挡土墙设计

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如图所示,路肩式挡土墙为M7.5墙砌片石圬工,墙与地基土之间的摩擦系数为μ = 0.4 ,地基为粘性土质,容许承载力 [σ]土 = 200 kN/m3,墙身圬工容重 γ圬 = 22 kN/ m3,墙后填土γ土 = 17 kN/ m3, φ = 35°,墙背摩擦角 δ =

φ, 墙高 H = 5m ,试确定该墙尺寸。(墙身截面 [σ]圬 =

1300 kPa, [τ]圬 = 210 kPa)

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解:

⑴ 求破裂角根据θ

根据已知条件得:

墙背倾斜角 α = arctan0.25 = 14°02′

Ψ = φ + δ﹣α = 35°+ 23°20′﹣14°02′= 44°18′

假设破裂面交于荷载分布范围以外,计算边界条件系数:

A0 =

H2 =

×52 = 12.5

B0 =

H2tanα﹣l0h0 =

×52×0.25﹣3.5×3.4 =﹣8.775

tanθ =﹣tanψ ±

=﹣0.9759 + 0.8115 =﹣0.1644

故破裂角θ =﹣9°20′

计算结果显然与原假设不符合。故重新假设破裂面交于荷载分布

范围内,计算边界条件系数:

A0 =

H(H+2h0) =

×5×(5+2×3.4) = 29.5

B0 =

H(H+2h0) tanα + dh

=

×5×(5+2×3.4)×0.25+0.15×3.4 = 7.885

tanθ =﹣tan ±

=﹣0.9759 +

= 0.7529

故破裂角θ = 36°58′

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校核假定: Htanθ = 5×0.7529 = 3.76m

Htanα + d = 5×0.25 + 0.15 = 1.4m

Htanα + d + l0 = 5×0.25 + 0.15 + 3.5 = 4.9m

故: Htanθ + d < Htanθ < Htanα + d + l0

即破裂面交于荷载分布范围内,与假设符合。

⑵ 求土压力系数λa、土压力Ea及土压力作用点

λa = (tanθ﹣tanα)

= (0.7529﹣0.25) ° ′ °

° °

= 0.1575

式中,λa 计算使用的tanθ 值,是通过前面试算后确定的。

Ea = γ(A0tanθ﹣B0)

= 17(29.5×0.7529﹣7.885)

= 76.28 kN/m

Ex = Ea cos(δ﹣α)

= 76.28×cos(23°20′﹣14°02′)

= 75.28 kN/m

Ey = Ea sin(δ﹣α)

= 76.28×sin(23°20′﹣14°02′)

= 12.33 kN/m

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h1 =

﹣ =

﹣ 0.3m

h2 = H﹣h1 = 5﹣0.3 = 4.7m

Zx = (H3+3h0h22)/(3(H2+2h0h2))

= (53+3×3.4×4.72)/(3(52+2×3.4×4.7))

= 2.05m

Zy = B+ Zx tanα = 1.45 + 2.05×0.25 = 1.96 m

⑶ 如需绘制应力图形求土压力时,还应计算:

σ0 = γh0λa = 17×3.4×0.1575 = 9.1 kPa

σh = γHλa = 17×5×0.1575 = 13.39 kPa

Ea = σ0 h2 +

σh×H = 9.1×4.7 +

= 76.25 kN/m

1、挡土墙稳定性验算

⑴抗滑稳定性验算:

应满足: (0.9G +γ Q1Ey) μ + 0.9Gtanα0 ≥ γ Q1Ex

(0.9G +γ Q1Ey) μ + 0.9Gtanα0

= (0.9×159.5 +1.4×12.33) 0.4 + 0

= 64.32 kN/m ≤ γ Q1Ex = 1.4×75.28 = 105.39 kN/m

故不满足抗滑稳定性要求。

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⑵抗倾覆稳定性验算:

应满足: 0.9G ZG + γ Q1(Ey Zx﹣Ex Zy) μ > 0

0.9G ZG + γ Q1(Ey Zx﹣Ex Zy) μ

= 0.9×159.5×1.35 + 1.4×(12.33×2.05﹣75.28×1.96)

= 22.61 kN > 0

故满足抗倾覆稳定性要求。

2、基底应力及偏心矩验算

⑴基底合力偏心矩:

ZN = (G ZG + Ey Zy﹣Ex Zx)/(G + Ey)

= (159.5×1.35+12.33×1.96﹣75.28×2.05)/ (159.5+12.33)

= 0.50 m

e =

﹣ZN = 1.45/2﹣0.50 = 0.225 m >

= 0.18 m

故基底合力偏心矩不满足要求。

⑵基底边缘最大、最小压应力设计值:

当e = 0.225 m≤

= 0.24 m时,

N1 =γ GG +γ Q1Ey﹣W = 0.9×159.5+1.4×12.33 = 160.81 kN

pmax =

(1 +

=

(1 +

214.15 kPa < 1.2[σ]土 = 240 kN/m3

pmin =

(1﹣

=

(1﹣

7.65 kPa < 1.2[σ]土 = 240 kN/m3

故基底压应力设计值满足要求。

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3、墙身截面强度验算

⑴强度计算

NG = 159.5 + 5×0.25×5×17/2 = 212.625 kN

NQ1 = Ey = 12.33 kN

αk =

=

= 0.776

αkARk/γk = 0.776×1.45×7500/2.31 = 3652 kN

Nj = γ0(γGNG + γ Q1 NQ1 +∑γ Qi ψ Ci NQi)

= 1.3×(0.9×212.625+1.4×12.33+0)

= 271.21 kN < αkARk/γk = 3652 kN

故满足要求。

⑵稳定计算

弯曲平面内的纵向翘曲系数:

Ψk =

=

= 6.90

=

= 0.842

=

= 3076.03 kPa

Nj = 229.49 kN <

= 3076.03 kPa

故满足要求。

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⑶正截面直接受剪时验算

根据规范,M7.5墙砌片石一般不用设计抗剪强度和设计弯曲抗拉强度,故不需要验算其正截面直接受剪。

综上所述,挡土墙的抗滑稳定性不满足要求,应增加抗滑稳定性。可采取如下措施:设置向内倾斜的基底,取基底倾角为 ° ′,即为1: 3。此外,还应验算通过墙踵的地基水平面I-I的滑动稳定性。

下图为挡土墙的排水布置: