Shear Strength - Introduction and Shear StrengthTesting

Shear Strength Theory

Strength of different materials

Steel

Tensile strength

Concrete

Compressive strength

Soil

Shear strength

Presence of pore waterComplexbehavior

Introduction

Shear strength defined as the maximum stress that a soil can sustain before failure occurs– Not a failure of soil particles – Failure by relative movement of particles– Depends on the normal stress acting on any

plane within the soil

failure plane does not pass through particles

particles move relative to each other

Embankment

Strip footing

Soils generally fail in shear

At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength.

Failure surface

Mobilized shear resistance

Shear failure mechanism

The soil grains slide over each other along the failure surface.No crushing of individual grains.

failure surface

At failure, shear stress along the failure surface () reaches the shear strength (f).

Why is shear strength important?

resistance to movement of soil block governed by strength on plane

movement on plane governed by weight of soil block

Seattle, 2014

The 1993 landslide at Holbeck Hall, North Yorkshire.

ultimate bearing capacity of foundations

ultimate bearing capacity of foundations

Retaining wall design - lateral earth pressures

At failure, shear stress along the failure surface () reaches the shear strength (f).

Introduction

In order to define the shear strength of a soil, you must be able to:– Define the state of stress in a soil mass; and– Establish a relationship between shear and

normal stress at failure – define a FAILURE CRITERION

H = σxsinα

C

B A

N = σα × 1

T = τα × 1

Area = 1

Area = 1cosαV = σycosα

α

Area = 1sinα

+τ+σ

Sign conventions

Forces acting on plane (AC) through point in soil mass:

Define the state of stress in a soil mass:

Summing forces in horizontal and vertical directions gives:

Dividing by area gives:

Solving for σα and τα yields:

Square and add these equations, gives us the equation of the Mohr circle of stress

Define the state of stress in a soil mass:

Introduction

What is shear strength?– Related to three components:

• Frictional resistance to sliding• Cohesion and adhesion • Interlock

Frictional resistance to sliding

T

NN

RR

α

α

R

T

N

At equilibrium,

T = RsinαN = Rcosα

Dividing T/N gives, Rsinα/Rcosα = tanα

Hence T = Ntanα

Dividing by area, gives

= σntanα

Frictional resistance to sliding

As is increased, sliding will be imminent when a limiting value of α is reached, this is ϕ, therefore at failure - f = σntanϕ

tan ϕf

σn

frictionFriction + interlock

Usually represented as…Mohr-Coulomb equation

Tanϕ’f

σn

‘cohesion’

Failure envelope:

f = c’ + σn’tan ϕ’

note that c’ and ϕ’ are curve fitting parameters

Shear Failure in Soils

Defined previously as:– the maximum stress that a soil can sustain before failure

occursCan also be considered as:

– the resistance to deformation by continuous shear displacement of soil particles along surfaces of rupture

Not purely a function of peak shear stresses

Shear Failure in Soils

Engineers need to be able to define the nature and extent of stress and deformation (strain) at the time of failure

For frictional soils the most appropriate theory is the Mohr Theory of Rupture

Mohr’s theory indicates that there is a critical combination of shear and normal stresses acting on a plane

Shear Failure in Soils

So, failure will occur when a state of stress exists in the soil so that one point on the Mohr circle touches the ‘failure’ or ‘rupture’ line– i.e. failure line is a tangent to the circle at a

single point

If the shear strength is represented by = f():

Normal stress

Shear stress

compressiontension Mohr’s envelope: = f()

I IIIII

IV V

Any combination of stresses that falls below this line, represent the STABLE condition

Normal stress

Shear stress

compressiontension Mohr’s envelope: = f()

I IIIII

IV V

Circles I – IV touch (are tangent to) the envelope, indicating impending failure

Normal stress

Shear stress

compressiontension Mohr’s envelope: = f()

I IIIII

IV V

Since the envelope is curved, it is clear that the friction angle reduces with increasing confining stress ()

Normal stress

Shear stress

compressiontension Mohr’s envelope: = f()

I IIIII

IV V

The non-linearity however, is slight, and so it is more convenient to represent this envelope as a straight line

Normal stress

Shear stress

compressiontensionMohr-Coulomb envelope: = c + tan

I IIIII

IV V

Shear Failure in Soils

31

Shear stress

Normal stress

Shear strength is a function of the normal stress and soil properties (such as c and )

3 1

f

f =

f

limiting value of f is defined as

Shear Failure in Soils

32

Shear stress

Normal stress

Shear strength is a function of the normal stress and soil properties (such as c and )

3 1

f

f

The shear strength for a purely frictional soil can be written as

tanffS

Shear Failure in Soils

33

Shear stress

Normal stress

Shear strength is a function of the normal stress and soil properties (such as c and )

1

f

f

The shear strength for a purely frictional soil can be written as

tanffS

Soil elements at different locations

Failure surface

X X

X ~ failure

YY

Y ~ stable

’

'tan'' cf

Y

c

c

c

Initially, Mohr circle is a point

c+

The soil element does not fail if the Mohr circle is contained within the envelope

GL

Y

c

c

c

GL

As loading progresses, Mohr circle becomes larger…

.. and finally failure occurs when Mohr circle touches the envelope

1

1

33

If we consider an element of soil subjected to a combination of stresses (Principal stresses 1 > 3)

At some value, a plane of failure will develop in the element

1

1

1

1

3

3

3

Along this failure plane, a critical combination of shear stress and normal stress will develop

3 1

1

1

1

3

3

3

c

Along this failure plane, a critical combination of shear stress and normal stress will develop

Using a Mohr circle, we can analyse these stress conditions

The failure envelope is defined by:

tancf

3 1

1

1

1

3

3

3

c

The orientation of the failure plane, and hence point D can be established, since we know that the angle between the major principal plane and the failure plane is

D

3 1

1

1

1

3

3

3

c

The orientation of the failure plane, and hence point D can be established, since we know that the angle between the major principal plane and the failure plane is

D

3 1

1

1

1

3

3

3

c

Since the angle subtended from the centre to the failure point is 2 or:

902

2

D

3 1

1

1

1

3

3

3

c

which becomes:

902

2

245

D

3 1

1

1

1

3

3

3

c

which becomes:

902

2

245

’3 ’1

’

1

1

1

3

3

3

c’

C D O E

F

BA

It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses

’3 ’1

’

1

1

1

3

3

3

c’

C D O E

F

BA

It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses

We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2

We can see that OF is:

’3 ’1

’

1

1

1

3

3

3

c’

C D O E

F

BA

It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses

We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2

We can see that OF is:

'sin)('sin BOABOAOF

’3 ’1

’

1

1

1

3

3

3

c’

C D O E

F

BA

It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses

We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2

We can see that OF is:

'sin)('sin BOABOAOF

’3 ’1

’

1

1

1

3

3

3

c’

C D O E

F

BA

It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses

We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2

We can see that OF is:

'sin)('sin BOABOAOF

’3 ’1

’

1

1

1

3

3

3

c’

C D O E

F

BA

It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses

We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2

We can see that OF is:

'sin)('sin BOABOAOF

’3 ’1

’

1

1

1

3

3

3

c’

C D O E

F

BA

It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses

We know that CF is parallel to the failure plane, inclined at angle , and that this is equal to 45 + ‘/2

We can see that OF is:

Since OF is the radius of the circle:

'sin)('sin BOABOAOF

31 ''5.0 OF

’3 ’1

’

1

1

1

3

3

3

c’

C D O E

F

BA

It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses

If,

and

and with BO written as:

Which defines the centre of the circle

'sin)('sin BOABOAOF

31 ''5.0 OF

'cot' cAB

31 ''5.0 BO

’3 ’1

’

c’ C D O E

F

BA

It is useful to be able to define the Mohr-Coulomb failure criterion in terms of the Principal stresses

becomes:

Simplifying and solving for ‘3, we obtain:

'sin)('sin BOABOAOF

31

31

31

31

'''cot'2

''

''5.0'cot'

''5.0'sin

cc

'sin1

'cos'2

'sin1

'sin1'' 13

c

Since:

and:

we can rearrange this:

2

'45tan

'sin1

'sin1 2

2

'45tan

'sin1

'cos

'sin1

'cos'2

'sin1

'sin1'' 13

c

To give:

or:

2

'45tan'2

2

'45tan'' 2

13

c

2

'45tan'2

2

'45tan'' 2

31

c

2

'45tan'2

2

'45tan'' 2

13

c

2

'45tan'2

2

'45tan'' 2

31

c

These are alternative forms of theMohr-Coulomb failure criterion, written in terms of effective stress

Determination of Shear StrengthIntroduction

– Only considered theoretical considerations to this point

– Validity and value of these considerations is closely related to the parameters and observations from experimental and field studies

– Strength parameters are obtained from:

• direct shear test• triaxial test; and

Determination of Shear StrengthIntroduction

– Soils are neither continuous, homogenous nor isotropic

– Limited number of tests will only give approximations to the characteristics of the soil mass

– Samples should be representative• ideally undisturbed• sample quality is profoundly important

– Test conditions should be representative

Other laboratory tests include,Direct simple shear test, torsional ring shear test, plane strain triaxial test, laboratory vane shear test, laboratory fall cone test

Determination of shear strength parameters of soils (c, orc’’

Laboratory tests on specimens taken from representative undisturbed samples

Field tests

Most common laboratory tests to determine the shear strength parameters are,

1.Direct shear test2.Triaxial shear test

1. Vane shear test2. Torvane3. Pocket penetrometer4. Fall cone5. Pressuremeter6. Static cone penetrometer7. Standard penetration test

Laboratory tests

Field conditions

z vc

vc

hchc

Before construction

A representative soil sample

z vc +

hchc

After and during construction

vc +

Laboratory testsSimulating field conditions in the laboratory

Step 1Set the specimen in the apparatus and apply the initial stress condition

vc

vc

hchc

Representative soil sample taken from the site

0

00

0

Step 2Apply the corresponding field stress conditions

vc +

hchc

vc + Traxia

l test

vc

vc

Direct shear test

Determination of Shear StrengthDirect shear test

– usually carried out in a shear box

Determination of Shear Strength

Direct shear test– usually carried out in a shear box

shear force applied to one half

soil sample

other half is restrained

Determination of Shear Strength

Direct shear test– usually carried out in a shear box

shear force applied to one half

soil sample

other half is restrained

vertical load N applied through the top platen

Determination of Shear Strength

Direct shear test– usually carried out in a shear box– very simple arrangement– very rapid test– no facility to control drainage and no

means of measuring pore water pressures

– tend to be used on coarse grained soils• free-draining• excess pore pressures will dissipate

instantaneously, hence u = 0• total stress = effective stress

Direct shear testTest procedure

Porous plates

Pressure plate

Steel ball

Step 1: Apply a vertical load to the specimen and wait for consolidation

P

Proving ring to measure shear force

S

Direct shear test

Step 2: Lower box is subjected to a horizontal displacement at a constant rate

Step 1: Apply a vertical load to the specimen and wait for consolidation

P

Test procedure Pressure plate

Steel ball

Proving ring to measure shear force

S

Porous plates

Direct shear test

Analysis of test results

sample theofsection cross of Area

(P) force Normal stress Normal

sample theofsection cross of Area

(S) surface sliding at the developed resistanceShear stressShear

Note: Cross-sectional area of the sample changes with the horizontal displacement

Direct shear tests on sands

Dense sand/ OC clay

fLoose sand/ NC clayf

Dense sand/OC Clay

Loose sand/NC Clay

Cha

nge

in h

eigh

t of

the

sam

ple Exp

ansi

onC

ompr

essi

on

Shear displacement

Stress-strain relationship

Determination of Shear Strength

Direct shear test– For any soil material, a number of

tests will be carried out (>3) at different values of normal stress

– From each test, the maximum shear stress f is plotted against the corresponding value of ‘n

– The straight line plotted through the points is the effective stress failure envelope for that soil, and satisfies:

'tan'' nf c

f1

Normal stress = 1

Direct shear tests on sandsHow to determine strength parameters c and

She

ar s

tres

s,

Shear displacement

f2

Normal stress = 2

f3

Normal stress = 3

She

ar s

tres

s at

fai

lure

,

f

Normal stress,

Mohr – Coulomb failure envelope

Direct shear tests on sandsSome important facts on strength parameters c and of sand

Sand is cohesionless hence c = 0

Direct shear tests are drained and pore water pressures are dissipated, hence u = 0

Therefore, ’ = and c’ = c = 0

Direct shear tests on clays

Failure envelopes for clay from drained direct shear tests

She

ar s

tres

s at

fai

lure

,

f

Normal force,

’

Normally consolidated clay (c’ = 0)

In case of clay, horizontal displacement should be applied at a very slow rate to allow dissipation of pore water pressure (therefore, one test would take several days to finish)

Overconsolidated clay (c’ ≠ 0)

Interface tests on direct shear apparatusIn many foundation design problems and retaining wall problems, it is required to determine the angle of internal friction between soil and the structural material (concrete, steel or wood)

tan' af cWhere, ca = adhesion, = interface angle of friction

Foundation material

Soil

P

S

Foundation material

Soil

P

S

Advantages of direct shear apparatus

Due to the smaller thickness of the sample, rapid drainage can be achieved

Can be used to determine interface strength parameters

Clay samples can be oriented along the plane of weakness or an identified failure plane

Disadvantages of direct shear apparatus

Failure occurs along a predetermined failure plane

Area of the sliding surface changes as the test progresses

Non-uniform distribution of shear stress along the failure surface