支模架板模板(扣件式)计算书施工方案
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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》JGJ162-2008
2、《混凝土结构设计规范》GB 50010-2010
3、《建筑结构荷载规范》GB 50009-2012
4、《钢结构设计规范》GB 50017-2003
一、工程属性
二、荷载设计
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
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/m
q
1静=0.9×[γ
G
(G
1k
+(G
2k
+G
3k
)×h)×b] =
0.9×[1.2×(0.1+(24+1.1)×0.12)×1]=3.361kN/m
q
1活=0.9×(γ
Q
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=(γ
G (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
L2+0.117q
1活
L2=0.1×3.361×0.352+0.117×3.15×0.352=
0.086kN·m
M
2=max[0.08q
2
L2+0.213pL,0.1q
2
L2+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·m
M
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/mm2≤[f]=12.5N/mm2满足要求!
2、挠度验算
ν
max
=0.677ql4/(100EI)=0.677×3.112×3504/(100×4500×183083.333)=0.384mm
ν=0.384mm≤[ν]=L/250=350/250=1.4mm
满足要求!
五、小梁验算
11k2k3k1k1k
+(G
2k +G
3k
)×h)+1.4×0.7×Q
1k
]×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,q
1静=0.9×1.2×(G
1k
+(G
2k +G
3k
)×h)×b=0.9×1.2×(0.3+(24+1.1)×0.12)×0.35=1.252kN/m
q
1活=0.9×1.4×Q
1k
×b=0.9×1.4×2.5×0.35=1.103kN/m
q
2=0.9×1.2×G
1k
×b=0.9×1.2×0.3×0.35=0.113kN/m
p=0.9×1.4×Q
1k
=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
L2+0.125q
1活
L2=0.125×1.252×12+0.125×1.103×12=
0.294kN·m
M
2=max[0.07q
2
L2+0.203pL,0.125q
2
L2+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·m
M
3=max[q
1
L
1
2/2,q
2
L
1
2/2+pL
1
]=max[2.354×0.22/2,0.113×0.22/2+3.15×0.2]
=0.632kN·m
M
max =max[M
1
,M
2
,M
3
]=max[0.294,0.647,0.632]=0.647kN·m
σ=M
max
/W=0.647×106/54000=11.989N/mm2≤[f]=15.44N/mm2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=1.472kN
V
2=0.625q
2
L+0.688p=0.625×0.113×1+0.688×3.15=2.238kN
V
3=max[q
1
L
1
,q
2
L
1
+p]=max[2.354×0.2,0.113×0.2+3.15]=3.173kN