1 1. INTRODUCTION to SOIL MATERIALS 1.1 Definitions Soil: • Uncemented Or Weakly Cemented Accumulation of Mineral and Organic

1. INTRODUCTION TO SOIL MATERIALS

1.1 Definitions

Soil: · uncemented or weakly cemented accumulation of mineral and organic particles and sediments found above the bedrock, or · any unconsolidated material consisting of discrete solid particles with fluid or gas in the voids

Rock: · indurated (consolidated by pressure or cementation ) material requiring drilling, blasting, brute force excavation

The dividing line between soil and rock is arbitrary; the same material may sometimes be either classified as “very soft rock” or “very hard soil”, depending on who classifies the material or what the application is. To a geologist “our” soil is drift or unconsolidated material.

Whereas we are concerned with soil to the depth of bedrock, soil scientists (pedology) and agricultural scientists (agronomists) are concerned with only the very uppermost layers of soil.

Soil Mechanics: (ASTM) the application of the laws and principles of mechanics and hydraulics to engineering problems dealing with soil as an engineering material.

Geotechnical Engineering: the application of civil engineering technology to some aspect of the earth, therefore including soil and rock as engineering materials. It combines the basic physical sciences, geology, pedology with hydraulic, structural, transportation, construction, environmental and mining engineering.

Soil mechanics is a subset of geotechnical engineering.

1.2 Origin of Soils

Soil is a three phase system of: - solid particles - pore fluid - pore gas

Most solid particles are mineral fragments that originated from the disintegration of rocks by physical or chemical action, often referred to as weathering.

Physical Weathering: erosion due to freezing & thawing, abrasion from glaciers, temperature changes, and the activity of plants and animals.

Chemical Weathering: decomposition due to oxidation, reduction, carbonation, and other chemical processes.

Introduction to Soils 1 CivE 381

Exceptions: Peat (organic) and shell deposits

Soils at a particular site can be:

- Residual or weathered in place (most common in tropical locations), or

- Transported by the action of: Glaciers (glacial) Moving water (fluvial) Wind (aeolian) Settling out in salt water (marine) Settling out in fresh water (lactustrine) Due to gravity movement downslope (colluvial) (most common in temperate regions)

1.3 Main Types of Soils

Granular: gravel, sand, (silt)

Cohesive: (silt), clay

Organic: marsh soil, peat, coal, tar sand

Man-Made: mine tailings, landfill waste, ash, aggregates

Soils can vary from 102 to 10-3 mm in diameter.

Naturally occurring soils are usually a mixture of two or more of the above components. (e.g., silty-sand, clayey-silt, clay with gravel)

In addition, the void space between the slid particles may be filled with either pore fluid gas.

1.4 The Unique Nature of Soil Material

· highly variable - properties vary widely from point to point within the soil mass - more heterogeneous rather than homogeneous - large variations over small distances

Introduction to Soils 2 CivE 381 · nonlinear stress-strain response

· nonconservative (i.e. inelastic) - soils “remember” what has happened to them in the past - stress history is very important - soil behaviour is quite different whether normally consolidated or overconsolidated (CivE381)

· anisotropic - different properties in different directions - primarily a result of depositional and loading history

· mulitphase system (soil, water, and air)

· empirical application in design - empirical - based on experience – what we can see / what we can measure - good design - combination of art, science and common sense

The behavior of soil in situ is often governed by soil fabric, weak layers and zones, and other defects in the material. It is therefore essential that the successful geotechnical engineer develops a feel for the soil behavior.

Generally we idealize the behavior using applied mechanics concepts, and then apply engineering judgement (based on our own experience and the experience of others) to come up with a final solution.

Because the soil is so complex, it is difficult to deal with as an engineering material. It is necessary to be able to CLASSIFY the soil based on ENGINEERING BEHAVIOUR.

Introduction to Soils 3 CivE 381 2. MASS - VOLUME RELATIONSHIPS AND DEFINITIONS

2.1 Soil System

Soil normally consists of a two- or three- phase system:

1) Solid mineral particles - quartz, feldspars, carbonates, mica / clay minerals, organic matter, - plus garbage, tailings, slag, etc.

2) Pore fluid - normally water - could be oil, bitumen - could be leachate

3) Pore gas - normally air - could be methane (landfill, pipeline)

- often excess CO2 in tropics, radon

2.2 Phase Diagram

For quantifying the properties of a soil, a series of definitions and terminology has developed to describe the three phase system – best illustrated with the use of a phase diagram.

VA MA=0 Air VV VW Water MW

VT Solid MT VS MS

· provides an easy means to identify both what is know and the relationship between known and desired quantities

· we usually measure the total volume VT, the mass of water MW, and the mass of solids MS

· we may then calculate the rest of the values and the mass volume relationships that we need. Most relationships are independent of sample size and are often dimensionless.

Introduction to Soils 4 CivE 381 2.3 Volumetric Relationships

Void Ratio, e: V e = V VS [1] VV = volume of voids VS = volume of solids

· Expressed as a decimal · Typically: Sands 0.4 < e < 1.0 very loose sand e » 0.8 Clays 0.3 < e < 1.5 soft clay e > 1 organic clays e > 3 · Empirically determined that much of soil behavior is related to e As e decreases density increases As e decreases strength increases As e decreases permeability decreases

Porosity, n: V n = V VT [2] VV = volume of voids VT = volume total

· Expressed as a decimal or percentage (usually decimal)

By substituting equation [1] into [2] we can show that: e n = 1+ e [2a] and n e = 1- n [2b] e.g., for a very loose sand with e=0.8,

Introduction to Soils 5 CivE 381 Degree of Saturation, S:

V S = W ´ 100 (%) VV [3] VW = volume of water VV = volume of voids

· Expressed as a percentage · Tells us the percentage of the total volume of voids that contain water · Range is from 0 to 100% S = 0 % soil is completely dry S = 100 % soil is saturated (i.e. pore spaces are completely filled with water)

2.4 Mass Relationships

Density, r: · Expressed as g/cm3, kg/m3 or Mg/m3 (=g/cm3)

Density of Solids, rS: MS = mass of solids VS = volume of solids

MS rS = VS [4]

Density of Water, rW:

M W 3 3 o r W = = 1.0g / cm = 1.0Mg / m at 4 C VW [5] MW = mass of water VW = volume of water

Bulk Density(also termed moist, wet, or total density), r: M r = T VT [6] MT = mass total VT = volume total

Introduction to Soils 6 CivE 381 Saturated Density, rsat:

S = 100%, therefore VA = 0

M T rsat = VT [7]

Similar to bulk density except that the sample must have S = 100% e.g. saturated soil below the water table

Dry Density, rd:

S = 0%, therefore MW = 0

M S rd = VT [8]

Buoyant (Submerged) Density, r’:

r¢ = rSAT - r W [9]

2.5 Weight Relationships

The relationships just defined in terms of masses (or densities) can be expressed in terms of weights and are called unit weights.

Unit Weight, g: g = r´g [10] g = acceleration due to gravity = 9.81 m/s2

· typically expressed in kN/m3 e.g. if r = 2100 kg/m3, g = 2100 kg/m3 ´ 9.81 m/s2 = 20601 kg×m = 20.6 kN / m3 s2×m3

Introduction to Soils 7 CivE 381 2.6 Basic Tests

Moisture Content, w: ASTM D2216

M w = W ´100(%) M S [11] · Expressed as a percentage · The amount of water present in a soil relative to the mass of dry soil. · See Bowles Experiment #1, pages 15-17.

Specific Gravity, Gs: ASTM D854

g r Gs = S = S g W r W [12] · Note that the Canadian Foundation and Engineering Manual (1992) terms this ratio as the relative density of the solid phase with respect to water and uses the symbol Dr. · See Bowles Experiment #7, pages 71-78 · Defined as the weight of soil divided by the weight of an equal volume of water at 20oC · Gs is found using a sample of soil and a pycnometer, which gives the average specific gravity of the materials from which the soil particles are made. · Typically 2.6 to 2.8 for the solid minerals in soil · Often Gs < 1 for organic particles

2.6 Useful Relationships

e S = w Gs [13]

æ G S + eS ö r = ç ÷ r w è 1+ e ø [14] set S = 1 for rsat set S = 0 for rd

Introduction to Soils 8 CivE 381 2.7 Typical Values

TABLE 1. Summary of typical values of porosity, void ratio, water content, saturated density and saturated unit weight.

SOIL n e w rsat gsat (%) (%) kg/m3 kg/m3 LOOSE, UNIFORM SAND 46 .85 32 1890 18.5 LOOSE, MIXED GRAINED SOIL 40 .67 25 1986 19.45 DENSE WELL GRADED SAND 30 .43 16 2163 21.2 HARD OR DENSE GLACIAL TILL 20 .25 9 2323 22.8 SOFT CLAY 55 1.2 45 1762 17.3 STIFF CLAY 37 0.6 22 2067 20.2 SOFT ORGANIC CLAY 75 3.0 110 1426 13.9 PEAT (VERY COMPRESSIBLE) 94 17 1000 10.2

TABLE 2. Summary of specific gravities for minerals and soils.

Specific Gravities of Minerals Specific Gravities of Soils

Quartz 2.65 Sand 2.65 K-Feldspars 2.65 - 2.57 Silty Sand 2.66 - 2.68 Na-Ca-Feldspars 2.62 - 2.76 Silt 2.67 - 2.68 Calcite 2.72 Silty Clay 2.70 - 2.72 Dolomite 2.85 Clay 2.70 - 2.80 Muscovite 2.7 - 3.1 Biotite 2.8 - 3.2 Gs > 2.80 - likely metals present Chlorite 2.6 - 2.9 Gs < 2.70 - likely organics present Pyrophyllite 2.84 Serpentine 2.2 - 2.7 Average Gs for sand = 2.65 Kaolinite 2.61 a Average Gs for well mixed soil = 2.70 2.64+/-0.02

Halloysite (2 H 2O) 2.55 Illite 2.84 a 2.60 - 2.86 Montmorillonite 2.74 a 2.75 - 2.78 Attapulgite 2.3

a Calculated from crystal structure.

Introduction to Soils 9 CivE 381 2.8 Example Problems

A saturated soil sample (S = 100%) has a water content of 42% and a specific gravity of 2.70. Calculate the void ratio, porosity, bulk unit weight, and bulk density.

A cylinder of soil has a volume of 1.15´10-3 m3, a mass of 2.290 kg and Gs of 2.68. The mass of solid obtained by drying is 2.035 kg. Calculate: r, g, wn, e, n, and S.

Introduction to Soils 10 CivE 381 3. GRAIN SIZE AND GRAIN SIZE DISTRIBUTION

3.1 Coarse Grained versus Fine Grained Soils

· convenient dividing line is the smallest grain that is visible to the naked eye

· with the Unified Soil Classification System (USCS) the division corresponds to a particle size of 0.075 mm.

Particles larger than this size are called coarse-grained, while soils finer than this size are called fine-grained.

Table 3.1 Textural and Other Characteristics of Soils

Soil Type Gravels, Sands Silts Clays

Grain size: - Coarse grained - Fine grained - Fine grained - Can see individual - Cannot see - Cannot see grains by eye individual grains individual grains

Characteristics: - Cohesionless - Cohesionless - Cohesive - Nonplastic - Nonplastic - Plastic - Granular - Granular

Effect of water on Relatively engineering unimportant Important Very Important behaviour: (exception: loose saturated granular materials and dynamic loading)

Effect of grain size distribution on Important Relatively Relatively engineering Important Unimportant behaviour:

Introduction to Soils 11 CivE 381 3.2 Grain Size Distribution

We are interested in both the particle size and the distribution of the particle sizes.

Sieve tests and hydrometer tests are used to define the distribution of grain sizes.

The range of particle sizes varies from 200 mm > D > 0.002 mm (i.e. by orders of magnitude) hence when we examine the particle size distribution we plot on a logarithmic scale.

Classification of soils according to particle sizes varies slightly between different classification systems.

The Unified Soil Classification System (USCS) is one commonly used classification system.

In describing the size of a soil particle, we can use either a dimension or a name that has been arbitrarily assigned to a certain size range. Classification from the USCS is described below:

Type Grain Size (mm)

Boulders > 300 Cobbles 75 to 300 Gravel 4.75 to 75 Sand 0.075 to 4.75 Silt < 0.075 Clay

0.001 0.01 0.1 1 10 100 1000 Particle Size (mm)

Introduction to Soils 12 CivE 381

Particle size distribution obtained by shaking a dry sample of soil through a series of woven-wire square-mesh sieves with successively smaller openings.

Since soil particles are rarely perfect spheres, particle diameter (or size) refers to an equivalent particle diameter as found from the sieve analysis. We will use the U.S. Standard Sieves. The sieve sizes are summarized in Table 3.2.

Table 3.2 U.S. Standard Sieve Sizes and their Corresponding Opening Dimension

Sieve No. Sieve Opening (mm) 3" 75 1.5" 38 0.75" 19 0.375" 9.5 #4 4.75 #10 2.00 #20 0.85 #40 0.425 #60 0.25 #100 0.15 #140 0.106 #200 0.075

Sand Gravel Fines (Silt, Clay) Boulders

F M C F C Cobbles

0.001 0.01 0.1 1 10 100 1000 Particle Size (mm)

Nested sieves are used for soils with grain sizes larger than 75 :m. For finer soils (silts and clays) the hydrometer test is used.

Introduction to Soils 13 CivE 381 Procedure for Soil Analysis

1. A soil sample is separated by passing it through the nest of sieves. 2. Determine the weight of soil retained on each sieve. 3. Calculate the percent of weight finer than each particle size. 4. Plot the grain size distribution as Percent Finer Than as the ordinate (y-axis) versus the log of the Particle Size as the abscissa (x-axis).

Calculation for Grain Size Analysis

Sieve Mass of Mass of Sieve Mass Cumulative % Cumulative % Passing Opening Sieve and Soil Retained Mass Retained Retained mm g g g g A B

Where T = total mass of dry sample

Typical Grain Size Curves

Introduction to Soils 14 CivE 381 Parameters Describing the Grain Size Distribution

1. Effective particle size, D10: - Denotes the grain diameter (in mm) corresponding to 10% passing by mass. - Controls flow for coarse grain soils.

2. Coefficient of Uniformity, Cu:

D Cu = 60 D10 where: D60 = grain diameter (in mm) corresponding to 60% passing by mass and, D10 = grain diameter (in mm) corresponding to 10% passing by mass.

(Note: if D60 = D10, Cu = 1, all particles between 10% and 60% are the same size).

3. Coefficient of Curvature, Cc:

D 2 Cc = 30 D10 D60 where: D30 = grain diameter (in mm) corresponding to 30% passing by mass.

For well graded soils,

CU > 4 for gravel

CU > 6 for sand

1 < CC < 3

If CU and CC do not meet both of the criteria above, the soil is poorly graded.

Well graded: good representation of particle sizes over a wide range; gradation curve is generally smooth.

Poorly graded: either excess or a deficiency of certain sizes, or most of the particles about the same size. (i.e. uniform soil)

Gap graded: a proportion of grain sizes within a specific range is low (it is also poorly graded).

Introduction to Soils 15 CivE 381 Introduction to Soils 16 CivE 381 3.3 Density Index of Granular Soil

· Also referred to as relative density

Definition: ID = emax - e x 100% emax - emin

where: emax = maximum void ratio corresponding to the loosest state, emin = minimum void ratio corresponding to the densest state, and e = void ratio of the sample.

Loosest State, ID = 0%, obtained by: · Sifting or funneling dry sand into narrow rows in a box · Gentle settling in a water column · If very fine, dumped in a damp, bulked state and submerged from below

Densest State, ID = 100%, obtained by: · Prolonged vibration at 20 - 30 cycles / sec under light static load in dry state · If very uniform sand, tamped lightly after dumping thin layers

Field Measurement: STD penetration test, "N" values · 63.5 kg (140 lb) hammer dropping 76.2 cm (30") · Count number of blows per ft to drive 2" sampler 61 cm

ID 0 15 35 65 85 100 Very Loose Compact Dense Very Loose Dense N 28o 30o 36o 45o

3.4 Application of Grain Size Distribution

1. Estimation of Coefficient of Permeability, k, in Sands and Gravels

An empirical correlation between PSD and permeability has been developed 2 k = c (D10) cm/s where 100 < c < 150 Developed by Hazen for uniform, loose, clean sands and gravels.

2. Frost Heave Susceptibility

Frost heaving occurs if water may be drawn towards the freezing front in soils from below, forming lenses of ice. Whether or not water may be drawn to the freezing front is largely governed by the pore size, which is a function of the grain size distribution of the

Introduction to Soils 17 CivE 381 soil. The pore sizes may be sufficiently small to allow capillary action of the pore water up to the freezing front, and sufficiently large to have a high enough permeability to allow the water to migrate fast enough.

Silts combine sufficiently high suction and permeability to maximize ice lens production, hence road base material, for example, is usually specified to have not more than 3% silt size particles to alleviate frost heave beneath roads.

A soil is frost susceptible if > 3% pass 0.02 mm.

3. Selection of Fill Material

· Used to specify material for concrete aggregate (sand & gravel), road base material · Used to examine and develop borrow pits.

4. Geotechnical Processes

· Used to evaluate soil drainage. · Used to examine the likely effectiveness of grouting or soil freezing techniques for soil stabilization.

5. Design of Protective Filters

Piping ratio: D 15(FILTER) < 4to5 D85(SOIL) Prevents the protected soil from moving through the filter. D 15(FILTER) > 4to5 D15(SOIL) Ensures that the filter is large enough to improve the situation. It may be necessary to place a number of filter materials in series to avoid piping.

Introduction to Soils 18 CivE 381 NATURE OF COHESIVE SOILS

Clay Water System

“Clay” refers to both a specific sheet size ( < 2 mm) and specific minerals (sheet silicates) that are somewhat similar to mica. The cohesive properties of natural soils are normally related to the presence of clay minerals (e.g., kaolinite, illite, monmorillonite, chlorite and vermiculite).

- all clay mineral are negatively charged

- hydrated cations (+ve) are attracted to -ve clay particles forming a double layer

The double layer (or bound water) is the main reason that the engineering behaviour of clayey soils are strongly influenced by the presence of water.

Atterberg Limits

Since water plays an important role in the behaviour with a significant clayey fraction, a range of water content has been defined that correlate strongly with the engineering properties of fine grained soils. The Atterberg limits are water contents that bracket different behavioural states for the soil.

INCREASING WATER CONTENT w (%)

shrinkage limit, ws plastic limit, wP liquid limit, wL natural water content, wn The range of water content over which a fine grained soil behaves as a plastic is defined as the Plasticity Index:

IP =

The Plasticity Index provides an important indication of soil properties and may indicate its composition. It is used in the classification of fine grained soils.

Also define the Liquidity Index as:

IL =

Relationship of Mineralogy to Atterberg Limits

Clay Mineral wL wP IP CEC (%) (%) (%) (meq/100g) kaolinite illite Na+ - montmorillonite Ca++ - montmorillonite

CEC = cation exchange capacity

NOTES:

1. High wL = montmorillonite = trouble

2. Na+ mont. MUCH more troublesome than Ca++ mont. CLASSIFICATION OF SOILS

UNIFIED SOIL CLASSIFICATION SYSTEM (USCS) Read: “An Introduction to Geotechnical Engineering”, Holtz and Kovacs pp. 47-64. ON RESERVE: UA Cameron Flr1 SciTec Reserve, CALL NUMBER: TA 710 H75 1981

1) MAIN SOIL TYPE PREFIX

COARSE GRAINED ( < 50% PASSES No. 200 SIEVE) GRAVEL ( < 50% PASSES No. 4 SIEVE) G SAND ( > 50% PASSES No. 4 SIEVE) S

FINE GRAINED ( > 50% PASSING No. 200 SIEVE) SILT M CLAY C ORGANIC O

PEAT Pt

2) SUBDIVISIONS SUFFIX

FOR GRAVEL AND SAND WELL GRADED (Cu > 4 and 1 < Cc < 3), CLEAN W POORLY GRADED (Cu Ý 4 and 1 Û Cc Û 3), CLEAN P APPRECIABLE FINES ( > 12% PASSES No. 200 SIEVE) M or C

FOR SILTS AND CLAYS (use plasticity chart)

LOW PLASTICITY (wL < 50%) L HIGH PLASTICITY (wL > 50%) H

see Example 3.1 H&K

COMPACTION OF SOILS

Compaction is the densification of soil by the application of mechanical energy.

Reasons for Compaction:

Road Subgrade - strong at small deflections - ultimate strength usually not a problem

Road Embankment - strong at ultimate strength for overall stability

Homogeneous Dam - strong and impervious

Dam Core - low permeability (relatively impervious) and usually weak - strength derived from shells of dam

Clay Liner - low permeability (relatively impervious) for municipal and toxic solid waste disposal

Main Factors Influencing the Compaction of Soils

4) 2

RELATIONSHIP BETWEEN DRY DENSITY AND WATER CONTENT

The relationship between the dry density (or unit weight) and water content of a soil is measured in the laboratory with the compaction test. Here a soil sample mixed to a certain water content is compacted in a cylinder of known volume. The dry density of the soil can be computed by measuring both the total mass of the soil and the water content.

• compact soil in layers • for each layer, drop known mass a certain height with a specified number of blows per layer

Standard Proctor Modified Proctor Number of Layers 3 5 Height of Fall 0.3048 m 0.4572 m Mass of Hammer 2.495 kg 4.536 kg Energy 593 kJ/m3 2694 kJ/m3 3

19 . . . 18 .

6 8 10 12 14 16 18 w

Typical Compaction Curve for Silty Clay

maximum dry density rd max optimum water content wopt

Explanation of shape:

- below wopt there is a water deficiency • get

- near wopt the clay particles are lubricated

- above wopt there is excess water • some of the compactive effort is taken by • also water takes up spaces that could be occupied by 4

Relationship between dry density, water content and degree of saturation can be calculated viz,

Note that:

- no data points should lie to the right of the zero air void curve

- complete saturation is never achieved, even at high water contents

Source: Holtz and Kovacs (1981)

Modified Proctor test has a greater compactive effort (CE) than the Standard Proctor.

• as CE 8,

• as w 8

• 5

COMPACTION AND STRENGTH OF COHESIVE SOILS

s 600

400 Strength (kPa) 200

w e %

w 6

COMPACTION AND PERMEABILITY OF CLAYEY SOILS

• hydraulic conductivity decreases as moulding water content

• huge decrease in k

• minimum k occurs 2 to 4% above the optimum water content

•

• if compacted wet of optimum the method of compaction influences k

e.g., at 4% wet of optimum, 100 times difference in k

! reason for lower k related to clay particle structure:

flocculated

dispersed 7

FIELD PLACEMENT OF CLAYEY BARRIER FOR WASTE CONTAINMENT

Compacted clay liners are commonly used as barriers in waste containment facilities (e.g., municipal solid waste landfills) to minimize the movement of contaminants from the facility.

MUNICIPAL SOLID WASTE LANDFILL

SOLID WASTE

Gravel (16 - 32 mm) Geotextile Perforated Pipe Gravel (50 mm) Geotextile

! since lowest values of k achieved with kneading compaction, make great effort in the field to repetitively knead the soil with many passes of pad-foot, club-foot or wedge-foot rollers

! kneading action breaks up clods and interclod macropores

! also compact in lifts with pad-foot compactor with feet long enough to penetrate through the lift being compacted into the underlying lift

! minimum thickness (normally) of 0.9 m (six lifts of 0.15 m) to minimize the risk of defects in a layer having a significant impact on performance - probability of cracks lining up is very small if compacted in more than four layers

! need to consider potential for clay-leachate interaction may not be a great problem clayey soils with low activity

! important that the liner not be permitted to: dessicate - freeze - 8

FIELD MEASUREMENT OF DENSITY

Having defined the necessary dry density for the soil, determined the type of compactor and determined the lift height, it is subsequently necessary to monitor the density of the field compacted soil to ensure that the soil is performing as expected and that contractor is performing the work as required.

The easiest and probably most common method of determining the soil density is with the use of a Nuclear Density Meter. Other methods include sand cone and rubber balloon tests.

Source: Bardet (1997) Nuclear Density Meter -non–destructive -measures both moisture content and bulk density directly -gamma radiation is used for density determination -neutron radiation is used for moisture content determination -radiation is sent out from an emitter and scattered radiation is counted by a detector. -calibration against compacted materials of known density and water content is necessary

Sand Cone and Balloon Density: Steps 1. Excavate a hole in the compacted fill at the desired sampling elevation. 2. Record the mass of soil removed for the hole. 3. Determine the water content of the soil removed. 4. Measure the volume of the hole using sand cone, balloon or other method. 5. Calculate bulk density knowing Mt and Vt. 6. Calculate the dry density knowing the bulk density and the water content. STEADY STATE SEEPAGE

TYPES OF PROBLEMS how fast and where water flows though soils - rate of leakage from an earth dam - movements of contaminants in subsurface rate of settlement of foundations - related to how fast water flows in soils the stability of earth structures - water influences the strength of soils

NATURE OF FLOW

flow from A to B - not in a straight line - not at a constant velocity - rather winding path from pore to pore

Flow occurs through the interconnected pores • isolated voids do not exist in an assemblage of spheres - regardless of packing density - gravels, sands, silts, and even most clays - probably no isolated voids - unless cemented

• some geologic materials (e.g., many crystalline rocks) have a high total porosity - most of which are interconnected

- effective porosity ne - percentage of interconnected pore space - contaminants may move very fast in fractured rock ONE DIMENSIONAL FLOW - DARCY’S LAW

Classical experiment performed by H. Darcy in the 1850's to study the flow properties of water through a sand filter bed.

It was experimentally found that:

Q = where:

Q = total volume of water collected per unit time - flow rate [ L3 / T ] k = experimentally derived constant [ L / T ]

h3 = height above datum of water rise in standpipe inserted at the top of the sand [ L ] h4 = height above datum of water rise in standpipe inserted at the base of the sand [ L ]

L = length of sample [ L ]

A = cross-sectional area of the sample container [ L2 ] Define gradient between any two points a and b as:

iab = where Dh is the difference in total head between points a and b.

Therefore Darcy’s Law can be written as:

Q =

Flow per unit area is given by:

where: v is the Darcy flux - volume of water that flows through a unit area per unit time - units with dimensions of [ L3 ] × [ L-2 ] × [ T-1 ] = [ L / T ] - same units as velocity - often called Darcy velocity but is actually a flux - fictitious velocity but useful

Consider the flow between points 1 - 2 and 3 - 4. Continuity of flow requires that:

Q 1-2 = Q 3-4

where: vs is the seepage velocity - also termed groundwater velocity and average linearized groundwater velocity - also fictitious quantity given tortuous flow path, but again useful quantity and n is the porosity

e.g., for sands, n .0.3 NATURE OF HEADS

h = hv + hp + z where: hv = velocity head

hp = pressure head - height to which liquid rises in a piezometer above that point

- pore pressure u = hp × ãw

z = elevation head - vertical distance from datum to point

h = total head [ L ]

** Water flows from high total head to low total head. **

Note: Total head is always measured relative to some datum. Since flow depends on the gradient (or change in head over a given distance) the choice of the position of datum is not important - however, choosing a datum (and clearly defining it) is of paramount importance.

Example 1. Example 2.

Example 3. PHYSICAL INTERPRETATION OF DARCY’S PROPORTIONALITY CONSTANT

Reflecting back on Darcy’s experiment, the proportionality constant k may be expected to be a function of the soil and the fluid.

k % ãw

k % 1 / µ

k % d2 where:

ãw = unit weight of water

µ = viscosity d = mean grain diameter of sand k = Darcy’s proportionality constant

Define k as the hydraulic conductivity - contains properties of both the porous medium and the fluid - units [ L / T ] - characterises the capacity of a porous medium to transmit water at a specific temperature - also referred to as the coefficient of permeability - hydraulic conductivity is most frequently used in ground water or hydrogeology literature - permeability used in petroleum industry where the fluids of interest are oil, gas and water

Darcy’s proportionality constant can be expressed as:

k = ki ãw µ

ki is defined as the intrinsic permeability - contains properties of the porous medium only - units [ L2 ] - characterises the capacity of a porous medium to transmit any fluid

Both the unit weight and the viscosity of water can change with temperature. For practical purposes of groundwater flow these changes are small; we ignore these effects (unless the temperatures approach 0°C), so we treat k as a soil property, independent of other effects. HYDRAULIC CONDUCTIVITY

The hydraulic conductivity is influenced by a number of factors including:

- effective porosity - grain size and grain size distribution - shape and orientation of particles - degree of saturation - clay mineralogy

Approximate range in values of hydraulic conductivity (Whitlow 1995).

k (m/s)

102 - 101 - 1 - 10-1 - 10-2 - 10-3 - 10-4 - 10-5 - 10-6 - 10-7 - 10-8 - 10-9 - 10-10 - HYDRAULIC CONDUCTIVITY AND CLAY MINERALOGY

In general, the higher the specific surface and cation exchange capacity, the greater amount of bound water and the lower the hydraulic conductivity value.

Clay Mineral Edge View Thickness Specific CEC Probable k (nm) Surface (meq/100g) (m/s) (km2/kg) kaolinite 50 - 2000 0.015 - 5 10-7 - 10-10

illite 30 .08 - 25 10-9 - 10-11

montmorillonite 3 100 - 100 10-10 - 10-15

Why is this so?

Implications for clayey barriers for waste containment: kaolinite - would have to be very pure to obtain low k because of low CEC - valuable as a pottery clay illite - probably best barrier clays - fairly inactive, no interlayer expansion or contraction - yield low k barrier if constitute about 20% of well graded soil montmorillonite - obtain the lowest hydraulic conductivity - susceptible to interlayer expansion and contraction - may get huge increase in k - BAD - most temperamental of the clay minerals LABORATORY MEASUREMENT OF HYDRAULIC CONDUCTIVITY

See Whitlow Sections 5.5, 5.6, 5.7

Constant Head Test Falling Head Test

FIELD MEASUREMENT OF HYDRAULIC CONDUCTIVITY

See Whitlow Section 5.9 ONE DIMENSIONAL FLOW PROBLEMS

The engineered barrier systems in modern municipal solid waste landfills provide excellent examples of the use of one dimensional flow problems to solve seepage problems.

TWO DIMENSIONAL STEADY STATE SEEPAGE

Several assumptions are required to derive the equation governing two dimensional steady state seepage.

• the soil is completely saturated • there is no change in void ratio of the porous medium • the hydraulic conductivity is isotropic • Darcy’s law is valid • the water is incompressible

Consider the flow of water into an element with dimensions dx and dy and unit width in the z direction.

Continuity of flow requires that, Analytical solutions can be obtained to Laplace’s equation for problems involving only simple boundary conditions.

Alternatively, either numerical or graphical solutions may be used.

e.g., finite difference e.g., flow nets finite element - SeepW ® - GMS Seep2D ®

Numerical solutions may be highly dependent upon the refinement of the finite-difference grid or finite-element mesh. For transient analysis suitable refinement of the time step is also important. Such numerical methods should be considered incorrect until proven correct.

Define two characteristics of flow:

1) Equipotential lines - EP - lines of constant total head

2) Flow lines - FL - lines parallel to the direction of flow

If we draw a flow net with constant head difference between EPs for flow through a homogeneous, isotropic porous medium then: - EP z FL - get curvilinear squares - don’t have to all be the same size - be able to fit circle tangent to all sides

7 Flow net for Darcy’s Apparatus 6

0 To solve for the flow Q use Darcy’s law: - calculate Q based on one square then multiply by the number of flow tubes

ÄQ = k Äh a L

ÄQ = flow in one flow channel (per unit width) Äh = total head drop across a pair of EPs L = distance over which head drop takes place a = distance between adjacent flow lines.

Common Boundary Conditions

equipotential line impermeable boundary

line of constant pressure

- sat. soil in contact with air

- hp = 0 - ˆ h = z - line of variable but known head

Steps in Drawing a Flow Net

1) Define and clearly mark a datum. 2) Identify the boundary conditions (EP, FL, LCP). 3) Draw intermediate equipotentials and flow lines. - draw coarse mesh with a few EPs and FLs 4) Verify the coarse mesh is correct. - Are the boundary conditions satisfied ? - Are all flow tubes continuous ? - Are EPs z FLs ? only if isotropic medium - Mostly “squares” ? 5) Add additional EPs and FLs for suitable refinement of the flow net. 6) Calculate desired quantities of flow and heads. Example: Steady State Seepage Beneath a Sheet Pile Wall Flow Beneath a Dam Seepage Through an Earth Dam EFFECTIVE STRESS

The compressibility and strength of soils is governed by the effective stresses.

Any deformation or mobilization of shearing resistance of a soil is associated with the soil skeleton since: - water is incompressible - water cannot support shear stresses

Terzaghi showed experimentally that for saturated soils:

óN = ó - u where: sN = effective stress s = total stress u = pore pressure

The effective stress principle is accurate provided that:

• point to point contact area between soil particles is small

ó = óN + u 1 - Ac A

• the compressibility of the soil particles is small and the strength of individual particles is large

Effective stress can be thought of as the force carried by the soil skeleton divided by the total area of the soil element (including the area of pore water).

Effective stress is an empirical concept that works well.

-1- CALCULATING EFFECTIVE STRESSES

• the total vertical stress sv within a particular soil layer is equal to the total weight per unit area of material above that point

where gi is the bulk unit weight of layer i with thickness di.

• pore pressure is equal to: u = hp × ãw

where: hp = pressure head

gw = unit weight of water

• vertical effective stress can then be found as: óvN = óv - u

• horizontal effective stress shN is equal to: óhN = K óvN

where: K = coefficient of lateral earth pressure = óhN / óvN

- horizontal stresses develop from the resistance to lateral movement - K is a function of soil type and stress history

- for conditions of zero lateral strain, Ko is used

Ko = “at rest” coefficient of lateral earth pressure - K is typically between 0.3 to 0.8 for normally consolidated clays - K is typically between 0.3 to 0.5 for normally consolidated sands

Some published correlations for Ko:

Ko . 0.95 - sin fN Brooker and Ireland (1965) where: f N = angle of internal friction

sin f N Ko . (1 - sin f N) ( OCR ) Kulhawy and Mayne (1990) where: OCR = overconsolidation ratio

-2- Example 1

-3- Example 2. Typical River Crossing with Artesian Conditions

-4- CAPILLARY AND SOIL SUCTION

The concept of soil suction is fundamental when considering the mechanical behaviour of

Total suction arises from two components: - matric suction, and - osmotic suction

ø = ( ua - u ) + ð where: y = total soil suction

ua = pore-air pressure u = pore-water pressure

( ua - u ) = matric suction p = osmotic suction

Matric suction is associated with the capillary phenomenon arising from the surface tension of water.

• interaction of surface molecules causes a condition analogous to a surface subjected to tension

• the capillary phenomenon is best illustrated by considering the rise of a water surface in a capillary tube

Consider a small glass tube inserted into water under atmospheric conditions:

wetting causes curvature liquid meets glass tube at angle á

-1- • water rises up a small tube resulting from a combination of the surface tension of a liquid and the tendency of some liquids ro wet surfaces which they come into contact with

Vertical equilibrium of the water in the tube requires that

2 2 ð r Ts cos á = ð r hc ñw g where: r = radius of capillary tube

Ts = surface tension of water . 73 dynes / cm - different in different liquids

a = contact angle - for any liquid, Ts 9 as temp 8 hc = capillary height g = gravitational acceleration.

solve for hc

Rs = radius of curvature ( r ÷ a)

For pure water and clean glass, a . 0, giving:

The radius of the tube is analogous to the pore radius in the soil. The smaller the pore radius, the greater the capillary rise.

• common to assume the effective pore size is 20% of effective grain size D10

Note that highly variable pore size and pore distribution complicate the capillary phenomenon in soils. However, useful qualitative deductions can be made from the glass tube analogy.

hc (m) Loose Dense Coarse sand 0.03 - 0.12 0.04 - 0.15 Fine sand 0.3 - 2.0 0.4 - 3.5 Silt 1.5 - 10 2.5 - 12 Clay > 10

-2- Consider several points in the capillary system that are in hydrostatic equilibrium

• weight of water column transferred to tube through the contractile skin

• for a soil with a capillary zone, this results in an increased compression on the soil skeleton

• matric suction increases the shear strength of an unsaturated soil

Re-examining the Water Table

Contact Moisture

Partially Saturated by Capillarity Saturated by

VADOSE ZONE Capillarity

Ground Water PHREATIC ZONE

-3- STRESSES IN AN ELASTIC MASS

Loading on the Surface of a Homogeneous Isotropic Semi-Infinite Mass

(a) Point Loading

3Pz3 Vertical Stress sz = 2pR 3

P é3r 2 z (1- 2v)R ù Radial Stress s = - - r 2 ê 3 ú 2pR ë R R + z û

P (1 - 2v) é R z ù Tangential stress s = - q 2 ê ú 2pR ëR + z R û

3Pr z2 Shear stress trz = 2pR5

(b) Uniformly Loaded Strip

P Vertical stress s = [ a + sin a cos (a + 2d) ] z p

P Horizontal stress s = [ a - sin a cos (a + 2d) ] x p

2p Horizontal stress s = n a y p

p Shear stress t = sin a sin (a + 2d) xz p

2 (c) Uniformly Load Circle

On axis, at depth z,

é 3 ù æ 1 ö 2 Vertical stress s = p ê 1 - ç ÷ ú z ê ç 2 ÷ ú è1 + (a / z) ø ëê ûú

é ù p 2 (1 + n)z z3 Horizontal stresses s = s = ê 1+ 2n - + ú r q ( ) 1 3 2 ê 2 2 2 2 2 2 ú ë (a + z ) (a + z ) û

For locations other than on the axis, see contour plot.

Increment in vertical stress (Dsv = Dqv) beneath a circular footing with radius R and subject to uniform vertical pressure Dqs on uniform, isotropic elastic half-space.

(d) Uniformly Loaded Rectangle

Vertical stress sz beneath the corner of a rectangle is given by Fadum’s chart. For points other than the corner, sz may be obtained by superposition of rectangles.

5 (e) General Shapes

Vertical stress sz may be obtained by use of the Newmark chart.

(f) Linear Superposition

For linear elastic problems solutions may be added or subtracted to solve problems involving more complex geometry.

For example:

6 SETTLEMENT OF SOILS

When soils are subjected to loads (e.g., construct a building or an embankment) deformation will occur.

The design of foundations for engineering structures requires that the magnitude and rate of settlement be known.

The total settlement ST is given by:

ST = Si + S + Ss where:

Si = immediate or distortion settlement - normally estimated using elastic theory

- judicious selection of stiffness parameters (E, n) over appropriate stress range

S = primary settlement in fine grained soils

- arises from the time dependent process of consolidation

- consolidation is the dissipation of excess pore pressure

- occurs because of changes in effective stress

Ss = secondary compression - arises from changes in void ratio at constant effective stresses

- also termed as creep

When a soil is loaded settlement occurs because of water and air squeezing out from the voids. This results in a decrease in void ratio, and hence settlement.

-1- MECHANICAL ANALOGY OF CONSOLIDATION

a) Initial conditions where:

total stress = sv

pore pressure = uo

effective stress = sv - uo = soN

b) Apply increase in total stress Ds with valve (V) closed. Then we

have: total stress = sv + Ds

pore pressure = uo + Du

effective stress = ( sv + Ds ) - ( uo + Du )

= sv - uo = soN

ˆ DsoN = 0 - no change in effective stress - no compression of the spring

- ST = Si S = 0

c) Open valve (V). Water allowed to flow out of the sample. ˆ Du 9 as t 8 as t 6 4, Du 6 0, then:

total stress = sv + Ds

pore pressure = uo

effective stress = ( sv + Ds ) - uo = sfN

here, DsN = sfN - soN = Ds

For real soil materials, the compression of the spring is analogous to a decrease in void ratio arising from a change in effective stresses

Consolidation is a time dependent process since it involves the flow of water from the pores. - consolidation is the dissipation of excess pore pressure

-2- STRAIN INTEGRATION

Recall the axial deformation d of a column with stiffness E and cross-sectional area A, subject to axial load P.

Likewise for a soil material subject to increases in effective stress, the settlement (vertical displacement) may be found be integrating the vertical strain, viz:

where:

Dez = change in vertical strain because of a change in sN D = thickness of compressible layer n = number of sub-layers

Dzi = thickness of sub-layers

Dez

Layer 1 Dz1

Layer 2 Dz2 .

Dzi

Layer n Dzn

The number (n) and thickness (Dzi) of sub-layers depends on the function to be integrated

-3- For conditions of one-dimensional strain, the change in volume strain Dev is equal to the change in vertical strain Dez.

BEFORE AFTER

D D - De S = ò Dez dz = ò dz 0 0 1+ eo

Now, need to express the relationship between void ratio and effective stress to calculate S because of change in sN.

-4- OEDOMETER TEST

In the laboratory we can measure the change in height of a sample (and thereby calculate the change in void ratio) for a certain effective stress. This test is called a consolidation test and is performed in an oedometer which permits one-dimensional strain.

- apply load

- initially all of the load is transferred into excess pore pressure

- drainage permitted by porous stones

- excess pore pressure will dissipate and effective stresses will increase and the soil will settle

- monitor the change in height of the sample until most of the pore pressures have dissipated (achieve 90% consolidation)

- apply next load increment

-5- Vertical Strain and Void Ratio Versus Effective Stress

Calculate the vertical strain or void ratio from the measurements of change in height of the sample. Can either plot these results on a linear or logarithmic scale of effective stress.

Dev = mv Ds¢

Note that:

- mv is not constant - depends on stress level

- mv decreases as sN increases i.e. soil becomes stiffer

Soils are normally strain hardening materials. - that is to say, they become stiffer as they are loaded

-6- Vertical Strain and Void Ratio Versus Logarithm of Effective Stress

- same data as previous plot now plotted versus the logarithm of effective stress

Note that: - apparently get a straight line - simplifies calculations - since log scale, still represents strain hardening behaviour

-7- Experimental results from an oedometer test are plotted with void ratio(e) versus the log of effective stress (sN): “ e log sN ” plot

0.8

0.7

0.6

0.5 Void Ratio e

0.4

0.3 1 10 100 1000 10000 Effective Stress s' (kPa)

where: eo = initial void ratio corresponding to soN

soN = initial effective stress - current in situ effective stress

spN = preconsolidation stress - maximum effective stress experienced by soil

Cc = compression index - slope of compression line on e log sN plot - typical values: NC medium sensitive clays 0.2 to 0.5 Leda Clay 1 to 4 Peats 10 to 15

Ccr = recompression index - slope of recompression line on e log sN plot

- Ccr < Cc

Cs = swelling index - slope of expansion line on e log sN plot

- Cs . Cs

-8- STRESS HISTORY OF SOILS

Soils have a “memory”, that is to say they remember the effective stresses that they have previously experienced. Represent the stress history with the over consolidation ratio OCR, where: OCR =

If the current vertical effective stress is equal to the preconsolidation stress:

0.8

0.7 NORMALLY CONSOLIDATED 0.6

0.5 Void Ratio e

0.4

0.3 1 10 100 1000 10000 Effective Stress s' (kPa)

If the current vertical effective stress 0.8 is less than the preconsolidation stress: 0.7

0.6

OVER CONSOLIDATED 0.5 Void Ratio e

0.4

0.3 1 10 100 1000 10000 Effective Stress s' (kPa)

-9- Determination of spN using Cassagrande’s graphical procedure:

1) Plot laboratory data on e vs log sN graph. This laboratory data must be corrected for errors arising from sample disturbance to get the field curve.

2) Select the point of minimum radius: point A.

3) Draw a horizontal line through A.

4) Draw a tangent to the curve at point A.

5) Bisect the angle between the horizontal line and the tangent through point A.

6) The intersection of the extension of the straight line portion of the compression curve with

the bisector line is the preconsolidation stress spN.

-10- Prediction of field e - log sN curves with Schmertmann’s Procedure NC Soils:

1) Find preconsolidation stress spN using Cassagrande’s method.

If the soil is normally consolidated NC ( soN . spN ) then

2) extend horizontal line from eo to spN

3) extend the laboratory compression line to intersect with 0.42 eo

4) connect this point with spN

-11- Prediction of field e - log sN curves with Schmertmann’s Procedure OC Soils:

1) Find preconsolidation stress spN using Cassagrande’s method.

If the soil is over consolidated OC ( svoN < spN ) then

2) extend horizontal line from eo to svoN

3) extend the laboratory compression line to intersect with 0.42 eo

4) find Ccr from unload - reload loop

5) from ( eo , svoN ) construct line parallel to unload - reload loop to find the void ratio

corresponding to spN

6) connect this point at spN with 0.42 eo

-12- Typical e - log sN curves:

If consolidation tests are conducted on many samples from different depths can construct profiles like the one shown below. This is typical of a stiffer “crust” material that has been preconsolidated. The material below 20 m is normally consolidated.

-13- Calculation of Primary Settlement

Two ways to calculate the change in vertical strain.

1) Elastic Model

Dez = mv DsN

mv = coefficient of volume decrease

- must use mv for appropriate stress range

2) Strain Hardening Model

- use e vs log sN plot to calculate De - implicitly models strain hardening behaviour of soil - depends on stress history - three cases a) for NC soil,

0.8

0.7

0.6

0.5 Vo id Ratio e

0.4

0.3 1 10 100 1000 10000 Effective Stress s' (kPa)

-14- b) for OC soil with sNf < sNp,

0.8

0.7

0.6

0.5 Void Ratio e

0.4

0.3 1 10 100 1000 10000 Effective Stress s' (kPa)

c) for OC soil with sNf > sNp,

0.8

0.7

0.6

0.5 Void Ratio e

0.4

0.3 1 10 100 1000 10000 Effective Stress s' (kPa)

-15- CONSOLIDATION

Consolidation is an important mechanism involving the flow of water through the soil leading to time dependent settlements. • process of consolidation involves the dissipation of excess pore pressure. • decrease in pore pressures result in increases in effective stresses. • increase in effective stresses lead to settlement.

Field Behaviour Under One-Dimensional (1D) Conditions

u (kPa) u (kPa) u (kPa) u (kPa) 0 50 100 150 200 0 50 100 150 200 0 50 100 150 200 0 50 100 150 200 0

z (m) 5

Initial Conditions Load rapidly applied, Some time after load Long time after load applied, applied,

-1- MECHANICAL ANALOGUE FOR CONSOLIDATION

-2- Governing Differential Equation for Consolidation

Assumptions: 1. soil is saturated and homogeneous 2. water and soil particles are incompressible 3. Darcy’s law is valid 4. one dimensional strain 5. k remains constant 6. change in volume results from change in void ratio and Me/MsN remains constant 7. total stress remains constant after application.

Terzaghi’s equation of consolidation is:

Du = excess pore pressure t = time z = depth below top of consolidating layer cv = coefficient of consolidation cv = k

mvgw

Define non-dimensional parameters:

Drainage Path Ratio, Z = z / H

H = length of drainage path

Time Factor, T = cv t H2

-3- substituting into the governing differential equation with s constant for t > 0 gives:

The solution to this equation for a layer of thickness z with two way drainage (ie. Z = 2) with boundary conditions:

at t = 0, Du = Duo for 0 # Z # 2

for t > 0, Du = 0 at Z = 0 and Z = 2 is:

¥ 2 Du 2 Du = å o (sin MZ) e-M T m=0 M

where: M = p / 2 ( 2m + 1 )

Useful to define another dimensionless parameter that represents the proportion of excess pore that has dissipated at a particular point in the deposit, viz:

Degree of Consolidation, Uz = eo - e = Duo - Du = 1 - Du

ef - eo Duo Duo

where:

eo = initial void ratio corresponding to soN e = void ratio at time t, e = f(t)

ef = final void ratio corresponding to sfN

Duo initial excess pore pressure Du = pore pressure at time t, Du = f(t)

The solution to the consolidation equation can be expressed graphically as shown in Figure C1.

-4- FIGURE C1.

FFIGURE C2. Average degree of consolidation.

-5- Example: A 4 m thick layer of clay is subject to rapid application of surface load from 5 m of fill (g = 20 kN/m3). Calculate the excess pore pressure and the effective stress at the mid-point of the clay layer: (a) initially, and (b) after 4 months.

SILT SILT

CLAY CLAY

SAND SAND

Initially

soN = so - uo

so = uo =

soN =

Immediately after fill placement, t=0

s = u = sN =

-6- After 4 months

- use consolidation theory to solve for excess pore pressure after 4 months - since silt and sand are much more permeable than clay, there is two way drainage for the clay

H = length of drainage path

Drainage Path Ratio, Z = z / H

Time Factor, T = cv t H2

Now solve for Du,

Du = Duo (1 - Uz) = =

s = u = sN =

Note: - this is at one point in the clay layer - look at another point (B), say 0.2 m below the top of the clay after 4 months at point B,

Z = z / H

ˆ Du = Duo (1 - Uz) =

soN = sN =

-7- Since

DsAN > DsBN

Need some way of averaging Du with depth to obtain the average degree of consolidation for the entire layer.

Average Degree of Consolidation

The average degree of consolidation U for a layer is given by:

D Du dz U = consolidation settlement at time t = St ò0 t = 1 - D total final consolidation settlement S Du dz ò0 o

- assuming: s constant with time,

mv constant with depth and time

Various solutions have been obtained for the average degree of consolidation. Figure C2 gives U for three cases where there is a linear variation in stress increment with depth.

-8- Example: Find the consolidation settlement of the 4 metre thick clay deposit 4 months after the fill is placed. 100 kPa Step 1: Find the total final settlement.

S = å De z Dzi i=1 3 m SILT Use just one sublayer here.

De De = z 3 1+ eo g = 16.3 kN/m wn = 70%, Gs = 2.72 4 m s = 2 oN CLAY cv = 1.26 m /yr sfN =

Cc = 1.055 Since OCR =1 , NC soil. OCR = 1

æ s¢f ö De = - Cc log10 ç ÷ è s¢o ø SAND

Find eo using: e S = w Gs

Total final consolidation settlement is:

S =

After 4 months,

Z =

T =

U =

S t = 4 mo = U × S =

-9- How to find the Coefficient of Consolidation cv Using Taylor’s Method

1) Plot change in height of the sample measured during consolidation test versus the square root of time. This is done for each load increment.

2) Fit straight line through the initial part of the compression curve.

3) The straight line is extended back to t=0 to find Ro.

4) Draw a second line from Ro with a slope 1.15 times larger than the line from step 2.

5) The intersection of this line with the compression curve is defined as t90.

6) Calculate cv using:

2 2 T Hdr T90 Hdr cv = = t t 90

-10- MOHR CIRCLE IN SOIL MECHANICS

Mohr circle of stress is a graphical representation of the state of stress at a point at equilibrium.

- used extensively to plot strength results

Sign Convention:

- compressive forces and stresses are taken as positive (change in normal sign convention because tension is rare in soil mechanics)

- positive shear stresses produce clockwise moments about a point just outside the element

- clockwise angles are positive

-1- The stability of an existing slope can be assessed by comparing the disturbing forces (self weight) with the strength of the soil mobilized along a potential failure surface. Consider the stresses acting a point along the potential failure surface below.

These stresses can be plotted on the Mohr circle.

-2- Now suppose we want to know the stresses for this same point but oriented in a different direction (e.g., on a potential failure plane). It is useful to define the pole of the Mohr circle.

If a line is drawn from a point on the circle (representing a state of stress) parallel to the plane on which the stress state exists it will intersect the circle at another point on the circle which is known as the pole, or origin of planes.

Any line drawn through the pole will intersect the circle at a point which represents the state of stress on a plane inclined at the same orientation in space as the line.

Once the pole is known, the stresses on any plane can be readily determined by drawing a line through the pole parallel to the plane in question; the stress on the plane will be the coordinates where the line intersects the circle.

How to find the pole:

1) Start from a known magnitude [ie. coordinates (s, t) ] and orientation of stress. Go to that point on the Mohr circle.

2) Draw a line through the point of known stress with the same orientation in space as the plane on which those stresses act.

3) The pole is where this line intersects the Mohr circle.

-3- t

Principal Stresses

- s1 and s3 are the respective maximum and minimum normal stresses on the Mohr circle

- note that the shear stress is equal to zero along the major and minor principal planes

-4- Example1. Given:

a) Find the normal and shear stresses acting on a plane inclined at 30o to the horizontal. b) Find the major and minor principal stresses. c) Determine the orientation of the major and minor principal planes. d) Find the maximum shear stress and the orientation of the plane on which it acts.

-5- Example 2. Given:

a) Find the magnitude of the normal and shear stresses on the horizontal plane. b) Find the magnitude and orientation of the principal stresses. c) Show the orientation of the planes of maximum and minimum shear.

-6- STRENGTH OF SOILS - MOHR-COULOMB FAILURE CRITERION

Failure or yield of a soil material occurs when the shear stresses exceed the shear strength.

Soil materials generally fail because of excess shear stresses. Uniform compressive stresses (ie. s1

= s3 ) alone will only tend to change the volume of the soil. Non-uniform compressive stresses

(e.g., s1 > s3 ) induce shear stresses in the soil.

The shear strength of soil is defined as the shear stress acting on the failure plane at failure.

This is commonly expressed using the Mohr-Coulomb failure criterion,

tff = cN + sNff tan fN

where: tff = the shear stress acting on the failure plane at failure - the shear strength of the soil

sNff = the normal effective stress acting on the failure plane at failure

cN = cohesion [ M T-2 L-1 ]

fN = angle of internal friction

Notes: - strength is governed by the effective stresses

- cN and fN are not unique material parameters - vary on many factors including stress range

-7- Consider a triaxial compression test on a medium sand.

-cylinder of sand

- subject to: vertical pressure s1

radial pressure s3

- increase s1 with s3 held constant - increase s1 until failure

-8- SHEAR STRENGTH OF SAND

Sands are termed as cohesionless or frictional materials since cN=0

Major factor influencing the shear strength of sands:

1. Void Ratio or relative density; 2. Pressure range of consideration; 3. Particle shape; 4. Grain mineralogy; 5. Grain size distribution; 6. Water; 7. Intermediate principal stress; and 8. Stress history.

Of these factors, void ratio is the single most important factor. Generally, the lower the void ratio (higher density) the higher the shear strength.

Direct Shear Testing of Sands t

Dilatancy: define the angle of dilatancy Explanation of Volume Change Behaviour for Sands

Dense Sand

If a sand is dense, the only way shearing can occur is for grains to move apart.

Therefore dense sands when sheared to failure exhibit an tendency for volume increase.

Loose Sand

If a sand is loose, during shearing the grains move closer together.

Therefore loose sands when sheared to failure exhibit an tendency for volume decrease. Example: A sample of loose sand is know to have a friction angle fN = 30o. It is tested in direct shear under a normal stress of 200 kN/m2. Determine the shear strength, the maximum shear stress and the major and minor principal stress at failure.

s Example: Direct shear tested were conducted on a sample of compacted sand. Determine the peak and ultimate friction angles based on the results that were recorded.

Test 1 2 3 4 Normal force (N) 110216324432 Ultimate shear force (N) 66 131195261 Peak shear force (N) 85 170253340

100

Shear Stress (kPa) 20

0 0 20 40 60 80 100 120 140 160 180 200 Normal Stress (kPa) SHEAR STRENGTH OF CLAY

The shear strength of clays depend on: - the effective stresses at failure - void ratio - stress history - mineralogy

Failure Envelopes for Clay t

s t

s Consider the Triaxial Test

- initial effective stress soN

- apply all around cell pressure W

either with drainage gives: - consolidation OR without drainage gives: - volume changes - no volume change

CONSOLIDATED TEST UNCONSOLIDATED TEST

Du = 0 Du = W

s1N = soN + W s1N = soN DsN = 0

s3N = soN + W s3N = soN

slowly without without with drainage drainage drainage

DRAINED UNDRAINED UNDRAINED TEST TEST TEST

Du = 0 Du = Dus Du = Dus s1N = soN + W + Z s1N = soN + W + Z - Dus s1N = soN + Z - Dus

s3N = soN + W s3N = soN + W - Dus s3N = soN- Dus

CD CU UU Consolidated Drained Triaxial Test (CD Test)

- soil is allowed to consolidate to a given effective hydrostatic stress scN with full drainage - the soil is loaded to failure very slowly so that no excess pore pressures develop

1. Changes in total stress equal the change in effective stress. - we can draw the Mohr circle for effective stresses at failure 2. Volume of the sample is allowed to change. - the void ratio will change during the test

Stress Path: locus of stress points on a given plane (normally the failure plane, but not always) TSP - Total Stress Path ESP - Effective Stress Path

s Example: A consolidated drained triaxial test was conducted on a normally consolidated clay. 2 The results were: s3 = 276 kN/m 2 (s1 - s3)f = 276 kN/m Determine: a) the angle of friction fN b) the inclination of the failure plane

c) the normal sffN and shear stress tff on the failure plane at failure

d) the normal stress snN on the plane of maximum shear stress tmax e) explain why shear failure did not take place on the plane of maximum shear stress. t

s Consolidated Undrained Triaxial Test (CU Test) With Pore Pressure Measurements

- soil is allowed to consolidate to a given effective hydrostatic stress scN with full drainage

- the drainage system is closed off

- the soil is loaded to failure relatively quickly - since drainage is prevented, excess pore pressures develop

- the pore pressures are measured

1. Control the applied (total) stresses and measure the pore pressures. - effective stresses can be calculated - can draw the Mohr circle for effective stresses at failure

2. Volume of the sample is not allowed to change during shearing. - the void ratio will NOT change during the test

Dus = B ( s3 + A ( s1 - s3 ) ) Example: CU Typical stress-strain and volume change versus strain curves for CD triaxial compression tests at the same effective confining stress.

Some examples of CD analyses for clays. Consolidated Undrained Triaxial Test (CU Test) With Pore Pressure Measurements

- soil is allowed to consolidate to a given effective hydrostatic stress scN with full drainage - the drainage system is closed off - the soil is loaded to failure relatively quickly - since drainage is prevented, excess pore pressures develop - the pore pressures are measured

1. Control the applied (total) stresses and measure the pore pressures. - effective stresses can be calculated - can draw the Mohr circle for effective stresses at failure

2. Volume of the sample is not allowed to change during shearing. - the void ratio will NOT change during the test

Normally Consolidated t

s Dus = excess pore pressure due to shear failure

- excess pore pressures may be (+)ve or (-)ve

- occur because sample wants to change volume but not allowed to (since drainage is prohibited)

- because there is no volume change, the tendency towards volume change induces Dus

- if the volume tries to decrease, water wants to squeeze out of the pores but can’t - develop (+)ve Du - NC - effective stresses are less than the total stresses - ESP lies to the left of the TSP

- if the volume tries to increase, wants to draw water into the pores but can’t - develop (-)ve Du - OC - effective stresses are greater than the total stresses - ESP lies to the right of the TSP

For saturated soils, S=100%

Dus = Ds3 + A ( Ds1 - Ds3 )

- Skempton’s pore pressure equation

For typical soft NC clay, A - 1/3 to 3/4 stiff OC clay, A - 1/3 to -1/2 Typical stress-strain, excess pore pressure vs. strain and ratio of major to minor principal effective stresses for normally and overconsolidated clays in consolidated undrained triaxial compression test. Overconsolidated t

s Example: A clay soil is known to have an effective stress envelope with cN=10 kPa and fN=25E. A series of four consolidated undrained (CU) tests were performed and the total stress Mohr circles at failure as shown. a) What is the excess pore pressure at failure for test #4 ? b) Is the clay normally consolidated or overconsolidated ? Why ? c) What is the shear strength for test #4 ? d) What is Skempton’s A factor at failure for test #4 ?

e) What is the shear strength of a sample with sNff = 500 kPa ?

f) What is the shear strength of a sample with a cell pressure s3 of 300 kPa ?

g) A clay was accidentally consolidated to s3 = 300 kPa. The technician then reduced the cell pressure to 200 kPa without drainage and ran an undrained test. What would the undrained strength of this clay be ? h) What would be the pore pressure at failure for part (g) ?

t (kPa) 300

200

100

100 200 300 400 500 600 s (kPa) Example cont’d

t (kPa) 300

200

100

100 200 300 400 500 600 s (kPa) Some examples of CU analyses for clays. Unconsolidated Undrained (UU) Triaxial Test

s t

s Some examples of UU analyses for clays. Typical stress-strain and volume change versus strain curves for CD triaxial compression tests at the same effective confining stress.

Some examples of CD analyses for clays. Consolidated Undrained Triaxial Test (CU Test) With Pore Pressure Measurements

1. Control the applied (total) stresses and measure the pore pressures. - effective stresses can be calculated - can draw the Mohr circle for effective stresses at failure

2. Volume of the sample is not allowed to change during shearing. - the void ratio will NOT change during the test

Normally Consolidated t

s Dus = excess pore pressure due to shear failure

- excess pore pressures may be (+)ve or (-)ve

- occur because sample wants to change volume but not allowed to (since drainage is prohibited)

- because there is no volume change, the tendency towards volume change induces Dus

- if the volume tries to decrease, water wants to squeeze out of the pores but can’t - develop (+)ve Du - NC - effective stresses are less than the total stresses - ESP lies to the left of the TSP

- if the volume tries to increase, wants to draw water into the pores but can’t - develop (-)ve Du - OC - effective stresses are greater than the total stresses - ESP lies to the right of the TSP

For saturated soils, S=100%

Dus = Ds3 + A ( Ds1 - Ds3 )

- Skempton’s pore pressure equation

e) What is the shear strength of a sample with sNff = 500 kPa ?

f) What is the shear strength of a sample with a cell pressure s3 of 300 kPa ?

t (kPa) 300

200

100

100 200 300 400 500 600 s (kPa) Example cont’d

t (kPa) 300

200

100

100 200 300 400 500 600 s (kPa) Some examples of CU analyses for clays. Unconsolidated Undrained (UU) Triaxial Test

s t

s Some examples of UU analyses for clays. Unconfined Compression (UC) Triaxial Test

- zero confining stress s3 = 0

- increase s1 to failure

- unconfined compressive strength, qc = s1f - equal to the diameter of the Mohr circle

- undrained shear strength = cu = qc / 2

- NOTE that strength of the soil is still controlled by the effective stresses - convenient to express strength in terms of total stresses here FIELD MEASUREMENT OF SHEAR STRENGTH

Field Vane Standard Penetration Test (SPT) Cone Penetration Test (CPT) EARTH PRESSURES AND RETAINING STRUCTURES

Introduction

The analysis of the pressures exerted by the ground against an engineering structure has been of paramount interest dating back to the time of Coulomb in 1776. Considerations of earth pressures are essential to the successful design of many engineering structures including bridges, retaining walls, tunnels; therefore, it is of concern in nearly all civil engineering projects. The subject is a vast one with a remarkable number of publications on various aspects of earth pressure theory and its application to real engineering situations. A wide variety of approaches have been available to solve these problems. However, because of complexities involved in this problem, all methods involve certain simplifying assumptions and none of them present a rigorous representation of the soil-structure interaction at failure. Many of the accepted analyses of the past are now being challenged as a result of a more complete understanding of the behaviour of soils when subjected to stress and strain.

The Complexity of Earth Pressure

A glance at many of the older handbooks of civil engineering and indeed at some modern textbooks for structural design would lead the uninitiated to believe that earth pressure can be calculated by simple formulas comparable to those for stress and deflection of steel or concrete members. One would be lead to conclude that the pressure exerted by the soil was unique for each type of soil, that the pressure was the same regardless of the type of structure or the problem, and that the pressure could be calculated with precision to two or three significant figures. Unfortunately none of these beliefs are correct. Earth pressure, in the broadest sense of the word, denotes forces and stresses that occur either in the interior of an earth mass or on the contact surface of soil and structure. Its magnitude will be determined by the physical properties of the soil, the physical interactions between soil and structure, value and character of absolute and relative displacements and deformations. Knowledge of the stress-strain and strength properties of soils is fundamental to solving these soil-structure interaction problems.

Classification of Earth Pressure Problems

Earth pressure problems can be separated into three main classes of problems. First, the case of an earth mass at rest where no deformations or displacements occur. This condition is strictly fulfilled in the infinite half space at rest only. This case is mainly of theoretical interest; it gives also the starting point for more practical problems. In the problems of the second group, the horizontal forces in the earth masses are to be determined. Here we have retaining wall problems, sheet piling, braced excavations, etc. Relative displacement between soil and structure occurs causing the soil either to expand or to contract. In the first case (ie. soil expansion) we have an active pressure, and in the second case (ie. soil contraction) a passive pressure. The most common example for this group is the retaining wall yielding around the bottom or pressed against the earth mass. There are also cases where at the same time compression and expansion in

-1- different parts of the mass occur. Problems where the vertical force prevails form the third group. These are the problems of foundations: stresses, deformations, failure of soil beneath foundation structures. Problems of buried structures and rock pressures also belong to this group.

Types of earth pressure problems

-2- Fundamental Concepts

Classical earth pressure theory is reviewed and the major assumptions that are made are discussed. A clear distinction is necessary between methods of:

1) allowable stress distributions, 2) limit analysis, and 3) complete deformation analysis.

Initial State of Stresses

Coefficient of earth pressure at rest (in terms of effective stresses)

s'h Ko = s'v

Typically for normally consolidated soils, Ko N.C. sands 0.4 to 0.5 N.C. clays 0.5 to 0.75

From field and lab tests for N.C. soils (Jaky 1948),

Ko » 1- sin f'

For over consolidated soils, Ko typically lies between 1 to -2.5 depending on soil type. The value of Ko is bounded by passive failure. Ko for O.C. soils increases with over-consolidation ratio

OCR (sNp ÷ sNvo) where sNp is the past maximum vertical effective stress and sNvo is the current vertical effective stress at a point within the ground. A relationship between Ko and the OCR has been reported by Brooker and Ireland (1965). The CFEM (1992) suggests the use of:

0.5 Ko = (1- sin f')OCR

as a first approximation for over consolidated soils.

For an elastic medium with Poisson’s ratio n and zero lateral strain (ie. ex = ey = 0)

n K = o 1- n

-3- Limiting Equilibrium (Rankine Theory)

A body of soil is in a state of plastic equilibrium if the state of stress at every point of the ground is on the verge of failure. Rankine (1857) investigated the stress conditions corresponding to the states of plastic equilibrium that can be developed simultaneously throughout a semi-infinite mass of soil acted on by no other force other than gravity. Consider the fictitious case of a large mass of cohesionless soil containing a thin embedded rigid wall of infinite depth. It is assumed that the wall does not influence the initial state of stress in the ground. The changes in stress at two points, A and B are considered for movements of the thin wall.

As wall displaces to the right,

- Point A experiences with corresponding in sx

- Point B experiences with corresponding in sx

Limiting equilibrium is reached by either:

sxN until induced shear stresses lead to failure 3N or

-4- sxN until induced shear stresses lead to failure 4

-5- Solution for Active and Passive Pressures (Rankine 1856)

s1' = s3 ' N f' + 2c' N f' where: æ f'ö N = tan2 ç45o + ÷ f' è 2 ø

-6- -7- What are the magnitudes of sxAN and sxPN ?

For cohesionless soils,

Active,

sxAN

Passive,

sxPN

-8- Example 1: Estimate the earth pressure acting on the wall.

3 2 m g1 = 17 kN/m f'1 = 34

5 m

3 3 m g2 = 19 kN/m

f'2 = 36

At Point A

sv =

At Point B

sv =

just above B, sNh =

just below B, sNh =

At Point C

sv =

u =

sNv =

sNh =

-9- What is the total horizontal force acting on the wall, and where does the resultant act?

-10- Example 2: Find the factor of safety against sliding and rotational failure for the gravity retaining wall shown.

1.5 m

5 m

3 m

-11- Effect of Sloping Ground Surface

Active earth pressure, sNhA =

Passive earth pressure, sNhP =

cosb - cos2 b - cos2 f¢ KA = cosb + cos2 b - cos2 f¢

cosb + cos2 b - cos2 f¢ KP = cosb - cos2 b - cos2 f¢

-12- Active and Passive Earth Pressure Coefficients - Cohesive Soils

Drained Response For cohesive soils with cN and fN

Active,

sxAN

Passive,

sxPN

Undrained Response

For cohesive soils - undrained response fN = 0, cu

Active,

sxA

depth of tension crack =

ie. the net pressure for z = Hc is zero. This is the theoretical maximum height of a vertical slope that can stand unsupported under short term conditions. The unsupported cut may only be stable for a very short time, due to seepage forces and softening by precipitation.

-13- SLOPE STABILITY slope failure - the downslope movement of a soil mass occurring along a failure surface

For a uniform soil without planes of weakness, then the failure surface is close to a circle.

For a homogeneous slope with f = 0 we have equilibrium along a circular sliding surface.

-1- GENERAL CASE c, f - METHOD OF SLICES

- shear strength is not constant - must integrate

F =

- to solve this equation for the factor of safety, F, we need to know the correct normal stress distribution.

- any method of analysis can be used for a slip circle provided that it correctly represents the overall statics for the problem.

Procedure

1. Divide the soil into a number of slices.

-2- 2. Look at a typical slice.

t =

S =

where: N is the normal force acting on the base of the slice, u is the pore pressure acting on the base of the slice.

Taking moments about the centre of the circle gives,

Must determine the force N. Resolving forces perpendicular to the slip surface gives,

It is often assumed that: N =

This implies that the resultant of the inner slice forces Xn, En acts parallel to the base of the slice. This is considered to be a conservative assumption.

-3- Determination of Pore Pressure u

May need to construct a flow net to determine the pressure head acting on the base of each slice.

General Notes

- with the method of slices we consider moment equilibrium

- we ignore considerations of vertical and horizontal equilibrium

- method tends to give conservative solutions for uniform clays without planes of weakness

-4- -5- Perform Tabular Calculation

Slice No. W l a c f u N1 N2

S =

N1 = W sin a N2 = [W cos a - u l ] tan f + c l

F = ( S N2 ) / ( S N1 ) where: W = weight of slice [kN/m] c = cohesion intercept [kN/m2] f = friction angle [degrees] 2 u = pore pressure = hpgw [kN/m ] a = angle between base of slice and horizontal [degrees] l = length of slip surface segments measured along base of slice [m]

Notes: 1) The slice weights W are calculated based on the dimensions of the slices and the unit weights of the soils within them. W can be calculated using:

W = b S ( gj hj ) where: b = width of slice

gj = unit weight of soil j hj = height of soil layer j where it is subtended by slice - measured at the centre of the slice

2) The values of c and f for each slice correspond to the type of soil at the bottom of each slice. For short term (ie. undrained) conditions use c = cu and f = 0. For long term (ie. drained) conditions use c = cN and f = fN.

3) The value of u for each slice is the average value at the middle of the slice.

4) The base length l and the base angle a are measured on a scale drawing of the slope.

-6- 5) Normally slices are drawn so that the base of each slice is in only one type of soil. Slices need not be of equal width. Example 1:

Slice No. W l a c f u N1 N2 1 119 4.35 -12 20 20 0 -25 129 2 288 4.00 0 20 20 0 0 185 3 410 4.08 11 20 20 0 78 228 4 480 4.38 24 20 20 0 195 247 5 464 5.03 37 20 20 0 279 235 6 368 6.83 54 20 20 0 298 215 7 44 4.00 70 20 20 0 41 85 S = 1325 867

-7- N1 = W sin a N2 = [W cos a - u l ] tan f + c l

F = ( S N2 ) / ( S N1 ) = 1.53

STABILITY CHARTS Taylor (1949) has prepared charts for the simple case of:

The factor of safety F depends on:

1) slope angle 2) angle of friction 3) c, g, and H

Define stability number N =

Use of Taylor’s Charts

- Taylor’s charts do not take account of pore pressures - used for short-term stability calculations (ie. total stress analysis)

CASE 1: Toe failure.

CASE 2: Rigid layer at elevation of toe. f > 10 f = 10 f = 5 f = 0

CASE 3: Circle passes below toe. f = 10 - toe failure

f = 5 - almost toe failure

-8- f = 0 - circle passes below toe - F depends on layer depth

- for f = 0 and n < 4.5, use T2

-9- Taylor’s Chart 1 Taylor’s Chart 2

-10- Example 2. What is the short-term factor of safety for the slope considered in Example 1? Take cu = 30 kPa.

What angle would the slope have to be to get F = 1.3 ?

Example 3. A wide excavation was made with a slope of 1V:1H in a material with unit weight g=18.8 kN/m3. Estimate the factor of safety for a depth of excavation of 13 m. The average undrained shear strength along the failure surface is 50 kPa.

Example 4. For the excavation in Ex. 3, estimate the factor of safety if a strong stratum exists at a depth of 13 m.

Example 5. Estimate the factor of safety and location of the critical failure circle when the rigid layer is 7.8 m below the toe of the slope.

Example 6. If a clay has undrained shear strength of 50 kPa and unit weight 20 kN/m3, find the maximum depth a vertical trench can be excavated.

-11- FIELD MONITORING OF SLOPE MOVEMENTS

-12-