偏心受压构件承载力计算例题

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,1,、,某偏心受压柱，截面尺寸,b,h,=300400 mm,，,采用,C20,混凝土，,HRB335,级钢筋，柱子计算长度,l,o,=3000 mm,，,承受弯矩设计值,M=,150kN.m,，,轴向压力设计值,N=,260kN,，,a,s,=,a,s,=40mm,，,采用对称配筋。求纵向受力钢筋的截面面积,A,s,=,A,s,。,偏心受压构件承载力计算,【,解,】,f,c,=9.6N/mm,2,=1.0,f,y,=,f,y,=300N/mm,2,，,b,=0.55,（,1,）,求初始偏心距,e,i,e,o,=,M,/,N,=15010,6,/26010,3,=577mm,e,a,=ma,x,（,20,，,h,/30,）,=ma,x,（,20,，,400/30,）,=20mm,e,i,=,e,o,+,e,a,=577+20=597mm,（,2,）,求偏心距增大系数,=3000/400=7.5,5,，应按式（,4.3.1,）计算。,取,1,=1.0,取,2,=1.0,（,3,）判断大小偏心受压,为大偏心受压。,（,4,）求,A,s,=,A,s,则有,A,s,=,A,s,=,=1235mm,2,（,5,）验算配筋率,A,s,=,A,s,=1235mm,2,0.2%,bh,=02%300400=240mm,2,，,故配筋满足要求。,（,6,）验算垂直弯矩作用平面的承载力,l,o,/,b,=3000/300=108,=0.992,N,u,=0.9,f,c,A,+,f,y,（,A,s,+,A,s,）,=0.90.992,9.6300400+300,（,1235+1235,）,=1690070N,N,=260,kN,故垂直弯矩作用平面的承载力满足要求。每侧纵筋选配,4 20,（,A,s,=,A,s,=1256mm,2,），,箍筋选用,8250,，,如图。,2,、,某矩形截面偏心受压柱，截面尺寸,b,h,=300mm500mm,，,柱计算长度,l,0,=2500mm,，,混凝土强度等级为,C25,，,纵向钢筋采用,HRB335,级，,a,s,=,a,s,=40mm,，,承受轴向力设计值,N,=1600kN,，,弯矩设计值,M,=180kNm,，,采用对称配筋，求纵向钢筋面积,A,s,=,A,s,。,【,解,】,f,c,=11.9N/mm,2,，,f,y,=300N/mm,2,=0.55,=1.0,=0.8,1,求初始偏心距,e,i,e,0,=,e,a,=（20，）=max(20,)=20mm,e,i,=,e,0,+,e,a,=112.5+20=132.5mm,2,求偏心距增大系数,l,0,/,h,=55，故,=1.0,3,判别大小偏心受压,h,0,=,h,-40=500-40=460mm,x,=,=448.2 mm,b,h,0,=0.55460=253 mm,属于小偏心受压构件。,4,重新计算,x,e=,e,i,+-,a,s,=1.0132.5+-40=342.5mm,=,=0.652,=0.652460=299.9mm,5,求纵筋截面面积,A,s,、,A,s,A,s,=,A,s,=,=1375mm,2,6,验算垂直于弯矩作用平面的承载力,l,0,/,b,=2500/300=8.338,=0.999,N,u,=0.9,（,A,s,+,A,s,）,f,y,+,Af,c,=0.90.999,（,1375+1375,）,300+30050011.9,=2346651N,N,=1600kN,故垂直于弯矩作用平面的承载力满足要求。每侧各配,2 22,（,A,s,=,A,s,=1520mm,2,），,如图所示。,