120mm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、《混凝土结构设计规范》GB 50010-20103、《建筑结构荷载规范》GB 50009-20124、《钢结构设计规范》GB 50017-2003一、工程属性模板设计平面图模板设计剖面图(模板支架纵向)模板设计剖面图(模板支架横向)四、面板验算面板类型覆面木胶合板面板厚度t(mm)13面板抗弯强度设计值[f](N/mm2)12.5面板抗剪强度设计值[τ](N/mm2) 1.4面板弹性模量E(N/mm2)4500面板计算方式三等跨连续梁算。

W＝bh2/6=1000×13×13/6＝28166.667mm3，I＝bh3/12=1000×13×13×13/12＝183083.333mm4承载能力极限状态q 1＝0.9×max[1.2(G 1k +(G 2k +G 3k )×h)+1.4×Q 1k ,1.35(G 1k+(G 2k +G 3k )×h)+1.4×0.7×Q 1k ]×b=0.9×max[1.2×(0.1+(24+1.1)×0.12)+1.4×2.5，1.35×(0.1+(24+1.1)×0.12)+1.4×0.7×2.5] ×1=6.511kN/mq 1静=0.9×[γG (G 1k +(G 2k +G 3k )×h)×b] =0.9×[1.2×(0.1+(24+1.1)×0.12)×1]=3.361kN/mq 1活=0.9×(γQ Q 1k )×b=0.9×(1.4×2.5)×1=3.15kN/m q 2＝0.9×1.2×G 1k ×b ＝0.9×1.2×0.1×1=0.108kN/m p ＝0.9×1.4×Q 1k ＝0.9×1.4×2.5＝3.15kN 正常使用极限状态q ＝(γG (G 1k +(G 2k +G 3k )×h))×b=(1×(0.1+(24+1.1)×0.12))×1＝3.112kN/m 计算简图如下：1、强度验算M 1＝0.1q 1静L 2+0.117q 1活L 2＝0.1×3.361×0.352+0.117×3.15×0.352＝0.086kN ·mM 2＝max[0.08q 2L 2+0.213pL，0.1q 2L 2+0.175pL]＝max[0.08×0.108×0.352+0.213×3.15×0.35，0.1×0.108×0.352+0.175×3.15×0.35]＝0.236kN ·mM max ＝max[M 1，M 2]＝max[0.086，0.236]＝0.236kN ·m σ＝M max /W ＝0.236×106/28166.667＝8.375N/mm 2≤[f]＝12.5N/mm2满足要求！2、挠度验算νmax＝0.677ql4/(100EI)=0.677×3.112×3504/(100×4500×183083.333)=0.384mmν＝0.384mm≤[ν]＝L/250＝350/250＝1.4mm满足要求！五、小梁验算11k2k3k1k1k+(G2k +G3k)×h)+1.4×0.7×Q1k]×b=0.9×max[1.2×(0.3+(24+1.1)×0.12)+1.4×2.5，1.35×(0.3+(24+1.1)×0.12)+1.4×0.7×2.5]×0.35=2.354kN/m因此，q1静＝0.9×1.2×(G1k+(G2k +G3k)×h)×b=0.9×1.2×(0.3+(24+1.1)×0.12)×0.35=1.252kN/mq1活＝0.9×1.4×Q1k×b=0.9×1.4×2.5×0.35＝1.103kN/mq2＝0.9×1.2×G1k×b=0.9×1.2×0.3×0.35＝0.113kN/mp＝0.9×1.4×Q1k＝0.9×1.4×2.5＝3.15kN计算简图如下：1、强度验算M 1＝0.125q 1静L 2+0.125q 1活L 2＝0.125×1.252×12+0.125×1.103×12＝0.294kN ·mM 2＝max[0.07q 2L 2+0.203pL，0.125q 2L 2+0.188pL]＝max[0.07×0.113×12+0.203×3.15×1，0.125×0.113×12+0.188×3.15×1]＝0.647kN ·mM 3＝max[q 1L 12/2，q 2L 12/2+pL 1]=max[2.354×0.22/2，0.113×0.22/2+3.15×0.2]＝0.632kN ·mM max ＝max[M 1，M 2，M 3]＝max[0.294，0.647，0.632]＝0.647kN ·mσ=M max /W=0.647×106/54000=11.989N/mm 2≤[f]=15.44N/mm 2 满足要求！ 2、抗剪验算V 1＝0.625q 1静L+0.625q 1活L ＝0.625×1.252×1+0.625×1.103×1＝1.472kNV 2＝0.625q 2L+0.688p ＝0.625×0.113×1+0.688×3.15＝2.238kNV3＝max[q1L1，q2L1+p]＝max[2.354×0.2，0.113×0.2+3.15]＝3.173kNVmax ＝max[V1，V2，V3]＝max[1.472，2.238，3.173]＝3.173kNτmax =3Vmax/(2bh)=3×3.173×1000/(2×40×90)＝1.322N/mm2≤[τ]=1.78N/mm2满足要求！3、挠度验算q＝(γG (G1k+(G2k+G3k)×h))×b=(1×(0.3+(24+1.1)×0.12))×0.35＝1.159kN/m挠度,跨中νmax＝0.521qL4/(100EI)=0.521×1.159×10004/(100×9350×243×104)＝0.266mm≤[ν]＝L/250＝1000/250＝4mm；悬臂端νmax ＝ql14/(8EI)=1.159×2004/(8×9350×243×104)＝0.01mm≤[ν]＝2×l1/250＝2×200/250＝1.6mm 满足要求！六、主梁验算q 1＝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)×0.12)+1.4×1.5，1.35×(0.5+(24+1.1)×0.12)+1.4×0.7×1.5]×0.35＝1.989kN/mq1静＝0.9×1.2×(G1k+(G2k+G3k)×h)×b＝0.9×1.2×(0.5+(24+1.1)×0.12)×0.35＝1.328kN/mq 1活＝0.9×1.4×Q1k×b＝0.9×1.4×1.5×0.35＝0.661kN/mq 2＝(γG(G1k+(G2k+G3k)×h))×b=(1×(0.5+(24+1.1)×0.12))×0.35＝1.229kN/m承载能力极限状态按二等跨连续梁，Rmax ＝1.25q1L＝1.25×1.989×1＝2.486kN按悬臂梁，R1＝1.989×0.2＝0.398kNR＝max[Rmax ，R1]＝2.486kN;正常使用极限状态按二等跨连续梁，R'max ＝1.25q2L＝1.25×1.229×1＝1.537kN按悬臂梁，R'1＝q2l1=1.229×0.2＝0.246kNR'＝max[R'max ，R'1]＝1.537kN;计算简图如下：主梁计算简图一主梁计算简图二2、抗弯验算主梁弯矩图一(kN·m)主梁弯矩图二(kN·m)σ=M/W=0.752×106/4120=182.593N/mm2≤[f]=205N/mm2 max满足要求！3、抗剪验算主梁剪力图二(kN)τmax =2Vmax/A=2×5.104×1000/384＝26.586N/mm2≤[τ]=125N/mm2满足要求！4、挠度验算主梁变形图一(mm)跨中νmax=1.402mm≤[ν]=1000/250=4mm悬挑段νmax＝1.027mm≤[ν]=2×200/250=1.6mm 满足要求！5、支座反力计算承载能力极限状态图一支座反力依次为R1=6.09kN，R2=7.08kN，R3=7.968kN，R4=3.723kN图二支座反力依次为R1=4.84kN，R2=7.59kN，R3=7.59kN，R4=4.84kN七、扣件抗滑移验算c c满足要求！八、立柱验算l=h=1500mmλ=l/i=1500/16=93.75≤[λ]=150满足要求！2、立柱稳定性验算根据《建筑施工模板安全技术规范》JGJ162-2008，荷载设计值q 1有所不同： 小梁验算q 1＝0.9×[1.2×(0.5+(24+1.1)×0.12)+1.4×0.9×1]×0.35 = 1.724kN/m 同上四～六步计算过程，可得：R 1＝5.282kN ，R 2＝6.583kN ，R 3＝6.91kN ，R 4＝4.197kN λ=l 0/i=1500.000/16=93.75 查表得，φ1=0.641 不考虑风荷载:N=Max[R 1,R 2,R 3,R 4]+0.9×γG ×q×H=Max[5.282,6.583,6.91,4.197]+0.9×1.2×0.15×3=7.396kNf=N/(φ1A)＝7.396×103/(0.641×384)=30.047N/mm 2≤[σ]=205N/mm 2 满足要求！ 考虑风荷载:M w ＝0.9×γQ φc ωk ×l a ×h 2/10＝0.9×1.4×0.9×0.051×1×1.52/10＝0.013kN ·mN w=Max[R 1,R 2,R 3,R 4]+0.9×γG ×q×H+M w /l b =Max[5.282,6.583,6.91,4.197]+0.9×1.2×0.15×3+0.013/1=7.409kNf=N w /(φ1A)+M w /W＝7.409×103/(0.641×384)+0.013×106/4120=33.255N/mm 2≤[σ]=205N/mm 2 满足要求！九、高宽比验算根据《建筑施工扣件式钢管脚手架安全技术规范》JGJ130-2011 第6.9.7：支架高宽比不应大于3H/B=3/7.8=0.385≤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11、受冲切承载力计算根据《混凝土结构设计规范》GB50010-2010第6.5.1条规定，见下表h t 0 u m =2[(a+h 0)+(b+h 0)]=1200mm F=(0.7βh f t +0.25σpc，m)ηu m h 0=(0.7×1×0.737+0.25×0)×1×1200×100/1000=61.908kN≥F 1=7.409kN满足要求！2、局部受压承载力计算根据《混凝土结构设计规范》GB50010-2010第6.6.1条规定，见下表c ，βcβl =(A b /A l )1/2=[(a+2b)×(b+2b)/(ab)]1/2=[(600)×(600)/(200×200)]1/2=3，A ln =ab=40000mm 2F=1.35βc βl f c A ln =1.35×1×3×6.902×40000/1000=1118.124kN≥F 1=7.409kN 满足要求！。