斜道计算书计算依据：1、《建筑施工扣件式钢管脚手架安全技术规范》JGJ130-20112、《建筑地基基础设计规范》GB50007-20113、《建筑结构荷载规范》GB50009-20124、《钢结构设计规范》GB50017-2003一、基本参数二、荷载参数搭设示意图：平面图立面图三、纵向水平杆验算计算简图如下：水平杆布置方式承载力使用极限状态q=(1.2×(0.038+G kjb×l b/(m+1 ))+ 1.4×G kq×l b/(m+1))×cosθ=(1.2×(0.038 +0.35×1/(2+1 ))+ 1.4×3×1/(2+1))×0.949 =1.505kN/m正常使用极限状态q'=((0.038+G kjb×l b/(m+1 ))+G kq×l b/(m+1))×cosθ=((0.038 +0.35×1/(2+1 ))+3×1/(2+1))×0.949 =1.096kN/m计算简图如下：1、抗弯验算M max=0.1q(l a/cosθ)2=0.1×1.505×(1.5/0.949)2= 0.376kN·mσ=M max/W=0.376×106/5080 = 74.016 N/mm2≤[f]=205N/mm2满足要求！2、挠度验算νmax =0.677q'(l a/cosθ)4/(100EI)=0.677×1.096×(1500/0.949)4/(100×206000×121900)=1.844mm≤[ν] = min[l a/cosθ/150，10]= min[1500/0.949/150，10]=10mm满足要求！3、支座反力计算承载力使用极限状态R max= 1.1×ql a/cosθ=1.1×1.505×1.5/0.949=2.617kN正常使用极限状态R max'= 1.1×q'l a/cosθ=1.1×1.096×1.5/0.949=1.906kN四、横向水平杆验算承载力使用极限状态F1=R max/cosθ=2.617/0.949=2.758kNq=1.2×0.038=0.046 kN/m正常使用极限状态F1'=R max'/cosθ=1.906/0.949=2.008kNq'=0.038kN/m计算简图如下：1、抗弯验算σ=M max/W=0.924×106/5080 = 181.89 N/mm2≤[f]=205N/mm2 满足要求！2、挠度验算νmax=2.856mm≤[ν] = min[l b/150，10]= min[1000/150，10]=6.667mm 满足要求！3、支座反力计算承载力使用极限状态R max= 2.781kN五、扣件抗滑承载力验算横向水平杆：R max=2.781KN≤Rc =0.8×8=6.4kN纵向水平杆：R max=2.617/0.949/2=1.379KN≤Rc =0.8×8=6.4kN满足要求！六、荷载计算立杆静荷载计算1、立杆承受的结构自重荷载N G1k每米内立杆承受斜道新增加杆件的自重标准值g k1＇g k1＇=(l a/cosθ+(l a/cosθ)×m/2)×0.038×n/2 /(n×H)=(1.5/0.949+(1.5/0.949)×2/2)×0.038×2/2 /(2×6)=0.01 kN/m单内立杆：N G1k=(g k+ g k1＇)×(n×H-H1)=(0.35+0.01)×(2×6-5) =2.52KN双内立杆：N GS1k =(g k+ g k1＇+0.038)×H1=(0.35+0.01+0.038)×5=1.99KN每米中间立杆承受斜道新增加杆件的自重标准值g k1＇g k1＇=(l a/cosθ+(l a/cosθ)×m/2)×0.038/H=(1.5/0.949+(1.5/0.949)×2/2)×0.038/6=0.02 kN/m单中间立杆：N G1k=(2×g k-0.038+ g k1＇)×(n×H-H1)=(2×0.35-0.038+0.02)×(2×6-5)=4.774KN双中间立杆：N GS1k =(2×g k+ g k1＇)×H1=(2×0.35+0.02)×5=3.6KN2、立杆承受的脚手板及挡脚板荷载标准值N G2k每米内立杆承受斜道脚手板及挡脚板荷载标准值g k2＇g k2＇=[G kjb×(l a/cosθ)×l b/2+G kdb×(l a/cosθ)]×(n/2 )/(n×H)=[0.35×(1.5/0.949)×1/2+0.14×(1.5/0.949)]×(2/2)/(2×6)=0.041kN/m单内立杆：N G2k=g k2＇×(n×H-H1)=0.041×(2×6-5) =0.287KN双内立杆：N GS2k =g k2＇×H1=0.041×5=0.205KN每米中间立杆承受斜道脚手板及挡脚板荷载标准值g k2＇g k2＇=[G kjb×(l a/cosθ)×l b/2+G kdb×(l a/cosθ)]/H=[0.35×(1.5/0.949)×1/2+0.14×(1.5/0.949)]/6=0.083k N/m单中间立杆：N G2k=g k2＇×(n×H-H1)=0.083×(2×6-5) =0.581KN双中间立杆：N GS2k =g k2＇×H1=0.083×5=0.415KN立杆施工活荷载计算N Q1k=[G kq×(l a/cosθ)×l b/2]×n j=[3×(1.5/0.949)×1/2]×2 =4.742 kN七、立杆稳定性验算1、立杆长细比验算立杆计算长度l0=kμh=1×1.5×1.8=2.7m长细比λ= l0/i =2700/15.8=170.886≤210满足要求！轴心受压构件的稳定系数计算：立杆计算长度l0=kμh=1.155×1.5×1.8=3.118m长细比λ= l0/i =3118/15.8=197.342查《规范》JGJ130-2011表A.0.6得，φ=0.1862、立杆稳定性验算不组合风荷载作用下的单立杆轴心压力设计值：单立杆的轴心压力设计值：单内立杆：N1=1.2×(N G1k+ N G2k)+ 1.4×N Q1k=1.2×(2.52+ 0.287)+ 1.4×4.742=10.007KN 单中间立杆：N2=1.2×(N G1k+ N G2k)+ 1.4×N Q1k=1.2×(4.774+ 0.581)+1.4×4.742=13.065KNN=max{N1，N2}=13.065KNσ= N/(φA) =13065/(0.186×489)=143.644N/mm2≤[f]=205 N/mm2满足要求！双立杆的轴心压力设计值:双内立杆：N S1=1.2×(N GS1k+ N GS2k)+ N1=1.2×(1.99+ 0.205)+ 10.007=12.641KN双中间立杆：N S2=1.2×(N GS1k+ N GS2k)+ N2=1.2×(3.6+ 0.415)+ 13.065=17.883KNN=max{N s1，N s2}=17.883KNσ= (K S×N S)/(φA) =(0.6×17883)/(0.186×489)=117.97N/mm2≤[f]=205 N/mm2满足要求！组合风荷载作用下的单立杆轴向力：单立杆的轴心压力设计值：单内立杆：N1=1.2×(N G1k+ N G2k)+ 0.9×1.4×N Q1k=1.2×(2.52+ 0.287)+0.9×1.4×4.742=9.343KN单中间立杆：N2=1.2×(N G1k+ N G2k)+ 0.9×1.4×N Q1k=1.2×(4.774+ 0.581)+0.9×1.4×4.742=12.401KNN=max{N1，N2}=12.401KNM w=0.9×1.4M wk=0.9×1.4w k l a h2/10=0.9×1.4×0.25×1.5×1.82/10=0.153 kN·mσ=N/(φA)+M w/W=12401/(0.186×489)+0.153×106/5080=166.462N/mm2≤[f]=205 N/mm2 满足要求！双立杆的轴心压力设计值:双内立杆：N S1=1.2×(N GS1k+ N GS2k)+ N1=1.2×(1.99+ 0.205)+ 9.343=11.977KN双中间立杆：N S2=1.2×(N GS1k+ N GS2k)+ N2=1.2×(3.6+ 0.415)+ 12.401=17.219KNN=max{N s1，N s2}=17.219KNM w=0.9×1.4M wk=0.9×1.4w k l a h2/10=0.9×1.4×0.25×1.5×1.82/10=0.153 kN·mσ=K S×N S/(φA)+M w/W=0.6×17219/(0.186×489)+0.153×106/5080=143.707N/mm2≤[f]=205 N/mm2满足要求！八、连墙件承载力验算lw k a长细比λ=l0/i=600/15.8=37.975，查《规范》JGJ130-2011表A.0.6得，φ=0.896(N lw+N0)/(φAc)=(5.67+2)×103/(0.896×489)=17.506N/mm2≤0.85 ×[f]=0.85×205N/mm2=174.25N/mm2满足要求！扣件抗滑承载力验算：N lw+N0=5.67+2=7.67kN≤0.8×12=9.6kN满足要求！九、立杆地基承载力验算单立杆的轴心压力设计值：单内立杆：N1=(N G1k+ N G2k)+ N Q1k=(2.52+ 0.287)+ 4.742=7.549KN单中间立杆：N2=(N G1k+ N G2k)+ N Q1k=(4.774+ 0.581)+ 4.742=10.097KN 双立杆的轴心压力设计值:双内立杆：N S1=(N GS1k+ N GS2k)+ N1=(1.99+ 0.205)+ 7.549=9.744KN双中间立杆：N S2=(N GS1k+ N GS2k)+ N2=(3.6+ 0.415)+ 10.097=14.112KN N=max{N1，N2，N s1，N s2}=14.112KN立杆底垫板平均压力P=N s/(k c A)=14.112/(0.4×0.25)=141.12kPa≤f g=160kPa 满足要求！。