)}/1.2gkР=0.281×489×0.205-{1.2×0.22+2.25+0.09×1.186×489/5.08}/1.2×0.1248Р =25.51/1.2×0.1288=170 mР则脚手架搭设高度限值[H]=HS/(1+0.001HS)=170/(1=0.001×170) =145mР所以脚手架满足稳定性要求Р3、纵向水平杆,横向水平杆计算Р① 强度验算РM纵=1.2MσK+1.4∑MQKР=1.2×0.08×0.35×1.05÷2×1.22+1.4×3×1.05÷2×1.22×0.08=0.422KN/mРM横=ql2/8=(0.35×1.2×1.2+3×1.2×1.4)1.052/8=0.764KnmРMMAX/W=0.764×103/5.08=150N/mm2<f=205N/mm2满足抗弯强度要求Р② 挠度验算:РV纵=0.677ql4/100EIР=0.677(1.2×0.35×1.05÷2+1.4×3×1.05/2)×1.24×103/100×2.06×105 ×12.19=1.3mmРV横=qal3(-1+4a2/l2+3a3/l3)24EIР=6.93×0.03×1.053 ×(-1+4×0.32÷1.052+3×0.33÷1.053)/24×2.06×105×12.19=Р2mmР则Vmax=2mm<[V]=10mm 满足挠度要求Рa).纵、横向水平杆与立杆连接时,其扣件的抗滑承载力验算РR横=ql(1+a/l)2/2Р=(1.2×0.35×1.2+1.4×3×1.2)×1.05×(1+0.3/1.05)2/2Р=3.4KNРR纵=0.6qlР=0.6(1.2×0.35×1.05/2+1.4×3×1.05/2)×1.2Р=1.8KNРR=3.4KN<8.00KNР满足扣件抗滑承载力要求Р故可按设计要求搭设脚手架。
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