geology & geoengineering

Subject CIV3247 Geoengineering 1.1 Topic 1 : Geology and Geoengineering

Department of Civil Engineering, Monash University

Edition Date: 1/2000

TOPIC 1

GEOLOGY AND GEOENGINEERING

TABLE OF CONTENTS PREVIEW.................................................................................................................... 2

Introduction .............................................................................................................. 2 Objectives................................................................................................................. 2

PREFACE .................................................................................................................... 2 Origin ................................................................................................................... 2 Geology and Pedology.......................................................................................... 2 Geomorphology, landforms and landscapes .......................................................... 3 Particulate Mechanics ........................................................................................... 3

GEOLOGY AND CIVIL ENGINEERING................................................................... 3

CONSTRUCTIONAL AND DESTRUCTIONAL PROCESSES.................................. 4 Emerged Coastal Plains ............................................................................................ 4 Tectonic landforms ................................................................................................... 4 Volcanoes and Intrusions .......................................................................................... 5 Zones of Deposition.................................................................................................. 6 Metamorphism.......................................................................................................... 7 Weathering ............................................................................................................... 7 Mass wasting ............................................................................................................ 8 Erosion ..................................................................................................................... 9 Karsts ....................................................................................................................... 9

GEOLOGICAL CYCLE............................................................................................... 9

GEOLOGICAL TIME.................................................................................................. 9

REFERENCES AND FURTHER READING............................................................. 10

REVIEW QUESTIONS.............................................................................................. 10




PREVIEW Introduction All infrastructure and mining projects interact with the ground. The behaviour of soil and rock and their interaction with the infrastructure are therefore vitally important to the design, maintenance, safety and life of the project. Properties of these materials can vary immensely. They exhibit a wide range of behaviour and as a result each major site must be treated in some ways as unique. The design of foundations, tunnels, retaining walls, slopes etc are therefore site specific. Experience gained from other sites must be applied with care and with a thorough knowledge of soil behaviour. A great deal of information relating to the engineering behaviour of rock and soil can be determined through knowledge of the material constituents and of its formation. Topic 1 looks at the many geological processes that form rock and soil. As will become apparent, water has a significant influence on these processes and on the engineering properties of rock and soil. Objectives • To gain knowledge of the various constructive and destructive geological processes

that lead to rock and soil formation • To understand the roles played by the crust, mantle and core and plate tectonics • To obtain an appreciation of geological time and a knowledge of the relative ages • To appreciate the importance of geology to civil engineering construction PREFACE Origin Rocks can be igneous, sedimentary or metamorphic in origin. Their properties are closely related to the minerals that make up the rock, their formation process and weathering condition. Rocks are generally in their strongest state when fresh, and gradually become weaker and softer the greater the weathering. All rock deposits contain joints, faults and other discontinuities that significantly change the behaviour and the properties of the mass. These discontinuities dominate the behaviour of the rock mass. Most soils are formed by physical and chemical weathering of rocks, the process either happening in-situ, leading to residual soils, or involving transport processes, by air, ice or water, followed by deposition in a range of environments, e.g. aeolian, lacustrine, marine, alluvial, glacial. The properties of a soil reflect the material of its origin, its mode of transportation, its depositional environment and its subsequent history. Geology and Pedology Geology is the scientific study of the earth's crust. Pedology is the scientific study of soil (Gk. Pedon - ground), not necessarily for utilitarian purposes.




Geomorphology, landforms and landscapes Geomorphology is the study of the classification, description, nature, origin and development of present landforms. A landform is any physical form or feature of the earth’s crust having a characteristic shape and produced by natural causes; e.g. plain, mountain, slope, dune, plateau. All together, the landforms make up the surface configuration of the earth. Landscape is an assemblage of landforms, generally that can be seen in a single view. Landscapes are being transformed continuously by constructional and destructional processes. Particulate Mechanics One feature that differentiates soil and rock from other engineering materials is the non-continuum nature of the medium, considered at any scale. All ‘soil and rock’, comprise an assemblage or skeleton of individual particles or grains and their behaviour under mechanical stress cannot be described adequately by conventional theories of continuum stress analysis (although it is often used). In addition, the interaction of the particles with water further complicates their mechanical behaviour, and for these two main reasons a new branch of engineering mechanics, known as geomechanics - has developed as the scientific basis of geotechnical engineering. Unfortunately, soil and rock are often considered as two different groups of materials, rather than a continuous spectrum. This has seen the development of two almost separate disciplines in geomechanics : rock mechanics and soil mechanics, both with their own separate classification systems, testing procedures, analysis and design methods. This separation can cause considerable problems when designing in weathered materials that lie somewhere between a "soil" and a "rock". In this subject we will attempt to marry the two approaches. The terms soil and rock will be retained for clarity. The different disciplines have emerged as rock is seen to be essentially an intact material that is intersected by discontinuities (joints, faults etc). Although the intact rock blocks are made up of grains, they are generally strongly cemented together. As the strength of these bonds is (usually) significantly stronger than the strength of the discontinuities, it is the discontinuities that control behaviour of the rock mass. As a result, most rock mechanics problems deal with the mechanics of discontinua. Soil on the other hand is a particular material with only at best very weak bonds between grains. Although discontinuities may exist, they tend to be of a similar strength to the soil. Soil mechanics therefore deals with continuum mechanics, treating the soil as a (usually) homogeneous assemblage of particles. GEOLOGY AND CIVIL ENGINEERING Engineering structures rest on/in the earth’s surface. This surface (landscape) comprising soil and rock is continuously being modified by various constructional and destructional processes. An understanding of the processes and resulting landforms an give us some vital clues to the nature and properties of the underlying materials. Civil Engineering design can accommodate almost any ground conditions which are correctly assessed and understood.




& REQUIRED

Waltham Chapter 1 : pp. 2 - 3

CONSTRUCTIONAL AND DESTRUCTIONAL PROCESSES Constructional • Emerged coastal plains • Tectonic landforms • Volcanoes and Intrusions • Oceans and Floodplains • Metamorphasis Destructional • Weathering (physical & chemical) • Mass wasting (landslips) • Erosion • karsts Emerged Coastal Plains Can be formed through

• major vertical earth movements or epeirogenic movements • isotasy • eustacy – fluctuating sea level (ice ages); e.g. Brighton coastal plain, Tarra

delta, parts of Gippsland plains Tectonic landforms Convection currents within the mantle cause plates to move, creating disturbances along the plate boundaries. These disturbances result in tectonic landforms such as mountains, fault scarps, garbens, tilted blocks, folds etc. Movement of plates results in fracturing (faults and joints) and folding of the rock that forms the surface. Faults are a brittle form of failure, while folds (synclines and anticlines) are ductile (plastic) behaviour. Joints often form due to folding action. Joints :

• are fractures or discontinuities along which no movement has occurred • also form during cooling and along bedding in sedimentary rocks • are weaknesses that affect engineering properties of the rock mass and

therefore must be accounted for in design and construction • are pervasive

Faults :

• joints along which movement has occurred • often contain broken rock (breccia), gauge and slickensides




• types include normal (tension), reverse (compression) and tear • are often significant weaknesses and sources of water that can be of

significant concern to civil engineering works Fractures control the engineering properties of the rock mass. Fracture characteristics that may be important include roughness, aperture, filling, orientation, persistence and weathering. Joints, faults, folds and other geological structures are covered in more detail in Topic 4.

& REQUIRED


Volcanoes and Intrusions Formation processes for igneous rocks : extrusive and intrusive. Volcanoes :

• located at plate boundaries (along with earthquake zones) • result in surface pyroclastic and extrusive igneous rocks • pyroclastic : particles thrown into air during eruption that settle to form ash,

tuff and agglomerate • magma extruded to the surface to form extrusive igneous rocks (lava); e.g.

rhyolite (acidic), andesite & basalt (basic). Cool quickly on the earth’s surface generating fine grained rocks. Often contain open vesicules due to contact with water during cooling. The acidic rocks have comparatively low viscosity and flows poorly. The basic rocks are less viscous and can flow on very gentle slopes over vast areas – e.g. basalt plains of Victoria

Intrusions :

• formed when magma solidifies below the crust. Later exposed by erosion of the earths surface.

• includes batholyths, dykes and sills • cool very slowly thereby generating large grains • include granite, diorite, gabbro, porphyry and dolerite (type depends on

acidity) Minerals :

• quartz, feldspar, muscovite, biotite, mafics (hornblends, augite, olivine)

& REQUIRED

Waltham Chapter 10 : pp. 20 – 21 Chapter 2 : pp 4 - 5




Zones of Deposition Floodplains (alluvial valleys) :

• deposition by streams – size of particle depends on stream velocity • alluvium characterised by rapid changes in materials • floods accelerate erosion, move particles down stream, finer particles left on

the flood plain where currents are lower. • Deltas and coastal deposition (wave action)

& REQUIRED

Waltham Chapter 14 : pp. 28 – 29

Oceans :

• sediment is washed down off the land by rivers and streams and is deposited in the sea. The thickness of sediment builds up over millions of years forming sedimentary rocks.

• subsequent earth movements lift the sea floor above the surface; erosion and transportation then tranform the surface exposing the older rocks below.

Lithification is the process of forming sedimentary rocks from transported soil. Includes three processes :

1. Cementation : filling of voids by mineral cements (silica, iron oxides, calcite or clay)

2. Recrystallization : small scale solution and redeposition of minerals – e.g. in limestone

3. Consolidation : decrease in void ratio due to weight of overburden

Sedimentary rocks : • Usually characterised by bedding (resulting from deposition process).

Bedding often forms a plane of weakness (discontinuity). • Clastic or Non-Clastic • Clastic : Rudaceous (coarse grained e.g. conglomerate, breccia)

Arenaceous (medium grained e.g. sandstone) Argillaceous (fine grained e.g. siltstone, mudstone, shale)

• Non- Clastic : Carbonates (limestone) Non-carbonates (e.g. flint, chert, coal, lignite, ironstone, salt and gypsum)

• Minerals : primarily quartz, but also muscovite, kaolonite, illite, smectite, calcite, dolomite, gypsum, liminite, pyrite.




& REQUIRED

Waltham Chapter 3 : pp. 6 - 7 Chapter 4 : pp. 8 - 9 Chapter 15 : pp. 30 - 31 Chapter 16 : pp. 32 - 33 Chapter 17 : pp. 34 - 35

Metamorphism Igneous and sedimentary rock can undergo metamorphasis when exposed to high temperature (up to about 600oC) and/or pressures (around 500 MPa at 20 km depth). Changes occur in the solid state, with the resulting type of rock depending on the original rock and the temperature and pressure conditions. Types of metamorphism include :

• Regional : involves high temperature and pressure and occurs in mountainous area due to collisions of plate boundaries

• Thermal or contact : involves high temperature only – baking of rock by igneous intrusions

Metamorphic rocks :

• Include marble (from limestone); quartzite (sandstone), hornfel, slate, schist gneiss (clay/mudstone/siltstone depending on type and degree of metamorphism); granites show little change

• Main minerals : quartz, feldspar muscovite, biotite, chlorite, epidote, calcite, kaolinite, limonite

Metamorphism causes :

• recrystalisation (e.g. limestone to marble, sandstone to quartzite) • growth of new minerals (e.g. clay minerals to mica to feldspar and mafics

e.g. basalt to greenstone) • directional pressure causing growth of grains in direction perpendicular to

maximum pressure – foliation (e.g. gneiss) • planar weaknesses – parallel micas cause cleavage and schistosity (e.g. slate

and schist).

& REQUIRED


Weathering Landscapes broken down by chemical and physical processes and erosion




Physical :

• includes temperature changes (freezing and thawing, thermal expansion), crystal growth, pressure, plant roots, burrowing animals

• cause disintegration of parent material and facilitates chemical weathering Chemical :

• always in water • includes hydration, hydrolysis, oxidation, reduction, carbonation and

exchange • examples include oxidation of Fe to form limonite, deposited in joints,

inhibits groundwater flow; hydrolysis of feldspars to form clay (kaolin) – forms infill for joints.

Understanding weathering is vital to knowledge of engineering behaviour Usually more weathered near surface and along joints (why ?). In general the more weathered a rock, the lower the strength. Weathering pattern depends on rock type and fracture pattern. Rocks classified according to weathering status : Fresh (F), Slightly (SW), Moderately (MW), Highly (HW), Completely (CW), Soil. Soil that has weathered from a rock insitu without being transported is called a residual soil.

& REQUIRED


Mass wasting Gravitational movement of weathered rock down slope without aid of water or wind (landslips). Transported material called colluvium. Often triggered by mans activity or by water. Can involve very small to immense volumes of material. Several mechanisms including sliding, toppling, unravelling, slumping, flow.




Controlled by discontinuities (joints, bedding, schistocity, faults etc).

& REQUIRED

Waltham Chapter 32 : pp. 64 – 65 Chapter 33 : pp. 66 – 67 Chapter 34 : pp. 68 – 69 Chapter 35 : pp. 70 - 71

Erosion Sheet erosion – by water flowing down valley sides – severe when vegetation removed and geological materials uncemented Stream erosion – materials brought downslope by mass wasting and sheet erosion are transported by streams. Erosion by streams cause meanders etc. Karsts Forms by dissolution of limestone – limestone is only common rock soluble in water – dissolved by carbon dioxide in rain water. Form highly variable ground conditions. Formation of sink holes – when buried karsts lead to surface subsidence.

& REQUIRED

Waltham Chapter 29 : pp. 58 – 59

GEOLOGICAL CYCLE Continuous cycle of weathering, erosion, transportation, sedimentation to break rock down to soil and then cementation, recrystallization, consolidation, metamorphism or melting to form soil into rock. See page 7 of Waltham. GEOLOGICAL TIME An understanding of geological time is important to determine the history of rocks and the likely subsurface conditions. The stratigraphic column divides geological time into periods (Holocene, Pleistocene, Pliocene … etc). These names are also used to identify




the rock. e.g. the bedrock of Melbourne – a siltstone is often referred to as the “Silurian”. See Waltham page 19. REFERENCES AND FURTHER READING Fookes REVIEW QUESTIONS 1. How old is the surface of the earth as we see it today. What are the main factors that

have contributed to its present form ? 2. Why do the plates move ? 3. Why are areas that are prone to earthquakes and volcanoes close to plate boundaries

? 4. What is the difference between isotasy and eustacy ? 5. Why are granite and basalt the most abundant types of igneous rock? 6. How are igneous rocks classified ? 7. At a depth of 10 km, what temperature (approximately) is required to melt rock ?

What about at the surface ? 8. What is the difference between intrusive and extrusive igneous rocks ? What

characteristic makes them easily identifiable ? 9. What are the main clay minerals and what are their characteristics ? How do you

think these charactersitics would affect engineering structures built on these materials ?

10. What is the most common mineral ? What elements does it contain ? 11. What do we mean by the term rock mass ? 12. Why is rock more highly weathered closer to the surface and along joints ? 13. What type of soil does basalt weather down to ? What about sandstone and granite ? 14. What is the difference in origin between colluvium, alluvium and a residual soil ?

How do you think their appearance and properties would differ ? 15. What are the main characteristics of a glacial till ? 16. How might you identify an alluvium gravel ? 17. What is the difference between a reducing and oxidising environment with regard to

rock weathering. 18. Can you name the periods of the stratigraphic column in order from youngest to

oldest and the time periods they represent ? 19. What problems would be associated with building a dam and reservoir on

limestone?

Subject CIV3247 Geoengineering 2.1 Topic 2 : Three Phase Model, Water and Effective Stress



TOPIC 2 – Three Phase Model, Water

and Effective Stress

TABLE OF CONTENTS PREVIEW 2.2

Introduction ........................................................................................................... 2.2 Objectives.............................................................................................................. 2.2

PREFACE 2.2 CLAY – WATER INTERACTION 2.3

Particle Size and Shape.......................................................................................... 2.3 Influence of mineralogy......................................................................................... 2.3

GROUNDWATER AND SEEPAGE 2.4 SOIL FABRIC – QUANTITATIVE APPROACH – 3 PHASE MODEL 2.5

Some must know definitions : ............................................................................ 2.6 EFFECTIVE STRESSES 2.7

Vertical Insitu Stress .............................................................................................. 2.9 REVIEW QUESTIONS 2.10




PREVIEW Introduction In the Level 2 subject - Introduction to Geoengineering - the emphasis was on treating soil as a granular material without any recognition of the important role played by water. This was done so that basic soil mechanics principles could be grasped more readily. Most engineering soils and rocks however contain water in the pore spaces and variations in the water regime of a soil (and rock) are often the dominating influence on its engineering behaviour. This subject treats soil (and rock) more realistically as a 3-phase material, viz solid grains, pore water and pore air (or gas). The influence of water is investigated through pore water pressure and effective stress, seepage (only briefly) and clay – water interaction. Objectives • To understand the effect that water has on the properties of clay and the role

mineralogy has in this. • To understand the three phase model of soil and rock and to be able to use it to

determine simple properties such as water content, void ratio, porosity, degree of saturation etc.

• To introduce the effective stress principle and to be able to use this principal to determine effective vertical in situ stresses within the ground.

PREFACE Understanding the influence that water has on the behaviour of soil and rock is of vital importance. Most failures in geomechanics involve water in some way or another. For example, most landslips are triggered by water. The strength and deformation properties of soil and rock change significantly with water content. Expansive clays swell and shrink with wetting and drying, often leading to large settlements. Water provides the medium for transporting pollutants through the ground. Water also introduces a time dependency into soil behaviour, with short term behaviour being considerably different to that of long term behaviour. For example, can you explain why a trench dug in clay will be stable for several months and then suddenly cave in, whereas attempting to dig the same unsupported trench in sand is a useless endeavour as the sand caves in immediately. This lecture briefly looks at some of the ways in which water influence soil and rock behaviour. The principle of effective stress will be introduced here, but expanded on in future lectures. Seepage and its role in environmental engineering will be dealt with in CIV3248 Groundwater and Environmental Engineering. However, to fully grasp some of the concepts covered in this subject it will be necessary to read material that will be covered in CIV3248 and contained in Holtz and Kovacs.




CLAY – WATER INTERACTION Particle Size and Shape Soil and rock are made up of individual particles or grains that can be cemented (e.g. rocks) or uncemented (e.g.. soil). Particles of soil may consist of a single mineral (e.g. quartz sand), while granite boulders consist of mineral grains of feldspar, quartz, mafics and biotite. For soil, particle size can vary from Boulders (>200 mm diameter.) down to Clays (<2µm). Most coarse soil grains are observed to be blocky or sub-spheroidal in shape, with freshly created particles being more angular than those which have been subjected to long-term weathering. For engineering purposes and correlations it is sufficient to give each particle a single ‘effective diameter’, the full physical significance of which depends somewhat on its method of measurement. On the other hand clays are seen, under electron microscopic examination, to be largely two dimensional in nature, frequently lamellar or plate shaped and often with striking geometric symmetry. Needle and tubular (rolled lamellae) particles also occur relatively frequently. Although clay particle size and shape are relevant to understanding the interaction of clays with water, they are not normally investigated in soil engineering practice. Influence of mineralogy The lamellar shapes of clay particles reflect the fundamental difference between clays (cohesive) and sands/gravels/cobbles (cohesionless), namely a difference in mineralogy. The cohesionless (coarse grained) soils have particles consisting of primary minerals found in the parent rock, particularly silica for sand grains, while the cohesive (fine grained) soils comprise particles which are the products of chemical weathering, and alteration of these primary minerals to produce so-called secondary minerals, mainly hydrous aluminosilicates. X-ray investigations have revealed the regular crystal structure of clay minerals, in which two basic layer structures - tetrahedral (silica) and octahedral (alumina) sheets - are packed together in a variety of ways to produce a given clay mineral. Substitution of cations with a valency less than that of silicon or aluminium, into the crystal structure produces variant mineral types and contributes to an electrical charge imbalance at the crystal surface. To varying degrees, clay particles behaving as though their faces carry a negative charge, while their edges carry a lesser positive charge. Double Layer : The surface of the clay particle is rich in oxygen atoms that form hydrogen bonds with the closest water atoms. This gives the water a structure that modifies its properties, and such water is known as the adsorbed water layer. Electrical neutrality of the system is provided by cations embedded in this water layer. The attachment of the cations is rather weak, and because they can be replaced by other cations in the free water, they are referred to as exchangeable cations. This double layer in which the adsorbed water close to the particle surface is greatly modified in properties, controls the ease with which particles may move relative to each other and




giving rise to the plasticity of clay soils. The primary factor here is the thickness of the adsorbed water layer that is affected by the valency of the cations adsorbed, and the cation concentration in the free water. In some minerals such as Montmorillonite, water may also penetrate between the crystal layers, causing excessive swelling,. As the interaction of water and cations with the particle is a surface phenomenon, its effects will be more noticeable with a greater particle surface area. As particle size decreases, the specific surface, or surface area per unit mass, increases greatly. The laminar shape further increases the surface area, explaining why clay properties are much more influenced by available water than cohesionless soils.

& REQUIRED

Holtz and Kovacs Chapter 4 : pp. 77 - 107

GROUNDWATER AND SEEPAGE Water exists in the ground. The term “water table” is often used to describe the upper most surface of the water in the ground. For example, consider a container full of dry sand. If we were to add water to the sand it would fill the voids between the sand grains up to a certain height from the bottom of the container ie they would be saturated. The surface that forms the interface between air-filled and water-filled voids is called the water table. This however is a simplistic view. In fine-grained soils, capillary forces “pull” water above this level, such that soil above the water table can still be close to saturated. Plants rely on this effect to obtain water from the soil. A more accurate term is the “piezometric surface” which defines the level at which the pore water pressure is zero (relative to atmospheric pressure). The pressure in the water above the piezometric surface is negative, and can be at a level of tens of MPa of tension (or suction). For stationary ground water the piezometric surface is horizontal and below it the pore pressure increases at a constant rate of 9.81 kPa per metre depth, referred to as a hydrostatic pressure distribution.

& REQUIRED

Waltham Chapter 18 : pp. 36 – 37 Holtz and Kovacs Chapter 6 : pp. 166 - 190

However, if some disturbance to this occurs such as dewatering by pumping, or addition of a load to the ground, then the ground water will flow, and the pressure distribution will not be hydrostatic. The movement of water through soil and rock is called seepage. The majority of seepage in rock is through the discontinuities, whereas in soil it is




through the pores. We assume that seepage is laminar and that it is governed by Darcy’s Law.

Here seepage velocity v m/s is the total volume flow rate of water q m3/s flowing through cross-sectional area A m2 normal to the flow, under a hydraulic gradient i, through a soil with hydraulic conductivity or “permeability”, k m/s. Permeability varies significantly both within the same soil and from soil to soil – with many orders of magnitude separating clay (10-9 - 10-11 m/s) and sand (10-6 - 10-3 m/sec). Hydraulic gradient is the rate of change of total head h in the direction of flow. Total head is identical to that considered in pipe flow, except that the velocity head is so small that it can be neglected without error. For small laboratory samples, change in elevation head will be small, and there the hydraulic gradient is given by just the pore pressure gradient. Note that under hydrostatic conditions, the total head is constant everywhere, and hydraulic gradient is zero, the changes in pressure head cancelling the changes in elevation head. For in rock, most water travels through joints and other discontinuities, the intact rock between the joints being much less permeable. The theory developed for seepage through homogeneous and isotropic soil is not really applicable for such flow in rock, but nevertheless, seepage theory is still extensively used with apparently good success for these materials. Are real soils homogeneous and isotropic? Seepage will be dealt with in second semester in the subject CIV3248 Groundwater and Environmental Engineering. Nevertheless to fully appreciate soil the material presented in this subject it will be necessary to carry out the suggested reading.

& SUGGESTED


SOIL FABRIC – QUANTITATIVE APPROACH – 3 PHASE MODEL The soil fabric, or the arrangement of grains, is often important and sometimes central to explaining soil behaviour. However, in routine geotechnical engineering the soil or rock is reduced to a model comprising three lumped component phases, with all the grains, all the water and all the air collected into single discrete layers.

It is imagined that all the grains in a block of soil or rock are ‘melted’ together and allowed to resolidify into a single solid block. The total original pore space - the space between the grains in the real fabric - is called the void space and is represented by the collected water and air components, and denoted by e. For calculation purposes it is convenient to choose a lump of soil containing a solids volume, Vs, equal to 1 and a cross sectional area of the soil block A=1, so that heights on the model are then equal to volumes (See Figure 1).

kiAvAq ==




Figure 1 : Volumetric relationships

Some must know definitions: (Fundamental quantities: Vv = volume of voids, Vs = volume of solids, Vw = volume of water, VT = total volume, Mw = mass of water, Ms = mass of solids, Ma = mass of air, very small so Ma = 0)

Void ratio, e s

v

VV

e = (usually used for soil)

Specific volume s

t

VV

v = = 1 + e

Porosity, n %100×=T

v

VV

n (usually used for rock)

Water content, w %100×=s

w

MM

w

Degree of Saturation, S %100×=v

w

VV

S

Bulk density, ρ T

ws

T

T

VMM

VM +

==ρ

Solid density, ρs s

ss V

M=ρ (range of 2.3 to 2.8, mean 2.65 t/m3)

Density of water, ρw 1≈=w

ww V

Mρ t/m3

A = 1

AIR

WATER

SOLIDS

VA

VW

VS = 1

VV = e




Dry density, ρd T

sd V

M=ρ (the reciprocal of total volume / solid mass)

Unit weight, γ kN/m3 or kPa/m gργ = where g =9.81m/s2 ( Unit weight of water is γω = 9.81 kN/m3 or kPa/m ) Unit weight may be dry, bulk, saturated etc depending on density value used. Density is measured in tonnes per metre cubed (t/m3) and unit weight in kN/m3. Don’t confuse the two. Using the 1+e model, relationships between the various parameters ρ, ρd, ρs,γ, γd, γs, w, SR, e, n, may be readily calculated. For example :

ee

n+

=1

n

ne

−=

1

es

d +=

1ρ

ρ ( )

( )ews

++

=11ρ

ρ

e

ewssaturated +

+=

1ρρ

ρ S

we

w

s

ρρ

= ρd = ρ /(1 + w)

& REQUIRED


@

Activity 2.1 Work through example problems 2.2 to 2.6 (pp. 16 – 24) in Holtz and Kovacs

EFFECTIVE STRESSES The most important interaction of water with soil (and to a lesser extent with rock) is the role it plays in influencing stress transfer in soil (and rock) deposits. For a dry soil (in which the pores are filled with highly compressible air,) any applied load is immediately transferred to the soil grains, causing an immediate change in dimensions, strength, or stiffness. If the voids are however completely full of water (which is relatively incompressible), the situation is changed dramatically. Applied loads or stresses are then initially carried partly by the soil skeleton and partly by changes in the pore water pressure. Where the soil skeleton is compressible, the pore space full of water ie saturated, and the pore water cannot leave the voids, then the pore water pressure will carry the full change in applied load.




The Terzaghi concept of an effective stress implies that for any applied stress, although part is resisted by pore pressure change, the remainder increases the stresses at the points of contact of the soil grains and is effective in changing the mechanical behaviour of the soil skeleton. Terzaghi formulated his effective stress law as follows.

Effective stress is a measure of the load transfer between particles, and for an uncemented soil cannot be negative ie in tension. (But pore pressure can be negative.) It is not necessary for stress change to have occurred for the effective stress law to be valid. It also applies for the static state in soil deposits that have reached steady state equilibrium with the water table present. Drained versus Undrained For cases of additional applied load caused by engineering operations a further complication arises, namely the effect of time. Clean sands and gravels are so free draining that any pore pressure increases from an applied load will dissipate fully as the load is being applied, so total and effective stresses will be essentially the same. At the other extreme, clays drain very slowly and months or years are necessary for construction induced pore pressures to dissipate noticeably. Hence for clays we have the concept of immediate undrained loading for the initial state and long term, fully drained loading for ultimate equilibrium. Behaviour in both cases is controlled by the effective stress law.

& REQUIRED


The piston and spring analog shown below demonstrates these actions.

THE EFFECTIVE STRESS LAW

σ′ = σ - u

where σ = total or externally applied stress u = pore water pressure σ′ = effective stress (this form of the equation is for soils with SR = 100%)




Vertical In situ Stress The total vertical stress at a depth, d, in a soil or rock mass is obtained by summing the weights of the materials above that point. Usually we divide the soil or rock mass into layers of height, hi, and sum the weights of each layer (as determined from bulk density,

ρi, the gravitational constant, g, and layer height); i.e. If the water table is at depth, z, below the surface, then the pore pressure at depth d is given by

where γw = ρwg. (Note g = 9.81 m2/sec). The effective vertical stress, according to the effective stress law is

The term buoyant weight, γ' is used to describe the effective weight of the soil when completely saturated:

γ' = γ - γw

@


∑ρ=σn

1ii

dv gh

w)zd(u γ−=

uvv −σ=σ′

∑=n

ihd1

TOTAL STRESS applied externally to the sample. (Utilised in all equilibrium calculations)

Water-filled space –PORE PRESSURE, measured by standard pressure gauges

Elastic spring – soil skeleton. Load carried is the EFFECTIVE STRESS

Piston with drain hole. UNDRAINED (no hole or a very small hole) then no water loss, no spring compression and no change in effective stress, pore water carries all the applied load. DRAINED (A) a very large hole, water drains quickly, spring compresses and carries all the applied load. (B) a small hole, and after sufficient time, water drains, spring compresses and carries all the applied load.

Closed-end cylinder

PISTON AND SPRING ANALOG




REVIEW QUESTIONS 1. Describe the structures of Kaolinite and Montmorillonite. How do their structures

affect the properties of these clay minerals? 2. What do we mean by specific surface and how does it affect interaction between

clay minerals and water? 3. Holtz and Kovacs : p 108 ; Questions 4-4, 4-5 and 4-7. 4. Holtz and Kovacs : pp : 41 – 45; Questions 2-1, 2-3 to 2-32 5. What is meant by the term “hydrostatic pressure” ? The Burnley Tunnel at it

deepest point is approximately 60 m below the ground surface. If the water table is at 1m depth, what is the water pressure acting on the tunnel if hydrostatic conditions are assumed ?

6. Holtz and Kovacs : p 274; Questions 7-14 to 7-17

Subject CIV3247 Geoengineering 3.1 Topic 3 : Soil and rock classification and the continuous spectrum



TOPIC 3 – Soil and rock classification and the

continuous spectrum

TABLE OF CONTENTS PREVIEW................................................................................................................. 3.2


PREFACE ................................................................................................................. 3.2

ROCK CLASSIFICATION ....................................................................................... 3.3

SOIL CLASSIFICATION ......................................................................................... 3.3 Particle Size........................................................................................................... 3.3 Particle Shape ........................................................................................................ 3.4 Unified Soil Classification ..................................................................................... 3.5 Strength, Consistency and Density......................................................................... 3.5

REVIEW QUESTIONS............................................................................................. 3.6




PREVIEW Introduction The engineering properties of soil and rock vary enormously depending on origin, stress history, mineral composition, degree of weathering, water content etc. Unlike structural engineers working with steel, we are not operating with only one material, but many different materials. Like the structural engineer, we must have a good knowledge of the properties and likely behaviour of the material. The identification and classification of soil and rock is therefore vitally important so that the engineer has a “feel” for the material he is working with and so that he can apply his past experience, or just as importantly, experience by others, with similar materials. Objectives • To appreciate the diversity of soils and rocks and be able to identify the main types

of each • To understand and be able to use the various classification systems used, and have

knowledge of the various tests used to in the classification process • To appreciate that soil and rock are not distinctly different materials, but are part of

one continuous spectrum in which behaviour gradually changes. PREFACE A scientifically based classification system is absolutely essential for clear and unambiguous contract writing and for the sharing of geotechnical knowledge and experience. Unfortunately no such system exists for both soil and rock, and many systems exists for soil or rock alone. This situation has arisen because of the way in which soil and rock mechanics have developed. One of the major geotechnical problems that faced early civil engineers was providing suitable foundations for structures in relatively weak soils. As the structural loads were often relatively small compared to those imposed by structures today, founding on rock was not usually perceived as a problem. As a result, civil engineers dealt mainly with soil behaviour and soil mechanics. On the other hand, rock mechanics arose out of the mining industry, where the most common problems arose in hard rocks, usually due to discontinuities or high stresses. As a result, soil and rock mechanics have developed as largely two distinct disciplines and have developed their own classification systems, tests etc. This can become a problem when dealing with weathered rocks at the surface, as these materials sit outside the normal envelope of experience, and the application of experience from soil or rock mechanics can lead to widely diverging results. Currently, the two disciplines remain largely separate, but the civil engineer must have a good appreciation of both areas. For example, when designing in a fissured clay, it is important that the orientation and strength of the fissures (discontinuities) be considered. This may often not be appreciated by an engineer dealing only in soil mechanics. Hopefully with time, the distinction between the two areas (soil and rock mechanics) will fade, and classification systems, identification procedures, testing etc will become uniform.




The purpose of this topic is to describe the various classification systems used for soil and rock. As these systems are currently different, the distinction between soil and rock will be maintained. ROCK CLASSIFICATION Due to the diverse characteristics of rock and soil, it is important that a uniform method of classification is adopted. Rock is generally classified according to origin (igneous, sedimentary, metamorphic) and grain size. Generic names such as granite, basalt, siltstone, sandstone, marble etc are used as the basis of this classification. When these materials weather to a soil, the name of the parent material is usually lost, and the soil is classified as either clay, silt, sand or gravel depending on particle size. However, on geological maps the name of the parent material is retained.

& REQUIRED

Waltham Chapter 2 : pp. 4 –5 Chapter 4 : pp. 8 – 9 Chapter 5 : pp. 10 – 11 Chapter 13 : pp. 26 - 27

For engineering purposes, rock is usually also classified on the basis of the strength of the intact blocks and on their weathering condition. Descriptors for strength range from extremely low to extremely high. Rock strength can be assessed through Unconfined Compressive Strength testing, Point Load index testing or by observation. Weathering condition is usually based on observation with descriptors ranging from residual soil, through extremely, highly, moderately and slightly weathered fresh (see Waltham pg 27). As previously discussed discontinuities also play a very important role in rock mass behaviour. A consistent method for describing discontinuities is therefore necessary. The Iternational Society for Rock Mechanics has published a recommended method for classifying rock joints. This method takes into account various factors including discontinuity roughness, alteration, block size, persistence, infill etc. SOIL CLASSIFICATION Particle Size Early soil classifications were based on the most readily observable and measurable parameter of a particle - its size. At one extreme Ayers Rock, or that part of it visible above ground level, could be regarded as part of a very large particle, while all clay and most silt particles are too small to be differentiated with the naked eye. The normal engineering range is from Boulders (>200 mm dia.) down to Clays (<2µm). Soil particle size analyses are usually presented as % by weight of particles finer than a specific size vs particle size in mm (log scale). Examples of particle size distributions




are shown on pages 74 to 76 of Holtz and Kovacs. Note that soil contains a range of particle sizes, so names such as clayey sand or sandy clay etc are necessary. Soil classification according to particle size is based on regular alternation of the numbers 2 and 6.

Soil Type Maximum Size Minimum Size

Boulders >200 200 mm Cobbles 200 mm 60 mm Gravel 60 mm 2 mm Sand 2 mm 0.06 mm Silt 0.06 mm 0.002 mm Clay 0.002 mm

The gravel, sand and silt components may also be subdivided into coarse, medium and fine fractions, again based on 2 - 6 alternation. This size classification is also used to distinguish mudstones from siltstones and sandstones etc.

& REQUIRED

Holtz and Kovacs Chapter 2 : pp. 25 – 33

@

Activity 3.1 Work through example problem 2.7 (pg. 31) in Holtz and Kovacs

Particle Shape Most coarse soil grains are observed to be blocky or sub-spheroidal in shape , with freshly created particles being more angular than long-term weathered ones. For engineering purposes and correlations it is sufficient to give each particle a single ‘effective diameter’, the full physical significance of which depends somewhat on its method of measurement. On the other hand clays are observed, under microscopic examination, to be largely two dimensional in nature, frequently lamellar or plate shaped and often with striking geometric symmetry. Needle and tubular (rolled lamellae) particles also occur relatively frequently. Although clay particle size and shape are relevant to understanding the interaction of clays with water, they are not normally investigated in soil engineering practice.




Unified Soil Classification The widely accepted Unified Soil Classification, based on principles put forward by A. Casagrande, uses particle size distribution and plasticity of fines to give soils a two letter classification, e.g. SW for “well graded sand”, CH for “clay of high plasticity”. In marginal cases double classifications may be used, e.g. CL-ML. Numerical values of plasticity are obtained from the Atterberg Limits tests for the Liquid Limit, wL, and the Plastic Limit, wP, giving the universally used Plasticity Index, IP = wL - wP .

& REQUIRED

Holtz and Kovacs Chapter 2 : pp. 34 – 41 Chapter 3 : pp. 47 - 76

Figure 1 : Casagrande’s plasticity chart

Strength, Consistency and Density Fine grained soils (clays and fine silts) are classified on the basis of consistency, while coarse grained soils (coarse silts, sands and gravels) on density. Consistency is usually related to the “undrained” shear strength of the soil and uses the descriptors very soft (undrained shear strength < 12 kPa), soft (12 to 25 kPa), firm (25 to 50 kPa), stiff (50 to 100 kPa), very stiff (100 to 200 kPa) and hard (>200 kPa). Density is quantified through density index or the “N” value obtained from Standard Penetration Testing (SPT). Descriptors range from very loose (N<4), loose (4 to 10), medium dense (10 to 30), dense (30 to 50) and very dense (>50). The two systems overlap in the hard soil extremely low strength rock range, which can often lead to confusion.

0 20 40 60 80 100

CL-ML

60

40

20

0

IP

WL

CH

CL

ML

OH or MH

A line IP = 0.73(wL - 20)

U line IP = 0.9(wL - 8)

OL or ML




It should be emphasised that the strength of soil and rock can be significantly affected by water. The above descriptors are used for soil and rock in their saturated state.

@

Activity 3.2 Work through example problem 3.1 and 3.2 (pp. 62 & 67) in Holtz and Kovacs

REVIEW QUESTIONS 1. Holtz and Kovacs : from page 41 on; Problems 2-2, 2-33 to 2-37. 2. Holtz and Kovacs : from page 72 on; Problems 3-1 to 3-6. 3. Describe the point load index and the uniaxial compressive strength tests on rock.

Subject CIV3247 Geoengineering 4.1 Topic 4: Geological structures and mapping



TOPIC 4

Geological structures and mapping.

TABLE OF CONTENTS PREVIEW 4.2


PREFACE 4.2 ROCK STRUCTURE 4.3 Horizontal limits to rock type………………………..………………………………4.3 Vertical limits to rock type…………………………………………………………..4.3 Planar features……………………………………………………………………….4.4

Orientation of a plane……………………………………………………………….4.5 Non-Planar Structures………………………………………………………………4.5

GEOLOGICAL MAPS 4.6

The base map……………………..…………………………………………………4.6 Geological boundaries………….…………………………………………………...4.6 Map symbols…………………….……………………… ………………………..4.6 Vertical cross sections………………………………………………………………4.6 The legend…………………………………………………………………………..4.7 Publishing details…………… …………………………………………………….4.8

REVIEW QUESTIONS4.7




PREVIEW Introduction The utilisation of soil or rock as a structural element, in say a foundation or a tunnel, is influenced not only by the type of material, but by its extent before another material is encountered, and the layering and fracturing that both may contain. The dimensions of the extent of a given material type are highly variable, ranging from millimetres to kilometres, and an important aspect of site evaluation is the awareness of this value. Both the material type and its state are the result of the past events at the site, and for sedimentary material particularly, the effect of the material being placed in layers becomes part of the rock, as seen in bedding planes, that is, the rock has structure. With subsequent ground movement, these once horizontal planes may be tilted, fractured or faulted, adding more variables to the rock structure. The geological map is one way of recording the type and structure of geoengineering materials in a region. The usual convention is to show in plan the boundaries of the surface exposures of the different materials present, with symbols to indicate the major structures. This is an objective presentation, in that at any point on the plan, all informed observers would see the material type indicated. Geological maps may show one or more vertical section along designated bearings. These sections are usually supported by data from bore holes, supplemented by inferences based on a theory of the material formation, and so are more subjective than the geological plan. The third significant feature of the map is the legend that provides a geological classification of the material. This groups materials according to their age and material type, often with a specific formation name. Geological maps exist for all the earths surface, although the scale may be coarse in some areas. When evaluating the geoengineering characteristics of any site the geological map is one of the first pieces of available data. At the very least, it will show the range of conditions possible at the site. Objectives • To understand the nature of the structures to be found in soil and rock, and have

knowledge of the basic types of rock structure. • To understand the nature of the information provided by a geological map, to know

how to read this information, and be aware of its limitations. PREFACE The structural properties of soil and rock depend on both the material and its condition, and for rock, the behaviour of the intact rock piece may be quite different for the larger rock mass containing defects such as joint planes or other structural features. Another aspect of a site is the landform, the result of applying the energy of the weather to the rock type and its structure. Hence, from knowledge of landform, weather and rock type, the structure may be inferred.




Each site for development involving extensive construction must be investigated separately. Information provided by the geological map will make the investigation process more efficient by suggesting what is probably there, and the field work needed to confirm it. This lecture looks at the major types of structure found in rock at a macro level rather than the mineralogical texture, and how information regarding rock type and its structure for a given area is presented on a geological map. ROCK STRUCTURE Horizontal limits to rock type The horizontal extent of a specific rock formation is first limited by the amount of material originally created. The basins in which sedimentary rock is formed are finite, but of variable size, as are the batholiths of plutonic rock. Erosion may remove some of the rock to expose underlying material, or deposition may cover part of it with a younger rock. An understanding of the regional geological events may indicate the nature of the boundaries. The slope of the interface may be steep or flat depending on the way the boundary was formed. Vertical limits to rock type The stress effects of a building are negligible at depths below about three times its width, and so site information below depths of more than 200m is rarely needed. For tunnels and caverns, basements and deep foundations, the depth of interest will extend for about three times the width below their deepest point. As for the horizontal extent, the vertical extent of a rock is determined by the process of formation. Because the Civil Engineering depth of interest is relatively small, some rock formations are effectively infinitely deep. A notable exception is the Basalt found in many areas around Melbourne, where in many places it may be less than a few metres thick. (Why?) Conformity Where a sedimentary rock is being formed and the nature of the material placed changes without other interruption, this is a conformable boundary, and is seen in the sedimentary layers in Melbourne Mudstone where sandstone and claystone layers are found in contact with each other. If there is a significant time gap between the two activities then the boundary is an Unconformity. Nonconformity: The creation process of one rock has ceased. With or without a time interval, a different rock type is placed in contact with the older. The Lysterfield Hills to the east of Melbourne are the result of granodiorite intruding into the Melbourne Mudstone, and the contact between the igneous granodiorite and sedimentary mudstone is nonconformable. Other boundaries include the Angular Unconformity where there is a steep slope between rock types, and Disconformity where there is parallel strata on both sides of the surface, but a time gap has allowed erosion on the surface of the older rock.




Planar features Planar features can include bedding planes created during the rock formation, joints due to tensile failure in stressed rock, and faults where significant sliding has occurred. Where minor plutonic activity has forced magma forced into existing joints, Sills and Dykes, are formed in horizontal and vertical joints, respectively. Bedding planes Bedding planes are visual features in some sedimentary rock caused variations in the sedimentary process from time to time, resulting in material of different texture or chemical composition being placed in succession. The planes indicate the surface of deposition, usually in water, but sometimes in air, and can become surfaces of weakness or differential weathering. Older rock is more likely to be tilted from tectonic activity, as seen in the sandstones of the Grampians Range in western Victoria. The younger Hawksbury sandstones around Sydney show minimal tilting. Unless the tilting is extreme, younger material will overlie older. Joints Joints are the most common structural feature of most rocks, and have a major influence on the strength of the rock mass. They are usually caused by tensile stresses. For example when liquid basalt cools and the shrinkage is restrained by the supporting material tensile stress generated can form a distinctive set of vertical joints in a hexagonal pattern. An example of this can be seen at the Organ Pipes National Park on the west side of Melbourne. Joints are generally found as a “set” of near parallel fracture planes at a spacing varying from millimetres to meters. Several sets of joints with different orientations may be present in the rock mass. Joints may be closed, open or filled with other material, fresh (unweathered) or weathered, and rough or smooth. Their persistence or lateral extent may be from tenths to tens of meters. Faults Faults, in contrast to joints, have experienced a shear displacement on the rupture surface. They are much less common than joints but can have a major influence on the character a site. The length of a fault may vary from meters to hundreds of kilometres, and the “throw” or shear displacement from tenths to hundreds of meters. The formation of Port Phillip Bay has been caused by movement of a pair of north - south faults, Selwyn’s near Dromana, and Rowsley near Bacchus Marsh, creating a trough (or garben). The reverse action, of uplift between two faults, forms a ridge (or horst). An example of this is the higher land of the Mornington Peninsular, bounded by Selwyn’s Fault in the west and Tyabb Fault in the east. The movement can involve the crushing of a band of material to form a zone of weakened material. The fault zone may consist of large broken pieces (Breccia), large voids (allowing a ready flow of water), or may be filled with finely ground material (gouge). Most of the large faults near Melbourne are inactive, as seen by the degree of erosion and the infrequency of earthquakes. This is not the case for the faults around the rim of the Pacific Ocean. Various terms used to define fault geometry include:




Foot wall - rock below the fault plane Hanging wall - rock above the fault plane Throw - relative vertical movement across the fault Heave - relative vertical movement across the fault Normal faults are due to tensile stress resulting in the hanging wall moving down relative to the foot wall. A reverse fault is due compression, resulting in the hanging wall moving up relative to the foot wall. Transcurrent, or tear faults occur where shear deformation produces horizontal relative movement. The San Andreas fault in California, USA is a notorious example of this type. Fault movements can cause rocks of different age to be found adjacent to each other. Orientation of a plane The attitude of a plane of weakness relative to the stress direction will determine the effect of the plane on rock behaviour. A plane has two degrees of freedom in that it may be rotated independently about vertical and horizontal axes, and so two directions are required to specify a particular plane. The strike of a plane is the intersection line of the plane with a horizontal plane, and this is often an observable feature. However, it is a clumsy parameter in that for a given strike, additional data is needed to indicate whether it slopes to the left or right The dip orientation values are more concise, the dip being the line of maximum downward slope within the plane. The dip is always normal to the strike, and the plane is defined uniquely by the dip direction and the dip angle. The dip direction is the direction of the projection of the dip on a horizontal plane, and the dip angle is that between the dip and the dip direction.

Inclined plane

Strike

Horizontal plane

Dip Dip direction

Figure 1: Angles defining a plane

The bearing of the dip direction is given by one of two conventions. It can be a single number as the clockwise angle from north usually in degrees, or as [reference direction] [angle] [reference direction]. A bearing of 345 degrees is also N15W or W75N. Note that “north” might be true north, grid north or magnetic north, and should identified.




Non-planar structures Under some conditions, horizontal compressive stress will cause the rock to buckle or fold. The scale of this folding may range from tenths of a meter to hundreds of kilometres. Folding concave up is a syncline, and concave down an anticline. It is also possible for the axis of the fold to be inclined to the horizontal. Where the upper portion of a syncline has been eroded away, it leaves the exposure of an inlier of younger rock surrounded by older rock. Locally, the bedding planes that were formed at the horizontal become inclined. The Melbourne Mudstones between Kew and Warrandyte have been folded into a series of syncline and anticlines, with the fold axes being just east of north. Folds may be as small as a metre or may form synclinal basins hundreds of kilometres wide.

& REQUIRED

Waltham Chapter 6: pp. 12 – 13 Geological structures

GEOLOGICAL MAPS Geological maps are produced at a variety of scales showing the relationship of geological features over different areas. For Australia the common scales of government maps ranges from 1:5,000,000 to 1:25,000. Much of Australia is mapped at 1:250,000, covering one degree of latitude and one and a half degrees of longitude (110 km x 130 km). Maps at 1:50,000 or 1:63,360 (inch to a mile) are available for some areas. The base map The base map on which the geological data is superimposed will contain a selection of topographical features such as height contours, streams and water bodies, and manmade features such as roads, townships and quarries. The Australian or other national map grid may be shown, as well as latitude and longitude and north points. Geological boundaries The surface exposure of different formations or rock types may be shown by both colouring and a letter symbol. These boundaries are determined by surface traverse supplemented by interpretation of aerial photography Map symbols Major planar features such as folds, faults, bedding planes and dykes can be indicated by a symbol, as may other features such as metamorphism where it occurs in localised bands. Vertical cross sections Vertical sections may be shown for typical portions of the map sheet. The section gives an indication the relative order in which materials will be found at depth, their typical relative thickness, and the nature of the interface slope (very steep or very flat?) On very large scale maps it is possible to infer the shape of the subsurface interface or the




structure (stratum) contours from the plan of the rock outcrop and the ground surface contours.

& REQUIRED

Waltham Chapter 7: pp.14 - 15 Geological maps and sections

The legend Maps produced by the Department of Minerals and Energy, Victoria have a legend set out as a table, with each row referring to the one material with the youngest material at the top and the oldest at the bottom. The left most columns can show the geological ages or eras. Moving right, the next columns can indicate the type of formation process ie sedimentary, igneous or metamorphic. Here within the appropriate row and column the colour code and letter symbol for the material may be given. Further to the right can be the Group name eg Brighton Group or Newer Volcanics, then the formation, bed or seam name eg Red Bluff Sands. Not every formation belongs to a group, nor might a group be divided into formations. The right most column will contain a list of the major components of the formation or group and the letter symbol repeated. The letter symbol may be formed with the first letter giving the formation age, and the remaining two an abbreviation of the formation name or rock type eg Dgl for Devonian Lysterfield Granodiorite, found at Cardinia Reservoir and Churchill National Park. Publishing details The publishers and authors of the map have an agenda. If the map was prepared for mining development then the rock formation where minerals were expected may receive detailed treatment, and other areas only cursory treatment. Geology is a dynamic discipline and the way of interpreting the data will change with time. Thus it is important to see who prepared the map and when it was published

& REQUIRED

Waltham Chapter 8: pp.16-17 Geological map interpretation

CONCLUSION Surprises during construction are often due to ignorance rather than error, and a thorough assessment of all the available geological data, especially the relevant geological map sheet, is a good start to a successful project. As well as the map there is the real world, and the geoengineer must be a good observer, and be able to make an initial assessment of any landscape from observation.




REVIEW QUESTIONS Obtain a copy of a 1:250,000 geological sheet and answer the following questions. 1. Who published this map, when, and who prepared it? 2. From the location of this sheet, what is the type of climate (rainfall and temperature)

and type of vegetation expected across this sheet? 3. What are the main topography features in this area? 4. Identify the highest point on the map by a selection of its name, map coordinates,

latitude and longitude, or distance and bearing from a significant named feature. 5. What is the approximate length of the major drainage feature shown on the sheet? 6. What is the youngest major rock type exposed on this sheet? 7. What is the oldest major rock type exposed on this sheet? 8. What and when was the geological activity that formed the bulk of the rock found in

this area? 9. What soil / rock type in this area would create the greatest problem for a Civil

engineering development? 10. What is the major structural feature shown on this sheet, and what effect has it had

on the distribution of material?

Subject CIV3247 Geoengineering 5.1 Topic 5: Site Investigation



TOPIC 5

Site Investigation

TABLE OF CONTENTS PREVIEW …………………………………………………………….... .2


PREFACE ……………………………………………………………..…2

Risk management…………………………………………………………………………..3

SITE INVESTIGATION STRATEGY………………………………… 4

Stages of Site Investigation…………………………………………………………...4 Definition of investigation objective………………………………………………… 4 Evaluation of existing data……………………………………………………………4 Planning and costing of further work…………………………………………………5 Site inspection………………………………………………………………………...6 Subsurface sampling…………………………………………………………… …..7 In situ testing………………………………………………………………………….8 Laboratory testing………………………………………………………………… ..9 Reporting… ………………………………………………………………………...10 Performance check………………………………………………………………… 10

Conclusion…………………………………………………………… 10 REVIEW QUESTIONS………………………………………………..10




PREVIEW Introduction Site Investigation is the process of obtaining the necessary and sufficient information concerning a site, especially but not exclusively, about the subsurface conditions, that will allow the confident design and economic construction of a given project. Sites vary concerning area and location. A small site at a location surrounded by extensive recent development may require minimal site investigation. A large site in a remote area with no previous construction may need a detailed site investigation. It might seem that the construction of a conventional house in a Melbourne suburb, where 1000’s have been constructed previously would need minimal site investigation. However, the most common litigation issue involving home construction is its foundations. The need for a site investigation for the construction of a petrochemical plant on the estuary of a large river is more self evident. Site investigation is an expensive activity, and all that might be produced physically is a written report plus a few litres of soil and rock samples. Along with all costs, the client wishes to keep this cost to a minimum. A site investigation is successful when it allows the project to be completed on time and under budget. Failure to achieve these goals can be due to an inadequate site investigation. It has been suggested that the client pays for the site investigation whether it is performed or not. Site investigation is a data collection operation, and its efficiency improves when maximum use is made of the existing data, and the collection of irrelevant data is kept to a minimum. This efficiency is assisted by using a purposeful strategy of the “assumed site hypothesis”. An educated guess is made about the site, based on the available data, and the field work is restricted to the confirmation of this hypothesis. In most cases the guess is good and a very economical site investigation results. There will be instances when the site differs from what was expected, and more field work will be needed to confirm a revised hypothesis. This is better than an uncoordinated set of activities to see just what might be found. Objectives • To understand the need for, and process of, site investigation. • The ability to plan a site investigation for a given project, on a given site. • To have a knowledge of the techniques available for use in site investigation in

terms of their benefits, costs and limitations PREFACE To commence the design of the geotechnical components of any project it is necessary to have a knowledge to the significant materials beneath a site in terms of their horizontal and vertical extent, and their structural properties of strength and stiffness. In addition, a knowledge is needed of the ground water regime and its relations with the




surface hydrology. There may be small amounts of material, either very weak or very strong which may or may not have been found by the site investigation, with varying relevance to the particular project. In planning a site investigation, both the location of the site and the proposed development must be known. A report about several sites to evaluate the most suitable will be different from that used for the final design at the chosen site. The investigation for 50 kilometres of rural road will be different from that for a 50 storey building on a 60 m square site. Risk management A frequent complaint of construction contractors is that the site investigation has been a waste. This may be because the report agrees with the contractors preconceived notion of the site, his ability to properly interpret the report, or the report did not get it completely right. Rarely would a site investigation retrieve more than a cubic metre of sample. A common early site activity is to move many tens of cubic metres of material and the amount of available site information increases accordingly. In a well-planned site investigation, the new data available during construction would be readily incorporated into the work. Contractual relevance Often the site investigation report is included as part of the contract documentation, and it is important for the report writer to know the consequences of the report. An important issue is who carries the risk when the actual conditions are more difficult than anticipated. Is it the contractor who must complete the project at a reduced profit? An astute contractor may then pad his price to allow for this. Or does the client assume responsibility for the cost of what he will ultimately own? The Observational Method This is a contractual process which acknowledges that site investigation performed at a reasonable cost is less than perfect or complete, and that further information will become available during construction. The investigation identifies the significant performance parameters and these are monitored during construction. Where these parameters are found to differ significantly form expected levels the contract makes provision for the appropriate work.

& REQUIRED

Waltham Chapter 19: pp.38 – 39 Site Investigation

SITE INVESTIGATION STRATEGY The basis of an effective and efficient site investigation is to make all the activity purposeful. By forming an initial hypothesis about the site conditions, further action can be limited to confirming (or refuting) the hypothesis. Where the initial hypothesis is based on extensive existing data and personal experience, the subsequent field work to




confirm it may be direct and limited. Even a wrong hypothesis should indicate the briefest fieldwork necessary to confirm the correct hypothesis. Stages of Site Investigation The stages of site investigation include

• Definition of investigation objective • Evaluation of existing data • Planning and costing of further work • Site inspection • Drilling and sampling • In situ testing • Laboratory testing • Reporting (Has the investigation objective been achieved • Inspection during construction • Inspection of the project in service

DEFINITION OF INVESTIGATION OBJECTIVE The required site investigation will depend on the stage of the project in question (preliminary report / final design report), the scale of the project ($K100 - $M2000), the lateral extent (site 100m x 100m / 50m x 100 km) and the project function (500 MW generator foundations / shipping container storage yard). The size, height and weight of structure and proposed excavations will indicate the type of report required. EVALUATION OF EXISTING DATA Sources of existing data include

• Maps – topographical, geological, from any source and covering the area of interest

• Aerial photographs • Investigation reports from neighbouring sites • History of the area • Personal experience

Maps Topographical maps will indicate the nature of the terrain, making some subsurface conditions more likely and others less likely. Geological maps will indicate the regional type of soil or rock, and the process of its formation, again limiting the range of materials that can be expected at the site. Maps of the area produced for any purpose may provide some insight into the nature of the materials present. Aerial photographs The most useful are taken with a vertical camera and in “stereo” pairs with 60% overlap so that a three dimensional image may be formed. With the necessary resources this image can be quantified to produce contour information. The image of any site will vary with the height of the camera (plane /satellite), time of day, time of year, the year taken,




and the type of film emulsion. Landform, drainage, vegetation and land use can be seen at a variety of scales. In some instances the older photographs may yield relevant information, such as the operation of a landfill site now covered over.. Investigation reports from neighbouring sites In areas of extensive development, it is probable that detailed reports exist for neighbouring sites, and the effort of retrieving them can be well rewarded. History of the area Certain prior activities on the site, such as mining, can have an influence on subsurface conditions. An awareness of the area history will identify such activities, Personal experience The personal conduct of investigations at adjacent sites will allow many insights into the materials at the site in question.

& REQUIRED

Waltham Chapter 20: pp40 – 41 S.I. Desk study

PLANNING AND COSTING OF FURTHER WORK The assessment of the available data will allow the development of an initial hypothesis of greater or lesser reliability depending on the nature and extent of this data. Given a total budget for the project and the influence of ground conditions on its successful construction, the field work budget range will be evident. The minimum activities needed to confirm the hypothesis must be identified and costed, as will those for laboratory work and report preparation. The date when the information is required must be known and faster work in a shorter time will be more expensive. Another significant financial issue is the nature of the site. A benign site where the materials are uniform and strong will involve a much lower site investigation cost than that at more challenging site where the material distribution is complex and many of these are very weak. The graph below indicates the nature of these costs, but the catch is that the nature of the site may not be known before the site investigation is undertaken.

Total Project cost $

Site Investigation Cost $

Challenging site

Benign site

Optimum cost




SITE INSPECTION When the clients has accepted the proposed budget then the engineer responsible for the production of the site investigation report should walk across the site. If the remoteness of the site and / or a small budget does not allow this then a suitable technician must observe and report on the site. Feature of importance that might be seen include

• Local topography. • Vegetation. • Drainage, water bodies. • Rock outcrops, cuttings, excavations. • Regional setting (all of above). • Performance of neighbouring structures. • Access, availability of services. • Geologist traverse • Anecdotal - afternoon in local pub?

Where the site is very steep and / or densely vegetated, the drilling contractor would need to consider this in selecting his equipment and submitting his price. Anecdotal information should always be treated with caution until confirmed from other independent sources. At regional view of the site is important for understanding the previous activities, both geological and recent, that have occurred there. At all sites the presence of buried services must be determined before drilling. Their rupture by any drilling operation is both dangerous and expensive to repair. Geological Inspection For a very large project such as the construction of a large dam, the geology traverse would be the subject of a separate report. As well as considering the materials present and geological processes, this could involve measuring joint plane spacing and orientation, evaluation of land slip potential, and geophysical testing such as seismic refraction and electrical resistivity. SUBSURFACE SAMPLING It is essential to obtain specimens of the significant subsurface materials, firstly for clear identification of what is present, and secondly, if the specimen is suitable, to be used for determining the structural properties of this material. The activities involved in subsurface sampling include

• Identifying where to sample • Exposing the sampling site • Collecting the sample • Identification and transport to the testing laboratory.




Location of sampling points It is desirable to obtain sufficient samples to be able to establish the nature and distribution of the significant materials beneath the site. Horizontally the sampling points would be spread uniformly across the site, except where there was some evidence of a major vertical interface between different types of material. Vertically samples should be taken of every significant layer and at intervals no greater than about 2 m. The sampling should extend down to a depth three times the width of the deepest footing, below the deepest footing level expected. Sampling sites will also be influenced by the ease of access, but this should not prevent sampling at significant points. Sampling over water or on very steep sites is possible but costly. Exposing the sampling site The material at depth to be sampled can be uncovered by any convenient excavation technique. Under crisis conditions a hand shovel or power backhoe can be used, but it is generally more efficient to use a site investigation drilling rig. These are smaller than the units developed for the mining and quarry industries. The latter are optimised for only drilling holes in hard materials, while the former must operate in a variety of materials, sometimes in the one hole, and have the facilities for taking specialised samples of a range of materials. The most basic unit is than hand auger, to be considered for a few shallow holes at inaccessible sites. The drill rig has the ability to push or pull the vertical drill rod over about 2m, and rotate it in either direction at arrange of speeds and torques. Drill rods of one to two metre lengths couple together to allow hole up to 80m to be drilled. For firm to stiff clays and moist sands that allow the hole to form unsupported, the quickest drilling technique is the continuous flight auger that cuts the material and clears it from the hole in the one action. Very soft clays and saturated sands must be provided with some support to prevent them from collapsing. The support can be provided by steel casing that is advanced as the hole is deepened, or by the pressure from a “drilling mud” slurry. The latter is easier to use but can result in contamination of the sample. The continuous flight auger can be awkward in this situation, and just the cutter on plain drill rod is used, with a fluid flush used to remove the cuttings from the hole. With steel casing either air or water forced down the centre of the hollow drill rod can be used as the flushing fluid, and an appropriate pump must be available. With drilling mud, the circulation of the mud through a sediment tank removes the cuttings. In the harder materials, either a “rock roller” bit is needed, or coring with a diamond studded cylindrical bit is used. A fluid flush is needed to remove the cuttings, and in the latter case, water is needed to keep the diamonds cool. Because these activities are somewhat glamorous, highly visible and expensive, site investigation is sometimes referred to as “drilling”. This is a only one of the necessary components of the data collection process, and does of itself provide some information from the character of the cuttings and the behaviour during the drilling.




Collecting the sample Have exposed the subsurface material to the required level and ensured that the surface at the bottom of the hole is free from cuttings and is undisturbed, soil samples can be taken by forcing a steel tube into the undisturbed soil, and with care a sample will be retained in the tube. For firm to stiff clays a seamless steel cylinder 1 mm thick, 50mm to 100mm diameter, and 300mm to 500mm long can be attached to the end of drill rod and slowly forced into the soil by the drill rig. This collects a sample with minimum disturbance, and once the ends are sealed to prevent moisture loss, the sample, protected by the steel tube may be taken to a laboratory for testing to determine the structural properties. These thin walled tube samples are commonly referred to as “undisturbed” as the disturbance level can be of low significance. For sands the resistance to penetration of the thin walled tube sampler is great, and sample loss is common. In this case it is more common to hammer a split spoon sample into the sand. The sample is disturbed but not contaminated with surrounding material. The sampler can be disassembled and the specimen removed and stored in a water tight container, the sampler being reassembled for further use. Where the hammer, sampler, and technique follow a standard process, this operation gives a “Standard Penetration Test” or SPT value that correlates well with the in situ properties of the sand. Where coring is used with the harder materials, a well designed core tube permits the collection of the core. The core when removed from the coring tube should be stored in plastic tubes in padded steel boxes to minimise further disturbance to the sample. A useful feature is to place material such as styrene where any core has been lost. Why? When sampling in the groundwater care is needed to ensure that the water does not damage the sample. Soft clay requires special samplers such as the piston sampler or Bishop sampler.

& REQUIRED

Waltham Chapter 21: pp 42 – 43 Site Investigation Boreholes

Identification and transport to the testing laboratory During the drilling operation, a record or “ field log” must be kept of all activities including the hole location, depth, time and type of all samples collected. The samples themselves should be marked with indelible pen, the tube number recorded or one or more robust tags attached securely. Why? As well as who, where, when, how, and why, and the location of samples, the drilling log should record where changes in material occur as indicated by the cutting or the operation of the drill. As recorded will be the depths at which ground water is encountered or changes in ground water flow occurs.




The samples should be stored out of the weather until they are transported to the laboratory without delay. Often a hole will be retained so that water level depths can be observed at a later period. This should be capped to prevent Bart from falling in. Otherwise it should be backfilled IN SITU TESTING Although laboratory testing can impose precise conditions on a sample and yield meaningful parameters, there is always doubt concerning the possible disturbance to the sample, and more importantly, it representativeness. The mass behaviour of rock and soil is influenced by the defects such as joints or thin seams of weak material, and the very nature of these defects makes them difficult to sample. The alternative is to undertake an in situ test where the test volume is large and undisturbed. A practical difficulty is that it is hard to control the field test conditions, and so even though the sample is relevant, the parameters obtained can have an uncertain meaning. In situ testing is of two kinds, one where some mechanical loading is applied to the soil, the other where a geophysical characteristic such as electrical resistivity is measured. Mechanical Tests Many of the mechanical tests including those listed below are performed in the borehole and may be part of the drilling operation.

Vane shear - undrained cohesive strength of soft clay Hammering - Standard Penetration Test (64 kg) Dynamic cone (9kg) -

pavements. . Lateral presssure tests - Pressuremeter in bore hole, Camkometer, Flat plate dilatometer Other mechanical tests required the development of a significant vertical force, requiring dead weight (kentledge) or ground anchors to provide the reaction, and include

Static (Dutch) Cone penetrometer Trial footing / pile tests, Californian Bearing Ratio

Usually some empirical correlation parameter is needed to relate Geophysical tests These tests involve large volumes of material and can be very useful for detecting the interface between different materials, but their ability to give any quantitative characteristic is poor.

& REQUIRED

Waltham Chapter 22: pp 44–45 S.I.Geophysical Surveys




Laboratory testing Disturbed samples may be subjected to identification tests such as grading analysis and Atterberg limits. The undisturbed samples may be used for these tests, and a representative subset would be used in triaxial and consolidation tests, because the are expensive and of appreciable duration. Sufficient tests should be done to establish confident structural characteristics for the significant materials. Reporting It may be necessary to report to the client during the drilling operation if it the information from this suggests that the actual subsurface conditions differ significantly from the initial hypothesis, and major changes are needed for the filed testing program. Although the contents of the final report will reflect the nature of the project and the site, most reports would cover the following issues in the following order. Introduction - the project , the environment (geology, hydrology, ecology) Data collection (summary) who, when how Bore logs - amalgamation of site and laboratory data, plan of hole locations Table of laboratory and in situ test results Description of subsurface materials and their distribution. Design recommendations – conclusions References and acknowledgements – who helped directly or indirectly Performance check Does the report satisfy the performance objective? Is more work needed, and is there money and time for this work? What is the risk associated with these identified unknowns? CONCLUSION Another view of the site inspection process is that it is there to reduce the number of surprises during construction. Often the geotechnical consultant will be engaged to prepare the site investigation report and then their involvement ceases unless a serious unresolved geotechnical issue arises during construction. Similarly, on completion of construction, it is rare for monitoring of the structure to occur when it is in service, unless it is a very sensitive structure such as a power station, failure has catastrophic consequences such as a large earth dam, or an ordinary structure suffers some structural distress. A successful site investigation minimises the cost and duration, and safety, of a construction project. REVIEW QUESTIONS 1. The building where you live, and those of all your immediate neighbours are to be demolished so that a supermarket can be developed on the site. Outline the site investigation process required for this development.




2. If you can get permission from the owner of the building where you live, and the surface soil is exposed, collect soil samples from the corners of the block. Use a spade to dig down 0.5 m and collect an unmixed sample of about 0.5 l, and seal it in a robust plastic bag wit suitable labelling. Produce a short report on your sampling activity indicating the site and sample locations, Ease / difficulty of excavation, soil water condition, and a description of the soil retrieved, and comments on the uniformity of the site. 3. As a drilling contractor, make a recommendation on how you would obtains soil samples needed for the construction of an overpass at the Princes Highway – Wellington Road intersection.

Subject CIV3247 Geoengineering 6.1 Topic 6 : Engineering Geology of Melbourne and the Yarra Valley



TOPIC 6

ENGINEERING GEOLOGY OF MELBOURNE AND THE YARRA VALLEY



REQUIRED READING ............................................................................................ 6.2

Subject CIV3247 Geoengineering 6.2 Topic 6 : Engineering Geology of Melbourne and the Yarra Valley



PREVIEW Introduction The engineering geology of Melbourne and the Yarra Valley is of paramount importance to the practice of geotechnical engineering in Melbourne. In this topic, the main aspects of the geological setting of Melbourne and the geology of the Yarra Valley are described. Objectives • To gain knowledge of geological setting of Melbourne including statigraphy,

structure, hydrogeology and geomorphology • To understand the geological development of the Yarra Valley and how it has

influenced the development of Melbourne. REQUIRED READING There are many papers and text that describe the geological setting of Melbourne. The following papers from the the book “Engineering Geology of Melbourne” edited by Peck, Neilson, Olds and Seddon; Balkema (Pubs) have been included on the CD for your benefit. Consider them to be required reading. 1. “Outline of the stratigraphy of the Melbourne region” by Archbold 2. “Geological structure” by Granger 3. “The hydrogeology of the Melbourne region” by Lane, Lakey and Leonard 4. “Geomorphology of the Melbourne region” by Joyce 5. “Silurian and Lower Devonian” by Sanders 6. “The Mount dandenong Volcanics” by Birch and Wilson 7. “Older Volcanics – Geology” by Anderson 8. “Brighton Group – Geology” by Kenley 9. “The newer volcanics” by Dahlhaus & Rourke 10. “Geology of the Yarra Delta” by Neilson

Subject CIV3247 Geoengineering 7.1 Topic 7 : Consolidation and the Oedometer Test



TOPIC 7 – Consolidation and the Oedometer Test

TABLE OF CONTENTS PREVIEW.................................................................................................................... 2


PREFACE .................................................................................................................... 2 Components of Deformation..................................................................................... 2 Consolidation............................................................................................................ 3

THE OEDOMETER TEST........................................................................................... 4

CONSOLIDATION SETTLEMENT............................................................................ 7

SEDIMENTARY ROCK FORMATION...................................................................... 9

REVIEW QUESTIONS................................................................................................ 9




PREVIEW Introduction The determination of ground movements is one very important aspect of designing structures, excavations etc in soil and rock. After all, if the structure becomes unserviceable or cannot fulfil its function due to excessive movements, then it has effectively failed even though the ultimate limit state has not been reached. Soil and rock undergo both vertical and lateral movements due to construction activities within these materials. For example, loads imparted by buildings to foundations will cause predominantly vertical movements or settlement, excavations will cause stress relief that will generate both lateral and vertical movements of the ground. Such movements may be localised (e.g. due to a footing) or occur over a large area (e.g. from addition of fill over extensive areas or by dewatering or oil and gas removal). For the current topic we will confine the discussion to vertical settlements over a large area. Such settlements may be considered to be one-dimensional. The total settlement that occurs can be split into 3 components, immediate or elastic settlement; consolidation settlement and secondary compression or creep. The last two components are time-dependent. In clays the consolidation and creep settlements dominate. This topic deals with consolidation settlements. Creep settlements are covered in Topic 8. Objectives • To understand the process of consolidation and how it affects soil properties • To become familiar with normally consolidated and over-consolidated behaviour,

and the properties that help estimate consolidation settlement in these materials • To be able to interpret the Oedometer test, be aware of its limitations and to apply

the results to the estimation of settlement PREFACE Components of Deformation As with any material, when a load is applied to soil/rock (e.g. by a footing), the soil/rock will deform. Deformation occurs from :

1. deformation of the soil/rock grains 2. compression of air and water in the voids 3. drainage of water and air from the voids (with reduction in void volume). For saturated clay, the first two components of deformation are small and can be ignored. Since clay is relatively impermeable, the water can only drain very slowly from the voids. Therefore, since the porewater is relatively incompressible compared to the clay skeleton, the porewater initially carries the any increase in applied load a (total stress change) and an increase in porewater pressure results. This change is added to the pore pressure present before the load increase occurred. At the instant of load application the clay is undrained, but water slowly squeezes out because the pore pressure is no longer in equilibrium. With this flow the porewater pressure decreases




back towards the preload value (often hydrostatic). The soil skeleton now carries an increasing proportion of the applied load, and is able to deform as the water flow from the voids allows a smaller void volume. Ultimately the clay is in the drained state and its density increases. This time dependent process is called consolidation. Consolidation represents the deformation (volume change) that occurs between the undrained and fully drained states.

For saturated sand, the same consolidation process occurs. However, as the permeability of sand is much greater, the process occurs almost instantaneously. As a result, consolidation is usually ignored, and only drained parameters are considered.

For saturated rock, consolidation also occurs. However, it is not as dominant as it is soil (and becomes even less so as the rock gets stronger and harder). In strong rock, the void ratio (and therefore the amount of water in the voids) and compressibility of the skeleton are small, and the first two components of deformation dominate.

For unsaturated soil and rock, consolidation still occurs, but the compression of air in the voids allows the process to occur at a much more rapid rate. Consolidation Consolidation occurs when soil is subjected to an increase in effective stress; wherein water is expelled (or drains) from the soil causing a reduction in soil volume. This may be due to an increase in applied total stress from: • foundation loading • placement of fill • deposition of soil (leading to sedimentary rock formation) More subtly, a reduction of the environment pore pressure will initiate consolidation. Ground water extraction by pumping, leading to a lowering of the water table, is the major cause of regional ground settlement world wide.

Swelling (the opposite of consolidation) occurs when there is a decrease in effective stress and water moves into the soil, increasing the soil volume. For example : • due to excavation or erosion reducing the total stress in the remaining soil • due to raising of the water table and increasing the pore pressure.

& REQUIRED


Consolidation can therefore be defined simply as the time dependent removal of water from soil by drainage brought about by effective stress changes. As soil consolidates it reduces in volume and becomes stronger; both have important ramifications with respect to the stability and deformation of structures built in or on soil. (For example the bearing capacity and settlement of footings for a building). Consolidation should not be confused with compaction. Compaction is the reduction in volume of soil by the compression and removal of air from the soil by mechanical means.




& SUGGESTED


THE OEDOMETER TEST When a soil layer is loaded vertically over a large area (e.g. by filling), the displacement of the soil by consolidation around the centre of the fill can be assumed to occur only in the vertical direction; i.e. in one dimension only. (Why ?) We can simulate this in the laboratory by the oedometer test (see Figure 1).

Figure 1 : Schematic of floating ring Oedometer Apparatus An undisturbed soil sample (diameter = 75 mm, height, Ho = 20 mm) is carefully trimmed and placed into a rigid (steel) confining ring. Porous stones of slightly smaller diameter (to eliminate side friction) placed on top and bottom of the sample confine the soil but allow water to pass freely into and out of the soil. The assembly is placed in the oedometer base, the loading plate placed in position and the base filled with water, ensuring that the top porous stone is completely covered. A sequence of loading increments is then applied to the specimen, and measurements of deformation with time are taken during each increment. Figure 2 shows a typical displacement versus time plot for one load increment from 10 to 20 kPa. The height of the sample and the void ratio at the start of the increment were 17.25 mm and 2.15 respectively.

Figure 2 : Displacement – time results from Oedometer test on clay

Soil specimen

Porous stone

Loading Plate

Load

Porous stone

Water Confiningring

Base

00.5

11.5

22.5

3

0 500 1000 1500

Time (min)

Dis

plac

emen

t (m

m)

.




As the sample is loaded, water is forced from the voids, resulting in a reduction in volume, height and void ratio. As the cross-sectional area of the sample is constant and volume change occurs through void reduction, the void ratio of the sample is directly proportional to the sample height. Each load increment is maintained until equilibrium is reached, with little or no further deformation; i.e. at end of primary consolidation, which is defined as the stage where all excess pore pressures have dissipated. (Note that creep (or secondary consolidation) will cause further displacements to occur and will therefore cause problems with the selection of this point). For the example above, consolidation has ceased at approximately 500 minutes with a displacement of 2.4 mm. Standard methods for determining the end of primary consolidation will be dealt with in Topic 8. At the end of primary consolidation, the final load applied can be interpreted in terms of effective stress (since all excess pore water pressures have dissipated). The load and deformation data at the end of primary consolidation for each increment can be plotted on the one graph to give the overall stress/strain behaviour of the soil. This can be carried out in a variety of ways: plotting percent consolidation or vertical strain against final effective consolidation stress, p’

f, or more usually as final void ratio, ef, against log p’

f. Convention dictates that the effective consolidation stress is plotted on a log axis.

( )( )e ee H H

Hf ii f i

i= +

+ −1

Where (i) and (f) subscripts refer to initial and final values for the increment respectively. For the time vs displacement plot shown in Figure 2, p’

f, =20kPa, Hf

=17.22mm and ei = 2.15, then Hf - Hi = -2.4mm and hence ef =1.71. The graph so obtained (by plotting ef against log p’

f. for each load increment) is essentially the stress-strain relationship for the soil. A typical result from an oedometer test on soft clay is shown in Figure 3. The soil specimen is initially at a point to the left of A. At point A the load has been increased to 2.5 kPa. The load is gradually increased through points B (5 kPa) to point C (80 kPa) where the soil is unloaded to point D (5 kPa) and then reloaded to point E (160 kPa).

0.5

1

1.5

2

2.5

3

1 10 100 1000

Effective consolidation stress, p ' (kPa)

Voi

d ra

tio, e

AB

CD

E

Figure 3: Typical e - log p curve for clay.




Note the following : • the approximate straight line behaviour between points AB, BC, CD and CE. The

lines AB and CD have similar slopes, as do lines BC and CE. Point B is an indication of the maximum vertical overburden stress that this sample has experienced sometime in the past, (approx. 5 kPa in this case) and is called the preconsolidation pressure. It is denoted by the symbol p’c. Due to the curved transition between AB and BC, the determination of preconsolidation pressure can be rather subjective. However, standard techniques are available for estimating this value (see required reading).

• Soils have a “memory” of the stress and other changes that have occurred during their history, and these changes are preserved in the soil structure (Casagrande, 1932). Point B represents the point at which the soil is loaded beyond what it has experienced in the past and the structure of the soil starts to break down and deform “plastically”. If the load is removed (e.g. point C to D) the soil rebounds “elastically”. If load is once again applied, the soil will behave in an elastic manner until it once again reaches it maximum past stress (point C in this case) at which point plastic behaviour will recommence.

• Disturbance during sampling can drastically affect the shape of the e – log p’ curve. It is very important that undisturbed samples are collected for testing.

• The initial flatter portion of the curve (section AB) is termed the reconsolidation (or recompression) curve, and the part after point B is called the virgin compression (or consolidation) curve (BCE).

• The soil is called normally consolidated (N/C) when the existing effective vertical overburden pressure lies on the virgin compression curve.

• The soil is called overconsolidated (O/C) if the preconsolidation pressure is greater than the existing overburden pressure. This may occur because of removal of overburden by erosion or excavation, change in pore water pressure from changes in water table (pumping) or artesian pressure or desiccation due to surface drying. Recent soils tend to be N/C, older (and residual) soils tend to be O/C.

• The soil is called underconsolidated if it has been recently deposited and is not yet in equilibrium; i.e. it is still consolidating, with pore water pressures greater than the hydrostatic.

• The overconsolidation ratio (OCR) is the ratio of the preconsolidation stress to the existing vertical effective overburden stress. That is,

For normal consolidated soils : OCR =1 For overconsolidated soils : OCR >1 For underconsolidated soils : OCR <1

• The slope of the virgin compression line (BCE) is called the compression index, Cc and is given by :

• Cc is independent of stress level. Typical values range considerably. However Terzaghi and Peck (1967) suggested the following correlation with liquid limit (LL) for clays of low to medium sensitivity:

′′

−=

1

2

21c

pp

log

eeC




• The slope of the reconsolidation line (AB or DC) or recompression index, Cr, and

the swelling line (CD) or swelling index, Cs, can be determined in a similar manner, the only differences being that they refer to different portions of the curve. Cr is usually assumed to be approximately 5 to 10 % of Cc with typical values in the range 0.015 to 0.035. The lower values are for clays of lower plasticity and low OCR.

• For non-linear portions of the curve (especially for O/C clays), we define the coefficient of volume change, mv, which can be determined from :

• Eoed is the constrained or oedometric modulus and is a measure of the one-dimensional stiffness of the soil. Note that mv is dependent on stress level, it is not a fundamental soil constant. Typical values vary from 0.4 to 0.8 MPa-1 for (N/C) soft clays to 0.1 to 0.2 MPa-1 for (O/C) stiff clays

• The parameters Cc, Cr and mv are useful for determining consolidation settlements. However, we must make allowance for three dimensional effects where appropriate.

• The behaviour of soil changes markedly from N/C to O/C behaviour

& REQUIRED


@


CONSOLIDATION SETTLEMENT Consolidation settlements in the field result from a change in the void ratio of the soil due to a change in effective stress. Such effective stress changes can occur through the placement of a surcharge (e.g. fill or foundations) or the lowering of the water table. If this settlement occurs over a large area, then the settlement can be considered to be one-dimensional and therefore simulated reasonably accurately by the oedometer test. The settlement, s, can be calculated from (Holtz and Kovacs pp. 309-310)

oo

Hee

s+∆

=1

( )( )( ) oedoo

v Eeppee

edpde

m1

11 12

21 ≈+′−′

−≈

+−

=

)10(009.0 −= LLC c




The values of Cc and Cr or mv define the change in void ratio (or volume of voids) with effective stress change. For a normally consolidated soil (referring to Figure 4 and remembering that p is plotted on a log scale)

i.e. the change in void ratio is simply the stress change multiplied by the slope of the virgin compression line. Settlement is therefore given by :

′′

+=

1

2

1

1 log1 p

pC

eH

s c

For an over consolidated soil (referring to Figure 5), the change in void ratio will depend on whether or not the stress change takes the soil past its preconsolidation pressure and into the normally consolidated range.

′′

=−=∆1

212 log

pp

Ceee c

e

p′c p′

e1

e2

p′1 p′2

Cc

∆e

∆p′

Figure 4 : Settlement of normally

consolidated soil

e

p′c p′

e1

e2

p′1 p′2

Cc

∆e

∆p′

e

p′c p′

e1

e2

p′1 p′2

Cc∆e1

∆p′

∆e2

Cr

cppp ′≤′∆+′1 cppp ′≥′∆+′1

Figure 5 : Settlement of over consolidated clay




For cppp ′≤′∆+′1 the soil remains in the over-consolidated range and

1

2

1

1

1

1

1

1 log1

log1 p

pe

HC

ppp

eH

Cs rr ′′

+=

′′∆+′

+=

For cppp ′≥′∆+′1 the soil becomes normally consolidated and both Cr and Cc are used to estimate settlement :

c

cc

r pp

eH

Cpp

eH

Cs′′

++

′′

+= 2

1

1

11

1 log1

log1

Sometimes the degree of overconsolidation or the change in stress may vary throughout a layer. In such a case you will need to divide the layer into a number of sublayers (of the same or differing height), calculate the settlement of each sub-layer in turn and then sum the settlements to obtain the overall consolidation settlement. This is best done using a spreadsheet. Refer to Topic 18 for further details.

& REQUIRED


@


SEDIMENTARY ROCK FORMATION Consolidation is one important component of the formation of sedimentary rocks such as siltstone and sandstone. These rocks are essentially reconsolidated and cemented soils. Many behave as highly overconsolidated soils. REVIEW QUESTIONS 1. Briefly explain the difference between consolidation and compaction. 2. Do you understand what is meant by the terms over consolidated, normally

consolidated, preconsolidation pressure, virgin compression line, recompression line and unconsolidated.

3. Does sand experience consolidation settlement. How would we estimate it ? 4. What do we mean by the term excess pore water pressures ? Can excess pore water

pressures be negative ? If so, what form of construction activity would generate such pressures ?

5. Holtz and Kovacs : from page 367 on; Problems 8-1 to 8-20, 8-23, 8-24, 8-27 and 8-29.

Subject CIV3247 Geoengineering 8.1 Topic 8 : Time Rate of Consolidation



TOPIC 8 – Time Rate of Consolidation



PREFACE ................................................................................................................. 8.2

TERZAGHI’S ONE-DIMENSIONAL CONSOLIDATION EQUATION. ................ 8.2

DETERMINATION OF CV ....................................................................................... 8.4 Displacement vs log(time) – Casagrande (1938) .................................................... 8.4 Displacement vs squareroot (time) – Taylor (1948)................................................ 8.5

SECONDARY COMPRESSION............................................................................... 8.6

REVIEW QUESTIONS............................................................................................. 8.7




PREVIEW Introduction In Topic 7, the time dependent behaviour of consolidation was introduced. The oedometer test was also introduced as one method to determine the likely volume changes that resulted from the consolidation process. The oedometer test can also tell us how long it will take for consolidation to occur. This is an important practical problem, because it is essential that we know how fast a structure will settle. Objectives • To understand Terzaghi’s one-dimensional consolidation theory and be able to apply

it to practical problems • To be able to interpret oedometer tests to determine properties for estimating the

rate at which consolidation occurs • To differentiate between primary and secondary consolidation and be able to

estimate secondary consolidation settlement. PREFACE Consolidation results from the dissipation of excess pore water pressures which generates movement of pore water within the soil. The amount of water that is squeezed out is directly proportional to the amount of excess pore water pressure that is dissipated. It follows then, that the rate of consolidation is directly related to the rate of excess pore pressure dissipation. TERZAGHI’S ONE-DIMENSIONAL CONSOLIDATION EQUATION. As introduced in Topic 2 and will be dealt with in the subject CIV3248 Groundwater and Environmental Engineering in Semester 2, the movement of pore water is governed by Darcy’s law. Darcy’s law tells us that the quantity of flow depends on the hydraulic gradient and the permeability of the soil. By equating the volume change of the soil due to water egress with the volume change of the soil due to change in effective stress, it is possible to derive the following differential equation which governs one dimensional consolidation. Here u is the pore water pressure, cv is the coefficient of consolidation (a soil “property”), t is time and z denotes the position where u is determined - u is a function of both z and t.

This equation was first suggested as a model of consolidation by Terzaghi in 1925. It can also be derived in three dimensions, but for most practical applications the one-dimensional equation is usually assumed. Note that the rate of consolidation is governed by the coefficient of consolidation not the permeability. cv depends on the permeability, k, and compressibility, mv, of the soil. This implies that for a soil and rock with the same permeability, the pore pressures will dissipate faster in the rock because it has lower compressibility (or greater stiffness)

vwv m

kc

γ=

tu

zu

c 2

2

v ∂∂

=∂∂




Terzaghi’s consolidation equation can be solved using analytical or numerical techniques. The solution obtained depends on the boundary conditions. For the oedometer test, with a sample of height, 2H, the boundary conditions are : 1. Complete drainage at top and bottom of sample; i.e. u = 0 at z = 0 and z = 2H 2. The initial excess pore water pressure ∆u = uI is equal to the applied stress

increment ∆σ; i.e. when t = 0, ∆u = uI = ∆σ = (σ′2−σ′1). Note the length of the longest drainage path is half the sample height, H, since drainage can occur to both the top and bottom of the sample.

The solution is obtained as a Fourier series which can be conveniently expressed in the following form :

Where Uz is the degree of consolidation at time t at depth z, and T is a non-dimensional time factor. Uz and T are given by :

where Hdr is the length of the longest drainage path. The solution is shown graphically in Figure 1.

Figure 1 : Time rate of consolidation curves for two way drainage This diagram can be used to determine the degree of consolidation throughout the layer. For example, the degree of consolidation, Uz, at mid height of an oedometer test for T=0.2 (Point A) is approximately 23%. However at the same time (and time factor) at other locations, Uz is different. At z/H=0.5, Uz = 44% and at z/H=0.1, Uz = 86%.

0

12

1

2

112

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Consolidation ratio, U z

Z =zH

T=0T=0.05

0.100.15

0.30

0.40

0.50

0.70

0.80 0.848

0.90

0.60

0.20

A

( )TfHz

f1U 20n

1z ∑∞

=

−=

i12

1

21

1z u

u1

eeee

U −=σ′−σ′σ′−σ′

=−−

= 2dr

v Ht

cT =




Note that other boundary conditions have different solutions. These can be obtained from most reference books that deal with consolidation. In most settlement calculations we are interested in the average degree of consolidation of the entire layer rather than the point values given above. This can be calculated by finding the area under the appropriate T curve in the above diagram. For problems in which initial excess pore pressure is constant throughout the consolidating layer, the following values apply. Uav 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95 1 T 0.008 0.031 0.071 0.126 0.197 0.287 0.403 0.567 0.848 1.163 ∞

Some useful approximations (Cassagrande, 1938; Taylor, 1948) are:

Uav > 60% T = 1.78 − 0.933log(100−Uav%)

2

2

100%

44%60

==< av

avavU

UTUππ

In terms of settlements : where s(t) is the settlement at any time, t, and sc is the total primary consolidation settlement at t100.

& REQUIRED

Holtz and Kovacs Chapter 9 : pp. 376 – 395 Appendix b-2 : pp. 683 - 690

@


DETERMINATION OF CV The coefficient of consolidation can be determined from the oedometer displacement and time data or from the consolidation phase of a triaxial test. Only the oedometer test will be dealt with. The displacement and time data need to be plotted in the form of displacement vs square root (time) or displacement vs log(time). Displacement vs log(time) – Casagrande (1938) 1. Determine end of primary consolidation tp (or t100) and d100 by plotting tangents to

curve- see below.

cav s)t(sU =




2. Determine true origin of test; i.e. d0 and t0 by the following process : Choose any two times t1 and t2 where t2 = 4 t1. Determine d2-d1, the settlement increment over the period t2-t1. d0 is located distance d2-d1 above d1.

3. d50 (or U=50%) is located midway between d0 and d100 . t50 can be read directly of the log(time) axis.

4. Determine cv from

Displacement vs squareroot (time) – Taylor (1948) 1. Draw straight line through

initial data points. Project line back to zero time to establish d0.

2. Draw second line from d0 with abscissas 1.15 times larger than that of the first line.

3. The intersection of this second line and the laboratory curve defines d90 and t90.

4. Determine cv from

50

2dr50

2dr

v tHT

tTH

c ==

90

2dr90

2dr

v tHT

tTH

c ==

0

0.5

1

1.5

2

2.5

3

0.1 1 10 100 1000 10000

Log time (mins)

Dis

pla

cem

ent

(mm

)

U=100 %

t p

Primary consolidation

Secondarycompression

t 1

d 1d 2

t 2 = 4t 1

U=0 %

U=50 %

t 50 =13.6 min

∆logt

∆e

Figure 2 : Displacement versus log time

0

0.5

1

1.5

2

2.5

3

0 10 20 30 40

Square root time, t0.5 (mins)

Dis

pla

cem

ent (

mm

)

d0.15d

t 90

t 100

d 90

0.1 d 90

Primaryconsolidation

Secondarycompression

Figure 3 : Displacement versus square root time




Note : 1. cv determined from root time method is usually slightly greater than that determined

from the log time method. 2. cv is not constant, but depends on stress level. 3. For pressures less than the preconsolidation pressure, consolidation occurs quite

rapidly, and cv can be relatively high. However, interpretation can be quite difficult, as the displacement time graphs do not have the “classical” shapes shown earlier. Typical values of cv for soft clays range from 0.1 to 0.5 m2/year and gradually increase with OCR. Due to soil fabric, roots, small sample size, etc; cv measured in the laboratory tend to be less than those measured in the field.

& REQUIRED


@

Activity 8.2 Work through example problem 9.9 (pg. 403) in Holtz and Kovacs

SECONDARY COMPRESSION Secondary compression or creep occurs at a much slower rate than primary consolidation and differs from primary consolidation in that it occurs at a constant effective stress. Nevertheless, creep settlements can be very large. One method of estimating creep settlement is through the secondary compression index Cα.

where ∆e is the change in void ratio between times ta and tb, and ∆logt =log tb - log ta. Cα is usually estimated over one log cycle of time; e.g. from ta = 100 to tb =1000 (or from 1000 to 10000 etc), then ∆logt = 1 and Cα = ∆e. See Figure 2. The modified secondary compression index, Cαε, , giving strain rather than void ratio change, can also be defined as

te

Clog∆∆

=α

peC

C+

=1

ααε




Where ep is the void ratio at the start of the linear portion of the e vs logt curve. (e0 is often used with little loss of accuracy). Both Cα and Cαε can be determined from the slope of the straight line portion of the displacement vs log(time) curve following the end of primary consolidation. Typical values of Cα /Cc :

organic soft clays : 0.05 +/- 0.1 inorganic soft clays : 0.04 +/- 0.1 sands : 0.015 – 0.03

& REQUIRED


@

Activity 8.3 Work through example problem 9.10 – 9.12 (pp. 410 - 423) in Holtz and Kovacs

REVIEW QUESTIONS 1. Which properties of a soil govern the rate at which excess pore water pressures

dissipate ? 2. For the following construction activities, plot on the same graph the expected

variation of pore water pressure (on the vertical axis) with time (on the horizontal axis) for a point in the soil immediately beneath the construction activity. Assume that the initial pore water pressure is the same in all cases and that all activities cause the same local magnitude change in pore water of ∆u. ∆u can be positive or negative depending on the activity. (a) Addition of extensive fill overlying a deep deposit of clay (b) Addition of extensive fill overlying a deep deposit of sand (c) Addition of extensive fill overlying a shallow layer of clay overlying sand (d) Footing on deep deposit of clay (e) Excavation in deep deposit of clay (f) Lowering of ground water table in clay (g) Raising of ground water table in clay

3. Holtz and Kovacs : from page 423 on; Problems 9-1 to 9-36

Subject CIV3247 Geoengineering 9.1 Topic 9 : Shear Strength of Soil and Rock



TOPIC 9 – Shear Strength of Soil and Rock



PREFACE ................................................................................................................. 9.2 The Triaxial Test ................................................................................................... 9.2 Influence of Confining Pressure............................................................................. 9.3

MOHR-COULOMB YIELD CRITERION................................................................ 9.4

HOEK AND BROWN CRITERION (1980) .............................................................. 9.6

JOHNSTON CRITERION......................................................................................... 9.6

HOEK AND BROWN (1997) CRITERION.............................................................. 9.7

INFLUENCE OF ANISOTROPY ............................................................................. 9.9

REFERENCES AND FURTHER READING.......................................................... 9.10

REVIEW QUESTIONS........................................................................................... 9.10




PREVIEW Introduction Strength is important for the prediction of performance under load. For example, without knowledge of the strength of soil to be used in an embankment, it is not possible to determine a safe batter angle. Soil and rock are significantly stronger in shear than in tension. Normally, tensile strength cannot be relied upon and is assumed to be zero. Shear strength is therefore of major interest. Topics 9, 10 and 11 cover shear strength of soil and rock. This topic introduces the general shear strength behaviour of these materials and describes a number of common strength criteria used in geomechanics. Topics 10 and 11 investigate the influence of pore pressures on strength behaviour. Objectives • To understand the basic principles of triaxial testing, principal and Mohr circle

stress plots and strength envelopes • To gain knowledge of the common failure criteria used in geomechanics • To understand the influence of discontinuities on the mass strength behaviour of soil

and rock. PREFACE In all forms of engineering science, it is important to be able to predict the strength of the materials involved so that an adequate factor of safety may be maintained against material failure. Geotechnical engineering is no exception. Unfortunately, unlike man-made materials, e.g. steel, the strength of soil and rock vary over many orders of magnitude - from a few kPa for very soft soils to greater than 300 MPa for very hard rocks. Even two adjacent samples of the sample soil or rock from the one location can have significantly different strength properties. Strength also varies with test method, sample condition (disturbed or undisturbed, saturated or partially saturated etc.) test rate, operator experience and so on. The Triaxial Test In most cases we are interested in the shear strength of soil and rock. One common method of determining shear strength is by triaxial testing. This test involves loading a cylindrical sample in compression. A confining stress, σ3, (usually constant) is applied to the curved surface of the sample, and the sample is brought to failure by gradually increasing the axial stress, σ1. The stresses applied to the soil/rock sample are illustrated in Figure 1. The confining stress models the insitu stresses present in the ground. Note that the intermediate

σ3

σ1

Figure 1 : Triaxial testing of geomaterials




principal stress, σ2 = σ3. It is common when testing rock to carry out this test under unconfined conditions, σ3 = 0. This is called a uniaxial (or unconfined) compressive strength test or UCS.

Influence of Confining Pressure By plotting the principal stresses at failure (σ1 vs σ3) we can see the influence of confinement on strength for various geomaterials – see Figure 2. This is called a principal stress plot. The line or curves shown on Figure 2 are called strength envelopes – 3 are plotted, one each for soil, hard soil or soft rock and hard rock. If we view the principal stress plot as two dimensional space, then the rock or soil is at yield (failure) when its stress state plots on the strength envelope. If it plots beneath the strength envelope then it is considered to be “elastic”. The soil or rock cannot exist in a stress state above (outside of) the strength envelope. Three dimensional stress states are handled by including a third principal stress axis (σ2) at right angles to σ1 and σ3. The strength envelopes shown in Figure 2 (and 3) are indicative of shape only. Soil strength envelopes are usually close to linear, becoming more non-linear (and close to parabolic) for hard rock. The intercept on the σ1 axis is the UCS, while the intercept on the σ3 axis is the uniaxial tensile strength (Note : compressive stresses are positive). The stresses at failure can also be plotted in terms of a Mohr-circle or shear stress, τ, versus normal stress, σ, plot. These envelopes are constructed by drawing a tangent to the Mohrs circles at failure. As with the principal stress plot, the strength envelope for soil is usually close to a straight line with τ intercept c (or cohesion) and slope tanφ (where φ is called the internal angle of friction). These two parameters can therefore be used to describe the strength envelope for soil and hence are known as strength properties. The non-linear envelopes of hard soil/soft rock and hard rock can also be described by mathematical equations referred to as strength criteria. In this case different strength properties (other than c and φ) are needed to define the strength envelope.

soft soil and sand

σ1

σ3

hard rock

UCS

hard soil andsoft rock

Figure 2 : Principal stress plot

soft soil and sand

τ

σ

hard rock

hard soil andsoft rock

σ1σ3

Figure 3 : Mohr-circle plot




31 σσ MI +=

& REQUIRED

Revision Holtz and Kovacs Chapter 10 : pp. 431 - 449

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Activity 9.1 - Revision Work through example problems 10.1 to 10.5 (pp. 436 – 445) in Holtz and Kovacs

MOHR-COULOMB YIELD CRITERION This is by far the most commonly applied failure criterion in geomechanics and is given by the expression

where τ is the shear stress at failure, σ is the normal stress, c is the cohesion and φ is the angle of friction. This criterion clearly describes a straight line in the shear stress-normal stress plane as shown above. For this criterion, the relationship between c and φ, the uniaxial compressive strength, UCS, and the uniaxial tensile strength, σt, are

It follows that the Mohr-Coulomb criterion gives the following expression for the ratio of uniaxial compressive to tensile strength

One of the principal problems associated with the construction of this criterion from actual test results, particularly in the case of results for rock that often show a fair degree of scatter, is that a best fit envelope must be fitted as a tangent to a number of Mohr circles, rather than as a straight line through a number of specific points. This is not a particularly easy task to achieve by eye nor is it amenable to the use of fitting techniques such as the method of least squares. Therefore, there may be some considerable merit in using a plotting technique that gives specific points such as a plot in the principal stress plane of σ1 and σ3 shown earlier. By constructing a line of best fit through the failure points, it is possible to define an intercept, I, and a slope, M.

φστ tan + c =

φφ

sin-1cos.c2

= UCS φφ

σsin+1cos.c2

- = t

φφ

σ sin-1sin+1

- = q

t

u




31 σσ MUCS +=

The relationship between the same criterion but in different plot format can be derived by consideration of a Mohr's circle plot. From the geometry of the circle, it may be shown that

or It follows that the intercept on the σ1 axis is given by UCS and the slope by:

It follows that the cohesion and the angle of friction may be calculated from the following

and

Another useful form of the Mohr-Coulomb criterion is the normalised principal stress plane. This is useful because it allows a comparison of trends in strength for materials of quite different strength values. This form is very similar to the principal stress plot except the axes are divided (or normalised) by the uniaxial compressive strength, UCS, to give

In soil engineering, the linear Mohr-Coulomb criterion is normally considered to be a reasonable representation of actual strength variations. Therefore, its wide-spread use is well justified. The variations of rock strength, however, tend to show a non-linear relationship when plotted in the τ - σn or σ1 - σ3 planes, especially for hard rock. Despite this, because of the relative simplicity of a linear criterion, Mohr-Coulomb it is often used for rocks for a limited stress range. The Mohr-Coulomb criterion is also very useful and practical to describe the behaviour of joint behaviour, and peak and residual strengths, particularly when dealing with various forms of slope stability problems.

& REQUIRED


φφ

σφφ

σsin-1sin+1

+ sin-1cosc2

= 31

φ−φ+

=sin1sin1

M

φφ−

=cos2

)sin1(qc u

1M+1M-

= sin φ

1 + M= n3in σσ




HOEK AND BROWN CRITERION (1980) In 1980, Hoek and Brown suggested an empirical strength criterion for rock masses (as distinct from intact rock). This criterion has been applied in many instances for hard rock engineering problems but does show some inconsistencies in some areas, particularly when applied to softer rocks. The basic criterion is given by

σ1n = σ3n + (mi σ3n + s)1/2

where m and s are constants that depend on the properties of the rock and on the extent to which it has been broken before being subjected to failure stresses. For intact rock, it has been suggested that s = 1. This reduces the criterion to:

σ1n = σ3n + (mi σ3n + 1)1/2

Hoek and Brown fitted this criterion to a range of test results for various forms of rock and as a result of this, values of m appeared to be a function of the rock type and varies between 7 and 25. It would appear that for hard rocks, this criterion gives reasonable agreement between predicted and measured results. It follows that it should be able to predict the strength of hard rocks on the basis of a knowledge of the rock type (to select m) and its uniaxial compressive strength. One the major shortcomings of the Hoek and Brown criterion is that it is of parabolic shape and therefore must predict a parabolic envelope for all intact materials. As it is known that soft rocks have strength envelopes that have much less curvature (i.e. they are approaching the relatively straight Mohr-Coulomb envelope that is more applicable for soils), it would appear that the Hoek and Brown criterion may only apply to a limited strength range of rocks. On the other hand, the Hoek and Brown criterion has a major contribution to make with regard to the prediction of the strength of rock masses that contain a range of defects (that can significantly decrease the strength of the mass below the intact value). Although the suggestions that were made must be considered a little tentative because of the very small amount of reliable test data that was available for the purposes of strength correlations, they do represent a simple method of predicting the strength of a rock mass. JOHNSTON CRITERION In their work on the soft rock Melbourne mudstone, Johnston and Chiu (1984) found that strength variations were intermediate to the two extremes (straight line and parabolic) with strength predicted by a curved envelope but not as extreme as a parabolic envelope. This is perhaps not too surprising when considering the general properties of soft rocks which are intermediate to soils and hard rocks. Further work by Johnston (1985) on a range of geotechnical materials showed that there was a general progression from a linear envelope for soft clays, through increasingly more curved relationships for stronger materials until a parabolic relationship was found for hard rocks. The criterion describing this progression is given by:




where M is a material constant depending on the strength of the material and its type, and B is another constant which depends on strength alone. The parameter S is similar in form to the s parameter used in the Hoek and Brown criterion and is equal to 1 for an intact material. Therefore for intact materials, the criterion reduces to

By fitting this criterion to a wide range of geotechnical materials ranging from soft clays through to hard rocks, values of B and M were suggested. It was found that for soft clay, that B approaches 1, and therefore the criterion reduces o the Mohr-Coulomb criterion. For hard rock, M increases, and B reduces to about 0.5 to produce a parabolic envelope very similar to that given by Hoek and Brown (1980). HOEK AND BROWN (1997) CRITERION Hoek and Brown (1997) revisited their 1980 criteria to address some of the criticisms made by Johnston amongst others. Using rock mass classification as a basis they introduced the concept of the Geotechnical Strength Index or GSI. Their new generalised criterion for rock mass strength can be stated as

Where m, s and a are all related to GSI by

For GSI >25 (i.e. good quality rock masses) a = 0.5 For GSI <25 (i.e. rock masses of very poor quality) s = 0 This empirical criterion relies on the assessment of the GSI from the two charts, one of which is shown in Figure 4. Although this process is subjective, it does provide a consistent basis on which to assess mass strength. It should be noted however that the criterion is isotropic, and should not be used where mass is likely to be anisotropic due to preferred orientation of joints etc.

−

=28

100GSIexpmm ib

200GSI

65.0a −=

σσ S + BM

= n3

B

n1

σσ 1 + BM

= n3

B

n1

−

=9

100GSIexps

a3

b31 sUCS

mUCS

+

σ+σ=σ




Figure 4 : Chart used to estimate GSI (after Hoek and Brown, 1997)




INFLUENCE OF ANISOTROPY Due to a number of features such as bedding, grain orientations, schistosity, etc., many rock types show preferred planes of weakness. It follows that rocks often show strength anisotropy which can have a pronounced effect on measured strength. This can be demonstrated by the following simple example. Consider a specimen of rock that has a plane of weakness inclined at an angle β to the line of action of the minor principal stress (or to the horizontal in the case shown in Figure 5). Let the shear strength of this plane be represented by the Mohr-Coulomb criterion so that at failure:

τ = cω + σ tan φω

where cω and φω are the cohesion and angle of friction of this plane of weakness. From a consideration of the geometry of the Mohr's circle, the general shear stress and the normal stress on any plane within the sample are given in terms of the principal stresses, σ1 and σ3, as

By rearranging these equations, failure in terms of the principal stresses may be expressed as

From this equation, it may be demonstrated that the minimum deviator stress required to cause failure is given by :

(σ1 - σ3)min = 2(cω + σ3 tan φ) [ (1 + tan2 φω)1/2 + tan φω ] and this occurs when

The above equations also show that for β approaching 90o or φω, the deviator stress approaches infinity. This implies that irrespective of the applied stresses, the sample will not fail on the plane of weakness for β < φω. However, although the rock substance of the sample may be much stronger than the plane of weakness, it cannot take infinite stress. As a result failure must occur through the rock substance and this can be determined from (for a Mohr Coulomb material) :

β

σσ

σσσ 2cos

2-

+ 2+

= 3131 β

σστ 2sin

2-

= 31

ββφφσ

σσ2sin)cot tan- (1

tan2 + c2 = - 3

31

2 + 45 = φ

β ω

σ1

β σ3

Figure 5 : Rock sample with inclined joint




30 60 90 β

σ1

σ3

Figure 6 : Variation of strength with confining pressure and joint inclination

τ = cs + σ tan φs where cs and φs are the cohesion and angle of friction of the rock substance. This effectively provides an upper limit to the strength of the rock as illustrated in Figure 6 for various confining pressures.

REFERENCES AND FURTHER READING Hoek and Brown (1980) Hoek and Brown (1997) Johnston (1985) Johnston and Chiu (1984) REVIEW QUESTIONS 1. Plot the strength envelopes in both Mohr circle and principal stress plots for the

following soils and rocks a) clay with c = 5 kPa, φ = 25o b) sand with c = 0 kPa, φ = 35o c) intact rock with UCS = 3 MPa, mi = 14 d) rock mass with GSI = 45, mi = 14 and UCS = 5 MPa e) rock mass with GSI = 15, mi = 14 and UCS = 5 MPa f) Rock sample of part c) but containing a joint inclined at 30o to minor principal

stress direction.

For each case determine the deviator stress at failure for σ3 = 0 and σ3 = 1 MPa. 2. For Q1 parts c), d) and e), determine the tangent values of c and φ at a confining

stress of 500 kPa. 3. Holtz and Kovacs : from page 485 on; Problems 10-1 to 10-23

Subject CIV3247 Geoengineering 10.1 Topic 10 : Soil Shear Strength - Undrained



TOPIC 10 – Soil Shear Strength - Undrained

TABLE OF CONTENTS PREVIEW............................................................................................................... 10.2

Introduction ......................................................................................................... 10.2 Objectives............................................................................................................ 10.2

PREFACE ............................................................................................................... 10.2

COULOMB’S EQUATION..................................................................................... 10.2

LABORATORY STRENGTH TESTING................................................................ 10.3

PRESENTATION OF RESULTS............................................................................ 10.4

THE φU = 0 CASE................................................................................................... 10.4 Skempton’s B parameter ...................................................................................... 10.5

UNDRAINED STRENGTH - DEPTH RELATIONSHIP........................................ 10.6

UNSATURATED SOILS........................................................................................ 10.6





PREVIEW Introduction In Topic 9, some common yield criteria for soil and rock were introduced. The strength behaviour of soil in particular is very sensitive to pore water pressures and the rate at which they dissipate. When clay is subjected to an increase in load e.g. by a footing, failure (if it is to occur) will usually occur soon after loading. However, excavations in clay may stand for months or years before collapsing. For the footing example, the short term strength of the clay is important, whereas for the excavation problem, the long term strength governs. The difference in behaviour is due to the pore water pressure response. This topic deals with shear strength under conditions where excess pore water pressures do not have time to dissipate ; i.e. undrained. Objectives • To be able to correctly interpret the results of a quick undrained triaxial test to

determine strength and deformation properties • To understand the behaviour of soil subjected to undrained loading conditions • To introduce unsaturated soil behaviour PREFACE Soil does not have a strength : because it is granular, its tensile strength is usually taken as zero and, when tested in compression, the failure which results is initiated by shear stresses, so soil strength is nearly always considered in terms of shear strength. To a lesser degree the same is true of soft rocks, but hard rocks require special treatment. Even when considering only shear strength any one soil has a wide range of strengths depending on loading patterns, water drainage conditions, past stress history and current environment. To simplify matters strength behaviour is divided into two major categories governed by the state of pore water drainage, i.e. undrained and drained COULOMB’S EQUATION Although the ‘immediate’ or ‘undrained’ condition does not usually involve considerations of pore water pressure, it is necessary to digress briefly to discuss the general shear strength equation, to permit explanation of undrained behaviour. Coulomb, in 1773, tested soils in direct shear and postulated the behavioural equation now known as the Mohr-Coulomb failure criterion (see Topic 9)

Coulomb’s Equation s = τf = c + σ tanφ

where s = shear strength, τf = shear stress at failure c = soil cohesion, σ = normal stress on failure plane and φ = angle of shearing resistance In effective stress terms this becomes :-

s = τf = c′+ (σ - u) tanφ′




Coulomb was unaware of effective stresses, yet the form of the equation has proved very satisfactory for effective stress calculations using the effective normal stress σ′ = (σ - u) and the effective stress strength parameters c′ and φ′ . Soil strength data are usually presented on a τ - σ plot, either as Mohr circles or stress paths (see later). The cohesion is that part of the soil strength which is independent of applied stress and has its origins in the electro-chemical interactions of the double layer. The friction is that part which is proportional to applied stress and has its origins in the sliding and rolling of particle upon particle. Depending on how a test is interpreted both c and φ may have ‘apparent’ values very different from their ‘real’ ones. LABORATORY STRENGTH TESTING The two most common tests are the direct shear test and the triaxial compression test, with the latter being more suited to clays and the former to sands. Clay behaviour is

more relevant to undrained or short term conditions so here we will concentrate on the triaxial test. In this test a cylindrical sample of soil, L = 2D, is surrounded with an impermeable latex membrane, placed in a pressure vessel and subjected to an all-round pressure, σ3, then rapidly failed by increasing the vertical or axial stress, σ1 . Failure is assumed to have occurred when the principal stress difference (σ1 - σ3) or deviator stress reaches a peak value or constant state or when the axial strain εax= ∆L/Lo exceeds say 20%, Lo being the initial length of the sample. The loading to failure typically takes about ten minutes and during the whole procedure no water is permitted to drain from the specimen and no attempts are made to measure the pore pressure developed in the soil.

σ

τ

c φ

• • σ1 σ3

(σ1 - σ3 )

invalid tensile stress region

Figure 1 : Mohr coulomb failure envelope

∆L

Axial Load - P

Cell Pressure σ3

Latex Membrane

Soil Specimen

O-ring Seal

Figure 2 : Triaxial test set-up for quick undrained test




& SUGGESTED


PRESENTATION OF RESULTS During the test readings are taken of Axial Load, P, and Axial Compression, ∆L. P may have to be corrected for the ‘blow-out effect’ of the cell pressure acting on the lower end of the loading ram. Graphs are prepared of (σ1 - σ3) versus axial strain in %. An area correction is also applied to account for lateral bulging of the specimen as the length decreases, the total volume remaining constant for undrained samples with S=100. Young’s modulus, Eu can be estimated from the “linear” portion of the (σ1 - σ3) versus axial strain graph. Some subjectivity may be required in choosing the “linear” portion of the curve. As the test is undrained, the modulus so determined is the elastic undrained Young’s modulus. As the sample undergoes no volume change (saturated samples only) then by definition Poisson’s ratio, νu = 0.5. THE φU = 0 CASE A practically very important case occurs when saturated clay specimens sharing the same stress history - say natural soil samples taken from the same depth at a site - are tested in undrained compression using a range of cell pressures, σ3, and the results are plotted as Mohr circles on a total stress plot (because pore pressures and hence effective stresses are usually not known). Figure 2 presents an idealised case with no sample variability, extremely rare in practice, but any clay with S=100% - whether or not it came from below a field water table - will exhibit φ = 0 behaviour in unconsolidated, undrained compression. The value of cu may then be used in total stress analyses of short term or immediate stability of foundations, slopes, etc.

(σ1 - σ3) = P/A For S = 100% A = Ao / (1 - ε )

where Ao = initial sample cross sectional area; ε = ∆L / Lo and Lo = initial sample length

For S < 100% A = Ao (1 + v ) / ( 1 - ε ) where v = volumetric strain = ∆V / Vo and Vo = initial sample volume (assumes volume increase is positive)




Reasons for φu=0. If two identical specimens of pore space saturated clay are placed in triaxial cells for undrained testing and subjected to cell pressures σ3 of 100 kPa and 200 kPa respectively, what will be the difference between their effective stresses? The soil skeleton is several orders of magnitude more compressible than

water so that, when an external stress increment is applied, almost all of it will be resisted by an increase in pore pressure, as in the undrained state the skeleton cannot compress to mobilise more resistance. Therefore ∆u = ∆σ3 , and from the effective stress law, ∆σ3′ = 0, so both specimens have the same effective stresses and therefore the same strengths (σ1 - σ3)f, ensuring a φu = 0 result. The actual pore pressures in the samples after applying the cell pressures will depend on the initial pore pressure (negative, i.e. less than atmospheric) and the pore pressures at failure, uf, for our two specimens would differ by 100 kPa. If values of uf were known, both specimens would plot as identical effective stress Mohr’s circles and no common tangent could be drawn. The very useful shear parameters cu and φu are therefore not fundamental soil parameters, but merely products of the method of testing and interpretation. Skempton’s B parameter The pore pressure response of soil can be determined using Skempton’s A and B parameters. At this stage we will consider only the B parameter which relates the change in pore water pressure to the change in all-round stress; i.e.

∆u = B ∆σ

For saturated soils, B is close to one and hence if a saturated soil is subjected to an all-round stress increase of ∆σ1 = ∆σ2 = ∆σ3 = ∆σ = 10 kPa, pore water pressures will (initially) increase by 10 kPa; i.e. ∆u = 10 kPa and hence ∆σ′ = ∆σ - ∆u = 10 – 10 = 0. That is the soil will feel no effective stress change, until the excess pore water pressures can dissipate.

In clay it can take a significant period of time for excess pore water pressures to dissipate, and hence the clay only feels a very gradual change in effective stress. In free draining sand, the response is considered to be instantaneous. For stiffer materials such as rock, B <1, and hence not all of a hydrostatic stress change is taken initially by the pore water.

Pore pressure parameter B (Skempton) For undrained all-round compression of a soil

∆u = B ∆σ and for SR = 100%, B ≅ 1, ∆u =∆σ, ∆σ′=0

τ

σ spec.1 spec.2 spec.3

φu = 0

c =cu

Mohr-Coulomb envelope

Figure 2 : UU test results




& REQUIRED


@

Activity 10.2 Work through example problems 11.12 and 11.13 (pp. 567 – 570) in Holtz and Kovacs

UNDRAINED STRENGTH - DEPTH RELATIONSHIP In a uniform deposit of N.C. clay cu often varies linearly with depth, particularly if the water table is close to the ground surface. The strength of the very near surface clay is often stronger as a result of desiccation (drying crust and large negative pore pressures). For O.C. clays the picture is more complex but often, at least over a limited depth range, cu may be taken as constant, with something less than the average of all test values for the soil layer being used in design.

UNSATURATED SOILS Many engineering soils exist with degrees of saturation less than 100%, e.g. compacted soils for earth fills, where S may be in the 80 - 90% range. For these soils the B parameter is significantly less than 1 and even under undrained conditions an applied load will cause some increase in effective stress, leading to measurable friction angles.

Figure 3 : Strength / Depth for

N.C. clay.

cu

depth

∇ drying crust

cu / γ ′z = const = cu / p

τ

σ

limit for S = 100%

Figure 4 : Undrained behaviour

of partially saturated soil




At increasing levels of applied stress the degree of saturation increases, as air is forced into solution in the pore water, causing φu to decrease continuously and approach zero asymptotically at high cell pressures. For such soils :

τ = c + k1 σk2 At very high pressures the soil reaches 100% saturation and φu becomes 0. The τ-σ plot may be linearised over a relevant stress range. REVIEW QUESTIONS 1. Why don’t we usually measure the undrained strength of sand ? 2. Show that the undrained Young’s modulus is equal to the slope of the deviator stress

versus axial strain curve from an undrained triaxial test. 3. Explain why excavations in clay may stand for several months and then suddenly

collapse, whereas in sand they collapse almost immediately. 4. Explain why the undrained strength of normally consolidated clays increases

linearly with depth. 5. Briefly describe the various methods for determining the undrained strength of a

clay and the limitations of each. 6. For what engineering design purposes do we use undrained strength ? 7. Briefly explain why Skempton’s B value decreases below 1 as soil/rock become less

compressible. 8. Explain briefly why you can build sand castles out of wet sand, but not out of dry

sand. 9. Do the deformations measured in an undrained triaxial test contain any component

of consolidation settlement? Briefly explain your answer.

Subject CIV3247 Geoengineering 11.1 Topic 11 : Soil Shear Strength - Drained



TOPIC 11 – Soil Shear Strength - Drained



PREFACE ............................................................................................................... 11.2

DRAINED STRENGTH TESTS ............................................................................. 11.2

DIRECT SHEAR TEST .......................................................................................... 11.3

TRIAXIAL COMPRESSION TEST........................................................................ 11.4





PREVIEW Introduction In contrast to the material presented in Topic 10 which was concerned mainly with undrained strength parameters derived from total stress τ - σ plots, this topic considers effective stress strength parameters derived from drained tests (D) before considering the third main test type, the consolidated undrained test (CU). Although in the CU test the compression to failure is carried out under undrained conditions, pore pressures are measured during the test so that effective stress τ - σ plots may be drawn and the resulting parameters c′ and φ′ are considered to be slightly conservative approximations to the truly drained parameters cd and φd . The advantage is that for clays the CU test may be completed much more rapidly than the D test. Objectives • To be able to correctly interpret the results of consolidated undrained and drained

triaxial tests to determine strength and deformation properties • To understand the behaviour of soil subjected to drained and undrained loading

conditions • To understand and be able to interpret the results of direct shear tests • To understand the importance of Skempton’s pore pressure parameters and be able

to apply them to practical situations to estimate pore pressure response • To understand stress paths and their application PREFACE Although the estimation of undrained strength is important in some applications, its relevance can only be understood in terms of effective stress and pore pressure response. The undrained strength is not a true property of the soil but a manifestation of the way in which we test the soil and interpret the test results. The true strength of the soil should be measured in terms of effective strength, giving rise to effective or “drained” strength parameters. The term “drained” refers to the dissipation of excess pore water pressures, not to the complete drainage of water out of the sample. “Drained” tests are carried out a slow enough rate so that excess pore water pressures do not build up, but dissipate at the same rate at the total stress being applied. That is, the soil “feels” the total stress change as an increase in effective stress and not as an increase in pore water pressure. DRAINED STRENGTH TESTS Drained strength may be measured conveniently in either the direct shear box or the triaxial test. The shear box is suited to both cohesionless soils (sands, gravels) and clays while the triaxial test is best suited to clays, as the preparation of cylindrical specimens is difficult for cohesionless soils. An additional advantage for the shear box is that the maximum length of drainage path is quite short, so that tests may be completed quickly. Where pore pressure information is required the triaxial test is universally used.




DIRECT SHEAR TEST A specimen of square cross section (normally 60 x 60 mm, L x L) and thickness approximately 1/3 L, is placed inside a metal box split into two halves across its mid-plane. A vertical normal load, P, is applied by a convenient mechanism and a motor drive pushes the lower half of the box, while the top half reacts against a load cell which measures the shearing force, T. The box usually sits inside a water bath for sample saturation. Test Procedure. For a drained test the load P is applied and the sample allowed to consolidate to say

95% degree of consolidation. A steadily increasing shear displacement is then applied at a rate consistent with near complete drainage and ∆L, T and ∆H are measured. For cohesive soils an area correction is necessary for σn and τ. A single test gives values of σn and τ at failure (peak or large displacement) and two or more tests are required for a τ - σ envelope to determine cd and φd.

σn = N/A = N/60(60 - ∆L) (N/mm2) τ = T/A = T/60(60 - ∆L) Note that for many silts and all sands and coarser soils cd is 0 and the area correction is not required for correct φd determination. For an undrained test on clay, the test needs to be carried out quite quickly, so

that pore water pressures do not have time to dissipate. This is actually very difficult to achieve as there will always be some dissipation of excess pore water pressure due to the very short drainage path. The results of undrained direct shear tests should therefore always be treated with caution.

& REQUIRED


@

Activity 11.1 Work through example problems 10.6 and 10.7 (pp. 460 – 463) in Holtz and Kovacs

P

T T

L

∆L

Figure 1 : Direct shear test (schematic)

σ′

φd

τ •

•

spec. 2

spec. 1 cd

Figure 2 : Drained direct shear - Coulomb failure envelope




TRIAXIAL COMPRESSION TEST More complex equipment is required for drained triaxial tests than for undrained tests. The specimen must have a porous disc at least at its lower end and often at both ends. A pressurisable burette (or volume change device) measures water flow to or from the specimen and a pressure transducer is needed for the CU test to monitor changes in pore water pressure during compression without significant volume change.

The use of a back pressure, ubp, is recommended wherever possible. A ‘back pressure’ is simply a controlled water pressure fed through the burette and into the porous disc/s at the end/s of the specimen. Its main functions are to assist in saturating the specimen (where relevant) and to provide an elevated initial pore pressure for specimens where the pore pressure is expected to decrease during compression and could become too negative to measure.

Figure 3 : D and CU triaxial tests Drained Test - The specimen is first back pressurised with a low difference between cell pressure, σ3, and back pressure, ubp, such that (σ3 - ubp) = initial estimated effective stress in specimen. The cell pressure and back pressure are then increased, in equal increments, until the specified value of back pressure has been reached. A “B – test” is then carried out to determine if the sample is close to fully saturated. This is done closing the back pressure valves and increasing the cell pressure by ∆σ. The pore pressure increase due to the rise in cell pressure is measured and the B value determined (see Topic 10). If B > 0.95 then the sample is considered to be close to fully saturated and the test can commence. The first main test phase is the consolidation phase, in which cell pressure σ3 is increased until the desired (σ3-ubp)=σ3′ has been achieved and the specimen is then allowed to consolidate. When at least 95% consolidation has occurred the second main phase, the drained compression phase, is commenced. The specimen is compressed vertically at a rate compatible with continued almost complete drainage, i.e. no significant pore pressures can build up in the specimen. Readings are taken of vertical external load, P, axial compression, ∆L, and volume of water outflow/inflow, ∆V, until either a peak strength or excessive deformation is reached. The area correction for calculating (σ1 -σ3) at any stage was given in Topic 10. Note that (σ1-σ3)=(σ1′-σ3′) for any type of triaxial test, or indeed any stressed soil element. Typical responses measured during drained triaxial testing for normally consolidated and over consolidated soils are shown in Figure 4. In general, loose or soft soils display a ductile deviator stress versus axial strain response. Failure usually occurs by bulging

Burette

Specimen

Membrane

Valve

To read outPore pressure

transducer

ubp

σ3

( )σ3- ubp = σ3




and throughout the test the sample expels water (i.e. contracts in volume). That is the soil is in a denser state at the end of the shearing stage than it was at the beginning. For dense or stiff soil samples, the stress strain behaviour is more brittle, often displaying significant softening after peak. The samples generally fail along a distinct inclined shear plane. The angle of inclination of the shear plane depends on the frictional angle of the soil. The sample may contract at the start of shearing, but then takes on water as it dilates (due to soil grains sliding over one another). The sample is therefore less dense at the end of the test. The strength parameters are obtained from a τ - σ plot using either Mohr circles or stress paths. A stress path is a graph of p = (σ1′+σ3′)/2 on the x axis against q = (σ1′-σ′3)/2 on the y axis and the stress path failure envelope yields parameters d (< c) and ψ (<φ). The stress path follows the path to failure taken by the sample. It is simply the trajectory taken by the top of the Mohrs circle. The strength (as distinct from the strength parameters) of the sample will depend on the stress path taken.

Figure 5 : Drained failure envelopes for c-φ soils

Figure 4 : Drained triaxial test results

(σ1-σ3)

+∆V

-∆V

axial strain %

axial strain %

dense, stiff

loose, soft

dense, stiff

loose, soft

•

•

stress path

σ3′ (σ1′ - σ3′) 45

•

p, (σ1′+σ3′)/2,σ′

q, (σ1′-σ3′)/2

d c

ψ φ

c cotφ = d cotψ

sinφ = tanψ NOTE For cohesionless soils and NC clays cd = 0 and all strength is ‘frictional’

stress path




The drained Young’s modulus and Poisson’s ratio can be estimated from the linear portion of the stress strain curve and the same portion on the volume change versus strain curve. Consolidated Undrained Test - The test procedure is exactly the same as for the drained test up to the end of the consolidation phase. Then the drainage to the burette is closed off, locking the ubp into the specimen as its initial pore pressure for the undrained compression phase. The axial compression is then applied, at a much faster rate than for a drained test on the same soil, and the pore water pressure is recorded in place of the volume change, which of course is now zero for a saturated specimen. Failure may be taken as either the peak strength, where this clearly occurs, or the maximum σ1′/σ3′, the principal effective stress ratio, in other cases. Results are presented as plots of (σ1-σ3), σ1′/σ3′ and ∆u versus εaxial% plus either Mohr circle or stress path τ- σ failure envelopes. Typical stress and pore pressure versus axial strain curves are shown in Figure 6 for stiff/dense (A –ve) and soft/loose (A+ve) soils. The A value refers to the second of the Skempton pore pressure parameters that relates pore pressure change to shear stress. Dense/stiff materials show a decrease in pore pressure as loading increases where soft/loose soils show an increase. Skempton’s general equation governing pore pressure development can be expressed as

( )[ ]∆ ∆σ ∆σ ∆σu B A= + −3 1 3 The value of A depends on soil type and stress path. Typical values are listed on page 603 of Holtz and Kovacs. The pore pressure generated during loading therefore depends on the mean stress change as well as the change in shear stress (or deviator stress). This equation can be used to predict pore pressure changes during undrained loading.

& REQUIRED

Holtz and Kovacs Chapter 11 : pp. 599 – 605 Appendix b-3 : pp. 691 -700

Figure 6 : Consolidated undrained triaxial test results

(σ1-σ3) σ1′/σ3′

axial strain %

axial strain %

A -ve

A +ve

σ1′/σ3′

+ ∆u

- ∆u

A +ve

A -ve




@

Activity 11.2 Work through example problem 11.14 (pp. 603) in Holtz and Kovacs

The failure envelope interpretation is more complex than for the drained test and the initial effective stress state, σ3′ (= σ1′), is commonly plotted as a point Mohr circle, p, the effective isotropic consolidation stress, on the σ axis. The effective stresses at failure are σ3′f and σ1′f and the excess pore pressures generated by shearing at failure is ∆uf.

Simple geometric calculations lead to the following expression relating the strength of the soil to its effective strength parameters and initial confining stress level p :

φφφσσ

′−+′+′′

=′−′

sin)12(1sincos

2

)( 31

f

f

Apc

For the test shown Skempton’s A parameter is about +0.7, i.e. 0.7 of the circle diameter plots to the left of point p. For A -ve, all of the circle will plot to the right of p. Note again that c′=0 for sands and NC clays.

& REQUIRED

Holtz and Kovacs Chapter 10 : pp. 473 - 484 Chapter 11 : pp. 490 - 639

Start of test 45+φ′/2

•

• •

• • •

τ (σ1-σ3)/2

failure point stress path

total stress

p OP

σ3′f σ3

σ′ (σ1′+σ3′)/2

σ1′f

ubp ∆uf

uf

∆uf = Af (∆σ1-∆σ3) = Af (σ1-σ3)

Figure 7 : CU test effective stress failure envelopes

End of test




@


REVIEW QUESTIONS 1. Use elastic theory to derive an equation for estimating Poisson’s ratio from the

volume change and axial strain measured in a drained triaxial test. 2. Are the Young’s modulus and Poisson’s ratio determined from a CUPP test drained

or undrained parameters ? Give reasons for your answer. 3. Holtz and Kovacs : from page 640 on; Problems 11-1 to 11-105.

Subject CIV3247 Geoengineering 12.1 Topic 12 : Soil Slope Stability



TOPIC 12 – Soil Slope Stability



PREFACE ............................................................................................................... 12.2

FACTOR OF SAFETY, F ....................................................................................... 12.3

TIME EFFECTS...................................................................................................... 12.3

SHAPES OF FAILURE SURFACES ...................................................................... 12.3

METHOD OF ANALYSIS...................................................................................... 12.4

BISHOP’S SIMPLIFIED METHOD OF SLICES.................................................... 12.5

PARTIAL SUBMERGENCE .................................................................................. 12.5

OTHER SOLUTIONS............................................................................................. 12.6

WEDGE METHODS............................................................................................... 12.6

TENSION CRACKS ............................................................................................... 12.7





PREVIEW Introduction Landslips/slope instability is a natural on-going phenomenon (see Topic 1) that involves slopes moving to a more stable condition. It is often man’s activity that brings about the instability by increasing slope angles and changing groundwater conditions. Since many urban areas, infrastructure projects (e.g. roads, railways, dams etc) and mining projects involve excavation, filling or other changes to the local topography and groundwater levels, instability is of significant concern. There are many factors that influence stability, the major ones being soil and rock strength, discontinuities and planes of weakness, water and external loading. The mechanism via which failure occurs also varies depending on these conditions. As a result, there are many forms of analyses. This topic concentrates on instability of soil slopes. Topic 13 deals with instability of rock slopes. Objectives • To understand the factors that affect slope stability and to be able to quantify their

influence in terms of a factor of safety • To understand and be able to implement Bishop’s modified method for vertical

slices • To understand the difference between slopes that are excavated and filled, and

which parameters to apply to determine stability in the long and short term. PREFACE Slope stability problems in soil and rock engineering may involve heights from 1m to several 1000m, with volumes of slipping material from a few m3 to several 100 x 106

m3, i.e. from failure of a small trench to mountain landslides. Not all such instabilities can be successfully analysed currently, but many slopes in the general range of engineering significance have become amenable to analysis with the development of improved mathematical techniques and comprehensive computer programs. Whereas a few decades ago the analysis by hand of a single possible failure mechanism could take an engineer half a day, now many tens of thousands of mechanisms may be investigated in that time and in much greater detail. Slopes of engineering interest may be either natural or man-made, terrestrial or submarine, stress increasing (e.g. embankments) or stress reducing (excavations), reinforced or unreinforced and statically or seismically loaded. The most common mode of failure is by mass movement of a body of soil or rock, either slowly or rapidly, though liquefaction and flow slides are also important. Analyses are usually more concerned with estimation of a Safety Factor, F, against total failure, than with pre-failure deformations, though these may also be estimated when required, using finite element or difference methods and, where sufficient data is available, probabilities of failure may also be calculated. Soil and rock slope




behaviours share some common aspects, but also exhibit significant differences and this topic covers only the matters in common, leaving much of rock slope behaviour for specialist studies (see Topic 13). FACTOR OF SAFETY, F Many definitions of F are possible, but the one almost universally used in practice is a safety factor on shear strength, i.e. τ = s/F = c/F + σ(tanφ)/F where τ = shear stress on slip surface s = available shear strength c, φ = appropriate shear parameters

σ = normal stress (total or effective as appropriate) This criterion must be satisfied at all points along a failure/slip surface or mechanism, implying that F is constant, though τ and s will vary from point to point. In practice distributions of τ are difficult to calculate and F is computed indirectly from moment equilibrium for each element of a failure mechanism. c/F and (tanφ)/F are referred to as the ‘developed cohesion’ and the ‘developed friction’ respectively, i.e. that proportion of the available cohesion and friction which must be called into play to just provide stability or equilibrium for the slope. Cohesion and friction are usually given the same degree of development, 1/F, though this matter is arguable. TIME EFFECTS Slopes are subject to the same considerations concerning time effects as all other geotechnical structures. This means that slopes in sands or free draining soils will be analysed using drained or effective stress parameters, while slopes in clays may be treated as either ‘immediate’ or ‘long term’ problems depending on circumstances. A temporary cut in clay would be analysed as an immediate or total stress problem without any pore pressure input, while a slope meant to last for decades could be either a short or long term problem, depending on whether or not it is ‘built-up’ or ‘excavated’, also taking into account likely future changes in ground water conditions. Most modern analysis methods handle all cases with appropriate choice of strength parameters and pore pressure data input. SHAPES OF FAILURE SURFACES Before an analytical method can be applied the shape of the slipping soil mass must be defined. This shape may be circular, cycloidal, log spiral, near linear, general non-linear, or combination surfaces, all of these agreeing reasonably well with measurements of specific failures. For reasons of simplicity and mathematical tractability, combined with acceptable accuracy, the circular slip was the first to be used widely.




x

circularslipsurface

R

W

S

Figure 1 :Circular slip surface - method of slices The weight of an element of soil, such as W for the vertical slice shown in Figure 1, exerts a destabilising or overturning moment about the circle centre O, while the shear strength S at the base of the slice provides a stabilising or restoring moment. The balance of these moments summed over the whole sliding mass controls the stability of the slope.

& REQUIRED

Waltham Chapters 32 to 34 : pp. 64 – 69

METHOD OF ANALYSIS Most commonly used methods of slope stability analysis adopt the relatively brute force method of limit equilibrium, i.e. a free body of soil is isolated by assuming a failure surface, based on past experience, and all body and boundary forces are evaluated. Equations of equilibrium for moments and vertical and horizontal forces are then used to solve for any unknowns, a process that usually involves one or more assumptions to make a solution feasible. A value of F is then calculated from the equation above, but is valid only for the failure surface postulated. To reveal the minimum and therefore critical value of F it is necessary to analyse a large number of different failure surfaces - a daunting task pre computer! Various researchers have developed methods of increasing complexity based on the limit equilibrium approach, most of them being extensions of the popular Bishop’s Simplified Method of Slices (itself developed from the Swedish Method of Slices). For slips where the failure surface is approximately circular Bishop’s method is still quite accurate and will be outlined below.




BISHOP’S SIMPLIFIED METHOD OF SLICES

In this method the slope is divided as in Figure 1 into a number of vertical slices, usually at least 15 - 20. The forces on a typical slice are shown in Figure 2. Provided a large number of slices is used the base of each slice may be approximated by a straight line and the slice weight may be calculated accurately as W = γbh. The total unit weight is used for both total and effective stress analyses as it is equilibrated against total boundary forces. The E forces include both soil and water components, but the X forces are soil shear forces only. The U force is the water force on the base of length l and is derived from flow net estimates of pore pressure.

By resolving forces vertically to determine N′ values, and summing moments of each slice about the circle centre, Bishop was able to derive his formula :

where As F appears in the formula for mα it cannot be calculated explicitly from Bishop’s equation and an iterative approach is necessary. A spreadsheet tabular solution is efficient and usually 3 iterations suffice to give F to two decimal places, which is more than conventional practice can justify. The above formula is for an effective stress analysis but is also suitable for a total stress analysis by using cu for c′, φu for φ′ and zero pore pressures. PARTIAL SUBMERGENCE As shown in the Figure 3 a slope may be partially submerged, with an external free water level above the toe of the slope. Bishop showed how this may be handled for total and effective stress analyses by drawing a construction line horizontally into the slope at the level of the external water and using γtotal for soil above this line and γ ′ for soil below this line, leading to an ‘effective weight’, Weff, for each slice. For effective

( )[ ]

∑∑

α

φ′−+′= α

sinWm1

tanubWbcF

Ftansin

cosmφ′α

+α=α

b n

n+1

Xn

En W

En+1

Xn+1 l T

N = N′ + U

h

β

α

∇

∇ phreatic surface

slope surface

sinα = x/R

Figure 2 : Force diagram for nth slice




stress analyses the actual pore pressure at the base of each slice must be reduced by ∆u = γw x (height difference between construction line and base of slice), i.e. uexcess = u - ∆u Bishop’s equation then becomes :

If this method is used any external water forces on the slope face must then be ignored. Note that where water seeps into the slope uexces is negative, increasing the factor of safety

OTHER SOLUTIONS Most computer packages offer a range of methods with names such as Spencer’s Method, Morgenstern-Price Method, General Limit Equilibrium Method, etc and these are generally extensions of the Bishop method to include a more accurate treatment of the side forces between slices, the Xn forces being effectively ignored in Bishop’s solution. They also permit general non circular surfaces and some form of search for the critical safety factor. Increasingly they include treatment of slope reinforcing and nailing. WEDGE METHODS For profiles containing, for example, thin weak layers, the failure surface is likely to be highly non-circular and Bishop-style analyses may become inaccurate. Recently there has been an increase in interest in multi-wedge solutions for such problems and several such programs have shown that they may be applied also to circular and near circular failure mechanisms with accuracy equal to or better than the conventional vertical slice approaches. Sarma’s multi-wedge method is regularly included in modern packages and Monash University has produced two very different wedge analyses - GWEDGEM (the General WEDGE Method) and EMU (the Energy Method Upper-bound).

( )[ ]

∑∑

α

φ′−+′= α

sinWm1

tanbuWbcF

eff

exceff

h = ∆u/γw

phreatic surface

∇ use γ

use γ ′

ignore external water

Figure 3 : Treatment of partial submergence




TENSION CRACKS When the soil at the slope crest is cohesive it may be wise to include a tension crack in the analysis. From earth pressure theory (see Topic 15), the maximum depth of crack is given by zo= 2c/γ√KA where KA = (1-sinφ)/(1+sinφ) and the worst case occurs when the crack is full of water after a rain storm, the water exerting a destabilising horizontal force of 1/2 γw zo

2 on the vertical side of the crack. This force gives rise to an additional overturning moment in the analysis.

& SUGGESTED

The relevant chapters on slope stability contained in any of the following books : Bowles, J.E., Foundation analysis and design. McGraw-Hill Das, B.M., Principles of geotechnical engineering. PWS-Kent Tomlinson, M.J., Foundation design and construction. Pitman

REVIEW QUESTIONS 1. What factors affect the stability of soil slopes ? Which of these has the greatest

affect for a given soil slope ? 2. For the free body diagram shown in Figure 2, show that the factor of safety for a

single slice is given by

3. What are the limitations of the simplified Bishop Method of vertical slices? What simplifying assumptions does it make ?

4. In an undrained analysis, why is the solution for factor of safety independent of the position of the piezometric surface ?

5. Determine the depth of tension crack and the extra disturbing force due to water in the tension cracks for the following soils : a) stiff clay : c′ = 10 kPa, φ′=27o, γ = 18 kN/m3, water table at the surface b) stiff clay : cu = 35 kPa, φu=0o, γ = 18 kN/m3, water table at the surface c) dense sand : c′ = 0 kPa, φ′=37o, γ = 20 kN/m3, water table at the surface d) stiff clay : cu = 35 kPa, φu=0o, γ = 18 kN/m3, water table at 1m depth e) stiff clay : c′ = 10 kPa, φ′=27o, γ = 18 kN/m3, water table at 1m depth

6. For the stiff clay in Question 5(a), investigate the influence of depth of water in the tension crack on disturbing force.

7. What affect does partial submergence have on the safety factor of a slope? 8. If you observed water flowing out from a slope of marginal stability would you be

concerned ? Why ? 9. Why is failure more likely in slope inundated by flood water after the water has

receded ?

( )[ ]

α

φα

sin

1tan

Wm

ubWbcF

′−+′=

Subject CIV3247 Geoengineering 13.1 Topic 13 : Rock Slope Stability



TOPIC 13 – Rock Slope Stability



PREFACE ............................................................................................................... 13.2

DISCONTINUITY ORIENTATION....................................................................... 13.4

STEREOGRAPHIC PROJECTION......................................................................... 13.5 The hemispherical projection of a plane ............................................................... 13.5

BASIC CONSTRUCTIONS FOR HEMISPHERICAL PROJECTION.................... 13.7 Plotting a line....................................................................................................... 13.7 Plotting a plane, its pole and its dip vector ........................................................... 13.7 Defining the line of intersection of two planes ..................................................... 13.8 Determining the angle between two lines in a plane ............................................. 13.9 Plotting of discontinuity orientation data.............................................................13.10

KINEMATIC ANALYSIS OF ROCK SLOPES .....................................................13.11 Circular failure....................................................................................................13.11 Plane failure........................................................................................................13.11 Wedge failure .....................................................................................................13.13 Toppling failure ..................................................................................................13.14

ANALYSIS OF SLIDING BLOCKS......................................................................13.15 Without water .....................................................................................................13.15 With water ..........................................................................................................13.15 With reinforcement .............................................................................................13.16

LANDSLIDES AND SLOPE STABILISATION ...................................................13.16

REFERENCES AND FURTHER READING.........................................................13.16

REVIEW QUESTIONS..........................................................................................13.17




PREVIEW Introduction Topic 12 presented an introduction in slope instability and concentrated on soil slopes. Such slopes can be reasonably modelled assuming failure occurs on a circular slip surface. The failure of rock slopes on the other hand is controlled not by the strength of the intact rock, but by the orientation, position strength etc of the many discontinuities that are contained in the rock mass. For this reason, there are several forms of failure that must be considered. For each of these, it is necessary to have a good knowledge of the 3D geometry of the discontinuities and how they influence the stability of the slope. This topic will briefly introduce the various mechanisms via which rock slopes can fail, describe a simple method for describing the 3D geometry of discontinuities and the simplistic rules used to determine if the discontinuities can form a kinematic mechanism of failure. Objectives • To identify the various forms of rock slope failure mechanism and the part played

by discontinuities in forming these mechanisms • To develop skills in using stereographic projection to analyse the discontinuities

within a rock mass; qnd • To be able to use these skills to assess the kinematic stability of rock slopes. PREFACE Almost everyone is affected by the stability of rock slopes. They exist in road and rail cuttings, dam abutments, quarries, open pit mines and tunnel portals, etc... Rock falls and slides from natural and excavated slopes constitute a major source of costs, delays, injuries and fatalities in mining and construction and in the normal environment of many communities. In conjunction with the safety, economic and environmental aspects of the excavation of a rock slope an extensive list of geological parameters must be considered. The degree of certainty required of the geological investigation for a rock slope varies widely; e.g. a rock slope adjacent to a major dam abutment must be designed with a substantial margin of safety, whereas an opencut mine would be uneconomic if its slopes were designed by the same criteria. This is reflected by the large number of deep open pit mines that have experienced major rock slides at some stage in their history. Recognition that uncertainty must exist in all but the most trivial problems is the first step in rock (and soil) slope engineering. It is usually better in the design of rock (and soil) slopes to work from the general to the particular. In this way, time is not wasted on detailed studies of small sites where subsequent investigations have shown that an entire mountain side to be unstable; on the other hand, the working practitioner is likely to deal with many small scale problems for every major one encountered.




A review of small rock slides shows that one general concept of stability analysis cannot satisfy all possible cases. Rock material can vary from a homogeneous unbroken mass to a heterogeneous fractured one with endless combinations of joints, faults, block patterns, orientations, zoning and fissuring. Depending on the nature of the rock mass

itself, the height and steepness of the slope, and the ground water conditions, a number of widely differing potential modes of failure may be identified that is plane sliding, wedge sliding, circular failure, toppling or any combination of these (see Figure 1). Hence each site must be treated individually and specialised methods of analysis adopted according to these modes of failure. The behaviour of a rock slope where normal stresses are quite low is dominated by the presence of discontinuities, such as

joints, faults, bedding, foliation etc. Hence most practical rock slope designs are based on the discontinuum approach. Elasticity approaches to slope stability are at present severely limited because our knowledge of the mechanical properties of rock masses is so inadequate that the choice of material properties becomes a matter of guess-work; i.e. what is the meaning of E and ν in a highly discontinuous medium. However, with the further development of numerical methods such as the finite element method incorporating joint elements and the discrete element methods, an improvement in analysis may be possible. The influence of discontinuities on the stability of steep slopes is shown in Figure 2. If one initially ignores the unstable slopes it can be seen from this figure that if joints are favourable to the slope face, a fairly clear line can be drawn for the

Figure 2 : Discontinuities and slope stability (after Hoek and Bray)

Figure 1 : Main failure modes of rock slopes (after Hoek and Bray)




maximum slope height versus slope angle of successfully excavated slopes. Based on this “experienced” curve, if one decided to excavate a 300 m high slope at an angle of 65°, a very comprehensive investigation would be necessary to show that there would be no risk of inducing a massive slope failure. However, while slopes may be stable at steep angles, and heights of several hundred metres, many flatter slopes fail with only heights of tens of metres as evidenced by the

solid points in the figure above. The reason is attributed to unfavourable orientations of controlling discontinuities in the slope, and the material properties existing on those discontinuities. For example, the variation in the critical height of a drained vertical slope containing a planar discontinuity with the inclination of the discontinuity is illustrated below. The discontinuity has the following properties - cohesion = 96 kPa, friction angle = 20°, bulk density= 2.56 t/m3. Figure 3 shows that under these circumstances, the rock slope can be theoretically of infinite height for dips of less than about 28° and more than 83° but reduces to a maximum height of approx. 30 m for dip angles of 55°.

The following sections describe some of the more common methods for estimating the stability of rock slopes. The different types of failure (plane sliding, wedge sliding, circular and toppling) require different methods of analysis. DISCONTINUITY ORIENTATION

The orientation of a discontinuity is simply the attitude of the discontinuity in space. It has components of dip direction (or azimuth) measured

clockwise from true north, and

dip the angle between the horizontal plane and the line of the steepest declination of the plane of weakness.

Figure 3 : Slope height versus discontinuity (after Hoek and Bray, 1981)

DipdirectionDip

N α

strike = αdip direction = α + 90

Figure 4 : Orientation of a discontinuity




Orientation is usually expressed as a three digit number followed by a two digit number in the following format - dip direction/dip; e.g. 035°/27° implies a dip direction of 35° clockwise from north and a dip of 27° from the horizontal. Other terms which are often used to define the orientation of discontinuities are : strike the trace of the intersection of the plane of weakness with a horizontal

reference plane. plunge is the dip of the line of intersection of two planes of weakness. trend is the direction of the horizontal projection of a line, measured clockwise from

true north. Although, the definitions above have been used in the context of planes of weakness, they apply to any plane. STEREOGRAPHIC PROJECTION The orientation of discontinuities can have a large influence on the stability of engineering and mining works. As engineers, we have to find some way of quantifying this influence. Unfortunately, since discontinuities are 3 dimensional, visualising the intersection of discontinuities with each other and with rock slopes or excavations can be very difficult. It follows that any analysis attempted in three dimensions will be even more difficult. Stereographic projection is a graphical method of representing and analysing three-dimensional features such as planes and lines in two dimensions. It is used widely in assessing the stability of rock slopes and underground openings. There are several very good references on stereographic projection including Goodman (1980), Hoek and Bray (1977) and Priest(1985). The hemispherical projection of a plane Consider a sphere (called the reference sphere) which is free to move anywhere in space. This sphere can be centred on any inclined plane (such as a discontinuity) as illustrated in Fig. 5. The line formed by the intersection of the inclined plane and the surface of the sphere will form a circle (called a great circle) as indicated in Fig. 5. A line drawn through the centre of the sphere that is perpendicular to the inclined plane will intersect the reference sphere at two diametrically opposing points which are called poles (of the plane). We can split the reference sphere into two hemispheres (one upper and one lower) by passing a horizontal reference plane Figure 5 : The reference sphere




through the centre of the reference sphere. The pole and the great circle in the upper hemisphere are repeated in the lower hemisphere. This implies that we only have to consider one of the hemispheres. In rock mechanics, the lower hemisphere is preferred and hence this form of projection is called the lower hemispherical projection.

The great circle and the pole can now be projected back onto the horizontal reference plane. This is achieved by connecting all points on the great circle and the pole with the zenith of the sphere as shown in Fig. 6. The projection of the great circle and the pole are given by the intersection of these connecting lines with the horizontal plane. This type of projection is commonly known as the stereographic or equal angle projection. Another type of projection which is commonly in use is called the equal area projection. However, we will

not be using the equal area projection in this course. The plot obtained from a stereographic projection is called a stereoplot. To facilitate the plotting of planes and their poles we use a stereonet (see Fig. 7). The great circles on the stereonet (the one running North to South) represent planes of constant dip. The stereonet also contains a number of small circles which are the circles centred around the North and South points. The angle between any two points on a great circle can be determined by counting the small circle graduations along the great circle between the two points. Some of the basic constructions which are needed to use stereographic projection are described below.

@

Activity 13.1 Make up your own stereonet by gluing the supplied net onto a piece of cardboard and pushing a drawing pin through the centre of the net. Locate the pin by pushing through from the front side of the net and then reverse the pin (so that the pointy end is on the same site as the net) and hold in place using sticky tape.

Figure 7 : Equal angle stereonet

Figure 6 : Stereoprojection of a great circle




BASIC CONSTRUCTIONS FOR HEMISPHERICAL PROJECTION Plotting a line This construction is needed to represent the orientations of lines in space such as the intersection of two planes, the normal to a plane (or pole) and force vectors. Recall that a line is usually defined by its plunge (to the horizontal) and its trend (to North). A line with a plunge of 55° and a trend of 27° is usually represented by 027/55. On a stereoplot, a line is represented by a single point. The process via which the location of this point is obtained is as follows. This process is also illustrated by way of an example; plot the line 027/55. 1. Place a drawing pin at the centre of your stereonet. Then place a piece of tracing

paper over your stereonet such that the tracing paper covers the entire stereonet. The drawing pins acts as a centre of rotation such that you can rotate your tracing paper relative to the stereonet. Mark the North point on the tracing paper.

2. Using the graduations around the edge of the stereonet, count of the trend of the line (i.e. 27° clockwise from North for this example - see Fig. 8a) and mark a point at this location.

3. Rotate the tracing paper about the drawing pin until this mark lies on the East-West diameter of the stereonet (see Fig. 8b).

4. Count from the point on the perimeter of the stereonet, along the E-W diameter the required plunge angle (55° for this example). Mark the point with some convenient notation (Fig. 8c).

5. Rotate the tracing paper such that the North points are re-aligned.

Plotting a plane, its pole and its dip vector This construction is needed to plot discontinuities such as faults, joints, bedding planes, etc. Recall that a plane is located by its dip direction and dip. It is generally represented

Fig. 8 : Plotting the line 027/55




by two numbers separated ba a slash - viz dip direction/dip. A dip vector is a line in the dip direction (or at 90° to the strike) and plunging at the dip angle. On a stereoplot, a plane is represented by a great circle and its pole by a point. To construct these, carry out the following instructions. Again the process is also illustrated by an example (Fig. 9) - plot the pole, dip vector and great circle for the plane 130/50. 1. Place a drawing pin at the centre of your stereonet. Then place a piece of tracing

paper over your stereonet such that the tracing paper covers the entire stereonet. Mark the North pint on the tracing paper.

2. Using the graduations around the edge of the stereonet, count of the dip direction of the plane clockwise from North (130° in this case - Fig. 9a) and mark a point at this location.

3. Rotate the tracing paper about the drawing pin until this mark lies on the East-West diameter of the stereonet.

4. Count from the point on the perimeter of the stereonet, along the E-W diameter the required dip angle (50° in this example - Fig. 9b). Mark this point as the dip vector. Trace in the great circle which passes through this point.

5. Plot the pole of the plane by counting a further 90° along the E-W diameter. Mark this point with some appropriate notation (Fig. 9b).

6. Rotate the tracing paper such that the North points are aligned (Fig. 9c).

Defining the line of intersection of two planes Two planes intersect in a line, which is represented by a point in a stereoplot. This procedure is required to determine the lines of intersection of two or more sets of discontinuities. To determine the trend and plunge of the line of intersection of two planes carry out the following steps. As an example find the trend and plunge of the line of intersection of two planes 130/50 and 250/30. 1. Plot the great circle and pole of each plane by the method described above

(Fig. 10a). The line of intersection of the two planes is represented by the point of intersection of the two great circles.

2. Rotate the tracing paper until the point of intersection of the two planes lies on the E-W diameter of the stereonet (Fig. 10b). The plunge is obtained by counting the number of degrees along the E-W axis from the perimeter (either the 90° or 270°

Figure 9 : Plotting the plane 130/50




point) - in this example 21°. With the tracing in this position, the poles of the two planes lie on the same great circle (Fig. 10b). This provides an alternative way of locating the line of intersection.

3. Rotate the tracing so that the North points are re-aligned. 4. Draw a straight line through the centre of the net and the point of intersection of

the two great circles to the perimeter of the net (Fig.10c). This line is the trend of the line of intersection - measured at 201° clockwise from North.

Determining the angle between two lines in a plane This can also be used to determine the angle between two planes if the planes have been defined by their poles. Find the angle between lines with orientations 240/54 and 140/40. 1. Plot the projection of the two lines as described above (Fig. 11a). 2. Rotate the tracing until these two points lie on the same great circle of the

stereonet. The dip direction and dip of the plane containing these two lines can be ascertained by carrying out the steps outlined in 6.3.2. above in reverse - in this case 200° and 60° respectively.

3. The angle between the two lines is found to be 64° and is obtained by counting the small circle divisions between the two points along the great circle (Fig. 11b).

Figure 10 : Plotting the line of intersection of two planes 130/50 and 250/30

Figure 11 : Determining the angle between the two co-planar lines 240/54 and 140/40.




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Activity 13.2 Using your own stereonet, work through the constructions above.

Plotting of discontinuity orientation data

The stereoplot is extremely useful for presenting orientation data and determining stability of rock slopes and excavations on a kinematic basis (see later). However, a stereoplot would become very crowded if for each location every discontinuity that was mapped had its pole, dip vector and great circle plotted onto the stereoplot. As a result, the poles of the discontinuity are the only points plotted. Such a plot is illustrated in Fig. 12. The discontinuity data is plotted as three distinct sets of points representing joints, bedding planes and a fault.

To aid interpretation, a contour plot of pole concentrations for bedding planes and joints is usually drawn up. Faults are usually treated independently. The contour plot for this data is included in Fig. 13. It is fairly obvious now that there are two major joint sets and one preferred orientation of the bedding planes. The central orientations of the two major joint sets are 347/22 and 352/83, and that of the bedding planes is 231/81. In most cases, due to the methods of mapping, the data obtained from a mapping exercise is biased. Corrections must be carried out before analysing the data to eliminate this bias. Appropriate techniques will be described later.

Figure 12 : Pole stereoplot of discontinuity data

Figure 13 : Contour plot of discontinuity data




KINEMATIC ANALYSIS OF ROCK SLOPES A kinematic analysis uses stereographic projection techniques to determine whether blocks of rock are free to move; i.e. many steep rock slopes are stable even though they contain steeply dipping discontinuities of weak strength. They are stable because the potentially unstable rock blocks are unable to move along a discontinuity as other rock blocks are in their way. If these rock blocks were removed, the slope would fail. The kinematic analysis is so called because it only accounts for the possible motion of the rock blocks and ignores all forces that are involved. As such it is a very quick and useful method for identifying potentially unstable rock slopes. A more sophisticated analysis can then be used to analyse the potentially unstable slopes. A summary of the types of slope failure and the associated stereoplots representing the structural conditions likely to give rise to these failures is illustrated in Figure 14. Note that it is important to include the cut face of the rock slope in the stereoplot as failure can only occur as a result of movement towards the free face created by the cut. Each type of failure is considered separately below.

Circular failure This type of failure can arise in overburden soil, waste rock dumps, and highly weathered or highly fractured rocks where failure is not dominated by geological features. As such the stereoplot shows a random scatter of poles with no identifiable structural pattern (see Fig. 14). In such cases a kinematic analysis is not very useful. Plane failure Plane failure is a fairly rare occurrence in rock slopes as the geometric configuration (of discontinuities and excavations) applicable for this type of failure only occasionally occurs in rock slopes. The geometric considerations for sliding on a single plane are: • the plane on which sliding occurs must strike approximately parallel (within +/-20°)

to the slope face,

Figure 14 : Failure modes and stereoplots




• the failure plane must exit in the slope face; i.e. the dip of the discontinuity must be less than the dip of the slope face,

• as will be demonstrated in the next section, the dip of the failure plane must be greater than the angle of friction of the discontinuity,

• the slope must be long enough such that end conditions are negligible or alternatively, “release” surfaces which provide negligible resistance to sliding must be present at some point within the rock mass.

Recall that a plane is plotted as a great circle on a stereoplot and can be represented by its pole N and dip vector D (see Fig 15a). Any block that is able to slide on a single plane surface, will move down the slope parallel to the dip of the weakest plane; i.e. parallel to the dip vector D. If the rock slope has a dip of angle, α then a rock block will slide if the dip vector D points into the free space of the excavation and the dip of the discontinuity on which sliding is to occur is less than a (Fig. 15b). Both of these conditions are satisfied on a stereoplot (Fig. 15c) if the dip vector of the discontinuity falls within the region above the rock slope great circle; i.e. the shaded region in Fig. 15c. For the dip vectors plotted in Fig. 15c, D1will allow sliding and D2 will not. These rules are also applicable for wedge failure.

Another condition can also be invoked for plane sliding and that is that the discontinuity must strike within approximately +/-20° of the slope face. This further restricts the possibilities of plane sliding to the shaded area shown in Fig. 15d. These rules can also be used to determine the steepest angle that a rock slope can be cut into a rock mass that contains a particular discontinuity set. This process is illustrated in Fig. 16. If the discontinuity set is represented by the dip vector D1, then the steepest safe slope with a specified dip direction (on a kinematic basis only) that can be excavated, is given by the dip of the great circle passing through D1 and the point on the

Figure 15 : Kinematic analysis of plane failure




circumference of the stereoplot representing the dip direction of the cut slope (of strike 1, 2 3 etc). The analysis described thus far has only accounted for the possible motion of the rock slope and has not accounted for the strength of the discontinuities, block size, forces

such as gravity and pore pressure etc.. There are constructions which can be made on the stereoplot which can account for some of these influences. However, we will only look at how we can include the frictional resistance of the discontinuity into this kinematic analysis. Sliding (due to self weight only) can only occur if the surface of sliding dips at a steeper angle than the joint friction angle φj, (assuming no joint cohesion). Hence if we draw a circle of radius 90-φj at the centre of the stereonet (called the friction circle) as shown in Fig. 17, the shaded area

outside this circle contains all lines plunging at angles less than φj. All dip vectors, D, that fall within this shaded area will not fail by sliding unless other forces, e.g. water pressure are present. A similar argument and result can be applied to toppling.

Wedge failure The most common form of failure in rock slopes is wedge failure. This occurs when the discontinuities strike across the slope and sliding takes place along the line of intersection of discontinuities (see Fig. 3). Due to the three dimensional aspects of these types of failures, analysis can be extremely complicated. However, a simple kinematic analysis can reduce the number of slopes that will need to be analysed by more sophisticated methods.

Figure 17 : Friction circles for sliding (left) & toppling (right)

Figure 16 : Steepest slope angle for sliding




The kinematic analysis is essentially the same as for plane failure, except in this case, the slope will not fail if the line of intersection of the two discontinuities forming the wedge at a lower angle than the friction angle of the discontinuity surface, i.e. if the points of intersection I on a stereoplot lie within the shaded area outside the friction circle as shown in Fig. 8, the slope will be stable. Note however, that if I plots outside the shaded region it does not necessarily mean that the slope will fail, only that it is able to fail. In these cases a more sophisticated analysis may be required. Toppling failure Unlike the previous mechanisms discussed which describe sliding failures, toppling failure involves the rotation of columns or blocks of rock about some fixed base. Goodman and Bray have described three different toppling modes, flexural, block

block-flexural toppling - each of these are illustrated in Fig. 18. In general, toppling failure occurs in rock masses which have well developed, steeply dipping discontinuities. The separate rock blocks break in flexure as they bend forward (Fig. 18). Sliding or undermining of the toe of the slope by erosion initiates the toppling process and it gradually progresses back into the rock mass, forming deep wide tension cracks in the process. It can be difficult to recognise a toppling failure from the toe of the slope. As indicated above, for toppling to occur there must be sliding along discontinuities. By a simple analysis, it can be shown that if the discontinuities have an angle of friction of φj, slip will only occur if the normals, N, to the discontinuities are inclined at a less steep angle than φj. That is, for a slope of dip α, containing discontinuities of dip, δ, toppling failure will occur if (90 - δ) + φj < α. On a stereographic projection, this means that toppling can occur only if pole, N, lies more than φj

degrees below the cut slope. Also, since the discontinuities must strike approximately parallel to the cut slope, say within +/- 30°, toppling can only occur if the pole of the

Figure 18 : Types of toppling failure.




discontinuity falls within the shaded area shown in Fig. 19. This area is bounded by a great circle φj degrees below the cut slope and striking parallel to it, the horizontal great circle (the outer edge of the stereoplot) and two small circles perpendicular to the strike of the cut and 30° from the centre of the net.

ANALYSIS OF SLIDING BLOCKS Without water Consider the problem of a block sitting on an inclined surface (Fig. 20). The block can move either by sliding or toppling or a combination of sliding and toppling. Only sliding will be considered here.

For the case of sliding only, and for a factor of safety of F = 1, consideration of force equilibrium gives

φαα tancossin ⋅+= WcAW where α is the slope angle, φ is the angle of friction between the block and the slope, c is the cohesion between the block and the slope, A is the base area of the block and W is the weight of the block.

If c = 0, then α = φ. Note that the values of c and φ depend on the magnitude of normal force. This leads to non-linear behaviour and more complications. However, for the time being we will assume that c and φ are constant. With water The prediction of water pressure is a rock mass is even more difficult than it is for soils, and considerable judgement must be used. Consider a similar configuration as that above, but this time include a water pressure distribution as shown in Fig. 21. Again for F=1

Figure 19 : Kinematic analysis of toppling

αW

W cosα

W sinα

R

Figure 20 : Sliding rock block




( ) φαα tancossin UWcAVW −+=+

where V and U are the total forces on the block due to water pressure and may be estimated from the following equations.

αγ ω cos21

bhU =

αγ ω cos21 2hV =

With reinforcement Again, consider a similar configuration, except this time introduce a rock anchor (Fig. 22). Assume that a tension force T is imposed into the rock anchor. As in the previous two cases, for F = 1

φβαφα tan)sincos(cossin TUWcATVW +−+=−+

For minimum T, calculate dT/dβ, which leads to β = φ. Alternatively,

( )βα

φβαcossin

tansincosTVWTUWcA

F−++−+

=

Note: F can be increased by increasing T and/or reducing U and V by dewatering.

LANDSLIDES AND SLOPE STABILISATION

& REQUIRED

Waltham Chapters 35 to 36 : pp. 70 – 73

REFERENCES AND FURTHER READING Goodman, R.E. 1980. Introduction to Rock Mechanics. John Wiley & Sons, New York. Hoek, E. and Bray, J.W. 1977. Rock Slope Engineering, Revised Second Edition, Institution of Mining and Metallurgy. Priest, S.D. 1985. Hemispherical Projection Methods in Rock Mechanics. George Allen and Unwen, London.

αW

W cosα

W sinαR

VU

V

Figure 21 : Sliding rock block with water

αW

W cosα

W sinαR

VU

V

Tβ

Figure 21 : Sliding rock block with water and rock anchor




REVIEW QUESTIONS 1. Determine the trend and plunge of the lines of intersection of two rock joint sets

with dip direction/dip of 146/59 and 266/36. What is the pitch of this line in each of the planes. (Ans. 219/26, 31o from SW end of plane 1, 49o from southern end Plane 2)

2. Two lines of trend/plunge 124/68 and 227/32 are known to be coplanar. Determine the obtuse angle between the lines. What is the trend and plunge of the line that bisects the obtuse angle between the two lines ? What is the orientation of the plane containing the two lines. What is the apparent dip of this plane in the direction 197o? (Ans. 115o, 068/21, 150/70, 62o)

3. A length of core, from a borehole with an axis of trend/plunge 143/68 contains a discontinuity plane of dip direction/dip 204/47. It is known that the core has rotated through a clockwise angle (looking down the axis) of 140o during retrieval from the hole. What will be the apparent dip direction/dip of the discontinuity plane as the core emerges from the borehole ? (Ans. 330/82)

4. A rock mass contains the following recurrent discontinuities : bedding strikes N32oE, dips 75oN58oW, φ = 25o; joint set 1 dip direction/dip 090/65, φ = 28o; joint set 2 horizontal, , φ = 20o. As part of a feasibility study you are asked to determine the maximum safe slopes for a circular quarry which is to be formed in this rock mass. Carry out your analysis at 15o intervals around the quarry.

5. For the rock mass described in the previous question, determine the best orientation for a highway cut through a ridge in this rock.

6. The block shown is required to have a safety factor of 1.2 against sliding. If the anchor is inclined at β = φ, what prestress, T, should be put into the anchor cable ? Investigate the influence of angle β on the required tension in the cable.

7. Calculate the factor of safety for the slope shown if the tension crack is half full of water. Investigate the change in factor of safety as the depth of water in the tension crack varies.

35ο

V

Tβ

3m

4m

Water filledcrack

Rockγ = 20 kN/m3

φ = 40o, c = 0

γ = 25 kN/m3

9m

14.8m

30o

30m

30oweak bedding planec′=50 kPaφ′ = 30o

Subject CIV3247 Geoengineering 14.1 Topic 14 : Earth Pressures in Soil and Rock



TOPIC 14 – Earth Pressures in Soil and Rock



PREFACE ............................................................................................................... 14.2

PLASTIC EQUILIBRIUM STATES....................................................................... 14.2

EFFECTS OF DISPLACEMENT............................................................................ 14.4

SLOPING SOIL ...................................................................................................... 14.5

c - φ MATERIALS .................................................................................................. 14.5

FINITE SIZE OF PLASTIC ZONE ......................................................................... 14.5

INFLUENCE OF WATER ...................................................................................... 14.6

WALL ROUGHNESS............................................................................................. 14.6

COULOMB’S WEDGE SOLUTION ...................................................................... 14.6

GENERAL WEDGE METHOD.............................................................................. 14.7

TABULATED SOLUTIONS................................................................................... 14.7

INSITU STRESSESS IN ROCK ............................................................................. 14.7






PREVIEW Introduction The stresses within the ground and those imposed by the ground can have a significant influence on the performance of structures supporting the ground, e.g. tunnels and underground openings, retaining walls etc. It is usual to assume that the principal insitu stresses are vertical and horizontal (although this is not always the case as changes to insitu stresses occur as a result of tectonic movements etc.). While the vertical stress can usually be estimated reasonably easily and accurately, the estimation of horizontal stress is much more problematic. The relative magnitudes of vertical and horizontal stresses also influence design and often govern the mode of failure. This topic introduces fundamental earth pressure theory. The topic will focus mainly on earth pressure theory applied to soil, but will also make some comments on earth pressures in rock. Objectives • To gain knowledge on fundamental earth pressure theory • To understand the concepts of active and passive earth pressure and earth pressure at

rest and to be able to calculate their values. • To understand the limitations of classical earth pressure theory and be aware of the

range of solutions that are available. PREFACE Earth pressures exerted by soils on structures cannot be calculated without an understanding of plastic equilibrium states in soils, i.e. what are the stresses in a soil mass on the point of shear failure. These stresses may then be used to calculate forces on structures such as retaining walls, but there is a major difference in comparison to above-ground structures where the applied loads are estimates of what are actually expected to be applied. With sub-surface structures the actual earth loads depend greatly on the deformations which are allowed to occur and in some cases the design loads may be significantly less than the loads the structure is carrying. Hopefully what follows will throw some light on this paradox! PLASTIC EQUILIBRIUM STATES For a semi-infinite body of soil with a horizontal ground surface the vertical stress at any depth, z, is simply the overburden pressure σv=γz (using γ or γ′ as appropriate). The lateral stress is then σh = Kγz = Kσv . K is a lateral earth pressure coefficient with a reasonably large range of values from about +0.2 to about +3.0 and in this range of values three special cases emerge, viz:




(1) Ko - the coefficient of earth pressure at rest

If the soil remains undisturbed the condition is described as at rest or zero lateral yield. For an ideally elastic soil Ko = ν/(1-ν) so that a Poisson’s Ratio of 0.3 would imply Ko = 0.43. Actual soils have typical values of :

Normally consolidated clay 0.45 ≤ Ko ≤ 0.7 Over consolidated clay 0.6 ≤ Ko ≤ 3.0 Sands and gravels 0.35 ≤ Ko ≤ 0.5

Rocks can have significantly higher and lower values of Ko due to tectonic movements, stress relief and erosion. There are some cases where Ko has been measured close to zero.

& REQUIRED


@

Activity 14.1 Work through example problem 7.9 (pp. 226) in Holtz and Kovacs

(2) KA - the coefficient of ACTIVE earth pressure

This coefficient defines the lowest possible lateral pressure in the mass, with the whole soil body on the point of shear failure. If, as a thought experiment, a lateral stress release is allowed to occur throughout the depth of the soil mass the lateral stress at any point will decrease from Koγz to KAγz, at which point shear failure occurs and any further lateral expansion takes place at constant stress KAγz.

From simple geometry KA = (1-sinφ)/(1+sinφ). Active pressures in rock are a function of discontinuity strength and orientation. Ka corresponds to conditions of normal faulting where the major principal stress is the vertical stress and failure occurs by horizontal extension.

(45+φ/2)

τ

FAILURE

FAILURE

σ3f′ σ1′ σH′ = Κ0σ1′ φ′

• • •

Ξ

Ξ

σ ′

σ3f′= KAσ1′

Figure 1 : Active limit equilibrium state




(3) KP - the coefficient of PASSIVE earth pressure

This coefficient defines the maximum possible lateral pressure in the soil mass. In this case the lateral pressure is caused to increase steadily so that the original lateral stress σ3′ = Koσ1′ increases, passing through σ1′ to become the new major principal stress at KPγz, when shear planes form at (45 - φ/2) to the horizontal. It may be shown that KP = (1+sinφ)/(1-sinφ), i.e. KP = 1/KA. Attempts to continue to compress the soil laterally take place at constant stress KPγz. Passive pressures in rock are a function of discontinuity strength and orientation. Kp corresponds to conditions for reverse faulting, with the vertical stress being the minor principal stress and failure occurs via horizontal compression

Note that in the ACTIVE case the soil movement is assisted by gravity, but in the PASSIVE case the soil movement is against gravity, hence the large difference in the two limiting pressures. KA and KP as defined above are known as the RANKINE (1857) earth pressure coefficients. EFFECTS OF DISPLACEMENT Very small lateral movements are sufficient to mobilise the fully active state, while large movements are required to mobilise the fully passive state. This has practical significance in that frequently an additional safety factor is applied to passive resistance to limit deformations.

(45-φ/2)

τ

FAILURE

FAILURE

σ3′

σ1f′ σΗ′

φ′ • • •

Ξ

Ξ

σ ′

σHf′= KPσv′ = σ1f′

Figure 2 : Passive limit equilibrium state

σv′ = σ3f′

1 100

Away from backfill W all movement

Pressure on wallC

A

B p o

pP

pA

∆x/H %Towards backfill

∆x

H

Figure 3 : Effect of lateral displacement on earth pressure




SLOPING SOIL Rankine also produced a closed form solution for the case of a semi-infinite cohesionless soil mass with ground surface sloping at angle β to the horizontal.

p K z zA A= =− −

+ −γ γ β

β φ β

β φ βcos

cos (sin sin )

cos (sin sin )

2 2

2 2

For β=0 (horizontal ground) the Rankine solution implies that there are no shear stresses on any vertical plane, while for sloping ground the resultant stress on vertical planes is parallel to the surface, i.e. inclined at angle β above the horizontal. These are matters of some significance. c - φ MATERIALS Bell (1915) extended the Rankine solution for horizontal soil surface to the case of a c - φ soil, leading to :

p z cA =−+

−−+

γφφ

φφ

( sin )( sin )

( sin )( sin )

11

211

i.e. pA = KA γz - 2c√KA

(This equation had actually been derived by Coulomb in 1776! but is now frequently referred to as the Rankine-Bell solution.) This equation implies that at the ground surface the lateral stress is -2c√KA, but as soil has very low tensile strength a tension crack is assumed to open to the depth at which the theoretical pressure becomes

positive, i.e. to depth zo where zcKo

A

=2

γ.

FINITE SIZE OF PLASTIC ZONE The solutions given above were all for a semi-infinite mass of soil. To be of use in retaining wall design they must also be applicable to finite shear zones, as illustrated in the Figure 4. If we imagine part of the soil, to depth H, to the left of a vertical plane to be removed and replaced by a wall structure, if the wall is allowed to displace to the left under the action of the lateral earth pressures a Rankine active shear zone will develop behind the wall and the stresses within this zone will be exactly the same as for the semi-infinite mass. This principle provides the basis for retaining wall design. If the wall is pushed into the soil (e.g. bulldozer blade) then a passive shear zone will form.

wall movement

elastic zone

plastic zone

Figure 4 : Limited zone of plastic equilibrium




INFLUENCE OF WATER If a water table is present in the soil behind a wall effective soil stresses must be used for calculating earth pressures. The effective weight is reduced by buoyancy to γ ′, leading to low values of KAγ′z, but the water exerts a full hydrostatic pressure of 9.81z at any depth, unmodified by the KA factor, causing large destabilising horizontal forces on the wall. Therefore it is always good practice to provide adequate wall drainage. WALL ROUGHNESS Application of Rankine solutions to retaining walls implies that the wall back is ‘perfectly smooth’, as there must be no shear stresses on vertical planes. In practice most walls are to some extent ‘rough’ and shear stresses must be mobilised as the soil slips with a downwards component relative to the wall. For a c-φ soil the soil/wall interface is characterised by two similar parameters cw (≤ c) and δ(≤ φ), the ‘wall adhesion’ and ‘angle of wall friction’ respectively. For soil on concrete δ is frequently taken as 2/3 φ, while cw is either ignored or taken as cw = c up to a cut-off limit of 50 kPa. For a normal stress of σn at a point on a wall back the maximum shear stress which may be mobilised is given by τmax = cw + σn tanδ. For a slip plane passing through soil only the equation becomes τmax = c + σn tanφ. COULOMB’S WEDGE SOLUTION

Coulomb, in his masterly 1776 paper, introduced wedge solutions to allow for rough wall effects, opening the way to realistic retaining wall design. In a trial wedge analysis a potential failure plane is assumed and a force triangle drawn, as shown in the figure below. Once PA has been determined from the force triangle various other slopes of failure surface, θ, are analysed until the one giving the maximum value of PA is found. For

purely frictional backfill Müller-Breslau (1906) derived an analytical solution for the geometry as shown in Figure 6, eliminating the need for trial graphical solutions. Note that in all wedge solutions it is not necessary for the soil inside the sliding wedge to be in a state of plastic equilibrium. The wedge slides as an intact body and only the single slip surface need be on the point of shear failure.

• max

W

θ δ

PA φ

• • PA

WR

PA

R

Figure 5 : Coulomb wedge analysis

Soil φ , γ

H

β

θ α

W

PA δ

Figure 6 : Coulomb analytical solution (Müller-Breslau)




When retaining walls are used to support rock masses, it is important that the orientation and strength of the discontinuities are considered in the analysis. For example, if a vertical cut is to be made into a very strong rock mass, then the rock mass will not need to be supported if there are no joints dipping into the excavation and if toppling is not an issue. However, if for example, continuous discontinuities exist with a strength of φ = 30o at a dip of 60o (= 45 + φ/2) into the excavation, then the full active earth pressure for φ = 30o will need to be supported; i.e. the rock mass is no stronger in this regard to a sand with a friction angle of φ = 30o. For shallower and steeper discontinuity inclinations, lower values of earth pressure need to be supported. The calculation of these pressures is based on a wedge analysis similar to that described by Coulomb. GENERAL WEDGE METHOD The Coulomb wedge method is readily extendable to more complex, general problems involving seeping water, layered backfills, surcharge loadings, cw and δ, and tension cracks. A more complex vector polygon must be drawn, but otherwise the principles remain the same. TABULATED SOLUTIONS The previous U.K. Code for Earth Retaining Structures (CECP2-1951) contained useful tables of earth pressure coefficients for use in the formula:

pAN = KA ∑ γeff z - KAC.c

where pAN is the normal earth pressure on the back of a vertical wall, with horizontal ground surface, and the KA and KAC coefficients include the effects of wall roughness. Graphed solutions for a number of realistic cases are also available in the current U.K. Code BS8002:1994 or in greater detail in books by Caquot and Kerisel (1948) and Kerisel and Absi (1990). The graphs and tables also include passive pressures.

& REQUIRED

Chapter on earth pressure theory in any book dealing with foundation engineering (e.g. Das).

INSITU STRESSESS IN ROCK Insitu stresses in rock masses not only result from the weight of over-lying materials, but are also significantly influenced by tectonic actions, confinement and loading history (or as more commonly referred to as stress history). In situ stresses may vary from almost zero to values which approach the uniaxial strength of the rock. The in situ stress can have a large influence on the behaviour of rock masses and hence is an important parameter to be estimated in design. For example, in regions of low in situ




stress, rock blocks may fall from the surface of excavations as the low stresses allow joints to open. On the other land, in regions of high in situ stresses, floor heave in excavations is likely and there may even be catastrophic results from a sudden and violent release of stored energy. In situ stresses are compressive. Although some situations may theoretically arise where stresses should be tensile, these have never been measured. In rock mechanics, much more emphasis is placed on the measurement of in situ stress than in soil mechanics. This is mainly due to the fact that the in situ stress in rock is comparatively much greater than it is in soil and as a result has a much greater influence on the design process. This is particularly true for horizontal in situ stresses. In soil (and rock) the horizontal in situ stress varies between Ka and Kp (active and passive values of K respectively - see Section 10.3) times the vertical stress, and in general is limited to a value of approximately 3 times larger than the vertical stress. However, in rock the horizontal in situ stress can be many times greater than the vertical in situ stress (which arises due to much greater values of Kp). This becomes very important when designing and constructing structures in or on rock. For example, when dealing with underground excavations the following comments are applicable: • the excavation has its longest dimension aligned with the highest principal stress • if the in situ stress is very high there may be a need to change the shape of the

excavation to minimise stress concentrations • cracks propagate in a direction approximately perpendicular to the minor (most

tensile) principal stress and hence there may be a need to design the excavations so that cracks don’t run into one another

• pressure tunnels can be constructed and operated in rock without lining if the in situ stress is greater than internal fluid pressure (there may be a need however to line the tunnel for other reasons such as pollution from seepage etc.).

• be wary of in situ stresses once they reach about 25% of the unconfined compressive strength as

- minimum stress concentration around a tunnel is approximately 2 - cracking begins in UCS tests at approximately qu/2.0 Other examples include: • when excavating with explosives on the surface, more efficient yields can be

obtained if the excavation is aligned with the minor principal stress direction. • measurement of in situ stress can be used as an early warning device of earthquakes

e.g. Newcastle earthquake • rock properties depend on stress level ⇒ accurate measurement of properties

depends on a knowledge of the in situ stress. REFERENCES AND FURTHER READING 1. BS8002 : 1994 British Standard : Code of practice for Earth retaining structures 2. Das, B.M. 1999. Principles of Foundation Engineering, Fourth Edition, ITP 3. Clayton, C.R.I and Milititsky, J. 1986. Earth pressure and earth-retaining structures,

Surrey University Press. 4. Most other books on foundation engineering.




REVIEW QUESTIONS 1. The stability of swimming pool walls in areas where the water table is high can be

of concern on emptying the pool. Why ? 2. A 5m high retaining wall is supporting medium dense sand with properties of c′ = 0,

φ′ = 30o, γbulk = 16.5 kN/m3 above the water table and γsat = 19.3 kN/m3 below the water table. The water table behind the retaining wall is at 2.5m depth. Determine the at rest, active and passive pressure distributions on the wall and the resultant forces for each case. Briefly explain the situations (e.g. types of wall) in which you would use the at rest, active and passive distributions that you have derived.

3. A 6m high retaining wall is to support a soil with unit weight γ = 17.4 kN/m3, soil strength parameters of c′ = 15 kPa, φ′ = 26o. Determine the Rankine active force per unit length of the wall both before and after the tensile crack occurs, and determine the line of action of the resultant in both cases.

4. Draw a free body diagram of the forces acting on a soil wedge behind a retaining wall. Assume that the wall has a height, H, wall friction, δ, soil internal friction angle of φ and unit weight γ. For different wedge angles, α, determine analytically the active force per metre acting on the wall. For assumed values of H, δ etc, plot these values on a graph of force versus wedge angle and confirm that your optimum solution agrees with the published solution.

5. A 10m high retaining wall supports a rock mass which has joints dipping at 60o into the excavation. The joints have a friction angle of 40o and the rock mass has a unit weight of γ = 24 kN/m3. Determine the active force per unit length acting on the wall for a smooth wall (δ = 0o) and a rough wall (δ = 30o). If the joints dip at 30o rather than 60o, what will be the active force on the wall?

6. A swimming pool wall is 5m high. The water level inside the pool is at the surface. The water table in the ground is at 2m depth. What is the net water force acting on the wall when the pool is full? How does this value change if the pool is emptied ?

Subject CIV3247 Geoengineering 15.1 Topic 15 : Earth Retaining Structures



TOPIC 15 – Earth Retaining Structures



RETAINING WALL TYPES .................................................................................. 15.2

DESIGN METHOD ................................................................................................ 15.3

MODES OF FAILURE ........................................................................................... 15.3

DEFORMATION CONSIDERATIONS.................................................................. 15.3

GRAVITY WALL DESIGN.................................................................................... 15.4

BEARING CAPACITY........................................................................................... 15.5

OVERALL SLOPE FAILURE ................................................................................ 15.5

CANTILEVER WALLS.......................................................................................... 15.5

ADDITIONAL LOADINGS ................................................................................... 15.6

OTHER WALL TYPES .......................................................................................... 15.6






PREVIEW Introduction The simplest way to hold two areas of soil (or rock) with differing ground levels in a stable state is to join them with a slope of appropriate angle, however this is very wasteful of usually valuable surface area. A vertical boundary between the two soils is therefore commonly required and for most soils the provision of some type of earth retaining structure or retaining wall is necessary. The design of retaining walls remains one of the least satisfactory aspects of modern Geoengineering, partly because of the ill-defined nature of the forces actually acting on such structures and this problem has created difficulties in adopting partial factor limit state design methods, though the new U.K. Code BS8002:1994 makes a bold attempt. The earlier Code, CECP2-1951 used a single lumped safety factor approach which is still popular in practice and the first Australian Code on retaining structures should be published soon. In this Topic, several aspects of geomechanics are combined to enable the analysis and design of retaining walls. There are many types of retaining walls available, the design and analysis of each one requiring an understanding of the interaction between the soil/rock and the wall. Emphasis will again be on walls used to retain soil, however much the same principles apply to rock. Objectives • to apply fundamental earth pressure theory to the design of retaining walls • to understand the possible modes of failure of retaining walls and to be able to

design against the occurrence of these modes • to appreciate the important role played by water RETAINING WALL TYPES New types or sub-types of retaining walls continue to emerge to take advantage of new technology, but most may be listed in three main categories, viz ♦ Gravity walls - including mass concrete, masonry, rock walls, crib walls, gabions ♦ Cantilevered walls - including reinforced concrete, diaphragm walls, soldier pile

walls, sheet piling ♦ Reinforced earth - including strip and grid reinforced walls, soil nailing, rock

dowelling Of course there is some overlap between these categories but the classification is useful. Choice of the best type for a specific site is based on considerations of cost, aesthetics, foundation conditions, height difference, cut vs fill, availability of materials, material to be retained, size of work area and scale of project.




DESIGN METHOD Once possible types have been selected and strength parameters measured for both retained soil or backfill and foundation subsoil, earth pressure theory is used to estimate design forces acting on the structure. The safety factor, or some other index of stability, is then calculated for the likely modes of failure and compared with acceptable values established by accumulated experience. A degree of iteration is usually necessary as the initial dimensions of the wall have to be assumed before calculations can start and if the wall proves to be either unsafe or overconservative in design these dimension must be revised. Attention is also given to general matters such as drainage, surface finishes, etc. Computer programs are available for much of this work, but there is also need for regular human intervention. MODES OF FAILURE The common modes of ‘failure’ are

♦ Sliding on base ♦ Overturning about toe ♦ Bearing capacity failure ♦ Overall slope failure ♦ Excessive deformation and settlement ♦ Structural failure ♦ Loss of toe resistance (sheet piles)

Not all modes need be investigated for all walls, as some may usually be ruled out as not relevant. DEFORMATION CONSIDERATIONS As indicated in the previous lecture, the earth pressures are controlled by the deformations that are possible. For example if a wall is constructed then backfilled, if no lateral movement is permitted the earth pressure on the wall back would be close to the Ko value. If the wall is then allowed to translate, a small movement would drop the pressure to the KA or active value. In reality, the wall would move sufficiently until the reduced earth pressure is in equilibrium with restoring forces, particularly base resistance, and there it would stop. For DESIGN it is usually assumed that at failure the ACTIVE pressure is mobilised, even though this is the lowest possible pressure the wall can experience. Shear failure in the soil behind a wall does not imply failure of the wall! Passive resistance of the soil in front of the toe of a wall is not always included in the equilibrium equations for practical reasons, but when it is it is usually reduced by a factor of 2 because of the large deformations needed for its full mobilisation. For a basement wall, where both top and base levels are prevented from translating by the floor structure, at least a Ko stress level should be used. Braced excavation walls experience unusual deformation patterns and special approaches are required.




GRAVITY WALL DESIGN As an introductory example of the process a gravity wall will be considered, followed by extensions to more complex designs.

PA and PP are calculated by the methods of Topic 14. PP should be ignored if the depth of soil on the passive side is less than (0.6 + 0.1H) to allow for possible future excavation in front of the wall. U1 and U2 are calculated from a flow net or other suitable method and W is the weight of the wall. The effective soil base force is :

N′= W + PA vert - U2 (+ PP vert)

The available base shear force, S, may then be estimated using the Coulomb shear strength equation and the soil/base shear parameters cbase and δbase. Many designers ignore cbase

because of the possibility of construction disturbance of this layer and take δbase as equal to φ (foundation soil) for cast-in-situ walls and 2/3 φ for precast walls. Hence

S = N′ tanδ′base ( +c′ B) where B = base width

F against sliding failure is the ratio of Restoring Force to Motivitating Force, i.e. FS = S/(PAhor + U1). Normally FS should be between 1.5 to 2.0 depending on factors such as the consequences of failure, quality of site investigation, confidence in contractor, etc. F against overturning failure is the ratio of Restoring Moment to Overturning Moment, taken about the toe point, A, but N′ is not included, as if rotation is about to occur the N′ force will have moved across to the toe.

MOT = PA hor yPA - PA vert xPA + U1 yU1 (+ U2 xU2 ?)

MR = WxW + (PP hor yPP ?)

FOT = MR/MOT and again should be between 1.5 and 2.0. Both safety factors are lumped safety factors with the same degree of reliability implied on all parameters and it can be argued that some of the terms could appear as either restoring or disturbing forces, with appropriate changes of algebraic sign. If FS is too low four approaches are possible : increase base width B slope the wall base use a base key reduce water pressures

yPA

xU2

PA

U1

W

PP

N′ U2

S • A

xPA

yU1

xW

xN

H

Figure 1 : Gravity wall analysis




FOT is usually expected to be acceptable if the resultant force on the base falls within the middle third. BEARING CAPACITY

The wall base exerts contact stresses on the foundation soil and the potential for bearing capacity failure must be checked. Forces involved are shown below : Vector R is the resultant of all forces on the wall back plus the wall weight, and its eccentricity at the base/soil interface may be found by taking moments about the toe. Standard bearing capacity theory may then be applied to determine the safety factor against bearing failure. Remedial measures include widening the base or using piled foundations.

OVERALL SLOPE FAILURE In this mode of failure the retaining wall and a considerable volume of soil surrounding it move as a unit, as for a typical slope stability failure, and the normal methods of slope stability analysis apply. Care must however be taken to exclude slip circles that pass through the wall (except where the wall is segmented or of inadequate construction). This is usually only a problem for thick deposits of soft clays. CANTILEVER WALLS For walls of medium height concrete cantilever walls are often used because of their efficient use of materials. Using the virtual wall back concept they may sometimes be analysed in a similar fashion to gravity walls, frequently using the simple Rankine solutions. If the angle θ shown in Figure 3 is ≤ (45 + φ/2) then a fully active Rankine zone abcda will develop as the wall yields to the left to produce the necessary stress release and there are zero shear stresses on the virtual wall back bd. For stability analyses the composite concrete and soil wall daghijbed is treated as a smooth backed gravity wall with Rankine pressures applied. If the soil surface adc slopes at angle β the Rankine solution for a sloping soil mass may be used

heel

c

x

R v

R h

R

toe e

Figure 2 : Contact forces at base

c

e f h i

j

θ

virtual back

a d g

b

Figure 3 : Cantilever wall - virtual wall back concept




with an effective wall friction angle δ = β on the vertical plane. If the heel width ef is too small to meet the θ criterion then aghijba may be taken as the composite wall, with forces on ab calculated from a general wedge or other appropriate analysis, using δ = φ on the sloping virtual wall back ab. ADDITIONAL LOADINGS Several other sources of stresses on walls often need to be considered. The main one is surcharge loading where loads of point, line and area nature are applied to the soil surface or the wall itself. Elastic stress analysis solutions are available for point and line loads and a uniform surcharge may be included as an additional height of soil above the real surface, with ∆h = q/γ, where q = uniform surcharge stress. Some codes recommend that all walls be designed for a minimum surcharge of 10 kPa. Another important source of stress is soil compaction behind non-yielding walls and several semi-empirical methods are available for estimating values. OTHER WALL TYPES The embedded walls sub-group of cantilever walls, which rely heavily on passive soil resistance for their stability, and the currently very popular reinforced and soil nailed walls require specialised analyses of their own, which may be found in most modern texts. Small walls for domestic and similar purposes are well treated in commercial data sheets.

& REQUIRED

Chapter on retaining walls in any book dealing with foundation engineering (e.g. Das).

REFERENCES AND FURTHER READING 1. BS8002 : 1994 British Standard : Code of practice for Earth retaining structures 2. Das, B.M. 1999. Principles of Foundation Engineering, Fourth Edition, ITP 3. Clayton, C.R.I and Milititsky, J. 1986. Earth pressure and earth-retaining structures,

Surrey University Press. 4. Most other books on foundation engineering. REVIEW QUESTIONS 1. Why is drainage an import aspect of retaining wall design? 2. Describe briefly the different modes of failure of a gravity retaining wall. 3. The ground surface immediately behind a retaining wall is subjected to a surcharge

loading of 10 kPa. If the soil behind the retaining wall has strength properties of c′=0 and φ′=30o, determine the active pressure imposed on the wall by the surcharge load at a depth of 1m, 2m and 5m.

4. What influence does the backfill (material properties, extent of etc) behind a retaining wall have on the pressures that act on the wall?

Subject CIV3247 Geoengineering 16.1 Topic 16 : Tunnels



TOPIC 16 Tunnels



REQUIRED READING .......................................... 16.¡Error! Marcador no definido.

Subject CIV3247 Geoengineering 16.2 Topic 16 : Tunnels



PREVIEW Introduction Tunnelling and underground space development is an important aspect of geotechnical engineering. With the increased cost of land in urban areas, underground space utilisation and development are becoming much more viable. This topic introduces preliminary aspects of tunnel design. The main emphasis is on the design of temporary lining using classification systems such as Q and RMR. Objectives • To gain knowledge of the problems associated with tunnelling in hard and soft

ground • To be able to apply the Q and RMR rock classification systems to the design of

temporary linings for tunnels • To become familiar with the various tunnel construction techniques and to be aware

of methods of tunnelling through bad ground

& REQUIRED

Waltham Chapter 37 : pp. 74 – 75 Chapter 38 : pp. 76 – 77

& SUGGESTED

Papers by Leca and Kaiser (on this CD)

Subject CIV3247 Geoengineering 17.1 Topic 17 : Shallow Foundations



TOPIC 17 – Shallow Foundations



BEARING CAPACITY........................................................................................... 17.2 Appropriate strength parameters .......................................................................... 17.3

Fine grained cohesive soils (slow draining) : eg. clay....................................... 17.3 Coarse grained granular soils (fast draining) : eg. Sand. ................................... 17.3

Other Considerations ........................................................................................... 17.3 Rigidity............................................................................................................ 17.3 Influence of adjacent footings .......................................................................... 17.4 Position of ground water table .......................................................................... 17.4 Soil strength increasing with depth................................................................... 17.4 Stiff soil overlying soft layer ............................................................................ 17.4 Soft soil overlying stiff layer ............................................................................ 17.5 Eccentric loading ............................................................................................. 17.5 Footings with moments.................................................................................... 17.5

Bearing Pressures ................................................................................................ 17.5 Gross Pressure ................................................................................................. 17.5 Net pressure ..................................................................................................... 17.6

Computer Analyses.............................................................................................. 17.6

SETTLEMENT ....................................................................................................... 17.6 Components of Settlement ................................................................................... 17.6 Total Settlement................................................................................................... 17.7 Methods for estimating settlement........................................................................ 17.7 Corrections for Rigidity and Depth .....................................................................17.10 Poulos and Davis Elastic Method........................................................................17.11

SHALLOW FOUNDATIONS ON ROCK..............................................................17.11

REFERENCES AND FURTHER READING.........................................................17.12

REVIEW QUESTIONS..........................................................................................17.12




PREVIEW Introduction Many structures rely on shallow foundations or footings to transmit structural loads to the ground. The design of such footings must ensure that the ground can safely carry the load without failing (usually in shear) (ultimate limit state) and free from excessive deformations (serviceability limit state). The term excessive is relative as the amount of allowable settlement that different structures can sustain varies considerably. The geotechnical design of shallow foundations therefore includes two main steps; determination of bearing capacity and of deformation (usually vertical settlement). Both topics were introduced in CIV2271 Introductory Geoengineering. This topic builds on that knowledge and supplements the information contained in the CIV2271 notes. Objectives • To be able to determine the bearing capacity of foundations on soil and rock • To be able to estimate settlements on these foundations • To understand the importance of short and long term behaviour and to know which

one will be critical to the performance of a foundation • To understand the difference between net and gross bearing pressure and when they

should be applied BEARING CAPACITY Refer to CIV2271 notes Chapter 7 – Bearing Capacity for introductory information. In the CIV2271 notes, Terzaghi’s bearing capacity equation and Meyerhof’s bearing capacity factors for a surface strip footing subjected to a vertical load were introduced.

γγBNqNcNq qcu 21

++=

where c = cohesion, q = effective stress at the level of the bottom of the foundation, γ = unit weight of soil, B = width (or diameter) of foundation and Nc, Nq, Nγ are the bearing capacity factors which depend on the friction angle of the soil. These factors need to be modified for footing shape and depth, and load inclination as well as a number of other influences. The full bearing capacity equation is given by :

γγγγγ NFFBFNFFqFNFFcFq idsqqiqdqsccicdcsu 21

++=

where Fcs, Fqs, Fγs are shape factors, Fcd, Fqd, Fγδ are depth factors and Fci, Fqi, Fγι are load inclination factors (see CIV2271 notes). This equation shows that bearing capacity depends on the soil strength and the geometry of the footing. The question arises as to which strength parameters should be used to estimate bearing capacity.




Appropriate strength parameters The choice of strength parameters for bearing capacity calculations depends largely on excess porewater pressure considerations. The generation and subsequent dissipation of excess porewater pressures depends on the soil type and the time taken to complete construction. Fine grained cohesive soils (slow draining) : eg. clay Usually use undrained parameters (cu and φu (=0)) since the time taken for construction is usually shorter than the time required for the dissipation of excess porewater pressures. That is, immediately after loading of the footing, all load will be carried by the porewater. This will generate a local rise in porewater pressure, which will slowly dissipate (in many clays, it may take several years for excess porewater pressures to dissipate completely). As the excess pore pressures dissipate (usually some time after construction has ceased), the soil will consolidate and become stronger. However at the end of construction, little dissipation of porewater pressure is likely to have occurred and the strength of the soil will not have changed significantly from its value before construction. To ensure safety, the undrained properties of the soil are therefore adopted. Coarse grained granular soils (fast draining) : eg. Sand. Due to their free draining nature, no excess porewater pressures can be generated (except perhaps under very rapid earthquake loading situations where increases in pore pressure may lead to liquefaction). Hence it is the effective or drained strength parameters which are usually used ( c’ (=0) and φ’.) Note however that the friction angle of sand depends on amongst other things, the density of the sand. Given the difficulty of obtaining undisturbed samples of sand, it is very difficult to carry out representative strength tests in the laboratory. Instead, insitu tests (e.g. the standard penetration test or SPT) are usually adopted. Unfortunately, for most of these tests, friction angle cannot be determined directly but instead must be inferred from empirical correlations. Such methods introduce increased uncertainty with regard to strength parameters. Other Considerations Rigidity It is usually accurate enough to assume that strips and pad footings are rigid. The contact pressure under rigid footings is reasonably uniform. However, as footings become larger, e.g. rafts, they tend to become more flexible, and contact pressures less uniform. Local contact pressures under footings can become very high (e.g. under columns) and may exceed the bearing capacity. Bending moments in the raft can also exceed design specifications. Special care must be taken with raft foundations and these will be dealt with at Level 4.




Influence of adjacent footings The influence of adjacent footings on one another is unclear. Usual procedure is ♦ when centre to centre spacing S > 3B (B is the width of the footing) then influence

can be ignored. ♦ For S < 3B, adjust shape factors to account for overall layout of the footings. (Has

very little influence though). Position of ground water table Can have a significant influence. Usually adopt the highest likely location. Choose the appropriate values of q and γ. Three possibilities : ♦ For GWT at surface, use γ’ ♦ For GWT greater than 2B use γ ♦ For 0 < GWT < 2B linearly interpolate between γ and γ’ Soil strength increasing with depth Meyerhof’s equations are appropriate for soil deposits of constant strength. However, most soil deposits (especially soft normally or lightly over-consolidated soils) increase in strength with depth, which can lead to over-estimates of bearing capacity if the average strength is adopted. Skempton suggests that the average strength can be used if it is within +/- 50% of the strength of the soil at a depth of 0.67B below the footing. Engineering judgement needs to be applied for other situations. Stiff soil overlying soft layer

Usually occurs when a stronger granular soil or compacted fill overlies a layer of soft clay, or in cases where a crust may form on the surface of a soft clay. Failure usually occurs by punching through the stiff soil. Nevertheless, in the absence of more appropriate solutions, the bearing capacity equation for generalized shear failure is usually adopted.

For a footing at depth, d, in a stiff layer of thickness, d + Z, the following steps are followed : 1. Calculate the bearing capacity for the soft layer (with q = γο(d +z) where γo is the

unit weight of the stiff soil). 2. Increase this value by a factor 1/η, where η is the ratio of the stress at the surface of

the soft layer divided by the contact stress at the base of the footing. η can be determined from appropriate tables or graphs (e.g. see 2271 settlement notes).

Care must be taken however, because if the stiff layer is very thick, the bearing capacity can be very large. Obviously, the bearing capacity cannot be greater than the value obtained for the stiff layer on its own.

B Stiff soil

Soft soil

Z

d




Soft soil overlying stiff layer Failure occurs by the soft soil being squeezed out from beneath the footing. Using plasticity theory it is possible to derive the following theoretical solutions :

For a strip footing of width B for Z/B<0.5 qu net = {B/2Z + π + 1}c for Z/B>0.5 normal bearing capacity equation

For circular foundation of diameter B for Z/B<0.17 qu net = {B/3Z + π + 1}c for Z/B>0.17 normal bearing capacity equation

Eccentric loading In many footings, the column load is applied eccentrically to the footing. This results in non-uniform pressure being applied to the soil, with local pressures being greater than the average pressure. Meyerhof (1953) suggested a practical solution to this problem by adjusting the footing dimensions until the column load becomes a central load. The new footing dimensions, L’ and B’, for bearing capacity calculations

are therefore L’ = L-2eL and B’ = B-2eB where L and B are the footing’s original dimensions and eL and eB are load eccentricities in the L and B directions respectively. Footings with moments Columns can apply both a moment, M, and axial loading, P, to a footing. The moment is handled by moving the axial load to an eccentricity, e, such that M = Pe. The above solution for eccentric loading can then be employed. Bearing Pressures A footing will be safe in bearing if the allowable net (rather than gross) bearing pressure applied by the footing is less than the bearing capacity of the soil (reduced by an appropriate amount by the strength reduction factor – see CIV2271 notes). Unfortunately structural engineers work with gross bearing pressures, so some care must be taken when communicating recommendations for allowable bearing pressures. Gross Pressure The gross pressure exerted is the total force (including the weight of the footing) acting at the base of the footing divided by the total plan area of the footing.

B

Soft soil

Stiff soil

Z

P e

qmax

e

P P

M




Net pressure The net bearing pressure for bearing capacity is that component of the gross pressure which requires mobilization of the soil strength to support the foundation. It depends primarily on foundation geometry and is usually independent of the position of the water table. Several different cases can be identified : ♦ For a dry excavation with a footing at depth d below the surface, qnet = qgross - γd (γ

is the unit weight of the soil). ♦ For a flooded excavation qnet = qgross - γ’d; and ♦ For an excavation which is much wider than the footing qnet = qgross. As is explained in the chapter on settlements, the net bearing pressure used for bearing capacity may be different from that used for settlement calculations. Computer Analyses Complex bearing capacity problems can be solved using several computer based methods including: ♦ limit equilibrium slope stability programs such as GWEDGEM in which the footing

is modeled as a flat slope with a surcharge loading. Solutions can be obtained relatively quickly with only minimal input data. Program must be able to handle surcharge loads and non-circular failure surfaces.

♦ Non-linear finite element programs. These programs are very sophisticated and require considerable experience to obtain reasonable solutions. Considerable effort is required to generate input data, and solution times can be substantial. Not generally recommended for design, but useful as a research tool.

♦ Other numerical programs such as FLAC. Can suffer from the same problems as the finite element programs, but tend to require less input data and smaller computation times. Again require considerable experience to use.

SETTLEMENT An introduction to the settlement of shallow footings is provided in the CIV2271 notes. This topic extends the analysis of settlement of footings. Components of Settlement Settlement of soil (and rock) is time dependent and consists of three separate components (see Topic 7) :

1. Initial or elastic settlement, ρι , occurs immediately on application of the load and results from the elastic compression of the soil. Elastic settlement (as the name implies) is completely reversible; i.e. on removal of the load the soil rebounds to its original position. It is time independent. Exhibited by all soil and rock.

2. Primary consolidation settlement, ρc , is a time dependent process that results from water being squeezed out of the voids due to effective stress changes in the




soil. It is only partly reversible. The amount and rate of settlement varies with soil type. In clay, drainage of excess pore water pressures and therefore primary consolidation settlement, occur very slowly. As described in Topic 7, the oedometer test can be used to determine the amount and rate of consolidation. Consolidation settlement is usually the most dominant component of settlement for most types of clay and can be many times larger than the initial settlement. In sand, drainage of excess pore pressures and therefore primary consolidation settlement, occur very rapidly. It is difficult to separate primary consolidation from elastic settlement. Therefore it is usual for primary consolidation to be included as part of the initial settlement. As a result, the initial settlement of sand is not completely recoverable. In rock, consolidation settlements tend to be small and are usually ignored or, like sand, included with initial settlement .

3. Secondary compression or creep, ρσ , is the time dependent (and very slow)

process that follows primary consolidation. It occurs at constant effective stress (i.e. no drainage of pore water occurs) and is irreversible. Depending on stress level, it can be significant in clay (usually soft clay for civil engineering applications), sand and rock (deep underground mining applications). In field measurements, it can be very difficult to separate primary consolidation and secondary compression settlements.

Total Settlement The total settlement, ρΤ , is the sum of the above three components: i.e.

ρ ρ ρ ρT i c s= + + Methods for estimating each component follow. Methods for estimating settlement Initial settlement is usually calculated using elastic solutions (e.g. see Elastic Solutions by Poulos and Davis). For a shallow footing founded on a layer of “elastic” soil, the initial settlement of the footing of width, B, can be calculated from

( )ρ

υρ I

E

Bqgrossi

21−=

where E and ν are the Young’s modulus and Poisson’s ratio of the soil respectively, qgross is the gross bearing pressure and Iρ is a settlement influence factor which depends on foundation shape, flexibility and layer depth. Values of Iρ for a rigid rectangular footing of length, L, founded on the surface of an infinitely deep “elastic” soil layer are listed in the CIV2271 notes on settlement of




shallow footings. Values of Iρ for other footing types and layer depths can be found in any one of a number of reference books. The choice of E and ν depend on the soil. For clay, no drainage has occurred (initially that is) and the undrained values Eu and νu (=0.5) should be used. For sand, drainage occurs rapidly. Hence the values that are adopted should reflect the drained behaviour of the material; i.e. use the drained values E′ and ν′. Rock is usually treated in the same way as sand. Young’s modulus and Poisson’s ratio for clay and rock can be determined from the appropriate triaxial test. Values for sand are usually estimated from correlations with simple field tests such as the Standard Penetration Test (or SPT). Primary consolidation settlement can be calculated from the results of the oedometer test. For normally consolidated clay (p′o = p′c):

′

′∆+′+

=o

o

o

orc p

ppe

HC log

1ρ

where p′c is the preconsolidation pressure, Ho is the initial layer thickness, p′o is the initial effective vertical stress, eo the initial void ratio, ∆p′ is the change in effective stress and Cc is the compression index. For overconsolidated clay (p′o < p′c) : If p′o+∆p′ < p′c then

′

′∆+′+

=o

o

o

orc p

ppe

HC log

1ρ

else, if p′o+∆p′ > p′c then

c

o

o

oc

o

c

o

orc p

ppe

HC

pp

eH

C′

′∆+′+

+′′

+= log

1log

1ρ

where Cr is the recompression index. For situations involving swelling (or effective stress decrease) the amount of swelling can be estimated from the first of the above two equations, but with Cr replaced by Cs. Alternatively, for non-linear e - logp′ responses, the following equation using the coefficient of volume change, mv, can be used.

pHm ovc ′∆=ρ

The implementation of these equations changes from application to application. As described below, in order to estimate the consolidation settlement beneath a shallow footing, each soil layer must be divided into a number of sub-layers. This is because the effective stress change felt by the soil will vary with depth. Immediately below the




footing, the soil will experience the full net bearing pressure applied by the footing. However, the soil at greater depth will only experience a portion of the net bearing pressure. There are a number of analytical solutions that can be used to determine the variation of stress beneath a uniformly loaded area (which approximates a shallow footing) on an infinitely deep soil layer. The Boussinesq equation for the vertical stress, σz, at depth, z, below a uniformly loaded circular area of radius, a is given by :

( ) 23

21

11

+

−=

zaq

zσ

The vertical stress at depth, z, below the corner of a uniformly loaded rectangular footing of length, L, and width, B, on an infinitely deep layer is given by

+−+++

+++++

+++++

= −

112

tan12

112

41

2222

221

22

22

2222

22

nmnmnmmn

nmnm

nmnmnmmn

qz

πσ

where m = B/z and n = L/z. To determine the stress below the centre of the footing, split the footing into quarters and use superposition. That is, determine the stress beneath the corners of a footing with dimensions L/2 x B/2. The stress below the centre of a footing L x B will be then be 4 times the stress below the corner of a footing of L/2 x B/2. Similar solutions exist for other footing shapes and for finite layers (refer to standard references). To estimate the consolidation settlement beneath the footing, carry out the following steps : 1. Divide the soil layer into a number of sub-layers. 2. Determine the initial effective overburden stress, po′ and the change in effective

stress, ∆p′ at the mid-height of each sub-layer. 3. Use the appropriate equation given above to determine the consolidation settlement

of each layer. Note that Ho in this case will be the thickness of each sub-layer (which may vary from layer to layer). Also, if using the equation involving mv, note that mv will vary with stress level, and the appropriate value of mv should be used.

4. Sum the settlements from all layers to determine the overall primary consolidation settlement.

This method is particularly suited to implementation on a spreadsheet. The process is somewhat simplified for situations in which the effective stress change is approximately uniform throughout the layer. This will arise when fill is placed over a large area or the water table is lowered (See Topic 7). In such cases, sub-layering is not required unless the properties of the layer vary with depth.




Through experience, the above method has been found to give reasonable estimates of primary consolidation settlement for N/C clays but over-estimates settlement for O/C clays, especially when the loaded area is small compared to the thickness of the clay layer. By considering 3D consolidation effects, Skempton and Bjerrum (1957) suggested that the settlement estimated from the above equation should be reduced by a factor µ given by :

)1( AA −+= αµ

Where A is Skempton’s pore pressure parameter and α is a factor that depends only on geometry. Values for circular (diameter B) and square (width B) footings are :

z/B 0 0.25 0.5 1 2 4 10 ∞ α 1.0 0.67 0.5 0.38 0.30 0.28 0.26 0.25 α 1.0 0.74 0.53 0.37 0.26 0.20 0.14 0.0

where z is the thickness of the consolidating layer. Typical values for A and µ are : A µ Sensitive soft clay 1 ~1 N/C clay 0.5 – 1 0.9 – 1 O/C clay 0.25 – 0.5 ~0.5 Heavily O/C sandy clay <0.25 0.25 – 0.3 Secondary compression or creep settlement for a layer of thickness Ho from time ti to tf can be estimated using the following equations:

+=

i

fo

ps t

tH

eC

log1

αρ

or

=

i

fos t

tHC logαερ

where ep is the void ratio of the soil at the end of primary consolidation (eo is often used instead). Corrections for Rigidity and Depth If it is assumed that a uniform pressure exists over the footing then the footing is assumed to be flexible. Most footings however are closer to being rigid and hence a correction needs to be made. The total settlement of a rigid footing is approximately 0.8 of that estimated for the centre of a flexible footing.




Total settlements also reduce as the footing founding depth increases. For a footing of width, B, founded at depth, d, below the surface, the following correction factors, DF, are appropriate :

d/B 0 0.5 1 2 3 4 5 >10 DF 1.0 0.95 0.89 0.86 0.84 0.82 0.81 0.8

Note that correction factors apply to total settlements. Poulos and Davis Elastic Method Poulos and Davis proposed that the soil could be modelled as a two-phase elastic material. The initial (undrained) settlement, ρi, is estimated as usual using elastic theory and undrained values of Young’s modulus, Eu, and Poisson’s ratio, νu. The total final settlement, ρT (ignoring creep settlements) is also estimated using the same elastic theory, but replacing the undrained parameters with the drained parameters, E′ and ν′. The consolidation settlement, ρc, is then the difference between the total final and initial undrained settlements; i.e.

iTc ρρρ −=

This type of model is particularly suited to computer analysis using numerical methods such as finite element, finite difference and boundary element techniques. It should be noted, that because consolidation occurs between the undrained and drained states of the soil, E′ < Eu. Care must also be taken in selecting the values of E′ and ν′ especially when dealing with soft clays that undergo significant volume change during loading. The Poisson’s ratio adopted must reflect this volume change, otherwise excessively large horizontal movements may be predicted. Values much lower than 0.3 and perhaps as low as 0.05 may need to be used. Parameters should be determined from laboratory tests (usually drained triaxial tests) that follow a similar stress path to that experienced by the soil in the field. SHALLOW FOUNDATIONS ON ROCK The design of shallow foundations on rock is usually carried out using the same techniques as applied to footings on soil. However care must be taken to use representative rock mass parameters (rather than intact rock values) which can be estimated using rock mass classifications systems such as GSI and RQD. In cases where preferential jointing is observed, rock mechanics principles should be applied to the design to account for the possibility of failure or deformation along these discontinuities.




REFERENCES AND FURTHER READING 1. Das, B.M. 1999. Principles of Foundation Engineering, Fourth Edition, ITP –

Chapters 3 & 4. 2. Most other books on foundation engineering. REVIEW QUESTIONS 1. Explain why you would use different parameters to determine the bearing capacity

of a footing and the stability of a cut in the same clay. 2. Why is likely that bearing capacity will be overestimated for shallow foundations

situated on normally consolidated clay deposits if the average strength of the clay is adopted ?

3. Derive the equations for net bearing pressures of dry, flooded and wide excavations listed on page 17.6.

4. When estimating elastic settlements, why should gross rather than net pressure be used ?

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