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高聳立柱三面翻廣告牌設(shè)計(jì)畢業(yè)論文(參考版)

2025-06-30 17:00本頁(yè)面
  

【正文】 ω * βl * fcc * Al = *3***160000 = 5847223N  Ab = (400+2*400)^2 = 1440000mm  Ab = (d + 2 * c) ^ 2  混凝土局部受壓面積 Al = d ^ 2 = 160000mm 樁局部受壓承載力計(jì)算:  ≥ F = 1215000N,滿足要求。 βl = Sqr(Ab / Al) = Sqr(27040000/7840000) =   c = Min{Cx, Cy, bx, by} = Min{1200,1200,2800,2800} = 1200mm  承臺(tái)在柱下局部受壓時(shí)的計(jì)算底面積按下列公式計(jì)算:  局部荷載設(shè)計(jì)值 F = 1215000N    βx = / (λx + ) = (+) =   ax = 2000 = 20002800/2400/2 = 400mm  (基礎(chǔ)規(guī)范 )   承臺(tái)斜截面受剪承載力按下列公式計(jì)算:  扣除承臺(tái)及其上填土自重后 Y 方向斜截面的最大剪力設(shè)計(jì)值:    βy = / (λy + ) = (+) =   ay = 2000 = 20002800/2400/2 = 400mm  (基礎(chǔ)規(guī)范 )  承臺(tái)斜截面受剪承載力按下列公式計(jì)算:  扣除承臺(tái)及其上填土自重后 X 方向斜截面的最大剪力設(shè)計(jì)值: 承臺(tái)斜截面受剪承載力計(jì)算:  = 2982444N ≥ = 1755518N,滿足要求。 [β1x * (c2 + a1y / 2) + β1y * (c1 + a1x / 2)] * βhp * ft * ho  樁內(nèi)邊緣到承臺(tái)外邊緣的水平距離:  (基礎(chǔ)規(guī)范 )  λ1y = a1y / ho = 400/(1200110) =   Y 方向上自樁內(nèi)邊緣到最近柱邊的水平距離:  (基礎(chǔ)規(guī)范 )  λ1x = a1x / ho = 400/(1200110) =   X 方向上自樁內(nèi)邊緣到最近柱邊的水平距離:  Nl ≤ [β1x * (c2 + a1y / 2) + β1y * (c1 + a1x / 2)] * βhp * ft * ho  Nl = Nmax = 1595925N (2)、角樁對(duì)承臺(tái)的沖切計(jì)算:  = 28631467N ≥ Fl = 1215000N,滿足要求。 2 * [βox * (bc + aoy) + βoy * (hc + aox)] * βhp * ft * ho  (基礎(chǔ)規(guī)范 )  λoy = aoy / ho = 400/(1200110) =   Y 方向上自柱邊到最近樁邊的水平距離:  (基礎(chǔ)規(guī)范 )  λox = aox / ho = 400/(1200110) =   X 方向上自柱邊到最近樁邊的水平距離:  Fl ≤ 2 * [βox * (bc + aoy) + βoy * (hc + aox)] * βhp * ft * ho  Fl = 1215000N (1)、柱對(duì)承臺(tái)的沖切計(jì)算:  (As = 9450)   As,min = 9360mm ρ = %  ①號(hào)筋 Asx = 1353mm Myct = Max{MyctL, MyctR} = = (+)*() = 柱右邊緣 MyctR = (N3 + N6) * (Sa )  = (+)*() = 柱左邊緣 MyctL = (N1 + N4) * (Sa )  (As = 9450)  As,min = 9360mm ρ = %  ②號(hào)筋 Asy = 8147mm Mxct = Max{MxctU, MxctD} = = (++)*()/2 = 柱下邊緣 MxctD = (N1 + N2 + N3) * (Sb bc) / 2 = (++)*()/2 = 柱上邊緣 MxctU = (N4 + N5 + N6) * (Sb bc) / 2  N6 = *() =   N4 = *() =   N2 = *() =   Ni = γz * (Qik Qgk)  Qgk = Gk / n = =   = ≤ = 1320kN  Q6k = (Fk + Gk) / n Mxk * Yi / ∑Yi ^ 2 + Myk * Xi / ∑Xi ^ 2  = (10916*4/2)/(6*4^2/4)  = ≤ = 1320kN  Q4k = (Fk + Gk) / n Mxk * Yi / ∑Yi ^ 2
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