Transcript Shear Strength in Soils

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
steel
plates:
probably
think
of
Shearing
Pins
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
shearshear
forcestress,
causes
Thethe
internal
τ isfailure,
simply
then
the shear
results,
the shear
force, stress
T actingthat
on the
failure
plane
divided
by strength
the area, of
A of
the
τf is the
shear
the
failure
plane:
material.
T
So shear forcesτ are
 those that tend
A
to cause shear failure.
Area, A
The Shear
Force that
acts
the
Since
the external
force
is on
acting
failure plane
resisted
by the
parallel
to theisfailure
plane,
the
strength
of the
material.
internal
strength
of the
material is
thought of as its internal friction, F.
This is the material’s reaction to the
external shear force, T.
Vector
addition
gives
thean
resultant
to
and
overcome
thereby
the
cause
friction
the
object
force,
F
to
on
that
must
be
applied
to
object
of
The
tangent
of
the
friction
angle,

vector,
R
which
acts
at
an
angle
of

plane
where
move.
the
object
known
weight,
W isrests
is the
theWRT
ratiothe
of
F
to
W
which
also
normal
to
the
plane.
The object’s weight vector, W acts
known as the coefficient of friction.
normal problems
to the failure
plane.
Friction
in mechanics
determine the external force, T
W
W

SHEAR
STRENGTH
If
the
soil
is
loadedof(yet
Consider and element
soilsober):
within a
largeSOILS
soil mass:
IN
Soil Surface
The loading of a material that
undergoes shear failure is not always
parallel to the failure
plane.
Soil Element
Soil Mass
Bedrock
σ1
For
This
visual
is the
simplicity
major principal
we replace
stress
the
The load transmits stress to the
distributed
with an equivalent
distribution,load
designated
σ1 due to
element by inter-particle contacts.
point
the load.
load.
σ1
σ2
σ3
σ3
σ2
σ1
The
thethe
element
will react
Since
Thesoil
element
webelow
assume
squeezed
soil
vertically
is isotropic,
will
with
of
equalpressure
magnitude
but
tend
the aconfining
tostress
bulge horizontally
lateral
to which
will be
the
the
same
allupwards
directions
so σis2 = σσ23
soil directed
reactsinwith
confining
pressures
soand
it too
allowing
us designated
to
view principal
it in σ21.dimensions.
and
σ3 in the
other
directions.
The
friction
force
on
this
failure
But
what
has
this
got
to
do
with
Soil
undergoes
shear
failure
when
one
For this to happen, a failure plane
plane
isSHEAR
overcome
by the
external
STRENGTH?
portion
moves
relative
to
the
rest.
SHEAR
FAILURE!
develops within the soil.
forces and viola:
σ1
2-D
Θ
σ1
1.
Direct
Shear
Test
The
angle
of
internal
friction,

The
Shear
at failure,
τtests
,
There
are 3Stress
basic laboratory
f is
Triaxial
Compression
Test
the2.
pressure
required
overcome
that
can be
performed
on soil of
characterizes
the
sheartostrength
the
friction
on
the
surface
of
the
3.
Unconfined
Compression
Test
samples
to
evaluate
the
shear
the soil and is one of its shear
failure plane
(a.k.a.
Shear
Strength).
strength parameters.
parameters:
strength
σf
Θ
τf

Rf
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
to direction of applied load).
DIRECT SHEAR TEST
This
forces
failure
tohorizontal)
occur
on a
Then
the
top
and
base
are
pushed
The
prepared
sample
is
placed
A
normal
to
the
A
shear
and
a(90
abox
normal
top
asoil
has
base
extension
load
three
piston
The
horizontal
force
is parts:
increased
horizontal
plane
between
the
in opposite
load
is applied
in
thedirections
to
box.
the soil.
until
the
sample
shears
in
two:
top and base:
The procedure is repeated two more
times using successively heavier
normal loads.
DIRECT SHEAR TEST
In
the means
CV504 labs,
inside
dimensions
This
the the
failure
plane
has anof
the shear
box
are
60
mm
by
60
mm.
2
area of 3600 mm .
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.
The
For
shear
this
reason,
force
required
the of
soil’s
toshear
shear
The
shear
strength
the
soil
therefore
not constant
but
thestrength
sample increases
isischaracterized
in proportion
by
changes
with
the
confining
to
the
normal
load.pressure.
shear strength
parameters:
(c,).
DIRECT SHEAR TEST
The
equation
of
Coulomb’s
failure
envelope:
The
τ
axis
intercept
is
the
apparent
cohesion,
c ofpoints:
the soil.
Fitting
a
best
fit
line
through
these
Plotting
the
shear
stress
versus
normal
stress:
The slope angle of this line is the
angle
τ
=
c
+
σ
tan
.
we have
an
estimate
of
Coulomb’s
failure
envelope
n
f
of internal friction,  of the
soil.
Shear Stress,
τ (kPa)
First Test
Second Test
Third Test
τf
τf

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
Specimens
specimen
are
is then
weighed
placed
and
in a
The
specimen
is
mounted
between
2 platens
Preparation
varies
with
material
properties
and
then
inserted
into
a latex
sleeve.
from
sampled
soil.
dimensions
measured
chamber.
first.
(clay
vsplexiglas
sand vs
cohesive
granular).
length
diameter
TRIAXIAL COMPRESSION TEST
Onceathe
cell chamber
is filled
withthe
water,
the air
release
For
drained
test
drain
valve
is
Then
the
is
placed
on
the
The
specimen
is
mounted
on
the
pedestal
For
The
an
assembly
undrained
is
then
test,
mounted
the
drain
on
valve
the
Water
is
forced
into
the
cell
with
the
supply
valve is closed and the cell pressure is increased
of
the
chamber
base
asthe
shown.
valve
open
asdesired
well
as
the
air
release
opened
and
pore
water
collected.
to
the
value
for
test. valve.
compression
islocked
closed.
testing
machine.
base
and
into
place.
loading ram
air release valve
plexiglas chamber
water supply for
cell (confining)
pressure
loading cap
latex sleeve
drainage
or pore
specimen
water
pressure
measurement
porous disc
pedestal
TRIAXIAL COMPRESSION TEST
The
Major
Principal
Stress,
σ
, is
the
Then
agoal
vertical
axial
load
is
applied
to
the
1known
The
cell
pressure,
σ
,
is
also
as
But
how
can
we
find
τ
and
σ
from
σ
The
effect
of
the
cell
pressure
on
The
is
to
simulate
the
stresses
3
f
f
Enter
Christian
Otto
Mohr:
combination
of creating
the deviator
stress and 1
loading ram
compressive
the
Minor
Principal
Stress.
the
specimen
is
illustrated
below:
confining
the
specimen
in
the
ground.
and
σ
?
cell
pressure:
stresses or the deviator
stress
∆σ :
3
∆ σ
σ3
σ3
σ1  Δσ  σ3
σ3
Plan View of
Specimen
∆ σ
Side View of
Specimen
Source:
“commons.wikimedia.org”
TRIAXIAL COMPRESSION TEST
Herr Mohr
in Germany on
forwas
anyborn
material,
1835-10-08
and
was
a
renowned
the internal shear and normal
Civil Engineer and professor until
stresses acting on ANY plane
In his
other
he
death words,
on 1918-10-02.
within the material,
Whilediscovered
contemplating the
symmetry of
MOHR’S
caused by external stresses or loads
his name, Otto started tinkering
CIRCLE
.
canthe
be determined
using
with
properties of
the acircle
trigonometric
transformation
when he discovered
that... of
the external stresses.
TRIAXIAL COMPRESSION TEST
Shear Stress, τ (kPa)
If
you
plot
σ1 got
and
on
the
σnone
axis
Ultimately,
the
test
ends
when
shear
failure
During
the
test,
this
circle
starts
as
point
at
Remember
plot
of
Stress
versus
then
fit
then
one
you’ve
circle
through
Mohr’s
these
circle!
points
The
point
of
tangency
ofaσShear
the
circle
and
failure
3
σoccurs
and then
grows
tothe
the
right
as
axial
stress,
anddefines
the
circle
has
become
tangent
Normal
Stress?
shear
strength,
τf
3envelope
∆σ increases
but
σ3stress,
remains
to
the
failure
envelope.
and
normal
σconstant.
f.
τf

c
σ3
∆σ
σ1
∆σσ∆σ
f
σ1
σ1
Normal Stress, σn(kPa)
TRIAXIAL COMPRESSION TEST
If one line cannot be drawn tangent to all three
As
with
most
lab
measurements,
the
ideal
(one
This
means
that
you
need
to
perform
the
test
But
how
can
you
be
sure
one
of
them
isn’t
bogus?
A
third
test
at
yet
another
cell
pressure
would
But
Geometrically,
how
do
we
find
you
need
the
failure
at
least
envelope
two
circles
from
ina
circles, a best fit is made as long as one circle
line
tangent
to on
allcompression
three
circles)
is
difficult
at
least
twice
the
same
material
but
at
help
to
confirm
the
validity
of
the
failure
order
triaxial
to
define
a
line
tangent
test?
to
both.
is not out to lunch compared to the others.
Shear Stress, τ (kPa)
to
achieve.
different
cell pressures.
envelope.

c
Normal Stress, σn(kPa)
TRIAXIAL COMPRESSION TEST
Instead
of
doing
this
graphically,
we
can
use
Remember
the
deviator
stress,
∆σ
=
σ
σand
So
the
radius
of
the
Mohr’s
Circle,
R
is
half
Thefor
Centre
Mohr’s
Circle,
C is then:
then
eachof
test,shear
the
shear
strength,
Once
we
have
the
strength
parameters,
1 τf the
3,
trigonometry to
find equations
for τf and
diameter
or:
which
is the
diameter
the
Circle.
normal
stress,
σ of
can
beMohr’s
found.
 and
c defining
the
failure
envelope,
Shear Stress, τ (kPa)
f
σf using the angle of the
failure plane, Θ
and the values of σ1 and σ3
σ1  σ 3
R
2
σ1  σ 3
C
2
specimen
Θ
failure plane

c
C
Θ
σ3
R
R
σ1
Normal Stress, σn(kPa)
TRIAXIAL COMPRESSION TEST
ABC
=&90
so
= 90
- the
and
DBC
=
To
follow
trig
label
vertices:
’s
EBC
BCFthe
areACB
bothwe
isosceles
& EBF
is 90.
= 180 =180
– 2Θ –= 2(9090 -  Θ) = 2Θ
 EFB = 90 – DCB
Θ & BCF

θ  45 
2
Shear Stress, τ (kPa)
Rearranging:

B
τf

A

c
E
2Θ
Θ
σ3
D
σf
σ1  σ 3
2
σ  σ3
C 1
2
R
C
F
σ1
Θ
Normal Stress, σn(kPa)
TRIAXIAL COMPRESSION TEST
Shear Stress, τ (kPa)
 you
can
find Θ
InSo…
DBC,
sideside
BD
is
same
asand
τf . 
Also
in knowing
DBC,
DCthe
= Rcos(180-2Θ)
1
using
Θ
and
the
σ
&
σ
values
for
each
trial,

1
3
 σf = C  f– Rcos(180-2Θ)
or
C
+
Rcos(2Θ)
Rsin180  2θ   σ1  σ3 sin2θ
τf and σf can be found
for each trial.
2
1
 f  σ1  σ3 sin2θ (Eqn. 4.3)
2
1
1
 f  σ1  σ3   σ1  σ3 cos2θ (Eqn. 4.4)
2
2
B
τf

A

c
E
2Θ
Θ
σ3

2
σ  σ3
R 1
2
σ  σ3
C 1
2
θ  45  
D
σf
C
F
σ1
Θ
Normal Stress, σn(kPa)
TRIAXIAL COMPRESSION TEST
What
happens
when
the
pore
water
is
not
Typically,
the
deviator
stress
at
failure
is a
As
the
external
pressure
increases,
the
internal
And,
thefinal
apparent
cohesion,
cu willisbe
the same
Therefore,
the
failure
envelope
typically
One
word
on
nomenclature…
The normal
for each trial will
then
allowedstress,
to drainσf(UNDRAINED
TEST)?
Shear Stress, τ (kPa)
pore
water
pressure
(acting
in
the
opposite
All
stress
symbols
used
in cell
fairly
constant
for
each
different
forAlleach
trial
and
equal
to
the
shear
horizonal
line
and

=
0.
symbols
used
u in DRAINED
be
σ
+
c
direction stress
to the
external)
increases
tonot
match
3
u tests are
UNDRAINED
pressure.
strength,
τprimed…σ
teststrivialize)
are
usually
feffect. 1’,σ3’,σf’
(and
the
primed…σ ,σ ,σ and  indicating
1 3 f that fthey are in
and f’ indicating
(Thethey
radiiare
areinallterms
the same)
that
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 = τ f
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
The
Mohr’s
circle
continues
to
grow
until
failure
Instead
of
calling
it
the
deviator
stress,
σ,
itof
is
If
a
q
does
not
maximize
before
15%
axial
load
starts
at
0
and
increases
Because
σ
=
0
and
q
is
the
diameter
IfThe
the
q
does
maximize
before
15%
strain,
This
is
analogous
to
the
circle
becoming
u
3
uf
u
The
point
of tangency
of the circle
and
occurs
either
when
the
specimen’s
shear
strain
is
reached
then
the
q
at
15%
called
the
unconfined
compressive
stress,
q.u.
steadily
as
in
the
triaxial
compression
test.
the
circle,
the
shear
strength,
τ
and
thenstrength
the
maximum
q
value
is
used
as
q
u
tangent
to
the
failure
envelope
when
f
u defines
uf
failure
envelope
the
shear
is
reached
or
15%
strain.
strain is used to define the
Shear Stress, τ (kPa)
normal
stress
at failure,
σf are both
shear
failure
occurs.
strength,
τ
and
normal
stress,
σfof
.
f
unconfined
compressive
strength
estimated to be half of quf.
the specimen, quf
c = τf
qu qu
qu σf qu
qu ququf
u
Normal Stress, σn(kPa)