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. Shearing Pins can be used to fasten together two steel plates:
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…
How do these fail?
Each pin has sheared into two pieces.

3. So shear forces are those that tend to cause shear failure.
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, τf is the shear strength of the material.
The internal shear stress, τ 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 material’s reaction to the external shear force, T. F T

5. to overcome the friction force, F on the plane where the object rests
Vector addition gives the resultant vector, R which acts at an angle of  WRT the normal to the plane.
and thereby cause the object to move.
The tangent of the friction angle,  is the ratio of F to W which is also known as the coefficient of friction.
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 object’s 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:
SHEAR STRENGTH IN SOILS
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

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

8. σ1 σ2 σ3 σ3 σ2 σ1
The element squeezed vertically will tend to bulge horizontally to which the soil reacts with confining pressures σ2 and σ3 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 σ2 = σ3 allowing us to view it in 2 dimensions.
The soil below the element will react with a stress of equal magnitude but directed upwards so it too is designated σ1.

9. But what has this got to do with SHEAR STRENGTH?
The friction force on this failure plane is overcome by the external forces and viola:
For this to happen, a failure plane develops within the soil.
But what has this got to do with SHEAR STRENGTH?
Soil undergoes shear failure when one portion moves relative to the rest.
SHEAR FAILURE! σ1
2-D Θ σ1

10. The angle of internal friction, 
The Shear Stress at failure, τf, 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  σf Rf τf Θ

11. Can be performed on all types of soil, moist or dry.
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. The horizontal force is increased until the sample shears in two:
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
A normal (90 to the horizontal) load is applied to the soil.
The prepared soil sample is placed in the box.
A shear box has three parts:
and a normal load piston
a base
a top extension
The procedure is repeated two more times using successively heavier normal loads.

13. This means the failure plane has an area of 3600 mm2.
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 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 soil’s shear strength is characterized by shear strength parameters: (c,).

14. DIRECT SHEAR TEST
The τ axis intercept is the apparent cohesion, c of the soil.
The equation of Coulomb’s failure envelope: τf = c + σntan .
Fitting a best fit line through these points:
The slope angle of this line is the angle of internal friction,  of the soil.
Plotting the shear stress versus normal stress:
we have an estimate of Coulomb’s failure envelope
First Test
Second Test
Third Test τf τf
Shear Stress, τ (kPa) τf  c
Normal Stress, σn(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.
Specimens are weighed and dimensions measured first.
Preparation varies with material properties (clay vs sand vs cohesive granular).
The specimen is then placed in a plexiglas chamber.
The specimen is mounted between 2 platens and then inserted into a latex sleeve.
diameter
length

17. TRIAXIAL COMPRESSION TEST
Then the chamber is placed on the base and locked into place.
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.
For a drained test the drain valve is opened and pore water collected.
Water is forced into the cell with the supply valve open as well as the air release valve.
The assembly is then mounted on the compression testing machine.
For an undrained test, the drain valve is closed.
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
drainage or pore water pressure measurement
latex sleeve
specimen
porous disc
pedestal

18. Source: “commons.wikimedia.org”
TRIAXIAL COMPRESSION TEST
Enter Christian Otto Mohr:
But how can we find τf and σf from σ1 and σ3 ?
The Major Principal Stress, σ1, is the combination of the deviator stress and cell pressure:
The goal is to simulate the stresses confining the specimen in the ground.
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 ∆σ :
The cell pressure, σ3, is also known as the Minor Principal Stress.
∆ σ σ3 σ3 σ3
∆ σ
Plan View of Specimen
Side View of Specimen
Source: “commons.wikimedia.org”

19. In other words, he discovered MOHR’S CIRCLE.
TRIAXIAL COMPRESSION TEST
Herr Mohr was born in Germany on and was a renowned Civil Engineer and professor until his death on
for any material,
the internal shear and normal stresses acting on ANY plane within the material,
In other words, he discovered MOHR’S 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 σ1 and σ3 on the σn axis
During the test, this circle starts as one point at σ3 and then grows to the right as axial stress, ∆σ increases but σ3 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
The point of tangency of the circle and failure envelope defines the shear strength, τf and normal stress, σf.
then you’ve got a Mohr’s circle!
Remember the plot of Shear Stress versus Normal Stress?
Shear Stress, τ (kPa) τf  c σ3 ∆σ σf ∆σ ∆σ σ1 σ1 σ1
Normal Stress, σn(kPa)

21. But how can you be sure one of them isn’t bogus?
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 isn’t 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.
A third test at yet another cell pressure would help to confirm the validity of the failure envelope.
This means that you need to perform the test at least twice on the same material but at different cell pressures.
But how do we find the failure envelope from a triaxial compression test?
Shear Stress, τ (kPa) c 
Normal Stress, σn(kPa)

22. TRIAXIAL COMPRESSION TEST
Instead of doing this graphically, we can use trigonometry to find equations for τf and σf using the angle of the failure plane, Θ and the values of σ1 and σ3
Remember the deviator stress, ∆σ = σ1 - σ3, which is the diameter of the Mohr’s Circle.
then for each test, the shear strength, τf and normal stress, σf can be found.
The Centre of the Mohr’s Circle, C is then:
So the radius of the Mohr’s Circle, R is half the diameter or:
Once we have the shear strength parameters,  and c defining the failure envelope,
Shear Stress, τ (kPa) Θ
specimen
failure plane R c C R R Θ  σ3 σ1
Normal Stress, σn(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 – Θ & BCF =180 – 2(90- Θ) = 2Θ
DCB = 180 – 2Θ = 90 - 
Rearranging:
Shear Stress, τ (kPa) B Θ τf  c 2Θ F E A Θ  D C
Normal Stress, σn(kPa) σ3 σf σ1

24. TRIAXIAL COMPRESSION TEST
In DBC, side BD is the same as τf . 
Also in DBC, side DC = Rcos(180-2Θ)
So… knowing  you can find Θ and
 σf = C – Rcos(180-2Θ) or C + Rcos(2Θ)
using Θ and the σ1 & σ3 values for each trial, τf and σf can be found for each trial.
Shear Stress, τ (kPa) B Θ τf  R c 2Θ F E A Θ  D C
Normal Stress, σn(kPa) σ3 σf σ1

25. One final word on nomenclature…
TRIAXIAL COMPRESSION TEST
And, the apparent cohesion, cu will be the same for each trial and equal to the shear strength, τf
Typically, the deviator stress at failure is fairly constant for each different cell pressure.
One final word on nomenclature…
What happens when the pore water is not allowed to drain (UNDRAINED TEST)?
Therefore, the failure envelope is typically a horizonal line and u = 0.
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.
All stress symbols used in DRAINED tests are usually primed…σ1’,σ3’,σf’ 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 primed…σ1,σ3,σf and f indicating that they are in terms of TOTAL STRESS and the shear strength parameters are denoted (u,cu)
The normal stress, σf for each trial will then be σ3 + cu
(The radii are all the same)
Shear Stress, τ (kPa)
u  0
cu = τf σf σf σf
Normal Stress, σn(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
If the qu does maximize before 15% strain, then the maximum qu value is used as quf.
Instead of calling it the deviator stress, σ, it is called the unconfined compressive stress, qu.
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, τf and normal stress, σf.
This is analogous to the circle becoming tangent to the failure envelope when shear failure occurs.
Because σ3 = 0 and quf is the diameter of the circle, the shear strength, τf and normal stress at failure, σf are both estimated to be half of quf.
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
The Mohr’s circle continues to grow until failure occurs either when the specimen’s shear strength is reached or 15% strain.
Shear Stress, τ (kPa)
c = τf qu qu qu σf qu qu qu
quf qu
Normal Stress, σn(kPa)