# SHEAR STRENGTH IN SOILS

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```					SHEAR STRENGTH

In general, the shear strength of
any material is the load per unit area
or pressure that it can withstand
before undergoing shearing failure.
Shearing Pins can be used to fasten
When you hear “Shear Failure” you
With high enough plate forces in
together two Shearing Pins
probably think of steel plates: or
opposite directions…
Bolts.
How do these fail?

Each pin has sheared into two pieces.
The failure plane for metals will be
parallel to the external shear forces.
If the shear force causes is simply
The internal shear stress, τ failure,
then the shear stress thatthe failure
the shear force, T acting on results,
planethe shear strength ofof the
τf is divided by the area, A the
failure plane:
material.
So shear forcesτ are T those that tend

A failure.
to cause shear
Area, A
The Shear Force that acts acting
Since the external force is on the
failure to the failure plane, the
parallel plane is resisted by the
strength of of material.
internal strength thethe material is
thought of as its internal friction, F.
This is the material’s reaction to the
external shear force, T.
and thereby cause theforce, to of
the friction resultant
toVector addition gives to an objecton
overcome of the friction angle,
that must be appliedthe object F 
The tangent        acts object rests
vector, R whichmove.at an angle of 
where W
is the planeknowntothe which is also
F weight, W
the ratio ofnormal to the plane.
The WRT the weight vector, W acts
object’s
known as the coefficient of friction.
normal to the failure plane.
Friction problems in mechanics
determine the external force, T
 W
W
SHEARisSTRENGTH a
If the soil loaded (yet within
Consider and element of soilsober):
IN SOILS
large soil mass:
Soil Surface

undergoes shear failure is not always
plane.
parallel to the failureElement
Soil
Soil Mass

Bedrock
σ1

This is the major principal stress
For visual simplicity we replace the
The load transmits stress to the
distribution, designated σ1 due to
element by inter-particle contacts.
σ1

σ2

σ3                  σ3

σ2
σ1

The element the element will will
The soil belowsqueezed verticallyreact
Since we assume the soil is isotropic,
tend confining of equal magnitude the
the to bulge lateral pressure will but
with a stress horizontally to which be
reacts all directions it so is
the directed upwards soandtoo σ2 = σ2
soil same inwith confining pressures σ3
allowing us designated σ1dimensions.
other it in 2 .
and σ3 in theto view principal directions.
The friction shear on this do with
But what has this got to failure
Soil undergoes force failure when one
For this to happen, a failure plane
SHEARrelative to external
plane is overcome by thethe rest.
STRENGTH?
SHEAR FAILURE!
portion moves within the soil.
develops
forces and viola:
σ1

2-D
Θ

σ1
1. Direct Shear at failure, τ is
TheShear 3 basicTest friction,, 
There are of internal
The angle Stress laboratory tests
f
thethat can be performed Test of
2.pressure required to on soil
Triaxial the shear strength
characterizes Compressionovercome
samples to evaluate its of the
the friction on the surfaceshear
3. Unconfined one of the shear
the soil and is Compression Test
failure plane (a.k.a. Shear Strength).
strength parameters:
strength parameters.

σf
Rf
Θ                 τf
DIRECT SHEAR TEST
Can be performed on all types of
soil, moist or dry.
Measures shear stress at failure
on failure plane for various
normal stresses.
Failure plane is controlled (parallel
DIRECT SHEAR TEST
Then shear a failuresample parts:a
A forces andextension pushed
normal (90 to forceoccur on
to
The the a box soil three is placed
horizontal base piston
ATheand top top base horizontal)
normalthe is
in is applied between
plane box.
horizontal in the to the soil. two:
until opposite directionsin
top and base:

The procedure is repeated two more
times using successively heavier
DIRECT SHEAR TEST
This CV504 the failure plane has an
In the means labs, the inside dimensions of
the shear box are 60 mm by .60 mm.
area of 3600 mm2
The shear force at failure (maximum)
and normal load, both in Newtons are
divided by this plane area to find the
shear stress at failure and the
normal stress in MPa.
For this force the of theshear
The shear strength soil’s shear
The shearreason,required tosoil
strengthincreases in proportion
sample is not constant but
the therefore ischaracterized by
with the confining pressure.
changes to the normal load. (c,).
shear strength parameters:
DIRECT SHEAR TEST
Fitting the shear the apparent failure envelope:
The equation fit stress versus these stress:
The τ axisa bestof Coulomb’sline normal points:
The slope angle of through is the the soil.
Plotting intercept is linethis cohesion, c of angle
= Coulomb’s .
τf ofc + σntanof the soil.
we have an estimate friction,  failure envelope
of internal
First Test         Second Test   Third Test
τ (kPa)

τf
τf
Shear Stress,

τf


c

Normal Stress, σn(kPa)
TRIAXIAL COMPRESSION TEST
Can be performed on all types of soil,
moist or dry and can consolidate
sample to in situ conditions by
tracking pore water pressures.
Measures vertical stress applied to
soil sample and confining pressure.
Shear stress on failure plane must be
calculated from principal stresses.
TRIAXIAL COMPRESSION TEST
Cylindrical specimens are prepared
The specimen is then placed
Specimens are weighed properties
Preparation varies with material andin a
The specimen is mounted between 2 platens
from measured first.
(clay vs sand sampled soil.
plexiglas chamber.
dimensions vs cohesive granular).
and then inserted into a latex sleeve.

diameter

length
TRIAXIAL COMPRESSION TEST
Then undrainedthencelldrain ison the
Once the cell chamberthe
a is forced is with is on with therelease
the is is intotest, pressurepedestalis
Foran drainedand the cellmounted increased
The specimen
The assembly
Water              test water,the air on the
filled      placed valve
the
Forvalve is closed mounted the drain valve
the            supply
valve open andpore base theplace.
of to and is testing shown.
base chamber water collected.
openedtheas well closed.asmachine.
the desired value into test.
locked air
compression as the for release valve.
air release valve

plexiglas chamber

cell (confining)                   latex sleeve
drainage or pore
pressure                         specimen
water pressure
measurement
porous disc
pedestal
TRIAXIAL COMPRESSION TEST
The cellvertical axialσ3Stress, Mohr: onσ1
Butcombination ofthe cellthe σ1f is to and
Then effect Principal ,τis also known as
The aMajor we finddeviatorstresses
Thegoal
howpressure, load and σ , from
applied the
TheEnteris to simulate ispressure the
can of the Otto stress
Christian f
thethe Minor isdeviator stress ∆σ
cellPrincipal the below:
confining the thepressure: Stress. :
specimenandillustrated ground.
σ3 ?
stresses or specimen in
∆ σ
σ3

σ3
σ1  Δσ  σ3

σ3
∆ σ
Plan View of     Side View of            Source:
Specimen         Specimen        “commons.wikimedia.org”
TRIAXIAL COMPRESSION TEST
Herr Mohr was born in Germany on
for any material,
and was a normal
1835-10-08 shear and renowned
the internal
Civil Engineer and professor until
stresses acting on ANY plane
other words, he
In hiswithin the 1918-10-02.
death on
material,
discovered MOHR’S
While contemplating the symmetry of
caused by external stresses or loads
his name, Otto started tinkering
CIRCLEusing a
can be properties
.
with the determinedof the circle
trigonometric transformation of
when he discovered that...
the external stresses.
TRIAXIAL COMPRESSION TEST
DuringIf of tangencyends on theseoneversus
Ultimately, plottestthrough circle points at
Remember you’ve1 and σ the thecircle!point
The pointyou thethis got a 3starts as nandfailure
then the one circlecircleMohr’s shear failure
fit test, plot of Shear Stress
then      σ    of when σ axis
and then the circleStress? strength, τf
σoccurs anddefines thehas become tangent
Normal right
3envelope grows to the shear as axial stress,
and normal remains constant.
∆σ increases but σ3stress, σ .
to the failure envelope.
f
Shear Stress, τ (kPa)

τf



c

σ3   ∆σ ∆σ ∆σ 1
σf    σ     σ1   σ1   Normal Stress, σn(kPa)
TRIAXIAL COMPRESSION TEST
If one line cannot be drawn tangent to all three
Butwith mostbest measurements,pressure would
As how canwelab sure oneto them isn’t bogus?
Thishow do youyou need at least the circles in
two the (one
A third testthat you is madeofperformidealcircle
Geometrically, find another celllong as one test
But means at yet the failure envelope from a
be need
circles, a      fit        as
atorder triaxial the line circles) is others.
help toout to lunchthree tangent to both.at
tangentdefinethe same materialfailure
to to on a validity to the but
twice all compared test?
lineleast confirm compression of thedifficult
is not
to cell pressures.
different achieve.
envelope.
Shear Stress, τ (kPa)

   c
Normal Stress, σn(kPa)
TRIAXIAL COMPRESSION TEST
Remember doing this graphically,RC = half the
of the shear stress, parameters,
Mohr’s Circle, we canf σ3
Instead have of the the shear strength, τ use
Once for each the Mohr’s Circle,
The Centre test,
then we                                  1 - and
trigonometry to find equations for τf and
diameter the
which normaldiameter of or:be found.Circle.
is c defining the failure envelope,
 and the stress, σ can Mohr’s
σf using the angle of the failure plane, Θ
f
and the values of σ1 and σ3
Shear Stress, τ (kPa)

σ1  σ3                        σ1  σ 3
R                             C
2                              2

specimen
Θ

failure plane


c                   C        Θ   R         R
σ3                         σ1   Normal Stress, σn(kPa)
TRIAXIAL COMPRESSION TEST
ABC = 90 so trig we label  & EBF is = 
EBC & BCF are both = 90 - the vertices:
’sTo follow the ACB isoscelesand DBC 90.
Θ & BCF – 2Θ = 90 - 
 EFB = 90 – DCB = 180 =180 – 2(90- Θ) = 2Θ

Rearranging:              θ  45 
Shear Stress, τ (kPa)



2

B
τf
σ1  σ3
R

2              Θ
σ  σ3
C 1
c        E        Θ
2Θ
F        2
A                  

D      C                 Normal Stress, σn(kPa)
σ3       σf                   σ1
TRIAXIAL COMPRESSION TEST
In DBC, sideside DC = sameΘ and. 
So… DBC,  you can find as τf
Also in knowing BD is theRcos(180-2Θ)
1
Θ and the σ1 & σ3 values + each trial,
using C – Rcos(180-2Θ) or Cfor Rcos(2Θ)
 σf =  f  Rsin180  2θ  σ1  σ3 sin2θ
τf and σf can be found for each trial.
2
Shear Stress, τ (kPa)

1
 f  σ1  σ3 sin2θ (Eqn. 4.3)
2
1            1
 f  σ1  σ3   σ1  σ3 cos2θ (Eqn. 4.4)
2            2

B                     θ  45 
τf                                                                      2
σ σ
R 1 3

2             Θ
σ  σ3
C 1
c          E        Θ
2Θ
F         2
A                  

D      C             Normal Stress, σn(kPa)
σ3       σf               σ1
TRIAXIAL COMPRESSION TEST
Typically, the deviatorincreases, the is same
What happens when the pore water internal
As the external pressureenvelopeat be thenot a
And, thefinal word on stress is failure is
One apparent cohesion, cu will typically
Therefore, the failure nomenclature…
allowed to drain for each trial TEST)?
The normal stress, σf(UNDRAINED will then
All stress equal in usedshear
the in
fairly constant and symbols = 0.opposite
pore water pressure (actingdifferent cell
for each to in DRAINED
for each trial line and u the
horizonal σ + c
All stress external)used
direction to the symbols increases tonot
be 3
UNDRAINEDu tests are match
pressure.
strength, feffect.
tests are usuallyτprimed…σ1’,σ3’,σf’
the
(and trivialize),σ and  indicating
primed…σ ,σ
Shear Stress, τ (kPa)

and f’ indicating that fthey are in
1 3 f
(The radii are all the same)
that they are in terms of TOTAL
terms of EFFECTIVE STRESS
STRESS and the shear strength
and the shear strength
parameters are denoted (u,cu)
parameters are denoted (’,c’).
u  0
cu = τf

σf   σf   σf       Normal Stress, σn(kPa)
UNCONFINED COMPRESSION TEST

Is performed mainly on cylindrical,
moist clay specimens sampled
from bore holes.

Measures vertical stress applied to soil
sample with no confining pressure.
Shear stress on failure plane is
determined similarly to undrained
triaxial compression test.
UNCONFINED COMPRESSION TEST
InsteadIfuais u σ3maximize ufto and before 15%is
If Mohr’sdoes of0not deviator stress, strain,
The of point it tangencyis grow diameter
This calling starts q 0of theuntil failure
q circle the maximize circle
q does and at
Becauseanalogous to the the15% σ,and
circle becoming
The either when the specimen’s shear
occurs in is reached then the stress, q .
tangent to reacheddefinesused at 15%
strain
thenstrengththethe compressive qu shear.
calledthe circle, the shear strength,asftest.
the                  compression and
triaxialor 15% the
the maximum u value is strain. qufu
envelope
failure is is used to define the
strain                        at normal σf are σ .
normal stressandfailure, stress,both
shear failure occurs.
strength, τcompressive strength fof
unconfined f to be half of q .
Shear Stress, τ (kPa)

estimated                uf
the specimen, quf

c = τf

qu qu    qu σf qu   qu ququ
uf      Normal Stress, σn(kPa)

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