Appendix 1附件1
Calculation of the Formworks模板计算书
1、Side Formwork Construction侧模施工
1.1、设计说明
Design description: using site procesd wood formwork, face plate is plywood of 15mm, condary keel is timber of 50mm×100mm (the material is northeast larch) with 250mm space in between. Main keel is the timber of 80mm×200mm as modeling with the min. height no less than 150mm. 2 main keel t up with spacing of 700mm, 250mm as bottom and 255mm as upper side of slab.
侧模采用现场加工木模板，面板为15厚胶合板；次龙骨为50mm×100mm木方（材质为东北落叶松），间距250mm；主龙骨使用80mm×200mm木方做造型木（材质为东北落叶松），造型木中心最小高度不小于150mm。主龙骨设置两道，间距700mm，距底部250mm和上侧255mm.
1.2、Computational Checking of Secondary Keel次龙骨验算
cuckoo
1）Load and Combination of Load荷载及荷载组合
a．side pressure on the form for concrete混凝土对模板的侧压力
t0=200/(25+15)=5h (即混凝土的温度按25℃计算)
F1=0.22γc t0β1β2V1/2=0.22×25×5×1.2×1.15×21/2 =53.67KN/m2
F2=γcH=25×1.2=30KN/m2 （取此值做强度验算）
(take this value for computational checking of strength )
b．load of concrete pouring混凝土倾倒荷载： 4KN/m2
c．load of concrete vibrating混凝土振捣荷载： 4KN/m2
largo
combination of load荷载组合：1.2×30+1.4×(4+4)=47.2KN/m2
line load化为线荷载：q=47.2×0.25=11.8KN/m
2）Computational Checking of Flexural Strength抗弯强度验算
Mmax =11.8×0.7^2×(1-4×0.252肚皮舞服装/0.72长大成人2)/8=0.52KN·m (建筑施工手册表Construction Manual 2-10)
Wn =1/6bh2 =1/6×50×1002 =250000/3
σm = M/Wn =0.52×106 /(250000/3)=6.24N/mm2≤ fm =17 N/mm2
Flexural Strength meets the requirement抗弯强度满足要求。
3）Computational Checking of Shear Strength抗剪强度验算
Vmax =1/2×11.8×0.7=4.13KN
A=50×100=5000mm2nd的过去式
т=VS/Ib=3V/2A=3×4.13×1000/2×5000=0.826 N/mm2≤fv =1.6 N/mm2
Flexural Strength meets the requirement抗剪强度符合要求
4）Computational Checking of deflection挠度验算
[ω]= 0.7×1000×1/400=1.75mm
Actual deflection obtained if calculation bad on supporting beam of two ends
挠度计算按照两端简支梁计算则有实际挠度
ωmax=qlbandicoot4/384EI=11.8×1高一英语必修12004 /384×10000×(50×100mimia3/12)=1.529mm
ω=1.529mm< [ω] =1.75mm
Deflection meets the requirement挠度满足要求
To sum up, the condary keel meets the design requirement
综上所述，次龙骨强度满足要求。
1.3、Calculation of Main Keel主龙骨计算
Calculation of main keel bad on three-span continuous beam under concentrated load, which is the longitudinal transmitting force of condary keel of P=11.8×0.7=8.26KN. The distance between M16 bolts is 600mm. Calculation drawing is as follows:
1）主龙骨的计算按照集中荷载作用下的三跨连续梁计算，集中荷载为次龙骨的纵向传递力，其大小为：P=11.8×0.7=8.26KN，计算时按照最不利截面计算，M16螺栓间距为600mm。
2）Computational Checking of Flexural Strength抗弯强度验算
Mmax =0.3PL=0.3×8.26×0.6=1.4868KN·m
Wn =1/6bh2 =1/6×80×2002 =533333
σm = M/Wn =1.4868×106 /533333=2.79N/mmmogic2≤ fm =17 N/mm2
3）Computational Checking of Shear Strength抗剪强度验算
Vmax=1.267×8.26=10.465KN
A=80×200=16000mm2
т=VS/Ib=3V/2A=3×10.465×1000/(2×16000)=0.981N/mm2≤fv =1.6 N/mm2
4）Computational Checking of Deflection Strength挠度验算
私人教练培训ωmax=1.883FL3/(100EI)=1.883×8.26×1000×6003/(100×10000×2003×80/12)=0.629mm
ωmax=0.629mm < [ω]=600×1/400=1.5mm
To sum up, the main keel meets the design requirement
综上所述主龙骨满足设计要求。
1.4、Computational Checking of Pull-Bolt Strength对拉螺栓强度验算
Using M16 bolt with allowable tension stress [N]=24.5KN
对拉螺栓使用M16螺栓。容许拉力[N]=24.5KN
N=47.2KN/m2×0.7m×0.6m=19.824KN<[N]=24.5KN
Pull-bolt strength meets the requirement.
对拉螺栓强度满足要求。
