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九江碧桂园高支模工程专项施工方案（59P）.doc1???????????
Q1k=1.5kN/m2
CECS431-2016标准下载 q1=0.9max[1.2(G1k+ (G3k+G2k)×h)+1.4Q1k,1.35(G1k+(G3k+G2k)×h)+1.4×0.7Q1k]×b=0.9max[1.2×(0.5+(1.1+24)×0.12)+1.4×1.5,1.35×(0.5+(1.1+24)×0.12)+1.4×0.7×1.5]×0.3=1.7kN/m
q1?=0.9×1.2(G1k+ (G3k+G2k)×h)×b=0.9×1.2×(0.5+(1.1+24)×0.12)×0.3=1.14kN/m
q1?=0.9×1.4×Q1k×b=0.9×1.4×1.5×0.3=0.57kN/m
q2=(G1k+ (G3k+G2k)×h)×b=(0.5+(1.1+24)×0.12)×0.3=1.05kN/m
????????
??????,Rmax=(1.143q1?+1.223q1?)L=1.143×1.14×0.9+1.223×0.57×0.9=1.79kN
????,R1=q1l=1.7×0.3=0.51kN
R=max[Rmax,R1]=1.79kN;
??,R'=1.35kN,R''=1.35kN
????????
??????,Rmax=1.143q2L=1.143×1.05×0.9=1.08kN
????,R1=q2l=1.05×0.3=0.32kN
R=max[Rmax,R1]=1.08kN;
??,R'=0.81kN,R''=0.81kN
??????:
?????(kN·m)
Mmax=0.43kN·m
σ=Mmax/W=0.43×106/4490=95.2N/mm2=[f]=205N/mm2
?????(kN)
Vmax=2.67kN
τmax=2Vmax/A=2×2.67×1000/424=12.57N/mm2?[τ]=125N/mm2
?????(mm)
νmax=0.67mm
νmax=0.67mm?[ν]=2.25mm
?????:l01=kμ1(hd+2a)=1×1.386×(1500+2×200)=2633.4mm
??????:l02=kμ2h =1×1.755×1500=2632.5mm
λ=l0/i=2633.4/15.9=165.62?[λ]=210
???????!
2????????
?????:l01=kμ1(hd+2a)=1.115×1.386×(1500+2×200)=2936.241mm
λ1=l01/i=2936.241/15.9=184.669,???,φ1=0.211
Mw=0.92×1.4ωklah2/10=0.92×1.4×0.18×0.9×1.52/10=0.04kN·m
Nw=0.9[1.2ΣNGik+0.9×1.4Σ(NQik+Mw/lb)]=0.9×[1.2×(0.5+(24+1.1)×0.12)+0.9×1.4×1]×0.9×0.9+0.92×1.4×0.04/0.9=4.04kN
f= Nw/(φA)+ Mw/W=4042.92/(0.21×424)+0.04×106/4490=54.4N/mm2=[f]=205N/mm2
Mw=0.92×1.4ωklah2/10=0.92×1.4×0.18×0.9×1.52/10=0.04kN·m
Nw=0.9[1.2ΣNGik+0.9×1.4Σ(NQik+Mw/lb)]=0.9×[1.2×(0.75+(24+1.1)×0.12)+0.9×1.4×1]×0.9×0.9+0.92×1.4×0.04/0.9=4.26kN
f= Nw/(φA)+ Mw/W=4261.62/(0.21×424)+0.04×106/4490=56.84N/mm2=[f]=205N/mm2
???????,??????N=4.04kN=[N]=30kN
????????????p=N/(mfA)=4.26/(1×0.1)=42.62kPa=fak=70kPa
（五）羽毛球场屋面板模板（扣件式）计算书
??«????????????»5.2.1"?????????"???,????,?????????????,????????,?1m?????????????:
W=bh2/6=1000×15×15/6=37500mm3,I=bh3/12=1000×15×15×15/12=281250mm4
q1=0.9max[1.2(G1k+ (G3k+G2k)×h)+1.4Q1k,1.35(G1k+ (G3k+G2k)×h)+1.4×0.7Q1k]×b=0.9max[1.2×(0.1+(1.1+24)×0.12)+1.4×2.5,1.35×(0.1+(1.1+24)×0.12)+1.4×0.7×2.5] ×1=6.51kN/m
q2=0.9×1.2×G1k×b=0.9×1.2×0.1×1=0.11kN/m
p=0.9×1.3×Q1K=0.9×1.4×2.5=3.15kN
Mmax=max[q1l2/8,q2l2/8+pl/4]=max[6.51×0.32/8,0.11×0.32/8+3.15×0.3/4]= 0.24kN·m
σ=Mmax/W=0.24×106/37500=6.33N/mm2=[f]=15N/mm2
q=(G1k+(G3k+G2k)×h)×b=(0.1+(1.1+24)×0.12)×1=3.11kN/m
ν=5ql4/(384EI)=5×3.11×3004/(384×10000×281250)=0.12mm=[ν]=l/400=300/400=0.75mm
?[L/la]??=[6000/900]??=6,?????????,???????????300mm,??????????,??????:
q1=0.9max[1.2(G1k+(G3k+G2k)×h)+1.4Q1k,1.35(G1k+(G3k+G2k)×h)+1.4×0.7Q1k]×b=0.9×max[1.2×(0.3+(1.1+24)×0.12)+1.4×2.5,1.35×(0.3+(1.1+24)×0.12)+1.4×0.7×2.5]×0.3=2.02kN/m
??,q1?=0.9×1.2(G1k+(G3k+G2k)×h)×b=0.9×1.2×(0.3+(1.1+24)×0.12)×0.3=1.07kN/m
q1?=0.9×1.4×Q1k×b=0.9×1.4×2.5×0.3=0.94kN/m
q2=0.9×1.2×G1k×b=0.9×1.2×0.3×0.3=0.1kN/m
p=0.9×1.4×Q1k=0.9×1.4×2.5=3.15kN/m
M2=0.077q2L2+0.21pL=0.077×0.1×0.92+0.21×3.15×0.9=0.6kN·m
M3=max[q1L12/2,q2L12/2+pL1]=max[2.02×0.32/2,0.1×0.32/2+3.15×0.3]=0.95kN·m
Mmax=max[M1,M2,M3]=max[0.14,0.6,0.95]=0.95kN·m
σ=Mmax/W=0.95×106/83330=11.39N/mm2=[f]=15.44N/mm2
V1=0.607q1?L+0.62q1?L=0.607×1.07×0.9+0.62×0.94×0.9=1.11kN
V2=0.607q2L+0.681p=0.607×0.1×0.9+0.681×3.15=2.2kN
V3=max[q1L1,q2L1+p]=max[2.02×0.3,0.1×0.3+3.15]=3.18kN
Vmax=max[V1,V2,V3]=max[1.11,2.2,3.18]=3.18kN
τmax=3Vmax/(2bh0)=3×3.18×1000/(2×100×50)=0.95N/mm2?[τ]=1.78N/mm2
q=(G1k+(G3k+G2k)×h)×b=(0.3+(24+1.1)×0.12)×0.3=0.99kN/m
??νmax=0.632qL4/(100EI)=0.632×0.99×9004/(100×9350×4166700)=0.11mm?[ν]=l/400=900/400=2.25mm
???νmax=qL4/(8EI)=0.99×3004/(8×9350×4166700)=0.03mm=[ν]=l1/400=300/400=0.75mm
1???????????
Q1k=1.5kN/m2
q1=0.9max[1.2(G1k+ (G3k+G2k)×h)+1.4Q1k,1.35(G1k+(G3k+G2k)×h)+1.4×0.7Q1k]×b=0.9max[1.2×(0.5+(1.1+24)×0.12)+1.4×1.5,1.35×(0.5+(1.1+24)×0.12)+1.4×0.7×1.5]×0.3=1.7kN/m
q1?=0.9×1.2(G1k+ (G3k+G2k)×h)×b=0.9×1.2×(0.5+(1.1+24)×0.12)×0.3=1.14kN/m
q1?=0.9×1.4×Q1k×b=0.9×1.4×1.5×0.3=0.57kN/m
q2=(G1k+ (G3k+G2k)×h)×b=(0.5+(1.1+24)×0.12)×0.3=1.05kN/m
????????
??????,Rmax=(1.143q1?+1.223q1?)L=1.143×1.14×0.9+1.223×0.57×0.9=1.79kN
????,R1=q1l=1.7×0.3=0.51kN
R=max[Rmax,R1]=1.79kN;
??,R'=1.12kN,R''=1.12kN
????????
??????,Rmax=1.143q2L=1.143×1.05×0.9=1.08kN
????,R1=q2l=1.05×0.3=0.32kN
R=max[Rmax,R1]=1.08kN;
??,R'=0.68kN,R''=0.68kN
??????:
?????(kN·m)
Mmax=0.56kN·m
σ=Mmax/W=0.56×106/4490=124.51N/mm2=[f]=205N/mm2
?????(kN)
Vmax=3.12kN
τmax=2Vmax/A=2×3.12×1000/424=14.71N/mm2?[τ]=125N/mm2
?????(mm)
νmax=0.47mm
??νmax=0.47mm?[ν]=900/400=2.25mm
???νmax=0.2mm?[ν]=150/400=0.38mm
?????:l01=kμ1(hd+2a)=1×1.386×(1500+2×200)=2633.4mm
??????:l02=kμ2h =1×1.755×1500=2632.5mm
λ=l0/i=2633.4/15.9=165.62?[λ]=210
???????!
2????????
?????:l01=kμ1(hd+2a)=1.217×1.386×(1500+2×200)=3204.848mm
λ1=l01/i=3204.848/15.9=201.563,???,φ1=0.179
Mw=0.92×1.4ωklah2/10=0.92×1.4×0.18×0.9×1.52/10=0.04kN·m
Nw=0.9[1.2ΣNGik+0.9×1.4Σ(NQik+Mw/lb)]=0.9×[1.2×(0.5+(24+1.1)×0.12)+0.9×1.4×1]×0.9×0.9+0.92×1.4×0.04/0.9=4.04kN
f= Nw/(φA)+ Mw/W=4042.92/(0.18×424)+0.04×106/4490=62.48N/mm2=[f]=205N/mm2
Mw=0.92×1.4ωklah2/10=0.92×1.4×0.18×0.9×1.52/10=0.04kN·m
Nw=0.9[1.2ΣNGik+0.9×1.4Σ(NQik+Mw/lb)]=0.9×[1.2×(0.75+(24+1.1)×0.12)+0.9×1.4×1]×0.9×0.9+0.92×1.4×0.04/0.9=4.26kN
f= Nw/(φA)+ Mw/W=4261.62/(0.18×424)+0.04×106/4490=65.36N/mm2=[f]=205N/mm2
???????,??????N=4.04kN=[N]=30kN
????????????p=N/(mfA)=4.26/(1×0.1)=42.62kPa=fak=70kPa
（六）大堂屋面板模板（扣件式）计算书
??«????????????»5.2.1"?????????"???,????,?????????????,????????,?1m?????????????:
W=bh2/6=1000×15×15/6=37500mm3,I=bh3/12=1000×15×15×15/12=281250mm4
q1=0.9max[1.2(G1k+ (G3k+G2k)×h)+1.4Q1k,1.35(G1k+ (G3k+G2k)×h)+1.4×0.7Q1k]×b=0.9max[1.2×(0.1+(1.1+24)×0.1)+1.4×2.5,1.35×(0.1+(1.1+24)×0.1)+1.4×0.7×2.5] ×1=5.97kN/m
q2=0.9×1.2×G1k×b=0.9×1.2×0.1×1=0.11kN/m
p=0.9×1.3×Q1K=0.9×1.4×2.5=3.15kN
Mmax=max[q1l2/8,q2l2/8+pl/4]=max[5.97×0.32/8,0.11×0.32/8+3.15×0.3/4]= 0.24kN·m
σ=Mmax/W=0.24×106/37500=6.33N/mm2=[f]=15N/mm2
q=(G1k+(G3k+G2k)×h)×b=(0.1+(1.1+24)×0.1)×1=2.61kN/m
ν=5ql4/(384EI)=5×2.61×3004/(384×10000×281250)=0.1mm=[ν]=l/400=300/400=0.75mm
q1=0.9max[1.2(G1k+(G3k+G2k)×h)+1.4Q1k,1.35(G1k+(G3k+G2k)×h)+1.4×0.7Q1k]×b=0.9×max[1.2×(0.3+(1.1+24)×0.1)+1.4×2.5,1.35×(0.3+(1.1+24)×0.1)+1.4×0.7×2.5]×0.3=1.86kN/m
??,q1?=0.9×1.2(G1k+(G3k+G2k)×h)×b=0.9×1.2×(0.3+(1.1+24)×0.1)×0.3=0.91kN/m
q1?=0.9×1.4×Q1k×b=0.9×1.4×2.5×0.3=0.94kN/m
q2=0.9×1.2×G1k×b=0.9×1.2×0.3×0.3=0.1kN/m
p=0.9×1.4×Q1k=0.9×1.4×2.5=3.15kN/m
M2=0.077q2L2+0.21pL=0.077×0.1×0.62+0.21×3.15×0.6=0.4kN·m
M3=max[q1L12/2,q2L12/2+pL1]=max[1.86×0.32/2,0.1×0.32/2+3.15×0.3]=0.95kN·m
Mmax=max[M1,M2,M3]=max[0.06,0.4,0.95]=0.95kN·m
σ=Mmax/W=0.95×106/83330=11.39N/mm2=[f]=15.44N/mm2
V1=0.607q1?L+0.62q1?L=0.607×0.91×0.6+0.62×0.94×0.6=0.68kN
V2=0.607q2L+0.681p=0.607×0.1×0.6+0.681×3.15=2.18kN
V3=max[q1L1,q2L1+p]=max[1.86×0.3,0.1×0.3+3.15]=3.18kN
Vmax=max[V1,V2,V3]=max[0.68,2.18,3.18]=3.18kN
τmax=3Vmax/(2bh0)=3×3.18×1000/(2×100×50)=0.95N/mm2?[τ]=1.78N/mm2
q=(G1k+(G3k+G2k)×h)×b=(0.3+(24+1.1)×0.1)×0.3=0.84kN/m
??νmax=0.632qL4/(100EI)=0.632×0.84×6004/(100×9350×4166700)=0.02mm=[ν]=l/400=600/400=1.5mm
???νmax=qL4/(8EI)=0.84×3004/(8×9350×4166700)=0.02mm=[ν]=l1/400=300/400=0.75mm
1???????????
Q1k=1.5kN/m2
q1=0.9max[1.2(G1k+ (G3k+G2k)×h)+1.4Q1k,1.35(G1k+(G3k+G2k)×h)+1.4×0.7Q1k]×b=0.9max[1.2×(0.5+(1.1+24)×0.1)+1.4×1.5,1.35×(0.5+(1.1+24)×0.1)+1.4×0.7×1.5]×0.3=1.54kN/m
q1?=0.9×1.2(G1k+ (G3k+G2k)×h)×b=0.9×1.2×(0.5+(1.1+24)×0.1)×0.3=0.98kN/m
q1?=0.9×1.4×Q1k×b=0.9×1.4×1.5×0.3=0.57kN/m
q2=(G1k+ (G3k+G2k)×h)×b=(0.5+(1.1+24)×0.1)×0.3=0.9kN/m
????????
??????,Rmax=(1.143q1?+1.223q1?)L=1.143×0.98×0.6+1.223×0.57×0.6=1.08kN
????,R1=q1l=1.54×0.3=0.46kN
R=max[Rmax,R1]=1.08kN;
??,R'=0.81kN,R''=0.81kN
????????
??????,Rmax=1.143q2L=1.143×0.9×0.6=0.62kN
????,R1=q2l=0.9×0.3=0.27kN
R=max[Rmax,R1]=0.62kN;
??,R'=0.46kN,R''=0.46kN
??????:
?????(kN·m)
Mmax=0.12kN·m
σ=Mmax/W=0.12×106/4490=27.06N/mm2=[f]=205N/mm2
?????(kN)
Vmax=1.08kN
τmax=2Vmax/A=2×1.08×1000/424=5.09N/mm2?[τ]=125N/mm2
?????(mm)
νmax=0.03mm
??νmax=0.03mm?[ν]=600/400=1.5mm
???νmax=0.02mm?[ν]=150/400=0.38mm
?????:l01=kμ1(hd+2a)=1×1.386×(1500+2×200)=2633.4mm
??????:l02=kμ2h =1×1.755×1500=2632.5mm
λ=l0/i=2633.4/15.9=165.62?[λ]=210
???????!
2????????
?????:l01=kμ1(hd+2a)=1.217×1.386×(1500+2×200)=3204.848mm
T／CECS 537-2018 城镇污水处理厂污泥隔膜压滤深度脱水技术规程 λ1=l01/i=3204.848/15.9=201.563,???,φ1=0.179
Mw=0.92×1.4ωklah2/10=0.92×1.4×0.18×0.6×1.52/10=0.03kN·m
Nw=0.9[1.2ΣNGik+0.9×1.4Σ(NQik+Mw/lb)]=0.9×[1.2×(0.5+(24+1.1)×0.1)+0.9×1.4×1]×0.6×0.6+0.92×1.4×0.03/0.6=1.63kN
f= Nw/(φA)+ Mw/W=1630.61/(0.18×424)+0.03×106/4490=27.62N/mm2=[f]=205N/mm2
Mw=0.92×1.4ωklah2/10=0.92×1.4×0.18×0.6×1.52/10=0.03kN·m
Nw=0.9[1.2ΣNGik+0.9×1.4Σ(NQik+Mw/lb)]=0.9×[1.2×(0.75+(24+1.1)×0.1)+0.9×1.4×1]×0.6×0.6+0.92×1.4×0.03/0.6=1.73kN
f= Nw/(φA)+ Mw/W=1727.81/(0.18×424)+0.03×106/4490=28.9N/mm2=[f]=205N/mm2
???????,??????N=1.63kN=[N]=30kN
江西省建设工程施工仪器仪表台班费用定额（2017版含增值税和非增值税版）.pdf ????????????p=N/(mfA)=1.73/(1×0.1)=17.28kPa=fak=70kPa