清华大学出版社 土力学地基基础(陈希哲第四版)

σ zA = 2.0 * (α1 − α2 ) * po = 2.0 * (0.1350− 0.0947) * po σ zA / σ zo = 2.0 * (0.1350 − 0.0947) / 4 * 0.1034 = 19.487%
L / b = 6 / 5 = 1.2, z / b = 10 / 5 = 2, α 1 = 0.0947
b b
σ
z
= α ⋅ p o = 0 . 306 * 100 = 30 . 6 kPa
z = 3 .0b
x z α = f ( , ) = f ( 0 ,3 . 0 ) = 0 . 208 b b
σ
z
= α ⋅ p o = 0 . 208 * 100 = 20 . 8 kPa
b/2
b/2
σ min = 50 kPa
在第2层土内中心点下的沉降差
s 21 =
α
7 b 1α 1 ( 0 − 7 b 1 ) − b 1α 1 ( 0 − b 1 ) E s1
= z f (0, ) = b1
⋅ po
1 ( 0 − 7 b1 )
7 b1 f (0, ) b1
条形平均附加应力系数查不到，借用矩形面积上 均布荷载角点下的平均附加应力系数
习 题
土木工程专业（建筑工程方向）
3.1（P134）
σ co = 0 σ ca = γ 1 ⋅ h1 = 18.0 *1.5 = 27.0kPa
′ σ cb = σ ca + γ 2 ⋅ h2 = 27.0 + (19.4 −10) * 3.6 = 60.84kPa ′ σ cc上 = σ cb + γ 3 ⋅ h3 = 60.84 + (19.8 − 10) *1.8 = 78.48kPa ′ σ cc下 = σ cc上 + γ ω ⋅ hω = 78.48 + 10 * (3.6 + 1.8) = 132.48kPa
6.0m
14.0m 5.0m
A 基础中心点下
5.0m
L / b = 7 / 5 = 1 . 4 , z / b = 10 / 5 = 2 . 0 , α = 0 . 1034
σ zo = 4 ⋅ α ⋅ po = 4 * 0.1034 po
M点下：
L / b = 20 / 5 = 4.0, z / b = 10 / 5 = 2, α 1 = 0.1350
设 E s1 ≈ 2 E s 2
s 22
b1 ⋅ p o b1 ⋅ p o = 2 . 168 * = 4 . 336 0 .5 E s1 E s1
b1 ⋅ po b1 ⋅ po ∆ 2 = (4.336 − 1.1306) ⋅ = 3.2054 ⋅ Es1 Es1
直接按条形荷载计算 （计算点位于大边下）： po=300kPa, b=6.0m, z=9.0m的均布荷载 x/b=6.0/6.0=1.0, z/b=9.0/6.0=1.5, α1=0.211, σz1=0.211*300=63.3kPa po=200kPa, b=6.0m, z=9.0m的三角形荷载 x/b= 6.0/6.0=1.0, z/b=9.0/6.0=1.5, α2=0.0.13, σz2=0.13*200=26.0kPa σz= σz1+ σz2 =63.3+26.0=89.30kPa （计算点位于小边下）： x/b=- 6.0/6.0=-1.0, z/b=9.0/6.0=1.5, α2=0.09, σz2=0.09*200=18.0kPa σz= σz1+ σz2 =63.3+18.0=81.30kPa
s 22 =
α
7 b 1α
2 ( 0 − 7 b1 )
− b 1α
2 ( 0 − b1 )
E s2
= z f (0, ) = 2 b1
⋅ po
2 ( 0 − 7 b1 )
7 b1 f (0, ) = 0 . 444 2 b1
s 22
b1 ⋅ p o 7 * 0 .444 − 0 .94 = ⋅ b1 ⋅ p o = 2 .168 * Es2 Es2
3.0m
6.0mБайду номын сангаас0.25m
P=2400kN
100 200
500 9.0m
po=200kPa, b=3.0m, z=9.0m的均布荷载 x/b=0.5, z/b=3.0, α3=0.198, σz3=0.198*200=39.6kPa po=100kPa, b=3.0m, z=9.0m的三角形荷载 x/b=-0.5, z/b=3.0, α4=0.10, σz4=0.10*100=10.0kPa σz= σz1+ σz2 –σz3-σz4= 82.0+48.0-39.6-10.0=80.4kPa
p
o1
=
N 1 − 20 b1
⋅ d
1
po2
N2 2 ⋅ N1 = − 20 ⋅ d 2 = − 20 ⋅ d 1 = p o 1 b2 2 ⋅ b1
在第1层土内中心点下的沉降差
s11 =
z * α 1( 0 − b1 ) E s1
⋅ po
s12 =
z * α 2 ( 0 −b1 ) E s1
⋅ po
6.0m 0.25m
3.0m 100 500
9.0m po=500kPa, b=3.0m, z=9.0m的均布荷载 x/b=0.5, z/b=3.0, α3=0.198, σz3=0.198*500=99.0kPa po=100kPa, b=3.0m, z=9.0m的三角形荷载 x/b=0.5, z/b=3.0, α4=0.10, σz4=0.10*100=10.0kPa σz= σz1+ σz2 –σz3-σz4= 123.0+75.0-99-10.0=89.0kPa
σ o = 100kPa
50
σ z (kPa )
z = 0 .25 b z = 0 . 5b z = 1 .0 b z = 2 .0 b z = 6 .0b
z
48.1 41.0 27.5 15.3 10.4
3.5(P135)
e
0.97 0.965 0.96 0.955 0.95 0.945 0.94 0.935 0.93 0.925 0.92 0
z b1 α 1( 0 − b1 ) = f ( 0 , ) = f ( 0 , ) = 0 . 807 b1 b1 z b1 α 2 ( 0 − b1 ) = f ( 0 , ) = f (0, ) = 0 . 940 2 ⋅ b1 2 ⋅ b1
s 11 =
s 12 =
α 1(0−b
E s1
α
1
α 1( 0 − 7 b )
1
z 7 b1 = 4 f (10 , ) = 4 f (10 , ) = 4 * 0 .0692 = 0 .2768 0 .5b1 0 .5 ⋅ b1
s 21
b1 ⋅ p o 7 * 0 .2768 − 0 .807 = ⋅ b1 ⋅ p o = 1 .1306 * E s1 E s1
3.8(P135)
N1
2N1
d1 b1 b1 γ 1 = 20kN / m3 粉土 −1
a1− 2 = 0.25MPa
2b1
6b1 粘 土
γ 2 = 19 kN / m 3
a1− 2 = 0.50 MPa −1
问两基础的沉降量是否相同？何故？通过整d和b，能否 使两基础沉降量接近？说明有几种方案，并给出评介。
当第四层为强风化岩时：
′ σ cc = σ cb + γ 3 ⋅ h3 = 60.84 + (19.8 − 10) *1.8 = 78.48kPa
天然地面
sc(kPa)
o
27.0
1.5m 3.6m
地下水
素填土 =18.0kN/m 粉土 =18.0kN/m
60.84
a
b
1.8m
中砂土 =19.8kN/m
78.48 132.48
c Z
坚硬整体岩石层
天然地面
sc(kPa)
o
27.0
1.5m 3.6m
地下水
素填土 =18.0kN/m 粉土 =18.0kN/m
60.84
a
1.8m 强风化岩石
中砂土
=19.8kN/m
78.48
b c Z
3.2(P135) 解：σ c = ∑γ i ⋅ hi = 20.1*1.1 + (20.1 −10) * (4.8 −1.1) = 59.48kPa
i =1 n
3.3(P135)
1 1 解： po = (σ max + σ min ) = (150 + 50) = 100kPa 2 2
z =0
x z α = f ( , ) = f ( 0 ,0 ) = 1 .0 b b
σ
z
= α ⋅ p o = 1 . 0 * 100 = 100 . 0 kPa
e～p曲线
中压 缩性 土
50
100
150
200
250
300
p
a1− 2 =
e1 − e2 0.952 − 0.936 = = 0.00016kPa −1 = 0.16 MPa −1 p2 − p1 100
1+ e1 1+ 0.952 Es1−2 = = = 12.2MPa a1−2 0.16
3.6(P135)
3.0m
6.0m 0.25m
P=2400kN
100 200
500 9.0m
po=200kPa, b=9.0m, z=9.0m的均布荷载 x/b=0.5, z/b=1.0, α1=0.410, σz1=0.41*200=82.0kPa po=300kPa, b=9.0m, z=9.0m的三角形荷载 x/b=-0.5, z/b=1.0, α2=0.16, σz2=0.16*300=48.0kPa