2020
Advanced Geotechnical
Engineering
Neal and Procter
Lecture Notes
Edition 2
TABLE OF CONTENTS
•
Course Objectives
•
Syllabus
•
Course Evaluation
•
Provisional Program (presentation may change sequentially)
•
References
•
Terms and Definitions
Content of Syllabus
Chapter 1 Strength of Soils
Page 1.1
Chapter 2 Earth Pressure
Page 2.1
Chapter 3 Stability of Slopes
Page 3.1
Chapter 4 Bearing Capacity
Page 4.1
Chapter 5 Consolidation
Page 5.1
Chapter 6 Permeability
Page 6.1
Laboratory program
A series of soil tests will be conducted during the early part of the semester that will complement the theory
learned in class time. Your dedicated technician will laboratory manuals to be used as a guideline and will include
safety instructions as per the current Health and Safety Act.
The student must familiarize him/herself with the contents of the laboratory manual as it contains important
information to successfully complete the soil tests. A date will be announced by the technician for the submission
of reports that will be marked and submitted to your Faculty. The written report must contain all results and a
summary of the soil test. A poster is required to be submitted at the end of the final submission.The final mark is
compiled alongside theory questions in the form of a quiz on each soil test. These quizzes are submitted on-line on
the e-learning platform known as Blackboard. The link is on the University's web page. A combination of quiz
work and written reports is compiled into a final mark. A minimum of 50% is required to pass the practical
module.
iv
Suggested reading material
Elements of Soil Mechanics - G.N. Smith 7'11 Edition
Solving Problems in Soil Mechanics - B.H.C. Sutton 2nd Edition
Soil Mechanics - RF. Craig Slh Edition
Highway Materials, Soils and concrete - Harold N. Atkins 3rd
Edition Principals of Geotechnical Engineering - Braja M. Das 2nd
Edition Basic Soil Mechanics - R Whitlow 2nd Edition
Soil Mechanics, SI Version - William T Lambe and
Elements of Foundation Design - GN Smith and EL Pole
Problems in Engineering Soils - PL Capper, WF Cassie, JD Geddes
Examination subjects for Civil Engineers, Soil Mechanics - MJ
Smith
-x-
Chapter 1
Shear Strength of Soils
Neal and Procter
Advanced Geotechnical
Engineering
OBJECTIVES
Upon completion ohhis chapter you must be able to:
Understand the difference between shear strength and stresses in soils.
Define Coulomb's Law, angle ofinternal friction and cohesion.
Descnbe the shear box test, and determine its shear parameters.
Describe the triaxial test, and determine its shear parameters.
Name the different types oftriaxial tests.
Understand the principal ofeffective stress.
Descnbe the unconfined compression test, as well as the shear vane test.
1.2
Shear Strength
Every building or structure which is founded in or above the earth imposes loads on the soil which in turn
is directed to the foundations. The stresses set up in the soil causes deformation of the soil (because soils
are elastic), and can occur in three ways:
•
By elastic deformation of the soil particles
•
By the change in volume of the soil resulting from the expulsion of fluid (water and or gas) from
the voids between the solid particles.
•
By the slippage of the soil particles, against each another, which may lead to the sliding of one
body of soil relative to the surrounding mass
The first of these is negligible for most soils at the usual levels of stress which occur in practice. The
second is known as consolidation, and will be dealt with in a later chapter.
The third is the process known as shear failure and occurs when shear stresses set up in the soil mass
exceed the maximum shear resistance i.e. its shear strength.
Shear strength definition: The shear strength of a soil mass is the internal resistance per unit area
(maximum shear stress) that the soil mass can provide in resisting failure and sliding along any plane
inside it (see the image below indicating the sliding surface under failure)
When applying a load on a confined mass of soil it will fail when the soil particles start to move relative
to each other. This failure is a result of sliding or shearing along the shear plane, and will be the shear
failure, and the maximum resistance of the soil to shearing is the shear strength. The soil at this stage is
in a plastic state and will not support any additional load. Therefore we can regard the strength of a soil as
the shearing resistance of that particular soil.
1.3
Shear Stttll1(th
There are a few factors governing shearing resistance:
•
Grain Size (large or small)
•
Grain Shape (round or cubic etc.)
•
Soil Structure and Density
•
Water Content and void ratio
One must understand the nature of shearing resistance in order to analyse soil stability
problems such as bearing capacity, slope stability. and lateral oressure on earth retaining
structures.
(a)
(c)
(d)
(f)
Simplified aamples of mechanisms of failurr in soils: fa) foundation.
(b) retaining wall, (c) bradng in Ut:t/1'ation. (d) dttp =avation. (t) cutting, (/)
embankment or earth dam
The stresses induced along any plane in a soil mass can be converted to:
•
Stresses perpendicular to the.plane, called the normal stress, denoted by the
Greek symbol "sigma" Un
•
Stresses along the plane, is called shear stress, and denoted by the Greek symbol
. . .. . .. .
''tau" t:
A
•.
1.4
. . .. .
SOJL l(J,4SS
Sltear Strength
1.2
SHEAR STRENGTH PARAMETERS OF A SOIL
Those properties ofa soil which determine its shear strength are called the shear strength
parameters. Although the shear strength of a soil depends on a large number of factors,
it is generally considered that the main contributors are:
•
Internal Friction
•
Cohesion
Shear Strength of Soils; Cohesion
• Cohesion (C), 1s a measure ot the forces lhal
cement panicles of sois
: �� =:��:::::��a��n ,A
:����;
1.2.1 Internal Friction
-iiillll":"']
I
I�--�-)
l
'-------"'-5
This is a resistance to movement between the individual particles. Resistance in
this case is a result of sliding of particles on each other or/and the interlocking of
particles. You can measure the internal friction in degrees, it is called the angle
of internal friction, and is denoted by the Greek word "phi" ¢ . Internal friction
is usually influence by grain size (coarse or fine), grain shape, soil structure and
density and water content.
1.2.2 Cohesion (or Apparent Cohesion)
This is the force tending to hold particles together as a solid mass, without the
application of any external forces. The reason for this is:
•
Attraction between molecules
•
Surface tension of water between the grains.
For Coarse-grained soils the cohesion strength depends very little on molecular
attraction and surface tension. For Fine-grained soils the molecular attraction is
so great in comparison to the grain mass, that cohesion becomes, a factor in the
shearing strength of the fine grained soils. It is mainly observed in silts and clays.
Cohesion is denoted by "C' and is expressed in stress units (kPa). Cohesion is
influence by grain size, state of packing of particles (molecular attraction) and
water content of sample
1.5
Shear Strength
1.3
COULOMBS LAW
A relative simple equation and theory relating the shear strength of a soil to the applied
normal stress was first suggested by Coulomb in 1773. Coulombs equation states the
relationship between shear stress, friction and cohesion.
c
where
'!'
c
a
+
=
=
=
atan </J
Shear Stress along a particular plane
Apparent Cohesion
Effective stress normal to the plane of failure
=
Angle of internal friction
¢
(Shear stress = Cohesion + Normal stress x TAN of angle of internal friction)
When represented graphically, this equation yield a straight line:
shear
stress
at
failure
'T't
normal stress Or,
Maximum shear stress related to normal stress Jrom shearbox tests
(Coulomb envelope)
From the above it is evident that the value ofthe "apparent cohesion " (C) is represented
by the intercept on the shear (vertical) axis, and that the "angle of internal friction" ( ¢)
is in fact the slope of the line. The straight line thus plotted is known as the "Coulomb
line". Any combination ofnormal stress and shear stress on a plane in the soil mass which
gives a plot below the Coulomb line indicates that failure will not occur along that plane.
In accordance with Terzaghi's effective stress concept, namely that the pore water
1.6
Shear Strength
pressure, µ, does not contribute towards shear strength, it is now generally accepted that
shear strength must be related to effective stress. In that case the equation become:
c' +
(J'
1
tan¢'
where (J' ' = effective normal stress
and c' and
¢' = shear strength parameters with respect to effective stress
The shear strength parameters c' and <fl would have constant values for a given soil,
providing that the void ratio, density and pore pressure also all remained constant for
different values ofnormal stress. It is extremely doubtful whether or not this ever occurs
in practice, either in the field or during laboratory tests. It is usual, therefore, to refer
measured values to the type oftest and conditions under which they were measured.
In any sample, the applied pressure (P) will be restricted or opposed by granular :friction
and pressure. This is called effective stress( </J' ). The other part ofthe resistance will be
the pressure developed in the water, called the pore water pressure(µ).
p = p' + µ
and
(in terms ofpressure)
a - a ' + µ (in terms ofstress)
effective stress
a -,_
-a - µ
effective pressure P ' = p -
]. 7
µ
Shear Strenmh
DETERMINATION OF THE SHEAR STRENGTH OF A SOIL BY MEANS OF
1.4
.
THE SHEAR BOX TEST
1.4.1 S c o ~ e
In principle the shear box test is a "angle of friction" test, in which one portion
of soil is made e along another by the action of a steadily increasing
horizontal shearing force, while a constant load is applied normal to the plane of
movement.
1.
Shear box machine with electric motor and multi-speed driver unit, and
load ring for measuring horizontal shear force.
2.
Shear box body in two halves, the upper half fitted with a "swan-neck"
yoke, and clamping and lifting screws.
&
mint U o ~ ~ l l e ~ t l m
b .n.o:&C*
domping scr*rr
C-holes lor clom@ng
8Cr.W.
L-nohs falifting 8er.w
(b)
Details of 60 mm s h e a r k fa) p h view, @I section rhrouph box
3.
Shear box carriage, watertight, running on roller bearings.
4.
Loading pad, porous plates, perforated grid plates and retaining plate.
5.
Loading yoke, weight hanger and lever-am, for applying the pressure to
the specimen.
6.
Stop watch.
Components of the direct shear box
Shear Strength
EXAMPLE
A sandy soil has been tested in an undrained 100x100 shear box. The results below are provided to
assess the apparent cohesion and angle of friction. A gauge attached to the proving ring provides the
deflection reading. The proving has been calibrated to reflect 20 N per division. Assume g = 9.81 m/s.
Plot the values on a graph. Remember to use the same scale divisions on X and Y axis with zero at the
origin.
Test Nr
Loading (kg)
1
2
3
4
40
76
114
150
Maximum deflection readings at
shear (divisions)
18
30
40
49
SOLUTION TABLE
Test Nr
Compressive
Stress (kN/m²)
1
2
3
4
0
Cohesion…………………………
Angle of friction………………..
1.13
Shear Stress
(kN/m²)
Shear Strength
IMAGES OF TYPICAL LABORATORY TEST APPARATUS OF THE DIRECT SHEAR BOX
Take note of the following
Usually only relatively slow drained tests are performed in a shear box
test. Clays should be tested at a rate that will prevent excess pore water
pressure building up (speed rate of <0.5 mm/min). Sands and gravels can
be performed quickly (speed rates up to 1.5 mm/min)
Tests on sands and gravels are usually performed dry. Moist or saturated
conditions do not significantly affect the drained strength.
Voids in the sample should be determined. You will need to determine the
dry density, moisture content and specific gravity. By comparing the voids
during the test one can judge whether the results are suspect or not.
Reminder of void calculations
Source: www.home.iitk.ac.in
1.14
Shear Strength
Class Example
An undisturbed soil sample has been extracted and tested in a small undrained shear box,
measuring 60 x 60 mm in cross section. The laboratory results from this test have been tabulated
below. You are required to do the following:
1.1
Complete the tabulated table accurately. [The proving ring has a calibration of 2.5 N per
division].
1.2
Produce a neat Coulomb failure envelope on graph paper.
1.3
Indicate accurately on the graph what the cohesion and angle of friction will be.
1.4
Classify the soil according to the strength parameters ONLY.
1.5
The contractors plan to place a vertical load of 300 kN on a 1 m x 1 m foundation at the
same sample extraction depth. Would this procedure be recommended? Give clear
motivation for your answer
Total Vertical
Load (kg)
Total
Vertical
Load (N)
Compressive
Stress
kN/m²
33.1
Horizontal
Load
(divisions)
144
66.2
218
99.3
292
132.4
364
1.15
Horizontal
load (kN)
Shear
Stress
kN/m²
Shear Strength
1.5
TYPES OF SHEAR BOX TESTS
Drainage ofthe soil sample can either be allowed or prevented during the test, depending
on whether porous or solid end plates are used. This will of course have an influence on
the values obtained for the shear strength parameters. It is of utmost importance that the
reported values for c and <P be therefore referred to the particular test conditions, e.g. if
an undrained test was performed, a subscript "u" is used to indicate that these values of
cu and <Pu refer to undrained conditions,
For drained tests sufficient time must be allowed after application of any loads to permit
complete drainage to zero excess pore pressure. For coarse grained soils (sands and
gravels) this occurs immediately, but for clayey soils it may take hours. Since there is no
increase in pore pressure, the applied stresses are all equal to the effective stresses with
the result that the parameters obtained during a drained test refer to effective stress, and
are denoted by c' and <P' respectively. No measure ofpore pressure is possible in the shear
box test.
'Three types of soil can be tested:
shear
stress
shear
stress
shear
stress
+ (,.,,........... ,.,,...� ------(a)
Normal
compressive
stress·
(b)
Normal
compressive
stress
(c)
Normal
compressive
stress
(a) Frictional soil (sands); c = 0. (b) Cohesive soil (clays); � = 0.
(c) Cohesive-frictional soil (silts); c - � soils
1.20
Source: GP Raymond
Shear Strength
Class Example
A shearbox test was performed in the laboratory on a soil sample with box dimensions of 100 x 100 mm.
The test was performed under undrained conditions and displacement was measured up to failure (i.e the
dial gauge displayed a reverse reading). Analyze the test data in Table 1 and do the following:
1.1
Provide the shear stress against shear distance curves on graph paper.
1.2
Determine the maximum shear stress at 15% of the shear box length. Indicate clearly where the
shear stress occurred.
1.3
Calculate the normal stress on each sample.
1.4
Determine the strength parameters and use the correct units of measurement.
1.5
A structure is to be built on the soil that was tested with an expected normal stress of 250 kPa and
soil shear stress of 125 kPa. Will the soil be capable of supporting the structure, and state the
reasons for your answer?
Table 1 Shear box Data
Shear box
Displacement
(mm)
2
4
6
8
10
12
14
16
Applied Force,
N
Force at failure
on Sample 1
Force at
failure on
Sample 2
Force at failure
on Sample 3
255
343
431
586
691
758
808
978
1090
984
1187
1444
1161
1395
1692
1258
1566
1804
1290
1634
1909
1268
1562
1873
2000
3000
4000
1.22
Effective
area under
shear , m²
Shear Stress,
kN/m²
Sample 1
Shear Stress,
kN/m²
Sample 2
Shear Stress,
kN/m²
Sample 3
Shear Strength
0
Displacement (mm)
Maximum shear sample 1=
Maximum shear sample 2=
Maximum shear sample 3=
1.23
Shear Strength
USING A LEVER ARM
A lever arm is essentially used to increase the stresses on a specimen. It is often the case when vertical
loading cannot achieve the desired stress due to a heavy mass incapable of being supported by the
hanger below. In these cases a simple beam is attached and an increased moment is applied. To
calculate the new increased stress the lever arm’s beam ratio is utilised. An example is given below.
Let’s assume the following masses on a shear box of 100x100 mm in effective size.
Lever arm ratio
Mass of hanger
Mass of lever arm beam
Mass on the hanger
Applicable stress, kN/m²
1*9.81*5/1000/0.01 = 4.905
2*9.81*2.5/1000/0.01= 4.905
10*9.81*5/1000/0.01= 49.050
1:5
1 kg
2 kg
10 kg
Total Stress
58.86 kN/m²
If a structure will experience 100 kPa at the footing depth where the samples were removed then the
above masses will be inadequate in simulating the stresses in a laboratory. Typical loading is 50%,
100% and 150% of the stresses expected on site. Let’s assume that 100 kPa is required on the footing.
We will need to calculate which mass, M, placed on a lever arm, will be required. Using the same
scenario above the formula would look like this:
100 = [(‫*ܯ‬9.81*5+1*9.81*5+2*9.81*2.5)/1000]/0.01
M =18.4 kg
So the “test bracket” of masses to be used in the shear box test would be:
1. 18.4/2 =9.2 kg
2. 18.4 kg
3. 18.4*1.5 =27.6 kg
Shear Strength
SELF EVALUATION EXERCISE NR 3
An undrained shearbox test was done on a soil sample. The shearbox is 60 x 60 mm. The test results in
the laboratory are recorded in the table below. The lever arm ratio to be applied is 1:5
Tst Nr
Test 1
Applied
Weights,
kg
8
Shear Stress at
Failure
kN/m²
100
Test 2
15
140
Test 3
22
180
a) Use the graph paper below to determine the strength parameters.
b) What type of soil can this be described as?
1.30
Shear strength
1.7 TRIAXIAL TESTING
Why conduct a triaxial test?
The triaxial test is one of the most versatile and widely performed geotechnical laboratory tests,
allowing the shear strength and stiffness of soil and rock to be determined for use in geotechnical
design. Advantages over simpler procedures, such as the direct shear test, include the ability to control
specimen drainage and take measurements of pore water pressures. Primary parameters obtained from
the test may include the angle of shearing resistance ϕ΄, cohesion c΄, and undrained shear strength cu,
although other parameters such as the shear stiffness G, compression index Cc, and permeability “k”
may also be determined. The illustration below gives an example of the engineering application of the
test – here triaxial compression provides strength information at the top of a cut slope, whilst triaxial
extension allows parameters for soil elements at the slope base to be determined. (Dr Sean Rees,
Geotechnical Specialist at GDS Instruments).
STRESS STATE DURING A TRIAXIAL TEST
The stresses applied to a soil or rock specimen when running a triaxial compression test are displayed
in the sketch below. The confining stress σc is applied by pressurising the cell fluid surrounding the
specimen – it is equal to the radial stress σr, or minor principal stress σ3. The deviator stress q (σ1 - σ3)
is generated by applying an axial strain εa to the soil – the deviator stress acts in addition to the
confining stress in the axial direction, with these combined stresses equal to the axial stress σa, or
major principal stress σ1. The stress state is said to be isotropic when σ1 = σ3, and anisotropic when σ1
≠ σ3.
The soil is sheared by applying an axial strain εa to the test specimen at a constant rate through
upward (compression) or downward (extension) movement of the load frame platen. This rate, along
with the specimen drainage condition, is dependent on the type of triaxial test being performed.
1.32
Shear Strength
_C<Z.11 t>Yessvvrz.
= 0-
.3
= l'l'"vino-t" p,,incipaJ
sk'1'e&S
tn�o< principal
.Sh-es&'
= O"� + (c,; -0"'?>)
= o-,
Generally the principal stresses a 1 and a3 are known and it is required to find the stresses
an and 't'. The Mohr circle must be constructed to solve this graphically. The principal
stresses a 1 and a3 are plotted on the horizontal axis and a semi-circle constructed across
them. If a soil sample is stressed to failure and the Mohr circle drawn, the straight line
representing Coulomb's equation will be tangential to the circle. This principal is used
in the triaxial test to find the shear strength parameters of a soil.
MOHR CIRCLE
J
Oi
1.33
Shear Strength
1.8
DETERMINATION OF THE SHEAR STRENGTH OF A SOIL BY MEANS OF
THE TRIAXIAL COMPRESSION TEST
1.8.1 Scope
The triaxial apparatus is possibly the most widely used and most versatile means
ofobserving the shear strength characteristics ofsoils. A cylindrical sample ofsoil
is enclosed in a pressurised chamber which subjects the sample to compressive
stresses in three mutually perpendicular directions. The vertical compressive
stresses is then increased in excess of the horizontal stresses until eventually the
soil fails in shear, or strain to such a point that excessive deformation results.
1.8.2 Apparatus
1.
A motorised triaxial load frame with detachable load ring.
2.
Triaxial cell, capable of sustaining an internal water pressure up to
1OOOkN/m2 with perspex body and base plate.
3.
Constant pressure system for maintaining cell pressures up to 1000 kN/m2
at a constant level.
4.
Latex rubber membranes, perspex end caps, rubber 0-rings, stopwatch
and a suction membrane stretcher.
1.8.3 Method
1.
A recompacted clay sample will be provided, and must be measured and
weight before the test commence.
2.
The Latex rubber membrane and end caps must then be fitted as follows.
membrane
stretcher
rubber
membrane
(a)
(b)
(cl
1.34
(e)
(1)
Shear Strength
ofthe compressive stress has been passed or a strain of 20% (about 15
mm) reached. The test should be continued until three or four consecutive
8.
readings show a decreasing stress.
When it is clear that failure ofthe specimen has occurred, stop the motor,
and wide the machine platen down by hand. Reduced the pressure in the
cell to zero. Dismantle the cell and sketch the mode of failure of the
specimen. Also take a sample for moisture content determination.
9.
The test is repeated on different samples ofthe same soil, each time under
a different value ofthe cell pressure.
1.8.4 Calculations
The Minor principal stress: � = cell pressure
The Major principal stress: 01 = o3 + Applied stress through ram
U1
=
U3
+
Force
Area
For each set oftest results a different Mohr circle can be constructed. The best
common tangent to these circles represents the failure envelope (Coulomb line),
and the strength parameters c and </J are determined.
failure envelope
.assumed to have
the equation
t,
c' + a� tan q,'
=
er,
Obtaining the Mohr-Coulomb envelope
1.36
a'n
Shear Strength
,,
Soil with cohesion
Cohesion less soil
1.8.5 Notes
1.
As the specimen shortens under the load, its diameter will increase,
producing a barrelling effect. In some cases, the specimen will shear
cleanly along a well defined plane; in others a number ofpartly developed
shear planes at approximately similar angles will be seen. In barrelling
there is no define failure point, and failure stress is taken as 20% ofstrain.
Example, if a sample is 75 mm long then 20% of75 mm is 15 mm.
r------1
(al
(bl
r------,
I
I
(cl
Types of failure in the triaxial test
(a) Brittle shear failure (b) Partial shear failure (c) Plastic yieldina failure
2.
During application ofthe vertical load the sample experiences shortening
in length with a corresponding increase in diameter. In addition, when
drainage is allowed, the volume ofthe sample will be reduced. The strain
dial gauge indicates the change in length of the sample. To record the
volume change during a drained test, a drainage connection is made from
the top ofthe specimen. A porous ceramic disc is placed between the soil
and the top (perforated) end-cap and a plastic tube connected between the
end-cap and drainage outlet in the cell base. A glass burette (measuring to
1.37
Shear Strength
0, I ml) is then connected with a plastic tube to the external outlet valve.
In the case ofa saturated test specimen, the volume ofthe water collected
in the burette represents the volume change (L\ V) in the specimen. In the
case of an undrained test, L\V may be taken as zero.
Change in Volume:
Axi;,I slraln �.
voturne change .:i V tCO te$t t
No change in volume:
v
A =
H - �H
Where:
A
=
Area at failure (mm2 or m2)
Ao
=
Original area (mm2 or m2)
=
Volume (at start oftest)(mm3 or m3)
v
H
L\H
L\V
3.
=
=
=
Height or length (mm or m)
Strain (Original height - final height)
Volume change (ml or m3)
The angle at which the sample shear is called the angle ofinclination(8).
This angle can be determined as follows:
(}
= 45° + ¢
2
Theoretical
failure plane
for H-piles and
lagging
Theoretical
failure plane
for sheeting
moment
45°- cj>/2
(a
1.38
(b)
Shear Strength
EXAMPLE:
The following data was obtained after three undrained triaxial tests (UU) were performed on
representative undisturbed samples:
Sample Nr Minor Stress kN/m² Axial Load At Failure, divisions on gauge
75
1
76
2
116
160
3
250
157
•
•
Find by graphical means the apparent cohesion and the angle of internal friction.
Determine graphically the angle of inclination.
Note
NB. Cell pressure is also referred to as Minor Principal Stress (ߪଷ ,)
Each sample is 72 mm long and 36 mm in diameter.
Each gauge division = 1,5N/div
SOLUTION:
Axial Loads:
Sample Nr
Test 1
Axial Load in kN
Test 2
Test 3
Axial Stress:
Area of Sample (A) = .........................mm² = ...................m² (store the calculator answer and recall later)
Axial Stress = Load / Area
Sample Nr
Test 1
Major Principle Stress*
Test 2
Test 3
*Major Principal Stress(ߪଵ ,) = Axial Stress + Cell Stress
Minor Principal Stress(ߪଷ ) = Cell Stress
1.39
So the Major Principal :
Test I: ..................................kNlm2 + 50kIWm2 = ...........................Wlm2
Test 2: ..................................kNlm2 + 150kN/m2 = ............................kN/m2
Test 3: ..................................kNlm2 + 250kN/m2 = ............................Wlm2
From the graph: c
=
4
=
Shear Strength
Class Example
Three undrained triaxial tests were completed on representative undisturbed samples.
Each sample is 75 mm long and 35 mm in diameter. Vertical deformation after testing
was recorded as 6.1 mm in each core. The volume change is zero. You are required to do
the following:
2.1
Complete the tabulated table accurately. [The proving ring has a calibration of
1.35 N per division].
2.2
Determine graphically the apparent cohesion and angle of friction.
2.3
Determine the angle of inclination on the graph.
Test
Nr
Load,
divisions
1
Cell
pressure
kN/m²
200
2
400
245
3
600
293
Axial Load,
kN
216
1.41
Deviator Stress
kN/m²
Principal or
Major Stress
kN/m²
Shear Strength
Site conditions will determine the type of test most applicable in a laboratory to simulate the
application of stresses expected. A summary of the test conditions has been tabulated below.
•
•
•
In the UU test the water content of the test specimen is not permitted to change during the
application of the confining pressure or during the loading of the specimen to failure by
increasing the deviator stress. The UU test is usually applicable only to soils that are not
free draining, i.e., to soils having a permeability less than 1.0 x 10-4 cm/sec (cohesive soils)
and normally consolidated clays, ie OCR =±1
All specimens must be completely saturated before application of the deviator stress in the
CU test. In general, it is preferable to saturate the soil after the specimens have been
prepared, encased in membranes, and placed within the compression chamber, using back
pressure. An example would be where an embankment is being excavated and some
consolidation can be expected with slow drainage.
The CD Test is usually done only on pervious (generally granular/cohesion-less) soils.
Drainage is fairly free flowing and permeability of above 1.10 cm/sec. Examples would be in
the upper layers of a road surface in which a network of drains have been laid; or clayey
embankments with drainage channels.
A summary of the stress application in each test type is provided below.
1.49
Shear Strength
EXAMPLE:
The following results refer to a series of triaxial tests on specimens of saturated clay, initially
38mm in diameter and 76 mm long:
Type of Test
Undrained
Drained
All-Round
Axial Load
Axial Defor-
Volumetric
Pressure (kPa)
(N)
mation (mm)
Change (ml)
200
222
9,83
No change
400
215
10,06
No Change
600
226
10,28
No Change
200
467
10,81
-6,6 ml
400
848
12,26
-8,2 ml
600
1265
14,17
-9,5 ml
Determine the value of the shear strength parameters with respect to:
(a).
Total stress
(b).
Effective stress
1.52
Shear Strength
SOLUTION:
With respect to total stress: (Using undrained results)
(a).
/). v
v
Actual Area
w
2
(m )
(kN)
w a)
A (kP
0
*1,3026 x 10-3
222 x 10-3
170
370
0,132
0
1,3066 x 10-3
215 x 10-3
164
564
0,135
0
1,3111 x 10-3
226 x 10-3
172
772
11L
-
-
200
0,129
400
600
U3
(kPa)
L
/). v
v
1- ­
*A
11L
1 -­
L
1- 0
1l
2
[
]
03
0
)
,
1,3026 x 10 -3
8
4(
9
1 0,12
With respect to effective stress: (Using drained results)
(b).
u/
(kN)
w a)
A (kP
*1,2208 x 10-3
467 x 10-3
382
582
0,0951
1,2236 x 10-3
848 x 10-3
693
1093
0,1102
1,2403 x 10-3
1265x 10-3
1020
1620
11L
-
-
/). v
v
Actual Area
w
2
(m )
200
0,1422
0,0766
400
0,1613
600
0,1864
u/
(kPa)
L
/). v
v
1-­
11L
1-­
L
- 0,0766
]
1- 0,1422
1
2
4(0,038 )[
1[
1.53
1,2208 x 10-3
Shear Strength
I I
I'
I I I
'I
I I
I I
T.
''
I I
'
J....:....i__
�'
"
�f-----�1--r..........i..;......:...;......,.��,
, �t...�,�,,.......;..c.��'--+-4--.�'---+
�i""'i-i--�+-!'+-1-+-!�+..:.-l--;��..;.--''-'-+--'--�i��· ......-+-.,......��.............�............----..............._............................__,_........._,_..._.4...;_t++·q--t+-:·
:--::::=
.::.:::.:
..........
·-.-
•j-r·
-,
Results:
Undrained:
Drained:
<Pu =
85 kPa
0
C' = 20kPa
°
</J' = 26
1.54
Shear Strength
EXAMPLE I CLASS TUTORIAL
The following are results of three tests done on the same soil sample to determine its strength
parameters:
1
2
3
Initial length (mm)
75,15
74,85
75,21
Sample diameter (mm)
38
38
38
Length at failure (mm)
73,9
73,1
72,75
Cell pressure (kPa)
150
250
350
Volume change (ml)
- 0,8
1,6
2,9
Pore pressure (kPa)
100
100
100
Load at failure (N)
92,4
148,4
205,8
Sample
(a).
What type of test was done?
(b).
Use the above results and determine the strength parameters of the soil.
SOLUTION:
1.55
Shear Strength
Class Example
The results as recorded below were completed on a soil sample to determine the strength parameters.
Complete the table in the spaces provided for all critical loads and stresses. Use the graph paper provided
to accurately display the failure curve.
Criteria
Original Diameter, mm
Sample 1
50
Sample 2
50
Sample 3
50
Original Length, mm
100
100
100
Length at failure, mm
95
93
92
Cell Pressure, kPa
200
400
600
Pore Water Pressure at
failure, kPa
Axial load at failure, N
100
100
100
469
645
821
Change in volume, ml
-5
-7
-10
Area, m²
Axial Load at failure,
kN
Axial Stress, kN/m²
Effective Minor Stress,
ken/m² ( ߪ ଷ ሻ
Effective Major Stress,
kN/m² (ߪ ଵ ሻ
2.1
What type of test was completed above?
Consolidated Drained Triaxial Test
(2)
2.2
Use the above results and determine C’ and ф’.
1.60
(4)
Shear Strength
Class Example
The results of an Unconfined Compressive Strength test are displayed in the figure below. A
spring with a calibrated torque of 8 Nmm was used.
3.1.
Determine the unconsolidated undrained shear strength of the sample. Show all
calculations. Answer to be in kN/m²
3.2.
What would the undrained cohesion value be?
Calculations
1.67_a
(in kN/m²
)
Shear Strength
1.14
THE SHEAR VANE TEST
In soft sensitive clays it is difficult to obtain samples that have only a slight degree of
disturbance and in-situ shear tests are usually carried out by means of a shear vane test.
The apparatus consists of a 75mm diameter vane, with four small rectangular blades
150mm long at the end of a steel rod. For stiff soils a smaller vane, 50mm diameter and
1 OOmm high may be used. The vanes are pushed into the clay 0,5m ahead of the boring
to eliminate disturbing effects, and the undrained strength of the clay is obtained from the
relationship with the torque necessary to turn the vane. The rate of turning the rods,
throughout the test, is kept within the range 6 - 12 ° per minute to determine the
undisturbed shear strength. If the handle is turned slowly, at the specified rate, an
undisturbed sample reading can be obtained. After the soil did shear, rapid turning should
start to get a disturbed sample reading.
The torque head is at the top of the rods. This is standard practice for most site
investigation work but for deep bores it is now possible to use apparatus in which the
torque motor is mounted down near to the vane, in order to remove the whip in the rods.
The undrained shear strength is obtained by equating the applied torque to the shear
resistance moment:
(
dJ
d2
4
d
T - Cu l[hdx -+
2 21[-X -
Hence cu
(1
T
1 ')
1[ -hd + d
2
6
2
=
3
1
T
1 )
(
2 l[d h+ 3d
1.68
2
Shear strength
EXAMPLE
A vane tests was completed inside a trial hole containing cohesive soils. A hand vane was used. The
diameter of the vane was 75 mm and the height was 150 mm. Rapid and slow rotations were applied
yielding results of 70 Nm and 100 Nm respectively. Determine the undrained shear strength of the
soil. (remember that slow rotations provide undisturbed values while rapid rotations provide disturbed
values).
SOLUTION
1.70
Shear Strength
SELF EVALUATION EXERCISE 4
A vane used to test a deposit of very soft alluvial clay, required a torque of 67,5 Nm. The
dimensions of the vane D = 80 mm and H = 150 mm.
Determine the value of the undrained shear strength of the clay.
SOLUTION:
1.71
Chapter 2
Earth Pressure
Neal and Procter
Advanced Geotechnical
Engineering
2.8
Earth Pressure
ADDITIONAL THEORY FOR SOLDIER PILES
SELECTING A SOLDIER PILE SECTION FOR A
SOLDIER PILE AND LAGGING WALL WITH ROCK SOCKETS
The bending moment in a soldier pile varies with depth and the material in which it is embedded.
The maximum bending moment (Mmax ) expected to occur in the soldier pile is used to size the
pile. The Mmax for a cantilevered soldier pile wall with some or all of its embedment in rock (i.e.,
rock sockets) is traditionally dependant on the elevation of the rock surface assumed during
design.
2.21
Earth Pressure
During construction, it is likely that the actual rock surface will vary somewhat from the
elevation(s) assumed in design. This occurs because the amount of subsurface information
available to the designer is generally insufficient to precisely define the rock profile. During
construction, if the actual rock elevation is found to be lower than the assumed rock elevation,
the soldier pile section specified on the Contract Plans is often no longer adequate. When this
occurs, the first step is for the wall designer to review the assumptions from the original analysis
and compare them to the actual site conditions (i.e. soil conditions, ground water elevation,
surcharge loads, etc.). A re-analysis with the revised assumptions may prove the soldier pile
section shown on the Contract Plans is still adequate. If a re-analysis indicates the section is
insufficient, possible remedies and their associated consequences are as follows:
i.
Remedy- Increase the section modulus of the soldier pile by either ordering new steel
or welding steel cover plates to the flanges of the existing soldier piles.
Consequence- Delays, orders-on-contract, claims.
ii. Remedy- Reduce the factors of safety in the original design.
Consequence- Not acceptable for permanent walls or critical temporary walls.
iii. Remedy- Change the wall design by adding anchors.
Consequence- R.O.W. restrictions, additional design, delays, requires specialty Contractor
and equipment, orders-on-contract, claims.
iv. Remedy- Change the wall design by reducing the soldier pile spacing.
Consequence- If decreasing the pile spacing is an option, payment will be required for the
additional quantities of drilling and soldier piles.
The most effective way to ensure safety and avoid costly delays and orders-on-contract is to
specify a soldier pile section that is able to accommodate likely variations in rock elevation
encountered during construction.
2.22
Earth Pressure
2.23-a
Earth Pressure
Lateral Earth Pressure—key points
2.23-b
Earth Pressure
2.33
Earth Pressures
WORKED EXAMPLE
A cantilevered sheet pile wall is to support the side of an excavation in sand 10 m deep. The sand is
fully drainable with a deep water table (not significant on the effect against the sheet pile). The unit
weight of the sand soil is 19 kN/m3 and the phi angle of friction is 30°. Use a safety factor against
sliding of 1.2 and determine the minimum depth the sheet pile should be installed at.
10
SOLUTION
Pp
d/3
(10+d)/3
d
Pa
Kp =
Ka = = 0.333
=3
Active Pressures & Thrusts
Pressure (pa)=Ka* (h+d)
= 0.333*19*(10+d)
=6.333 (10+d) kN/m²
Passive Pressures & Thrusts
Pressure (pp) = Kp* *d
=3*19*d
=57d kN/m²
Thrust (PA)=0.5*Pa*(h+d)
=0.5*6.333
=0.5*6.333 (10+d)*(h+d)
(10+d)*(h+d)
=3.1665(10+d)² kN/m run
P.O.A =(
=(10+d)
10+d)/3 above the base
Thrust (PP) = 0.5*pp*d
=0.5*57d*d
=28.5 d² kN/m run
P.O.A =d/3 above the base
Now determine the minimum
minimum depth “d” by taking moments about the base of the sheet pile.
pile.
Remember to apply the safety factor to the passive thrusts
Ie Clockwise
Clockwise moments = antianti-clockwise moments
1/FOS * PP * D/3 = PA * (10+d)/3
So… 1/1.2
1/1.2 (28.5d²) * d/3 = 3.1665(10+d)² *(10+d)
10+d)/3
7.916
7.9167d3 = 1.0555(1000+300d+30d²+d
1.0555(1000+300d+30d²+d3)
7.916
7.9167d3 = 1055.5+316.65d+31.665d
1055.5+316.65d+31.665d2+1.0555 d3
0 = 1055.5+316.65d+31.665d
1055.5+316.65d+31.665d2 -6.8617d3
So by cubic equation, d =10.445 m*
m*
•
A graphical solution is shown on the next page
2.52
Earth Pressure
Figure 1 Google Images: http://kshitija.wordpress.com/2006/06/23/
Figure 2 Sheet pile profile shapes http://www.ce-ref.com/sheet-pile.htm
2.55
Earth Pressures
Class Example
The sheet pile shown below retains soil with the following properties: δ= 18.5 kN/m3; φ = 25°; and C
= 0 kN/m2. The planned tie rod tension @ T is not expected to exceed 750 kN.
1.1 Reproduce the sketch and write in the P.O.A on the sketch, ie at h1,h2,h3 (alongside the arrows)
1.2 Solve for “d” (depth of sheet pile). Use any method available to you.
1.3 Determine the final horizontal forces.
1.4 Determine the spacing of the tie rods, to the nearest 0.01m.
•
Use only 40% of passive thrust as F.O.S.
Show all calculations. Hint: Write up all the applicable forces in terms of F * h,( ie F1 * h1)
2.64
Earth Pressures
Figure 1A Solid Core rock anchored; B deformed tensile steel
Figure 2 Hydraulic drill rigging for rock bolts
Figure 3Installed rock bolt
2.66
Earth Pressure
A rock bolt is a long anchor bolt, for stabilizing rock excavations, which may be used in tunnels
or rock cuts. It transfers load from the unstable exterior, to the confined (and much stronger)
interior of the rock mass.
Rock bolts were first used in mining starting in the 1890s, with systematic use documented at the
St Joseph Lead Mine in the US in the 1920s. Rock bolts were applied to civil tunneling support
in the US and in Australia, starting in the late 40s. Rock bolts were used and further developed,
starting in 1947, by Australian engineers who began experimenting with four meter long
expanding anchor rock bolts while working on the Snowy Mountains Scheme.
Examples of rock bolts with “Multiple corrosion Protection”
2.68
Chapter 3
Stability of Slopes
Neal and Procter
Advanced Geotechnical
Engineering
STABILITY OF SLOPES
3.1 INTRODUCTION
The stability analysis of slopes plays a very important role in civil engineering. Stability analysis is used
in the construction of transportation facilities such as highways, railroads, airports and canals; the
development of natural resources such as surface mining, refuse disposal, and earth dams; as well as
many other human activities involving building construction and excavations. Failures of slopes in these
applications are caused by movements within the human-created fill, in the natural slope, or a
combination of both.
The assessment of the stability of an earth slope against failure is an important criterion in the design of
engineering projects where slopes are formed by excavation or by fill embankments. A fairly common
engineering failure is slipping of an embankment or cutting, and considerable research has been carried
out into the causes of such failures.
Water is frequently the cause of earth slips, either by eroding a sand stratum, lubricating shale or
increasing the moisture content of clay, and hence decreasing the shear strength. Failure of a slope
normally takes the form of a slip along one or more rupture surfaces somewhere within the soil mass.
When a slip in a clay soil occurs it is generally found to be along a circular arc, and therefore this shape is
assumed when studying the stability of a slope.
3.3
This method is more
accurate!
Stability of Slopes
Slice Nr
Applicable Area, m²
Weight of
slice, W, kN
Deflection N-Force, kN
angle,∞ ° W*cos∞
T-Force, kN
W*sin∞
1
2
3
4
5
Totals
3.21
෍ܰ =
෍ܶ =
Stability of Slopes
Slice Nr
Applicable Area, m²
Weight of
slice, W, kN
Deflection N-Force, kN
angle,∞ ° W*cos∞
T-Force, kN
W*sin∞
1
2
3
4
5
Totals
3.23
෍ܰ =
෍ܶ =
Chapter 4
Bearing Capacity
Neal and Procter
Advanced Geotechnical
Engineering
Note: Reinforcement does not
prevent settlement but rather
reduces cracking and torsional
movement of the concrete
Note: In clays the
footing width is
preferably built
narrower than that in
sandy soils. Heaving
clays that have
access to wider
footings may lead to
increased cracking
See Paige-Green or Kesslers'
model for bearing capacity on page
4.25
Bearing Capacity in Shallow Foundations
Further research by Meyerhof and Vesic has led to the following expressions for the bearing
capacity coefficients:
Nc = Cot (Nq -1)
Note that in saturated clays where the friction
angle is often zero, this value is never taken as
absolute. For example the friction angle is slightly
more, more likely >0.05°. Otherwise the bearing
capacity factors cannot be calculated
Nq = tan² (45+ /2
N
1.8
1
4.6.3 Shape Factors
Terzaghi's basic equation is only valid for a long strip foundations. For other shapes like
rectangular, square and / or circular foundations, the shape of the failure surface is three
dimensional rather than two dimensional, as for strip foundations. Thus the bearing capacity
is altered. The basic form of the equation remains the same, but the shape factors are
introduced:
Square footing:
Circular footing:
.
.
.
.
.
Rectangular footing:
.
.
Where L is the length of the footing.
Alternatively recent research by De Beer and Vesic resulted in expressions for the shape
factors. So Terzaghi's basic bearing capacity equation can be modified by shape factors.
0.5
Where , and are shape factors founded in the following table, and should be
substituted in the formula above.
4.13
Bearing Capacity of foundations
5.
IN-SITU INSTRUMENTATION TO ASSESS BEARING CAPACITY
ESABLISHING STRENGTH PARAMETER RELATIONSHIPS USING THE DCP
5.1.1
Equipment Required
•
•
•
•
5.1.2
A Dynamic Cone Penetrometer (Figure 1), complete with replacement tips, two
25 cm spanner wrenches, a 10 mm x 250 mm breaking rod and one 6 mm Allan
wrench.
A stool or sturdy box for the person responsible for holding the instrument
perpendicular to the surface. The end gate of a vehicle may be used instead.
A jack designed to retrieve the rod after the test
Operation
Normally three people are needed to complete the test. One person stands on the stool and holds
the apparatus by the handle while the second person lifts the drop weight. The third observes the
readings and records them on the appropriate form. The steel rule attached to the guide foot is
placed through the slot in the hand guard. The foot is placed on the surface to be tested and the
cone tip passed through the guide hole. The entire apparatus is then held by the handle
perpendicular to the surface. The technician observes the reading on the rule at the top of the
hand guard and records this as the Zero Reading of DCP (Figure 1).The drop weight is then
raised to its maximum height and released. It is extremely important to gain maximum height for
each drop but care must be taken not to strike the weight against the handle. Doing so would
cause the instrument to withdraw and results would be in question. The readings are taken with
each blow of the weight. If the penetration rate is below 20mm/blow, the frequency of readings
may be decreased to:
•
•
•
one for every two blows with readings from 10-20 mm
one for every five blows with readings from 5-9 mm
one for every ten blows with readings from 2-4 mm.
Note that penetration depths less than 1 mm and exceeding 20 blows is considered as refusal.
The test depth is determined by the site engineer. Normally, readings are taken to 1.0 m below
the contact with the subgrade in road works. No test should be less than 1.0 m from surface.
For foundations assessments where new dwellings are to be built the DCP is useful in
determining the existing bearing capacity and relative density, %DR (see 5.1.4 below). It is
4.21
Bearing Capacity of foundations
suggested by practitioners in industry that the upper unconfined (overburden) soil be removed,
say upper 400 mm, and then proceed to determine the DN (mm/blow) for the DCP. It’s important
the DCP penetrate through to the bottom of the expected pressure bulb below the foundations.
5.1.3
Californian Bearing Ratio (CBR) and Unconfined Compressive Strength (UCS).
Paige and Green (2009) refer in their report to the DN of the DCP having a relationship with the
CBR and UCS. Correlations are shown below.
If DN > 2 mm/blow CBR = 410 x DN-1.27
If DN < 2 mm/blow CBR = (66.66 x DN2) - (330 x DN) + 563.33
UCS = 15 x CBR 0.88
or, UCS = 2900 x DN-1.09
4.22
(5.1)
(5.2)
(5.3)
(5.4)
Bearing Capacity of foundations
Various other models are available for converting the DCP penetration rate to the CBR. These
are shown below in Table 2.6 and refer to models utilizing the 60° cone only.
Table 5.1 CBR Conversion models
5.1.4
Relative Density (%DR)
The relative density is a useful parameter for describing the consistency of the sand (see Figure
3). Research by Coduto (2001) to correlate the DCP’s DN to the % DR led to the development of
the following formula:
%DR = 189.93/DN 0.53
(5.5)
In a laboratory the % can be calculated from measurement and is calculated using the
formula below.
% 100
(5.6)
A good correlation was achieved by research with the DCP to determine the % , yielding a
correlation co-efficient R² = 0.98. Figure 2 below illustrates the correlation work achieved.
4.23
Bearing Capacity of foundations
Figure 2 % DR vs DCP mm/blow (Coduto, 2001)
Figure 3
Relative states of compaction are defined above in Figure 3 and is referenced from AS 1726.
Meyerhof (1959) suggested that the % DR could also be correlated with the internal angle of
friction, for normally consolidated sand, as shown below.
(5.7)
4.24
Bearing Capacity of foundations
5.2
Correlations for Bearing Capacity
Definition: The soils bearing capacity is its ability to withstand a vertical load placed on
its foundation without failing due to shear and in some cases, uncontrolled settlement.
Ultimate bearing capacity (qu) is the minimum gross pressure at the base of the
foundation at which soil fails in shear. The DCP has been used in the past to relate its DN
to the in-situ bearing capacity and three models are presented below.
5.2.1
Model 1 (Kessler)
For analysis of shallow foundations an estimate of bearing capacity can be made from the
equation adapted from the Portland Cement Association (PCA, 1955) showing the
relationship between bearing capacity and CBR.
Bearing capacity (kPa)= 26.16*CBR 0.664
(5.8)
The CBR value is determined from the DN number as: CBR = 292 / DN 1.12 (Kessler,
2010)
It should be noted that this equation is not suited for soils classified as CL or for clays
with very low CBR values (<10%).
In South Africa (Kleyn, 1970) suggested using Log CBR = 2.628 - 1.273 log(DN). This
formula has been used for many successful years of road rehabilitation assessments.
5.2.2
Model 2 (Paige-Green)
The DCP penetration rate has also been correlated in foundations, predicting the bearing
capacity of soils for founding structures. This method provides a general indication and
should not replace conventional testing, but can be a useful addition to extend the results
of other tests using a cheap in situ test method. One such model is:
Bearing capacity (kPa) = 3426.8 DN-1.0101
5.2.3
(5.9)
Model 3 (DN converted SPT “N” value)
Attempts in the past have been made to predict the SPT “N” value by inverting the DN of
the DCP. The number of blows is recorded to drive a SPT split spoon sampler and its
cone through 305 mm and is usually done at the bottom of a pre-drilled bore hole. If the
DCP’s DN number is divided into the SPT depth (305 mm), and ignoring moisture
influence, a DCP correlated “N” value is obtained, i.e. DCP DN of 45mm/blow can
4.25
Bearing Capacity of foundations
correlated to 305/45 = 6.7 (unadjusted for a water table). The SPT method has been
adopted by the ASTM in their standards (ASTM, D1586-64).
With the DCP “N” value an attempt can be made at using Terzaghi’s classic bearing
factors for shallow foundations to assess the bearing capacity (Terzaghi’s, 1967). For
strip footings in NC sand Terzaghi’s classic formula would read as shown below:
Bearing capacity (kPa) 0.5 (5.10)
Where
z= depth from NGL to the base of the foundation (m)
C =cohesion (Aeolian sands are cohesion less)
= unit weight of the sand (kN/m3
Nc; Ny and Nq are the classic bearing capacity factors
B= width of the foundation (m)
A more direct correlation has been used before to correlate the DCP’s DN to the SPT “N”
value and is shown below (MnRoad, 2002).
log !1.05 2.03 log $%
(5.11)
Where,
DCP DN = average penetration rate of the DCP in mm/blow (in firm ground)
SPTN = number of SPT blows to drive through 305 mm (in firm ground)
It should be noted that the above equation is valid only for SPT blows > 30 per 305 mm.
5.2.4
Angle of friction vs DCP DN
Mohammadi, et al (2007) proposed that a good correlation exists between the DCP’s DN
and the internal angle of friction of sandy soils. The formula for this correlation proposed
through their work is shown below.
& = 52.16 / (DN) 0.13
(5.12)
Where, DN is in mm/blow
4.26
Chapter 5
Consolidation
Neal and Procter
Advanced Geotechnical
Engineering
Consolidation
Before construction the layer below ground will carry the overburden load. It’s important to evaluate
these soils to a depth beyond 4*B, where B is the planned width of the foundation (in this case a strip
footing). After construction these layers (in which the foundation are set) will bear the loading (Qu)
plus some of the overburden stresses. Loading above caused tremendous consolidation in the footings
below (figure 5.2 below)
When dealing with relatively thick layers it is normal to evaluate stresses through the layer as a
whole. Stresses are calculated to the centre of the layer in question, that is, where the most settlement
will occur. In-situ apparatus are used to determine the expected settlement. And if possible bearing
capacity is measured (the use of devices such as a bearing plate or SPT, DMT or DCP are common).
So when evaluating effective stresses we consider stresses:
•
•
Before construction to the centre of the layer
After construction to the centre of the layer (taking into account shape influence-Fig 5.3)
Fig 5.2 Consolidation of soils below the footings
Fig 5.3 (a) below strip footings and (b) below square footings
5.5
Consolidation
5.3 ESTIMATION OF VERTICAL STRESSES BELOW A SOLID MASS OF SOIL
It’s important to evaluate stresses at any point in a soil-mass due to external vertical loading and
especially so before an attempt is made in the prediction of settlements of buildings and other
structures. Many have developed equations to compute stresses at any point in a soil mass. They also
have made assumptions on the basis of the theory of elasticity. According to elastic theory: “constant
ratios exist between stresses and strains” in non-elastic soil the elastic theory may be assumed to
hold so long as the stresses induced in the soil mass are relatively small (Terzaghi, K., Peck, R.B. and
Mesri, G., 1996) Since the stresses in the subsoil of a structure having adequate factor of safety
against shear failure are relatively small in comparison with the ultimate strength of the material, the
soil may be assumed to behave elastically under such stresses.
When a load is applied to the soil surface, it increases the vertical stresses within the soil mass. The
increased stresses are greatest directly under the loaded area, but extend indefinitely in all directions.
Many formulas based on the theory of elasticity have been used to compute stresses in soils. They are
all similar and differ only in the assumptions made to represent the elastic conditions of the soil mass.
The formulas that are most widely used are the Boussinesq and Westergaard formulas. These
formulas have been integrated to give stresses below uniform strip loads and rectangular loads
For ease of use a chart to read off the
influence factors is provided below. Both
Boussinesq and Westergaards’ values can be
read off. This guideline will only follow
Boussinesq as it is more widely used. For his
values to be true he assumed soil to be elastic,
isotropic, homogenous and semi-infinite.
Boussinesq stress influence, IB can be written
down as follows: = ∗
/
Consider the sketch bottom right. An arbitrary
point, P, can chose at a distance, r, below the
point load, Q. If r/z is calculated below the
point, r/z =0, giving an IB value of 0.48.
The formula to calculate the vertical stress according to
Boussinesq is:
= ² ∗ 5.5a
Consolidation
CACULATING VERTICAL STRESS USING BOUSSINESQ METHOD
Here follows an example of how to use Boussinesq in determining the vertical stress below the soil. In
this guideline only the Boussinesq method will be used.
A point load is placed on the ground surface. The soil below is homogenous with no variation. The
moisture is also evenly distributed. Calculate the vertical stresses (a) at a depth of z = 4 m below the
point load, (b) at a distance r = 3 m , i.e. z, at the same depth. Use Boussinesq equation in both cases.
SOLUTION
Boussinesqs’ equation is = ² ∗
Where
=
/
/
(a) Notice that r/z =0 (when the point calculated is directly below the load, z =0), IB =3/2π = 0.48
(Or deter mine this from the graph, although this method is a little less accurate-see previous
page)
= 0.48*Q/z² = 0.48*1000/4*4 = 30 kN/m²
(b) When r/z = ¾ =0.75
=
/ . / = 0.156
= 0.156*1000/4*4 = 9.8 kN/m²
So it can be seen that the further the stress point moves laterally away from below the
point load, the less the vertical stress becomes.
BOUSSINESQS’ STUDY ON DEFORMATION
The general expression for the elastic deformation of a uniformly loaded plate resting on the surface
of a uniform, homogeneous, isotropic, semi-infinite elastic half-space can be obtained from the
solution presented by Boussinesq or a general theory of elasticity text and has the form:
A = BC/D
Where:
s = deformation
q = applied foundation stress
B =foundation width
E = Young's modulus
I = influence factor
The influence factor “I” is included to account for the shape of the foundation and the thickness of the
compressible zone. Values of “I” which are often used with the above Equation were first presented
by Steinbrenner (1934) and are reproduced by Terzaghi (1943), Lambe and Whitman (1969), and
Bowles (1988).
5.5b
Consolidation
Stresses in a Saturated Soil
To have a good understanding of what the stresses below ground are in a soil, imagine you are
standing in your favourite swimming pool which, at first, is empty .Memorize the level of the stresses
(pressure) you feel under your feet. Now, stay there in the bottom of the swimming pool and ask
someone to start filling it. As the water level rises, you will feel the pressure under your feet
progressively decreasing, until you are completely immerged. Now that you are completely
immerged, the level of the water may rise higher and higher, without affecting the pressure under your
feet.
Buoyancy contributes to the effect of less pressure on the surface of an object that is submerged. So
when calculating effective stress below ground that is affected by a water table, we need to subtract
the unit of water, which is 9.81 kN/m3. Let’s consider the example below.
Calculate the effective stresses before and after construction to the middle of the bottom layer. The
load on the foundation is 1000 kN. The water table is very deep in the summer months but increasing
to 20 meters BGL in the winter. Assume the footing shape to be square and extends 2 meters into the
2nd layer. Calculate the stress influence to the centre of the second layer.
1. Consider stresses before and after construction (to the centre of the 2nd layer)
Before construction during the summer months
Calculations , units = kN/m²
22*10 =220
18*(30/2) =270
=490
After construction during the summer months
Layer Nr
Calculations , units = kN/m²
1
0
2
18*13 =234
Vertical Stress (by Boussinesq, r/z =0.48)
1000*0.480/(15-2)²=2.84 [to centre of clay layer]**
Total stresses
=1000+2.84+234 = 1236.84
Before construction during the winter months
Layer Nr
Calculations , units = kN/m²
1
22*10 =220
2
18*10+(20-9.81)*(15-10) =231
Total stresses
=451
After construction during the winter months
Layer Nr
Calculations , units = kN/m²
1
0
2
18*(10-2)+(20-9.81)*(15-10) =195
Vertical Stress (by Boussinesq, r/z =0.48)
1000*0.480/(15-2)²=2.84 [to centre of clay layer] **
Total stresses
1000+195+2.84 = 1197.84
Layer Nr
1
2
Total stresses
**Note: if the point load was located at the surface, the full depth would then be used as z
5.6
Consolidation
Pre-consolidation Pressure (Soil has memory)
OCR=σ'P/σ'VO,
Where,
σ'P=pre-consolidation stress (determined by Casagrande's method),
σ'VO= current overburden stress (pre-determined and based on site conditions)
•
OCR>1,over consolidation
•
OCR=1,normal consolidation
•
OCR<1,under consolidation
A graphical solution is used here to determine where σ'P can be plotted.
Casagrande procedure is as follows:
1. Choose by eye the point of minimum radius (or maximum curvature) on the consolidation curve
(point A in Fig.8.6).
2. Draw a horizontal line from point A.
3. Draw a line tangent to the curve at point A.
4. Bisect the angle made by steps 2 and 3.
5. Extend the straight line portion of the virgin compression curve up to where it meets the bisector
line obtained in step 4. The point of intersection of these two lines is the pre-consolidation stress
(point B on the figure above)
5.10_a
Consolidation
2.
The following can now be calculated:
Void ratio after swelling period = ef = Gs*mf, [where mf is the final moisture content]
(Since the soil is saturated, Sr = 1.0)
The void ratio for each stress increment is given by:
e1 =
(The ratio
h1(1 + ef )
−1
hf
h1(1 + ef )
is a constant which can be calculated using the "final" values as obtained
hf
in the test above. It can just as well be calculated using any other value of void ratio, say the
initial e1, provided the corresponding thickness H is known).
3.
The results are presented in the form of a compression curve e vs log P. (Log P is used to
get a straight line). The settlement can now be calculated as follows:
S=
e0 − e1
*h
1 + e0
5.18
Example of consolidation calculation
Consolidation
A soil sample was removed from a clayey layer below ground where a high rise apartment block
is to be built. A consolidation settlement test was conducted in an Oedometer cell in the
laboratory (results below). The clay layer extends from 8.6 m to 12.9 m and above the clay layer,
only sandy gravel is encountered. A water table was logged at 6.3 m below ground level. Above
the water table the sandy gravel has a dry density of 1925 kg/m3 and below the water table a
saturated density of 2010 kg/m3. The clay’s saturated density is 1745 kg/m3. The foundations for
the building will extend to 6 m below ground and exert a line load of 1350 kN/m2.
Note: The initial thickness of the specimen was 26 mm, and at the end of the test, the moisture
content was 28.95%. Assume Gs = 2.7 g/cm3, g = 9.81 m/s² and Sr =1.
Pressure,
kPa
0
30
60
120
240
480
960
1920
0
Dial gauge,
mm
6
start
5,12
4,86
4,63
4,38
4,13
3,90
3,65
4,16
end
You are required to do the following:
4.1.
Determine the expected pressures at the founding levels before and after construction.
Show all calculations and use the standard unit format of kPa.
4.2.
Determine the void ratios at each loading increment and present these values in the table
provided below.
4.3.
Establish a graph on the graph paper provided and plot the compression curve.
4.4.
Indicate clearly the final void ratios, before and after construction. Note: Choose an
appropriate scale to three decimal places.
4.5.
Calculate the clay’s predicted consolidation settlement due to the erection of the building
and increase in stress. Round up the final answer to the nearest mm.
4.6.
Determine from data available from the NHBRC (National Home Builders Registration
Council) how this site may classified, with reference to the differential settlement being
assumed to be 50%.
5.19
~nsoIldabbn
EXAMPLE 1 CLASS TUTORIAL
A site investigation showed that the water table is 2,7 m below the surface. The saturated density
of the clay is 2120 kg/m3fiom 5,8 m to 7,4 m below the ground surface. The material on top of
the clay layer has a bulk density of 1980 kg/m3above the water table and a saturated density of
2060 kg/m3below the water table.
It is planned to erect a building, of which the foundation will exercise a pressure of 1220 kPa at
a depth of 3,5 m,on this site. The following are results obtained fiom a consolidation test done
on a sample fiom this clay layer:
Pressure
kPa
100
200
400
800
1600
Thickness after 24 hours
mm
26,40
26,17
25,93
25,71
25,47
After completion of the test, the pressure was removed and the sample was allowed to expand
whilst it was kept saturated. After expansion the thickness of the sample was 26,73 mm and the
moisture content was 24,1%. Determine the expected settlement.
SOLUTION:
Consolidation
5.9
RATE OF CONSOLIDATION
Settlement in sands and gravels take place in a short time, usually as construction proceeds, and
these rarely cause major problems. But in clay soils, because of their low permeability, settlement
can take place over much longer periods, i.e. months, years, decades, even centuries, after
completion of construction. Estimates of the rate of settlement, and time within which settlement
will be virtually complete, are therefore important factors in foundation design. The rate of
consolidation depends on the rate at which water is expelled under the applied pressure. It is thus
evident that the rate will be influenced by those factors which affect the flow of water through the
soil, the most important of which is the permeability of the soil and its compressibility. Other
factors which play a role are the distribution of the effective stress through the layer and the
length of the drainage path.
•
Degree of consolidation, (Uv)
Uv =
•
Settlement after time" t"
Final Settlement
Co-efficient of Consolidation,
(Cv)
Cv =
Where:
πh²
T*4
CV - Coefficient of consolidation
h = length of drainage path in meter (check if drainage path is in both directions or
not, if so divide “h” by 2; see sketch above)
t =time for complete consolidation (in seconds) assuming that consolidation rate is
kept constant (this will be obtained from the graph)
When the layer under consideration may drain freely from both its upper and lower surfaces, it is
said to be an open layer, and the length of the drainage path H, is equal to half the thickness. If
free drainage can only occur at one boundary, the layer is said to be half-closed, and the length of
drainage path equates the thickness.
5.33
Consolidation
The co-efficient of consolidation can be calculated graphically after recording values for settlement vs
time. The Y-axis values can be strain gauge reading, voids, or actual settlement. Common values are
the sample thicknesses corresponding to the time intervals. The X-axis values are the square root of
interval time, that is, the time intervals when the strain gauges were read.
Both Casegrande and Taylor (1948) have produced graphical methods and both will be demonstrated
below.
Casegrande Method (simplified version)
•
•
•
Begin by recording the dial gauge (or specimen height) readings on the X-axis vs the function
of the logarithm of time on graph paper (see pg 5.38 a)
Draw tangents to the primary settlement and after 24 hours to find the point of intersection.
Read the time interval as real time. The example on pg 5.38a is 12 minutes)
This reading equates to 100% consolidation.
Taylor’s Method
•
•
If you plot change in height (or dial gage reading) as a function of the square root of time, it
should plot as a straight line as long as U < 60%. If you extrapolate the linear portion of the
laboratory consolidation curve back toward the origin, you can determine the dial reading R0
corresponding to 0% consolidation (i.e., the dial reading after the immediate settlement has
occurred). Taylor noted that if you extrapolate the linear portion of the laboratory
consolidation curve forward (beyond 60% consolidation) it gives you a T90 that’s about 15%
too low (see figure below). So the tangent on the primary settlement section of the curve is
extended by this amount. The example below reveals a T90 time reading of 0.93²=0.86
minutes
As can be seen in the example on pg 5.38a the reading for time should be squared to reveal
the actual time for 90% consolidation.
Fig 5.5
5.35
Consolidation
Here follows some more guidelines on assessing the rate of consolidation using Taylors Method
1. Using the Oedometer data from your laboratory test construct an accurate graph, plotting
the settlement (y-axis) vs the square root of time intervals (x-axis). See the previous page for
an example of this.
2. You might see some initial settlement on the top of the graph. This can be ignored as it
signifies some early quick settlement due to air voids collapsing.
3. Draw a primary tangent to the primary curve and extend this to the bottom of the graph to cut
the x-axis. See an example below.
4. Extend the primary tangent by 15% (see previous page). Read off the Consolidation Time
where the extended line cuts the primary settlement curve (arrow above).
5. Now use Taylor’s formula to calculate the rate of consolidation, Cv.
∗
2
/
The T-factor can be chosen from the table below and depends on what type of consolidation is
desired, ie 50% consolidation =0.197, 90% = 0.848
5.35a
Laboratory Consolidation Test and Analysis
Data obtained from one increment in a conventiional multi increment Consolidation Test
Sample Diameter (mm)
= 50.00
Initial Sample tickness (mm)
= 20.00*
Dial gauge reading at the start of the test (mm)
= 0.000
Dial gauge reading at the start of the
= 3.744
current increment (mm)
Initial Moisture content of the sample %
= 43.43
Specific gravity of the particles
= 2.20
Current load increment is from 60 kN/m2 to 120 kN/m2
Date
Time
Time
Elapsed
(min)
26/06/2009
2:00 PM
0.00
0.00
0.25
Dial Reading
Root
Time
(min 1/2 )
Major
Minor
Dial
Reading
(mm)
Settlement
(mm)
3.6
72
3.744
0.000
0.50
3.8
28
3.856
0.112
0.50
0.71
3.8
42
3.884.
0.140
1.00
1.00
3.8
62
3.924
0.180
2.00
1.41
3.8
94
3.988
0.244
4.00
2.00
4.0
38
4.076
0.332
8.00
2.83
4.0
94
4.188
0.444
15.00
3.87
4.2
44
4.288
0.544
30.00
5.48
4.2
84
4.368
0.624
60.00
7.75
4.4
6
4.412
0.668
120.00
10.95
4.4
21
4.442
0.698
1440.00
37.95
4.4
58
4.516
0.772
* Sample thickness is 20 mm at the start when the load is 0 kN/m2. At the beginning of the load
increment 60 – 120 kN/m2, sample has already settled 3.744 mm and the sample thickness is
16.256 mm.
Determine Cv using Taylors and Casegrande's graphical methods.
5.38
Solution:
Consolidation
5.38_a
Specimen Calculations
Coefficient of Volume Compressibility(mv)
mv =
∆h
0.772
=
= 7.91 x 10-4 m2/kN
∆σ × H 60 × 16.256
Coefficient of Consolidation (Cv)
a) By the Square Root Time (Taylor’s) Method
t 90 = 4.0 min 1 / 2
t 90 = 16 min
2
T ×d2
⎛ 16.256 / 2 ⎞
2
2
C v = 90
= 0.848 × ⎜
⎟ = 3.5 mm min = 1.84 m /year
t 90
16
⎝
⎠
b) By the Log (Time) – Casagrande’s Method
δ 0 = 0.05mm
δ 50 =
δ 100 = 0.65mm
0.05 + 0.65
= 0.35 mm
2
t 50 = 4.68mm
2
Cv =
T50 × d 2
⎛ 16.256 / 2 ⎞
2
2
= 0.197 × ⎜
⎟ = 2.78 mm min = 1.46 m /year
t 50
4
.
68
⎝
⎠
5.38_b
CLASS EXAMPLE ON RATE OF CONSOLIDATION
The readings that follow were taken for an increment of vertical stress of 50 kPa in an Oedometer test
on a saturated clay sample of diameter 75 mm and thickness original of 20 mm. Drainage was
permitted from both the top and the bottom boundaries.
•
•
Determine the coefficient of consolidation ( Cv) using the Taylors’ root time method.
Determine how long the clay will take to reach 90% consolidation.
•
Determine how the long clay will take to settle below the structure if the full thickness of the clay
is 4 m.
Hint: Calculate the final height, h, of the clay after 24 hours when assessing the Cv value.
௛
‫ܥ‬௩ =0.848*(ଶ )²/ܶଽ଴
√ Root time, min
20
19.8
Settlement, mm
19.6
19.4
19.2
19
18.8
18.6
18.4
18.2
18
0
20
40
60
Length of time for
settlement ...............................
5.40
80
100
120
Consolidation
Class Example
The results obtained from a consolidation test on a clay sample for a pressure increment of 100 - 200
kN/m2 are shown below.
Thickness of sample (mm)
Time (min)
25,200
0
25,141
¼
25,108
1
25,075
2¼
25,046
4
24,985
9
24,922
16
24,865
25
24,827
36
24,809
49
24,800
64
24,797
81
24,780
1440
4.1
Provide a settlement vs time graph on the graph paper provided and show clearly how the
time value is determined using Casagrande’s graphical method. (3)
4.2
How long (in years) would a layer of this clay, 6 m thick and drained on both sides, take
to reach 40% primary settlement? (3)
5.42
Consolidation
NHBRC Site Classification
5.43
Chapter 6
Permeability
Neal and Procter
Advanced Geotechnical
Engineering
OBJECTIVES
Upon completion of this chapter you must be able to:
•
Apply the various laws of conductivity
•
Calculate the various stresses upon soil strata of different types
•
Evaluate the co-efficient of permeability and identify the soil type
•
Execute a laboratory test suitable for the soil type
•
Determine the drawdown after a pump out test
6.2
6. PERMEABILITY
6.1
INTRODUCTION - OCCURRENCE OF SUBSURFACE WATER
Less than I per cent of the earth's water occurs as liquid fresh water associated with land
masses, the rest is either saline water in the oceans and seas, or water vapour in the
atmosphere. This land-based water is derived from rainfall and subsequently flows under
the influence of gravity. On the surface streams, rivers and lakes are formed, but since
soils consist of solid particles surrounded by interconnecting pore spaces it is possible for
water to flow through, or be held in store by, a soil stratum. Water below the surface of
the ground is known as subsurface water. The depth of penetration and movement of of
this percolating water depends on the permeability of the soil strata.
vadose
water
--
phreatic
water
ll !/i
-- --- --... ---
(
/JercIoJ C
ati·o
n
- -_
f
'
1
- - 't7 wat
e �abt
�.:.!:_
e
/Jhrear;---...::;;: - -.
c flow
-.
---
river or
lake
..........................................,
Occurrence of groundwater
From an engineering point of view, groundwater in soil exists in two distinct zones
namely; vadose water and phreatic water, separated by the water table. The water table
is the level at which the pore water pressure is equal to the atmospheric pressure, in other
words the level at which water will stand in a borehole.
6.3
Pmneability
6.2
CAPILLARY WATER
Capillary water is held above the water table by surface tension which is the attractive
force exerted at the interface of surface between materials. The total regime of subsurface
water may be divided into four zones as shown below:
Ground surface
closed face
open face
partially saturated with
percolating water only
��;tt�!��t=
_____,;;z_�x. capillary rise
EXCAVATION
Groundwater zones
•
Below the water table, a zone saturated with water subject to normal hydrostatic
•
•
•
pressure.
Immediately above the water table, a zone saturated with capillary water.
Followed by a zone partially saturated with capillary water.
Finally, the top zone where water is either percolating downwards or evaporating.
The height to which water will rise under capillary action depends on the size of the
intergranular pore spaces, the smaller the pores the bigger the rise. Heights of capillary
rise can quite significant, ranging from almost being negligible for gravel to values
exceeding I Om for clayey soils.
6.3
WATER TABLE
Since soils are made up ofmasses of discrete particles, the pore spaces all interconnected,
so that water may pass from zones of high pressure to zones of low pressure. The level
6.4
Permeabilitv
at which the porewater pressure is equal to that of the atmosphere is called the water
table or phreatic surface. Below the water table the soil is assumed to be fully saturated,
although it is likely that, due to the presence of small bodies of entrapped air, the degree
of saturation will be marginally below 100%. The level of the water follows
approximately the topographic features of the ground surface, the level also fluctuates
with climatic conditions (rainfall, drought).
In the case of unconfined bodies of groundwater, the water table correspond to free water
surface, such as will be revealed in a borehole or standpipe inserted into the ground. A
local water table, occurring above the normal water table, called a perched water table
may be caused by the presence of an isolated impermeable layer.
excavation
Normal and perched water tables
A acquifer is a water bearing stratum. An unconfined acquifer is one where the water
surface is free to fluctuate and there is consequently no build-up of porewater pressure
beyond the normal hydrostatic pressure.
A confined acquifer is one situated between impermeable strata, in which case the
porewater pressure may develop beyond that corresponding to the top of the water level
in the acquifer.
6.5
Permeability
6.4
EFFECTIVE AND NEUTRAL STRESS
When an external stress is applied to a soil mass that is saturated with porewater, the
immediate effect is an increase in the pore pressure. This produces a tendency for the
porewater to flow away through adjoining voids, with the result that the pore pressure
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, namely:
•
Effective stress ( a?
This is the stress transmitted thru the points of
contact of the solid particles (intergranular). It is this stress component that is
effective in controlling both volume change deformation and the shear strength
of the soil.
Interpretation of effective stress.
•
Pore water pressure (u)
This is the pressure exerted by the mass of
water in the voids below the water table. Water is able to transmit normal stress,
but not shear stress, and is therefore ineffective in providing shear resistance. For
this reason, the pore water pressure is sometimes referred to as neutral stress.
Suppose, for example, we require the stress as a result of the soil's own weight at a level
z - z which is h meter below the surface. Assume the water table is below z - z , and take
the density of the soil as p kg/m3 . Consider the weight of a column of soil of area A:
6.6
Permeability
I Dens�
I
I
I
I
I
I
fw I
h
�·--c·.13A
ARE.A
Stress at level h
Total stress
u
=
=
=
=
=
=
=
(7
where
(7
d
u
Weight of soil + Area
(Inass x gravity) + Area
(p x vol x g) + A
(p x A x h x g)
pgh = rh
A
Ysat X h
=
=
(rsub + rw )xh
(rsub xh) + (rw xh)
d + u
=
=
total stress normal to the plane under consideration
effective stress between the soil particles at the plane
=
under consideration
pore water pressure at that level
6.7
Permeability
EXAMPLE:
A borehole on a building site is shown in the figure. Calculate the effective stress at the bottom
of the hole (clay):
(a).
(b).
Under the conditions shown.
If the water table is lowered by 2,4 m by pumping,. but assuming that the sand remains
saturated with capillary water up to the original level of the water table.
. . .. ..
·....
p =. 1q30 ke1 ,.,?
SAND : G-:..= ;;, "1:Z
.... -....
...
w. ..:::·. :.. L
�
-:;
-:-;--
·:·. ..·:.:.
:.·
..
5o+uro.kd sa.nd. Hint:
Sketch the effective I total stress profiles
SOLUTION:
6.9
Psat
-
Gs + e
1
+ e
x Pw
Permeability
6.5
FLOW OF WATER THROUGH SOILS
6.5.1 Darcy's Law
The flow of water through soils under saturated conditions will occur from a
higher to a lower water level, and flow will proceed as long as there is a
difference in head.
-
-- - -
""'J!adierii, i = llH
t::.L
-
llH
area = A
q
One-dimensional flow in soil
Water will flow through a soil ifthere is a hydraulic gradient. This flow of water
through a fully saturated soil takes place in accordance with Darcy's empirical
law which states that the rate of flow through a column of saturated soil is
proportional to the hydraulic gradient and the cross sectional area transverse to
the direction offlow.
In symbols
Q
=
Where Q
volume of water flowing per unit time (m3 Is)
X-sectional area
k
=
=
=
i
=
hydraulic gradient (
A
kiA
coefficient of permeability
6.17
Head causing flow
Length of flow path
�)
Permeabilizy
6.5.2 Coefficient of Permeability (k)
The capacity of a soil to allow water to pass through it is termed permeability.
The coefficient ofpermeability (k) is a measure of the permeability a soil and is
defined as the flow rate per unit area under a hydraulic gradient of unity.
The dimensions of k are those of velocity:
k
Q
iA
[m3 ][m]
- mis
[s][m][m2 ]
While k has the dimensions of velocity, it must not be seen as the true seepage
velocity ofthe water through the voids ofthe soil, but as the velocity which would
produce the same flow rate Q if the water flowed through the whole area A
instead of through the voids only.
Since the coefficient ofpermeability is a measure of the ability of the soil to allow
water to percolate through it, it is obvious that the value of k depends mainly on
the size of the pores and thus is a function of:
•
The particle size distnbution
•
The shape and orientation of the soil particles
•
The pore size distnbution (degree of compaction)
•
The degree of soil saturation
•
The porosity of the soil
•
In clay, soil structure (presence of fissures)
The range ofvalues ofk is extremely large, extending from zero in the case of clay
tp perhaps 1 Om/s or even more for coarse grained gravels. The following table
shows the range of average values for k for various soils and also indicates
potential drainage conditions:
6.18
COEFFICIENT OF PERMEABILITY
cm/s (log scale)
102
Drainage property
101
{
Application in earth
dams and dikes
{
Types of soil
1.0
10·1
10·2
Good drainage
10·>
10·•
Clean gravel
.
Constant Head Permeameter:
little experience required.
Reliable;
little experience
required
Computation:
From the grain size distribution
(e.g.. Hazen's formula). Only
applicable to clean. cohesionless
sands and gravels
Indirect determination
of coefficient of
permeability
I
I
10
1.0
10·•
"'Impervious'" soils e.g.,
homogeneous clays
below zone of
weathering
I
I
Failing !lead Permeameter;
Range of unstable permeability;•
much experience necessary for
correct interpretation
1
10·•
10·
I
0
Constant head test in triaxial cell;
reliable with experience and no leaks
Lirly reli,Je;
considerable experience necessary
(do in triaxial cell)
Computations:
frofn consolidation
tests; expensivt labor·
atory equipment and
Horizontal Capillarity Test:
Very little experience necessary; especially useful for rapid testing of a large number of samples
in the field without laboratory facilities.
1
10·•
Pra�tically im�rvious
(Note: Considerable experience
also required in this range.)
Direct testing of soil in its original position (e.g., well points).
If properly conducted, reliable; considerable experience required.!
Direct determin.ition
of coefficient of
permeabilitV
Poor drainage
10·•
Very fiM sands, organic ind inorganic
silts, mixtures of sand, silt, and clay,
glacial till, stratified clay deposits, etc.
I
I
I
""Impervious" soil; which are ,,;odified by
the effect of vegetation and weathering;
fissured, we�thered clays; fractured OC clays
Clean sands,
clean sand and gravel mixtures
I
I
10·5
I
!
I
Impervious sections of earth dams and dikes
Pervious sections of dams and dikes
I
I
Permeability
10· 5
ience required.
considerable exper10·1
10·•
10·•
10"9
The table above gives typical values of k for different soil types. the following
are some further noteworthy facts regarding the value of k:
•
•
In general, the smaller the particles, the smaller the average sizes of the
pores and the lower the value of k.
The presence of a small % of fines in a coarse grained soil will result in
a significant reduction of the value of k.
•
In clays, the presence of fissures will yield the large mass permeable
although the k-value of the clay representing the stratum might be
negligible. In such cases a field determination of k yields more realistic
values than laboratory tests which are conducted on small sample.
The following test methods can be used to determine k:
•
Coarse grained soils
Constant head permeameter
•
•
Fine grained soils
Falling head permeameter
In-situ soils
Pumping out test
6.19
Permeability
6.6.3. Apparatus
1.
Permeameter cell, fitted with loading piston, perforated plates, flow tube
connection, wire gauze, etc.
manometer
,._-�-+1+-tubes
�------piston
,------- locking nut
. �--top connection
upper
perfora ted---1Htt':!>
plate
manometer·---+---�
connection
, glands
1(3 sets of 3)--...
I,
l
'-
--
about uniform gravel
40mmJ or glass balls
. ·. .
: .. :..: : ·
"'""--i--- ac rylic plastic
cell body
H
. .
· .· · · ..
· -· -· '"7'"tt--+-- sample
-ffi't--.
___ _J_L �' . : : .. · .. ·
3 tie rods
�---- wire gauze
�·.·._·p:·.··
------..i
,,,,.---,.-- wire gauze
1
about uniform gravel
40mmJ or glass balls
0-ri ng sea1--t;±e___::;:;:::;:::�[o)=
supporting feet
Details of constant head permeameter cell
2.
. Glass piezometer tubes mounted on a stand with a scale marked in mm.
3.
Rubber tubing for water flow, and measuring cylinder.
4.
Constant head reservoir, with a means of adjusting the water level, and
stopwatch.
6.6.4 Method
1.
The constant head apparatus and the sample will be already setup before
you enter the laboratory.
2.
Water under constant head is allowed to percolate vertically through the
sample in order to saturate the sample completely and to expell all
entrapped air from the apparatus.
6.21
manometer
panel
-- ------
11+----+--manometer
tubes
adjustable
constant
head
tonk
-
-
to
drain
Perm�ability
water
supply
.
.
i
-
outlet
reservoir
�-----0
-
.
b
"•
.�:=::;:::
-!-: :
permeometer
cell
timer
3.
General arrangement for constant head permeability test (downward
flow)
Once constant flow conditions have been reached, and all entrapped air
have been removed, the test can start. To start a test run, start the timer at
the instant the measuring cylinder is placed under the outlet flow. Record
the clock time at which the first run is started.
4.
Read the levels of the water in the manometer tubes. When a certain time
has elapsed, stop the stopwatch and remove the cylinder from the outlet
flow.
5.
Noted the time and measure the volume of water.
6.
Repeat the test at different time intervals.
6.6.5 Calculations
1.
The following information was recorded:
•
Time for each test (t)
•
•
Volume of water (Q)
Diameter of sample(d)
6.22
Permeablllty
EXAMPLE:
In a constant head permeameter test the following results were obtained:
Duration of test
4 minutes
Quantity of water collocted
300ml
Head difference in manometer
50mm
Distance between manometer tappings
JOO mm
Diameter of test sample
JOO mm
Determine the coefficient of permeability of the material.
SOLUTION:
6.24
Permeability
6.7
THE DETERMINATION OF THE PERMEABILITY OF FINE GRADED
SOIL (CLAYS AND SILTS) BY MEANS OF THE FALLING HEAD
PERMEAMETER.
6.7.1 Scope
The falling head procedure is used for measuring the permeability of soils of
intermediate and low permeability (less than 104 mis) i.e. silts and clays. In the
falling head test a relatively short sample is connected to a standpipe which
provides both the head ofwater and the means ofmeasuring the quantity of water
flowing through the sample. Several standpipes ofdifferent diameters are available
from which can be selected the diameter most suitable.
6.7.2. Applications
Knowledge of the permeability characteristics of soil is required for many
construction projects in which drainage is an important feature. The following are
some of the main applications:
•
Excavations in water bearing ground i.e to estimate the quantity of water
likely to flow into the excavation, whether groundwater lowering is
feasible, for the deign of sheet pile walls, etc.
•
Earth dams i.e. to determine the different types of soil, including filter
zones, of which an earth dam is constructed need to be known in order to
•
estimate the likely quantity of seepage flow through the dam.
Seepage pressures i.e. water percolating through any porous material
exerts a pressure, known as the seepage pressure. This seepage pressure
affect the stability of earth structures such as embankments and cuttings,
•
earth and concrete dams, retaining walls etc.
To determine the drainage possibility of highway and airfield bases and
subbases.
•
To design graded filters
6.27
Permeability
6.7.3. Apparatus
1.
Permeameter cell, comprising of a cell body, 100 mm diameter and 130
mm long. Perforated base plate with straining rods and wing nuts. Top
clamping plate. Connecting tubes !��ttings.
inlet
�
::,inc:, clip
1·
r ubber
tubing
wing nuts
flat r uDber
sealing rings
3�e-,cdS�
L
sample
--1--wire gauze
1---+--cutting edge
feet
perforated baseplate
water outlet
tnr-ougn pertorations
Derails of falling head permeamerer cell
2.
Standpipe panel fitted with glass standpipe tubes of different diameters
with a scale mark in mm next to it. Each standpipe must have a valve at
its base and connecting tubing.
3.
A source of distilled water.
4.
Stop-clock or stopwatch
5.
Immersion tank, with overflow
6.7.4. Method
1.
The soil specimen, at the appropriate density, is placed inside the cylinder
of known diameter. In the case of fine grained soils, undisturbed samples
are normally used, and the containing cylinder may be the sample tube
itself. The falling head apparatus will be already setup before you enter
the laboratory.
6.28
rl
·
Permeability
from
water supply
standpipe tubes
(different diameters)
1--
0
General arrangement forJailing head permeability test
2.
Water is allowed to percolate through the sample until the sample and
container are completely saturated. This must usually be done overnight.
After saturation, check that no air is entrapped in the pipes.
3.
Before the test can start, steady flow conditions must first be check.
4.
Let water flow from the water supply to fill up the standpipe to a high h1
(or even higher) and close the screw clip to stop the water from the water
supply. As soon as the water level reaches the level hl> start the stopwatch
immediately. The time t required for the level in the standpipe to drop
from h1 to h2 is observed and noted.
5.
The standpipes can be re-filled for a repeat run by opening the screw clip
to let water into the standpipe. This test can also be repeat by using
different diameter standpipes, and taking readings at different time
intervals.
6.29
Permeability
6. 7.5. Calculations
1.
The following information was recorded:
•
Area ofstandpipe (a)
•
Area ofsample (A)
•
Length ofsample([)
•
Time for each test (t)
•
Height at beginning oftest (h1 )
•
2.
Height at end oftest (h2)
Calculate the coefficient ofpermeability:
I
a Ln !!J...
At
hi_
k =
Where:
a
A
I
t
h1
=
=
=
Cross-sectional area ofstandpipe(mm2 )
Cross-sectional area ofsample(mm2 )
Length ofsample(mm)
Time for test (min or sec)
Height ofwater in standpipe at beginning
oftest(mm)
Height ofwater in standpipe at end oftest
h2
(mm)
6.30
Permeabilitv
6.8
THE DETERMINATION OF THE PERMEABILITY CHARACTERISTICS OF
AN IN-SITU SOIL IN THE FIELD BY MEANS OF THE FIELD PUMP OUT
TEST.
6.8.1 Scope
Because of the problems associated with the reliability of laboratory tests, as
discussed in the pervious section, field methods should be used when
permeability values are of significant importance. Comprehensive multiple-well
pumping tests can be expensive to carry out, but offer a high level of reliability
and is effective up to a depth of about 45 m.
6.8.2 Method
A casing of about 400 mm diameter is driven to bedrock or to impervious
stratum. Observation wells of at least 35 mm diameter are put down on radial
lines (r1 etc.) from the casing, and both the casing and the observation wells are
perforated to allow easy entrance of water. The test consi�t of pumping water out
from tp.e central casing at a measured rate (q), and observing the resulting
drawdown in ground water level by means of the observation wells. At least four
observation wells, arrange in two rows at right angles to each other should be
q
used.
pumping
· · ·· well
observation wells
------,'
r
Pumping test in an unconfined aquifer
6.34
·-
•.
- - - ·-
impermeable stratum
Permeability
6.8.3. Calculations
Calculate the coefficient of permeability by means of the field pump out test:
7Ln(){)
k -
Where
1r(1zi-iz;)
q
=
=
Quantity of water collected in time t
Radius from pumping well to nearest observation
well(m)
=
Radius from pumping well to furthest observation
well(m)
=
Nearest observation well. Distance from the
impermeable stratum (bottom) to top of
drawndown water table(m).
=
Furthest observation well. Distance from the
impermeable stratum (bottom) to top of
drawndown water table (m).
6.35
Permeability
EXAMPLE:
A 9,15 m thick layer of sandy soil overlies an impermeable rock stratum. Ground water level is
at a depth of 1,22 m below the top of the soil. Water was pumped out of the soil from a central
pumping well at a rate of 5680 kg/min and the drawdown of the water table was noted in two
observation wells. These two wells were on a radial line from the centre of the main well at
distances of 3,05 and 30,5 m respectively. During pumping the water level in the well nearest to
the pump was 4,57 m below ground level and in the furthest well 2, 13 m below ground level.
Determine an average value for the permeability of the soil in mis.
SOLUTION:
6.36
EXAMPLE 1 CLASS TUTORIAL
A site investigationhas shown that the water table was approximately 3,5 rn below the relatively
horizontal soil surface. A layer ofpermeablematerial extended to a depth of 25,4 m, below which
a layer of relatively impermeable clayey material was encountered. The plane separating the two
types of material is almost horizontal. It is decided to determine the coefficient ofpermeability of
the permeable material by means of the pumping out test. Calculate the coefficient of permeability
if the pumping out rate is 305 litres per hour, r, = 8 m, r2 = 13 m, h, = 9,3 m and h2 = 11,8 m,
SOLUTION:
Permeabi(ify
SELF EVALUATION EXERCISE 2
1.
During a constant head permeameter test a flow of 173 ml was measured in 15 min. The
sample was 100 mm in diameter, and the head difference of61 mm was measured between
tapping points 200 mm apart. Calculate the coefficient of permeability in mm/s.
(0,24 mm/s)
2.
When testing a cylindrical soil sample's coefficient of permeability in a constant head
permeameter, it was found that 17,4 cm3 of water flowed through the sample in 45
seconds. The sample was 190 mm long, and 60 mm in diameter. The water level in the
reservoir was 850 mm above the outlet for the duration of the test. Calculate the soil's
coefficient of permeability.
3.
(0,031 mm/s)
The following particulars refer to a falling head permeameter test:
Length of sample
=
Diameter of sample
Diameter of standpipe
=
Height readings
Time interval between readings
=
160 mm
60 mm
8 mm
1200 mm; 990 mm; 810 mm
10 min.
Determine whether stable flow conditions existed, and calculate the coefficient of
(9,11 x 104 mm/s)
permeability of the soil.
4.
In a field pump out test water is pumped from a bore-hole at a constant rate of 328
liters/min. When stable conditions were established, it was observed that the water table,
initially 2,35 m below ground level, was at depths of 4,5 m and 2,8 min two observation
wells at distances of 15 m and 60 m from the bore hole. The layer of soil, from the surface
to an impermeable rock stratum, is 10 m thick. Calculate the soil's coefficient of
(1,12 x 104 mm/s)
permeability.
5.
All observations in problem 4 apply. the only difference is that the top 5 m of the soil is
clay with negligible permeability. Calculate the permeability of the soil between the clay
(1,42 x 104 mm/s)
and the impenneable rock stratum.
6.38
Permeabilltv
EXAMPLE:
Excavations is being carried out in a soil with porosity n = 0,35 and grains relative density
Gs = 2,65. A 1,25 m layer of soil is subjected to an upward seepage head of 1,85 m. What depth
of soil would be required above the soil to give a FoS of 2 against piping?
(Assume that the coarse sand has the same porosity and grain relative density as the soil, and that
there is negligible head loss in the sand)
SOLUTION:
6.44
Permeability
6.12
FLOW NETS
6.12.1 Seepage through soil
In water retaining dams, unless the foundations continue down to impervious
rock, a steady flow of water is set up under the structure owing to the difference
in head. This may lead to an undesirable amount of leakage and, with upward
flow of water on the downstream side, dangerous quicksand conditions may
occur, with possible subsequent failure of the dam. This seepage can be studied
by the use offl.ow nets.
6.12.2. Purpose of flow nets
A flow net is a pictorial representation of the flow pattern of water when passing
through a pervious material. It is used for the solution of practical seepage
problems i.e. the estimation of the amount of water lost due to seepage through
earth embankments or underneath retaining structures. It can also be employed
to predict instability of foundations due to seepage, as well as the possibility of
piping.
6.12.3. Components of a flow net
A flow net consists of two sets of orthogonal lines depicting the flow of wat_er
through a pervious medium. It is made up offlow lines and equipotential lines
crossing each other at right angles.
E
i
Permeability
Theflow lines represents the path offlow ofthe water through the soil, the spaces
between them being called flow channels. There are an infinite number offlow
lines, the path of which never cross (laminar flow). They makes smooth curves
;.---------;.=-=�-;;,;�_;.
----------
when changing direction, and they are approximately parallel to each other.
---------- pile
wall
----------
Equipotenti1I/
lines
I
I
1/Sjj/
�
lmpennabll
stratum
t
I
(bl
(al
The equipotential lines join points ofequal pressure head. Water will only flow
ifthere is a pressure head, say H, and this is dissipated as water flows through the
soil. On each flow line, therefor, there will be a point where the pressure head has
dissipated to (say) 1/eH. All these points where the pressure head are equal will
fall on the same equipotential line.
6.12.4. The use of a flow net to estimate seepage loss
The use ofa flow net may be illustrated by referring to the following figure. This
show a typical example of a flow net. In this case a water retaining structure is
founded on a permeable stratum which is underlain by an impermeable rock
layer. The length of the wall is L.
-..
E
;;
-..
Q
Flow net (scale I : 250)
6.48
s
Pumeabllity
·············•·
·-·······-··-······
Equipotential drop
Flow path
""•• ••••••••
•·•··-· •
• ••••••••••- • ••
• • •-•o
,,..,,,
=
=
Nwnber of flow paths (in this case N1 = 5)
Nwnber of equipotential drops (in this case Ne = 8)
Consider one "square" ABCD with side length say /:
= ki A
The flow rate from AD to BC is:
llQ
f).H
= k - - (Ix L)
1
= kllHL
But llH
llQ
=
H
Ne
H
= k-L
Ne
This is thus the flow rate from AD to BC, i.e. for one flow path. The total flow
rate from PQ to RS is then obtained by multiplying with the nwnber offlow paths
�:
N
Total seepage loss Q - kHL _i_
Ne
Where
Q
k
H
L
N1
Ne
=
=
=
Total seepage loss
Coefficient of permeability
=
Water height causing flow
Wall length
=
Nwnber of flow paths
=
Nwnber of equipotential drops
6.49
Permeabilitv
EXAMPLE:
Study the following figure, and use the information given, and then determine the estimated
seepage loss in liters/dag.
Coefficient of permeability of soil
Wall length
90m
Wall height
lOm
1.3 x 10-6 m/s
...
...
-
Flow net (scale 1 : 250)
6.50
PermeabiQ
EXAMPLE:
Consider the example on page 6.50(wall length 90m; coefficient of permeability 0,0013 d s ) .
It was decided to decrease the loss of water, so sheet piles were driven in at the toe of the dam
to a depth of 5,s m and an impervious apron was constructed at the heel, 6m wide.
What will be the loss of water per day when these measures are taken?
SOLUTION:
/A\ug\~ug\\~w\w\'Cx\~\.c/n~m~~~\~~~~~\~
Flow net (scale 1 :250)
Pumeabllity
EXAMPLE I CLASS TUTORIAL
The following figure shows the cross section of a line of sheet piles driven to a depth of 7 m into
a stratum of homogeneous sandy soil which has a thickness of 12 m and is underlain by a
impermeable stratum. From an original depth of 5,5 m the water level on one side of the piles is
reduced by pumping to a depth of 0,5 m. Draw a flow net for the seepage conditions and from
it determine the quantity (m3 /day) seepage under the piles per meter run.
SOLUTION:
��--�����VIL...:.;.;::;.._
����������������----..,--�
•••••••••••••••••••••••••••••••••••••••••••••••••••••••
,I\.
V Sheet-pile wallWL
5,5 rr.
.,
,'
'"
H=5m
j 0,5 m
, I\.
7m
sandy soil
12m
\/
Impermeable stratum
6.56
Permeab@y
SELF EVALUATION EXERCISE:
1.
A falling head permeameter test carried out on a sample of the permeable soil in the
figure below, produced the following results:
Length of sample 101,6 mm; diameter of sample 73 mm; initial head 1352 mm; final
head 352 mm; time interval 147 sec; diameter of standpipe 5 mm.
Construct the flow net for the conditions shown and estimate the seepage loss in
liters/day/lOm length of wall.
SOLUTION:
. . ... ... . . . ..' •.......
• ••
1·2m
1·5m
.•• .••
• .• • . . . . . . •. .
•
.
•
• • •• •• • •
·:• .• . • •. . •:
.• . •
•
•
•
••
•
•
•
••
•• •
•• •
•
•• ••
••
• •• •
4 ••
•
••
•
••
..,E
..• ..
..
••• • . .. .
• •
. .. . .. . ...• .. .. . . . .•. .• ..• ..
.
.
.
.
.
.
. . . .. . .
. . . . .. .. .. .. . .
• •
•• • • •• t•
•
•
•
�
Impermeable rock
6.57
Permeabllitv
2.
A concrete dam. with a base length 24,4 m, holds back water to a height ofl 2,2 m above
its base. The dam. which is 457 m wide, sits on a stratum of soil 18,3 m thick which
overlies impermeable rock. The base ofthe dam is horizontal and, at the downstream end,
a sheet pile cut-off has been driven to a depth of6,lm
The coefficient of permeability of the soil is 4 x 10-7 mis. Determine the seepage loss
through the soil, in m3 /day, if there is a head of 1,52 m of water above the base on the
[± 536 m3 /day]
downstream side.
SOLUTION:
6.58
Chapter 7
Site Investigation and Remedial
Foundation Support
Neal and Procter
Advanced Geotechnical
Engineering
7.1
Foundation Support & Underpinning
Foundation Underpinning
1.
Introduction
Underpinning is the structural process by which the load superimposed by an existing
foundation is transferred to a new bearing point, usually lower than that of the existing.
Before demolition or underpinning is commenced, the following matters should be
attended too (suggested in the Civil Engineering Code of practice). Public liability is
always a contentious issue and all areas of accountability must be investigated.
2.
Process for site investigation
1. Serve notice on owners of adjoining properties, of your intention to proceed, and
give details of proposed temporary shoring, propping and barricades etc.
2. Carry out a detailed survey of the building to be underpinned before commencing
work. A detailed list should be compiled, listing all cracks and defects in this, and
adjoining buildings, and the list should be supported by photographs, which are
dated, and signed before a notary. (And preferably acknowledged by all owners.)
3. Apply glass tell-tales (slivers of glass ± 150 x 75 mm in size), fixed in positions
with steel glue, to indicate both lateral and vertical movement, and should be
provided over al visible cracks. Check levels; referenced to a suitable datum,
should be established; and monitored regularly.
4. Work should be under the continuous control of an experienced person.
5. Carry out an investigation of the sub soil beneath the building to be underpinned
(particularly if settlement has taken place), to establish whether a suitable bearing
can be established. DCP’s can be used or DMT’s, which are more accurate and
provide more insightful information about the actual conditions, deep below the
foundations, without disturbing the strata.
7.2
Foundation Support & Underpinning
Underpinning
6. Reduce loading on the footing to be underpinned, as far as possible. This might
involve the removal of superimposed floor loads, or the introduction of raking
shores etc.
7. Suspension of plumb bobs provides a useful and simple safety device.
3.
Excavations
The process involves excavating under an existing foundation and the building up of a
new supporting structure from a lower level, to the underside of the existing foundation –
the object being, to transfer the load as mentioned previously.
This may be necessary for the following reasons:
1. If the existing footing is insufficient, possibly due to uneven settlement, uneven
loading imposed, etc.
2. To increase the load bearing capacity of the existing foundation.
3. To facilitate the lowering of level of the adjacent ground level e.g. to provide for
the construction of a new basement.
4. To provide for a new structure underneath an existing structure e.g. to provide a
new basement below an existing structure
Because the safety of the existing building is of paramount importance, underpinning
must be carried out in stages and with great care. As the exposed footing has to span
the distance of the support removed and to avoid settlement, the underpinning stages
should be kept short, not exceeding 1.2 m in face width. Stages should be carried out
in the specified sequence formulated by the site engineer, so that no two stages are
worked on consecutively.
7.3
Foundation Support & Underpinning
Underpinning
The length of the stages will be determined by the following:
1. the length, height and thickness of the wall to be underpinned i.e. the load to
be transformed
2. the width of existing footing
3. the general condition of the wall and the footing
4. the superimposed loading on the wall and thus also the footing
5. the nature of the ground below the footing
6. estimated spanning ability of the existing footing
The length of the stages is normally ±1.2 m long, with the restriction that the sum total of
the unsupported lengths should not exceed 25% of the total wall length. It is furthermore,
undesirable to excavate for and construct a leg immediate adjacent to a stage that has just
been constructed.
7.4
Foundation Support & Underpinning
Underpinning
A suitable sequence of stages is indicated above, where stages are numbered in the order
in which they are constructed. No two adjacent legs have been handled one after the
other, and at no time is the length of wall being underpinned in excess of a quarter (25%)
of the total length.
The new supporting walls may be constructed of brick or concrete. Concrete is easily
adaptable for use in confirmed positions. Either material requires some form of vertical
key for bonding into successive sections to ensure that the whole underpinned structure
forms one homogenous unit. Toothings (gaps) are left in brickwork and in the case of
concrete, steel rods or dowels can be used.
Sequence of operations:
1.
Excavations for stages are carried down to the new foundation level.
2.
If necessary, planking and stetting is inserted to support the excavated faces.
3.
The new concrete footing is laid, the necessary dowels to achieve continuity, being
bedded in the ends.
4.
The new brickwork or concrete is constructed. Brickwork is generally built in
English bond, using strong mortar mix, with toothings left for bonding purposes.
7.5
Foundation Support & Underpinning
Underpinning
5.
The gap between this new work is filled with a weak concrete mix (or soil),
progressively as the wall is taken up.
6.
After a period of approximately 12 hours, the gap at the top of the new wall is
firmly wedged so that weight is taken up on new foundation, and the dry strong
grout mix firm tamped in.
7.
The existing projecting concrete footing is now cut is now cut off flush.
8.
Planking and stetting is removed.
9.
Excavated soil is returned and rammed, if it does not form part of a future basement
excavation.
The condition of the existing footing and the proximity of the new adjacent structure will
determine whether or not it is to be removed. A new basement wall against the existing
will necessitate its removal. If the footing is deformed it must be cut away completely,
and the next work taken up to the bottom of the existing brickwork.
4.
Piled Underpinning
This method of underpinning is used in the following circumstances:
If the presence of ground water prevents the use of the above mentioned methods
(problems will be encountered in excavating to the desired depth). A satisfactory solid
foundation is at a considerable depth below the base to be underpinned. It is probably too
expensive or impracticable to attempt the normal method, to such as depth and where a
vibration free system is required (in lieu of a sheet piled system in this case).
7.6
Foundation Support & Underpinning
Underpinning
The arrangement of piles (and beams), depends on the condition of the existing structure
and the necessity of avoiding disturbance. The most common method is a system of
jacked in piles, positioned immediately under the existing foundation. The system
consists of short lengths of precast concrete
pile, jacked into the ground, until a suitable
subsoil bearing is reached.
Sequence of operations
1. The excavation is carried down to
and
under
the
footing
to
be
underpinned (in short face widths)
2. The first length of pile (with pointed shoe) is positioned and a steel bearing plate
placed against the soffit of the footing. The hydraulic jack is then positioned
between the top of the pile and the bearing plate.
3. The initial pile length is jacked to its full length, and the jack removed.
4. A steel dowel is inserted in the continuous bore of the pile, and grouted solid, a
further length of pile is added, and the jack and bearing plate repositioned.
5. The process continues and the pressure (the resistance to jacking) is monitored
continuously. Once a suitable bearing resistance is achieved, jacking ceases.
(Design safety margin is usually 50%)
6. The jack and base plate are removed and the top of the pile is wedged and pinned
to the soffit of the footing in the usual manner.
7.8
Foundation Support & Underpinning
Underpinning
A variation is the use of a steel jacking mandrel which forces down a concrete
shoe, followed by a series of tubular steel sections- the completed tubes removed
thereafter.
As mentioned, jacked pile underpinning causes no vibration, is virtually noiseless and
requires little headroom for jacking. The expense of the method however, limits its use.
5.
Underpinning with Needles and beams
Where access is available on both sides of the wall (e.g., free standing boundary wall or
empty house dwelling), it is possible to sink pairs of piles at intervals along the walls, one
on either side, to be connected by a “needle” or beam, which passes through the wall.
Reinforced concrete or steel joists are used for needles. Steel joists occupy less space, but
are more difficult to wedge and pin effectively.
An alternative layout is to stagger the piles on each side o the wall. This reduces the
number of piles to be used (the expensive element of this form of underpinning),
although each is required to support an increased load. Needles are longer and thus of a
greater cross sectional dimension.
Where it is not possible to work within the building (owners still occupy the dwelling), or
gain access to both sides of an isolated wall, groups of two or more piles, along the
outside of the building, can be used to support cantilevered needles, or support beams.
7.9
Foundation Support & Underpinning
Underpinning
When walling is weak or when stages of greater lengths than described above are to be
opened up, it may be necessary to provide temporary needles, passing through the wall,
or under the existing foundation.
Needles (timber or steel), should be of sufficient size to avoid deflection, and should be
kept wedged up firmly by means of folding wedges and jacks.
Supports must be kept well clear of the working area of the underpinning and of course,
as was the case with a dead shore, taken down onto a solid bearing.
7.10
Foundation Support & Underpinning
Underpinning
When needles pass through the wall above foundation level, secondary needles (or
springing pieces), are hung from the main needles, to carry that section of wall, and
possibly the existing foundation, below the level of the main needles.
The above described methods of underpinning and final pinning up, results in negligible
movement due to settlement, but in large and heavy buildings, a system pre-compressing
the soil is adopted. So, the soil is compacted before the load is applied, and settlement is
avoided.
In this method, the strip footing is cast on the bottom of the excavation, and on maturity,
a hydraulic jack is fixed between this footing and the underside of the existing work over.
A variation is to establish precast concrete beams on the top of the jack, and pinned up to
the soffit of the existing structure. On hardening of the pinning grout, load is applied by
the jacks, and bearing capacity is achieved. Spaces between the jacks in adjacent sections
are bricked up and pinned. When this work is set, jacks are removed and the spaces left
by the jacks built up.
7.11
Foundation Support & Underpinning
Underpinning
6.
Grout Treatment of Soil as Underpinning (see also notes under Chemical
Grouting)
The bearing resistance of certain soils can be improved by grout injections and is used as
an underpinning method when soil conditions are appropriate. An example would be the
stabilization of the sandy silts and marine clays below the foundation of the Leaning
Tower of Pisa. This procedure included soil extraction (to reposition the tilt to near
vertical) as well and grout stabilization.
References used in compilation of notes
1. Building in Expansive Soils. The South African Builder. October 1962 (reprint)
2. Building Houses on Expansive Soils. Tom McGoldrick. Watermeyer Legge
Piesold & Uhlman ,CE
3. SA Construction World. April 1990
4. Frankipile. “Guide to Piling and Foundations Systems”
5. National Department of Housing. Generic Specification, GFSH-2. Sep 2002
7.12