Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show. Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.

1. 1
Geotechnical Engineering–II [CE-321]
BSc Civil Engineering – 5th Semester
by
Dr. Muhammad Irfan
Assistant Professor
Civil Engg. Dept. – UET Lahore
Email: mirfan1@msn.com
Lecture Handouts: https://groups.google.com/d/forum/geotech-ii_2015session
Lecture # 24
6-Dec-2017

2. 2
Practice Problem #4











 




sin1
sin1
2
45tan2
o
a
aK
aaa KczK  2
c’ = 0
’ = 35°
 = 18 kN/m3
c’ = 0
’ = 30°
WT
 = 19 kN/m3
sat = 21 kN/m3
c’ = 0
’ = 32°
sat = 20 kN/m3
4m
4m
2m
4m
Determine the total active force per meter acting on the wall
along with its point of application.
q = 50 kPa

3. 3
Practice Problem #5











 




sin1
sin1
2
45tan2
o
a
aK
aaa KczK  2
c’ = 10 kPa
’ = 35°
 = 18 kN/m3
WT
c’ = 50
’ = 0°
sat = 20 kN/m3
4m
4m
4m
Determine the total active force per meter acting on the wall
along with its point of application.
q = 50 kPa
c’ = 20 kPa
’ = 19.5°
sat = 21 kN/m3

4. 4
Practice Problem #6











 




sin1
sin1
2
45tan2
o
a
aK
aaa KczK  2
c’ = 50 kPa
’ = 10°
 = 18 kN/m3
10 m
A retaining wall of 10 m height retains a cohesive soil.
Determine the active force with respect to various possibilities
of tension crack.
a
c
K
c
z



2

5. 5
RANKINE THEORY
ACTIVE PRESSURE -- SUMMARY --











 




sin1
sin1
2
45tan2
o
a
aK
aaa KczK  2
a
c
K
c
z



2

6. 6
COULOMB’S EARTH PRESSURE
THEORY
ASSUMPTIONS
1. The soil is homogeneous and isotropic.
2. Soil has both cohesion and friction (c- soil).
3. Rupture surface as well as backfill surface is planar.
4. There is friction between wall and soil.
5. Failure wedge is a rigid body undergoing translation.
Coulomb (1776)

7. 7
BENEFITS OF ASSUMPTIONS
-- DIFFERENCE BETWEEN THEORY AND REALITY --

8. 8
BENEFITS OF ASSUMPTIONS
-- DIFFERENCE BETWEEN THEORY AND REALITY --
Theoretical Earth Pressure Actual Earth Pressure

9. 9
COULOMB’S ACTIVE EARTH PRESSURE
a
b
q
180aq
ab
A
B
C
D
qb
W
W = Weight of soil wedge ABC
𝑊 = 1
2∙𝐴𝐶∙𝐵𝐷∙1∙𝛾 ⋯⋯⋯(1)
𝐴𝐶
sin(𝛼 + 𝛽)
=
𝐴𝐵
sin(𝜃 − 𝛽)
∆𝑨𝑩𝑪
Using law of sines
𝐴𝐶 =
𝐴𝐵
sin(𝜃 − 𝛽)
∙ sin(𝛼 + 𝛽)
𝐴𝐶 =
𝐻
sin 𝛼 ∙ sin(𝜃 − 𝛽)
∙ sin(𝛼 + 𝛽)
H

10. 10
COULOMB’S ACTIVE EARTH PRESSURE
a
b
q
180aq
ab
A
B
C
D
qb
W
W = Weight of soil wedge ABC
𝑊 = 1
2∙𝐴𝐶∙𝐵𝐷∙1∙𝛾 ⋯⋯⋯(1)
𝐴𝐶 =
𝐻
sin 𝛼 ∙ sin(𝜃 − 𝛽)
∙ sin(𝛼 + 𝛽)
∆𝑨𝑩𝑫
𝐵𝐷
sin(180 − (𝛼 + 𝜃))
=
𝐴𝐵
sin 90
∵ sin(180 − 𝛼 + 𝜃 ) = sin(𝛼 + 𝜃)
𝐵𝐷
sin(𝛼 + 𝜃)
=
𝐴𝐵
sin 90
𝐵𝐷 = sin(𝛼 + 𝜃) ∙
𝐴𝐵
1
𝐵𝐷 = sin(𝛼 + 𝜃) ∙
𝐻
sin 𝛼
H

11. 11
COULOMB’S ACTIVE EARTH PRESSURE
a
b
q
180aq
ab
A
B
C
D
qb
W
W = Weight of soil wedge ABC
𝑊 = 1
2∙𝐴𝐶∙𝐵𝐷∙1∙𝛾 ⋯⋯⋯(1)
𝐴𝐶 =
𝐻
sin 𝛼 ∙ sin(𝜃 − 𝛽)
∙ sin(𝛼 + 𝛽)
𝐵𝐷 = sin(𝛼 + 𝜃) ∙
𝐻
sin 𝛼
𝑊 = 1
2 ∙
𝛾𝐻2
𝑠𝑖𝑛2 𝛼
∙
sin(𝛼 + 𝛽) ∙ sin(𝛼 + 𝜃)
sin(𝜃 − 𝛽)
Eq. 1 →
H

12. 12
COULOMB’S ACTIVE EARTH PRESSURE
a
b
q
180aq
ab
A
B
C
D
qb
W
W
R

R = Resultant of shear and normal forces
acting on failure plane
R
(q)
(ad)180(adq)
d
Pa
𝛿 = 2
3 𝜙
Our Goal:
Determine active force (Pa) on the wall.
 Draw force polygon of the system.
Pa
d = angle of wall friction
(𝑅𝑒𝑠𝑜𝑛𝑎𝑏𝑙𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛)

13. 13
COULOMB’S ACTIVE EARTH PRESSURE
a
b
q
180aq
ab
A
B
C
D
qb
W
W
R

R
(q)
(ad)180(adq)
d
Pa
𝑃𝑎
sin(𝜃 − 𝜙)
=
𝑊
sin[180 − 𝛼 − 𝛿 + 𝜃 − 𝜙 ]
Applying sine law on force polygon
Pa
𝑃𝑎
sin(𝜃 − 𝜙)
=
𝑊
sin(𝛼 − 𝛿 + 𝜃 − 𝜙)
Replacing value of ‘W’
𝑃𝑎 =
1
2
∙
𝛾𝐻2
𝑠𝑖𝑛2 𝛼
∙
sin(𝛼 + 𝛽) ∙ sin(𝛼 + 𝜃) ∙ sin(𝜃 − 𝜙)
sin(𝜃 − 𝛽) ∙ sin(𝛼 − 𝛿 + 𝜃 − 𝜙)

14. 14
COULOMB’S ACTIVE EARTH PRESSURE
a
b
q
180aq
ab
A
B
C
D
qb
W
W
R

R
(q)
(ad)180(adq)
d
Pa
As designers, we want to determine max.
value of Pa
Pa
𝑃𝑎 =
1
2
𝛾𝐻2 ∙
𝑠𝑖𝑛2(𝛼 + 𝜙)
𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 +
sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽)
sin(𝜙 − 𝛿) ∙ sin(𝜙 + 𝛽)
2
To determine critical value of b for max. Pa,
we have 𝑑𝑃𝑎
𝑑𝛽
= 0

15. 15
COULOMB’S ACTIVE EARTH PRESSURE
a
b
q
180aq
ab
A
B
C
D
qb
W
R
d
Pa
𝐾 𝑎 =
𝑠𝑖𝑛2(𝛼 + 𝜙)
𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 +
sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽)
sin(𝜙 − 𝛿) ∙ sin(𝜙 + 𝛽)
2
Since,
𝑃𝑎 =
1
2
∙ 𝛾𝐻2
∙ 𝐾 𝑎
𝑃𝑎 =
1
2
𝛾𝐻2
∙
𝑠𝑖𝑛2
(𝛼 + 𝜙)
𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 +
sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽)
sin(𝜙 − 𝛿) ∙ sin(𝜙 + 𝛽)
2

16. 16
COULOMB’S ACTIVE EARTH PRESSURE
a
b
q
180aq
ab
A
B
C
D
qb
W
R
d
Pa
𝑃𝑎 =
1
2
𝛾𝐻2
∙
𝑠𝑖𝑛2
(𝛼 + 𝜙)
𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 +
sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽)
sin(𝜙 − 𝛿) ∙ sin(𝜙 + 𝛽)
2
For a vertical wall face and horizontal
levelled ground
𝛼 = 90° , 𝑎𝑛𝑑 𝛽 = 0°
𝑃𝑎 =
1
2
𝛾𝐻2 ∙
1 − sin 𝜙
1 + sin 𝜙
Above equation is reduced to
i.e. same as Renkine’s Solution

17. 17
CONCLUDED
REFERENCE MATERIAL
Principles of Geotechnical Engineering – (7th Edition)
Braja M. Das
Chapter #13
Essentials of Soil Mechanics and Foundations (7th Edition)
David F. McCarthy
Chapter #17
Geotechnical Engineering – Principles and Practices – (2nd Edition)
Coduto, Yueng, and Kitch
Chapter #17