脚手架稳定性计算
05-15
脚手架稳定性计算
有局部脚手架的搭设高度过高,达到60m ,底部30m 为双立管,现对
其稳定性进行计算。
材料参数 A=4.89cm2 W=5.08cm3
I=12.09cm4 I=1.58cm
L 0=kμh=1.155×1.5×1.8=3.1185
λ= L0/I=3.1185×102/1.58=197.4
查表 ψ=0.186
G K =hi (gk1+gk3)+n1l a g k2
g k1=0.1307KN/m
g k2=0.331 KN/m
g k3=0.0614 KN/m
G K =hi (gk1+gk3)+n1l a g k2
=
风荷载体型系数
μs =1.0ψ, 其中ψ=1.2 则μs =1.2
风压高度变化系数μδ =0.62
W k =0.7×μδ×μs ×W 0
=0.7×0.62×1.2×0.55
=0.286KN/m2
M W =0.85×1.4×W k ×la ×h 2/10
=0.85×1.4×0.286×1.8×1.82/10
=0.198KN.m
考虑风荷载组合
σ= N /Ψa+ MW /W
=20.05×103/0.186×4.89×1.5×100+0.198×106/5.08×103
=147+39
=186N/mm2
∴稳定性满足要求
以上公式右端就是支架的材料抗力R d (=φAf ),当立杆为φ48×3.5
普碳钢管,其A =489mm 2,f =205N/mm2,则有:
R d =100.245φ (kN )