shear strength of soils final (complete)
TRANSCRIPT
Shear Strength of Soils
CE2112 Soil MechanicsNational University of Singapore
Department of Civil and Environmental Engineering
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
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
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
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.
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?)
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
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
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
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
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
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
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
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
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
Shear Strength of Soils : Stress-Strain Response
Stress-Strain Curves - 1
Shear Strength of Soils : Stress-Strain Response
Stress-Strain Curves - 2
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
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
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
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
Shear Strength of Soils : The Mohr Circle Method
The Mohr Circle MethodThe Mohr Circle Method
Mohr Circle
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
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
Mohr Circle – Failure PlaneShear Strength of Soils : General Stress State on Any Plane
σ1τf
σ3
fθf
σnf
Y τ Dτf
OX
σnA B
2θf
Cφ
σ3
σnf
σ3
σ1
Mohr Circle - 3
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°
Cα
σ3
σnθ
σ3
σ1
Mohr Circle - 4
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
Relationship between φ and θf
Shear Strength of Soils : The Mohr Circle Method
p φ f
Y τ Dτf
Oσn
A2θf
Cφ
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°
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
Cφ
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°
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
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
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
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
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
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)?
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
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.
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
• 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
• 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
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
• 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
• 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
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
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)
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)
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)
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
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
N ΔVShear Strength of Soils : Direct Shear Test
N
Tδ
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
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
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
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
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
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
• 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
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.
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
Triaxial Shear Test
Shear Strength of Soils : Triaxial Shear Test
Triaxial Shear Test
Triaxial Shear Test - 6
Triaxial Shear TestShear Strength of Soils : Triaxial Shear Test
Triaxial Shear Test
water
AB
Triaxial Shear Test - 7
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
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
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
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
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
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
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.
Fδ
• 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
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
Lab Vane Shear Test
Shear Strength of Soils : Lab Vane Shear Test
Lab Vane Shear Test
Other Shear Strength Tests - 2
Lab Vane Shear Test
Shear Strength of Soils : Lab Vane Shear Test
Lab Vane Shear Test
Blade or Spindle
Other Shear Strength Tests - 3
Shear Strength of Soils : Torvane
Torvane
Other Shear Strength Tests - 4
Miniature Cone Penetrometer
Shear Strength of Soils : Miniature Cone Penetrometer
Miniature Cone Penetrometer
Other Shear Strength Tests - 5
Pocket Penetrometer
Shear Strength of Soils : Pocket Penetrometer
Pocket Penetrometer
Other Shear Strength Tests - 6
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
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
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
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
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
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’.
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.
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
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.
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
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
• 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
• _______________ 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
• 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
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)
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.
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
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
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.) =
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.) =
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.) =
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
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
Rφ
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
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
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
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
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
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
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?
Shear Strength of Soils : Mohr Coulomb Criterion in terms of Principal Stresses
τ
τff σ−σ
φ σ1f
σ3f
θf
τff
2f3f1 σσ
Rφ
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
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
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
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
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)
Direct Shear TestShear Strength of Soils : The Direct Shear Test
Schematic of the Shear Box ApparatusN ΔV
δ
Tδ
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)
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
Direct Shear Test on Dense SandShear Strength of Soils : The Direct Shear Test
N ΔV
Tδ
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
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
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
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
(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°
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
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
Triaxial TestShear Strength of Soils : The Triaxial Test
cell pressure
cell pressure
Triaxial Test - 2
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
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
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=
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
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
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
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
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
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
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.
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
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)
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.
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
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
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.
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
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=
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=
vσ
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
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
Empirical correlation of A-Parameter and OCRShear Strength of Soils : Pore Pressure Parameters
Pore Pressure Parameters 3
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
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
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
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.
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
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
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
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
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
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°
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
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=
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.
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
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
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.
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.
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.
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
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
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.
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
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
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
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