high thick masonry structural chiefly for defense —often used in plural b : a masonry fence around a garden, park, or estate

2 Retaining Walls - Applications
high
way

3 Retaining Walls - Applications
basement wall
High-rise
building

4
Retaining Walls - Applications
Road
Train
Metros and Subways

TYPES
GRAVITY WALLS
RETAINING WALLS

TYPES
RETAINING WALLS
CANTILEVER

TYPES
RETAINING WALLS
COUNTERFORT

TYPES
RETAINING WALLS
COUNTERFORT
School of Civil Engineering-AIT
SIIT-Thammasat University

TYPES
RETAINING WALLS
BUTTRESS

PARTS
CANTILEVER RETAINING WALLS
STEM
or
Wall Slab
TOE HEEL
KEY
BACKFILL
FRONT

Lateral Earth Pressure
(R.P. Weber)
(R.P. Weber)
??
??

Water Pressure and Soil Pressure
Consider “at-rest” (geostatic) condition
Consider hydrostatic
condition
Anisotropic
sz
sz ≠ sz sx
>
Isotropic
sx
sx

y
gy

EARTH PRESSURES
 PRESSURE AT REST
 ACTIVE EARTH PRESSURE
 PASSIVE EARTH PRESSURE

PRESSURE AT REST
RIGID

At Rest Earth Pressure
One common earth pressure coefficient for the “at rest”
condition in granular soil is:
Ko = 1 – sin(φ)
Where: Ko is the “at rest” earth pressure coefficient and φ is
the soil friction value.
K0gh
h
gz
K0gz
o
o K
h
E 2
2
1
g

z
h/3

EARTH PRESSURES

ACTIVE EARTH PRESSURE
s3 = s1 . tan2 (45-/2)-2c.tan (45-/2)
RANKINE ACTIVE EARTH PRESSURE

Ka = tan2 (45 - /2)
sa = sv . Ka – 2cKa
RANKINE ACTIVE EARTH PRESSURE
s3 = s1 . tan2 (45-/2)-2c.tan (45-/2)
sa = sv . tan2 (45-/2)-2c.tan (45-/2)

28
Active Stress Distribution (c ≠ 0)
-
γ
c ≠ 0
Φ
dry soil
H
Ka γH
zo
2 c (Ka)1/2 Ka γH – 2 c (Ka)1/2
=
Find zo:
Ka γzo – 2 c (Ka)1/2 = 0
Zo = 2c / γ (Ka)1/2
Pa = ?
_

Note :
z = 0  sv = 0 ; sa = -2cKa
z = H  sv = gH
The tensile stress decreases with depth and becomes zero at a depth z = zc
or
gzcKa – 2cKa = 0
and
a
c
K
c
z
g
2
 zc = depth of tensile crack

RANKINE ACTIVE EARTH PRESSURE FOR INCLINED BACKFILL
  1
cos
sin
cos
8
cos
4
cos
cos
cos
4
sin
cos
2
cos
2
cos
1
'
:
2
2
2
2
2
2
2
2














































 


g

g





g

 z
c
z
c
z
c
K
where
a












2
2
2
2
cos
cos
cos
cos
cos
cos
cos
Ka
Ka
.
H
.
.
Pa 2
2
1 g

(for granular soil, c = 0)
For c- soil

g
g
s cos
'
a
a
a zK
zK 


PASSIVE EARTH PRESSURE

RANKINE PASSIVE EARTH PRESSURE

sp= sv . tan2(45+/2) + 2c . tan (45+/2)

Kp = tan2 (45 + /2)
sh = sv . Kp + 2cKp

40
Passive Stress Distribution (c ≠ 0)
-
γ
c ≠ 0
Φ
dry soil
H
Kp γH 2 c (Kp)1/2 Kp γH + 2 c (Kp)1/2
=
Pp = ?
+
Kp = tan2 (45 + /2)
sh = sv . Kp + 2cKp

Relation among three earth pressures
-△ +△
+△
-△
E
o
△a △p
Ea
Eo
Ep

Example 1

Example 2
h=5m
g1=17kN/m3
c1=0
1=34o
g2=19kN/m3
c2=10kPa
2=16o
h
1
=2m
h
2
=3m
A
B
C

A
B
C
h=5m
h
1
=2m
h
2
=3m
0
1
1 
 a
aA zK
p g
kPa
K
h
p a
aB 4
.
10
1
1
1 ＝
上 g

kPa
K
c
K
h
p a
a
aB 2
.
4
2 2
2
2
1
1 ＝
－
下 g

kPa
K
c
K
h
h
p a
a
aC 6
.
36
2
)
( 2
2
2
2
2
1
1 


 g
g
m
kN
Ea /
6
.
71
2
/
3
)
6
.
36
2
.
4
(
2
/
2
4
.
10 ＝





10.4kP
a
4.2kPa
36.6kP
a
Solution:

48
Active Stress Distribution (c = 0)
γ
c = 0
Φ
dry soil
H
σa‘ = Ka σv’ – 2 c (Ka)1/2
σa‘ = Ka σv’ 0
σa‘ is the stress distribution
Pa is the force on the wall (per foot of wall)
How is Pa found?
Pa = ?
? - What is this value

49
Passive Stress Distribution (c = 0)
γ
c = 0
Φ
dry soil
H
0
σp‘ is the stress distribution
Pp is the force on the wall (per foot of wall)
How is Pp found?
Pp = ?
? - What is this value
σp‘ = Kp σv’ – 2 c (Kp)1/2
σp‘ = Kp σv’

50
Stress Distribution - Water Table (c = 0)
H1
Ka γ H1
or
Pa = Σ areas = ½ Ka γH1
2 + Ka γH1H2 + ½ Ka γ’H2
2 + 1/2γwH2
2
H2
Ka γ H1 Ka γ’ H2
Ka (γ H1 + γ’ H 2)
Effective Stress Pore Water Pressure
Pa
γw H2

51
Stress Distribution With Water Table
H1
Ka γ H1
or
H2
0
Ka γ H1 Ka γ’ H2
Ka (γ H1 + γ’ H 2)
Effective Stress Pore Water Pressure
Pa
γw H2
Why is the water pressure considered separately? (K)

Assumptions:
-Fill material is
granular soil
- Friction of
wall and fill
material is
considered
- Soil failure
shape is plane
(BC1, BC2 …)
Pa = ½ Ka . g . H2
 
2
2
2
)
sin(
).
sin(
)
sin(
).
sin(
1
sin
.
sin
)
(
sin
Ka



























COULOMB ACTIVE EARTH PRESSURE

COULOMB’S ACTIVE EARTH PRESSURE WITH A SURCHARGE ON THE
BACKFILL

 
2
2
2
)
sin(
).
sin(
)
sin(
).
sin(
1
sin
.
sin
)
(
sin
Kp





























Pp = ½ Kp . g . H2
COULOMB PASSIVE EARTH PRESSURE

STABILITY
OVERTURNING
SLIDING
BEARING

OVERTURNING
Highway Loading (Surcharge)

OVERTURNING
Overturning Forces
No Surcharge Here
Full Surcharge Here
Active Pressure
Soil+Surcharge

OVERTURNING
Restoring Forces
No Passive
Pressure
Weight of Soil
Weight of Wall
Weight of Soil
(with care)

OVERTURNING
Restoring Moment
Overturning Moment
FOS vs OT =
A FOS = 2 is considered sufficient

SLIDING
Sliding Forces
No Surcharge Here
Full Surcharge Here
Active Pressure
Soil+Surcharge
H1

SLIDING
Resisting Forces
H2 +  S V
=Coeff of Friction
No Surcharge Here
H2
Vs1
Vs2
Vc1
Vc2 Vc3
Resisting Forces

SLIDING without KEY
Passive Earth Pressure Force+ S V
Active Earth Pressure Force
FOS vs Sliding =
A FOS = 1.5 is considered sufficient

SLIDING with KEY
Sliding Forces
No Surcharge Here
Active Pressure
Soil+Surcharge

Resisting Forces
No Surcharge Here
H
Vs1
Vs2
Vc1
Vc2 Vc3
SLIDING with KEY

Find Vertical forces
acting in front and
back of key
No Surcharge Here
Vs2
Vc1
Vc2
Vs1
Vc3
RESULTANT
Active Pressure
Soil+Surcharge
SLIDING with KEY

Determine Pressure
Distribution Under Base
B
B/2
e A=B
S=B2/6
2
6
B
Ve
B
V

2
6
B
Ve
B
V

V
x
SLIDING with KEY

Determine Force in Front of KEY
B
x1
P1 P2
y1
y2
y3
y3=y2+(y1-y2) (B-x1)/B
P1=(y1+y3) x1/2
P2=V-P1
SLIDING with KEY

When Pressure Distribution Under
Base is Partially Negative
B
B/2
e
2
6
B
Ve
B
V

2
6
B
Ve
B
V

V
SLIDING with KEY

B
x
e
2
6
B
Ve
B
V

2
6
B
Ve
B
V

V
3x
2V
3x
Determine P1 and
P2 once again
P1 P2
SLIDING with KEY

Total Sliding Force = H1
Total Resisting Force = P1 tan    P2 + H2
Force in Front of Key
Internal Friction of Soil
Passive Earth
Pressure Force
Force on and
Back of Key
Friction b/w Soil, Concrete
Active Earth Pressure Force
SLIDING with KEY

BEARING
1. No surcharge on heel
2. Surcharge on heel
There are two possible critical conditions

BEARING
No Surcharge on Heel
Vs2
Vc1
Vc2
Vs1
Vc3
RESULTANT
Active Pressure
Soil+Surcharge
This case has been dealt already

BEARING
Surcharge on Heel
Vs2
Vc1
Vc2
Vs1
Vc3
RESULTANT
Active Pressure
Soil+Surcharge
Vs
DETERMINE THE PRESSURE
DISTRIBUTION UNDER BASE SLAB

Determine Pressure
Distribution Under Base
B
B/2
e A=B
S=B2/6
2
6
B
Ve
B
V

2
6
B
Ve
B
V

V
x

Compare Pressure with
Bearing Capacity
B
2
6
B
Ve
B
V

2
6
B
Ve
B
V

2
6
B
Ve
B
V

FOS vs Bearing =
Allowable Bearing
Max Bearing Pressure

ALTERNATELY
B
2
6
B
Ve
B
V
 2
6
B
Ve
B
V

FOS vs Bearing =
Allowable Bearing
Max Bearing Pressure
2V/3x
2V/3x
3x

BENDING OF WALL

DESIGN OF STEM
CRITICAL SECTIONS
Active Pressure
Soil+Surcharge
Critical Section
Moment
Critical Section Shear
d

DESIGN OF STEM
Design Moment
=1.6 (H1 y1 + H2 y2)
H1=Ca s h
Surcharge = s N/m2
H2=0.5 Ca gs h2
y1 y2
h

DESIGN OF STEM
Design Shear=1.7(H '1+H '2)
H'1=Ca s (h-d)
Surcharge = s N/m2
H'2=0.5 Ca gs (h-d)2
d
h













 


2
2
1
7
.
1
h
d
h
H
h
d
h
H

DESIGN OF TOE SLAB
CRITICAL SECTIONS
Critical Section
Moment
d
Critical Section (Shear)

DESIGN OF TOE SLAB
Design Loads
1.6Soil Pressure
0.9 Self Wt
0.9 Soil in Front
(may be neglected)

TOE : DESIGN MOMENT
1.6(0.5 T y3) T/3
+1.6(0.5 T y1) 2T/3
-0.9 wc T2/2
-0.9 ws T2/2
T
y1
y3

TOE : DESIGN SHEAR
1.6(0.5 Ts) y3 Ts/T
+1.6(0.5 T y1-0.5 d [y1/T] d)
-0.9 wc Ts
-0.9 ws Ts
Ts=T-d
y1
y3

DESIGN OF HEEL SLAB
CRITICAL SECTIONS
Critical Section
Moment & Shear
TENSION FACES

DESIGN OF HEEL SLAB
DESIGN LOADS
1.6s + 1.2 gs +1.2 gc
Soil Pressure Neglected

MAIN REINFORCEMENT
Minimum 75 mm Clear Cover

ACI CODE
SECONDARY STEELS
ACI Minimum SLAB
ACI 14.3.3
ACI 14.3.2

DRAINAGE
Sand + Stone Filter
Weepers
Or
Weep Holes

DRAINAGE
Drainage Pipes f 100-200 mm @ 2.5 to 4 m

DRAINAGE (Alternate)
Perforated Pipe
Suited for short walls

END OF PART III
End of Part III

Active and Passive Limit Conditions
Ka = Coefficient of Active
Earth Pressure
(Wall Moving Away from
Backfill)
(small sx)
Active Failure Condition movem
ent
Active
Failure
Wedge
(45+/2
)
Kp = Coefficient of
Passive Earth Pressure
(Wall Moving Toward
Backfill)
Passive Failure Condition movem
ent
Passive
Failure
Wedge
(45 -
/2)































2
'
45
tan
'
sin
1
'
sin
1
..
'
sin
1
'
sin
1
'
'
2 




s
s
a
x
z
K
so
Pole
Point
Active
Failure
45
/2
45
/2
Rankine Active Failure Surface

Pole
Point
Rankine Passive Failure Surface






























2
'
45
tan
'
sin
1
'
sin
1
..
'
sin
1
'
sin
1
'
'
2 




s
s
p
z
x
K
so
45
/2
Passive
Failure

movem
ent
Passive
Failure
Consider Mohr’s Circles… sx decreases until failure
sx increases until failure

movement…
Active Failure
at Ka
Passive Failure
at Kp
Stationary (at
rest)
Movement
toward
backfill
Movement
away
from backfill
Ka < K0< Kp

101
• We can calculate σv’
• Now, calculate σh’ which is the horizontal stress
• σh‘/ σv‘ = K
• Therefore, σh‘ = Kσv‘ (σV‘ is what?)
σv’
σh’
H

102
• There are 3 states of lateral earth pressure
Ko = At Rest
Ka = Active Earth Pressure (wall moves away from soil)
Kp = Passive Earth Pressure (wall moves into soil)
Passive is more like a resistance
σv
σh
z
H

103
At Rest Earth Pressure
At rest earth pressure occur when there is no wall rotation such as
in a braced wall (basement wall for example)
Ko can be calculated as follows:
Ko = 1 – sin φ for coarse grained soils
Ko = .44 + .42 [PI / 100] for NC soils
Ko (oc) = Ko (NC) (OCR)1/2 for OC soils
σv
σh
z
H

AT REST EARTH PRESSURE
q
sv
sh
z sv = g . z + q
v
h
K
s
s

At rest, K = Ko
Jaky, Broker and Ireland  Ko = M – sin ’
Sand, normally consolidated
clay  M = 1
Clay with OCR > 2  M =
0.95
Sherif and Ishibashi  Ko =  +  (OCR – 1)
 = 0.54 + 0.00444
(LL – 20)
 = 0.09 + 0.00111
Broker and Ireland
Ko = 0.40 + 0.007 PI , 0  PI  4
Ko = 0.64 + 0.001 PI , 40  PI 

high thick masonry structural chiefly for defense —often used in plural b : a masonry fence around a garden, park, or estate

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high thick masonry structural chiefly for defense —often used in plural b : a masonry fence around a garden, park, or estate