钢筋混凝土结构设计原理
解:已知K=1.20, fc=14.3N/mm2, fy=f'y=360N/mm2, a=a'=50mm, h0=500-50=450mm
l0/h=2500/500=5h0/30=450/30=15mm, 故按实际偏心距e0=500mm
ηe0=500mm>0.3h0=0.3h0=0.3*450=135mm, 故按大偏心受压计算
ξ=KN/fcbh0=1.20*400000/(14.3*500*450)=0.149 x= ξh0=0.149*450=67.13mm ρmin*bh0=0.2%*500*450=450mm2
As及A's各选用4F20+4F20 (As=A's=1256mm2)