模板支撑体系计算书计算依据:1 、《建筑施工模板安全技术规》JGJ162-20082 、《建筑施工扣件式钢管脚手架安全技术规》JGJ 130-20113 、《混凝土结构设计规》GB 50010-20104 、《建筑结构荷载规》GB 50009-20125 、《钢结构设计规》GB 50017-2003-、工程属性、荷载设计三、模板体系设计设计简图如下:平面图本图梁侧支撐构遥仅作示意，具体详见衆側模樋设计四、面板验算面板类型覆面木胶合板 面板厚度t(mm)142面板抗弯强度设计值[f](N/mm )15 2面板抗剪强度设计值[T ](N/mm )1.52面板弹性模量E(N/mm)5400取单位宽度b=1000mm 按三等跨连续梁计算:W = bh 2/6=1000 X 14X 14/6 = 32666.667mm, I = bh 3/12=1000 X 14X 14X 14/12 = 228666.667mmq 1 = 0.9 X max[1.2(G ik +(G k +G k ) X h)+1.4Q 2k , 1.35(G ik +(G k +G k ) X h)+1.4 书 cQk ]X b=0.9 X max[1.2 X (0.1+(24+1.5) X 0.9)+1.4 X 2, 1.35 X (0.1+(24+1.5) X0.9)+1.4 X 0.7 X 2] X 1 = 29.77kN/mV1<------------ 1 ---------------------- cyjLTc----------- «JLT IT立面图q 備=0.9 X 1.35 X [G ik+(G k+G k) X h] X b= 0.9 X 1.35 X [0.1+(24+1.5) X 0.9] X 1= 28.006kN/mq 1活= 0.9 X 1.4 X 0.7 X Q k X b = 0.9 X 1.4 X 0.7 X 2X 1 = 1.764kN/mq 2= [1 X (G1k+(G k+G k) X h)] X b= [1 X (0.1+(24+1.5) X 0.9)] X 1 = 23.05kN/m 计算简图如下：1 、强度验算M max= 0.1q1 静L2+0.117q1 活L2= 0.1 X 28.006 X 0.1 2+0.117 X 1.764 X 0.1 2=0.03kN • m(r = MUW= 0.03 X 106/32666.667 = 0.92N/mn i< [f] = 15N/mrm满足要求！2 、挠度验算v max= 0.677q2L4/(100EI)=0.677 X 23.05 X 1004/(100 X 5400X 228666.667)= 0.013mmC [ v ] = L/250 = 100/250 = 0.4mm满足要求！3 、支座反力计算设计值(承载能力极限状态)R 1=R=0.4q1 静L+0.45q1 活L=0.4 X 28.006 X 0.1+0.45 X 1.764 X 0.1 = 1.2kNR 2=R=1.1q1 静L+1.2q1 活L=1.1 X 28.006 X 0.1+1.2 X 1.764 X 0.1 = 3.292kN 标准值(正常使用极限状态)R 1 =R' =0.4q 2L=0.4 X 23.05 X 0.1 = 0.922kNR 2=R'=1.1q2L=1.1 X 23.05 X 0.1 = 2.536kN五、小梁验算承载能力极限状态：梁底面板传递给左边小梁线荷载：5左=R/b=1.2/1 = 1.2kN/m梁底面板传递给中间小梁最大线荷载：4仲=Max[F2,R3]/b =Max[3.292,3.292]/仁3.292kN/m梁底面板传递给右边小梁线荷载：5右=F4/b=1.2/1 = 1.2kN/m小梁自重：q2= 0.9 X 1.35 X (0.3-0.1) X 0.3/3 =0.024kN/m梁左侧模板传递给左边小梁荷载43左=0.9 X 1.35 X 0.5 X(0.9-0.12)=0.474kN/m梁右侧模板传递给右边小梁荷载= 0.9 X 1.35 X 0.5 X(0.9-0.12)=0.474kN/m梁左侧楼板传递给左边小梁荷载44左=0.9 X Max[1.2 X (0.5+(24+1.1) X 0.12)+1.4 X 2, 1.35 X (0.5+(24+1.1) X 0.12)+1.4 X 0.7 X 2] X (0.5-0.3/2)/2 X 1= 1.105kN/m梁右侧楼板传递给右边小梁荷载= 0.9 X Max[1.2 X (0.5+(24+1.1) X 0.12)+1.4 X 2, 1.35 X (0.5+(24+1.1) X 0.12)+1.4 X 0.7 X 2] X ((1-0.5)-0.3/2)/2 X 1= 1.105kN/m 左侧小梁荷载q左= +q2+q a左+中左=1.2+0.024+0.474+1.105=2.803kN/m中间小梁荷载q中= q 仲+ q 2=3.292+0.024=3.317kN/m右侧小梁荷载q右=4佑+q2+q a右+q4右=1.2+0.024+0.474+1.105=2.803kN/m小梁最大荷载q=Max[c左,q 中,q 右]=Max[2.803,3.317,2.803]=3.317kN/m正常使用极限状态：梁底面板传递给左边小梁线荷载：5左’=R'/b=0.922/1 = 0.922kN/m梁底面板传递给中间小梁最大线荷载：q仲’=Max[R',R訂/b =Max[2.536,2.536]/仁2.536kN/m梁底面板传递给右边小梁线荷载：q i右’=R//b=0.922/1 = 0.922kN/m小梁自重：q2' = i x (0.3-0.1) X 0.3/3 =0.02kN/m梁左侧模板传递给左边小梁荷载q3左'=1X 0.5 X (0.9-0.12)=0.39kN/m梁右侧模板传递给右边小梁荷载q a右'=1X 0.5 X (0.9-0.12)=0.39kN/m梁左侧楼板传递给左边小梁荷载q4左'=[1 X (0.5+(24+1.1) X 0.12)] X(0.5-0.3/2)/2 X 1=0.615kN/m梁右侧楼板传递给右边小梁荷载q4右'=[1 X (0.5+(24+1.1) X 0.12)] X ((1-0.5)-0.3/2)/2 X 1=0.615kN/m左侧小梁荷载q左' = q佐'+q J+q 3左'+q 4左'=0.922+0.02+0.39+0.615=1.947kN/m 中间小梁荷载q中'= q 仲'+ q 2‘=2.536+0.02=2.556kN/m右侧小梁荷载q右' = q佑'+q J+q 3右'+q 4右'=0.922+0.02+0.39+0.615=1.947kN/m小梁最大荷载q'=Max[q 左',q 中',q 右']=Max[1.947,2.556,1.947]=2.556kN/m为简化计算，按简支梁和悬臂梁分别计算，如下图：i、抗弯验算M ma尸max[0.125ql i2, 0.5ql 2] = max[0.125 x3.317 x O.52, 0.5 x 3.317 x0.3] =0.149kN • m(T =MUW=0.149X 106/32667=4.569N/mn i< [f]=11.44N/mm 2满足要求！2 、抗剪验算V max= max[0.5ql 1, ql 2] = max[0.5 x 3.317 x 0.5 , 3.317 x 0.3] = 0.995kN2T ma=3Vn ax/(2bh 0)=3 X 0.995 X 1000/(2 x 40x 70) = 0.533N/mn i<[T ]=1.232N/mm2满足要求！3 、挠度验算V 1 = 5q'l 14/(384EI) = 5x 2.556 x 5004/(384 x 7040x 114.333 x 104) = 0.258mm w [ V ] = 11/250 = 500/250 = 2mm4 4 4V2= q'l 2/(8EI) = 2.556 x 300 /(8 x 7040x 114.333 x 10) = 0.322mmC [ V ]=2l 2/250 = 2x 300/250 = 2.4mm满足要求！4 、支座反力计算承载能力极限状态R ma=[qL1,0.5qL 1+qL2]=max[3.317 x 0.5,0.5 x 3.317 x 0.5+3.317 x0.3]=1.824kN同理可得：梁底支撑小梁所受最大支座反力依次为R i=1.542kN,R2=1.824kN,R3=1.824kN,R4=1.542kN正常使用极限状态R maX=[q'L i,0.5q'L i+q'L2〕=max[2.556 X 0.5,0.5 X 2.556 X 0.5+2.556 X0.3]=1.406kN同理可得：梁底支撑小梁所受最大支座反力依次为R i'=1.071kN,R 2'=1.406kN,R J=1.406kN,R /=1.071kN六、主梁验算L.a24tN l.«4kN 1.542kMJ L 痂L 中J L垣J Lr r 1 、抗弯验算主梁弯矩图(kN • m)(T =MUW=0.141 X 106/4120=34.284N/mm = [f]=205N/mm 2满足要求！2 、抗剪验算主梁剪力图(kN)V max = 3.366kNT maX =2Vn ax /A=2 X 3.366 X 1000/384 = 17.532N/mn i < [ T ]=125N/mn n满足要求！3、挠度验算S. 661 kN 3.6«lkM 3.661 khl 3.«61kM 工昭lkN主梁变形图(mm)v max= 0.056mm^ [ v ] = L/250 = 334/250 = 1.336mm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Ri=0.295kN, F2=3.661kN, R=3.661kN, F4=0.295kN正常使用极限状态支座反力依次为R i'=0.224kN，R'=2.701kN，R'=2.701kN，F4'=0.224kN七、2号主梁验算P = max[F2, F3]=Max[3.661 , 3.661]=3.661kN , P = max[F2', R，] = Max[2.701 , 2.701]=2.701kNJ L 咖L 咖Lr 1 、抗弯验算V ma尸2.38kN2 号主梁弯矩图（kN • m）（T =ML/W=0.641 X 106/4120=155.509N/mmk满足要求!2 、抗剪验算2号主梁剪力图(kN)S. 661 kN3.6«lkM3.661 khl3.«61kM工昭lkNT ma =2Vn aJ A=2 X 2.38 X 1000/384 = 12.394N/mm < [ T ]=125N/mn i满足要求！3 、挠度验算2号主梁变形图(mm)v max = 1.533mm^ [ v ] = L/250 = 1000/250 = 4mm满足要求！4 、支座反力计算极限承载能力状态支座反力依次为 R i =4.942kN , F 2=7.871kN , R=7.871kN , F 4=4.942kN 立柱所受主梁支座反力依次为 艮=7.871/1=7.871kN ，F 3=7.871/1=7.871kN八、纵向水平钢管验算计算简图如下:O.MEkJN 0.255-kM O.^Skhl CL 曲秋N 0.295WU OJ晒kN JL 咖L 咖Lr 1 、抗弯验算V ma尸0.192kN纵向水平钢管弯矩图（kN • m）c = ML/W= 0.052 X 106/4120 = 12.531N/mm= [f] = 205N/mrm满足要求!2 、抗剪验算纵向水平钢管剪力图(kN)T ma尸2V max/A=2 X 0.192 X 1000/384 = 0.999N/mm k [ T ] = 125N/mrm 满足要求！3 、挠度验算纵向水平钢管变形图(mm)v max= 0.127mm^ [ v ] = L/250 = 1000/250 = 4mm满足要求！4 、支座反力计算支座反力依次为R i=0.398kN, F2=0.634kN, R=0.634kN, F4=0.398kN同理可得：两侧立柱所受支座反力依次为R=0.634kN, R=0.634kN九、可调托座验算1 、扣件抗滑移验算两侧立柱最大受力N= max[R,R4] = max[0.634,0.634] = 0.634kN< 1 X 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！2 、可调托座验算可调托座最大受力N= max[R, R B] = 7.871kN < [N]=30kN满足要求！十、立柱验算、长细比验算1 o=h=15OOmm入=l o/i=15OO/16=93.75 < [入]=150长细比满足要求！查表得，©二0.6412 、风荷载计算2 2M w= 0.9 X© c X 1.4 Xw k X I a X h/10 = 0.9 X 0.9 X 1.4 X 0.29 X 1 X 1.5 /10 =0.074kN • m3 、稳定性计算根据《建筑施工模板安全技术规》JGJ162-2008,荷载设计值5有所不同：1) 面板验算q 1 = 0.9 X [1.2 X (0.1+(24+1.5) X 0.9)+1.4 X 0.9 X 2] X 1 = 27.162kN/m2) 小梁验算q 1= max{1.098+0.9 X 1.2 X [(0.3-0.1) X 0.3/3+0.5 X (0.9-0.12)]+0.9 X [1.2 X(0.5+(24+1.1) X 0.12)+1.4 X 0.9 X 1] X max[0.5-0.3/2 , (1-0.5)-0.3/2]/2 X 1, 3.01+0.9 X1.2 X (0.3-0.1) X 0.3/3}=3.032kN/m同上四〜八计算过程，可得：R 1 = 0.574kN ,艮=7.003kN , 7.003kN, R = 0.574kN立柱最大受力Nv= max[R+Ni 1, R, R, Ri+N*2]+0.9 X 1.2 X 0.15 X (31.2-0.9)+M W\ b= max[0.574+0.9 X [1.2 X (0.5+(24+1.1) X 0.12)+1.4 X 0.9 X 1] X (1+0.5-0.3/2)/2 X 1,7.003 , 7.003 , 0.574+0.9 X [1.2 X (0.5+(24+1.1) X 0.12)+1.4 X 0.9 X 1] X (1+1-0.5-0.3/2)/2 X 1]+4.909+0.074/1 = 11.985kNf = N/( © A)+M/W= 11985.215/(0.641 X 384)+0.074 X 106/4120 = 66.653N/mn i< [f] = 205N/mrn满足要求！十一、高宽比验算根据《建筑施工扣件式钢管脚手架安全技术规》JGJ130-2011第6.9.7 :支架高宽比不应大于3H/B=31.2/20=1.56 V 3满足要求，不需要进行抗倾覆验算！十二、立柱支承面承载力验算F 1=N=11.985kN1 、受冲切承载力计算GB50010-20K第6.5.1条规定，见下表根据《混凝土结构设计规》可得：B h =1, f t =0.858N/mm,n =1, h o =h-2O=1OOmmu m =2[(a+h o )+(b+h o )]=15OOmmF=(0.7 B h f t +O.25(T pc ,m ) n U m h O =(O.7 X 1 X O.858+O.25 X O) X 1 X 15OO X 1OO/1OOO=9O.O9kN > R=11.985kN满足要求！2 、局部受压承载力计算根据《混凝土结构设计规》GB50010-201C 第6.6.1条规定，见下表可得：f c=7.488N/mm c=1,B i=(A b/A i)1/2=[(a+2b) x (b+2b)/(ab)] 1/2=[(1000) x (1350)/(100 x1/2 2450)] =5.477, A in =ab=45000mmF=1.35 B 詡l fA=1.35 x 1 x 5.477 x 7.488 x 45000/1000=2491.568kN > R=11.985kN满足要求！。