water in soil - ipb university

Soil Mechanics Asep Sapei

1

WATER IN SOIL

SOIL WATER

- Soils are permeable material water being free to

flow through the interconnected pores between the solid

particles

- Two distinct zones: Phreatic zone and vadose zone

- Water tables o The level at which the pore water pressure is equal to that

of the atmosphere

o In unconfined, the water table corresponds to the free

water surface normal and perched water table


2

- Capillary water o Is held above water table by surface tension

o d

Th

w

cγ

αcos4=

o For soil : 10

30ed

hc = ; ed10 : effective size

PERMEABILITY

- In one dimension, water flows through a fully saturated

soil in accordance with Darcy’s law

v = ki

q = Aki

v : flow velocity, m/s

q : volume of water flowing per

unit time, m3/s

k : coefficient of permeability,

m/s

i : hydraulic gradient = ∆h/∆L

A : cross sectional area, m2


3

- The values of k for different types of soils

- Permeability Determination

o Laboratory

� Constant head permeability

test

• For coarse grained soils

• Aht

QL

Ai

qk ==


m/s

Q : volume of water collected in

time t, m3

L : Distance between manometer

taping, m

A : cross section area of sample,

m2

h : different in manometer level,

m

t : running time, s


4

� Falling head permeability test

• For fine sand, silt and

clays

•

=

2

1log3.2h

h

At

aLk


m/s

L : Length of sample, m

A : cross section area of sample,

m2

a : cross section area of tube, m2

t : elapsed time from h1 – h2, s

o In situ test

� Constant head borehole test • For confined soil stratum

• The lower end of the borehole should

not less than 5 d (d: internal diameter

of the casing) from either the top or

bottom of the stratum • Water is allowed to flow under

constant head into the stratum

through the bottom of borehole

• The rate of flow (q) required to

maintain the constant water level is

measured

• )75.2/( dhqk =

� Variable head borehole test • For unconfined soil stratum

• The rate of flow from the stratum

into the cased borehole is measured

by observing time for the water in the

borehole changes from h1 to h2

•

=

2

1ln11 h

h

t

dk

π


5

- Permeability of stratified soil

� Horizontal flow The head loss between the entry and exit faces will be the same

h1 = h2 = h3 = h The hydraulic gradients are the same : i1 = i2 = i3 = i

Total flow : ikAqqqq hh =∆+∆+∆= 321 ; Akiq =∆ ; BDA =

( ) ikDDDBikBDikBDikBD h321333222111 ++=++

321

332211

DDD

kDkDkDkh

++

++=

� Vertical flow

The rate of flow will be

the same through each layer

∆q1=∆q2=∆q3=qv The head lost in each layer : h1, h2 and h3 The hydraulic gradient in each layer : i1=h1/D1 i2=h2/D2 i3=h3/D3

The total flow, L

hAkiAkq vvv ==

Total head lost : 321 hhhh ++= and 321 DDDL ++=

Also 1

11

Ak

qDh = , etc

321

3

3

2

2

1

1

DDD

Ak

qD

Ak

qD

Ak

qDAk

q

v

v++

++

=

So

3

3

2

2

1

1

321

kD

kD

kD

DDDkv

++

++=


6

FLOW NET

- Two dimensional flow of seepage

Continuity of equation: Quantity flowing into element = quantity flowing out of element

dydxdzz

vvdydzdx

x

vvdydvdydv z

xx

xzzzx

∂

∂++

∂

∂+=+

Therefore 0=∂

∂+

∂

∂

z

v

x

v zx

Darcy : z

hkvand

x

hkv zx

∂

∂−=

∂

∂−=

Introducing potential function φ(x,z) and flow function ψ(x,z),

xzz

hkvand

zxx

hkv zx

∂

∂−=

∂

∂=

∂

∂−=

∂

∂=

∂

∂=

∂

∂−=

ψφψφ

Satisfies as Laplace equation

Differentiation of that equation:

The function of ψ(x,z): dzvdxvdzz

dxx

d xz +−=∂

∂+

∂

∂=

ψψψ


7

If ψ(x,z) constant, then dψ = 0 and x

z

v

v

dx

dz=

The function of φ(x,z): dzvdxvdzz

dxx

d zx +=∂

∂+

∂

∂=

φφφ

If φ(x,z) constant, then dφ=0 and z

x

v

v

dx

dz−=

Constant φ is called equipotential and constant ψ is called

flow lines. A graphical construction of equipotential and

flow lines is called a flow net

- Flow nets construction rule and boundary condition


8

o Square field : the areas bounded by equipotentials and flow lines must be as near as square as possible

o Right-angle intersection: the intersection of an equipotential with flow line must occur 900

o Impermeable boundary: since no flow takes place across an impermeable boundary. A boundary is a flow line

o Permeable boundary : a submerged permeable boundary along which the head is constant will be equipotential

o Phreatic surface: along a phreatic surface, the pore pressure u = 0, so it is a flow line

o Seepage surface: a seepage surface occurs where the phreatic surface intersects tangentially with the ground surface

- Seepage quantity

Flow net may be used for determination of seepage

quantities and pressure o A constant difference in head (∆H) : intervals between

adjacent equipotentials

o A constant flow quantity (∆q) : interval between adjacent flow

lines

o Total head lost: H=∆H x no. of equipotential drops=∆HxNd

o Total seepage flow: q=∆q x no. of flow intervals=∆qxNf

o Darcy: total flow, q = k∆HNf =d

f

N

NkH

- Example 1 The cross section of a line of sheet piling driven to a depth of 6 m

into a stratum of homogeneous sandy soil which has a thickness of

8.6 m and is underline by an impermeable stratum. From the

original depth of 4.5 m the water level on one side of the piles is

reduced by pumping to a depth of 0.5 m.

Solution - Impermeable boundary: along the sheet piling BEC and along

impermeable stratum FG. BEC and FG are flow lines


9

- Permeable boundary: along AB the pressure head is a constant

4.5 m and along CD the pressure head is 0.5 m. AB and CD are

equipotentials

- Trial a flow line HJ near the piling. The lines must start at

right angle to equipotential AB and follow a smooth curve round

the bottom of pilling

- Trial equipotential lines are then drawn between the flow lines

BEC and HJ, intersecting both flow lines at right angles and

forming curvelinear squares. If necessary the position of HJ

should be altered slightly so that a whole number of squares is

obtained between BH and CJ


10

- The procedures is continued by sketching the estimated line of

flow KL

- The procedur is repeated until the boundary FG is reached.

- At the first attempt, the last flow line drawn will be

inconsistent with FG. Then the entire flow net is adjusted, and

the last flow line should be consistent with FG as figure

- The number of flow lines drawn is 4 – 5 flow channels

- For the flow net shown: Nd = 12 and Nf = 4.3

- The total volume of water flowing :

q = kh(Nf/Nd) = k x 4 x (4.3/12) = 1.44 k m3/s

- Total head at point P :

hp=h(nd/Nd)=4x(10/12)=3.33 m

- Pore pressure at P can be calculated by Bernoulli’s theorem:

up = g(hp-zp) ; g : gravitational acceleration


11

Example 2

The section through a dam is shown as figure. Determine the

quantity of seepage under the dam and the distribution of uplift

pressure on the base of the dam. k=2.5 X 10-5 m/s

Solution

Total head loss =h=4.00 m

The figure comprises 4.7 flow channel (Nf) and 15 equipotential

interval (Nd)

Seepage quantity = q = kh(Nf/Nd)=2.5x10-5x4.00x(4.7/15)

= 3.1x10-5 m3/s per m dam

ui = g(hi-zi) Point h (m) z (m) h-z (m) u (kN/m2)

1 0.27 -1.80 2.07 20.3

2 0.53 -1.80 2.33 22.9

3 0.80 -1.80 2.60 25.5

4 1.07 -2.10 3.17 31.1

5 1.33 -2.40 3.73 36.6

6 1.60 -2.40 4.00 39.2

7 1.87 -2.40 4.27 41.9

7.5 2.00 -2.40 4.40 43.1


12

SEEPAGE THROUGH EARTH DAMS AND EMBANKMENTS

- Unconfined the upper boundary of seepage is

phreatic surface as top flow line

- The basic shape of phreatic surface is parabola, but

need modification due to inconsistencies at entry and

exit surface

- Correction method proposed by Casagrande Correction factors for earth dam flow net

β(o) 30 60 90 120 150 180

∆a/a 0.36 0.32 0.26 0.18 0.10 0

o The parabola is assumed to start at D

o The directrix (EH) is located by striking an arc radius DF

(DE=DF)

o All point on a parabola are equidistant from EH and point

F; FG=GH, and for all points X, XX’=FX

o For exit tangential to downstream face

� The exit point K is changed to point J


13


14

STRESS IN SOIL

ELASTICITY AND PLASTICITY

- From A typical stress (σ) –strain (ε) curve

Brittle material: cast iron, high carbon steel, concrete, hard rock, etc.

Ductile material : structural steel, aluminum, other alloy

o Stiffness : result of the relationship between the stress

and the strain

o Material is said to be elastic when the same amount of

strain caused at the same level of stress regardless of

whether the material has been loaded or unloaded

o If the deformation is not fully recoverable upon unloading is

called plastic

o If the stress strain curve is a straight line Hookean

materials. The slope of the stress-strain curve

modulus of elasticity or Young’s modulus

ε

σ

∂

∂=E

o If the stress strain curve is not a straight line non-

Hookean materials (some non-ferrous metals, plastic, soil)

The slope E varies with stress and quoted as the tangent or

secant value


15

Tangent, ε

σ

∂

∂=

''

E

Secant, ε

σ

∆

∆=

''

secE

Prime (‘) denote effective

stress

- Shear Stress (τ) and Strain (γ)

The slope of shear stress-shear strain modulus of rigidity

or shear modulus

)1(2 υγ

τ

+=

∂

∂=

EG ; υ : Poisson’s ratio (= 0.5 for

saturated clay and 0.33 for sand)

THE STRESS-STRAIN CHARACTERISTICS OF SOIL

Soils behave like other solids when subject to changes in loading, but

there are significant differences

- Soils cannot sustain tension


16

- When loaded, soil will generally undergo a change in volume or an

increase in pore fluid pressure

- Saturated soils can only undergo a change in volume as porewater

is squeezed out. The rate of water loss is controlled by

permeability

- Some (hard) soils will exhibit brittle failure by shearing, while

others will simply distort plastically

- Once a shear slip has occurred the problem changes from one of

solid mechanics to one of rigid body mechanism

CONTACT PRESSURE

Is the intensity of loading transmitted from the underside of a

foundation to the soil

Flexible footing

Rigid footing on cohesive soil

Rigid footing on cohesionless soil


17

STRESS IN A SOIL MASS DUE TO APPLIED LOADING

Assumption: - Soil mass is a semi-infinite elastic half-space

- Soil is homogenous and isotropically elastic

a) Stresses due to a Vertical Point Load

Boussinesq:

5

3

2

3

R

Pzz

πσ =∆

+

−−=∆

zR

R

R

zr

R

Pr

)21(3

2 3

2

2

υ

πσ

+−−=∆

zR

R

R

z

R

P)12(

2 2υ

πσθ

z

r

R

rPzzrz σ

πσ ∆==∆

5

2

2

3

0=∆=∆ rz θθ ττ

Can be written:

pz Iz

P.

2=∆σ

Ip = influence factor

=

2/5

2

5

)/(1

1

2

3

2

3

−=

zrR

z

ππ


18

b) Stresses due to a Long Line Load

- have length, but not breadth

- uniform load along the length

( )222

32

zx

zQz

+=∆

πσ

( )222

22

zx

zxQx

+=∆

πσ

( )222

22

zx

xzQxz

+=∆

πσ

c) Stresses due to a Strip Load

- the length is very long compared with its breadth

- uniform load

- two-dimensional


19

[ ])2(cossin βαβββπ

σ +−=∆q

z

[ ])2(cossin βαβββπ

σ ++=∆q

x

[ ])2(cossin βαββπ

σ +=∆q

xz

d) Stresses due to a Triangular strip load

−=∆ αβ

πσ 2sin

c

xqz

−−

−−+=∆

222

22

ln2sinzcx

zx

c

z

c

xqx ββ

πσ

−+=∆ αβ

πσ

c

zqxz

22cos1


20

e) Stresses due to a uniform-loaded circular area

The load of small element =

Q x rdθdr

∫ ∫

+=∆

π

π

θσ

2

0 0

2/5

22 )/(1

1

2

3a

zzrz

drqrd

−−=

2/3

2)/(1

11

zaq

= q(A+B)

A and B : partial influence

Factor


21

f) Stresses due to a uniformly loaded rectangular area

Stress beneath one corner :

Rz qI=∆σ ; IR : influence factor dependent on length (L),

breadth (B) and depth (z)

+++

+++

++

++

+++

++= −

1

12tan

1

2

1

12

4

12222

221

22

22

2222

22

nmnm

nmmn

nm

nm

nmnm

nmmnIR

π

zBm /= zLn /=

IR can be determined by using table or Fadum’s influence chart


22


23

g) Stresses due to trapezoidal load

- Such dike, road, embankment etc

qIz =∆σ ; I : influence factor

Osterberg’s influence chart


24

h) Newmark’s influence chart

- By graphical means

- xqxIeredfieldsofno qz )cov .(=∆σ

Iq : the chart influence value Nqz 002.0=∆σ

- The loading structure is drawn (usually on tracing paper) with

the scale length AB = z


25

Distribution of stress – Pressure bulbs

- I f equal values of vertical stress are plotted on a cross

section a diagram pressure bulb

SETTLEMENT DUE TO ELASTIC COMPRESSION

Vertical surface displacement of a soil layer of infinite depth due to

uniform load :

ρυ IE

qBSi )1( 2−=

q : intensity of contact pressure

B : least lateral dimension (breadth or diameter)

υ : Poisson’s ratio

E : modulus of elasticity

Iρ : influence factor for vertical displacement


26

For cases where the layer thickness is less than 2B and where υ ≈

0.5 (Janbu et al, 1956)

)1( 210 υµµ −=

E

qBSi


27

Example

1. A continuous strip footing of breadth 3 m carries a uniform load

of 50 kN/m2. Calculate the vertical stress at a point 3 m from

the center line and 5.0 m below the footing?

Answer:

[ ])2cos(sin βαββπ

σ ++=∆q

z

α = arc tan (1.5/5) = 0.327 rad

α+β = arc tan (4.5/5) = 0.733 rad

β = 0.733 – 0.327 = 0.406 rad

[ ] 2/55.9)406.0327.02cos(406.0sin406.050

mkNxz =++=∆π

σ

2. A rectangular foundation transmits a uniform contact pressure of

120 kN/m2. Determine the vertical stress induce by this loading

:

a. At the depth of 10 m below point A

b. At the depth of 5 m below point B

Answer:

a.

∆σz(A) =∆σz(1) +∆σz(2) +∆σz(3)

+∆σz(4)

∆σz(A) =q (IR(1) +IR(2) +IR(3)

+IR(4))


28

For z = 10 m, use chart or table to find IR

Rectangle B/z L/z IR

1

2

3

4

1.0

1.0

0.5

0.5

0.5

2.0

2.0

0.5

0.1202

0.1999

0.1350

0.0840

0.5391

∆σz(A) =120 x 0.5391 = 65 kN/m2

b.

∆σz(A) =q (IR(1) -IR(2) -

IR(3) +IR(4))

For z = 5 m, use chart or table to find IR

Rectangle B/z L/z IR

1

2

3

4

3.8

3.8

0.8

0.8

6.2

1.2

6.2

1.2

0.2480

0.2171

0.1850

0.1684

∆σz(A) =q (IR(1) -IR(2) -IR(3) +IR(4)) = 120(0.2480-0.2171-

0.1850+0.1684)

= 1.7 kN/m2


29

SOIL STRENGTH

� In soil, the induced stress is as shear stress Therefore

Soil strength usually changed with SHEAR STRENGTH

� The shear strength is the maximum value of the shear stress that

may be induced within its mass before the soil yield

� If at a point on any plane within a soil mass the shear stress

become equal to the shear strength of the soil, failure will occur

� Analyzed base on Friction model

'lim tanφµ NNT it ==

T : tangential component

N : normal component

µ : coefficient of friction

φ’ : angle of internal friction or angle of friction

� Peak, ultimate and residual stress

THE MOHR-COULOMB FAILURE CRITERION


30

� Mohr-Coulomb equation:

φστ tan+= c

τ : shear stress

c : (apparent) cohesion

σ : normal stress

φ : angle of internal friction

� According to Terzaghi in term of effective stress:

'tan''' φστ += c

� In term of major (σ1) and minor (σ3) stress :

2

'45

2cos)''(2

1)''(

2

1

2sin)''(2

1

3131

31

φθ

θσσσσσ

θσστ

+=

−++=

−=

o

)''(2

1'cot'

)''(2

1

'sin

31

31

σσφ

σσφ

++

−=

c

Therefore 'cos'2'sin)''()''( 3131 φφσσσσ c++=−

Or


31

)2

'45tan('2)

2

'45(tan'' 002

31

φφσσ +++= c

TYPE OF SHEAR STRENGTH TEST

1. Unconsolidated Undrained Tests or Quck undrained tests

In this test, no consolidation and no drainage of pore water are

allowed. For fully saturated soils, the increase in pore pressure

will be equal to the increase of total stress, and so no increase in

effective stress

This test refers to the total stress

2. Consolidated-Undrained tests

The test specimen is first allowed to consolidated under

conditions of constant isotropic stress and full drainage. After

which the axial load is increase with no dranage allowed

This test are used to obtain the effective stress parameters

3. Consolidated-Drained Tests

The test specimen is first allowed to consolidated under

conditions of constant isotropic stress and full drainage. When

the consolidation stage is complete, the axial load is increase at a

rate low enough to ensure that no increase in pore pressure

This test obtained the effective stress parameter

LABORATORY SHEAR STRENGTH TEST

1. The Direct Shear Test

- A rectangular prism

is carefully cut from

a soil sample and

fitted into a square

metal box (usually

60 mm x 60 mm)

- A vertical load is then applied to the speciment by means of

a static weight hanger


32

- The soil is sheared by applying a horizontal force at a

constant rate of strain

- The magnitude of shearing force is measured (usually use

proving ring)

- The procedure is repeated

on four or five specimens of

the same soil

Volume change during test:

dilation for dense soil and

contraction for loose soil

Example:

A shear box test was carried out on a sandy clay. The results as follow:

Normal load (N) 108 202 295 390 484 576

Shear load at failure (N) 172 227 266 323 374 425

Area of shear plane : 60 mm x 60 mm. Determine the cohesion and angle of

friction for the soil !

Answer:

Area of shear plane : 60x60x10-4 m2

Normal stress = normal load/area of shear plane

Shear stress at failure = shear load at failure/area of shear plane

Normal stress, σ’ (kN/m2) 30.0 56.1 81.9 108.3 134.4 160.0

Shear stress, τ’(kN/m2) 47.8 63.1 73.9 89.7 103.9 118.1

Shear strength envelope

Cohesion=c’=33 kN/m2

Angle of friction = φ’=280

2. TRIAXIAL

COMPRESSION TEST


33

- Is the most widely used

shear strength test

- The test is carried out on

a cylindrical specimen of

soil having a

hight/diameter of 2:1.

The sizes are 76 x 38

mm or 100 x 50 mm

- Reading are taken: the

change in length of the

specimen, the axial load and pore pressure (optional)

The test is repeated at least three time using the same

condition specimen and difference cell pressure

- When shear failure occurs, the vertical compressive stress is

termed peak or ultimate deviator stress


34

- Type of failure

- Mohr-Coulomb envelope A minimum of three circles will be required to give a reliable result

- Area and Volume change

0

00

/1

/1

ll

VVAA

∆−

∆−=

Ao : original cross section area of specimen

Vo : original volume of specimen

lo :original length of specimen ∆V : change in volume

∆l : change in length

In the case of undrained test, ∆V = 0, 0

0

/1 ll

AA

∆−= or

a

AA

ε−=

1

0

εa : axial strain = ∆l/lo


35

Example A drained triaxial compression test carried out on three specimens of the

same soil yielded the following results:

Test no. 1 2 3

Cell pressure (kN/m2) 100 200 300

Ultimate deviator stress (kN/m2) 210 438 644

Draw the shear strength envelope and determine the shear strength

parameters, assuming that the pore pressure remains constant during the

axial loading stage

Answer:

Minor principal stress, σ3 = cell pressure

Major principal stress, σ1 = cell pressure + deviator stress

Since u = 0, σ1’=σ1 and σ3’=σ3

Test no. 1 2 3

σ3’ (kN/m2) 100 200 300

σ1’ (kN/m2) 310 638 944

Mohr circles:

From figure:

C’ = 0

φ’ = 31 o

3. UNCONFINED COMPRESSION TEST

- Triaxial compression with cell pressure equal to zero (σ3 = 0)

- The test is only applicable where φu = 0 (fully saturated non fissured

clays)

- Mohr circle :


36

- Mohr Coulomb at different type of test

Undrained and drained test at the same cell pressure

Effective and total stress


37

IN SITU SHEAR STRENGTH TEST

1. The shear vane test

- for soft silts and clays

- Determine the undrained shear strength

+=

342

2

2d

xdd

hdxcT u ππ

Hence

+

=

+

=

dhd

T

dhd

Tcu

3

1

2

1)

6

1

2

1( 232 ππ

2. The cone penetration test

- Have no direct relation with soil

shear parameter

- Use for identified relative soil

strength, density, root penetration

rate etc.

- Expressed by Cone Index,

CI = load/cone base area, kN/m2, Pa

Load = weight of penetrometer +

applied load

- Penetration rate 1 cm/sec


38

Example:

A shear vane had a diameter of 75 mm and a length of 150 mm use for a test.

The average torque recorded after slow and rapid rotations were 65 and 26 Nm.

Determine the undrained shear strength and its sensitivity

Answer:

Undisturbed undrained strength,

+

=

+

=−

075.03

1150.0075.0

2

1

1064

3

1

2

1 2

3

2 ππ

x

dhd

Tcu =

41.4 kN/m2 Remoulded undrained strength,

+

=

+

=−

075.03

1150.0075.0

2

1

1016

3

1

2

1 2

3

2 ππ

x

dhd

TcuR = 10.3 kN/m2

Sensitivity, St = cu/cuR = 41.4/10.3 = 4.0

ESTIMATES OF SHEAR STRENGTH PARAMETERS FROM INDEX

TEST

- Usually using plasiticity index (Ip) and liquidity index (IL)

IP = wL-wP ; IL = (w-wP)/IP wL : liquid limit; wP : plastic limit

- Scholfield and Wroth (1978) : )6.4exp(170 Lu Ic −= kN/m2

- Skempton and Bjerrum (1957) : P

vo

u Ic

37.011.0'

+=σ

- Ladd et al. (1977) : ( )( )

8.0

'/

'/o

ncvou

ocvouR

c

c=

σ

σ

- Correlation: Lu Ic 0.22.0log +=

- Massarch (1979): PIK 42.044.00 +=

cu : undrained shear strength, cuR : remoulded undrained shear strength, Ro : overconsolidation ratio, σvo’ : effective overburden stress

Example: Obtain an estimate for remoulded shear strength of soil which has wL

= 37 %, wP = 19 % and w = 23 %

Answer: using correlation : log cuR = 0.2+2.0 IL; IL = (w-wP)/IP

= 0.2+2.0x (37-23)/(37-19)

cuR = 58 kN/m2


39

SOIL CONSOLIDATION

� Consolidation

- Is the gradual reduction in volume of a fully saturated soil of

low permeability due to drainage of some of the pore water,

the process continuing until the excess pore water pressure

set by an increase in total stress has completely dissipated

- One-dimensional consolidation : Terzaghi’s model (1943)

- Consolidation settlement : is the vertical displacement of the surface corresponding to the volume change at any stage of consolidation process

- A soil is said to be fully consolidated : when its volume remains constant under a constant state of stress

- A soil is said to be normally consolidated : at the present time in a state corresponding to its final consolidation pressure

- A soil is said to be over consolidated : when its present-day overburden pressure is less than its final consolidation pressure was

sometime in the past

� The volume change – compressibility


40

ooo e

e

H

H

V

V

+

∆=

∆=

∆

1

Therefore, the change in thickness :

oo

e

eHH

+

∆=∆

1

� The one-dimensional consolidation test (Oedometer test)

- Soil specimen size

usually 75 mm in

diameter and 15-20

mm in thickness

- A vertical static load

is then applied through

a lever system

- The soil thickness will

reduced, and

measured by using displacement dial gauge

- Readings are continue until the specimen is fully consolidated

(usually 24 hrs: at 6, 9, 15 and 30 sec, 1, 1.5, 2, 3, 5, 7, 10, 15,

20, 30, and 40 mnts, 1, 1.5, 2, 3, 6 and 24 hrs)

- Further increments of load are then applied (double from the

previous load) up to 6 different load

(example: 0.1, 0.2, 0.4, 0.8, 1.6, 3.2, 6.4 and 12.8 kg/cm2)

- After full consolidation reach under the final load, the load is

removed and the sample is allowed to swell

- Data and calculation:

� Water content after swelling period : w1

� Void ratio after swelling period : e1 = w1Gs (Sr = 1)

� Thickness at the end of stage = h1

� Thickness at start of stage = ho

� Void ratio at end of stage = e1

� Change of thickness = ∆h

� Change in void ratio, )1( 11

eh

he −

∆=∆

� Void ratio at start of stage, eeeo ∆−= 1


41

� Suppose a stratum of clay thickness Ho, therefore

Consolidation settlement, oo

oc H

e

eeHs

+

−=∆=

1

1

� Consolidation Parameters

1. Compression index (Cc)

From the e vs σ’ or the e vs log σ’

The slope of the the straight normal consolidation curve is

referred to as the Cc

)'/'log('log 01

10

σσσ

eeeCc

−=

∆

∆=

2. Preconsolidation stress (po)

Casagrande (1936) suggested an empirical graphical method based

on the e vs log σ’ curve a. Select point P which is the

point of maximum curvature

between A and B

b. Draw two lines passing point P.

one is a tangent to the curve

(TPT) and other (PQ) is

parallel to the stress axis

c. Divide the angle QPT into 2

same angles by line PR

d. The point of intersection S

gives an approximate value for

preconsolidation stress


42

3. Coefficient of volume compressibility (mv)

- Represents the change in unit volume that results from a unit

increase in effective stress

- mv is not constant for a given soil depends on the level of

effective stress

- H

Hmv

'σ∆

∆=

But 01 e

e

H

H

+

∆=

∆, so:

01

1.

' e

emv

+∆

∆=

σ

Where 'σ∆

∆e is the slope of the e vs σ’ curve

Example: The following readings were obtained from an oedometer test on a specimen of

saturated clay. The load being held constant for 24 hr before the addition of

the next increment

Applied stress (kN/m2) 0 25 50 100 200 400 800

Thickness (mm) 19.60 19.25 18.98 18.61 18.14 17.68 17.24

At the end of the last period the load was removed and sample allowed to

expand for 24 hrs, at the end of which time its thickness wa 17.92 mm and its

water content found to be 31.8 %. The specific gravity of soil was 2.66

a) Plot the e vs σ’ curve and determine the coefficient of volume

compressibility for an effective stress range 220-360 kN/m2

b) Plot the e vs log σ’ curve and determine the compressibility index and

preconsolidation pressure

c) Obtain the value for consolidation settlement for 4 m thick layer of the

clay when the average effective stress changes from 220-360 kN/m2

Answer:

Since Sr = 1.0, e1 = w1Gs = 0.318x2.66 = 0.842

Change in void ratio, )1( 00

eh

he +

∆=∆

During swelling stage : 070.0)842.1(92.17

68.0==∆e

During 400-800 stage : 045.0)772.1(24.17

44.0−=

−=∆e


43

a) the e vs σ’ curve

from the curve:

σo’ = 220, e = 0.858

σ1’ = 360, e = 0.825

01

1.

' e

emv

+∆

∆=

σ

127.0858.1)220360(

10)825.0858.0(3

=−

−=

x

xm2/MN

b) the e vs log σ’ curve

cc= 153.0log200-log800

0.772-0.864=

c) at the end of stage, σ1’ = 360 kN/m

2

mv = 0.13 m2/MN

Sc = 0.13x10-3(360-220)4

x103= 73 mm


44

4. Coefficient of Consolidation (Cv)

- Assumptions on consolidation (Terzaghi, 1925): a) The soil is fully saturated and homogenous

b) Both the water and the soil particles are incompressible

c) Darcy’s law of water applies

d) The change in volume is one dimensional in the direction of applied

stress

e) The coefficient of permeability in this direction remains constant

f) The change in volume corresponds to the change in void ratio and

∂e/∂σ’ remains constant

- Coefficient of Consolidation , wv

vm

kC

γ= ; in m2/year

Degree of consolidation, oof

oz

u

uu

ee

eeU

−=

−

−= 0

Time factor, 2

d

tCT v

v =

eo : initial void ratio, e : void ratio after t, ef : final void ratio, uo : initial

excess pore pressure, u : excess pore pressure after time t, t : time,

d : length of drainage path

- Relationship between average degree of consolidation (U) and

time factor (Tv)


45

- Cv determination

1. The log time method (due to Casagrande) a) Plot log t vs compression from oedometer test

b) Select two points P and Q on the first part of curve for which the

values of t ratio of 1 : 4

c) Set the vertical difference PF = PQ. Point F corresponds to U0

d) On the final part of curve, determine point E which corresponds to

U100

e) Divide vertical difference FE into 2 equal part (point H). Point H

corresponds to U50. And determine t50 by projecting point H on to

the curve


46

f) Calculate Cv, 50

250

t

dTCv =

Example:

One of the loading stages in a consolidation test results:

Time (min) 0.00 0.04 0.25 0.50 1.00 2.25 4.00 6.25 9.00

Change in

thickness (mm) 0.00 0.121 0.233 0.302 0.390 0.551 0.706 0.859 0.970

--------------------------------------------------------------------------------------------------------

Time (min) 12.25 16.00 25.00 36.00 64.00 100 360 1440

Change in

thickness (mm) 1.065 1.127 1.205 1.251 1.300 1.327 1.401 1.482

At the end, the thickness of the specimen was 17.53 mm, the stress had been

raised by 100 kN/m2 , Gs = 2.70 and the water content was 24.7 %. Determine

(a) the coefficient of consolidation and (b) the coefficient of volume

compressibility !

Answer:

(a)

Plot log t vs change in thickness

Select tP = 0.25 min and ∆hP=0.233 mm

tQ=1.00 min and ∆hQ=0.390 mm

So ∆hF= 0.233-(0.390-0.233) = 0.076 mm

Determine point E, and ∆hE = 1.224 mm

Then ∆h50= (1.224-0.076)/2 + 0.076 = 0.650 mm


47

Now, determine t50 from the curve. Found that log t50=0.525 and t50 = 3.35 min

From curve Tv vs U, T50 = 0.197

Average thickness of the specimen =17.53 + 1.482/2 = 18.27 mm

Length of drainage, d=18.27/2=9.14 mm

91.435.3

14.9197.0 2

50

250 ===

x

t

dTCv mm2/min

(b)

Final void ratio, e1=w1Gs = 0.247x2.70 = 0.667

Initial thickness, ho = 17.53 + 1.472 = 19.00 mm

Change in void ratio, 130.000.19

)667.01(482.1)1( =

+=+

∆=∆

xe

h

he o

o

Initial void ratio, eo = 0.667+0.130 = 0.797

723.0797.1100

10130.0

1

1.

'

3

0

==+∆

∆=

x

x

e

emv

σm2/MN

2. Square root of time method (Taylor’s method) a) Plot root of time (min) vs sample thichness (mm) or compression

(mm)

b) Draw the best straight line

through the point in the first 60

%

c) Next drawn a straight line with

abscissa 1.15 times those the

first straight line

d) This line intersect the curve at

point C which correspond to U90

e) Determine t90

f) Calculate Cv, 90

290

t

dTCv =

Example:

Using the previous example, determine Cv by using root time

Answers:

Plot root t vs the change of thickness

Draw the best straight line through the point in the first 60 %. This line

intersect the thickness axis at point F which correspond to Uo (with ∆ho =

0.078 mm)


48

Next drawn a straight line with abscissa 1.15 times those the first straight line.

This line intersect the curve at point C which correspond to U90

From the plot: root t90 = 3.79, and t90 = 14.36 min

From Tv chart, T90 = 0.848

93.436.14

14.9848.0 2

90

290 ===

x

t

dTCv mm2/min


49

EARTH PRESSURE EARTH PRESSURE

- Pressure in soil at

horizontal/lateral direction,

so also named lateral earth

pressure

- The magnitude of earth

pressure is depends on: o The shear strength

o The lateral strain condition

o The vertical effective stress, o The state of equilibrium

- Related to retaining structure problems such as retaining wall

- Some terminology: o A body is said to be in a state of elastic equilibrium when a small change

(increase or decrease) in stress acting upon it produces a corresponding

and reversible change in strain.

o Irreversible strains are caused if the stress is increased beyond the

yield point

o In a state of plastic equilibrium, irreversible strain is taking place at

constant stress

- The strain states relating to earth pressure calculations fall into

three categories:

o At rest state : elastic equilibrium with no lateral strain taking place

o Active state : plastic equilibrium with lateral expansion taking place

o Passive state : plastic equilibrium with lateral compression taking place

EARTH PRESSURE AT REST


50

- The soil is in elastic equilibrium if the stress state in soil mass is

still below the Mohr-Coulomb failure envelope

- Under natural conditions of depositions there is a negligible

amount of horizontal strain, although some lateral contraction

may occur upon loading The soil at rest state

- Horizontal effective stress:

'' 0 vh K σσσσσσσσ ====

Where Ko : coefficient of earth pressure at rest and σv’: vertical

effective stress

For normally consolidated soil Jaky (1944) proposed:

'sin10 cK φφφφ−−−−====

Where φc’ : the critical angle of friction

Range values for K0

Type of soil K0

Loose sand

Dense sand

Normally Cons. clay

Over Cons. clay

Compacted clay

0.45-0.6

0.3-0.5

0.5-0.7

1.0-4.0

0.7-2.0

RANKINE’S THEORY

- Rankine’s theory (1857) considers the state of stress in a soil

mass when the condition of plastic equilibrium has been reach

- Rankine’s original derivation assumed a zero value of the shear

strength parameter c’

- If there is a movement of the wall away from soil, σx decrease to

a minimum value such that of plastic equilibrium develops.

Horizontal stress σx’ is must be the minor principal stress (σ3’),

and vertical stress σz’ is the major principal stress (σ1’)


51

)'cot'2''( 2/1

)''( 2/1'sin

31

31

φφφφσσσσσσσσ

σσσσσσσσφφφφ

c++++++++

−−−−====

'cos'2)'sin1(')'sin1(' 13 φφφφφφφφσσσσφφφφσσσσ c−−−−−−−−====++++

'sin1

)'sin1('2

'sin1

'sin1''

2

13 φφφφ

φφφφ

φφφφ

φφφφσσσσσσσσ

++++

−−−−−−−−

++++

−−−−==== c

++++

−−−−−−−−

++++

−−−−====

'sin1

'sin1'2

'sin1

'sin1'' 13 φφφφ

φφφφ

φφφφ

φφφφσσσσσσσσ c

[[[[ ]]]])2/'(45tan 02 φφφφ−−−− can be substituted for (1-sinφ’)/(1+sinφ’)

Overburden pressure σ1’ at the depth z : σ1’=γz

The horizontal stress for the above condition : active pressure

(pA)

At active state:

If 'sin1

'sin1

φφφφ

φφφφ

++++

−−−−====AK , is active pressure coefficient,

The equation above become AAA KczKp '2−−−−==== γγγγ


52

If the horizontal stress = the active pressure in the active

Rankine state

- If the wall is moved against the soil mass, value of σx will increase

until a state of plastic equilibrium reach

Horizontal stress σx’ is become σ1’ and vertical stress σz’ is

become σ3’, σ3’ = γz

Horizontal stress in this case: passive pressure

The equation become:

++++

−−−−−−−−

++++

−−−−====

'sin1

'sin1'2

'sin1

'sin1'' 31 φφφφ

φφφφ

φφφφ

φφφφσσσσσσσσ c

If 'sin1

'sin1

φφφφ

φφφφ

−−−−

++++====PK , is passive pressure coefficient,

The equation above become PPP KczKp '2−−−−==== γγγγ

When c’=0, triangular distribution are obtained. When c’ > 0, pA =

0 at a particular depth z0

There fore: AK

cz

γγγγ

'20 ====

The total active thrust (PA) for a vertical wall surface of height

H:


53

∫∫∫∫ −−−−−−−−−−−−========H

zoAAAA zHKczHKdzpP ))(('2)(

2

10

20

2γγγγ

20 )(

2

1zHK A −−−−==== γγγγ ; if z0 = 0;

2

2

1HKP AA γγγγ====

The force PA acts at a distance of 1/3(H-z0) above the bottom of

the wall

The total passive thrust (PP) for a vertical wall surface of height

H:

HKcHKdzpP PP

H

PP )('22

1 2

0

++++====∫∫∫∫==== γγγγ

The two components of PP act at distance of H/3 and H/2

respectively above the bottom of the wall


54

Additional pressures due to uniformly distributed surcharge

The vertical stress, σz become: qzz ++++==== γγγγσσσσ , resulting additional

pressure of KAq in active cases and KPq in passive cases, with

being constant at any depth

The corresponding forces on a vertical wall of height H are KAqH

and KPqH, and act at H/2

Soil below water table

- In fully drained condition, active and passive pressure in term

of effective

- The hydrostatic pressure γwz is as additional pressure

Example 1. Calculate the resultant active thrust on a vertical smooth retaining wall of

height 5.4 m. The water table is below the base of the wall. Soil properties:

c’=0, φ’=300, and γ=20 kN/m3.

Answer:

333.030sin1

30sin1

'sin1

'sin1====

++++

−−−−====

++++

−−−−====

φφφφ

φφφφAK

At the depth of 5.4 m:

σh’=KAγz=0.333x20x5.4 = 36 kN/m2

Resultant of active thrust: 20 )(

2

1zHKP AA −−−−==== γγγγ =1/2x36.0x5.4=97.2kN/m2

This will acts at a height of H/3 above the base = 1.8 m

2. A retaining wall with a smooth vertical back of height 7.0 m retains soil

having an unsurcharged horizontal surface. The soil properties are c’=0,

φ’=320, γsat=20 kN/m3 and γ=18 kN/m3.

Determine the distribution of horizontal stresses on the wall and also the

magnitude and position of the resultant horizontal thrust when the

groundwater at 3.0 m below the surface

Answer:


55

Active case, so 307.032sin1

32sin1

'sin1

'sin1====

++++

−−−−====

++++

−−−−====

φφφφ

φφφφAK

At z=3.0 m :

Vertical total stress : σv=γz=18x3=54.0 kN/m2

Pore water pressure : u=0

Vertical effective stress : σv’=σv-u=54.0 kN/m2

Horizontal effective stress, σh’=KAσv’=0.307x54.0=16.59 kN/m2

Horizontal total stress : σh=σh’+u = 16.59 kN/m2

At z = 7.0 m

Vertical total stress : σv=(γz)1+(γz)2= (18x3)+(20x4) =134.0 kN/m2

Pore water pressure : u=γw(z-zw)=9.81(7.0-3.0)=39.24 kN/m2

Vertical effective stress : σv’=σv-u=134.0-39.24=94.76 kN/m2

Horizontal effective stress, σh’=KAσv’=0.307x94.76=29.12 kN/m2

Horizontal total stress : σh=σh’+u = 68.36 kN/m2

The resultant horizontal thrust :

Ph = P1+P2+P3+Pw;

=1/2x16.59x3+16.59x4+1/2x(29.12-16.59)x4.0+1/2x9.81x4.02

= 194.8 kN/m2

The line of action oh Ph found by equating moment to the base

Phħ= P1x5.0+P2x2.0+P3x4/3+Pwx4/3

ħ=(24.9x5.0+66.4x2.0+25.1x4/3+78.5x4/3)/194.8 = 2.03 m


56

COULOMB’S THEORY

- Coulomb’s theory (1776) involves consideration of stability, as a

whole, of the wedge of soil between a retaining wall and a trial

failure plane

- The friction between the wall and the adjacent soil is taken into

account, denoted by δ

- The shape of the failure surface is curved near the bottom of the

wall

- Active case

2

2

1HKP AA γγγγ====

Where

[[[[ ]]]]

2

)sin(

)sin()sin()sin(

sin/)sin(

−−−−

−−−−++++++++++++

−−−−====

ββββαααα

ββββφφφφδδδδφφφφδδδδαααα

ααααφφφφααααAK

The point of application is assumed at the distance of H/3 above

the base of the wall

- Passive case

Angles between W and P is (1800-α+δ) and between W and R is

(θ+φ)

2

2

1HKP PP γγγγ====

Where

[[[[ ]]]]

2

)sin(

)sin()sin()sin(

sin/)sin(

−−−−

++++++++−−−−−−−−

++++====

ββββαααα

ββββφφφφδδδδφφφφδδδδαααα

ααααφφφφααααPK

RETAINING WALL


57

Type of retaining wall:

- Gravity walls Depend largely upon their own weight for stability.

Have wide base and usually a rigid construction

- Embedded walls Consists of vertical driven or placed sheets or piles that may be anchored,

tied or propped


58

- Reinforced earth Wall material may be in situ rock or soil into which reinforcement is inserted

STABILITY OF THE WALL

- Gravity and cantilever walls

o They are liable to rotational or translational movement

o Lateral pressure calculation use Rankin or Coulomb’s theory

o Basic condition:

� The base pressure at the toe < the allowable bearing capacity

Pressure under the base, )6

1(2B

e

B

Rp v ±±±±====

Where R : the resultant force acting on the base wall

e : the eccentricity < B/6

� The safety factor against sliding between the base and the

underlying > 1.5 or >2.0

The factor of safety against sliding h

vs

R

RF

δδδδtan==== ,

δ is the angle of friction between the base and the underlying soil


59

� The safety factor against overturning > 2.0

∑∑∑∑

∑∑∑∑====

).., (

).., sing(

hOT

Rgemomentsgoverturnin

WgemomentsstabiliF

Example:

Check the stability of the reinforced concrete retaining wall shown in figure.

The soil is a silty gravel sand with a horizontal surface and no surcharge. The

water table lies below the base. The soil properties: c’=0, φ’=400 and γ=20

kN/m3. Concrete properties: γ=24 kN/m3

Answer:

KA=(1-sin40)/(1+sin40)=0.217

2

2

1HKP AA γγγγ==== =1/2x0.217x20x8.02= 138.9 kN/m = Rh

Vertical force : W1 = 24x0.3x7.25 = 52.2 kN/m

W2 = 24x5.2x0.75 = 93.6 kN/m

Ws = 20x3.2x7.25 = 464.0 kN/m

Total = 609.8 kN/m

Sliding:

68.39.138

40tan8.609tan============

h

vs

R

RF

δδδδ (stable)

Overturning: overturning moment : Mo = 138.9x8/3=370 kNm/m

Stabilising moment, Ms =52.2x1.85+93.6x2.6+464(2.0+3.2/2)

= 2010 kNm/m

FOT = 2010/370=5.43 (stable)


60

The maximum base pressure:

Position of base resultant, x = (2010-370)/609.8=2.689 m

eccentricity, e=1/2x5.2-2.689=-0.089 (on the right side)

129)2.5

089.061(

2.5

8.609)

61(

22====++++====++++====

x

B

e

B

Rp v kN/m2


61

SLOPE STABILITY

Soil mass movement may take place as the result of a shear failure along an

internal surface or when a general decrease in effective

stress particles causes full or partial liquefaction

Type of soil mass movement

Slope failure

are usually precipitated by a variation in condition, such as a

change in rainfall, drainage, loading or surface stability (such

as removal of vegetation)

may occur immediately after construction or may develop

slowly

TRANSLATIONAL SLIDE ON AN INFINITE SLOPE

- Is as a plane translational movement at a shallow depth parallel to

a long slope

- Usually cause by sudden increase in pore pressure


62

- Undrained infinite slope

Forces:

Weight of the element, cbzW ββββγγγγ cos ====

Normal reaction on the slip plane, cWN ββββcos====

Tangential force down the slope, cWT ββββsin====

Shear resistance force up the slope, bR ττττ==== Where βc : critical angle of slope

τf : undrained shear strength of soil = cu

E1=E2 cancel out

For limiting equilibrium: R-T=0

Therefore, cccu zbWbc ββββββββγγγγββββ sincossin ========

z

cor

z

c uccc

u

γγγγββββββββββββββββ

γγγγ

2sin2 2sin

2

1cossin c ============

This stability condition only valid for 00 < βc < 450, i.e. 2cu/γz ≤ 1

Critical depth, c

uc

cz

ββββγγγγ 2sin

2====

Factor of safety, ββββ

ββββ

2sin

2sin cF ====

Where β is actual angle of slope


63

Drained infinity slope

Under drain condition 'tan 'tan φφφφσσσσττττφφφφσσσσττττ nfnf or ========

Forces:

Weight of the element, cbzW ββββγγγγ cos ====

Normal reaction on the slip plane, cWN ββββcos====

Tangential force down the slope, cWT ββββsin====

Pore pressure force on the slip plane, cwhbu ββββγγγγ 2cos====

Shear resistance force up the slope, bR 'ττττ====

When c’=0, for limiting equilibrium R=T

'tan)coscos('tan' 2 φφφφββββγγγγββββφφφφ cwc hbWNR −−−−========

'tancos)( 2 φφφφββββγγγγγγγγ cw bhz −−−−====

ccc zbWT ββββββββγγγγββββ sincossin ========

Then: 'tancos)(sincos 2 φφφφββββγγγγγγγγββββββββγγγγ cwcc bhzzb −−−−====

'tan

tan

φφφφ

ββββ

γγγγ

γγγγγγγγ c

z

whz====

−−−− and 'tan1tan φφφφ

γγγγ

γγγγββββ

−−−−====

z

wc

h

Factor of safety : ββββ

φφφφ

γγγγ

γγγγ

tan

'tan1

−−−−====

z

whF

ββββγγγγ 2coshu w==== ; so ββββ

φφφφ

ββββγγγγ tan

'tan

cos1

2

−−−−====

z

uF

Special cases:

a) Dry sand or gravel


64

h = 0, so tan βc = tan φ’

b) Groundwater level coincident with the slip of plane

Coarse grain soil: c’=0 and h=0, so tan βc = tan φ’

c) Groundwater level below the slip plane

'tan1tan φφφφγγγγ

γγγγββββ

++++====

z

swc

h; hs : distance of GWL below the slip plane

d) Waterlogged slope with steady parallel seepage The upper flow line is coincident with ground surface. h = z

'tan'

tan 'tan1tan φφφφγγγγ

γγγγββββφφφφ

γγγγ

γγγγββββ ====

−−−−==== c

wc or

e) c'=0

ββββββββγγγγ

φφφφββββγγγγγγγγφφφφ

cossin

'tancos)(''tan''2

z

hzc

T

NbcF w−−−−++++

====++++

====

Example: 1. Calculate the factor of safety relating to the undrained stability of a long

slope of 35o, if at a depth of 1.8 m a weak layer of cohesive soil occurs. The

soil has cu=24 kN/m2 and γ=18.5 kN/m3

Answer:

44.18.15.18

24222sin ============

x

x

z

cuc γγγγ

ββββ

Factor of safety, 52.1)352sin(

44.1

2sin

2sin

0============

xF c

ββββ

ββββ

2. Calculate the factor of safety of the waterlogged slope (slope=140) against

failure along a slip plane parallel to the surface at a depth of 4 m, if at this

depth there is a thin layer of cohesive soil with the following

properties:c’=12 kN/m2, φ’=240 and γ=20 kN/m3.

Answer:

55.114cos14sin420

24tan14cos)481.9420(12

cossin

'tancos)(' 22

====−−−−++++

====−−−−++++

====xxx

xxx

z

hzcF w

ββββββββγγγγ

φφφφββββγγγγγγγγ


65

ROTATIONAL SLIP

- Type of failure surface

- Stability :

1. Short-term (end-of-construction)

- undrained condition (τ=cu)

- referred to total stress

2. Long-term (long after the construction)

- Related to effective stress

- UNDRAINED STABILITY-TOTAL STRESS ANALYSES (φu=0)

Disturbing moment = Wd

Resistance: length of arc AB=Rθ

Shear resistance force along AB=cuRθ

Shear resistance moment = cuR2θ


66

The factor of safety, Wd

Rc

momentdisturbing

momentceresisshearF u θθθθ2

tan ========

Effect of crack

- crack reduced length of AB

- If crack filled with water

hydrostatic force Pw

2

2

1oww zP γγγγ====

- cw

cu

yPWd

RcF

++++====

θθθθ2

Multy-layer

Submerged slope

γ’=γsat-γw

Location of the most critical

circle

...

...)(2

++++++++

++++++++====

BBAA

BuBAuA

dWdW

ccRF

θθθθθθθθ


67

Taylor’s stability number method (1948)

For: - homogeneous soil

- Using total stress analysis

- Ignoring the possibility of tension crack

Wd

RLcF u====

L∝H and W∝ γH2, so: L=K1H, W=K2γH2

Then dKH

HRKcF u

22

1

γγγγ====

The stability number, HF

c

RK

dKN u

γγγγ========

1

2

Hence HN

cF u

γγγγ====


68

(a) φu = 0 (b) φu > 0

- DRAINED STABILITY-EFFECTIVE STRESS ANALYSES-THE

METHOD OF SLICES

The force acting on a slice of length 1 m will be as follows:

1. The total weight of the slice, bhW γγγγ====

2. The normal force on the base, ulNlN −−−−======== '' σσσσ

3. The shear force induced along the base, ααααsinWT ====

4. The normal interslice force, E1 and E2

5. The tangential interslice force, X1 and X2

The number of slices > 5

At the point of limiting equilibrium:

The total disturbing moment = the moment of the total mobilized

shear force along AB

∑∑∑∑ ∑∑∑∑ ∑∑∑∑======== RWlRF

lRf

m ααααττττ

ττττ sin then ∑∑∑∑

∑∑∑∑====

αααα

ττττ

sinW

lF

f

'tan' φφφφσσσσττττ nf c ++++==== and 'tan'' φφφφττττ Nlclf ++++==== ,

so

∑∑∑∑

∑∑∑∑ ∑∑∑∑++++====

αααα

φφφφ

sin

'tan''

W

NlcF

For homogeneous soil:

∑∑∑∑

∑∑∑∑++++====

αααα

φφφφ

sin

''tan'

W

NLcF AC , where LAC=arc length AC=θR


69

FELLENIUS’METHOD

It is assumed : E1 = E2 and X1 = X2 So ulWN −−−−==== ααααcos' , and

∑∑∑∑

∑∑∑∑ −−−−++++====

αααα

ααααφφφφ

sin

)cos('tan'

W

ulWLcF AC

W sin α and W cos α can be determined graphically for each slices.

This solution underestimates the factor of safety (5-20%)

BISHOP’S SIMPLIFIED METHOD

It is assumed X1 = X2 but E1 ≠ E2

For equilibrium along the base of the

slice:

F

NLcWl

FW

f 'tan''sinsin0

φφφφαααα

τττταααα

++++−−−−====−−−−====

So: ∑∑∑∑

∑∑∑∑ ++++====

αααα

φφφφ

sin

)'tan''(

W

NlcF

For equilibrium in a vertical direction:

ααααττττ

αααααααα sincoscos'0 lF

ulNWf

−−−−−−−−−−−−====

ααααφφφφ

αααααααα sin))'tan''

(coscos'F

NlculNW

++++−−−−−−−−−−−−====

αααα

φφφφαααα

αααααααα

sin'tan

cos

cossin'

'

F

ullF

cW

N

++++

−−−−−−−−==== ; ααααsecbl ====

Then: [[[[ ]]]]

∑∑∑∑++++

−−−−++++

∑∑∑∑====

F

ubWbc

WF

'tantan1

sec'tan)('

sin

1

φφφφααααααααφφφφ

αααα


70

Example:

Determine the factor of safety in terms of effective stress for the slope in

figure in respect of the trial circle shown. The soil properties are : c’=10

kN/m2, φ’=280, γ=18 kN/m3.

Answers:

Graphically, the soil slip mass has

been divided into slices of width 3

m 8 slices

The average height (h) of each

slice is scaled of the diagram,

the weight of each slice:

hxhxhbW 0.54318 ============ γγγγ kN/m

the length of the chord at the

base of each slice is scaled of the

diagram

the pore pressure : xlxhu w 81.9====

A triangle of force is drawn at the

base of each slice to obtain values

of N and T

N=Wcosα and T=Wsinα

The calculation :

LAC=θR=91.47x(π/180)x18.58=29.7 m

05.1814

105228tan7.2910

sin

''tan' 0

====++++

====∑∑∑∑

∑∑∑∑++++====

xx

W

NLcF AC

αααα

φφφφ


71

A cutting in a cohesive soil has a slope angle of 350 and a vertical height of 8 m.

The hard rock layer found at the depth of 4 m below the floor of the cutting.

The soil has cu=40 kN/m2, φu=0 and γ=18 kN/m

3. determine the factor of

stability using Taylor’s method

Answer:

H= 8 m, DH=12 m D=12/8=1.5

From Taylor’s chart: when D=1.5, β=350 and φu=0 N=0.168 and n=0.6

65.1818168.0

40============

xxHN

cF u

γγγγ

The break-out point (nH) = 0.6x8 = 4.8 m

A slope is excavated to a depth of 8 m in a deep layer of saturated clay of unit

weight 19 kN/m3. cu=65 kN/m2 and φu=0. determine the factor of safety for

the trial failure as shown in figure.

Answer:

From figure:

Cross sectional area ABCD = 70 m2

Weight of soil mass = 70x19 = 1330

kN/m

The centroid ABCD is 4.5 m from O,

angle AOC=89.50, radius OC=12.1 m

and the arc length ABC=18.9 m

48.25.41330

9.181.1265============

x

xx

Wd

RLcF u


72

STABILITY OF EARTH DAMS

- The factor of safety of both slopes must be determined for the

most critical conditions

- The most critical condition:

o The upstream slope: - at the end of construction

- during rapid drawn of the reservoir level

o The down stream slope: - at the end of construction

- during steady seepage when the

reservoir

full

- It is influenced dominantly by pore water pressure

- At end of construction o Excess pore water pressure

o An effective stress analysis is preferable

o Factor safety > 1.3

The pore pressure at any point: 1σσσσ∆∆∆∆++++====∆∆∆∆++++==== Buuuu oo

Where B is pore pressure coefficient (1σσσσ∆∆∆∆

∆∆∆∆==== uB )

Then pore pressure ratio h

Bh

ur ou

γγγγ

σσσσ

γγγγ1∆∆∆∆

++++====

If it is assumed that the increase in total major principal stress is equal to

fill pressure, so Bh

ur ou ++++====

γγγγ

- Steady seepage o It is analyzed in term of effective stress

o Pore pressure being determined from the flow net

o Factor of safety > 1.5

- Rapid drawdown

o Will result in a change of the pore pressure distribution


73

o Factor of safety > 1.2

The pore pressure before drawdown at a typical point P on a potential failure

surface : )'( hhhu wwo −−−−++++==== γγγγ

For drawdown depth exceeding hw: wwhγγγγσσσσ −−−−====∆∆∆∆ 1

And the change of pore pressure : wwhBBu γγγγσσσσ −−−−====∆∆∆∆====∆∆∆∆ 1

Therefore the pore pressure at P immediately after drawdown:

}')1({ hBhhuuu wwo −−−−−−−−++++====∆∆∆∆++++==== γγγγ

Hence :

}'

)1(1{h

hB

h

h

h

ur w

sat

w

satu −−−−−−−−++++========

γγγγ

γγγγ

γγγγ

water in soil - ipb university

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