扣件式梁模板支撑计算书(梁截面300mm×750mm)

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扣件式梁模板支撑计算书(高支模)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、荷载设计三、模板体系设计设计简图如下: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29 面板弹性模量E(N/mm2) 9000[f](N/mm2)取单位宽度1000mm,按三等跨连续梁计算,计算简图如下:W=bh2/6=1000×12×12/6=24000mm3,I=bh3/12=1000×12×12×12/12=144000mm4 q1=0.9max[1.2(G1k+ (G2k+G3k)×h)+1.4Q2k,1.35(G1k+(G2k+G3k)×h)+1.4×0.7Q2k]×b=0.9max[1.2×(0.1+(24+1.5)×0.75)+1.4×2,1.35×(0.1+(24+1.5)×0.75)+1.4×0.7×2]×1=25.122kN/mq1静=0.9×1.35×[G1k+(G2k+G3k)×h]×b=0.9×1.35×[0.1+(24+1.5)×0.75]×1=23.358kN/m q1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)×0.75]×1=19.225kN/m1、强度验算M max=0.1q1静L2+0.117q1活L2=0.1×23.358×0.12+0.117×1.764×0.12=0.025kN·mσ=M max/W=0.025×106/24000=1.059N/mm2≤[f]=29N/mm2满足要求!2、挠度验算νmax=0.677q2L4/(100EI)=0.677×19.225×1004/(100×9000×144000)=0.01mm≤[ν]=l/400=100/400=0.25mm满足要求!3、支座反力计算设计值(承载能力极限状态)R1=R4=0.4 q1静l +0.45 q1活l=0.4×23.358×0.1+0.45×1.764×0.1=1.014kNR2=R3=1.1 q1静l +1.2 q1活l=1.1×23.358×0.1+1.2×1.764×0.1=2.781kN标准值(正常使用极限状态)R1'=R4'=0.4 q2l=0.4×19.225×0.1=0.769kNR2'=R3'=1.1 q2l=1.1×19.225×0.1=2.115kN五、小梁验算为简化计算,按四等跨连续梁和悬臂梁分别计算,如下图:q1=max{1.014+0.9×1.35×[(0.3-0.1)×0.3/3+0.5×(0.75-0.12)]+0.9max[1.2×(0.5+(24+1.1)×0.12)+1.4×2,1.35×(0.5+(24+1.1)×0.12)+1.4×0.7×2]×max[0.6-0.3/2,(1.2-0.6)-0.3/2]/2×1,2.781+0.9×1.35×(0.3-0.1)×0.3/3}=2.841kN/mq2=max[0.769+(0.3-0.1)×0.3/3+0.5×(0.75-0.12)+(0.5+(24+1.1)×0.12)×max[0.6-0.3/2,(1.2-0.6)-0.3/2]/2×1,2.115+(0.3-0.1)×0.3/3]=2.135kN/m1、抗弯验算M max=max[0.107q1l12,0.5q1l22]=max[0.107×2.841×0.62,0.5×2.841×0.12]=0.109kN·mσ=M max/W=0.109×106/64000=1.71N/mm2≤[f]=15.44N/mm2满足要求!2、抗剪验算V max=max[0.607q1l1,q1l2]=max[0.607×2.841×0.6,2.841×0.1]=1.035kNτmax=3V max/(2bh0)=3×1.035×1000/(2×60×80)=0.323N/mm2≤[τ]=1.78N/mm2满足要求!3、挠度验算ν1=0.632q2l14/(100EI)=0.632×2.135×6004/(100×9350×2560000)=0.073mm≤[ν]=l/400=600/400=1.5mmν2=q2l24/(8EI)=2.135×1004/(8×9350×2560000)=0.001mm≤[ν]=l/400=100/400=0.25mm满足要求!4、支座反力计算梁头处(即梁底支撑小梁悬挑段根部)承载能力极限状态R max=max[1.143q1l1,0.393q1l1+q1l2]=max[1.143×2.841×0.6,0.393×2.841×0.6+2.841×0.1]=1.948kN同理可得,梁底支撑小梁所受最大支座反力依次为R1=R4=1.924kN,R2=R3=1.948kN正常使用极限状态R'max=max[1.143q2l1,0.393q2l1+q2l2]=max[1.143×2.135×0.6,0.393×2.135×0.6+2.135×0.1]=1.464kN同理可得,梁底支撑小梁所受最大支座反力依次为R'1=R'4=1.467kN,R'2=R'3=1.464kN六、主梁验算主梁类型钢管主梁材料规格(mm) Ф48×3可调托座内主梁根数 2 主梁弹性模量E(N/mm2) 206000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) 10.78 主梁截面抵抗矩W(cm3) 4.49 主梁自重忽略不计,因主梁2根合并,则抗弯、抗剪、挠度验算荷载值取半,计算简图如下:1、抗弯验算主梁弯矩图(kN·m)σ=M max/W=0.137×106/4490=30.586N/mm2≤[f]=205N/mm2满足要求!2、抗剪验算主梁剪力图(kN)V max=1.844kNτmax=2V max/A=2×1.844×1000/424=8.696N/mm2≤[τ]=125N/mm2满足要求!3、挠度验算主梁变形图(mm)νmax=0.033mm≤[ν]=l/400=600/400=1.5mm满足要求!4、扣件抗滑计算R=max[R1,R3]=0.185kN≤1×8=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!同理可知,右侧立柱扣件受力R=0.185kN≤1×8=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!七、立柱验算长细比验算顶部立杆段:l01=kμ1(h d+2a)=1×1.386×(1500+2×180)=2577.96mm非顶部立杆段:l02=kμ2h =1×1.755×1500=2632.5mmλ=l0/i=2632.5/15.9=165.566≤[λ]=210长细比满足要求!1、风荷载计算M w=0.9×1.4×ωk×l a×h2/10=0.9×1.4×0.22×0.6×1.52/10=0.037kN·m2、稳定性计算根据《建筑施工模板安全技术规范》公式5.2.5-14,荷载设计值q1有所不同:1)面板验算q1=[1.2×(0.1+(24+1.5)×0.75)+0.9×1.4×2]×1=25.59kN/m2)小梁验算q1=max{0.933+(0.3-0.1)×0.3/3+[1.2×(0.5+(24+1.1)×0.12)+0.9×1.4×1]×max[0.6-0.3/2,(1.2-0.6)-0.3/2]/2×1,2.556+(0.3-0.1)×0.3/3}=2.576kN/m同上四~六计算过程,可得:R1=0.17kN,R2=6.732kN,R3=0.17kN顶部立杆段:l01=kμ1(h d+2a)=1.185×1.386×(1500+2×180)=3054.883mmλ1=l01/i=3054.883/15.9=192.131,查表得,φ1=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[0.17+[1.2×(0.5+(24+1.1)×0.12)+0.9×1.4×1]×(1.2+0.6-0.3/2)/2×1.2,6.732,0.17+[1.2×(0.5+(24+1.1)×0.12)+0.9×1.4×1]×(1.2+1.2-0.6-0.3/2)/2×1.2]+0.037/1.2=6.795kNf=N/(φA)+M w/W=6794.663/(0.195×424)+0.037×106/4490=90.515N/mm2≤[f]=205N/mm2满足要求!非顶部立杆段:l02=kμ2h =1.185×1.755×1500=3119.512mmλ2=l02/i=3119.512/15.9=196.196,查表得,φ2=0.188立柱最大受力N w=max[R1+N边1,R2,R3+N边2]+0.15×(10-0.75)+M w/l b=max[0.17+[1.2×(0.75+(24+1.1)×0.12)+0.9×1.4×1]×(1.2+0.6-0.3/2)/2×1.2,6.732,0.17+[1.2×(0.75+(24+1.1)×0.12)+0.9×1.4×1]×(1.2+1.2-0.6-0.3/2)/2×1.2]+1.388+0.037/1.2=8.151kNf=N/(φA)+M w/W=8150.978/(0.188×424)+0.037×106/4490=110.59N/mm2≤[f]=205N/mm2满足要求!八、可调托座验算由"主梁验算"一节计算可知可调托座最大受力N=max[R2]×2=7.374kN≤[N]=30kN 满足要求!九、立杆基础底面的平均压力应满足下式的要求:Pk = Nk/A ≤fg式中:Pk ——立杆基础底面处的平均压力标准值(kPa);Nk ——上部结构传至立杆基础顶面的轴向力标准值(kN);A——基础底面面积(m2);fg——地基承载力特征值(kPa),应按《建筑施工扣件式钢管脚手架安全技术规范》JGJ 130-2011第5.5.2 条规定采用。