避雷针计算书

一、风荷载计算：避雷针整体高度为H+H+H二2+9+9二20m>8m 取0=2123zH管径为申402x93ywd2=1x0.35x0.422=0.0617>0.015z0取卩=0.6sw d二2x0.6x1x0.35x0.42二0.18KN/mkzsz0H管径为0299x92ywd2=1.25x0.35x0.2992=0.0391>0.015z0取卩=0.6sw=0卩ywd二2x0.6x1.35x0.35x0.299二0.16KN/mkzsz0H=2m（避雷针针尖）采用070x5。风荷载统一取w=0.2KN/mk采用MIDAS对该20m避雷针建模，将风荷载以集中活荷载作用于每段杆件中点。NM+ANM+AYWnX18500 40700000+11106 1.15x1068130=40N/mm2<80N.： <f=215N//mm2 /mm2按《变电所建筑结构设计技术规定》（NDGJ96-92）10.2.2条规定，管壁应力不宜超过80%m2（在荷载标准值作用下）。满足要求。二、挠度：独立避雷针挠度：H/100=20000/100=200mm环形构件受弯强度设计值:受弯：402=44.7<3806038060=177.02299=受弯：402=44.7<3806038060=177.02299=33.2<3806038060=177.02受压：402二44.7<2410024100215二112二33.2<2410024100215二112广f=215N/mm2受压：402二44.7<2410024100215二112二33.2<2410024100215二112广f=215N/mm2长度系数:H20.0000863K=―^• = =041I H1 0.000215 °2H1 'N•I!10x0.000215 …n= • 1——= =1.121H2 N•I22.16x0.0000863T2・ 1卩=3.12 |i=3.3721=T=200.57卩xH1 1i1=爲笄=295.6轴心受压稳定系数：九=®・Z=沖•工=2.05>0.215n2 兀E3.14 206000尢n1九if=295.6i'215忘飞~E=3.14•飞206000=3.04>0.215+九2)+n+aX+九2一4X2n查《变电构架设计手册》表A-2得：a2=°965,a3=0・3009=1.189=1.121弯矩作用下钢管截面整体稳定：G299x9)钢管：N'=兀2EA/(.1X213.142x206000x8200/(1.1x295.62)=1732762EyN 卩M 10000 1x1700000010000173276.2+ mx x = 10000173276.29AYW(1-0.8上)1.1x82001.15x577130x(1-0.8xx N' (=28.1N/mm2<f=215N/mm2G299x9)满足要求G402x9)钢管：N'=n2EA/(.l九213.142x206000x8723/(1.1x200.572)=4003757EyN 卩M 22200 1x5700000022200)400375.7丿+ mx 22200)400375.7丿9AyW(1-0.8-^)1.18x111061.15x1068130x(1-0.8xx N' (=50N/mm2<f=215N/mm2G402x9)满足要求压弯构件局部稳定计算G299x9）钢管N2Afc*MCIfbN2Afc*MCIfb100008200x21517x106x29928628.09x104x215=0.143<1G402x9)钢管MC2+—MC2+—AfIb22200157x106x40228723x215 21469.37x104x215=0.262<1满足要求挠度计算:采用midasgen计算，避雷针在风荷载作用下且考虑重力二阶效应，顶点位移&H/100满足要求。三、法兰计算法兰螺栓计算：M=17KN•m N=10KNV=3KNNbtmaxMYZY2i<NbntM=1.7<R=149.5时，取管中心为旋转轴Y=237Y=205Y=1181 2 3工Y2=2372+2xCo52+II82L168067INbtmaxML+N=17°°°%237+10=12.75KN「n 2x16806712螺栓采用6.8级M20Nb=哎fb=245x300=73500N=73.5KN>N=12.75KNt4ttmax满足要求2.法兰板计算l=117l二70xyl70斗= l117x0.6取0=0.0747Nb 1max11=1.6KN/mm2117x70xyM=0ql2=0.0747x1.6x1172=1636Nx板厚t=16mmQ235 f=215Mpa f=125Mpar5满足要求3.肋板计算：肋板厚度t=6mm正应力：=正应力：=37.2N/mm2<f=215N/mm25Nbb5x12750x35max—=——h21 1002x6NNb 1maxht100x6-12750-21.25N/mm2<f-125N/100x6r满足要求4.焊缝计算h=6mm=160Mpa竖向：NbNbQ— 1maxfhlew12750 、_0.7x6x2x(100-2x6)_17.25N/mm2<pfw二1.22x160二195.2N/mm2ft水平：NNbT— t