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1. NED UNIVERSITY OF ENGINEERING &TECHNOLOGY, KARACHI-75270

2. 2

3. syllabus INTRODUCTION TO SOIL MECHANICS ORIGIN AND FORMATION OF SOIL CLAY MINERALOGY COMPOSITION & PHYSICAL PROPERTIES OF SOIL SOIL CLASSIFICATION 3

4. syllabus SOIL COMPACTION SOIL HYDRAULICS SEEPAGE THEORY THE CONCEPT OF EFFECTIVE STRESS IN SOIL STRESS DISTRIBUTION IN SOIL FROM SURFACE LOADS CONSOLIDATION THEORY SHEAR STRENGTH OF SOIL 4

5. LIST OF BOOKS Soil Mechanics : Basic Concepts and Engineering Applications by A. Aysen Soil mechanics : concepts and applications by William Powrie Advanced soil mechanics by Braja M. Das Soil mechanics by T. William Lambe , Robert V. Whitman Soil mechanics in engineering practice by Karl Terzaghi 5

6. LIST OF BOOKS elements of soil mechanics by Ian Smith Soil Mechanics and Foundations by Dr. B.C. Punmia Soil mechanics by Robert F. Craig Soil Mechanics & Foundation Engineering by Purushothama Raj Soil Mechanics and Foundations by Muniram Budhu 6

7. LIST OF BOOKS Fundamentals of geotechnical engineering by Braja M. Das Principles of Geotechnical Engineering by Braja M. Das Geotechnical engineering of dams by Robin Fell, Patrick MacGregor , David Stapledon Geotechnical engineering : principles and practices of soil by V. N. S. Murthy 7

8. Prerequisite knowledge and/or skills Students should have a basic understanding of the following: basic tools for stress, strain and strength analysis to design structures, predict failures and understand the physical properties of materials. 8

9. Prerequisite knowledge and/or skills analyze and design structural members subjected to tension, compression, torsion and bending using fundamental concepts of stress, strain, elastic behavior and inelastic behavior . conduct themselves professionally and with regard to their responsibilities to society, especially with respect to designing structures to prevent failure. 9

10. Prerequisite knowledge and/or skills a fundamental knowledge and understanding of the mechanics of fluid at rest and in motion by describing and observing fluid phenomena and by developing and using the principles and laws for analyzing fluid interactions with natural and constructed systems . 10

11. Course objectives Develop technical competence in basic principles of soil mechanics and fundamentals of application in engineering practice. (Outcomes b, e, k) Ability to list the salient engineering properties of soils and their characteristics and describe the factors which control these properties. (Outcomes c) 11

12. Course objectives Ability to apply laboratory methods of determining the properties of soils. (Outcomes a, c, d, k) Ability to identify common situations when the soil becomes a factor in an engineering or environmental problem. (Outcomes a, c, i, k) Capable of performing basic analytical procedures in these situations to obtain the engineering quantity desired given the formuli , tables, and the soil properties and understand their limitations. (Outcomes a, c, d, k ) 12

13. Relationship of course to undergraduate degree program objectives and outcomes This course serves students in a variety of engineering majors. The information below describes how the course contributes to the college's educational objectives. 13

14. Relationship of course to undergraduate degree program objectives and outcomes Soil Mechanics refers to the application of the mechanics of materials and fluids to describe the behavior of soils. The course covers soil mechanics along with GeoEngineering in a less rigorous and more qualitative manner. Upon completion of the course, students should have an understanding of the properties of soils, the standard analyses performed by a GeoEngineer , and the qualitative aspects of soil behavior which are used in the analysis of problems in GeoEngineering . 14

15. Relationship of course to undergraduate degree program objectives and outcomes Within the Geological Engineering Program, this course helps provide key educational outcomes as listed below: a. an ability to apply knowledge and principles of mathematics, science, and engineering to geological engineering problems. This includes differential equations, calculus-based physics, chemistry, and geological science topics that emphasize geologic processes, the identification of minerals and rocks, geophysics, and field methods. This also includes engineering science topics such as statics, properties/strength of materials, and geomechanics . 15

16. Relationship of course to undergraduate degree program objectives and outcomes b . an ability to design and conduct experiments, as well as to analyze and interpret data c. an ability to design a system, component, or process to meet desired needs within realistic constraints such as economic, environmental, social, political, ethical, health and safety, constructability, and sustainability. This requires exposure to topics such as surface and near-surface natural processes, the impacts of construction projects, disposal of wastes, and site remediation . 16

17. Relationship of course to undergraduate degree program objectives and outcomes d . an ability to function on multi-disciplinary teams e. an ability to identify, formulate, and solve geological engineering problems in space and time. This includes the knowledge of the physical and chemical properties of earth materials, surface water, ground water and their distribution. f. an understanding of professional and ethical responsibility. 17

18. Relationship of course to undergraduate degree program objectives and outcomes g . an ability to communicate effectively h. the broad education necessary to understand the impact of engineering solutions in a global, economic, environmental, and societal context. i. a recognition of the need for, and an ability to engage in life-long learning j. a knowledge of contemporary issues k. an ability to use the techniques, skills, and modern engineering tools necessary for engineering practice. 18

19. Engineering nomenclature Science : is the study of the physical and natural world using theoretical models and data from experiments and observations. OR Knowledge gained through experiments and observations. Technology : technology is the practical application of science. Engineering is the application of scientific methods to solve discrete problems. 19

20. 20

21. INTRODUCTION TO SOIL MECHANICS 21

22. What is soil and what is rock…? To an engineer, Soil is the un-aggregated and un-cemented deposits of minerals and organic particles covering the earth crust. To an engineer, ROCK is the aggregated and cemented deposits of minerals and organic particles covering the earth crust. 22

23. What is soil and Rock mechanics…? Soil Mechanics is a branch of engineering mechanics that describes the properties and behavior of soils to the force field of their physical environment. Rock Mechanics is a branch of engineering mechanics that describes the properties and behavior of Rocks to the force field of their physical environment. 23

24. What is soil and mechanics…? 24

25. What is foundation engineering…? Foundation engineering is the application of soil mechanics and rock mechanics in the design of foundation elements of structures . 25

26. Geomechanics Geomechanics involves the geologic study of the behavior of soil and rock. The two main disciplines of Geomechanics are soil mechanics and rock mechanics. 26

27. Geotechnical engineering…?? Geotechnical engineering is the branch of civil engineering concerned with the engineering behavior of earth materials. Geotechnical engineering uses principles of soil mechanics and rock mechanics to investigate subsurface conditions and materials; determine the relevant physical/mechanical and chemical properties of these materials; evaluate stability of natural slopes and man-made soil deposits; assess risks posed by site conditions; design earthworks and structure foundations; and monitor site conditions, earthwork and foundation construction . 27

28. Geotechnical engineering…?? 28

29. ground Soil foundation interaction

30. typical Geotechnical Project construction site Geo-Laboratory for testing Design Office for design & analysis soil samples soil properties design details

31. bed rock firm ground Shallow Foundations for transferring building loads to underlying ground mostly for firm soils or light loads

32. Shallow Foundations

33. bed rock weak soil Deep Foundations P I L E for transferring building loads to underlying ground mostly for weak soils or heavy loads

34. 34 Deep Foundations Driven timber piles, Pacific Highway

35. Retaining Walls for retaining soils from spreading laterally Road Train ` retaining wall

36. Earth Dams for impounding water soil reservoir clay core shell

37. Concrete Dams reservoir soil concrete dam

38. Concrete Dams Three Gorges Dam, Hong Kong

39. Concrete Dams

40. Earthworks Roadwork, Pacific Highway preparing the ground prior to construction

41. Construction hazard an unwelcome visitor at an earthwork site A dead Anaconda python (courtesy: J. Brunskill) What does it have to do with Geo?#!

42. 42 Geofabrics used for reinforcement, separation, filtration and drainage in roads, retaining walls, embankments… Geofabrics used on Pacific Highway

43. Reinforced Earth Walls using geofabrics to strengthen the soil

44. 44 Soil Nailing steel rods placed into holes drilled into the walls and grouted

45. 45 Sheet Piles Resist lateral earth pressures Used in excavations, waterfront structures, ..

46. 46 Sheet Piles used in temporary works

47. 47 Sheet Piles interlocking sections

48. Cofferdam sheet pile walls enclosing an area, to prevent water seeping in

49. Landslides

50. Sheet Piles sheets of interlocking steel or timber driven into the ground , forming a continuous sheet ship warehouse sheet pile

51. Shoring propping and supporting the exposed walls to resist lateral earth pressures

52. Tunneling

53. Blasting For ore recovery in mines

54. Ground Improvement Impact Roller to Compact the Ground

55. Ground Improvement Sheepsfoot Roller to Compact Clay Soils

56. Ground Improvement Smooth-wheeled Roller

57. Ground Improvement Big weights dropped from 25 m, compacting the ground. Craters formed in compaction

58. Environmental Geomechanics Waste Disposal in Landfills

59. Instrumentation to monitor the performances of earth and earth supported structures. to measure loads, pressures, deformations, strains ,…

60. Soil Testing Cone Penetration Test Truck – Lavarach Barracks, Townsville

61. Soil Testing More Field Tests Standard Penetration Test Vane Shear Test

62. Soil Testing Triaxial Test on Soil Sample in Laboratory

63. Soil Testing Variety of Field Testing Devices

64. Soil Testing laboratory staff

65. Typical Safety Factors Type of Design Safety Factor Probability of Failure Earthworks 1.3-1.5 1/500 Retaining structures 1.5-2.0 1/1500 Foundations 2.0-3.0 1/5000

66. Some unsung heroes of Civil Engineering… … buried right under your feet. foundations soil exploration tunneling

67. Great Contributors to the Developments in Geotechnical Engineering Hall of fame

68. Karl Terzaghi 1883-1963 C.A.Coulomb 1736-1806 WJM Rankine 1820-1872 A.Casagrande 1902-1981 L. Bjerrum 1918-1973 A.W.Skempton 1914- G.F.Sowers 1921-1996 G.A. Leonards 1921-1997

69. Geotechnical Engineering Landmarks challenges

70. Leaning Tower of Pisa Our blunders become monuments!

71. Hoover Dam, USA Tallest (221 m) concrete dam

72. Petronas Tower, Malaysia Tallest building in the world

73. and…. Monuments

74. Some Suggestions Attend the lectures Develop a good feel for the subject It takes longer to understand from the lecture notes It is practical, interesting and makes lot of sense.

75. Some Suggestions Work in groups

76. Some Suggestions Thou shall not wait till the last minute

77. Exams My mama always said, “Exam is like a box of chocolates; you never know what you are gonna get”

78. ORIGIN AND FORMATION OF SOIL

79. To an agriculturist, soil is the substance existing on the earth’s surface, which grows and develops plants. To a geologist, soil is the material in a relatively thin surface zone within which roots occur, and rest of the crust is termed as rock irrespective of hardness. To an engineer, soil is the un-aggregated and un-cemented deposits of minerals and organic particles covering the earth crust . 79 ORIGIN AND FORMATION OF SOIL

80. 80 Soil Formation Parent Rock Residual soil Transported soil in situ weathering (by physical & chemical agents) of parent rock weathered and transported far away by wind, water and ice.

81. 81 Residual Soils Formed by in situ weathering of parent rock

82. Transported Soils 82

83. 83 Transported Soils Transported by : Special name: wind “Aeolian” sea (salt water) “Marine” lake (fresh water) “Lacustrine” river “Alluvial” ice “Glacial”

84. Geological cycle of soil formation 84

85. weathering 85

86. upheaval 86

87. transportation 87

88. deposition 88

89. Parent Rock Formed by one of these three different processes Igneous Sedimentary Metamorphic formed by cooling of molten magma (lava) formed by gradual deposition, and in layers formed by alteration of igneous & sedimentary rocks by pressure/temperature e.g., limestone, shale e.g., marble e.g., granite

90. Rock types There are three major type of rocks: 90

91. 91 Rock cycle

92. Igneous rocks formation Igneous rocks form when magma (molten matter) such as that produced by exploding volcanoes cools sufficiently to solidify. Igneous rocks can be coarse-grained or fine-grained depending on weather cooling occurred slowly or rapidly. Relatively slow cooling occurs when magma is trapped in the crust below the earth surface (such as at the core of mountain range), while more rapid cooling occurs if the magma reaches the surface while molten lava flow. 92

93. Igneous rocks formation In coarse-grained igneous rocks the most common is granite, a hard rock rich in quartz, widely used as a construction material and for monuments. Most common in fine-grained igneous rocks is basalt, a hard dark coloured rock rich in ferromagnesian minerals and often used in road construction. Being generally hard, dense, and durable, igneous rocks often make good construction materials. Also, they typically have high bearing capacities and therefore make good foundation material. 93

94. 94 Igneous rocks formation

95. 95 Igneous rocks formation

96. Igneous rocks formation 96

97. There are over 700 types of igneous rocks identified and most of them are formed below the surface of the earth. Based on their mode of occurrence, igneous rocks are either intrusive or plutonic rocks and extrusive or volcanic rocks. 97 Igneous rocks formation

98. Sedimentary rocks formation Sedimentary rocks are formed when mineral particles, fragmented rock particles and remains of certain organisms are transported by wind, water, and ice and deposited, typically in layers, to forms sediments. Over a period of time as layers accumulate at a site; pressure on lower layers resulting from the weight of overlying strata hardens the deposits, forming sedimentary rocks. Sedimentary rocks comprise the great majority of rocks found on the earth’s surface. 98

99. Sedimentary rocks formation Additionally deposits may be solidified and cemented by certain minerals (e.g., silica, iron oxides, calcium carbonate). Sedimentary rocks can be identified easily when their layered appearance is observable. The most common sedimentary rocks are shale, sandstone, limestone, and dolomite. The strength and hardness of sedimentary rocks are variable, and engineering usage of such rocks varies accordingly. 99

100. Relatively hard shale makes a good foundation martial. Sandstones are generally good construction materials. Limestone and dolomite, if strong, can be both good foundation and construction materials. Clastic sedimentary rocks The fragments of pre-existing rocks or minerals that make up a sedimentary rock are called clasts . Sedimentary rocks made up of clasts are called clastic ( clastic indicates that particles have been broken and transported). Clastic sedimentary rocks are primarily classified on the size of their clasts . 100 Sedimentary rocks formation

101. 101 Sedimentary rocks formation

102. 102 Sedimentary rocks formation

103. 103 Sedimentary rocks formation

104. Metamorphic rocks formation When sedimentary or igneous rocks literally change their texture and structure as well as mineral and chemical composition, as a result of heat, pressure, and shear; metamorphic rocks are formed. Metamorphic rocks are much less common at the earth’s surface than sedimentary rocks Metamorphic rocks can be hard and strong if un-weathered. 104

105. Metamorphic rocks They can be a good construction material like marble is often used for buildings and monuments but foliated metamorphic rocks often contain planes of weakness that diminish strength. Metamorphic rocks sometimes contain weak layers between very hard ones. The most common metamorphic rocks are gneiss, schist, slat, quartzite and marble. 105

106. 106

107. 107

108. 108

109. Identification of rock types Igneous: A tough, frozen melt with little texture or layering; mostly black, white and/or gray minerals; may look like granite or like lava. Sedimentary: Hardened sediment with layers (strata) of sandy or clayey stone; mostly brown to gray; may have fossils and water or wind marks. Metamorphic: Tough rock with layers (foliation) of light and dark minerals, often curved; various colours; often glittery from mica. 109

110. Identification of rock types 110 Examine the specimen closely. Begin by looking for differences in texture like solid, porous, crystallized or grainy elements. Pour vinegar over the porous or smooth rock. If it sizzles, you have a common sedimentary rock known as limestone or possibly a metamorphic rock known as marble. If not, you will need to further examine the rock for answers. Note the arrangement of crystals, the colour of its holes or the size of any particles. Small particles are usually sedimentary rock like sandstone, while large particles are another sedimentary rock called conglomerate. and schist.

111. Identification of rock types 111 Pay special attention to rocks with holes, as they can fall into either of two categories. Light-coloured holes represent a sedimentary rock called tufa , while dark holes are surprisingly found to be igneous rock such as pumice. Check the surface of the rock. Is it glassy, shiny or dull? Observe the layers within a layered rock. Although sedimentary rock is known for its layered characteristics, several types of metamorphic rock also reveal layers such as slate and schist.

112. Igneous rocks 112 Andesite Basalt Diorite Gabbro Granite

113. Sedimentary rocks 113 Seventy percent of all the rocks on earth are sedimentary rocks Limestone Peat Gypsum Rock salt Sandstone Shale

114. Sedimentary rocks 114 Siltstone Coal Conglomerate Sandstone Conglomerate Limestone

115. Metamorphic rocks 115 Gneiss Marble Quartzite Schist Slate Phyllite

116. 117

117. CLAY MINERALOGY

118. Clay minerology 119

119. background 120

120. background 121

121. 122 Atomic structure

122. Elements of Earth 123

123. Elements of Earth 124

124. 125 Elements of Earth

125. 126 Elements of Earth 12500 km dia 8-35 km crust % by weight in crust O = 49.2 Si = 25.7 Al = 7.5 Fe = 4.7 Ca = 3.4 Na = 2.6 K = 2.4 Mg = 1.9 other = 2.6 82.4%

126. Elements of Earth Geotechnical engineers are interested mainly in the top 100 metres of the earth crust. As you can see from the table, 82% of the elements are oxygen, silicon and aluminium . 127

127. CLAY MINERALOGY Clay minerals are very tinny crystalline substances evolved primarily from chemical weathering of certain rock forming minerals. All clay minerals are very small, colloidal sized crystals and they can only be seen with an electron microscope. Most of the clay minerals have sheet or layered structures. A few have elongated tubular or fiber structures. 128

128. CLAY MINERALOGY Soil masses generally contain mixture of several clay minerals named for the predominating clay mineral with varying amounts of non clay minerals. Clay minerals are complex aluminium silicates composed of two basic units : Silica tetrahedron, and Aluminium octahedron 129

129. 130 Basic Structural Units 0.26 nm oxygen silicon 0.29 nm aluminium or magnesium hydroxyl or oxygen Clay minerals are made of two distinct structural units. Silicon tetrahedron Aluminium Octahedron

130. CLAY MINERALOGY The particular way, in which these are stacked, together with different bonding and different metallic ions in the crystal lattice, constitute the different clay minerals. 131

131. silica tetrahedron Each tetrahedron unit consists of four oxygen atoms surrounding a silica atom. Together oxygen and silicon account for 80 % of the earth crust. This must mean that crustal minerals are mainly combinations of these two elements. 132

132. silica tetrahedron 133

133. tetrahedron 1.Silicate Clays

134. 1.Silicate Clays silica tetrahedron

135. 136   silica tetrahedron

136. octahedron 137

137. octahedron

138. octahedron The octahedral unit consists of six hydroxyl units surrounding an aluminium (or Magnesium) atom. The octahedral layer is composed of cations in octahedral coordination with oxygen. The combination of the aluminium octahedral units forms a Gibbsite . If the main metallic atoms in the octahedral units are Magnesium these sheets are referred to as Bruce sheets. 139

139. 140 B or G octahedron http://www.luc.edu/faculty/spavko1/minerals /

140. gibbsite The octahedral unit is called gibbsite if the metallic atom is mainly aluminium. Chemical Formula for gibbsite is Al ( OH) 3 , Aluminium Hydroxide. Gibbsite is an important ore of aluminium and is one of three minerals that make up the rock Bauxite. Bauxite is often thought of as a mineral but is really a rock composed of aluminium oxide and hydroxide minerals such as gibbsite, boehmite , AlO (OH) and diaspore , HAlO2, as well as clays, silt and iron oxides and hydroxides. 141

141. alumina octahedron

142. brucite The octahedral unit is called brucite if the metallic atom is mainly magnesium. Chemical Formula of brucite is Mg (OH) 2 , Magnesium Hydroxide. Brucite is a mineral that is not often used as a mineral specimen but does have some important industrial uses. It is a minor ore of magnesium metal and a source of magnesia. It is also used as an additive in certain refractory . 143

143. tetrahedral sheet The tetrahedral sheet is composed of tetrahedron units. Each unit consists of a central four-coordinated atom (e.g. Si, Al, or Fe) surrounded by four oxygen atoms that, in turn, are linked with other nearby atoms (e.g. Si, Al, or Fe), thereby serving as inter-unit linkages to hold the sheet together. Sheet of horizontally linked, tetrahedral-shaped units; serve as one of the basic structural components of silicate clay minerals. 144

144. tetrahedral sheet 145

145. tetrahedral sheet 146

146. tetrahedral sheet 147

147. tetrahedral sheet 148 Several tetrahedrons joined together form a tetrahedral sheet.

148. Octahedral sheet 149

149. Octahedral sheet 150

150. Octahedral sheet 151

151. Clay minerals Kaolinite Illite Montmorillonite Vermiculite Smectite 152

152. kaolinite The chemistry of Kaolinite is Al 2 Si 2 O 5 (OH) 4 , Aluminium Silicate Hydroxide and the skeletal is shown in the figure. Kaolinite's structure is composed of silicate sheets (Si 2 O 5 ) bonded to aluminium oxide/hydroxide layers (Al 2 (OH) 4 ) called gibbsite layers. The silicate and gibbsite layers are tightly bonded together with only weak bonding existing between these silicate/gibbsite-paired layers (called s-g layers). 153

153. kaolinite The weak bonds between these s-g layers cause the cleavage and softness of this mineral. The Kaolinite structural unit consists of alternating layers of silica tetrahedra with tips embedded in an aluminum (gibbsite) octahedral unit. This alternating of silica and gibbsite layers produces what is some times called a 1:1 basic unit. The resulting flat sheet unit is about 7 Å thick and extends infinitely in other two dimensions. 154

154. kaolinite 155 Kaolinite is used in the production of ceramics, as filler for paint, rubber and plastics and the largest use is in the paper industry to produce a glossy paper such as is used in most magazines .

155. 156 kaolinite Different combinations of tetrahedral and octahedral sheets form different clay minerals: 1:1 Clay Mineral (e.g., kaolinite, halloysite ):

156. Kaolinite Si Al Si Al Si Al Si Al joined by strong H-bond  no easy separation 0.72 nm Typically 70-100 layers joined by oxygen sharing Kaolinite is used for making paper, paint and in pharmaceutical industry.

157. kaolinite 158

158. Click here, for 3D view of Kaolinite kaolinite

159. Illite The chemistry of Illite is KAl 2 [(OH) 2 |AlSi 3 O 10 ], Potassium aluminium silicate hydroxide hydrate mineral of the clay group. The Illite clay mineral consists of an octahedral layer of gibbsite sandwiched between two layers of silica tetrahedral. This produces a 1:2 mineral with the additional difference that some of the silica positions are filled with aluminium atoms and potassium ions are attached between layers to make up the charge deficiency. 160

160. Illite This bonding results in a less stable condition than for Kaolinite, and thus the activity of Illite is greater. Illite is found in a wide variety of environments, the materials of this series form by either weathering or hydrothermal alteration of muscovite. Illite can be formed from kaolin , if potassium is added. 161

161. Illite 162

162. 163 Illite Si Al Si Si Al Si Si Al Si 0.96 nm joined by K + ions fit into the hexagonal holes in Si-sheet

163. Illite 164

164. Montmorillonite The chemistry of Montmorillonite is (Na, Ca ) (Al, Mg) 6 (Si 4 O 10 ) 3 (OH) 6 - nH 2 O, Hydrated Sodium Calcium Aluminium Magnesium Silicate Hydroxide. Montmorillonite is a member of the general mineral groups the clays. It swells in water and posses high cation -exchange capacities. It typically forms microscopic or at least very small platy micaceous crystals. The water content is variable, and in fact when the crystals absorb water they tend to swell to several times their original volume. 165

165. Montmorillonite This makes montmorillonite a useful mineral for several purposes. It is the main constituent in a volcanic ash called bentonite , which is used in drilling muds. The bentonite gives the water greater viscosity ("thickness" of flow), which is very important in keeping a drill head cool during drilling and facilitating removal of rock and dirt from within a drill hole. 166

166. Montmorillonite Another important use of Montmorillonite is as an additive to soils and rocks. The effect of the Montmorillonite is to slow the progress of water through the soil or rocks. This is important to farmers with extended dry periods, engineers of earthen dams or levees or perhaps to plug up old drill holes to prevent leakage of toxic fluids from bottom levels to higher aquifers used for drinking water. 167

167. Montmorillonite As a mineral specimen, Montmorillonite does not get much consideration. Usually, pure samples of Montmorillonite are massive, dull and not very attractive. However, as with all minerals, there are those exceptional specimens that defy the norm. Montmorillonite has been found as attractive pink inclusions in quartz crystals, and these make for interesting specimens. 168

168. Montmorillonite 169 G or B Si Si (a) Schematic of basic unit (b) Montmorillonite cluster G G G     Na, Ca, -nH 2 O bond Na, Ca, -nH 2 O bond

169. 170 Montmorillonite Different combinations of tetrahedral and octahedral sheets form different clay minerals : 2:1 Clay Mineral (e.g., Montmorillonite, Illite)

170. 171 Montmorillonite Si Al Si Si Al Si Si Al Si 0.96 nm joined by weak van der Waal’s bond  easily separated by water also called smectite ; expands on contact with water

171. Montmorillonite 172

172. 173 Summary - Montmorillonite Montmorillonite have very high specific surface, cation exchange capacity, and affinity to water. They form reactive clays. Bentonite (a form of Montmorillonite) is frequently used as drilling mud. Montmorillonite have very high liquid limit (100+), plasticity index and activity (1-7).

173. Vermiculite Vermiculite is a 2:1 clay, meaning it has 2 tetrahedral sheets for every one octahedral sheet. It is a limited expansion clay with a medium shrink-swell capacity. Vermiculite has a high cation exchange capacity at 100-150 meq /100 g. Vermiculite clays are weathered micas in which the potassium ions between the molecular sheets are replaced by magnesium and iron ions. 174

174. CLAY MINERALS 175

175. Soil fabric The arrangement and organisation of particles and other features within a soil mass is termed its structure or fabric. This includes bedding orientation, stratification, layer thickness, the occurrence of joints and fissures, the occurrence of voids, artefacts, tree roots and nodules, the presence of cementing or bonding agents between grains. 176

176. Soil fabric The soil fabric is the brain it retains the memory of the birth of the soil and subsequent changes that occur. Two common types of soil fabric flocculated and dispersed are formed. flocculated structure : If a flocculated structure, formed under a salt environment, many particles tend to orient parallel to each other. When a flocculated structure formed under fresh water, many particles tend to orient perpendicular to each other. 177

177. Soil fabric Dispersed structure : A dispersed structure is the result when a majority of the particles orient parallel to each other. 178

178. 179 Clay Fabric Flocculated Dispersed edge-to-face contact face-to-face contact

179. 180 Clay Fabric Electrochemical environment (i.e., pH, acidity, temperature, cations present in the water) during the time of sedimentation influence clay fabric significantly . Clay particles tend to align perpendicular to the load applied on them.

180. Soil fabric 181

181. Defuse double layer Diffuse double layer ( DDL ) is an ionic structure that describes the variation of electric potential near a charged surface, such as clay, and behaves as a capacitor. The surface charges on fine-grained soils are negative. These negative surface charges attract cations and the positively charged side of water molecules from surrounding water. 182

182. Defuse double layer Consequently , a thin film or layer of water, called adsorbed water, is bonded to the mineral surfaces. The thin film or layer of water is known as the diffuse double layer. The largest concentration of cations occurs at the mineral surface and decreases exponentially with distance away from the surface. 183

183. Defuse double layer 184

184. Defuse double layer 185

185. 186 Cation Concentration in Water + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + cations cation concentration drops with distance from clay particle - - - - - - - - - - - - - - clay particle double layer free water

186. 187 Adsorbed Water - - - - - - - - - - - - - - A thin layer of water tightly held to particle; like a skin 1-4 molecules of water (1 nm) thick more viscous than free water adsorbed water

187. 188 Clay Particle in Water - - - - - - - - - - - - - - free water double layer water adsorbed water 50 nm 1nm

188. 189 A Clay Particle Plate-like or Flaky Shape

189. 190 Isomorphous Substitution substitution of Si 4+ and Al 3+ by other lower valence (e.g., Mg 2+ ) cations results in charge imbalance (net negative) + + + + + + + _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ positively charged edges negatively charged faces Clay Particle with Net negative Charge

190. Cation exchange capacity 191 Cation -exchange capacity is defined as the degree to which a soil can adsorb and exchange cations . Cation -a positively charged ion (NH 4 + , K + , Ca 2+ , Fe 2+ , etc...) Anion-a negatively charged ion (NO 3 - , PO 4 2- , SO 4 2- , etc...)

191. 192 Cation Exchange Capacity (c.e.c) capacity to attract cations from the water (i.e., measure of the net negative charge of the clay particle) measured in meq / 100g (net negative charge per 100 g of clay) milliequivalents known as exchangeable cations The replacement power is greater for higher valence and larger cations. Al 3+ > Ca 2+ > Mg 2+ >> NH 4 + > K + > H + > Na + > Li +

192. Specific surface 193 Specific surface area " SSA " is a property of solids which is the total surface area of a material per unit of mass, solid or bulk volume, or cross-sectional area. The specific surface area of natural particles is an important parameter to quantify processes such as mineral dissolution and sorptive interactions in soils and sediments.

193. 194 Specific Surface surface area per unit mass (m 2 /g) smaller the grain, higher the specific surface e.g., soil grain with specific gravity of 2.7 10 mm cube 1 mm cube spec. surface = 222.2 mm 2 /g spec. surface = 2222.2 mm 2 /g

194. 195 A Comparison Mineral Specific surface (m 2 /g) C.E.C (meq/100g) Kaolinite 10-20 3-10 Illite 80-100 20-30 Montmorillonite 800 80-120 Chlorite 80 20-30

195. 196 Identifying clay minerals

196. 197

197. 198 Others… X-Ray Diffraction (XRD) Differential Thermal Analysis ( DTA ) to identify the molecular structure and minerals present to identify the minerals present

198. 199 Scanning Electron Microscope common technique to see clay particles qualitative plate-like structure

199. 200 Casagrande’s PI-LL Chart montmorillonite illite kaolinite chlorite halloysite

200. Soil types Bentonite is a clay formed by the decomposition of volcanic ash with a high content of montmorillonite . It exhibits the properties of clay to an extreme degree. Varved Clays consist of thin alternating layers of silt and fat clays of glacial origin. They possess the undesirable properties of both silt and clay. The constituents of varved clays were transported into fresh water lakes by the melted ice at the close of the ice age . 201

201. Soil types Kaolin , China Clay are very pure forms of white clay used in the ceramic industry. Boulder Clay is a mixture of an unstratified sedimented deposit of glacial clay, containing unsorted rock fragments of all sizes ranging from boulders, cobbles, and gravel to finely pulverized clay material. 202

202. Soil types Calcareous Soil is a soil containing calcium carbonate. Such soil effervesces when tested with weak hydrochloric acid. Marl consists of a mixture of calcareous sands, clays, or loam. Hardpan is a relatively hard, densely cemented soil layer, like rock which does not soften when wet . Boulder clays or glacial till is also sometimes named as hardpan.. 203

203. Soil types Caliche is an admixture of clay, sand, and gravel cemented by calcium carbonate deposited from ground water. Peat is a fibrous aggregate of finer fragments of decayed vegetable matter. Peat is very compressible and one should be cautious when using it for supporting foundations of structures. Loam is a mixture of sand, silt and clay . 204

204. Soil types Loess is a fine-grained, air-borne deposit characterized by a very uniform grain size, and high void ratio . The size of particles ranges between about 0.01 to 0.05 mm. The soil can stand deep vertical cuts because of slight cementation between particles. It is formed in dry continental regions and its color is yellowish light brown . 205

205. Soil types Shale is a material in the state of transition from clay to slate. Shale itself is sometimes considered a rock but, when it is exposed to the air or has a chance to take in water it may rapidly decompose. 206

206. 207 Summary - Clays Clay particles are like plates or needles. They are negatively charged. Clays are plastic; Silts, sands and gravels are non-plastic. Clays exhibit high dry strength and slow dilatancy.

207. Future tasks Organic clay Black cotton soil humic acid 208

208. Organic clay ORGANIC clay is defined as a clay with sufficient organic content to influence the soil properties [ASTM, D2487]. Organic clays are normally characterized by high plasticity, high water content , high compressibility, low permeability and low shear strength . These characteristics make organic clays unsuitable for engineering construction unless a suitable improvement method is applied. 209

209. Humic acid Humic acid is a principal component of humic substances, which are the major organic constituents of soil. Humic acid has the average chemical formula C 187 H 186 O 89 N 9 S 1 and is insoluble in strong acid (pH = 1). 210

210. Humic acid 211

211. Humic acid 212

212. Humic acid powder 213

213. Black cotton soil 214

214. lime treated clay Lime can be used to treat soils in order to improve their workability and load-bearing characteristics in a number of situations. 215

215. COMPOSITION & PHYSICAL PROPERTIES OF SOIL

216. Soil Composition While a nearly infinite variety of substances may be found in soils, they are categorized into four basic components: minerals, organic matter, air and water . M ineral matter obtained by the disintegration and decomposition of rocks; Organic matter, obtained by the decay of plant residues, animal remains and microbial tissues ; 217

217. Soil Composition Water , obtained from the atmosphere and the reactions in soil (chemical, physical and microbial); Air or gases, from atmosphere, reactions of roots, microbes and chemicals in the soil 218

218. SOIL PHASE RELATIONSHIPS Soil mass is generally a three phase system. It consists of solid particles (i.e., minerals and organic matters), liquid and gas. The phase system may be expressed in terms of mass-volume or weight-volume relationships. 219

219. Phase diagram of soil 220

220. Phase diagram of soil 221

221. Volumetric Ratios There are three volumetric ratios that are very useful in geotechnical engineering and these can be determined directly from the phase diagram: 222

222. Volumetric Ratios Voids ratio Degree of saturation 223

223. Volumetric Ratios Porosity 224

224. Volumetric Ratios Percentage air voids:   225

225. Volumetric Ratios Air content: Air content, ac is the ratio of the volume of air voids to the volume of voids.   226

226. Mass-volume relationship Unit weight Unit weight of solid 227

227. Soil composition Saturated unit weight Submerged unit weight 228

228. index properties of soil Index properties are the properties of soil that help in identification and classification of soil. Water content, Specific gravity, Particle size distribution, In situ density (Bulk Unit weight of soil), Consistency Limits and relative density are the index properties of soil. These properties are generally determined in the laboratory. In situ density and relative density require undisturbed sample extraction while other quantities can be determined from disturbed soil sampling. 229

229. Index Properties of soil The following are the Index Properties of soil. 1 . Water content 2 . Specific Gravity 3. In-situ density 4 . Particle size distribution 5 . Consistency limits 6 . Relative Density 230

230. Index Properties of soil Water content 231

231. Index Properties of soil Specific Gravity : The ratio of the specific weight of a substance to the specific weight of water is called specific gravity . This number indicates how much heavier or lighter a material is than water. In soils, SG refers to the mass of solid matter of a given soil sample as compared to an equal volume of water. 232

232. Relative density bottle (RD-Bottle) 233

233. Specific Gravity 1 ft 3 Soil Solids Water W = 187.2 lbs W= 62.4 lbs SG = 187.2/62.4 = 3.0 1 ft 3

234. General Ranges of SG For Soils Sand 2.63 – 2.67 Silt 2.65-2.7 Clay & Silty Clay 2.67-2.9 Organic Soils <2.0

235. Specific Gravity For example: A material with SG of 2 is twice as heavy as water (2x 62.4 lbs /ft 3 ) = 124.8 lbs /ft 3

236. In-situ density   237

237. Density of water 238

238. In-situ density test methods Sand Cone Method Rubber Balloon Method Nuclear Density Method 239

239. Sand cone apparatus 240

240. In-situ density by Sand cone apparatus 241

241. In-situ density by Sand cone apparatus 242

242. In-situ density by Sand cone apparatus 243

243. Rubber balloon apparatus 244

244. Nuclear density method A nuclear density gauge is a tool used in civil construction and the petroleum industry , as well as for mining and archaeology purposes. It consists of a radiation source that emits a directed beam of particles and a sensor that counts the received particles that are either reflected by the test material or pass through it. By calculating the percentage of particles that return to the sensor, the gauge can be calibrated to measure the density and inner structure of the test material. 245

245. Nuclear density apparatus 246

246. Consistency limits Liquid limit Plastic limit Shrinkage limit 248

247. Liquid limit apparatus 249

248. Liquid limit apparatus 250

249. example No, of blows Can no. mass of can, M1 (g) Mass of the can+wet soil, M2 (g) mass of the can +dry soil, M3 (g) moisture content, w(%) 30 25 15.4 32.36 30.35 13.44 16 20 15.6 40.2 36.96 15.17 9 10 15.39 42.65 38.75 16.70 5 33 15.85 52.9 47.25 17.99 251

250. solution 252

251. Plastic limit 253

252. Plastic limit 254

253. Shrinkage limit 255

254. Cracks due to shrinkage 256

255. Atterberg limits 257

256. Use of Atterberg limit 258

257. Use of Atterberg limit 259

258. Free swell index of soil Free Swell Index Degree of expansiveness LL PL SL <20 Low 0.50 0-35% >17% 20-35 Moderate 40-60% 25-50% 8-18% 35-50 High 50-75% 35-65% 6-12% >50 Very high >60% >45% <10% 260

259. Relative density in which e max = void ratio of sand in its loosest state having a dry density of p dmin e min void ratio in its densest state having a dry density of p dMax e = void ratio under in-situ condition having a dry density of pd   261

260. Maximum void ratio 262

261. Minimum void ratio 263

262. Interrelationships Void ratio (e) in terms of porosity (  ) i.e. e = f (  ) 264

263. Interrelationships Void ratio (e) in terms of porosity (  ) i.e. e = f (  )   265

264. Interrelationships Porosity in terms of void ratio   266

265. Interrelationships 267

266. Exercise Determine the void ratio of a soil whose porosity is 32%. A laboratory test reveals that the void ratio of a certain soil is 1.234. What is the porosity of this soil? [ Ans : e = 0.471,  =55.3%] 268

267. Interrelationships Prove that   269

268. Interrelationships   270

269. Interrelationships Dry unit weight in terms of specific gravity, void ratio and unit weight of water. i.e , 271

270. Interrelationships 272

271. Interrelationships Dry unit weight as a function of in-situ unit weight and water content : 273

272. Interrelationships 274

273. Interrelationships Dry unit weight in terms of specific gravity, water content, degree of saturation and specific weight of water i.e., 275

274. Interrelationships 276

275. Interrelationships Specific weight in terms of specific gravity, void ratio, and degree of saturation 277

276. Interrelationships 278

277. Interrelationships Saturated unit weight in terms of specific gravity, voids ratio and specific weight of water i.e., 279

278. Interrelationships 280

279. Interrelationships Submerged unit weight in terms of specific gravity, voids ratio and unit weight of water i.e. 281

280. Interrelationships 282

281. Interrelationships Prove that 283

282. Interrelationships 284

283. Interrelationships 285

284. example The weight of a chunk of moist soil sample is 45.6 lb. the volume of the soil chunk measured before drying is 0.40 ft 3 . After being dried out in an oven, the weight of dry soil is 37.8 lb. the specific gravity of soil is 2.65. Determine water content, unit weight of moist soil void ratio, porosity, and degree of saturation. 286

285. example The dry density of a soil is 1.78 g/cm 3 . It is determined that the void ratio is 0.55. What is the moist unit weight if S = 50 %? If S = 100 %? Assume G = 2.76 287

286. example For a soil in natural state, given e = 0.8, w = 24%, and G s = 2.68, Determine the moist unit weight, dry unit weight, and degree of saturation. 288

287. Exercise -I An oven tin containing a sample of moist soil was weighed and had a mass of 37.82 g; the empty tin had a mass of 16.15 g. After drying, the tin and soil were weighed again and had a mass of 34.68 g. Determine the void ratio of the soil if the air void content is (a) zero (b) 5 % take G s = 2.70. [ Ans : (a) e = 0.456, (b) e = 0.533] 289

288. Exercise-II The moist mass of a soil specimen is 20.7 kg. The specimen’s volume measured before drying is 0.011 m 3 . The specimens dried mass is 16.3 kg. The specific gravity of solids is 2.68. Determine void ratio, degree of saturation, wet unit mass, dry unit mass, wet unit weight, and dry unit weight. 290

289. Exercise-III An undisturbed soil sample has e = 0.78, w = 12%, and G s = 2.68. Determine wet unit weight, dry unit weight, degree of saturation, and porosity. Ans: =16.49 kN/m 3 , =14.72 kN/m 3 , S r = 40.9 %,  = 44.0 %] 291

290. Exercise-IV A sample of dry sand having a unit weight of 16.5 kN /m 3 and specific gravity of 2.70 is placed in the rain. During the rain the volume of the sample remains constant but the degree of saturation increases to 40 %. Determine the unit weight and water content of the soil after being in the rain. 292

291. Exercise-V For a given soil, the in situ void ratio is 0.72 and Gs = 2.61. Calculate the porosity, dry unit weight ( lb /ft 3 , kN /m 3 ), and the saturated unit weight. What would the moist unit weight be when the soil is 60% saturated? 293

292. Exercise-VI A 100 % saturated soil has a wet unit weight of 120 lb /ft 3 . The water content of this saturated soil was determined to be 36 %. Determine void ratio, and specific gravity of solids. [ Ans : e =1.04, G s =2.88] 294

293. Exercise-VII A proposed earth dam will contain 50,000,000 m 3 of earth. Soil to be taken from a borrow pit will be compacted to a void ratio of 0.78. The void ratio pf soil in the borrow pit is 1.12. Estimate the volume of soil that must be excavated from the borrow pit. 295

294. Exercise-VIII A fibre reinforced cemented sandy soil sample of 50 mm diameter and 100 mm height is to be prepared for the subsequent testing of its engineering properties. The sample is to be compacted to a targeted dry density of (15 +RN/10) kN /m 3 at an optimum moisture content of 10%. If the soil is added with 5% cement and 0.25% fibre. 296

295. Exercise-VIII Total volume, Volume of solid, Volume of voids, Volume of water, Volume of air voids, Void ratio, Porosity. Determine the amount of sand, cement, fibre and water to be mixed for sample preparation. 297

296. Exercise-VIII Hint:   298

297. Soil gradation

298. Soil gradation Soil gradation is very important to geotechnical engineering. It is an indicator of other engineering properties such as compressibility, shear strength, and hydraulic conductivity. In a design, the gradation of the in situ or on site soil often controls the design and ground water drainage of the site. A poorly graded soil will have better drainage than a well graded soil because there are more void spaces in a poorly graded soil . 300

299. Soil gradation When a fill material is being selected for a project such as a highway embankment or earthen dam, the soil gradation is considered. A well graded soil is able to be compacted more than a poorly graded soil. These types of projects may also have gradation requirements that must be met before the soil to be used is accepted . 301

300. Soil gradation Grain size distribution Hydrometer Analysis 302

301. 303 Grain Size Distribution In coarse grain soils …... By sieve analysis Determination of GSD: In fine grain soils …... By hydrometer analysis Sieve Analysis Hydrometer Analysis hydrometer stack of sieves sieve shaker

302. Sieve analysis 304 Sieve Stack

303. Sieve analysis 305 Sieve Shaker

304. 306 Soil Dispersion Mixer Hydrometer analysis

305. What is a Hydrometer? Device used to determine directly the specific gravity of a liquid Consists of a thin glass tube closed at both ends Large bulb contains lead shot to cause the instrument to float upright in liquid. Scale is calibrated to indicate the specific gravity of the liquid.

306. Hydrometer analysis 308

307. Measure Sample Collect 50 g of fine from mechanical sieving procedure 1 ft 3

308. Required Test Equipment Mixer Scale Hydrometer Jar Scale Deflocculating Solution

309. Deflocculating Stage Add sample to 125 ml of 40g /L sodium hexometaphosphate (deflocculating solution) Allow to soak for at least 12 hours

310. Mix sample with spatula to dislodge settled particles

311. Sample Preparation Pour sample into mixing cup Use distilled water to rinse beaker

312. Add distilled water to the soil in mixer cup to make it about two-thirds full. Mix using mixer for two minutes Sample Preparation

313. Seal the jar or simply place hand over end and rotate about 60 times. Set the jar on the bench and record the time. This is time (t=0) on your data sheet. Take a hydrometer reading and temperature reading at prescribed intervals. After each reading the hydrometer is put into the transparent cylinder Starting the Test

314. 316 Definition of Grain Size Boulders Cobbles Gravel Sand Silt and Clay Coarse Fine Coarse Fine Medium 300 mm 75 mm 19 mm No.4 4.75 mm No.10 2.0 mm No.40 0.425 mm No.200 0.075 mm No specific grain size-use Atterberg limits

315. 317 General Guidance Coarse-grained soils: Gravel Sand Fine-grained soils: Silt Clay NO.200 0.075 mm Grain size distribution C u C c PL, LL Plasticity chart 50 % NO. 4 4.75 mm Required tests: Sieve analysis Atterberg limit LL>50 LL <50 50%

316. example Plot the gradation curve and determine the PSD parameters. 318 Sieve No. 4 10 20 50 80 100 120 200 Pan Sieve opening (mm) 4.3 2.0 0.9 0.3 0.2 0.2 0.1 0.1 0.0 Mass retained (grams) 2.0 2.5 18.4 124.0 476.6 192.2 10.1 72.2 68.0 percent retained 0.2 0.3 1.9 12.8 49.3 19.9 1.0 7.5 7.0 cumulative % retained 0.2 0.5 2.4 15.2 64.5 84.4 85.5 93.0 100.0 Cumulative % passing 99.8 99.5 97.6 84.8 35.5 15.6 14.5 7.0 0.0

317. example 319

318. example 320 < 1% 93% 7% Gravel Sand Fine

319. Soil grading 321

320. well graded soil A well graded soil is a soil that contains particles of a wide range of sizes and has a good representation of all sizes from the No. 4 to No. 200 sieves. A well graded gravel is classified as GW while a well graded sand is classified as SW. 322

321. poorly graded soil A poorly graded soil is a soil that does not have a good representation of all sizes of particles from the No. 4 to No. 200 sieve. Poorly graded soils are either uniformly graded or gap-graded. A poorly graded gravel is classified as GP while a poorly graded sand is classified as SP. Poorly graded soils are more susceptible to soil liquefaction than well graded soils . 323

322. Psd parameters the Coefficients of Uniformity: Calculating the coefficients of uniformity and curvature requires grain diameters. The grain diameter can be found for each percent of the soil passing a particular sieve. This means that if 40% of the sample is retained on the No. 20 sieve then there is 60% passing the No. 20 sieve . 324

323. gap-graded soil A gap-graded soil is a soil that has an excess or deficiency of certain particle sizes or a soil that has at least one particle size missing. An example of a gap-graded soil is one in which sand of the No. 10 and No. 40 sizes are missing, and all the other sizes are present . 325

324. Gradation curves 326

325. Psd parameters the Coefficients of Uniformity : The coefficient of uniformity , C u is a crude shape parameter and is calculated using the following equation: where D 60 is the grain diameter at 60% passing, and D 10 is the grain diameter at 10% passing.   327

326. Psd parameters The Coefficients of Curvature : The coefficient of curvature , C c is a shape parameter and is calculated using the following equation:   328

327. Psd parameters 329

328. Criteria for Grading Soils The following criteria are in accordance with the Unified Soil Classification System: For a gravel: C u > 4 & 1 < C c < 3 If both of these criteria are not met, the gravel is classified as poorly graded or GP. If both of these criteria are met, the gravel is classified as well graded or GW . 330

329. Criteria for Grading Soils For a sand : C u ≥ 6 & 1 < C c < 3 If both of these criteria are not met, the sand is classified as poorly graded or SP. If both of these criteria are met, the sand is classified as well graded or SW. 331

330. Psd parameters Fineness modulus ( FM ): An empirical factor obtained by adding the total percentages of a sample of the aggregate retained on each of a specified series of sieves, and dividing the sum by 100 .   332

331. Fineness modulus The fineness modulus is an empirical factor that gives a relative measure of the proportions of fine and coarse particles in an aggregate. It is a value widely used to indicate the relative fineness or coarseness of an aggregate . 333

332. Fineness modulus Soil Type   Fineness Modulus       Fine sand   2.20 to 2.60 Medium sand   2.60 to 2.90 Coarse sand   2.90 to 3.20 Fine aggregate 2.70 to 3.0 2.0 to 4.0 Coarse aggregate 6.50 to 8.0 334

333. Group Index he group index was devised to provide a basis for approximating within-group evaluations. Group indexes range from 0 for the best subgrade soils to 20 for the poorest . The higher the value of GI the weaker will be the soil and vice versa. Thus, quality of performance of a soil as a subgrade material is inversely proportional to GI. 335

334. 336 Group Index For Group A-2-6 and A-2-7 The first term is determined by the LL The second term is determined by the PI In general, the rating for a pavement subgrade is inversely proportional to the group index, GI. use the second term only F200: percentage passing through the No.200 sieve

335. example U.S. sieve size Size opening (mm) Weight retained (g) ¾ in. 19.0 3/8 in. 9.50 158 No.4 4.75 308 No. 10 2.00 608 No. 40 0.425 652 No. 100 0.150 224 No. 200 0.075 42 Pan -- 8 337 An air-dry soil sample weighing 2000g is brought to the soils laboratory for mechanical grain size analysis. The laboratory data are as follows: Plot a grain size distribution curve for this soil sample. And with the help of distribution curve determine Cc and Cu. as well.

336. solution U.S. sieve size Size opening (mm) Weight retained (g) Percentage retained Cumulative percentage retained Percentage passing ¾ in. 19.0 100 3/8 in. 9.50 158 7.9 7.9 92.1 No.4 4.75 308 15.4 23.3 76.7 No. 10 2.00 608 30.4 53.7 46.3 No. 40 0.425 652 32.6 86.3 13.7 No. 100 0.150 224 11.2 97.5 2.5 No. 200 0.075 42 2.1 99.6 0.4 Pan -- 8 0.4 100 -- 338

337. solution 339

338. Identification and CLASSIFICATION of Soils

339. Identification and CLASSIFICATION of Soils Introduction Field identification of soils Engineering classification of soils Purpose of soil classification Unified soil classification system AASHTO soil classification system Textural soil classification system 341

340. introduction It is necessary to have a standard language for a careful description and classification of a soil. Soil classification is the arrangement of soils into various groups or sub groups so as to express briefly the primary material characteristics without detailed descriptions. 342

341. 343 introduction Classifying soils into groups with similar behavior, in terms of simple indices, can provide geotechnical engineers a general guidance about engineering properties of the soils through the accumulated experience . Simple indices GSD, LL, PI Classification system (Language) Estimate engineering properties Achieve engineering purposes Use the accumulated experience Communicate between engineers

342. Field identification of soils Field identification of soil is of great importance for civil engineering. Sometimes the lack of time and facilities makes laboratory soil testing impossible in construction . Even when laboratory tests are to follow , field identification tests must be made during the soil exploration . Experience is the greatest asset in field identification; learning the technique from an experienced technician is the best method of acquiring the skill. 344

343. Field identification of soils Advantages of field identification:- •It is very economical. •It can be carried out in short duration of time. •No pre-setting is required Disadvantages of field identification:- •It is just an approximation. •You cannot completely rely on it. •Experienced people are required. 345

344. Visual Examination Colour of soil :- Visual examination should establish the colour, grain size , grain shapes (of the coarse-grained portion), some idea of the gradation, and some properties of the undisturbed soil. It helps in Unified classification of soil. Colour is often helpful in distinguishing between soil types, and with experience , one may find it useful in identifying the particular soil type. 346

345. Visual Examination Compressive strength :- Knowing the consistency index (consistency) of the soil we can have the value of compressive strength by field determination . However , this value is an approximate value and can be used as a guideline . It can be tabulated as follows:- 347

346. Visual Examination 348

347. Types of soil Sand, silt, and clay are the basic types of soil. Most soils are made up of a combination of the three. The texture of the soil, how it looks and feels, depends upon the amount of each one in that particular soil. The type of soil varies from place to place on our planet and can even vary from one place to another in your own backyard. 349

348. Types of soil Loam is a rich soil consisting of a mixture of sand and clay and decaying organic materials. Loam is composed of sand , silt , and clay in relatively even concentration (about 40-40-20% concentration respectively ). 350

349. 351 Organic Soils Highly organic soils- Peat (Group symbol PT) A sample composed primarily of vegetable tissue in various stages of decomposition and has a fibrous to amorphous texture, a dark-brown to black color, and an organic odor should be designated as a highly organic soil and shall be classified as peat, PT. O rganic clay or silt( group symbol OL or OH): “The soil’s liquid limit (LL) after oven drying is less than 75 % of its liquid limit before oven drying.” If the above statement is true, then the first symbol is O. The second symbol is obtained by locating the values of PI and LL (not oven dried) in the plasticity chart.

350. Purpose of soil classification 1 . Provides a concise and systematic method for designating various types of soil. 2. Enables useful engineering conclusions to be made about soil properties. 3. Provides a common language for the transmission of information. 4. Permits the precise presentation of boring records and test results. 352

351. 353 Classification Systems Two commonly used systems: Textural Classification of soil Unified Soil Classification System (USCS). American Association of State Highway and Transportation Officials ( AASHTO ) System

352. Soil texture triangle 354

353. examples Example 1 . What is the classification of a soil sample with 18% sand, 58% silt, and 24% clay? Entering the left axis at 18%, the bottom axis at 58%, and the right axis at 24%, and moving to the intersection point, the soil's classification is Silty Clay Loam . 355

354. examples Example 2 . What is the classification of a soil sample with 47% sand, 32% silt, and 21% clay? Entering the left axis at 47%, the bottom axis at 32%, and the right axis at 21%, and moving to the intersection point, the soil's classification is Clay Loam . Example 3 . What is the classification of a soil sample with 32% gravel, 38% sand, 22% silt, and 8% clay? 356

355. General Classification Granular Materials (35% or less passing No. 200 sieve) Silt Clay Materials (More than 35% passing No. 200 sieve) Group Classification A-1 A-2       A-7 A-1-a A-1-b A-3 A-2-4 A-2-5 A-2-6 A-2-7 A-4 A-5 A-6 A-7-5 A-7-6 Sieve Analysis %   2.00 mm (No .10) 50 max - - - - - - - - - - 0.425 mm (No. 40) 30 max 50 max 51 min - - - - - - - - 0.075 mm (No. 200) 15 max 25 max 10 max 35 max 35 max 35 max 35 max 36 min 36 min 36 min 36 min Characteristics of fraction passing   Liquid Limit - - 40 max 41 min 40 max 41 min 40 max 41 min 40 max 41 min Plasticity Index 6 max N.P 10 max 10 max 11 min 11 min 10 max 10 max 11 min 11 min 1 Usual types of significant constituent materials Stone fragments, gravel, and sand Fine sand Silty or clayey gravel and sand Silty soils Clayey soils General ratings as subgrade Excellent to good Fair to poor 357

356. example For the given sieve analysis data classify the soil according to the AASHTO Classification System. 358 Sieve No. 4 10 20 50 80 100 120 200 Pan Sieve opening (mm) 4.3 2.0 0.9 0.3 0.2 0.2 0.1 0.1 0.0 Mass retained (grams) 2.0 2.5 18.4 124.0 476.6 192.2 10.1 72.2 68.0 percent retained 0.2 0.3 1.9 12.8 49.3 19.9 1.0 7.5 7.0 cumulative % retained 0.2 0.5 2.4 15.2 64.5 84.4 85.5 93.0 100.0 Cumulative % passing 99.8 99.5 97.6 84.8 35.5 15.6 14.5 7.0 0.0

357. 359

358. 360 < 1% 93% 7% Gravel Sand Fine

359. solution 361

360. 362

361. 363

362. 364 Plasticity Chart (Holtz and Kovacs, 1981) LL PI H L The A-line generally separates the more claylike materials from silty materials, and the organics from the inorganics. The U-line indicates the upper bound for general soils. Note: If the measured limits of soils are on the left of U-line, they should be rechecked.

363. 365 Borderline Cases (Dual Symbols) For the following three conditions, a dual symbol should be used. Coarse-grained soils with 5% - 12% fines. About 7 % fines can change the hydraulic conductivity of the coarse-grained media by orders of magnitude . The first symbol indicates whether the coarse fraction is well or poorly graded. The second symbol describe the contained fines. For example: SP-SM, poorly graded sand with silt .

364. 366 Borderline Cases (Dual Symbols) Fine-grained soils with limits within the shaded zone. (PI between 4 and 7 and LL between about 12 and 25). It is hard to distinguish between the silty and more claylike materials. CL-ML: Silty clay, SC-SM: Silty, clayed sand. Soil contain similar fines and coarse-grained fractions. possible dual symbols GM-ML

365. 367 Borderline Cases (Summary) (Holtz and Kovacs, 1981)

366. 368 Definition of Grain Size Boulders Gravel Sand Silt-Clay Coarse Fine 75 mm No.4 4.75 mm No.40 0.425 mm No.200 0.075 mm No specific grain size-use Atterberg limits

367. 369 General Guidance 8 major groups: A1~ A7 (with several subgroups) and organic soils A8 The required tests are sieve analysis and Atterberg limits. The group index, an empirical formula, is used to further evaluate soils within a group (subgroups). The original purpose of this classification system is used for road construction (subgrade rating). A4 ~ A7 A1 ~ A3 Granular Materials  35% pass No. 200 sieve Silt-clay Materials  36% pass No. 200 sieve Using LL and PI separates silty materials from clayey materials Using LL and PI separates silty materials from clayey materials (only for A2 group)

368. Classification of soil based on sensitivity The degree of disturbance of undisturbed clay sample due to remoulding can be expressed as:   370 Classification of soil on the basis of St (After Skempton and Northey )

369. Exercise-I An air-dry soil sample is brought to the soils laboratory for mechanical grain size analysis. The laboratory data is given in the table. Plot a grain size distribution curve for this soil sample. And with the help of distribution curve determine Cc and Cu. as well. Also determine the percentage of gravel, sand, silt, and clay particles present in the sample, by using USC system. 371

370. Exercise-I U.S. sieve No. Size opening (mm) Mass retained (g) No.4 4.75 0.0 No. 10 2.00 40.0 No. 20 0.085 60.0 No. 40 0.425 89.0 No. 60 0.250 140.0 No. 80 0.177 122.0 No. 100 0.150 210.0 No. 200 0.075 56.0 Pan -- 12.0 372

371. Exercise-II A borrow pit site soil is to be checked for its suitability for using as a subgrade and embankment material for an earthen dam. The results of a sieving analysis are tabulated herein in Table 1. The liquid and plastic limit values were found to be (30+RN/10) % and (20+RN/10) % respectively. Classify the soil according to AASHTO and USCS Classification system. 373

372. Exercise-II Using these data, check the suitability of the soil to be used as subgrade or Earthfill dam embankment material. Suggest suitable type of soil improvement method if required for both purposes. 374

373. Exercise-II 375

374. Exercise-III 376

375. 377 References Main References: Das, B.M . (1998). Principles of Geotechnical Engineering , 4th edition, PWS Publishing Company. (Chapter 3) Holtz, R.D. and Kovacs, W.D . (1981). An Introduction to Geotechnical Engineering , Prentice Hall. (Chapter 3) Others: Santamarina , J.C., Klein, K.A ., and Fam , M.A. (2001). Soils and Waves , John Wiley & Sons, LTD.

376. SOIL COMPACTION

377. Soil compaction Introduction Compaction test Saturation (zero air void) line Laboratory compaction tests Field compaction 379

378. introduction Compaction of soil may be defined as the process by which the soil particles are artificially rearranged and packed together into a state of closer contact by mechanical means in order to decrease its porosity and thereby increase its dry density. This is usually achieved by dynamic means such as tamping, rolling, or vibration. The process of compaction involves the expulsion of air only. 380

379. introduction The following are the important effects of compaction : Compaction increases the dry density of the soil, thus increasing its shear strength and bearing capacity through an increase in frictional characteristics; Compaction decreases the tendency for settlement of soil ; and, Compaction brings about a low permeability of the soil. 381

380. Solids Water Air Solids Water Air Compressed soil Load Soil Matrix g soil (1) = W T1 V T1 g soil (2) = W T1 V T2 g soil (2) > g soil (1) introduction

381. Factor Affecting Soil Compaction : 1- Soil Type 2- Water Content ( w c ) 3- Compaction Effort Required (Energy) Why Soil Compaction : 1- Increase Soil Strength 2- Reduce Soil Settlement 3- Reduce Soil Permeability 4- Reduce Frost Damage 5- Reduce Erosion Damage Types of Compaction : (Static or Dynamic) 1- Vibration 2- Impact 3- Kneading 4- Pressure Water is added to lubricate the contact surfaces of soil particles and improve the compressibility of the soil matrix

382. Soil compaction The response of a soil to compaction depends on both the method of compaction and the moisture content. All other things being equal, the amount of compaction is greatest at an “ optimum moisture content”; the terminology comes from engineers. In general, soils compacted “wet-of-optimum” are more liable to either swell or to collapse when wetted, having higher permeability, a lower air entry value, greater ultimate strength at the moulding moisture content, and greater rigidity both at the moulding moisture content and when swollen. 384

383. Soil compaction Moulding moisture content, however, seems to have little effect on strength when swollen; judging from tests on foundry sands, dry strength increases with the moulding moisture content. There are reports of failure by piping erosion of earth dams that had been compacted dry-of-optimum. 385

384. Compaction test To determine the soil moisture-density relationship and to evaluate a soil as to its suitability for making fills for a specific purpose, the soil is subjected to a compaction test . Proctor (1933) showed that there exists a definite relationship between the soil moisture content and the dry density on compaction and that, for a specific amount of compaction energy used. 386

385. Compaction test There is a particular moisture content at which a particular soil attains its maximum dry density. Such a relationship provides a satisfactory practical approach for quality control of fill construction in the field. 387

386. Proctor compaction 388

387. Dry and wet of optimum The water content at compaction is also sometimes specified because of its effect on soil fabric, especially for clays. Clays compacted dry of optimum have a flocculated fabric, which generally corresponds to higher permeability, greater strength and stiffness, and increased brittleness. Conversely, clays compacted wet of optimum to the same equivalent dry density tend to have a more oriented or dispersed fabric, which typically corresponds to lower permeability, lower strength and stiffness, but more ductility. 389

388. Dry and wet of optimum 390 Effect of compacted water content on soil fabric for clays ( Coduto , 1999).

389. Zero air void line A line showing the relation between water content and dry density at a constant degree of saturation S may be established from the equation: 391

390. Zero air void line If one substitutes S = 100% and plots the corresponding line, one obtains the theoretical saturation line, relating dry density with water content for a soil containing no air voids. It is said to be ‘theoretical’ because it can never be reached in practice as it is impossible to expel the pore air completely by compaction. 392

391. Moisture Content Dry Density g d max Compaction Curve for Standard Proctor (OMC) g d max (OMC) Zero Air Void Curve S r < 100% Zero Air Void Curve S r =100% Zero Air Void Curve S r = 60% Compaction Curve for Modified Proctor Zero air void line

392. Compaction efforts The amount of compaction ( energy applied per unit of volume) is called compaction efforts. Increase in the compaction efforts results in an increase in the maximum dry density and decrease the OMC. In laboratory compaction efforts are applied through Standard Proctor Test or Modified Proctor Test. In both the cases the compaction energy is given as: 394

393. Compaction efforts The degree of compaction is not directly proportional to compaction efforts and dry density doesn’t increase indefinitely. When the soil is initially loose, the compaction increases the dry density, but further compaction beyond certain point doesn’t increase the density . 395

394. Water Content Dry Density Effect of Energy on Soil Compaction Higher Energy ZAV Increasing compaction energy Lower OWC and higher dry density In the field increasing compaction energy = increasing number of passes or reducing lift depth In the lab increasing compaction energy = increasing number of blows Effect of compaction efforts

395. Soil Compaction in the Lab : 1- Standard Proctor Test 2- Modified Proctor Test 3- Gyratory Compaction Standard Proctor Test Modified Proctor Test Gyratory Compaction

396. Soil Compaction in the Lab : 1- Standard Proctor Test w c1 w c2 w c3 w c4 w c5 g d1 g d2 g d3 g d4 g d5 Optimum Water Content Water Content Dry Density g d max Zero Air Void Curve S r =100% Compaction Curve 1 2 3 4 5 (OWC) 4 inch diameter compaction mold. (V = 1/30 of a cubic foot) 5.5 pound hammer 25 blows per layer H = 12 in Wet to Optimum Dry to Optimum Increasing Water Content g dry = g wet W c 100 % 1+ g ZAV = G s g w W c G s 1+ S r

397. Compaction of unsaturated soils 399 Dry of optimum Optimum moisture content Wet of optimum liable to either swell or to collapse when wetted, having higher permeability

398. Compaction characteristics 400 Some cohesionless soils exhibit two peaks in the compaction curve; one at very dry conditions, where there are no capillary tensions to resist the compaction effort, and the other at the optimum moisture content, where optimum lubrication between particles occurs.

399. 401 Compaction characteristics

400. 402 Compaction characteristics

401. 1- Rammers 2- Vibratory Plates  3- Smooth Rollers 4- Rubber-Tire 5- Sheep foot Roller 6- Dynamic Compaction Field Compaction

402. Field Soil Compaction Because of the differences between lab and field compaction methods, the maximum dry density in the field may reach 90% to 95%. Moisture Content Dry Density g d max (OMC) ZAV 95% g d max

403. Example: The laboratory test for a standard proctor is shown below. Determine the optimum water content and maximum dry density. If the G s of the soil is 2.70 , draw the ZAV curve. Solution: g dry = g wet W c 100 % 1+ Volume of Proctor Mold (ft 3 ) 1/30 1/30 1/30 1/30 1/30 1/30 Weight of wet soil in the mold (lb) 3.88 4.09 4.23 4.28 4.24 4.19 Water Content (%) 12 14 16 18 20 22 Volume of Mold (ft 3 ) 1/30 1/30 1/30 1/30 1/30 1/30 Weight of wet soil in the mold (lb) 3.88 4.09 4.23 4.28 4.24 4.19 Water Content (%) 12 14 16 18 20 22 Wet Unit Weight (lb/ft 3 ) 116.4 122.7 126.9 128.4 127.2 125.7 Dry Unit Weight (lb/ft 3 ) 103.9 107.6 109.4 108.8 106.0 103.0 g ZAV = G s g w W c G s 1+ S r

404. g dry max Optimum Water Content

405. example A proctor compaction test was conducted on a soil sample, and the following observations were made : If the volume of the mold used was 950 cm 3 and the specific gravity of soils grains was 2.65. 410 Water content, percent 7.7 11.5 14.6 17.5 19.7 21.2 Mass of wet soil, g 1739 1919 2081 2033 1986 1948

406. example Make necessary calculations and draw, (i) compaction curve and (ii) 80% and 100% saturation lines . 411

407. Checking Soil Density in the Field: 1- Sand Cone (ASTM D1556-90) 2- Balloon Dens meter The same as the sand cone, except a rubber balloon is used to determine the volume of the hole 3- Nuclear Density (ASTM D2292-91 ) Nuclear Density meters are a quick and fairly accurate way of determining density and moisture content.  The meter uses a radioactive isotope source (Cesium 137) at the soil surface (backscatter) or from a probe placed into the soil (direct transmission).  The isotope source gives off photons (usually Gamma rays) which radiate back to the mater's detectors on the bottom of the unit.  Dense soil absorbs more radiation than loose soil and the readings reflect overall density.  Water content (ASTM D3017) can also be read, all within a few minutes.  A small hole (6" x 6" deep) is dug in the compacted material to be tested.  The soil is removed and weighed, then dried and weighed again to determine its moisture content.  A soil's moisture is figured as a percentage.  The specific volume of the hole is determined by filling it with calibrated dry sand from a jar and cone device.  The dry weight of the soil removed is divided by the volume of sand needed to fill the hole.  This gives us the density of the compacted soil in lbs per cubic foot.  This density is compared to the maximum Proctor density obtained earlier, which gives us the relative density of the soil that was just compacted.

408. Nuclear Density Sand Cone

409. Compaction Specifications: Compaction performance parameters are given on a construction project in one of two ways: 1- Method Specification detailed instructions specify machine type , lift depths , number of passes , machine speed and moisture content .  A "recipe" is given as part of the job specifications to accomplish the compaction needed.  2- End-result Specification Only final compaction requirements are specified (95% modified or standard Proctor). This method, gives the contractor much more flexibility in determining the best, most economical method of meeting the required specs. 

410. Exercise-I An embankment for a highway is to be constructed from a soil compacted to a dry unit weight of 18 kN /m 3 . The clay to be trucked to the site from a borrow pit. The bulk unit weight of the soil in the borrow pit is 17 kN /m 3 and its natural moisture content is 5%. Calculate the volume of clay from the borrow pit required for 1 cubic meter of embankment. Assume Gs =2.7. 416

411. Exercise-I (Continue) Estimate the amount of water to be added per cubic meter of the soil, if the optimum moisture content to achieve the targeted dry density of 18 kN /m 3 is 7%. 417

412. Exercise-II Soil from a borrow pit to be used for construction of an embankment as shown in Figure gave the following laboratory results when subjected to the ASTM D 698 Standard Proctor test: Maximum dry unit weight = 118.5 lb /ft 3 , Optimum moisture content = 12.5% The soil on the borrow pit site is found to have: Maximum dry unit weight = (110 +RN/10) lb /ft 3 , Moisture content = (5 +RN/10) %. 418

413. Exercise-II Determine the following for 1 Km Road section: Amount of soil to be excavated from borrow pit. Amount of water to be added in the embankment. 419

414. Exercise-III The contractor, during construction of the soil embankment, conducted a sand-cone in-place density test to determine whether the required compaction was achieved. The following data were obtained during the test: Weight of sand used to fill test hole and funnel of sand-cone device = 845 g. Weight of sand to fill funnel = 323g. 420

415. Exercise-II Unit weigh of sand = 100 lb /ft 3 . Weigh of wet soil from test hole = 600g. Moisture content of soil from test hole = 17%. Based on the contract, the contractor is supposed to attain the 95% compaction. Will you approve the contractor’s work? 421

416. Exercise-III Condition The combined weight of a mold and the specimen of compacted soil it contains is 9.0 lb. The mold’s volume is 1/35ft 3 The mold’s weight is 4.35 lb. The specimen’s water content is 12%. What is dry unit weight of the specimen? 422

417. Soil hydraulics

418. Soil hydraulics 424

419. Soil hydraulics 425 Baluchistan 2012

420. permeability 426

421. permeability Measurement of soil/rock sample permeability is of significant importance. For instance, Pumping water, oil or gas into or out of a porous formation Disposing of brine wastes in porous formations Storing fluids in mined caverns for energy conversions Predicting water flow into a tunnel 427

422. 428

423. Geological formations of soil Aquifer - An aquifer is an underground layer of water-bearing permeable rock or unconsolidated materials (gravel, sand, or silt) from which groundwater can be usefully extracted using a water well . 429

424. Geological formations of soil 430 AQUIFER

425. Geological formations of soil Aquitard -Geological formation that may contain groundwater but is not capable of transmitting significant quantities of it under normal hydraulic gradients. May function as confining bed. 431

426. Geological formations of soil 432 AQUITARD

427. Aquifuge - a formation which has no interconnected openings and hence cannot absorb or transmit water. 433 Geological formations of soil

428. Aquiclude - a formation which , although porous and capable of absorbing water , does not permit its movement at rates sufficient to furnish an appreciable supply for a well or spring. 434 Geological formations of soil

429. 435 Geological formations of soil

430. 436 Geological formations of soil

431. aqueduct -An aqueduct is a water supply or navigable channel constructed to convey water. In modern engineering, the term is used for any system of pipes, ditches, canals, tunnels, and other structures used for this purpose . 437 Geological formations of soil

432. 438 Geological formations of soil

433. Darcy’s law 439 Sand filter h u/s d/s q q

434. 440

435. Coefficient of PERMEABILITY 441

436. 442 Coefficient of permeability versus particle size ( Novac 1982)

437. Flow velocity 443 This velocity is an average velocity, since it represents flow rate divided by gross cross-sectional area of the rock mass/ soil. This area however includes both solid soil material and voids. Since water or oil moves only through the voids, the actual (interstitial) velocity is,

438. Flow velocity 444 v v v s q

439. example In a rock permeability test, it took 160.0 minutes for 156 cm 3 of water to flow through a rock sample, the cross-sectional area of which was 50.3 cm 2 . The voids ratio of the rock sample was 0.45. Determine the average and actual velocity of the water through the rock sample. 445

440. Permeability tests 446 Field permeability tests Laboratory Permeability tests Falling head permeability test Constant head permeability test

441. Aquifer recuperation 447

442. Aquifer recuperation (confined) 448 Aquifer H r 2 r r 1 h 1 h h 2 Impermeable layer

443. 449 From Darcy’s law Aquifer recuperation

444. 450 Integrating for whole of the cross-section Aquifer recuperation

445. 451 Aquifer recuperation

446. 452 Aquifer recuperation (Unconfined)

447. Unconfined aquifer Consider a pumping well is located in an unconfined, homogeneous aquifer; in this case the piezometric surface lies within the aquifer and water can be drawn from the hole thickness h. thus From Darcy’s law For very small portion with distance dr and height dh the discharge will be 453

448. Unconfined aquifer 454

449. Unconfined aquifer 455

450. scavenger well A well located between a good well (or group of wells) and a source of potential contamination, which is pumped (or allowed to flow) as waste to prevent the contaminated water from reaching the good well. 456

451. 457 scavenger

452. Injection wells An Injection well is a device that places fluid deep underground into porous rock formations, such as sandstone or limestone, or into or below the shallow soil layer. These fluids may be water, wastewater, brine (salt water), or water mixed with chemicals. Injection wells have a range of uses that include long term (CO 2 ) storage, waste disposal, enhancing oil production, mining, and preventing salt water intrusion. 458

453. Injection wells 459

454. Constant head permeability test 460

455. 461 From Darcy’s law Constant head permeability test

456. Example-1 A constant head permeability test was carried out on a cylindrical sample of sand 4 in. in diameter and 6 in. in height. 10 in 3 of water was collected in 1.75 min, under a head of 12 in. Compute the hydraulic conductivity in ft /year and the velocity of flow in ft /sec. 462

457. Falling head permeability test 463

458. 464 From Darcy’s law falling head permeability test With negative sign used to indicate a falling head. The flow of water into the specimen is therefore The flow of water through and out of the specimen is

459. Falling head permeability test 465

460. Falling head permeability test 466

461. Falling head permeability test Where a = area of cross section of burette A = area of cross section of sample L = length of sample t = time elapsed ( t 1 -t 2 ) h 1 = initial head of burette at time t 1 h 2 = final head of burette at time t 2 467

462. example The hydraulic conductivity of a soil sample was determined in a soil mechanics laboratory by making use of a falling head permeameter. The data used and the test results obtained were as follows : diameter of sample = 2.36 in, height of sample = 5.91 in, diameter of stand pipe = 0.79 in, initial head h Q = 17.72 in. final head h l = 11.81 in. Time elapsed = 1 min 45 sec. Determine the hydraulic conductivity in ft /day. 468

463. solution 469

464. solution 470

465. 471

466. Example-2 A sand sample of 35 cm 2 cross sectional area and 20 cm long was tested in a constant head permeameter . Under a head of 60 cm, the discharge was 120 ml in 6 min. The dry weight of sand used for the test was 1 120 g, and Gs = 2.68. Determine (a) the hydraulic conductivity in cm/sec, ( b) the discharge velocity, and (c) the seepage velocity. 472

467. Pinhole test Pinhole tests differentiates between dispersive and non dispersive clays. For dispersive clay the water flowing through the specimen carries a cloudy colored suspension of colloids, whereas water running through ordinary, erosion resistant clays is crystal clear . This test is used for evaluating clay soils for erodibility by flowing water through a small hole that is drilled through the compacted specimen. 473

468. Pinhole test under hydraulic heads (H) ranging between 50 and 1020 cm. Dispersibility is assessed by observing effluent colour and flow discharge through the hole, by visual inspection of the hole after the completion of the test . Prior to each pinhole experiment the soil is air-dried at room temperature ( 20°C ) for 3-4 days and is sieved at a mesh size of 1.25 cm diameter. 474

469. dispersive soil A dispersive soil is structurally unstable. In dispersive soils the soil aggregates – small clods – collapse when the soil gets wet because the individual clay particles disperse into solution . Using dispersive clay soils in hydraulic structures, embankment dams, or other structures such as roadway embankments can cause serious engineering problems if these soils are not identified and used appropriately . 475

470. dispersive soil 476

471. Pinhole test 477

472. Pinhole test 478

473. Seepage theory

474. Equation of continuity d z d x d y q q +  q x z y

475. Equation of continuity

476. Equation of continuity

477. Equation of continuity

478. Equation of continuity

479. Equation of continuity Three dimensional equation of continuity for isotropic and homogeneous soil. Two dimensional equation of continuity for isotropic and homogeneous soil.

480. 486

481. The concept of effective stress

482. introduction 488 The pressure transmitted through grain to grain at the contact points through a soil mass is termed as intergranular or effective pressure. It is known as effective pressure since this pressure is responsible for the decrease in the void ratio or increase in the frictional resistance of a soil mass.

483. introduction The importance of the forces transmitted through the soil skeleton from particle to particle was recognized in 1923 when Terzaghi presented the principle of effective stress , an intuitive relationship based on experimental data. The principle applies only to fully saturated soils and relates the following three stresses: 489

484. introduction The total normal stress (  ) on a plane within the soil mass, being the force per unit area transmitted in a normal direction across the plane, imagining the soil to be a solid (single-phase) material; The pore water pressure (u), being the pressure of the water filling the void space between the solid particles; The effective normal stress ( ` ) on the plane, representing the stress transmitted through the soil skeleton only. 490

485. introduction Effective stress (σ') acting on a soil is calculated from two parameters, total stress (σ) and pore water pressure ( u) according to : The principle of effective stress is the most important principle in soil mechanics. Deformations of soils are a function of effective stresses, not total stresses. The principle of effective stresses applies only to normal stresses and not to shear stresses.   491

486. Spring analogy The concept of effective stress in soil is analogue to the spring analogy which is expressed in the following figure

487. Ping pong balls 493

488. During consolidation 494  remains the same (=q) during consolidation . u decreases (due to drainage) while ’ increases transferring the load from water to the soil.   GL saturated clay q kPa A  u  ’   u  ’ q

489. During consolidation 495  remains the same (=q) during consolidation. u decreases (due to drainage) while ’ increases transferring the load from water to the soil. GL saturated clay q kPa A   u  ’    u  ’ q

490. 496

491. NO FLOW condition 497

492. Flow FROM TOP TO BOTTOM 498

493. Flow from BOTTOM TO TOP 499

494. Example-I Calculate the effective stress for a soil element at depth 5 m in a uniform deposit of soil, as shown in Figure. Assume that the pore air pressure is zero. 500

495. solution 501

496. solution 502

497. Example-II A borehole at a site reveals the soil profile shown in Figure. Plot the distribution of vertical total and effective stresses with depth. Assume pore air pressure is zero. 503

498. Example-II 504

499. solution 505

500. solution 506

501. solution 507

502. solution 508

503. Example-III The soil profile shown in figure, determine the present effective overburden pressure at the midheight of compressible clay layer. 509

504. Example-IV The depth of water in a well is 3 m. Below the bottom of the well lies a layer of sand 5 meters thick overlying a clay deposit. The specific gravity of the solids of sand and clay are respectively 2.64 and 2.70. Their water contents are respectively 25 and 20 percent. Compute the total, inter-granular and pore water pressures at points A and B shown in Figure. 510

505. Example-IV 511

506. solution 512

507. solution 513

508. Example-V A trench is excavated in fine sand for a building foundation, up to a depth of 13 ft. The excavation was carried out by providing the necessary side supports for pumping water. The water levels at the sides and the bottom of the trench are as given Fig . Examine whether the bottom of the trench is subjected to a quick condition if Gs = 2.64 and e = 0.7. If so, what is the remedy? 514

509. Example-V 515

510. Exercise-I A uniform layer of sand 10 m deep overlays bedrock. The water table is located 2 m below the surface of the sand which is found to have a voids ratio e = 0.7. Assuming that the soil particles have a specific gravity Gs = 2.7 calculate the effective stress at a depth 5 m below the surface. 516

511. Exercise-Ii A stratum of sand 2.5 m thick overlies a stratum of saturated clay 3 m thick. The water table is 1 m below the surface. For the sand, Gs = 2.65, e = 0.50 and for the clay G = 2.72, e = 1.1. Calculate the total and effective vertical stresses at depths of 1 m, 2.5 m and 5.5 m below the surface assuming that the sand above the water table is completely dry. 517

512. Exercise-III In a deep deposit of clay the water table lies 3 m below the soil surface. Calculate the effective stresses at depths of 1 m, 3 m and 5 m below the surface. Assume that the soil remains fully saturated above the water table.   518

513. Exercise-IV Plot the distribution of total stress, effective stress, and pore-water pressure with depth for the soil profile shown in Figure. Neglect capillary action and pore air pressure. 519

514. Exercise-V A clay layer 3.66 m thick rests beneath a deposit of submerged sand 7.92 m thick. The top of the sand is located 3.05 m below the surface of a lake. The saturated unit weight of the sand is 19.62 kN /m 3 and of the clay is 18.36 kN /m 3 . Compute (a) the total vertical pressure, (b) the pore water pressure, and (c) the effective vertical pressure at mid height of the clay layer. 520

515. Exercise-V 521

516. Exercise-VI A large excavation is made in a stiff clay whose saturated unit weight is 109.8 lb /ft 3 . When the depth of excavation reaches 24.6 ft , cracks appear and water begins to flow upward to bring sand to the surface. Subsequent borings indicate that the clay is underlain by sand at a depth of 36.1 ft below the original ground surface. What is the depth of the water table outside the excavation below the original ground level? 522

517. Exercise-VI 523

518. reference 524

519. CONSOLIDATION THEORY

520. introduction Consolidation may be defined as the” Time rate of compression of soil under static loading resulting to the dissipation of pore water pressure”. It is the gradual reduction in volume of a fully saturated soil of low permeability due to drainage of some of the pore water, the process continuing until the excess pore water pressure set up by an increase in total stress has completely dissipated; 526

521. introduction the simplest case is that of one-dimensional consolidation, in which a condition of zero lateral strain is implicit. The process of swelling, the reverse of consolidation, is the gradual increase in volume of a soil under negative excess pore water pressure . 527

522. CONSOLIDATION THEORY A general theory for consolidation, incorporating three-dimensional flow vectors is complicated and only applicable to a very limited range of problems in geotechnical engineering. For the vast majority of practical settlement problems, it is sufficient to consider that both seepage and strains take place in one direction only; this usually being vertical. 528

523. CONSOLIDATION THEORY One-dimensional consolidation specifically occurs when there is no lateral strain, e.g. in the Oedometer test. One-dimensional consolidation can be assumed to be occurring under wide foundations. 529

524. CONSOLIDATION THEORY 530

525. Terzaghi 1-d consolidation equation 531 x z y dx dy dz  Q in  Q out

526. Terzaghi one-dimensional equation To derive the equation for time rate of settlement using an element of the soil sample of thickness dz and cross-sectional area of dA = dxdy , we will assume the following:  The soil is saturated, isotropic and homogeneous Darcy’s law is valid Flow only occurs vertically The strains are very small 532

527. Terzaghi one-dimensional equation 533

528. Terzaghi one-dimensional equation Since the change in volume of the soil (  V) is equal to the change in volume of pore water expelled (  V w ), which is equal to the change in volume of the voids (  V v ) therefore , 534

529. Terzaghi one-dimensional equation 535

530. Terzaghi one-dimensional equation 536

531. Terzaghi one-dimensional equation 537 From Darcy’s law

532. Terzaghi one-dimensional equation 538 By Partial Differentiation with respect to depth z gives

533. one-dimensional equation 539 As there is no change in the volume of solid therefore, As we know that

534. Terzaghi one-dimensional equation 540 As we know that the change in total vertical pressure is equal to the change in pore water pressure. i.e.,  u =  v we can write

535. Terzaghi one-dimensional equation 541 where Therefore,

536. 542 TERZAGHI 1-D consolidation equation

537. 543 TERZAGHI 1-D consolidation equation

538. SOLUTION TO TERZAGHI EQUATION 544 Where

539. SOLUTION TO TERZAGHI EQUATION 545

540. DEGREE OF CONSOLIDATION The ratio, expressed as a percentage, of the amount of consolidation at a given time within a soil mass, to the total amount of Consolidation obtainable under a given stress condition. It is the ratio of the settlement occurred at a particular time and depth to the total expected settlement. This parameter can be expressed as 546

541. DEGREE OF CONSOLIDATION 547

542. DEGREE OF CONSOLIDATION 548

543. SOLUTION TO TERZAGHI EQUATION 549

544. SOLUTION TO TERZAGHI EQUATION 550

545. U-T v relation 551

546. U-  T v relation 552

547. U-  T v relation 553

548. U-  T v relation 554

549. U-  T v relation 555

550. U-  T v relation 556

551. U-  T v relation 557

552. U-  T v relation 558 b/a =0.15

553. U-  T v relation 559

554. Consolidation test The consolidation or Oedometer test is used to determine the compressibility characteristics of saturated undisturbed or remoulded soil. 560

555. Consolidometer

556. 562

557. Compressibility characteristics Compressibility is the degree to which a soil mass decreases in volume when supporting a load. A soil is a particulate material, consisting of solid grains and void spaces enclosed by the grains . The voids may be filled with air or other gas, with water or other liquid, or with a combination of these. The volume decrease of a soil under stress might be conceivably attributed to: 563

558. Compressibility characteristics Compression of the solid grains; Compression of pore water or pore air; Expulsion of pore water or pore air from the voids , thus decreasing the void ratio or porosity . 564

559. Compressibility characteristics Compressibility characteristics of soils forms one of the important soil parameters required in design considerations. Compression index, Cc, which is the slope of the linear portion of void ratio, e vs. logarithm of effective pressure p(log p) relationship, is extensively used for settlement determination. The e–log p is most often assumed to be linear at higher pressure range and hence Cc is taken as a constant. 565

560. Swelling characteristics The swelling behavior of expansive soils often causes unfavorable problems, such as differential settlement and ground heaving. 566

561. Obtaining t 90 from the experimental results

562. 571

563. CONSOLIDATION PARAMETERS k = c v m v  w

564. EXAMPLE-1 For the following given information of an Oedometer test calculate the consolidation parameters of the soil. Moisture content, % w= 32 Specific gravity, G = 2.7 Diameter, mm D = 74.94 Height, mm H =19.22

565. EXAMPLE-1

566. EXAMPLE-1

567. TIME FOR 90% CONSOLIDATION t 90 From the graph

568. From the graph

569. Total vertical displacement after each stage

570. Final void ratio

571. COMPRESSION CURVE

572. COEFFICIENT OF CONSOLIDATION C V Coefficient of consolidation C v Coefficient of volume compressibility m v

573. COEFFICIENT OF PERMEABILITY, k

574. ANALYSIS OF THE LABORATORY MEASUREMENTS FOR EACH STAGE Loading stage 1 2 3 Pressure  `z[kPa] 50 100 200 Initial void ratio, e 0.864 0.756 0.701 Initial height, H [mm] 18.81 17.715 17.162 Height change,  H 1.095 0.553 0.569 Void ratio change,  e 0.108 0.055 0.056 Final void ratio, e f 0.756 0.701 0.645 t 90 [min] 11.56 C v [m 2 /in] 6.49  10 -6 m v [m 2 /MN] 1.158 K [m/s] 12.275  10 -10

575. An 8 ft clay layer beneath a building is overlain by stratum of permeable sand and gravel and is underlain by impermeable bedrock. The total expected total settlement for the clay layer due to the footing load is 2.5 in. the coefficient of consolidation ( c v ) is 2.68  10 -3 in 2 /min. How many years it will take for 90 % of total expected consolidation settlement to take place. Compute the amount of consolidation settlement that will occur in one year? How many years will it take for consolidation settlement of one inch to take place? EXAMPLE-2

576. EXAMPLE-2 Sand and gravel Impermeable bedrock Clay 8.0 ft S t = 2.5 in c v = 2.68  10 -3 in 2 /min t years = ? for U = 90% S 1 year = ? for t = 1 year t = ? for S = 1 in

577. For T V = 0.15, U = 43 % EXAMPLE-2

578. EXAMPLE-2 For U = 40 %, T V = 0.126

579. EXAMPLE-3 A foundation is to be constructed on a site where the soil profile is as shown in the figure, the coefficient of consolidation C V = 4.96  10 -6 m 2 /min. How long it will take for half the expected consolidation settlement to take place if the clay layer is underlain by (a) permeable sand and gravel? (b) Impermeable bedrock?

580. EXAMPLE-3 Elev. 185.6 m Elev. 192 m Elev. 198 m Elev. 200 m Water table Sand and gravel Unit weight = 19.83.0 kN/m 3 Clay Unit weight = 17.10 kN/m 3 Water table Elev. 195.5 m

581. EXAMPLE Elev. 185 m Elev. 190 m Elev. 198 m Elev. 200 m Water table Sand and gravel Unit weight = 20 kN/m 3 Clay Unit weight = 17 kN/m 3 Water table Elev. 195 m

582. EXAMPLE C V = 4.96  10 -6 m 2 /min H dr = H/2 = 6.4/2 = 3.2 m (two way drainage) t =? U = 50 % For U = 50 %, T v = 0.196

583. FIELD CONSOLIDATION LINE The modified curve of the logarithm of vertical effective stress versus void ratio (e- log  `) is called the field consolidation line. There are two methods for determining the field consolidation line, one for normally consolidated clay, and the other for over consolidated clay. In the case of normally consolidated clay, determination of the field consolidation line is fairly simple. However for over consolidated clay, finding the field consolidation line is somewhat difficult.

584. FIELD CONSOLIDATION LINE In the case of normally consolidated clay, with the given (e- log  `) curve develop from the laboratory test, the point on the (e- log  `) curve corresponding to 0.4 e o is determined let point f . a straight line connecting points a (point a is the point designated by a pressure of  ` and void ratio of e o ) and f gives the field consolidation line for normally consolidated clay. NORMALLY CONSOLIDATED CLAY

585. FIELD CONSOLIDATION LINE Normally consolidated clay e log  ` NORMALLY CONSOLIDATED CLAY

586. FIELD CONSOLIDATION LINE a Normally consolidated clay e e o  ` o log  ` NORMALLY CONSOLIDATED CLAY

587. FIELD CONSOLIDATION LINE a Normally consolidated clay e e o  ` o log  ` NORMALLY CONSOLIDATED CLAY

588. FIELD CONSOLIDATION LINE a Normally consolidated clay e e o  ` o 0.4 e o log  ` f NORMALLY CONSOLIDATED CLAY

589. FIELD CONSOLIDATION LINE a Normally consolidated clay e e o  ` o 0.4 e o log  ` f NORMALLY CONSOLIDATED CLAY

590. FIELD CONSOLIDATION LINE a Normally consolidated clay e e o  ` o 0.4 e o log  ` f af is called field consolidation line NORMALLY CONSOLIDATED CLAY

591. COMPRESSION INDEX a e e 1 log  ` 1 e 2 log  ` f log  ` 2 The slope of the field consolidation line is called compression index

592. COMPRESSION INDEX

593. EXAMPLE-4 When the total pressure acting at midheight of a consolidating clay layer is 200 kN/m 2 , the corresponding void ratio of the clay is 0.98. When the total pressure acting at the same location is 500 kN/m 2 , the corresponding void ratio decreases to 0.81. Determine the void ratio of the clay if the total pressure acting at midheight of the consolidating clay layer is 1000 kN/m 2 .

594. MAXIMUM past PRESSURE e  ` ( log scale)

595. MAXIMUM past PRESSURE e  ` ( log scale)

596. MAXIMUM past PRESSURE g e  ` ( log scale)  Point of maximum curvature

597. MAXIMUM past PRESSURE g h i e  ` ( log scale) Horizontal line Tangent line

598. MAXIMUM OVERBURDEN PRESSURE g h j e  ` ( log scale) i Bisector line

599. MAXIMUM OVERBURDEN PRESSURE k g h j e  ` ( log scale) i

600. MAXIMUM past OVERBURDEN PRESSURE g h j e  ` ( log scale)  ` m k i

601. Example-5 611 The soil profile shown in figure, determine the present effective overburden pressure at the midheight of compressible clay layer.

602. NORMAL AND OVER CONSOLIDATED CLAY e e log  ` log  `

603. NORMAL AND OVER CONSOLIDATED CLAY e e o e e o  ` o  ` o log  ` log  ` a b a

604. NORMAL AND OVER CONSOLIDATED CLAY a e e o e e o  ` o  ` o log  ` log  ` b a

605. NORMAL AND OVER CONSOLIDATED CLAY b a a b e e o e e o  ` o  ` o log  ` log  `

606. NORMAL AND OVER CONSOLIDATED CLAY a b e e o e e o  ` o  ` o log  ` log  ` Normally consolidated clay Over consolidated clay b a

607. NORMAL CONSOLIDATION LINE

608. Example-6 During a consolidation test, a sample of fully saturated clay 3 cm thick is consolidated under a pressure increment of 200 kN/m 2 . When equilibrium is reached, the sample thickness is reduced to 2.60 cm. The pressure is then removed and the sample is allowed to expand and absorb water. The final thickness is observed as 2.8 cm ( ft ,) and the final moisture content is determined as 24.9%. If the specific gravity of the soil solids is 2.70, find the void ratio of the sample before and after consolidation . 618

609. Example-6 619

610. Example-8 A soil sample has a compression index of 0.3. If the void ratio e at a stress of 2940 Ib /ft 2 is 0.5, compute (i) the void ratio if the stress is increased to 4200 Ib /ft 2 , and (ii) the settlement of a soil stratum 13 ft thick. 620

611. Example-9 A 2.5 cm thick sample of clay was taken from the field for predicting the time of settlement for a proposed building which exerts a uniform pressure of 100 kN/m2 over the clay stratum. The sample was loaded to 100 kN/m2 and proper drainage was allowed from top and bottom. It was seen that 50 percent of the total settlement occurred in 3 minutes. Find the time required for 50 percent of the 621

612. Example-9 total settlement of the building, if it is to be constructed on a 6 m thick layer of clay which extends from the ground surface and is underlain by sand. 622

613. Example-10 A laboratory sample of clay 2 cm thick took 15 min to attain 60 percent consolidation under a double drainage condition. What time will be required to attain the same degree of consolidation for a clay layer 3 m thick under the foundation of a building for a similar loading and drainage condition ? 623

614. Exercise-1 The soil profile at a site for a proposed office building consists of a layer of fine sand 10.4 m thick above a layer of soft normally consolidated clay 2.0 m thick. Below the soft clay there is a deposit of coarse sand. The ground water table was observed at 1.0 m below the ground level. The void ratio of the sand is 0.75 and the water content of the clay is 45 %. 624

615. Exercise-1 (Continue) The building will impose a vertical stress increase of 150 kPa at the mid height of the clay layer. Estimate the primary consolidation settlement of the clay. Assume the soil above the water table to be saturated. Take C c = 0.45 and G s = 2.67. 625

616. Stress distribution in soil

617. Stress distribution in soil 627

618. Stress distribution in soil 628

619. Westergaard equation 629

620. Westergaard equation 630 where  = Vertical stress increment at depth z Q = Concentrated load  = Poisson’s ratio (ratio of lateral strain to axial stress in a material) z = depth

621. Westergaard equation If Poisson’s ratio is zero 631

622. Boussinesq Equation 632 This equation also gives stress q as a function of both the vertical distance z and horizontal distance r. for low r/z ratio; the Boussinesq equation gives higher values of q than the Westergaard equation. The Boussinesq equation is more widely used. Boussinesq’s equation considers a point load on a semi-infinite, homogeneous, isotropic, weightless, elastic half-space.

623. example A concentrated load of 1000 kN is applied at the ground surface. Compute the vertical pressure (i) at a depth of 4 m below the load, (ii) at a distance of 3 m at the same depth. Use Boussinesq's equation . 633

624. example 634 Q = 1000 kN z = 4 m ( a) r = 0  =? ( b) r = 3 m  =?

625. example A concentrated load of 250 tons is applied to the ground surface. Determine the vertical stress increment due to this load at a depth of 20 ft: (a) directly below the ground surface and (b) 16 ft from the line of the concentrated load using Westergaard and Boussinesq equations. 635

626. example Data Q = 250 tons z = 20 ft (a) r = 0  =? (b) r = 16 ft  =? 636 Westergaard method Boussinesq method

627. Vertical pressure below a uniform load Analysis of stress distribution resulting from uniform loaded surface area is generally more complicated than those resulting from concentrated loads. Two methods are discussed here Approximate method Method based on Elastic theory 637

628. Approximate method The approximate method based upon the assumption that the area of stress below a concentrated load increases with depth with the slope 2:1. Accordingly, stress at a depth z is given by : 638 Vertical pressure below a loaded surface area (uniform load)

629. Approximate method 639

630. Example-1 A 10-ft by 15-ft rectangular area carrying a uniform load of 5000 lb/ft 2 is applied to the ground surface. Determine the vertical stress increment due to this load at a depth of 20 ft below the ground surface by the approximate method. 640

631. Example-1 Data Q = 5000  (10  15) = 750,000 lb B = 10 ft L = 15 ft z =20 ft  = q = ? 641

632. Example-2 Determine vertical soil pressure using 2:1 method: Given:- Footing: 8 feet x 4 feet rectangular footing Column load = 25 kips Requirement : - Determine vertical soil pressure at 6’ below bottom of footing 642

633. Example-2 643

634. solution 644

635. Exercise-3 A concentrated load of 45000 Ib acts at foundation level at a depth of 6.56 ft below ground surface. Find the vertical stress along the axis of the load at a depth of 32.8 ft and at a radial distance of 16.4 ft at the same depth by (a) Boussinesq, and (b) Westergaard formulae for  = 0. Neglect the depth of the foundation. 645

636. Exercise-4 A rectangular raft of size 30 x 12 m founded at a depth of 2.5 m below the ground surface is subjected to a uniform pressure of 150 kPa . Assume the center of the area is the origin of coordinates (0, 0). and the corners have coordinates (6, 15). Calculate stresses at a depth of 20 m below the foundation level by the methods of (a) Boussinesq, and (b) Westergaard at coordinates of ( 0, 0), (0, 15), (6, 0) (6, 15) and (10, 25). Also determine the ratios of the stresses as obtained by the two methods. Neglect the effect of foundation depth on the stresses. 646

637. 647 Exercise-4

638. solution 648

639. solution 649

640. STRIP LOADS The state of stress encountered in this case also is that of a plane strain condition. Such conditions are found for structures extended very much in one direction, such as strip and wall foundations , foundations of retaining walls, embankments, dams and the like. For such structures the distribution of stresses in any section (except for the end portions of 2 to 3 times the widths of the structures from its end) will be the same as in the neighbouring sections, provided that the load does not change in directions perpendicular to the plane considered. 650

641. Method based on elastic theory 651

642. Stress due to point load 652

643. 653

644. 654

645. Newmark’s chart 655

646. example A 20- ft by 30- ft rectangular foundation carrying a uniform load of 6000 lb / ft 2 is applied to the ground surface. Determine the vertical stress increment due to this uniform load at a depth of 20 ft below the center of the loaded area. 656

647. example 657

648. Influence factors 658 Influence factors for vertical stress increase due to a point load (Craig, 1997)

649. r/z I r/z I r/z I 0.0 0.478 0.8 0.139 1.6 0.020 0.1 0.466 0.9 0.108 1.7 0.016 0.2 0.433 1.0 0.084 1.8 0.013 0.3 0.385 1.1 0.066 1.9 0.011 0.4 0.329 1.2 0.051 2.0 0.009 0.5 0.273 1.3 0.040 2.2 0.006 0.6 0.221 1.4 0.032 2.4 0.004 0.7 0.176 1.5 0.025 2.6 0.003 659

650. Influence factors 660 Influence coefficients for points under uniformly loaded circular area

651. Influence factors 661 Influence coefficients for points under uniformly loaded rectangular areas

652. 662

653. Example-1 A rectangular footing as shown in Figure exerts a uniform pressure of 500 kN /m 2 . Determine the vertical stress at the center of the footing for a depth of 2 m. 663

654. Example-1 664

655. Example-2 A rectangular L-shape footing as shown in Figure, exerts a uniform pressure of 500 kN /m 2 . Determine the vertical stress at point A for a depth of 2 m. Use table for influence factor. 665

656. Example-2 666 B

657. Example-3 A water tank is supported by a ring foundation having outer diameter of 8m and inner diameter of 6m. The uniform load intensity on the foundation is 200 kN /m 2 . compute the vertical stress caused by the water tank at a depth 4 m below the centre of the foundation 667

658. Example-7 A column of a building carries a load of 1000 kips. The load is transferred to sub soil through a square footing of size 16 x 16 ft founded at a depth of 6.5 ft below ground level. The soil below the footing is fine sand up to a depth of 16.5 ft and below this is a soft compressible clay of thickness 16 ft. The water table is found at a depth of 6.5 ft below the base of the footing. 668

659. Example-7 The specific gravities of the solid particles of sand and clay are 2.64 and 2.72 and their natural moisture contents are 25 and 40 percent respectively . The sand above the water table may be assumed to remain saturated. If the plastic limit and the plasticity index of the clay are 30 and 40 percent respectively, estimate the probable settlement of the footing. 669

660. Example-7 670

661. Do like this 671

662. Don’t do like this 672

663. Mohr’s Circle of Stress

664.  1  1  3  3  3  3 Stresses in Soil

665. Stresses in Soil pressure  1 pressure  1 zero pressure zero pressure pressure  1 pressure  1

666. Stresses in Soil  1  1 direct stress  1 shear stress  =  1  

667. Stresses in Soil  1  1  1    

668. Stresses in Soil  1  1   1    

669. Stresses in Soil  1  1    1  

670. Stresses in Soil  1  1  1    

671. Stresses in Soil  1  1    1  

672. Stresses in Soil  1  1  = 0  = 0  1    1

673. Stresses in Soil  1  1    1    

674. Stresses in Soil  1  1    1    

675. Stresses in Soil  1  1  1    

676. Stresses in Soil  1  1  1    

677. Stresses in Soil  1  1  1    

678. Stresses in Soil  1  1  1   1  

679. Stresses in Soil  1  1  1  = 0  1  

680. Stresses in Soil  1  1   Mohr’s Circle of Stress  1 Question : What is the stress at this point? Answer : This circle

681. Stresses in Soil pressure  1 pressure  1 pressure  3 pressure  3  1  1  3  3

682. Stresses in Soil  1  1 direct stress  =  1 shear stress  =  1    3  3  3  3

683. Stresses in Soil  1  1  1      3  3  3  3

684. Stresses in Soil  1  1   1      3  3  3  3

685. Stresses in Soil  1  1    1    3  3  3  3

686. Stresses in Soil  1  1  1      3  3  3  3

687. Stresses in Soil  1  1    1    3  3  3  3

688. Stresses in Soil  1  1  3  = 0  1   1   3  3  3  3

689. Stresses in Soil  1  1    1      3  3  3  3

690. Stresses in Soil  1  1    1      3  3  3  3

691. Stresses in Soil  1  1  1      3  3  3  3

692. Stresses in Soil  1  1  1      3  3  3  3

693. Stresses in Soil  1  1  1      3  3  3  3

694. Stresses in Soil  1  1  3   1    3  3  3  3

695. Stresses in Soil  1  1  3   1    1  3  3  3  3  3

696. Stresses in Soil  1  1  Mohr’s Circle of Stress   1  3  3  3 Question : What is the stress at this point? Answer : This circle

697. Stresses in Soil  1  1  Mohr’s Circle of Stress   1  1  1  3  3  3  3  3  3  3

698. Soil with Cohesion and Friction   = c +  tan  Mohr’s Circle of Stress c c  Soil fails when Mohr’s circle touches these lines   1  3

699. (c, ) and ( 3 ,  1 ) relationship

700. (c, ) and ( 3 ,  1 ) relationship

701. (c, ) and ( 3 ,  1 ) relationship ccot  ( 1 - 3 )/2 ( 1 - 3 )

702. (c, ) and ( 3 ,  1 ) relationship

703. (c, ) and ( 3 ,  1 ) relationship

704. (c, ) and ( 3 ,  1 ) relationship

705. Coefficient of lateral earth pressure, K For cohesionless soils, c = 0 And

706. RELATION B/W  AND   + 90 + 180 - 2 = 180 2 = 90 +  = 45 +  /2 = ( /4 +  /2)

707. Maximum shear stress  max

708. Shear stress at failure  f F ` F C 2 

709. Normal Stress  n  c F( n ,  f )  max

710. Pole point The pole on Mohr’s circle identifies a point through which any plane passing will intersect the Mohr’s circle at a point that represents the stresses on that plane. The pole point is determined from drawing a line parallel to the face of the element, which the stresses are acting.

711. Pole point O C A(  y ,  yx ) B (  x ,  xy )  3  1   D E

712. Pole point O C A(  y ,  yx ) B (  x ,  xy )  3  1   D E Pole

713. Pole point O C A(  y ,  yx ) B (  x ,  xy )  3  1   D E  max Pole

714. Pole point O C A(  y ,  yx ) B (  x ,  xy )  3  1   D E  max Pole A` B`

715. Radius of Circle From the geometry of the figure AA` =  yx The radius of the circle can be determined as:

716. Centre of circle The location of the centre of Mohr circle in terms of stress components can be located as:

717. Maximum shear Therefore the maximum shear stress can also be expressed in terms of stress components as:

718. Major and Minor principal stresses

719. 731

720. SHEAR STRENGTH OF SOIL

721. SHEAR STRENGTH OF SOIL Shear strength is a term used in soil mechanics to describe the magnitude of the shear stress that a soil can sustain. The shear resistance of soil is a result of friction and interlocking of particles, and possibly cementation or bonding at particle contacts. 733

722. Shear strength of soil 734

723. Shear strength of soil 735

724. STRENGTH THEORIES FOR SOILS A number of theories have been proposed for explaining the shearing strength of soils. Of all such theories, the Mohr’s strength theory and the Mohr-Coulomb theory, a generalisation and modification of the Coulomb’s equation, meet the requirements for application to a soil in an admirable manner. 736

725. Laboratory tests for determination of shear strength parameters Direct shear test Triaxial test Simple shear test Plane strain triaxial test Torsional ring shear test Unconfined Compression Test Laboratory Vane Shear Test 737

726. Motor drive Load cell to measure Shear Force Normal load Rollers Soil Porous plates Top platen Measure relative horizontal displacement, dx vertical displacement of top platen, dy Shear Box Test

727. Shear Box Test

728. The limiting shear stress (soil strength) is given by t = c + s n tan f where c = cohesion (apparent) f = friction angle t s n Mohr-Coulomb failure criterion

729. Normal and shear stress 741  n  1 1 2 2 3 3 4 4 5 5

730. 742 t s n Shear parameters (c, )

731. 743 t s n Shear parameters (c, )

732. 744 t s n Shear parameters (c, )

733. 745 t s n Shear parameters (c, )

734. 746 t c f Shear parameters (c, ) s n

735. Maximum shear and shear at failure (  max ,  f )

736. 748 t f Maximum shear and shear at failure (  max ,  f ) s n

737. 749 t f s n Maximum shear and shear at failure (  max ,  f )

738. 750 t f s n  max  f  n Maximum shear and shear at failure (  max ,  f )

739. example The stresses at failure on the failure plane in a cohesionless soil mass were: Shear stress = 4 kN/m 2 ; normal stress = 10 kN/m 2 . Determine the resultant stress on the failure plane, the angle of internal friction of the soil and the angle of inclination of the failure plane to the major principal plane. 751

740. Graphical solution The procedure is first to draw the σ-and τ-axes from an origin O and then, to a suitable scale, set-off point D with coordinates (10,4), Joining O to D , the strength envelope is got. The Mohr Circle should be tangential to OD to D . DC is drawn perpendicular to OD to cut OX in C , which is the centre of the circle. With C as the centre and CD as radius, the circle is completed to cut OX in A and B . 752

741. 753

742. exercise The following results were obtained in a shear box text. Determine the angle of shearing resistance and cohesion intercept : Normal stress (kN/m 2 ) 100 200 300 Shear stress (kN/m 2 ) 130 185 240 754

743. solution 755

744. example Clean and dry sand samples were tested in a large shear box, 25 cm × 25 cm and the following results were obtained : Normal load (kN) 5 10 15 Peak shear load (kN) 5 10 15 Ultimate shear load (kN) 2.9 5.8 8.7 Determine the angle of shearing resistance of the sand in the dense and loose states. 756

745. solution The value of  obtained from the peak stress represents the angle of shearing resistance of the sand in its initial compacted state; that from the ultimate stress corresponds to the sand when loosened by the shearing action . The area of the shear box = 25 × 25 = 625 cm 2 . = 0.0625 m 2 . Normal stress in the first test = 5/0.0625 kN/m 2 = 80 kN/m 2 757

746. solution Similarly the other normal stresses and shear stresses are obtained by dividing by the area of the box and are as follows in kN/m 2 : Normal stress, σ 80 160 240 Peak shear stress,  max 80 160 240 Ultimate shear stress,  f 46.4 92.8 139.2 758

747. solution 759

748. 760 Practical worked example

749. 761 Practical worked example

750. 762 Practical worked example

751. example The following data were obtained in a direct shear test. Normal pressure = 20 kN/m 2 , tangential pressure = 16 kN/m 2 . Angle of internal friction = 20°, cohesion = 8 kN/m 2 . Represent the data by Mohr’s Circle and compute the principal stresses and the direction of the principal planes. 763

752. solution 764

753. Unconfined compression test 765

754. 766 Unconfined compression test

755. 767 Unconfined compression test

756. TRIAXIAL TEST

757. TRIAXIAL TEST Cell pressure Pore pressure and volume change Rubber membrane Cell water O-ring seals Porous filter disc Confining cylinder Deviator load Soil

758. s r s r = Radial stress (cell pressure) s a = Axial stress F = Deviator load s r STRESSES IN TRIAXIAL SPECIMEN

759. s r s r = Radial stress (cell pressure) s a = Axial stress F = Deviator load s r From equilibrium we have STRESSES IN TRIAXIAL SPECIMEN

760. 772

761. Example-1 A series of shear tests were performed on a soil. Each test was carried out until the sample sheared and the principal stresses for each test were : Test No. ( kN/m 2 ) ( kN/m 2 ) 1 200 600 2 300 900 3 400 1200 773

762. solution 774

763. Example-2 From a direct shear test on an undisturbed soil sample, the following data have been obtained. Evaluate the undrained shear strength parameters. Determine shear strength, major and minor principal stresses and their planes in the case of specimen of same soil sample subjected to a normal stress of 100 kN/m 2 . : s n (kN/m 2 ) 70 96 114 t (kN/m 2 ) 138 156 170 775

764. Example-2 776

765. Example-3 A cylindrical sample of saturated clay 4 cm in diameter and 8 cm high was tested in an unconfined compression apparatus. Find the unconfined compression strength, if the specimen failed at an axial load of 360 N, when the axial deformation was 8 mm. Find the shear strength parameters if the angle made by the failure plane with the horizontal plane was recorded as 50°. 777

766. Example-3 778

767. Triaxial testing procedure Flushing Saturation ramp B-check Consolidation Shearing

768. Saturation ramp

769. B check

770. Consolidation

771. Shearing

772. 784

773. Type of Triaxial tests Depending on the nature of loading and drainage condition, triaxial tests are conducted in three different ways Unconsolidated Undrained test ( UU -test) Consolidated Undrained test (CU-test) Consolidated Drained test ( CD-test)

774. example A cylindrical sample of soil having a cohesion of 80 kN/m 2 and an angle of internal friction of 20° is subjected to a cell pressure of 100 kN/m 2 . Determine: (i) the maximum deviator stress ( s 1 - s 3 ) at which the sample will fail, and (ii) the angle made by the failure plane with the axis of the sample. 788

775. 789 example

776. example A normally consolidated clay was consolidated under a stress of 3150 lb /ft 2 , then sheared undrained in axial compression. The principal stress difference at failure was 2100 lb /ft 2 , and the induced pore pressure at failure was 1848 lb /ft 2 . Determine (a) the Mohr-Coulomb strength parameters , in terms of both total and effective stresses analytically, (b) compute ( s 1 / s 3 ), and ( s ’ 1 / s ’ 3 ), and (c) determine the theoretical angle of the failure plane in the specimen. 790

777. example The following results were obtained at failure in a series of consolidated-undrained tests, with pore pressure measurement, on specimens of saturated clay. Determine the values of the effective stress parameters c’ and ’ by drawing Mohr circles. 791

778. Stress-strain behaviour

779. Stress-strain behaviour A graph that describes the relationship between stress and strain. Stress-strain graphs indicate the elastic and plastic regions for a given material. Stress - stress in soil/rock is the body's reaction to a change due to external agencies that requires a physical adjustment or response. The intensity of stress is expressed in units of force divided by units of area. 793

780. Stress-strain behaviour Deviator stress- is t he difference between the major and minor principal stresses in a triaxial test. Mean effective stress 794

781. Stress-strain behaviour 795 Strain - is the deformation of a physical body under the action of applied forces. is defined as the ratio of the change in dimension of the soil/rock mass to the original dimension.

782. Stress-strain behaviour 796

783. 797  a q

784. e a (%)

785. Dilatancy 804

786. YIELDING DURING ISOTROPIC COMPRESSION

787. THE MODE OF FAILURE

788. Transitional Failure mode 807

789. Failure mode 808

790. Strain localization 809

791. Shear band pattern 810

792. Shear band pattern

793.  1  1  3  3  n  f  s Shear band pattern

794. Shear band pattern 813

795. Illustration of progressive failure 814 (Klein et al., 2001)

796. Brittleness index (I B ) 815

797. Portaway sand 816 F A I L U R E SUMMARY

798. Factors affecting the mode of failure Initial relative density Confining pressure Type and amount of reinforcement 817

799. Effect of initial relative density 818

800. Effect of initial relative density 819

801. Effect of initial relative density 820

802. Effect of initial relative density 821

803. Effect of confining pressure 822

804. Effect of confining pressure 823

805. Effect of cement content 824

806. Effect of cement content 825

807. Effect of cement content 826

808. Effect of cement content 827

809. Effect of cement content 828

810. Effect of cement content 829

811. SPECIMEN PREPARATION

812. SAND CEMENT PROPORTION The following is the procedure for the determination of the void ratio of a cemented specimen, Determine The specific gravity of sand G sand The specific gravity of cement G cement The dry mass of specimen M dry i.e., the mass of solid M solid Specimen dimensions i.e., height H and diameter D

813. AVERAGE SPECIFIC GRAVITY The average specific gravity of the specimen G (taking cement content = C %)

814. INITIAL VOID RATIO The volume of solid (i.e., sand + cement) V solid The total volume of the specimen V total The volume of voids V voids

815. INITIAL VOID RATIO The initial void ratio of the specimen e

816. CONSTANT DRY DENSITY WITH INCREASING CEMENT CONTENT The mass of sand M sand and mass of cement M cement can be calculated as follow:

817. SAND CEMENT MIXING BUCKET Once dry sand and cement mixed thoroughly, water of required percent must be added and mixed to get a uniform and consistent sand-cement paste.

818. SAND CEMENT MIXING The targeted dry unit weight of the material for a standard size specimen can be maintained by varying the weight of the moist sample using the following equation. The required weight of mixture should be taken for specimen preparation

819. MEMBRANES Rubber membranes: (a) Latex; (b) Neoprene.

820. MATERIALS

821. SPECIMEN PREPARATION ACCESSORIES

822. USE OF TRANSPARENCY SHEET

823. CEMENTED SPECIMEN PREPARATION Grine, K. and Glendinning, S. (2007) Geotech Geol Eng 25, 441–448 Grine & Glendinning (2007)

824. Bradshaw and Baxter (2007)

825. Ismail, M. A., Joer , H. A., and Randolph, M. F., (2000) Sample Preparation Technique for Artificially Cemented Soils,” Geotechnical Testing Journal , 23( 2), , pp. 171–177. Ismail, et al (2000)

826. Ladd, R. S. (1978) Preparing test specimens using undercompaction," Geotechnical Testing Journal, 1(1) , 16-23. Ladd, (1978)

827. Ladd, (1978)

828. Bradshaw, A. S. and Baxter, C. D. P. (2007) Sample preparation of silts for liquefaction testing, Geotechnical Testing Journal, 30(4), 1-9. Bradshaw and Baxter (2007)

829. Membrane Puncture remediation

830. UNCEMENTED SPECIMEN PREPARATION

831. Uncemented specimen preparation

832. Uncemented specimen preparation

833. Uncemented specimen preparation

834. Uncemented specimen preparation

835. Uncemented specimen preparation

836. Uncemented specimen preparation

837. THE END

838. FINAL YEAR PROJECT LIST

839. Final year project list Development of Soil Map for Karachi city. Use of Geosynthetics as reinforced subgrade materials. Use of Geosynthetics as drainage and filter materials. Groundwater and Seepage modelling using DrainMod Mechanical behaviour of reinforced soil using Matlab 862

840. ASSIGNMENTS

841. 865