合肥厂房安全检测鉴定专业办理单位
很多顾客都会经常问到一个问题,标准厂房的承受重量是多少?那么严格讲是活荷载,如果货物长期堆放,且不移动的话,在堆放时轻拿轻放,可以考虑按恒荷载衡量能否放置此重量的货物,如若移动,则必须按活荷载考虑。若按一般厂房设计楼板能承受标准荷载是3.5kn/m2。厂房放置设备,要看放置设备本身重量及设备运行频率产生的动荷载决定,同时建议提供结施图及设备安装资料.经结构工程师计算审核后方可做出决定。
根据检测得出的系统偏差,在通过计算,确定是否需要对楼板承重系统进行处理,进而通过理论计算分析提出建议,保证厂房结构的安全稳定。
Many customers often ask a : what is the weight of astandard factory building? So strictly speaking is a live load. Ifthe goods are stacked for a long time and do not move, they can betaken lightly when stacking, and the weight can be measured byconstant load. If they move, it is necessary to consider the liveload. According to the general plant design, the floor canwithstand standard load of 3.5kn/m2. Plant placement e on the dynamic load on the weight of the e itselfand the operating frequency of the e. At the same time, itis recommended to provide the data of the drawing and installationof the e. The decision can be made by the structuralengineer.
According to the system deviation measured, it is determinedwhether the floor load-bearing system should be treated bycalculation, and then the suggestion is put forward throughtheoretical calculation and analysis to ensure the safety andstability of the structure of the factory building.
二、合肥厂房安全检测鉴定专业办理单位
采用MorGain 结构快速设计程序 V2004.15.1162.0。
3.3 验算过程
3.3.1 地基承载力特征值
⑴ 地基承载力特征值 fak = 85kPa 基础宽度的地基承载力修正系数 ηb = 0
基础埋深的地基承载力修正系数 ηd = 1 基础底面以下土的重度 γ = 18kN/m?
基础底面以上土的加权平均重度 γm = 11.7kN/m? 基础底面宽度 b = 2.2m
基础埋置深度 d = 1.65m
当 b = 2.2m < 3m 时,按 b = 3m
⑵ fa = 85+0*18*(3-3)+1*11.7*(1.65-0.5) = 98.5kPa
修正后的地基承载力特征值 fa = 98.5kPa
⑶ 天然地基基础抗震验算时,地基土抗震承载力按《建筑抗震设计规范》
(GB 50011-2001)(式 4.2.3)调整:
faE = ξa * fa = 1*98.5 = 98.5kPa
3.3.2 基本资料
⑴ 柱子高度 hc = 650mm (X 方向) 柱子宽度 bc = 450mm (Y 方向)
⑵ 基础底面宽度 b = 3000mm (X 方向) 底面长度 l = 2200mm (Y 方向)
基础根部高度 H = 500mm 端部高度 h1 = 200mm
⑶ 柱边基础截面面积
X 轴方向截面面积 Acb = h1 * b + (b + hc + 2*0.05) * (H - h1) / 2 = 1.16m?
Y 轴方向截面面积 Acl = h1 * l + (l + bc + 2*0.05) * (H - h1) / 2 = 0.85m?
冲切验算
⑴ 基底净反力 pj
pmax = γz * 138 = 186.3kPa
pj = pmax - G / A = 186.3-353.7/6.6 = 132.8kPa
⑵ X 方向(b 方向)
ab = Min{bc + 2Ho,l} = Min{0.45+2*0.45,2.2} = 1.350m
amx = (bc + ab) / 2 = (0.45+1.35)/2 = 0.900m
Flx = pj * Alx = 132.8*1.41 = 187.8kN
0.7 * βhp * ft * amx * Ho / γRE = 0.7*1*1271*0.9*0.45/0.85 = 423.9kN
≥ Flx = 187.8kN,满足要求。