water in soil - universiti teknologi...

Influence of water in soil

Prepared by:

Dr. Hetty

Muhammad Azril

Fauziah Kassim

Norafida

Copyright Dr. Hetty et. al. 2014

C:/Users/User/Desktop/HD/My Passport/Soil Mechanics/E-BOOK Soil mechanics.pdf

C:/Users/User/Desktop/HD/My Passport/Soil Mechanics/E-BOOK Soil mechanics.pdf

Important things

How water flow in soil?

How fast water flow in soil?

Does water effect soil condition?

Can water permit in geotechnical

structures (Dam, Retaining Wall and etc.)?

Why do we need to know water flow?


Introduction

From an engineering mechanics point of view, groundwater in soil may be one of two types, occurring in two distinct zones separated by the water table or phreatic surface.

Two zones of water in soil are:-

a) Phreatic water or gravitational water which;

-subject to gravitational forces

-saturates the pore spaces in the soil below the water table

-has an internal pore pressure greater than atmospheric pressure

-tends to flow laterally

(b) Vadose water, which may be:

-transient percolating water, moving downwards to join the phreatic water below the water table

-capillary water held above the water table by surface tension forces (with internal pore pressure less than atmospheric)


Water zones in soil

Fig. 1: Water zones in soil

Source: Whitlow, 2001Copyright Dr. Hetty et. al. 2014

Water Flows through Soil (1 D)

As water flows through pipes it will also

flow to any mediums that permit

Water could flow through soil medium

that can be explained by the Bernoulli’s

equations


Bernoulli Equation

ht= hz + u + v2 = hz + hp

w 2g

where hz = position or elevation head

u = pressure head to pore pressure

w

v2 = velocity head

2g

Water will flow through soil if there exists a difference

of head within 2 points Copyright Dr. Hetty et. al. 2014

HpA

HpB

▲Ht A-B = hTA- hTB

= (ZA+hpA)- (ZB +hpB)Copyright Dr. Hetty et. al. 2014

Darcy Law for saturated flow

In saturated conditions, one-dimensional flow is governed by Darcy’s Law, which states that the flow velocity is proportional to the hydraulic gradient:

v α i or v = ki

where v = flow velocity

k = the flow constant or coefficient of permeability

i = the hydraulic gradient = ∆ h

∆ L

∆h = difference in total head over a flow path length ( ∆L)


Coefficient of Permeability

The permeability of a soil is a measure of its capacity

to allow the flow of water through the pore spaces

between solid particles.

The degree of permeability is determined by applying a

hydraulic pressure gradient, i in a sample and

measuring the consequent rate of flow of water.

The coefficient of permeability is expressed as velocity,

(mm/s2)



The value of k is used as a measure of the resistance to flow by the soil, and it is affected by several factors:

a) The porosity of the soil

b) The particle size distribution

c) The shape and the orientation of soil particles

d) The degree of saturation/presence of air

e) The type of cation and thickness of adsorbed layers associated with clay minerals (if present)

f) The viscosity of the soil water, which varies with temperature


Permeability

“the larger a soil’s void space, the greater

will be its permeability. Conversely, the

smaller the void space, the lesser will be

it’s permeability”

Therefore, coarse soil > fine soil on

permeability value.



k value (m/s) Type of soil Note

102 Clean gravels Very good drainage

101

1

10-1

10-2 Clean sands

Gravel-sand mixtures10-3 Good drainage

10-4

10-5 Very fine sands

Silts and silty sands

Poor Drainage

10-6

10-7 Clay silts (>20% clay)

10-8 Practically impervious

10-9

Source: Roy Whitlow (2001) Copyright Dr. Hetty et. al. 2014

Question

Could you estimate the coefficient of

permeability if you are given a sandy soil

sample?

The answer

It ranges from 10-4 to 10-5 m/s


Determination of k Laboratory

- Falling head

- Constant Head

In situ ( On field Test)

- Steady state Pumping - confined aquifer/unconfined aquifer

-Borehole test- Rising head, Variable head, Constant Head


Lab vs. In situ (Advantages)

Lab

1. Reliability is

questionable

Insitu

1. Reliable

Lab vs. In situ (Disadvantages)

• Lab

1. Simple

2. Cheap (Cost)

• Insitu

1. Complex

2. Expensive


Laboratory Constant Head

Usually use for coarse-grained soils

The equation use to calculate is k = Q L

Aht

Falling Head

Usually use for cohesive soils

The equation use to calculate is k = 2.303 aL log10 h1

At h2Copyright Dr. Hetty et. al. 2014

Constant Head TestThe total volume of water collected may be expressed as

Q = Avt = A(ki)t = Ak(h/L)t

where Q = volume of water collected

A = area of cross section of the soil specimen

t = duration of water collection

h = head loss

or the equation can be substituted into

Q = A(kh)t

L

L = length of specimen

or

k = QL

AhtCopyright Dr. Hetty et. al. 2014

Constant Head Apparatus

Fig. 2: Constant Head apparatus (M. Das, 2000)


Constant Head Apparatus

Fig. 3: Constant Head apparatus


Example 1

A test from Constant Head Test give this value :-

L = 18 mm ; A = area of specimen = 3.5 mm2

h = constant head difference = 28 mm

Water collected in a period of 3 minutes = 21.58 mm3

Solution :-

•k = QL

Aht

Given ; Q= 21.58 mm3, L = 18 mm, A = 3.5 mm2, h = 28 mm and t = 3 min

K = (21.58)(18) = 0.022mm/s

(3.5)(28)(3)(60)Copyright Dr. Hetty et. al. 2014

Example 2Determine the time required from the coefficient of

permeability,k =0.0986 cm/s according to following

data:

Quantity of water discharge during test = 250 cm3

Length of specimen = 11.43 cm

Head (difference between manometer level) = 5.5 cm

Diameter of specimen = 10.16 cm


Solution :-

k = QL

Aht

Given ; k =0.0986 cm/s Q= 250 cm3, L = 11.43 mm, d

= 10.16 cm, h = 5.5 cm

A = ∏ x 10.162/4 = 81.07 cm2

t = (250)(11.43) = 65 s

(81.07)(5.5)(0.0986)


Example 3During a test using a constant-head permeameter, the following

data were collected. Determine the average value of k

Diameter of sample = 100 mm

Distance between manometer tapping points = 150 mm

Quality collected in 2 min

(1x103mm3)

541 503 509 474

Difference in manometer

level (mm)

76 72 68 65


Solution

Cross-sectional area of sample = 1002 x ∏/4 = 7854 mm2

Flow time = t = 2 x 60 = 120 s

k = (Q)(150) = 0.159 Q/h mm/s

(7854)(h)(120)

Average k = (k1+k2+k3+k4) / 4 = 1.15 mm/s


Falling Head Test

The specimen is first saturated with water.

Water is then allowed to move through the soil specimen under a falling-head condition while the time required for a certain quantity of water to pass through the soil specimen is determine and recorded.

Water is insert through the burette ( (a) cross section area of burette) for several time (t)


Falling Head

The rate of flow of the water through the specimen at any time t can be given by

qin = -a(dh) qout = k(h/L)A

(dt)

where q = flow rate L = length specimen

a = cross-sectional area of the standpipe

A = cross-sectional area of the soil specimen

Equating qin and qout gives

-a(dh/dt) = kA(h/L) k = (aL/At) In (h1/h2)

or

k = 2.303 aL log10 h1 At h2

h1 = hydraulic head of

beginning

h2 = hydraulic head of endCopyright Dr. Hetty et. al. 2014

Falling Head Test Set Up

Fig. 4: Schematic of falling permeability setup ( Liu and Evett,

2002)


Falling Head Apparatus


Example 1

In a laboratory, a falling head permeability test was

conducted on a silty soil. The following data were

obtained:

Length of specimen = 15.80 cm

Diameter of specimen = 10.16 cm

Cross section area of burrette = 1.83 cm2

Hydraulic head at beginning h1 = 120.0 cm

Hydraulic head at end h2 = 110.0 cm

Time requirement = 20.0 min

Determine the k value for the soil sample? Copyright Dr. Hetty et. al. 2014

Solution :

k = 2.303 aL log10 h1

At h2

A = ∏ (10.16 cm)2/4 = 81.07 cm2

k = 2.303 (1.83)(15.80) log10 (120)

(81.07)(1200) (110)

k = 2.58 x 10-5 cm/s


Example 2

A falling head permeameter mould of 75 mm

diameter and 150 mm height is filled with

soil height.the internal diameter of burette

tube is 5 mm. Determine the time ( in

seconds) required for the hydraulic head to

fall from 1300 mm to 50 mm if the

coefficient of permeability of soil in the

mould is 0.0215 mm/s?


Given ; k = 0.0215 mm/s

D = 75 mm, L = 150 mm

Internal diameter of tube, d = 5 mm

h1 = 1300 mm , h2 = 50 mm

Solution :

Specimen cross-section area, A = ∏(752)/4

= 4418.44 mm2

Tube cross-sectional area, a = ∏ (52)/4

= 19.64 mm2

0.0215 = 2.303 (19.64)(150) log10 (1300)

(4418.44)(t) (50)

t = 2.303 (19.64)(150) log10 (1300)

(4418.44)(0.0215) (50)

t = 101 s


Example 3Sampel tanah liat berdiameter 100mm sepanjang 150mm diuji

kebolehtelapanya di makmal , set data ujikaji kebolehtelapan

turus menurun adalah seperti jadual berikut:-

Diameter

paip

(mm)

Aras dalam paip(mm)

Bacaan awal Bacaan akhir

Perbezaan masa

(s)

5.00 1200

800

800

400

82

149

9.00 1200

900

700

900

700

400

177

169

368

Dapatkan purata bagi nilai k tanah ini.


Given;

Diameter sample = 100 mm, L = 150 mm

Diameter pipe = 5 mm dan 9 mm

Solution:

A = ∏ (1002)/4 = 7855 mm2

a1 = ∏ (52)/4 = 19.64 mm2; a2 = ∏ (92)/4 = 63.63

For a1

k = 2.303 (19.64)(150) log10 h1

(7855)t h2

= 0.864 log10h1

t h2Copyright Dr. Hetty et. al. 2014

For a2

k = 2.303 (63.63)(150) log10 h1

(7855)t h2

= 2.798 log10h1

t h2

a h1 h2 t k =

1 1200 800 82 0.011176

800 400 149 0.006469

2 1200 900 177 0.016476

900 700 169 0.017191

700 400 368 0.008313

kpurata = 0.011925 mm/s


In Situ Permeability Test

Permeability determined in a laboratory may not be truly

indicative of the in situ permeability.

Thus, field tests are generally more reliable because the

test are performed on the undisturbed soil.

Other reason are :-

Soil stratification.

Overburden stress.

Location of the groundwater table.

Several field methods for evaluating permeability such as

pumping, borehole and tracer tests.


Pumping Test

Pumping tests involve the measurement of a pumped quantity from a well together with observations in other wells of the resulting drawdown of the groundwater level.

A steady state is achieved when at a constant pumping rate, the levels in the observation wells also remain constant.

The pumping rate and the levels in two or more observation wells are then noted.

The analysis of the results depends on whether the aquifer is confined or unconfined.


Confined Aquifer

Through an impermeable layer and then a

permeable layer (an aquifer) to another

impermeable layer.

If water is pumped from the well at a

constant discharge (q), flow will enter the

well only from the aquifer and the

piezometric surface will be drown toward

the well as shown on figure.


Confined Aquifer (Cont.)

An equilibrium condition will be reached at some time after pumping begins.

The piezometric surface can be located by auxiliary observation wells located at distances r1 and r2 from the pumping well.

The piezometric surface is located at distance h1 above the top of the aquifer at point r1 from the pumping well and at distance h2 at point r2.


Unconfined Aquifer

The piezometric surface lies within the

aquifer.

The analysis of this type of well is the

same as that for the confined aquifer.


Steady state pumping test (confined

aquifer)

Source: Whitlow (2001)Copyright Dr. Hetty et. al. 2014

Confined Aquifer

k = q ln (r2/r1)

2D(h2-h1)‘

Unconfined Aquifer

k = q ln (r2/r1)

(h22-h1

2)


Steady state pumping test

(unconfined aquifer)

Source: Whitlow (2001)Copyright Dr. Hetty et. al. 2014

Example 1

A pumping test was performed in a well penetrating a confined aquifer to evaluate the coefficient of permeability of the soil in the aquifer. When equilibrium flow was reached, the following data were obtained:

q = 750 l/min

water level (h1 & h2) = 5m & 6 m

Distance from the well (r1 & r2) = 20m & 60m

Thickness of aquifer = 6m


Solution :

k = q ln (r2/r1)

2D(h2-h1)

q = 750 l/min = 0.0125 m3/s

D = 6m; h1 = 5m; h2 = 6m; r1 = 20m; r2 = 60m

k = (0.0125m3/s) ln (60m/20m) = 0.0137

2(6m)(6m-5m) 37.7

= 0.00036m/s


Example 2

Some conditions as in Example 1, except that the well is

located in an unconfined aquifer.


Solution:

k = q ln (r2/r1)

(h22-h1

2)

q = 750 l/min = 0.0125 m3/s

h1 = 5m; h2 = 6m; r1 = 20m; r2 = 60m

k = (0.0125) ln (60/20) = 0.0137

(62-52) 34.54

= 0.00397 m/s


Horizontal and Vertical flow in stratified soils

Fig.6 a: Horizontal Flow b: Vertical flow

Source: Whitlow (2001)


Horizontal flow

Horizontal flow, occur when the water

flow are tangential to strata

The head lost between the entry and exit

faces will be the same for each layer:

# h1 = h2 = h3 = h

Hence the hydraulic gradients are the

same:

# i1 = i2 = i3 = i


Horizontol flow

Starting with Darcy law (q=Aki) the flows in the layer will be:

∆q1 = A1k1i1 ∆q2= A2k2i2 ∆q3 = A3k3i3

which Ā= B (D1 +D2+ D3) and total flow qH = ∆q1+ ∆q2+ ∆q3 = Ā kHi

kH = average horizontal coefficient of permeability

= D1k1+ D2k2+ D3k3

D1+D2+D3


Vertical flow Vertical flow is a flow which normal to strata. The rate

of flow will be the same through each layer.

#∆q1 = ∆q2= ∆q3 = ∆qv

The head lost in each layer will be h1,h2 and h3 giving

hydraulic gradients or

#i1=h1/D1 i2 = h2/D2 i3=h3/D3


Vertical flow

From Darcy law (q=Aki)

The total flow, qv = kvĀi but the total head lost ∑h =

h1+h2+h3 and L = D1+D2+D3

Therefore kv = D1+ D2+ D3

D1 + D2 +D3

k1 k2 k3


Question

A stratified soil consists of sand and silts.

The sand are generally 150mm thickness and

have a permeability k= 6.5 x 10 -1 mm/s, the

silts layer are 1.8m and the k is 2.5 x 10-4

mm/s. Assuming flow conditions are

isotropic determine horizontal and vertical

flow?


Capillary water in Vadose zone

Capillary water is held above the water table by surface tension which the attractive force exerted at the interface or surface between materials in different physical states, i.e. liquid/gas, solid/liquid

For example, a water/ air surface exhibits an apparently elastic molecular skin due to the sub-surface water molecules (which are more dense than air) exerting a greater attraction than air molecules. Similarly, water is attracted towards a solid interface because of the greater density and therefore attraction of the solid


Capillary water

Fig.7 Capillary water in capillary tube



Capillary rise

Capillary rise, hc = 4T cos / w d, As an approximation for soils put T = 0.000074 kN/m w = 9.81 kN/m3 = 0 and d ed10 (where d10 = effective size)

Giving hc 4 x 0.000074 x 106 = 30

ed10 x 9.81 ed10

This estimate may improved to allow for the effect of grading and shape characteristics, such as irregularity and flakiness (Terzaghi and Peck)

hc = C

ed10

where C=a value between 10 and 40 mm2


Capillary rise in soil



Effect of particle size on capillary rise


Soil Profile


Total stress, pore pressure and effective stress

Stress or intensity of loading is the load per unit area. The fundamental definition of stress is the ratio of the force ∆P acting on a plane.

Total stress () is the stress carried by the soil particles and the liquids and gases in the void.

= ∑ (h x )

When an external stress is applied to a soil mass that is saturated with pore water the immediate effect is an increase in the pore pressure. This produces a tendency for the pore water to flow away through adjoining voids, with the result that the pore decreases and the applied stress is transferred to the granular fabric of the soil. At a given time after application, therefore, the applied total stress will be balanced by two internal stress components.


Effective stress Effective stress (’): This is the stress transmitted through

the soil fabric via inter-granular contacts. It is this stress component that is effective in controlling both volume change deformation and the shear strength of the soil since both normal stress and shear stress is transmitted across grain to grain contacts. Terzaghi (1943) showed that, for a saturated soil, effective stress may be defined quantitatively as the difference between the total stress and the pore pressure:

’ = - u

As a conclusion effective stress () is the stress carried by the soil particles and pore pressure (u) is the pressure of water held in soil pores.

Note: ’ = sat -w where,

’ is submerge unit weight,

sat is saturated unit weight

w is water unit weight


Pore pressure

Pore pressure (u): This is the pressure induced in the fluid (either water, or vapor and water) filling the pores. Pore fluid is able to transmit normal stress, but not shear stress, and is therefore ineffective in providing shear resistance. For this reason, the pore pressure is sometimes referred to as neutral pressure.

Pore pressure (u) = wh

While pore pressure for capillary zone ur = -Sr wh/ 100


b = 18kN/m3 Sand

b =19.0 kN/m3 Silty Clay

sat = 20 kN/m3 Clay

A

B

C

D

Ground surface

1m

1m

2 m


Example


d = 18kN/m3 Dry Sand

b =19.0 kN/m3 S= 50% Silty Clay

sat = 22 kN/m3 Clay

A

B

C

D

Ground Surface

1m

0.5m

2 m


Exam question

Soil profile shown in following figure. You are

required to determine and plot stresses

distribution at point A,B,C dan D (Total stress,

pore water pressure and effective stress)

A-B : 3 m , B-C: 2m C-D: 4m

Determine actual maximum capillary rise

according Terzaghi dan Peck if C is 40mm2 .


Gs= 2.7, e = 0.6 Dry sand d10 = 0.2mm

d = 14.03 kN/m3 Silty Clay w = 20% d10 =

0.05mm (capillary zone) e = 0.85 Gs = 2.65

sat = 22 kN/m3 Clay

w = 55%

A

B

C

D

Ground surface


Exam question

Determine stresses distribution σ, and σ’ on

each soil profile level if the GWL rise up to

ground surface. γb dan γsat of each soil layer as

following:-

A-B: 3 m γb = 16 kN/m3 dan γsat = 18kN/m3

B-C: 2m γb = 17kN/m3 dan γ sat = 19kN/m3

C-D: 4m γb = 18kN/m3 dan γsat = 20kN/m3


Flow net

1. Streamlines Y and Equip. lines are .

2. Streamlines Y are parallel to no flow

boundaries.

3. Grids are curvilinear squares, where

diagonals cross at right angles.

4. Each stream tube carries the same flow.


Flow Net in Isotropic Soil

Portion of a flow net is shown below


Flow Net in Isotropic Soil

The equation for flow nets originates from

Darcy’s Law.

Flow Net solution is equivalent to solving

the governing equations of flow for a

uniform isotropic aquifer with well-defined

boundary conditions.


Method of Drawing

1. Draw to a convenient scale the cross

sections of the structure, water elevations,

and aquifer profiles.

2. Establish boundary conditions and draw

one or two flow lines Y and equipotential

lines F near the boundaries.


Method of Drawing

3. Sketch intermediate flow lines and equipotential lines by smooth curves adhering to right-angle intersections and square grids. Where flow direction is a straight line, flow lines are an equal distance apart and parallel.

4. Continue sketching until a problem develops. Each problem will indicate changes to be made in the entire net. Successive trials will result in a reasonably consistent flow net.


Method of Drawing

5. In most cases, 5 to 10 flow lines are

usually sufficient. Depending on the no.

of flow lines selected, the number of

equipotential lines will automatically be

fixed by geometry and grid layout.

6. Equivalent to solving the governing

equations of GW flow in 2-dimensions.


Example of Flow net


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