最全某某工程施工结构梁拆除方案正式版

五、主要施工方法及措施 6 七、各项保证措施 9 一、工程概况3.15m。根据建立单位的要求，撤除B区地下室首层〔地下室顶板〕相应部位〔撤除部位详见二、编制依据.z.-平安员预算劳资员本钱员测量组平安员预算劳资员本钱员测量组质检员试验组内业组材料组机械队安装队装饰队土建队工程施工组织构造网络图.z.绝一切重大平安质量事故；杜绝机械设备事故，杜，我们将采取施工全封施工，在确保工程质量，最大限度减少对环境的施工准备用途造破碎的混凝土建筑垃圾建筑垃圾明施工台把1.2作业条件准备.z.-5、现场准备危险区域，施工前应发出告示，通报施工考前五、主要施工方法及措施1)支承架体:梁截面400mm*800mm及以上按梁跨度方向立柱间距450mm,梁两侧立柱间mm杆。可杆、水平拉杆、剪刀撑应采用Ø48.3×3.0mm钢管，用扣件与钢管立.z.-mm。可。本工程所有部位立柱接长全部采用对接扣件连接，严禁搭接，接头位置要梁交接9、梁撤除时将梁分段剔凿，梁按照1.5m每段剔凿，剔凿宽度不得小于150mm，将需要才可逐段切断钢筋，每切割一段后及时使用塔吊吊离施工作业.z.-梁、板〕按梁、板相应跨度方向不少于2/3搭无误后才能进展，根据施工现场实际了确保其他楼栋的施工正常进展，故在撤除时安排专人值班，以消除撤除施工平安，则需轻轻敲打，避过钢筋剔除周围混凝土，尽量防止损坏钢得损坏周边钢筋。撤除的楼板与保存的楼板混凝土别离后，才空缺，应设置临时围护栏，围护栏设置应醒清理和成品保护容。由工长主持上.z.-照ISO9001质量保证体系的要求，落实施工过程控制的责任，适应程序的具体工程经理为质量保证第一人，工程总工把关，工程师、技术员、各工种工长亲据施工方案、工艺标准、质量方案，明确质量管理重点和管理措施，对各分包七、各项保证措施制度。灭火器材和其它相应的防护楚作业围平安状态，提钢筋处于绝缘状态。所有电器.z.-作。杜绝高空坠落等均按平安操作业；立体穿插作业作好防护措施，必要品火器等防火灭火器材，动火前及八、雨季施工.z.-。九、成品保护措施工完成的主体构造使用木模板进展格挡，防止剔凿下来的砼。7-.z.GkkN/m3)Gk(kN/m3)GkkN/m3)QkQk(kN/m2)2时Q1k(kN/m2)22500.z.- m00mm4q＝γ×ma*[1.2(G+(G+G)×h)+1.4Q，1.35(G+(G+G)×h)+1.4ψQ]×b=0.9101k2k3k2k1k2k3kc2kq＝0.9×1.35×[G+(G+G)×h]×b＝0.9×1.35×[0.1+(60+3.75)×1]×1＝1静1k2k3kq＝0.9×1.4×0.7×Q×b＝0.9×1.4×0.7×2×1＝1.764kN/m1活2kq＝[G+(G+G)×h]×b＝[0.1+(60+3.75)×1]×1＝63.85kN/m21k2k3kM＝0.107qL2+0.121qL2＝0.107×77.578×0.1752+0.121×1.764×0.1752＝ma*1静1活σ＝M/W＝0.261×106/54000＝4.829N/mm2≤[f]＝15N/mm2ma*ν＝0.632qL4/(100EI)=0.632×63.85×1754/(100×10000×486000)＝0.078mm≤[ν]＝l/250ma*2R=R=0.393ql+0.446ql=0.393×77.578×0.175+0.446×1.764×0.175＝5.473kN151静1活.z.-R=R=1.143ql+1.223ql=1.143×77.578×0.175+1.223×1.764×0.175＝15.895kN241静1活R=0.928ql+1.142ql=0.928×77.578×0.175+1.142×1.764×0.175＝12.951kN31静1活R'=R'=0.393ql=0.393×63.85×0.175＝4.391kN152R'=R'=1.143ql=1.143×63.85×0.175＝12.772kN242R'=0.928ql=0.928×63.85×0.175＝10.369kN320小梁截面惯性矩I(cm4)三等跨连续梁q＝ma*{5.473+0.9×1.35×[(0.3-0.1)×0.7/4+0.5×(1-0.16)]+0.9×ma*[1.2×1mq＝ma*{4.391+(0.3-0.1)×0.7/4+0.5×(1-0.16)+(0.5+(60+2.75)×0.16)×ma*[0.6-0.7/2，2M＝ma*[0.1ql2，0.5ql2]＝ma*[0.1×15.938×0.452，0.5×15.938×0.12]＝0.323kN·mma*1112σ＝M/W＝0.323×106/4490＝71.879N/mm2≤[f]＝205N/mm2ma*V＝ma*[0.6ql，ql]＝ma*[0.6×15.938×0.45，15.938×0.1]＝4.303kNma*1112τ＝2V/A=2×4.303×1000/424＝20.298N/mm2≤[τ]＝125N/mm2ma*ma*ν＝0.677ql4/(100EI)＝0.677×12.807×4504/(100×206000×107800)＝0.16mm≤[ν]＝l/2501211.z.-ν＝ql4/(8EI)＝12.807×1004/(8×206000×107800)＝0.007mm≤[ν]＝2l/250＝2×100/2502222R＝ma*[1.1ql，0.4ql+ql]＝ma*[1.1×15.938×0.45，0.4×15.938×0.45+15.938×0.1]ma*111112同理可得，梁底支撑小梁所受最大支座反力依次为R＝R＝3.884kN，R＝R＝15247.889kN，R＝6.432kN3R'＝ma*[1.1ql，0.4ql+ql]＝ma*[1.1×12.807×0.45，0.4×12.807×0.45+12.807×0.1]ma*212122同理可得，梁底支撑小梁所受最大支座反力依次为R'＝R'＝3.349kN，R'＝R'＝15246.339kN，R'＝5.154kN3Ф48×320主梁截面惯性矩I(cm4)主梁截面抵抗矩W(cm3)计算σ＝M/W＝0.211×106/4490＝47.06N/mm2≤[f]=205N/mm2ma*.z.-V＝5.556kNma*τ＝2V/A=2×5.556×1000/424＝26.208N/mm2≤[τ]＝125N/mm2ma*ma*ν＝0.07mm≤[ν]＝l/250＝400/250＝1.6mmma*支座反力依次为R=0.402kN，R=14.594kN，R=14.584kN，R=0.4kN12342两侧立柱最大受力R＝ma*[R,R]＝ma*[0.402,0.4]＝0.402kN≤0.85×8=6.8kN4可调托座最大受力N＝ma*[R，R]＝14.594kN≤[N]＝30kN23mmФ48×3立柱截面面积A(mm2).z.-mm架自重标准值q(kN/m)0.15顶部立杆段：l=kμ(h+2a)=1×1.386×(600+2×200)=1386mm011d022λ=l/i=2632.5/15.9=165.566≤[λ]=2100M＝γ×1.4×ψ×ω×l×h2/10＝1.4×0.9×0.182×0.45×1.52/10＝0.023kN·mw0cka根据"建筑施工扣件式钢管脚手架平安技术规"JGJ130-2011，荷载设计值q有所不同：1q＝[1.2×(0.1+(60+3.75)×1)+1.4×0.9×2]×1＝79.14kN/m1q＝ma*{5.466+1.2×[(0.3-0.1)×0.7/4+0.5×(1-0.16)]+[1.2×(0.5+(60+2.75)×0.16)+1.41mR＝0.401kN，R＝14.574kN，R＝14.563kN，R＝0.4kN1234顶部立杆段：l=kμ(h+2a)=1.155×1.386×(600+2×200)=1600.83mm011dλ＝l/i＝1600.83/15.9＝100.681，查表得，φ＝0.5881011立柱最大受力N＝ma*[R+N，R，R，R+N]+M/l＝ma*[0.401+[1.2×w1边1234边2wb+1.4×0.9×2.5]×(1.2+1.2-0.6-0.7/2)/2×0.9]+0./1.2＝14.593kNf＝N/(φA)+M/W＝14593.286/(0.588×424)+0.×106/4490＝63.705N/mm2≤[f]＝wNmm2m022.z.hηηhηη-λ＝l/i＝3040.537/15.9＝191.229，查表得，φ＝0.1972022立柱最大受力N＝ma*[R+N，R，R，R+N]+1.2×0.15×(7-1)+M/l＝w1边1234边2wbNf＝N/(φA)+M/W＝15673.286/(0.197×424)+0.×106/4490＝192.812N/mm2≤[f]＝wNmm2F=N=15.673kN1计规"GB50100-2010第6.5.1条规定，见下表F1ftσpc,mum012中反力设计值1.0-3.5N/㎜2围h0/2均值作用面积形状的影响系数s=0.5+as×h/4U20m.z.-用面积为矩形时的长边与短边尺寸比βs拟，βs不宜大于4：当βs<2时取βs=2，当面积为圆形时，取βs=2sas在本工程计算中为了平安和简化计算起见，不考虑上式中σpc,m之值，将说明可得：β=1，f=1.57N/mm2，η=1，h=h-20=100mm，ht0u=2[(a+h)+(b+h)]=1200mmm00F=(0.7βf+0.25σ)ηuh=(0.7×1×1.57+0.25×0)×1×1200×100/1000=131.88kN≥htpc，mm0F=15.673kN1计规"GB50100-2010第6.6.1条规定，见下表部荷载或局部压力设计值1.4-1取值受压的计算底面积，按本规第6.6.2条确定βl=(Ab/Al)1/2F1fcAlnAlAbcclbllblF=1.35ββfA=1.35×1×3×16.7×40000/1000=2705.4kN≥F=15.673kNclcln1.z.-L)L)7Gk(kN/m3)GkkN/m3)QkQk(kN/m2)2时Q1k(kN/m2).z.-梁两侧有板，梁板立柱共用(A)lamm900lbmm1200hmm1500混凝土梁居梁两侧立柱中的位置居中mm600梁底增加立柱根数2加立柱布置方式按梁两侧立柱间距均分mm400,800梁底支撑小梁根数5每纵距附加梁底支撑主梁根数0mm200构造外表的要求构造外表隐蔽 面板厚度(mm)mm00mm4q＝γ×ma*[1.2(G+(G+G)×h)+1.4Q，1.35(G+(G+G)×h)+1.4ψQ]×b=0.9101k2k3k2k1k2k3kc2kq＝0.9×1.35×[G+(G+G)×h]×b＝0.9×1.35×[0.1+(60+3.75)×0.8]×1＝1静1k2k3k.z.-q＝0.9×1.4×0.7×Q×b＝0.9×1.4×0.7×2×1＝1.764kN/m1活2kq＝[G+(G+G)×h]×b＝[0.1+(60+3.75)×0.8]×1＝51.1kN/m21k2k3kM＝0.107qL2+0.121qL2＝0.107×62.×0.0752+0.121×1.764×0.0752＝0.039kN·mma*1静1活σ＝M/W＝0.039×106/54000＝0.714N/mm2≤[f]＝15N/mm2ma*ν＝0.632qL4/(100EI)=0.632×51.1×754/(100×10000×486000)＝0.002mm≤[ν]＝l/250＝ma*2.3mmR=R=0.393ql+0.446ql=0.393×62.×0.+0.446×1.764×0.＝1.889kN151静1活R=R=1.143ql+1.223ql=1.143×62.×0.+1.223×1.764×0.＝5.484kN241静1活R=0.928ql+1.142ql=0.928×62.×0.+1.142×1.764×0.＝4.472kN31静1活152242320小梁截面惯性矩I(cm4)三等跨连续梁q＝ma*{1.889+0.9×1.35×[(0.3-0.1)×0.3/4+0.5×(0.8-0.16)]+0.9×ma*[1.2×1.z.-q＝ma*{1.506+(0.3-0.1)×0.3/4+0.5×(0.8-0.16)+(0.5+(60+2.75)×0.16)×ma*[0.6-0.3/2，2M＝ma*[0.1ql2，0.5ql2]＝ma*[0.1×5.574×0.92，0.5×5.574×0.22]＝0.452kN·mma*1112σ＝M/W＝0.452×106/4490＝100.561N/mm2≤[f]＝205N/mm2ma*V＝ma*[0.6ql，ql]＝ma*[0.6×5.574×0.9，5.574×0.2]＝3.01kNma*1112τ＝2V/A=2×3.01×1000/424＝14.199N/mm2≤[τ]＝125N/mm2ma*ma*ν＝0.677ql4/(100EI)＝0.677×4.396×9004/(100×206000×107800)＝0.879mm≤[ν]＝l/2501211ν＝ql4/(8EI)＝4.396×2004/(8×206000×107800)＝0.04mm≤[ν]＝2l/250＝2×200/250＝2222R＝ma*[1.1ql，0.4ql+ql]＝ma*[1.1×5.574×0.9，0.4×5.574×0.9+5.574×0.2]＝ma*111112同理可得，梁底支撑小梁所受最大支座反力依次为R＝R＝5.447kN，R＝R＝152419kN，R＝4.446kN3R'＝ma*[1.1ql，0.4ql+ql]＝ma*[1.1×4.396×0.9，0.4×4.396×0.9+4.396×0.2]＝ma*212122N.z.-同理可得，梁底支撑小梁所受最大支座反力依次为R'＝R'＝4.355kN，R'＝R'＝15244.352kN，R'＝3.539kN3Ф48×320主梁截面惯性矩I(cm4)主梁截面抵抗矩W(cm3)计算σ＝M/W＝0.423×106/4490＝94.191N/mm2≤[f]=205N/mm2ma*V＝6.598kNma*τ＝2V/A=2×6.598×1000/424＝31.12N/mm2≤[τ]＝125N/mm2ma*ma*ν＝0.215mm≤[ν]＝l/250＝400/250＝1.6mmma*支座反力依次为R=1.403kN，R=14.598kN，R=14.585kN，R=1.402kN1234.z.-2两侧立柱最大受力R＝ma*[R,R]＝ma*[1.403,1.402]＝1.403kN≤0.85×8=6.8kN14可调托座最大受力N＝ma*[R，R]＝14.598kN≤[N]＝30kN23mmФ48×3mm支架自重标准值q(kN/m)顶部立杆段：l=kμ(h+2a)=1×1.386×(600+2×200)=1386mm011d022λ=l/i=2632.5/15.9=165.566≤[λ]=2100M＝γ×1.4×ψ×ω×l×h2/10＝1.4×0.9×0.182×0.9×1.52/10＝0.046kN·mw0cka根据"建筑施工扣件式钢管脚手架平安技术规"JGJ130-2011，荷载设计值q有所不同：1q＝[1.2×(0.1+(60+3.75)×0.8)+1.4×0.9×2]×1＝63.84kN/m1.z.-q＝ma*{1.892+1.2×[(0.3-0.1)×0.3/4+0.5×(0.8-0.16)]+[1.2×(0.5+(60+2.75)×0.16)+1.41R＝1.439kN，R＝14.913kN，R＝14.9kN，R＝1.438kN1234顶部立杆段：l=kμ(h+2a)=1.155×1.386×(600+2×200)=1600.83mm011dλ＝l/i＝1600.83/15.9＝100.681，查表得，φ＝0.5881011立柱最大受力N＝ma*[R+N，R，R，R+N]+M/l＝ma*[1.439+[1.2×w1边1234边2wb.4×0.9×2.5]×(1.2+1.2-0.6-0.3/2)/2×0.9]+0./1.2＝14.952kNf＝N/(φA)+M/W＝14951.547/(0.588×424)+0.×106/4490＝70.314N/mm2≤[f]＝wNmm2m022λ＝l/i＝3040.537/15.9＝191.229，查表得，φ＝0.1972022立柱最大受力N＝ma*[R+N，R，R，R+N]+1.2×0.15×(7-0.8)+M/l＝w1边1234边2wbmaf＝N/(φA)+M/W＝16067.547/(0.197×424)+0.×106/4490＝202.704N/mm2≤[f]＝wNmm2F=N=16.068kN1计规"GB50100-2010第6.5.1条规定，见下表.z.hηηhηη-ftftσσpc,m1.0-3.5N/㎜2pc,mh0/2um处板垂直截面的最不利周长。012012作用面积形状的影响系数sβs=0.5+as×h/4U20m拟，βs不宜大于4：当β=0.5+as×h/4U20m在本工程计算中为了平安和简化计算起见，不考虑上式中σpc,m之值，将说明可得：β=1，f=1.57N/mm2，η=1，h=h-20=100mm，ht0u=2[(a+h)+(b+h)]=1200mmm00F=(0.7βf+0.25σ)ηuh=(0.7×1×1.57+0.25×0)×1×1200×100/1000=131.88kN≥htpc，mm0F=16.068kN1计规"GB50100-2010第6.6.1条规定，见下表-.z.部荷载或局部压力设计值1.4-1取值受压的计算底面积，按本规第6.6.2条确定βl=(Ab/Al)1/2F1fcAlnAlAbccβ=(A/A)1/2=[(a+2b)×(b+2b)/(ab)]1/2=[(600)×(600)/(200×200)]1/2=3，A=ab=40000mm2lbllnF=1.35ββfA=1.35×1×3×16.7×40000/1000=2705.4kN≥F=16.068kNclcln166-.z.Gk(kN/m3)GkkN/m3)QkQk(kN/m2)2时Q1k(kN/m2)25.z.-0 m00mm4q＝γ×ma*[1.2(G+(G+G)×h)+1.4Q，1.35(G+(G+G)×h)+1.4ψQ]×b=0.9101k2k3k2k1k2k3kc2kq＝0.9×1.35×[G+(G+G)×h]×b＝0.9×1.35×[0.1+(60+3.75)×0.8]×1＝1静1k2k3kq＝0.9×1.4×0.7×Q×b＝0.9×1.4×0.7×2×1＝1.764kN/m1活2kq＝[G+(G+G)×h]×b＝[0.1+(60+3.75)×0.8]×1＝51.1kN/m21k2k3kM＝0.107qL2+0.121qL2＝0.107×62.×0.1252+0.121×1.764×0.1252＝0.107kN·mma*1静1活σ＝M/W＝0.107×106/54000＝1.984N/mm2≤[f]＝15N/mm2ma*ν＝0.632qL4/(100EI)=0.632×51.1×1254/(100×10000×486000)＝0.016mm≤[ν]＝l/250＝ma*2.z.-R=R=0.393ql+0.446ql=0.393×62.×0.125+0.446×1.764×0.125＝3.148kN151静1活R=R=1.143ql+1.223ql=1.143×62.×0.125+1.223×1.764×0.125＝9.14kN241静1活R=0.928ql+1.142ql=0.928×62.×0.125+1.142×1.764×0.125＝7.454kN31静1活152N242kN320小梁截面惯性矩I(cm4)三等跨连续梁q＝ma*{3.148+0.9×1.35×[(0.3-0.1)×0.5/4+0.5×(0.8-0.12)]+0.9×ma*[1.2×1mq＝ma*{2.51+(0.3-0.1)×0.5/4+0.5×(0.8-0.12)+(0.5+(60+2.75)×0.12)×ma*[0.6-0.5/2，2M＝ma*[0.1ql2，0.5ql2]＝ma*[0.1×9.171×0.452，0.5×9.171×0.22]＝0.186kN·mma*1112σ＝M/W＝0.×106/5080＝36.556N/mm2≤[f]＝205N/mm2ma*V＝ma*[0.6ql，ql]＝ma*[0.6×9.171×0.45，9.171×0.2]＝2.476kNma*1112τ＝2V/A=2×2.476×1000/489＝10.127N/mm2≤[τ]＝125N/mm2ma*ma*.z.-ν＝0.677ql4/(100EI)＝0.677×7.326×4504/(100×206000×121900)＝0.081mm≤[ν]＝l/2501211ν＝ql4/(8EI)＝7.326×2004/(8×206000×121900)＝0.058mm≤[ν]＝2l/250＝2×200/250＝2222R＝ma*[1.1ql，0.4ql+ql]＝ma*[1.1×9.171×0.45，0.4×9.171×0.45+9.171×0.2]ma*111112同理可得，梁底支撑小梁所受最大支座反力依次为R＝R＝2.776kN，R＝R＝15244.539kN，R＝3.705kN3R'＝ma*[1.1ql，0.4ql+ql]＝ma*[1.1×7.326×0.45，0.4×7.326×0.45+7.326×0.2]ma*212122同理可得，梁底支撑小梁所受最大支座反力依次为R'＝R'＝2.46kN，R'＝R'＝15243.626kN，R'＝2.949kN3Ф48×320主梁截面惯性矩I(cm4)主梁截面抵抗矩W(cm3)计算.z.-σ＝M/W＝0.194×106/4490＝43.234N/mm2≤[f]=205N/mm2ma*V＝3.198kNma*τ＝2V/A=2×3.198×1000/424＝15.086N/mm2≤[τ]＝125N/mm2ma*ma*ν＝0.088mm≤[ν]＝l/250＝400/250＝1.6mmma*支座反力依次为R=0.46kN，R=9.632kN，R=9.623kN，R=0.46kN12342两侧立柱最大受力R＝ma*[R,R]＝ma*[0.46,0.46]＝0.46kN≤0.85×8=6.8kN4可调托座最大受力N＝ma*[R，R]＝9.632kN≤[N]＝30kN23.z.-mmФ48×3mm支架自重标准值q(kN/m)顶部立杆段：l=kμ(h+2a)=1×1.386×(600+2×200)=1386mm011d022λ=l/i=2632.5/15.9=165.566≤[λ]=2100M＝γ×1.4×ψ×ω×l×h2/10＝1.4×0.9×0.182×0.45×1.52/10＝0.023kN·mw0ckaq＝[1.2×(0.1+(60+3.75)×0.8)+1.4×0.9×2]×1＝63.84kN/m1q＝ma*{3.153+1.2×[(0.3-0.1)×0.5/4+0.5×(0.8-0.12)]+[1.2×(0.5+(60+2.75)×0.12)+1.41RkN，R＝9.642kN，R＝9.633kN，R＝0.46kN1234顶部立杆段：l=kμ(h+2a)=1.155×1.386×(600+2×200)=1600.83mm011dλ