施工组织设计下载简介
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xxx关幼儿园结构工程专项施工方案静载线荷载q1=0.9×1.35bG4k=0.9×1.35×0.5×29.87=18.146kN/m
活载线荷载q2=0.9×1.4×0.7bQ3k=0.9×1.4×0.7×0.5×2=0.882kN/m
NB/T 42108-2017 家用和类似用途低压电路用的连接器件汇流排σ=Mmax/W=0.104×106/(1/6×500×122)=8.701N/mm2≤[f]=14.74N/mm2
作用线荷载q=bS正=0.5×29.87=14.935kN/m
ν=0.677ql4/(100EI)=0.677×14.935×233.3334/(100×8925×(1/12×500×123))=0.466mm≤[ν]=l/400=233.333/400=0.583mm
小梁上作用线荷载q=bS承=0.233×38.056=8.88kN/m
小梁弯矩图(kN·m)
Mmax=0.215kN·m
σ=Mmax/W=0.215×106/21.33×103=10.075N/mm2≤[f]=13.5N/mm2
小梁剪力图(kN·m)
Vmax=2.324kN
τmax=3Vmax/(2bh0)=3×2.324×1000/(2×80×40)=1.09N/mm2≤[τ]=1.35N/mm2
小梁上作用线荷载q=bS正=0.233×29.87=6.97kN/m
ν=0.726mm≤[ν]=1.5mm
(规范中缺少相关计算说明,仅供参考)
连续梁中间集中力取小P值;两边集中力为小梁荷载取半后,取P/2值。
Rmax=4.572kN
P=Rmax/2=2.286kN
R’max=3.589kN
P’=R’max/2=1.794kN
长边柱箍弯矩图(kN·m)
长边柱箍剪力图(kN)
M1=0.118kN·m,N1=3.812kN
Rmax=4.572kN
P=Rmax/2=2.286kN
R’max=3.589kN
P’=R’max/2=1.794kN
短边柱箍弯矩图(kN·m)
短边柱箍剪力图(kN)
M2=0.206kN·m,N2=4.692kN
M/Wn=0.206×106/(4.12×103)=49.881N/mm2≤[f]=205N/mm2
长边柱箍变形图(mm)
短边柱箍变形图(mm)
ν1=0.019mm≤[ν]=l/400=0.792mm
ν2=0.056mm≤[ν]=l/400=1.065mm
N=4.692×2=9.384kN≤Ntb=17.8kN
N=4.692×2=9.384kN≤26kN
10.3板模板(扣件式)计算书
2、《建筑施工扣件式钢管脚手架安全技术规范》JGJ 130-2011
楼板面板应搁置在梁侧模板上,本例以三等跨连续梁,取1m单位宽度计算。
W=bh2/6=1000×15×15/6=37500mm3,I=bh3/12=1000×15×15×15/12=281250mm4
q1=0.9×max[1.2(G1k +(G2k+G3k)×h)+1.4×Q1k ,1.35(G1k +(G2k+G3k)×h)+1.4×0.7×Q1k]×b=0.9×max[1.2×(0.1+(24+1.1)×0.1)+1.4×2.5,1.35×(0.1+(24+1.1)×0.1)+1.4×0.7×2.5] ×1=5.969kN/m
q1静=0.9×[γG(G1k +(G2k+G3k)×h)×b] = 0.9×[1.2×(0.1+(24+1.1)×0.1)×1]=2.819kN/m
q1活=0.9×(γQQ1k)×b=0.9×(1.4×2.5)×1=3.15kN/m
q2=0.9×1.2×G1k×b=0.9×1.2×0.1×1=0.108kN/m
p=0.9×1.4×Q1k=0.9×1.4×2.5=3.15kN
q=(γG(G1k +(G2k+G3k)×h))×b =(1×(0.1+(24+1.1)×0.1))×1=2.61kN/m
M1=0.1q1静L2+0.117q1活L2=0.1×2.819×0.32+0.117×3.15×0.32=0.059kN·m
M2=max[0.08q2L2+0.213pL,0.1q2L2+0.175pL]=max[0.08×0.108×0.32+0.213×3.15×0.3,0.1×0.108×0.32+0.175×3.15×0.3]=0.202kN·m
Mmax=max[M1,M2]=max[0.059,0.202]=0.202kN·m
σ=Mmax/W=0.202×106/37500=5.388N/mm2≤[f]=15N/mm2
νmax=0.677ql4/(100EI)=0.677×2.61×3004/(100×10000×281250)=0.051mm
ν=0.051mm≤[ν]=L/250=300/250=1.2mm
q1=0.9×max[1.2(G1k+ (G2k+G3k)×h)+1.4Q1k,1.35(G1k +(G2k+G3k)×h)+1.4×0.7×Q1k]×b=0.9×max[1.2×(0.3+(24+1.1)×0.1)+1.4×2.5,1.35×(0.3+(24+1.1)×0.1)+1.4×0.7×2.5]×0.3=1.855kN/m
因此,q1静=0.9×1.2×(G1k +(G2k+G3k)×h)×b=0.9×1.2×(0.3+(24+1.1)×0.1)×0.3=0.91kN/m
q1活=0.9×1.4×Q1k×b=0.9×1.4×2.5×0.3=0.945kN/m
q2=0.9×1.2×G1k×b=0.9×1.2×0.3×0.3=0.097kN/m
p=0.9×1.4×Q1k=0.9×1.4×2.5=3.15kN
M1=0.125q1静L2+0.125q1活L2=0.125×0.91×12+0.125×0.945×12=0.232kN·m
M2=max[0.07q2L2+0.203pL,0.125q2L2+0.188pL]=max[0.07×0.097×12+0.203×3.15×1,0.125×0.097×12+0.188×3.15×1]=0.646kN·m
M3=max[q1L12/2,q2L12/2+pL1]=max[1.855×0.252/2,0.097×0.252/2+3.15×0.25]=0.791kN·m
Mmax=max[M1,M2,M3]=max[0.232,0.646,0.791]=0.791kN·m
σ=Mmax/W=0.791×106/83330=9.487N/mm2≤[f]=15.444N/mm2
V1=0.625q1静L+0.625q1活L=0.625×0.91×1+0.625×0.945×1=1.16kN
V2=0.625q2L+0.688p=0.625×0.097×1+0.688×3.15=2.228kN
V3=max[q1L1,q2L1+p]=max[1.855×0.25,0.097×0.25+3.15]=3.174kN
Vmax=max[V1,V2,V3]=max[1.16,2.228,3.174]=3.174kN
τmax=3Vmax/(2bh0)=3×3.174×1000/(2×50×100)=0.952N/mm2≤[τ]=1.782N/mm2
q=(γG(G1k +(G2k+G3k)×h))×b=(1×(0.3+(24+1.1)×0.1))×0.3=0.843kN/m
挠度,跨中νmax=0.521qL4/(100EI)=0.521×0.843×10004/(100×9350×416.67×104)=0.113mm≤[ν]=L/250=1000/250=4mm;
悬臂端νmax=ql14/(8EI)=0.843×2504/(8×9350×416.67×104)=0.011mm≤[ν]=2×l1/250=2×250/250=2mm
1、小梁最大支座反力计算
q1=0.9×max[1.2(G1k +(G2k+G3k)×h)+1.4Q1k,1.35(G1k+(G2k+G3k)×h)+1.4×0.7×Q1k]×b=0.9×max[1.2×(0.5+(24+1.1)×0.1)+1.4×1.5,1.35×(0.5+(24+1.1)×0.1)+1.4×0.7×1.5]×0.3=1.542kN/m
q1静=0.9×1.2×(G1k +(G2k+G3k)×h)×b=0.9×1.2×(0.5+(24+1.1)×0.1)×0.3=0.975kN/m
q1活=0.9×1.4×Q1k×b=0.9×1.4×1.5×0.3=0.567kN/m
q2=(γG(G1k +(G2k+G3k)×h))×b=(1×(0.5+(24+1.1)×0.1))×0.3=0.903kN/m
按二等跨连续梁,Rmax=1.25q1L=1.25×1.542×1=1.928kN
按悬臂梁,R1=1.542×0.25=0.386kN
R=max[Rmax,R1]=1.928kN;
按二等跨连续梁,R'max=1.25q2L=1.25×0.903×1=1.129kN
按悬臂梁,R'1=q2l1=0.903×0.25=0.226kN
R'=max[R'max,R'1]=1.129kN;
主梁弯矩图一(kN·m)
主梁弯矩图二(kN·m)
σ=Mmax/W=0.671×106/4120=162.925N/mm2≤[f]=205N/mm2
τmax=2Vmax/A=2×3.885×1000/384=20.235N/mm2≤[τ]=125N/mm2
跨中νmax=1.071mm≤[ν]=1000/250=4mm
悬挑段νmax=0.874mm≤[ν]=2×250/250=2mm
支座反力依次为R1=5.755kN,R2=6.328kN,R3=7.097kN,R4=3.956kN
支座反力依次为R1=4.817kN,R2=6.751kN,R3=6.751kN,R4=4.817kN
按上节计算可知,可调托座受力N=7.097kN≤[N]=30kN
顶部立柱段:l01=kμ1(hd+2a)=1×1.386×(1500+2×200)=2633mm
非顶部立柱段:l0=kμ2h =1×1.755×1800=3159mm
λ=max[l01,l0]/i=3159/16=197.438≤[λ]=210
q1=1×[1.2×(0.5+(24+1.1)×0.1)+1.4×0.9×1]×0.3 = 1.462kN/m
同上四~六步计算过程,可得:
R1=5.453kN,R2=6.398kN,R3=6.726kN,R4=4.564kN
l01=kμ1(hd+2a)=1.155×1.386×(1500+2×200)=3041.577mm
λ1=l01/i=3041.577/16=190.099
查表得,φ=0.199
N1 =Max[R1,R2,R3,R4]=Max[5.453,6.398,6.726,4.564]=6.726kN
f= N1/(ΦA)=6726/(0.199×384)=88.018N/mm2≤[f]=205N/mm2
Mw=1×γQφcωk×la×h2/10=1×1.4×0.9×0.051×1×1.82/10=0.021kN·m
N1w =Max[R1,R2,R3,R4]+Mw/lb=Max[5.453,6.398,6.726,4.564]+0.021/1=6.747kN
f= N1w/(φA)+ Mw/W=6747/(0.199×384)+0.021×106/4120=93.39N/mm2≤[f]=205N/mm2
l0=kμ2h =1.155×1.755×1800=3648.645mm
λ=l0/i=3648.645/16=228.04
查表得,φ1=0.14
N=Max[R1,R2,R3,R4]+1×γG×q×H=Max[5.453,6.398,6.726,4.564]+1×1.2×0.15×4.5=7.536kN
f=N/(φ1A)=7.536×103/(0.140×384)=140.179N/mm2≤[σ]=205N/mm2
Mw=1×γQφcωk×la×h2/10=1×1.4×0.9×0.051×1×1.82/10=0.021kN·m
Nw=Max[R1,R2,R3,R4]+1×γG×q×H+Mw/lb=Max[5.453,6.398,6.726,4.564]+1×1.2×0.15×4.5+0.021/1=7.557kN
H/B=4.5/2.9=1.552≤3
满足要求,不需要进行抗倾覆验算 !
十、立柱支承面承载力验算
F1=N=7.557kN
um =2[(a+h0)+(b+h0)]=1200mm
F=(0.7βhft+0.25σpc,m)ηumh0=(0.7×1×0.737+0.25×0)×1×1200×100/1000=61.908kN≥F1=7.557kN
2、局部受压承载力计算
可得:fc=6.902N/mm2,βc=1,
βl=(Ab/Al)1/2=[(a+2b)×(b+2b)/(ab)]1/2=[(600)×(600)/(200×200)]1/2=3,Aln=ab=40000mm2
F=1.35βcβlfcAln=1.35×1×3×6.902×40000/1000=1118.124kN≥F1=7.557kN
10.4梁、板、柱模板支撑系统示意图
10.5楼梯模板支撑系统示意图
NB/T 14017-2016 页岩气录井技术规范10.6云关幼儿园结构施工阶段平面布置图
10.7云关幼儿园基础与主体结构施工进度计划
10.8模板工程及支撑体系安全验收表
DG/TJ08-2012-2018标准下载模板工程及支撑体系安全验收表
注:1、模板工程由项目经理、项目技术负责人、劳务分包项目经理、班组长和监理单位相关人员进行验收。
2、模板工程及支撑体系安全验收根据施工方案要求进行分段验收,工程管理部负责组织验收工作。