梁模板(600、1300)

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梁模板(轮扣式,梁板立柱共用)计算书计算依据:1、《建筑施工脚手架安全技术统一标准》GB51210-20162、《建筑施工模板安全技术规范》JGJ162-20083、《混凝土结构设计规范》GB 50010-20104、《建筑结构荷载规范》GB 50009-20125、《钢结构设计规范》GB 50017-2003一、工程属性每纵距内附加梁底支撑主梁根数0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t(mm) 12 面板抗弯强度设计值[f](N/mm2) 12.5 面板抗剪强度设计值[τ](N/mm2) 1.4 面板弹性模量E(N/mm2) 4500W=bh2/6=1000×12×12/6=24000mm3,I=bh3/12=1000×12×12×12/12=144000mm4q1=γ0×max[1.2(G1k+(G2k+G3k)×h)+1.4Q1k,1.35(G1k+(G2k+G3k)×h)+1.4ψc Q1k]×b=1.1×max[1.2×(0.1+(24+1.5)×1.8)+1.4×3,1.35×(0.1+(24+1.5)×1.8)+1.4×0.7×3]×1=71.544kN/mq1静=γ0×1.35×[G1k+(G2k+G3k)×h]×b=1.1×1.35×[0.1+(24+1.5)×1.8]×1=68.31kN/mq1活=γ0×1.4×0.7×Q1k×b=1.1×1.4×0.7×3×1=3.234kN/mq2=[1×(G1k+(G2k+G3k)×h)]×b=[1×(0.1+(24+1.5)×1.8)]×1=46kN/m计算简图如下:1、强度验算M max=0.1q1静L2+0.117q1活L2=0.1×68.31×0.152+0.117×3.234×0.152=0.162kN·m σ=M max/W=0.162×106/24000=6.759N/mm2≤[f]=12.5N/mm2满足要求!2、挠度验算νmax=0.677q2L4/(100EI)=0.677×46×1504/(100×4500×144000)=0.243mm≤[ν]=min[L/150,10]=min[150/150,10]=1mm满足要求!3、支座反力计算设计值(承载能力极限状态)R1=R4=0.4q1静L+0.45q1活L=0.4×68.31×0.15+0.45×3.234×0.15=4.317kNR2=R3=1.1q1静L+1.2q1活L=1.1×68.31×0.15+1.2×3.234×0.15=11.853kN标准值(正常使用极限状态)R1'=R4'=0.4q2L=0.4×46×0.15=2.76kNR2'=R3'=1.1q2L=1.1×46×0.15=7.59kN五、小梁验算梁底面板传递给左边小梁线荷载:q1左=R1/b=4.317/1=4.317kN/m梁底面板传递给中间小梁最大线荷载:q1中=Max[R2,R3]/b =Max[11.853,11.853]/1= 11.853kN/m梁底面板传递给右边小梁线荷载:q1右=R4/b=4.317/1=4.317kN/m小梁自重:q2=1.1×1.35×(0.3-0.1)×0.45/3 =0.045kN/m梁左侧模板传递给左边小梁荷载q3左=1.1×1.35×0.5×(1.8-0.4)=1.04kN/m梁右侧模板传递给右边小梁荷载q3右=1.1×1.35×0.5×(1.8-0.4)=1.04kN/m梁左侧楼板传递给左边小梁荷载q4左=1.1×Max[1.2×(0.5+(24+1.1)×0.4)+1.4×3,1.35×(0.5+(24+1.1)×0.4)+1.4×0.7×3]×(0.475-0.45/2)/2×1=2.361kN/m梁右侧楼板传递给右边小梁荷载q4右=1.1×Max[1.2×(0.5+(24+1.1)×0.4)+1.4×3,1.35×(0.5+(24+1.1)×0.4)+1.4×0.7×3]×((0.95-0.475)-0.45/2)/2×1=2.361kN/m左侧小梁荷载q左=q1左+q2+q3左+q4左=4.317+0.045+1.04+2.361=7.762kN/m中间小梁荷载q中= q1中+ q2=11.853+0.045=11.898kN/m右侧小梁荷载q右=q1右+q2+q3右+q4右=4.317+0.045+1.04+2.361=7.762kN/m小梁最大荷载q=Max[q左,q中,q右]=Max[7.762,11.898,7.762]=11.898kN/m正常使用极限状态:梁底面板传递给左边小梁线荷载:q1左'=R1'/b=2.76/1=2.76kN/m梁底面板传递给中间小梁最大线荷载:q1中'=Max[R2',R3']/b = Max[7.59,7.59]/1= 7.59kN/m梁底面板传递给右边小梁线荷载:q1右'=R4'/b=2.76/1=2.76kN/m小梁自重:q2'=1×(0.3-0.1)×0.45/3 =0.03kN/m梁左侧模板传递给左边小梁荷载q3左'=1×0.5×(1.8-0.4)=0.7kN/m梁右侧模板传递给右边小梁荷载q3右'=1×0.5×(1.8-0.4)=0.7kN/m梁左侧楼板传递给左边小梁荷载q4左'=[1×(0.5+(24+1.1)×0.4)]×(0.475-0.45/2)/2×1=1.318kN/m梁右侧楼板传递给右边小梁荷载q4右'=[1×(0.5+(24+1.1)×0.4)]×((0.95-0.475)-0.45/2)/2×1=1.318kN/m左侧小梁荷载q左'=q1左'+q2'+q3左'+q4左'=2.76+0.03+0.7+1.318=4.808kN/m中间小梁荷载q中'= q1中'+ q2'=7.59+0.03=7.62kN/m右侧小梁荷载q右'=q1右'+q2'+q3右'+q4右' =2.76+0.03+0.7+1.318=4.808kN/m小梁最大荷载q'=Max[q左',q中',q右']=Max[4.808,7.62,4.808]=7.62kN/m为简化计算,按二等跨连续梁和悬臂梁分别计算,如下图:1、抗弯验算M max=max[0.125ql12,0.5ql22]=max[0.125×11.898×0.452,0.5×11.898×0.22]=0.301kN·mσ=M max/W=0.301×106/42667=7.059N/mm2≤[f]=15.444N/mm2满足要求!2、抗剪验算V max=max[0.625ql1,ql2]=max[0.625×11.898×0.45,11.898×0.2]=3.346kNτmax=3V max/(2bh0)=3×3.346×1000/(2×40×80)=1.569N/mm2≤[τ]=1.782N/mm2满足要求!3、挠度验算ν1=0.521q'l14/(100EI)=0.521×7.62×4504/(100×9350×170.667×104)=0.102mm≤[ν]=min[l1/150,10]=min[450/150,10]=3mmν2=q'l24/(8EI)=7.62×2004/(8×9350×170.667×104)=0.096mm≤[ν]=min[2l2/150,10]=min[400/150,10]=2.667mm满足要求!4、支座反力计算承载能力极限状态R max=max[1.25qL1,0.375qL1+qL2]=max[1.25×11.898×0.45,0.375×11.898×0.45+11.898×0.2]=6.693kN同理可得:梁底支撑小梁所受最大支座反力依次为R1=4.366kN,R2=6.693kN,R3=6.693kN,R4=4.366kN正常使用极限状态R max'=max[1.25q'L1,0.375q'L1+q'L2]=max[1.25×7.62×0.45,0.375×7.62×0.45+7.62×0.2]= 4.286kN同理可得:梁底支撑小梁所受最大支座反力依次为R1'=2.704kN,R2'=4.286kN,R3'=4.286kN,R4'=2.704kN六、主梁验算主梁类型钢管主梁截面类型(mm) Φ48×2.8主梁计算截面类型(mm) Φ48×2.8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 主梁截面抵抗矩W(cm3) 4.25主梁弹性模量E(N/mm2) 206000 主梁截面惯性矩I(cm4) 10.19 可调托座内主梁根数 2 主梁受力不均匀系数0.6受集中力为Ks×Rn,Rn为各小梁所受最大支座反力1、抗弯验算主梁弯矩图(kN·m)σ=M max/W=0.17×106/4250=40.052N/mm2≤[f]=205N/mm2满足要求!2、抗剪验算主梁剪力图(kN)V max=3.527kNτmax=2V max/A=2×3.527×1000/398=17.723N/mm2≤[τ]=125N/mm2满足要求!3、挠度验算主梁变形图(mm)νmax=0.012mm≤[ν]=min[L/150,10]=min[250/150,10]=1.667mm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=0.092kN,R2=3.395kN,R3=6.86kN,R4=3.362kN,R5=0.252kN 立杆所受主梁支座反力依次为P1=0.092/0.6=0.154kN,P2=3.395/0.6=5.659kN,P3=6.86/0.6=11.433kN,P4=3.362/0.6=5.603kN,P5=0.252/0.6=0.421kN七、可调托座验算两侧立杆最大受力N=max[R1,R5]=max[0.092,0.252]=0.252kN≤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[P2,P3,P4]=11.433kN≤[N]=30kN满足要求!八、立杆验算h max=max(ηh,h'+2ka)=max(1.2×1800,600+2×0.7×500)=2160mmλ=h max/i=2160/16=135≤[λ]=150长细比满足要求!查表得:φ=0.3712、风荷载计算M wd=γ0×φc×γQ×Mωk=γ0×φc×γQ×(ζ2×ωk×l a×h2/10)=1.1×0.6×1.4×(1×0.021×0.45×1.82/10)=0.003kN·m3、稳定性计算P1=0.154kN,P2=5.659kN,P3=11.433kN,P4=5.603kN,P5=0.421kN梁两侧立杆承受楼板荷载:左侧楼板传递给梁左侧立杆荷载:N边1=1.1×max[1.2×(0.5+(24+1.1)×0.4)+1.4×3,1.35×(0.5+(24+1.1)×0.4)+0.7×1.4×3]×(0.9+0.475-0.45/2)/2×0.45=4.887kN右侧楼板传递给梁右侧立杆荷载:N边2=1.1×max[1.2×(0.5+(24+1.1)×0.4)+1.4×3,1.35×(0.5+(24+1.1)×0.4)+0.7×1.4×3]×(0.9+0.95-0.475-0.45/2)/2×0.45=4.887kNN d=max[P1+N边1,P2,P3,P4,P5+N边2]+1.1×1.35×0.15×(4.41-1.8)=max[0.154+4.887,5.659,11.433,5.603,0.421+4.887]+0.581=12.014kNf d=N d/(φA)+M wd/W=12014.2/(0.371×398)+0.003×106/4250=82.071N/mm2≤[f]=205N/mm2满足要求!九、高宽比验算根据《建筑施工脚手架安全技术统一标准》GB51210-2016 第8.3.2条: 支撑脚手架独立架体高宽比不应大于3.0H/B=4.41/30=0.147≤3满足要求!十、架体抗倾覆验算支撑脚手架风线荷载标准值:q wk=l'a×ωfk=0.9×0.566=0.509kN/m:风荷载作用在支架外侧竖向封闭栏杆上产生的水平力标准值:F wk= l'a×H m×ωmk=0.9×1.5×0.153=0.207kN支撑脚手架计算单元在风荷载作用下的倾覆力矩标准值M ok:M ok=0.5H2q wk+HF wk=0.5×4.412×0.509+4.41×0.207=5.864kN.m参考《规范》GB51210-2016 第6.2.17条:B2l'a(g k1+ g k2)+2ΣG jk b j≥3γ0M okg k1——均匀分布的架体面荷载自重标准值kN/m2g k2——均匀分布的架体上部的模板等物料面荷载自重标准值kN/m2G jk——支撑脚手架计算单元上集中堆放的物料自重标准值kNb j——支撑脚手架计算单元上集中堆放的物料至倾覆原点的水平距离mB2l'a(g k1+ g k2)+2ΣG jk b j=B2l'a[qH/(l'a×l'b)+G1k]+2×G jk×B/2=302×0.9×[0.15×4.41/(0.9×0.9)+0.5]+2×1×30/2=1096 .5kN.m≥3γ0M ok =3×1.1×5.864=19.352kN.M满足要求!十一、立杆支承面承载力验算11、受冲切承载力计算根据《混凝土结构设计规范》GB50010-2010第6.5.1条规定,见下表h t 0u m =2[(a+h 0)+(b+h 0)]=1320mm F=(0.7βh f t +0.25σpc ,m )ηu m h 0=(0.7×1×0.829+0.25×0)×1×1320×130/1000=99.579kN ≥F 1=12.014kN 满足要求!2、局部受压承载力计算根据《混凝土结构设计规范》GB50010-2010第6.6.1条规定,见下表c cβl=(A b/A l)1/2=[(a+2b)×(b+2b)/(ab)]1/2=[(600)×(600)/(200×200)]1/2=3,A ln=ab=40000mm2F=1.35βcβl f c A ln=1.35×1×3×8.294×40000/1000=1343.628kN≥F1=12.014kN 满足要求!。