OUTLINE
Applied Geotechnics
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Introduction
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Lateral Earth Pressu res – At rest – Active
RETAINING WALLS
A/Prof Hadi Khabbaz Email:
[email protected] Room CB11.11.224 CB11.11.224
– Passive
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Rankine’ Rankine’s s Theory Theory
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Coulomb’s Theory
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Types of Retaining Walls
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Design of Gravity Walls
http://www.geofffox.com/MT/archives/2005/05/12/the-collapse-along-the-henry-hudson-parkway.php
A retaining wall holding back a steep hill gave way, allowing the hill to tumble onto the Henry Hudson Parkway and its access road. Debris was piled up at least 7.5 m deep and you could see partially buried cars at the edge of the slide area. This all took place in the shadow of the George Washington Bridge.
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This retaining wall collapsed after an unknown person drove a stolen vehicle into it in May 2013. The City Council accepted a bid to fix the wall for for $34 K.
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A retaining wall, which has collapsed on top of cars, is seen after heavy rains caused by Typhoon Sanba in Yeosu, about 460 km (286 miles) south of Seoul September 17, 2012. Tens of thousands of people were forced to evacuate and hundreds of sea and air passenger services were cancelled as the powerful typhoon Sanba made landfall in southern South Korea on Monday, local Yonhap news agency reported.
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Retaining Wall Collapse in Sydney
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Lateral Earth Pressures
Lateral Earth Pressure at Rest Estimation of Ko:
In normally consolidated clays and granular soils: Ko = 1 – sin
The conditions in the bed of a river consist of 1m of water over sandy clay: 0 - 1m Water 1 - 10m Clay (overconsolidated) t = 20 kN/m3 = 30
OCR = 2
In over-consolidated soils: Ko = (1 – sin )
Example 1
OCR0.5
Determine the total horizontal stress (h) at a point in the soil 5 m below the surface of the water in the river, which is 4 m below the top of the soil layer.
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Rankine’s Theory
Solution of the Example 1m
Assume that the wall is frictionless;
Applicable to horizontal soil surface;
4m Sandy clay
OCR=2
s = 20 kN/m 3
′ = 30
The vertical stress may then be calculated in the usual way: v = t z.
The vertical and horizontal stresses throughout the retained soil mass are the principal stresses:
The normal stress acting on the wall will thus be a principal stress; The retained soil is assumed to be at failure everywhere and the horizontal stress can be calculated from the failure criterion.
Types of Analyses
Rankine’s Theory Active Earth Pressure
Drained:
Clays: long time after construction, sand and gravel: always.
Passive Earth Pressure
No excess pore pressure in the soil.
Use drained strength parameters in an effective stress analysis, effective stresses, , drained cohesion and friction angle, c & ;
In the presence of water, the water pressure must be applied to the wall.
Types of Analyses
Undrained:
Clays, immediately after construction.
Example 2
Find distribution of active earth pressure on the wall and the resultant horizontal force acting per metre run of the wall.
There may be excess pore pressure in soil;
Sand
Use undrained strength parameters in a total stress analysis, total stresses, , undrained cohesion, c u and undrained
3m
t = 16 kN/m 3 =30o
friction angle, u (zero for saturated normally consolidated clay).
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Example 3
Solution of Example 2
Find distribution of active earth pressure on the wall.
2m
Sand t = 21 kN/m3 = 30o, c = 0
2m
Clay t = 20 kN/m3 = 0, c = 30kPa
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Solution of Example 3
Tension Crack
Limitations in Rankine’s Theory
In a total stress analysis, Rankine’s theory indicates negative stress in cohesive soils under a low vertical stress. Soil particles cannot provide tensile strength. Cracks may develop in the tensile zone, which may be filled with water if soil is submerged. Water in Depth of tension crack: zc the crack
Assume that the wall is frictionless;
In reality no retaining wall is smooth.
w zc
Applicable to vertical walls.
Applicable to horizontal soil surface;
No surcharge
The method can be modified for non-vertical walls.
Application to inclined surface is possible, but not simple.
With surcharge
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Rankin’s Equations Chu (1991) Frictionless soil-wall interface
General Formulas for Lateral Earth Pressures
Rankin’s Equations
Coulomb’s Limit Equilibrium Equations Poncelet (1840)
Coulomb’s Limit Equilibrium Equations
Different Types of Retaining Walls
Passive earth pressure is overestimated . The error is small if
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Types of Retaining Walls
Choice of the walls:
Masonry or concrete gravity walls
Cantilever walls
Counterfort and buttress walls
Sheet piling
Anchored walls
Gabions
Others
(e.g. crib retaining walls, Geosynthetic reinforced soil walls)
Types of Retaining Walls
Conterfort and Buttress Walls
Gravity walls
Rely on their weight for stability Massive concrete or masonry walls Cantilever walls Common height: 3-4m Gabions walls Rarely > 8m Counterfort and buttress
Embedded walls
Rely on passive pressure on embedded section for stability Sheet piling Slurry concrete walls Soldier Beam and Lagging
Hybrid system
Cost effective for temporary excavations May show significant ground movement
Anchored walls Reinforced soil walls 40
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Modes of Failure
Construction of shoring using soldier beams and lagging
Modes of failure
Sliding
Rotation
Bearing failure
Global failure
Excessive
Excessive
settlements deformation
http://arconcretecorp.com/Quickstart/ImageLib/P7270144.JPG 43
Photo taken from: Guney Olgan (2003)
Separation of the facing panels due to vertical distortion and spilled backfill material
Design of Retaining Walls Using a single global factor of safety:
Design of Retaining Walls
F = 1.5 – 2.0 for sliding/overturning failure 2.0 – 4.0 for bearing capacity failure
Using partial factors of safety:
Strength parameters decreased by reduction factors; Applied loads increased by load factors;
Limit resisting force > Ultimate disturbing force Limit resisting moment > Ultimate disturbing moment
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Retaining Walls
Retaining Walls
Design for strength:
Overturing failure
Design for strength:
Moments are usually calculated around the toe of the foundation of the wall.
Factor of safety depends mainly on the passive pressure.
Moment due to active earth pressure is normally a function of (height) 3. It becomes increasingly large when the wall height is large.
It is normal to ignore 0.5 to 1m of top soil in front of the wall to reduce the possibility of failure due to inadvertent or careless excavation.
Action forces Resistance forces
Action forces Resistance forces
Retaining Walls
Sliding failure
Design for strength:
It is normal to ignore 0.5 to 1m of top soil in front of the wall to reduce the possibility of failure due to inadvertent excavation.
Shear key
Retaining Walls Design for strength:
Global failure
A suitable method shall be employed to check the overall stability of the soil and wall together against general global failure. Methods of analyses will be discussed later.
Bearing failure
Eccentricity and inclination of the resultant force on the foundation reduces the vertical capacity of the foundation.
Factor of safety depends mainly on the passive pressure.
Action forces Resistance forces
Retaining Walls
Design for strength:
Sliding failure
Retaining Walls
Design for strength:
Structural failure Sections of wall and foundation shall be checked against structural failure.
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Example 4 Design of a Gravity Wall
Solution of Example 4 Using a globe Factor of Safety approach
Design based on a single global factor of safety 1.5m
Determine the stability of the concrete gravity wall shown in the figure.
Sand =30o
2.5m
t=18kN/m3
0.5m
t=25kN/m3 = 2/3
a) Check for Sliding
b) Check for Overturning
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c) Check for Bearing Capacity using Hansen’s theory
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Bearing Capacity Equation
Hansen’s Bearing Capacity Factors
Terzaghi’s equation:
Hansen’s equation:
For ≠ 0: qu = cNcscdcicgcbc + qoNqsqdqiqgqbq + 0.5BNsdigb
For = 0: qu = 5.14 cu (1 + sc + dc - ic – bc - gc) + qo
Factors
Shape Factors
s
sc , sc , sq , s
Depth Factors
d
dc , dc , dq , d
Inclination Factors
i
ic , ic , iq , i
Ground Factors
g
gc , gc , gq , g
Base Factors
b
bc , bc , bq , b 64
c) Check for Bearing Capacity using Hansen’s theory
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Example 5 Design of a Gravity Wall A concrete cantilever wall is shown in the following figure. Determine the value of B to reach a minimum global factor of safety of 2 for overturning stability of the wall. Groundwater table is 5m below the base of the wall. The effect of passive force due to soil should be included in your calculations. (NOTE: no need to determine the factor of safety for sliding stability or bearing capacity strength.)
Solution of Example 5
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Resisting Moments
4 2 1 3
Australian Standard
AS 4678-2002 “Earth Retaining Structures” specifies a limit state design approach for earth retaining walls. All loads are increased by appropriate load factors.
Load factors: 1.5 for live loads, minimum live load of 5 kPa. 1.0 for hydrostatic pressures. 1.25 to soil unit weight of disturbing forces 0.8 to soil unit weight of resisting forces
Australian Standard
Australian Standard AS 4678-2002 “Earth Retaining Structures”
All strength properties are reduced by appropriate strength reduction factors.
Factors vary depending upon the uncertainty associated with soil parameters
Load factors:
Based mainly on AS 1170: Load factor
Strength
Stability
Serviceability
Dead load of wall and contained soil
1.25
0.8
1
Dead load of earth pressure behind wall
1.25
1.25
1 1
0.8
0.8
Factors are different for c and , refer to
Dead load of fill in front of wall Water pressures on either side of wall
1
1
1
tables in the handout.
Live load on top of wall and contained soil
1.5
0
0.7 or 0.4*
Live load on backfill behind wall
1.5
1.5
0.7 or 0.4*
0
0
0
Live load on fill in front of wall
* 0.7 for long term case and 0.4 for short term case
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Australian Standard
Australian Standard
Strength reduction factors:
Cohesion, c, and friction angle, , are reduced by a material strength factor, , so that the design cohesion, c*, and the design friction angle, *, are: c* = uc c * = tan-1( u tan( ) ) Strength reduction factor
Fill class I (98%)
Fill class II (95%)
Uncontrolled fill
Design action effects need to be calculated for all possible modes of failure (S*).
The design resistance effects (R*) is calculated for all possible modes of failure by further reducing the resistance effects by a structural classification factor, n. n varies from 0.9 to 1.1 The design action effects must be less than the design resistance effects:
Insitu soil
Drained parameters, c , 0.90
Strength
u
0.95
0.75
0.85
uc
0.90
0.75
0.50
0.70
Serviceability
u
1.0
0.95
0.90
1.00
uc
1.0
0.85
0.65
0.85
n
Undrained parameters, cu Strength
u
0.6
Serviceability
uc
0.9
0.5
0.3
0.5
0.8
0.5
0.75
R* ≥ S*
Australian Standard
Structural classification factor
The Australian standard AS4678 takes into account the consequence of failure of the retaining wall in design.
Strength analysis
Q = 1.5
Q = 1.5
G = Dead load (DL) factor Q = Live load (LL) factor
Structure Type
Consequence of failure
n
G = 1.25 1
2
3
Where failure would result in significant damage or loss of life
0.9
Failure results in moderate damage and loss of services
1.0
Failure results in minimal damage and loss of access
1.1
G = 1.25 Q=0
G = 0.8
Australian Standard
Stability analysis
Q=0
Q = 1.5
Australian Standard
Serviceability analysis
G = Dead load (DL) factor
G = Dead load (DL) factor
Q = Live load (LL) factor
Q = Live load (LL) factor
G = 0.8
G = 0.8
Q = 0.7
G = 1.0 G = 1.25
Q=0
Q = 0.4
G = 1.0 Q=0
G = 1.0
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Example 6 Design of a Cantilever Wall 1.6m
0.4
Design based on Australian Standard AS 4678:
1.5m
Minimum Surcharge 5 kPa
Sand
Structure type: 2
4.4m
= 35o
Uncontrolled Fill
t=
0.6m
t=
25 kN/m3
18 kN/m3
0.4m
=
Retaining Walls
Check for Overturning (Structure Type: 2)
S* = 35 + 173 - 1.4 S* ≈ 207 k N.m
pmr
Gravity Forces
1 x 388 > 207 kN.m pmr Check for Sliding
(Structure Type: 2)
The mass of soil above the toe (front of the wall) is ignored. i.e. 3.7 kN.m pmr
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Check for Bearing Capacity
Zone
Force (kN)
Arm (m)
Moment (kN.m)
1
0.4 x 4.6 x 31.25 = 57.5
0.05
-2.9
2
0.4 x 3.5 x 31.25 = 43.8
0
0
3
1.5 x 4.6 x 22.5 = 155.2
1
-155.2
4
7.5 x (1.5 + 0.4) = 14.2
0.8
-11.4
5 0.2 x 1.6 x 14.4 = 4.6 0.95 +4.4 ---------------------------------------------------------------------------------------------------Total: V = 277.9 kN pmr -160.3 kN.m pmr ---------------------------------------------------------------------------------------------------Moments due to active force
due to Passive force
M = -160.3 + (35 + 173) - 1.4 = 46.3 kN.mpmr
THANK YOU &
V = 277.9 kN pmr , H = (104 + 14) –7 = 111 kN pmr e = M / V = 46.3 / 277.9 = 0.17 m B = B – 2e = 3.5 –0.34 = 3.16 m
GOOD LUCK
Complete the solution for the bearing capacity similar to the previous example.
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