(完整版)ansys钢筋混凝土梁的建模方法约束方程法

用拘束方程法模拟钢筋混凝土梁结构问题描述P位移载荷hE 1bL建立钢筋线建立混凝土单元对钢筋线节点以及混凝土节点之间建立拘束方程后施加拘束以及位移载荷混凝土的应力云图混凝土开裂fini/clear,nostart/config,nres,5000/filname,yuesufangcheng5jiamihunnintu/prep7/title,rc-beamb=150h=300a=30l=2000displacement=5!定义单元种类et,1,solid65et,2,beam188et,3,plane42!定义截面种类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!定义资料属性，混凝土资料属性mp,ex,1,24000tb,conc,1,1,9tbdata,,0.4,1,3,-1!纵向受拉钢筋mp,ex,2,2e5tb,bkin,2,1,2,1tbdata,,350!横向箍筋，受压钢筋资料属性mp,ex,3,2e5tb,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*alesize,all,,,20lesize,all,,,20lesize,all,50lesize,all,50kgen,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,1lmesh,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会集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,btype,3amesh,all!拉伸成混凝土单元type,1real,3mat,1extopt,esize,30extopt,aclear,1vext,all,,,,,-lalls!建立拘束方程cmsel,s,hoopingreinforcementcmsel,a,longitudinalnsll,s,1ceintf,,ux,uy,uzallsel,all!界限条件拘束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!施加外面荷载/solunsel,allnsel,s,loc,y,hnsel,r,loc,z,-1000d,all,uy,-displacementalls!求解nlgeom,onnsubst,200outres,all,allneqit,100pred,onallselsolvefinish/post1allselplcrack,0,1plcrack,0,2!时间历程后办理/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