× 2× (18 − 9) = 12 × 10 −7 m 3 / s⏺4-8、某建筑场地的地层分布均匀,第一层杂填土厚 1.5m, γ= 17kN / m 3 ;第二层粉质黏⏺⏺土厚 4m,γ= 19kN / m 3 , G= 2.73 ,ω = 31% ,地下水位在地面下 2m 深处;第三层⏺⏺淤泥质黏土厚 8m , γ= 18.2kN / m 3 , G= 2.74 , ω = 41% ;第四层粉土厚 3m ,⏺⏺γ = 19.5kN / m 3 ,G= 2.72 ,ω = 27% ;第五层砂岩未钻穿。试计算各层交界处的竖向⏺自重应力σ c ,并绘出σ c 沿深度分布图。

解:(1)求γ '⏺' WS − VS γ Wγ (WS − VS γ W )γ (GS ⋅ γ W − γ W )γ ⋅ γ W (GS − 1)γ (GS − 1)⏺γ = =V W=WS + WW=GS γ W+ ωGS γ W=Gs(1 + ω )⏺⏺由上式得:γ '= 9.19kN / m 3 ,γ '= 8.20kN / m 3 ,γ '= 9.71kN / m 3 ,⏺(2)求自重应力分布σ c1 = γ 1 h1 = 1.5 × 17 = 25.5kPa⏺σ = γ h + γ h‘= 25.5 + 19 × 0.5 = 35.0kPa⏺⏺σ c 2= σ c水+ γ ’(4 − h ' ) = 35.0 + 9.19 × 3.5 = 67.17kPa⏺σ = σ + γ ’h = 67.17 + 8.20 × 8 = 132.77 kPaσ = σ + γ ’h = 132.77 + 9.71 × 3 = 161.90kPaσ 4不透水层 = σ c4 + γ W (3.5 + 8.0 + 3.0) = 306.9kPa9、基础埋深为 2m,底面尺寸为 4m×2m。试求基底平均压力p和边缘最大压力pmax,并绘出沿 偏心方向的基底压力分布图。解:(1)全力的偏心距 e(F + G )⋅ e = F ×1.31⏺1.31× 680e = 680 + (4 × 2 × 2 × 20)= 0.891m⏺⏺(2) pmaxF + G ⎛= ⎜1 ±6e ⎞⎟⏺minA ⎝ l ⎠⏺⎛ e ⎞ ⎛ × ⎞⏺因为 ⎜1 ± 6⎟ = ⎜1 ± 60.891⎟ = 1 ± 1.337 出现拉应力⏺⎝ l ⎠ ⎝ 4 ⎠⏺故需改用公式 pmax =2(F + G )2(680 + 4 × 2 × 20)== 301kPa⏺3b⎜ l− e ⎟