SHEAR STRENGTH

1
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.

2
How do these fail?

• When you hear Shear Failure you probably think
  of Shearing Pins or Bolts.

Shearing Pins can be used to fasten together two
steel plates
With high enough plate forces in opposite
directions
Each pin has sheared into two pieces.
3
The failure plane for metals will be parallel to
the external shear forces.
If the shear force causes failure, then the shear
stress that results, tf is the shear strength of
the material.
The internal shear stress, t is simply the shear
force, T acting on the failure plane divided by
the area, A of the failure plane

• So shear forces are those that tend to cause
  shear failure.

T
Area, A
4

• The Shear Force that acts on the failure plane is
  resisted by the strength of the material.

Since the external force is acting parallel to
the failure plane, the internal strength of the
material is thought of as its internal friction,
F. This is the materials reaction to the
external shear force, T.
F
T
5

• to overcome the friction force, F on the plane
  where the object rests

that must be applied to an object of known
weight, W
The tangent of the friction angle, ? is the ratio
of F to W which is also known as the coefficient
of friction.
and thereby cause the object to move.
Vector addition gives the resultant vector, R
which acts at an angle of ? WRT the normal to the
plane.
The objects weight vector, W acts normal to the
failure plane.
Friction problems in mechanics determine the
external force, T
R
W
W
?
F
T
6
SHEAR STRENGTH IN SOILS
Consider and element of soil within a large soil
mass
If the soil is loaded (yet sober)
LOAD
Soil Surface

• The loading of a material that undergoes shear
  failure is not always parallel to the failure
  plane.

Soil Element
Soil Mass
Bedrock
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LOAD
s1

• The load transmits stress to the element by
  inter-particle contacts.

This is the major principal stress distribution,
designated s1 due to the load.
For visual simplicity we replace the distributed
load with an equivalent point load.
8
s1
s2
s3
s3
s2
s1

• The soil below the element will react with a
  stress of equal magnitude but directed upwards so
  it too is designated s1.

The element squeezed vertically will tend to
bulge horizontally to which the soil reacts with
confining pressures s2 and s3 in the other
principal directions.
Since we assume the soil is isotropic, the
confining lateral pressure will be the same in
all directions and so s2 s3 allowing us to view
it in 2 dimensions.
9

• But what has this got to do with SHEAR STRENGTH?

For this to happen, a failure plane develops
within the soil.
The friction force on this failure plane is
overcome by the external forces and viola
Soil undergoes shear failure when one portion
moves relative to the rest.
SHEAR FAILURE!
s1
2-D
T
s1
10

• The Shear Stress at failure, tf, is the pressure
  required to overcome the friction on the surface
  of the failure plane (a.k.a. Shear Strength).

The angle of internal friction, ? characterizes
the shear strength of the soil and is one of its
shear strength parameters.
There are 3 basic laboratory tests that can be
performed on soil samples to evaluate the shear
strength parameters

• 1. Direct Shear Test
• 2. Triaxial Compression Test
• 3. Unconfined Compression Test

?
sf
Rf
tf
T
11
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).
12
DIRECT SHEAR TEST
The horizontal force is increased until the
sample shears in two
This forces failure to occur on a horizontal
plane between the top and base
Then the top and base are pushed in opposite
directions
The prepared soil sample is placed in the box.
A normal (90? to the horizontal) load is applied
to the soil.
A shear box has three parts
a base
a top extension
and a normal load piston

The procedure is repeated two more times using
successively heavier normal loads.
13
DIRECT SHEAR TEST
In the CV504 labs, the inside dimensions of the
shear box are 60 mm by 60 mm.
This means the failure plane has an 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.
The shear force required to shear the sample
increases in proportion to the normal load.
The shear strength of the soil therefore is not
constant but changes with the confining pressure.
For this reason, the soils shear strength is
characterized by shear strength parameters (c,?).
14
DIRECT SHEAR TEST
Fitting a best fit line through these points
The t axis intercept is the apparent cohesion, c
of the soil.
The equation of Coulombs failure envelope
tf c sntan? .
Plotting the shear stress versus normal stress
The slope angle of this line is the angle of
internal friction, ? of the soil.
we have an estimate of Coulombs failure envelope
First Test
Second Test
Third Test
tf
tf
Shear Stress, t (kPa)
tf
?
c
Normal Stress, sn(kPa)
15
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.
16
TRIAXIAL COMPRESSION TEST
Cylindrical specimens are prepared from sampled
soil.
Preparation varies with material properties (clay
vs sand vs cohesive granular).
Specimens are weighed and dimensions measured
first.
The specimen is mounted between 2 platens and
then inserted into a latex sleeve.
The specimen is then placed in a plexiglas
chamber.
diameter
length
17
TRIAXIAL COMPRESSION TEST
Then the chamber is placed on the base and locked
into place.
For a drained test the drain valve is opened and
pore water collected.
Once the cell is filled with water, the air
release valve is closed and the cell pressure is
increased to the desired value for the test.
The assembly is then mounted on the compression
testing machine.
For an undrained test, the drain valve is closed.
Water is forced into the cell with the supply
valve open as well as the air release valve.
The specimen is mounted on the pedestal of the
chamber base as shown.
loading ram
air release valve

plexiglas chamber
loading cap
water supply for cell (confining) pressure
latex sleeve
drainage or pore water pressure measurement
specimen
porous disc
pedestal


18
TRIAXIAL COMPRESSION TEST
Enter Christian Otto Mohr
The goal is to simulate the stresses confining
the specimen in the ground.
The Major Principal Stress, s1, is the
combination of the deviator stress and cell
pressure
But how can we find tf and sf from s1 and s3 ?
The effect of the cell pressure on the specimen
is illustrated below
Then a vertical axial load is applied to the
loading ram creating compressive stresses or the
deviator stress ?s
The cell pressure, s3, is also known as the Minor
Principal Stress.
? s
s3
s3
s3
? s
Plan View of Specimen
Side View of Specimen
Source commons.wikimedia.org
19
TRIAXIAL COMPRESSION TEST
Herr Mohr was born in Germany on 1835-10-08 and
was a renowned Civil Engineer and professor until
his death on 1918-10-02.
for any material,
the internal shear and normal stresses acting on
ANY plane within the material,
In other words, he discovered MOHRS CIRCLE.
While contemplating the symmetry of his name,
Otto started tinkering with the properties of the
circle when he discovered that...
caused by external stresses or loads
can be determined using a trigonometric
transformation of the external stresses.
20
TRIAXIAL COMPRESSION TEST
If you plot s1 and s3 on the sn axis
During the test, this circle starts as one point
at s3 and then grows to the right as axial
stress, ?s increases but s3 remains constant.
Ultimately, the test ends when shear failure
occurs and the circle has become tangent to the
failure envelope.
then fit one circle through these points
then youve got a Mohrs circle!
The point of tangency of the circle and failure
envelope defines the shear strength, tf and
normal stress, sf.
Remember the plot of Shear Stress versus Normal
Stress?
Shear Stress, t (kPa)
tf
?
c
s3
s1
s1
s1
?s
?s
sf
?s
Normal Stress, sn(kPa)
21
TRIAXIAL COMPRESSION TEST
If one line cannot be drawn tangent to all three
circles, a best fit is made as long as one circle
is not out to lunch compared to the others.
But how can you be sure one of them isnt bogus?
As with most lab measurements, the ideal (one
line tangent to all three circles) is difficult
to achieve.
Geometrically, you need at least two circles in
order to define a line tangent to both.
This means that you need to perform the test at
least twice on the same material but at different
cell pressures.
A third test at yet another cell pressure would
help to confirm the validity of the failure
envelope.
But how do we find the failure envelope from a
triaxial compression test?
Shear Stress, t (kPa)
c
?
Normal Stress, sn(kPa)
22
TRIAXIAL COMPRESSION TEST
Instead of doing this graphically, we can use
trigonometry to find equations for tf and sf
using the angle of the failure plane, T and the
values of s1 and s3
Remember the deviator stress, ?s s1 - s3, which
is the diameter of the Mohrs Circle.
then for each test, the shear strength, tf and
normal stress, sf can be found.
The Centre of the Mohrs Circle, C is then
So the radius of the Mohrs Circle, R is half the
diameter or
Once we have the shear strength parameters, ? and
c defining the failure envelope,
Shear Stress, t (kPa)
specimen
failure plane
R
c
R
C
R
T
?
s3
s1
Normal Stress, sn(kPa)
23
TRIAXIAL COMPRESSION TEST
To follow the trig we label the vertices
?ABC 90? so ?ACB 90 - ? and ?DBC ?
?s EBC BCF are both isosceles ?EBF is 90?.
? ?EFB 90 T ?BCF 180 2(90- T) 2T
?DCB 180 2T 90 - ?
Rearranging
Shear Stress, t (kPa)
B
tf
?
c
2T
F
E
T
A
?
D
Normal Stress, sn(kPa)
C
s3
s1
sf
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TRIAXIAL COMPRESSION TEST
In ?DBC, side BD is the same as tf . ?
Also in ?DBC, side DC Rcos(180?-2T)
So knowing ? you can find T and
using T and the s1 s3 values for each trial,
tf and sf can be found for each trial.
? sf C Rcos(180?-2T) or C Rcos(2T)
Shear Stress, t (kPa)
B
tf
?
R
c
2T
F
E
T
A
?
D
Normal Stress, sn(kPa)
C
s3
s1
sf
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TRIAXIAL COMPRESSION TEST
And, the apparent cohesion, cu will be the same
for each trial and equal to the shear strength, tf
What happens when the pore water is not allowed
to drain (UNDRAINED TEST)?
Typically, the deviator stress at failure is
fairly constant for each different cell pressure.
One final word on nomenclature
As the external pressure increases, the internal
pore water pressure (acting in the opposite
direction to the external) increases to match
(and trivialize) the effect.
Therefore, the failure envelope is typically a
horizonal line and ?u 0?.
All stress symbols used in DRAINED tests are
usually primeds1,s3,sf and ?f indicating
that they are in terms of EFFECTIVE STRESS and
the shear strength parameters are denoted (?,c).
All stress symbols used in UNDRAINED tests are
not primeds1,s3,sf and ?f indicating that they
are in terms of TOTAL STRESS and the shear
strength parameters are denoted (?u,cu)
The normal stress, sf for each trial will then be
s3 cu
(The radii are all the same)
Shear Stress, t (kPa)
?u ? 0?
cu tf
sf
sf
sf
Normal Stress, sn(kPa)
26
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.
27
UNCONFINED COMPRESSION TEST
Instead of calling it the deviator stress, ?s,
it is called the unconfined compressive stress,
qu.
If the qu does maximize before 15 strain, then
the maximum qu value is used as quf.
The axial load starts at 0 and increases steadily
as in the triaxial compression test.
The point of tangency of the circle and failure
envelope defines the shear strength, tf and
normal stress, sf.
If a qu does not maximize before 15 strain is
reached then the qu at 15 strain is used to
define the unconfined compressive strength of the
specimen, quf
This is analogous to the circle becoming tangent
to the failure envelope when shear failure occurs.
Because s3 0 and quf is the diameter of the
circle, the shear strength, tf and normal stress
at failure, sf are both estimated to be half of
quf.
The Mohrs circle continues to grow until failure
occurs either when the specimens shear strength
is reached or 15 strain.
Shear Stress, t (kPa)
c tf
qu
qu
qu
qu
qu
qu
qu
sf
quf
Normal Stress, sn(kPa)