新建商业项目(中城上城四号地施工二期)商业项目工程超限模板支撑专项施工方案(共108页,有配图,计算图)

新建商业项目(中城上城四号地施工二期)商业项目工程超限模板支撑专项施工方案(共108页,有配图,计算图)
仅供个人学习
反馈
文件类型:.zip解压后docx
资源大小:4.97M
标准类别:施工组织设计
资源属性:
下载资源

施工组织设计下载简介

内容预览随机截取了部分,仅供参考,下载文档齐全完整

新建商业项目(中城上城四号地施工二期)商业项目工程超限模板支撑专项施工方案(共108页,有配图,计算图)

νmax=5ql4/(384EI)=5×3.865×2004/(384×10000×341333.333)=0.024mm

ν=0.024mm≤[ν]=L/400=200/400=0.5mm

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)×0.15)+1.4×2.5,1.35×(0.3+(24+1.1)×0.15)+1.4×0.7×2.5]×0.2=1.508kN/m

因此GB 38507-2020 油墨中可挥发性有机化合物(VOCs)含量的限值,q静=0.9×1.2×(G1k +(G2k+G3k)×h)×b=0.9×1.2×(0.3+(24+1.1)×0.15)×0.2=0.878kN/m

q1活=0.9×1.4×Q1k×b=0.9×1.4×2.5×0.2=0.63kN/m

q2=0.9×1.2×G1k×b=0.9×1.2×0.3×0.2=0.065kN/m

p=0.9×1.4×Q1k=0.9×1.4×2.5=3.15kN

计算简图如下:

M1=0.125q1静L2+0.125q1活L2=0.125×0.878×0.92+0.125×0.63×0.92=0.153kN·m

M2=max[0.07q2L2+0.203pL,0.125q2L2+0.188pL]=max[0.07×0.065×0.92+0.203×3.15×0.9,0.125×0.065×0.92+0.188×3.15×0.9]=0.579kN·m

M3=max[q1L12/2,q2L12/2+pL1]=max[1.508×0.252/2,0.065×0.252/2+3.15×0.25]=0.79kN·m

Mmax=max[M1,M2,M3]=max[0.153,0.579,0.79]=0.79kN·m

σ=Mmax/W=0.79×106/54188=14.57N/mm2≤[f]=15.444N/mm2

V1=0.625q1静L+0.625q1活L=0.625×0.878×0.9+0.625×0.63×0.9=0.848kN

V2=0.625q2L+0.688p=0.625×0.065×0.9+0.688×3.15=2.204kN

V3=max[q1L1,q2L1+p]=max[1.508×0.25,0.065×0.25+3.15]=3.166kN

Vmax=max[V1,V2,V3]=max[0.848,2.204,3.166]=3.166kN

τmax=3Vmax/(2bh0)=3×3.166×1000/(2×45×85)=1.242N/mm2≤[τ]=1.782N/mm2

q=(γG(G1k +(G2k+G3k)×h))×b=(1×(0.3+(24+1.1)×0.15))×0.2=0.813kN/m

挠度,跨中νmax=0.521qL4/(100EI)=0.521×0.813×9004/(100×9350×230.297×104)=0.129mm≤[ν]=L/400=900/400=2.25mm;

悬臂端νmax=ql14/(8EI)=0.813×2504/(8×9350×230.297×104)=0.018mm≤[ν]=2×l1/400=2×250/400=1.25mm

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)×0.15)+1.4×1.5,1.35×(0.5+(24+1.1)×0.15)+1.4×0.7×1.5]×0.2=1.301kN/m

q1静=0.9×1.35×(G1k +(G2k+G3k)×h)×b=0.9×1.35×(0.5+(24+1.1)×0.15)×0.2=1.036kN/m

q1活=0.9×1.4×0.7×Q1k×b =0.9×1.4×0.7×1.5×0.2=0.265kN/m

q2=(γG(G1k +(G2k+G3k)×h))×b=(1×(0.5+(24+1.1)×0.15))×0.2=0.853kN/m

承载能力极限状态

按二等跨连续梁,Rmax=1.25q1L=1.25×1.301×0.9=1.464kN

按悬臂梁,R1=q1l1=1.301×0.25=0.325kN

主梁2根合并,其主梁受力不均匀系数=0.6

R=max[Rmax,R1]×0.6=0.878kN;

正常使用极限状态

按二等跨连续梁,R'max=1.25q2L=1.25×0.853×0.9=0.96kN

按悬臂梁,R'1=q2l1=0.853×0.25=0.213kN

R'=max[R'max,R'1]×0.6=0.576kN;

计算简图如下:

主梁计算简图一

主梁弯矩图一(kN·m)

σ=Mmax/W=0.305×106/4490=68.03N/mm2≤[f]=205N/mm2

主梁剪力图一(kN)

τmax=2Vmax/A=2×2.195×1000/424=10.354N/mm2≤[τ]=125N/mm2

主梁变形图一(mm)

跨中νmax=0.263mm≤[ν]=900/400=2.25mm

悬挑段νmax=0.075mm≤[ν]=2×250/400=1.25mm

5、支座反力计算

承载能力极限状态

支座反力依次为R1=3.465kN,R2=3.998kN,R3=3.998kN,R4=3.465kN

按上节计算可知,可调托座受力N=3.998/0.6=6.663kN≤[N]=30kN

l0=h=1500mm

λ=l0/i=1500/15.9=94.34≤[λ]=150

2、立柱稳定性验算

q1=0.9×[1.2×(0.5+(24+1.1)×0.15)+1.4×0.9×1]×0.2 = 1.148kN/m

同上四~六步计算过程,可得:

R1=3.059kN,R2=3.529kN,R3=3.529kN,R4=3.059kN

λ=l0/i=1500.000/15.9=94.34

查表得,φ1=0.634

不考虑风荷载:

N =Max[R1,R2,R3,R4]/0.6+0.9×γG×q×H=Max[3.059,3.529,3.529,3.059]/0.6+0.9×1.2×0.15×17.03=8.64kN

f=N/(φ1A)=8.64×103/(0.634×424)=32.141N/mm2≤[σ]=205N/mm2

Mw=0.9×γQφcωk×la×h2/10=0.9×1.4×0.9×0.076×0.9×1.52/10=0.017kN·m

Nw =Max[R1,R2,R3,R4]/0.6+0.9×γG×q×H+Mw/lb=Max[3.059,3.529,3.529,3.059]/0.6+0.9×1.2×0.15×17.03+0.017/0.9=8.659kN

f=Nw/(φ1A)+Mw/W=8.659×103/(0.634×424)+0.017×106/4490=35.998N/mm2≤[σ]=205N/mm2

H/B=17.03/15=1.135<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=8.659kN

1、受冲切承载力计算

um =2[(a+h0)+(b+h0)]=3120mm

F=(0.7βhft+0.25σpc,m)ηumh0=(0.7×1×0.737+0.25×0)×1×3120×580/1000=933.573kN≥F1=8.659kN

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=8.659kN

7.4、 250楼板支撑计算书

板模板(扣件式)计算书

设计简图如下:

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。

承载能力极限状态

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)×0.25)+1.4×2.5,1.35×(0.1+(24+1.1)×0.25)+1.4×0.7×2.5] ×1=10.035kN/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

正常使用极限状态

q=(γG(G1k +(G2k+G3k)×h))×b =(1×(0.1+(24+1.1)×0.25))×1=6.375kN/m

计算简图如下:

M1=q1l2/8=10.035×0.152/8=0.028kN·m

M2=q2L2/8+pL/4=0.108×0.152/8+3.15×0.15/4=0.118kN·m

Mmax=max[M1,M2]=max[0.028,0.118]=0.118kN·m

νmax=5ql4/(384EI)=5×6.375×1504/(384×10000×341333.333)=0.012mm

ν=0.012mm≤[ν]=L/400=150/400=0.375mm

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)×0.25)+1.4×2.5,1.35×(0.3+(24+1.1)×0.25)+1.4×0.7×2.5]×0.15=1.538kN/m

因此,q1静=0.9×1.2×(G1k +(G2k+G3k)×h)×b=0.9×1.2×(0.3+(24+1.1)×0.25)×0.15=1.065kN/m

q1活=0.9×1.4×Q1k×b=0.9×1.4×2.5×0.15=0.472kN/m

q2=0.9×1.2×G1k×b=0.9×1.2×0.3×0.15=0.049kN/m

p=0.9×1.4×Q1k=0.9×1.4×2.5=3.15kN

计算简图如下:

M1=0.125q1静L2+0.125q1活L2=0.125×1.065×0.92+0.125×0.472×0.92=0.156kN·m

M2=max[0.07q2L2+0.203pL,0.125q2L2+0.188pL]=max[0.07×0.049×0.92+0.203×3.15×0.9,0.125×0.049×0.92+0.188×3.15×0.9]=0.578kN·m

M3=max[q1L12/2,q2L12/2+pL1]=max[1.538×0.22/2,0.049×0.22/2+3.15×0.2]=0.631kN·m

Mmax=max[M1,M2,M3]=max[0.156,0.578,0.631]=0.631kN·m

σ=Mmax/W=0.631×106/54188=11.644N/mm2≤[f]=15.444N/mm2

V1=0.625q1静L+0.625q1活L=0.625×1.065×0.9+0.625×0.472×0.9=0.865kN

V2=0.625q2L+0.688p=0.625×0.049×0.9+0.688×3.15=2.195kN

V3=max[q1L1,q2L1+p]=max[1.538×0.2,0.049×0.2+3.15]=3.16kN

Vmax=max[V1,V2,V3]=max[0.865,2.195,3.16]=3.16kN

τmax=3Vmax/(2bh0)=3×3.16×1000/(2×45×85)=1.239N/mm2≤[τ]=1.782N/mm2

q=(γG(G1k +(G2k+G3k)×h))×b=(1×(0.3+(24+1.1)×0.25))×0.15=0.986kN/m

挠度,跨中νmax=0.521qL4/(100EI)=0.521×0.986×9004/(100×9350×230.297×104)=0.157mm≤[ν]=L/400=900/400=2.25mm;

悬臂端νmax=ql14/(8EI)=0.986×2004/(8×9350×230.297×104)=0.009mm≤[ν]=2×l1/400=2×200/400=1mm

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)×0.25)+1.4×1.5,1.35×(0.5+(24+1.1)×0.25)+1.4×0.7×1.5]×0.15=1.433kN/m

q1静=0.9×1.35×(G1k +(G2k+G3k)×h)×b=0.9×1.35×(0.5+(24+1.1)×0.25)×0.15=1.235kN/m

q1活=0.9×1.4×0.7×Q1k×b =0.9×1.4×0.7×1.5×0.15=0.198kN/m

q2=(γG(G1k +(G2k+G3k)×h))×b=(1×(0.5+(24+1.1)×0.25))×0.15=1.016kN/m

承载能力极限状态

按二等跨连续梁,Rmax=1.25q1L=1.25×1.433×0.9=1.612kN

按悬臂梁,R1=q1l1=1.433×0.2=0.287kN

主梁2根合并,其主梁受力不均匀系数=0.6

R=max[Rmax,R1]×0.6=0.967kN;

正常使用极限状态

按二等跨连续梁,R'max=1.25q2L=1.25×1.016×0.9=1.143kN

按悬臂梁,R'1=q2l1=1.016×0.2=0.203kN

R'=max[R'max,R'1]×0.6=0.686kN;

计算简图如下:

主梁计算简图一

主梁计算简图二

主梁弯矩图一(kN·m)

主梁弯矩图二(kN·m)

σ=Mmax/W=0.517×106/4490=115.252N/mm2≤[f]=205N/mm2

主梁剪力图一(kN)

主梁剪力图二(kN)

τmax=2Vmax/A=2×3.314×1000/424=15.633N/mm2≤[τ]=125N/mm2

主梁变形图一(mm)

主梁变形图二(mm)

跨中νmax=0.789mm≤[ν]=900/400=2.25mm

悬挑段νmax=0.557mm≤[ν]=2×200/400=1mm

5、支座反力计算

承载能力极限状态

支座反力依次为R1=4.422kN,R2=6.001kN,R3=6.323kN,R4=3.561kN

支座反力依次为R1=3.981kN,R2=6.173kN,R3=6.173kN,R4=3.981kN

按上节计算可知,可调托座受力N=6.323/0.6=10.539kN≤[N]=30kN

l0=h=1500mm

λ=l0/i=1500/15.9=94.34≤[λ]=150

2、立柱稳定性验算

q1=0.9×[1.2×(0.5+(24+1.1)×0.25)+1.4×0.9×1]×0.15 = 1.268kN/m

同上四~六步计算过程,可得:

R1=3.914kN,R2=5.464kN,R3=5.597kN,R4=3.524kN

λ=l0/i=1500.000/15.9=94.34

查表得,φ1=0.634

不考虑风荷载:

N =Max[R1,R2,R3,R4]/0.6+0.9×γG×q×H=Max[3.914,5.464,5.597,3.524]/0.6+0.9×1.2×0.15×7.4=10.528kN

f=N/(φ1A)=10.528×103/(0.634×424)=39.164N/mm2≤[σ]=205N/mm2

Mw=0.9×γQφcωk×la×h2/10=0.9×1.4×0.9×0.076×0.9×1.52/10=0.017kN·m

Nw =Max[R1,R2,R3,R4]/0.6+0.9×γG×q×H+Mw/lb=Max[3.914,5.464,5.597,3.524]/0.6+0.9×1.2×0.15×7.4+0.017/0.9=10.547kN

f=Nw/(φ1A)+Mw/W=10.547×103/(0.634×424)+0.017×106/4490=43.021N/mm2≤[σ]=205N/mm2

H/B=7.4/20=0.37<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=10.547kN

施工方案包括的内容 1、受冲切承载力计算

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=10.547kN

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广州地铁一号线东站施工方案,Aln=ab=40000mm2

F=1.35βcβlfcAln=1.35×1×3×6.902×40000/1000=1118.124kN≥F1=10.547kN

©版权声明
相关文章