ansys模擬鋼管混凝土的極限承載能力

1、薄壁鋼管混凝土柱軸心受壓性能的實(shí)例分析建立的單元類型都為solid45號(hào)單元，因?yàn)橛?jì)算較為簡單的緣故。構(gòu)件參數(shù)d=200，t=5，l=840，fy=265.8e6，fc=27.15e6，ec=21700e6，es=210000e6。一、建立鋼管混凝土實(shí)體模型并對(duì)此模型進(jìn)行網(wǎng)格劃分鋼管混凝土單元單元二、材料應(yīng)力應(yīng)變關(guān)系的定義 鋼管應(yīng)力應(yīng)變關(guān)系 混凝土材料的應(yīng)力應(yīng)變關(guān)系（采用隨動(dòng)硬化模型）三、施加位移約束和荷載ANSYS加載方式有兩種：一種是位移加載，一種是力加載。本文采用的是位移加載方式，采用力加載法在接近應(yīng)力峰值時(shí)會(huì)遇到很大的收斂問題，而采用位移加載法比較容易收斂，且得等到的荷載位移曲線較好在

2、底部施加固定端約束，同時(shí)在頂面施加在面方向的移動(dòng)，UX、UZ，在頂面的所用節(jié)點(diǎn)進(jìn)行耦合，使其在沿著柱的方向上有相同的位移。然后是施加位移，在頂面所有節(jié)點(diǎn)施加位移荷載，UY=0.003大概是這么多。四、求解定義完求解步之后，設(shè)置分析選項(xiàng)后，就可進(jìn)行有限元的計(jì)算。五、檢查結(jié)果鋼管的馮米澤斯應(yīng)力整體的馮米澤斯應(yīng)力節(jié)點(diǎn)位移圖六、荷載應(yīng)變?nèi)糖€計(jì)算節(jié)點(diǎn)力的時(shí)候是通過頂面節(jié)點(diǎn)Y方向節(jié)點(diǎn)力求和操作來求解。!鋼管與混凝土完全粘接情況下的抗壓承載性能，使用面荷載加載方式。fini/clear/prep7wprota,90,csys,4d=0.2l=0.840t=0.005n=d/tfy=265.8e6fc=2

3、7.15e6ec=21700e6es=210000e6p=2000e3!混凝土單元及材料參數(shù)et,1,45mp,ex,1,ec mp,prxy,1,0.2 tb,kinh,1,13 !混凝土選用的是hongnestad模型tbpt,0.001,2.17e+07tbpt,0.0011,2.30e+07tbpt,0.0012,2.41e+07tbpt,0.0013,2.50e+07tbpt,0.0014,2.58e+07tbpt,0.0015,2.64e+07tbpt,0.00154,2.66e+07tbpt,0.00154,2.66e+07tbpt,0.00176,2.72e+07tbpt,0.

4、002,2.66e+07tbpt,0.00246,2.50e+07tbpt,0.00292,2.34e+07tbpt,0.00338,2.18e+07!鋼管單元及材料參數(shù)et,2,solid45mp,ex,2,2.1e11mp,prxy,2,0.3tb,biso,2,1,2tbtemp,0tbdata,fy,0cyl4,(d-2*t)/2,-l cyl4,d/2,(d-2*t)/2,-l vglue,all csys,0wpstyl,defawpstyle,0!劃分網(wǎng)格mat,1type,1vsweep,1mat,2type,2vsweep,3!施加約束條件asel,s,loc,y,0da,all,allallsel,allasel,s,loc,y,lnsla,s,1cp,1,uy,allsfa,all,1,pres,-p/(3.14*(d*d)/4)allsel,allfinish!進(jìn)入求解層、設(shè)置求解選項(xiàng)/soluautots,onnsubst,64outres,all,allsolvefini!進(jìn)入post1后處理*get,rtime,active,0,set,time!進(jìn)入post26,繪制荷載-位移曲線（節(jié)點(diǎn)的豎向應(yīng)變與荷載曲線）/post26nsol,2,71,u,yprod,3,2,1/lprod,4,1,p/1e3/axlab,x,strain/a