用ANSYS建立钢筋混凝土梁模型
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ANSYS 理论基础一、钢筋混凝土模型1、Solid65单元——模拟混凝土和岩石等抗压能力远大于抗拉能力的非均匀材料开发的单元,可以模拟混凝土中的加强钢筋(或玻璃纤维、型钢等);普通8节点三维等参元,增加针对混凝土材料参数和整体式钢筋模型;基本属性:——可以定义3种不同的加固材料;——混凝土具有开裂、压碎、塑性变形和蠕变的能力;—-加强材料只能受拉压,不能承受剪切力。
三种模型:分离式模型——把混凝土和钢筋作为不同的单元来处理,各自划分单元,或钢筋视为线单元(杆件link-spar8或管件pipe16,20);钢筋和混凝土之间可以插入粘结单元来模拟界面的粘结和滑移;整体式模型——将钢筋分布于整个单元中,假定混凝土和钢筋粘结很好,并把单元视为连续均匀材料;组合式模型—-分层组合式:在横截面上分成许多混凝土层和若干钢筋层,并对截面的应变作出某些假设(如应变沿截面高度为直线);或采用带钢筋膜的等参单元。
2、本构模型线性弹性、非线性弹性、弹塑性等;强度理论——Tresca、V on Mises、Druck —Prager等;3、破坏准则单轴破坏(Hongnested等)、双轴破坏(修正的莫尔库仑等)、三轴破坏(最大剪应力、Druck—Prager等),三参数、五参数模型;混凝土开裂前,采用Druck—Prager屈服面模型模拟塑性行为;开裂失效准则,采用William-Warnke五参数强度模型.4、基本数据输入混凝土:ShrCf-Op—张开裂缝的剪切传递系数,0~1ShrCf—Ol—闭合裂缝的剪切传递系数,0。
9~1UnTensSt—抗拉强度,UnCompSt—单轴抗压强度,(若取-1,则以下不必要)BiCompSt—双轴抗压强度,HydroPrs—静水压力,BiCompSt—静水压力下的双轴抗压强度,UnCompSt-静水压力下的单轴抗压强度,TenCrFac—拉应力衰减因子。
加固材料(材料号、体积率、方向角)二、其他材料模型在Ansys中,可在Help菜单中查阅各种不同单元的特性.例1、矩形截面钢筋混凝土板在中心点处作用-2mm的位移,分析板的受力、变形、开裂(采用整体模型分析法).材料性能如下:1、混凝土弹性模量E=24GPa,泊松比ν=0。
ANSYS在钢筋混凝土梁热分析中的应用【摘要】在火灾荷载的条件下,钢筋混凝土构件内部的温度场分布,对火灾后的构件能否继续使用,具有重要的作用。
ANSYS作为大型有限元软件,在有限元分析中得到了普遍的应用.本文首先从混凝土梁截面热分析入手,然后进行混凝土构件梁整体热分析,从而比较两者在热分析中的误差,从而得出ANSYS 在热分析中方法及思路。
【关键词】ANSYS;热分析;钢筋混凝土梁Reinforced concreted beam in the application of thermal analysis with ANSYS【Abstract】With the fire load conditions, the inside temperature field distribution of concrete beam has an important role on the components. As large-scale finite element software, the finite element analysis has gained widespread application. Comparing the thermal analysis of concrete beam section with the overall thermal analysis of concrete beams, and then draw the differences and similarities, which take thermal analysis in ANSYS in the methods and ideas.【Key words】ANSYS;Thermal analysis;Reinforced concrete beam1. 前言组成钢筋混凝土梁构件的材料,在火灾荷载作用下,其热工性能和力学性能会产生明显的变化,变形也会明显增大,由于构件在受火时,体积膨胀、截面温度不均匀分布,都会使截面产生自平衡的温度应力和构件弯曲变形[1]。
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$wpoff,,,0.125+0.06$vsbw,all$numcmp,allcm,zhuzi,volu!!!!!!!!!!赋钢筋单元属性并划分网格!!!!!!!!!elemsiz=50e-3lsel,s,loc,x,-0.16$lsel,r,loc,z,0.078$cm,waizj1,line$latt,4 ,3,3,0$lesize,all,elemsizlsel,s,loc,x,-0.16$lsel,r,loc,z,0.16$cm,waizj2,line$latt,4,3,3,0$lesize,all,elemsizlsel,s,loc,x,-0.11$lsel,r,loc,z,0.16$cm,waizj3,line $latt,4,3,3,0$lesize,all,elemsizlsel,s,loc,x,0.16$lsel,r,loc,z,0.078$cm,ywaizj1,line$latt,4 ,3,3,0$lesize,all,elemsizlsel,s,loc,x,0.16$lsel,r,loc,z,0.16$cm,ywaizj2,line $latt,4,3,3,0$lesize,all,elemsizlsel,s,loc,x,0.11$lsel,r,loc,z,0.16$cm,ywaizj3,line $latt,4,3,3,0$lesize,all,elemsiz$cmsel,s,waizj1,line$$cmsel,a,waizj2,line$$cmsel,a,waizj3,l ine$$cmsel,a,ywaizj1,line$$cmsel,a,ywaizj2,line$$cmsel,a,ywaizj 3,line$$cm,zj,line!节点外域箍筋lsel,s,tan1,y !对称外定义箍筋$lsel,u,loc,x,-2.25,-0.161 $lsel,u,loc,x, 0.161,2.25$lsel,u,loc,y,0.125-0.015$lsel,u,loc,y,-0.125+0.015 $lsel,u,loc,y,-0.295,0.295 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0.125$lsel,u,loc,y,-0.125$Lsel,r,loc,x,-0.16$Lsel,r,loc,z,0,0.16$cm,ssgj3,line$lsel,s,tan1,y !对称外定义箍筋$lsel,u,loc,x,-2.25,-0.161$lsel,u,loc,x, 0.161,2.25$lsel,u,loc,y,0.125-0.015$lsel,u,loc,y,-0.125+0.015$lsel,u,loc,y,-0.125-0.06$lsel,u,loc,y,0.125+0.06$lsel,u,loc,y,-0.095,0.095$lsel,u,loc,y,0.251,1.5$lsel,u,loc,y,-1.5,-0.251$lsel,u,loc,y, 0.125$lsel,u,loc,y,-0.125$Lsel,r,loc,x,0.16$Lsel,r,loc,z,0,0.16$cm,ssgj4,line$$cmsel,s,ssgj1,line$$cmsel,a,ssgj2,line$cmsel,a,ssgj3,line$cmsel,a,ssgj4,line$$cm,guj3,line$$cmsel,s,guj1,line$$cmsel,a,guj3,line$$cm,guj,line$latt,5,4 ,3,0$lesize,all,elemsiz$cmsel,s,guj,line$cmsel,a,zj,line$cm,gj,lineCmsel,s,gzg,volu$VSEL,r,LOC,Y,-0.125+0.015,0.125-0.015 $VSEL,r,LOC,z,0,0.003$cm,fb,VOLUCmsel,s,gzg,volu$VSEL,u,LOC,Y,-0.125,0.125$cm,GZhu,VOLU Cmsel,s,gzg,volu$$cmSEL,U,fb,VOLU$VSEL,r,LOC,Y,-0.125+0.015 ,0.125-0.015$cm,mianban,VOLUCmsel,s,gzg,volu$$cmSEL,U,fb,VOLU$$cmSEL,U,gzhu,VOLU$$cmSEL ,U,mianban,VOLU$$cm,yy,VOLU!!!!!!!!!!!!!!!!!!!!!!!! Mesh钢筋!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!cmsel,s,gj,line$lmesh,all $lsel,all$/eshape,1$eplot !钢筋划分完毕!!!!!!!!!!!!!!!!!!!! mesh 钢筋混凝土柱!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Cmsel,s,zhuzi,volu $esize,50e-3$vatt,2,1,2,0$MSHAPE,0,3D$MSHKEY,1$VMESH,allallsel,all!!!!!!!!!!!!!!!!!! Mesh 翼缘!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Cmsel,s,yy,volu$$LSEL,s,length,,1.8LESIZE,all, , ,12vatt,1,1,1,0$esize,50e-3$$MSHAPE,0,3D$MSHKEY,1$VMESH,all$allsel,all!!!!!!!!!!!!!!!!!! Mesh fb!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Cmsel,s,fb,volu$vatt,1,1,1,0$esize,50e-3$$MSHAPE,0,3D$MSHKEY,1$VMESH,all$allsel,all!!!!!!!!!!!!!!!!!! Mesh mianban gzhu!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Cmsel,s,mianban,volu$Cmsel,a,gzhu,volu$vatt,3,5,1,0$esize,50e-3$$MSHAPE,0,3D$MSHKEY,1 $VMESH,all$allsel,all$nummrg,node$numcmp,all!定义远距离加载点$ET,4,TARGE170ET,5,CONTA174R,6keyopt,4,2,1$keyopt,5,2,2$keyopt,5,4,2$keyopt,5,12,5 asel,s,loc,x,2$nsla,s,1$real,6$type,5$esurfnmax=ndinqr(0,14)+1$n,nmax,2,,type,4$tshap,pilo$e,nmaxR,7keyopt,4,2,1$keyopt,5,2,2$keyopt,5,4,2$keyopt,5,12,5 asel,s,loc,x,-2$nsla,s,1$real,7$type,5$esurfnmax=ndinqr(0,14)+1$n,nmax,-2,,type,4$tshap,pilo$e,nmaxR,8keyopt,4,2,1$keyopt,5,2,2$keyopt,5,4,2$keyopt,5,12,5 asel,s,loc,y,1.5$nsla,s,1$real,8$type,5$esurfnmax=ndinqr(0,14)+1$n,nmax,0,1.5,type,4$tshap,pilo$e,nmaxAllsel,all!施加约束和荷载Allsel,allAsel,s,loc,z,0Da,all,symm$D,8665,uy,,,,,uz$D,8666,uy,,,,,uz$Asel,s,loc,y,-1.5Da,all,allAllsel,all/solu$ANTYPE,0$OUTRES,ALL,ALL$$pred,off$neqit,30$lnsrch,on$nlgenom,off$Cnvtol,f,,0.05$kbc,0$eqslv,sparse,1.0E-5,2 $p=0.01$time,1$nsubst,50,,10$Asel,s,loc,y,1.5f,8667,fy,-582000$$allsel,all$OUTRES,ALL,ALL $ALLSEL,ALL $LSWRITE,1TIME,2$D,8667,Ux,19*p$$kbc,0$ NSUBST,100,1000000,1$AUTOTS,o n$OUTRES,ALL,ALL $ALLSEL,ALL$LSWRITE,2$Lssolve,1,2$save$finish/post1SET,LAST$SET,PREVIOUS*GET,RTIME,ACTIVE,0,SET,TIME !获取最终收敛时间Pldisp$plnsol,s,x !显示变形图与X方向应力分布图Plnsol,epel,x !X弹性应变分布图Plnsol,eppl,x !x塑性应变分布图Plnsol,epto,x !X方向总应变分布图Plnsol,nl,srat !显示塑性区域/device,vector,off$plcrack,Cmsel,s,gzgEslv,s/POST26nsel,s,loc,x,0.975NUMVAR,200$NSOL,2,8667,U,x$nsel,s,loc,y,-1.55$RFORCE,3,8667,F,xcmsel,s,gj,line$Esll,s$/output,Z1,csv$prod,4,2,,,,,,1000$Prod,5,3,,,,,,1/1000PRVAR,4,5!!! 钢梁应力的显示/post1$Cmsel,s,gzg$cmsel,a,FUBAN$vsel,r,loc,z,0,0.0045$vsel,r,l oc,y,-0.125+0.006,0.125-0.006$vsel,r,loc,x,-0.175,0.175$eslv,s$nsle,splnsol,s,eqvPlnsol,epel,x !X弹性应变分布图Plnsol,eppl,x !x塑性应变分布图Plesol,epto,1 !X方向总应变分布图/post1$Cmsel,s,gzg$$eslv,s$nsle,s$plnsol,s,eqv/post1$Cmsel,s,zhuzi$eslv,s$nsle,splnsol,s,eqv$Cmsel,s,zhuzi$cmsel,a,FUBAN$vsel,r,loc,z,0,0.0045$vsel,r ,loc,y,-0.125+0.006,0.125-0.006$vsel,r,loc,x,-0.175,0.175$e slv,s$nsle,splesol,s,eqv/post1$Cmsel,s,zhuzi$eslv,s$nsle,splnsol,s,eqv/post1SET,LAST$SET,PREVIOUS*GET,RTIME,ACTIVE,0,SET,TIME !获取最终收敛时间Pldisp$plnsol,s,x !显示变形图与X方向应力分布图Plnsol,epel,x !X弹性应变分布图Plnsol,eppl,x !x塑性应变分布图Plnsol,epto,eqv !X方向总应变分布图Plnsol,nl,srat !显示塑性区域/post1Cmsel,s,gj$esll,sETABLE,rebar,ls,1PLETAB,rebar/post1$wpstyl,defa$$path,23,2,30,120$ppath,1,,0.185,0.125,0$ppath ,2,,0.975,0.125,0pdef,myseqv,s,eqv$pdef,mysy,s,1plpath,myseqv,mysyPLPAGM,myseqv,1,blankPLPAGM,myseqv,1,nodeprpath,myseqv,,xg,yg,zg/device,vector,on$plcrack,Cmsel,s,gzgEslv,snsel,s,loc,x,0.975NUMVAR,200$NSOL,2,9989,U,Y$nsel,s,loc,y,-1.55$RFORCE,3,9989,F,Y/output,Z1,csv$prod,4,2,,,,,,-1000$Prod,5,3,,,,,,-1/1000 PRVAR,4,5/POST26nsel,s,loc,x,0.975NUMVAR,200$NSOL,6,10678,U,Y$nsel,s,loc,y,-1.55$RFORCE,7,10678,F,Y/output,Z2,csv$prod,8,6,,,,,,1000$Prod,9,7,,,,,,1/1000!后处理阶段/post1SET,LAST$SET,PREVIOUS*GET,RTIME,ACTIVE,0,SET,TIME !获取最终收敛时间Pldisp$plnsol,s,x !显示变形图与X方向应力分布图Plnsol,epel,x !X弹性应变分布图Plnsol,eppl,x !x塑性应变分布图Plnsol,epto,eqv !X方向总应变分布图Plnsol,nl,srat !显示塑性区域esel,s,type,,2cmsel,s,gj,line$esll,sETABLE,rebar,ls,1PLETAB,rebar!混凝土裂缝的显示!外混凝土$Cmsel,s,zhuzi,$vsel,r,loc,z,0.0625+0.001,0.175$vsel,r,lo c,y,-0.125+0.006,0.125-0.006$eslv,s$$/dscale,,off$/device,vector,on$plcrack,1,1 !第一条裂纹!内混凝土$Cmsel,s,zhuzi$vsel,r,loc,z,0,0.0625$vsel,r,loc,y,-0.125+ 0.006,0.125-0.006$eslv,s$/dscale,,off$$/device,vector,on$pl crack/post1!混凝土应力的显示$/gformat,defa !恢复缺省格式$Cmsel,s,zhuzi$eslv,s$plnsol,s,1 !主应力$Cmsel,s,zhuzi$eslv,s$plnsol,s,3 !最大压应力$Cmsel,s,zhuzi$eslv,s$plnsol,s,eqv !等效应力$Cmsel,s,gzg$Cmsel,a,fb$eslv,s$plnsol,s,eqv/post1!内混凝土矢量图$Cmsel,s,zhuzi,$vsel,r,loc,z,0,0.0625$vsel,r,loc,y,-0.125 +0.006,0.125-0.006$eslv,s$$/device,vector,off$/dscale,,off$$plnsol,s,eqv $Plvect,u$$Plvect,s$/vscale,,1$plvect,s,,,,vect !调整箭头大小$/gline,,-1$plvect,s,,,,vect ! 取消单元边界$/dscale,,off$$plnsol,s,1 !关闭结构变形$/dscale,defas$/gline,1 !恢复变形比例缺省,显示单元边界$/Device,vector,1 !设置矢量显示模式$/clabel,,5 !设置等高线上的文字间隔$/contour,,18,-16,,500 !设置等高线数目$Plnsol,s,x!外混凝土矢量图$Cmsel,s,zhuzi,$vsel,r,loc,z,0.0625+0.001,0.175$vsel,r,lo c,y,-0.125+0.006,0.125-0.006$eslv,s$/device,vector,off$/dscale,,off$$plesol,s,3$Plvect,u$$Plvect,s$/vscale,,1$plvect,s,,,,vect !调整箭头大小$/gline,,-1$plvect,s,,,,vect ! 取消单元边界$/dscale,,off$$plnsol,s,1 !关闭结构变形$/dscale,defas$/gline,1 !恢复变形比例缺省,显示单元边界$/Device,vector,1 !设置矢量显示模式$/clabel,,5 !设置等高线上的文字间隔$/contour,,18,-16,,500 !设置等高线数目$Plnsol,s,x!混凝土主压应变与等效应变曲线$Cmsel,s,zhuzi,$vsel,r,loc,z,0,0.175$vsel,r,loc,y,-0.125+ 0.006,0.125-0.006$vsel,r,loc,x,-0.135+0.03,0.175$eslv,s$nsle,s/POST26NUMVAR,200$$Esol,3,1080,26,s,eqv$esol,4,1080,26,epel,eqv$esol,5,1080,26,eppl,eqv$add,4,4,5$prod,5,3,,,,,,1/1000000$/axlab,x,strain-eqv$/axlab,y,sigx-eqv(MPa)$PRVAR,4,5/POST26NUMVAR,200$$Esol,3,1080,45,s,eqv$esol,4,1080,45,epel,eqv$esol,5,1080,45,eppl,eqv$add,4,4,5$prod,5,3,,,,,,1/1000000$/axlab,x,strain-eqv$/axlab,y,sigx-eqv(MPa) $PRVAR,4,5/POST26NUMVAR,200$$Esol,3,2706,46,s,eqv$esol,4,2706,46,epel,eqv$esol,5,2706,46,eppl,eqv$add,4,4,5$prod,5,3,,,,,,1/1000000$/axlab,x,strain-eqv$/axlab,y,sigx-eqv(MPa) $PRVAR,4,5/POST26NUMVAR,200$$Esol,3,3157,296,s,eqv$esol,4,3157,296,epel,eqv$esol,5,3157,296,eppl,eqv$add,4,4,5$prod,5,3,,,,,,1/1000000$/axlab,x,strain-eqv$/axlab,y,sigx-eqv(MPa) $PRVAR,4,5/POST26NUMVAR,200$$Esol,3,3157,291,s,eqv$esol,4,3157,291,epel,eqv$esol,5,3157,291,eppl,eqv$add,4,4,5$prod,5,3,,,,,,1/1000000$/axlab,x,strain-eqv$/axlab,y,sigx-eqv(MPa)$PRVAR,4,5/POST26NUMVAR,200$$Esol,3,3160,1169,s,eqv$esol,4,3160,1169,epel,eqv$esol,5,3160,1169,eppl,eqv$add,4,4,5$prod,5,3,,,,,,1/1000000$/axlab,x,strain-eqv$/axlab,y,sigx-eqv(MPa)$PRVAR,4,5! 剪应变与梁端荷载曲线$Cmsel,s,gzg$vsel,r,loc,z,0,0.0045$vsel,r,loc,y,-0.125+0. 006,0.125-0.006$eslv,s$nsle,s/POST26NUMVAR,200$$RFORCE,3,7150,F,Y$esol,4,6351,2661,epel,xy$esol,5,6351,2661,eppl,xy$add,4,4,5$prod,5,3,,,,,,-1/1000$PRVAR,4,5/post1$Cmsel,s,gzg$vsel,r,loc,z,0,0.0045$vsel,r,loc,y,-0.125+0. 006,0.125-0.006$$vsel,r,loc,x,-0.175,0.175$eslv,s$nsle,sSET,LAST$SET,PREVIOUS*GET,RTIME,ACTIVE,0,SET,TIME !获取最终收敛时间Pldisp$plnsol,s,x !显示变形图与X方向应力分布图Plnsol,epel,xy !Xy弹性应变分布图Plnsol,eppl,xy !xy塑性应变分布图Plnsol,epto,xy !Xy方向总应变分布图/post1Allsel,all!腹板剪力$Cmsel,s,gzg$Nslv,s,1$Nsel,r,loc,x,-0.175,0.175$Nsel,r,loc,z,-0.0045,0.0045 $Nsel,r,loc,y,0,0.05$Esln,s,1$Nsel,r,loc,y,0$fsum!内混凝土剪力Allsel,all$Cmsel,s,zhuzi$Nsel,r,loc,x,-0.175,0.175$Nsel,r,loc,z,-0.0625,0.0625$Nsel,r,loc,y,0,0.05$Esln,s,1$Nsel,r,loc,y,0$fsum/post1Allsel,all!外混凝土剪力$Cmsel,s,zhuzi$Nslv,s,1$Nsel,r,loc,x,-0.175,0.175$Nsel,u,loc,z,-0.0625+0.001,0.0625-0.001 $Nsel,r,loc,y,0,0.05$Esln,s,1$Nsel,r,loc,y,0$fsum!混凝土总剪力Allsel,all$Cmsel,s,zhuzi$Nsel,r,loc,x,-0.175,0.175 $Nsel,r,loc,y,0,0.05$Esln,s,1$Nsel,r,loc,y,0$fsum!总剪力/post1Allsel,all$Nsel,s,loc,x,-0.175,0.175 $Nsel,r,loc,y,0,0.05$Esln,s,1$Nsel,r,loc,y,0$fsum。
⼿把⼿教你学ansys--钢筋混凝⼟梁⼤家好,我是⽔哥。
⽔哥ansys使⽤经验三年多,既做过重⼤科研项⽬,也做过许多实际项⽬,对ansys的使⽤有⼀定的⼼得体会,本着分享经验的精神,今⽇以⼀个钢筋混凝⼟梁的建模求解过程来简单说明ansys的基本操作步骤。
(我的ansys14.0)总的说来,⽆论⼩项⽬还是⼤项⽬,总体过程⽆⾮定义单元、定义材料、物理模型、有限元模型、加载、约束、求解、查看这⼏个过程,和我们⼯程类的设计软件例如PKPM、SAP2000等基本过程都差不多,只不过最⼤的区别在于ansys的建模实在是蛋疼了⼀些。
废话不多说,以下⾯的⼀根悬臂钢筋混凝⼟梁来教新⼿如何快速进⼊ansys 的⼤堂,每⼀步都有GUI操作,完了之后会有相应部分的命令流,这⾥多说⼀句,⼀个ansys的使⽤⾼⼿必然是⼀个精通apdl编程的能⼿,所以我建议新⼿在学习的时候最好以apdl⼊⼿,GUI操作辅助,这样在学习的时候能节省⼤量时间,⽽不会浪费在GUI毫⽆意义的重复操作上。
此题如下:悬臂梁如下,梁宽200mm,梁顶有两根直径为16的钢筋,钢筋中⼼距梁边的距离为40mm,在梁端附近受集中⼒P=100KN的作⽤.要求对此悬臂梁进⾏完全线弹性分析,结果要显⽰主应⼒迹线。
材料参数:混凝⼟弹性模量为3000MPa,钢筋的弹性模量取200GPa,不考虑材料⾃重。
(建模时注意单位的协调性)⼀、题⽬解读与材料单元定义注意此题要求进⾏完全线弹性分析,此话的意义在于我们可以⽤除solid65 以外的其他实体单元。
在ansys单元中,solid65是专门⽤于模拟钢筋混凝⼟构件的单元,但⽤此单元模拟时,⼀般是需要考虑材料的⾮线性,也即是多⽤于构件的⾮线性分析,并且需要材料的本构模型。
此题要求我们做弹性分析,我们可以⽤link8单元模拟钢筋,⽤solid45单元模拟混凝⼟,注意⾼版本的ansys已经将许多低阶单元合并掉了,以ansys14为例,在link单元中只有180,⽽低阶的link8、link10等已被合并。
钢筋混凝土梁受压区温度裂缝分析与ANSYS仿真模拟钢筋混凝土梁受压区温度裂缝分析与ANSYS仿真模拟摘要:利用大型有限元分析软件ANSYS中的三维实体单元Solid65对钢筋混凝土梁进行热―结构耦合分析,模拟高温环境中预埋地脚螺栓对混凝土受热膨胀产生的约束作用。
计算结果显示,混凝土构件自身由于温度变化而产生的变形在受到外来阻力约束时,会导致混凝土裂缝的出现。
关键词:钢筋混凝土;温度应力;裂缝;非线性有限元;仿真中图分类:TU378文献标识码:A 文章编号:一般来说,不同季节和不同时间环境温度改变都会给材料带来“热胀冷缩”现象。
这种因为温度变化使材料产生的应力为温度应力。
对于室内环境温度较高或室外需要露天工作的混凝土构件,其温度变化较大所以产生的温度应力也较大,而这种温度应力对构件带来影响往往也是不可忽视的。
由于钢筋混凝土结构的性质复杂,材料非线性与几何非线性常同时存在,所以用传统的方法来分析和描述这种温度应力产生的变形则难度非常大[1]。
随着计算机处理能力的不断增强以及非线性有限元方法的日臻完善,有限元作为一个强有力的数值分析工具,在钢筋混凝土结构非线性分析中正显示着越来越大的实用性和方便性[2]。
目前,可以利用比较完善的特种单元来近似模拟混凝土或钢筋混凝土材料,在大型通用有限元软件ANSYS中,Solid65单元常被用来模拟钢筋混凝土等抗压能力远大于抗拉能力的非均匀材料。
另外ANSYS提供的热―结构耦合分析,可以将温度影响施加到建立的模型中进行分析,从而得出温度对结构产生的应力。
1工程描述由于水泥在煅烧过程将产生大量的粉尘和一定的热量,所以造成厂房内温度很高,环境恶劣。
电收尘器作为水泥工厂中常用的除尘设备,一般都是在烧成车间的梁上预埋钢板焊接或预埋高标号的地脚螺栓安装。
2002年,某水泥厂窑尾车间用于承受电收尘器的大梁,在使用一年后受压区靠近预埋螺栓的部位产生细小裂缝(见图1)。
为此建设单位邀请设计、施工、监理和部分专家对裂缝产生的原因和其对厂房的安全性,耐久性进行分析:1.1 出现裂缝的梁承受荷载、截面尺寸、配筋等均按国家规范和工艺要求设计,梁裂缝宽度及挠度均控制在规范允许范围内。
基于ANSYS的钢筋混凝土梁的非线性分析摘要:本文主要通过介绍混凝土的本构模型,利用solid65号单元阐述ansys如何实现钢筋混凝土梁的建模,开裂,破坏等受力性能。
关键词:混凝土,有限元,非线性the nonlinear analysis of reinforced-concrete beam based on ansysdang jianping(baotou construction engineering cost can administer station, baotou 014010)abstract:by introducing the concrete constitutive model,the paper expounds ansys how to realize the modeling, craze, destructive force performance of the reinforced concrete beam using solid65 element.keywords: concrete, finite element, nolinear1 solid65单元的材料属性ansys的solid65单元是专为混凝土,岩石等抗压能力远大于抗拉能力的非均匀材料开发的单元。
它可以模拟混凝土中的加强钢筋(或玻璃纤维,型钢等),以及材料的抗裂和压溃现象。
solid65单元最多可定义3种不同的加固材料,即此单元允许同时拥有四种不同的材料。
混凝土材料具有开裂,压碎,塑性变形和蠕变的能力;加强材料则只能受拉压,不能承受剪切力。
2 材料本构关系模型2.1 混凝土本构模型根据弹塑性理论建立混凝上的本构关系时,必须对屈服,条件流动法则、硬化法则即塑性模型三要素做出基本假定。
ansys弹塑性本构关系主要使用mises屈服准则或drucker-prager屈服准则。
2.2 混凝土破坏准则混凝土模型采用willam-warnke五参数破坏准则,破坏面通过以及在静水压力p下的来定义。
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,solid45 et,2,beam188et,3,plane42!定义截面类型sectype,1,beam,csolid,,0 secoffset,centsecdata,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,ecmp,prxy,1,0.2tb,kinh,1,,16tbpt,,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,,0.001790667,2.69e1 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,,0.0038,2.28e1tb,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,8,88,4 l,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,hooping reinforcement,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,0 nsel,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,-displacement alls!求解nlgeom,onnsubst,50outres,all,all neqit,50pred,oncnvtol,f,,0.05,2,0.5 allselsolvefinish/post1allsel/device,vector,1!时间历程后处理/post26nsel,s,loc,z,-l/2*get,Nmin,node,0,num,min nsol,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,min n0=Nmin rforce,5,Nmin,f,y*do,i,2,ndinqr(1,13) ni=ndnext(n0) rforce,6,ni,f,y add,5,5,6n0=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-beamb=150h=300a=30l=2000fcu=40ec=2.85e4displacement=5!定义单元类型et,1,solid65 et,2,beam188et,3,plane42!定义截面类型sectype,1,beam,csolid,,0 secoffset,centsecdata,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,ecmp,prxy,1,0.2tb,kinh,1,,16tbpt,,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,,0.001790667,2.69e1 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,2,2e5mp,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,hooping reinforcement,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,hooping reinforcement cmsel,a,longitudinal nsll,s,1ceintf,,ux,uy,uz allsel,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,-displacement alls!求解nlgeom,onnsubst,200outres,all,all neqit,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,min nsol,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,min n0=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。