当前位置:文档之家› 避雷针计算书

避雷针计算书

避雷针计算一.设计条件:1.计算依据《钢结构设计规范》GB50017-2003《变电站建筑结构设计技术规定》NDGJ96-92《建筑地基基础设计规范》GB 50007-2002《建筑结构荷载规范》GB 50009-2001(2006年版)《建筑抗震设计规范》GB 50011-2008《变电构架设计手册》2.独立避雷针荷载计算:H=35m,第一段高度h1=7300mm,采用钢管Φ580/Φ490x10,平均直径Φ535,N=9.5 kN第二段高度h2=7000mm,采用钢管Φ490/Φ390x8,平均直径Φ440,N=6 kN第三段高度h3=7000mm,采用钢管Φ390/Φ290x7,平均直径Φ340,N=5 kN第四段高度h4=7000mm,采用钢管Φ290/Φ190x6,平均直径Φ240,N=2.5 kN第五段高度h5=2400mm,采用钢管Φ152x4,N=0.5 kN第六段高度h6=1950mm,采用钢管Φ133x4,N=0.4 kN第七段高度h7=1600mm,采用钢管Φ114x4,N=0.3 kN第八段高度h5=1050mm,采用钢管Φ95x3,N=0.2 kN按各段高度及外径求得加权平均外径为:D=(7300×535+7000×440+7000×340+7000×240+2400×152+1950×133+1600×114+1050×95)÷(7300+7000×3+2400+1950+1600+1050)=339mm(实际取用364mm偏于安全)风荷载计算:按《建筑结构荷载规范》(GB 50009-2001)(2006版)查得ω0=0.60kN/m2,风荷载标准值:ωk=βz.μs.μz.ω0风振系数:单钢管柱(h>8m),βz=2.0风压高度变化系数μz:h=35m查《建筑结构荷载规范》(GB 50009-2001)表7.2.1(B类)插值得:μz=1.42+(1.56-1.42)×5÷(40-30)=1.49风荷载体型系数μs:μzω0.d2=1.49×0.60×0.3642=0.118>0.015,取μs=+0.6ωk=βz.μs.μz.ω0=2.0×0.6×1.49×0.60=1.073kN/m2作用于各段钢管的风荷载标准值:第一段钢管Φ580/Φ490x10,q1=ωk xD=1.073×0.535=0.574 kN/m第二段钢管Φ490/Φ390x8,q2=ωk xD=1.073×0.44=0.472 kN/m第三段钢管Φ390/Φ290x8,q3=ωk xD=1.073×0.34=0.365kN/m第四段钢管Φ290/Φ190x6,q4=ωk xD=1.073×0.24=0.258 kN/m第五段钢管Φ152x4,q5=ωk xD=1.073×0.152=0.163 kN/m第六段钢管Φ133x4,q6=ωk xD=1.073×0.133=0.143 kN/m第七段钢管Φ114x4,q7=ωk xD=1.073×0.114=0.122 kN/m第八段钢管Φ95x3,q8=ωk xD=1.073×0.095=0.102 kN/m二、内力分析各段钢管底风荷载标准值:1)剪力第八段钢管Q k8=0.102×1.05=0.107 kN第七段钢管Q k7=0.107+0.122×1.60=0.107+0.195=0.302 kN第六段钢管Q k6=0.302+0.143×1.95=0.302+0.279=0.581 kN第五段钢管Q k5=0.581+0.163×2.40=0.581+0.391=0.972 kN第四段钢管Q k4=0.972+0.258×7=0.972+1.806=2.778 kN第三段钢管Q k3=2.778+0.365×7=2.778+2.555=5.333 kN第二段钢管Q k2=5.333+0.472×7=5.333+3.304=8.637 kN第一段钢管Q k1=8.637+0.574×7.3=8.637+4.19=12.827 kN2)弯矩第八段钢管M k8=0.5×1.05×0.107=0.056 kNm第七段钢管M k7=0.056+0.107×1.6+0.5×1.6×0.195=0.056+0.171+0.156=0.383 kNm 第六段钢管M k6=0.056+0.107×(1.6+1.95)+0.156+0.195×1.95+0.5×1.95×0.279=0.056+0.38+0.156+0.38+0.272=1.244 kNm第五段钢管M k5=0.056+0.107×(1.6+1.95+2.40)+0.156+0.195×(1.95+2.40)+0.272+0.279×2.40+0.5×2.4×0.391=0.056+0.637+0.156+0.85+0.272+0.67+0.469=3.574 kNm 第四段钢管M k4=0.056+0.107×(1.6+1.95+2.40+7)+0.156+0.195×(1.95+2.40+7)+0.272+0.279×(2.40+7)+ 0.469+0.391×7+0.5×7×1.806=0.056+1.386+0.156+2.213+0.272+2.623+0.469+2.734+6.321=16.23 kNm第三段钢管M k3=0.056+0.107×(1.6+1.95+2.40+7+7)+0.156+0.195×(1.95+2.40+7+7)+0.272+0.279×(2.40+7+7)+ 0.469+0.391×(7+7)+6.321+1.806×7+0.5×7×2.555=0.056+2.135+0.156+3.578+0.272+4.576+0.469+5.474+6.321+12.642+8.943=44.622 kNm第二段钢管M k2=0.056+0.107×(1.6+1.95+2.40+7+7+7)+0.156+0.195×(1.95+2.40+7+7+7)+0.272+0.279×(2.40+7+7+7)+ 0.469+0.391×(7+7+7)+6.321+1.806×(7+7)+8.943+2.555×7+0.5×7×3.304=0.056+2.884+0.156+4.943+0.272+6.529+0.469+8.211+6.321+25.284+8.943+17.885+11.564=95.517 kNm第一段钢管M k1=0.056+0.107×(1.6+1.95+2.40+7+7+7+7.3)+0.156+0.195×(1.95+2.40+7+7+7+7.3)+0.272+0.279×(2.40+7+7+7+7.3)+ 0.469+0.391×(7+7+7+7.3)+6.321+1.806×(7+7+7.3)+8.943+2.555×(7+7.3)+11.564+3.304×7.3+0.5×7.3×4.19=0.056+3.665+0.156+6.367+0.272+8.565+0.469+11.065+6.321+38.468+8.943+36.537+11.564+24.119+15.294=171.862 kNm3)轴力第八段钢管N k8=0.2kN第七段钢管N k7=0.2+0.3=0.5kN第六段钢管N k6=0.5+0.4=0.9kN第五段钢管N k5=0.9+0.5=1.4kN第四段钢管N k4=1.4+2.5=3.9kN第三段钢管N k3=3.9+5=8.9kN第二段钢管N k2=8.9+6=14.9kN第一段钢管N k1=14.9+9.5=24.4kN三、钢管截面特性计算(按平均截面计算)第一段钢管Φ580/Φ490x10, 平均直径Φ535的截面特性W x=W y=π(d4-d41)/(32d)=3.141592×(5354-5154)÷(32×535)=2125061.3mm3 i x=i y=(d2+d21)0.5/4=(5352+5152)0.5÷4=185.7mm185.8A=π(d2-d21) /4=3.141592×(5352-5152) ÷4=16493.3 mm2第二段钢管Φ490/Φ390x8, 平均直径Φ440的截面特性I x=I y=π(d4-d41)/64=3.141592×(4404-4244)÷64=253366931.8mm4W x=W y=π(d4-d41)/(32d)=3.141592×(4404-4244)÷(32×440)=1151667.9mm3 i x=i y=(d2+d21)0.5/4=(4402+4242)0.5÷4=152.8mmA=π(d2-d21) /4=3.141592×(4402-4242) ÷4=10857.3 mm2第三段钢管Φ390/Φ290x8, 平均直径Φ340的截面特性I x=I y=π(d4-d41)/64=3.141592×(3404-3244)÷64=115031326.3mm4W x=W y=π(d4-d41)/(32d)=3.141592×(3404-3244)÷(32×340)=676654.9mm3 i x=i y=(d2+d21)0.5/4=(3402+3242)0.5÷4=117.4mmA=π(d2-d21) /4=3.141592×(3402-3242) ÷4=8344.1 mm2第四段钢管Φ290/Φ190x6, 平均直径Φ340的截面特性I x=I y=π(d4-d41)/64=3.141592×(2404-2284)÷64=30209536.1mm4W x=W y=π(d4-d41)/(32d)=3.141592×(2404-2284)÷(32×240)=251746.1mm3 i x=i y=(d2+d21)0.5/4=(2402+2282)0.5÷4=82.8mmA=π(d2-d21) /4=3.141592×(2402-2242) ÷4=5830.8 mm2第五段钢管Φ152×4截面特性I x=I y=π(d4-d41)/64=3.141592×(1524-1444)÷64=5095913.6mm4W x=W y=π(d4-d41)/(32d)=3.141592×(1524-1444)÷(32×152)=67051.5mm3 i x=i y=(d2+d21)0.5/4=(1522+1442)0.5÷4=52.3mmA=π(d2-d21) /4=3.141592×(1522-1442) ÷4=1859.8 mm2第六段钢管Φ133x4截面特性I x=I y=π(d4-d41)/64=3.141592×(1334-1254)÷64=3375252.6mm4W x=W y=π(d4-d41)/(32d)=3.141592×(1334-1254)÷(32x133)=50755.7mm3i x=i y=(d2+d21)0.5/4=(1332+1252)0.5÷4=45.6mmA=π(d2-d21) /4=3.141592×(1332-1252) ÷4=1621 mm2第七段钢管Φ114x4截面特性W x=W y=π(d4-d41)/(32d)=3.141592×(1144-1064)÷(32×114)=36728mm3i x=i y=(d2+d21)0.5/4=(1142+1062)0.5÷4=38.9mmA=π(d2-d21) /4=3.141592×(1142-1062) ÷4=1382.3 mm2第八段钢管Φ95x3截面特性I x=I y=π(d4-d41)/64=3.141592×(954-894)÷64=918345.5mm4W x=W y=π(d4-d41)/(32d)=3.141592×(954-894)÷(32×95)=193333.6mm3i x=i y=(d2+d21)0.5/4=(952+892)0.5÷4=32.5mmA=π(d2-d21) /4=3.141592×(952-892) ÷4=867.1mm2四、强度验算第一段钢管N/A+M x/(γx W x)=1.2×24.4×1000÷16493.3+1.4×171.862×1000000÷(1.15×2125061.3)=1.78+98.46=100.24N/m m2<215×0.7=150.5 N/mm2N/A-M x/(γx W x)=24.4×1000÷16493.3-1.4×171.862×1000000÷(1.15×2125061.3) =1.48-98.46=-96.98N/m m2<215×0.7=150.5 N/mm2第二段钢管N/A+M x/(γx W x)=1.2×14.9×1000÷10857.3 +1.4×95.517 ×1000000÷(1.15×1151667.9)=1.65+100.97=102.61N/m m2<215×0.7=150.5 N/mm2N/A-M x/(γx W x)= 14.9×1000÷10857.3 -95.517 ×1000000÷(1.15×1151667.9)=1.37-72.12=-70.75N/m m2<215×0.7=150.5 N/mm2第三段钢管N/A+M x/(γx W x)= 1.2×8.9×1000÷8344.1 +1.4×44.622 ×1000000÷(1.15×676654.9)=1.28+80.28=81.56N/m m2<215×0.7=150.5 N/mm2N/A-M x/(γx W x)= 8.9×1000÷8344.1 -44.622×1000000÷(1.15×676654.9)=1.07-57.34=-56.27N/m m2<215×0.7=150.5 N/mm2第四段钢管N/A+M x/(γx W x)= 1.2×3.9×1000÷5830.8 +1.4×16.23×1000000÷(1.15×251746.1) =0.8+78.48=79.28N/m m2<215×0.7=150.5 N/mm2N/A-M x/(γx W x)= 3.9×1000÷5830.8 -16.23×1000000÷(1.15×251746.1)=0.67-56.06=-55.39N/m m2<215×0.7=150.5 N/mm2第五段钢管N/A+M x/(γx W x)= 1.2×1.4×1000÷1859.8 +1.4×3.574×1000000÷(1.15×67051.5) =0.9+64.89=65.79N/m m2<215×0.7=150.5 N/mm2N/A-M x/(γx W x)= 1.4×1000÷1859.8-1.4×3.574×1000000÷(1.15×67051.5)=0.75-64.89=-64.14N/m m2<215×0.7=150.5 N/mm2第六段钢管N/A+M x/(γx W x)= 1.2×0.9×1000÷1621+1.4×1.244×1000000÷(1.15×50755.7)=0.67+29.84=30.51N/m m2<215×0.7=150.5 N/mm2N/A-M x/(γx W x)= 0.9×1000÷1621-1.4×1.244×1000000÷(1.15×50755.7)=0.56-29.84=-29.28N/m m2<215×0.7=150.5 N/mm2第七段钢管N/A+M x/(γx W x)= 1.2×0.5×1000÷1382.3+1.4×0.383×1000000÷(1.15×36728)=0.43+12.69=13.12N/m m2<215×0.7=150.5 N/mm2N/A-M x/(γx W x)= 0.5×1000÷1382.3-1.4×0.383×1000000÷(1.15×36728)=0.36-12.69=-12.33N/m m2<215×0.7=150.5 N/mm2第八段钢管设计值作用下:N/A+M x/(γx W x)= 1.2×0.2×1000÷1382.3+1.4×0.383×1000000÷(1.15×36728)=0.17+12.69=12.86N/m m2<215×0.7=150.5 N/mm2N/A-M x/(γx W x)= 0.2×1000÷1382.3-1.4×0.383×1000000÷(1.15×36728)=0.14-12.69=-12.55N/m m2<215×0.7=150.5 N/mm2设计值作用下:N/A+M x/(γx W x)= 1.2×0.2×1000÷1382.3+0.383×1000000÷(1.15×36728)=0.17+9.07=9.24N/mm2<80 N/mm2N/A-M x/(γx W x)= 0.2×1000÷1382.3-1.4×0.383×1000000÷(1.15×36728)=0.14-12.69=-12.55N/mm2<80 N/mm2五、稳定性验算第一段钢管1)平面内的稳定性等效长度计算系数 K=1+M 1/M 2=1+95.517÷171.862=1.556注:(M 1为钢管上部弯矩;M 2为钢管下部弯矩)λx =Kl/i x =1.556×7300÷185.7=61.17<150,查得x φ=0.8158147131)17.61.116493.3/(1206000141592.3)1.1/(2222'=⨯⨯⨯==x Ex EA N λπmkN m kN N N W M A N Ex x x x mx /215/92.10074.9818.2)8147131244002.18.01(2125061.315.11000000862.1710.14.13.16493815.0244002.1)8.01(φ'1x <=+=⨯⨯-⨯⨯⨯⨯⨯+⨯⨯=-+=γβσ 2)平面外的稳定性mkN m kN W M A N x x tx /215/43.8125.7918.22125061.30.11000000862.1710.14.17.03.16493815.0244002.1φφ1b x ≤=+=⨯⨯⨯⨯⨯+⨯⨯=+βη 第二段钢管1)平面内的稳定性等效长度计算系数 K=1+M 1/M 2=1+44.622÷95.517=1.467注:(M 1为钢管上部弯矩;M 2为钢管下部弯矩)λx =Kl/i x =1.467x7000÷152.8=67.21<150,查得x φ=0.7854442507)21.67 /(1.110857.3206000141592.3)1.1/(2222'=⨯⨯⨯==x Ex EA N λπmkN m kN N N W M A N Ex x x x mx /215/4.1033.10110.2)4442507149002.18.01(1151667.915.11000000 95.5170.14.1 10857.3785.0149002.1)8.01(φ'1x <=+=⨯⨯-⨯⨯⨯⨯⨯+⨯⨯=-+=γβσ 2)平面外的稳定性mkN m kN W M A N x x tx /215/37.8327.8110.29.15166710.11000000517.950.14.17.03.10857785.0149002.1φφ1b x ≤=+=⨯⨯⨯⨯⨯+⨯⨯=+βη 第三段钢管1)平面内的稳定性等效长度计算系数 K=1+M 1/M 2=1+16.23/44.622=1.36注:(M 1为钢管上部弯矩;M 2为钢管下部弯矩)λx =Kl/i x =1.36x7000÷117.4=81.09<150,查得x φ=0.7042345411)09.81 /(1.18344.1206000141592.3)1.1/(2222'=⨯⨯⨯==xEx EA N λπ mkN m kN N N W M A N Ex x x x mx /215/39.8257.8082.1)234541189002.18.01(9.76654615.11000000 622.440.14.1 1.3448704.089002.1)8.01(φ'1x <=+=⨯⨯-⨯⨯⨯⨯⨯+⨯⨯=-+=γβσ 2)平面外的稳定性mkN m kN W M A N x x tx /215/42.666.6482.19.6766540.11000000622.440.14.17.01.8344704.089002.1φφ1b x ≤=+=⨯⨯⨯⨯⨯+⨯⨯=+βη 第四段钢管1)平面内的稳定性等效长度计算系数 K=1+M 1/M 2=1+3.574÷16.23=1.22注:(M 1为钢管上部弯矩;M 2为钢管下部弯矩)λx =Kl/i x =1.22x7000÷82.8=103.14<150,查得x φ=0.563102222'104.3)563.0 /(1.18.8305206000141592.3)1.1/(⨯=⨯⨯⨯==x Ex EA N λπmkN m kN N N W M A N Ex x x x mx /215/91.7948.7843.1)104.339002.18.01(1.25174615.11000000 23.160.14.18.5830563.039002.1)8.01(φ10'1x <=+=⨯⨯⨯-⨯⨯⨯⨯⨯+⨯⨯=-+=γβσ 2)平面外的稳定性mkN m kN W M A N x x tx /215/37.6418.6319.11.2517460.1100000023.160.14.17.08.5830563.039002.1φφ1b x ≤=+=⨯⨯⨯⨯⨯+⨯⨯=+βη 根据上述结构计算,第五、第六、第七、第八段平面内及平面外都满足要求。

相关主题