Part6-lateralearthpressure.pdf

Part 6 :

Earth Pressure & Earth Retaining

Structures

1

2

Earth Retaining Structures

3

At Rest Earth Pressure

The ratio of the horizontal principal effective stress to the

vertical principal effective stress is called the lateral earth

pressure coefficient at rest (Ko)

For coarse-grained soils, (Jaky, 1944)

1 sin
o

K  

For fine-grained soils, (Massarsch,1979)

 

( ) ( )

%
0.44 0.42

100
o

o overconsolidated o normallyconsolidated

PI
K

K K OCR

 
   

 



4

Earth Pressure Concept

5

‘ ‘

‘ ‘

‘ ‘
2

‘

1 sin 1 sin
2

1 sin 1 sin

1 sin
tan 45

1 sin 2

2

 
 

 

 



 

    
    

    

   
     

   

 

a v

a

a v a a

c

K

K c K

Rankine Active Earth Pressure

(for smooth frictionless wall)

6

‘ ‘

‘ ‘

‘
2

‘

1 sin 1 sin
2

1 sin 1 sin

1 sin ‘
tan 45

1 sin 2

2

 
 

 

 



 

    
    

    

   
     

   

 

p v

p

p v p p

c

K

K c K

Rankine Passive Earth Pressure

(Smooth frictionless wall)

Note: 𝑘𝑝 =
1

𝑘𝑎

7

Slip Planes

Slip Planes within a soil mass near a retaining wall

8

Variation of active and passive lateral earth pressures

Variation of Stresses with Depth for cohesionless soils

9

Stresses due strip load

Variation of Stresses with Depth for water, surcharge, and strip load

10

Water pressure cancels out

Example 5-1 For the retaining wall shown, find:
a) Resultant active force and passive force

b) Overturning moment and resisting moment

𝑘𝑎 =
1 − 𝑠𝑖𝑛30

1 + 𝑠𝑖𝑛30
= 0.333, 𝑘𝑝=

1

𝑘𝑎
= 3

11

Active side

depth g’ ’ ka a(soil) a(surch.)

kN/m
3

kN/m
2

kN/m
2

kN/m
2

0 16 0 0.333 0 3.33

4 16 64 0.333 21.3 3.33

6 10.2 84.4 0.333 28.1 3.33

Passive side

depth g’ ’ kp p
m kN/m

3
kN/m

2
kN/m

2

0 16 0 3 0

1 16 16 3 48

3 10.2 36.4 3 109.2

area No. Force Aram moment

kN m kN.m

1 42.6 3.3 142.1

2 49.4 1.0 47.2

surcharge 20.0 3.0 60.0

Fa = 112.0 Mo = 249.2

3 24.0 2.3 56.0

4 157.2 0.9 136.8

Fp = 181.2 Mr = 192.8

active side

passive

side

For the retaining wall shown, find:

a) Resultant active force and passive force

b) Overturning moment and resisting moment

Example 5-2

𝑘𝑎 =
1 − 𝑠𝑖𝑛30

1 + 𝑠𝑖𝑛30
= 0.333, 𝑘𝑝=

1

𝑘𝑎
= 3

Water pressure cancels out

12

For the retaining wall shown, find:

a) Resultant active force and passive force

b) Overturning moment and resisting moment

Example 5-3
Active side

depth g’ ’ ka a(soil) a(surch.)

kN/m
3

kN/m
2

kN/m
2

kN/m
2

0 16 0 0.333 0 3.33

4 16 64 0.333 21.3 3.33

4 10.2 64 0.271 17.3 2.71

6 10.2 84.4 0.271 22.9 2.71

Passive side

depth g’ ’ kp p
m kN/m

3
kN/m

2
kN/m

2

0 16 0 3 0

1 16 16 3 48

1 10.2 16 3.69 59.0

3 10.2 36.4 3.69 134.3

area No. Force Aram moment

kN m kN.m

1 42.6 3.33 142.1

2 40.2 0.95 38.4

5 13.3 4.00 53.3

6 5.4 1.0 5.4

Fa = 101.6 Mo = 239.1

3 24.0 2.33 56.0

4 193.4 0.87 168.3

Fp = 217.4 Mr = 224.3

active side

passive

side

surcharge

Water pressure cancels out

13

Horizontal Stresses Due to Surface Strip Load

14

Active side

depth g’ ’ ka a(soil) a(surch.)

kN/m
3

kN/m
2

kN/m
2

kN/m
2

0 16 0 0.333 0 3.33

4 16 64 0.333 21.3 3.33

6 10.2 84.4 0.333 28.1 3.33

Passive side

depth g’ ’ kp p
m kN/m

3
kN/m

2
kN/m

2

0 16 0 3 0

1 16 16 3 48

3 10.2 36.4 3 109.2

area No. Force Aram moment

kN m kN.m

1 42.6 3.33 142.1

2 49.4 0.95 47.2

strip 34.2 3.96 135.4

Fa = 126.2 Mo = 324.7

3 24.0 2.3 56.0

4 157.2 0.9 136.8

Fp = 181.2 Mr = 192.8

active side

passive

side

Example 5-4
For the retaining wall shown, find:

a) Resultant active force and passive force

b) Overturning moment and resisting moment

𝑘𝑎 =
1 − 𝑠𝑖𝑛30

1 + 𝑠𝑖𝑛30
= 0.333, 𝑘𝑝=

1

𝑘𝑎
= 3

Water pressure cancels out

15

Active Case With Cohesive Backfill (C > 𝟎)

Active pressure distribution to
be considered

16

Active side

depth g’ ’ ka c’ a
kN/m

3
kN/m

2
kN/m

2
kN/m

2

0 17 0 0.406 15 -19.1

4 17 68 0.406 15 8.5

6 10.2 88.4 0.406 15 16.8

Passive side

depth g’ ’ kp c’ p
m kN/m

3
kN/m

2
kN/m

2
kN/m

2

0 16 0 2.46 15 47.1

1 16 16 2.46 15 86.4

3 10.2 36.4 2.46 15 136.6

area No. Force Aram moment

kN m kN.m

1 5.2 2.4 12.6

2 27.3 0.9 24.3

Fa = 32.5 Mo = 36.9

3 66.8 2.5 163.6

4 223.0 0.9 206.3

Fp = 289.8 Mr = 369.9

active side

passive

side

Example 5-5

𝑧𝑐 =
2𝑐′

𝛾 𝑘𝑎
=

2(15)

17 0.406

𝑘𝑎 =
1 − 𝑠𝑖𝑛25

1 + 𝑠𝑖𝑛25
= 0.406, 𝑘𝑝=

1

𝑘𝑎
= 2.46

For the retaining wall shown, find:

a) Resultant active force and passive force

b) Overturning moment and resisting moment

17

Active pressure at depth z =

Rankine Pressure with Sloping Surface

Resultant active force =

2 2

2 2

cos cos cos
cos

cos cos cos
a

K
  


  

 


 

Rankine active pressure

18

Coulomb’s Earth Pressure

21

2
a a

P K Hg

The resultant active force behind dry sand

can be calculated as

Coulomb’s limit equilibrium approach, obtained expressions for Ka where wall friction (d ) is present,
the wall face is inclined at an angle b, and the backfill is sloping at an angle 

19

Consider the retaining wall shown. Calculate the Coulomb’s
active force per unit length of the wall and the angle it makes
with the horizontal q. Assume d = (2

3
𝜙′)

𝑘𝑎 =
𝑠𝑖𝑛2(75 + 32)

𝑠𝑖𝑛2 75 sin(75 −
64
3
) 1 +

sin 32 +
64
3

sin(32 − 10)

sin 75 −
64
3

sin(10 + 75)

= 0.4683

Resultant active force = =
1

2
(0.4683)(18.5)(6.5)2 = 183 𝑘𝑁

𝜃 = 15 +
2

3
32 = 36.3

Example 5-6

20

Flexible Retaining Walls

21

Mode of Failure in Flexible Walls

22

Cantilever Walls

Mostly used to support sallow cuts (less than 4m)

Using the approximate pressure distribution will simplify the solution to find do . The actual depth of
embedment d  1.2do

A factor of safety is applied to the passive pressure.

23

Active side

depth g’ ’ ka a(soil) a(surch.)

kN/m
3

kN/m
2

kN/m
2

kN/m
2

0 16 0 0.333 0 3.33

4 16 64 0.333 21.3 3.33

L 10.2 64+10.2L 0.333 21.3+3.4L 3.33

Passive side

depth g’ ’ kp p
m kN/m

3
kN/m

2
kN/m

2

0 16 0 2 0

1 16 16 2 32

L 10.2 16+10.2L 2 32+20.4L

area No. Force Aram moment

kN m kN.m

1 42.6 L+4/3 42.6L+56.8

2 21.3L L/2 10.7L
2

3 1.7L
2

L/3 0.57L
3

4 13.3+3.33L 2+L/2 1.67L
2
+13.3L+26.6

5 -16.0 L+1/3 -16L-5.33

6 -32L L/2 -16L
2

7 -10.2L
2

L/3 -3.4L
3

SM  -2.83L
3
-3.63L

2
+39.9L+78.1 =0

L = 4 m

d = 1.2(1+4) = 6 m

active side

passive

side

Determine the required depth of embedment of the

piling to ensure a factor of safety of 1.5 with respect

to gross passive resistance

Example 5-7

24

Anchored Walls

25

Active side

depth g’ ’ ka a(soil) a(surch.)

kN/m
3

kN/m
2

kN/m
2

kN/m
2

0 16 0 0.333 0 3.33

4 16 64 0.333 21.3 3.33

L 10.2 64+10.2L 0.333 21.3+3.4L 3.33

Passive side

depth g’ ’ kp p
m kN/m

3
kN/m

2
kN/m

2

0 16 0 2 0

1 16 16 2 32

L 10.2 16+10.2L 2 32+20.4L

area No. Force Aram moment

kN m kN.m

1 42.6 2.17 92.3

2 21.3L L/2+3.5 10.65L
2
+74.5L

3 1.7L
2

2L/3+3.5 1.13L
3
+5.95L

2

4 13.3+3.33L 1.5+L/2 1.67L
2
+11.6L+20.0

5 -16.0 3.17 -50.7

6 -32L L/2+3.5 -16L
2
-112L

7 -10.2L
2

2L/3+3.5 -6.8L
3
-35.7L

2

Tie force(T) -T 0.0 0.0

SF  -8.5L
2
-7.37L+39.9 -T = 0 SM  -5.67L

3
-33.4L

2
-25.9L+61.6 =0

substitue L = 0.97 m L = 0.97 m

T = 24.8 kN/m d = (1+0.97)  2 m

active side

passive side

Example 5-8

26

Rigid Walls

27

Mode of Failure in Rigid Walls

28

Pressure Distribution acting on rigid walls

Based on Rankine Lateral Pressure (frictionless wall)

29

Example

Check the stability of the cantilever concrete

retaining wall against overturning, sliding and

bearing capacity. Assume concrete unit weight

of 23.58 kN/m3

FS (overturning)  2

FS (sliding)  1.5

FS (bearing capacity)  3

30

Forces Acting on the Wall

Forces that resist sliding are the passive force Pp plus the friction force between the foundation of the wall and the

soil underneath. The friction force can be calculated as follows

1 2
2 3

tan

For a unit length of wall,

tan

is the friction angle between the base of the wall and the soil ( to )

C is the cohesion bet

friction normal cs

base

friction friction base cs

soil

cs

C

A B

F A V BC

  d

 d

d 

 



  

1 2
2 3

ween the base of the wall and the soil ( to )
soil

C

31

Example

Check the stability of the cantilever concrete

retaining wall against overturning, sliding and

bearing capacity. Assume concrete unit weight

of 23.58 kN/m3

FS (overturning)  2

FS (sliding)  1.5

FS (bearing capacity)  3

32

Overturning Moments,

SMo

Resisting Moments,

SMR

Sliding forces,

SFS

Resisting force,

SFR
c

Pa

ph

pv

H’

H’/3
Pp

( )

R

overturning

o

M
FS

M





( )

R

sliding

S

F
FS

F





Forces Acting on the Wall

33

Sloped ground in the active side

34

Factor of safety against overturning

The passive force

effect on overturning

is usually ignored

35

Factor of Safety Against Sliding

Forces that resist sliding are the passive force Pp plus the friction force between the
foundation of the wall and the soil underneath. The friction force can be calculated as
follows

1 2
2 3

tan

For a unit length of wall,

tan

is the friction angle between the base of the wall and the soil ( to )

C is the cohesion bet

friction normal cs

base

friction friction base cs

soil

cs

C

A B

F A V BC

  d

 d

d 

 



  

1 2
2 3

ween the base of the wall and the soil ( to )
soil

C

36

2 6

R o
M MB B

e
V


  

 


(max)

(min)

6
1

toe

heel

V e
q

B B

 
  

 



max

(bearing capacity) 3u
q

FS
q

 