用ANSYS建立钢筋混凝土梁模型

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。P=5mm位移L=2000mm图1钢筋混凝土结构尺寸图H=300mmB=150mm、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(self”钢材的应力应变关系混凝土的弹性模量采用线弹性EX24752価24152他337523£Q；3C-：-2I3C2GC创0TEMP怯比"II2525tC；LOOCacQ2275-HJ30-f30-BOO建立钢筋线十',<._fp1<'.<1对钢筋线划

2、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一Qe-Z37.CBS112.031.SO.S5013£.1722£3.443 me. 70'*混凝土的应力云图25.155.3557135.36?11.3'7316.09522.4J.111S .62214.12319.644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:EPS混凝土单元钢筋单元力与位移曲线 ?3L& ；'£t3.311 u 二工.丁Pl155.5fc71

3、 e.-B.J1jB3t1： X； ib" -Qff3ue 二2233.343斗MTp1：hn合并节点时的命令流fini /clear, no start /con fig, nres,5000 /prep7/title,rc -beamb=150h=300 a=30 l=2000fcu=40ec=2.85e4dis pl aceme nt=10!定义单元类型et,1,solid45et,2,beam188et,3, pla ne42!定义截面类型sect yp e,1,beam,csolid,0 secoffset,ce nt secdata,8,0,0,0,0,0,0,0,0,0

4、sect yp e,2,beam,csolid,0 secoffset,ce nt 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,ecmp,p rxy,1,0.2tb,ki nh,1,16tbp t,0.000179067,5.10tbp t,0.000358133,9.67tbp t,0.0005372,1.37e1tbp t,0.000716267,1.72e1tbp t,0.000895333,2.01e1tbp t,0.0010744,2.26e1tbp t,0.001253467,2.44e1tbp t,0.001432533,

5、2.58e1tbp t,0.0016116,2.66e1tbp t,0.001790667,2.69e1tbp t,0.0019916,2.65e1tbp t,0.002393467,2.57e1tbp t,0.002795333,2.48e1tb pt,0.0031972,2.40e1tbp t,0.003599067,2.32e1tb pt,0.0038,2.28e1tb,co nc,1,1,9tbdata,0.4,1,3, -1!纵向受拉钢筋mp ,ex,2,2e5mp,p rxy,2,0.3tb,bki n,2,1,2,1tbdata,350!横向箍筋，受压钢筋材料属性mp ,ex,3

6、,2e5mp,p rxy,3,0.25tb,bki n,3,1,2,1tbdata,200!生成钢筋线k,k,bkge n,2,1,2,hk,a,ak,b-a,akge n,2,5,6,h-2*akge n,21,5,8,-100*do,i,5,84,1l, i,i+4*enddo*do,i,5,85,4l,i,i+1l,i,i+2*enddo *do,i,8,88,41.1.1- 11.1.1- 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,lon gitudi nal,li ne

7、ty pe,2mat,2sec nu m,1 lesize,all,50 lmesh,all allsd,all,uyd,all,ux!施加外部荷载/solunsel,all nsel,s,loc,y,h nsel,r,loc,z,-1000 d,all,uy,-dis placeme nt alls!求解nl geom ,onn subst,50 outres,all,alln eqit,50p red, oncn vtol,f,0.05,2,0.5 allselsolve finish/p ost1 allsel /device,vector,1!时间历程后处理/p ost26n sel,

8、s,loc, z,-l/2*get,N min,no de, 0,num,min n sol,2, nmi n,u,yprod,3,2, -1n sel,s,loc,y,0n sel,r,loc, z,0*get, Nnum,no de,0,co unt *get,Nmi n,no de,0, num ,min n0=Nminrforce,5,Nmi n,f,y*do,i,2, ndi nqr(1,13) ni=ndn ext (n0) rforce,6, ni,f,y add,5,5,6n0=ni*enddop rod,7,5,1/1000 /axlab,x,uy/axlab, y,p(kn

9、) xvar,3cmsel,u,lo ngitudi nalcm,ho oping rein forceme nt,li ne!箍筋，受压钢筋typ e,2mat,2sec nu m,2 lesize,all,50 lmesh,all /esha pe,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,1,2,4,3lsel,s,loc,y,0 lsel,a,loc,y,h lesize,all,8 lsel,all lsel,s,loc,x,0 lsel,a,loc,x,b lesize,all,10 typ e,3 amesh,all!拉伸成混凝土单

10、元typ e,1real,3mat,1ext op t,esize,20ext op t,aclear,1 vext,all,-l alls!合并节点num mrg,all numcmp ,all!边界条件约束n sel,s,loc,y,0n sel,r,loc, z,0d,all,uyd,all,uxn sel,s,loc,y,0n sel,r,loc, z,-lpl var,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155.350对钢筋线节点以及混凝土节点之间建

11、立约束方程身一£hK.u.a*隨严严严貸 ?SWW!?f *£鼻卜L ?VU”N SUTI, IRb后施加约束以及位移载荷r 进入求解器进行求解；钢筋单元的受力云图.01190517.78736+的91丄总苛5194.45233.3373L1+112混凝土的应力云图KN.1472163,55010.37913-7920.61217.20127.43324.02230.644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"kn】钢材的本构关系曲线SIGZPS钢筋的von mises应力35077.3S155.62233,37239.

12、992116.74194.4%272-2 刖:混凝土的应力,1287095,9951L.6633.0126.77914.54517.42520.31223,195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, no start /con fig, nres,5000 /prep7/title,rc -beam b=150 h=300a=30 1=2000 fcu=40 ec=2.85e4 dis pl aceme nt=5!定义单元类型et,1,solid65 et,2,beam1

13、88et,3, pla ne42!定义截面类型sect yp e,1,beam,csolid,0 secoffset,ce nt secdata,8,0,0,0,0,0,0,0,0,0sect yp e,2,beam,csolid,0 secoffset,ce nt 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,ecmp,p rxy,1,0.2tb,ki nh,1,16tbp t,0.000179067,5.10tbp t,0.000358133,9.67tbp t,0.0005372,1.37e1tbp t,0.000716267,1.

14、72e1tbp t,0.000895333,2.01e1tbp t,0.0010744,2.26e1tbp t,0.001253467,2.44e1tbp t,0.001432533,2.58e1tbp t,0.0016116,2.66e1tbp t,0.001790667,2.69e1tbp t,0.0019916,2.65e1tbp t,0.002393467,2.57e1tbp t,0.002795333,2.48e1tbp t,0.0031972,2.40e1tbp t,0.003599067,2.32e1tb pt,0.0038,2.28e1!tb,co nc,1,1,9!tbdat

15、a,0.4,1,3, -1!纵向受拉钢筋mp ,ex,2,2e5mp,p rxy,2,0.3tb,bki n,2,1,2,1tbdata,350!横向箍筋，受压钢筋材料属性mp ,ex,3,2e5mp,p rxy,3,0.25tb,bki n,3,1,2,1 tbdata,200!生成钢筋线k,k,bkge n,2,1,2,hk,a,ak,b-a,akge n,2,5,6,h-2*akge n,21,5,8,-100*do,i,5,84,1l, i,i+4*enddo*do,i,9,81,4l,i,i+1l,i,i+2*enddo *do,i,12,84,41.1.1- 11.1.1- 2*e

16、nddo!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,lon gitudi nal,li ne ty pe,2mat,2d,all,uxn sel,s,loc,y,0 n sel,r,loc ,z,-l d,all,uy d,all,ux!施加外部荷载/solunsel,all nsel,s,loc,y,h nsel,r,loc,z,-1000 d,all,uy,-dis placeme nt alls!求解nl geom ,on nsubst,200 outres,all,all neqit,10

17、0 p red, oncn vtol,f,0.05,2,0.5 allselsolve finish/p ost1 allselp lcrack,0,1p lcrack,0,2!时间历程后处理/p ost26n sel,s,loc,z,-l/2*get,Nmi n,no de,0, num ,min n sol,2, nmi n,u,yprod,3,2, -1n sel,s,loc,y,0n sel,r,loc, z,0*get, Nnum,no de,0,co unt *get,N min,no de, 0,num,min n0=Nminrforce,5,Nmi n,f,y*do,i,2,

18、ndi nqr(1,13) ni=ndn ext (n0) rforce,6, ni,f,y add,5,5,6 n0=ni*enddosec nu m,1 lesize,all,50 lmesh,all alls cmsel,u,lo ngitudi nalcm,ho oping rein forceme nt,li ne!箍筋，受压钢筋typ e,2mat,2sec nu m,2 lesize,all,50 lmesh,all /esha pe,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,1,2,4,3lsel,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 typ e,3 a