POLITEKNIK SULTAN HAJI AHMAD SHAH presents
GEOTECHNICAL ENGINEERING
The Writers & Editors Manisah binti Mohamad Emmy Liana binti Ayob Nur Bazilah binti Ishak Zakiah binti Hassan Nik Nor Asima binti Ariffin
Publications & Declaration All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical including photocopy, recording or any information storage and retrieval system, without permission in writing form from Civil Engineering Department, Polytechnic of Sultan Haji Ahmad Shah, Kuantan..
Acknowledgement We would like to include a special note of thanks to Portfolio E-Learning POLISAS for giving us chance to contribute in this project with guidance and countless help in every part of it production process. We would like to express our gratitude for the support and love of our family and friends.
Abstract This e-book has develop out of a series of lectures notes use in Geotechnical Engineering Course in Civil Engineering Department, POLISAS. It contains eight chapters that parallel to the syllabus of DCC30093 Geotechnical Engineering for Diploma of Civil Engineering in Polytechnic Curriculum. This e-book complete with basic fundamental of geotechnical engineering and enriched with variety of examples and tutorials that may help students to have better understanding in Geotechnical Engineering.
Chapters
Geotechnical Engineering
01
INTRODUCTION TO GEOTECHNICS
1 - 16
03
BASIC CHARACTERISTICS OF SOIL
37 - 61
05
FOUNDATION
69 - 88
07 SEEPAGE
101 - 133
02
SITE INVESTIGATION
17 - 36
04
SHEAR STRENGTH
62 - 68
06
EFFECTIVE STRESS & LATERAL EARTH PRESSURE
89 - 100
08
SLOPE STABILITY
134 - 157
TOPIC 1 INTRODUCTION TO GEOTECHNICS
OBJECTIVES To identify soil mechanics in general To classify the types of soil in Malaysia To explain soil formation process
INTRODUCTION
To Identify Soil Mechanics in General
TERMINOLOGIES Soil is natural resources. It is an un-cemented aggregate of mineral grains and decayed organic matter (solid particles) with liquid and gas in the empty spaces between the solid particles. Soil is used as construction materials and as support for structures on and within them. Soil mechanics is the branch of science that deals with the study of the physical properties of soil and the behaviour of soil masses subjected to various types of forces. Soil engineering is the application of the principles of soil mechanics to practical problems. Geotechnical engineering is the sub discipline of civil engineering that involves natural material found close to the surface of the earth. It includes the application of principles of soil mechanics and rock mechanics to the design of foundation, retaining structures, and earth structures.
PAGE 2
INTRODUCTION
Soil mechanics in general
Soil mechanics is a subset of geotechnical engineering, which involves the application of soil mechanics, geology and hydraulics to the analysis and design of geotechnical systems (Budhu, M., 2010)
The engineering aspect of soils
Three structural properties are of primary interest to soil engineers: a) The resistance of soil mass to change in volume with changes in load b) The ability of a soil mass to resist shear forces or lateral displacement under load c) The permeability of the soil mass when it affects the load volume change and shear characteristics
(Holtz, 1974)
PAGE 3
INTRODUCTION
To Classify the Types of Soil in Malaysia The formation of rock and soils are explained by its geological process. The earth crust is made up of in-place rock and unconsolidated sediments. The igneous, sedimentary and metamorphic are known as the parent rocks. The igneous rock is formed by solidification of molten magma that originated within the earth. The sedimentary rock is formed by the deposition of fragmented rock in a layer form. Metamorphic rock formed by the creation of new characteristics in term of mineral and structures in preexisting rock due to the increased temperature and pressure that is known as metamorphism process. Figure 1.1 below show the geological cycle in formation of rock.
Figure 1.1: Geology Cycle of Rock Formation.
PAGE 4
INTRODUCTION
PAGE 5
INTRODUCTION
PAGE 6
INTRODUCTION
PAGE 7
INTRODUCTION
PAGE 8
INTRODUCTION
Scan me for more information about the rock cycle
PAGE 9
INTRODUCTION
To Explain Soil Formation Process Soil is formed by the weathering process of rock which is the disintegration and the decomposition of rock and minerals near the earth surface through the action of natural or mechanical or chemical agent into smaller and smaller grains. Weathering is the process of chemical or physical decomposition of rock. There are several agents of weathering such as wind, water and glazier. Soil formation process can produce 4 types of soils that are; the organic soil or top soil, the residue soil, the transported soil and the marine soil.
PAGE 10
INTRODUCTION
Formation of soil Soil consist of solid particles with varying amounts of water, gases and other organic matters. The formation of soil through the action of natural or mechanicals factors and the reaction of the chemical agents.
Organic soil or top soil Organic soil or top soil found in the earth surface. The top soil: a) Not suitable for construction activity. The top soil should be removed (about 500mm) before construction. b) Contains organic materials that can give effect to the construction activities. Example: Peat soils.
PAGE 11
INTRODUCTION
Figure 1.2: The Removing of Top Soils Before Construction.
Residue soil • Have formed from mostly the weathering rock and remain at the location of their origin. • It occurs when the rate of weathering is higher than the rate at which soil is transported by agents of erosion. • If the rocks are igneous and metamorphosis, then the size of the resulting range is from silt to gravel. • Laterite was formed from limestone by the action of rainwater that decompose soluble rocks and leave the insoluble iron hydroxide and aluminium hydroxide.
Figure 1.3: The Residue Soils PAGE 12
INTRODUCTION
Transported soil
• Transported soil are those materials that have been moved from their place of origin by the transportation agents. • The transportation agents are water, glasier, wind and gravity.
Scan me for more information about the residue soil versus transported soil
Figure 1.4: The Transported Soils
Gravity transported soil • Under the action of gravity, soils can transport through short distances. • Soil masses and rock fragments collected at the foot of the cliffs or steep slopes had fallen from a higher elevation under the action of the gravitational force. Colluvial soils have been deposited by the gravity.
Water transported soil • Flowing water is one of the most important agents of transportation of soils. If the velocity of flow is large, then it carries a large quantity of soil either by suspension or by rolling along the bed. • The size of the soil particle carrier depends on the velocity of flow. PAGE 13
Wind transported soil
INTRODUCTION
• Some soil particles are transported by wind. •
In this phenomenon, we get wind transport soils, the particle size of the soil depends upon the velocity of the wind. Soils deposited by the wind called Aeolian deposits.
•
When wind speed is very fast then he takes a lot of soils one place to another place. If wind speed is slow then he takes some soil then all depend on the speed of winds
Glasier transported soils •
The glacier is large masses of ice form by the compaction of snow.
•
As the glaciers grow and move, they carry with them soils varying in size finegrained to huge boulders.
•
Soils get to mix with the ice and transport far away from their original position.
•
Drift is a general term uses for the deposits to make by glaciers directly or indirectly. Table 1.1: The Types of Transported Soils Based on Its Source of Transportation Agents.
(Source: civilnotes.com)
PAGE 14
Marine clay
INTRODUCTION
PAGE 15
INTRODUCTION
ASSESSMENT Try to assess your understanding on this chapter by answering this quiz. Good luck. https://forms.office.com/r/WKsaEVfh1v
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TOPIC 2 SITE INVESTIGATION
Site Investigation is the process of collecting information, assessment of the data and reporting potential hazards beneath a site which are unknown.
SITE INVESTIGATION
OBJECTIVES OF SITE INVESTIGATION The main objective of the site investigation has been outlined in BS 5930: 1991
1. SUITABILITY - To determine the suitability of the site and the environment for the
work which has been determined. (Proposed work)
2. DESIGN - To enable an adequate and economic design can be provided, including
temporary works design. 3. CONSTRUCTION METHOD - To plan the best method of construction and to
determine the sources of materials and surplus materials or waste. 4. EFFECT OF CHANGES - To determine the changes that will occur in the soil and
surrounding circumstances, either naturally or because of work. 5. SITE SELECTION - When there is a site options, to provide advice and guidance.
In addition, site investigation may also be necessary to: i.
Reporting on the work of security available.
ii.
Design of additional work to the work of the existing.
iii.
Investigate cases in which the failure occurred.
PAGE 18
SITE INVESTIGATION
WORK PROCEDURES IN SITE INVESTIGATION
1.
Preliminary and Site Exploration
2.
Program Planning and Scope of Site Investigation
3.
Soil Investigation/ Sampling Lab Test
4.
Preparation of Site Investigation Report
5.
Design and Review During Construction and Monitoring
A)
PRELIMINARY AND SITE EXPLORATION
Preliminary / Desk study- The collection of wide information related to building such as maps, drawings, geological details, utilities details and ownership of adjacent property. Site Exploration - An early examination of site by appropriate expert such as geologic and soil engineer to collect data on overall site layout, topography, basic geology, details of access and high restriction on site. PAGE 19
B)
SITE INVESTIGATION
PROGRAMME PLANNING AND SCOPE OF SITE INVESTIGATION
Program Planning:
1. Borehole - how many boreholes? Criteria to terminate BH 2. Field test - criteria & frequency 3. Sampling type - criteria & frequency 4. References Notes / Codes - for certain project such as bridge, road and building 5. Documentations & drawings
The scope of Site Investigations is: 1. Topography 2. Soil Profile 3. Ground Water Condition
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SITE INVESTIGATION
C)
SOIL INVESTIGATION AND SAMPLING LAB TEST
Investigation of detailed geology and sub-surface soil conditions using surface surveys, trial pits, boreholes, sounding, geophysical methods, as appropriate, survey of ground water conditions over a signification period.
D)
PREPARATION OF SITE INVESTIGATION REPORT
1. Details of geological study, including structures, stratigraphy, and mapping. 2. Results of borings including logs. 3. References for samples and stratigraphy interpretation upon request. 4. Comments and recommendations relating to the design and construction of the proposed works. 5. Recommendation relating to further investigations or testing, and to ongoing or post completion monitoring.
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E)
SITE INVESTIGATION
DESIGN AND REVIEW DURING CONSTRUCTION AND MONITORING
Monitoring during construction and maintenance period is required whether the expectations of the proceeding investigation have been realized. No one can ensure that the soil parameters used for design is most representative of the soil conditional at site unless the response is observed. Among the measurement made during the monitoring stage are: • Settlement • Displacement • Deformation • Inclination • Pore water pressure
DETERMINE IN-SITU SOIL TESTING a) JKR & Mackintosh Probe b) Standard Penetration Test (SPT) c) Cone Penetration Test (CPT) d) Plat Bearing Test
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1.
JKR AND MACKINTOSH PROBE
SITE INVESTIGATION
Introduction of Probe Mackintosh: •
Dynamic probing is a continuous soil investigation technique, which is one of the simplest soil penetration test.
•
Figure 2.1: Probe Mackintosh Testing It basically consists of repeatedly driving a metal tripped probe into the ground using a drop weight of fixed mass and travel.
•
Testing is carried out continuously from ground level to the final penetration depth.
•
The continuous sounding profiles enable easy recognition of dissimilar layers and even thin strata by the observed variation in the penetration resistance. Advantages of Probe Mackintosh:
•
• Lightweight • Portable penetrometer It is considerably faster and cheaper tool than boring equipment (especially for moderate depth penetration, for loose and soft soil) • Simple and economic testing method to gather preliminary data • Data can be used for shallow foundation design
PAGE 23
SITE INVESTIGATION
Objective
PROCEDURES OF JKR / MACKINTOSH PROBE TEST
https://youtu.be/Ce6lejk1Yto
PAGE 24
SITE INVESTIGATION
Mackontosh probe JKR VS SPT
Pictures
PAGE 25
SITE INVESTIGATION
Figure 2.2: Equipment and Procedures of Probe Mackintosh Testing
2.
• •
STANDARD PENETRATION TEST (SPT)
Figure 2.3: SPT Procedures The SPT is carried out at every 0.75 m vertical intervals in a borehole.
This can be increased to 1.5 m if the depth of borehole is large. •
Due to presence of boulders or rocks, it may not be possible to drive the sampler to a distance of 450 mm. In such a case, the N value can be recorded for the first 300 mm penetration.
PAGE 26
SITE INVESTIGATION
The boring log shows refusal and the test halted if: • 50 blows required for any 150 mm penetration • 100 blows required for any 300 mm penetration • 10 successive blows produce no advances
PAGE 27
SITE INVESTIGATION
THE PROCEDURES OF STANDARD PENETRATION TEST https://youtu.be/pvY0cm9dG9g
3.
CONE PENETRATION TEST
Cone penetration testing (CPT) is an in-situ test that is used to identify the soil type. In this test a cone penetrometer is pushed into the ground at a standard rate and data are recorded at regular intervals during penetration. A cone penetration test rig pushes the steel cone vertically into the ground. Static CPT are conducted to measure and evaluate characteristics of very soft and loose soils (soft sensitive silts and clays and some coarse cohesionless soils) where transportation of heavy equipment was not possible.It measures: 1.
3.
Stratification 2. Soil type Relative soil density and in-situ conditions 4. Shear strength parameter
PAGE 28
SITE INVESTIGATION
Pictures
THE PROCEDURES OF STATIC CPT https://youtu.be/Cvu9iBSnQYo
PAGE 29
SITE INVESTIGATION
4. DYNAMIC CONE PENETRATION TEST
•
•
The preliminary use is in cohesionless soils when static penetration test is difficult to perform. The aim is to determine the effort required to force a point through the soil and so obtain resistance value.
The cone is driven into the soil by blows of 65 kg hammer falling from a height of 750 mm. The blow count for every 30cm penetration is made to get continuous record of the variation of the soil consistency with depth.
•
•
The sufficient circulation of the bentonite slurry is necessary for elimination of the friction on the rods. (Bentonite slurry is not used when the investigation is required up to 6m only).
PAGE 30
Pictures
SITE INVESTIGATION
THE PROCEDURES OF DCP TEST https://youtu.be/U4pzkNKBEfk PAGE 31
5. PLATE BEARING TEST
•
SITE INVESTIGATION
The Plate Bearing Test (or Plate Loading Test) is an in-situ load bearing test of soil used for determining the ultimate bearing capacity of the ground and the likely settlement under a given load. •
Generally referred as 'Plate Load Test'
The test consists of loading a steel plate at foundation level and the settlement due to each load increment are recorded.
•
•
Thus, the ultimate bearing capacity of soil is obtained by dividing the total load on the plate. •
The plates used in the test are made of steel generally with dimensions of: 1. 25 mm thick and 150-722 mm 2. square plates with 305 mm x 305 mm
PAGE 32
SITE INVESTIGATION
PICTURES
Dial Gauge Reference B earn
Figure 2.4: Equipment for Plate Bearing Test
PAGE 33
SITE INVESTIGATION
THE PROCEDURES OF PLATE BEARING TEST https://youtu.be/6Z1JrhsdD1M PAGE 34
SITE INVESTIGATION
METHOD OF SOIL SAMPLE
1. Disturbed soil sample 2. Undisturbed soil sample
A)
DISTURB SAMPLE
The structure of the soil is disturbed to the considerable degree by the action of the boring tools or the excavation equipment
The disturbances can be classified in following basic types: 1.Change in the stress condition 2. Change in the water content and the void ratio 3. Disturbance of the soil structure 4. Chemical changes PAGE 35
B)
UNDISTURBED SAMPLE
SITE INVESTIGATION
It retains as closely as practicable the true in-situ structure and water content of the soil. For undisturbed sample the stress changes cannot be avoided. Undisturbed samples are generally taken by cutting blocks of soil or rock, or by pushing or driving tubes into the ground.
The following requirements are looked for: 1. No change due to disturbance of the soil structure, 2. No change in void ratio and water content, 3. No change in constituents and chemical properties.
PAGE 36
TOPIC 3 BASIC CHARACTERISTICS OF SOIL
OBJECTIVE At the end of the topic, student will be able to analyse the size of the soil particle by wet and dry sieve methods. Define plastic limit (PL), liquid limit (LL) and shrinkage limit (SL) clearly Compute the PL and LL using Casagrande & Cone Penetration test method Calculate the Plasticity Index (PI), Liquid Index (LI) and Activity Index (A)
BASIC CHARACTERISTICS OF SOIL (SOIL PHASE RELATIONSHIP)
BASIC CHARACTHERISTICS OF SOIL let's identify what's in a soil other than soil particles itself
Soil particle Air
Water
Figure 3.1: Soil Diagram
to get a picture, lets transfer into a soil phase
VOLUME
MASS
AIR
WATER
SOIL
Figure 3.2 : Soil Phase Relationship
PAGE 38
BASIC CHARACTERISTICS OF SOIL (SOIL PHASE RELATIONSHIP)
SOLVING PROBLEM
Question:
Solution:
PAGE 39
BASIC CHARACTERISTICS OF SOIL (SOIL PHASE RELATIONSHIP)
(2.65) ( 9.81 :�) G s Y--1 w =-------1 = 0.96 e)e =Ydry 13.24 � m
t) n = 1
e
+e
=1
0.96
+ 0.96 =
0·49
Vw mG 5 0.125(2.65) = ---- = 0.35 g) S = - = e 0.96 t{,
Where, 3 .4kg Vw. __ M·. w __ .,.,. Pw 1000� m
= 3.4x10- 3 m 3
From,
s = Vv.w V
So, .A =
3 3 3 Vv - Vw = (9.,71x10- - 3.4x10- )m = 0.34 V.T 18 , 3 X ]0-3 ill3
PAGE 40
BASIC CHARACTERISTICS OF SOIL (SOIL CLASSIFICATION METHOD)
INTRODUCTION • • •
Soil classification is a necessary method in Geotechnical Engineering to provide a conventional classification of types of soil for the purpose of describing the various materials encountered in site exploration. From this classification can determine whether the soil is gravel, sand, or silt. In site investigation specification BS5930:1981, soil particle classification shows in Table 3.1 Table 3.1: Soil particle size distribution table (BS 5390 : 1981) Very Coarse Soils G Gravel Course Soils
Fine Soils
Boulders Cobbles
S Sand M Silt C Clay
Course (C) Medium (M) Fine (F) Course (C) Medium (M) Fine (F) Course (C) Medium (M) Fine (F)
> 200 mm 60 – 200 mm 20 – 60 mm 6 – 20 mm 2 – 6 mm 0.6 – 2.0 mm 0.2 – 0.6 mm 0.06 – 0.2 mm 0.02 – 0.06 mm 0.006 – 0.02 mm 0.002 – 0.006 mm < 0.002 mm
PARTICLE SIZE ANALYSIS Hydrometer The Hydrometer method covers the quantitative determination of the particle size distribution in a soil from the coarse sand size to the clay size by means of sedimentation. The test is normally not required if less than 10 % of the material passes the 75 mm test sieve in a wet or dry sieving analysis. Distribution of soil particles having sizes less than 75 micron (Fine Grained soils) is often determined by a sedimentation process using a hydrometer to obtain the necessary data such as the borderline between clay and silt. Using this test the GSD or grain size distribution for soils containing appreciable amount of fines is obtained. Dry sieve Sieve analysis is used to determine the distribution of the larger grain sizes. The soil is passed through a series of sieves with the mesh size reducing progressively, and the proportions by weight of the soil retained on each sieve are measured.
PAGE 41
BASIC CHARACTERISTICS OF SOIL (SOIL CLASSIFICATION METHOD)
CHARACTERISTICS OF GRADATION The shape and position of the grading curve are used to identify some characteristics of the soil.
Figure 3.4 Grading curve Some typical grading curves are shown in Figure 3.1. The following descriptions are applied to these curves: W : Well-graded material U : Uniform material P : Poorly graded material or gap graded material C : Well graded with some clay F : Well graded with an excess of fines
COEFFICIENT OF UNIFORMITY (CU) AND COEFFICIENT OF CURVATURE (CC) Two parameters can be determined from the grain-size distribution curves of coarse-grained soil. These coefficients are D60 Uniformity coefficient, Cu = D10 Coefficient of curvature, Cc = Where,
D10 D30 D60
= = =
Per cent finer than 10% Per cent finer than 30 % Per cent finer than 60 %
PAGE 42
BASIC CHARACTERISTICS OF SOIL (SOIL CLASSIFICATION METHOD)
From the numbers of Cu and Cc parameter, type and grade of soil can be determined as; Cu > 4 Well graded gravel 1 ≤ Cc ≤ 3 Cu > 6 Well graded sand 1 ≤ Cc ≤ 3 If Cu or Cc does not fall in that range, so that the soil sample has poorly grade
Example 1 A sieve analysis of soil was tested in the laboratory, and the results of the laboratory test shows in Table 3.2. Determined the value of Cu and Cc, and conclude the type of that soil. Table 3.2: Sieve analysis result Sieve size (mm)
Mass retain (g)
5.0
11
2.36
18
1.18
24
0.6
21
0.3
41
0.212
32
0.15
16
0.063
22
Pan
20 Solution
From the data given, percentage of soil passing for each sieve calculated as Table 3.3 Table 3.3: Calculation for percentage passing Sieve size (mm)
Mass retained (g)
Mass passing (g)
% Passing
5.0
11
205 – 11 = 194
(194/205)100 = 95
2.36
18
194 – 18 = 176
(176/205)100 = 86
1.18
24
176 – 24 = 152
(152/205)100 = 74
0.6
21
152 – 21 = 131
(131/205)100 = 64 PAGE 43
BASIC CHARACTERISTICS OF SOIL (SOIL CLASSIFICATION METHOD)
0.3
41
131 – 41 = 90
(90/205)100 = 44
0.212
32
90 – 32 = 58
(58/205)100 = 28
0.15
16
58 – 16 = 42
(42/205)100 = 20
0.063
22
42 – 22 = 20
(20/205)100 = 10
Pan
20
20 – 20 = 0
(0/205)100 = 0
∑ = 205 A graph was plotted on semi-log paper from the result from Table 3.2. After graph was plotted, type and grad of soil can be determined. Figure 3.3: Sieve analysis graph
PAGE 44
BASIC CHARACTERISTICS OF SOIL (SOIL CLASSIFICATION METHOD)
From semi-log graph, the result were as follows: % Sand = % passing size of 2.35mm - % passing size of 0.06 mm = 83 % – 10 % = 73 % > 50% (Sand) D10
=
0.63,
D30
=
0.22,
Therefore ,
D60
=
0.52
D60 D10 0.52 = 0.063 = 8.25 > 6
The uniformity coefficient, Cu
=
The coefficient of gradation, Cc
= = =
D302 D10 D60 0.222 (.063)(0.52) 1.5 ( 1 < Cu < 3)
Hence, the sample is well-graded sand
Activity Test your understanding before proceeding further input. Please check your answers in the feedback on the next page A sieve analysis test was conducted at geotechnical laboratory and the data has shown below: Sieve size (mm) 37.5 20.0 10.0 5.0 2.36 1.18 0.6 0.3 0.212 0.15 0.063 Pan
Mass retained (g) 0 59 38 33 27 30 22 15 17 16 9 11
PAGE 45
BASIC CHARACTERISTICS OF SOIL (SOIL CLASSIFICATION METHOD)
Determine: i.
Plot the grading curve on semi log paper.
ii.
Calculate the value of uniformity coefficient (Cu) and coefficient of gradation (Cc)
iii.
Summarize type of the soil from grading curve, value of Cu and value of Cc.
Solution
Sieve Analysis % P a s s i n g
Sieve size (mm) SAND
GRAVEL
From semi-log graph: Per cent of gravel = 60 % > 50% D10 = 0.18 Cu = D60/D10
D30 = 0.95
D60 = 7.5
= 7.5/0.18 = 41.67
Cc = (D30)2/(D10D60) = (0.95)2(0.18x7.5) = 0.67 Cu > 4 and Cc < 1 Therefore, type of soil is well-graded gravel (GP)
PAGE 46
BASIC CHARACTERISTICS OF SOIL (SOIL CLASSIFICATION METHOD)
Self-Assessment Test your understanding before proceeding further input.
Question 1 List down types of the test will be conducted in particle size analysis. Question 2 List down different between wet sieve and dry sieve to determine the distribution of course and fine-grained soil. Question 3 According to the results of a sieve analysis in table below, plot the sieve analysis graph and construct a soil particle-size distribution curve Sieve size (mm) 4.75 2.00 0.850 0.425 0.250 0.180 0.150 0.075 Pan Determine
Mass of soil retained (g) 0 40 60 89 140 122 210 56 12
a. Effective size D10, D30 and D60 b. Uniformity coefficient, Cu c. Coefficient of curvature, Cc d. Classify the soil based on the graph
PAGE 47
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
INTRODUCTION The liquid limit can best be explained by referring to Figure 1. The specimen is very soft clay with high water content. The soil is liquid and very soft. It offers no shearing resistance and flows like liquid. As the water contents decrease, it becomes stiffer and starts to exhibit resistance to deformation. The soil is now in the plastic stage. The water content of the soil at this is called the liquid limit (LL). Liquid limit is the water content in soil at the point of transition from liquid to plastic.
Solid
Semi solid
Shrinkage limit SL
Plastic
Liquid
Liquid limit Plastic LL limit PL Figure 3.5 Atterberg’s Limit
Water content
As the water content of soil is reduced, the soil becomes semi-solid. The soil tends to crack when moulded. The water content of soil at this stage is termed the plastic limit (PL). Plastic limit is defined as the moisture content of soil at the same transition from plastic to semi-solid. As the water content is further reduced, the volume decreases until a stage is reached where, no further volume change occurs. At this stage, the soil has reached the solid state. The shrinkage limit (SL) is defined as the water content in soil where the soil changes from semi-solid state to solid state. Liquid limit (LL) Liquid limit is the minimum water content to the soil can be flows like liquid. Liquid limit can be obtained in the laboratory using Casagrande apparatus or the cone penetrometer Casagrande Method
Figure 3.6: Casagrande Apparatus for Liquid Limit PAGE 48
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Example 1
After a series of laboratory liquid limit test using Casagrande method, the following data were established for an organic soil. Given plastic limit is 32%. Determine liquid limit Test No
1
2
3
4
Mass of soil + can (g)
48.61
55.53
51.71
50.51
Mass of dry soil + can (g)
41.19
46.05
42.98
41.54
Mass of can (g)
17.33
17.41
17.45
17.36
Number of Blows
34
27
22
17
1
2
3
4
48.61
55.53
51.71
50.51
41.19
46.05
42.98
41.54
17.33
17.41
17.45
17.36
7.42
9.48
8.73
8.97
23.86
28.64
25.53
24.18
31.1
33.1
34.2
37.1
34
27
22
17
Solution: First, calculate the value of water content. Test No Mass of soil + can (g) W1 Mass of dry soil + can (g) W2 Mass of can (g) W3 Mass of water (W1 – W2) Mass of solid (W2 – W3) Water content W = Mw x 100 Ms Number of Blows
PAGE 49
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Second, plot a graph of water content versus number of blows
Value of moisture content that corresponds to 25 blows is taken as soil liquid limit (LL) From the graph. 25 blows = 33.8% of moisture content
Cone penetrometer method
To analyse, plot relationship between moisture content with cone penetration. Moisture content that corresponds to cone penetration of 20 mm is taken as soil liquid limit (LL).
Figure 3.7: Cone Penetration Apparatus PAGE 50
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Example 2
Table shows the results of a cone penetration test conducted on a cohesion soil sample. Determine the liquid limit of the soil. Average penetration (mm) 15 17.5 19 21 22.5 Average moisture content (%) 35 45 50 60 67 Solution: Plot a graph of penetration versus moisture content on a graph paper. Penetration versus Moisture Content Graph
Moisture content that corresponds to cone penetration of 20 mm is taken as soil liquid limit (LL). From the graph; 20 mm penetration = 55% of moisture content
PAGE 51
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Plastic limit determination (PL) Plastic limit is the minimum moisture content that the soil can be rolling into a 3.5 mm diameter thread until its starts to crack.
Figure 3.8: Plastic limit determination procedure
Example 3 The following results were recorded during Plastic limit test on a cohesive soil. Test no. Mass of can (g) Mass of wet soil + Mass of dry soil + can (g) can (g) 1 8.1 20.7 18.7 2 8.4 19.6 17.8 Calculate plastic limit of the soil Solution: Mw w1 = x100% Ms 20.7 − 18.7 w1 = x100% = 18.87% 18.7 − 8.1 Mw x100% Ms 19.6 − 17.8 w2 = x100% = 19.15% 17.8 − 8.4 w2 =
So the value of plastic limit is, PL =
18.87 + 19.15 = 19.01% 2
PAGE 52
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Shrinkage Limit (SL) Soil shrinks, as moisture is gradually lost from it. With continuing loss of moisture, a stage of equilibrium is reached at which more loss of moisture will result in no further volume change.. The moisture content, in percent at which the volume of the soil mass stops to change is defined as the shrinkage limit.
Figure 3.9: Shrinkage Limit Test: (a) Soil Pat Before Drying; (b) Soil Pat After Drying The shrinkage limit can be determined as; Shrinkage limit, SL Where, wi
= =
w
=
wi
=
m1 m2
= =
However,
Where,
Also,
Where,
w
Vi
=
=
wi (%) - w (%) (5.2) Initial moisture content when the soil is placed in the shrinkage limit dish Change in moisture content (that is, between the initial moisture content and the moisture content at the shrinkage limit)
m1 − m2 m2
x 100
(5.3)
Mass of wet the wet soil pat in the dish at the beginning of the test (g) Mass of the dry soil pat
(Vi - Vf )rw m2
x 100
(5.4)
Initial volume of the wet soil pat (that is, inside volume of the dish, cm3)
Vf
=
Volume of oven-dried soil pat (cm3)
rw
=
Density of water (g/cm3) PAGE 53
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Plasticity index (PI) The plasticity index (PI) is the difference between the liquid limit and the plastic limit of a soil, or PI
=
Liquid limit (LL) – Plastic limit (PL) PI 0 1-5 5-10 10-20 20-40 >40
(5.5)
Description Non plastic Slightly plastic Low plasticity Medium plasticity High plasticity Very high plasticity Burmister (1949) Example 4
Liquid limit and plastic limit test was conducted on a sample of non-organic silt soil. The results for liquid limit and plastic limit respectively 45% and 35%. Determine the plasticity index of that soil. Solution: Given: LL = 45% and PL = 35% Plasticity Index, PI
= = =
Liquid limit (LL) – Plastic limit (PL) 45 % - 35 % 10 % Liquid index (LI)
The relative consistency of a cohesive soil in the natural state can be defined by a ratio called the liquidity index, which is given by
(w) - (PL) (PI) Where w = in situ moisture content of soil. If LI > 1 ➔ these soils, when remoulded, can be transformed into a viscous to flow like a liquid Liquid index, LI
=
Soil deposits that are heavily over consolidated may have a natural moisture content less than plastic limit. PAGE 54
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Example 5 A sample of soil from site construction has a moisture content of 33%. The values of liquid limit and plastic limit test respectively 5% and 31%. Determine liquid index (LI) from the data given. Solution: Given : w = 33%; LL = 5%; PL = 31% Liquid index, LI
=
=
Initial moisture content (w) - Liquid Plastic (PL) Plasticity Index (PI) 33% − 31% 45% − 31%
=
0.142
Activity (A) Activity is used as an index for identifying the swelling potential of clay soil. This activity may be expressed as: Activity, A
=
Plasticity Index (PI ) % of clay size friction, by weight
(5.7)
Example 6 Liquid limit and plastic limit test was conducted on organic soil and the result as follow; Liquid limit = 45% Plastic limit = 35% From hydrometer test, the size of soil particle less than 2um is 20%. Determine activity for that soil. Solution Activity, A
=
= =
Plasticity Index (PI) % clay soil 45% − 35% 20% 0.5
PAGE 55
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Plasticity chart A plasticity chart is shown in figure 3.10. It is based on experimental data obtained worldwide. It was found that clay, silt, and organic soil line within a distinctive section of the graph. A line called A line separate the clay from silts and organic soil. From the graph below, it may be seen that clay lies above the A line while silt and organic soil lies below the A line.
Figure 3.10: Plasticity Chart Where,
CL CH ML MH OL OH
= = = = = =
Low plasticity clay High plasticity clay Low plasticity silt High plasticity silt Low plasticity organic soil High plasticity organic soil
Example 7 Liquid limit and plastic limit test were conduct on organic clay soil and the result shown below Liquid limit = 45% Plastic limit = 35% Determine Plasticity Index of the soil. Solution Plasticity Index, PI = = =
Liquid limit (LL) – Plastic limit (PL) 45 % - 35 % 10 %
PAGE 56
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
From the plasticity chart, type of soil is ML or OL
Self-Assessment Test your understanding before proceeding further input. Please check your answers in the feedback on the next page Question 1 Given that the liquid limit and plastic limit of soil is 58% and 30%. By using Plasticity Chart, classify the type of soil. Question 2 After a series of laboratory test, the following data was established for fine soil:
i. ii.
Liquid limit 48% Plastic limit 32% Clay content 24.2% Calculate the activity of the soil Determine the liquidity index of the soil when its natural moisture content is 39%
PAGE 57
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Question 3 The following results were recorded during a Casagrande Test and Plastic limit test on a cohesive soil. No, of blow (n) 98 70 55 35 18 Water content (%)
13
The result from the plastic limit test were Test no. Mass of can (g) 1 2
8.1 8.4
24
32
Mass of wet soil + can (g) 20.7 19.6
45
59.4
Mass of dry soil + can (g) 18.7 17.8
Determine the following: a. Plot a graph moisture content versus no of flow b. Calculate liquid limit, plastic limit and Plasticity Index of the soil Question 4 In a liquid limit test on a fine-grained soil using a cone penetrometer, the following results were recorded. Determine the liquid limit and plasticity index of the soil and classify it according to the British Soil Classification System. The plastic limit was found to be 25%. Cone penetration (mm) Water content %
15.9 32.6
17.7 42.9
19.1 51.6
20.3 59.8
21.5 66.2
Question 5 After a series of laboratory liquid limit test using Casagrande method, the following data were established for a non organic soil. Given plastic limit is 52%, water content 55% and percent of clay soil is 40%. Determine i. Liquid limit ii. Plasticity index iii. Liquid Index iv. Activity v. Describe the type of soil Test No Mass of soil + can (g) Mass of dry soil + can (g) Mass of can (g) Number of Blows
1 48.2 36.3 16.3 28
2 50.3 37.1 15.7 22
3 54.4 39.9 16.9 19
4 48.1 35.4 16.1 16
PAGE 58
BASIC CHARACTERISTICS OF SOIL (ATTERBERG LIMIT)
Answer Question 1 PI = 28% Type of soil = MH Answer Question 2 Activity of soil, AI = 0.66 Liquidity index = 0.44 Answer Question 3 Liquid limit (LL) = 54% Plastic limit = 19.01 Plasticity index of the soil = 34.99 Answer Question 4 P.I = 32 Type of soil is CH Answer Question 5 LL = 61% PI = 9% Liquid Index (LI) = 0.33 Activity (A) = 0.225 Type of soil = MH
PAGE 59
BASIC CHARACTERISTICS OF SOIL (COMPACTION)
COMPACTION OF SOIL
WHAT IS COMPACTION?
In geotechnical field, compaction is a process of densification of soil mass by reducing the air voids. WHAT'S THE OBJECTIVES?
This process is to improve the shear strength of soil by determine the suitable water amount at which can be figure out from the compaction graph. The suitable amount of water was known as Optimum Moisture Content. Refer to the Figure 3.
Figure 3.3 : The Relationship Between Dry Density and Water Content
TYPES OF COMPACTION METHOD IN LABORATORY STANDARD PROCTOR COMPACTION TEST
MODIFIED PROCTOR COMPACTION TEST
Hammer mass=2.5kg Hammer drop =300mm
Hammer mass=4.9kg Hammer drop=450mm
PAGE 60
BASIC CHARACTERISTICS OF SOIL (COMPACTION)
SOLVING PROBLEM
Question:
Solution:
PAGE 61
TOPIC 4: SHEAR STRENGTH
OBJECTIVE Define shear strength of soil and it parameters; cohesion and friction angle. Explain Coulomb and Mohr Circle. Apply the parameter of shear strength.
SHEAR STRENGTH
WHAT IS SHEAR STRENGTH OF SOIL? Shear strength is a force that resist the soil to failure in a shear. The shear came from the movement between a grain’s particle itself and the strength occur from a friction from that movement. The wrong selection of strength value will become disaster where the foundation can’t support the structure due to weak of soil strength.
DOES IT RELATED WITH THE SOIL STRENGTH? Yes, it's related. Coarse grain and fine grain have a different movement ability for a water through between the grain particle and can be stated as a drained and undrained of soil. Coarse grain consists of sand and gravel, while fine grain consists with a clay and sand. Generally, coarse grain has more ability for water through the particles and it’s called as a drained soil instead of fine coarse which is undrained. So that, soil strength is rely on these criteria of soil.
WHAT ABOUT PARAMETER FOR SHEAR STRENGTH?
Ø = friction angle C = cohesion
WHERE THESE PARAMETERS COME FROM? Both parameters will get from a Mohr-Columb Graph that resulted by Direct Shear Test or Triaxial Test. This terms will be explain deeper at the next page.
PAGE 63
SHEAR STRENGTH
INTRODUCTION TO MOHR-COLUMB METHOD Normally, this method is used to identify the critical material that will fail during or after load are subjected to the ground. There are 2 types of stresses that are used for this method analysis, which is vertical stress and horizontal stress.
HOW THIS STRESSES CAN BE RELATE WITH THIS METHOD? Imagine the square box of diagram was a soil just like a Figure 1.
Figure 4.1: The Stress Direction
re
at
th
e
so il st is ,
Lets place the Figure 1 under the ground and there's no load on the ground surface. The load came from the soil itself.
PAGE 64
SHEAR STRENGTH
Now, place a load. Lets say, a building at the ground surface
FAILURE PLANE AT MOHR GRAPH Basically, the failure plane for Mohr graph was differentiate according to the grain of soil. Refer to Figure 2 to understand the differentiate of it.
Figure 4.2: The position of failure plane Able to identify why the failure plane has a different position?
PAGE 65
SHEAR STRENGTH
Yes, it's because of the parameters
WHAT IF CONSIST BOTH COARSE AND FINE GRAIN?
re The failu going plane is e this to be lik
THE LABORATORY TEST RELATED WITH THIS PARAMETERS? DIRECT SHEAR TEST may use a disturbed soil sample
TRIAXIAL TEST may use undisturbed soil sample
PAGE 66
SHEAR STRENGTH
SOLVING PROBLEM DIRECT SHEAR TEST: QUESTION: Refer to table below. The raw data collected from a direct shear test and calibration factor given is 0.02 kN/div. The mould sample size was 60mmx60mm. Using the data given, determine the parameters.
SOLUTION:
PAGE 67
SHEAR STRENGTH
TRIAXIAL TEST: The table below showed a raw data collection from a triaxial test and the calibration factor given is 1.4 N/div. The undisturbed sample was used in this testing and diameter and length was 37.5mm diameter, 75mm respectively. Determine the parameters.
SOLUTION:
PAGE 68
TOPIC 5
FOUNDATION
OBJECTIVES To explain foundation in general
FOUNDATION
INTRODUCTION • The lowest part of a structure is often referred to as the foundation • Its function is to supports a structure i.e transfer load of a structure to the soil on which it is resting. • It is important that the foundation is properly designed to avoid overstressing the soil • Overstressing of the soil may lead to shear failure of the soil or may cause unnecessary settlement of the structure.
DEFINE FOUNDATION • Foundation refers to something that supports a structure, such as a column or wall, along with the load carried by the structure
Shallow foundation • The depth of foundation is less than or equal to the width of foundation. • A foundation is an important part of a structure • The stability of a structure depends the stability of the supporting soil.
Deep foundation • A pile foundation is a relatively long and slender member • The structural loads may be transferred to deeper firm strata (pile). • Pile may be subjected to vertical or lateral loads or combination of vertical and lateral loads. PAGE 70
FOUNDATION
Deep foundations transfer the load by taking into account the side friction between the soil and footing. The bearing capacity of deep foundation is increased because of side friction, in addition to usual bearing of the sub soil strata.
Shallow foundation (with most type as spread footing) actually spreads the load of the structure over a greater area by the footing pad and it rely on the direct contact pressure between footing and near to surface soil/rock
Defferent of foundation Wall of Column
More than 3m Are commonly used for smaller projects and when the top layer of soil can adequately handle the distribution of weight
Transfer the load down to layer of substrata bedrock to ensure structural integrity
SHALLOW FOUNDATION
Level of firmer stratum
DEEP FOUNDATION
PAGE 71
FOUNDATION
Types of Foundation In general, all foundations are divided into two categories, - shallow and deep foundations. The terms Shallow and Deep Foundation refer to the depth of the soil at which it is placed. Generally, if the width of the foundation is greater than the depth, it is labelled as the “Shallow Foundation”. If the width is smaller than the depth of the foundation it is called as “Deep Foundation.” However, deep foundation and shallow foundation can be classified as shown in the following chart.
In general, all foundations are divided into two categories, - shallow and deep foundations. The terms Shallow and Deep Foundation refer to the depth of the soil at which it is placed. Generally, if the width of the foundation is greater than the depth, it is labeled as the “Shallow Foundation”. If the width is smaller than the depth of the foundation it is called as “Deep Foundation.” However, deep foundation and shallow foundation can be classified as shown in the following chart.
TYPES OF FOUNDATION
Shallow foundation
Deep foundation
Isolated sperad footing Wall Footing or Strip footing Combined Footing Cantilever or Strap Footing Raft or Mat Foundation
Pile foundation Pier Foundation Caisson Foundation
PAGE 72
FOUNDATION
Shallow Foundation Isolated Spread Footing This is the most widely recognized and most straightforward shallow foundation type, as this is the most economical type. They are typically utilized for shallow establishments to convey and spread concentrated burdens caused, for instance, by pillars or columns. They are generally used for ordinary buildings (Typically up to five stories).
Isolated footing comprises a foundation directly at the base of the segment. Generally, every section has its footing. They straightforwardly transfer the loads from the column to the soil. It might be rectangular, square, or roundabout. It can comprise both reinforced or non-reinforced material. For the non-reinforced footing, however, the stature of the footing has to be more prominent to give the vital spreading of the load. They should possibly be utilized when it is sure beyond a shadow of a doubt that no differing settlements will happen under the whole structure. Spread footings are inadmissible for the orientation of large loads. It is given to lessen the twisting minutes and shearing powers in their primary areas. The size of the footing can be roughly calculated by dividing the total load at the column base by the allowable bearing capacity of the soil. The followings are the types of spread footing. i. ii. iii. iv. v. vi.
Single pad footing. Stepped footing for a column. Sloped footing for a column. Wall footing without step. Stepped footing for walls. Grillage foundation. PAGE 73
Types of Spread Footing
FOUNDATION
Pad Footing
Stepped column
Sloped column PAGE 74
Types of Spread Footing
FOUNDATION
Wall footing
Stepped Walls
Grillage Foundation
PAGE 75
FOUNDATION
Wall Footing or Strip footing Wall footing is also known as continuous footing. This type is used to distribute loads of structural or non- structural load-bearing walls to the ground in such a way that the load-bearing limit of the soil isn't outperformed. It runs along the direction of the wall. The width of the wall foundation is usually 2-3 times the width of the wall. The wall footing is a continuous slab strip along the length of the wall. Stone, brick, reinforced concrete, etc. are used for the construction of wall foundations. •
•
•
On account of block walls, the footing comprises a few courses of bricks, the least course being generally double the expansiveness of the wall above. On account of stone masonry walls, the counterbalances could be 15 cm, with the statues of the course as 30 cm. Along these lines, the size of footings is marginally more than that of the block divider footings. If the heap on the wall is substantial or the soil is of low bearing limit, this reinforced concrete foundation type can be given.
STRIP FOUNDATION
PAGE 76
FOUNDATION
Combined Footing The combined footing is very similar to the isolated footing. When the columns of the structure are carefully placed, or the bearing capacity of the soil is low and their footing overlap each other, combined footing is provided. It is fundamentally a blend of different footings, which uses the properties of various balances in a single-footing dependent on the necessity of the structure.
The foundations which are made common to more than one column is called combined footings. There are different types of combined footing, including slab type, slab and beam type, rectangular, raft, and strap beam type. They may be square, tee-shaped, or trapezoidal. The main objective is the uniform distribution of loads under the entire area of footing, for this is necessary to coincide with the centre of gravity of the footing area with the centre of gravity of the total loads.
COMBINED FOOTING
PAGE 77
FOUNDATION
Cantilever or Strap Footing Strap footings are similar to combined footings. Reasons for considering or choosing strap footing are identical to the combined one.
In strap footing, the foundation under the columns is built individually and connected by a strap beam. Generally, when the edge of the footing cannot be extended beyond the property line, the exterior footing is connected by a strap beam with interior footing.
CANTILEVER FOOTING
PAGE 78
FOUNDATION
Raft or Mat Foundation Raft or Mat foundations are used where other shallow or pile foundations are not suitable. It is also recommended in situations where the bearing capacity of the soil is inadequate, the load of the structure is to be distributed over a large area or structure is subjected continuously to shocks or jerks. Raft foundation consists of a reinforced concrete slab or T-beam slab placed over the entire area of the structure. In this type, the whole basement floor slab acts as the foundation. The total load of the structure is spread evenly over the entire area of the structure. This is called raft because, in this case, the building seems like a vessel that floats on a sea of soil.
RAFT FOUNDATION
PAGE 79
FOUNDATION
Deep Foundations 1. Pile Foundation Pile is a common type of deep foundation. They are used to reduce cost, and when as per soil condition considerations, it is desirable to transmit loads to soil strata which are beyond the reach of shallow foundations. The followings are the types of pile foundations :-
BASE ON FUNCTION OR USE
BASE ON METERIALS OR CONSTRUCTIO N METHOD
Sheet Piles Load Bearing Piles End Bearing Piles Friction Piles Soil Compactor Piles
Timber Piles Concrete Piles Steel Piles Composite Piles
PAGE 80
FOUNDATION
2. Pier Foundation Pier is an underground structure that transmits a more massive load, which cannot be carried by shallow foundations. It is usually shallower than piles. The pier foundation is generally utilized in multi-story structures. Since the base region is determined by the plan strategy for the regular establishment, the single pier load test is wiped out. Along these lines, it is increasingly well known under tight conditions.
PAGE 81
FOUNDATION
3. Caisson Foundation Caisson foundation is a watertight retaining structure used as a bridge pier, construction of the dam, etc. It is generally used in structures that require foundation beneath a river or similar water bodies. The reason for choosing the caisson is that it can be floated to the desired location and then sunk into place.
PAGE 82
FOUNDATION
Load transfer mechanism of foundation
Load transfer mechanism of shallow foundation da
The dropping of Zone I creates two zones of plastic equilibrium, II and III, on either side of the footing. Zone II is the radial shear zone whose remote boundaries bd and af meet the horizontal surface at angles (45° - 0/2), whereas Zone III is a passive Rankine zone. The boundaries de and fg of these zones are straight lines and they meet the surface at angles of (45° - 0/2). The curved parts cd and cf in Zone II are parts of logarithmic spirals whose centers are located at b and a respectively.
Load transfer mechanism of deep foundation
c) Load- settlement curves a) Single pile
b) Load Transfer curves
When the ultimate load applied on the top of the pile is Qu, a part of the load is transmitted to the soil along the length of the pile and the balance is transmitted to the pile base. The load transmitted to the soil along the length of the pile is called the ultimate friction load or skin load Qf and that transmitted to the base is called the base or point load Qb. PAGE 83
FOUNDATION
Example Transfer Load of deep foundation
▪
The base of an end bearing pile rests on a relatively firm (strong) soil such as rock, very dense sand or gravel and the base of pile bears the load of the structure
▪
The load of the structure is transmitted through the pile into this firm soil
▪
This type of piles transmits load of the structure by skin friction or cohesion between the soil and the embedded surface of pile
▪
Use if the firm soil is at a considerable depth which it becomes too expensive to use and bearing pile
▪
Predominate in clays and silts
▪
May be used to support a downward or an upward load
PAGE 84
FOUNDATION
Function of foundation Transfer the load to the soil layers safely
To prevent soil settlement
Prepare a flat surface for construction work Enhance / increase the stability of the building
Prevent the structure from vibration or erosion of sediment PAGE 85
FOUNDATION
Factors effecting the requirement of the foundation Structural requirement Internal condation of soil
Size of the project Strength of foundation (durability) Environmental factors Cost/economy Implementation of work PAGE 86
FOUNDATION
Criteria design of foundation -
Soil bearing capacity and depth of foundation
-
-
-
Soil bearing capacity is refer to the ability of soil to supporting the load from structure The safe bearing capacity is the maximum pressure which the soil can carry safely without risk of shear failure or excessive settlement The value of the safe bearing capacity is ratio between the net ultimate bearing capacity and suitable factor of safety The depth of the foundation is less than or equal to the width of the foundation Minimum depth requirement for JKR standard is 1.5m
Total settlement is the magnitude of downward movement while differential settlement is the difference in vertical movement between various locations of the structure and distorts the structure Limitations to total and differential settlement depend on the function and type of structure Total settlement should not exceed 50mm for most facilities A typical specification of total settlement for commercial buildings is 25mm -
Requirement of Factor of Safety
-
Settlement limit that is total Settlement and differential settlement
Factor of safety needed when uncertainty soil condition and inadequacies may occur in the construction Factor of safety apply for reduce the probability failure of foundation Factor of safety apply to protection of natural changes in soil shear strength A factor of safety of 2.5 to 3 is commonly applied to the value of qu (bearing capacity)
PAGE 87
FOUNDATION
Test your understanding before your move to the next input......
ACTIVITY You can go to next topic if you can answer all the question in this activity
QUESTION 1
Define the shallow and deep foundation?
State THREE types of shallow foundation?
QUESTION 3
QUESTION 2
Determine THREE criteria to the design of foundation?
PAGE 88
TOPIC 6 EFFECTIVE STRESS & LATERAL EARTH PRESSURE
OBJECTIVES To analyze soil stress, and determine changes in stress, calculate lateral earth pressure and calculate the active and passive pressure for cohesion and cohesion less soil.
Introduction to Stress in Soil
EFFECTIVE STRESS
TERMINOLOGIES Stress is an intensity of loading. Strain is the measurement of deformation. Loading is the mass that applied to soil. There are two types of loading that are; external load and internal load (self-weight of soil). Soil stresses are consisting of 3 types of soil stresses; vertical stress, horizontal stress and effective stress. Factors that affected soil stress; soil types, human activities, construction, soil water (ground water level) and loading activities. Stress Changes is cause by self-weight of soil, changes of water capillary and water level. Factors that affected soil stress; soil types, human activities, construction, soil water and loading. Pore Pressure is the water in the pores of a soil called pore water. The pressure within this pore water is called pore pressure, which usually expressed as U. Effective Stress is the difference between the total stress (vertical stress) and pore pressure. PAGE 90
EFFECTIVE STRESS
VERTICAL STRESS, σv The total vertical stress acting at a point below the ground surface is due to the weight of everything lying above: soil, water, and surface loading. Total stresses are calculated from the unit weight of the soil. Any change in vertical total stress (σv) may also result in a change in the horizontal total stress (σh) at the same point. Vertical stress, σv = Weight / Area (kN/m2) = ρgh Where, ρg = γ (unit weight of soil) So that, σv = γh
HORIZONTAL STRESS, σH
Horizontal stress is the product of horizontal stress coefficient (k) with the
vertical stress (σv) Horizontal stress, σh = kσv Where, k is determine using Rankine Formula: K = 1+ sin ø 1 – sin ø Where, ø = angle of soil friction
PORE PRESSURE, U Under hydrostatic conditions (no water flow) the pore pressure at a given point is given by the hydrostatic pressure: u = γw .h Where, h = depth below water table or overlying water surface γw = unit weight of water (1000kg/m3 or 9.81 kN /m3)
PAGE 91
EFFECTIVE STRESS, σv
EFFECTIVE STRESS
Ground movements and instabilities can be caused by changes in total stress (such as loading due to foundations or unloading due to excavations), but they can also be caused by changes in pore pressures (slopes can fail after rainfall increases the pore pressures). The difference between the total stress (vertical stress) and the pore pressure is called the effective stress Effective stress = Vertical Stress – Pore Pressure σ’= σv - U
Figure 6.1: The Principle of Effective Stress
Figure 6.2: The Soil Load Distribution Diagram of Effective Stress PAGE 92
MULTI LAYER VERTICAL STRESS, σv
EFFECTIVE STRESS
In a soil mass with three different soil layers, the vertical normal stress at X is
γ1h1 + γ2h2 + γ3h3
Figure 6.3: Overburden Stress at a Point in a Layered Soil
Example 1
A soil profile has a depth and unit weight as shown in the diagram below. Determine the vertical stress, pore pressure and effective stress at a depth of 10 meters.
Solution
PAGE 93
Example 2
EFFECTIVE STRESS
Plot the variation of total and effective vertical stresses, and pore water pressure with depth for the soil profile shown below
Solution
Graph Plotting
PAGE 94
EFFECTIVE STRESS
Tutorial 1
A 4m thick layer of sand is underlain by a layer of 6m stiff saturated clay. Ground water level is 1.5m from the surface layer. Unit weight of sand is 18.4 kN/m3, unit weight of saturated sand is 21.2 kN/m3 and clay 18.1 kN/m3. Calculate the total stress, pore water pressure and effective stress at 10m depth from the ground level. Plot the variation of total and effective vertical stresses, and pore water pressure with depth for the soil profile. Solution Depth (m) 0 1.5 4 10
σ (kN/m2) 0 18.4 x 1.5m = 27.6 27.6 + (21.2 x 2.5m) = 80.6 80.6 + (18.1 x 6m ) = 189.2
u (kN/m2)
σ’ (kN/m2)
0
0
0
27.6
9.81 x 2.5 = 24.53
56.07
9.81 x 8.5 = 83.39
105.82
Graph Plotting
PAGE 95
Tutorial 2
EFFECTIVE STRESS
On a certain site, a surface layer of sand is 4 m thick and this overlies a layer of clay 6 m thick. Ground-water level was located at 1.5 meters under the ground surface. Above the water table, unit weight of sand is 18.4 kN/m3, while the unit weight of saturated sand and clay are respectively 21.1 and 18.1 kN/m3. a) Calculate vertical stress, pore pressure and effective stress at a depth of 10 m under the ground surface. b) Draw the stress distribution for vertical stress, pore pressure and effective stress.
Tutorial 3 A surface layer of sand is 5 m thick and this overlies a layer of clay 4 m thick. Draw a stress distribution for vertical stress and effective stress against of these cases: ~ Ground Water level at 2 m from ground surface. ~ Ground water level (G.W.L) at the top of clay layer. Given: Unit weight of saturated sand, γsand = 20.9 kN/m3 Unit weight of dry sand, γdry = 17.4 kN/m3 Unit weight of clay, γclay = 17.8 kN/m3
PAGE 96
LATERAL EARTH
LATERAL EARTH PRESSURE Introduction to lateral earth pressure Lateral earth pressure is the pressure that soil exerts in the horizontal direction. The lateral earth pressure is important because it affects the consolidation behavior and strength of the soil and because it is considered in the design of geotechnical engineering structures such as retaining walls, basements, tunnels, deep foundations and braced excavations.
PAGE 97
LATERAL EARTH
Concept of active and passive pressure of soil The active state occurs when a retained soil mass is allowed to relax or deform laterally and outward (away from the soil mass) to the point of mobilizing its available full shear resistance (or engaged its shear strength) in trying to resist lateral deformation. That is, the soil is at the point of incipient failure by shearing due to unloading in the lateral direction. It is the minimum theoretical lateral pressure that a given soil mass will exert on a retaining that will move or rotate away from the soil until the soil active state is reached (not necessarily the actual in-service lateral pressure on walls that do not move when subjected to soil lateral pressures higher than the active pressure). While, the passive state occurs when a soil mass is externally forced laterally and inward (towards the soil mass) to the point of mobilizing its available full shear resistance in trying to resist further lateral deformation. That is, the soil is at the point of incipient failure by shearing, but this time due to loading in the lateral direction. Thus active pressure and passive resistance define the minimum lateral pressure and the maximum lateral resistance possible from a give mass of soil.
LATERAL PRESSURE IN COHESIONLESS SOIL, C=0 5.1.1
Rankine’s Rankine’s Theory – lateral earth pressure for stratum surface
Theory lateral
earth pressure for stratum surface
• A vertical and smooth wall was keeping the cohesionless soil that staying stratum out behind
the wall which is show in figure 4.1. Assume that unit weight of soil, and friction angle,
Vertical v ø is constant with depth. Vertical pressure, v distribution depth at soil pressure, is: distribution depth at soil
v = Z
• If the wall has been moved sufficiently to meet the active soil pressure, the v active = pressure, Z Pa in the Mohr circle in figure 6.1(a), v = Z is a major principles stress and P
The active pressure Mohr Circle
Pa = Ka Z Ka
Pa = 1/2 Ka Z²
=
1 2 1P−=sinK ,ZOr 1 + sin
Ka = tan2 (45 − ) 2
The active thrust behind the wall Surface failure
Ka Z Active Thrust Distribution PAGE 98
LATERAL EARTH
LATERAL PRESSURE IN COHESION SOIL, C = ? Rankine’s Theory for cohesion soil. What is cohesion soil? Is a clay soil or silt soil. Has a value of friction angle and shear strength of the soil. she
Pa
Ka Z surface failure
Active Thrust Distribution Diagram
The value of active pressure, Pa will remain positive as long as Z ≥ 2C √ 1 γ Ka
Z
The value of active pressure will be negative hence the soil Z ≤ 2C √ 1 γ Ka
The depth of tension crack friction angle and
Zc ≥ 2C √ 1 γ Ka
PASSIVE PRESSURE The passive pressure coefficient, 1 + sin Kp= 1 − sin
PAGE 99
LATERAL EARTH
By using Rankine’s theory, determine the total of active thrust acting on the retaining wall with 8 meter height. Given the unit weight of soil is 19kN/m³ and friction angle is Ø = 30 º
Solution Ka = 1 - sin 30º = 0.333 1 - sin 30º
At z = 0m Ø = 30º γ = 19 kN/m³
a = K a ( Z ) = 0.333 (19 x 0) = 0 kN/m2 At z = 8m
8m
a = Ka (Z ) = 0.333 (19 x 8) = 50.616 kN/m2
Pa
Pa = 1 (50.616x8) 2 = 202.464kN/m
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TOPIC 7 SEEPAGE
OBJECTIVES To explain seepage in general To describe the seepage theory in soil To apply the flow net through medium and specific structures To analyze the quantity and pore pressure of the seepage in sheet piles, concrete dam and earth dam using flow net method.
To Explain Seepage in General
SEEPAGE
Figure 7.1: Cofferdam at Montgomery Point Lock, USA (Courtesy: U.S. Army Corps of Engineers , 2004)
Terminologies Seepage can be defined as the flow of water through the soil mass. It can occur due to the existence of soil permeability Flow line is a line followed by particles of water during seepage occurs. Along the route they would suffer loss of pressure head. A flow line is a line along which a water particle will travel from upstream to the downstream side in the permeable stratum. Equipotential line is a line which connecting to other points with having the same equipotential forces. An equipotential line is a line along which the potential head at all points is equal. Flow net is a series of squares created by the intersection between the line of flow lines with the equipotential line. Network flow is a flow strips of water through a material and drawn in a scale. The pattern of approximate squares formed by these two sets of lines is known as a flow net.
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To Describe Seepage in General
SEEPAGE
Seepage can be defined as the flow of water through the soil mass. It can occur due to the existence of soil permeability. Seepage occurs when water flow through the soil, it can cause water strips formed. This thread will bring with it the details of structural unsustainability of land is the cause of soil failure. Process by which water flowed through pores among soil particles which is can occur because the existence of soil permeability. The important of learning seepage is to give the overview when a problem cause to any hydraulic structure that are built on land such as concrete dam, an earth dam and bedding dam.
Permeability and Seepage Permeability is a term to measure of how easily a fluid can flow through a porous medium. In geotechnical engineering, the porous medium are soils and the fluid is water at ambient temperature. Generally, the coarser the soil grains, larger the voids and larger the permeability. Therefore, gravels are more permeable than silts (N. Sivakugan, 2005).
Figure 7.2: Seepage Beneath (a) A Concrete Dam (b) A Sheet Pile (Source: N. Sivakugan, 2005)
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SEEPAGE
To Apply the Flow Net Through Medium and Specific Structure The quantity of water seepage under the structure can be determined by flow net method. Components present in flow net is composed of the flow line and the equipotential line. A flow net for an isometric medium is a network of flow lines and equipotential lines intersecting at right angles to each other. Groundwater flow can be studied using flow net illustrated by the graphical method. Figure 7.3, Figure 7.4, Figure 7.5 and Figure 7.6 below show the diagram of flow net under the sheet pile pieces, concrete dam and earth dam.
Figure 7.3: Flow net under the pile pieces.
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SEEPAGE
Figure 7.4: Flow Net for Seepage Around a Single Row of Sheet Pile
Figure 7.5: Flow Net Under a Dam with Toe Filter
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SEEPAGE
Figure 7.6: Seepage in an Earth Dam
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SEEPAGE
How to construct flow net for isotropic soil including sheet pile and concrete dam? (A) Construct Flow Net for Concrete Dam. 1. Use graph paper and draw Figure 7.7 (a)1 by using scale 1cm: 1m
2. Start by planning the number of flow lines to be drawn by marking Figure 7.1b
Figure 7.7 (a)
Figure 7.7 (b)
3. Next start the first sketch of the flow line (HJ). (Figure 7.7c) to be followed by subsequent flow lines, KL. Flow lines should be parallel to each other line by line. Then start drawing the line is the same building, by sketching the same disability under the pile sheet first. (Figure 7.7d)
Figure 7.7 (c)
Figure 7.7 (d)
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SEEPAGE
4. Then start drawing the line is the same building, by sketching the same
disability under the pile sheet (Make sure each line of intersection between the line of flow lines and equipotential to make a square corner.) 5. Draw the lines of the other efforts to ensure that each intersection of the lines and flow lines to attempt a square corner. (Figure 7.7e) 6. If this sketch has met all the criteria required, complete the line by making the line drawings directly as shown in (Figure 7.7f).
Figure 7.7 (e)
Figure 7.7 (f) PAGE 108
SEEPAGE
How to construct flow net for isotropic soil including sheet pile and concrete dam? B Construct Flow Net for Sheet Pile. 1. Use graph paper and draw Figure 7.8 (a)1 by using scale 1cm: 1m 2. Start by planning the number of flow lines to be drawn by marking Figure 7.8(b)
Figure 7.8 (a)
Figure 7.8 (b) PAGE 109
SEEPAGE
3. Next start the first sketch of the flow line (HJ) to be followed by subsequent flow lines, KL. Flow lines should be parallel to each other line by line. Then start drawing the line is the same building, by sketching the same disability under the pile sheet first. (Figure 7.8c) 4. Then start drawing the line is the same building, by sketching the same disability under the pile sheet (Make sure each line of intersection between the line of flow lines and equipotential to make a square corner.)
Figure 7.8 (c)
5. Draw the lines of the other efforts to ensure that each intersection of the lines and flow lines to attempt a square corner. (Figure 7.8d) 6. If this sketch has met all the criteria required, complete the line by making the line drawings directly as shown in (Figure 7.8e).
PAGE 110
SEEPAGE
Figure 7.8 (d)
Figure 7.8 (e)
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SEEPAGE
To Calculate Quantity of Seepage and Pore Pressure 1.
To Calculate the Quantity and Pore Pressure of the Seepage in Sheet Piles.
Figure 7.9(a)
Based on the example which shows the flow net has been drawn under a pile piece. If the soil permeability coefficient is 7.5 x 10-2 mm/s, determine the: i. water lost through seepage (Q) in m3/hr units. (Quantity of seepage) ii. The pore pressure (Up) in soil at point P and Q
PAGE 112
SEEPAGE
Solution 1) Step 1 is sketching the flow net. Take the lower as the datum.
Figure 7.9 (b)
From diagram Nf = 4 Ne = 8 Upstream and downstream head difference, H = 2m Given, k = 7.5 x 10-2 mm/s = (7.5 x 10-2 x 60 x 60 ) /1000 = 0.270 m/hour Therefore, the seepage flow rate (Quantity of seepage under sheet pile), Q = kH
Nf
Ne
= 0.270 x 2 x 4/8 = 0.270 m3/hour /m long. PAGE 113
SEEPAGE
Figure 7.9 (c)
i)
Pore water pressure at the point P (see Figure 7.9 c)
Piezometric tube shown at the same point P on the line marked equipotential ne = 7. Total head at point P is: hp
=
𝑛𝑛𝑒𝑒
𝑁𝑁𝑒𝑒
7
𝐻𝐻 = × 2 = 1.75m 8
the water level in the tube is 1.75 m above datum. Point P with distance 1.4m (0.5m + 0.9m) below the datum, i.e. the height-1.4m column. Pore water pressure at the point P can be calculated from Bernoulli's Theorem: Up = γw [ hp – (-zp ) ] = 9.81 x ( 1.4 + 1.75 ) = 30.9 kN/m2
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SEEPAGE
Pressure head at point Q Piezometric tube is shown at point Q on the same line marked equipotential ne = 2. Total head at point Q is: hq
=
𝑛𝑛𝑒𝑒
𝑁𝑁𝑒𝑒
2
𝐻𝐻 = × 2 = 0.5m 8
the water level in the tube is 0.5 m above the datum. Point Q is 2.0m (0.5m + 1.5m) below datum, namely the head height - 2.0m. Pore water pressure at the point Q can be calculated from Bernoulli's Theorem: Uq = γw [ hq - ( - zq ) ] = 9.81 x ( 0.5 + 2.0 ) = 24.53 kN/m2 2.
To Calculate the Quantity and Pore Pressure of the Seepage in Concrete Dam
Figure 7.10 (a)
Refer to Figure above, calculate i.
Seepage flow rate (Q) in units of length m3/hour/m under concrete dams if given k = 0.0015 mm / s.
ii.
Pore water pressure (Ux) at point X PAGE 115
SEEPAGE
Solution 1) Step 1 is sketching the flow net. Take the lower as the datum
i) Seepage flow rate From sketches the flow net, Nf = 4 Ne = 11
Figure 7.10(b)
Differential head of water upstream and downstream of the dam concrete is H = (3 m - 1 m) = 2m Given k = 0.0015 mm/s = [0.0015 x 60 x 60 ] x (1/1000) = 0.005 m/hour Thus the value of the seepage flow under the dam concrete is N Q = kH f Ne
= 0.005 x 2 x 4/11 = 0.004 m3/hour/m long
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SEEPAGE
Figure 7.10 (c)
ii. Pore water pressure at point X Refer to Figure 7.10 c. Piezometric tube is shown at point X on the same line marked disability nex = 9. Total head at point X is: hx
= =
𝑛𝑛𝑒𝑒
𝑁𝑁𝑒𝑒 9
11
𝐻𝐻
×2
= 1.64m the water level in the tube 1.64m above datum. Point X is 3m (1.5m + 0.5m +1 m) below the datum, i.e. with the head height = 3m. Pore water pressure at point X can be calculated from Bernoulli's Theorem: UX = γw [ hx - ( - zx ) ] = 9.81 x ( 1.64 + 3.0 ) = 45.52 kN/m2
PAGE 117
Earth Dam
SEEPAGE
• Soil dam or earthen dam is made of compacted earth and the rock-filled materials. It requires very large quantity of materials. • The materials are commonly used in earthen dam are clayey materials, black cotton soil, silty clay loam, sandy material, rock for pitching and rip-rap, seepage drain and masonry for filters, cement, steel lime and other materials for construction spillway and outlets.
Figure 7.11 : Review of Banasura Sagar Dam, Vythiri, India. The Second Largest Eathen Dam in Asia. (Source: Tripadvisor.com)
Figure 7.12 : Review of Hirakud Dam, Odisha, India. The World Longest Eathen Dam in Asia. (Source: Aseanrecords.World) PAGE 118
Soil Dam Failure
SEEPAGE
Figure 7.13 : Diagram of Soil Dam Failure. (Source: The Constructor)
Figure 7.14 : Various of Soil Dam Failures Right: Erosion Upstream Left: Erosion of Downstream
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SEEPAGE
Figure 7.15 : Various of Soil Dam Failures Right: Overtopping of Dam Left: Formation of Gullies due to Heavy Rainwater
3. To Calculate the Quantity and Pore Pressure of the Seepage in Earthen Dam (Soil Dam With Toe Filter)
PAGE 120
SEEPAGE
Solution
1. Step 1 is sketching the earth dam with the correct scale. Mark point A, B, C, D, H and S. Mark point G with the ratio HS= 0.3GS. Mark focal point, F at the top of toe filter. Refer figure 7.16 (a). 2. Use GF as radius and point G as a center point, draw a curve that intersect waterline and mark point I. Project down a line through point I and intersect at the toe filter. Mark the intersection point as E. A line IE known as directrix.
Figure 7.16 (a)
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SEEPAGE
1. Mark point O at the centre between E and F.
2. Draw a vertical line that parallel to directrix; X1, X2, X3, and X4 with the multiple distances; XX1, XX2, XX3, and XX4.(Refer figure 7.16 (b)). 3. Use distances; XX1, XX2, XX3, and XX4 as radius and point F as the center point, then draw a curve that intersect to vertical line X1, X2, X3, and X4 respectively. 4. Mark the intersection point A1, A2, A3, and A4. Connect point A1, A2, A3 and A4 to point G.
Figure 14 : The Construction Method of Flow Net for Soil Dam with Toe Filter
Figure 7.16 (b)
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SEEPAGE
Figure 7.16 (c)
5. Draw the correction line that perpendicular to BS and connect this line to point A4. This line known as TFL, Top Flow Line. (Refer Figure 7.16(d)).
Figure 7.16 (d)
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SEEPAGE
Figure 7.16 (e)
Draw the flow net in the boundary of OF and SA. (Refer Figure 7.16(e)).
Figure 14 (e)
Figure 7.16 (f) PAGE 124
SEEPAGE
From sketches the flow net, Nf = 4 Ne = 9 Differential head of water upstream and downstream of the dam concrete is H = 16m Given k = 6.2 x 10-3mm/s = [6.2 x 10-3 x 60 x 60 ] x (1/1000) = 0.0223 m/hour Thus the value of the seepage flow under the dam concrete is N Q = kH f Ne
= 0.0223 x 16 x 4/9 = 0.159 m3/hour/m long
ii. Pore water pressure at point A Refer to figure in the tutorial question. Piezometric tube is shown at point A on the same line marked disability ne A = 9. Total head at point X is: hA
= 4
𝑛𝑛𝑒𝑒
𝑁𝑁𝑒𝑒
𝐻𝐻
= × 16 9
= 7.1m
Pore water pressure at point X can be calculated from Bernoulli's Theorem: UA = γw [ zA - hA ) ] = 9.81 x ( 10.0 – 7.1) = 28.45 kN/m2
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SEEPAGE
6. To Calculate the Quantity and Pore Pressure of the Seepage in Earthen Dam (Soil Dam Without Toe Filter) Tutorial
Diagram below show the earth dam with the vertical flow. Construct the top flow line and flow net for the earthen dam without the toe filter.
7. Step 1 is sketching the earth dam with the correct scale. Mark point A, B, C, D, H and S. Mark point G with the ratio HS= 0.3GS. Mark focal point, F at the top of toe filter. Refer figure 7.17 (a).
8. Use GF as radius and point G as a center point, draw a curve that intersect waterline and mark point I. Project down a line through point I and intersect at the permeable layer. Mark the intersection point as E. A line IE known as directrix. Mark centrepoint O in between point E and F.
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SEEPAGE
Figure 7.17 (a)
9. Draw a vertical line that parallel to directrix; Y1, Y2, Y3, Y4 and Y5 with the multiple distances; YY1, YY2, YY3, YY4 and YY5.(Refer figure 7.17b). 10.Use distances; YY1, YY2, YY3, YY4 and YY5 as radius and point F as the center point, then draw a curve that intersect to vertical line Y1, Y2, Y3, Y4 and Y5 respectively. 11.Mark the intersection point A1, A2, A3, A4 and A5. (Refer figure 7.17 (c)).
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SEEPAGE
Figure 7.17 (b)
Figure 7.17 (c) PAGE 128
SEEPAGE
12. Join the intersection point A1, A2, A3, A4 and A5 with point G and O. Measure length of ( a + ∆ a ). Refer Figure 7.17 (d). 13. Determine the dam angle α, and use the given table to make an
interpolation to obtain the value of ∆ a. Project the line from ∆ a. and obtain the point M. MF is a straight line. 14.Construct the TFL and the flow net for the dam as shown in the Figure 7.17(e)
Figure 7.17 (d)
Figure 7.17 (e) (i) PAGE 129
SEEPAGE
Figure 7.17 (e) (ii)
PAGE 130
SEEPAGE
ASSESSMENT Try to assess your understanding on this chapter by answering this tutorial. Good luck. 1. Sheet Pile
Based on the flow network sketch given above, find: i.
seepage flow under the unit m3/jam/m pile long pieces if given k = 2.15 x 10-3 mm / s.
ii.
Pore water pressure at point Y.
PAGE 131
SEEPAGE
2) Concrete Dam
Based on the flow net sketch given above, find: i.
seepage flow under the concrete dam in the long m3/hour/m unit if given k = 2.15 x 10-3 mm / s.
ii.
Pore water pressure at point X.
PAGE 132
SEEPAGE
3) Earthen Dam
Based on the figure given above: i.
Construct the top flow line and sketch the flow net for the earthen dam above.
ii.
Calculate seepage flow under the concrete dam in the long m3/hour/m unit if given k = 6.2 x 10-3 mm / s.
iii.
Pore water pressure at point A.
PAGE 133
CHAPTER 8 SLOPE STABILITY
THIS TOPIC COVERS THE BASIC KNOWLEDGE OF SLOPE, TYPES OF SLOPE FAILURE, SLOPE STABILITY ANALYSIS, TOTAL STRESS METHOD, USING TOTAL STRESS ANALYSIS, TAYLOR AND SLICES CHART STABILITY METHODS. STUDENTS WILL ALSO BE EXPOSED TO VARIOUS CALCULATIONS FOR FACTOR OF SAFETY AND SLOPE STABILIZATION.
SLOPE STABILITY
8.1 Introduction Slopes are typically categorized in two types: natural and artificially-made slopes. Natural slopes are formed due to physical processes that include plate tectonics and weathering/erosion of rock masses that result in material deposition. Artificially-made slopes are established to facilitate infrastructure projects, ex., embankments, earth dams, road cuttings etc.
The stability of a slope is of critical importance in Geotechnical Engineering applications. A slope movement (also referred as a landslide) can lead to severe issues including infrastructure damage or/and casualties. Slope stability depends on the capability of the soil mass to withstand its gravitational forces, the additional loads acting on the slope, as well as potential dynamic loads (such as that of an earthquake).
PAGE 135
SLOPE STABILITY
8.2 Define Slope Inclined soil surfaces or soil slope occur in highway cuts and fill sections, earth dams, building platforms and so on. Whenever there are soil slopes, there are always the potential for the soil to slide from a higher location to a lower one creating landslides. Landslides occurs when the shear stress at the potential slip surface exceed the shear strength of the soil. The potential for the occurrence of landslides depends on several factors such as:
i) The geometry of the slope including the geometric configuration of the underlying strata.
PAGE 136
SLOPE STABILITY
ii) Water flow or seepage within the slope
iii) The material properties of the differing strata, including the unit weight, angle of friction and cohesion. PAGE 137
SLOPE STABILITY
iv) The additional loading on the slope by man
PAGE 138
SLOPE STABILITY
8.3 Slope Failure The most common and complete classification system of landslides is that provided by Varnes (1978), who introduces a system that requires the definition of the landslide material and the type of the movement induced. The ground materials are distinguished in 5 categories: Rock: An intact rockmass which used to be located at its initial position (i.e., it has not been eroded) before the movement occurred. Soil: Soil mass formed or transferred due to weathering and erosion of rocks. Soils consists of solid particles and voids filled with liquid and/or air, representing a three-phase system. Earth: The soil material in which more than 80% consist of particles smaller than 2 millimeters (upper limit of sand particles). Mud: The soil in which more than 80% consists of particles smaller than 0.06 millimeters (upper limit of silt particles). Debris: The soil material that has 20%-80% of particles that are bigger than 2 millimeters.
Figure 8.1: The Highland Towers collapse occurred on 11 December 1993 in Taman Hillview, Ulu Klang, Selangor, Malaysia from a major landslide caused by heavy rains that burst diversion pipes resulted in 48 deaths. PAGE 139
SLOPE STABILITY
Slope failures can be triggered by natural of human-induced causes, or a combination of the two. The natural causes of landslides include: gravitational forces that tend to destabilize the ground, water saturation, erosion, dynamic loads (e.g., earthquakes), the sudden uplift of the aquifer level, volcanic eruptions and freeze-thaw weathering cycles. The presence of water is one of the most common factors that triggers landslides. Water saturation can be caused as a result of heavy precipitation, snow melt or changes in the ground water level. Water saturation reduces the shear strength of soils. In particular, it decreases the normal effective stress that acts between the grains and hence, the frictional resistance is reduced. The Mohr-Coulomb failure criterion suggests that the shear strength of the ground is proportional to the normal effective stress as:
Where t is the shear strength, σn is the effective stress, σt is the total stress, u is the pore-water pressure, c is the cohesion, and φ the friction angle.
PAGE 140
SLOPE STABILITY
8.4 Types of Slope Failure Rotational slips – occurs characteristically in homogeneous soft rocks or cohesive soils; the movement taking place along a curved shear surface in such a way that the slipping mass slumps down near the top of the slope.
Figure 8.2: The rotational Holbeck Hall landslide landslide, in Scarborough North Yorkshire, England (June 1993) (photo from British Geological Survey) PAGE 141
SLOPE STABILITY
Lateral spreads are deformational phenomena caused by liquefaction, the process during which a saturated soil (usually sands) experiences loss of strength after a sudden change in its initial stress conditions. Therefore, the soil tends to behave more like a liquid than a solid. Such deformations occur on less steep slopes and are usually triggered by dynamic loads such as that of an earthquake. Lateral spreading is usually a progressive process that occurs mainly near shores, riverbanks, and ports where loose and saturated sandy soils exist. Infrastructure founded on those type of soils is prone to extensive damage
Figure 8.3 a) An illustration of a translational landslide (USGS, 2004) and b) Lateral spreading example caused by an earthquake in Pakistan (Independent)
PAGE 142
SLOPE STABILITY
Translation of slide failure – which may involve linear movement of rock blocks along bedding planes or of a soil layer lying near to the (sloping) surface.
Figure 8.4 a) Illustration of a translational landslide (USGS, 2004) and b) Co-seismic translational landslide triggered in Japan in 2016 (Highland and Bobrowsky, 2018 by Khang Dang and Kyoji Sassa)
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SLOPE STABILITY
Failure of flows – here the slipping mass is internally disrupted and moves partially or wholly as a fluid. Flows often occur in weak saturated soils when the pore pressure has increased sufficiently to produce a general loss of shear strenght
Fugure 8.5 a) Accumulated material from a debris flow in Cephalonia Island, Greece (GEER, 2020), b) Titled tree chunks as an indicator of soil creep (USRA by Tom McGuire), and c) La Conchita landslide in Ventura County, California (1995). A slump resulted in an earthflow downwards (Photo by Mark Reid, U.S. Geological Survey
PAGE 144
SLOPE STABILITY
Failure of falls – these are characterized by movement away from existing discontinuities, such as joins, fissures, steeply-inclined bedding planes, fault planes, etc. and within which the failure condition may be assisted or precipitated by the effects of water or ice pressure
Figure 8.6 a) Illustration of a rockfall (USGS, 2004), b) Rockfall in Central Pyrenees, Spain (Corominas, 2017)
PAGE 145
SLOPE STABILITY
8.5 Slope Stability and Safety Factors Slope stability is about stress and strength. Gravity and other factors combine to produce a “driving force” potentially capable of mobilizing a failure mass. Shear strength within the soil and/or rock mass provides a “resisting force” to help keep it in place. There are several factors affecting stability of slopes; Topography and its surrounding physical conditions Geological conditions such as the nature and the depth of its subsoil, degree of decomposition or location of fracture etc Shear strength of the slope-forming materials Surface and ground water condition External loading and surcharges Stability of slopes can be achieved by the following methods; construct surface drain system usually consists of surface channel, stepped or trapezoidal channel, catch pit or sand trap constructing gabion wall protection and treatment to rock slope such as scaling forming soil nail and rock bolts construct buttress support Meshing Shortcreting Turfing sloping slope provide Nylon Mesh for subsurface drainage and to reinforce the root of grass
Figure 8.7 Surface drainage system to stable the slope
PAGE 146
8.6 Factor Of Safety
SLOPE STABILITY
Factor of Safety can be defined as the ratio of the resisting force to the driving force. The Factor of Safety tells us, on a percentage basis, how much greater the resisting force is, or is likely to be, than the driving force. When the Factor of Safety approaches 1, the resisting and driving forces are balanced, and failure is assumed imminent with only a small reduction in strength or a slight increase in stress. The risk of slope failure can be minimized by placing an appropriate value for the factor of safety during the design work. Safety factors can be determined using 3 methods which are: i) Total stress analysis ii) Taylor’s stability number method iii) Fellenius’ method
Total Stress Analysis A total stress analysis may be applied to the case of a newly cut or newly constructed slope in a fully saturated clay (which is the condition immediately after construction) . In this analysis, certain assumption has been made such as; The undrained shear strength being τ = cu. the failure surface will take the cross-sectional form of a circular arc prevailing stability may be caused by the body weight of soil, W
PAGE 147
SLOPE STABILITY
0
d
w R d
w 0
19
-
= =
-
radius centroid di· tarrnce ·from 0 body v eig ht oentroid S'.
ctor angle
PAGE 148
SLOPE STABILITY
Figure shows the cross section of a slope together with a trial slip circle of radius, R and center, O. Instability tends to be caused due to the moment of the body weight, W of the portion above the slip circle. In the analysis, only be considered where the equilibrium of moment.
The tendency to move is resisted by the moment of the mobilised shear strength acting along the circular arc AB.
The values of W and d are obtained by dividing the shaded area into slices or triangular/rectangular segments and then taking area-moments about a vertical axis passing through the toe, or other convenient point
PAGE 149
SLOPE STABILITY
LE EXAMP
PAGE 150
SLOPE STABILITY
Taylor's Stability Number Method Taylor’s stability number method is determining the minimum factor of safety of slope in a homogeneous soil. Using a total stress analysis and ignoring the possibility of tension cracks, he produced a series of curves which relate a stability number (N) to the slope angle, β.
PAGE 151
SLOPE STABILITY
LE EXAMP
The slope is safe because the Factor of Safety > 1
PAGE 152
SLOPE STABILITY
Fellenius Method In this method, it is assumed that the interslice forces are equal and opposite and cancel each other out. It is now only necessary to resolve the forces acting on the base of the slice. Fellenius slice method assumes that the failure surface may occur as a circular arc and the subsequent failure to attempt a plane divided into several slices. This method is also known as a method Swedish.
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SLOPE STABILITY
Rsin a·+--0
R
b
i,.-- D
X X
�im1L
Fmm the analysis, the force.s acting on each slice are; L Force slice , \V = y b Z 2.. Normal force N 3. Wateo1· recomces, µ 4.. Shear for-ce, C 5.. Fo:rce:s , dge E and X
PAGE 154
SLOPE STABILITY
LE EXAMP For slope as shown in below, get a safety factor of slopes by using the method of slices. Divide into 6 slices and use parameter given as per following;
PAGE 155
Solution :
SLOPE STABILITY
Choose the right scale and on graph paper as per following;
PAGE 156
SLOPE STABILITY
The slope is safe because the Factor of Safety > 1
PAGE 157
LIST OF REFERENCES Basack & Purkayastha. (2009) Engineering Properties of Marine Clay from the Eastern Coast of India. Journal of Engineering Technology Research Vol.1. pp109-114
01
02
03
04
05
06
Braja M Das. (2010). Principles of Geotechnical Engineering (7th ed.). Cengage Learning. United States of America. Donald P. Coduto. (1999). Geotechnical Engineering (Principles and Practices). Prentice Hall, Inc. Upper Saddle River, New Jersey, US. Marcus M. Truitt. (1983). Soil Mechanics Technology. United States of America. Prentice Hall, Inc. Englewood Cliffs, New Jersey, US.
Muni Budhu. (2011). Soil Mechanics & Foundation (3rd ed.). John Wiley & Sons Inc.
N. Sivakugan. (2005). Permeability and Seepage. University of Houston.