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模板支撑体系计算书

模板支撑体系计算书计算依据: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、《建筑施工扣件式钢管脚手架安全技术规范》JGJ130-20114、《混凝土结构设计规范》GB 50010-20105、《建筑结构荷载规范》GB 50009-20126、《钢结构设计规范》GB 50017-2003一、工程属性风荷载参数:荷载系数参数表:设计简图如下:模板设计平面图模板设计立面图四、面板验算楼板面板应搁置在梁侧模板上,本例以简支梁,取1m单位宽度计算。

W = bh2/6=1000x13x13/6 = 28166.667mm3, I = bh3/12=1000x13x13x13/12 = 183083.333mm4承载能力极限状态% = 丫。

”1.35'5 +(G2k+G3k)xh)+1.4x/x(Q1k + Q2k)]xb=1x[1.35x(0.1+(24+1.1)x0.4)+1.4x0.9x2.5]x1=16.839kN/m正常使用极限状态q=(Y G(G1k +(G2k+G3k)xh))xb =(1x(0.1+(24+1.1)x0.4))x1 =10.14kN/m计算简图如下:W? TV?瞥HIP"」闻,,1、强度验算M max=q1l2/8 = 16.839x0.252/8 = 0.132kN-mo=M /W=0.132x106/28166.667 = 4.671N/mm2S[f] = 15N/mm2 max满足要求!2、挠度验算v max=5ql4/(384EI)=5x10.14x2504/(384x10000x183083.333)=0.282mmv=0.282mm<[v] = L/400=250/400 = 0.625mm满足要求!五、小梁验算q1 = 丫。

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+(5+%9)+1.4*限9比+Q2k)]xb=1x[1.35x(0.3+(24+1.1)x0.4)+1.4x0.9x2.5]x0.25=4.277kN/m因此,q1静=Y0x1.35x(G1k +(G2k+G3k)xh)xb=1x1.35x(0.3+(24+1.1)x0.4)x0.25 = 3.49kN/m q1活=Y0x1.4xW c x(Q1k + Q2k)xb=1x 1.4x0.9x2.5x0.25 = 0.787kN/m 计算简图如下:1、强度验算M1= 0.125q1静L2+0.125q1 活L2=0.125x3.49x0.92+0.125x0.787x0.92=0.433kN-mM2 = q1L12/2=4.277x0.32/2 = 0.192kN-mM max=max[M], M2] =max[0.433, 0.192] = 0.433kN-m gM max/W=0.433x106/42667=10.15N/mm2<[f]=15.444N/mm2 满足要求!2、抗剪验算V1= 0.625q1静L+0.625q1活L=0.625x3.49x0.9+0.625x0.787x0.9 = 2.406kNV2=q1L[=4.277x0.3 = 1.283kNV max=max[V], V2]=max[2.406, 1.283] =2.406kNT max=3V max/(2bh0)=3x2.406x1000/(2x40x80) = 1.128N/mm2<[T]=1.782N/mm2 满足要求!3、挠度验算q=(Y G(G1k +(G2k+G3k)xh))xb=(1x(0.3+(24+1.1)x0.4))x0.25 = 2.585kN/m挠度,跨中v max=0.521qL4/(100EI)=0.521x2.585x9004/(100x9350x170.667x104) = O.554mm<[v]=L/400=900/400=2.25mm;悬臂端v max= ql14/(8EI)=2.585x3004/(8x9350x170.667x104) = 0.164mm<[v]=2xl1/400 = 2x300/400= 1.5mm满足要求!六、主梁验算1、小梁最大支座反力计算q1 = Y0x[L35x(G ik +(G2k+G3k)xh)+1.4x%x(Q ik +Q2k)]xb=1x[1.35x(0.5+(24+1.1)x0.4)+1.4x0.9x2.5]x0.25=4.345kN/mq1 静=Y0x1.35x(G1k +(G2k+G3k)xh)xb = 1x1.35x(0.5+(24+1.1)x0.4)x0.25 = 3.557kN/mq1活=丫/1.4*限91k + Q2k)xb =1x1.4x0.9x2.5x0.25 = 0.787kN/mq2= (Y G(G1k +(G2k+G3k)xh))xb=(1x(0.5+(24+1.1)x0.4))x0.25 = 2.635kN/m承载能力极限状态按二等跨连续梁,R max=1.25q1L=1.25x4.345x0.9=4.888kN按二等跨连续梁按悬臂梁,R1= (0.375%静+0.437q1活)L +q1l1= (0.375x3.557+0.437x0.787)x0.9+4.345x0.3 = 2.814kN主梁2根合并,其主梁受力不均匀系数=0.6R=max[R max,RJx0.6 = 2.933kN;正常使用极限状态按二等跨连续梁,R'max = 1.25q2L= 1.25x2.635x0.9 = 2.964kN按二等跨连续梁悬臂梁,R'1= 0.375q2L +qj = 0.375x2.635x0.9+2.635x0.3 =1.68kNR'=max[R'max,R'Jx0.6 = 1.779kN;计算简图如下:2.933kN 2.933kN 2.933kN 2.933kN 2.933kN 2.933kN 2.933kN 2.933kN 2.933kN 2.933kN 2.933kN 2.933kN 2.933kN主梁计算简图一2、抗弯验算gM max/W=0.865x106/4490=192.748N/mm2s[f]=205N/mm2满足要求!3、抗剪验算主梁剪力图一(kN)T max=2V max/A=2x6.664x1000/424 = 31.436N/mm2<[i]=125N/mm2满足要求!4、挠度验算跨中v max=0.974mms[v]=900/400=2.25mm悬挑段v max=0.332mmS[v]=2x150/400=0.75mm满足要求!5、支座反力计算承载能力极限状态图一支座反力依次为R1=8.001kN,R2=11.064kN,R3=11.064kN,R4=8.001kN 七、可调托座验算按上节计算可知,可调托座受力N=11.064/0.6=18.44kNS[N] = 30kN 满足要求!八、立杆验算1、长细比验算l o=k|i(h+2a)=1x1.1x(1500+2x250)=2200mmX=l o/i=2200/15.9=138.365<[X]=230 满足要求!2、立杆稳定性验算考虑风荷载:l0=k|i(h+2a)=1.155x1.1x(1500+2x250)=2541mm九=l0/i=2541.000/15.9=159.811查表得,叼=0.277M wd=Y0xw w Y Q M wk二Y0xw w Y Q(C2w k l a h2/10)=1x0.9x1.4x(1x0.024x0.9x1.52/10)=0.006kN-m N d=Max[R1,R2,R3,R4]/0.6+1xy G xqxH=Max[8.001,11.064,11.064,8.001]/0.6+1x1.35x0.15x4=19.25kNf d=N d/(91A)+M wd/W =19.25x103/(0.277x424)+0.006x106/4490=165.266N/mm2<[o]=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=4/4=1<3满足要求!十、架体抗倾覆验算风荷栽fl制示甘国支撑脚手架风线荷载标准值:4.=1/、=0.9*0.111=0.1四加: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= 1a xH m X Wmk=0.9x 1.5x0.163=0.22kN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卜:okM ok=0.5H2q wk+HF wk=0.5x42x0.1+4x0.22=1.679kN.m参考《规范》GB51210-2016第6.2.17条:B21a(g k1+8卜/冯占应以g 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一支撑脚手架计算单元上集中堆放的物料自重标准值kNbj——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mB21a(g k1+ g k2)+2SG jk b j =B21a[qH/(1a x1b)+G1k H2xG jk xB/2=42x0.9x[0.15x4/(0.9x0.9)+0.5]+2x1x4/2=21.867kN. m>3y0M ok =3x1x1.679=5.038kN.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,见下表可得:P h=1,f t=0.992N/mm2,”=1,h0=h-20=380mm,u m =2[(a+h0)+(b+h0)]=2120mm F=(0.7Pf +0.25oh t pc)nu m h0=(0.7x1x0.992+0.25x0)x1x2120x380/1000=559.409kNNF1=19.25kNm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,见下表可得:f c=11.078N/mm2,氏=1,P i=(A b/A l)i/2=[(a+2b)x(b+2b)/(ab)]i/2=[(400)x(300)/(200x100)]i/2=2.449A =ab=20000mm2inF=1.35P c P l f c A ln=1.35x1x2.449x11.078x20000/1000=732.657kN>F1=19.25kN 满足要求!。

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