4. Earth Pressures
• At-rest – no movement of support
• Active pressure – support moves AWAY from soil
• Passive pressure – support moves INTO soil
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5. Lateral earth pressures
• The way that a wall is installed
affects the way pressures are
applied to it
• Unyielding wall – the horizontal
pressures in the soil before
excavation are retained
• Yielding wall – horizontal pressures
in supported soil are reduced
• Soil placed after wall is built –
method of placing soil will control
horizontal pressures
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9. Effective Stress
sv’ = sv - u
Effective
Stress
Total
Stress
(gh)
Porewater
Pressure
(gwhw)
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10. What are the total and effective stresses at the top of the clay
layer?
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sv = gh =
u = gwhw =
sv’ = sv - u =
sv =
u =
sv’ =
sv =
u =
sv’ =
sv =
u =
sv’ =
sv =
u =
sv’ =
3 x 19 = 57 kPa
10 x 1 = 10 kPa
57 – 10 = 47 kPa
Silty sand
Silty sand
Silty sand
Silty sand
Silty sand
g=19kN/m3
1
2
3
4
5
11. Answers
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sv =
u =
sv’ =
sv =
u =
sv’ =
sv =
u =
sv’ =
sv =
u =
sv’ =
sv =
u =
sv’ =
3 x 19 = 57 kPa
10 x 1 = 10 kPa
57 – 10 = 47 kPa
3 x 19 = 57 kPa
10 x 2 = 20 kPa
57 – 20 = 37 kPa
3 x 19 = 57 kPa
10 x 3 = 30 kPa
57 – 30 = 27 kPa
3 x 19 + 2 x 10 = 77 kPa
10 x 5 = 50 kPa
77 – 50 = 27 kPa
3 x 19 + 3 x 10 = 87 kPa
10 x 6 = 60 kPa
87 – 60 = 27 kPa
1
2
3
4
5
12. Q.1
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g=19kN/m3
Silty sand
Depth, h (m)
Vertical effective stress, sv’ (kPa)
h1 g = 2 x 19 = 38 kPa
h1 g + hw (g - gw ) =
2 x 19 + 1 x (19 - 10) = 47 kPa
h1
hw
Depth, h (m)Depth, h (m)
Porewater pressure, u (kPa)Total vertical stress, sv (kPa)
h2
h2 g = 3 x 19 = 57 kPa hw gw = 1 x 10 = 10 kPa
0
1
2
3
0
1
2
3
0
1
2
3
13. Earth Pressure Coefficients
K = s’h / s’v
s’h Horizontal effective stress
s’v Vertical effective stress
K can be either:
Ka Coeff. of Active Earth Pressure
Kp Coeff. of Passive Earth Pressure
Ko Coeff. of Earth Pressure at Rest
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14. Coefficient of Earth Pressure at Rest (K0)
Empirical formulae
Jaky’s formula (simplified) K0 = 1- sin f
For sloping ground K0 = {1- sin f}/ {1- sin b}
(where b is the slope angle)
The above equations are generally valid for normally
consolidated soils and lightly over-consolidated soils. Higher
values can be expected for heavily over-consolidated soils.
K0(OC) = (1- sin f) OCRsinf
(OCR = over-consolidation ratio)
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15. Typical K0 values
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Soil Type Typical Ko
Dense sand (not highly preconsolidated) 0.35-0.40
Loose sand 0.55-0.60
Soft clay 0.50-0.60
Overconsolidated London Clay 1.0-2.8
Compacted clay 1.0-2.0
Compacted sand 1.0-1.5
16. Earth pressure theories
Rankine theory (1862)
– Assumes failure in entire soil mass. Very simplified.
Coulomb wedge (1776)
– Assumes wedge failure for both active & passive cases.
Very simplified.
Numerical Analyses
– More versatile – can incorporate many factors
– Generally part of a more complete system analysis
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17. Rankine Coefficient of Active/Passive Earth Pressure
Ka = {1- sin (f)}/ {1+ sin (f)} = tan2(45 - f/2)
for horizontal ground surface and smooth wall
and
Kp = 1/ Ka = tan2(45 + f/2)
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In total stress, undrained
short-term conditions use
undrained cohesion, Cu and
fu = 0, therefore, Ka = Kp = 1
18. Earth Pressures – non-cohesive soil
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g=19kN/m3
Silty sand
Depth, h (m)
Vertical effective stress, sv’ (kPa)
h1 g = 2 x 19 = 38 kPa
h1 g + hw (g - gw ) =
2 x 19 + 1 x (19 - 10) = 47 kPa
h1
hw
Depth, h (m)
Porewater pressure, u (kPa)
h2
pw = hw gw = 1 x 10
= 10 kPa
Depth, h (m)
Horizontal effective stress, sh’ (kPa)
pa1’ = Kasv’ = Ka h1 g
= 0.33 x 38 = 12.5 kPa
pa2’ = Ka (h1 g + hw (g - gw )) =
0.33 x 47 = 15.5 kPa
Assume f’=30 then Ka = 0.33
pa’ = effective active pressure = Ka sv’
If we have a retaining wall supporting the
top 3m of soil what are the active earth
pressures acting on the wall?
0
1
2
3
0
1
2
3
0
1
2
3
19. Forces
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Depth, h (m)
Horizontal effective stress, sh’ (kPa)
P1 = ½ x pa1’ x h1 =
0.5 x 12.5 x 2 = 12.5 kN
P3 = ½ x (pa2’-pa1’) x hw =
0.5 x (15.5-12.5) x 1 = 3 kN
Depth, h (m)
Porewater pressure, u (kPa)
pa1’ = 12.5 kPa
pa2’ = 15.5 kPa
Depth, h (m)
Porewater pressure, u (kPa)
pw = 10 kPa
Depth, h (m)
Horizontal effective stress, sh’ (kPa)
P2 = pa1’ x hw = 12.5 x 1 = 12.5 kN
The forced are calculated
from the areas under the
graphs
Pw = ½ x hw x pw =
0.5 x 1 x 10 = 5 kN
0
1
2
3
0
1
2
3
0
1
2
3
0
1
2
3
h1
hw
20. Effective pressures
• Non-cohesive soils
• pa’ = Ka (gh – gwhw) = Ka sv’
• pp’ = Kp (gh – gwhw) = Kp sv’
• Cohesive soils
• pa’ = Ka (gh – gwhw) – 2c’ √Ka
• pp’ = Kp (gh – gwhw) + 2c’ √Kp
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Depth, h (m)
Horizontal effective stress, sh’ (kPa)
0
– 2c’ √Ka
Ka (gh – gwhw) – 2c’ √Ka
h0
c’ means
reduced
active
pressure
and increased
passive earth
pressure
as well as effectively
zero active earth
pressure over height
h0
21. Ref.: Geoguide 1
Example 5
Clay soils only,
depth of tension crack (long
term) –
zo or ho = 2c’ / g √Ka
depth of tension crack (short
term) –
zo or ho = 2Cu / g
as Ka=1
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Soil1Soil2 We assume a water pressure
over the depth of the tension
crack
29. Displacements Required to Develop Active &
Passive Pressures – AS4678-2002
Displacement (translation or rotation)
required for active conditions
Sand 0.001H
Clay 0.004H
Displacement required for passive
conditions
Sand – translation 0.05H
Sand – Rotation about base >0.1H
Greater movement is required to
activate passive resistance than
active.
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30. Passive Pressures
So, although the wall may move enough to generate active earth pressure
conditions, the movement may not be enough to fully develop passive
resistance.
Therefore, a reduction to Kp may be applied (i.e. use Kp/2 or Kp/3 in
analysis).
Should consider the effects
of unplanned excavations in
front of wall and reduction to
Kp (typically H/10 or 0.5m)
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43. Soil Properties - General
Ranges of soil properties are given in AS4678:2002 in the Appendices.
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44. Determination of Shear Strength Parameters
There are a number of different shear strengths used in
retaining wall design:
Peak Values: c’ & f’ – should be used normally
Residual Values: fr’ – should be used when previous shear
surface are present and may be reactivated
Critical State: f’crit - sliding, wall friction
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46. Determination of Shear Strength Parameters
Granular Soils:
f’max = 30 + A + B + C
f’crit = 30 + A + B
Ref: BS8002:1994 & AS4678:2002
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48. Wall Friction & Adhesion
BS8002 EC7 (Draft) CP2 Geoguide 1,
1st Ed
Geoguide 1,
2nd Ed
Wall friction, dw 2/3f’
75% of design shear
strength of soil (or
representative value
determined by test)
Clause 3.2.6
d = fcrit’ for rough
surfaces
d = 20 for smooth
surfaces
Clause 2.2.8
2/3 fcrit’
Clause 8.5.1(4)
dsoil/concrete = 20
Clause 1.4321
Design of base and
stem
d = 0
Sliding or tilting, d = f
Clause 1.435
dpassive = ½ dsoil/concrete
Clause 1.435
dmax = f’/2
Precast concrete
dmax = 2/3 f’
Cast insitu concrete
Section 2.7
f’/2 or backfill slope
angle, whichever is
smaller for virtual
back of wall
Section 5.11.2
Wall adhesion, cw 0
Clause 2.2.8
0
Clause 8.5.1(4)
0
(cohesionless soil)
0 0
Section 5.11.2
Base friction, db 2/3 f’
75% of design shear
strength of soil (or
representative value
determined by test)
Clause 3.2.6
d = fcrit’ for rough
surfaces
d = 20 for smooth
surfaces
Clause 2.2.8
f’design for cast insitu
concrete
2/3 f’design for smooth
precast foundations
Clause 6.5.3(8)
f for cast insitu
concrete
when precast
db = dw = 20
Clause 1.4922
d = 2/3 f’
without shear key
d = f’
with shear key
Section 2.6
d = 0.9 – 1 fcrit’
Rough surfaces
d = 0.8 – 0.9 fcrit’
Smooth surfaces
Section 5.12
Base adhesion, cb 0
Clause 2.2.8
0
Clause 6.5.3(8)
0
(cohesionless)
0 0
(cohesionless soil)
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49. Values of d from CP2
Material/Condition Values of d
Concrete/brick
Steel sheet piling coated with
tar/bitumen
Uncoated steel sheet piling
Structure subject to vibration or
downward movement
20
30
15
0
Notes: 1. In the case of reinforced concrete cantilever walls d can be considered as
being equal to f (Cl. 1.435).
2. Bitumen coating is usually applied to reduce skin friction and other
references state d = 0.
Note: in the tutorials and exam a “smooth wall” means d = 0.
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