shear strengthlibvolume3.xyz/.../shearstrengthofsoilpresentation1.pdf · 2014-12-25 · failure...

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.

When you hear “Shear Failure” you probably think of Shearing Pins or

Bolts. How do these fail?

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.

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

The failure plane for metals will be parallel to the external shear forces.

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:

Area, A

A

T=τ

If the shear force causes failure, then the shear stress that results, τf is the shear strength of the material.

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.

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, φ is the ratio of F to W which is also known as the coefficient of friction.

Friction problems in mechanics determine the external force, T

W W

and thereby cause the object to move.

The object’s weight vector, W acts normal to the failure plane.

Vector addition gives the resultant vector, R which acts at an angle of φ

WRT the normal to the plane.

φ

SHEAR STRENGTH IN SOILS

The loading of a material that undergoes shear failure is not always

parallel to the failure plane.

Consider and element of soil within a large soil mass:

Soil Surface

Bedrock

Soil Mass

If the soil is loaded (yet sober):

Soil Element

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

This is the major principal stress distribution, designated σ1 due to

the load.

For visual simplicity we replace the distributed load with an equivalent

point load.

σ1

The soil below the element will react with a stress of equal magnitude but

directed upwards so it too is designated 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.

σ1

σ3

σ1

σ3

σ2

σ2

But what has this got to do with SHEAR STRENGTH?

Soil undergoes shear failure when one portion moves relative to the rest.

For this to happen, a failure plane develops within the soil.

σ1

σ1

2-D

The friction force on this failure plane is overcome by the external

forces and viola:

Θ

SHEAR FAILURE!

The Shear Stress at failure, ττττffff, 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, φφφφ

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:

σf

1. Direct Shear Test

2. Triaxial Compression Test

3. Unconfined Compression Test

Θ ττττffff

φφφφ

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

A shear box has three parts:

a base a top extension and a normal load piston The prepared soil sample is placed in the box.

A normal (90° to the horizontal) load is applied to the soil.

Then the top and base are pushed in opposite directions

This forces failure to occur on a horizontal plane between the

top and base:

The horizontal force is increased until the sample shears in two:

The procedure is repeated two more times using successively heavier

normal loads.

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 soil’s shear strength is characterized by

shear strength parameters: (c,φ).

DIRECT SHEAR TEST Plotting the shear stress versus normal stress:

Sh

ear

Str

ess,

ττ ττ (kP

a)

Normal Stress, σn(kPa)

First Test

ττττffff

Second Test

ττττffff

Third Test

ττττffff

Fitting a best fit line through these points: we have an estimate of Coulomb’s failure envelope

The τ axis intercept is the apparent cohesion, c of the soil.

c

The slope angle of this line is the angle of internal friction, φ of the soil.

φφφφ

The equation of Coulomb’s failure envelope: ττττffff = c + σntanφ .

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.


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.

diameter

leng

th

The specimen is mounted between 2 platens and then inserted into a latex sleeve.

The specimen is then placed in a plexiglas chamber.

TRIAXIAL COMPRESSION TEST The specimen is mounted on the pedestal

of the chamber base as shown. Then the chamber is placed on the

base and locked into place. The assembly is then mounted on the

compression testing machine.

specimen

For an undrained test, the drain valve is closed.

For a drained test the drain valve is opened and pore water collected.

latex sleeve

porous disc

loading cap

pedestal

plexiglas chamber

loading ram

drainage or pore water pressure measurement

water supply for cell (confining)

pressure

air release valve

Water is forced into the cell with the supply valve open as well as the air release valve.

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 effect of the cell pressure on the specimen is illustrated below:

The goal is to simulate the stresses confining the specimen in the ground.

Then a vertical axial load is applied to the loading ram creating compressive

stresses or the deviator stress ∆σ :

Plan View of Specimen

The cell pressure, σ3, is also known as the Minor Principal Stress.

The Major Principal Stress, σ1, is the combination of the deviator stress and

cell pressure:

Side View of Specimen

σ3

σ3

σ3

∆ σ

∆ σ

31 σΔσσ +=

But how can we find τf and σf from σ1

and σ3 ? Enter Christian Otto Mohr:

Source: “commons.wikimedia.org”


for any material,

the internal shear and normal stresses acting on ANY plane

within the material,

caused by external stresses or loads

can be determined using a trigonometric transformation of

the external stresses.

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.

While contemplating the symmetry of his name, Otto started tinkering with the properties of the circle

when he discovered that...

In other words, he discovered MOHR’S

CIRCLE.


Remember the plot of Shear Stress versus Normal Stress?

If you plot σ1 and σ3 on the σn axis then fit one circle through these points

Normal Stress, σn(kPa) σ1111 σ3333

ττττffff

c

φφφφ

She

ar S

tres

s, τ

τ

τ

τ (kPa

)

then you’ve got a Mohr’s circle! 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.

∆σ σ1111 σ1111 ∆σ ∆σ

Ultimately, the test ends when shear failure occurs and the circle has become tangent

to the failure envelope.

The point of tangency of the circle and failure envelope defines the shear strength, τf

and normal stress, σf.

σffff


But how do we find the failure envelope from a triaxial compression test?

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.


c φ

She

ar S

tres

s, τ

τ

τ

τ (kPa

) But how can you be sure one of them isn’t bogus? A third test at yet another cell pressure would

help to confirm the validity of the failure envelope.

As with most lab measurements, the ideal (one line tangent to all three circles) is difficult

to achieve.

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.


Once we have the shear strength parameters, φ and c defining the failure envelope,

then for each test, the shear strength, τf and normal stress, σf can be found.


R

σ3333

c φ

She

ar S

tres

s, τ

τ

τ

τ (kPa

) Remember the deviator stress, ∆σ = σ1 - σ3,

which is the diameter of the Mohr’s Circle. So the radius of the Mohr’s Circle, R is half the

diameter or:

CCCC

σ1111

R

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

The Centre of the Mohr’s Circle, CCCC is then:

Θ

specimen

failure plane

Θ

2

σσ R 31 −=

2

σσ 31 +=CCCC

TRIAXIAL COMPRESSION TEST To follow the trig we label the vertices: ∆∆∆∆’s EBC & BCF are both isosceles & ∠∠∠∠EBF is 90°°°°.

Normal Stress, σn(kPa) σ3333

ττττffff

c φ

She

ar S

tres

s, τ

τ

τ

τ (kPa

) ∠∠∠∠ABC = 90° so ∠∠∠∠ACB = 90 - φ and ∠∠∠∠DBC = φ

∠∠∠∠DCB = 180 – 2Θ = 90 - φ

CCCC σ1111

∴ ∠∠∠∠EFB = 90 – Θ & ∠∠∠∠BCF =180 – 2(90- Θ) = 2Θ

Rearranging:

σffff

2Θ Θ

Θ 2

σσ R 31 −=

2

σσ 31 +=CCCC

A

B

D

E F

φ

245θ

φ−=

o

TRIAXIAL COMPRESSION TEST In ∆∆∆∆DBC, side BD is the same as τf . ∴

using Θ and the σ1 & σ3 values for each trial, τf and σf can be found for each trial.

Normal Stress, σn(kPa) σ3333

ττττffff

c φ

She

ar S

tres

s, τ

τ

τ

τ (kPa

) Also in ∆∆∆∆DBC, side DC = Rcos(180°-2Θ) So… knowing φ you can find Θ and

CCCC σ1111

∴ σf = CCCC – Rcos(180°-2Θ) or CCCC + Rcos(2Θ)

σffff

2Θ Θ

Θ 2

σσ R 31 −=

2

σσ 31 +=CCCC

A

B

D

E F

φ

245θ

φ−= o

( ) ( )sin2θσσ2

12θ180Rsin 31f −=−= oτ

( ) 4.3) (Eqn. sin2θσσ2

131f −=τ

( ) ( ) 4.4) (Eqn. cos2θσσ2

1σσ

2

13131f −++=σ

TRIAXIAL COMPRESSION TEST What happens when the pore water is not

allowed to drain (UNDRAINED TEST)? 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.

Typically, the deviator stress at failure is fairly constant for each different cell

pressure.


cu = τf

φu ≈ 0°

She

ar S

tres

s, τ

τ

τ

τ (kPa

) Therefore, the failure envelope is typically a

horizonal line and φu = 0°. And, the apparent cohesion, cu will be the same

for each trial and equal to the shear strength, τf

The normal stress, σf for each trial will then be σ3 + cu

(The radii are all the same)

σf σf σf

One final word on nomenclature… All stress symbols used in DRAINED

tests are usually primed…σ1’,σ3’,σf’ and τf’ indicating that they are in terms of EFFECTIVE STRESS

and the shear strength parameters are denoted (φ’,c’).

All stress symbols used in UNDRAINED tests are not

primed…σ1,σ3,σf and τf indicating that they are in terms of TOTAL STRESS and the shear strength parameters are denoted (φu,cu)

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.

The axial load starts at 0 and increases steadily as in the triaxial compression test.

Instead of calling it the deviator stress, ∆σ, it is called the unconfined compressive stress, qu.

The Mohr’s circle continues to grow until failure occurs either when the specimen’s shear

strength is reached or 15% strain.

Normal Stress, σn(kPa) qu qu

c = c = c = c = ττττffff

She

ar S

tres

s, τ

τ

τ

τ (kPa

)

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

If the qu does maximize before 15% strain, then the maximum qu value is used as quf.

qu qu qu qu qu

This is analogous to the circle becoming tangent to the failure envelope when

shear failure occurs.

The point of tangency of the circle and failure envelope defines the shear strength, τf and normal stress, σf.

σffff

UNCONFINED COMPRESSION TEST

quf

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.

shear strengthlibvolume3.xyz/.../shearstrengthofsoilpresentation1.pdf · 2014-12-25 · failure...

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