Lateral Earth pressure

Geo Technical Engineering

1. Active Earth Pressure
Cohesive Soil
H
c
2C cotα
Net pressure
γH cot
2
α-2C cotα
[ ]pressureNetCHHP
Cdzdzzp
dzCzdzppressureTotalNow
CZp
KCKp
a
H
O
H
O
a
H
O
H
O
a
a
aavaha
aag
aag
aag
aag
ss
cot2cot
2
1
)cot2cot(
)cot2cot(
cot2cot
2
22
2
2
2
-=
-=\
-=
-=
-==
ò ò
òò

2. •Practically crack do occur ,neglect the
negative pressure and consider the whole
positive pressure
aag
aag
a
g
g
cot)(2cot)(
2
1
)cot2cot(
tan
4
2
2
22,0
222
2
ZoHCZoH
dzCzP
C
K
C
H
ZcorZobewillthisP
H
Zo
a
a
c
a
---=
-=
=
÷
÷
ø
ö
ç
ç
è
æ
=\
=
ò

3. g
g
g
a
g
ag
g
a
g
ag
a
a
g
a
a
a
g
gag
a
g
aa
g
g
a
g
2
2
22
22
22
22
2
2
2
2
2
222
2
2
2
2
2
2
1
2
cot2
2
cot
2
1
4
cot2
2
cot
2
1
tan
tan4
cot2
tan
1
tan
4
2
1
cot
2
1
tan
2
2cottan
2
2
1
tan
2
C
KCHKHP
C
CH
C
HP
C
CH
C
HP
C
CH
C
HP
C
HC
C
HP
C
zoofvaluethePut
aaa
a
a
a
a
+-=
+--=
+--=
+-´-=
÷
÷
ø
ö
ç
ç
è
æ
--
ï
þ
ï
ý
ü
ï
î
ï
í
ì
÷
÷
ø
ö
ç
ç
è
æ
=
=
P
a

4. Lateral Earth Pressure
Cohesionless Soil
Circle shows the soil at rest position
Circle represents the active state
Circle represents the passive state
O A C
σ
v F σ
hp
φ
G
D
E
1
2
3
σ
ha
1
2
3
B

5. ( )
p
v
h
a
v
h
vhv
p
v
hp
p
p
a
a
K
Sin
Sin
K
Sin
Sin
c
ccc
CasePassiveCaseActive
OR
Sin
Sin
K
FGOSinOFRGandcircleofRadiusRGFEBFAs
OF
BF
OF
FE
BFOF
EFOF
OB
OE
K
pressureearthpassiveoftCoefficienK
casestatepassiveforSimilarly
Sin
Sin
a
K
Sin
OC
CD
andCDCBAC
OC
CB
OC
AC
CBOC
ACOC
OB
OA
K
pressureearthactiveoftcoefficienK
=
-
+
===
+
-
==
==
===
=+=+=
-
+
=
D===
-
+
=
-
+
===
=
+
-
=\
÷
ø
ö
ç
è
æ
===
+
-
=
+
-
==
=
f
f
a
s
s
f
f
as
s
assass
assass
aassaass
f
f
f
s
s
f
f
f
1
1
tan
1
1
tan
1
tantan
tantan)0(
)0(tan2tantan2tan
1
1
)()3(
1
1
3 # circle Ref..
1
1
,
1
1
2
2
22
3
2
31
2
31
2
31
2
31




6. Active Earth Pressure
Cohesive soil or C-φ Soil
Active
state
At rest
τ =c′+σ′tanφ
Y
φ
c
σ
ha
Zσ
vσ
n
τ
X
φ
Oc/tanφ
Z
o

Z
c

D
c
-2C cotα
γD cot
2
-2Ccotα
Or
γD cot
2
-2Ccotα

7. ( )
f
ff
ff
f
s
f
f
ff
f
s
ff
f
ff
f
s
ff
f
ss
f
f
s
s
Sin
SinSinc
Sin
Sin
Sin
Sinc
Sin
Sin
c
Sin
Sinc
Sin
Sin
c
XOAsXO
Sin
Sin
XO
Sin
Sin
XZZY
XZZY
ZYXZ
ZYXZ
XO
XO
v
v
v
vha
v
ha
+
---
+
÷
÷
ø
ö
ç
ç
è
æ
+
-
=
÷
÷
ø
ö
ç
ç
è
æ
-
+
-
+
÷
÷
ø
ö
ç
ç
è
æ
+
-
=
¢
-
÷
÷
ø
ö
ç
ç
è
æ
+
-
+
÷
÷
ø
ö
ç
ç
è
æ
+
-
=
þ
ý
ü
î
í
ì
=-
÷
÷
ø
ö
ç
ç
è
æ
+
-
+=
+
-
=
+
-
=
+
-
=
+
+
1
11(
tan1
1
1
1
1
tan1
1
tan1
1
tan1
1
tan
,
1
1
1
1
1
1
Contd…

8. [ ]
ahaa
vha
aaaavha
v
v
v
vha
KCCpDIf
CDC
KcKDKCK
eqsBell
Sin
Sin
C
Sin
Sin
CosSin
Sin
SinC
Sin
Sin
Sin
C
Sin
Sin
Cos
Sin
Sin
Sin
Sin
Cos
C
Sin
Sin
2cot2,0
cot2cotcot2cot
22
.'
1
1
2
1
1
1
1
12
1
1
1
cos2
1
1
tan
1
2
1
1
22
22
2
-=-===
-=-=
-=-=
®
+
-
-
÷
÷
ø
ö
ç
ç
è
æ
+
-
=
=+
+
-
-
÷
÷
ø
ö
ç
ç
è
æ
+
-
=
+
-
÷
÷
ø
ö
ç
ç
è
æ
+
-
=
ú
û
ù
ê
ë
é
=
÷
÷
ø
ö
ç
ç
è
æ
+
-
´´+
÷
÷
ø
ö
ç
ç
è
æ
+
-
=
as
aagaass
gss
f
f
f
f
s
jj
f
f
f
f
s
f
f
f
f
s
f
f
f
f
f
f
f
f
f
ss
Contd…

9. •OR
aaa
aavha
vha
vha
vha
hav
KCDKp
KCK
Sin
Sin
C
Sin
Sin
C
C
C
caseactiveIn
C
2
2
1
1
2
1
)1(
cot2cot
tan2tan
tan2tan
:
tan2tan
2
2
2
2
31
-=\
-=
+
-
-
+
-
=
-=
-=
+=
+=
g
ss
f
f
f
f
ss
aass
asas
aass
aass

10. g
f
a
gg
g
g
C
ZorDorD
If
C
K
C
D
KCDK
KCDK
DbewillwhatpIf
KCp
HorDIf
cco
a
aa
aa
a
aa
2
0
tan
22
2
20
?,0
2
,00
=
=
==
=
-=
==
-=
==

11. Passive Earth Pressure
Cohesive soil and C-φ soil
P
p
2C tanα
γD tan
2
α
τ =c+σ tanφ
Q
Oσ
ha
σ
v
P σ
hpσ
n
τ
CφX

12. aag
aass
ss
f
f
f
f
ss
ff
f
ss
f
f
s
s
tan2tan
tan2tan
2
1
1
2
1
1
tan1
1
)(
1
1
1
1
2
2
CDP
Cor
KCK
similarlyor
donepreviouslyAs
Sin
Sin
C
Sin
Sin
C
XOXO
Sin
Sin
XO
Sin
Sin
XPPQ
XPPQ
PQXP
PQXP
XO
XO
Now
p
vhp
ppvhp
vhp
vhp
v
hp
+=
+=
+=
þ
ý
ü
î
í
ì
-
+
+
÷
÷
ø
ö
ç
ç
è
æ
-
+
=
þ
ý
ü
î
í
ì
=-
÷
÷
ø
ö
ç
ç
è
æ
-
+
+=\
-
+
=
-
+
=
-
+
=
+
+


13. .2
2
tan2tan
tan2tan
tan2tan
2
1
tan2,0
2
2
31
22
ppp
ppvhp
vhp
p
p
KCKDPOr
KCK
C
COr
CHHP
CPDIf
+=
+=
+=
+=
+=
==
g
ss
aass
aass
aag
a

14. For Inclined Surface
β
P
a
bg
fbb
fbb
b
bg
fbb
fbb
b
cos
2
1
coscoscos
coscoscos
cos
cos
2
1
coscoscos
coscoscos
cos
2
22
22
2
22
22
HKP
K
HKP
K
pp
p
aa
a
=
--
-+
´=
=
-+
--
´=

15. Coulomb’s Earth Pressure Theory
(a) Trial Failure wedge (b)Force
A
B
F
C
Β-α
α
90-θ+α
90+θ-β
W D
β
H
θ
δ
P
a
φ
90-θ-δ
90+θ+δ-β+φ
Β-φ
W
Pa
F

16. Derivation of Coulomb’s Earth
Pressure Theory
Cohesionless Soil
Active case
Let AB (Fig.) be the back face of a retaining wall
supporting a granular soil, the surface of which is
constantly sloping at an angle α with the horizontal.
BC is a trial failure surface. In the stability
consideration of the probable failure wedge ABC,
the following forces are involved(per unit width of
the wall):

17. 1W, the weight of the soil wedge.
2F, the resultant of the shear and normal forces
on the surface of failure, BC
This is inclined at an angle of φ to the normal
drawn to the plane BC.
3P
a
, the active force per unit width of the wall.
The direction of P
a
is inclined at an angle δ is the
angle of friction between the wall and the soil.
The force triangle for the wedge is shown in Fig.
From the law of sines we have,

18. ()
)5(
)90((
sin
4
cos
)cos(
.
)90(.
cos
)90(
)3(.)()(
2
1
).(
)2(.
)90(
)(
)1(
)()90(
®
+-
=
-
®
-
=
-+=
-+=
®=
®
+-++
-
=
®
-
=
+-++
aqab
q
bq
bq
q
bq
g
fbdq
fb
fbfbdq
Sin
BC
Sin
AB
esoflawthefromAgain
H
Sin
H
SinABAD
But
BCADW
wedgetheofweighttheaFigFrom
W
Sin
Sin
P
or
Sin
P
Sin
W
a
a

19. ú
û
ù
ê
ë
é
+-++-
--
=
®
-
--
=
®
-
-
=
-
-
=
)90sin(.)sin(.cos
.)cos(.)cos(
2
1
)2.(exp
)7(
)sin(.cos
)cos(.)cos(
2
1
),3(.)6()4(.
)6(.
)sin(.cos
)cos(
.
)(
)cos(
2
2
2
2
fbdqabq
aqbq
g
abq
aqbq
g
abq
aq
ab
aq
HP
EqinWofressionthengSubstituti
HW
getweEqinandEqsngSubstituti
HAB
Sin
BC
Or
a

20. •In the preceding expression for P
a
, the values
of γ, H, θ, α, φ, and δ are constants, and β is
the only variable. To determine the critical
value of β for maximum P
a
we have
)10(
)cos(.)cos(
)sin(.)sin(
1)cos(.cos
)(cos
:
'
)9(
2
1
:
')7.(int
),8(.
)8(0
2
2
2
2
®
ú
û
ù
ê
ë
é
-+
-+
++
-
=
®=
®=
aqqd
affd
qdq
qf
g
b
b
a
a
aa
a
K
bygivenistcoefficien
pressureearthactivesCoulombisKWhere
HKP
asobtainedispressureearth
activesCoulombEqogsubtitutinis
ofiprelationshthewhenEqsolvingAfter
d
dP

21. •Note that when α=0
o
, θ=0
o
, and δ=0
o
•,Coulomb’s active earth pressure coefficient
becomes equal to (1-sinφ)/(1+sinφ)
,horizontal which is the same as the Rankine’s
earth pressure coefficient.
•The variation of the value of K
a
for retaining walls
with a vertical back (θ=0
o
) and horizontal backfill
(α=0
o
) is given in table. From this table, not that
for a given value ofφ the effect of wall friction is
to reduce somewhat the active earth pressure
coefficient.

22. Table Values of K
a
(Eq.10)

for θ=0
o
,α=0
o

23. C
β-α
α
90-θ+α
90+θ-β
W
D
β
H
θ
δ
P
p
φ
A
W
90-θ+α
β+φ
P
p
F
F
[180-(90-θ+δ)-(β+φ)]
Coulomb’s passive pressure
Trial failure wedge
Force polygon

24. Passive Case
•Fig. shows a retaining wall with a sloping
cohesionless backfill. The force polygon for
equilibrium of the wedge ABC for passive
state is shown in fig.(b). P
p
is the notation for
the passive force. In a procedure similar to the
one we followed in the active case, we get:
• P
p
= ½ K
p
γ H
2 --------------------(12)
• where K
p
= coefficient of passive earth
pressure for Coulomb’s case:
2
2
2
)cos()cos(
)sin()sin(
1)cos(cos
)(cos
ú
û
ù
ê
ë
é
--
+-
--
+
=
qaqd
afdf
qdq
qf
p
K

25. •For a frictionless wall with the vertical back face
supporting granular soil backfill with a horizontal
surface that is, α = 0
o
,θ = 90
o
and δ = 0
o
Eq. 12
becomes:
• K
p =1+sinφ/1-sinφ = tan
2
(45+φ/2)
•This is the same relationship that was obtained for the
passive earth pressure coefficient in Rankine’s case.
•The variation of k
p
with ф and δ for θ =0 and α = 0 is
given in Table. It is also observed from this table that
for given values of α and ф the value of K
p
increases
with the wall friction.

26. Table values of K
p
(Eq,12) for θ = 0 and α =0
δ(degree)
ф(deg)0 5 10 15 20
15 1.695 1.90 2.130 2.405 2.735
20 2.040 2.313 2.636 2.030 3.525
25 2.464 2.83 3.286 3.855 4.597
30 3.00 3.506 4.143 4.977 6.105
35 3.69 4.390 5.310 6.854 8.324
40 4.600 5.59 6.946 8.870 11.772

27. Example-1
•A retaining wall with a smooth vertical back retains
sand backfill for a depth of 6 m. The backfill has a
horizontal surface and has the following properties:
c′= 0, f′ = 28
o
; g=16 kN/m
3
; g
s a t
=20 kN/m
3
.
Calculate the magnitude of the total thrust
against the wall for the conditions given
below:
(a) back fill fully drained but the top of the wall
is restrained against yielding, (b) backfill fully
drained and the wall is free to yield and (c) wall
free to yield, water table at 3 m depth and there
is no drainage. Determine the point of
application of the resultant thrust for case (c)

28. Solution:
•(a) If the wall is restrained against yielding, the
lateral pressure that would develop against
the wall would be the earth pressure ‘at rest’.
Calculate coefficient of earth pressure at rest
K
o
using the following Eq.
• K
o
=1-sinf′ = 1-sin28
o
=0.53
•Hence
6 m
P
o
P
o
(a)
P
A
P
A
(b)

29. For this case,
Fig. shows the pressure distribution diagram for
this case.
Lateral pressure, p
o
=K
o
g. z
At z = 6 m, p
o
= o.53x16x6 =50.88 kN/m
2
Total thrust per meter length of the wall is
P
o
= 1/2x 50.88x6 = 152.64 kN
(b) For this case again, g= 16 kN/m
3
, but the
lateral pressure is the active earth pressure.
g=16 kN/m
3.

30. Fig. shows the pressure distribution diagram.
Above the water table, g =16 kN/m
3
and below
the water table,g = g′=g
sat
–g
w
=20- 9.8 =10.2
kN/m
3 .
The lateral thrust due to water is shown as
P
3
.
p
1
= K
a
g H
1
= 0.36 x 16 x 3 =17.28 kN/m
2


2
2
/68.103656.34
2
1
/56.3461636.0,6
36.0
sin1
sin1
,28
mkNP
mkNpmzAt
zKp
K
A
A
AA
a
o
=´´=\
=´´==
=
=
+
-
==¢
g
f
f
f

31. p
2
= K
a
g′ H
2
=0.36x10.2x3=11 kN/m
p
3
= g
w
H
2
9.8 x3 = 29.4 kN/m
2
The total thrusts which are shown in Fig.
are calculated below:
P
1
= ½ x17.28 x 3 = 25.92 kN
acting
at (3+1/3x3) = 4 m from base
P′ = 17.28x3 51.84 kN acting at 2/3 =
1.5 m from base
P
2
= 1/2x11x3 =16.5 kN acting at
1/3x3 =1 m from base
P
3
= 1/2x29.4x3 = 44.1 kN acting at
1/3x3=1 m from base.

32. 3 m
3 m
P
3
P′
P
1
P
2
P
1
P
2
P
3
•It can be seen that the lateral thrust due to water
contributes substantially to the total lateral
thrust.
Total thrust = P
1
+P′+P
2
+P
3
= 25.92+51.84+16.5+44.1
= 138.36 kN/m length.
The distance of the
resultant P from the base
of the wall can be obtained by taking moments
about the base.
mH 75.1
36.138
242
36.138
11.4415.165.184.51492.25
==
´+´+´+´
=

33. Example-2

34. •Δp
A
= K
A
. q
= 0.36 x 25 = 9 kN/m
2
(Fig)
Hence the increase in total lateral pressure
Δp
A =9 x 6 = 54 kN/m

length. The point of
application of the thrust is at a distance
1/2X6=3 m from the base.
Δp
A
Δp
A
=9

35. Example-3
•A retaining wall with a smooth vertical back is
10 m high and retains a 2-layer sand backfill
with the following properties
0-5 m depth: c′ = 0, f′=30
o
g = 18 kN/m
3
Below 5 m: c′ = 0, f′=34
o
g = 20 kN/m
3

Show the active earth pressure distribution,
assuming that the water table is well
below the base of the wall.

36. Solution:
•When the backfill consists of more than one
soil layer, the lateral pressure distribution for
each of the layers is worked out and a
combined diagram drawn. At the interface of
two layers, there will be a break in the
pressure distribution diagram, since there will
be two values of pressure- one value at the
base of the upper layer and another at the top
of the lower layer. For the lower layer, the
upper layer will act as a surcharge.

37. 2
1
2
2
1
/3090333.0.
/90518;5
0;0;0
:
283.0
34sin1
34sin1
333.0
30sin1
30sin1
mkNKp
mkNpressureverticalmz
ppressureverticalmz
layeruppertheforondistributipressureActive
layerlowwertheforK
layeruppertheforK
vAA
v
Av
o
o
A
o
o
A
=´==
=´==
===
=
+
-
=
=
+
-
=
s
s
s

38. •Active pressure distribution for the lower layer:
z = 5 m; vertical pressure σ
v
=90 kN/m
2
p
A
= K
A2
σ
v
=0.283 x 90 = 25.47 kN/m
2
z = 10 m; σ
v
=90 + 20x5 190 kN/m
2
p
A
= 0.283 x 190 = 53.77 kN/m
2

The active pressure distribution is shown in Fig.
In reality, there can not be a sudden change in
lateral pressure since shear stresses which develop
along the interface have not been considered. But
this does not introduce any serious error in the
magnitude and direction of the resultant thrust

39. c′=0,φ=30
o

γ= 18 kN/m
3
c′=0,φ=34
o

γ= 20 kN/m
3
5 m
5 m
30
25.47
53.77
Fig. Retaining wall and earth pressure diagram

40. Example-4
•A retaining wall, 8 m high, with a smooth
vertical back, retains a clay backfill with c
′= 15

kN/m
2
,f′=15
o
and g=18 kN/m
3
.
Calculate the total active thrust on the
wall assuming that tension cracks
may develop to the full theoretical
depth.

41. Solution:
•The active pressure at a depth z in a c-φ soil is
given by:

m
K
c
zatp
KczKpmkNpmzAt
KcpmkNpzAt
K
KczKp
A
oA
AAAA
AAA
o
o
A
AAA
17.2
588.018
1522
0
)2(/67.6123818588.0,8
)2(/23588.0152,0
588.0
15sin1
15sin1
sin1
sin1
2.
2
2
=
´
=
-
==
-==-´´==
-=-=´-==
=
+
-
=
+
-
=
-=
g
g
f
f
g



42. 8 m
_
2.17
5.83
23
61.67
•From Fig., it can be seen that the pressure is negative
or tensile upto a depth of 2.17 m.
The soil tends to break away from the wall and
tension cracks develop in the soil.
Hence, the resultant
active thrust is obtained
by determining the area
of the hatched portion
of the pressure diagram.
P
A
=1/2 x61.67x 5.83 = 179.8 kN/m

43. Example-5
An excavation was made in saturated soft clay (φ
u

= 0), with its sides more or less vertical. When the
depth of excavation reached 6 m, the sides caved
in. What was the approximate value of cohesion
of the clay soil? Take unit weight of clay = 20
kN/m
3
.
Solution:
The depth of tension crack z
o
in a c-φ
soil is equal to and the depth for which the
total net pressure is zero is equal to 2z
o
or .
A
K
c
g
2
A
K
c
g
4

44. •An excavation in such a soil should be able to
sustain its vertical faces upto a depth of
without any lateral support.
•Critical depth of cut in a φ
u
= 0 soil is therefore
equal to since K
A
= 1 for φ
u
= 0 .
A
K
c
g
4
g
d
c4
2
/30
4
206
4
6
6
4
mkNcor
c
u
u
=
´
==
=\
g
g

45. Example-6
•Fig.(a) shows a three layered backfill behind a 15
m high retaining wall with a smooth vertical
back. Draw the active earth pressure
distribution.
•Layer 1: (c = 0 soil)
41.0
25sin1
25sin1
)(:2
/2727.0100
/100520;5
0.
0;0
27.0
35sin1
35sin1
2
2
=
+
-
=-
=´=
=´==
==
==
=
+
-
=
o
o
A
A
v
vAA
v
o
o
A
KsoilcLayer
mkNp
mkNmz
Kp
pressureverticalz
K
f
s
s
s

46. H
1
=5m


H
2
=5m


H
3
=5m


Layer g c f
(kN/m
3
) (kN/m
2
)
1 20 0 35
2 18 20 25
3 18 35 0
2715.4
52.3
120
200
5x18x0.41=36.9
K
A
g.D
(a) (b)
Fig. (a) Retaining wall with backfill (b) Earth pressure diagram- Example-6

47. •z = 5 m; s
v
= 20 x 5 = 100 kN/m
2


s
v
= K
A
.s
v
– 2c √ K
A


= 0.41 x 100 -2 x 20 √0.41= 15.4 kN/m
2
z = 10 m; s
v
=100 + 18 x 5 =190 kN/m
2

s
v
=0.41 x 190 – 2 x 20 √0.41= 52.3 kN/m
2

Or 15.4 + 18 x 5 x 0.41= 15.4+36.9=52.3
Layer -3: K
A
=1 (f
u
=0 soil)
z=10 m; s
v
=190 kN/m
2

p
A
=p
v
-2c
u
= 190-2 x 35 =120 kN/m
2
[:
.
p
A
= k
A
s
v
-2c
u
√K
A
where K
A
=1]