云题海 - 专业文章范例文档资料分享平台

当前位置:首页 > 立柱三面翻广告牌设计计算书

立柱三面翻广告牌设计计算书

  • 62 次阅读
  • 3 次下载
  • 2026/4/30 4:21:55

Q2k = (Fk + Gk) / n + Mxk * Yi / ∑Yi ^ 2 = 285.2+(10916*4/2)/(6*4^2/4) = 1194.9kN ≤ 1.2Ra = 1320kN

Q3k = (Fk + Gk) / n + Mxk * Yi / ∑Yi ^ 2 + Myk * Xi / ∑Xi ^ 2 = 285.2+(10916*4/2)/(6*4^2/4)+(980*2)/(4*2^2) = 1317.4kN ≤ 1.2Ra = 1320kN

Q4k = (Fk + Gk) / n - Mxk * Yi / ∑Yi ^ 2 - Myk * Xi / ∑Xi ^ 2 = 285.2-(10916*4/2)/(6*4^2/4)-(980*2)/(4*2^2) = -747.0kN ≤ 1.2Ra = 1320kN Q5k = (Fk + Gk) / n - Mxk * Yi / ∑Yi ^ 2 = 285.2-(10916*4/2)/(6*4^2/4) = -624.5kN ≤ 1.2Ra = 1320kN

Q6k = (Fk + Gk) / n - Mxk * Yi / ∑Yi ^ 2 + Myk * Xi / ∑Xi ^ 2 = 285.2-(10916*4/2)/(6*4^2/4)+(980*2)/(4*2^2) = -502.0kN ≤ 1.2Ra = 1320kN

每根单桩所分配的承台自重和承台上土自重标准值 Qgk: Qgk = Gk / n = 811.2/6 = 135.2kN

扣除承台和其上填土自重后的各桩桩顶相应于荷载效应基本组合时的竖向力设计值:

Ni = γz * (Qik - Qgk)

N1 = 1.35*(1072.4-135.2) = 1265.2kN N2 = 1.35*(1194.9-135.2) = 1430.6kN N3 = 1.35*(1317.4-135.2) = 1595.9kN

N4 = 1.35*(-747-135.2) = -1190.9kN N5 = 1.35*(-624.5-135.2) = -1025.6kN N6 = 1.35*(-502-135.2) = -860.2kN (2)、X 轴方向柱边的弯矩设计值:(绕 X 轴)

柱上边缘 MxctU = (N4 + N5 + N6) * (Sb - bc) / 2 =

(-1190.9+-1025.6+-860.2)*(4-2.8)/2 = -1846.0kN·m 柱下边缘 MxctD = (N1 + N2 + N3) * (Sb - bc) / 2

= (1265.2+1430.6+1595.9)*(4-2.8)/2 = 2575.0kN·m

Mxct = Max{MxctU, MxctD} = 2575.0kN·m ②号筋 Asy = 8147mm

ζ = 0.031 ρ = 0.15%

47Φ16@110 (As

ρmin = 0.15% As,min = 9360mm= 9450)

(3)、Y 轴方向柱边的弯矩设计值:(绕 Y 轴)

柱左边缘 MyctL = (N1 + N4) * (Sa - 0.5hc)

= (1265.2+-1190.9)*(2-2.8/2) = 44.6kN·m

柱右边缘 MyctR = (N3 + N6) * (Sa - 0.5hc)

= (1595.9+-860.2)*(2-2.8/2) = 441.5kN·m

Myct = Max{MyctL, MyctR} = 441.5kN·m ①号筋 Asx = 1353mm

ζ = 0.005 ρ = 0.02%

47Φ16@110 (As

ρmin = 0.15% As,min = 9360mm= 9450)

2、承台受冲切承载力计算: (1)、柱对承台的冲切计算:

扣除承台及其上填土自重,作用在冲切破坏锥体上的冲切力设计值: Fl = 1215000N

柱对承台的冲切,可按下列公式计算:

Fl ≤ 2 * [βox * (bc + aoy) + βoy * (hc + aox)] * βhp * ft * ho (基础规范 8.5.17-1) X 方向上自柱边到最近桩边的水平距离:

aox = 2000 - 0.5hc - 0.5d = 2000-2800/2-400/2 = 400mm λox = aox / ho = 400/(1200-110) = 0.367

X 方向上冲切系数 βox = 0.84 / (λox + 0.2) (基础规范 8.5.17-3) βox = 0.84/(0.367+0.2) = 1.482 Y 方向上自柱边到最近桩边的水平距离:

aoy = 2000 - 0.5bc - 0.5d = 2000-2800/2-400/2 = 400mm λoy = aoy / ho = 400/(1200-110) = 0.367

Y 方向上冲切系数 βoy = 0.84 / (λoy + 0.2) (基础规范 8.5.17-4) βoy = 0.84/(0.367+0.2) = 1.482

2 * [βox * (bc + aoy) + βoy * (hc + aox)] * βhp * ft * ho = 2*[1.482*(2800+400)+1.482*(2800+400)]*0.967*1.433*1090 = 28631467N ≥ Fl = 1215000N,满足要求。

(2)、角桩对承台的冲切计算:

扣除承台和其上填土自重后的角桩桩顶相应于荷载效应基本组合时的竖向力设计值:

Nl = Nmax = 1595925N

承台受角桩冲切的承载力按下列公式计算:

Nl ≤ [β1x * (c2 + a1y / 2) + β1y * (c1 + a1x / 2)] * βhp * ft * ho (基础规范 8.5.17-5)

X 方向上自桩内边缘到最近柱边的水平距离:

a1x = 2000 - 0.5hc - 0.5d = 2000-2800/2-400/2 = 400mm λ1x = a1x / ho = 400/(1200-110) = 0.367

X 方向上角桩冲切系数 β1x = 0.56 / (λ1x + 0.2) (基础规范 8.5.17-6)

β1x = 0.56/(0.367+0.2) = 0.988

Y 方向上自桩内边缘到最近柱边的水平距离:

a1y = 2000 - 0.5bc - 0.5d = 2000-2800/2-400/2 = 400mm λ1y = a1y / ho = 400/(1200-110) = 0.367

Y 方向上角桩冲切系数 β1y = 0.56 / (λ1y + 0.2) (基础规范 8.5.17-7)

β1y = 0.56/(0.367+0.2) = 0.988

桩内边缘到承台外边缘的水平距离:

c1 = c2 = Sc + 0.5d = 600+400/2 = 800mm

搜索更多关于: 立柱三面翻广告牌设计计算书 的文档
  • 收藏
  • 违规举报
  • 版权认领
下载文档10.00 元 加入VIP免费下载
推荐下载
本文作者:...

共分享92篇相关文档

文档简介:

Q2k = (Fk + Gk) / n + Mxk * Yi / ∑Yi ^ 2 = 285.2+(10916*4/2)/(6*4^2/4) = 1194.9kN ≤ 1.2Ra = 1320kN Q3k = (Fk + Gk) / n + Mxk * Yi / ∑Yi ^ 2 + Myk * Xi / ∑Xi ^ 2 = 285.2+(10916*4/2)/(6*4^2/4)+(980*2)/(4*2^2) = 1317.4kN ≤ 1.2Ra = 1320kN Q4k = (Fk + Gk) / n - Mxk * Yi / ∑Yi ^ 2 - Myk * Xi / ∑Xi ^ 2 = 285.2-(10916*4/2)/(6*4^2/

× 游客快捷下载通道(下载后可以自由复制和排版)
单篇付费下载
限时特价:10 元/份 原价:20元
VIP包月下载
特价:29 元/月 原价:99元
低至 0.3 元/份 每月下载150
全站内容免费自由复制
VIP包月下载
特价:29 元/月 原价:99元
低至 0.3 元/份 每月下载150
全站内容免费自由复制
注:下载文档有可能“只有目录或者内容不全”等情况,请下载之前注意辨别,如果您已付费且无法下载或内容有问题,请联系我们协助你处理。
微信:fanwen365 QQ:370150219
Copyright © 云题海 All Rights Reserved. 苏ICP备16052595号-3 网站地图 客服QQ:370150219 邮箱:370150219@qq.com