ansys模擬鋼管混凝土的極限承載能力

...wd......wd......wd...薄壁鋼管混凝土柱軸心受壓性能的實例分析建設的單元類型都為solid45號單元，因為計算較為簡單的緣故。構件參數(shù)d=200，t=5，l=840，fy=265.8e6，fc=27.15e6，ec=21700e6，es=210000e6。一、建設鋼管混凝土實體模型并對此模型進展網(wǎng)格劃分鋼管混凝土單元單元二、材料應力應變關系的定義鋼管應力應變關系混凝土材料的應力應變關系〔采用隨動硬化模型〕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力峰值時會遇到很大的收斂問題，而采用位移加載法比擬容易收斂，且得等到的荷載—位移曲線較好在底部施加固定端約束，同時在頂面施加在面方向的移動，UX、UZ，在頂面的所用節(jié)點進展耦合，使其在沿著柱的方向上有一樣的位移。然后是施加位移，在頂面所有節(jié)點施加位移荷載，UY=0.003大概是這么多。四、求解定義完求解步之后，設置分析選項后，就可進展有限元的計算。五、檢查結果鋼管的馮米澤斯應力整體的馮米澤斯應力節(jié)點位移圖六、荷載—應變?nèi)糖€計算節(jié)點力的時候是通過頂面節(jié)點Y方向節(jié)點力求和操作來求解。!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=27.15e6ec=21700e6es=210000e6p=2000e3!混凝土單元及材料參數(shù)et,1,45mp,ex,1,ecmp,prxy,1,0.2tb,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,,0.002,2.66e+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,,,,-lcyl4,,,d/2,,(d-2*t)/2,,-lvglue,allcsys,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!進入求解層、設置求解選項/soluautots,onnsubst,64outres,all,allsolvefini!進入post1后處理*get,rtime,active,0,set,time!進入post26,繪制荷載-位移曲線〔節(jié)點的豎向應變與荷載曲線〕/post26nsol,2,71,u,yprod,3,2,,,,,,1/lprod,4,1,,,,,,p/1e3/axlab