ページ3:

EX. 11.8
a clay
正常壓密的沉陷量?
b.過壓密的沉陷量?
Jc = 200 kN
C. J c
=
150.
m²
N
m²
2 m
4 m
3.5 m
Ar 100 kN/m²
Ydry 14 kN/m³
Groundwater table
Year 18 kN/m³
Yat 19 kN/m³
Void ratio, e 0.8
☐ Sand ☐ Clay
Figure 11.24
65=2.686
00=0.178
a. Jo
= 2
C₁ = 0.0356
2 r dry 4 (r sat, sand -rw) + 3.5 (rsat, clay-rw)
+
= 2:14 + 4 ( 18-981) +1.75 (19-981)
76.8
IN
m²
So= Ce log (to +st) H
1 + e.
0.178
1+08
-log (-
· 76.8 +100 )
3.5
76.8
b
To + of
C
76 + 100 < 200
So
=
C
( Jo' to t ) H
=
log (Jets)
0.1250
0.0356
108
-log (.
0.025 m
768+100)
76.8
3.5

ページ4:

C. Jo to σ > σo'
76 +100
>150
Sc=
Ite
0.0356
1+08
)
So be log (To /H + be by (ots)
·log (158) 4.5+
768
log H
0.178 76.8 +100 ).
-log (
1408
0.02 + 0.025 0.045 m
=
3.5

ページ5:

EX 11.10
Example 11.10
A soil profile is shown in Figure 11.26a. Laboratory consolidation tests were
conducted on a specimen collected from the middle of the clay layer. The field
consolidation curve interpolated from the laboratory test results is shown in
Figure 11.26b. Calculate the settlement in the field caused by primary consoli-
dation for a surcharge of 60 kN/m² applied at the ground surface.
05' 60 LN
kN/m²
1.12
1.10
(a)
1.08
Groundwater table
4m
Void ratio, e
苕 1.06
1.04
8 m
1.1
中點
Yat 18 kN/m³
1.02
Clay Rock
Jo' = 4 (rsat-rw) - 32.76 KN
To' + 00' = 92.76.
3
m²
Sc =
I+lo
H = 0.21m
1.076
1.045
6e = 0.055
92.76
1.00
32.76
70
Pressure or (log scale) (kN/m²)
(b)
100

ページ6:

₤
Primary consolidation % k f
正常
7
£ <
<
secondary consolidation 次魘宏沉陷
超額孔係水壓都打出去,e≠0
有效應力 'const.
So C₂ Hly (2)
=
Ca
Ca
Sc=
oe
T+lo
Ca secondary compression index
ep void ratio of the end of
primary consolidation
H
ep = e.. se
-
e primary

ページ7:

The rate of consolidation
-M²Tu
[sin (M²)]
M
Sih
Har
1-way [zdr
m=0
u =
m = 0
M = (1) (2m+1)
Tv = Cut
Har
U₂ = 1-2
Uz |-
2-way
·U1%).
Tu
50
0.197
0.403
90
0.848
Z
Hdr
U,超額孔隙水壓
excess
pore
Water
pressure.
M:體積壓縮係數
To:時間因子(無光)
高度
Har:排水路徑
Cv.
:
壓密係數
Uz:壓密完成度
degree of consolidation
U: 平均壓密完成度(%)
Uz pg pg t by u
:
uo: 初袷u
(空間)
Tv
0.1
0.2
03
Uz

ページ8:

EX: 11:12
111-02
U:=1-2sin M
Mz)
M H
m=0
= ?
Determine the degree of consolidation Uatz=H/3 and T = 0.3.
=
M+ (1) (2m+1)
把公式用到的敉找出來
m
35
OFN
I
2
忽略
570
2
2
M
Z
Hdz
Tv
2
3元
Z
T
M
Hdz
TV
Ma
3
0.3
571
Sin
MZ
Hdz
0.5
0.5
""
e-M²T
0.48 0.0013
= 0
→
0
0.3036 0.0005
= 0
→0
20.304
Uz= -0.304 = 0.6959
"

ページ9:

11.13
·Uz -50%
Hdri
ti
=
:
0.025
2
m
3×60+155
Tv = Cut
Har
·U₁₁ = 50%
Hdr2=3m
tz = ? day.
ti
tz
Hdri
Hdr?
tz =
ti
Hdri
X
Hdrz
195
=
0.025
× 3
=
130 day

ページ10:

EX. 11.14 U-10%, to? day
·U 1%)
Tv
50
0.197
0403
90
0.848
Tv = Cut
Har
Cv=
ti
t3
Tv, Hari
TV3 Har
195
t3
10. 197 (0.025)²
0.403:32
t3 = 266
266 day

ページ11:

EX. 11.15
Hdr = 3 m
U1%)
Tv
50
0.197
U
=
90%
no
0.403
t = 75 day
90
0.848
C₁ = ?
Tv = Cut
0.848
Har
Cv 75x 24x60²
(3)²
Cv = 2.94 - 107 m²
2

ページ12:

EX. 11. 1b.
J₁ = 200 IN
e=e0 = 1.22
m
=
4
k-061×10 day
Har = 4m
To to σ = 400
kN
e=0.98
m²
U = 60%
t: ? day
=
oe 1.22-0.98-024
0σ = 400-200
= 200
kN
be
m²
0.24
av
200
Mv
Ite
Heo
171.22
·C₁ =
mvrw
T 60=
Cutbo
Hdr²

ページ13:

Mohr-Coulomb failure criterion 土壤破壞
Shear Failure
2日; = 90° + 中
A:= 45° +$/2
20
10
20
0->
111(3)
總應力:
If =
剪力強度 shear stress
If = C + J tan 中
有效應力:
C = 凝聚力 cohesion
中:(總應力)摩擦角
Tf = C' + O' tan d'
丁: 破壞面的正向應力 normal stress
Mohr's Circle
I
J = J + U
(σ; (4)
试验
J₁ = √ 3 tan² (45 + 1 )
Jo'
Ji
+2c' tan (45+2)
5.(上下)有效最大主應力
b
←
圍一圈(共三軸)
⑤:(左右)有效最小主應力

ページ14:

direct shear test 直接剪力試驗
振張
'd' tan" (71)
零沙
剪位物
鬆沙
剪位移
髭沙
壓縮
triaxial shear test 三軸試驗
03
t
J₁ = σ3 (od) failure
axial deviater.
wator
membrane
→ 膜
Soil
排水→沒有以
CD Consolidated-Drained
壓密
UC = 0 ←
√3
03:微壓
UC:超額孔练水厘
ˇ
B =¨
↑
J
uc
J
(01-05')
2
△d:固體釉差
sin p' =
p = sin" (01-0

ページ15:

EX. 12.3
CD
(有效應力)摩擦角
√3 = 140 LN
中?
052 = 104
m²
JcN
m²
J₁ = √3 + ood
p = sin
0
=
Ex. 12.4
=
Ji-'
51 +53)
0 = 7
CD = σ3 = J₂
140+104-244
J₁ = J₁'
=
Sin
45' + I = 52.86°
45+1
'
!!) J & J f = ?
e) When Z Max, σ₁ = ?
T
(σ, 2+)
74°20
(1)幾合:(0
-I
244-140
244+140
土
-) = 15.71°
1100
20
180-20-74
0:圆心的值-a值
7
T
=
:
(104+344) - (344-104) cos 74° = 116.73
$15 sin 74°
x
Zo (σ Z) = ( 244-104) Sin 74° - 49.59
=
(2)
0' = 11)
51+53
140+244
=
=192
2
2

ページ16:

Ex. 12.6 shear strength parameters (p'. C'- ?)
kN
第一顆莫爾圓
·J₂ = 70 m²
,
=
EN
0
JJ = 130
m²
m²
=
△JJ:233.5
EN
m²
第二顆莫爾圓
Ji
=
√3 + 00d
·CD→ J3 = Jj
J₁ = J₁'
091 = εD
01470+30=200 √3 = 10
J₁ (2) = 160+723.5
03(2)=160
P₁ = 2 (tan'
-
45°
03-550)
p' = 20°
+)
c' = = 20
C = √1 - σ3' tan² (45 +
2 tan (451)
KN
m²

ページ17:

CU Consolidated - Undrained
壓密
A = oud
不排水
Ji - oud = J₁'
·σ3 - old = σ3'
.有效應力
總應力
EX. 12.8
√3-105 L
=
Jd=70
ond
=
m²
P = ?
IN
'm²
N
50
m'
J₁ = √3 + od
J₁ = J, -old
·J₁ = σ3 - oud
=
=
P
175
=125
=
55
-
Sin
Sin
51 +53
=
14.5°
(8-5) 22.9°
=

ページ18:

EX 12.9 CU
A = 0.81
0 = 28°
KN
√3 = 70
A
Loud
m²
岩,
ord = ?
old A od 0.810σd
=
J₁ = √3 + ord = 10 + o Jd
Ji'= J, -ould
=
70+ 0.19.02
=
70-0.81002
J₁ = √3 - old
p' = sin'
28
=
sin" |
ord.
+31
ofd
140 -0.62002).
140-0.62002
001=509
KN
k
m²
= Sin 28° = 0.47

ページ19:

[UU Unconsolidated-Undrained 通常用黏土
不壓密
不排水
u = B03 + Ā (01-03)
m".
7p C = Cu
=
003 = 011c
2Cu
己。
✓
3=Uc.
Cu:不排水剪力强度
每個試體莫爾圓好都一樣
J₁' Joud
)
=
J₁ = 03 - oud
Cu:凝聚力(莫爾圓直徑)
裂縫深度

ページ20:

Lateral earth pressure
5 水平有效應力
靜態
K:土壓力常數
at rest 19 P. == KorH²
k
ˊ
垂直有效應力
k。
ka
Ja'
=
kp
=
passive
被動
Jo'
active主動
Passive
壓
Active

ページ21:

Ex. 13. the location of resultant force x =?
Po
OUR = 1.5
合力作用位置
σ (kN/m²)
u (kN/m²)
3 m
1.5 m
B
c'=0
d' = 35°
y
15.7 kN/m³
Sand
(a)
Groundwater table
c=0
d' = 35°
Yt
19.2 kN/m³
1.5
4.5
25.34
3
2
H
4
39.92
4.5
14.72
z (m)
(b)
z (m)
JuH
(c)
F: D = *25.34 × 3 = 38.01
© 25.34 × 1.5= 38.01
x
Figure 13.6
③ (39.92 -25.34)×1.5=10.935
z
9 × 14.72 1.5 11.04
Σ x x x
:
EF!
=
×
=
9 (1.5 +≤׳) +© (½½³) + ( 15 )
N
98 km
m.
t
+® | |_)

ページ22:

EX. 13.2
Sand
y-18 kN/m
6'34'.<=0
OCR-2
Clay
You - 19 kN/m
IL-36. Pl. 14, OCR-3
G.W.T
6
20
40
60
80
100
20
40
60
0
T
0
T
T
σ'(kN/m²)
2
area
(46.8.
66.38
2
area
4
88.33
+
6
4
19.62
u(kN/m²)
=Po=267.93.
KN
r = rsat-rw
z (m)
Figure 13.7
(b)
'sin &'
Ko, sand = (1-sin p³) (OCR) = 0.65
ock=
Ko, clay.
=
Z=4
Co’過去最大
現在
(0.44+0.42 (PL)) (OCK)
Jo
▼z (m)
(c)
0.5
=
0922
Jo-u
=
rh
h-0 = 18×4-72
z =
b
Jo'不變,但Ch'變了!
Th', sand = ko, sand To' = 46.8
Th', clay Ko clay To=
=
=6638
·Jo' = 18 x 4 + (1998) 2 = 904
(19-98)2=904
0
Thi clay
83.33
=
u = 28w = 1962

ページ23:

Rankne's theory 朗令
active
1.假設沒有f个
1.假設擋土牆的延伸
ka
passive
Ta'
To'
Ja Ja'
=
tan² (45)
1-sing'
It sing'
2
kp
=
tan 145 +
5+)
It sing'
1-sinf)
如果不是沙土,要考慮凝聚
Ja = Kar H = 2 c'√ ka
7

ページ24:

EX. 13. An 6-m-high retaining wall is shown in Figure 13.21a. Determine:
=
a. Rankine active force per unit length of the wall and the location of the
resultant
b. Rankine passive force per unit length of the wall and the location of the
resultant
Ta'
Jo'
=
tan² 145-1
1-sing'
Itsing!
= 0.26
rz
=
16 x 6
=
96
a.
Ka
·Jó
=
Ja ka σ = 0.26.96 24.96
= Jo'
=
P₁ = ^ area = x 24 9 6 x6-9488 kN
=
Normally consolidated sand (OCR 1)
active
m
passive
Z = 6m
16 kN/m³
6'-36"
c-0
(a)
6 m
74.88 kN/m
24.96 kN/m².
(b)
2 m
6 m
369.6 kN/m²
(c)
1108.8 kN/m
2-2m
b. ki
kp
=
=
A
Jo'
kpσ
area =
It sing'
tan² (45+ 2)
=
= 3.85
1-sing
3.8596
=
369.6
+ x 3696×6 - 1108.8
N
m

ページ25:

Ex. 13.8
A frictionless retaining wall is shown in Figure 13.23a. Determine:
a. The active force P after the tensile crack occurs
b. The passive force P
q= 10 kN/m²
-6.09 kN/m²
4 m
y= 15 kN/m³
6'=26°
c=8 kN/m²
(a)
4-z 2.96 m.
(b)
=1.04 m
17.31 kN/m²-
-153.6 kN/m²
51.2 kN/m²
(c)
Ka =
1-sin p
Itsin &'
=
0.39
Ja = KarH - 2 c'√ ka
z=0, Ja= 0.39 (10) - 2×8 √0.39
-6.09
Z=4.
Ja
=
0.39 (15x4+10) -2×8 √0.39 17.31
:
X
4-x
6.09 1231
= X = 1.04
KN.
Pa
=
17.31 (4-1.04)
=25.62
m