2(3 ) ==7.67×106 3 33 9.5×10×102==1.548 / <0.25 =0.25×1.0×11.9 =2.975 / 2
截面尺寸满足要求
又因为
=1.24 / 2>0.7 =0.889 / 2
故需按计算配置受扭钢筋
（3）①抗扭箍筋
=2×(140+390)=1060mm
=140×390=54600 2
取ξ=1.2
1 0.35 9.5×106 0.35×1.43×7.67×106
===0.292 2/ 1.2√ξ
取箍筋直径为Φ10（A=78.5 2）则得箍筋间距为
78.5S≤=268.8 取s=250mm = 总
=2×78.5 1.27=0.314%> ， =0.28=0.28×=0.132%
1.2×270×0.292×1060 1 ===278.568 2 ②抗扭纵筋
梁上下端分别配受扭纵筋
140 =278.568×=36.792 2 选用2根直径为5mm的HRB400级钢筋 =39 2 梁两侧面分别配受扭纵筋
390 =278.568×=102.492 2 选用2根直径为8.2mm的HRB400级钢筋 =106 2
最小配筋率
1.27 , =0.6√=×=0.3% 上下面纵筋最小配筋量
140 , =0.3%×200×450×=35.58 2<39 2 梁侧面纵筋最小配筋量