用ANSYS建立钢筋混凝土梁模型

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。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B=150mmH=300mmL=2000mmP=5mm位移图1钢筋混凝土结构尺寸图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合并节点时的命令流:fini/clear,nostart/config,nres,5000/prep7/title,rc-beamb=150h=300a=30l=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,,0.0038,2.28e1tb,conc,1,1,9tbdata,,0.4,1,3,-1!纵向受拉钢筋mp,ex,2,2e5mp,prxy,2,0.3tb,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,,bkgen,2,1,2,,,hk,,a,ak,,b-a,akgen,2,5,6,,,h-2*akgen,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,8,88,4l,i,i-1l,i,i-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,2mat,2secnum,1lesize,all,50lmesh,allallscmsel,u,longitudinalcm,hoopingreinforcement,line!箍筋，受压钢筋type,2mat,2secnum,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,1,2,4,3lsel,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,1real,3mat,1extopt,esize,20extopt,aclear,1vext,all,,,,,-lalls!合并节点nummrg,allnumcmp,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,-ld,all,uyd,all,ux!施加外部荷载/solunsel,allnsel,s,loc,y,hnsel,r,loc,z,-1000d,all,uy,-displacementalls!求解nlgeom,onnsubst,50outres,all,allneqit,50pred,oncnvtol,f,,0.05,2,0.5allselsolvefinish/post1allsel/device,vector,1!时间历程后处理/post26nsel,s,loc,z,-l/2*get,Nmin,node,0,num,minnsol,2,nmin,u,yprod,3,2,,,,,,-1nsel,s,loc,y,0nsel,r,loc,z,0*get,Nnum,node,0,count*get,Nmin,node,0,num,minn0=Nminrforce,5,Nmin,f,y*do,i,2,ndinqr(1,13)ni=ndnext(n0)rforce,6,ni,f,yadd,5,5,6n0=ni*enddoprod,7,5,,,,,,1/1000/axlab,x,uy/axlab,y,p(kn)xvar,3plvar,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曲线钢筋的vonmises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。020406080100120024681012合并节点约束方程约束方程法命令流:fini/clear,nostart/config,nres,5000/prep7/title,rc-beamb=150h=300a=30l=2000fcu=40ec=2.85e4displacement=5!定义单元类型et,1,solid65et,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,1,1,9!tbdata,,0.4,1,3,-1!纵向受拉钢筋mp,ex,2,2e5mp,prxy,2,0.3tb,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,,bkgen,2,1,2,,,hk,,a,ak,,b-a,akgen,2,5,6,,,h-2*akgen,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,12,84,4l,i,i-1l,i,i-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,2mat,2secnum,1lesize,all,50lmesh,allallscmsel,u,longitudinalcm,hoopingreinforcement,line!箍筋，受压钢筋type,2mat,2secnum,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,1,2,4,3lsel,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,1real,3mat,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,0nsel,r,loc,z,0d,all,uyd,all,uxnsel,s,loc,y,0nsel,r,loc,z,-ld,all,uyd,all,ux!施加外部荷载/solunsel,allnsel,s,loc,y,hnsel,r,loc,z,-1000d,all,uy,-displacementalls!求解nlgeom,onnsubst,200outres,all,allneqit,100pred,oncnvtol,f,,0.05,2,0.5allselsolvefinish/post1allselplcrack,0,1plcrack,0,2!时间历程后处理/post26nsel,s,loc,z,-l/2*get,Nmin,node,0,num,minnsol,2,nmin,u,yprod,3,2,,,,,,-1nsel,s,loc,y,0nsel,r,loc,z,0*get,Nnum,node,0,count*get,Nmin,node,0,num,minn0=Nminrforce,5,Nmin,f,y*do,i,2,ndinqr(1,13)ni=ndnext(n0)rforce,6,ni,f,yadd,5,5,6n0=ni*enddoprod,7,5,,,,,,1/1000/axlab,x,uy/axlab,y,p(kn)xvar,3plvar,7