、 real,3 mat,1 extopt,esize,30 extopt,aclear,1 vext,all,,,,,-l alls !建立约束方程 cmsel,s,hooping reinforcement cmsel,a,longitudinal nsll,s,1 ceintf,,ux,uy,uz allsel,all !边界条件约束 nsel,s,loc,y,0 nsel,r,loc,z,0 d,all,uy d,all,ux nsel,s,loc,y,0 nsel,r,loc,z,-l d,all,uy d,all,ux !施加外部
19、荷载 /solu nsel,all nsel,s,loc,y,h nsel,r,loc,z,-1000 d,all,uy,-displacement alls !求解 nlgeom,on nsubst,200 outres,all,all neqit,100 pred,on cnvtol,f,,0.05,2,0.5 allsel solve finish /post1 allsel plcrack,0,1 plcrack,0,2 !时间历程后处理 /post26 nsel,s,loc,z,-l/2 *get,Nm
20、in,node,0,num,min nsol,2,nmin,u,y prod,3,2,,,,,,-1 nsel,s,loc,y,0 nsel,r,loc,z,0 *get,Nnum,node,0,count *get,Nmin,node,0,num,min n0=Nmin rforce,5,Nmin,f,y *do,i,2,ndinqr(1,13) ni=ndnext(n0) rforce,6,ni,f,y add,5,5,6 n0=ni *enddo prod,7,5,,,,,,1/1000 /axlab,x,uy /axlab,y,p(kn) xvar,3 plvar,7 Welcome To Download !!! 欢迎您的下载，资料仅供参考！
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