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CEGEOEN Module 0 Introduction to Geotechnical Engineering 1 Dr. Mary Ann Q. Adajar  Introduction to Geotechnical Engineering  Geotechnical engineering is the subdiscipline of civil engineering that involves natural materials found close to the surface of the earth.  I...

CEGEOEN Module 0 Introduction to Geotechnical Engineering 1 Dr. Mary Ann Q. Adajar  Introduction to Geotechnical Engineering  Geotechnical engineering is the subdiscipline of civil engineering that involves natural materials found close to the surface of the earth.  It includes the application of the principles of soil mechanics and rock mechanics to the design of foundations, retaining structures, and earth structures. 3 SOILS  defined as the uncemented aggregate of mineral grains and decayed organic matter (solid particles) with liquid and gas in the empty spaces between the solid particles. 4 To an engineer, soil is a material that can be:  built on: foundation to buildings, bridges  built in: tunnels, culverts, basements  built with: roads, runways, embankments, dams  supported: retaining walls, quays All engineering projects that come in contact with earth require the expertise of geotechnical engineers. 5 Geotechnical Engineering Applications  Design of shallow and deep foundations for buildings and offshore structures  Design of earth retaining structures such as basement walls and bridge abutments, road and railway construction  Stability analysis of hillsides and riverbanks, and engineered slopes such as earth dams, highway cuts, embankments, and dikes for flood protection 6 Earth‐related topics:  Geo-environmental engineering deals with preventing and remediating the effects of industrial and agricultural wastes  Geosynthetics and ground improvement provide engineered solutions for projects in weak soils and soft ground  Soil‐structure interaction of buried lifelines and tunnels 7 The analytical tools we use to perform engineering analyses are known as soil mechanics and rock mechanics.  SOIL MECHANICS, also called geotechnique or geotechnics or geomechanics, is the application of engineering mechanics to the solution of problems dealing with soils as a foundation and as a construction material.  Every application of soil mechanics involves uncertainty because of the variability of soils—their stratification, composition, and engineering properties. in Geotechnical Engineering  MARVELS OF CIVIL ENGINEERING 9 Beijing National Stadium world’s largest steel structure Bailong Elevator in Zhangjiajie, China The highest and heaviest outdoor elevator in the world that is 330 m high and can carry 50 people in one trip  MARVELS OF CIVIL ENGINEERING 10 Eurotunnel Tunnel start from England and end in France. The length is 31 miles and 23 miles of which is in the sea. Palm Islands Biggest artificial islands off the coast of Dubai. Made from 85,000,000 cubic meters of sand Along with concrete supports, this sand was used to shape the complex foundation of a palm tree and support the hotel. 11  A satisfactory foundation design requires the proper application of soil mechanics principles, accumulated experience, and good judgment.  The stability and life of any structure—a building, an airport, a road, dams, levees, natural slopes, power plants—depend on the stability, strength, and deformation of soils.  Thus, successful civil engineering projects are heavily dependent on geotechnical engineering.  Our blunders become monuments! 12 Leaning Tower of Pisa Soil conditions need to be known in advance in order to properly support structures during their entire life time. 13  The foundation pressures exceeded the bearing capacity of the underlying clay foundations. 14 GEOTECHNICAL LESSONS FROM FAILURES Causes of failures on structures:  inadequate site and soil investigations  unforeseen soil and water conditions;  natural hazards;  poor engineering analysis, design, construction, and quality control;  damaging postconstruction activities;  and usage outside the design conditions 15  When failures are investigated thoroughly, we obtain lessons and information that will guide us to prevent similar types of failure in the future 16  Lessons learned from these failures : the importance of soil investigations, soils tests, and the effects of rate of loading  Great Contributors to the 17 Developments in Geotechnical Engineering GEOENG1 18  “The father of soil mechanics”  He formulated the Theory of Consolidation, which is considered as the one of the most significant milestones in civil engineering.  The year 1925 was considered as the “birth” of geotechnical engineering as a widely recognized discipline, for that was when Terzaghi published the first comprehensive book “The Mechanics of Earth Construction Karl Terzaghi Based on Soil Physics”. 1883-1963 GEOENG1   A French physicist who is best 19 remembered for his work electricity and magnetism. However, he also made contributions in other fields including the computation of lateral earth pressures.  He was the first to define soil strength using both cohesion and friction, the first to consider wall friction, and the first to Charles Augustin analytically search for the Coulomb orientation of the most critical 1736-1806 failure plane. GEOENG1 20  A disciple of Terzaghi and a professor at Harvard University, he made many contributions to the analysis of soft clays, soil composition and classification, and seepage through earth structures. Arthur Casagrande 1902-1981 GEOENG1 21  He was an English civil engineer internationally recognised, along with Karl Terzaghi, as one of the founding fathers of the engineering discipline of soil mechanics.  He has made major contributions to soil mechanics on the fundamentals of the effective stress, pore pressures in clays, bearing capacity, and slope Alec Wesley Skempton stability. 1914 - 2001 GEOENG1 Some unsung heroes of Civil Engineering… foundations soil exploration tunneling 22 … buried right under your feet.  CEGEOEN Module 1 Weight-Volume Relationship of Soil 1 Dr. Mary Ann Q. Adajar  Soil: A 3-Phase Material Air Water Solid The Mineral Skeleton Solid Particles Volume Voids (air or water)  Three-Phase Diagram Air Water Solid Mineral Skeleton Idealization: Three Phase Diagram  Three - Phase System Va Air Wa~0 Vv Vw V Water Ww W Vs Solid Ws Volume Weight  Fully Saturated Soils VW = VV Water V = VV + VS V = VW + VS Solid W = WW + WS Mineral Skeleton Fully Saturated  Dry Soils Va = VV Air V = VV + VS Wa = 0 W = WS Solid Mineral Skeleton Dry Soil 8 Partly Saturated Soils Air VV = Va + VW Water V = VV + VS Wa = 0 Solid W = WW + WS Mineral Skeleton Partly Saturated Soils VOID RATIO and POROSITY Void ratio (e) is defined as the ratio of the volume of voids to the volume of solids.  Volume Components: Vv  Volume of Solids = Vs Void Ratio, e = (1)  Volume of Water = Vw Vs  Volume of Air = Va  Volume of Voids = Va + Vw = Vv Porosity (n) is defined as the ratio of the volume of voids to the total volume. Vv Porosity, n(%) = 100% (2) V 9 Relationship between void ratio and porosity  From (1) : 𝑉𝑣 = 𝑒𝑉𝑠  Since V = Vv +Vs  Then (2) can be written as: 𝑒𝑉𝑠 𝑒𝑉𝑠 𝑉𝑠(𝑒) 𝑛= = = 𝑉𝑣 + 𝑉𝑠 𝑒𝑉𝑠 + 𝑉𝑠 𝑉𝑠(𝑒 + 1) e n= (3) 1+ e n e= (4) 1− n 10 DEGREE OF SATURATION or SATURATION RATIO The degree of saturation (S) is defined as the ratio of the volume of water to the volume of voids. Volume Components:  Volume of Solids = Vs  Volume of Water = Vw  Volume of Air = Va  Volume of Voids = Va + Vw = Vv Vw Degree of Saturation , S (%) = 100% (5) VV For dry state: VW = 0 S = 0 For fully saturated state: VW = VV S = 100% 11 Specific Gravity Specific gravity (Gs) is defined as the ratio of the unit weight of a given material to the unit weight of equal volume of water. Weight of a Substance Specific Gravity = Weight of an Equal Volume of Water s Ws GS = = (6)  w Vs w Density of a Substance Specific Gravity = Density of Water s Ms Gs = = (7)  w Vs  w 12 Specific Gravity 13 14 Expected Value for GS (Bowles, 1978) Type of Soil GS Sand 2.65 – 2.67 Silty sand 2.67 – 2.70 Inorganic clay 2.70 – 2.80 Soils with mica or iron 2.75 – 3.00 Organic soils Variable, but maybe under 2.00 WATER CONTENT or MOISTURE CONTENT Moisture content (w) is referred to as water content and is defined as the ratio of the weight of water to the weight of solids in a given volume of soil. Weight Components: Weight of Solids = Ws Weight of Water = Ww Weight of Air ~ 0 Ww Water Content , w(%) = 100% (8) Ws MW w(%) =  100% (9) MS 15 16 DENSITY and UNIT WEIGHT Density = mass per unit volume Unit Weight = weight per unit volume Density of water, ρw = 1000 kg/m3 ρw = 1 g/cm3 Unit weight of water, γw = 9.81 kN/m3 γw = 62.4 lb/ft3 17 Soil’s Unit Weight Moist Unit Weight or Bulk Unit weight - unit weight of a soil when void spaces of the soil contain both water and air. W Ws + Ww  = = (10) V Vs + VV (not fully saturated) Vv = Vw + Va 18 Soil’s Unit Weight Saturated unit weight - unit weight of a soil when all void spaces of the soil are completely filled with water, with no air. W Ws + Ww  sat = = (11) V Vs + Vv Fully saturated condition: Vv = Vw  Soil’s Unit Weight 19  Dry unit weight - unit weight of a soil when all void spaces of the soil are completely filled with air, with no water Ws d = (12) V  The relationship of unit weight, dry unit weight, and moisture content can be derived from: and 𝑊𝑠 = 𝛾𝑑 𝑉 𝛾 𝛾𝑑 = (13) 1+𝑤 20 Soil’s Unit Weight  Submerged (Buoyant) unit weight or Effective Unit Weight) - the difference between the saturated unit weight and the unit weight of water  ' =  sat −  w (14) 21 Typical Unit weights RELATIONSHIPS among Unit Weight, Void Ratio, Moisture Content, and Specific Gravity Consider a volume of soil in which the volume of the soil solids, Vs = 1 Fig. 1.3 Three separate phases of a soil element with volume of soil solids equal to 1 Ws = Gs w Ww = wGs w  Using the definitions of unit weight and dry unit weight (17) (18) (19) Since volume of water is 𝑊𝑤 𝑤𝐺𝑠𝛾𝑤 𝑉𝑤 = = = 𝑤𝐺𝑠 𝛾𝑤 𝛾𝑤 The degree of saturation: 𝑉𝑤 𝑤𝐺𝑠 (20) 𝑆= = x 100% or 𝑆𝑒 = 𝑤𝐺𝑠 𝑉𝑣 𝑒 If the soil sample is saturated, that is, the void space are filled with water (Fig. 1.4), the relationship for saturated unit weight : (21) For soil sample where the void spaces are not filled with water, the moist unit weight (γ) can be written as:  w (Se + Gs )  = (22) 1+ e 25 Relative Density, Dr The term relative density is commonly used to indicate the in-situ denseness or looseness of granular soil. It is defined as: emax − e Dr = x100% emax − emin (23) where e − void ratio (natural void ratio) e min − Minimum void ratio (soil in very dense condition) e max − Maximum void ratio (soil in very loose condition) 26 Relative Density, Dr  By using the definition of dry unit weight given in Eq. (18), we can express relative density in terms of maximum and minimum possible dry unit weights. Thus, (24) 27 Relative Density, Dr 28 Relative Compaction Another term used in regard to the degree of compaction of coarse-grained soils is relative compaction, Rc Degree of compaction is also sometimes expressed in terms of an index called relative compaction (RC) defined as follows: d Rc = (25)  d (max)  Derivations of Eqns. (17) to (20) if Vs is not 29 assumed as 1 𝑊𝑠 𝐺𝑠 = 𝑊𝑠 = 𝐺𝑠𝑉𝑠 𝛾𝑤 𝑉𝑠 𝛾𝑤 𝑊𝑤 𝑊𝑤 𝑤= 𝑊𝑠 = 𝑊𝑤 = 𝑤𝐺𝑠𝑉𝑠 𝛾𝑤 𝑊𝑠 𝑤 𝑉𝑣 𝑉 = 𝑉𝑣 + 𝑉𝑠 𝑒= 𝑉𝑣 = 𝑒𝑉𝑠 𝑉 = 𝑒𝑉𝑠 + 𝑉𝑠 = 𝑉𝑠(1 + 𝑒) 𝑉𝑠 𝑊 𝑊𝑠 + 𝑊𝑤 𝐺𝑠𝑉𝑠 𝛾𝑤 + 𝑤𝐺𝑠𝑉𝑠 𝛾𝑤 𝐺𝑠𝑉𝑠 𝛾𝑤 (1 + 𝑤) 𝐺𝑠𝛾𝑤 (1 + 𝑤) 𝛾= = = = = (17) 𝑉 𝑉𝑠(1 + 𝑒) 𝑉𝑠(1 + 𝑒) 𝑉𝑠(1 + 𝑒) 1+𝑒 𝑊𝑠 𝐺𝑠𝑉𝑠 𝛾𝑤 𝐺𝑠𝛾𝑤 𝛾𝑑 = = = 𝑉 𝑉𝑠(1 + 𝑒) 1 + 𝑒 (18) 𝐺𝑠𝛾𝑤 𝑒= −1 (19) 𝛾𝑑 30 𝑊𝑤 𝑊𝑤 𝑤𝐺𝑠𝑉𝑠 𝛾𝑤 𝛾𝑤 = 𝑉𝑤 = = 𝑉𝑤 𝛾𝑤 𝛾𝑤 𝑉𝑊 𝑤𝐺𝑠𝑉𝑆 𝑆= = 𝑉𝑉 𝑒𝑉𝑠 𝑆𝑒 = 𝑤𝐺𝑠 (20)  CEGEOEN Module 2 PLASTICITY AND SOIL STRUCTURE Particle Size & Shape 1 Dr. Mary Ann Q. Adajar 2 Individual soil particles in a soil can have different; Sizes and Shapes… These characteristics have a significant effect on its engineering behavior 3 Particle Size Classification According to ASTM, particles larger than 3 inches (75mm) in diameter are known rock fragments. Smaller particles are known as soil, and are classified to their ability to pass through certain size sieves.   Gravels are pieces of rocks with occasional particles of quartz, feldspar, 4 and other minerals.  Sand particles are made of mostly quartz and feldspar. Other mineral grains also may be present at times.  Silts are the microscopic soil fractions that consist of very fine quartz grains and some flake-shaped particles that are fragments of micaceous minerals.  Clays are mostly flake-shaped microscopic and submicroscopic particles of mica, clay minerals, and other minerals.  Non-clay soils can contain particles of quartz, feldspar, or mica that are small enough to be within the clay classification. Hence, it is appropriate for soil particles smaller than 2 microns (2μm), or 5 microns (5μm) as defined under different systems, to be called clay-sized particles rather than clay. 5 ASTM D422 Standard Test Method for Particle-Size Analysis of Soils  This test method covers the quantitative determination of the distribution of particle sizes in soils.  The distribution of particle sizes larger than 75 µm (retained on the No. 200 sieve) is determined by sieving – Sieve Analysis  The distribution of particle sizes smaller than 75 µm is determined by a sedimentation process, using a hydrometer to secure the necessary data – Hydrometer Method 6 Particle Size Distribution Sieve Analysis Grain-size analysis of soils containing relatively large particles is accomplished using sieves. It is an apparatus having openings of equal size and shape through which grains smaller than the size of the opening will pass, while larger grains are retained. 7 Particle Size Distribution Hydrometer Analysis The hydrometer method is based on Stoke’s law, which says that the larger the grain size, the greater its settling velocity in a fluid. ASTM Standard Sieves (per ASTM D422) Sieve Opening Size Identification (in) (mm) 3 inches 3.00 76.2 2 inches 2.00 50.8 1½ 1.50 38.1 1 inch 1.00 25.4 ¾ inch 0.75 19.0 3/8 inch 0.375 9.52 #4 0.187 4.75 #8 0.0929 2.36 8 #10 0.0787 2.00 ASTM Standard Sieves (per ASTM D422 and E100) Sieve Opening Size Identification (in) (mm) #16 0.0465 1.180 #20 0.0335 0.850 #30 0.0236 0.600 #40 0.0167 0.425 #50 0.0118 0.300 #60 0.00984 0.250 #100 0.00591 0.150 #140 0.00417 0.106 9 #200 0.00295 0.075 10 The larger sieves are identified by their opening size (3” to 3/8”) Ex. A ¾ -inch sieve will barely pass a ¾ inch diameter sphere. Smaller sieves are numbered, with the number indicating the openings per inch (#4 to #200) Ex. #8 sieves has 8 openings per inch or 64 per square inch. ASTM Particle Size Classification (per ASTM D2487) Sieve Size Particle Diameter Soil Classification Passes Retained (in) (mm) on 12 in >12 >350 Boulder Rock 12 in 3 in 3-12 75-350 Cobble Fragments 3 in ¾ in 0.75-3 19-75 Coarse Soil Gravel ¾ in #4 0.19-0.75 4.75-19 Fine Gravel #4 #10 0.079-0.19 2-4.75 Coarse Sand #10 #40 0.016-0.079 0.425-2 Medium Sand #40 #200 0.0029-0.016 0.075- Fine Sand 0.425 #200 95% of (modified) maximum (b) >95% of (modified) maximum dry dry unit weight unit weight and w within 2% of wopt  CEGEOEN Module 4 PERMEABILITY (One-dimensional Flow 1 of Water through Soils) Dr. Mary Ann Q. Adajar 2 Flow of Water in Soils Karl Terzaghi once wrote… “… in engineering practice, difficulties with soils are almost exclusively due not to the soils themselves, but to the water contained in the voids. On a planet without any water there would be no need for soil mechanics….” 3 Specific water-related geotechnical issues The effect of water on the behavior and engineering properties of soils and rocks The potential for water flowing into excavations The potential for pumping water through wells or other facilities The effect of water on the stability of excavations and embankments 4 Specific water-related geotechnical issues The resulting uplift forces on buried structures The potential for seepage-related failures, such as piping The potential for transport of hazardous chemicals along with the water 5 HYDROLOGY Hydrology is the study of water movements across the earth… it includes assessments of rainfall intensities, stream flows, and lake water levels, known as surface water hydrology and studies of underground water, known as ground water hydrology 6 Ground Water Hydrology Subsurface water encompasses all underground water, virtually all of which is located within the soil voids or rock fissures Ground water table (also called phreatic surface) is one of the most important informations to describe and understand subsurface water. 7  Groundwater table can be located by installing observation wells.  Soil profiles represent the groundwater table as a line marked with triangle 8 Ground Water Hydrology Ground water table often changes with time, depending on the: ✓ season of the year, ✓recent patterns of rainfall, ✓irrigation practices, ✓pumping activities,  Subsurface water may be divided into two 9 sections: 1. Phreatic zone 2. Vadose zone 10 Ground Water Hydrology Phreatic Zone – (zone of saturation) the portion below the ground water table. ❑This water is subjected to a positive pressure as a result of the weight of the overlying water ❑Most of the subsurface water is in the phreatic zone 11 Ground Water Hydrology Vadose Zone – (zone of aeration) the portion above the water table. ❑This water has a negative pressure… ❑And is held in place by capillary action and other forces present in the soil 12 Ground Water Hydrology Technically, only the water in the phreatic zone is true ground water. However, we often use the term ground water to describe all the subsurface water 13 Aquifers, Aquiclude and Aquitard Aquifers – some soils such as sands and gravels that can transmit large quantities of groundwater. Aquicludes – other soils such as clays that transmit water very slowly Aquitards – intermediate soils, such as silty sand that pass water at a slow-to-moderate rate Perched groundwater condition can occur when an aquiclude separates two acquifers. 14 aquiclude aquifer 15 Confined and Unconfined Aquifers Unconfined aquifer – is one in which the bottom flow boundary is defined by an aquiclude, but the upper flow boundary (the ground water table) is free to reach its own natural level Confined aquifer – is one in which both the upper and the lower flow boundaries are defined by aquicludes. This type of aquifer is similar to a pipe that is flowing full. 16 17 Artesian Condition Artesians – Most confined aquifers are also artesians, which means that the water at the top of the aquifer are under pressure Artesian wells – people often drill wells into artesian confined aquifers, because the water will rise up through the aquiclude without pumping. If the water pressure is high enough, artesian wells deliver water all the way to the ground surface without pumping  REVIEW OF FLUID MECHANICS 18 Head and Pore Water Pressure Flow rate, q – the quantity of water that passes through a pipe per unit time q = vA (discharge volume per unit of time) v = rate of flow A = cross-sectional area 19 Head and Pore Water Pressure Piezometer – A vertical tube with one end attached to the pipe and the other open to the atmosphere. If the flow rate through the pipe remains constant, and the piezometer is sufficiently tall, the water level on the piezometer remains stationary 20 Head and Pore Water Pressure Pitot tube – it is similar to a piezometer except that the tip is pointed upstream and receives a ramming effect from the flowing water The water in the pitot tube is slightly higher.  Energy 21 Potential energy – due to its elevation above the datum Strain energy- due to its pressure in the water Kinetic energy- due to its velocity  Heads 22 Head is the energy divided by the acceleration due to gravity, g Elevation Head, hz – difference in elevations between the datum and the point being considered. It describes the potential energy at that point 23 Heads Pressure Head, hp – difference in elevations between the point and the water level in a piezometer attached to the pipe. It describes the strain energy 24 Heads Velocity Head, hv – difference in elevations between the piezometer and the pitot tube and describes the kinetic energy. It is related to the velocity, v, and the acceleration due to gravity, g 2 v hv = 2g  According to Bernoulli’s equation, the total head at a point in water under motion can be given by the sum of the pressure, 25 velocity, and elevation heads. Total head , h = hz + h p + hv 26 Application to Soil and Rock Although velocity is very important in pipes and open channels, the velocity of flow in soil is much lower, so the velocity head is very small (less than 5mm). Thus, we can neglect it for practical soil seepage problems Total head , h : h = hz + h p hz = elevation head hp = pressure head 27 Head Loss and Hydraulic Gradients, i Groundwater flow through the soil pores is driven by a hydraulic gradient , i which is defined as the rate of change of total head with distance dh h i= = dl l where; dh − change in total heads = Head loss dl − dist. the water travels  What is the hydraulic gradient, i between points A and B? 28 hA = 3.62 + 3.01 + 0.50m = 7.13m hB = 4.28 + 1.61 + 0.50m = 6.39m h = hA − hB = 7.13 − 6.39 = 0.74m h 0.74m i= = = 0.0037 l 200m  Pore Water Pressure, u 29 -The pressure in the water within the soil voids. -Also called gage pressure (difference between the absolute water pressure and atmospheric pressure). -For points below the ground water table, the pore water pressure is u =  whp where; u − pore water pressure h p − pressure head  w − unit weight of water  Pore Water Pressure, u 30 If the PWP is due solely to the force of gravity acting on the pore water, we call this hydrostatic condition, and the associated PWP is the hydrostatic PWP, uh Therefore, the PWP is u = uh= wzw where: u = pore water pressure uh = hydrostatic pore water pressure γw = unit weight of water zw = depth from the ground water table to the point  Pore Water Pressure, u 31 Above the GWT, we normally consider the PWP to be zero In reality, surface tension effects between the water and the solid particles produce negative PWP above the GWT, and this negative pressure is called soil suction.  Ground Water Flow Conditions 32 Laminar Flow – Sometimes flows in a smooth orderly fashion. This flow pattern occurs when the velocity is low. Turbulent Flow – the water swirls as it moves. This happens when the velocity is high. Turbulent flow consumes much more energy, and produces more head loss 33 Ground Water Flow Conditions One-dimensional flow – the velocity vectors are parallel and of equal magnitude. The water always moves parallel to some axis and through a constant cross section.  Ground Water Flow Conditions 34 Two-dimensional flow – when the velocity of the flow conditions are confined to a single plane, but vary in direction and magnitude within that plane.  Ground Water Flow Conditions 35 Three-dimensional flow – the most general condition. It exists when the velocity vectors vary in the x, y, and z directions.  Steady and Unsteady Flow 36 Steady-state flow – a system has reached equilibrium. The flow pattern has been established and is not in the process of changing. When this condition exists, the flow rate Q, remains constant with time. Unsteady Condition (transient flow) – it occurs when the PWP, GWT locations, flow rate, or other characteristics are changing, perhaps in response to a change in the applied head. Unsteady flow varies with time.  PERMEABILITY (One-Dimensional Flow) 37  Soils are permeable due to the existence of interconnected voids through which water can flow from points of high energy to points of low energy.  The study of the flow of water through permeable soil media is important in soil mechanics.  It is necessary for estimating the quantity of underground seepage under various hydraulic conditions, for investigating problems involving the pumping of water for underground construction, and for making stability analyses of earth dams and earth-retaining structures that are subject to seepage forces. 38 One of the major physical parameters of a soil that controls the rate of seepage through it is hydraulic conductivity, otherwise known as the coefficient of permeability. The ease with which the groundwater can flow through the soil pores is quantified by the Hydraulic Conductivity ( or coefficient of permeability), k 39 DARCY’S LAW Darcy (1856) published a simple relation between the discharge velocity and the hydraulic gradient: where: v = discharge velocity = ki i = hydraulic gradient k = coefficient of permeability. It has units of velocity (m/s or cm/s)  The quantity of water flowing through the soil in unit time is q,and can be given by: 𝑞 = 𝑘𝑖𝐴 Where: A is the cross-section of the soil perpendicular to the direction of flow. 40 Seepage Velocity The actual velocity of water called the seepage velocity,vs through the void spaces is greater than v  Seepage Velocity 41 The seepage velocity, vs, is the rate of movement of an element of water through a soil ki vs = where; ne vs − seepage velocity k − hydraulic conductivity i − hydraulic gradient ne − effective porosity sandy soils : ne = n( porosity) clayey soils : ne = 0.10(if no test data is available)  Validity of Darcy’s law 42  Darcy’s law given by equation, ν = ki, is true for laminar flow through the void spaces.  The flow of water through all types of soil (sand, silt, and clay), the flow is laminar and Darcy’s law is valid.  For coarse sands, gravels, and boulders, turbulent flow of water can be expected  Darcy’s Law implies that the discharge velocity bears a linear relation with the hydraulic gradient. 43  Permeability is often determined by laboratory tests; and to be representative of the soil’s in situ permeability, tests should be performed on “undisturbed sample”  In the case of granular soils, the recommended procedure is to perform permeability tests on three soil samples, with each specimen having a different void ratio.  A relationship between void ratio and permeability can be established for a given soil by plotting a graph of permeability on a logarithmic scale versus void ratio on the arithmetic scale.  (b) 10-2 44 10-3 k (cm/sec) 10-4 trendline F= 0 10-5 F=10 F=30 F=50 F=70 -6 100.6 0.7 0.8 0.9 1 1.1 void ratio, e k vs void ratio, e  e9.74  ( k = 3.52 10−3 ) exp ( −0.043F )    1+ e   Table 4.1. Classification of Soil according to their 45 Coefficient of Permeability (Terzaghi, 1996) Soil Coefficient of Degree of Permeability k Permeability (cm/s) Gravel Over 10-1 cm/sec High gravel, clean 10-1-10-3 cm/sec Medium sand, fine sand Sand, dirty sand, 10-3-10-5 cm/sec Low silty sand Silt, silty clay 10-5-10-7 cm/sec Very Low Clay Less than 10-7 Practically cm/sec impermeable 46 Relationships for Hydraulic Conductivity on Granular Soil  For fairly uniform sand (that is, sand with a small uniformity coefficient), Hazen (1930) proposed an empirical relationship for hydraulic conductivity in the form: where: 𝑐𝑚 c = a constant that varies from 1.0 to 𝑘 = 𝑐(𝐷10 )2 1.5 𝑠𝑒𝑐 D10 = the effective size, in mm  More recently, Chapuis (2004) proposed an empirical relationship for k where: e = void ratio D10 = the effective size, in mm 47 Relationships for Hydraulic Conductivity—Cohesive Soils  Samarasinghe et al. (1982) conducted laboratory tests on New Liskeard clay and proposed that, for normally consolidated clays, 𝑒𝑛 𝑘=𝐶 1+𝑒 where C and n are constants to be determined experimentally. 48 Factors affecting the coefficient of permeability 1. Shape and size of the soil particles. Sample A Sample B 49 2. Void ratio  Permeability increases with increase in void ratio. 3. Degree of saturation  Permeability increases with increase in degree of saturation. 4. Composition of soil particles  For sands and silts this is not important; however, for soils with clay minerals this is one of the most important factors. Permeability depends on the thickness of water held to the soil particles 50 5. Soil structure  Fine-grained soils with a flocculated structure have a higher coefficient of permeability than those with a dispersed structure. 6. Viscosity of the fluid 7. Density and concentration of the fluid 51 Determination of coefficient of permeability in the laboratory  Two standard laboratory tests are used to determine the hydraulic conductivity (k) of soil— 1. constant-head test 2. falling-head test An indirect determination of k can be obtained from consolidation test  Constant-head test 52 The constant-head test is suitable for more permeable granular materials. The soil specimen is placed inside a cylindrical mold, and the constant-head loss h of water flowing through the soil is maintained by adjusting the supply. The outflow water is collected in a measuring cylinder, and the duration of the collection period is noted. The value of k is calculated as: where: QL k= Q = volume of water collected A = area of cross section of the hAt soil specimen t = duration of water collection L = length of specimen h = difference in head  Falling-head test 53  The falling-head permeability test is more suitable for fine- grained soils. The soil specimen is placed inside a tube, and a standpipe is attached to the top of the specimen. Water from the standpipe flows through the specimen. The initial head difference h1 at time t = 0 is recorded, and water is allowed to flow through the soil such that the final head difference at time t = t is h2. The value of k can be calculated as: aL h1 k = 2.303 log a = cross-sectional area of the standpipe At h2 A = cross-sectional area of the soil specimen L = length of specimen t = time required for the head to drop from h1 to h2  Equivalent Hydraulic Conductivity in 54 Stratified Soil Many natural soils, especially alluvial soils, contain horizontal stratifications that reflect the history of deposition. The hydraulic conductivity in some layers is often much greater than in others, so ground water flows horizontally much more than vertically 55  Equivalent Hydraulic Conductivity in 56 Stratified Soil In a stratified soil deposit where the hydraulic conductivity for flow in a given direction changes from layer to layer, an equivalent hydraulic conductivity can be computed to simplify calculations. We need to determine two values of k: The equivalent horizontal hydraulic conductivity, kh(equiv.) The equivalent vertical hydraulic conductivity, kv(equiv.)  FLOW IN HORIZONTAL DIRECTION IN 57 STRATIFIED SOIL If kh1 , kh2 , kh3 , …, are the coefficients of permeability of layers 1, 2, 3, …, respectively. = kh1H1 + kh 2 H 2 +... + khi H i kh ( equiv.) =  k H hi i H kh ( equiv.) H1 + H 2 +... + H i i  FLOW IN VERTICAL DIRECTION IN 58 STRATIFIED SOIL kv1 , kv2 , kv3 ,…, be the coefficients of permeability for flow in the vertical direction. H1 + H 2 +... + H i kv ( equiv.) = H1 H 2 Hi + +... + k v1 k v 2 kvi kv ( equiv.) =  H i H    k i  vi   CEGEOEN Module 5 SEEPAGE (TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS) 1 Dr. Mary Ann Q. Adajar 2  Flow of water through soils is called seepage.  Seepage takes place when there is difference in water levels on the two sides of the structure such as a dam or a sheet pile as shown in Fig. 3  Whenever there is seepage, it is often necessary to estimate the quantity of the seepage, and permeability becomes the main parameter here.  Problems on the calculation of flow of water through soil can be solved: a. Analytical Method – use of Laplace Equation of Continuity b. Graphical Method - use of flownets (a trial-and- error method)  FLOWNETS 4  A set of flow lines and equipotential lines is called flow net  A flow line is an imaginary line that traces the path that a particle of ground water would follow as it flows through an aquifer.  Equipotential lines are lines of constant total head 5  Flow nets are constructed for the calculation of groundwater flow and the evaluation of heads in the media.  To complete the graphic construction of a flow net, one must draw the flow lines and equipotential lines in such a way that 1. The equipotential lines intersect the flow lines at right angles. 2. The flow elements formed are approximate squares (curvilinear squares).  FLOWNETS 6  A set of flow lines and equipotential lines is called flow net. Flow line Equipotential line 7 (Impermeable layer) (permeable layer) (Approximate square) 8 in Isotropic Soil Nf = Nd = Qi = kH/Nd Qt = NfQi Qt = kH(Nf/Nd) 9 in Isotropic Soil 10 in Isotropic Soil Nf =Nd = Qt = kH(Nf/Nd) 11  If flow elements in the flow nets were constructed as rectangles, the width-to-length ratio of the flow nets must be constant, that is: b1 b2 b3 = = =... = n l1 l2 l3 Nf Rate of seepage per unit length, Qt = kH n Nd  SUMMARY: 12 FLOWNETS IN ISOTROPIC SOILS  The rate of seepage through each flow channel: H Qi = k Nd Where: H = head difference between the upstream and downstream sides Nd = number of potential drops (or head drops) H h = Δh = head loss between each Nd potential drop 13 FLOWNETS IN ISOTROPIC SOILS  The total rate of flow through all channels: HN f Qt = k N f = number of flow channel Nd  Total flow rate through all flow channels with rectangular flow nets: Nf b1 b2 b3 Qt = kH n = = =... = n Nd l1 l2 l3  Heads and Hydraulic Gradient 14  Consider the base of the standpipe as the point 15 considered:  The height of water in the standpipe is the pressure head, hp  The elevation head, hz of a point is the height above the datum line to the base of the standpipe.  The height above the datum to the water level in the standpipe is the total head, h. If the point considered is above the datum: hp = h – hz hp If the point considered is below the datum: hp = h + hz 16  If there is seepage: Pressure head, hp = h – head loss + hz H head loss, h = Nd h  Hydraulic gradient, i= L Δh = head loss L = flow length which can be measured directly from flow net  UPLIFT FORCE UNDER A 17 HYDRAULIC STRUCTURES  Flow nets can be used to determine the uplift pressure at the base of a hydraulic structure.  The uplift pressure at any point j is calculated as follows: 1. Select a convenient datum (ex. downstream water level). 2. Determine the total head at j: 𝐻𝑗 = 𝐻𝑈𝑃 − 𝑁𝑑 𝑗 Δℎ Where: HUP = head at the upstream level Δh = head loss between each potential drop (Nd)j = number of potential drops at point j  3. Subtract the elevation head (hz) at point j from the 18 total head (Hj) to get the pressure head (hp)j ℎ𝑝 = 𝐻𝑈𝑃 − 𝑁𝑑 𝑗 Δℎ − ℎ𝑧 𝑗 If elevation head is below the datum, hz is negative If elevation head is above the datum, hz is positive If there is no seepage, the pressure head is the height to the free-water table. 4. The pore water pressure is: u j = (h p ) j  w 5. The uplift force per unit length due to groundwater flow: n Where: u = average j Pw =  u j x j porewater pressure over an j =1 interval Δxj  CEGEOEN Module 5 SEEPAGE (TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS) 1 Dr. Mary Ann Q. Adajar 2  Flow of water through soils is called seepage.  Seepage takes place when there is difference in water levels on the two sides of the structure such as a dam or a sheet pile as shown in Fig. 3  Whenever there is seepage, it is often necessary to estimate the quantity of the seepage, and permeability becomes the main parameter here.  Problems on the calculation of flow of water through soil can be solved: a. Analytical Method – use of Laplace Equation of Continuity b. Graphical Method - use of flownets (a trial-and- error method)  FLOWNETS 4  A set of flow lines and equipotential lines is called flow net  A flow line is an imaginary line that traces the path that a particle of ground water would follow as it flows through an aquifer.  Equipotential lines are lines of constant total head 5  Flow nets are constructed for the calculation of groundwater flow and the evaluation of heads in the media.  To complete the graphic construction of a flow net, one must draw the flow lines and equipotential lines in such a way that 1. The equipotential lines intersect the flow lines at right angles. 2. The flow elements formed are approximate squares (curvilinear squares).  FLOWNETS 6  A set of flow lines and equipotential lines is called flow net. Flow line Equipotential line 7 (Impermeable layer) (permeable layer) (Approximate square) 8 in Isotropic Soil Nf = Nd = Qi = kH/Nd Qt = NfQi Qt = kH(Nf/Nd) 9 in Isotropic Soil 10 in Isotropic Soil Nf =Nd = Qt = kH(Nf/Nd) 11  If flow elements in the flow nets were constructed as rectangles, the width-to-length ratio of the flow nets must be constant, that is: b1 b2 b3 = = =... = n l1 l2 l3 Nf Rate of seepage per unit length, Qt = kH n Nd  SUMMARY: 12 FLOWNETS IN ISOTROPIC SOILS  The rate of seepage through each flow channel: H Qi = k Nd Where: H = head difference between the upstream and downstream sides Nd = number of potential drops (or head drops) H h = Δh = head loss between each Nd potential drop 13 FLOWNETS IN ISOTROPIC SOILS  The total rate of flow through all channels: HN f Qt = k N f = number of flow channel Nd  Total flow rate through all flow channels with rectangular flow nets: Nf b1 b2 b3 Qt = kH n = = =... = n Nd l1 l2 l3  Heads and Hydraulic Gradient 14  Consider the base of the standpipe as the point 15 considered:  The height of water in the standpipe is the pressure head, hp  The elevation head, hz of a point is the height above the datum line to the base of the standpipe.  The height above the datum to the water level in the standpipe is the total head, h. If the point considered is above the datum: hp = h – hz hp If the point considered is below the datum: hp = h + hz 16  If there is seepage: Pressure head, hp = h – head loss + hz H head loss, h = Nd h  Hydraulic gradient, i= L Δh = head loss L = flow length which can be measured directly from flow net  UPLIFT FORCE UNDER A 17 HYDRAULIC STRUCTURES  Flow nets can be used to determine the uplift pressure at the base of a hydraulic structure.  The uplift pressure at any point j is calculated as follows: 1. Select a convenient datum (ex. downstream water level). 2. Determine the total head at j: 𝐻𝑗 = 𝐻𝑈𝑃 − 𝑁𝑑 𝑗 Δℎ Where: HUP = head at the upstream level Δh = head loss between each potential drop (Nd)j = number of potential drops at point j  3. Subtract the elevation head (hz) at point j from the 18 total head (Hj) to get the pressure head (hp)j ℎ𝑝 = 𝐻𝑈𝑃 − 𝑁𝑑 𝑗 Δℎ − ℎ𝑧 𝑗 If elevation head is below the datum, hz is negative If elevation head is above the datum, hz is positive If there is no seepage, the pressure head is the height to the free-water table. 4. The pore water pressure is: u j = (h p ) j  w 5. The uplift force per unit length due to groundwater flow: n Where: u = average j Pw =  u j x j porewater pressure over an j =1 interval Δxj  CEGEOEN Module 6 STRESSES IN A SOIL MASS 1 Dr. Mary Ann Q. Adajar Prepared by: MQAdajar 2  An important function in the study of soil mechanics is to predict the stresses and strains imposed at a given point in a soil mass due to certain loading conditions  This is necessary to estimate settlement and to conduct stability analysis of earth and earth-retaining structures, as well as to determine stress conditions Prepared by: MQAdajar 3 The “kissing” silos. (Bozozuk, 1976, permission from National Research Council of Canada.) These silos tilt toward each other at the top because stresses in the soil overlap at and near the internal edges of their foundations. The foundations are too close to each other. Prepared by: MQAdajar 4 Leaning Tower of Pisa Stresses in soil need to be known in advance in order to design the foundation that can properly support the structure during their entire life time. Prepared by: MQAdajar  NORMAL STRESSES AND STRAINS 5 A normal stress is the load per unit area on a plane normal to the direction of the load. Pz Px Py z = x = y = xy yz xz Normal stresses compress or elongate a material. Prepared by: MQAdajar 6 A normal strain is the change in length divided by the original length in the direction of the original length. z x y = y z = x = z x y The volumetric strain is: v =  x +  y +  z Prepared by: MQAdajar  SHEAR STRESSES AND SHEAR STRAINS 7 A shear stress is the load per unit area on a plane parallel to the direction of the shear force. F = xy Shear stresses distort a material Prepared by: MQAdajar 8 Shear strain is a measure of the angular distortion of a body by shearing forces. x  zx = tan −1 z x For small strains, tan γzx = γzx ,  zx = Prepared by: MQAdajar z  Material Responses to Normal Loading 9 and Unloading For uniaxial loading : Change in vertical stress P  z = A Vertical strain is Where: Ho is the original length Radial strain is ro is the original radius + for compression - for expansion Prepared by: MQAdajar 10  The ratio of the radial (or lateral) strain to the vertical strain is called Poisson’s ratio, ν Prepared by: MQAdajar 11 The elastic modulus or initial tangent elastic modulus (E) is the slope of the stress– strain line for linear isotropic material. The tangent elastic modulus is the slope of the tangent to the stress–strain point under consideration. The secant elastic modulus is the slope of the line joining the origin (0, 0) to some desired Linear and nonlinear stress–strain stress–strain point. curves of an elastic material Prepared by: MQAdajar  Elastoplastic Material 12  Some materials—soil is one of them—do not return to their original configurations after unloading. They exhibit a stress–strain relationship similar to that depicted in Fig. 6.6, where OA is the loading response, AB the unloading response, and BC the reloading response. The strains that occur during loading, OA, consist of two parts—an elastic or recoverable part, BD, and a plastic or unrecoverable part, OB. Such material behavior is called elastoplastic. Part of the loading response is elastic, the other plastic. 13 Material Response to Shear Forces  A typical response of an elastoplastic material to simple shear is shown in Figure. The initial shear modulus (Gi) is the slope of the initial straight portion of the shear stress (𝜏𝑧𝑥 ) versus shear strain (𝛾𝑧𝑥 ) curve. The secant shear modulus (G) is the slope of a line from the desired shear stress–shear strain point to the origin of the 𝜏𝑧𝑥 versus 𝛾𝑧𝑥 plot.   The stress at a point is defined by components 14 acting in several directions z  x = normal stress acting on the x-face (x-face is perpendicular to the x-axis) 𝜎𝑧 = normal stress acting on the z-face 𝜏𝑥𝑧 = shearing stress acting on the x-face in x the z-direction 𝜏𝑧𝑥 = shearing stress acting on the z-face in the x-direction 𝜏𝑥𝑧 = 𝜏𝑧𝑥 plane, direction   Some problems may require computation of the 15 stresses on other planes. σn τn σn = normal stress and τn = shear stress on a plane that makes an angle θ Prepared by: MQAdajar 16  Sign Convention:  Compressive normal stresses are taken as positive.  Shear stresses are positive if they tend to produce a counterclockwise rotation Prepared by: MQAdajar  STRESSES ON A PLANE ORIENTED AT AN 17 ANGLE  FROM HORIZONTAL PLANE θ is positive for σn τn counterclockwise orientation measured from x-axis σn = normal stress on a plane that makes an angle θ z +x z −x n = + cos 2 +  xz sin 2 2 2 τn = shear stress on a plane that makes an angle θ z −x n = sin 2 −  xz cos 2 Prepared by: MQAdajar 2 18 MAJOR AND MINOR PRINCIPAL STRESSES:  = inclination of plane of maximum or minimum normal stresses (principal planes) 2 xz tan 2 = z −x σ1 = maximum normal stress or major principal stress z +x z −x  2 1 = +   + ( xz )2 2  2  σ3 = minor normal stress or minor principal stress z +x z −x  2 3 = −   + ( xz ) 2 2  2  Prepared by: MQAdajar 19  max = maximum shear stress z −x  2  max =   + ( xz ) 2  2  1 −  3  max =  2 Prepared by: MQAdajar 20 STRESSES ON PLANE ORIENTED AT AN ANGLE FROM THE PRINCIPAL PLANES σn = normal stress on a plane that makes an angle θ with the principal planes. 1 +  3 1 −  3 n = + cos 2 2 2 τn = shear stress on a plane that makes an angle θ 1 −  3 n = sin 2 2 Prepared by: MQAdajar  The maximum and minimum 21 normal stresses occur at 90o apart. Maximum and minimum normal σ1 stresses occur on planes of zero z shearing stress. σ3 Maximum and minimum normal θ stresses are called the principal x stresses and the planes on which σ3 they act are called principal σ1 planes. The planes of maximum shearing stress are inclined at 45o with the principal planes Prepared by: MQAdajar 22 MOHR’S CIRCLE FOR STRESS STATES We can obtain these stresses using a graphical representation called a Mohr’s circle, which was developed by the German engineer Otto Mohr (1835-1918). The radius of the circle to any point on its circumference represents the axis directed normal to the plane whose stress components are given by the coordinates of the points Prepared by: MQAdajar  z 23 τzx σz τ τxz x σx σx x  max τxz τzx ( x , xz ) σz z R  xz 2 3 C  1 C = center of the circle x −z σ x +z 2 C= ( z ,− zx ) 2 R = radius of the circle  max z −x  2 R=   + ( xz ) 2 σ1 = maximum normal stress  2  σ1 = C + R σ3 = minimum normal stress σ3 = C - R τmax = maximum in-plane shearing stress Prepared by: MQAdajar 24  The angle between the radii to selected points on Mohr’s circle is twice the angle between the normal to the actual plane represented by these points.  The rotational sense of this angle corresponds to the rotational sense of the actual angle between the normal to the plane. Counterclockwise is positive, clockwise is negative.  The σ1 and σ3 occur at 90o apart in actual plane, 180o apart in Mohr’s circle Prepared by: MQAdajar  CEGEOEN Module 7 IN-SITU STRESSES: Geostatic Stresses Stresses in Saturated Soils 1 Dr. Mary Ann Q. Adajar Prepared by: MQAdajar  SOURCES OF STRESS IN THE GROUND 2  To evaluate the stresses at a point in the ground, we need to know the locations, magnitudes, and directions of the forces that cause them.  Two broad categories of sources of stress in the ground: GEOSTATIC STRESSES – are those that occur due to the weight of the soil above the point being evaluated. INDUCED STRESSES – are those caused by external loads such as structural foundations, vehicles, or fluid in a storage tank. Prepared by: MQAdajar 3 TOTAL AND EFFECTIVE STRESSES  The stress, σ is called the total stress  In geostatic condition, the total stress, σ at a point is the saturated unit weight of the soil and the unit weight of water above it. 4 The total stress, σ can be divided into two parts:  A portion is carried by water in the continuous void spaces called porewater pressure.  The rest of the total stress is carried by the soil solids at their point of contact. This is called the effective stress, σ’. The resistance or reaction to σ is provided by a combination of the stresses from the solids, called effective stress (σ ′), and from water in the pores, called porewater pressure (u). The equilibrium equation is:  =  '+u so that effective stress, 𝜎′ = 𝜎 − 𝑢 5 GEOSTATIC STRESSES Prepared by: MQAdajar  1. Stresses in Saturated Soil without Seepage 6  The total stress at the elevation of point A can be obtained from the saturated unit weight of the soil and the unit weight of water above it.  = H w + (H A − H ) sat Where: σ = total stress at the elevation of point A γw = unit weight of water γsat = saturated unit weight of soil H = height of water table from the top of the soil column HA = distance between point A and the water table. Prepared by: MQAdajar 7  The porewater pressure at pt. A: 𝑢 = 𝐻𝐴 𝛾𝑤  Effective stress at pt. A: '=  − u  ' = H w + (H A − H ) sat  − H A w  ' = (H A − H )( sat −  w )  ' = (Height of the soil column)   ' Where: γ’ = γsat – γw = submerged unit weight of soil The effective stress at any point A is independent of the depth of water H, above the submerged soil. Prepared by: MQAdajar  at A : Total stress :  A = H1 w 8 Porewater Pressure : u A = H1 w Effective stress :  A' =  A − u A = 0 at B : Total stress :  B = H1 w + H 2 sat Porewater Pressure : uB = ( H1 + H 2 ) w Effective stress :  B' =  B − uB = H 2 ' at C : Total stress :  C = H1 w + z sat Porewater Pressure : uB = ( H1 + z ) w Effective stress :  C' =  C − uC = z ' Prepared by: MQAdajar 9 In summary: The total stress at any point in the soil mass is due to the weight of the soil and water above it. Effective stress is the force per unit area carried by the soil skeleton. The effective stress in the soil mass controls its volume change and strength. Increasing the effective stress induces soil to move into a denser state of packing. Prepared by: MQAdajar  Example No. 1 10 a.) A soil profile is shown in Fig. Calculate the total stress, pore water pressure and effective stress at points A, B, and C. Prepared by: MQAdajar 11 b.) How high should the water table rise so that the effective stress at C is 190 kN/m2. Assume γsat = 19.25 kN/m2 for both layers Prepared by: MQAdajar 12 EFFECTS OF SEEPAGE If water is seeping, the effective stress at any point in a soil mass will differ from that in the static case. It will increase or decrease depending on the direction of seepage. Prepared by: MQAdajar  2. Stresses in Saturated Soil with Upward Seepage 13 at A : Total stress :  A = H1 w Porewater Pressure : u A = H1 w Effective stress :  A' =  A − u A = 0 at B : Total stress :  B = H1 w + H 2 sat Porewater Pressure : uB = ( H1 + H 2 + h ) w Effective stress :  B' =  B − uB  B' = H 2 ( sat −  w ) − h w = H 2 '−h w Prepared by: MQAdajar  at C : 14 Total stress :  C = H1 w + z sat h Porewater Pressure : uB = ( H1 + z + z ) w H2 Effective stress :  C' =  C − uC  = z ( sat −  w ) − h ' C z w H2 h  'C = z '− z w H2 h Note that is the hydraulic gradient i H2 caused by the flow. Prepared by: MQAdajar  'C = z '−iz w 15 From 1. Stresses in soil without seepage,  C' = z ' From 2. Stresses in soil with upward seepage,  'C = z '−iz w  The effective stress at a point located at depth z measured from the surface of a soil layer is reduced by an amount izγw because of upward seepage.  If the rate of seepage and hydraulic gradient gradually are increased, a limiting condition will be reached at which the effective stress becomes zero.  If the effective stress becomes zero, the soil loses its intergranular frictional strength and behaves like a viscous fluid. The soil at this state is called static liquefaction. Prepared by: MQAdajar 16 WHEN EFFECTIVE STRESS (σ’C )= 0  'C = z '−iz w = 0 ' i= = icr = critical hydraulic gradient w ' icr = w  ' =  sat −  w = Gs w + e w −w = (Gs − 1) w 1+ e 1+ e  ' Gs − 1 icr = =  w 1+ e Prepared by: MQAdajar 17  Static liquefaction occurs when icr is reached.  The soil stability is lost.  Events connected to static liquefaction are: ✓ Boiling occurs when the upward seepage force exceeds the downward force of the soil. ✓ Piping (or tunneling) refers to the subsurface “pipe- shaped” erosion that initiates near the toe of dams and similar structures. ✓ Quicksand is the existence of a mass of sand in a state of static liquefaction. ✓ Heaving occurs when seepage forces push the bottom of an excavation upward. Prepared by: MQAdajar  3. Stresses in Saturated Soil with Downward Seepage 18 at A : Total stress :  A = H1 w Porewater Pressure : u A = H1 w Effective stress :  A' =  A − u A = 0 at B : Total stress :  B = H1 w + H 2 sat Porewater Pressure : uB = ( H1 + H 2 − h ) w Effective stress :  B' =  B − uB  B' = H 2 ( sat −  w ) + h w = H 2 '+ h w Prepared by: MQAdajar  at C : 19 Total stress :  C = H1 w + z sat h Porewater Pressure : uB = ( H1 + z − z ) w H2 Effective stress :  C' =  C − uC  = z ( sat −  w ) + h ' C z w H2  'C = z '+iz w Prepared by: MQAdajar  SEEPAGE FORCE 20 The effect of seepage is to increase or decrease the effective stress at a point in soil. seepage force = iz w A Seepage force per unit volume = i w Prepared by: MQAdajar  Example No. 2 21  A 9-m thick layer of stiff saturated clay is underlain by a layer of sand. The sand is under artesian pressure. Calculate maximum depth of cut H that can be made in the clay. γsat = 18 kN/m3 9m 3.6 m 3m γsat = 16.5 kN/m3 Prepared by: MQAdajar  Example No. 3 Consider the upward flow of water through a layer of sand in a 22 tank as shown in Fig. For sand, the following are given: Void ratio, e = 0.52, specific gravity = 2.67 a.) Calculate the total stress, pore water pressure, and effective stress at points A and B. b.) What is the upward seepage force per unit volume of soil? Prepared by: MQAdajar  CEGEOEN Module 7 IN-SITU STRESSES: Geostatic Stresses Safety of Hydraulic Structures, Capillarity 1 Dr. Mary Ann Q. Adajar Prepared by: MQAdajar 2 GEOENG2 3 GEOENG2  SEEPAGE AND TUNNELING (PIPING) 4  Seepage thru earth dams occurs when the water seeps through the tiny soil pores and finds its way into some bigger cracks. This starts a process of “tunneling” where a tiny crack becomes larger and larger as the water starts moving through it and carrying the surrounding soil particles away with it. Eventually the crack widens to the point where the water comes rushing GEOENG2through the levee and crumbles the entire structure.  SAFETY OF HYDRAULIC 5 STRUCTURES AGAINST PIPING  When “piping”, “heaving” or “quicksand” occurs, the soil has no bearing capacity, hence it can not support structures. GEOENG2 6 Piping or heaving originates in the soil mass when hydraulic gradient i is greater than or equal to the critical hydraulic gradient, icr, ' icr = w Gs − 1 icr = 1+ e GEOENG2 7 Heaving in Soil Due to Flow around Sheet Piles  TERZAGHI (1922) conducted some model tests with a single row of sheet piles as shown in Figure and found that the failure due to heaving (or piping) takes place within a distance of D/2 from the sheet piles (D is the depth of penetration of the sheet pile). GEOENG2 8  In order to prevent failure, the weight (W) of the soil prism in the zone must be greater than the uplifting force due to seepage GEOENG2  Factor of 9 Safety against heaving: W FS = 3 U Where: W = submerged weight of soil in the heave zone per unit length of sheet pile U = uplifting force caused by seepage on the same volume of soil GEOENG2 10 H = Total head loss HT h2 D D 2 GEOENG2  STEPS: 11 1. Estimate the average pressure head (h1) at point P along the base (a-b) of the soil prism of unit thickness. D The variation of pressure over the base is considered to be parabolic. ha + 2hb h1 = 3 hb h1 ha = H T − (h )a ha U hb = H T − (h )b GEOENG2 12 2. Determine the actual seepage pressure head (hs) to be dissipated through the soil prism. hs = h1 − h2 D 3. Estimate the uplifting force, U D U = A whs =  whs hb 2 h1 ha D U where : A = (1) Surface area of 2 the base of soil prism GEOENG2  The average hydraulic gradient 13 across the prism, hs iav = D D U =  w (Di av ) = D  wiav D 1 2 2 2 4. Calculate the submerged weight of the soil prism. hb h1 1 2 ha W = D ' U 2 GEOENG2  5. Calculate the Factor of Safety: 14 1 2 ′ 𝑊 𝐷 𝛾 𝐹𝑆 = = 2 𝑈 1 𝐷ℎ 𝛾 2 𝑠 𝑤 D ' ' icr FS = or FS = or FS = hs w iav w iav GEOENG2  TERZAGHI’s Alternate Method 15 for Flow around a Sheet Pile D ' FS = Co w (H1 − H 2 ) (Das, 2014) GEOENG2  HARZA (1935) investigated the safety of hydraulic structures 16 against piping. The factor of safety (FS) against piping: icr FS = iexit Where:  ' Gs − 1 icr = =  w 1+ e iexit is the maximum exit gradient which can be determined from the flow net. h H Where: Δh = head loss between the last iexit = h = two equipotential lines. L N d L = the length of the flow element A factor of safety of 3 is considered adequate for GEOENG2 the safe performance of the structure.  Harza also presented 17 charts for the maximum exit gradient of dams constructed over deep homogeneous deposits. h iexit =C B GEOENG2  Example #1 18 The sheet pile arrangement shown in Figure is to be examined for adequacy. Determine the factor of safety against piping failure using a.)Terzaghi’s method b.) Harza’s method e = 0.61 Gs = 2.67 γsat = 20 kN/m3 k = 2.6 x 10-5 m/s GEOENG2 Impermeable 19 GEOENG2  PREVENTION OF PIPING 20  To increase the factor of safety against failure, several methods were recommended: 1. Placing filter material over the danger zone. GEOENG2 21 2. Lengthening the flow lines, by driving the sheet pile deeper or by installing sheet piles at one or both ends of a concrete dams. Filter GEOENG2 22 3. Lengthening the flow lines at concrete dams by constructing upstream or downstream concrete aprons. GEOENG2 23 CAPILLARITY 24 Capillarity (or capillary action) is the upward movement of a liquid into the vadose zone, which is above the level of zero hydrostatic pressure ✓ this upward movement occurs in porous media ✓ It is the result of the surface tension between the water and the media. Capillarity 25 26 The theoretical height of the capillary rise, hc, in a glass tube of diameter, d, at a temperature of 200C is: 0.03 hc = d where; hc − height of the capillary rise(m) d − diameter of glass tube(mm) 27 ✓ Capillary rise in soils is more complex because soils contain an interconnected network of different-size pores. ✓ Using d = 0.2D10 generally produces good results for sands and silts. ✓ The height of the capillary rise in these soils is approximately: where; 0.15 hc = hc − height of the capillary rise( m) D10 D10 − grain diameter(mm) with 10% passing  EFFECT OF CAPILLARY RISE 28  Water surface exposed to the atmosphere is under tension, called capillary tension. 1 hc  d The smaller the capillary tube diameter, the larger the capillary rise. GEOENG2 29  In soils, water also rises above the ground water table because of surface tension. The speed of rise depends on the soil types: a.) In clay, capillary rise is slow due to very small pore size as well as the presence of water bonded to the clay particles. b.) In sand and silty sand, the rise depends on the: pore size, particle shape and distribution density, original water content, viscosity of water GEOENG2 30 GEOENG2  EFFECTIVE STRESS IN THE ZONE OF 31 CAPILLARY RISE  The effects of capillary rise on pore pressure are: 1. An increase in the density of the soil within the region; this increases the total stress (σ). 2. The pore pressure, u is negative throughout the region. The value at a point in a layer of fully saturated soil by capillary rise is u = − h w If partial saturation is caused by capillary action: Where:  S  h = height of the point under u = −  wh consideration measured from the  100  ground water table. S = degree of saturation in percent GEOENG2 32 3. The pore water pressure due to capillary rise varies linearly with depth, becoming zero at the ground water table. 4. The capillary action has no effect on the pore water pressure below the ground water table. GEOENG2  Example #2 33  A soil profile is shown in Fig. Given H1 = 1.83m, H2 = 0.91m, H3 = 1.83m. Plot the variation of total stress, pore water pressure, and effective stress with depth. GEOENG2  CEGEOEN Module 8 IN-SITU STRESSES: INDUCED STRESSES Part 1 Dr. Mary Ann Q. Adajar 1 GEOENG2  INDUCED STRESSES – are those caused by external loads such as structural foundations, vehicles, or fluid in a storage tank.  Construction of a foundation causes changes in the stress, usually a net increase.  The net stress increase in the soil depends on the load per unit area to which the foundation is subjected, the depth below the foundation at which the stress estimation is desired, and other factors.  It is necessary to estimate the net increase of vertical stress in soil that occurs as a result of the construction of a foundation so that settlement can be calculated.  1. STRESSES CAUSED BY A POINT LOAD 4  Boussinesq’s solution for normal stresses due to point load (P) applied at the soil surface. GEOENG2 5 Eq. (8.1) GEOENG2 6 Table 8.1  2. VERTICAL STRESS CAUSED BY A 7 VERTICAL LINE LOAD  The Figure shows a vertical line load of infinite length that has intensity q/unit length on the surface Eq. (8.2a) OR Eq. (8.2b) Eq. (8.2c) GEOENG2 8 Table 8.2 GEOENG2 9 3. VERTICAL STRESS CAUSED BY A HORIZONTAL LINE LOAD Eq. (8.3) GEOENG2 10 Table 8.3 GEOENG2  Example #1 11 An inclined line load with a magnitude of 10 kN/m is shown in Figure. Determine the increase of vertical stress (Δσz) at point A due to the line load. GEOENG2  CEGEOEN Module 8 IN-SITU STRESSES: INDUCED STRESSES Part 2 Dr. Mary Ann Q. Adajar 1 GEOENG2  4. VERTICAL STRESS CAUSED BY A VERTICAL STRIP 2 LOAD (FINITE WIDTH AND INFINITE LENGTH) GEOENG2 3 Table 8.4 GEOENG2  Table 8.4 4 GEOENG2  Table 8.4 5 GEOENG2  Table 8.4 6 GEOENG2  Table 8.4 7 GEOENG2  EXAMPLE #2 8  Given:  B = 4m;  q = 100 kN/m2  For point A: z = 1m; x = 1m Determine the Vertical stress Δσz GEOENG2  2 x 2(1) 2 z 2(1) 9 = = 0.5 = = 0.5 B 4 B 4 Using Table 8.4:  z = 0.902 q  z = 0.902(100) = 90.2kN/m 2 Table 8.4 GEOENG2  5. Linearly Increasing Vertical Loading 10 on an Infinite Strip GEOENG2 11 Table 8.5 GEOENG2  6. Vertical Stress due to Embankment 12 Loading GEOENG2 13  Simplified form: Fig. 8.6b GEOENG2  14 Fig. 8.6b GEOENG2  EXAMPLE #3 15  An embankment is shown in Figure. Determine the stress increase under the embankment at points A1 and A2. GEOENG2  7. Vertical Stress Below the Center of a 16 Uniformly Loaded Circular Area  GEOENG2 17 The variation of σz/q with z/R is given in Table 8.6: Table 8.6 GEOENG2  8. Vertical Stress at Any Point below a 18 Uniformly Loaded Circular Area Δσz at any point A located at a depth z at any distance r from the center of the loaded area can be given as: Where: A’ and B’ are functions of z/R and r/R (refer to Tables 8.7 and 8.8) GEOENG2 19 Table 8.7 GEOENG2  Table 8.7 20 r/R GEOENG2  Table 8.8 21 GEOENG2 22 Table 8.8 r/R GEOENG2  9. Vertical Stress Caused by a Rectangularly 23 Loaded Area  Case 1 Vertical stress below the corner of a uniformly loaded flexible rectangular area GEOENG2 24 GEOENG2  The variation of I3 with m and n is shown in Table 8.9 25 Table 8.9 GEOENG2  m 26 GEOENG2 27 Case 2 Vertical stress increase below the center of a rectangular area GEOENG