Soil Mechanics Lecture Notes h672a

Introduction to Soil Mechanics

Dr. Hop Minh Nguyen

3.1.

Introduction

People may often think about civil engineering as the field related to the design and construction of the superstructures above the ground level. What is not easy to be seen is the foundation system below the building; the construction of this part is, however, a crucial component of any construction project. Failure in the foundation design and construction could easily lead to dire consequences as it can affect the stability of the whole structure. Foundation can be considered as a common type of loading impact on the original ground condition. Civil engineers need to study soil mechanics to be able to understand and predict the behaviour of soils in response to any change to its in-situ (original) condition, which would happen as the result of the construction activities. Applications for soil mechanics theories are very diverse, which include (among others) foundation design, stability analysis of natural and man-made slopes, design of retaining structures and embankments, tunnel design, etc. An important factor that makes soils apart from other construction materials is: soils are natural materials. Further than that, soils have history, which would explain why soil is now at its current location and condition. Would it have been formed here or it has been formed at some where else and then transported to the current place? Soil history (normally in of hundreds or even thousands of years) makes soil at each location on the earth unique and affects significantly its behaviour. Understand what happened in the past helps the engineers to predict what could happen when the material is subjected to new loading conditions. Soil behaviours are complex; working with soils involves many uncertainties, you will learn later that the application of a factor of safety (FOS ≈ 2 ÷ 3) is a common practice in the design of earth structures. The idea is that the capacity of the material is purposely reduced, so the design would be based on a weaker material and the solution is on the safe side. We all live on the outmost layer of the earth that is called the crust. The crust is mainly composed of rocks. When the rock surface exposes to the environment, it will be subjected to the weathering processes. For examples, frequent large variations of temperature could lead to rock cracking and ultimately to complete disintegration; or water entering the cracks from the surface could be subjected to freezing and thawing cycle and the volume expansion could also break rock into small fragments. These two processes (not exclusive) are classified as physical weathering. In addition, there also have chemical weathering, in which water and carbon dioxide in the atmosphere form carbonic acid that could react with the chemical substances within the composition of rocks. These chemical processes result in the creation of new finer materials with the mineral composition different than the mineralogy of the parent rocks. According to Craig (2004), soil is defined as “any uncemented or weakly cemented accumulation of mineral particles formed by the weathering of rocks”. Soil particles resulted from physical weathering of rock are normally in bulky form, with size greater than 2 µm and having the same mineral composition of the parent rock. Typical soils are silt, sand, gravel, cobble and boulder. Soils resulted from chemical weathering, on the other hand, are clayey soils, which normally compose of platy clay mineral particles with size < 2 µm. As platy particles having high specific area (i.e. high ratio of surface area per unit mass), soil structure is affected significantly by surface forces. When clays are mixed with water, the platy particles carrying negative charge on their surfaces attract cations and hydrogen atoms of the water molecules. The water held onto the clay particles by attraction force is called double-layer water. The innermost 2

layer of water next to the clay surface is called absorbed water, which has higher viscosity than the free water away from the clay particles. Absorbed water can move freely in parallel to the particle surface but is restricted in the normal direction. The existence of the double layer water and the ability of platy particles sliding relatively to each other explain for the plasticity property of clay when it is mixed with water.



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3.2.

Soil classification & phase diagram:

Figure 2.1: Particulate nature of soil Figure 2.1 shows a diagram of an idealized soil sample. As you can see, soil is not a continuum but an assembly of soil particles. This statement becomes obvious when you observe a small heap of sand put on your hand. The particulate nature of soil, however, still holds true for clays, although the very small particles are not visible to naked eyes. Soil sample always contains a lot of internal voids; if there is no water inside the void space, soil is dry as for layers of coarse sand or gravel above the underground water table. Soils below underground water level could have the void space fully filled by water; in this case, soil is fully saturated. Soils with small particle size (clay, silt, fine sand) immediately above the water table can be fully saturated to partially saturated, because water can rise above the steady water table due to the capillary effect. As soil particles are considered as incompressible solid, the total volume of soil particles in a sample is always constant. Consequently, volume change in soil is actually equal to change in the void volume that occurs when the particles rearrange under the new loading condition.

Figure 2.2: Soil particle sizes (Atkinson, 1993) Particle size of natural soils varies greatly from the “micrometre” range for clay particles up to as large as half metre for boulder. The size of 0.06 mm is considered as the boundary between coarse particle and fine particle. Coarse soil particles are visible to naked eyes, whereas the particles of fine soils are only visible with the help of hand lens and/or microscope. Coarse soils in nature are normally mixed with fine material, which is the dirt (clean material, however, can have all the particles in the same size range, e.g. clean sand). In general, soil containing 35% and greater of fine particles is referred as fine soil, coarse soils are soils with less than 35% fine content. With no magnifying glass, it can be difficult to differentiate between clay and silt as both are fine particles. Here, the distinction can be based on the plasticity property of clay mineral particles. Silt particles can be fragments from broken sand particles and hence similar to sand, there is no bonding between

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them so we can easily brush silt particles off our hand. On the other hand, clay mineral can absorb water and becomes sticky on your finger, which will need to be washed off. The existence of even a small content of clay mineral in soil could have an out of proportion effect to the overall behaviour of the sample. Based on the plasticity characteristic of clay mineral, soils are often classified as cohesive soil and cohesionless soil. Cohesionless soils are coarse soils (sand, gravel, sandy gravel, etc.), which mainly consist of discrete particles with no bonding so it is also called granular soil. Cohesive soils are fine soils with significant plastic behaviour resulted from the interaction between the clay mineral content and the water inside the pore space. The cohesive strength comes from surface forces, which is more dominant for small particles, and the negative pore water pressure due to capillary tension. •

Classification of granular (coarse) soils based on particle size distribution (PSD):

Relative proportion of different particle sizes by mass within a soil sample can be determined by doing sieve test. Before the test, any cobble or boulder (> 63 mm) will be removed, their mass should be estimated; the total mass (M) of the remained soil sample is then weighted. The mass of soil with particles smaller than 63 mm (D0) is let ing through a series of successively smaller sieves (D1 > D2 > …> Di > ..). The mass of material retained on each sieve size (i.e. for particle size larger than the opening size of the mesh) is weighted as well as the residual mass ing through the last (smallest) sieve (size ≈ 63 µm). Dividing the retained mass (mDi) on each sieve (Di) by the total mass (M) that we have determined at the beginning of the test, we can calculate the percentage of the particles that are greater than the sieve size Di and smaller than the greater sieve size immediately above it (Di-1). The percentage of material smaller than a size Di can be obtained by subtracting the total mass (M) by the sum of mass of all materials larger than Di and then divided by M, which can be given in the following form:

m D ≤ Di M

(%) =

M − ∑ (m D1 + ... + m Di −1 + m Di ) × 100 % M

(2.1)

Size distribution for particles smaller than 63 µm can be obtained via hydrometer analysis, which is based on the sedimentation of particles in water. Further details can be found in the text book of soil mechanics (e.g. Das, Advanced soil mechanics, 2008).

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Example 2.1: Test results on particle size distribution of soil

Craig (2004)

The first column in the table shows the sieve size (Di) in BS standard. “Percentage smaller” is the value of

mD≤ Di

M

in Eq. 2.1 for the corresponding sieve size. The particle size distributions of four

soils A, B, C and D are plotted as follows:

The particle size corresponding to any specific value of “Percentage smaller” on the vertical axis can be read off from the PSD graph. For example, particle size at 10% smaller of Soil B can be obtained by drawing a horizontal line crossing the vertical axis at 10% and intersect the PSD curve of Soil B at approximately near the boundary between fine and medium sand. Furthermore, a vertical line drawing at the boundary size between gravel and sand intersects the PSD of Soil A at around 23% on the vertical axis, which would mean Soil A consists of about 23% sand and 77% gravel, and hence the sample is gravel dominant (according to the table below, Soil A is a very sandy GRAVEL).

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Based on the percentage of different particle sizes from the PSD graph, coarse soils consisting of more than one type of material can be classified as follows: Table 2.1 Classification of composite coarse soils (Craig, 2004)

The characteristics of a coarse soil can be reflected through the PSD. The shape and the slope of the curve can be described in of two simple parameters defined as follows:

D60 D10

(2.2)

D302 D60 D10

(2.3)

The coefficient of uniformity (CU): CU =

The coefficient of curvature (CZ): CZ =

Here, D10, D30 and D60 are the particle sizes (abscissa) corresponding to 10%, 30% and 60% values on the vertical axis (coordinate) of the PSD graph. A soil is referred as well-graded if we have CU ≥ 4 and 1 ≤ CZ ≤ 3. A soil having CU ≤ 2 is considered uniform, which means its particles are more or less of similar size. •

Classification of fine soils based on plasticity characteristic:

The behaviour of fine soils depends very much on the amount of water inside the pore space. Clays with no or very little water are brittle, which could crumble between your fingers. Adding enough water, we now can easily deform the sample into any shape without cracking. However, if too much water is added, clay can flow like a viscous liquid under gravity.

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Figure 2.3: Consistency of cohesive soils (Das, 2008) Water content or moisture content is defined as the ratio of the mass of water over the mass of the solid phase (i.e. soil particles). Figure 2.3 describes different states of a cohesive soil depending on the value of moisture content. Liquid limit is the boundary moisture content when soil stops behaving as a liquid with a decrease in water content. Compared to the liquid state, soil in plastic state would develop some resistances against shearing deformation and hence the sample could hold its shape. As water content continues to decrease, clay sample will its plastic limit when it moves into the semisolid state with brittle failure, and soil becomes crumbly as the result. Shrinkage limit is reached once the volume of the sample becomes constant regardless of any further decrease of water content, and soil is now in solid state. Cohesive soils at the field are normally in plastic state, the range of water content for soil existing in the plastic state is called plasticity index, PI, which is equal to the difference between PL (plastic limit) and LL (liquid limit) or PI(%) = LL(%) – PL(%). The values of PL, LL and PI for cohesive soil can be determined from standard laboratory tests. Since plasticity of clay is due to the existence of the double-layer water, the plasticity index is affected by the capacity of soil in absorbing water, which depends on the amount of clay mineral in the sample and the type of the clay mineral. In general, the finer the mineral, the higher the specific area and the higher plasticity index. Skempton (1953) found that the PI of a soil is a linear function of the percentage by mass of the clay mineral particles (< 2µm) within the sample; this can be described by the following equation:

A=

PI % by mass of clay mineral particles inside soil sample

(2.4)

Here, A is a constant called activity which depends on the type of the clay mineral. For example, the value of A for illite mineral is 0.5 – 1.0 > A = 0.5 for kaolinite. Note that the specific area of illite (100 m2/g) is greater than kaolinite (10 m2/g), and this would imply that illite mineral can absorb more water than kaolinite mineral. We need to do standard tests on soil to determine the value of PL, LL and PI; fine soil can then be classified as clay or silt depending on the value of PI (see Fig. 2.4). In general, PI of clay will be greater than PI of silt (in Fig. 2.4, clay (C) is above the A-line whereas silt (M) is below). Furthermore, based on the value of LL (wL in Table 2.2), soils are further divided into smaller groups with low, intermediate, high plasticity, etc.

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Figure 2.4: Plasticity chart in BS standard (Craig, 2004)

Table 2.2 and letter symbols used for classification of soils (Craig, 2004)

Some examples on the use of the letter symbols in Table 2.2: GW (well graded gravel), GCL (clayey gravel with clay of low plasticity), MIS (sandy silt with silt of intermediate plasticity). •

Phase diagram and phase relationships:

Figure 2.5: Phase diagram for soil Soil is multi-phase material, which consists of soil particles as the solid phase, pore water as the liquid phase, and pore air as the gas phase. Fully saturated soils only consist of two phases: solid and liquid. As the behaviour of soil will depend on the relative proportions of its individual phases, the properties of a soil sample can be studied based on an idealized phase diagram as shown in Fig. 2.5. Here, while the letter symbols M and V indicate mass and volume; the subscripts a, w, s specify the mass and volume of each individual phase: air, water and solid, respectively; Vv is the volume of 9

void, which is the sum of air volume and water volume. Using the phase diagram, several soil properties can be defined as follows: Void ratio (e) and porosity (n):

e=

volume of void Vv volume of void Vv e n ; n= = ; n= ;e= = total volume V volume of solid V s 1+ e 1− n

Degree of saturation (Sr):

Sr =

volume of water Vw = volume of void Vv

Dry soil: Sr = 0 (void is full of air); fully saturated soil: Sr = 1 (void is full of water); otherwise, 0 < Sr < 1. Water content or moisture content (w):

w=

mass of water M w Ww = = mass of solid M s Ws

Weight and mass: weight is the downward force acting on a mass due to earth’s gravity (g = 9.81 m/s2), the dimension of weight is the dimension of force (e.g., N, kN). W = mass * gravity = M * g

(1 N = 1 kg * 1 m/s2)

Specific gravity (Gs): is the ratio between the mass (or weight) of a material and the mass (or weight) of the same volume of water. For the solid phase of soil,

Gs =

ρ Ms = s Vs ρ w ρ w

where ρs and ρw are the densities (mass per unit volume) of solid (soil particle) & water, respectively. Unit weight of soil (γ): weight of a unit volume of soil. For dry soil, γ = γdry; for saturated soi: γ = γsat; for the general case of natural soil (0 ≤ Sr ≤ 1):

γ = γ dry =

G + eS r total weight W W s + W w G s γ wV s + γ wVv S r = = = =γw s total volume V V s + Vv V s + eV s 1+ e

G +e Ws Gγ V Gγ weight of solid (because S r = 1) = = s w s = s w ; γ sat = γ w s 1+ e total volume Vs + Vv Vs + eVs 1+ e

Here, γw is the unit weight of water, γw = ρw * g = 1000 (kg/m3) * 9.81 (m/s2) = 9.81 (kN/m3) Relationship between e, w, Gs and Sr: eSr = wGs ; or e = wGs for fully saturated soil. 10

Example 2.2: A saturated sample of clay was extracted from the ground and weighted. It was then placed in an oven for 24 hours to dry and weighted again afterwards. Determine the void ratio and saturated unit weight knowing that specific gravity of soil particle Gs = 2.7, and: Wet mass of soil = 275 g Dry mass of soil = 196 g Answer: M = 275 g, Ms = 196 g → Mw = M – Ms = 275 – 196 = 79 g Vs = Ms/ρs = Ms/ (Gsρw) = 196/(2.7*1) = 72.6 cm3 Soil is fully saturated: Vv = Vw = Mw/ ρw = 79/1 = 79 cm3 Void ratio: e = Vv/Vs =79/72.6 = 1.09 Saturated unit weight: γsat = γw(Gs+e)/(1+e) = 9.81(2.7+1.09)/(1+1.09) = 17.8 kN/m3

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3.3.

Effective stress & drainage

Figure 3.1: Normal and shear stresses acting at the bottom of a block sitting on a horizontal surface Figure 3.1 shows a block sitting on top of a flat surface. Due to its weight, the block will apply onto the surface a normal stress that is equal to its self-weight divided by the area. If the block and the surface are made of hard materials, the block will not deform as well as will not sink into the surface under reasonable combination of its self-weight and any extra vertical loads applied on top of the block. If the block tends to move horizontally, there will have shear stress acting horizontally against that movement and can be calculated by dividing the horizontal force over the area. The block will start sliding only when the applied horizontal force is greater than the maximum frictional resistance. The stress condition in Fig. 3.1 can be an approximation of the stress condition at the point between soil grains. For the general stress range in engineering applications, soil particles can be considered as incompressible, which means the grains will not deform. Furthermore, there is shearing resistance at the point, which gives shear strength to the soil skeleton. When the applied stress exceeds the strength of the soil grain skeleton, there will have slippage occurs at many points, which leads to rearrangement of the soil particles and a new structure will be formed. Volume of the soil sample, consequently, could change during this process. The water inside the pores between the soil grains is considered incompressible, which means water does not change volume. However, unlike the soil skeleton, water has no shearing resistance and hence can not help to restrict the distortion of the soil sample. As both soil grains and water are incompressible, any attempt to reduce the volume of a saturated sample is not possible if the pore water is not allowed to be vented out. You will learn later that this condition is called undrained loading condition. Only when water is allowed to escape from the pore then the volume of saturated sample can change and this is called drained loading condition.

Figure 3.2: Discrete points and average pressure acting at the bottom of a soil column Figure 3.2 shows the internal situation inside an idealized soil sample with particles each other at discrete points. In reality, there is countless number of points as there are easily millions of soil particles in even a small geotechnical problem. In addition, the stress condition at each point is not the same, and hence it is not practical to define the stress condition at a point inside a soil body based on the local condition surrounding the point. Instead, we will need to find the stress value in an average manner. Figure 3.2 shows a soil column in a similar situation to the solid block in Fig. 3.1. As we will not be able to count the exact number of points that a real soil column would have with the horizontal surface, we assume that the weight of the column is 12

distributed evenly over the whole bottom area of the column. The average vertical pressure can be determined as:

weight of the column cross sectional area of the column γV γAh = = = γh A A

σv =

The total vertical stress at depth h below the ground surface can be calculated as:

σv = γh

(3.1)

where γ is the unit weight of natural soil. For dry soil, γ = γdry; similarly γ = γsat for saturated soil. If the water table is at the middle of the column, we need to apply different γ for soil above and below the water table.

Figure 3.3: A soil column from the ground

Example 3.1: A container of 2.5 m2 cross-sectional area (A = 2.5 m2) and 3 m deep (h = 3m) is filled up with dry sand with a voids ratio of 0.6 (e = 0.6). Determine the total vertical normal stress on the base of the container. Ignore wall side friction. Answer: Dry unit weight of sand: γd = (γwGs) / (1+e) = 10 *2.7 / (1 + 0.6) = 16.9 kN/ m3 Total weight: W = γdV = γdAh = 16.9 (2.5 · 3) = 126.7 kN Total vertical normal stress (assumed acting uniformly on the base): σv = W/A = 126.7 / 2.5 = 50.7 kN/m or: σv = Ahγd/ A = γdh = 16.9*3 = 50.7 kN/m2 If the soil is now saturated (i.e. all the voids are filled up with water): We need to calculate the saturated unit weight of the soil: γsat = γw(Gs + e)/(1+e) = 20.6 kN/m3 Total vertical normal stress: σv = γsath = 20.6*3 = 61.8 kN/m2

Similar to the calculation of the total vertical stress in Eq. 3.1, water pressure at depth hw below the underground water table can be obtained as follows:

σ w = γ whw

(3.2)

Here, the void spaces are assumed interconnected, and pore water is continuous; the pore water pressure defined in Eq. 3.2 corresponds to hydrostatic condition, which means the pressure is the 13

same in all directions. We have discussed in the previous part that the behaviour of a soil sample depends on the stress condition at the points. We, however, have not discussed how the stress condition at that points can be affected by the pressure of water inside the pore and vice versa. This is a complex coupled process that sets soil apart from other construction materials, and truly a fundamental of soil mechanics that civil engineers must understand.

Figure 3.4: The combination of interparticle forces and pore water pressure in equilibrium with the external load applied at the boundary of a soil sample Figure 3.4 shows an idealized soil element subjected to external load at the boundary. In a stable situation, all soil grains should be at rest, and forces are in equilibrium. Considering the equilibrium of forces on any arbitrary horizontal plane ing through the element, we can see that the selfweight of soil and any extra applied load will be transmitted partly through interparticle forces; the rest of the load can be equilibrated by pore water pressure. Now if there occurs a small perturbation leading to sliding movement at several points, particles will rearrange and voids collapse. As void tends to reduce its volume, water pressure inside the void will increase. At the same time, when particles rearrange, some s will be lost and the corresponding forces will be released. Note that during this process, the sum of the soil’s self-weight and any extra load will maintain the same. This means a decrease in forces could lead to an increase in pore pressure; the total force, meanwhile, is kept constant. In soil mechanics, such a situation can be conveniently explained in of the conceptual effective stress. In 1925, Terzaghi suggested the principles of effective stress for saturated soil in the following form: σ = σ’ + u

(3.3)

where: σ is the total stress; σ’ is the effective stress that represents (in an average sense) the stress carried by the soil skeleton; lastly, u is the pore water pressure (PWP). It is important to note that only the total stress, σ, and pore water pressure, u, are two measurable physical quantities. The effective stress, σ’, is always obtained as the difference between σ and u. In theory, effective stress can be calculated if you know the magnitude and direction of all interparticle forces; however, this is impractical and perhaps unnecessary since we often need to consider millions of particles for real soils. The application of the effective stress has been validated in many engineering problems and it is well considered as one of the most important concepts in geotechnical engineering. Equation 3.3 can be rewritten as: σ’ = σ - u

(3.4)

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Based on Eq. 3.4, if σ is kept constant, an increase in u will lead to a decrease in σ’. Refer to Fig. 3.4, an increase in the positive pore water pressure will tend to push soil particles apart from each other and hence decrease the forces between the particles. If the PWP is large enough to become equal to σ, then σ’ will become zero and particles can float inside water. Since there is no normal stress at , there will have no shearing resistance and the shear strength of soil is zero. If you stand on top of such sand, you will immediately sink into the ground because the soil does not have any strength to your weight. If there is a flow of water, the particles can easily go with the flow and be washed away. The condition of zero effective stress often happens for sand and is called quick sand. The value of PWP in Eq. 3.4 can go to negative, which implies water pressure is in tension. Negative PWP can increase effective stress by pulling particles to each other, and hence increase the shearing resistance at as the frictional force is proportional to normal force. This improvement effect, however, is not permanent. Negative PWP tends to suck water into the void space, and once more water is added, negative PWP and its improvement effect will disappear. An excellent example is the case of sand castle. Sand castle is built with damp sand that has negative PWP so it stands, when water is gradually added to a sand castle, the castle will collapse at some points.

Example 3.2: With the water table at 3m underground, draw the variation of the following with depth: [1] total vertical stress (σv) [2] pore water pressure (u), and [3] effective vertical stress (σ’v). Determine the values of σv, u & σ’v at level X.

Answer: Region A: At h = 0m, σv = 0 kPa u=0 At h = 3m, σv = (20)(3) = 60 kPa u=0 σ’v = σv – u = 60-0 = 60 kPa Region B: At h = 5m, σv = 60 + 20(2) = 100 kPa u = 10(2) = 20kPa σ’v = σv – u = 100-20 = 80 kPa Region C: At h = 8m, σv = 100 + 17(3) = 151 kPa u = 10(5) = 50kPa 15

σ’v = σv – u = 151-50 = 101 kPa At level X: At h = 6.5m, σv = 100 + 17(1.5) = 126 kPa u = 10(3.5) =35kPa σ’v = σv – u = 126-35 = 91 kPa Note: the unit weight of fine sand above the water table was taken as saturated because water can rise above the WT due to capillary effect. The amount of capillary rise for coarse sand and gravel is negligible, whereas it can be a few metres for silt and fine sand.

We have discussed that the relationship between effective stress and pore water pressure is a coupled process, which means any change in one parameter will affect the other and vice versa. We will now discuss what would happen to the effective stress and PWP for an increase in the total stress. According to Eq. 3.3, we should have: ∆σ = ∆σ’ + ∆u

(3.5)

When there is a large enough increase ∆σ of the total stress and the present soil skeleton is not able to sustain the higher load, particles need to rearrange to form a new structure that could resist the load. However, since the soil element in the ground is constrained by other elements surrounding it as well as the voids are fully occupied by water for saturated soil, particle rearrangement will not take place unless pore water can escape from the void. In this situation, firstly, PWP will increase by the same increase amount of the total stress, i.e. ∆u = ∆σ; and then, as water going out of the void, ∆u decreases and the corresponding load is transferred to the soil skeleton and effective stress is increased by ∆σ’. Eventually, PWP will return to its original hydrostatic value; at that point, ∆σ’ = ∆σ and ∆u = 0.

Figure 3.5: Dissipation of excess pore water pressure with time (Whitlow, 1995) The increase of PWP over its hydrostatic condition is called excess pore water pressure (∆u), the dissipation of excess PWP can be described graphically in Fig. 3.5. The soil element is modelled as a combination of a spring system immersed with water inside a watertight container; water is only allowed to go out through a valve. Here, the spring system represents the stiffness of the soil skeleton; the valve reflects the permeability of soil. Highly permeable soil means a larger valve. The dissipation of excess PWP can be explained by using the spring analogy. When we increase the vertical load in Fig. 3.5, the load is instantly transferred to water because the stiffness of water is much higher than the stiffness of the spring. Since water pressure is increased, it is pumped out through the valve. When water goes out, the piston goes down, and the load is gradually transferred to the spring system. The whole system will come to new equilibrium when the load is equilibrated 16

by the reaction force from the spring system, water returns to hydrostatic condition and water is no longer pumped out of the container. For soils with low permeability, e.g. clays, the dissipation of excess PWP will take years to complete, so we often need to check the stability of earth structure in short term based on the assumption of no dissipation of excess PWP taking place and this is called undrained analysis. Dissipation of excess PWP occurs quickly for sand and soil with high permeability, which means the applied load will be transferred to the soil skeleton during the construction process and we need to do the analysis in drained condition. Note that during earthquake, sand tends to be consolidated by ground vibration, which occurs very rapidly that excess PWP in sand can not be dissipated and sand can fail in undrained condition. This type of failure of sand is called liquefaction. In summary, drained or undrained depends on the relative relationship between the rate of loading and the permeability of soil. The same saturated soil could behave as undrained in one rapid loading condition but become drained in another slower mode of loading.

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3.4.

Permeability & one dimensional seepage

Soil is a porous material with interconnected voids that can form continuous ageways for water ing through it. There exists, however, two different types of water flow in soil: steady state flow and transient flow. Steady state flow is satisfied when pore water pressure at any point along a flow line is a constant with time and so is the flow rate. Effective stress also does not vary and soil grains can be considered in static condition (i.e. the grains do not move); there is, consequently, no volume change. On the other hand, transient flow occurs when there is volume reduction leading to excess pore water pressure, which is then dissipated towards any nearby free drainage boundary. In this case, excess pore water pressure and flow rate will decrease with time and eventually becomes zero when the system reaches a new equilibrium. Effective stress will increase at the end. The process of transient water flow linked to soil compression is called consolidation in soil mechanics. In this section, we will, however, only focus on the steady state water flow in soil, which is called steady state seepage or just seepage. Like the waterfall, water in soil flows from a point of high energy to a point of low energy. According to Bernoulli’s theorem, the total head at a point is defined as:

H=

v2 u + +z 2g γ w

(4.1)

where H is the total head, hv = v2/2g is the velocity head, hw = u/γw is the pressure head, and finally z is the elevation head. The velocity of seepage flow, v, is very small, and hence hv can be neglected from the calculation of the total head for seepage flow. To measure water pressure at a point under the ground, we can insert a small diameter tube (called standpipe or piezometer) into the ground; the bottom end of the tube is located at the depth where we would want to measure the PWP. Water is only allowed to enter the tube from its end; the value of PWP at the bottom end of the tube is equal to the pressure due to the weight of the rising water column or u = hw*γw. The pressure head, hw, defined in Eq. 4.1 is equal to the height of the water column that rises inside the standpipe. Elevation head, z, is the distance from an arbitrary horizontal datum to the point.

Figure 4.1: Total head loss in seepage flow Figure 4.1 shows a one dimensional water flow ing through a saturated soil sample placed in a small pipe, the flow is said in one dimension as we assume it only occurs in parallel to the pipe but 18

not in the direction normal to it. Water flows from point A to point B as HA > HB, the difference ∆H = HA - HB describes a loss in energy, which is due to friction between soil grains and water; ∆H is called total head loss. Note that the horizontal datum can be chosen arbitrarily because it always affects the values of zA and zB by the same amount; the value of ∆H is unchanged. The flow from A to B in Fig. 4.1 will continue as long as the values of HA and HB are maintained. If there is no water going into and out of the pipe, then water level on the A side will go down, water level on the B side will go up; seepage flow will stop once HA = HB or ∆H = 0. The flow rate and the outflow velocity of seepage flow in Fig. 4.1 can be described by Darcy’s law in the following form: q = Aki

(4.2)

where q is the volume of water ing through the cross-sectional area, A, of a soil sample in a unit of time, k is the coefficient of permeability of soil, i is the hydraulic gradient defined as:

i=

∆H L

(4.3)

∆H is the total head difference and L is the distance between two sections at A and B in Fig. 4.1. The outflow velocity can be calculated as:

v=

q ∆H = ki = k A L

(4.4)

Idealised diagram:

Since:

Qin = Qseepage = Qout vA = vsAvoids Avoids << A vs >> v

Figure 4.2: Comparison between seepage velocity and outflow velocity The outflow velocity in Eq. 4.4 corresponds to the whole cross-sectional area of the soil sample. We should, however, note that seepage flow inside the sample only es through the very small area of void space. Since the volume of water is the same, seepage velocity is much greater than the outflow velocity as shown in Fig. 4.2. The coefficient of permeability, k, of a soil can be determined in the laboratory (e.g. constant head test or falling head test) or can be determined at the field (e.g. well pumping test). For sands, k can be obtained based on empirical correlation, for example, Hazen showed that:

k = 10−2 D102 (m/s)

(4.5)

19

where D10 (mm) is the particle size at 10% finer read from the PSD curve of the sample. The typical value of k for different types of soil is: (m/s)

Figure 4.3: Coefficient of permeability for different soils (Craig, 2004)

Example 4.1: Effective stress affected by upward and downward seepage. Calculate the effective stress at the bottom of a sand column in the following cases, knowing that unit weight of dry sand, γdry = 16.9 kN/m3, and saturated unit weight γsat = 20.6 kN/m3. Answer: Effective stress at the bottom of the sand column in each case can be calculated as follows: Case (3): saturated soil – upward seepage Case (1): dry soil σv = 20.6*2 = 41.2 kPa σv = 16.9*2 = 33.8 kPa u=0 u = 10*3 = 30 kPa σ’v = 33.8 kPa σ’v = 41.2 – 30 = 11.2 kPa Case (2): saturated soil – hydrostatic condition σv = 20.6*2 = 41.2 kPa u = 10*2 = 20 kPa σ’v = 41.2 – 20 = 21.2 kPa

Case (4): saturated soil – downward seepage σv = 20.6*2 = 41.2 kPa u = 10*1 = 10 kPa σ’v = 41.2 – 10 = 31.2 kPa

It should be noted in the previous example that the effective stress in the case of downward seepage (4) is greater than the hydrostatic pressure (2). This is due to the effect of a seepage force acting in the direction of the flow. The seepage force that drives water flow through soil can be calculated as: 20

J = γ w × ∆H × A

(4.6)

If we divide J by the volume of the soil sample, we can obtain the value of seepage force per unit volume of soil, that is:

j=

J ∆Hγ w A = = iγ w V AL

(4.7)

The seepage force can be related to the pressure transferred from water to soil particles via frictional drag; the magnitude of seepage force depends on the hydraulic gradient, i, of the flow. As seepage force acts in the direction of flow, downward seepage will increase the value of effective stress and vice versa. As the result of this, there exists a critical condition for an upward flow into a hydrostatic soil-water system. When the flow is strong enough, the pressure due to seepage force can be equal to the effective stress of soil in the original hydrostatic condition but acts in opposite direction, and hence the effective stress is reduced to zero. The hydraulic gradient in this situation is called critical hydraulic gradient (ic). For the soil sample in Example 4.1, ic can be determined as: •

Original effective stress in hydrostatic condition: σ’v = γsat*2 - γw*2 = (γsat - γw)*2

•

Seepage pressure due to upward flow:

•

ps = σ’v →

ps = jV / A = icγ w L = icγ w * 2

icγ w * 2 = (γ sat − γ w ) * 2 → ic =

γ sat − γ w 20.6 − 10 = = 1.06 > the value of i in γw 10

case 4: i = (2-1)/2 = 0.5 The critical hydraulic gradient that could cause quick condition (σ’ = 0) in soil can be determined as:

ic =

γ sat − γ w γw

(4.8)



21

3.5.

Two dimensional seepage & flow net

We have discussed in the previous section that one dimensional flow is driven by a difference in of total head. One dimensional flow is most suitable to describe vertical flows at the field (i.e. upward and downward flows) where we only have the variation of the total head in the vertical direction. Total head on a horizontal line, on the other hand, is a constant, and hence there is no horizontal water flow (the same situation as in Example 4.1). If there is a variation of total head in both horizontal and vertical directions, water flow will occur in two dimensions. In this section, we will discuss the derivation of a governing equation for two dimensional flows. Furthermore, the construction of a graphical solution (called flow net) for two dimensional seepage flows will be presented. Using flow net, we can calculate the value of PWP (u) at any point under the ground. This has an important role as we need to be able to determine the value of effective stress (σ’=σ-u) in order to assess the stability of the earth structure, which has seepage flow going under or through it.

Figure 5.1: In-flow and out-flow seepage velocities of a soil element in two dimensions Soil is assumed fully saturated, homogeneous and has constant permeability in all directions, k = kx = kz. The volume of water going into the soil element in Fig. 5.1 can be calculated as: Qin = vinA = vxAx + vzAz = vx*∆z*1 + vz*∆x*1 Similarly, the out-flow volume of water is: Qout = voutA = (vx + δvx)Ax + (vz+ δvz)Az = (vx + δvx)*∆z*1 + (vz+ δvz)*∆x*1 Since Qin = Qout, we have: Qout - Qin = 0 or δvx*∆z + δvz*∆x = 0 ⇔

δv x ∆x

+

δv z ∆z

=0

For ∆x → 0 and ∆z → 0, we have:

∂v x ∂v z + =0 ∂x ∂z

(5.1)

22

From Darcy’s law, we have vx = kx(∂H/∂x) and vz = kz(∂H/∂z). Substituting these to Eq. 5.1, we have:

kx

∂2H ∂2H ∂2H ∂2H + = 0 → + = 0 as k x = k z = k k z ∂x 2 ∂z 2 ∂x 2 ∂z 2

(5.2)

Equation 5.2 is Laplace’s equation and is the governing equation for two dimensional steady seepage flows. The Laplace’s equation in Eq. 5.2 can be expressed in of two conjugated functions: the potential function (φ) and the stream function (ψ). •

The potential function φ(x,z) is defined such that:

∂φ ∂H ∂φ ∂H = vx = −k and = vz = −k ∂x ∂z ∂x ∂z

(5.3)

The incremental form of φ(x,z) is in the following form:

dφ =

∂φ ∂φ dx + dz = v x dx + v z dz ∂x ∂z

if dφ = 0, we have

dz v =− x dx vz

(5.4)

Integrating Eq. 5.3, we obtain:

φ ( x , z ) = − kH ( x , z ) + C

(5.5)

If the potential function φ(x,z) in Eq. 5.5 is given a constant value, e.g. φ1, then it represents a curve, every points on this curve will have the same value of total head, H1. The potential function defines a family of curves with each curve is associated with a constant total head (or equipotential); and hence it is called equipotential line. •

The stream function ψ(x,z) is defined such that:

∂ψ ∂H ∂ψ ∂H = vx = −k and − = vz = −k ∂z ∂x ∂x ∂z

(5.6)

The incremental form of ψ(x,z) is in the following form:

dψ =

∂ψ ∂ψ dx + dz = − v z dx + v x dz ∂x ∂z

(5.7)

If the stream function ψ(x,z) is given a constant value, e.g. ψ1, then it represents a curve. And because we have dψ = 0 for a constant value of ψ, the curve has the following characteristic:

23

− v z dx + vx dz = 0 →

dz v z = dx v x

(5.8)

Equation 5.8 means the tangent of the curve specifies the direction of seepage velocity vector at that point, and hence it is a flow line. The stream function defines a family of the travel paths (flow lines) that water moves along through soil in seepage flow. Furthermore, according to Eq. 5.4 and Eq. 5.8, the flow line and equipotential line will intersect each other at right angle. Finding solution for seepage flow can be done by searching for the exact form of the two potential and stream functions. The exact analytical solution, however, is usually not available for practical problems with complex soil and boundary conditions. For simple problems, an approximate solution can be found graphically by drawing the flow net. Flow net is a net constructed by a set of curvilinear equipotential lines intersected perpendicularly by a set of curvilinear flow lines. The area between any two adjacent equipotential lines and two flow lines forms a “curvilinear square”. The accuracy of the solution depends on how good these requirements can be met by the flow net. Drawing a flow net normally requires several trials until the best solution can be found.

Example 5.1: Draw a flow net for the estimation of seepage flow occurring under a dam as shown in the figure below. Answer: Start drawing the flow net by identifying from the boundary conditions of the problem any obvious flow lines and equipotential lines. In general, any boundary along impermeable materials can be flow lines as water cannot penetrate these surfaces. In the problem below, two flow lines (i.e. blue lines) can be identified as: FL1) running along the base of the dam and the two sides of the sheet pile wall, and FL2) running along the boundary between soil and the underlying impermeable material. Similarly, two equipotential lines (i.e. the red lines) can be identified: EPL1) the ground surface on the reservoir side of the dam, any point on this line will have the same total head (H = 18 + h) and EPL2) the ground surface on the downstream side of the dam, H = 18 m. After defining some initial flow lines and equipotential lines, the flow net can be further developed as follows: a) Draw a new flow line (FL3) next to the first FL1, the line should cross the EPL1 and EPL2 at right angles. b) The two flow lines FL1 and FL3 forms a flow channel. Divide the area of this first flow channel into several squares by adding a suitable number of new equipotential lines. c) Extend all new EPL beyond the FL3. When do this, you should keep in mind that you need to form new squares between FL3 and the next FL4 flow line, and hence the EPLs should oriented accordingly.

24

d) Draw the next flow line FL4. Space the FL4 so that squares are formed at each area between two adjacent EPLs and bounded by FL3 and FL4. If you can not have all the squares as you want, readjust the relative position of FL4 and the EPLs in previous steps to get better results. e) Continue the process and that the EPLs and the FLs should cross at right angles until you have the complete flow net with the all best squares as you can. Note that the flow net only gives you a rough estimation for seepage flow. It is, however, more useful to help you “feel” the problem.

h

EPL2

FL1

EPL1

Datum

FL2

z=0

FL1

EPL1

EPL2

EPL3 EPL4 5

FL3

10 6 7 8 9

FL2

FL1

FL1

FL3

FL3 FL4

FL4 FL5 FL2

FL2

25

Figure 5.2: Constant head drop & constant flow rate in a flow net Figure 5.2 shows a small part of a flow net comprising of three flow lines and three equipotential lines. Considering the flow along a flow channel, the same flow rate should be transferred through Box 1 and then Box 2, which means: QBOX1 = QBOX2 = Q1 where:

QBOX1 = k1 A1 (H1-H2)/L1 QBOX2 = k2 A2 (H2-H3)/L2

Now if only these equipotential lines and flow lines are drawn to make the box a SQUARE, i.e. A1=L1, A2=L2 and assuming k is a constant: Q1 = k(H1-H2) = k(H2-H3) Consequently, H1-H2 = H2-H3 → the head drop between any two adjacent equipotential lines (∆Hi) is the same. If the total number of equipotential lines is NEPL, the total number of head drops (∆Hi) occurring in a flow net is Nd = NEPL -1. The equipotential drop ∆Hi can be obtained as:

∆H i =

∆H ∆H = Nd N EPL − 1

(5.9)

If we know the value of the total head at the boundary EPLs (e.g., in Example 5.1, the total head of EPL1 and EPL2 can be determined from the boundary conditions) and the value of the total head loss ∆H, then based on the flow net, we can calculate ∆Hi and the total head Hi of each EPL. Since the total head of each equipotential line (Hi) can be determined, the pressure head at any point A on a EPL can be calculated as (hw)A = (Hi)A – zA. Here, zA is the elevation head or the vertical distance from the chosen datum to point A. The value of pore water pressure at point A is equal to: uA = γw * (hw)A

(5.10) 26

In a similar way, if we consider Box 1 and Box 3, then: QBOX1 = k1 A1 (H1-H2)/L1 QBOX3 = k3 A3 (H1-H2)/L3 Since A1=L1, A3=L3 and assuming k is a constant. QBOX1 = k(H1-H2) = k∆Hi QBOX3 = k(H1-H2) = k∆Hi So we have: QBOX1 = QBOX3; or Q1 = Q2. Consequently, the flow rate is the same for each flow channel in a flow net and can be determined as:

Qi = k∆H i = k

∆H Nd

(5.11)

Using the flow net, the total flow rate of seepage flow can be calculated as the summation of the flow rate of each flow channel:

Q = ∑ Qi = N f k

N ∆H N −1 = k∆H f = k∆H FL Nd Nd N EPL − 1

(5.12)

where Nf is the total number of flow channels in a flow net, Nf = NFL – 1.



27

3.6.

Uniform settlement of soil

We have discussed the dissipation of excess pore water pressure that would lead to an increase in effective stress. Furthermore, as water leaving the pore space, the sample’s volume reduces and this process is called consolidation. The flow rate of the transient flow and the compression rate of the soil sample will decrease with time. Here, we will not discuss the time scale of the consolidation process but rather focus our discussion on the initial (original, before loading) state and the “final” state of the sample under the new loading condition. The word “final” is used here to indicate the final stage of the consolidation process when the excess PWP has been completely dissipated. Volume reduction of soil would lead to settlement on the ground surface or settlement at the base of the foundation system. It is one of the major objectives in foundation design to keep the settlement in control. Excessive settlement will cause damages to the building or at least reduce its serviceability. Unequal settlement between different parts of the same building is of particular importance as it can cause tilting (think about the famous Pisa tower in Italy as an example!) or structural failure of the load bearing components (e.g. beam, column) of the superstructure.

Figure 6.1: Volume change due to an increase in effective stress Figure 6.1 shows a diagram of a fully saturated soil sample subjected to an increase in the effective vertical stress. As the solid particles and water are incompressible, the volume change is equal to the volume of water squeezed out of the void space. The volume change in relative to the initial volume (or the volumetric strain) can be calculated as:

εv =

V0 − V ∆V ∆v ∆e = = = V0 V0 v0 1 + e0

(6.1)

where εv is the volumetric strain; V0, e0 and ν0 are the initial volume, initial void ratio and initial specific volume (ν = 1 + e), respectively; V, e and ν are the corresponding parameters for the sample after the application of the stress increment. From Eq. 6.1, εv has positive value when V < V0 in the case of compressive behaviour; for the reverse case of V > V0, we have εv < 0 and the behaviour is dilative. Figure 6.2 compares a soil sample in the initial condition (V0) and in its deformed state (the shaded area, V) after an increment in the vertical stress. The two deformed samples in Fig. 6.2b and 6.2c have the same volume V, although lateral strain is involved in only one case (6.2b). Consolidation settlement of soil is normally calculated based on the assumption of no lateral strain (6.2c), which is true when the width of the loading area is significantly larger than the thickness of the soil layer. As the deformation only occurs in the vertical direction in Fig. 6.2c, it is called one 28

dimensional settlement. One dimensional settlement can also be reflected as the uniform settlement of horizontal soil strata during the sedimentation process, in which the weight of the overlying material tends to compress the underlying layer. a)

c)

b)

Figure 6.2: Compressive behaviour with and without lateral strain For the case of one dimensional settlement, Eq. 6.1 can be rewritten as follows:

εv =

∆V ∆H ∆e = = V0 H 0 1 + e0

(6.2)

where H and H0 are the current and the initial height of the sample; ∆H (=H0 – H) is commonly referred as the vertical settlement (or just settlement). From Eq. 6.2, the settlement of soil can be determined if we know the initial thickness (H0) of the layer (the “field-equivalent” to the sample’s height) and the volumetric strain (= change in void ratio/initial void ratio). The change in void ratio, on the other hand, depends on the stress increment and the properties of soil. The relationship between stress and strain for soil is very complex, and usually needs to be determined from experiments. Tests are conducted on undisturbed soil samples extracted from the ground, and on samples reconstituted from natural soil in the laboratory; soil parameters used in geotechnical design can then be obtained based on the measured stress-strain behaviour from the tests.

Figure 6.3: A typical experimental relationship between void ratio and effective vertical stress Figure 6.3 shows a typical experimental relationship of e-σ’v obtained for one dimensional compression. We can see that an increase in vertical stress (∆σ’ = σ’B - σ’A) would compress the sample and lead to a reduction in void ratio (∆e = eB - eA). The stress-strain behaviour of soil in Fig. 6.3 is highly non-linear. The slope of the tangent to the curve decreases at higher stress and at smaller void ratio, which means soil in denser state is less compressible. Using the compression curve, we can predict the one dimensional compressive behaviour of soil in engineering application,

29

of which the applied load and the in-situ stress condition should fall within the range of (σ’A,σ’B). Note that if the applied load could not fit in the stress range in the experiment, the prediction is unreliable; and you should be aware of this problem. Considering a small stress increment in Fig. 6.3, the compression curve can be approximated as a straight line, the change in void ratio can be calculated as:

∆e = av ∆σ 'v

(6.3)

where av is the slope of the curve. Equation 6.3 can also be written as:

εv =

∆e a = v ∆σ 'v = mv ∆σ 'v 1+ e 1+ e

(6.4)

Here mv is called the coefficient of volume compressibility and can be interpreted as the volumetric strain over an increased unit of stress. From Eq. 6.2 and 6.4, if we know the original thickness of the soil layer H, the vertical settlement of the soil layer can be calculated as:

∆H = mv ∆σ 'v H

(6.5)

The values of mv have been obtained for different soil types are as follows: Table 6.1: Typical values of mv (Smith & Smith, 1998)

Example 6.1: From the measured stress-strain curve in the figure below, estimate the vertical settlement of a 25 metres thick layer of the same soil due to an increase in vertical stress of 100 kPa knowing that the unit weight of soil is 20 kN/m3 and assuming there is no water involved.

σv (kN/m2) Answer: Consider the middle point of the layer, the original vertical stress is: 30

σ1 = γh = 20 * 12.5 = 250 (kPa) New vertical stress due to the 100 kPa stress increment:

σ2 = σ1 + ∆σ = 250 + 100 = 350 (kPa) According to the stress-strain curve, the values of void ratio corresponding to σ1 and σ2 are: e1 = 0.666 and e1 = 0.658; the coefficient of volume compressibility can be calculated as:

mv =

e1 − e2 1 0.666 − 0.658 1 × = × = 4.8 × 10 −5 (m 2 /kN) 1 + e1 ∆σ v 1 + 0.666 100

The vertical settlement of the 25 metres thick layer can be determined as:

∆H = mv ∆σ 'v H = 4.5 ×10−5 ×100× 25 = 0.12 (m) = 12 (cm)



31

3.7.

Oedometer test & measurement of soil stiffness



Figure 7.1: Cross section of a standard oedometer Oedometer test is a common laboratory test used to determine the stress-strain behaviour in one dimensional compression; Fig. 7.1 diagrammatically shows the cross section of a standard oedometer. A small and thin soil sample (generally 75 mm in diameter and 20 mm in thickness) is prepared inside a rigid steel ring, which can prevent lateral expansion of the sample during compression. Two porous stones are placed at the top and the bottom of the sample to serve as the free drainage boundaries for the dissipation of excess PWP. The vertical load is applied onto the loading plate on top of the sample. The whole sample is immersed in water inside an open cell to make sure the sample maintains fully saturated during the test. Loading is applied in stages; the range of the applied stress in the experiment should cover the expected stress range at the site. As the load is applied, excess PWP is generated inside the sample and then gradually dissipated when water is squeezed out through the two porous plates. This process normally takes time depending on the permeability of the tested soil. Each loading step is usually kept for 24 hours; the vertical movement is monitored by dial gauge. Reading is made more frequently at the beginning of the each loading step as the compressive rate is higher in this range (i.e. the compression curve is steeper). Reading is continued until there is no further change can be observed before the next loading stage can be applied.

Figure 7.2: Loading and unloading in oedometer test Figure 7.2 shows the typical stress-strain behaviour of soil from oedometer test; the curve is similar to Fig. 6.3 but the loading condition here is more complex. Soil sample is compressed from a relatively loose initial state (point A) until the curve reaches point B. The sample is now unloaded by removing the applied load and returning the effective stress to its original value. The unloading stress-strain curve follows a new branch to point C instead of going back to point A. The change in void ratio ∆e(B→C) is much smaller than ∆e(A→B), which means volume reduction is mostly 32

irreversible. For the sample at point C, the value of σv is not the maximum stress value that the sample has ever experienced. Instead, σvB is the maximum historical stress, and it is called preconsolidation stress or σp/c. For a stress condition on the BC branch, σv < σp/c, soil is referred as overconsolidated; and the BC branch is called the overconsolidation line. Soil on the A→B branch is normally consolidated as the current stress is also the maximum stress that soil has experienced. The reverse way from B→A is not possible in oedometer test, and point A will never be reached again. The ratio between σp/c/σv is referred as the overconsolidation ratio (OCR), for normally consolidated soil OCR = 1, and OCR > 1 for overconsolidated soil. If the sample at point C is now recompressed, the curve will go back to point B on the same BC branch and then it will move on the extension part of the AB curve as if the sample has been continuously loaded from A to D. Figure 7.2 shows that the sample is subjected to two more loadingunloading cycles until the test is terminated at point F. The ABDF part corresponds to the loading stages and is called the normal compression line (NCL), whereas BC, DE and FG are three overconsolidation lines representing the unloading and recompression stages. Comparing the behaviour of soil on the NCL with the behaviour on the overconsolidation line under a same stress increment, we can see that soil has much higher volume reduction on the NCL. This has an important engineering implication. If we could somehow preload the ground temporarily to a high enough stress and then unload it before the construction of the real structure, then the vertical settlement of the real structure later will be much less because a large part of the compression curve would occur on the overconsolidation line. In this way, we convert a normally consolidated soil to an overconsolidated soil; this technique is called preloading and is a method of ground improvement.

Figure 7.3: Compression curve in semi log plot It is a more common practice that the oedometer test result is plotted with the abscissa in logarithm scale as shown in Fig. 7.3. In this way, we could cover a much wider stress range. The loading stage, now, can be represented by a straight line. The unloading and reloading stages also follow another straight line. The slope of the overconsolidation line is smaller than the slope of the NCL, which describes a smaller compressibility for overconsolidated soil. The preconsolidation stress can be defined at the intersection of these two lines as shown in the figure. From Fig. 7.3, two parameters can be derived for the calculation of the vertical consolidation settlement; one is called the compression index, Cc, which is simply the slope of the NCL, the other parameter is the swelling index, Cs, or the slope of the overconsolidation line. Note that the overconlidation lines are in parallel (see Fig. 7.2) so only one value of Cs is obtained from the test results. Furthermore, unlike the parameter mv that varies according to the stress value, Cc and Cs are not affected by the applied 33

stress and hence it can be used in different engineering applications. The change in void ratio for a soil sample going from point A to point B can be obtained as follows:

σ  σ  ∆e = ∆eA + ∆eB = Cs log10  p / c  + Cc log10  B  σ   σA   p/c 

(7.1)

Here the total void ratio change ∆e is separated into two parts: ∆eA when soil is overconsolidated, and ∆eB when soil is normally consolidated as the applied stress exceeds the preconsolidation stress. The parameters Cs and Cc can then be used for the calculation of ∆eA and ∆eB, respectively. The consolidation settlement can be determined as:

εv =

 σ  σ  ∆H ∆e Cs C log10  p / c  + c log10  B  = = σ  H 1 + eA 1 + eA  σ A  1 + eA  p/c 

(7.2)

 σ  σ  Cs C log10  p / c  H + c log10  B  H σ  1 + eA 1 + eA  σA   p/c 

(7.3)

Hence,

∆H =

Note that if there is no information on the preconsolidation stress, we should assume soil is normally consolidated. This is a safe assumption, as settlement will be greater when soil is compressed along the NCL.

Example 7.1: A 6 metres thick clay layer with the water table at the ground surface is subjected to a stress increment of 150 kN/m2. If the preconsolidation stress is 100 kN/m2, estimate the overall settlement knowing that γ = 20 kN/m3, Cc = 0.3, Cs = 0.06, and e = 1.0.

Answer: Consider the middle point of the clay layer:

σ1 = γh = 20 * 3 = 60 (kPa) ; u1 = γwh = 10 * 3 = 30 (kPa) σ’1 = 60 – 30 = 30 (kPa) New vertical stress due to the 150 kPa stress increment: 34

σ’2 = σ’1 + ∆σ = 30 + 150 = 180 (kPa) Change in void ratio due to the stress increment:  σ   σ ' 100    180  ∆e = C R  log10 p / c  + CC  log10 B  = 0.06 log  + 0.3    σ ' σ 30    100   A  p/c   = 0.06 × 0.52 + 0.3 × 0.25 = 0.031 + 0.075 = 0.106

Vertical consolidation settlement of the clay layer can be determined as:

(

)

 = 6 0.106 ∆H = H  ∆e = 6 × 0.053 = 0.32 (m) = 32 (cm)  1+1  1 + eA 



35

3.8.

Shear strength criterion for soils & Mohr’s circle diagram

In this section, we will discuss the failure criteria for soils. In general, soil can fail in both tension and shear modes, although shear failure is the more common type of failure encountered in geotechnical problems. Shear failure occurs when different parts of a soil mass sliding relatively to each other. Sliding movements usually occur within a thin band of the soil body, which is called the failure plane (or shear band). Tensile failure, on the other hand, is often associated with fracture and opening cracks, which is more common for rocks. The shear strength of soil comes from the accumulation of frictional resistance at the points as soil is an assembly of particles. This means shear strength is a function of effective stress; a change in pore water pressure to some extents could trigger shear failure in soil. In addition, compact soil to denser state increases the number of points and the interparticle forces, and this could lead to higher shear strength. For clays, attractive surface forces could also contribute to the shear strength of soil.

Figure 8.1: Frictional resistance (Smith & Smith, 1998) Figure 8.1 shows a stress diagram of a block resting on a horizontal surface, which is about to move as it is being subjected to a horizontal force, H. In Fig. 8.1a, the block is in static condition and the weight is counterbalanced by the vertical reaction force. In Fig. 8.1b, a horizontal force H is applied to the block. If H is not too large, it can be offset by the friction force acting at the interface and the block stays still. The reaction force, R, is now inclined and make an obliquity angle α with the vertical direction. The frictional resistance is not limitless but is capped by a maximum force defined as: F = µ W = tan φ W

(8.1)

where µ (= tanφ) is the friction coefficient, and φ is called the friction angle. These two parameters are affected by the interfacial properties and can be determined by experiment. When the applied H exceeds the maximum frictional force in Eq. 8.1, the block starts sliding relatively to the surface and we have α = φ. Figure 8.1 can be used as an approximation for the stress condition at the points between soil particles. At the point, H and W can be interpreted as the shear and normal forces. Also, µ is a material property, which depends on particle shape, surface roughness, mineralogy, etc…. Sliding may occur at s, which encourages rearrangement of particles. When sliding is oriented along a preferred direction, a localized failure plane can be formed and much larger relative movement can be accommodated. The onset of a failure plane formation usually marks the overall failure of a soil sample or of a soil mass. In design, we need to

36

make sure the stress condition of the earth structure in its normal working condition is far from the stress condition that can trigger the formation of a shear band. The normal and shear components of stress acting on a plane ing through a particular point in a soil mass will depend on the orientation of that plane. From the experiment, people measured the orientation of the shear band and obtained a consistent angle between the failure plane and the horizontal direction, the angle can be considered as an intrinsic property of soil. Based on this observation, we could think that the combination of the stress components on the failure plane should probably be the most critical one among all other possibilities. Furthermore, if soil is a frictional material, the ratio of shear stress over normal stress on the failure plane should satisfy the limiting condition in Eq. 8.1, that is τ/σn = tanφ. a)

b)

Figure 8.2: Representation of stress condition at a point using Mohr’s circle (Smith & Smith, 1998) Figure 8.2b shows a small soil element in a soil mass. If the whole soil mass is in static equilibrium, so is a small element within it. Knowing the principal stresses σ1 and σ3 acting on the horizontal and vertical planes in Fig. 8.2b, the shear and normal stress on a plane making an angle θ with the horizontal direction can be calculated based on the condition of force equilibrium as:

τ=

σ1 − σ 3

sin 2θ 2 σ n = σ 3 + (σ 1 − σ 3 ) cos 2 θ

(8.2)

The collection of all stress points satisfying Eq. 8.2 forms a circle as shown in Fig. 8.2a. The circle is named Mohr’s circle after the person who originally developed the method, the radius and the centre of the circle can be given as:

S=

σ1 + σ 3 2

;R =

σ1 − σ 3 2

(8.3)

where S and R are the abscissa of the circle’s centre and the circle’s radius, respectively. The X-axis and the Y-axis represent the normal stress and shear stress components. A Mohr’s circle can be drawn if we know the stress components on two orthogonal planes, which form two symmetrical stress points collinear with the circle’s centre (e.g., point A and point B in Fig. 8.2a). The Mohr’s circle represents the locus of the stress condition on all possible planes ing through a common point in 37

a soil mass. The common point can be represented on a Mohr’s circle by the pole point. In Fig. 8.2a, if we draw a vertical line at point A to depict the plane that σ3 acting upon and a horizontal line at point B to depict the plane that σ1 acting upon, then these two lines crossing each other at point A. Point A, hence, is the pole point of the Mohr’s circle in Fig. 8.2a. Using Mohr’s circle to find the stress components on the inclined plane in Fig. 8.2b, we can draw a plane ing through the pole point A and making an angle θ with the horizontal direction. The plane intersects the circle at point D; the stress components can then be obtained as the coordinates at the point and actually give the same results as in Eq. 8.2. For the stress condition at point D, connecting the point with the origin, we have:

tan α =

DE τ = OE σ n

(8.4)

From Fig. 8.1, we know that if α = φ, sliding can occur. In other words, using Mohr’s circle, we need to search for the failure plane as a plane with maximum obliquity angle α. a)

b)

Figure 8.3: Failure envelope and orientation of the failure plane (Smith & Smith, 1998) In Fig. 8.3a, we can see that the angle of obliquity α is maximum when OD becomes the tangent to the circle. If the stress circle corresponds to the stress condition at the failure state of soil, the line OD can be referred as the failure envelope and the angle DOA is referred as the friction angle of the material, φ. For a soil sample with vertical stress σ1 greater than horizontal stress σ3, the failure plane is formed at an angle of (45° + φ/2) to the horizontal plane as shown in Fig. 8.3b. Note that the shear stress on the failure plane is smaller than the maximum shear stress, which means soil fails in of maximum stress obliquity but not in of maximum shear stress.

c Figure 8.4: Determination of failure envelope from experimental data (Smith & Smith, 1998)

38

The failure envelope of soil is usually determined from experiment. Samples are prepared and sheared at different confining pressures (i.e. σ3 as shown in Fig. 8.3b). The effect of confining pressure (σ3) is similar to the effect of the normal force W in Fig. 8.1b. Larger H is required to move the block at larger W; the friction angle, however, does not change as long as the interfacial properties remain the same. The stress circles at the failure condition for samples with different σ3 are shown in Fig. 8.4. The failure envelope can be determined as the best common tangent drawn to all three stress circles, of which the equation can be given as:

τ = c + σn tanφ

(8.5)

Equation 8.5 is called Mohr-Coulomb failure criterion for soil. Here, c is the cohesion, which describes the shear strength of soil at zero normal stress; we usually have c > 0 for clays. Note that shear strength of soil is a function of the effective stress but not of the total stress, hence, Eq. 8.5 needs to be rewritten in of effective stress as:

τ = c'+σ 'n tanφ '

(8.6)

Here, c’ and φ’ are the effective cohesion and effective friction angle, and they are usually referred as the shear strength parameters of soil. Note that shear stress, τ, is the same in of both effective stress and total stress. For example, considering the shear stress acting on a 45° plane (i.e. at the apex of the circle), we have τ’ = (σ’1 - σ’3)/2 = (σ1 - u - σ3 + u)/2 = (σ1 - σ3)/2 = τ.

Figure 8.5: Failure envelope in of effective stress and total stress (Aysen, 2002) Figure 8.5 shows the total stress Mohr circle and the effective stress Mohr circle for a soil sample at failure condition. The two circles are separated by the value of PWP, u. Note that the effective stress circle is on the left, which means u is positive as σ’ = σ - u. In addition, the two circles have the same radius because R’ = (σ’1 - σ’3)/2 = (σ1 - u - σ3 + u)/2 = (σ1 - σ3)/2 = R. Different tangential lines can be drawn to each circle, and hence we have two sets of shear strength parameters (c,φ) and (c’,φ’). Geotechnical engineers should always use the effective strength parameters (c’,φ’) in stability analysis of earth structures. The total strength envelope is only applicable when there is no measurement of PWP, and also for the case of fully saturated clays in undrained loading condition (further details will be given in Section 10).



39

3.9.

Measurement of soil strength in laboratory: Direct shear test

Shearing plane

Figure 9.1: Diagram of a direct shear box (Craig, 2004) Direct shear test is a very common standard test in soil mechanics, in which the sample is sheared along a predetermined horizontal plane. As the shear plane is predetermined, the test is most suitable to measure the stress-strain behaviour at the interface between soil and other materials (e.g., concrete foundation, steel sheet pile wall, etc.) or when there is a weak layer of soil exists inside a stronger soil mass so the shearing movement leading to failure would likely to occur in this weak layer. In the more general cases that the failure plane would develop along the plane of maximum stress obliquity, then direct shear test is not a realistic presentation of the situation and we need to refer to a more versatile test (e.g., triaxial test). Figure 9.1 shows a diagram of a direct shear box, a thin soil sample (about 60x60x20 mm3) was prepared inside a split shear box. For fully saturated samples, the shear box is put inside an outer box filled with water. Two porous plates are placed at the top and bottom of samples to provide drainage boundaries to allow for water to drain out of the sample. For dry samples, porous plates can be replaced by grooved thin metal plates, which can help to minimize slippage at the interface. Normal force in of dead weight is applied to the loading plate on top of the sample. The upper half and the low half of the split box can be slightly separated to eliminate friction between the upper and the lower boxes, and then shear displacement is applied so the two halves is forced to move in opposite directions. During the shearing process, vertical and horizontal displacements as well as shear force are usually monitored by dial gauges and load transducer. Based on the measurement, the following parameters could be determined from the test: •

Normal stress: σ = N/A

•

Shear stress: τ = T/A

•

Shear strain: γ = ∆l/h0

•

Vertical strain: εver = ∆h/h0

•

Volumetric strain: εvol = εver = ∆h/h0

•

A: area of the shear box (& soil sample)

•

h0: initial thickness of soil sample

40

Samples of different densities can be prepared and tested at different confining pressures (by changing the value of N applied on top of the sample). Furthermore, different loading conditions can be considered. For sands, because of the high permeability, sands are normally sheared in drained or dry conditions, the shearing rate is about 1 mm/min. For clays, because of very low permeability, clays can be tested in either undrained or drained conditions. For undrained condition, the sample is sheared quickly (shearing rate is about 1 mm/min), so that there is not enough time for water to drain out and the excess PWP can not be dissipated. For drained condition, a much slower shearing rate is required such that no excess PWP is generated during shearing, and it could take a few days to complete the test. The direct shear test results on a few typical soils in different loading conditions are as follows: Direct shear test results on sand: a)

b)

Figure 9.2: Direct shear test results on sand (Das, 2007) Figure 9.2 plots the results of direct shear test on sand sample of different densities (loose, medium dense and dense). We can see that the stress-strain curve of dense sample in Fig. 9.2a (small void ratio) comprises of two parts, an initial part where stress increases with horizontal displacement until a maximum stress value is obtained. This type of behaviour is called strain hardening, and the maximum stress is called peak stress (or peak strength). After the peak stress, stress decreases until the curve reaches a horizontal plateau at large displacement; this type of behaviour is called strain softening. As the samples become looser (dense→medium→loose), the softening behaviour becomes less prominent and eventually disappears in loose sample. Furthermore, the effect of initial densities is suppressed at large displacement as all samples approach the same stress value. When this occurs, the samples are at the ultimate condition, and the ultimate stress value is called residual strength. Shear displacement can be increased further without any change in the ultimate stress values since the two upper and lower halves behave like two rigid blocks sliding relatively to each other. Figure 9.2b plots the corresponding volumetric behaviour to the stress-strain curves in Fig. 9.2a. Dense and medium dense samples show some little initial compression (volume reduction) and then the samples dilate (volume expansion) until the end of the test. Loose sample, on the other hand, does not dilate but compress all the way to the ultimate state.

41

a)

b)

Figure 9.3: Maximum dilatancy and ultimate (or critical) void ratio (Craig, 2004) We can see in Fig. 9.2a that the strength of the dense sample is higher than of the loose sample (peak stress > ultimate stress), and this is due to the effect of interlocked sand particles in dense sample. Since there is not enough void space to allow for free rearrangement of the particles in dense sample, the particles need to climb over each other and this leads to volume expansion or dilatancy behaviour of dense sand. The extra energy required in this process is reflected in the increased shear stress of the stress-strain curve. The peak stress is obtained at the maximum rate of dilatancy (dεv/dγ) as shown in Fig. 9.3a. Volume expansion does not happen for loose sample as there is more than enough void space inside the sample to allow for free rearrangement of particles. As shearing movement is applied, the granular structure of loose sample can even collapse leading to volume reduction. Both dense and loose samples will approach the same void ratio at the ultimate state (Fig. 9.3b), and the samples can be continually sheared at constant volume (ecv), and constant shear stress (τcv). a)

b)

Figure 9.4: Effect of confining pressure on the stress-strain behaviour of sand (Craig, 2004) The test results in Fig. 9.2 and 9.3 are for samples of different initial densities but subjected to the same confining pressure. Figure 9.4, on the other hand, shows the test results on replica samples at different confining pressures. Increase confining pressure would suppress dilatancy behaviour of the sample. Under large enough confining stress, dense sample could have only strain-hardening behaviour; whereas under small enough confining pressure, loose sample can dilate. Note that very high confining stress could cause particle breakage. As particles break, new degrees of freedom are created for particle rearrangement, and volume expansion is reduced. The overall behaviour is very complex as all different mechanisms could happen at the same time, and particle breakage is still an on-going research topic in soil mechanics. The stress values at the peak and at the ultimate state can be plotted in the (τ,σ’) space, the Mohr-Coulomb failure envelopes can be obtained by adding the best fit lines to the data points as shown in Fig. 9.5. The intercept and the slope of the line give you the value of cohesion and friction angle. Note that for clean sand, cohesion is usually zero so the line es through the origin. The peak strength should correspond to higher friction angle (φ) as compared to the friction angle obtained at the ultimate state (φcv). 42

Figure 9.5: Mohr-Coulomb failure envelopes from direct shear test (Das, 2008) Direct shear test results on saturated clays in undrained condition: Figure 9.6 shows the test results for saturated clay in undrained loading condition. There is no volume change because water can not drain out of the sample. Instead, there would have change in pore water pressure. However, since PWP is not measured in standard direct shear test, the only test result is the stress-strain curve in of total stress. Figure 9.6 plots the stress curve for undisturbed sample and remoulded sample. Undisturbed sample is the soil sample collected from the site and carefully preserved in its original condition. Remoulded sample is usually created from the soil of undisturbed samples after the test. Natural soil structure and interparticle bonding are completely destroyed in remoulded samples. Remoulded sample behaves as strain hardening with a small ultimate strength of about 5 kPa. The peak strength of undisturbed soil is much higher (cu = 40 kPa). The residual strength of undisturbed sample (cult = 25 kPa) is also above the strength of remoulded sample. Undrained test is very common for clays as it reasonably represents the condition of soil during the construction period. The consolidation process of clays occurs over much longer time beyond the construction period, so drained test is only suitable for long term stability analysis.

Figure 9.6: Direct shear test results on saturated clay in undrained condition (Bardet, 1997) Direct shear test results on saturated clays in drained condition: Figure 9.7a plots the stress curve of an overconsolidated clay sample under drained test. Note that the test may take a few days to complete due to the low permeability of clay and to ensure there is no excess PWP is generated during shearing. The curve resembles the result of dense sand in 43

Fig. 9. 2a, which comprises of two parts: an initial hardening curve and then followed by a strain softening behaviour. If density plays an important role in granular soils, then the same level of importance can be assigned to the role of overconsolidation ratio in clays. Overconsolidated clays (OCR > 1) behave similarly to dense sand with the existence of a peak stress and the dilatancy behaviour, whereas normally consolidated clays (OCR = 1) behave like loose sand with compressive volume change during shear. Furthermore, Fig. 9.7b shows that the Mohr-Coulomb envelope of OC clay has a positive cohesion (c > 0). For NC clay, the envelope es through the origin and so c = 0.

b)

a)

Figure 9.7: Direct shear test results on saturated clay in drained condition (Das, 2006)



44

3.10. Measurement of soil strength in laboratory: Triaxial test Figure 10.1 describes the stress condition at a point under the ground, which requires knowing all the stress components on three orthogonal planes. Since soil deposition occurs vertically, the vertical axis is usually the symmetrical axis for soil, so we could have: σx = σy, where σx and σy are the normal stresses acting on the horizontal plane. Any new load applied on the ground surface (e.g. the point load P in Fig. 10.1) would lead to changes in the stress components at the point. The stress increments will depend on the characteristics (area, distribution, etc.) of the applied loads, soil properties and the location of the point in relative to the loading area. It is obvious that the stress condition as shown in Fig. 10.1 is not the stress condition in direct shear test. Stress condition at a point under the ground is better represented in triaxial test.

Figure 10.1: Stress condition at a point under the ground surface (Das, 2006) The cross section of a triaxial test apparatus is shown in Fig. 10.2a. A cylindrical soil sample (typical size of 38 mm diameter and 76 mm height) was set upon a fixed pedestal. The sample is covered by impermeable rubber membrane, which is sealed at the top and the bottom by O-rings. Water, hence, can only be drained out of the sample through the porous plates at its two end faces. The whole system of sample + loading cap + pedestal was put inside a transparent chamber. The chamber is fully filled with water, and the water (outside the membrane) can be pressurized to provide confining effect around the sample (referred as cell pressure). Inside the membrane, water can flow in to saturate the sample through a small conduit inside the pedestal (referred as pore pressure). There is a drainage valve installed on this water age to block or to allow water going in and out of the sample. Vertical load can be applied through a piston attached to the loading cap on top of the sample. The following parameters can be monitored continuously during a shear test: cell pressure (σ2=σ3 or σr), vertical displacement (→strain, ε1 or εa), vertical load (→stress, σ1 or σa), and either the volume of water drained out of the sample (εv, if the valve is left open) or the value of pore water pressure (u, if the valve is closed). Figure 10.2b describes the stress condition of a sample in triaxial test, which includes an equal all-around pressure coming from the pressurized water inside the chamber, and an axial stress together with its reaction as the result of the vertical applied load. The effect of the cell pressure is similar to the confining effect of the surrounding soil to a soil element inside the ground. The vertical stress can be interpreted as the extra load from the construction activities.

45

a)

b)

Figure 10.2: Diagram of triaxial test apparatus and the simulating stress condition (Craig, 2004) There are three common types of shear test with the triaxial apparatus: •

Consolidated – Drained test (CD test): once the sample is set upon the pedestal and the chamber is filled with water, the cell pressure is increased gradually in stages until it reaches a predetermined value (σc) that is usually chosen corresponding to the stress condition at the site. The drainage valve is then opened to allow for water to drain out of the sample, the volume of which is monitored and is equal to the reduction volume of the soil sample. When the consolidation phase is complete, vertical stress is applied slowly (σ1 is increased) while the drainage valve is kept open to let water going out. The sample is sheared at constant confining pressure (σ3 = σc is kept constant) under the effect of deviatoric stress (σd = σ1 - σ3) and no excess PWP is generated during the shearing process.

•

Consolidated – Undrained test (CU test): the consolidation phase is similar to the CD test, the shearing phase, however, is carried out under undrained condition. This means the drainage valve is closed and water can not drain out. The sample’s volume does not change, and excess PWP is built up inside the sample. The value of PWP is measured during the shearing phase.

•

Unconsolidated – Undrained test (UU test): as soon as the confining pressure is applied, the sample is sheared. The drainage is closed and no volume change occurs during the experiment. The test is done very quickly, and there is not enough time for excess PWP to be distributed evenly within the sample, PWP is generally not measured in the test.

Figure 10.3: Changes in the dimensions of soil sample during triaxial test (Bardet, 1997) 46

The values of stress and strain during the triaxial test can be determined as: •

During the consolidation phase:

Vc = V0 − ∆Vc ; ε v =

∆Vc V0

 1 ∆Vc   H c = H 0 1 −  3 V0   1 − ∆Vc / V0  V  Ac = c = A0  Hc  1 − ∆Vc / 3V0  1 (σ 1 − σ 3 ) = 0 as σ 1 = σ 3 = σ c 2 1 s = (σ 1 + σ 3 ) = σ c 2 u = 0 at the end of the consolidation phase t=

•

During the shearing phase:

1 − εv  V  = Ac  1 H − ε 1   V = Vc − ∆V ; H = H c − ∆H A=

ε1 =

∆H ∆V ; εv = = ε1 + 2ε 3 Hc Vc

1 [(σ 1 − u ) − (σ 3 − u )] = 1 (σ 1 − σ 3 ) = t = P 2 2 A 1 s = (σ 1 + σ 3 ) = t + σ 3 2 1 s ' = [((σ 1 − u ) + (σ 3 − u ))] = s − u 2

t'=

Figure 10.4: Typical results of CD and CU tests on clays (Craig, 2004) Figure 10.4c and 10.4d describe the behaviour of NC clay and OC clay under drained condition. We can see that the behaviour of NC clay is purely strain-hardening with compressive volumetric strain, whereas for OC clay it has significant dilatancy with strain softening occurs after the peak stress. Compare with the undrained behaviour in Figs. 10.4a,b, compressive volumetric strain (in Figs. 10.4c,d) would tend to increase PWP (u>0, Figs. 10.4a,b); volume expansion, on the other hand, 47

decreases PWP. Increasing the OCR value, then OC clay behaves in a similar way to denser sand; volume expansion could even lead to negative PWP. If the drainage valve is opened at the end of the undrained test on OC clay, the negative PWP can suck water into the sample. The peak stress and the ultimate stress values can be used to determine the strength envelope as similar to Figs. 9.5 and 9.7. We will, however, focus more on the strength envelope obtained in UU test.

Figure 10.5: Strength envelope from UU test on saturated clay (Das, 2006) We have known that consolidation could make soil a stronger material as the sample is in denser state. In UU test, consolidation does not take place as drainage is not allowed regardless of how large the confining pressure is. Consequently, it would be reasonable to expect the strength of a soil sample in UU test does not depend on the value of the confining pressure (σc). In Fig. 10.5, the stress circle P represents the total stress condition at failure of soil sample 1, of which the corresponding effective stress circle can be given as circle Q. The two circles P and Q are distanced from each other by the value of excess PWP (∆ud). Note that since there is no measurement of PWP in UU test, circle Q shown here is only for the demonstration purpose but can be impractical in real test. Now if we consider another sample 2 created by subjecting sample 1 to a higher confining pressure σc2, then its total stress condition at failure can be described by circle R. The two circles P and R are separated from each other by the confining pressure increment ∆σ3 =∆σc = σc2 - σc1, which should be equal to an increase in of PWP (∆uc = ∆σc = ∆σ3) because in undrained condition, any change in confining pressure will be countered by an equivalent change of pore water pressure. The effective stress circle at failure of sample 2 is still circle Q as: σ’3R = σ3R – uR = σ3R – (us + ∆uR) = σ3R – (0 + ∆ud + ∆uc) = σ3P – ∆ud = σ’3P, where us is the background static PWP that is negligible in normal triaxial test procedure. Note that the circle P and R have the same radius, which means there is no strength improvement due to the increase in confining pressure (∆σc). Drawing a tangent to the two Mohr circles P and R, we can obtain a horizontal strength envelope in of total stress, which means the friction angle in UU test is zero (φu = 0). The effective strength envelope shown in Fig. 10.5 is not generally obtained from the test results of UU test but can be an overlaying line borrowed from the test results of CD or CU test on the same sample. The discontinuous line in Fig. 10.5 represents the total strength envelope for only sample 1, which is not trustworthy as we know that the true total strength envelope should have φu = 0. The interception of the horizontal total strength envelope with the vertical axis is referred as the undrained shear strength of saturated clays, cu.

48

b)

a)

Figure 10.6: Pore water pressure coefficient, B, under isotropic stress change (Craig, 2004) To calculate effective stress, we need to know the value of PWP. However, we don’t always have the facility for the measurement of PWP at the field and it would be useful if there is an empirical relationship to describe the response of PWP to a change in the total stress values in undrained condition. Based on the work by Skempton (1954), the excess PWP can be estimated as:

∆u = B[∆σ 3 + A(∆σ1 − ∆σ 3 )]

(10.1)

where A and B are called pore water pressure coefficients. Parameter A relates the PWP response to a change in deviatoric stress (∆σd = ∆σ1 - ∆σ3), whereas parameter B relates the PWP response to a change in isotropic stress state (∆σc). Figure 10.6a shows a soil specimen subjected to an equal allaround pressure (i.e. σ1 = σ2 = σ3 = σc), the isotropic stress state is then increased by the same amount ∆σ3 = ∆σc in all directions. In triaxial test, this can be done during the consolidation phase when we close the drainage valve and increase the cell pressure. After leaving some times to ensure PWP has been evenly distributed within the sample, we can check the new value of PWP, if the initial PWP is zero, then this reading is equal to the excess PWP, ∆u. Parameter B can then be determined as:

B = ∆u / ∆σ 3

(10.2)

Figure 10.6b describes the variation of parameter B as a function of soil saturation ratio (Sr). For fully saturated sample, B = 1, which means the increase of total stress is completely balanced by an equivalent increase of PWP. However, for partially saturated soil, B < 1 and this is because the air contained inside the void space is more compressible than water.

a)

b)

Figure 10.7: Pore water pressure coefficient, A, under deviatoric stress change (Craig, 2004) 49

Figure 10.7a shows a soil sample in the shearing phase, which is being subjected to an increase in vertical stress ∆σ1, the horizontal total stresses remain the same or ∆σ2 = ∆σ3 = 0. In undrained condition, pore water pressure should change accordingly, and the excess PWP can be estimated as:

∆u = A × B × ∆σ 1

(10.3)

If the sample is fully saturated, B = 1 and ∆u = A*∆σ1. Figure 10.7b plots the variation of parameter A obtained at the failure condition (Af) for fully saturated clays with different OCR values. Highly overconsolidated clays (OCR > 4) have Af < 0, which means the increase in vertical stress could generate negative PWP. This can be explained by the dilatancy of OC clays as shown in Fig. 10.4b and d. The total excess PWP generated in both consolidation and shearing phases of a triaxial test can be estimated as the sum of the excess PWP defined in Eqs. 10.2 and 10.3, which is formulated in Eq. 10.1. The values of parameters A and B for a soil can be determined from laboratory test; Eq. 10.1 can then be used to estimate (predict) the excess PWP resulted from the construction activities at the site.



50

3.11. Lateral earth pressure Retaining wall is one of the most common structures in geotechnical engineering. The main function of the structure is to retain the soil mass behind its back. The displacement of the wall needs to be strictly controlled to ensure the stability and serviceability of the infrastructures and/or buildings constructed behind it. To analyse the stability of the wall itself, we need to know the value of the lateral pressure acting on the wall due to the weight of the retained soil mass. We have, so far, only discussed about the vertical stress at a point under the ground but not really mention about the corresponding lateral pressure. And that is because the lateral stress acting on a soil element within a horizontal soil strata is always balanced by the lateral stress acting on the elements next to it and the forces cancel each other out. Consequently, there is no lateral strain within the ground was involved during the sedimentation process (the process in which underlying soil layer is compressed vertically by the weight of soils accumulated right above it). The lateral pressure corresponding to the no lateral strain condition is referred as at-rest lateral pressure, which is normally defined relative to the vertical stress as:

σ 'h = K0 ×σ 'v

(11.1)

where K0 is referred as coefficient of at-rest earth pressure. For the cases of non zero lateral strain, depending on the direction of the strain (tensile or compressive) that the lateral stress will decrease or increase as compared to the at-rest value in Eq. 11.1.

Figure 11.1: Active and ive cases of lateral earth pressure behind a retaining wall (Craig, 2004) Figure 11.1 shows a vertical wall retaining a soil mass behind a back. If the wall is in static condition, the soil mass is at rest. The vertical stress a depth z below the ground surface can be calculated by Eq. 3.1; the horizontal stress can be calculated by using Eq. 11.1 (we will discuss the value of K0 later). Now, if there is a movement of the wall, the lateral earth pressure will change accordingly. In the case that the wall moves away from soil, the soil mass will expand laterally and the lateral stress will decrease; the vertical stress meanwhile remains the same. The reduction of lateral stress will continue (and the corresponding stress circle will grow in size) until a point when the Mohr circle touches the failure envelope, and the shearing plane can be formed along the plane of maximum stress obliquity. In this case, the lateral pressure at the failure condition of the soil mass is referred as active earth pressure.

51

a)

b)

c)

Figure 11.2: Mohr circle, failure planes & resultant thrust of active pressure (Smith & Smith, 1998) Figure 11.2a plots the Mohr circle at failure in active condition. The discontinuous circle is an example of the initial at-rest stress condition; as lateral stress decreases, the circle increases in size until it touches the failure envelope. Vertical and horizontal stresses are the major and minor principal stresses, respectively. Point A is the pole point as it is at the intersection of two principal stress planes. Based on the stress circle, there are two sets of the failure planes can be identified, which make an angle of (45° - φ/2) with the vertical direction as shown in Fig. 11.2b. Similar to Eq. 11.1, the active pressure can be defined as:

σ 'h = Ka ×σ 'v

(11.2)

where Ka is the coefficient of active earth pressure. Using Fig. 11.2a, Ka can be determined as:

Ka =

σ '3 OA OC − AC OC − DC 1 − DC OC 1 − sin φ ' φ'   = = = = = = tan 2  45° −  σ '1 OB OC + CB OC + DC 1 + DC OC 1 + sin φ ' 2 

(11.3)

As Ka is a constant, the horizontal stress increases linearly with depth z in similar way to the vertical stress. The distribution of lateral stress on the wall is a triangle diagram as shown in Fig. 11.2c. The total horizontal force can be obtained by taking integration of σ’h with depth or (because of the linear variation) simply as the area of the triangle stress diagram, which means:

Pa =

1 γh 2 × h × K a γh = K a 2 2

(11.4)

The horizontal active thrust es through the centroid of the stress diagram, which is two thirds of the total height below the ground surface as shown in Fig. 11.2c. The opposite case of active pressure is ive pressure, in which the wall in Fig. 11.1 moving against the retained soil mass. When this happens, the soil mass is compressed laterally and hence, lateral stress is increased and becomes greater than vertical stress. The lateral compression will continue until the Mohr circle touching the failure envelope. However, the horizontal stress, here, is the major principal stress, and the vertical stress is the minor principal stress.

52

a)

b) Failure planes in the case of active pressure

c) Failure planes in the case of ive pressure

Figure 11.3: Mohr circle, failure planes of ive pressure (Smith & Smith, 1998) Figure 11.3a plots the Mohr circle for ive pressure, which is actually Fig. 11.2a rotated by 90 degree as now the horizontal stress becomes the major principal stress and vertical plane is the major principal stress plane. The failure planes of ive pressure now form an angle of (45° - φ/2) with the horizontal direction instead of the vertical direction as for the case of active pressure. Figure 11.3a and 11.3b compare the failure planes of the two opposite cases of active and ive pressures, which describes clearly the 90 degree rotation. Using Fig. 11.3a, the coefficient of ive pressure can be determined as:

Kp =

σ '1 OA OC + DC 1 + DC OC 1 + sin φ ' φ'   = = = = = tan 2  45° +  σ '3 OB OC − DC 1 − DC OC 1 − sin φ ' 2 

(11.5)

The ive pressure can be defined as:

σ 'h = K p × σ 'v

(11.6)

And the ive thrust is (the shape of the stress diagram and the point of application are similar to Fig. 11.2c):

Pp =

a)

1 γh 2 × h × K pγh = K p 2 2

(11.7)

b)

Figure 11.4: Development of active and ive pressures with lateral strain

53

Figure 11.4a shows a very common situation of a retaining wall, in which the wall is embedded in the soil with a shallow soil layer placed in front of its toe at the opposite side with the retained mass. If the wall in Fig. 11.4a moves to the right, then soil in front of the wall is compressed and the lateral stress would increase until it reaches the ive condition. Soil behind the wall’s back, on the other hand, expands laterally and would eventually satisfy the active condition. As the ive pressure acts in the opposite direction with the movement of the wall, it could help to resist the movement and improve the wall’s stability. Figure 11.4b describes the development of the active and ive pressures with the magnitude of lateral strain. We can see that the active pressure is obtained at smaller absolute strain as compared to the ive condition, which means soil behind the wall’s back will fail earlier than soil at the front of the wall. Furthermore, the actual ive failure plane is a curve instead of a straight line as it was assumed in the theoretical derivation (Eq. 11.5).

Figure 11.5: Lateral pressure due to surcharge (Craig, 2004) It is very often that there is construction of infrastructure and/or building behind the retaining wall. The load (is called surcharge) coming from these structures will contribute to the lateral force acting on wall. If the surcharge pressure is distributed uniformly as shown in Fig. 11.5, the correspond active (or ive) pressure can be calculated as σ’h = Ka*q (or σ’h = Kp*q). The lateral thrust es through the centroid of the rectangle stress diagram, which is at the middle point of the wall (H/2). The lateral stress due to surcharge is calculated separately and added to the lateral earth pressure due to the weight of the soil mass. We have discussed the equations used to calculate active (Eq. 11.3) and ive (Eq. 11.5) lateral pressures. Finally, we need to determine the value of lateral pressure at rest. The coefficient of atrest pressure can be determined using the following empirical relationships: •

NC soils:

K0 = 1 − sinφ'

(11.8)

•

OC soils:

K0 = (1 − sinφ ' )OCRsinφ '

(11.9)

where OCR is the value of the overconsolidation ratio. For example, Fig. 11.6 describes a typical relationship between K0 and the OCR value of a clayey soil. We can see that for OCR greater than about 3, K0 > 1 that means the vertical stress is smaller than the horizontal stress. On the other hand, K0 < 1 for NC soil, which means σ’h < σ’v.

54

Figure 11.6: Coefficient of at-rest lateral earth pressure (Craig, 2004)



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3.12. Stability of retaining walls The derivation of the coefficient of lateral pressures in Eq. 11.3 and 11.5 are based on the Rankine’s theory of earth pressure. In the Rankine’s method, it is assumed the principal stress directions are the same everywhere as well as the failure condition is attained at every point within the soil mass. For problems with more complex geometry, the principal stress directions would change at each location. Furthermore, there is often only one major failure surface, the soil masses at two sides of the surface tend to slide relatively against each other like rigid blocks, of which the soil inside the block is much less deformed as compared to soil within the shear band.

a)

b)

Figure 12.1: Coulomb’s wedge and earth pressure acting on an inclined wall (Smith & Smith, 1998) Figure 12.1 shows a retaining wall with the back inclined at an angle ψ with the horizontal direction. The ground surface is also inclined at an angle β as shown in Fig. 12.1b. In addition, there is shear stress acting at the interface between the wall and soil due to friction, which results in the total thrust P making an angle δ with the normal to the back of the wall. In this case, the vertical and horizontal directions are no longer the principal stress planes. The total thrust can be determined using Coulomb’s wedge theory by considering the force equilibrium of the soil wedge between the wall and the failure plane. In general, we don’t know the exact location of the failure plane, and need to search for the direction that yields the maximum thrust on the wall (the worst scenario for the stability of the wall). In Fig. 12.1a, the line BC represents the plane inclined at the friction angle, φ, of soil. Soil below this plane is considered stable as the stress obliquity on any plane below the plane BC is smaller than the maximum obliquity. Consequently, we should search for the failure plane above the plane BC. For example, the forces acting on the wedge ABD in Fig. 12.1a are as follows: •

The self-weight of soil (W): is equal to the volume (area in two dimensions) of the soil wedge multiplied by the unit weight of soil. W always acts vertically.

•

The active reaction of the wall on the soil wedge (Pa): inclines at angle δ to the normal of the back of the wall; the angle δ is the friction angle of the interface between the wall and the soil mass. Pa points upward as the soil wedge tends to slide downward relatively to the wall in the case of active pressure.

•

The reaction R acting on the failure surface: inclines at angle φ below the normal of the failure plane as the wedge tends to slide downward. The angle φ is the friction angle of the soil mass. 56

Figure 12.2: Force diagram for the wedge ABD in Fig. 12.1a (Smith & Smith, 1998) Knowing the inclination angles of the wall, the failure plane and of the ground surface, the direction of all force vectors can be determined and they should form a closed force diagram as shown in Fig. 12.2. As we know the values of the force diagram’s angles and the value of the selfweight force W, the magnitudes of the other two forces Pa and R can be calculated using the law of cosines in trigonometry. The plane BD is just one trial failure plane, we should repeat the calculation for a numbers of selected trial planes to search for the maximum active thrust, Pa.

Figure 12.3: Construction of Culmann line to search for the maximum thrust (Smith & Smith, 1998) Figure 12.3 describes a graphical method suggested by Culmann (1866), which can be used to search for the maximum total thrust. The reaction force Pa is determined for four different trial failure planes. The force diagrams of the four cases are drawn overlying on each other, the distance Xdi is proportional to the area (the weight) of each soil wedge. The length eidi represents the magnitude of the reaction force Pa, the locus of all points ei is a smooth curve called Culmann line, which describes the variation of the total thrust with the direction of the failure plane. The maximum thrust can be found be drawing a vertical line tangential to the Culmann line, the maximum length ed obtained at the tangential point represents the maximum total thrust. The angle e3Xe is equal to the angle EBG, so the corresponding failure plane can be defined (i.e. line BG in Fig. 12.3a).

57

Figure 12.4: Application point of the total thrust using the Culmann line (Smith & Smith, 1998) The point of application of the total thrust in the Coulomb’s wedge theory can be determined as shown in Fig. 12.4. A line can be drawn ing through the centroid of the failure wedge and in parallel to the failure plane; the intersection of the line with the wall is the application point of the total thrust. For a typical problem as shown in Fig. 12.1b, we have an analytical solution for the Coulomb’s theory of earth pressure. The coefficients of earth pressures can be given as: •

Active pressure: σ 'h = Ka ×σ 'v

1 γh 2 Pa = × h × K aγh = K a 2 2

  sin (ψ − φ )     sinψ Ka =    sin (ψ + δ ) + sin (φ + δ )sin (φ − β )    sin (ψ − β ) •

2

(12.1)

ive pressure: σ 'h = K p × σ 'v

Pp =

1 γh 2 × h × K pγh = K p 2 2

  sin (ψ + φ )     sinψ Kp =    sin (ψ − δ ) − sin (φ + δ )sin (φ + β )    sin (ψ − β )

2

(12.2)

The earth pressure based on the analytical solution of Coulomb’s theory in Eq. 12.1 and 12.2 can be assumed to have a triangular distribution as similar to the Rankine’s theory. The application point of the total thrust is given in Fig. 12.5 for different retaining walls. Note that while the active wedge tends to slide downward, the ive wedge tends to move upward relatively to the wall. The reaction force P acting at the wall interface and the reaction force R acting on the failure plane, hence, should be oriented in the way that resists this type of movement of the ive wedge. 58

Figure 12.5: Application point of the active thrust for the analytical solution (Smith & Smith, 1998) In the previous paragraphs, we discussed the derivation of lateral earth pressures using the Coulomb’s wedge theory. The Coulomb’s theory not only provides a solution for complex geometry problems but also shed some lights on how we could find the solution for an engineering problem. In the case of active thrust, it was assumed that as the wall is moving away, a failure soil wedge will be formed behind the wall. In other words, we need to assume a failure mechanism. The job of the engineer is then to design the structure so that such a scenario will not happen for the real structure. To come up with a failure mechanism, we can study the failure of similar real structures in the past, or conduct test on small scale physical model of the real structure, or even do full scale field test, etc. In the design process, we need to consider all the driving forces that may cause instability, and ensure their impacts are less than the effects of the resistance forces that help to stabilize the structure. Furthermore, a factor of safety is required, which means the structure is designed with the allowable load as only a fraction (normally in the range of 30%∼70% depending on the problem) of the estimated maximum resistance capacity. And this is to cover for any uncertainties and to ensure the serviceability of the structure in its normal working condition (the deformation level is within the acceptable limit so it does not affect the function of the structure). Figure 12.6 introduces several failure scenarios of a gravity wall. Gravity walls are usually massive and heavy walls as the stability comes from the self-weight of the wall; the frictional resistance at the base of the wall helps to resist against the lateral thrust of the retained soil mass.

Figure 12.6: Failure modes of gravity retaining wall (Smith & Smith, 1998) The wall in Fig. 12.6a fails as the whole soil mass surrounding it fails along a slipping surface; this type of failure is usually assessed in of slope stability analysis, which means another problem of bigger scale. Figure 12.6b describes the bearing capacity failure, which happens when the overturning moment of the lateral thrust causes higher stress concentration below the toe of the 59

wall. The stress concentration could be over the bearing capacity of the soil beneath the wall; and when the soil foundation fails, the wall could suddenly sink to the ground (not evenly) and topple over. Figure 12.6c shows the horizontal sliding failure mode of the wall when the friction resistance at the base is less than the lateral thrust. To avoid this, we can increase the weight of the wall (increase W in Eq. 8.1), increase the roughness of the base (increase µ in Eq. 8.1) or embed the wall into the ground in order to generate the ive resistance of soil in front of its toe. One also needs to check the overturning failure mode to ensure the ratio between the overturning moment and the restoring moment is within a safe limit (in of a factor of safety). Finally, the level of deformation of the wall and the ground needs to be controlled to avoid failure of the structural components of the wall and of the structures built behind it. Structural failures (cracks, etc.) could cause deterioration of the design capacity and lead to progressive failure of the whole structure, or at least reduce their serviceability for smaller damages. The overall stability of gravity wall needs to be checked against: •

Overturning failure about the wall’s toe:

FS overturnin g =

where:

•

∑M ∑M

R

(12.3)

O

FSoverturning is the factor of safety against overturning failure; the minimum value of FS should be in the range of 1.5 – 2.0. ΣMO is the summation of the moments that tend to overturn the wall about its toe, and ΣMR is the summation of the moments that oppose the overturning movement.

Sliding failure at the wall’s base:

FS sliding =

∑F ∑F

R

(12.4)

D

where:

•

FSsliding is the factor of safety against basal sliding failure; the minimum value is about 1.5. ΣFD is the summation of the horizontal driving forces, and ΣFR is the summation of the horizontal resistance forces.

Bearing capacity failure of soil beneath the wall’s base:

FSbearing =

where:

qu qmax

(12.5)

FSbearing is the factor of safety against bearing capacity failure; FS = 3 is generally required. qu is the ultimate bearing capacity of soil beneath the wall (the calculation of qu is usually covered in foundation design). qmax is the maximum applied pressure, which can be determined as follows:

60

Figure 12.7: Bearing pressure due to gravity retaining wall (Smith & Smith, 1998) In Fig. 12.7, RV is the total vertical force (the sum of the weight of the wall, the weight of soil and any surcharge within the perimeter of the base); e is the eccentricity of the total vertical force, which is the distance between the symmetrical axis of the base and the total vertical force. The eccentricity can be calculated as:

e=

B ∑ M R − ∑ MO − 2 RV

(12.6)

where B is the width of the base. Depending on the value of the eccentricity (e), the distribution of bearing pressure can be separated into there cases as shown in Fig. 12.7. The maximum pressure in the distribution (qmax) should be checked against the ultimate bearing capacity of soil (qu).

61