斜屋面板模板(扣件式)计算书计算依据:1、《建筑施工模板安全技术规范》JGJ162-20082、《建筑施工扣件式钢管脚手架安全技术规范》JGJ 130-20113、《混凝土结构设计规范》GB50010-20104、《建筑结构荷载规范》GB 50009-20125、《钢结构设计规范》GB 50017-2003一、工程属性新浇混凝土楼板名称斜屋面新浇混凝土楼板板厚(mm) 140 新浇混凝土楼板边长L(m) 7.5 新浇混凝土楼板边宽B(m) 3.8 二、荷载设计三、模板体系设计设计简图如下:模板设计平面图模板设计剖面图(楼板长向)模板设计剖面图(楼板宽向)四、面板验算面板类型覆面木胶合板 面板厚度(mm)15 面板抗弯强度设计值[f](N/mm 2)15面板弹性模量E(N/mm 2)4500根据《建筑施工模板安全技术规范》5.2.1"面板可按简支跨计算"的规定,另据现实,楼板面板应搁置在梁侧模板上,因此本例以简支梁,取1m 单位宽度计算。
计算简图如下:W =bh 2/6=1000×15×15/6=37500mm 3,I =bh 3/12=1000×15×15×15/12=281250mm 4 1、强度验算q 1=0.9max[1.2(G 1k + (G 3k +G 2k )×h)+1.4Q 1k ,1.35(G 1k +(G 3k +G 2k )×h)+1.4×0.7Q 1k ]×b=0.9max[1.2×(0.1+(1.1+23)×0.14)+1.4×2.5,1.35×(0.1+(1.1+23)×0.14)+1.4×0.7×2.5] ×1=6.9kN/m q 2=0.9×1.2×G 1k ×b=0.9×1.2×0.1×1=0.11kN/m p =0.9×1.4×Q 1K =0.9×1.4×2.5=3.15kNM max =max[q 1l 2/8,q 2l 2/8+pl/4]=max[6.9×0.32/8,0.11×0.32/8+3.15×0.3/4]= 0.24kN·m σ=M max /W =0.24×106/37500=6.33N/mm 2≤[f]=15N/mm 2 满足要求! 2、挠度验算q =(G 1k +(G 3k +G 2k )×h)×b=(0.1+(1.1+23)×0.14)×1=3.47kN/mν=5ql 4/(384EI)=5×3.47×3004/(384×4500×281250)=0.29mm≤[ν]=l/400=300/400=0.75mm 满足要求!五、小梁验算因[B/l b]取整=[3800/800]取整=4,按四等跨连续梁计算,又因小梁较大悬挑长度为100mm,因此需进行最不利组合,计算简图如下:1、强度验算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+23)×0.14)+1.4×2.5,1.35×(0.3+(1.1+23)×0.14)+1.4×0.7×2.5]×0.3=2.14kN/m因此,q1静=0.9×1.2(G1k+(G3k+G2k)×h)×b=0.9×1.2×(0.3+(1.1+23)×0.14)×0.3=1.19kN/mq1活=0.9×1.4×Q1k×b=0.9×1.4×2.5×0.3=0.94kN/mM1=0.107q1静L2+0.121q1活L2=0.107×1.19×0.82+0.121×0.94×0.82=0.15kN·mq2=0.9×1.2×G1k×b=0.9×1.2×0.3×0.3=0.1kN/mp=0.9×1.4×Q1k=0.9×1.4×2.5=3.15kNM2=max[0.077q2L2+0.21pL,0.107q2L2+0.181pL]=max[0.077×0.1×0.82+0.21×3.15×0.8,0.107×0.1×0.82+0.181×3.15×0.8]=0.53kN·m M3=max[q1L12/2,q2L12/2+pL1]=max[2.14×0.12/2,0.1×0.12/2+3.15×0.1]=0.32kN·m M max=max[M1,M2,M3]=max[0.15,0.53,0.32]=0.53kN·mσ=M max/W=0.53×106/54000=9.89N/mm2≤[f]=12.87N/mm2满足要求!2、抗剪验算V 1=0.607q 1静L+0.62q 1活L =0.607×1.19×0.8+0.62×0.94×0.8=1.05kN V 2=0.607q 2L+0.681p =0.607×0.1×0.8+0.681×3.15=2.19kN V 3=max[q 1L 1,q 2L 1+p]=max[2.14×0.1,0.1×0.1+3.15]=3.16kN V max =max[V 1,V 2,V 3]=max[1.05,2.19,3.16]=3.16kNτmax =3V max /(2bh 0)=3×3.16×1000/(2×90×40)=1.32N/mm 2≤[τ]=1.39N/mm 2 满足要求! 3、挠度验算q=(G 1k +(G 3k +G 2k )×h)×b=(0.3+(23+1.1)×0.14)×0.3=1.1kN/m跨中νmax =0.632qL 4/(100EI)=0.632×1.1×8004/(100×7480×2430000)=0.16mm≤[ν]=l/400=800/400=2mm 满足要求!六、主梁验算主梁类型钢管 主梁材料规格(mm) Φ48×2.75 可调托座内主梁根数1 主梁弹性模量E(N/mm 2) 206000 主梁抗弯强度设计值[f](N/mm 2) 205 主梁抗剪强度设计值[τ](N/mm 2) 125 主梁截面惯性矩I(cm 4)10.04主梁截面抵抗矩W(cm 3)4.181、小梁最大支座反力计算 Q 1k =1.5kN/m 2q 1=0.9max[1.2(G 1k + (G 3k +G 2k )×h)+1.4Q 1k ,1.35(G 1k +(G 3k +G 2k )×h)+1.4×0.7Q 1k ]×b=0.9max[1.2×(0.5+(1.1+23)×0.14)+1.4×1.5,1.35×(0.5+(1.1+23)×0.14)+1.4×0.7×1.5]×0.3=1.82kN/mq 1静=0.9×1.2(G 1k + (G 3k +G 2k )×h)×b=0.9×1.2×(0.5+(1.1+23)×0.14)×0.3=1.26kN/m q 1活=0.9×1.4×Q 1k ×b=0.9×1.4×1.5×0.3=0.57kN/mq 2=(G 1k + (G 3k +G 2k )×h)×b=(0.5+(1.1+23)×0.14)×0.3=1.16kN/m 承载能力极限状态按四跨连续梁,R max =(1.143q 1静+1.223q 1活)L =1.143×1.26×0.8+1.223×0.57×0.8=1.7kN按悬臂梁,R1=q1l=1.82×0.1=0.18kNR=max[R max,R1]=1.7kN;正常使用极限状态按四跨连续梁,R max=1.143q2L=1.143×1.16×0.8=1.06kN 按悬臂梁,R1=q2l=1.16×0.1=0.12kNR=max[R max,R1]=1.06kN;2、抗弯验算计算简图如下:主梁弯矩图(kN·m)M max=0.4kN·mσ=M max/W=0.4×106/4180=94.93N/mm2≤[f]=205N/mm2 满足要求!3、抗剪验算主梁剪力图(kN) V max=3.18kNτmax=2V max/A=2×3.18×1000/391=16.29N/mm2≤[τ]=125N/mm2 满足要求!4、挠度验算主梁变形图(mm) νmax=0.47mm跨中νmax=0.47mm≤[ν]=800/400=2mm悬挑段νmax=0.27mm≤[ν]=120/400=0.3mm满足要求!七、立柱验算1、长细比验算顶部立杆段:l01=kμ1(h d+2a)=1×1.386×(1500+2×100)=2356.2mm非顶部立杆段:l02=kμ2h =1×1.755×1500=2632.5mmλ=l0/i=2632.5/16=164.53≤[λ]=210长细比满足要求!2、立柱稳定性验算顶部立杆段:l01=kμ1(h d+2a)=1.217×1.386×(1500+2×100)=2867.495mmλ1=l01/i=2867.495/16=179.218,查表得,υ1=0.223f=N/(υA)=0.9max[1.2×(0.5+(1.1+23)×0.14)+1.4×1,1.35×(0.5+(1.1+23)×0.14)+1.4×0.7×1]×0.8×0.8×1000/(0.22×391)=3.58×1000/87.19=41.02N /mm2≤[f]=205 N/mm2满足要求!非顶部立杆段:l02=kμ2h =1.217×1.755×1500=3203.752mmλ2=l02/i=3203.752/16=200.234,查表得,υ2=0.18非顶部立杆段:l02=kμ2h =1.217×1.755×1500=3203.752mmλ2=l02/i=3203.752/16=200.234,查表得,υ2=0.18f=N/(υA)=0.9max[1.2×(0.75+(1.1+23)×0.14)+1.4×1,1.35×(0.75+(1.1+23)×0.14)+1.4×0.7×1]×0.8×0.8×1000/(0.18×391)=3.77×1000/70.38=53.58 N/mm2≤[f]=205 N/mm2满足要求!八、扣件抗滑移验算按上节计算可知,扣件受力N=3.58kN≤R c=k c×8=0.85×8=6.8kN满足要求!九、立柱地基基础验算地基土类型碎石土地基承载力设计值f ak(kPa) 320立柱垫木地基土承载力折减系数m f0.4 垫板底面面积A(m2) 0.1 立柱底垫板的底面平均压力p=N/(m f A)=3.77/(0.4×0.1)=94.28kPa≤f ak=320kPa 满足要求!。