Geotechnical Design I

Lateral earth pressure and support

Dr. Y.M. Cheng

Department of Civil and Environmental Engineering
http://www.cee.polyu.edu.hk/~ceymcheng/
 Local requirement
Practice Note for APs, RSEs & RGEs – APP 57

An ELS plan submission is required where the excavation is:

• >4.5m deep and > 5m long
• liable to affect any road, structure, slope (>30), or water main (≥75mmø)
within the 45 line up from the base of the excavation to the ground surface

Contents of an ELS plan

Lateral support system, e.g. sheet piles
-Strutting (or tie-back) layout
-Construction sequence
-Supporting geotechnical documents
-Realistic estimates of ground movements
-Assessment of effects of excavation & dewatering on adjoining structures
-Monitoring proposal
 Earth pressure theory
Rankine Earth Pressure Theory (Lower bound theory)
Rankine lateral earth pressure is totally based on the Mohr-Coulomb criteria
(yield) and is the lower bound instead of the exact solution. It is however on
the safe side and is simple to be adopted for design. For a vertical wall with
no wall friction and wall adhesion and the backfill is level, Ka and Kp are
given by the classical equations as
1  sin  1  sin 
Ka  Kp 
1  sin  1  sin 
and the active and passive earth pressures are given by the Bell’s equations
(Mohr-Coulomb criterion) as
 a   v K a  2c K a  p   v K p  2c K p
 Rankine Earth pressure theory

 With a slope of angle  behind the wall, Ka and Kp by Rankine are given by

cos   cos 2   cos 2  cos   cos 2   cos 2 

Ka  Kp 
cos   cos   cos 
2 2
cos   cos 2   cos 2 

 These equations are not correct in that Ka is independent of  (=1) if  = 

and Kp decrease with 


Pp


Pa

 Rankine’s equations do not consider the failure mechanism and are not the
ULS solution.
 Rankine Earth pressure theory
Mazindrani and Ganjali
(1997) gives

2c ' 1  sin  '

Depth of tension crack zc 
 1  sin  '

Chu (1991) gives

sin  sin 'sin  a

 a  sin 1 ( )    2   tan 1 ( )
sin ' 1  sin 'cos a
 Coulomb earth pressure theory (Upper bound)

 Coulomb’s earth pressure is based on the assumption of an assumed failure

mechanism and is an upper bound solution. In general, a planar failure
surface is assumed and this method is easy to be used and is applicable to
more general problems, including earthquake loadings. The limitation of
Coulomb’s theory is that while the error in the active pressure is usually
small, the error in the passive pressure can be significant if the wall friction
is high. The equation based on the Coulomb theory for a cohesionless soil
can be derived from the following figure. The weight of soil wedge ABE is
H 2  sin(   ) 
W  A(1)  sin(   ) (1)
2 sin  
2
sin(    ) 

Failure wedge used in deriving the Coulomb equation for active pressure. Note  may be  and 0 <  < 180˚
 Coulomb earth pressure theory

A A

(a) Assumed conditions (b) force vectors may not (c) Force triangle to
for failure pass through point O; hence establish Pa
static equilibrium is not
satisfied
From triangular of forces, the active force Pa is
Pa

W W sin(    )
or Pa  (2)
sin(    ) sin(180         ) sin(180         )

From Eq. (2) Pa depends on the failure plane angle . If all other terms for
a given problem are constant, the value of Pa will be given by the greatest
possible value. Combining Eqs (1) and (2), we obtain
 H2  sin(   )  sin(    )
Pa  sin(   )   (3)
2sin 2   sin(    )  sin(180         )
 Coulomb earth pressure theory

 The maximum active wall force Pa is found from setting dPa/ d = 0 to give
H 2 sin 2 (   )
Pa   2
(4)
2  sin(   ) sin(   ) 
sin 2  sin(   ) 1  
 sin(   ) sin(   ) 
 If  =  = 0 and  = 90˚ (a smooth vertical wall with horizontal backfill), Eq.
(4) simplifies to
 H 2 (1  sin  )  H 2 
Pa    tan 2 (45  ) (5)
2 (1  sin  ) 2 2

which is also the Rankine equation for the active earth pressure.
Equation (5) takes the general form
sin 2 (   )
H 2 Ka  2
(6)
Pa  Ka  sin(   ) sin(   ) 
2 sin 2  sin(   ) 1  
 sin(   ) sin(   ) 

and Ka is a coefficient that considers , , , , but is independent

of  and H
 Coulomb earth pressure theory

Passive earth pressure is derived similarly except that the inclination at

the wall and the force triangle is as shown above. From the above figure,
the weight of the assumed failure mass is
 H 2 sin(   ) sin(   ) (7)
W sin(   ) 
2 sin 2  sin(    )
and from the force triangle, using sine law,
sin(    )
Pp  W  (8)
sin(180         )
 Coulomb earth pressure theory
Setting the derivative dPp/d = 0 gives the minimum value of PP as
H 2 sin 2 (   )
Pp   2
(9)
2  sin(   ) sin(   ) 
sin 2  sin(   ) 1  
 sin(   ) sin(   ) 

For a smooth vertical wall with horizontal backfill ( =  = 0 and 

= 90˚), Eq. (9) simplifies to

 H 2 1  sin   H 2 
Pp    tan 2 (45  ) (10)
2 1  sin  2 2

Eq. (9) can also be written as

H 2 sin 2 (   )
Pp  Kp Kp  2
(11)
2  sin(   ) sin(   ) 
sin 2  sin(   ) 1  
 sin(   ) sin(   ) 

Eq. (11) cannot be used if the term in “[ ]” in the denominator is negative.

 At-rest earth pressure
Consider a deposit of soil formed by sedimentation in thin layers over a wide area.
No lateral yield occurs as a result of the imposition of load upon it by the
deposition of successive layers above. The in situ horizontal effective earth
pressure h in such a soil is known as the ‘earth pressure at rest’. Terzaghi used
the concept of an earth pressure coefficient K0 which is defined as:
'h
K0  '
v
In principle, if it were possible to insert an embedded wall into the ground without
disturbance, the earth pressures on either side of the wall would remain the same as
in the ground before wall construction. Theoretical and experimental determinations
have shown that the value of the coefficient of earth pressure at rest (K0) lies
between the active and passive earth pressure coefficients, Ka and Kp. Jaky (1944)
developed the following theoretical equation for a granular material under first
loading, i.e. in the normally consolidated state where the vertical effective stress has
never been higher than at present: 2
 sin ' )
K 0  (1  sin ') 3  (1  sin ')
 sin ' )
Mayne and Kulhawy (1982) and Meyerhof (1976) give an empirical equation for
K0 which takes into account over-consolidation for a ground slope angle β as
K 0  (1  sin )OCRsin  K 0  (1  sin ) OCR (1  sin )
 Discussion on earth pressure theory
Rankine and Coulomb earth pressure represent the upper and lower limits of the
ultimate earth pressure, but they are the true solutions only for special cases. Yield
alone is not necessarily the ULS, hence Rankine theory is the lower bound. Once an
assumed failure mechanism is assumed, a more critical solution may be available,
hence Coulomb theory is the upper bound. The ultimate lateral pressure can be
obtained by the partial differential equations or by design figures.

For Kp, the Coulomb equation is poor and design figure/tables should be used. For
active equation, the difference between the Coulomb active pressure and the rigorous
solution is actually small. For Coulomb active pressure, as long as a failure surface is
chosen, it is either the true solution or less critical solution. If it is a less critical
solution, the force from failure should be lesser, hence we need to search for the
maximum active pressure.

For Coulomb passive pressure, as long as a failure surface is chosen, it is either the
true solution or less critical solution. If it is a less critical solution, then the soil mass
is less easy to fail or equivalently greater passive pressure must be exerted to generate
failure. Hence we need to search for the minimum passive pressure.
Furthermore, the failure surface may not be a planar surface, and we need to try all
possible failure shapes in the analysis (at least in theory).
 Discussion on lower and upper bound

Any other trial failure surface,

less easy to fail or less force to
support the soil mass.

Critical surface
Active pressure to resist
soil mass from failure
Why search for maximum active pressure ?
 Rigorous result for ULS (curved failure surface)
 Rankine and Coulomb are the two bounds to the rigorous ULS solution.
Rigorous formulation of Ka and Kp are governed by combining the yield
criterion and the equilibrium and formulate the governing equations along the
failure surface as the slip line equations by
p   x y 
 sin 2  2R   sin(  2  )  cos(  2  ) 0
S S  S S 

p   x y 
sin 2   2R   sin(  2  )  cos(  2  ) 0
S  S   S  S 
 

1   3 1  3  
p R  
2 2 4 2

These two partial differential equations can only be solved by iterative finite
difference method. Rigorous solution indicates that log-spiral curve is a close
approximation of the actual failure surface. Design figures are given in GEO
Guide 1 and detailed solutions are provided by the earth pressure tables (~200
pages) by Kerisel and Absi. The solution of the above two equations to earthquake
loading case have been achieved by Dr. Cheng and program KA.exe and KP.exe
can also include earthquake loading.
 Earth pressure table

Approximate solution program KAKP used in some consultant, and more

rigorous programs Ka, Kp by Dr. Cheng
 Approximate formulae from Euro Code 7
For walls with level back, Euro code 7 recommends Ka and Kp (normal component)
as in radian term

[1  sin sin(2m w  )] 2(mt  m w  tan [1  sin sin(2m w  )] 2(mt  m w  tan
Ka  e Kp  e
 sinsin(2m t  )  sinsin(2m t  )

1  sin  1 sin 
where m t = (cos 1 ( )    ) and m w = (cos 1 ( )    ) for passive and active
2  sin  2 sin 

Β is the slope behind the wall,  is the inclination of wall from vertical or 90-wall
angle (0 for normal sheet pile wall),  is wall friction. The error is small which is
good enough for design. For normal problem, the results are actually good. For Ka,
actually, Coulomb active pressure coefficient is acceptable, but Coulomb passive
pressure coefficient can be poor and over-estimate the passive pressure coefficient
a lot. If the total earth pressure coefficient is required, the value should be divided
by cos.

Note that in mt, if term inside cos-1 is negative, take the second quadrant. For
example, if β=15 and =30, then cos-1 should take 121.17.
Design soil parameters
Design soil parameters
Design loadings
Movement to mobilize active and passive pressure
Active and passive pressure from GeoGuide
Use of Passive pressure design figure from GeoGuide

/=1.0, correction factor =1.0

For =30,=15, =10

Kp=13, Rp=0.65 (neglect the –ve sign)
Kp=13*0.65=8.45
From program by Dr. Cheng solving the
PDE, Kp =6.324, Ka=0.379, hence
design figure is still not very good.
From Euro code, Kp=6.23, Ka=0.361,
so Euro code is quite good
Compaction pressure arising from fill compaction
Effect of Line load or point load on lateral pressure
Types of retaining wall

MH3, use when wall

is less than 7m

MHL2, L is the spacing between the

counterfort, reduce bending moment but
more cost on formwork and labourer
Other types of retaining structures

Hong Kong –Zhuhai-

Macau Bridge
Drainage design
 Types of retaining systems in HK
1. Sheet pile wall – versatile and cheap, water-tight, cannot avoid utility, but flexible,
usually for bottom up construction, e.g. Lok Fu station

2. Soldier pile wall – restricted in use, cheap, not water-tight, can avoid utility, usually for
bottom up construction, e.g. Diamond Hill station

3. Caisson wall – limited in use now, stiff wall, require little machine, can avoid utility,
flexible, dangerous to workers, usually for bottom up construction and a permanent
wall, e.g. Choi Hung station, Sun Plaza

4. Diaphragm wall – very common, stiff wall, require heavy machine and care, cannot
avoid utility, flexible, usually for bottom up construction and a permanent wall, e.g.,
Wan Chai, Central, Causeway, Sheung Wan, Times Square….

5. Secant pile wall – very few, require heavy machine and care, cannot avoid utility, very
expensive, usually for bottom up construction and a permanent wall, e.g. Mongkok,
Prince Edward

6. Pipe pile wall - versatile and cheap, not water-tight and require back grouting, can
avoid utility and flexible, usually for top-down up construction e.g. Nathan Centre

7. PIP wall – Jordan,Tsim Sha Tsui .., Chartered Bank

 Types of support systems

wale strut Anchor (tie-back support), e.g.

Post reduce Kornhill, Building after Fortress Hill
slenderness ratio station, Lok Fu station
Strut support
 Sheet pile wall

Cheap, water-tight wall, but flexible. Limited by vibration and cannot avoid utility.
 Secant pile wall

For secant pile wall, each unit cut into the adjacent one to form water-tight
interlocking joint. Expansive construction and require heavy machine
 Diaphragm Wall

hydrofraise

Second most popular in HK. Precaution is required for very loose fill with cobbles.
 Diaphragm wall construction
Construct this Final stage
panel first Second stage excavation
excavation

a) b) c)

d) e) f) g)
 Diaphragm wall construction
Construction Sequence of Diaphragm wall
-Guide wall Construction – support top soil, weight or reinforcement cage and
rotary grab
-Trenching by use of bentonite and grab or hydrofraise
-Trench Cleaning and stop ends fixing
-Reinforcement Cage lowering, use guide wall as temporary support
-Tremie concreting
-Withdrawal of Stop ends
Bentonite Slurry as a support fluid (affecting stability of slurry trench)
1. The bentonite suspension is montmorillonite group clay with exchangeable sodium
cations (Na+).
2. The action of bentonite in stabilizing the sides of bore holes is primarily due to the
thixotropic property of bentonite suspension
3. The bentonite suspension when undisturbed forms a jelly which when agitated becomes a
fluid again. In case of granular soils, the bentonite suspension penetrates into the sides
under positive pressure and after a while forms a jelly. The bentonite suspension gets
deposited on the sides of the hole resulting in the formation of a filter cake in contact with
soil against which the fluid pressure acts. – jelly sheet pile wall as separation
4. In case of impervious clay, the bentonite does not penetrate into the soil, but deposits
only a thin film on the surface of the hole.
5. Level of the supporting fluid must be above ground water table to maintain a net force
stabilizing the trench
 Caisson wall

Choi Hung station and also for the building behind Fortress Hill station,
retaining wall at Kornhill
 Soldier pile wall

Diamond Hill MTRC station

 Pipe pile/Soldier pile wall

Can avoid utility and is cheap, but exposed surface means cannot be constructed below water
table. The exposed surface must also be stable for several days before the lagging construction.
Pipe pile wall
Flexible construction, can avoid utility. Required drillin
then grouting to avoid water leakage in-between piles.

Cement grout
Pipe pile Grout pipe between
Pipe pile for sealing
Pipe pile wall (as soldier pile)
Soldier pile wall
 Grouted pile wall (PIP)

Jordan, Tsim
Sha Tsui,
Charter Bank

Back grouting at the pile/pile joint may be necessary.

Grouted pile wall (Jet grouting)

e.g. Singapore MRT

Typical support system Wale, continuous
beam, support wall

Strut, take axial

load, support wale
Typical support system
 Strut preloading

Sheung
Wan
Concourse
 Anchor system

Lok Fu
Station

Limitation :
Intrusion to outer
boundary is not
allowed
Bottom up / top down construction

Conventional, simpler and cheaper.

Construct from bottom upward. Longer
construction time if super-structure is
to be constructed. For deep excavation,
settlement can be large. Lok Fu, Wong
Tai Sin, Jordan, Tsim Sha Tsui MTR

Construction from top slab downward.

More expensive. Shorter construction
time if super-structure is to be
constructed. Lesser ground settlement
compared with bottom up construction.
Central, Wanchai, Causeway MTR
Top down construction
 Bottom up no strut construction

Circular lining
under normal
pressure q, axial
stress =2qr/2t=qr/t.
Take moment at
any section, axial
stress only, no net
tensile stress –
prestressing action

Strut is not required because of shape. Only axial stress on wall lining
Bottom up procedures for a MRT station in Taiwan

Basement layout
Bottom up construction
 Method of analysis

Frew and Plaxis are more common in HK nowaday, while Wallap has
been used in the past
 Deformation of wall with excavation

No matter top down or bottom up construction, there will be a

cantilever stage followed by supported stage

Note that strut is constructed

at the deflected location

Wall deflection
 Deformation of wall with excavation

v is usually taken as 0.5 to 1.0 of h for design Maximum deflection may

be below formation level
Deformation of wall with excavation

Due to stress release, there is

minor upheaval near to wall
Deformation of wall with excavation (measurement at Taiwan)
 Design of retaining wall (hand calculation)
Cantilever Case
For this case, a high pressure confined within a narrow range at the bottom of the
retaining wall is found. Most of the engineers will adopt the simplified approach by
taking moment at the bottom of the retaining wall. Since the lever arm of the high
pressure range is very small, the contribution of this high pressure range can be
neglected. The numerical example as shown below will illustrate the principle of the
simplified approach.

For =18 kN/m3 and =33 , Ka = 0.295, Kp = 3.392

Dry Condition (3m cantilever)

Take moment at bottom of the retaining wall

1 1  D 3
3
Ka D  3  KpD 3 Kp
3
    11.499 3m
6 6  D  Ka
 D+3 = 2.257D
 D = 2.39m + 20% say 2.86m

Please note that the 20% increase in depth of penetration does D Big point force
not affect all other remaining calculations. It is the general at bottom
practice to add 20% to the depth of penetration to compensate
for the high pressure range at the bottom of the retaining wall.
Notes in taking moment

1/3 height from base if moment to

base, and 2/3 from tip of triangle
pressure if moment to tip
 Design of retaining wall

Wet Condition
If water is present at a depth of 2m below ground and at the dredge level for the two
sides. sat=19.5, ’ = 19.5-9.81 = 9.69 kN/m3
For water pressure, it is only necessary to consider the difference of the pressure on
the two sides of the sheet pile. This pressure difference can be determined by the use
of flow-net construction, computer analysis or approximation. The net differences
are usually approximated to a triangular pressure as shown for ease of hand
calculation. The maximum value of the triangle is usually taken as wH or from
GEO guide Review of Design Methods of Excavation 1/90 as given below
2  HD 2D 1
3
u W   9.81 from GEO
2D  H 2D  1

D is the depth of sheet pile below water level inside 4

the excavation
6
The active pressure can be divided into 5
regions for analysis.
2 5
1 2 1 1 1 1 1 2 1
K a 22 (  1  D )  K a 2 ( D  1) 2  K a '( D  1)3  u (  D )  uD D  K p ' D 3
2 3 2 6 2 3 2 3 6
1 2 3 4 5 6
UDL, depth D+1 triangular load Lever arm=1/3+D
 Design water pressure difference
Actually, ground water table is lowered

Actual water pressure difference, if

ground water table draws down
outside retaining wall

This part can be taken as a

straight line
H

2  HD
u   W from GEO
2D  H
D
Water pressure difference
Design pressure from flow net, assuming
difference for simplicity ground water table
remains constant

At this tip, there is only 1 water pressure, hence the

difference must be zero
To avoid ground settlement, some engineers adopt toe grouting below diaphragm wall to form a cut-
off, then there will be no major seepage and hydrostatic pressure will be used in design. For deep
excavation, recharge well may be used outside the retaining wall.
Analysis of free surface seepage flow by FEM
Water table actually drop
outside retaining wall
More realistic seepage model
Water pressure difference across a retaining wall Recharge well

Submersible pump
inside excavation to
lower ground water
table

Submersible pump system

 Design of retaining wall
2D  1  D3
or 0.885(D  3)  0.409(D  1)  4.91
3 3
  D   6.54  5.478D3
2D  1  3  2D  1

Try D = 3.5m LHS = 257.3 RHS = 234.9

D = 3.8m LHS = 297.0 RHS = 300.
+20% D = 4.56m
In most cases, the depth of penetration D will be close to the cantilever
height which can be used as the trial for evaluation of D. (Alternatively,
many engineers use u = wH for simplicity which is on the safe side,
where H is the water head difference)
Determination of Maximum Bending Moment
Maximum bending moment occur at the point where the shear force is
zero. Assume it is located at a distance x below the dredge level.
1 1
K a   x  3  K p x 2
2
Dry case
2
2 2
 x3 Kp
     11.499 3m
 x  Ka
Pt. of zero shear x
 x  1.255m
1 1
M max   0.295  18  4.2553   3.392  18  1.2553  48.1 kNm/m
6 6
 Design of retaining wall

With water (use 3.8m depth of penetration in your calculation)

u = 8.67 kPa from GEO Guide as above
1 1
 0.295  18  2 2  10.621  x    0.295  9.691  x 
2

2 2
1 1 3 .8  x  1
  8.67  1   8.67  8.67  x   3.392  9.69 x 2
2 2 3. 8  2
25.6  10.62 x  1.4291  x   8.67 x  1.14 x 2 = 16.434 x 2
2
or
16.145 x 2  22.15 x  27.03  0 gives x = 2.15m
Maximum moment is given by active pressure take moment to point of zero
shear-passive pressure take moment to this point :
1 2  1
M max   0.295  18  22    3.15   10.62  3.152 / 2   0.295  9.69  3.153
2 3  6
1 1  1 2 1
  8.67    2.15   3.76  2.152 / 2   4.91 2.152   3.392  9.69  2.153
2 3  2 3 6

= 80.7 kNm/m Choose FSPII type sheet pile from Japan which is commonly
used in Hong Kong. Section modulus Z = 874 cm3/m
80.7 106
  92.3N/mm 2
 165N/mm2 (from BS449) O.K
874 103
 Free-Earth Support Method for Single Layer of Support
Assume water pressure to be hydrostatic for simplicity and it is on the safe side :
(u = 2x9.81 = 19.62 kPa)
Unit 1 has no contribution

After installation of first strut, and excavate, the

wall below the formation level can move, and
the big point force at bottom is released. The
design pressure can be taken as simple active and
passive pressure then for this case.

Take moment about support 2

3
for simplicity

2 3
D2  1 2 2 
15.93  D  2    1   0.295  9.69  2  D    D  2   1
 2  2  3 
1 4  1 D  1 2 
  19.62  2    1   19.62  D    3    3.392  9.69  D 2  D  3  (Passive)
2 3  2 3  2 3 
D  22  D  2 
or 15.93  D  2    2   1.43  2  D   D  2.333   45.78  9.810   3   16.434 D 2  D  3 
2  3  3  3 
 Free-Earth Support Method for Single Layer of Support

Try D = 2.5: LHS = 432.2 RHS = 479.3

D = 2.3: LHS = 400.7 RHS = 394.1 O.K.
D  2.3 + 20% = 2.76m (use 2.3m in later calculation)
1 1 Actually, this is wale load
Anchor Force   3  15.93  15.93  4.3   0.295  9.69  4.32
2 2 =Total active force – total
1 1
  19.62  4.3   3.392  9.69  2.32 = 74.1 kN/m passive force
2 2
1 1
Active Pressure above dredge level   15.93  3  15.93  21.65   19.62  2
2 2
= 81.1 kN/m > 74.1 kN/m
 point of zero shear is located above dredge level, say distance x from ground level
1 1 1
 15.93  3  15.93  x  3   0.295  9.69   x  3   9.81   x  3  74.1
2 2

2 2 2
6.33  x  3  15.93  x  3  50.21  0
2
x-3=1.83, so x=4.83m

1 1
Mmax=  15.93  3  2.83  15.93  1.832 / 2   0.295  9.69  1.833
2 6
1
  9.81  1.833  74.1  2.83  147.5kNm / m
6
 Fixed-earth Support Method for Multi-Layers of Support
Assumption: point of zero earth
pressure = point of zero moment
This assumption is approximate but
is commonly used in China and is
also used in Hong Kong.

Given D=10m for this example

2  2  10  9.81
u  17.84 KPa
20  2
10  x
RHS  18  3  9.69  2  x    0.295   17.84
10
 9.69  3.392  x  stress at LHS
 32.868x  21.647  2.859 x  17.84  1.784 x  x  1.242m
Take moment at hinge point
1 1
TA  4.242   15.93  3  4.242  15.93  3.2422 / 2   0.295  9.69  3.2423
2 6
1 2  1 2 1
  17.84  2    1.242   15.624  1.2422 / 2   2.22 1.2422    9.69  3.392 x1.2423
2 3  2 3 6

 TA  58.6kN / m Wale Load

 Fixed-earth Support Method for Multi-Layers of Support

Excavate 3m more, assume TA remains unchanged in 2nd stage (unreasonable, but

make this assumption) 2  5  7  9.81
u  36.14 kPa from GEOGUIDE
14  5
7x
RHS  18  3  9.69  5  x    0.295   36.14  9.69  3.392 x ( LHS )
7
30.22  2.859x  36.14  5.163x  32.868x  x  1.89m
1 1
Hence TA  7.89  TB  4.89   15.93  3  7.89  15.93  6.89 2 / 2   0.295  9.69  6.893
2 6
1 5  1 1
  36.14  5    1.89   26.38 1.89 2 / 2   9.76 1.89 2   9.69  3.392 x1.893
2 3  3 6

 TB  123.4 kN / m Wale Load

This process continue until all strut

loads are determined.

Strut Load= wale load x strut spacing

 Final stage

Average load on span CD = (30.22+38.8)/2+9.81x6.5

= 98.28
Design bending moment of sheet pile on span
CD=98.28x32/8 =110.57 kN-m/m

Check : quick sand and depth of penetration

 Check depth of penetration by force and moment

According to GEO publication 1/90 and DM 7

Re = [(40.23+41.66)/2+(47.65+50.45)/2] x 0.5 38.8
= 45 kN/m)
Lowest strut
1
Passive resistance   3.392  9.69  6 2  591.6 kN / m
2
1 6
Active force from = 45  41.66  6   0.295  9.69  62  50.45  D=6m
2 2
mid height to bottom
= 497.8 kN/m < 591.6 kN/m O.K
Mid-height between lowest
strut and formation
Alternatively, D can be estimated from moment balance by taking moment about
the lowest strut D (using pressure below lowest strut) as:
1 1 1
 38.8D  1   0.295  9.69  D  1  47.65  
2 3

2 3 2
1 D  1 2 
 50.45 D    1   3.392  9.69  D 2  D  1
2 3  2 3 

Try D = 3.5: LHS = 694.8 RHS = 671

D = 4.0: LHS = 879 RHS = 964 D  3.6m
Simple check of depth of penetration by hydraulic gradient

Piping is another critical condition to be checked in deep excavation. To

avoid the use of computer software or the construction of flownet, a
simplified method can be performed as
since icrit = (Gs-1)/(1+e) = ’/w  1.0

For piping, average hydraulic gradient 

9
i  0.43  I crit /1.5 OK
9  2 6

Note, in this approximate formula, the shortest flow path is used and the average
hydraulic gradient is used. The shortest flow path is adjacent to the wall.

Alternatively, computer program like SEEP, Modflow, Plaxis can be used for
seepage analysis
 Example on checking on piping by flownet construction

head difference, H = ai = 4 m
Sheet-pile penetration, he = 4.8 m
 bulk = 20.2 kN/m3
oe = 2.4 m

Calculate the buoyant weight of soil in

element heop
Wbuoy= (- w)he*oe
=(20.2-10.0)*4.8*2.4=118kN/m

Calculate the average head perpendicular to eo

At e, he = 2.00 m ; At b, hb = 1.50 m ; At c, hc = 1.38 m ; At d, hd = 1.20 m
At o, ho = 1.10 m ; average head hoe is 1.44m. Proportion of total head loss m=1.44/4=0.36.
Calculate the critical head at the base of element heop
hp  hoe  bulk   w
  hp  hoe  4.9m
4.8 w
4.9
hcrit   13.6  FS=13.6/4.0=3.4
m
Design figure for piping
Pressure envelope for strut load design

Difficult to be used, for layered

ground, presence of water.

Such pressure envelope is not

the true earth pressure, but is
the result based on measure
strut load and assume load
distribution calculation.
Ground settlement measurement by Peck
Alternate method for estimating settlement
 Computer modelling

Outward forces

Upward force
generate upheave

Common programs - Flac2D, Flac3D, Plaxis 2D, Plaxis 3D, Frew, WallFEA, Wallap

Excavation = removal of soil = apply stress opposite to the existing stress. For the wall,
outward forces are generated at the excavation region. For the formation level, upward
forces are generated. The upheave can be major for cities like Shanghai with soft clay,
 Computer modelling

Example programs -
WallFEA, Wallap

The Winkler model requires an estimation of the spring stiffness at the different levels on
both sides of the wall. These are usually based upon measured undrained shear strength (for
clays) or density and stress level (for sands and gravels). Computer programs that use this
form of analysis will normally give guidance based upon experience. In addition, the
limiting (active or passive) forces must be estimated at each level. Computer codes
generally do so by calculating active and passive pressures from either total or effective
strength parameters, as for simple rigid-plastic models.
 Sample results from Oasys

Wall moment from Oasys

 Sample results from Oasys

Strut load and earth pressure

 Sample Computer modelling using Flac3D
;------------------------------------------------------ ini szz 0 grad 0 0 -1.5e4
; Excavation in a saturated soil ini sxx 0 grad 0 0 -1.2e4
;------------------------------------------------------ ini syy 0 grad 0 0 -1.2e4
config fluid apply nstress 0 grad 0 0 -1.2e4 range x 0.0 4.0 y 3.9 4.1 z 0.0
; --- geometrical model --- 5.0
gen zone brick p1 12 0 0 p2 0 12 0 p3 0 0 12 size 12 12 12 rat 1 1 1 apply nstress 0 grad 0 0 -1.2e4 range x 3.9 4.1 y 0.0 4.0 z 0.0
group soil 5.0
group excavate range x 0 4 y 0 4 z 0 5 apply nstress -7.5e4 range x 0.0 4.0 y 0.0 4.0 z 4.9 5.1
group wal1 range x 4 5 y 0 5 z 0 7 ; --- setting ---
group wal2 range x 0 4 y 4 5 z 0 7 set gravity 0 0 10
group wall range group wal1 any group wal2 any ; --- initial state ---
; --- fluid flow model --- solve force 1 ; check initial equilibrium
model fl_iso ; --- histories ---
prop perm 1e-12 poro 0.3 set hist_rep 40
ini fdensity 1e3 hist fltime
ini fmod 2.0e9 ftens -1e-3 hist gp pp 0 0 6
model fl_null range group excavate hist gp xdis 4 0 0
model fl_null range group wall hist gp xdis 4 0 2
ini pp 0 grad 0 0 1e4 hist gp xdis 4 2 0
fix pp range z -0.1 0.1 hist gp zdis 0 0 5
fix pp range x -0.1 4.1 y -0.1 4.1 z 4.9 5.1 hist gp zdis 2 0 5
; --- mechanical model --- hist gp zdis 4 0 5
model elas hist gp zdis 2 2 5
prop bul 3.9e6 shea 2.8e6 hist gp zdis 4 2 5
model null range group excavate hist gp zdis 4 4 5
ini density 1.2e3 hist gp zdis 10 0 1
ini density 1.5e3 range group wall hist gp zdis 10 0 2
fix x range x -.1 .1 ;
fix x range x 11.9 12.1 ; --- excavation ---
fix y range y -.1 .1 set fluid off
fix y range y 11.9 12.1 ; apply pore pressure at walls
fix z range z 11.9 12.1 apply nstress 0 grad 0 0 -1.e4 range x 0.0 4.0 y 3.9 4.1 z 0.0 5.0
; initial total stresses apply nstress 0 grad 0 0 -1.e4 range x 3.9 4.1 y 0.0 4.0 z 0.0 5.0
 Computer modelling
apply nstress -5.e4 range x 0.0 4.0 y 0.0 4.0 z 4.9 5.1 set fluid on
solve ;force 1 fix pp 0 range x -0.1 4.1 y -0.1 4.1 z 4.9 5.1
save exc1.sav cyc 9000
; save exc3.sav
; --- drainage --- plot create excav
apply remove nstress plot set rot 200 0 195
def relaxsetup plot set cent 6 6 6
step0 = step plot set dist 39.18
end plot set magn 0.8
relaxsetup plot set plane ori 0 0 0 normal 0 -1 0
def relax plot add cont pp out on
rstep = step - step0 plot add block group lgra range group wall
if rstep < ncyc then plot add cont pp int 10000 max 110000 range x 0 4 y 0 4 z
relax=1.0-(float(rstep)/float(ncyc)) 57
else plot add flow plane
relax = 0.0 plot show
endif ret
end
set ncyc = 1000
Sample input command, to illustrate the use
apply nstress 0 grad 0 0 -1.e4 hist relax &
of command to generate a problem, carry out
range x 0.0 4.0 y 3.9 4.1 z 0.0 5.0
apply nstress 0 grad 0 0 -1.e4 hist relax & analysis. No graphical menu is required, and
range x 3.9 4.1 y 0.0 4.0 z 0.0 5.0 easier for complicated problem.
apply nstress -5.e4 hist relax &
range x 0.0 4.0 y 0.0 4.0 z 4.9 5.1
cyc ncyc
solve
save exc2.sav
;
; --- percolation ---
3D modelling