扣件抗滑移验算两侧立杆最大受力N=max[R1,R3]=max[1.433,1.433]=1.433kN≤0.85×8=6.8kNP1=2.388kN,P2=28.293kN,P3=2.389kN立杆稳定性验算中立杆最大受力Nw=max[P1+N边1,P2,P3+N边2]+γ0×1.3×每米立杆自重×(H-梁高)+Mw/lb=max[2.388+1.1×[1.3×(0.5+(24+1.1)×0.18)+1.5×0.9×3]×(0.9+0.6-0.6/2)/2×0.9,28.293,2.389+1.1×[1.3×(0.5+(24+1.1)×0.18)+1.5×0.9×3]×(0.9+1.2-0.6-0.6/2)/2×0.9]+1.1×1.3×0.15×(5.35-1)+0.054/1.2=29.271kNf=Nw/(φA)+Mw/W=29270.841/(0.482×424)+0.054×106/4490=155.253N/mm2≤[f]/γR=300/1=300N/mm2在最新的验算中考虑支撑架竖向斜支撑需要参考轴力设计值,要是立杆稳定性验算中Nw=max 为轴向力值那么 P1,P2,P3分别是什么?...阅读全文