Rankine’s lateral earth pressure ...
Embed Size (px)
Transcript of Rankine’s lateral earth pressure ...

Foundation Analysis LATERAL EARTH PRESSURE

INTRODUCTION
Vertical or nearvertical slopes of soil are supported by
retaining walls, cantilever sheetpile walls, sheetpile
bulkheads, braced cuts, and other similar structures. The
proper design of these structures requires an estimation of
lateral earth pressure, which is a function of several factors,
such as a) the type and amount of wall movement, b) the
shear strength parameters of the soil, c) the unit weight of the
soil, and d) the drainage conditions in the backfill.

INTRODUCTION
Lateral earth pressure is a function of wall movement (or
relative lateral movement in the backfill soil).

LATERAL EARTH PRESSURE AT REST
Consider a vertical wall of height H, as shown in Figure 7.3,
retaining a soil having a unit weight of . A uniformly
distributed load, q/unit area, is also applied at the ground
surface.
(No Lateral Movement)

LATERAL EARTH PRESSURE AT REST (No Lateral Movement)
The shear strength of the soil is,
= +
where,
c is the cohesion
is the effective normal stress
is the effective angle of friction
At any depth z below the ground surface, the vertical
subsurface stress and lateral earth pressure are expressed as,
0 = +
= 00 +
where,
u is the pore water pressure
0 is the coefficient of atrest earth pressure

LATERAL EARTH PRESSURE AT REST (No Lateral Movement)
For normally consolidated soil (Jaky, 1944)
0 = 1 For overconsolidated soil (Mayne and Kulhawy, 1982)
0 = (1 )
The total force, 0 , per unit length of the wall can now be obtained from the area of the pressure diagram as,
0 = 1 + 2 = 0 +1
220
The location of the line of action of the resultant force, 0, can be obtained by taking the moment about the bottom of the wall. Thus,
=1
2
+ 23
0
Note: If the surcharge = 0 and the pore water pressure = 0, the pressure diagram will be a triangle.

LATERAL EARTH PRESSURE AT REST (No Lateral Movement)
If the water table is located at a depth z < H, the atrest
pressure diagram will have to be somewhat modified.

PROBLEM SET 10
1. For the retaining wall shown in the figure below, determine
the lateral earth force at rest per unit length of the wall. Also determine the location of the resultant force. Assume OCR = 1.

LATERAL EARTH PRESSURE
ACTIVE AND PASSIVE
Based on assumptions of the intervening forces and the failure
mode, different theories have been developed.
These theories differ only in terms of the coefficient of lateral
earth pressure but operate with similar stress/pressure
equations.
The three widelyaccepted theories are the following: 1. Rankines
2. Coulombs
3. Logspiral

LATERAL EARTH PRESSURE
ACTIVE AND PASSIVE
Below are the comparison of the three theories and their
applicability.
Method Failure
Mode Wall Friction
Active Case Passive Case
based on experimentation and actual
failure observations
Rankine planar no wall friction poor estimate poor estimate
Coulomb planar considered good estimate (less) poor
estimate
Logspiral curved considered better
estimate
better
estimate

RANKINES LATERAL EARTH PRESSURE
RANKINES THEORY
z 'v
'h
A
B
Unit weight of soil =
Assumptions:
Vertical frictionless wall
Dry homogeneous soil
Horizontal backfill
' tan ' c' f

RANKINES LATERAL EARTH PRESSURE
ACTIVE EARTH PRESSURE
z
'h
A
B
If wall AB is allowed to move
away from the soil mass
gradually, horizontal stress
will decrease.
Plastic equilibrium in soil
refers to the condition
where every point in a soil
mass is on the verge of
failure.
This is represented by Mohrs
circle in the subsequent
slide.
Unit weight of soil =
' tan ' c' f
'v

RANKINES LATERAL EARTH PRESSURE
o Koo 'a c'
Based on the diagram:
'sin 1
'sin  1 )
2
'  (45 tan
'
' 2
0
aaK
aa0
2
0
K2c'  K '
)2
'  (45 tan 2c'  )
2
'  (45tan' '
a
' ' tan ' c' f
ACTIVE EARTH PRESSURE
is the Rankine active earth pressure coefficient

RANKINES LATERAL EARTH PRESSURE
ACTIVE EARTH PRESSURE

RANKINES LATERAL EARTH PRESSURE
aa0 K2c'  K'
z
aK2c' aK2c' 
K' a0
ACTIVE EARTH PRESSURE DISTRIBUTION
a
cK
cz
'2

RANKINES LATERAL EARTH PRESSURE
z 'v
'h
A
B
If the wall is pushed into the
soil mass, the principal stress
h will increase. On the verge of failure the stress
condition on the soil
element can be expressed
by Mohrs circle b.
The lateral earth pressure,
p, which is the major principal stress, is called
Rankines passive earth
pressure.
Unit weight of soil =
' tan ' c' f
PASSIVE EARTH PRESSURE

RANKINES LATERAL EARTH PRESSURE
Shea
r st
ress
Normal stress
' tan ' c' f
C
D
D
O A
'p Koo
b
a
o ' c'
pp0
2
0
K2c' K'
)2
' (45 tan 2c' )
2
' (45 tan' '
p
'sin 1
'sin 1 )
2
' (45 tan
'
'2
0
p
pK
PASSIVE EARTH PRESSURE
is the Rankine passive
earth pressure coefficient

RANKINES LATERAL EARTH PRESSURE
PASSIVE EARTH PRESSURE

RANKINES LATERAL EARTH PRESSURE
For cohesionless soil,
z
K z ppK2c'
ppv K z K p
PASSIVE EARTH PRESSURE DISTRIBUTION

RANKINES LATERAL EARTH PRESSURE
SPECIAL CASES
Submergence:
Inclined backfill:
Inclined but smooth back face of wall:

RANKINES LATERAL EARTH PRESSURE
SPECIAL CASES
Inclined backfill with c soil:

RANKINES LATERAL EARTH PRESSURE
ILLUSTRATIVE PROBLEM
A frictionless retaining wall is shown in Figure 12.22a. Determine:
a) The active force Pa after the tensile crack occurs.
b) The passive force Pp.
z
A
B
= 15 kN/m3
= 26o
c = 8 kN/m2
q = 10 kN/m2
H = 4 m
Figure 12.22a Frictionless retaining wall AB

RANKINES LATERAL EARTH PRESSURE
SOLUTION
a) The active force Pa after the tensile crack occurs.
Ka =1 sin
1 + sin=1 si n( 26)
1 + si n( 26)= .
a = Kav 2c Ka
v = z
At z = 0 m;
= q = 10 kN/m2
= 0.39(10) 2(8)( 0.39) = . kN/m2
At z = 4 m;
= 10 + 15(4) = 60 kN/m2
= 0.39(60) 2(8)( 0.39) = . kN/m2

RANKINES LATERAL EARTH PRESSURE
SOLUTION
A
B
= 15 kN/m3
= 26o
c = 8 kN/m2
q = 10 kN/m2
H = 4 m
Figure 12.22 (a) Frictionless retaining wall AB, and (b) active pressure distribution diagram
6.09 kN/m2
17.32 kN/m2
y = 1.04 m
4 y = 2.96 m
(a) (b)

RANKINES LATERAL EARTH PRESSURE
SOLUTION
6.09 kN/m2
17.32 kN/m2
y = 1.04 m
4 y = 2.96 m
(b)
From Figure 12.22b,
6.09
y=17.31
4 y; = . m, 4 = . m
Pa =1
217.31 2.96 = . kN/m
The active force Pa after the tensile crack occurs is equal to the area of the active pressure distribution diagram below point C, or
and the location is located at
z =1
32.96 = . m
C
Pa
z=0.99 m

RANKINES LATERAL EARTH PRESSURE
SOLUTION
b) The passive force Pp.
Kp =1 + sin
1 sin=1 + si n( 26)
1 si n( 26)= .
At z = 0 m;
= q = 10 kN/m2
= 2.56(10) + 2(8)( 2.56) = . kN/m2
At z = 4 m;
= 10 + 15(4) = 60 kN/m2
= 2.56(60) + 2(8)( 2.56) = . kN/m2
p = Kpv + 2c Kp
v = z

RANKINES LATERAL EARTH PRESSURE
SOLUTION
A
B
= 15 kN/m3
= 26o
c = 8 kN/m2
q = 10 kN/m2
H = 4 m
Figure 12.22 (a) Frictionless retaining wall AB, and (c) passive pressure distribution diagram
(a) (c)
51.2 kN/m2
51.2 kN/m2 153.6 kN/m2

RANKINES LATERAL EARTH PRESSURE
SOLUTION
(c)
From Figure 12.22c,
Pp = 51.2 4 +1
2153.6 4 = kN/m
The passive force Pp is equal to the area of the passive pressure distribution, or
and the location is calculated by taking summation moment at the base, or
512 z = 51.2 44
2+
1
2153.6 (4)
1
3(4)
51.2 kN/m2
51.2 kN/m2 153.6 kN/m2 z = . m
Pp
z=1.6 m

RANKINES LATERAL EARTH PRESSURE
FINAL ANSWERS
a) The active force Pa after the tensile crack
occurs has a magnitude of 25.62 kN per unit
length of the frictionless retaining wall and is
acting at 0.99 meters above the base.
b) The passive force Pp has a magnitude of 512 kN
per unit length of the frictionless retaining wall
and is acting at 1.6 meters above the base.

PROBLEM SET 10
2. Assume that the retaining wall shown in the figure below
can yield sufficiently to develop an active state. Determine
the Rankine active force per unit length of the wall and the
location of the resultant line of action.

PROBLEM SET 10
3. Assume that the retaining wall shown in the figure below
can yield sufficiently to develop passive state. Determine the
Rankine passive force per unit length of the wall.

COULOMBS LATERAL EARTH PRESSURE
ACTIVE EARTH PRESSURE

COULOMBS LATERAL EARTH PRESSURE
PASSIVE EARTH PRESSURE

COULOMBS LATERAL EARTH PRESSURE
where,
is the angle the back face is inclined with the horizontal
is the inclination of backfill with the horizontal
is the wall friction angle
is the angle of internal friction

PROBLEM SET 10
4. A retaining wall shown below has a height of 4.5 m. The unit
weight of soil is 16.5 /3. The angle of internal friction of soil is 36, the wall friction angle is 24, and soil cohesion is 0. The wall is supporting a horizontal backfill. 4.1 Compute the Coulombs active earth pressure coefficient.
4.2 Compute the Coulombs active force per unit length of wall.
4.3 Compute the location of the Coulombs active force from the bottom of
the wall.

PROBLEM SET 10
5. A vertical retaining wall has a height of 4 m and is
supporting a horizontal backfill. The unit weight of soil is 16.5 /3. The angle of internal friction of soil is 35, the wall friction angle is 20, and soil cohesion is 0. 5.1 Compute the Coulombs passive earth pressure coefficient.
5.2 Compute the Coulombs passive force per unit length of wall
perpendicular to the wall.
5.3 Compute the location of the Coulombs passive force from the bottom
of the wall.