Shear Strength of Soils Final (Complete)

176
Shear Strength of Soils CE2112 Soil Mechanics National University of Singapore Department of Civil and Environmental Engineering

Transcript of Shear Strength of Soils Final (Complete)

Page 1: Shear Strength of Soils Final (Complete)

Shear Strength of Soils

CE2112 Soil MechanicsNational University of Singapore

Department of Civil and Environmental Engineering

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Shear Strength of Soils : Introduction

What is Strength?• Power to Resist Force

Merriam-Webster Dictionary

Typical Material Strengths

Merriam-Webster Dictionary

Typical Material Strengths considered in Civil Engineering

• Concrete Strength• Steel Strength• Soil Strength

Introduction - 1

• Soil Strength

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Concrete StrengthShear Strength of Soils : Introduction

Concrete Strength• Concrete Cubes

• Typically cured for 28 days

• Apply axial load until the cube crushescube crushes

• The maximum stress (l d/ ) th t th t(load/area) that the concrete cube can sustain before it

h i k itcrushes is known as its ___________________ (e.g.

)compressive strength

28 d i t thIntroduction - 2

________________________ ).28 day compressive strength

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Steel StrengthShear Strength of Soils : Introduction

Steel Strength• Steel bar specimen

• Subject the steel bar specimen to tensile pulling at its two ends.

• Stress-strain curve looks like this:

yield strength

Introduction - 3

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Soil StrengthShear Strength of Soils : Introduction

Soil Strength• Triaxial Cell Equipment

Loading Piston

Triaxial Cell Equipment

• Cylindrical Soil Specimen, e.g. 76 mm (height) x 38 mm (diameter), typical

• Triaxial cell filled with controlled

(height) x 38 mm (diameter), typical aspect ratio ___. 2:1

• Triaxial cell filled with controlled pressurized _____.

V ti l t i li d t th il

water

• Vertical stress is applied to the soil specimen by controlling the ___________________________.movement of the loading piston

• Force on the loading piston and the movement of the piston is measured and

Introduction - 4

recorded. The data is processed to obtain the stress-strain curve.

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Two Important Aspects of Geotechnical ProblemsShear Strength of Soils : Introduction

p p(a) Deformation/Consolidation

• Covered by Prof Lee.y• This course provides you with some simple

tools to make preliminary settlement /

• Difficult and challenging topic in practicedeformation calculations

• Very experienced engineers may come upwith different predictions for deformation /settlementsettlement

• Material effects : variability, complex stress-strain relationship, elastic-plastic response

• Dependent on modulusstrain relationship, elastic plastic response

(gradient of stress-strain curve)

• For soft clays C and C (are these related to modulus?)

• Geometrical effects : 2D and 3DIntroduction - 5

• For soft clays, Cc and Cs(are these related to modulus?)

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Two Important Aspects of Geotechnical ProblemsShear Strength of Soils : Introduction

(b) Stability (or Failure)p p

• Dependent on Soil Strength• Soil strength is highly variable, even within

it• Collapse (may or may not happen)

same site

• Loss of Stability

• Loss of Functionality

• Excessive deformation

• Loss of Functionality

Introduction - 6

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Response of Soil to Load Applications

Shear Strength of Soils : Response of Soils to Load Applications

Response of Soil to Load Applications

When the applied loads are small:Soil stresses are low, deformations are smallStability of the soil structure typically not a problem

When the applied loads are large:Soil stresses are high, deformations are largeStability of the soil structure may become a problemFailure may occur

Introduction - 7

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Shear Strength of Soils : Response of Soils to Load Applications

In geotechnical engineering, FAILURE can have different ‘meanings’:

In extreme case, we have collapse.In more general cases, we have limiting conditions (not necessarily complete breakdown) beyond which the soil structure cannot be used.

Introduction - 8

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I t t h i l i i bl

Shear Strength of Soils : Response of Soils to Load Applications

In most geotechnical engineering problems,ground deformation is only considered afterwe have checked that the design is safewe have checked that the design is safeagainst failure.

In some cases, where deformation is not ani id i himportant consideration, we may not have tocheck for deformation/settlement, but we stillhave to check that the structure is safe againsthave to check that the structure is safe againstfailure.

To analyze or perform a failure check, we needto understand the material behavior of the soil.to understand the material behavior of the soil.

Introduction - 9

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If l k t il t i l th th

Shear Strength of Soils : Response of Soils to Load Applications

If we look at soil as a material, then there areinvariably two major aspects that one oftenneeds to be concerned with for any material:needs to be concerned with for any material:

A. StiffnessA. Stiffness

B. StrengthThe _______ of a material defines how much deformation the material will undergo given a

stiffness

certain stress (or, more appropriately in soils, stress increment).e.g. Young’s modulus, shear modulus, etc

Note: the compression index C is the of thereciprocalNote: the compression index Cc is the _________ of the soil modulus in 1-D compression

reciprocal

Introduction - 10

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Shear Strength of Soils : Hooke’s Law

Hooke’s LawH k ’ L i i l f id li iHooke’s Law is a simple way of idealizingthe behaviour of most materials, bysimply assuming that the strainincrement that the material undergoes islinearly

c e e t t at t e ate a u de goes srelated to the stress increment.

This simple law has been found to worki ll i h i lquite well with most materials, up to a

point.

Hookes’ Law - 1

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Hooke’s LawShear Strength of Soils : Hooke’s Law

Hooke s Law

2 )N

)

200fixed

(kN

/m2

linear

e F

(kN

150

100= F/

A

L cross-sectional E

Forc 100

50ress

σsectional area A

1

E

δ

Str

Elongation δ (mm)F

Strain ε = δ/LStrain ε = δ/L

σ = Eε Hookes’ Law - 2 where E is the Young’s modulus

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Th St th f M t i lShear Strength of Soils : Layman’s Definition

The Strength of Materials

In reality, no material can be made to sustainstresses infinitely. Beyond a certain point(which varies depending upon the material),the structure of a material breaks down andno further increase in stress is then possible.At this point, we often say that the materialhas _____, and the stress level at which thisstate is achieved is known as the ________ of

failedstrength

that material.

A G l L ’ D fi itiA General, Layman’s Definition

Strength of Materials - 1

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Shear Strength of Soils : Layman’s Definition

no further increase

perfectly plastic

no further increase in stress

strength

stress σ onset of yieldingyielding

b H k ’ Llinear

strain ε

obey Hooke’s Law

strain ε

A General, Layman’s Definition

Strength of Materials - 2

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Shear Strength of Soils : Stress-Strain Response

Stress-Strain Curves - 1

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Shear Strength of Soils : Stress-Strain Response

Stress-Strain Curves - 2

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Some Aspects of the Stress-Strain ResponseShear Strength of Soils : Stress-Strain Response

Some Aspects of the Stress Strain ResponseThe example stress-strain curves show that variousdefinitions of strength can be adopted.g pSteel:

• Clearly defined yield point, or limit ofy y p ,proportionality.

• Beyond this, material suffers increasingBeyond this, material suffers increasingdeformation without breaking up.

• Can define ____________ as the stressyield strength____________level at which yielding occurs.

y g

(limit of proportionality exceeded)

• Can define _______________ as thestress level at which the material finallybreaks up

ultimate strength

breaks up.(maximum strength attainable)

Stress-Strain Curves - 3

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E l St St i C f

Shear Strength of Soils : Stress-Strain Response

St lExample Stress-Strain Curve of _____Steel

ultimate strength

yieldyield strength

Stress-Strain Curves - 4

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Shear Strength of Soils : Stress-Strain Response

Example Stress-Strain Curve of Clay

perfectly plastic

no further increase in stress

ultimate

onset of

perfectly plasticultimate strength

yield stress σ onset of

yieldingstrength

obey Hooke’s Lawlinear

strain ε

Stress-Strain Curves - 5

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Stress-Strain Response of Soft SoilsShear Strength of Soils : Stress-Strain Response of Soft Soils

What if there is no perfectly plastic response observed?Stress Strain Response of Soft Soils

σ

triaxial compression

approximately linear modulus ε

initial modulus Ei

approximately linear modulus at very small strains

ε

Stress-Strain Curves - 6

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Shear Strength of Soils : Stress-Strain Response of Soft Soils

Stress-Strain Response of Soft Soils

In such cases, it may be more relevant todefine the strength as the stress level atwhich _________________________ isreached.

a certain limiting strain level

For most soft soils, this level is often seto ost so t so s, t s e e s o te setat ____ strain.20%

Stress-Strain Curves - 7

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Shear Strength of Soils : Stress-Strain Response of Soft Soils

Stress-Strain Response of Soft Soils

σ

triaxial compression

failure t thstrength

initial modulus E

approximately linear modulus at very small strains

ε

initial modulus Ei

20% failure strainat very small strains failure strain

Stress-Strain Curves - 8

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I ti il b l d d i

Shear Strength of Soils : Different Modes of Loading Soils

• In practice, soil can be loaded in manydifferent ways.e g constr ct a ne b ilding the soil beneath the b ildinge.g. construct a new building – the soil beneath the building

experiences an ________ in load, which may cause parts of

increase

the soil to reach the yield strength in loading

excavate for basement – the soil beneath the excavation experiences a ________ in load, which may cause parts of the soil

decreasey p

to reach the yield strength in unloading

• Different modes of loading give rise to_______ strengths!differentWe will discuss this in greater detail later.

Types of Strength - 1

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Th St th f S il

Shear Strength of Soils : Different Types of Strength

The Strength of Soils• In textbooks, soil reports, other literature, you may, p , , y y

encounter different types of soil strengths beingdiscussed.

• Most commonly, you would see the term _____________.Shear Strength

• For rocks sometimes you may come across the terms• For rocks, sometimes you may come across the terms___________________ and ______________.Tensile StrengthCompressive Strength

F il h d l f th• For soils, such as sands or clays, we focus on the_____________.Shear Strength

• As the name implies, the shear strength of a soilrefers to the strength mobilized by the soil undershearing conditionsshearing conditions.

Types of Strength - 2

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Different Modes of Strength Testing

Shear Strength of Soils : Different Modes of Strength Testing

Different Modes of Strength Testing

σaσa - σa

σaσa

σa - σa σa

Compressive Testing Tensile Testing Isotropic Testing

Do these load conditions result in shearing?

Types of Strength - 3

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Diff t M d f T ti

Shear Strength of Soils : Different Modes of Strength Testing

Different Modes of Testingσa

σ σbσb

σa

σa ≠ σb

⇒ Shear Loading

Types of Strength - 4

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Diff t M d f T ti

Shear Strength of Soils : Different Modes of Strength Testing

Different Modes of Testingσa

σ σbσb

σa

σa ≠ σb

⇒ Shear Loading

Types of Strength - 4

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Diff t M d f T ti

Shear Strength of Soils : Different Modes of Strength Testing

Different Modes of Testingσa

σ

τ

σb

σn

σa ≠ σb

⇒ Shear Loading

Types of Strength - 4

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Complex StressShear Strength of Soils : Complex Stress

p

O

• Consider a body that is acted upon by external forces

• Consider an arbitrary point O within the body.

General and Principal Stresses - 1

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Complex StressShear Strength of Soils : Complex Stress

σap

τc

τa

O σb

σcτb

• Consider a body that is acted upon by external forces

D t th t l f th t ti

• Consider an arbitrary point O within the body.

• Due to the external forces, the stress acting on anyplane that passes through O is generally inclined tothe normal to the plane.the normal to the plane.

• Such a stress has both a normal and a tangentialcomponent, and is known as a compound, or complex,

General and Principal Stresses - 1

component, and is known as a compound, or complex,stress.

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Principal PlaneShear Strength of Soils : Principal Planes

pτa

O σb

τb σnprincipal

• At any point O within the body, a plane that is actedupon by a only is known as a principalnormal stress

principal plane

upon by a ____________ only is known as a principalplane.

• On a principal plane there is no tangential or

normal stress

shear• On a principal plane, there is no tangential, or _____,stress present.

shear

• For a general body, only _____ principal planes canexist in a stressed mass.

three

General and Principal Stresses - 2

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Principal StressShear Strength of Soils : Principal Stress

• The normal stress acting on a principal plane isreferred to as a principal stress.

• At every point in a soil mass, the applied stresssystem that exists can be resolved into three principalstresses that are mutually orthogonal

• These are the _____ principal stress, σ1major

stresses that are __________________.mutually orthogonal

the ___________ principal stress, σ2

the _____ principal stress, σ3

intermediateminor

• Critical stress values and obliquities generally occuron the two planes normal to the intermediate plan, soth t th ff t f b i dthat the effects of σ2 can be ignored.

• Two dimensional solution is possible.

General and Principal Stresses - 3

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General Stress State on Any PlaneShear Strength of Soils : General Stress State on Any Plane

• Consider the major principal plane, acted upon by themajor principal stress σ1 , and the minor principalplane acted upon by the minor principal stress σ3plane, acted upon by the minor principal stress σ3 .

σ1τ

σ3

τθ

B id i th ilib i f th l t h it

σn

• By considering the equilibrium of the element shown, itcan be shown that on any plane, inclined at an angle ofθ to the direction of the major principal plane, there is aj p p p ,shear stress τ and a normal stress σn . The magnitudesof these stresses are :

θσ−σ 231 i( ) 2 θ=τ 2

231 sin( ) θσ−σ+σ=σ 2

313n cosGeneral and Principal Stresses - 4

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Shear Strength of Soils : The Mohr Circle Method

The Mohr Circle MethodThe Mohr Circle Method

Mohr Circle

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Mohr CircleShear Strength of Soils : The Mohr Circle Method

θσ−σ

=τ 22

31 sinθσ−σ

+σ+σ

=σ 222

3131n cos

These equations, which provide a complete (in two dimensions) description for the state of stress, describe a circle.

This graphical representation of the state of stress is known as the Mohr circle.

σ1 SIGN CONVENTION

S

σ3

τθθ

• Stresses will be considered positive when ___________.

• τ is positive when _______________.compressive

counterclockwise

σnθ• θ is measured _______________

from the direction of σ1 .counterclockwise

Mohr Circle - 1

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Mohr Circle – General PlaneShear Strength of Soils : General Stress State on Any Plane

σ1τθ

σ3

θθ

σnθ

Y τDτθ

OX

σnA B

2θC

α

σ3

σnθ

σ3

σ1

Mohr Circle - 2

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Mohr Circle – Failure PlaneShear Strength of Soils : General Stress State on Any Plane

σ1τf

σ3

fθf

σnf

Y τ Dτf

OX

σnA B

2θf

σ3

σnf

σ3

σ1

Mohr Circle - 3

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Mohr Circle – Maximum Shear StressShear Strength of Soils : The Mohr Circle Method

Mohr Circle Maximum Shear Stressσ1

τθ

σ3

θ45°

σnθ

Y τ Dτθ

OX

σnA B

90°

σ3

σnθ

σ3

σ1

Mohr Circle - 4

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Strength EnvelopesShear Strength of Soils : The Mohr Circle Method

Assume φ is constant.Mohr strength envelope

Y τ

OX

σnA BC

φφ Xφ

Circle B : tangential to the strength envelopeg g pcondition of incipient failure

Circle A : completely within the strength envelopequite stable

Mohr Circle - 5

quite stableCircle C : beyond the strength envelope

cannot exist

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Relationship between φ and θf

Shear Strength of Soils : The Mohr Circle Method

p φ f

Y τ Dτf

Oσn

A2θf

σ f

τf

OXA BC

σ3

σnf

σ1

A l DCO 180° 2θAngle DCO =Angle DOC =

A l ODC

180° – 2θf

φ

90°Angle ODC =

Sum of angles in ΔODCSum of angles in ΔODC

90°

Mohr Circle - 6

φ + 90° + 180° – 2θf = 180°

Hence θf = φ/2 + 45°

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Example 1O f il l i l f i ti l f d d th t t l

Shear Strength of Soils : The Mohr Circle Method (Example 1)

On a failure plane in a purely frictional mass of dry sand the total stresses at failure were:

shear = 3.5 kN/m2 ; normal = 10.0 kN/m2

Determine the resultant stress on the plane of failure, the angle of shearing resistance of the soil, and the angle of inclination of the failure plane to the major principal plane

Y τ D3.5

failure plane to the major principal plane.

OX

σnA B

2θf

10 XA BC10

Resultant stress = OD = 22 1053 +. = 10.6 kN/m2

Mohr Circle - 7

tan φ = Angle of shearing resistance φ : 3.5/10 = 0.35 ⇒ φ = 19.3°

Angle of inclination of failure plane θ = φ/2 + 45° = 54.6°

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Example 2Gi

40 kN/m2

B

Shear Strength of Soils : The Mohr Circle Method (Example 2)

30°

Given

20 kN/m220 kN/m2

B σ1

σ3

τθθ

20 kN/m20 kN/m2

B

3

σnθ

40 kN/m2

θ =Solution 2θ = 60°⇒Calculate the stresses on the plane B-B.

30°

Y τθ =Solution 2θ = 60⇒

τθ

30°

Oσn60°

4020

10

σ θ

θ

OX4020

10cos 60°

σnθ

Mohr Circle - 8

Normal stress on B-B σn =Shear stress on B-B τ =

30 + 10 cos 60° = 35 kN/m2

10 sin 60° = 8.7 kN/m2

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Example 3 20 kN/m2

Shear Strength of Soils : The Mohr Circle Method (Example 3)

30°

Given

40 kN/ 240 kN/ 2

B

σnθ

τθ

θθ40 kN/m240 kN/m2

B

σ1

σ3

θθ

- 60°

20 kN/m2B

Calculate the stresses on the plane B-B.

σ3

Y τθ = -60°Solution 2θ = -120°⇒

Oσn

20

10cos 60°

σnθOX120° 4020

10τθ

Mohr Circle - 9

Normal stress on B-B σn =Shear stress on B-B τ =

30 - 10 cos 60° = 25 kN/m2

-10 sin 60° = -8.7 kN/m2

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Shear Strength of Soils : The Mohr Circle Method (Method of Poles)

The Method of PolesThe Method of Polesfor Mohr Circle Solutionsfor Mohr Circle Solutions

Mohr Circle

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Example 1 40 kN/m2

Shear Strength of Soils : Method of Poles (Example 1)

30°

Given

20 kN/m220 kN/m2

B

20 kN/m220 kN/m2

B40 kN/m2

Calculate the stresses on the plane B-B (using the origin of planes or the method of pole).Solution (35, 8.7)

p )

a) Draw the Mohr circleb) Draw line A’-A’ through point (40 0) and

X B’

OP30°

4020

b) Draw line A -A through point (40,0) and parallel to plane on which (40,0) acts.

A’ A’c) Intersection of A’-A’ with Mohr circle at point (20 0) is the origin of planes OP

60°

10cos 60°point (20,0) is the origin of planes, OP.d) Draw line B’-B’ through OP parallel to

B-B.) R d di t f i t X h B’ B’

B’

Mohr Circle - 10 Shear stress τ = 8.7 kN/m2

e) Read coordinates of point X where B’-B’ intersects Mohr circle.

Normal stress on B-B σn = 35 kN/m2

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Example 2 20 kN/m2

Shear Strength of Soils : Method of Poles (Example 2)

30°

Given

40 kN/m2 40 kN/m2

B

40 kN/m2 40 kN/m2

B20 kN/m2

Calculate the stresses on the plane B-B (using the origin of planes or the method of pole).Solution

p )

a) Draw the Mohr circleb) Draw line A’-A’ through point (20 0) and

OP

30° 4020

10cos 60°

b) Draw line A -A through point (20,0) and parallel to plane on which (20,0) acts.

A’ A’c) Intersection of A’-A’ with Mohr circle at point (40 0) is the origin of planes OP 60°

B’

(25 8 7)

point (40,0) is the origin of planes, OP.d) Draw line B’-B’ through OP parallel to

B-B.) R d di t f i t X h B’ B’

60

XB’

Mohr Circle - 11 Shear stress τ =

(25, -8.7)

-8.7 kN/m2

e) Read coordinates of point X where B’-B’ intersects Mohr circle.

Normal stress on B-B σn = 25 kN/m2

B

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Example 340 kN/m2

Shear Strength of Soils : Method of Poles (Example 3)

Given40 kN/m

20 kN/m2

B60°

20 kN/m2

B B

40 kN/m2

Calculate the stresses on the horizontal plane B-B (using the origin of planes or the method of pole).

Where is the pole (origin of planes)?Where is the pole (origin of planes)?

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Example 3 40 kN/m2

Shear Strength of Soils : Method of Poles (Example 3)

Given20 kN/m2

B B60°

20 kN/m2

B B

40 kN/m2

Calculate the stresses on the horizontal plane B-B (using the origin of planes or the method of pole).

OPSolution (35, 8.7)

of planes or the method of pole).

a) Draw the Mohr circleb) Draw line A’-A’ through point (40 0) and

A’ XB’ B’

4020

b) Draw line A -A through point (40,0) and parallel to plane on which (40,0) acts.

A’c) Intersection of A’-A’ with Mohr circle at

point O is the origin of planes A’point Op is the origin of planes.d) Draw line B’-B’ through OP parallel to

B-B.) R d di t f i t X h B’ B’

Mohr Circle - 12 Shear stress τ = 8.7 kN/m2

e) Read coordinates of point X where B’-B’ intersects Mohr circle.

Normal stress on B-B σn = 35 kN/m2

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Example 4Gi 40 kN/ 2

Shear Strength of Soils : Method of Poles (Example 4)

Given 40 kN/m2

10 kN/m2

B B30°

10 kN/m2

B B

OP

20 kN/m2

Calculate the magnitude and direction of the principal stresses.

A’P

(20 10)

Solutiona) Draw the Mohr circleb) Draw line A’-A’ through point (20 10) and

A’B’

44.14

(20, 10)b) Draw line A -A through point (20,10) and parallel to plane on which (20,10) acts.

A’c) Intersection of A’-A’ with Mohr circle gives the origin of planes OP

15.8637.5°

gives the origin of planes, OP.d) Draw line B’-B’ through OP so that it

intersects the major principal stress at (44 1 0)

B’

(40, -10)

Mohr Circle - 13

at (44.1, 0).e) Measure the angle the line B’-B’ makes with the horizontal =

f) The direction of the principal stress =37.5°

52.5° to the horizontal.

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In fact the shear strength of soils come into play in

Shear Strength of Soils : Different Engineering Applications

In fact, the shear strength of soils come into play inalmost all types of geotechnical loading conditions,____________________________________________.even when the shearing mechanism is not obvious

slopeslope

mobilised potential weak

plane caused by soil shearing

shear strength

Basic Shear Strength - 1

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• In fact the shear strength of soils come into play in

Shear Strength of Soils : Different Engineering Applications

• In fact, the shear strength of soils come into play inalmost all types of geotechnical loading conditions,____________________________________________.even when the shearing mechanism is not obvious

retaining wall

steelretained

soilsteel struts

soil

mobilised shear strengthshear strength

Basic Shear Strength - 2

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• In fact the shear strength of soils come into play in

Shear Strength of Soils : Different Engineering Applications

• In fact, the shear strength of soils come into play inalmost all types of geotechnical loading conditions,____________________________________________.even when the shearing mechanism is not obvious

Load

Pile shaft QQs

Basic Shear Strength - 3

Pile base, Qb

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Shear Strength of Soils : Drained vs Undrained Strength

• The shear strength is one of the key input parametersthat the user has to specify in order to carry out ap y yrealistic geotechnical analysis.

• When carrying out a geotechnical analysis, it isimportant to distinguish between two types of shearstrengths for a given soil:strengths for a given soil:

_______________ vs. _________________Drained Strength Undrained Strength

For the same soil, drained and undrainedstrengths can be significantly different

Basic Shear Strength - 4

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• Within each type of shear strength (either drained ord i d) th diff t f h

Shear Strength of Soils : Different Modes of Testing

undrained), there are different _________ of shearstrength, depending on the ____________________________________.

measuresstrength testing equipment

and proceduredirect shearsimple shear

triaxial shearothers (lab vane test, pocket

penetrometer)

• For the same soil tested under drained (or undrained)condition, the shear strengths obtained from directh t t i l h t t t i i l t t ld

p penetrometer)

shear tests, simple shear tests or triaxial tests wouldshow some variation due to the _____________________________________________________________

different boundaryand load conditions associated with different test________.

• In this course, we would cover some of the morecommon laboratory test methods typically carried out

methods

common laboratory test methods typically carried outon soils.

• You will also have some hands-on practice on howth t t i d t i th E i t G2 l bthese tests are carried out in the Experiment G2 labsession.

Basic Shear Strength - 5

Page 56: Shear Strength of Soils Final (Complete)

• In our lecture we will cover the following tests:

Shear Strength of Soils : Different Modes of Testing

direct shear testIn our lecture, we will cover the following tests:

simple shear test

lab vane testtriaxial shear test

torvane test andExperiment G2

Th k thi b t th diff t th d f h

torvane test and pocket penetrometer

• The key thing about these different methods of shearstrength testing is that we apply an increasing shearload to the soil specimen to __________________continually deform ituntil it _______________________________.

• There are different ways to apply the shearing load.

fails in some predetermined manner

There are different ways to apply the shearing load.

• The most direct method is the .Direct Shear TestThe most direct method is the _______________.Direct Shear Test

Basic Shear Strength - 6

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Direct Shear TestShear Strength of Soils : Direct Shear Test

• Consider a soil mass

• We can load the soil mass such that it deforms asfollo sfollows:

F

F

failure shear plane• The value of F required to cause the failure shear planeThe value of F required to cause the failure shear plane

to develop is used to calculate the shear strength.Direct Shear Test - 1

Page 58: Shear Strength of Soils Final (Complete)

Schematic of the Shear Box Apparatus

Shear Strength of Soils : Direct Shear Test

Schematic of the Shear Box Apparatus

N ΔV

δ

T

• Oldest strength test? Coulomb probably used something similar• Shear box has two halves• Bottom half _____, top half _________fixed moveable• Apply ______ load N to soil specimen• Apply constant rate of _________________ to upper half of shear box

Measure T δ and ΔV as the specimen is sheared

normalhorizontal motion δ

Direct Shear Test - 2

• Measure T, δ and ΔV as the specimen is sheared

• τ = T/A, σ = N/A (where A is the nominal area of the specimen)

Page 59: Shear Strength of Soils Final (Complete)

Schematic of the Shear Box Apparatus

Shear Strength of Soils : Direct Shear Test

Schematic of the Shear Box Apparatus

N ΔV

δ

T

• Oldest strength test? Coulomb probably used something similar• Shear box has two halves• Bottom half _____, top half _________fixed moveable• Apply ______ load N to soil specimen• Apply constant rate of _______________ to upper half of shear box

Measure T δ and ΔV as the specimen is sheared

normalhorizontal motion

Direct Shear Test - 2

• Measure T, δ and ΔV as the specimen is sheared

• τ = T/A, σ = N/A (where A is the nominal area of the specimen)

Page 60: Shear Strength of Soils Final (Complete)

Schematic of the Shear Box Apparatus

Shear Strength of Soils : Direct Shear Test

Schematic of the Shear Box Apparatus

N ΔV

δ

T

• Oldest strength test? Coulomb probably used something similar• Shear box has two halves• Bottom half _____, top half _________fixed moveable• Apply ______ load N to soil specimen• Apply constant rate of _______________ to upper half of shear box

Measure T δ and ΔV as the specimen is sheared

normalhorizontal motion

Direct Shear Test - 2

• Measure T, δ and ΔV as the specimen is sheared

• τ = T/A, σ = N/A (where A is the nominal area of the specimen)

Page 61: Shear Strength of Soils Final (Complete)

Schematic of the Shear Box Apparatus

Shear Strength of Soils : Direct Shear Test

Schematic of the Shear Box Apparatus

N ΔV

δ

T

• Oldest strength test? Coulomb probably used something similar• Shear box has two halves• Bottom half _____, top half _________fixed moveable• Apply ______ load N to soil specimen• Apply constant rate of _______________ to upper half of shear box

Measure T δ and ΔV as the specimen is sheared

normalhorizontal motion

• Measure T, δ and ΔV as the specimen is sheared

• τ = T/A, σ = N/A (where A is the nominal area of the specimen)Direct Shear Test - 2

Page 62: Shear Strength of Soils Final (Complete)

Direct Shear Test

Shear Strength of Soils : Direct Shear Test

Direct Shear TestExamples of Laboratory Equipment for Direct Shear Test

circularsquare box

circular ring box

Direct Shear Test - 3

Page 63: Shear Strength of Soils Final (Complete)

N ΔVShear Strength of Soils : Direct Shear Test

N

Dense Sand

C t t t f th diff t l f N (N > N > N )• Carry out test for, say, three different values of N• For each value of N, apply motion to the upper half of shear box and

measure T and δC l l t T/A d l t δ f ll

(N3 > N2 > N1)

e = constant (dense sand)τ = T/A

N /A

• Calculate τ = T/A and plot τ vs δ as follows:

σn3 = N3/A

σn2 = N2/Aσ = N /A

shear strength orresponding to σn2 = N2/A

shear strength orresponding to σn3 = N3/A

τ2f

τ3f

σn1 = N1/A

δ

shear strength orresponding to σn1 = N1/A

to σn2 N2/A

Direct Shear Test - 4

τ1f

Page 64: Shear Strength of Soils Final (Complete)

Simple Shear TestShear Strength of Soils : Simple Shear Test

• Similar to Direct Shear, but instead of rigid box, it usesa series of reinforced membranes, or thin rings,stacked together so that the soil specimen deformedstacked together so that the soil specimen deformedas shown.

multiple shear failure planes

• Results in a more _________________ of shear stressthroughout the specimen

uniform distribution

• Interpretration of results similar to that of Direct ShearDirect Shear Test - 5

Page 65: Shear Strength of Soils Final (Complete)

Triaxial Shear TestShear Strength of Soils : Triaxial Shear Test

• Triaxial = _________ = ____________Tri + Axial Three + Axis• Triaxial Loading = Loading Applied in All Three Axis

• Usually carried out on cylindrical soil specimensσ

• Triaxial Loading = _____________________________.Loading Applied in All Three Axis

yσy

xz

σz σ

• Why do we need σ and σ ?

σz σx

Why do we need σx and σz ?In the field, a typical soil element (or particle) isacted upon by stresses from ___________all directionsFor ________________ soils, the specimen cannotstand unsupported in the absence of σx and σz

sandy or granularTriaxial Shear Test - 1

Page 66: Shear Strength of Soils Final (Complete)

In the field and are usually equal

Shear Strength of Soils : Triaxial Shear Test

• In the field, σx and σz are usually equal

σv

σvv

σx σx

σz

• Generally σ’ = σ’ = K σ’

• Ko is the _____________________________________coefficient of lateral earth pressure (at rest)

• Generally, σ x = σ z = _____Ko σ y

• For normally consolidated clay or loose to medium sandK 0 4 to 0 6

• For heavily consolidated clay or very dense sand

Ko ≈ ________0.4 to 0.6

Ko > __ may be 2 ~ 5 or even higher1Triaxial Shear Test - 2

Page 67: Shear Strength of Soils Final (Complete)

Example:

Shear Strength of Soils : Triaxial Shear Test

pWhat are the stresses acting on a soil element located 8 mbelow the ground surface? Assume water table is at theground surface. Take γbulk = 18 kN/m3 and Ko = 0.5.

At 8 m below ground,

ground surface. Take γbulk 18 kN/m and Ko 0.5.

total vertical stress σ = 8 m x 18 kN/m3 = 144 kN/m2total vertical stress σy = 8 m x 18 kN/m 144 kN/mpore water pressure u = 8 m x 10 kN/m3 = 80 kN/m2

effective vertical stress σy’ = 144 – 80 = 64 kN/m2

effective lateral stress σx’ = σz’ = Koσy’ = 0.5 x 64 kN/m2

= 32 kN/m2

total lateral stress σx = σz =Total Stresses Effective Stresses

32 + 80 = 112 kN/m2

144 kN/m2Total Stresses

64 kN/m2Effective Stresses

112 kN/m2

112 kN/m2

32 kN/m2

32 kN/m2

Triaxial Shear Test - 3

Page 68: Shear Strength of Soils Final (Complete)

Shear Strength of Soils : Triaxial Shear Test

• Consider two soil elements at different depths of 8mand 16m.

σvσ

8 m

σx

σv

16 m

σx

Triaxial Shear Test - 3a

Page 69: Shear Strength of Soils Final (Complete)

• Consider effective stresses from previous exampleShear Strength of Soils : Triaxial Shear Test

p p

64 kN/m2At 8 m depth At 16 m depth (same soil)

128 kN/m2

32 kN/m2 64 kN/m2

• Will the two soil elements (at 8 m depth and at 16 md th) h th h t th?

32 kN/m2 64 kN/m2

depth) have the same shear strength?No!! Everything being the same, the soil at the

larger depth will have a shear strengthlargerlarger depth will have a _____ shear strengththan the soil at the shallower depth.

larger

• This is due to the larger stresses, in all three directions,acting on the soil element at larger depth.

• Hence, the __________ of a soil element (due to thet ti th l t i ll 3 di ti

stress statestresses acting on the element in all 3 directions x, yand z) has an important effect on the shear strength.

Triaxial Shear Test - 4

Page 70: Shear Strength of Soils Final (Complete)

Shear Strength of Soils : Triaxial Shear Test

• Consider the soil element subjected to the uniformlydistributed load as shown below.

σv

σσx

• Consider two situations: (i) the soil response is

Triaxial Shear Test - 4a

( ) pundrained, (ii) the soil response is drained.

Page 71: Shear Strength of Soils Final (Complete)

It is important to incorporate the effect of ’ and ’

Shear Strength of Soils : Triaxial Shear Test

• It is important to incorporate the effect of σx’ and σz’when carrying out tests to measure shear strength.

• The test allows us to do this.triaxial shearThe ___________ test allows us to do this.triaxial shear

• In conventional triaxial testing, the soil specimens are.cylindrical_________.cylindrical

• The lateral stresses σx’ and σz’ are controlled byapplying all round the specimen.water pressureThis is called the ________ pressure,applying _____________ all round the specimen.water pressure

confining or the ___ pressure.cell

• Because water pressure is __________, the lateralstresses σx’ = σz’ in a triaxial test.

hydrostatic

• Note, however, that, while σx’ = σz’ , the lateral stressesσx’ and σz’ are _____________ equal to the confiningpressure applied by the water.

not necessarilyp pp yYou will understand why a bit later!

Triaxial Shear Test - 5

Page 72: Shear Strength of Soils Final (Complete)

Triaxial Shear Test

Shear Strength of Soils : Triaxial Shear Test

Triaxial Shear Test

Triaxial Shear Test - 6

Page 73: Shear Strength of Soils Final (Complete)

Triaxial Shear TestShear Strength of Soils : Triaxial Shear Test

Triaxial Shear Test

water

AB

Triaxial Shear Test - 7

Page 74: Shear Strength of Soils Final (Complete)

Isotropic Compression PhaseShear Strength of Soils : Triaxial Shear Test

• Note the two valves present in the setupBottom right: for controlling the cell water pressure (A)Bottom left: for controlling the inflow and outflow ofBottom left: for controlling the inflow and outflow of

water into the soil specimen (B)• After the specimen is placed in the cell and everythingp p y g

is set up, water is allowed to flow into and completelyfill the cell via the cell water pressure valve (A).

• The cell water pressure is adjusted to a value based on• The cell water pressure is adjusted to a value based onthe test requirement.

• By doing this, the soil specimen is subjected to an

At this point the soil is shear loading

y g , p j_________________________ equal to the cell waterpressure.all-round confining pressure

not subjected to

• If the soil specimen contains pore water,

• At this point, the soil is ______________ shear loadingyet.

excess poreIt is experiencing ________ loading.isotropic

not subjected to

If the soil specimen contains pore water, ________________________ will be generated as a result of thehydrostatic loading.

excess porewater pressure

Triaxial Shear Test - 8

Page 75: Shear Strength of Soils Final (Complete)

Isotropic Compression PhaseShear Strength of Soils : Triaxial Shear Test

• By opening or closing the porewater valve B (bottomleft side), you can choose whether to let the excessporewater pressure under hydrostatic loading dissipate:porewater pressure under hydrostatic loading dissipate:

If valve B is open, water can flow out: excess porewater p ppressure can dissipate, hence ____________ occurs.consolidation

If valve B is closed, water cannot flow out: excess porewater ppressure cannot dissipate, no ____________ occurs.consolidationWe call this an ______________ specimen.unconsolidated

• Note that if your soil specimen is a fine-grained materiallike clay, it may take for the consolidation tolongerlike clay, it may take ______ for the consolidation tooccur under the isotropic loading.

longer(could be 24 hoursor longer)

Triaxial Shear Test - 9

Page 76: Shear Strength of Soils Final (Complete)

Shear Loading PhaseShear Strength of Soils : Triaxial Shear Test

• To apply shear loading, you move the loading pistoneither down (or up) so that the soil specimen is( p) psubjected to ______________________ in axial load.an increase (or decrease)

• It looks like you are introducing just a compressive or atensile loading onto the soil specimen.

• However, it can be shown that, besides thecompressive or tensile component, there is now a

t d th_____ component, due toshearσ’x = σ’z ≠ σ’y

• This is how we subject the soil specimen to a shearloading in a triaxial test.g

Triaxial Shear Test - 10

Page 77: Shear Strength of Soils Final (Complete)

Shear Loading PhaseShear Strength of Soils : Triaxial Shear Test

• Note that, during the shear loading, the pore-watervalve B can be either open or closed.If the valve is open, then water _______________ ofthe specimen. The specimen response will be ______ .

can flow in or outdrained

If the valve is closed, then water _________________of the specimen. The specimen response will be

.

cannot flow in or out

undrained_________

• In this way, we can conduct triaxial tests to study boththe drained and undrained response of soils.p

• Triaxial test is more involved and complicated than the direct shear but also much more versatilethe direct shear, but also much more versatile

Triaxial Shear Test - 11

Page 78: Shear Strength of Soils Final (Complete)

The Use of Back-Pressure in Triaxial TestingShear Strength of Soils : Triaxial Shear Test

• backpressure helps to raise the saturation level to close to _____100%

• without backpressure, may be difficult to attain degree of saturation close to 100%, as

cell pressure

cell pressure

σ

pore pressure = backpressure u

,dissolved air may come out of solution

• Example: σcell p

cell pressure σcell = 100 kPa,backpressure u = 60 kPa,If sample allowed to consolidate under cell pressure, then effective stress after consolidation is:σ’v = σ’h = σcell – u = 100–60 = 40 kPa

AB

• water here is ___________• known as backpressure u

• If the required σ’v = σ’h value after consolidation is 80 kPa, and if we maintain the backpressure at u = 60 kPa then the required cell

pressurized

• hence the porewater pressure in the soil sample = backpressure u

60 kPa, then the required cell pressure is

σcell = σ’v + u = 80 + 60 kPa= 140 kPaTriaxial Shear Test - 12

Page 79: Shear Strength of Soils Final (Complete)

Classification of Triaxial Tests

Shear Strength of Soils : Triaxial Shear Test

Classification of Triaxial Tests according to Drainage Conditions

Before Shear During Shear SymbolDue to isotropic

Valve B Valve B

Unconsolidated Undrained UU

ue to sot op ccell pressure loading

Closed Closed

Consolidated Undrained CUOpen Closed

Consolidated Drained CDOpen Open

In experiment G2, we will be doing the _______.UU test

Triaxial Shear Test - 13

Page 80: Shear Strength of Soils Final (Complete)

In experiment G2: • Set the machine so that the F

Shear Strength of Soils : Triaxial Shear Test

piston moves down slowly and compresses the soil specimen at a constant rate.

• Measure the load F that the piston is applying on the soil specimen

L A

specimen

• Measure the compression δof the soil specimen

σcell • Process the results following

the method described in the manual to obtain the stress-manual to obtain the stress-strain curve of the form

stress σ

• Some further interpretation of the results

strain ε

to be carried out using Mohr’s circle approach.

Triaxial Shear Test - 14

Page 81: Shear Strength of Soils Final (Complete)

Shear Strength of Soils : Other Shear Strength Tests

Other Shear Strength Tests

• Quite easy and quick to use• Useful for providing quick (but

rough) estimates of undrainedrough) estimates of undrained shear strengths

• Not as Versatile or Accurate as the Triaxial TestTriaxial Test

Other Shear Strength Tests - 1

Page 82: Shear Strength of Soils Final (Complete)

Lab Vane Shear Test

Shear Strength of Soils : Lab Vane Shear Test

Lab Vane Shear Test

Other Shear Strength Tests - 2

Page 83: Shear Strength of Soils Final (Complete)

Lab Vane Shear Test

Shear Strength of Soils : Lab Vane Shear Test

Lab Vane Shear Test

Blade or Spindle

Other Shear Strength Tests - 3

Page 84: Shear Strength of Soils Final (Complete)

Shear Strength of Soils : Torvane

Torvane

Other Shear Strength Tests - 4

Page 85: Shear Strength of Soils Final (Complete)

Miniature Cone Penetrometer

Shear Strength of Soils : Miniature Cone Penetrometer

Miniature Cone Penetrometer

Other Shear Strength Tests - 5

Page 86: Shear Strength of Soils Final (Complete)

Pocket Penetrometer

Shear Strength of Soils : Pocket Penetrometer

Pocket Penetrometer

Other Shear Strength Tests - 6

Page 87: Shear Strength of Soils Final (Complete)

When Do We Mobilize the Strength of a Material?Shear Strength of Soils : Stress Dependency

• So far, we have looked at the stress-strainresponse of typical soils.

g

response of typical soils.

• The concept of ‘strength’ has been discussedin relation to the stress strain responsein relation to the stress-strain response.

• We have looked at ‘strength’ in terms of thegstrain required to mobilize it.

The stress strain response however does not• The stress-strain response, however, does nottell the complete picture

• It does not tell us the _____________________ thatwill result in the strength being mobilized.

combination of stresses

Strength Concepts Overview - 1

Page 88: Shear Strength of Soils Final (Complete)

When Do We Mobilize the Strength of a Material?Shear Strength of Soils : Stress Dependency

g

σ A

30 kPB

30 kPa

C

The stress-strain response, by itself, does not tell ushi h f th th i t A B C i lik l t h

ε

which of the three points, A, B or C, is likely to havethe strength of 30 kPa shown.

To estimate the strength at a point, we need to knowthe stress combination acting at that point, and howit affects the shear strength. This is known as the

Strength Concepts Overview - 2

it affects the shear strength. This is known as the______________.failure criterion

Page 89: Shear Strength of Soils Final (Complete)

SHEAR STRENGTHShear Strength of Soils : Shear Strength

The property that enables a material

SHEAR STRENGTH

The property that enables a materialto remain in equilibrium when itssurface is is known as itsnot levelsurface is ________ is known as itsshear strength.

not level

Think of _____, which cannot come to a state ofequilibrium when its surface is not level It has no

waterequilibrium when its surface is not level. It has noshear strength!

Shear Strength Concepts - 1

Page 90: Shear Strength of Soils Final (Complete)

SHEAR STRENGTHShear Strength of Soils : Shear Strength

SHEAR STRENGTH

A soil mass derives a large part of its shear strength from :frictional resistanceshear strength from _________________:frictional resistance

between particlesrelative sliding• _____________ between particles• ___________ between particles

relative slidinginterlocking

Shear Strength Concepts - 2

Page 91: Shear Strength of Soils Final (Complete)

F i tiShear Strength of Soils : Friction Concepts

FrictionW W WW W

H

W

H’

R

H

R’

H

R R

α φ

(a) No Horizontal Force Applied

(b) Horizontal Force Applied

(b) Horizontal Force Applied

block stationaryblock stationary

W = R cos α

sliding is imminent

W = R’ cos φR = WH = R sin α

φH’ = R’ sin φ

α = angle of obliquity φ = angle of frictionShear Strength Concepts - 3

Page 92: Shear Strength of Soils Final (Complete)

only achieves the value of whenφ

Shear Strength of Soils : Friction Concepts

• α only achieves the value of __ whensliding occurs.

φ

F i ti l i t i t t t d• Frictional resistance is not constant andvaries with the applied load untilmovement occurs

• The term tan φ is known as the _________coefficient

movement occurs.

_________.of friction

Note thatwhere N is the

l ti d'Htanφ 'H

• Note that normal reaction due to the weight WW

tan =φN

=

• Hence the friction angle φ is related to theratio of the normal force N to thehorizontal force H’

Shear Strength Concepts - 4

horizontal force H’.

Page 93: Shear Strength of Soils Final (Complete)

h i th h t ti th

Shear Strength of Soils : Friction Concepts

• Let H’ = τf A where τf is the shear stress acting on the contact surface when α = φ and A is the area of contact

• Let N = σn A where σn is the normal stress and A is the area of contact

AAtan f

στ

=φ f

στ

=

• Hence the tangent of the friction angle can alsob th ht f th ti f th h t t

Anσ nσ

be thought of as the ratio of the shear stress tothe normal stress (when sliding is imminent).

• In soil mechanics, when the ratio of the shearstress to the normal stress reaches the value oftan φ we say that the soil element has yielded

Shear Strength Concepts - 4

tan φ, we say that the soil element has yielded.

Page 94: Shear Strength of Soils Final (Complete)

The Mohr Coulomb Failure Criterion

Shear Strength of Soils : The Mohr-Coulomb Failure Criterion

The Mohr-Coulomb Failure Criterion

Otto Mohr (1835 – 1918): German civil engineerOtto Mohr (1835 1918): German civil engineerMohr Circle (1882)Early theory of strengthEarly theory of strength

Charles Augustin De Co lomb (1736 1806) F h h i i tDe Coulomb (1736 – 1806) : French physicist

Coulomb’s law: Charge ∝ 1/r2

Military engineerFrictionShear strength of soils

Mohr Coulomb - 1

Page 95: Shear Strength of Soils Final (Complete)

The Mohr-Coulomb Failure CriterionShear Strength of Soils : The Mohr-Coulomb Failure Criterion

The Mohr Coulomb Failure CriterionMohr’s failure criterion

Materials fail when the on the failureshear stressMaterials fail when the ___________ on the failure plane at failure reaches some unique function of the on that plane.

shear stress

normal stress____________ pτff = f(σff)

h tτ = shear stressσ = normal stress1st subscript f refers to the plane on which the stress acts, p pi.e., the failure plane

2nd subscript f refers to ‘at failure’

τff = shear strength of the material

Note that Mohr does not explicitly say anything about

Mohr Coulomb - 2

p y y y gfriction angle.

Page 96: Shear Strength of Soils Final (Complete)

The Mohr-Coulomb Failure CriterionShear Strength of Soils : The Mohr-Coulomb Failure Criterion

Mohr’s failure criterionMaterials fail when the shear stress on the failureMaterials fail when the shear stress on the failure plane at failure reaches some unique function of the normal stress on that plane.

τ B(σff , τff)

τff = f(σff)σ1f

A(σff , τff)

τff = f(σff)σ1f

τff

σ3f σff

τffσ3f

element B

σ1f

σ3fσ3f σffσ1f

element A

Mohr Coulomb - 3

σ

Page 97: Shear Strength of Soils Final (Complete)

How to Obtain the Mohr Failure EnvelopeShear Strength of Soils : The Mohr-Coulomb Failure Criterion

• Say we carry out a test to fail a soil element we know the principal stress or the stress state on specific planes at failurespecific planes at failurePlot the Mohr circle to represent the state of stress for this element

• Repeat tests to obtain failure for other stress states• Repeat tests to obtain failure for other ___________.

• Draw a line to the Mohr circles

stress states

tangent• Plot Mohr circles for these other stress states.

τ = f(σ )

• Draw a line _______ to the Mohr circles.

τ• This line is the Failure Envelope.

tangentσ1f

τff = f(σff)τσ3fσ3f σff

τff

σ1f

Mohr Coulomb - 4

σσ1fσ3f

Page 98: Shear Strength of Soils Final (Complete)

• The failure envelope expresses the functional relationship b t th h t d th l t t

Shear Strength of Soils : The Mohr-Coulomb Failure Criterion

between the shear stress τff and the normal stress σff at failure.

• Failure occurs only when the combination of shear andFailure occurs only when the combination of shear and normal stress is such that the Mohr circle is _______ to the Mohr failure envelope.

tangent

• Circle A lies below the Mohr failure envelope – element with such a stress state is _____.P t f Ci l B li b th M h f il l h

stable• Part of Circle B lies above the Mohr failure envelope – such a

stress state ___________. Material would ___ before reaching this stage of stress.

cannot exist fail

τff = f(σff)τ

reaching this stage of stress.

AB

Mohr Coulomb - 5 σ

Page 99: Shear Strength of Soils Final (Complete)

• _______________ of a circle with the failure envelope gives Point of tangencyShear Strength of Soils : The Mohr-Coulomb Failure Criterion

the stress conditions on the failure plane at failure.• Using the pole method, we can determine the angle of the

plane θ associated with the point of tangencyplane θf associated with the point of tangency. Mohr’s failure hypothesis

f ( f fThe point of tangency (of the Mohr circle with the failure envelope) defines the _____________________ in the element or the test specimen

angle of the failure plane

τff = f(σff)τ

element or the test specimen.σ1f θfτff f(σff)

σ3fσ3f σff

τff

θ

σ1f

Mohr Coulomb - 6

σOp

θf

Page 100: Shear Strength of Soils Final (Complete)

• In soil mechanics, we commonly draw only the top half of th M h i l

Shear Strength of Soils : The Mohr-Coulomb Failure Criterion

the Mohr circle.• There is a __________. Hence, there is a ‘__________’ failure

envelopebottom half bottom half

envelope.• According to Mohr’s failure hypothesis, it is equally likely

that a failure plane will form at an angle of .-θf

τff = f(σff)τ σ1f θf

that a failure plane will form at an angle of ___ .

θf

f

ff ff

σ3fσ3f

O θf

σ1f

σOp θf

θf

Mohr Coulomb - 7

Page 101: Shear Strength of Soils Final (Complete)

The Coulomb’s EquationShear Strength of Soils : The Mohr-Coulomb Failure Criterion

q• Involved in design of military works such as revetments

and fortress walls.• At that time, design mainly by rule of thumb – hence,

many structures failed.• Coulomb carried out experiments to determine the shear• Coulomb carried out experiments to determine the _____

________________. • He observed that there was

shearresistance of soils

a _________________ component of shear strengthHe observed that there was

a component of shear strength

stress-independent

stress dependenta _______________ component of shear strength

• _______________ component similar to sliding friction in solids Coulomb called this

stress-dependent

Stress-dependentthe angle of internalin solids. Coulomb called this _________________

_________.• component seemed to be related

the angle of internalfriction (φ)Stress-independent

Mohr Coulomb - 8

_________________ component seemed to be related to the intrinsic ___________ of the material. Stress-independent

cohesion (c)

Page 102: Shear Strength of Soils Final (Complete)

The Coulomb’s EquationShear Strength of Soils : The Coulomb’s Equation

The Coulomb s Equation

Coulomb’s equation isτf = σ tan φ + c

τf is the shear strength of the soilf g

σ is the applied normal stress

φ and c are the strength parameters of the soil.

This relationship gives a straight line

• Neither nor are inherent properties of the soil

This relationship gives a straight line.

φ cNeither __ nor __ are inherent properties of the soil.

• They are dependent on the conditions operative in the test

φ c

Mohr Coulomb - 9

test.

Page 103: Shear Strength of Soils Final (Complete)

The Mohr-Coulomb Failure CriterionShear Strength of Soils : The Mohr-Coulomb Failure Criterion

The Mohr-Coulomb Failure Criterion

• Reasonable to combine the Coulomb equation with the• Reasonable to combine the Coulomb equation with the Mohr failure criterion.

Approximate Mohr failure envelope by a straight lineApproximate Mohr failure envelope by ____________Equation for that line in terms of the Coulomb strength parameters (φ and c) can be written

a straight line

parameters (φ and c) can be written

• Mohr-Coulomb Criterion

τff = σff tan φ + c

Mohr Coulomb Criterion

Mohr Coulomb - 10

Page 104: Shear Strength of Soils Final (Complete)

Relationship between θ and φ

Shear Strength of Soils : The Mohr-Coulomb Failure Criterion

Relationship between θf and φ

From earlier slide

φ2

45fφ

+°=θ2

where θ is the failure angle measured relative to thewhere θf is the failure angle measured relative to the plane of the major principal stress

Mohr Coulomb - 11

Page 105: Shear Strength of Soils Final (Complete)

Factor of Safety (F.S.) in the Shear StrengthShear Strength of Soils : Factor of Safety

y ( ) g• Consider a soil element subjected to principal stresses

________ the stresses required to cause failureless than

τff = f(σff)τ

φ

Note that, if we know φ, then we know the potential failure plane even if the soil element

σσ3 σ1f

τfτff

θfσ1

phas not failed yet.

245f

φ+°=θ

is the mobilized shear resistance on the potential failure plane

σ3 1f1=σ3f

( )

τf is the mobilized shear resistance on the potential failure planeτff is the shear strength available (shear stress on the failure plane at failure) when σ3 is kept constant.

Mohr Coulomb - 12

( )( )appliedavailable

f

ff

ττ

) 3 p

Factor of Safety (F.S.) =

Page 106: Shear Strength of Soils Final (Complete)

Factor of Safety (F.S.) in the Shear StrengthShear Strength of Soils : Factor of Safety

y ( ) g• Consider a soil element subjected to principal stresses

________ the stresses required to cause failureless than

τff = f(σff)τ

φ

Note that, if we know φ, then we know the potential failure plane even if the soil element

σσ3 σ1f

τfτff

θfσ1

phas not failed yet.

245f

φ+°=θ

is the mobilized shear resistance on the potential failure plane

σ3 1f1=σ3f

( )

τf is the mobilized shear resistance on the potential failure planeτff is the shear strength available (shear stress on the failure plane at failure) when σ3 is kept constant.

Mohr Coulomb - 12

( )( )appliedavailable

f

ff

ττ

) 3 p

Factor of Safety (F.S.) =

Page 107: Shear Strength of Soils Final (Complete)

Factor of Safety (F.S.) in the Shear StrengthShear Strength of Soils : Factor of Safety

• If the minor principal stress changes, so that _______

f( )τ

σ3f > σ3

φ

y ( ) g

τff = f(σff)φ

τf

τff

θf θf

σσ3 σ1f

θfσ1 σ3f

θf

is the mobilized shear resistance on the potential failure plane

( )

τf is the mobilized shear resistance on the potential failure planeτff is the shear strength available (shear stress on the failure plane at failure) when σ3f > σ3

Mohr Coulomb - 13

( )( )appliedavailable

f

ff

ττ

) 3f 3

Factor of Safety (F.S.) =

Page 108: Shear Strength of Soils Final (Complete)

Shear Strength Independent of Normal StressShear Strength of Soils : Shear Strength Independent of Normal Stress

g p

• φ = __• Mohr failure envelope is _________

0horizontal

φ __

φ = 0τ

φ

τf

45°σf σ

• Such materials sometimes described as purely cohesive• Valid for special (laboratory testing) conditions• Failure ‘theoretically’ occurs on the plane45°

Such materials sometimes described as ______________p y

• Failure theoretically occurs on the ___ plane

• Shear strength is τf = __________ and the normal stress th th ti l f il l t f il i

45

(σ1f - σ3f)/2( + )/2

Mohr Coulomb - 14

on the theoretical failure plane at failure is __________(σ1f + σ3f)/2

Page 109: Shear Strength of Soils Final (Complete)

Mohr-Coulomb Criterion in Terms of Principal StressShear Strength of Soils : Mohr Coulomb Criterion in Terms of Principal Stresses

τ

τff σ−σ

φ σ1f

σ3f

θf

τff

2f3f1 σσ

σ1f

σ3f

σ3f

2θfcφc cot φ

D

σ3f

2f3f1 σ+σ

σ1fσff σ

• sin φ = R/D ⇒

σ−σ

=φsin 2f3f1

sin φ R/D ⇒

φi1

φ+σ+σ

φcot

sinc

2f3f1

• If c = 0, thenf3f1

f3f1

σ+σσ−σ

=φsin or φ−φ+

=σσ

sinsin

11

f3

f1

Mohr Coulomb - 15⎟⎠⎞

⎜⎝⎛ φ

+°=σσ

2452

f3

f1 tan ⎟⎠

⎞⎜⎝

⎛ φ−°=

σσ

245tan2

f1

f3and • Alternatively

Page 110: Shear Strength of Soils Final (Complete)

Mohr-Coulomb Criterion in Terms of Principal StressShear Strength of Soils : Mohr Coulomb Criterion in Terms of Principal Stresses

p

τ φ σ1f θf

τff

2f3f1 σ−σ

Rσ3f

σ3f

φc cot φ

D

σ1f2θfσ3f

f3f1 σ+σσ1fσff σ

c

• If c ≠ 0 then

D 2

• If c ≠ 0, then

+tan= sin + 1= cotc + 21f⎥⎤

⎢⎡ φπφφσ

2

4tan sin - 1

cotc + 3f

⎥⎦⎢⎣φφσ

Mohr Coulomb - 16

Page 111: Shear Strength of Soils Final (Complete)

How do concepts from Mohr Circle apply to Soil Mechanics?

Shear Strength of Soils : Mohr Circle Application to Soil Mechanics

In the field, a typical soil element (or particle) isacted upon by stresses from ___________all directions

How do concepts from Mohr Circle apply to Soil Mechanics?

σvσ

σv

σx σx

σzz

• Generally, σ’x = σ’z = Ko σ’y• Ko is the _____________________________________coefficient of lateral earth pressure (at rest)

• For normally consolidated clay or loose to mediumFor normally consolidated clay or loose to mediumsand

Ko ≈ 0.4 to 0.6• For heavily consolidated clay or very dense sand

Mohr Coulomb - 17

• For heavily consolidated clay or very dense sandKo > may be 2 ~ 5 or even higher1

Page 112: Shear Strength of Soils Final (Complete)

Example:

Shear Strength of Soils : Mohr Circle Application to Soil Mechanics

Example:What are the stresses acting on a soil element located 8 mbelow the ground surface? Assume water table is at theground surface Take γb lk = 18 kN/m3 and K = 0 5At 8 m below ground,ground surface. Take γbulk = 18 kN/m and Ko = 0.5.

total vertical stress σy = 8 m x 18 kN/m3 = 144 kN/m2total vertical stress σy 8 m x 18 kN/m /pore water pressure u = 8 m x 10 kN/m3 = 80 kN/m2

effective vertical stress σy’ = 144 – 80 = 64 kN/m2

effective lateral stress σx’ = σz’ = Koσy’ = 0.5 x 64 kN/m2

= 32 kN/m2

total lateral stress σx = σz =Total Stresses Effective Stresses

32 + 80 = 112 kN/m2

144 kN/m2Total Stresses

64 kN/m2Effective Stresses

112 kN/m2

112 kN/m2

32 kN/m2

32 kN/m2

Mohr Coulomb - 18

Page 113: Shear Strength of Soils Final (Complete)

T t l St Eff i S

Shear Strength of Soils : Mohr Circle Application to Soil Mechanics

144 kN/m2Total Stresses

64 kN/m2Effective Stresses

112 kN/m2

112 kN/m2

32 kN/m2

32 kN/m2

112 kN/m 32 kN/m

Can you plot the Mohr circles corresponding to the totaland effective stresses?

τ (kPa)

u = 80 kPaTotal StressesEffective

Stresses

Mohr Coulomb - 19

σ (kPa)32 64 112 144

Page 114: Shear Strength of Soils Final (Complete)

Shear Strength of Soils : Mohr Circle Application to Soil Mechanics

Consider the effective stress.

If the failure envelope is as shown on the plot, what canyou say about the soil state? Has it failed?

τ (kPa)failure

envelope

STABLE!

effective

Not failed yet!

σ (kPa)32 64

stresses φ

Mohr Coulomb - 20

Page 115: Shear Strength of Soils Final (Complete)

Shear Strength of Soils : Mohr Circle Application to Soil Mechanics

Assuming the lateral (or horizontal stress) remainsconstant, by how much should the vertical stress beincreased to cause the soil element to fail?

failureτ (kPa)

failure envelope

Mohr circle at failure

(kPa)32 64

φ Δσy

σ (kPa)32 64

known can calculate this?

Mohr Coulomb - 21

known can calculate this?

Page 116: Shear Strength of Soils Final (Complete)

Shear Strength of Soils : Mohr Coulomb Criterion in terms of Principal Stresses

τ

τff σ−σ

φ σ1f

σ3f

θf

τff

2f3f1 σσ

σ1f

σ3f

σ3f

2θfcφc cot φ

D

σ3f

2f3f1 σ+σ

σ1fσff σ

• sin φ = R/D ⇒

σ−σ

=φsin 2f3f1

sin φ R/D ⇒

φi1

φ+σ+σ

φcot

sinc

2f3f1

• If c = 0, thenf3f1

f3f1

σ+σσ−σ

=φsin or φ−φ+

=σσ

sinsin

11

f3

f1

Mohr Coulomb - 15⎟⎠⎞

⎜⎝⎛ φ

+°=σσ

2452

f3

f1 tan ⎟⎠

⎞⎜⎝

⎛ φ−°=

σσ

245tan2

f1

f3and • Alternatively

Page 117: Shear Strength of Soils Final (Complete)

Shear Strength Tests for SoilShear Strength of Soils : Shear Strength Tests for Soil

In trying to understand shear strength concepts for soil, it is important to note that

A typical soil element in the ground is subjected toA typical soil element in the ground is subjected to _______________________ , including shear stresses.When studying the shear strength of soil, we have to stresses from all directions

consider the stresses acting on the soil, that is the stress state of the soil. Different stress states, at different depths , por different locations, can result in ________ shear strengths.differentShear strengths at A B C and

σv

A

B

DShear strengths at A, B, C and D are likely to be ________.different

σv

σhσh

B

C

τ

Shear Strength Testing - 2σv

Page 118: Shear Strength of Soils Final (Complete)

In trying to understand shear strength concepts for soil, it Shear Strength of Soils : Shear Strength Tests for Soil

y g g pis important to note that

We also have to consider the or poreeffect of waterWe also have to consider the _____________, or pore pressure.

We have to consider if the soil behaviour is ‘ ’

effect of water

drainedWe have to consider if the soil behaviour is _______ or ‘_________’.

‘drained strength’ vs ‘undrained strength’

drainedundrained

drained strength vs undrained strength

________ stress vs _____ stresseffective total

σv u σ’v

σhσh u σ’huσ’h

Shear Strength Testing - 3

σv u σ’v

Page 119: Shear Strength of Soils Final (Complete)

Shear Strength of Soils : Shear Strength Tests for Soil

Hence, shear strength tests for soils should incorporate, to some extent, ,

the stresses acting on the soil from differentthe stresses acting on the soil from different directions

the effect of pore water and whether it leads to drained or undrained behaviour

Shear Strength Testing - 4

Page 120: Shear Strength of Soils Final (Complete)

Shear Strength Tests for SoilShear Strength of Soils : Shear Strength Tests for Soil

Some of the more common tests for determining the shearing strength of soils:

• Direct Shear TestTriaxial Test• Triaxial Test

• Hollow Cylinder Test Laboratory Tests

• Plane Strain Test• True Triaxial TestTrue Triaxial Test

• Vane Shear TestC P t t T t In Situ Tests• Cone Penetrometer Test

• Standard Penetration Test

In Situ Tests(in the natural or original position

Shear Strength Testing - 5

• Pressuremeter Test or place)

Page 121: Shear Strength of Soils Final (Complete)

Direct Shear TestShear Strength of Soils : The Direct Shear Test

Schematic of the Shear Box ApparatusN ΔV

δ

soil

• Oldest strength test? Coulomb probably used something similar• Shear box has two halves• Bottom half _____, top half _________fixed moveable• Apply ______ load N to soil specimen• Apply constant rate of _______________ to upper half of shear box

Measure T δ and ΔV as the specimen is sheared

normalhorizontal motion

Direct Shear Test - 1

• Measure T, δ and ΔV as the specimen is sheared

• τ = T/A, σ = N/A (where A is the nominal area of the specimen)

Page 122: Shear Strength of Soils Final (Complete)

Direct Shear Test

Shear Strength of Soils : The Direct Shear Test

Direct Shear TestExamples of Laboratory Equipment for Direct Shear Test

Direct Shear Test - 2

Page 123: Shear Strength of Soils Final (Complete)

Direct Shear Test on Dense SandShear Strength of Soils : The Direct Shear Test

N ΔV

T

Dense Sand

e = constant (dense sand)τ = T/A

τ

σn3 = N3/Aσn2 = N2/A

14

11 X

X14

11n2 2

σn1 = N1/A

δ

7

σ

X7

φσn3 > σn2 > σn1

Direct Shear Test - 3

δ σnσn1 σn2 σn3τ - δ plot (test results) Mohr diagram

Page 124: Shear Strength of Soils Final (Complete)

Key Points of the Direct Shear Test

Shear Strength of Soils : The Direct Shear Test

Key Points of the Direct Shear Test

• The failure plane is forced to be with this apparatushorizontalp _________ pp

• Initially, before test, horizontal plane is a ________ plane (no shear stress).principal

• After shearing stress is applied, the horizontal plane ______ be a principal plane.

cannot

• At failure, the horizontal plane is ___ a principal plane.

• of the principal planes must occur in the direct shear test

not

Rotation• ________ of the principal planes must occur in the direct shear test.

• If we want to obtain the principal stresses at failure, we will have to

Rotation

infer from the ___________________________.

• The _____ of the Mohr-Coulomb failure envelope affects the angle of

Mohr Coulomb failure envelope

slope

Direct Shear Test - 4

_____ p grotation of the principal planes.

p

Page 125: Shear Strength of Soils Final (Complete)

ExampleTh i iti l d f il diti i di t h t t h b l

Shear Strength of Soils : The Direct Shear Test Example Problem

The initial and failure conditions in a direct shear test are shown below.

σn = σ1o σnτff

principal planes at failure

σh = Ko σ1o = σ3o σh

ff

(i) initial conditions (ii) at failure

Plot the Mohr circles for both initial conditions and at failure. Find the i i l t t f il d th i l f t ti t f ilprincipal stresses at failure and their angles of rotation at failure.

Solutioni iti l f ilτ τ

pole φ

initial failure

pole

pτff

Direct Shear Test - 6

σσ3o σ1oσσ3f σ1fσff = σn

σ1fσ3f

Page 126: Shear Strength of Soils Final (Complete)

Example

Shear Strength of Soils : The Direct Shear Test Example Problem

pA direct shear test is run on a medium dense sandy silt, with the normal stress σn = 65 kPa, Ko = 0.5. At failure, the normal stress is still 65 kPa and the shear stress is 41 kPa.Draw the Mohr circles for the initial conditions and at failure and determine:

(a) The principal stresses at failure(a) The principal stresses at failure(b) The orientation of the failure plane(c) The orientation of the major principal plane at failure(d) The orientation of the plane of maximum shear stress at failure.( ) p

σn = σ1o = 65 kPa σn = 65 kPaσn σ1o 65 kPa

σh = Ko σ1o = 0.5 x 65 32 5 kP

n

σh

τff = 41 kPa

= 32.5 kPa

(i) initial conditions (ii) at failure

Direct Shear Test - 7

Page 127: Shear Strength of Soils Final (Complete)

(i) Draw the Mohr circle for initial condition with σ1 = __ kPa, and σ3 = 0.5 x __ = ____ kPa65 (ii) Locate the failure point σff = kPa, and τff = kPa. In the absence of other information,

65 32.5 65 41

Shear Strength of Soils : The Direct Shear Test Example Problem

(ii) Locate the failure point σff __ kPa, and τff __ kPa. In the absence of other information, the line drawn from this failure point to the origin represents the ______________.

(iii) The angle of friction is obtained as tan φ = _____, hence φ = __°. (iv) From the failure point (65, 41), draw a line perpendicular to the failure envelope to meet

65 41 failure envelope

41/65 32

the σ axis. The point of intersection is the ______ of the Mohr circle associated with that state of failure.

(v) Draw the Mohr circle for the failure state. The principal stresses at failure are obtained as = kPa and = kPa

centre

139 43as σ1f = ___ kPa, and σ3f = __ kPa.(vi) The failure plane is _________ (forced to be so in the direct shear test). (vii) Locate the pole P. ( iii) F th l P d li t th j i i l t Thi li t th

139 43 horizontal

(viii) From the pole P, draw a line to the major principal stress. This line represents the orientation of the __________________. Hence the orientation is about ____° to the horizontal.

(ix) From the pole P, draw a line to the point of maximum shear stress. Hence the

major principal plane 60.5

τ P φ = 32°

( ) p , porientation of the plane of max. shear stress is about __° to the horizontal.16

τff = 41 16°

Direct Shear Test - 8σ6532.5 43 139

60.5°

Page 128: Shear Strength of Soils Final (Complete)

Advantages of the Direct Shear Test

Shear Strength of Soils : Advantages / Disadvantages of The Direct Shear Test

Advantages of the Direct Shear Test• Inexpensive

• Fast

• Simple

• Works well for granular materials – conditiondrainedg _______

Disadvantages of the Direct Shear Test

• Difficult to control ________ – very difficult to perform for ______ soils

F il l f d t i th h i t l di ti t b

drainage clayey

• Failure plane forced to occur in the horizontal direction – may not be the ________ direction, or the same critical direction as occurs in the field.

weakest

• Serious stress _____________ at the sample boundaries, which may lead to highly ___________ stress conditions within the test specimen.

t ti f th i i l l d t

concentrationsnon-uniform

U t ll d

Direct Shear Test - 5

• ___________ rotation of the principal planes and stresses.Uncontrolled

Page 129: Shear Strength of Soils Final (Complete)

Triaxial Test

Shear Strength of Soils : The Triaxial Test

Triaxial Test• Direct shear test popular in the early days of soil

mechanicsmechanics• About 1930, Arthur Casagrande at M.I.T. began

research on cylindrical compression tests in anresearch on cylindrical compression tests in an attempt to overcome some of the serious disadvantages of the direct shear test.g

• Nowadays, such tests, called triaxial tests, are the more popular of the two.p p

• Based on the principle in which a cylindrical specimen of soil is either ___________ or ________compressed extendedp ___________ ________

• Not immediately obvious, but compressing or extending a soil specimen can lead to _____________

p

shear stresses

Triaxial Test - 1

g p _____________and ___________.shear failure

Page 130: Shear Strength of Soils Final (Complete)

Triaxial TestShear Strength of Soils : The Triaxial Test

cell pressure

cell pressure

Triaxial Test - 2

Page 131: Shear Strength of Soils Final (Complete)

Versatility of the Triaxial TestShear Strength of Soils : Versatility of The Triaxial Test

y• Much more complicated than the direct shear, but

also much more versatileD i f t b t ll d it ll• Drainage of pore water can be controlled quite well

can turn off drainage to examine _________ behaviour

undrainedbehaviourcan allow drainage to study _______ behaviour

• The pore pressure u can also be controlled.drained

usually σ1 is the ______ stress and σ3 is • The principal stresses σ1 and σ3 can be controlled

p p

verticaly 1 ______ 3the _________ (or confining) stressin most triaxial tests, the orientations of

horizontal

however, if required, triaxial testing allows us to the principal stresses do not change

Triaxial Test - 3

rotate principal planes in a controlled manner, e.g. triaxial ___________ vs triaxial ________compression extension

Page 132: Shear Strength of Soils Final (Complete)

Stresses in triaxial specimensShear Strength of Soils : Stresses in Triaxial Specimens

pF = Deviator load

σr (due to piston moving up or down)

σr σr = Radial stress (cell pressure)

= σc

A i l tdeviator stress

σa = Axial stress

σ σa rFA

= +From equilibrium we have

Triaxial Test - 3

Aaxial stress area of cylindrical

cross-section

Page 133: Shear Strength of Soils Final (Complete)

Stresses in triaxial specimensShear Strength of Soils : Stresses in Triaxial Specimens

Stresses in triaxial specimensF/A is known as the deviator stress, and is given the symbol q

( ) ( )31raq σ−σ=σ−σ= total stresses

The axial and radial stresses are principal stresses

Note that σa , σr or σ1 , σ3 are total stresses

In terms of effective stressesIn terms of effective stresses,

( ) ( )aσ 1σ 3σ

( ) ( )''σr

( ) ( )u'u'q ra +σ−+σ=

'' σσ effective stresses

( ) ( )u'u' 31 +σ−+σ=

'' σσ=

Triaxial Test - 3

ra σ−σ= effective stresses31 σ−σ=

Page 134: Shear Strength of Soils Final (Complete)

S i i i l i

Shear Strength of Soils : Stresses in Triaxial Specimens

Stresses in triaxial specimens

Increasing cell pressure σc while keeping q = 0 will result in

• volumetric compression if the soil is free to drain. The peffective stresses will increase and so will the strength.

• increasing pore water pressure if soil volume is constantincreasing pore water pressure if soil volume is constant (that is, undrained). As the effective stresses cannot change, it follows that Δu = Δσc

is required to cause failureIncreasing q___________ is required to cause failure.Increasing q

Triaxial Test - 3

Page 135: Shear Strength of Soils Final (Complete)

TRIAXIAL COMPRESSIONShear Strength of Soils : Triaxial Compression

Δσ i i t i d

During test

t t= q

downward load applied to move ram downwards

σcell

Δσaxial • σcell is maintained ________

• if _______________, Δσaxial is increased at a certain rate and the

σ1constant

stress-controlled= σaxial

= q

σcell =

increased at a certain rate and the downward movement of the piston is measuredσ2 = σ3

• if ______________, the piston is programmed to move downward at a certain rate and Δσaxial is measured

strain-controlled

σaxial = σ1 =Δσaxial =

axial

• in either case,Δσaxial + σcellprincipal stress

σ1 - σ3 = qσaxial σ1

• σaxial > σcell

σaxial σcellprincipal stress difference = q + σcell

Triaxial Test - 4

• TRIAXIAL COMPRESSIONspecimen is ___________compressed

Page 136: Shear Strength of Soils Final (Complete)

TRIAXIAL EXTENSIONShear Strength of Soils : Triaxial Extension

During test• σcell is maintained constant

upward load applied to move ram upwards

σcell

Δσaxial • if stress-controlled, Δσaxial due to upward ram load is increased at a certain rate and the upward

σ3 = σaxial= q

σcell =

certain rate and the upward movement of the piston is measured

• if strain-controlled, the piston is σ1

programmed to move upward at a certain rate and Δσaxial is measured

σaxial = σ3 =Δσaxial = • in either case,

σcell - Δσaxial

σ1 - σ3 = q

• σaxial < σcell= σcell – q

Triaxial Test - 5

• TRIAXIAL EXTENSIONspecimen is ________extended

Page 137: Shear Strength of Soils Final (Complete)

Strains in triaxial specimensShear Strength of Soils : Strains in Triaxial Specimens

Strains in triaxial specimensFrom the measurements of change in height, dh, and change in volume dV we can determine

Axial strain ε adhh

= −0

εa is +ve for shortening

Volume strain ε VdVV

= −0

εV is +ve for volume reduction

where h0 is the initial height and V0 is the initial volume

It is assumed that the specimens deform as right circularIt is assumed that the specimens deform as right circular cylinders. The cross-sectional area, A, can then be determined from

⎟⎟⎞

⎜⎜⎛ ε

⎟⎟⎟⎞

⎜⎜⎜⎛

vo

0o 1

- 1 A = dhVdV + 1

A =A

Triaxial Test - 3

⎟⎠

⎜⎝ ε⎟

⎠⎜⎜

⎝a

o

0

o - 1A

hdh + 1

A

Page 138: Shear Strength of Soils Final (Complete)

TRIAXIAL COMPRESSION vs TRIAXIAL EXTENSIONShear Strength of Soils : Triaxial Compression vs Triaxial Extension

• Triaxial compression and extension tests typically yield ________ ______________ of shear strengths.

• Example : Slip Plane associated with Bearing Capacity Failuredifferent values

slightly

σ1 = σv σ3 = σv

σ3 = σh σ1 = σhslip or failure

plane

σv > σh

triaxialσh > σv

τ

τ

• Shear strength applicable to different parts of the slip plane vary

triaxial compression triaxial

extension

τ

direct shear

• Shear strength applicable to different parts of the slip plane vary, depending on whether the dominant mode of failure deformation is (i) ___________, (ii) _________ or (iii) __________.I ti i ll l th h t th bt i d fcompression extension direct shear

Triaxial Test - 6

• In practice, engineers usually use only the shear strength obtained from __________________ tests.

• _______________ tests not commonly done.triaxial compression

Triaxial extension

Page 139: Shear Strength of Soils Final (Complete)

Cl ifi ti f T i i l T t

Shear Strength of Soils : Classification of Triaxial Tests

Classification of Triaxial Tests according to Drainage Conditions

Before Shear After Shear Symbol

Unconsolidated Undrained UUUnconsolidated Undrained UU

Consolidated Undrained CU

Consolidated Drained CD

Triaxial Test - 8

Page 140: Shear Strength of Soils Final (Complete)

Types of triaxial testShear Strength of Soils : Classification of Triaxial Tests

There are many test variations. Those used most in practice are:ypes o t a a test

• UU (unconsolidated undrained) test.UU (unconsolidated undrained) test.Cell pressure σcell applied without allowing drainage. Hence, no consolidation of soil is permitted. Then keeping cell pressure p p g pconstant, increase deviator load q to failure without drainage.

• CIU (isotropically consolidated undrained) testCIU (isotropically consolidated undrained) test.Drainage allowed during cell pressure application. Soil is given time to fully consolidate under cell pressure. After consolidation, y p ,drainage valve is closed. Then without allowing further drainage, increase q keeping σr or σcell constant as for UU test.

• CID (isotropically consolidated drained) test

Similar to CIU except that in the final shearing stage as

Triaxial Test - 8

Similar to CIU except that, in the final shearing stage, as deviator stress q is increased, drainage is permitted.

Page 141: Shear Strength of Soils Final (Complete)

Unconsolidated-Undrained (UU) TestD i l th h t t tl d fl f t

Shear Strength of Soils : Unconsolidated-Undrained (UU) Test

• Drainage valves ______ throughout test – ______________ in and out of the specimen. No backpressure applied.

• Applied confining pressure does not cause any .

closed no flow of water

consolidationpp g p y ____________• Pore water pressure not measured.• Specimen sheared undrained

(no change to existing effective stresses in the soil.)

• _____ test – specimen sheared to failure in 10 to 20 minsQuick

Page 142: Shear Strength of Soils Final (Complete)

Unconsolidated-Undrained (UU) TestD i l th h t t tl d fl f t

Shear Strength of Soils : Unconsolidated-Undrained (UU) Test

• Drainage valves ______ throughout test – ______________ in and out of the specimen. No backpressure applied.

• Applied confining pressure does not cause any .

closed no flow of water

consolidationpp g p y ____________• Pore water pressure not measured.• Specimen sheared undrained

(no change to existing effective stresses in the soil.)

• _____ test – specimen sheared to failure in 10 to 20 minsExample : Given two specimens of a certain Clay A

Specimen 1 : confining (cell) pressure = 50 kPa

Quick

τ

Specimen 1 : confining (cell) pressure = 50 kPaprincipal stress difference at failure = 70 kPa

Specimen 2 : confining (cell) pressure = 100 kPaprincipal stress difference at failure = 70 kPa

φ = 0τ

35 kP

principal stress difference at failure = 70 kPa

τf = 35 kPa

σ50 120100 170

Triaxial Test - 9

Note : because no consolidation takes place, the confining pressure has no effect on the effective stresses; the undrained strength τfis _________________ for all Clay A specimens tested in UU.the same (in theory)

Page 143: Shear Strength of Soils Final (Complete)

Consolidated-Undrained (CU) TestShear Strength of Soils : Consolidated-Undrained (CU) Test

Consolidated Undrained (CU) Test

• Drainage valves opened initially to allow the g p ysoil specimen to __________ under the cell pressure.

consolidatep

• Backpressure may be used to help attain degree of saturation ~ 100%degree of saturation ~ 100%

• After specimen has consolidated, _____ the closedrainage valve.

• Specimen sheared undrained• Specimen sheared undrained.

• _________________________ developed Excess pore water pressures

Triaxial Test - 10

_________________________ pduring shear are measured.

Page 144: Shear Strength of Soils Final (Complete)

ExampleShear Strength of Soils : Consolidated-Undrained (CU) Test Example

A series of undisturbed samples from a normally consolidated clay was subjected to consolidated undrained tests.

Cell pressure Principal stress difference Excess pore water(kPa) at failure (kPa) pressure at failure (kPa)

200 118 110400 240 220600 352 320

Plot the strength envelope of the soil (a) with respect to total stresses, and (b) with respect to effective stresses.

Assume the given cell pressures are ‘net’ or effective values, i.e., the back pressure u, if any has already been subtractedif any, has already been subtracted.

Triaxial Test - 11

Page 145: Shear Strength of Soils Final (Complete)

Solutiongiven given givencalculated calculated calculated

Shear Strength of Soils : Consolidated-Undrained (CU) Test Example

cell pressure principal stress difference σ1 at failure Δu at failure σ3' σ1'

σcell = σ3 (kPa) at failure Δσ, kPa = σcell + Δσ, kPa kPa = σ3 - Δu (kPa) = σ1 - Δu (kPa)200 118 318 110 90 208

given given given

200 118 318 110 90 208400 240 640 220 180 420600 352 952 320 280 632

Triaxial Test - 12

Page 146: Shear Strength of Soils Final (Complete)

Solutiongiven given givencalculated calculated calculated

Shear Strength of Soils : Consolidated-Undrained (CU) Test Example

cell pressure principal stress difference σ1 at failure Δu at failure σ3' σ1'

σcell = σ3 (kPa) at failure Δσ, kPa = σcell + Δσ, kPa kPa = σ3 - Δu (kPa) = σ1 - Δu (kPa)200 118 318 110 90 208

given given given

200 118 318 110 90 208400 240 640 220 180 420600 352 952 320 280 632

(kP )φ ’

τ (kPa)

φ

σ (kPa)200 400 640318 600 952 σ (kPa)200 400 640318 600 952

φ is the total stress friction angle.φ’ is the effective stress friction angle

Triaxial Test - 12The values of φ and φ’ can be obtained by direct measurement.φ is the effective stress friction angle.

Page 147: Shear Strength of Soils Final (Complete)

f3f1 σ−σφ

Shear Strength of Soils : Consolidated-Undrained (CU) Test Example

Alternatively, can use the formula f3f1

f3f1

σ+σσσ

=φsin

For the Mohr stress circles created when the cell pressure, σcell = σ3 = 200 kPa

f3f1

f3f1

σ+σσ−σ

=φsin = 0.228200318200318

+−

= ⇒ φ = 13.2°f3f1

f3f1

''''

'sinσ+σσ−σ

=φ = 0.3969020890208

+−

= ⇒ φ = 23.3°f3f1 '' σ+σ 90208 +

Only valid when both failure envelopes pass through _________.the origin

Triaxial Test - 13

Page 148: Shear Strength of Soils Final (Complete)

Pore Pressure ParametersShear Strength of Soils : Pore Pressure Parameters

Pore Pressure Parameters• In triaxial strength testing, there are two important pore

t l d th A d B tpressure parameters commonly used: the ________________.

Δ change in pore• The A parameter

A and B parameters

( )31

uAσ−σ

Δ=

change in pore water pressure

The A parameter

applied deviator stress

Stress ratioRelates the increase (or decrease) in pore water pressure ( )to the applied deviator stress σ1 – σ3

The A-parameter varies with ____________, as well as different soils(shearing)

the ___________________________________.magnitude of the applied deviator stressAt failure

fuΔ

Pore Pressure Parameters 1( )f31

ff

uAσ−σ

Δ=

Page 149: Shear Strength of Soils Final (Complete)

OVERCONSOLIDATION RATIOShear Strength of Soils : Overconsolidation Ratio

• The overconsolidation ratio of a clay iscomputed as

OVERCONSOLIDATION RATIO

computed as

OCR =soilthebydexperiencecurrentlystresseffectiveexisting

history its in soil theby dexperience stress effective maximumsoiltheby dexperiencecurrently stresseffectiveexisting

max

''σ

σ=

c'p= where p’ = preconsolidation pressure

• A normally consolidated clay is one in which thev'σ

= where p c preconsolidation pressure

• A normally consolidated clay is one in which the____________________ is the maximum that it hasever experienced in its history.existing effective stress

Overconsolidation Ratio OCR - 1

• A normally consolidated clay is has OCR = __1

Page 150: Shear Strength of Soils Final (Complete)

Pore Pressure ParametersShear Strength of Soils : Pore Pressure Parameters

Pore Pressure Parameters• A-parameter also varies with the stress history of the soil

value can be positive or negative

when shearing loose sand or normally consolidated clayswhen shearing loose sand or normally consolidated clays, _______ pore pressures develop, hence A-parameter is

.positivepositive_______.

when shearing dense sand or highly overconsolidated clays pore pressuressmaller positive or even negative

positive

clays, _____________________________ pore pressures develop, hence A-parameter ____________________

values.

smaller positive or even negativedecreases to smaller or

even negative____________g

Empirical relation between Af (at failure) and OCR was presented by Bishop and Henkel in 1962 (see next slide)presented by Bishop and Henkel in 1962 (see next slide).

Pore Pressure Parameters 2

Page 151: Shear Strength of Soils Final (Complete)

Empirical correlation of A-Parameter and OCRShear Strength of Soils : Pore Pressure Parameters

Pore Pressure Parameters 3

Page 152: Shear Strength of Soils Final (Complete)

Pore Pressure ParametersShear Strength of Soils : Pore Pressure Parameters

Pore Pressure Parameters

uΔ change in pore water pressure

• The B parameter

c

uBσΔ

Δ=

Stress ratio

water pressure

change in cell pressure

Stress ratioRelates the increase (or decrease) in pore water pressure to the change in applied cell pressure σ (isotropicIf B = 1, soil is _____________.to the change in applied cell pressure σc

fully saturatedIf B 0 il i

(isotropic compression)

Used in the consolidation stage of CU test as an indicator

If B = 0, soil is ___.dry

of the degree of saturation.Difficult to obtain B = 1 during saturation stage of CU test; in most cases, we require B to be at least ____.0.95

Pore Pressure Parameters 4

Page 153: Shear Strength of Soils Final (Complete)

Recall Previous ExampleShear Strength of Soils : Pore Pressure Parameters

A series of undisturbed samples from a normally consolidated clay was subjected to consolidated undrained tests.

Cell pressure Principal stress difference Excess pore water(kPa) at failure (kPa) pressure at failure (kPa)

200 118 110400 240 220600 352 320

cell pressure principal stress difference σ1 at failure Δu at failure Af-parameter(kP ) t f il Δ kP Δ kP kP Δ / Δσcell = σ3 (kPa) at failure Δσ, kPa = σcell + Δσ, kPa kPa = Δuf / Δσ1

200 118 318 110

400 240 640 220

600 352 952 320

Pore Pressure Parameters 5

Page 154: Shear Strength of Soils Final (Complete)

Recall Previous ExampleShear Strength of Soils : Pore Pressure Parameters

A series of undisturbed samples from a normally consolidated clay was subjected to consolidated undrained tests.

Cell pressure Principal stress difference Excess pore water(kPa) at failure (kPa) pressure at failure (kPa)

200 118 110400 240 220600 352 320

cell pressure principal stress difference σ1 at failure Δu at failure Af-parameter(kP ) t f il Δ kP Δ kP kP Δ / Δσcell = σ3 (kPa) at failure Δσ, kPa = σcell + Δσ, kPa kPa = Δuf / Δσ1

200 118 318 110 0.93

400 240 640 220 0.92

600 352 952 320 0.91

Pore Pressure Parameters 5

Page 155: Shear Strength of Soils Final (Complete)

Consolidated-Drained (CD) TestShear Strength of Soils : Consolidated-Drained (CD) Test

Consolidated Drained (CD) Test• Drainage valves opened initially to allow the soil

specimen to __________ under the cell pressure.consolidate• Backpressure may be used to help attain degree of

saturation ~ 100%• After specimen has consolidated, leave the drainage

valve ____.open (water can freely flow in or out)

• Specimen sheared drained.

• develop during shearNo excess pore water pressures• ____________________________ develop during shear.No excess pore water pressures

• Typically carried out for ______________ materials.granular (sandy)

• If carried out on clays/silts, have to apply the load very ______ (to allow drainage from the clay specimens) so slowly

CD - 1

as to minimize the excess pore pressures generated. Usually not carried out for clays/silts as it takes too long.

Page 156: Shear Strength of Soils Final (Complete)

What do we get get out of Consolidated-Drained (CD) Tests?Shear Strength of Soils : Consolidated-Drained (CD) Test

g g ( )

• Drained strength(s) (direct)

• Stress-strain curve(s) (direct)

• Drained or effective stress friction angle φ’ D i d ff ti t f il l

• Stress-strain curve(s) (direct)

(postprocess)

( t )• Drained or effective stress failure envelope (postprocess)

• Usually have to perform _____________ test (at diff t ll ) i d t bt i f i ti

more than onedifferent cell pressure) in order to obtain friction angle and failure envelopeIf l t t i f d th h t

• material has no cohesion

• If only one test is performed, then we have to assume

• material has no cohesion• failure envelope is a straight line

i d t d i th f i ti l d th

CD - 2

in order to derive the friction angle and the failure envelope

Page 157: Shear Strength of Soils Final (Complete)

Mohr Circle and Stress-Strain Curve for a CD TestAt Start of Test

Shear Strength of Soils : Consolidated-Drained (CD) Test

σcell

Δσaxial • σ1 = σ3 = σcellσ1

At Start of Test

During Test

= σaxial

σcell = σ2 = σ3 • Δσaxial increases ⇒During Test

σ1 increases• σcell remains constant

Δσaxial = σ1 - σ3

cell• Mohr circle expands with σ3 fixed• Note corresponding point on

τ τ =(σ σ )/2

final Mohr circle corresponding to failure

stress-strain curve

(σ1 - σ2)/2

C

DC

D

corresponding to failure

strength

σσ3 εσ1

A AB

CB

C

CD - 3Start of test

Before applying axial load

1

major principal stressat failure

Page 158: Shear Strength of Soils Final (Complete)

If we carry out several CD tests at different confining pressures σc

Shear Strength of Soils : Consolidated-Drained (CD) Test

τ τ =(σ1 - σ2)/2

strength for σcell = σc-C

φ ’

strength for σcell = σc-B

strength for σcell = σc-A

σc-A

εσ

σc-B σc-D

• We get three different strengths for three different σcell

σc-C

• Strength of the soil depends on the value of σ ll used in the test

c D

Strength of the soil depends on the value of σcell used in the test• σcell = confining stress • Hence, when discussing the strength of sands (or materials that exhibit

drained response) it is important to know the confining stress

• The friction angle φ’ (associated with the drained failure envelope) allows

drained response), it is important to know the confining stress• Say, if we have the three test results shown above, can we obtain the

shear strength of the same soil for another σcell , say, σc-D ?

CD - 4

• The friction angle φ (associated with the drained failure envelope) allows us to relate shear strength to the confining stress

• The friction angle is an important STRENGTH parameter in soil mechanics

Page 159: Shear Strength of Soils Final (Complete)

Friction Angle (Angle of Internal Friction) φ’Shear Strength of Soils : Angle of Internal Friction

• Note that φ’ is not a strength – it does not directly provide a strength value.• φ’ is a drained strength parameter – it controls the drained strength that

can be mobilized at different confining pressures.can be mobilized at different confining pressures.• When discussing shear strength of sands, we usually do so in terms of the

friction angle φ’.• We can apply the use of φ’ to situations outside of the triaxial (and direct

One common application of φ’

• We can apply the use of φ to situations outside of the triaxial (and direct shear) tests.

Estimate the shear strength (along a horizontal plane) of the dry sand at 5m and 10m depths in the ground. Take φ = 30° for the sand.

( ) P i t A

A

5 m

10 mdry sand

(a) Point AEffective vertical stress σ’v =

= 18 x 5 = 90 kN/m2

For a horizontal plane σ’ = σ’

γd h

B

y

φ = 30°

γd = 18 kN/m3

For a horizontal plane, σ v = σ n

The shear strength along this horizontal plane is τf = σ’n tan φ’ = 90 tan 30° = 52 kN/m2

(b) Point B

CD - 5

(b) Point BRepeat the calculations for depth of 10 m.The computed shear strength along the horizontal plane at this depth = ____ kN/m2104

Page 160: Shear Strength of Soils Final (Complete)

More on Friction Angle φ’

Shear Strength of Soils : Angle of Internal Friction

More on Friction Angle φ

• Using the friction angle φ’, the calculated shear g g φ ,strengths ________ with depth (for a soil that exhibits drained response).

increase

• This is due to the __________ of the soil, which causes the confining stress to increase with depth.

self-weight

• Hence, the drained strength of a soil generally increases with depth

causes the confining stress to increase with depth.

increases with depth.• ‘Drained’ strength ⇒ use with ________ stresseseffective

• φ’ also known as effective stress friction angle

CD - 6

Page 161: Shear Strength of Soils Final (Complete)

Typical Values of the Friction Angle φ’Shear Strength of Soils : Angle of Internal Friction

Friction Angle φ’

G ( )Gravel 40 – 55°Gravel (sandy)Sand

35 – 50°

Loose dry 28 – 34°Loose saturated 28 – 34°D d 36 45Dense dry 36 – 45°Dense saturated 36 – 45°

Silt or silty sandLoose 27 – 30°Dense 30 – 35°

Clay

CD - 7

General 19 – 27°Singapore marine clay ~22°

Page 162: Shear Strength of Soils Final (Complete)

ExampleShear Strength of Soils : Consolidated-Drained (CD) Test

pA conventional consolidated-drained (CD) triaxial test is conducted on a sand. The cell pressure is 120 kPa, and the applied deviator t t f il i 220 kPstress at failure is 220 kPa.

(a) Plot the Mohr circles for both the initial and failure stress conditions.(b) Determine φ (assume c = 0).

(c) Determine the shear stress on the failure plane at failure τff , and find the theoretical angle of the failure plane in the specimen.

(d) Determine the maximum shear stress at failure τmax , and the angle of the plane on which it acts. Calculate the available shear strength onthe plane on which it acts. Calculate the available shear strength on this plane and the factor of safety on this plane.

CD - 8

Page 163: Shear Strength of Soils Final (Complete)

σcell = σ3o = 120 kPa σcell =

Δσaxial = 220 kPa = (σ1 – σ3 )f σ1f = 340 kPa

σ1fσ3f = 120

σcell = σ3o = 120 kPa σcell = σ3f = 120 kPa

Initial Conditions At Failure

φ‘= 29°τ (kPa) f3f1

f3f1

σ+σσ−σ

=φ'sin

220i'φ460

arcsin'=φ

τffφ

σ−σ−

σ+σ=σ sin

22f3f1f3f1

ff

59.3°

45°4125230avail .'tan =φ=τ

P

ff

φ’

22°−= 628110230 .sin 4177.=

2f3f1 σ−σ

'tanφσ=τ ffff °= 6284177 .tan.796

σ (kPa)100 200 300 400 500

I iti l C diti σ = 120 kPa

Pσff 230

σ = 340 kPa

796.=

kPa110τ

CD - 9

Initial Condition (a single point)

σ3f = 120 kPa σ1f = 340 kPa

2f3f1 σ+σ

kPa110=τmaxkPa 4125avail .=τ

1411104125SF avail ././.. max ==ττ=

Page 164: Shear Strength of Soils Final (Complete)

Undrained Strength from CU Test and Drained Strength from

Shear Strength of Soils : Drained vs Undrained Strengths

g gCD Test – Starting from the Same Confining Pressure

τ =(σ1 - σ3)/2 Drained stress-strain curve for σcell = 600 kPa

Oft

Undrained stress-strain curve for σcell = 600 kPa

Drained strength sd

Often plotted as

q =( )

Undrained strength su

(σ1 - σ3)

εshear strain

I t tNote that, for the same soil subjected to the same initial stresses the drained strength is higher than the undrained

Important:

stresses, the drained strength is higher than the undrained strength.

Page 165: Shear Strength of Soils Final (Complete)

Undrained Strength from CU Test and Drained Strength from

Shear Strength of Soils : Drained vs Undrained Strengths

g gCD Test – Starting from the Same Confining Pressure

(kP )

φ ’

τ (kPa) Undrained Strength from CU Test

Δuf

Total Stress Mohr Circles

Effective Stress Mohr Circles

Undrained strength su

σ (kPa)200 400 800600 1000

Drained vs Undrained - 1

Page 166: Shear Strength of Soils Final (Complete)

Undrained Strength from CU Test and Drained Strength from

Shear Strength of Soils : Drained vs Undrained Strengths

g gCD Test – Starting from the Same Confining Pressure

(kPa)

φ ’

τ (kPa) Drained Strength from CD Test

Effective Stress Mohr Circles

Drained strength sd

200 400 800600 1000 σ (kPa)

Drained vs Undrained - 1

Page 167: Shear Strength of Soils Final (Complete)

Drained vs Undrained StrengthShear Strength of Soils : Drained vs Undrained Strengths

g• For engineering design purposes, whether we use the drained or

undrained strength depends on the soil type present.• If the problem involves mainly clays or silts (i.e., fine-grained

materials), then for the short-term response, the undrained shear strength su (also commonly denoted as cu) should be considered.g u ( y u)This type of analysis (using su or cu) implies that φ = 0.Total stresses (not effective stresses) are usually adopted for the

• For the long-term response of clays and silts, the drained strength sd should be used.

stress calculations when the undrained strength su or cu is used.

• The drained strength sd of clays and silts should be calculated using the friction angle φ’ and the effective confining stresses to

hi h th il ill b bj t d t i th l t

sd should be used.

which the soil will be subjected to in the long term..• For sands or gravels that can drain easily, there is little difference

between the short-term and long-term strength for mostbetween the short term and long term strength for most geotechnical problems, hence only the drained strength is considered using φ’ and the confining stresses.

Page 168: Shear Strength of Soils Final (Complete)

Drained vs Undrained StrengthShear Strength of Soils : Drained vs Undrained Strengths

g• However, in some situations, even for a granular material like

sand, we may still use the undrained strength su to characterize i b h i d id l di lik h k h kiits behavior under very rapid loading like earthquake shaking, when the time scale is too short to allow the excess pore pressures to dissipate.

• The difference between the drained and undrained condition is the generation of the excess pore pressure in the soils.

• Undrained strength is typically smaller than drained strength, due to the presence of excess pore water pressure, which tends to reduce the effective stresses.tends to reduce the effective stresses.

• For design purposes, the undrained shear strengths are commonly obtained by UU tests (experiment G2) or vane-shear tests. These are direct measurements of the undrained shear strength.

• The undrained shear strength can also be obtained by doing CUThe undrained shear strength can also be obtained by doing CU tests. The undrained shear strengths may be obtained directly during the tests, by plotting the Mohr circles or stress strain curves.

Page 169: Shear Strength of Soils Final (Complete)

Drained vs Undrained StrengthShear Strength of Soils : Drained vs Undrained Strengths

g• However, when doing CU tests, the main purpose is to obtain

the friction angle φ’ and c’.the friction angle φ and c .

• Once, we have the φ’ and c’ from the CU tests, we can calculate the undrained shear strengths and the drained shear strengths.

• Hence, theoretically, we can calculate the undrained shear strengths from φ’ and c’.

• However, this is not commonly done for hand calculations. Some computer software allows us to do undrained analysis

i ’ d ’using φ’ and c’.

• For the most part, it is more common for undrained shear t th t b bt i d di tl i UU t t hstrengths to be obtained directly via UU tests or vane-shear

tests, or obtain from correlations, or past experience.

Page 170: Shear Strength of Soils Final (Complete)

Useful Correlations for the Undrained Strength of ClaysShear Strength of Soils : Useful Correlations for Undrained Strength of Clays

• For a ___________________ claynormally consolidated

20su

Note that ’ is the at theeffective overburden stress

250s

vo

u .'

≈σ

• Note that σ’vo is the ________________________ at the depth of interest.

effective overburden stress

• This correlation implies that the undrained strength su

• Remember: it is a correlation. It provides useful

This correlation implies that the undrained strength su

___________________________.increases with depthestimates

• Other correlations:

Remember: it is a correlation. It provides useful _________ of the shear strength, but it is not exact!

220s

vo

u .'

≈σ

based on vane shear strengths (Bjerrum)

sp

vo

u I370110s

..'

+≈σ

(Skempton)

Drained vs Undrained - 5

Page 171: Shear Strength of Soils Final (Complete)

Correlation between s and Undrained Modulus E

Shear Strength of Soils : Useful Correlations for Undrained Modulus

Correlation between su and Undrained Modulus Eu

• For settlement or compressibility analyses,

Eu = 200 su to 500 su

• Rough correlation : useful guide in the absence of

u u u

other information.

• Depends on the location and the kind of clay present

• Can take Eu = 300 su as a first approximation

p y p

u u pp

Drained vs Undrained - 6

Page 172: Shear Strength of Soils Final (Complete)

Drained vs Undrained StrengthShear Strength of Soils : Drained vs Undrained Strengths

Drained vs Undrained Strength

If we want to obtain the undrained or drained shear strengths from the friction angle φ’ and cohesion c’, we can use the following few slides to do so.

Page 173: Shear Strength of Soils Final (Complete)

Relationship between drained strength sd and φ’Shear Strength of Soils : Relationship between Drained Strength and φ’

d

τ

τmax = sd

sd

φ’

sd

σσc = σ’c = σ’3f σ’1f

φ

σ’ s

• For a soil with c’ = 0σ c sd

'sins'

s

dc

d φ=+σ

'sin1'sin's cd φ−

φσ=⇒

Relationship between sd and φ’ - 1

Page 174: Shear Strength of Soils Final (Complete)

Relationship between drained strength sd and φ’-c’Shear Strength of Soils : Relationship between Drained Strength and φ’ - c’

Relationship between drained strength sd and φ cτ

sdφ’sd

sdc’

0

σσc = σ’c = σ’3f σ’1f

d

c’ cot φ’

σ’ sd 0• For a soil with nonzero φ’ and c’

σ c sd

'sinsd φ=

( ) ( ) 'sin'cot'c''sin1s cd φφ+σ=φ−⇒

sin'cot'c's cd

φ=φ+σ+

( )'sin1

'sincot'c's cd φ−φ

φ+σ=⇒

( ) ( )cd

Relationship between sd and φ’ - 2

Page 175: Shear Strength of Soils Final (Complete)

Relationship between undrained strength su and φ’Shear Strength of Soils : Relationship between Undrained Strength and φ’

τ φ’Mohr circle at failure if drained effective stress

M h i l f

sd (if drained test)

test carried outMohr circle for undrained failure uf

σσc = σ3f

test)

total stress Mohr circle

sususu

σ’3f

0• For a soil with c’ = 0, and noting that

total stress Mohr circle for undrained failure

s ( )ufff s2/uq/uA ==

uf3

u

s's'sin+σ

=φ ( )uff3u sus +−σ=( )uuf3u sAs2s +−σ=

( )( )fucu A21s's −+σ=

( )[ ] φσ=φ−− sin''sinA211s cfu

( ) ⎭⎬⎫

⎩⎨⎧

φ−−φ

σ='sinA211

'sin'sf

cuRelationship between su and φ’ - 1

Page 176: Shear Strength of Soils Final (Complete)

Relationship between undrained strength s and φ’

Shear Strength of Soils : Relationship between Undrained Strength and φ’ - c’

Relationship between undrained strength su and φτ φ’

uf

σ

sd (if drained test)sususu

c’

0

σσc = σ3f

For a soil with nonzero φ’ and c’

σ’3f

0• For a soil with nonzero φ’ and c’

( ) ⎬⎫

⎨⎧ φ'sin( ) ( ) ⎭

⎬⎫

⎩⎨⎧

φ−−φ

φ+σ='sinA211

sin'cot'c'sf

cu

Relationship between su and φ’ - 2