梁模板（扣件式）计算书计算依据：1、《建筑施工模板安全技术规范》JGJ162-20082、《建筑施工扣件式钢管脚手架安全技术规范》JGJ 130－20113、《混凝土结构设计规范》GB50010-20104、《建筑结构荷载规范》GB 50009-20125、《钢结构设计规范》GB 50017-2003一、工程属性二、荷载设计平面图立面图四、面板验算面板类型覆面木胶合板面板厚度(mm) 12面板抗弯强度设计值[f](N/mm2) 15 面板弹性模量E(N/mm2) 10000W＝bh2/6=1000×12×12/6＝24000mm3，I＝bh3/12=1000×12×12×12/12＝144000mm4q1＝γ0×max[1.2(G1k+(G2k+G3k)×h)+1.4Q2k，1.35(G1k+(G2k+G3k)×h)+1.4ψc Q2k]×b=0.9×max[1.2×(0.1+(24+1.5)×2.6)+1.4×2，1.35×(0.1+(24+1.5)×2.6)+1.4×0.7×2]×1＝82.44kN/mq1静＝0.9×1.35×[G1k+(G2k+G3k)×h]×b＝0.9×1.35×[0.1+(24+1.5)×2.6]×1＝80.676kN/mq1活＝0.9×1.4×0.7×Q2k×b＝0.9×1.4×0.7×2×1＝1.764kN/mq2＝[G1k+(G2k+G3k)×h]×b＝[0.1+(24+1.5)×2.6]×1＝66.4kN/m1、强度验算M max＝0.125q1L2＝0.125q1l2＝0.125×82.44×0.152＝0.232kN·mσ＝M max/W＝0.232×106/24000＝9.661N/mm2≤[f]＝15N/mm2满足要求！2、挠度验算νmax＝0.521q2L4/(100EI)=0.521×66.4×1504/(100×10000×144000)＝0.122mm≤[ν]＝l/250＝150/250＝0.6mm满足要求！3、支座反力计算设计值(承载能力极限状态)R1=R3=0.375 q1静l +0.437 q1活l=0.375×80.676×0.15+0.437×1.764×0.15＝4.654kN R2=1.25q1l=1.25×82.44×0.15＝15.458kN标准值(正常使用极限状态)R1'=R3'=0.375 q2l=0.375×66.4×0.15＝3.735kNR2'=1.25q2l=1.25×66.4×0.15＝12.45kN五、小梁验算小梁弹性模量E(N/mm2) 9600 小梁截面抵抗矩W(cm3) 53.33小梁截面惯性矩I(cm4) 213.33 验算方式三等跨连续梁q1＝max{4.654+0.9×1.35×[(0.3-0.1)×0.3/2+0.5×(2.6-0.16)]+0.9×max[1.2×(0.5+(24+1.1)×0.16)+1.4×2，1.35×(0.5+(24+1.1)×0.16)+1.4×0.7×2]×max[0.5-0.3/2，(1-0.5)-0.3/2]/2×1，15.458+0.9×1.35×(0.3-0.1)×0.3/2}=15.494kN/mq2＝max{3.735+(0.3-0.1)×0.3/2+0.5×(2.6-0.16)+(0.5+(24+1.1)×0.16)×max[0.5-0.3/2，(1-0.5)-0.3/2]/2×1，12.45+(0.3-0.1)×0.3/2}＝12.48kN/m1、抗弯验算M max＝max[0.1q1l12，0.5q1l22]＝max[0.1×15.494×0.4012，0.5×15.494×0.22]＝0.31kN·mσ＝M max/W＝0.31×106/53330＝5.811N/mm2≤[f]＝13N/mm2满足要求！2、抗剪验算V max＝max[0.6q1l1，q1l2]＝max[0.6×15.494×0.401，15.494×0.2]＝3.728kNτmax＝3V max/(2bh0)=3×3.728×1000/(2×50×80)＝1.398N/mm2≤[τ]＝1.4N/mm2 满足要求！3、挠度验算ν1＝0.677q2l14/(100EI)＝0.677×12.48×4014/(100×9600×2133300)＝0.107mm≤[ν]＝l1/250＝401/250＝1.604mmν2＝q2l24/(8EI)＝12.48×2004/(8×9600×2133300)＝0.122mm≤[ν]＝2l2/250＝2×200/250＝1.6mm满足要求！4、支座反力计算梁头处(即梁底支撑主梁悬挑段根部)承载能力极限状态R max＝max[1.1q1l1，0.4q1l1+q1l2]＝max[1.1×15.494×0.401，0.4×15.494×0.401+15.494×0.2]＝6.834kN同理可得，梁底支撑小梁所受最大支座反力依次为R1＝R3＝3.294kN，R2＝6.834kN正常使用极限状态R'max＝max[1.1q2l1，0.4q2l1+q2l2]＝max[1.1×12.48×0.401，0.4×12.48×0.401+12.48×0.2]＝5.505kN同理可得，梁底支撑小梁所受最大支座反力依次为R'1＝R'3＝2.888kN，R'2＝5.505kN六、主梁验算主梁抗弯强度设计值[f](N/mm2) 205 主梁抗剪强度设计值[τ](N/mm2) 125 主梁截面惯性矩I(cm4) 12.19 主梁截面抵抗矩W(cm3) 5.081、抗弯验算主梁弯矩图(kN·m)σ＝M max/W＝0.294×106/5080＝57.879N/mm2≤[f]=205N/mm2满足要求！2、抗剪验算主梁剪力图(kN)V max＝2.894kNτmax＝2V max/A=2×2.894×1000/489＝11.836N/mm2≤[τ]＝125N/mm2满足要求！3、挠度验算主梁变形图(mm)νmax＝0.077mm≤[ν]＝l/250＝500/250＝2mm满足要求！4、支座反力计算承载能力极限状态支座反力依次为R1=0.4kN，R2=12.622kN，R3=0.4kN七、可调托座验算荷载传递至立杆方式可调托座扣件抗滑移折减系数k c0.85 可调托座内主梁根数 1 可调托座承载力容许值[N](kN) 301、扣件抗滑移验算两侧立柱最大受力R＝max[R1,R3]＝max[0.4,0.4]＝0.4kN≤0.85×8=6.8kN单扣件在扭矩达到40～65N·m且无质量缺陷的情况下，单扣件能满足要求！2、可调托座验算可调托座最大受力N＝max[R2]＝12.622kN≤[N]＝30kN满足要求！八、立柱验算顶部立杆段：l01=kμ1(h d+2a)=1×1.386×(600+2×200)=1386mm非顶部立杆段：l02=kμ2h =1×1.755×1500=2632.5mmλ=l0/i=2632.5/15.8=166.614≤[λ]=210长细比满足要求！2、风荷载计算M w＝γ0×1.4×ψc×ωk×l a×h2/10＝1.4×0.9×0.156×0.401×1.52/10＝0.018kN·m3、稳定性计算根据《建筑施工扣件式钢管脚手架安全技术规范》JGJ130-2011，荷载设计值q1有所不同：1)面板验算q1＝[1.2×(0.1+(24+1.5)×2.6)+1.4×0.9×2]×1＝82.2kN/m2)小梁验算q1＝max{4.647+1.2×[(0.3-0.1)×0.3/2+0.5×(2.6-0.16)]+[1.2×(0.5+(24+1.1)×0.16)+1.4×0.9×1]×max[0.5-0.3/2，(1-0.5)-0.3/2]/2×1，15.412+1.2×(0.3-0.1)×0.3/2}=15.448kN/m 同上四～六计算过程，可得：R1＝0.4kN，R2＝12.596kN，R3＝0.4kN顶部立杆段：l01=kμ1(h d+2a)=1.155×1.386×(600+2×200)=1600.83mmλ1＝l01/i＝1600.83/15.8＝101.318，查表得，φ1＝0.58立柱最大受力N w＝max[R1+N边1，R2，R3+N边2]+M w/l b＝max[0.4+[1.2×(0.5+(24+1.1)×0.16)+1.4×0.9×1]×(1+0.5-0.3/2)/2×1，12.596，0.4+[1.2×(0.5+(24+1.1)×0.16)+1.4×0.9×1]×(1+1-0.5-0.3/2)/2×1]+0.018/1＝12.614kNf＝N/(φA)+M w/W＝12614.022/(0.58×489)+0.018×106/5080＝47.966N/mm2≤[f]＝205N/mm2满足要求！非顶部立杆段：l02=kμ2h =1.155×1.755×1500=3040.537mmλ2＝l02/i＝3040.537/15.8＝192.439，查表得，φ2＝0.195立柱最大受力N w＝max[R1+N边1，R2，R3+N边2]+M w/l b＝max[0.4+[1.2×(0.5+(24+1.1)×0.16)+1.4×0.9×1]×(1+0.5-0.3/2)/2×1，12.596，0.4+[1.2×(0.5+(24+1.1)×0.16)+1.4×0.9×1]×(1+1-0.5-0.3/2)/2×1]+0.018/1＝12.614kN立柱最大受力N w＝max[R1+N边1，R2，R3+N边2]+1.2×0.15×(7.25-2.6)+M w/l b＝max[0.4+[1.2×(0.75+(24+1.1)×0.16)+1.4×0.9×1]×(1+0.5-0.3/2)/2×1，12.596，0.4+[1.2×(0.75+(24+1.1)×0.16)+1.4×0.9×1]×(1+1-0.5-0.3/2)/2×1]+0.837+0.018/1＝13.451kNf＝N/(φA)+M w/W＝13451.022/(0.195×489)+0.018×106/5080＝144.554N/mm2≤[f]＝205N/mm2满足要求！。