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某公路桥墩基础采用独立方形基础，基础底面边长为 3m，基础埋深2.5m，拟建场地地下水位距地表 1.0m，地基土层分布及主要物理力学指标见下表。 假如对应于作用的准永久组合时基础底面处的压应力 p = 150kPa，沉降计算深度 znz_n取 6.2m，按按《公路桥涵地基与基础设计规范》(JTG 3363-2019)， 计算该基础中心的最终沉降量为多少?

层序土名层底深度(m)天然重度γ(kN/m3)\gamma(kN/m^3)地基承载力特征值fakf_{ak}压缩模量Es(MPa)E_s(MPa)
①填土1.0018.0
②粉质黏土4.5018.71607.5
③黏土15.019.31809.9
解答
l/b=1,z/b=2/3=0.6667⇒αˉ1=0.816+(0.858−0.816)/(0.7−0.6)×(0.7−0.667)=0.83 l/b=1,z/b=6.2/3=2.067⇒αˉ2=0.429+(0.446−0.429)/(2.1−2)×(2.1−2.067)=0.435 A1=0.83×2=1.66 A2=0.435×6.2−1.66=1.035 Eˉs=1.66+1.0351.667.5+1.0359.9=8.276MPa p0=150−18−1.5×(18.7−10)=118.95kPa<0.75×160=120 ψs=0.4+0.7−0.415−7×(15−8.276)=0.652 s=0.652×[118.957.5×1.66+118.959.9×1.035]=25.27mm l/b = 1, z/b = 2 / 3 = 0.6667 \Rarr \bar \alpha_1 = 0.816 + (0.858-0.816)/(0.7-0.6) × (0.7 - 0.667) = 0.83 \\~\\ l/b = 1, z/b = 6.2 / 3 = 2.067 \Rarr \bar \alpha_2 = 0.429 + (0.446-0.429)/(2.1-2) × (2.1 - 2.067) = 0.435 \\~\\ A_1 = 0.83 × 2 = 1.66 \\~\\ A_2 = 0.435 × 6.2 - 1.66 = 1.035 \\~\\ \bar E_s = \cfrac{1.66 + 1.035}{\cfrac{1.66}{7.5} + \cfrac{1.035}{9.9}} = 8.276 MPa \\~\\ p_0 = 150 - 18 - 1.5 × (18.7-10) = 118.95 kPa < 0.75 × 160 = 120 \\~\\ \psi_s = 0.4 + \cfrac{0.7 - 0.4}{15 - 7} × (15 - 8.276) = 0.652 \\~\\ s = 0.652 × \lbrack \cfrac{118.95}{7.5} × 1.66 + \cfrac{118.95}{9.9} × 1.035 \rbrack = 25.27mm