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天津碧桂园凤凰酒店工程施工方案11093121冯硕11建工一（57P）-公众号（建筑教程资料库）发布.doc

三、模板支撑体系设计

设计简图如下：

SL_T 780-2020 水利水电工程金属结构制作与安装安全技术规程（清晰无水印，附条文说明）模板设计剖面图(楼板长向)

模板设计剖面图(楼板宽向)

q1＝0.95×0.9max[1.2×(G1k+(G3k+G2k)×h)+1.4Q1k，1.35(G1k+ (G3k+G2k)×h)+1.4×0.7Q1k]×b=0.95×0.9max[1.2×(0.3+(1.1+24)×0.15)+1.4×2.5，1.35×(0.3+(1.1+24)×0.15)+1.4×0.7×2.5] ×0.5=3.58kN/m

q2＝λ×0.9×1.2×G1k×b＝0.95×0.9×1.2×0.3×0.5=0.15kN/m

p＝λ×0.9×1.4×Q1k＝0.95×0.9×1.4×2.5＝2.99kN

M1=q1l2/8=3.58×0.92/8=0.36kN·m

M2=q2l2/8+pl/4=0.15×0.92/8+2.99×0.9/4=0.69kN·m

M3=max[q1l2/2，q2l2/2+pl]=max[3.58×0.32/2，0.15×0.32/2+2.99×0.3]=0.9kN·m

Mmax＝max[0.36，0.69，0.9]= 0.9kN·m

σ＝Mmax/W＝0.9×106/10460＝86.49N/mm2≤[f]＝205N/mm2

q＝(G1k+(G3k+G2k)×h)×b=(0.3+(1.1+24)×0.15)×0.5=2.03kN/m

跨中νmax＝5ql4/(384EI)＝5×2.03×9004/(384×206000×475000)＝0.18mm≤1.5mm

悬臂端νmax＝ql4/(8EI)＝2.03×3004/(8×206000×475000)＝0.02mm≤1.5mm

因[B/lb]取整＝[5000/900]取整＝5，按四等跨连续梁计算，又因小梁较大悬挑长度为250mm，因此需进行最不利组合，计算简图如下：

q1＝0.95×0.9max[1.2(G1k+ (G3k+G2k)×h)+1.4Q1k，1.35(G1k+ (G3k+G2k)×h)+1.4×0.7Q1k]×b=0.95×0.9×max[1.2×(0.5+(1.1+24)×0.15)+1.4×2.5，1.35×(0.5+(1.1+24)×0.15)+1.4×0.7×2.5]×0.75/1＝5.53kN/m

因此，q1静＝0.95×0.9×1.2×(G1k+(G3k+G2k)×h)×b＝0.95×0.9×1.2×(0.5+(1.1+24)×0.15)×0.75/1＝3.28kN/m

q1活＝0.95×0.9×1.4×Q1k×b＝0.95×0.9×1.4×2.5×0.75/1=2.24kN/m

q2＝0.95×0.9×1.2×G1k×b＝0.95×0.9×1.2×0.5×0.75/1＝0.38kN/m

p＝0.95×0.9×1.4×Q1k/1＝0.95×0.9×1.4×2.5/1＝2.99kN

M1＝0.107q1静L2+0.121q1活L2＝0.107×3.28×0.92+0.121×2.24×0.92＝0.5kN·m

M2＝max[0.077q2L2+0.21pL，0.107q2L2+0.181pL]＝max[0.077×0.38×0.92+0.21×2.99×0.9，0.107×0.38×0.92+0.181×2.99×0.9]＝0.59kN·m

M3＝max[q1L12/2，q2L12/2+pL1]=max[5.53×0.252/2，0.38×0.252/2+2.99×0.25]＝0.76kN·m

Mmax＝max[M1，M2，M3]＝max[0.5，0.59，0.76]＝0.76kN·m

σ＝Mmax/W＝0.76×106/4490＝169.3N/mm2≤[f]＝205N/mm2

V1＝0.607q1静L+0.62q1活L＝0.607×3.28×0.9+0.62×2.24×0.9＝3.05kN

V2＝0.607q2L+0.681p＝0.607×0.38×0.9+0.681×2.99＝2.25kN

V3＝max[q1L1，q2L1+p]＝max[5.53×0.25，0.38×0.25+2.99]＝3.09kN

Vmax＝max[V1，V2，V3]＝max[3.05，2.25，3.09]＝3.09kN

τmax＝2Vmax/A=2×3.09×1000/424＝14.57N/mm2≤[τ]＝125N/mm2

q＝(G1k+(G3k+G2k)×h)×b/1=(0.5+(24+1.1)×0.15)×0.75/1＝3.2kN/m

跨中νmax＝0.632qL4/(100EI)=0.632×3.2×9004/(100×206000×107800)＝0.6mm≤[ν]＝l/400＝900/400＝2.25mm

悬臂端νmax＝qL4/(8EI)=3.2×2504/(8×206000×107800)＝0.07mm≤[ν]＝l1/400＝250/400＝0.62mm

顶部立杆段：l01=kμ1(hd+2a)=1×1.386×(1500+2×200)=2633.4mm

非顶部立杆段：l02=kμ2h =1×1.755×1800=3159mm

λ=l0/i=3159/15.9=198.68≤[λ]=210

长细比满足要求！

2、立柱稳定性验算

顶部立杆段：l01=kμ1(hd+2a)=1.155×1.386×(1500+2×200)=3041.577mm

λ1＝l01/i＝3041.577/15.9＝191.294，查表得，φ1＝0.197

Mw＝0.92×1.4×0.22×0.9×1.82/10＝0.07kN·m

Nw＝0.95×0.9×[1.2×(0.75+(24+1.1)×0.15)+0.9×1.4×1]×0.9×0.75+0.95×0.92×1.4×0.07/0.9＝3.94kN

f＝ Nw/(φA)+ Mw/W＝3939.54/(0.2×424)+0.07×106/4490＝63.07N/mm2≤[f]＝205N/mm2

非顶部立杆段：l02=kμ2h =1.155×1.755×1800=3648.645mm

λ2＝l02/i＝3648.645/15.9＝229.475，查表得，φ2＝0.139

Mw＝0.92×1.4×0.22×0.9×1.82/10＝0.07kN·m

Nw＝0.95×0.9×[1.2×(1.1+(24+1.1)×0.15)+0.9×1.4×1]×0.9×0.75+0.95×0.92×1.4×0.07/0.9＝4.18kN

f＝ Nw/(φA)+ Mw/W＝4181.93/(0.14×424)+0.07×106/4490＝86.86N/mm2≤[f]＝205N/mm2

按上节计算可知，可调托座受力N＝3.94kN≤[N]＝30kN

八、立柱地基基础验算

立柱底垫板的底面平均压力p＝N/(mfA)＝4.18/(1×0.1)＝41.82kPa≤fak＝70kPa

板模板（扣件式）组合钢模板计算书

2、《建筑施工扣件式钢管脚手架安全技术规范》JGJ 130－2011

三、模板支撑体系设计

设计简图如下：

模板设计剖面图(楼板长向)

模板设计剖面图(楼板宽向)

q1＝0.95×0.9max[1.2×(G1k+(G3k+G2k)×h)+1.4Q1k，1.35(G1k+ (G3k+G2k)×h)+1.4×0.7Q1k]×b=0.95×0.9max[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] ×0.5=3.2kN/m

q2＝λ×0.9×1.2×G1k×b＝0.95×0.9×1.2×0.3×0.5=0.15kN/m

p＝λ×0.9×1.4×Q1k＝0.95×0.9×1.4×2.5＝2.99kN

M1=q1l2/8=3.2×0.92/8=0.32kN·m

M2=q2l2/8+pl/4=0.15×0.92/8+2.99×0.9/4=0.69kN·m

M3=max[q1l2/2，q2l2/2+pl]=max[3.2×0.32/2，0.15×0.32/2+2.99×0.3]=0.9kN·m

Mmax＝max[0.32，0.69，0.9]= 0.9kN·m

σ＝Mmax/W＝0.9×106/10460＝86.49N/mm2≤[f]＝205N/mm2

q＝(G1k+(G3k+G2k)×h)×b=(0.3+(1.1+24)×0.12)×0.5=1.66kN/m

跨中νmax＝5ql4/(384EI)＝5×1.66×9004/(384×206000×475000)＝0.14mm≤1.5mm

悬臂端νmax＝ql4/(8EI)＝1.66×3004/(8×206000×475000)＝0.02mm≤1.5mm

因[B/lb]取整＝[5000/900]取整＝5，按四等跨连续梁计算，又因小梁较大悬挑长度为250mm，因此需进行最不利组合，计算简图如下：

q1＝0.95×0.9max[1.2(G1k+ (G3k+G2k)×h)+1.4Q1k，1.35(G1k+ (G3k+G2k)×h)+1.4×0.7Q1k]×b=0.95×0.9×max[1.2×(0.5+(1.1+24)×0.12)+1.4×2.5，1.35×(0.5+(1.1+24)×0.12)+1.4×0.7×2.5]×0.75/1＝4.95kN/m

因此，q1静＝0.95×0.9×1.2×(G1k+(G3k+G2k)×h)×b＝0.95×0.9×1.2×(0.5+(1.1+24)×0.12)×0.75/1＝2.7kN/m

q1活＝0.95×0.9×1.4×Q1k×b＝0.95×0.9×1.4×2.5×0.75/1=2.24kN/m

q2＝0.95×0.9×1.2×G1k×b＝0.95×0.9×1.2×0.5×0.75/1＝0.38kN/m

p＝0.95×0.9×1.4×Q1k/1＝0.95×0.9×1.4×2.5/1＝2.99kN

M1＝0.107q1静L2+0.121q1活L2＝0.107×2.7×0.92+0.121×2.24×0.92＝0.45kN·m

M2＝max[0.077q2L2+0.21pL，0.107q2L2+0.181pL]＝max[0.077×0.38×0.92+0.21×2.99×0.9，0.107×0.38×0.92+0.181×2.99×0.9]＝0.59kN·m

M3＝max[q1L12/2，q2L12/2+pL1]=max[4.95×0.252/2，0.38×0.252/2+2.99×0.25]＝0.76kN·m

Mmax＝max[M1，M2，M3]＝max[0.45，0.59，0.76]＝0.76kN·m

σ＝Mmax/W＝0.76×106/4490＝169.3N/mm2≤[f]＝205N/mm2

V1＝0.607q1静L+0.62q1活L＝0.607×2.7×0.9+0.62×2.24×0.9＝2.73kN

V2＝0.607q2L+0.681p＝0.607×0.38×0.9+0.681×2.99＝2.25kN

V3＝max[q1L1，q2L1+p]＝max[4.95×0.25，0.38×0.25+2.99]＝3.09kN

Vmax＝max[V1，V2，V3]＝max[2.73，2.25，3.09]＝3.09kN

τmax＝2Vmax/A=2×3.09×1000/424＝14.57N/mm2≤[τ]＝125N/mm2

q＝(G1k+(G3k+G2k)×h)×b/1=(0.5+(24+1.1)×0.12)×0.75/1＝2.63kN/m

跨中νmax＝0.632qL4/(100EI)=0.632×2.63×9004/(100×206000×107800)＝0.49mm≤[ν]＝l/400＝900/400＝2.25mm

悬臂端νmax＝qL4/(8EI)=2.63×2504/(8×206000×107800)＝0.06mm≤[ν]＝l1/400＝250/400＝0.62mm

顶部立杆段：l01=kμ1(hd+2a)=1×1.386×(1500+2×200)=2633.4mm

非顶部立杆段：l02=kμ2h =1×1.755×1800=3159mm

λ=l0/i=3159/15.9=198.68≤[λ]=210

长细比满足要求！

2、立柱稳定性验算

顶部立杆段：l01=kμ1(hd+2a)=1.155×1.386×(1500+2×200)=3041.577mm

λ1＝l01/i＝3041.577/15.9＝191.294，查表得，φ1＝0.197

Mw＝0.92×1.4×0.22×0.9×1.82/10＝0.07kN·m

Nw＝0.95×0.9×[1.2×(0.75+(24+1.1)×0.12)+0.9×1.4×1]×0.9×0.75+0.95×0.92×1.4×0.07/0.9＝3.42kN

f＝ Nw/(φA)+ Mw/W＝3418.05/(0.2×424)+0.07×106/4490＝56.83N/mm2≤[f]＝205N/mm2

非顶部立杆段：l02=kμ2h =1.155×1.755×1800=3648.645mm

λ2＝l02/i＝3648.645/15.9＝229.475，查表得，φ2＝0.139

Mw＝0.92×1.4×0.22×0.9×1.82/10＝0.07kN·m

Nw＝0.95×0.9×[1.2×(1.1+(24+1.1)×0.12)+0.9×1.4×1]×0.9×0.75+0.95×0.92×1.4×0.07/0.9＝3.66kN

f＝ Nw/(φA)+ Mw/W＝3660.44/(0.14×424)+0.07×106/4490＝78.02N/mm2≤[f]＝205N/mm2

按上节计算可知，可调托座受力N＝3.42kN≤[N]＝30kN

八、立柱地基基础验算

立柱底垫板的底面平均压力p＝N/(mfA)＝3.66/(1×0.1)＝36.6kPa≤fak＝70kPa

梁模板（扣件式）组合钢模板计算书

2、《建筑施工扣件式钢管脚手架安全技术规范》JGJ 130－2011

三、模板支撑体系设计

设计简图如下：

q＝0.95×0.9max[1.2(G1k+(G2k+G3k)×h)+1.4Q2k，1.35(G1k+(G2k+G3k)×h)+1.4×0.7Q2k]×b=0.95×0.9max[1.2×(0.1+(24+1.5)×0.8)+1.4×2，1.35×(0.1+(24+1.5)×0.8)+1.4×0.7×2]×0.6＝15.2kN/m

M1=ql2/8=15.2×0.82/8=1.22kN·m

M2=ql2/2=15.2×0.42/2=1.22kN·m

Mmax＝max[M1,M2]=max[1.22，1.22]= 1.22kN·m

σ＝Mmax/W＝1.22×106/11980＝101.52N/mm2≤[f]＝205N/mm2

q＝(G1k+(G3k+G2k)×h)×b=(0.1+(1.5+24)×0.8)×0.6=12.3kN/m

跨中νmax＝5ql4/(384EI)＝5×12.3×8004/(384×206000×543000)＝0.59mm≤[ν]＝l/400＝800/400＝2mm

悬臂端νmax＝ql4/(8EI)＝12.3×4004/(8×206000×543000)＝0.35mm≤[ν]＝l1/400＝400/400＝1mm

计算简图如下：

承载能力极限状态

正常使用极限状态

承载能力极限状态：

梁混凝土和面板传递给小梁q1＝[0.95×0.9max[1.2(G1k+(G2k+G3k)×h)+1.4Q2k，1.35(G1k+(G2k+G3k)×h)+1.4×0.7Q2k]×b]/2=[0.95×0.9max[1.2×(0.1+(24+1.5)×0.8)+1.4×2，1.35×(0.1+(24+1.5)×0.8)+1.4×0.7×2]×0.8]/2＝10.14kN/m

正常使用极限状态：

面板传递给小梁q1＝[(0.1+(24+1.5)×0.8)×0.8]/2＝8.2kN/m

小梁弯矩图(kN·m)

σ＝Mmax/W＝0.12×106/4490＝27.18N/mm2≤[f]＝205N/mm2

小梁剪力图(kN)

Vmax＝2.046kN

τmax＝2Vmax/A=2×2.046×1000/424＝9.65N/mm2≤[τ]＝125N/mm2

小梁变形图(mm)

νmax＝0.06mm≤[ν]＝l/400＝400/400＝1mm

4、支座反力计算

承载能力极限状态

R1＝0.19kN，R2＝2.98kN，R3＝2.98kN，R4＝0.19kN

正常使用极限状态

R1＝0.15kN，R2＝2.33kN，R3＝2.33kN，R4＝0.15kN

最大支座反力Rmax=2×0.19=0.38kN，0.38kN≤8.0kN

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！

顶部立杆段：l01=kμ1(hd+2a)=1×1.386×(1500+2×200)=2633.4mm

非顶部立杆段：l02=kμ2h =1×1.755×1500=2632.5mm

λ=l0/i=2633.4/15.9=165.62≤[λ]=210

长细比满足要求！

Mw＝0.92×1.4×ωk×la×h2/10＝0.92×1.4×0.22×1.2×1.52/10＝0.07kN·m

q1＝0.9×[1.2×(0.1+(24+1.5)×0.8)+0.9×1.4×2]×0.4＝9.76kN/m

q1＝0.95×0.9×[1.2×(0.1+(24+1.5)×0.8)+0.9×1.4×2]×0.8=18.55kN/m

同上计算过程，可得：

顶部立杆段：l01=kμ1(hd+2a)=1.155×1.386×(1500+2×200)=3041.577mm

λ1＝l01/i＝3041.577/15.9＝191.294，查表得，φ1＝0.197

f＝N/(φA)+Mw/W＝6.95×103/(0.197×424)+0.07×106/4490＝98.27N/mm2≤[f]＝205N/mm2

JTG 3830-2018标准下载 非顶部立杆段：l02=kμ2h =1.155×1.755×1500=3040.538mm

λ2＝l02/i＝3040.538/15.9＝191.229，查表得，φ2＝0.197

f＝N/(φA)+Mw/W＝7.22×103/(0.197×424)+0.07×106/4490＝101.5N/mm2≤[f]＝205N/mm2

由"立柱验算"一节计算可知可调托座最大受力N＝max[R2DB34／T 3584-2020 公路大厚度水泥稳定碎石基层.pdf，R3]＝5.33kN≤[N]＝30kN

八、立柱地基基础计算

立柱底垫板的底面平均压力p＝N/(mfA)＝7.22/(1×0.15)＝48.17kPa≤fak＝140kPa