Shear Strength Theory
Strength of different materials
Steel
Tensile strength
Concrete
Compressive strength
Soil
Shear strength
Presence of pore waterComplexbehavior
Introduction
Shear strength defined as the maximum stress that a soil can sustain before failure occurs– Not a failure of soil particles – Failure by relative movement of particles– Depends on the normal stress acting on any
plane within the soil
failure plane does not pass through particles
particles move relative to each other
Embankment
Strip footing
Soils generally fail in shear
At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength.
Failure surface
Mobilized shear resistance
Shear failure mechanism
The soil grains slide over each other along the failure surface.No crushing of individual grains.
failure surface
At failure, shear stress along the failure surface () reaches the shear strength (f).
Why is shear strength important?
resistance to movement of soil block governed by strength on plane
movement on plane governed by weight of soil block
Seattle, 2014
The 1993 landslide at Holbeck Hall, North Yorkshire.
ultimate bearing capacity of foundations
ultimate bearing capacity of foundations
Retaining wall design - lateral earth pressures
At failure, shear stress along the failure surface () reaches the shear strength (f).
Introduction
In order to define the shear strength of a soil, you must be able to:– Define the state of stress in a soil mass; and– Establish a relationship between shear and
normal stress at failure – define a FAILURE CRITERION
H = σxsinα
C
B A
N = σα × 1
T = τα × 1
Area = 1
Area = 1cosαV = σycosα
α
Area = 1sinα
+τ+σ
Sign conventions
Forces acting on plane (AC) through point in soil mass:
Define the state of stress in a soil mass:
Summing forces in horizontal and vertical directions gives:
Dividing by area gives:
Solving for σα and τα yields:
Square and add these equations, gives us the equation of the Mohr circle of stress
Define the state of stress in a soil mass:
Introduction
What is shear strength?– Related to three components:
• Frictional resistance to sliding• Cohesion and adhesion • Interlock
Frictional resistance to sliding
T
NN
RR
α
α
R
T
N
At equilibrium,
T = RsinαN = Rcosα
Dividing T/N gives, Rsinα/Rcosα = tanα
Hence T = Ntanα
Dividing by area, gives
= σntanα
Frictional resistance to sliding
As is increased, sliding will be imminent when a limiting value of α is reached, this is ϕ, therefore at failure - f = σntanϕ
tan ϕf
σn
frictionFriction + interlock
Usually represented as…Mohr-Coulomb equation
Tanϕ’f
σn
‘cohesion’
Failure envelope:
f = c’ + σn’tan ϕ’
note that c’ and ϕ’ are curve fitting parameters
Shear Failure in Soils
Defined previously as:– the maximum stress that a soil can sustain before failure
occursCan also be considered as:
– the resistance to deformation by continuous shear displacement of soil particles along surfaces of rupture
Not purely a function of peak shear stresses
Shear Failure in Soils
Engineers need to be able to define the nature and extent of stress and deformation (strain) at the time of failure
For frictional soils the most appropriate theory is the Mohr Theory of Rupture
Mohr’s theory indicates that there is a critical combination of shear and normal stresses acting on a plane
Shear Failure in Soils
So, failure will occur when a state of stress exists in the soil so that one point on the Mohr circle touches the ‘failure’ or ‘rupture’ line– i.e. failure line is a tangent to the circle at a
single point
If the shear strength is represented by = f():
Normal stress
Shear stress
compressiontension Mohr’s envelope: = f()
I IIIII
IV V
Any combination of stresses that falls below this line, represent the STABLE condition
Normal stress
Shear stress
compressiontension Mohr’s envelope: = f()
I IIIII
IV V
Circles I – IV touch (are tangent to) the envelope, indicating impending failure
Normal stress
Shear stress
compressiontension Mohr’s envelope: = f()
I IIIII
IV V
Since the envelope is curved, it is clear that the friction angle reduces with increasing confining stress ()
Normal stress
Shear stress
compressiontension Mohr’s envelope: = f()
I IIIII
IV V
The non-linearity however, is slight, and so it is more convenient to represent this envelope as a straight line
Normal stress
Shear stress
compressiontensionMohr-Coulomb envelope: = c + tan
I IIIII
IV V
Shear Failure in Soils
31
Shear stress
Normal stress
Shear strength is a function of the normal stress and soil properties (such as c and )
3 1
f
f =
f
limiting value of f is defined as
Shear Failure in Soils
32
Shear stress
Normal stress
Shear strength is a function of the normal stress and soil properties (such as c and )
3 1
f
f
The shear strength for a purely frictional soil can be written as
tanffS
Shear Failure in Soils
33
Shear stress
Normal stress
Shear strength is a function of the normal stress and soil properties (such as c and )
1
f
f
The shear strength for a purely frictional soil can be written as
tanffS
Soil elements at different locations
Failure surface
X X
X ~ failure
YY
Y ~ stable
’
'tan'' cf
Y
c
c
c
Initially, Mohr circle is a point
c+
The soil element does not fail if the Mohr circle is contained within the envelope
GL
Y
c
c
c
GL
As loading progresses, Mohr circle becomes larger…
.. and finally failure occurs when Mohr circle touches the envelope
1
1
33
If we consider an element of soil subjected to a combination of stresses (Principal stresses 1 > 3)
At some value, a plane of failure will develop in the element
1
1
1
1
3
3
3
Along this failure plane, a critical combination of shear stress and normal stress will develop
3 1
1
1
1
3
3
3
c
Along this failure plane, a critical combination of shear stress and normal stress will develop
Using a Mohr circle, we can analyse these stress conditions
The failure envelope is defined by:
tancf
3 1
1
1
1
3
3
3
c
The orientation of the failure plane, and hence point D can be established, since we know that the angle between the major principal plane and the failure plane is
D
3 1
1
1
1
3
3
3
c
The orientation of the failure plane, and hence point D can be established, since we know that the angle between the major principal plane and the failure plane is
D
3 1
1
1
1
3
3
3
c
Since the angle subtended from the centre to the failure point is 2 or:
902
2
D
3 1
1
1
1
3
3
3
c
which becomes:
902
2
245
D
3 1
1
1
1
3
3
3
c
which becomes:
902
2
245
’3 ’1
’
1
1
1
3
3
3
c’
C D O E
F
BA
It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses
’3 ’1
’
1
1
1
3
3
3
c’
C D O E
F
BA
It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses
We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2
We can see that OF is:
’3 ’1
’
1
1
1
3
3
3
c’
C D O E
F
BA
It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses
We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2
We can see that OF is:
'sin)('sin BOABOAOF
’3 ’1
’
1
1
1
3
3
3
c’
C D O E
F
BA
It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses
We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2
We can see that OF is:
'sin)('sin BOABOAOF
’3 ’1
’
1
1
1
3
3
3
c’
C D O E
F
BA
It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses
We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2
We can see that OF is:
'sin)('sin BOABOAOF
’3 ’1
’
1
1
1
3
3
3
c’
C D O E
F
BA
It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses
We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2
We can see that OF is:
'sin)('sin BOABOAOF
’3 ’1
’
1
1
1
3
3
3
c’
C D O E
F
BA
It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses
We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2
We can see that OF is:
Since OF is the radius of the circle:
'sin)('sin BOABOAOF
31 ''5.0 OF
’3 ’1
’
1
1
1
3
3
3
c’
C D O E
F
BA
It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses
If,
and
and with BO written as:
Which defines the centre of the circle
'sin)('sin BOABOAOF
31 ''5.0 OF
'cot' cAB
31 ''5.0 BO
’3 ’1
’
c’ C D O E
F
BA
It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses
becomes:
Simplifying and solving for ‘3, we obtain:
'sin)('sin BOABOAOF
31
31
31
31
'''cot'2
''
''5.0'cot'
''5.0'sin
cc
'sin1
'cos'2
'sin1
'sin1'' 13
c
Since:
and:
we can rearrange this:
2
'45tan
'sin1
'sin1 2
2
'45tan
'sin1
'cos
'sin1
'cos'2
'sin1
'sin1'' 13
c
To give:
or:
2
'45tan'2
2
'45tan'' 2
13
c
2
'45tan'2
2
'45tan'' 2
31
c
2
'45tan'2
2
'45tan'' 2
13
c
2
'45tan'2
2
'45tan'' 2
31
c
These are alternative forms of theMohr-Coulomb failure criterion, written in terms of effective stress
Determination of Shear StrengthIntroduction
– Only considered theoretical considerations to this point
– Validity and value of these considerations is closely related to the parameters and observations from experimental and field studies
– Strength parameters are obtained from:
• direct shear test• triaxial test; and
Determination of Shear StrengthIntroduction
– Soils are neither continuous, homogenous nor isotropic
– Limited number of tests will only give approximations to the characteristics of the soil mass
– Samples should be representative• ideally undisturbed• sample quality is profoundly important
– Test conditions should be representative
Other laboratory tests include,Direct simple shear test, torsional ring shear test, plane strain triaxial test, laboratory vane shear test, laboratory fall cone test
Determination of shear strength parameters of soils (c, orc’’
Laboratory tests on specimens taken from representative undisturbed samples
Field tests
Most common laboratory tests to determine the shear strength parameters are,
1.Direct shear test2.Triaxial shear test
1. Vane shear test2. Torvane3. Pocket penetrometer4. Fall cone5. Pressuremeter6. Static cone penetrometer7. Standard penetration test
Laboratory tests
Field conditions
z vc
vc
hchc
Before construction
A representative soil sample
z vc +
hchc
After and during construction
vc +
Laboratory testsSimulating field conditions in the laboratory
Step 1Set the specimen in the apparatus and apply the initial stress condition
vc
vc
hchc
Representative soil sample taken from the site
0
00
0
Step 2Apply the corresponding field stress conditions
vc +
hchc
vc + Traxia
l test
vc
vc
Direct shear test
Determination of Shear StrengthDirect shear test
– usually carried out in a shear box
Determination of Shear Strength
Direct shear test– usually carried out in a shear box
shear force applied to one half
soil sample
other half is restrained
Determination of Shear Strength
Direct shear test– usually carried out in a shear box
shear force applied to one half
soil sample
other half is restrained
vertical load N applied through the top platen
Determination of Shear Strength
Direct shear test– usually carried out in a shear box– very simple arrangement– very rapid test– no facility to control drainage and no
means of measuring pore water pressures
– tend to be used on coarse grained soils• free-draining• excess pore pressures will dissipate
instantaneously, hence u = 0• total stress = effective stress
Direct shear testTest procedure
Porous plates
Pressure plate
Steel ball
Step 1: Apply a vertical load to the specimen and wait for consolidation
P
Proving ring to measure shear force
S
Direct shear test
Step 2: Lower box is subjected to a horizontal displacement at a constant rate
Step 1: Apply a vertical load to the specimen and wait for consolidation
P
Test procedure Pressure plate
Steel ball
Proving ring to measure shear force
S
Porous plates
Direct shear test
Analysis of test results
sample theofsection cross of Area
(P) force Normal stress Normal
sample theofsection cross of Area
(S) surface sliding at the developed resistanceShear stressShear
Note: Cross-sectional area of the sample changes with the horizontal displacement
Direct shear tests on sands
Dense sand/ OC clay
fLoose sand/ NC clayf
Dense sand/OC Clay
Loose sand/NC Clay
Cha
nge
in h
eigh
t of
the
sam
ple Exp
ansi
onC
ompr
essi
on
Shear displacement
Stress-strain relationship
Determination of Shear Strength
Direct shear test– For any soil material, a number of
tests will be carried out (>3) at different values of normal stress
– From each test, the maximum shear stress f is plotted against the corresponding value of ‘n
– The straight line plotted through the points is the effective stress failure envelope for that soil, and satisfies:
'tan'' nf c
f1
Normal stress = 1
Direct shear tests on sandsHow to determine strength parameters c and
She
ar s
tres
s,
Shear displacement
f2
Normal stress = 2
f3
Normal stress = 3
She
ar s
tres
s at
fai
lure
,
f
Normal stress,
Mohr – Coulomb failure envelope
Direct shear tests on sandsSome important facts on strength parameters c and of sand
Sand is cohesionless hence c = 0
Direct shear tests are drained and pore water pressures are dissipated, hence u = 0
Therefore, ’ = and c’ = c = 0
Direct shear tests on clays
Failure envelopes for clay from drained direct shear tests
She
ar s
tres
s at
fai
lure
,
f
Normal force,
’
Normally consolidated clay (c’ = 0)
In case of clay, horizontal displacement should be applied at a very slow rate to allow dissipation of pore water pressure (therefore, one test would take several days to finish)
Overconsolidated clay (c’ ≠ 0)
Interface tests on direct shear apparatusIn many foundation design problems and retaining wall problems, it is required to determine the angle of internal friction between soil and the structural material (concrete, steel or wood)
tan' af cWhere, ca = adhesion, = interface angle of friction
Foundation material
Soil
P
S
Foundation material
Soil
P
S
Advantages of direct shear apparatus
Due to the smaller thickness of the sample, rapid drainage can be achieved
Can be used to determine interface strength parameters
Clay samples can be oriented along the plane of weakness or an identified failure plane
Disadvantages of direct shear apparatus
Failure occurs along a predetermined failure plane
Area of the sliding surface changes as the test progresses
Non-uniform distribution of shear stress along the failure surface