支撑、系杆计算

支撑、隅撑受压构件长细比限值：支撑、隅撑受拉构件长细比限值：kN/m 2一、工程概况

场地类别：II类

2204000.25工程地点：云南昆明风荷载体型系数按《建筑结构荷载规范》GB50009-2001取值抗震设防烈度：6度,设计基本地震加速度值为0.05g

12.3米：ωk2=μs μ

z3ω0=

X 6.50X P 3=0.24 6.503.5=

14.75X 1.417.621.4kN

X 0.50kN X 1.4+X +0.25+X 1.67X X P 2=0.24X P 1=0.24X 3.50 5.005.00=kN =1.413.51=+

7.0

0.26X 1.4X X 1.4X

6.5X 0.25X 5.00P 4=0.24X

7.00 2.30X 6.50X 0.30X =0.24=X 0.251.10=0.26

0.30X 6.5 1.13地面粗糙度:kN/m 2

基本风压：

类B

0.80kN/m 20.801. 屋面水平支撑计算

(屋面水平荷载由第一道屋面水平支撑承担)

二、厂房支撑、系杆计算简图

风荷载标准值ωk =μs μz ω0

0.30[λ1]=[λ2]=钢材采用Q235钢，

X 1.4

0.30X 1.051.4X 0.80X X （50年一遇）

10.5米 :ωk2=μs μz2ω0=10米以下:ωk1=μs μz1ω0=X 5.00支撑、系杆计算

kN/m 2

1.00N/mm2

215[σ]=f y

[σ]=Aa杆长mm mm

做上柱支撑，mm mm ==

[σ]=215纵向水平荷载mm

则λx =0.9L/i x0=mm mm ==

4.78*b/t*(1+l oy 2*t 2/13.5*b 4)=

查表得φx =

[σ]=N/mm2

纵向水平荷载L=220220

mm mm

角钢做中柱支撑，N/mm 2kN

Ncos α=(R 1-P 1)

斜杆所受力N=所以斜杆所受力N=R B /2cos β=21.78满足要求!

40.9副柱间支撑共同承担长肢相连4.302.2 B/F轴中柱支撑4号杆计算 6.26则R A =7880≤21.780.789

mm 2

215N=(R 1-P 1)/cos α=8.85=选用φ182542. 柱间支撑、系杆计算

故可知传至柱间支撑的荷载R 1+R 1'=13.62在风压力作用下传至柱间支撑的荷载 R 1=0.80.5=X 计算可得支座反力：R 1=21.78

kN R 3= 1.414.75kN X X 6.51.4+0.25kN P 5=0.24X 6.50X 5.00X X 1.4=1.4X 0.503.5

6.5 1.671.13选用i x =kN

X 5.00X 9850

≤13.510.25P 6=0.24X kN

R 2=43.57=

6sin β=选用i x =2L100x6 6.50Ab杆长

2.1 B/F轴上柱支撑2号杆计算

L Aa =9000X 5.00X X 1.4+0.25+X P 7=0.24X 3.50X 6.50 1.49850L Ab =kN

35.40

76.1/kN

(R 1+R 1')/

AB杆长419.38=则在风吸力作用下传至柱间支撑的荷载 R 1'=R 1 X N/mm 2圆钢做水平支撑:

A e =则σ=N/Ae=该力由90009.69=0.914

2386A=i y =40.9mm

所以斜杆所受力N=R A /sin β=mm

kN kN

=2120b 受力最大的斜杆：cos α=

bc杆长22.0

满足要求!

165l oy =

522031.7≤168mm

≤22.0

80t=b 2=/1500选用31.1

38.5sin θ=0.287

R C =15002L100x80x6

则λy =L/i y =5220≤mm2

＞0.54l oy /b

35220

i y =mm 2

31mm

=31.7mm

0.487

1500/b==

56t 18.675220=σ=N/(φx A)=N/mm 2满足要求!

16.67cos β=/400

[λ2]=mm

l oy =

186

≤l oy 4000L AB =31.0则λy =0.8L Ab /i y =则λx=L/i x ==

b /t =7880

193mm

则λx =0.5L Ab /i x =159≤4925/31.0=R B =/8.85≤1908

则λy =0.9L/i y0=1908t=L BC =≤4.06

b=100t=6λyz λy(1+0.475*b 4/(l oy 2*t 2)

σ=N/A=mm

[λ2]=40021204号杆长

i x0=

=1908

6.54kN A=2127mm

i y0=[λ1]=

≤6号杆长

8718

1732202.3 B/F轴下柱支撑6号杆计算L=334mm 2

11.0220220

400≤5220[λ1]=

[λ1]=[λ2]=[λ1]=[λ1]==≤kN

2151110.708

11.046=0.58l oy /b N/mm 2

N/mm 2

56x3=

==≤[σ]=N/mm 2

焊管做系杆，

[σ]=N/mm 2mm

220220

mm mm ==λy

220

[σ]=N/mm 2mm

220220

mm mm ==

220

[σ]=N/mm 2mm mm

做上柱支撑，

mm mm ==

≤N/mm 2

2.4 B/F轴屋顶系杆1号杆计算

43.3kN 查表得φx =0.19

21.06λ

208≤/t /43.3

t=69000σ=N/(φx A)=则λy =L/i y = 2.02N/mm 2b 24=

16.67该力由≤=100在风压力作用下传至柱间支撑的荷载 R 2=/满足要求!

=9000

mm 1250.570.80

(R 2+R 2')/

l oy 24.0

≤43.57=

0.820839

0.48l oy /b 2≤196=

短肢相连

/43.3

≤208≤/24.0=

≤125σ=N/(φx A)=38.8N/mm 2则λx=L/i x =3000Bb杆受拉力选用2L100x80x6215选用φ16.67故可知传至柱间支撑的荷载R 2+R 2'=Ee杆长L Ee =9000EF杆长付柱间支撑共同承担则λx=L/i x =3000则λ=L/i=则λy =0.8L Ab /i y =8568/

40.98568

2L100X6t b 1=σ=N/A φx =11.7≤≤λ

/t 13.339000/46.0

满足要求!

0.48l oy /b 2

mm 2

b 217.70kN 0.186

查表得φx =

27.2341225查表得φx =

则在风吸力作用下传至柱间支撑的荷载 R 2'=R 2 X 8.85kN

则λx =0.5L Ab /i x =R B =

A=满足要求!

46.0系杆Aa受力i=

5355查表得φx =0.262

λy(1+1.09*b 24/(l oy 2*t 2)=

165220

2127i y =则λy =L/i y =133X3

9000≤0.48l oy /b 2mm2

则R E =2.7 D轴上柱支撑8号杆计算

mm 2

2L100X80X6b=≤=

i x =31.0选用=

σ=N/(φx A)=208R c =

=Cc系杆受力A=2386≤mm

4.30/24.0kN

i x =

2127mm 20.19

b 1=100t=69000

l oy b 2/2.6 B/F轴中间系杆5号杆计算选用215100[σ]=

t=6≤0.58l oy /b l oy =

16.6750400≤b /t σ=N/A=

8.8

/31.0=173λ

λy (1+0.475*b 4/(l oy 2*t 2)=202sin β'=9000/10710=0.840

10710

mm 所以斜杆所受力N=R E /sin β'=5800Ef杆长

L Ef =

[λ2]=[λ2]=N/mm 2R A =A=31

i y =2.5 B/F轴中间系杆3号计算43.3

短肢相连

i x =8.85400

L EF ==209≤215λ

21.9N/mm 2≤[λ1]=[λ1]=21524.0kN

220

[λ1]=≤39

≤[λ1]=[λ1]=mm

400i y =40.9[λ2]=mm

[λ1]=[λ1]=[λ1]=215

纵向水平荷载mm

做中柱支撑，

则λx =0.9L/i x0=mm mm ==

4.78*b/t*(1+l oy 2*t 2/13.5*b 4)=

[λ1]=[σ]=N/mm 2

纵向水平荷载L=220220

mm mm

≤[σ]=N/mm 2

屋顶系杆受力kN

133X3A=

[σ]=N/mm 2mm

220220mm mm =λy

[σ]=N/mm 2mm

220

0.19

2.9 D轴下柱支撑12号杆计算满足要求!

17.70

σ=N/(φx A)=N/mm 2≤2084号杆长

220

≤

则λx=L/i x =

3000b 243.8λ

=2.8 D 轴中柱支撑10号杆计算t 16.67Gg系杆受力R G =

6.20kN

2.11 D 轴中间系杆9号杆计算/45.8[λ1]=[λ1]=满足要求!

220

125220

1225mm 2

查表得φx =2.10 D轴屋顶系杆7号杆计算kN

24.0N/mm 2[λ1]==

0.48l oy /b 2=

208mm

/24.0

/选用≤=满足要求!

b 1=100则λy =L/i y =σ=N/(φx A)=43.3

≤

39t=37.3/90006l oy 则λx=L/i x =30002L100X80X6Ff杆受力

则λ=L/i=选用φR F =

17.70R E =

43.3

查表得φx =≤i y =24.0≤131

6000/

45.8=

R G =焊管做系杆，

A=21272120mm 2

0.387

17.70短肢相连

i x =mm ≤5220t=σ=N/(φx A)=38.7N/mm 2≤λyz=λy(1+1.09*b24/(l oy 2*t2)=

165查表得φx =0.262

[λ1]=≤0.48l oy /b 2则λy =L/i y =b 2/t l oy =

13.33b 2=8065220

≤i y =长肢相连/31.1

168mm 2

则λx=L/i x =5220/31.7=

165A=则λy =0.9L/i y0=21.58kN

N/mm 26.20所以斜杆所受力N=31sin θ'=18

b mm

b=kN

3cos γ'=15001500=查表得φx =＞/t 22.0

==87173l oy =

kN 56t=R F =满足要求!

BC杆长0.54l oy /b /λ

选用

1908/

=0.708

L=1908

11.0mm

1908

所以斜杆所受力N=R F /2cos γ'=22.0

18.67L BC =i x0=56x3212012.52kN 12号杆长

5220mm

≤mm 2

111≤0.487

选用2L100X80X6

[λ1]=[λ1]=1500σ=N/(φx A)=77

2127/A=

5220

A=2127mm 2

43.3

=i y ==

选用2L100X80X6短肢相连

24.0i x =215215215i=mm

≤≤[λ1]=2.12 B/F轴中间系杆11号杆计算334220

≤[λ1]=

≤9000mm

=1250.287

[λ1]=

215220

220

[λ1]=i y0=i x =31.7mm

11.0

220mm mm =λy

[σ]=N/mm 2重量（kg/m)

长度（m）总量单方用钢量

65236072953153120925561039用钢量统计

σ=N/(φx A)=15.3N/mm 224

24圆钢Φ181(水平支撑轴线面积为1140m 2）

11.02

11.012.466

≤16.67满足要求!

t =

≤

l oy 39查表得φx =

0.19

208≤

b 2/t=6[λ1]=

220

≤2.466=208数量0.48l oy /b 2=

3.77

2.05

16.718.7416.7245.2218.74

1.143.59

2(柱间支撑轴线面积为3284.1m 2）

241681303.8

（长并）651.62柱支撑

L56X3

99.85(短并）2L100X80X62L100X6

2L100X62L100X80X611778.7

10.21圆管?133X3.0

9.618

215[λ1]=9000mm

则λy =L/i y =9000/43.3

b 1=100