1 ElasticPlasticElasto-Plastic 12Mac2012

7/29/2019 1 ElasticPlasticElasto-Plastic 12Mac2012

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Advance soil mechanics

Elastic, plastic and elasto-plastic behaviour

By

Assoc. Prof. Dr. Mohd.Jamaludin bin Md.Noor

Faculty of Civil Engineering

UiTM Malaysia


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2

At the end of this lecture the student should be able to answer these questions.

1. Plastic behaviour is considered as soil irrecoverable deformation while elastic is taken the

behaviour when deformation is recoverable. Illustrate using the relevant graphs that soil

behave elastically within the yield limit and elastic-plastic when the stress exceeded the yield

limit.2. Differentiate the meaning of yield and failure. Explain the meaning of failure at peak strength

and failure at critical state.

3. By referring to the changes undergo by the solid particles when being compressed explain how

the soil able to behave plastic and elastic manner. With the aid of v log p curve differentiate

the path which is elastic and plastic and show how their magnitude can be calculated.

4. The yield limit can be define by the pre-consolidation pressure in clay and the past maximum

deviator stress in sand. Explain these.5. Show that the yield limit is the past maximum mobilised shear strength whether it is clay or

sand. Explain how that mobilised shear strength governs the elastic and plastic behaviour of

sand.

6. Mobilised shear strength envelope define the yield limit or the yield locus. In other words the

mobilised shear strength represents a certain deformation parameter. Explain this.

7. With the aid of graphs of q-p and v-p explain how the different sizes of the yield locus is

achieved and relate them to the soil elastic and plastic behaviour.8. Using the behaviour of a specimen in CID triaxial test explain the principal governing factor that

influence the specimens deformation. Emphasize your answer by considering 3 identical

specimens sheared at different effective stress.

9. Explain the uniqueness of the critical state line (CSL) by the sketching the line in 3-dimensional

space of q-p-v. Differentiate the stress path that lead to the CSL when the test specimens are

normally consolidated, lightly consolidated and over-consolidated clay.

10. With the aid of sketches of the relevant graphs show that it is not the critical state strength thatgoverns soil volume change behaviour but it is in fact the mobilised shear strength.

Problem base / outcome base teaching approach.


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3

Confined uniaxial testing

machine known misleadingly as

triaxial apparatus.

Deviator stress, = q

Cell

pressure3

Stress variables

1 = + 3q = (1 - 3)p = [(1 + 23)/3] - u

Soil

specimen

Pore water

pressure

measurement

Specimen

volume

change unit

Load cell

attached

to ram

Filling with

de-aired and

de-ionised

water

eV

VVv

s

vs

1

Specific volume = actual

volume occupied by aunit volume of soil solids

v

v

p

Volumetric

strain

Net mean stress


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4

Triaxial test

Relationship between volumetric strain, p and triaxial shearstrain, q with axial strain, a. and radial strain, r.

rapvol or 2 Drained triaxial test

ar

rap

vol

Therefore

21

02

321

Undrained triaxial test

3

2 raq

Triaxial shear strain,

Quantities measured in

triaxial tests,

l = change in length of specimenv = change in volume of specimen

VVand

ll

pa

2

// llVVr

3

/VV

l

lq

and ?


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DEFINITION OF

Mean stress, p, Deviatoric or shear stress, q,

Volumetric strain, p, Shear strain, q.

3

321 p

Mean stress represented by space diagonal3

1

2p

q

21

2

13

2

32

2

212

1 q

Deviatoric stress or shear stress

Volume

change

Distortion

or

shearing

Volumetric

strain

Deviatoric

strain

p

q

321 p

2

12

13

2

32

2

213

2 q


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Triaxial condition/

Axisymmetric

31 2 p

313

2 q

32 32

3

2 '3'

1' p'

3

'

1 q3

1

2

Plane strain 02

3

'

3

'

2

'

1' p 2

12

13

2

32

2

212

1 q

31 p 21

31

2

3

2

13

2 q

TWO COMMON CONDITIONS OF STRESS AND STRAIN IN

GEOTECHNICAL ENGINEERING

Volumetric strain Shear strain


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7

BULK MODULUS, K, SHEAR MODULUS, G AND ELASTIC MODULUS, E.

''

1p

K

e

P

qG

e

q3

1

e

q

e

p

GK

qp

300''

Shear stress do not cause volume changes and net

mean stress do not cause shear deformation.

'213

''

E

K

'12'

'

E

G '622'3'KGGK

Effective bulk

modulus

Shear modulus

modulus

Poisson ratio as a

function of K and G

Shear stress only produce shear deformation andnet mean stress only produce volume changes.

Soil elastic response to stress


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Soil elasticity

Change in net mean stress, p produces no distortion q = 0 i.e. noshear strain.

Change in the distortional deviator stress, q produces no change in

volume, p = 0.

q

q

1

3Gq

a1

E

The initial linear sections of the stress-strain curve can be interpreted as

the elastic response of the soil to the imposed change of stress.

E = Youngs modulus

G = Shear modulus

Elastic constants

deduced from

conventional drained

triaxial compression

tests.

313

2 q


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9

Elastic-plastic behaviour of soil

0

100

200

300

400

500

600

700

800

900

0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00

Axial strain (%)

Deviatorstress(kPa)

Eff.stress 50 Eff.stress 100 Eff. stress 200 Eff. stress 300

Drained triaxial tests on compacted standard sand

The first loading follow a

curved load-deformation path

which cannot be retraced

when the load is removed.

Upon unloading at zero load

the soil is left with permanentcompressed height.

When the soil is reloaded to

load less than the previous

maximum, essentially the soil

shows an elastic response.

As soon as the past

maximum load is exceeded

the elastic response change

to plastic response where the

soil proceed with the curved

stress-strain behaviour.

When the soil behaves elastic and when it

behaves elastic-plastic ?


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Soil plastic response to stress

pet

e

p eq

p

pp

qElastic

volumetric

strain

Elastic

shear

strain

Plastic

volumetric

strain

Plastic

shear

strain

'

'

'

'

o

o

pp

p

o

o

p

vp

p

v

v

vp

p

v

v

'

'

p

pve

'

'

vp

pe

p

''

1p

K

e

P

qG

e

q3

1

qMvp

pMMvp

p

p

2

''

' 2222

22

qMMvp

pMvp

p

q

22

2

2222

4

''2

'

TOTAL strain Elastic strain Plastic strain

Cam clay

Cam clay


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What is meant by yield, failure and work hardening ?

Yields:

A material yields when its stress-strain behaviour changes from being

purely elastic to partly plastic OR when the deformation stops beingrecoverable upon unloading. This is often marked by an abrupt change in

the slope (i.e. stiffness) of stress-strain curve. Yielding does not necessarily

mean failure.

Failure:

In Mohr-Coulomb theory failure is the onset of mobilising the maximum

shear stress where the Mohr stress circle (i.e. representing normal and

shear stresses on a slip plane) touches the failure envelope.

In critical state the onset of critical state is considered as failure at which

deformation continues at constant stress ratio and volume.

Soil hardening or work hardening:

Between yield and failure as the soil deforms elastic-plastically in reaction

to loading, the shear strength increases. The capacity of the soil to sustain

an increasing stress with increasing plastic strain is known as soil

hardening.

The soil stress state can never lie outside the current yield locus. As

deviator stress increases the yield locus expands so that the current stressstate is alwa s l in on the current ield locus.


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Yield and

yield envelope

D

3 tests

Set up in triaxial

apparatus, all inequilibrium at

effective stress

state, A

Specimen 1: Subjected to isotropic compression under increasing cell pressure. A yield point, Y1 is observed where

the stiffness changes markedly in the plot of specific volume, v versus p.

Specimen 2: Subjected to one-dimensional loading (like in an oedometer) by controlling the cell pressure as the axial

stress is increased in such a way that lateral strain of the specimen does not occur. A yield point, Y2 is observedwhere the stiffness of the specimen changes sharply in the plot of specific volume, v versus vertical effective stress.

Specimen 3: Subjected to a conventional undrained compression test with pore water pressure measurement.

Yielding is observed at Y3 in the plot of deviator stress versus triaxial shear strain, q.

A Y1v

p

A Y2v

v

q

q

Y3

A

q

Y1

Y3Y2

p

A

Yield locus joining

the yield points

Specimen 1 Specimen 2 Specimen 3

Yield envelope

marks the points ofthe same specific

volume

3 increasing

Pre-consolidation

pressure

Like in an oedometer


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Yield surface deduced from triaxial

tests on undisturbed specimens of

clay from St. Louis, Canada(Tavenas, des Rosiers, Leroueil,

LaRochelle and Roy, 1979)

Yield curves deduced from

triaxial tests on undisturbed

Winnipeg clay taken at fourdifferent depths (Graham,

Noonan and Lew, 1983) pg.75

vc = preconsolidation pressure

Yield locus in qp spaceof natural soil


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Compression index, Cc & swelling index, Csin comparison to

the slope of normal compression line NCL, & slope of unloading and reloading line URL, .

'ln pvv Equation for

unloading and

reloading line URL

'ln pvv Equation for normal

compression lineNCL

NCL and URL become linear whenplotted with natural logarithmic scale

for the mean stress axis.

Results from oedometer tests are often

plotted on a semilogarithmic basis.

3.210ln

3.210ln

'

'

s

c

C

Cv and v are intercepts onthe lines at p=1.

One-dimensional compressionIsotropic compression

Equation for normalcompression phase

Equation for unloadingor swelling phase

'

10

' log vss Cvv

'

10

' log vcc Cvv

vc and vs are

intercepts forv=1.


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Elastic and plastic deformationDuring normal compression :-

Plastic deformation, p refers to the greater part of the deformationwhich is due to the slippage between the soil particles as the soilskeleton rearranges itself to accommodate higher loads. This

component of deformation is irrecoverable or plastic.

Elastic deformation, eis taking place along unload and reloading linewhere change in stress can be accommodated without the need for the

rearrangement of soil particles. Deformation is primarily due to the

distortion of the soil particles and can be recovered on unloading.

e p

Distortion of particles


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16 wu

'

1 200fES200

'

3

ELASTIC AND PLASTIC STRAINACCORDING TO ROTATIONAL MULTIPLE YIELD SURFACE FRAMEWORK

'minf

'

min 1mob

'

min 2mob

'

min 3mob

a

A

B

CKink = yield point = yield limit

= maximum mobilised shear strength

e

A B

e Elastic zone

Elastic-plastic zone

e


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Elastic-plastic model for soilChanges in stress within yield locus

1. Changes in the size of the yield locus are related to the change in volume.

2. Changes of stress within the yield surface are accompanied by purely elastic or recoverable

deformation.

3. All the stress states within the yield locus, have their corresponding point on an unloading-reloading line

(URL). This represents the combination of specific volume, v and mean effective stress, p that produce

elastic volumetric behaviour.

q

p

v

p

NCLURL

C

CAB

B

A

O

yl Soil response for stress changes from point A to B is elastic since the

stress changes is within the yield locus.

The corresponding volumetric change in the compression plane isrepresented by the path AB along the URL.

Any stress state within the yield locus will correspond to a point on a

single URL in the compression plane, v : p.

The route from A to B or the stress path is immaterial since the

response is elastic.

Expansion of yield locus is related to volume change and produce

corresponding change in URL.

Shortcoming

There is the possibility of elastic volume change due to changes in

deviator stress, q at constant mean effective stress, p such as stress

path BD. But this cannot be model in the compression plane since it

only caters for isotropic stress condition only. Therefore the model

cannot differentiate between one-dimensional compression and

isotropic compression.

D


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Elastic-plastic model for soilChanges in stress which causes the soil to yield

Base on the response of Winnipeg clay;

Whatever the value of preconsolidation pressure,

the shape of the current yield surface would be the

same, with only its size changing.Irrespective of the stress path that produce the new

yield surface, its shape remain the same. Note: this

just a convenient assumption but not a necessary

one, since the complexities that arise when yield

surface are allowed to change shape are great.

q

pv

p

A

A

B

M

B

M

O

yl 1

Kyl 2

K

L

Lvp

Yield locus yl 2 is assumed to

have the same shape as yl 1.

Yield locus, yl 1 is attained by the soil

through one-dimensional compression

which normally compressed to point A.

Point K can be established in the

compression plane lying on URL 1 passing

through point A.

po1 po2

'

1

'

2

'

1

'

2

'

1

'

2

ln

lnln

o

op

o

o

o

op

ppv

p

p

p

pv

Total change in volumewhen the mean effective

stress increase from A to B.

Part of the total change which isrecovered when mean effective

stress is reduce again from B to A.

''

o

op

ppv

'

'

'

'

o

o

pp

p

o

o

p

vp

p

v

v

vp

p

v

v

Plastic volumetric strain

Expansion of yield locus is

associated to volume change !

?

''

o

op

p

pv


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19

'

'

vp

pe

p

p

p

e

pp

'

'

''

o

op

vpp

vpp

Total volumetric strain

increment = sum elastic

and plastic components

Elastic volumetric strain'

'

p

p

v

e

Changes in elastic

specific volume

pe vvv Total changes inspecific volume

''

''

o

o

pp

ppv

'

'

'

'

o

o

pp

p

o

o

p

vp

p

v

v

vp

p

v

v

Plastic volumetric strain

Elastic and plastic

volumetric strain

Plastic

volumetric

strain

Elastic

volumetric

strain

Total

volumetric

strain

= +Plastic

change in

specific

volume

Elastic

change in

specific

volume

Total

change in

specific

volume

= +

?

Elastic and plastic change

in specific volume


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20

q

pv

p

A

A

B

M

B

O

yl 1

Q

yl 2P

vp

po1 po2

The difference between elastic and plastic volumetric

strain in response to three stress changes.

RS

Q

ve

Path PQ is directed towards the interior of the locus yl 1and therefore produces purely elastic response.

In the compression plane, the path moves up the URL 1.

The elastic change in volume is given by;

And the elastic volumetric strain is;

This isve vol. strain i.e. swelling, vol. increase.

This elastic process occurs with no change of po = po1.

'

'

p

pve

'

'

vp

pe

p

Path PR is vertically upwards at constant p. The new effective

stress state lies on a new larger yield locus yl 2. This yl 2 could

be obtained by further one-dimensional normal compression AB

in q: p plane.

Since there is no change in p then there is no elastic volumetric

strain. Therefore the volume change resulting from the change in

po ( i.e. po = po1) is purely plastic;

'

'

o

o

pp

pvp

p

v

v

The path PS is the stress path with the (1) same change in p as the

path PQ and (2) has the same expanded yield locus yl 2 as point R but

(3) has the same specific volume as point P (i.e. no change in volume).

This stress path PS, involves both elastic and plastic change in volume

since the overall change in volume is zero;

Or in terms of volumetric strain;

The elastic volumetric strain is negative i.e. swelling;

Note: +ve volumetric strain is compression.

The plastic volumetric strain is positive;

0 pe vvv 0 pp

e

pp

'

'

vp

pe

p

'

'

o

o

pp

p

vp

p

v

v

P

1. Note that even though the stress paths PR, PS and AB produce the same

yield locus yl 2, they have the same plastic but diff. elastic volumetric strain.

2. Expansion of the yield locus is associated with equal plastic volume change

(i.e. vertical distance between two url in compression plane) but the elastic

volume change would be different due to the different change in p.

3. Yielding can occur in undrained condition i.e. stress path PS with elastic and

plastic deformation balanced to give zero resultant volume change.

4. The condition of no overall volume change i.e. undrained condition does notstop elastic and plastic deformation from taking place.

Expansion of yield locus elastic + plastic vol. change

p elastic vol. change


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21

q

p

v

p

C

A

B

B

O

yl 1

yl 2

G

vp

po1 po2

H

ve

Expansion of yield locus elastic + plastic vol. change

p elastic vol. change

Ifp = 0 then purely plastic volume changei.e. no elastic vol. change

F

E

D

A

LK

NM

QP

HOW THE DIFFERENT IN p PRODUCE A DIFFERENT ELASTIC VOLUME CHANGE BUT THESAME PLASTIC VOLUME CHANGE

q

p

v

p

C

A

B

B

O

yl 1

yl 2

G

vp

po1 po2

H

ve

D

A

QP

Compare stress paths PQ and CD

1. Same plastic strain, vp.

2. Different in elastic strain,

ve

.

Expansion of the yield locus

can be achieved by one of the

stress paths AB, CD, EF, GH,

KL, MN and PQ.

The recoverable volume

change would be different

due to different changes in

net mean stress, p.

The irrecoverable volume

change would be the same

caused by each stress path

since the change in po (i.e.

indicate the size of the yield

loci) caused by each stress

path is the same. The

magnitude is given by the

vertical separation between

the url.


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22


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23

Stress path in qp spacea. Drained triaxial testb. Undrained triaxial test

'3'2'131' p

In triaxial test, (i.e. biaxial symmetry);'

3

'

2

'3'1 23

1' p

As 1 is increased uniaxially

'

3

'

1 q

'

3

'

3

'

3

'

1

'

3

'

1

3

1'

33

12

3

1'

qp

p

'3

3

1'

dpdq

dq

dp

In drained triaxial test; there is no pore pressure increase

q

P

'Mpq

C

DU

3

1

p1

p1 u1

O

OC represent isotropic compression and at C,

1 = 3 and u = 0

CD is the drained effective stress path.

Pore water pressure changes during

undrained triaxial compression

q

au

aIn undrained triaxial test

If u1 is the pore water pressure

increase during undrained triaxial

compression, then CU will be the

effective stress path.

0, since constant


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EXAMPLE (Response of soil as isotropic and elastic material)

A triaxial soil specimen of diameter 50mm and height 100mm is subjected to an axial

effective stress of 400kPa and radial effective stress of 100kPa. The resultant axial and

radial displacements are 0.5mm and -0.04mm respectively. Assuming the soil is an isotropic

and elastic material, calculate (a) the mean and deviatoric stress (b) the volumetric and

shear strains and (c) the shear, G, bulk, K and elastic, E moduli.

kPa400'1

SOLUTION

kPa100'3

kPap 2003

1002400

3

2 '3'

1'

kPaq 300100400'3'1

(a)

(b) To determine the volumetric and shear strain.

005.0100

5.01

L

zz

0016.0

25

04.03

r

rr

31 2 p

%18.00018.00016.02005.02 rze

p

313

2 q

%44.00044.00016.0005.032

32 rz

eq

(c)'

'

1p

K

e

P qG

e

q3

1

kPap

Ke

p

111,111

0018.0

200''

kPaq

Ge

q

727,220044.03

300

3'

'12'

'

E

G

'62

2'3'

KG

GK

4.0111,1116727,222727,222111,1113

'622'3' KGGK

kPa

GE

636,63

4.01727,,222'12'


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25

Cam clay model: Stress and strain relationship

'

'

vp

pep

qG

e

q '3

1

qMvp

pMMvp

p

p

2

''

' 2222

22

qMMvp

pMvp

p

q

22

2

2222

4

''2

'

Elastic volumetric

strain

Elastic deviatoric

strain

Plastic deviatoric

strain

Plastic volumetric

strain

q

p

G

vp

eq

e

p

'

'3/10

0'/

q

p

M

M

Mvpp

q

p

p

'

/42

2

' 222

22

22

'0

'

pp

v

opp


26/52

26

Cam clay model: Stress and strain relationship

po

q

p

2

'

op

2

'

oMp

p

q

p

p

Whenever there is a stress increase

beyond the current yield locus, plastic

strain is triggered, the elliptical yield locus

enlarged. This is the hardening of the soil.

Stress path confined within the yield locus

only associate with the elastic strain.

222

'

'

M

M

p

p

o

'p

q

Equation of ellipse

Size of ellipse is

controlled by po and the

shape controlled by M

q

ppMM op

q

p

p

2

'2

2

'222

'

0

'

p

p

v

op

p

'' op

p

o vpp

Hardening relationship

Magnitude of plastic volumetric strain

Cam clay model: Stress and M it d f l ti l t i t i


27/52

27

Cam clay model: Stress and

strain relationship in Drained

Triaxial Compression

'

0

'

p

p

v

op

p

Magnitude of plastic volumetric strain

'p

q

po

q

p

pq

p

p'3 pq Since the cell

pressure is

constant.

F

ABO

C

url 1

ncl

csl

url 2

v

p

Isotropically

consolidated to point

A, followed by

unloading to point B.

Then the specimen isshear to C and

progressively to F.

At F ultimate plastic

shear strain developedwith no plastic

volumetric strain.

ylF

Note:Plastic

volumetric strain isassociated with the

expansion of the

yield locus.

0pp

Mp

q

'

At F

0'

p

p

p

q

p

q

CSL

Note:

Yielding only

produces plastic

volumetric and

plastic shear strain

i.e. no elastic

strain.

? Cam clay model prediction of stress and strain response


28/52

28

qMMvp

pMvp

p

q

22

2

2222

4

''2

'

'3 pq 3/' qp or Substitute this intoabove equation.

qMMvp

q

Mvp

p

q

22

2

2222

4

'32

'

The plastic shear strain

can be calculated fromthis equation and thencethe stress-strain curve

can be predicted.

qMMvp

Mpq

2222

222

'3

122

Rearrange

Cam clay model prediction of stress and strain response

po

q

p

F

ABO

C

ylF

CSL

1

3

q

Plastic shear strain,p

q

Linear elastic stress path BC

within the yield locus

Note:

Yielding only

produces plasticvolumetric and


i.e. no elastic

strain.?

Note:

Yielding only

produces plasticvolumetric and


i.e. no elastic

strain.?


29/52

29

EXAMPLE (Cam clay model prediction of soil response to stress)A specimen of Cam clay is in equilibrium in a triaxial cell under effective stress; r = 200kPa

and a = 300kPa. The specimen was deforming plastically just before this effective stress

state was reached. Then the specimen is subjected to changes in effective stresses; r=-1

kPa and a = +4 kPa. Estimate the increments in axial and radial strains that will result.

Take the values of soil parameters;=0.26,

=0.07, N=3.52, M=0.85 and G=1500 kPa.

SOLUTION

kPa300'1 kPa200'

3

kPap 3.2333

2002300

3

2 '3'

1'

kPaq 100200300'3'1

kPap3

2124

3

12

3

1' '3

'

1

Given 1'

3 4'

1

kPaq 5)1(4'3'1

429.03.233

100'

p

q

To calculate total volumetric strain, p ppepp

To calculate total deviatoric strain, q pqeqq ''

vppe

p

qG

e

q '3

1

qMvp

pMMvp

p

p

2'

'' 22

22

22

qMMvppMvpp

q

22

2

2222

4

''2'

10.242.152.33.233ln26.052.3ln ' opNv

?1 a

?3 r

31 2 p

313

2 q

Solved for1 and 3.

To calculate total volumetric strainSolution continue


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30

31 2 p

313

2 q

po

q

p

p

v

kPap 3.233'

kPaq 100

kPap32'

kPaq 5

Plastic

deformation

A

B

To calculate total volumetric strain, p

To calculate total deviatoric strain, q

00009525.03.23310.2

3/207.0

'

'

vp

pep

qMvp

pMMvp

p

p

2

''

' 2222

22

00111.0515003

1

'3

1

q

G

e

q

qMMvp

pMvp

p

q

22

2

2222

4

''2

'

A

B

Plastic

deformation

along ncl

url

ncl

Solution continue

00199.029.4359.0000428.0

5429.02429.085.03.23310.2

07.026.0

667.0429.085.0429.085.03.23310.2

07.026.0

22

22

22

pp

00317.0836.6572.0000428.0

5429.085.0

429.04

429.085.03.23310.2

07.026.0

667.0429.02429.085.03.23310.2

07.026.0

22

2

22

22

pq

002085.000199.000009525.0 p

p

e

pp

00428.000317.000111.0 pq

eqq

31 2002085.0

313

200428.0

Then, solve for1 and 3.

Solution continue


31/52

31

31 2 p

313

2 q

31 2002085.0

313

200428.0

Then, solve for1 and 3.

Solution continue

3100642.0

31 2002085.0

Subtract

33004335.0

001445.03

004975.0

)001445.0(00642.0

00642.0

1

1

31

-ve radial strain = swell

+ve axial strain = compressed

MAGNITUDE OF PRIMARY CONSOLIDATION SETTLEMENT


32/52

32

MAGNITUDE OF PRIMARY CONSOLIDATION SETTLEMENT

For overconsolidated clays

Virgin consolidation

curve for

overconsolidated clay,

slope = Cc.=

compression index

eeo

Pressure, p

(log scale)pc

p

slope = Cs.= swell index

po

pCase 1Case 2

10009.0 LLCc(Skempton 1944)

p

CLAY

SAND

SAND

Hc

Depth

GWT

Hs2

Hs1

If po + p pc.

0

0

0

log1 p

pp

e

HCS s

Case 1

If po < pc < po + p .

c

c

o

cs

p

pp

e

HC

p

p

e

HCS 0

00

log1

log1

Case 2

po = average effective stress at the middle of clay layer.

wclaysatcwsandsatssanddrysH

HHp )()(2)(102

H

e

eS

01

or

since

0

0logp

pp

eCc

PRIMARY CONSOLIDATION OF NORMALLY CONSOLIDATED CLAY


33/52

33+

PRIMARY CONSOLIDATION OF NORMALLY CONSOLIDATED CLAY

A layer of fine sand of 10.4m thick overlies a 2m layer ofsoft normally consolidated clay. GWT is 3m belowground level. Assume that Gs for sand and clay is 2.7,the Cc for clay is 0.3, void ratio of sand is 0.76 and thewater content of clay is 43%. If the construction of a

building will impose an increase in the effective stress of140kPa at the centre height of the clay layer, determinethe primary consolidation settlement of the clay.

e

Log (kPa)

140kPa

eD= ?

Cc=0.3Cr

A

BC

D

ncl

? ?

eB= ?

What is known and unknown?

1. The clay is normally consolidated thence it will follow the path ABD and the gradient Cc is known.

2. The increase in stress due to load is 140kPa. 3. Initial void ratio, eo of clay (Need to calculate).

4. The initial effective stress at mid clay layer is required which correspond to point B in the graph. Thencethe unit weight of sand and clay is required to calculate the overlying pressure at the mid clay layer.

Effective

stress at

mid-depth of

clay layer

135.9135.9

135.9 275.9

0

0

0

log1 p

pp

e

HCS c

DETERMINATION OF C FROM TEST DATA & PRIMARY CONSOLIDATION


34/52

34

DETERMINATION OF Cc FROM TEST DATA & PRIMARY CONSOLIDATION

A saturated clay specimen was normally consolidated to a vertical stress of 200kPa in an oedometer and

the corresponding void ratio is 1.52. An increase in vertical stress by 150 kPa compresses the specimen

to a void ratio of 1.43. Determine the compression index Cc.

The specimen was unloaded to a vertical stress of 200kPa and the void ratio increased to 1.45.

Determine the slope of the recompression curve, Cr, i.e. recompression index.What is the overconsolidation ratio of the soil sample after being unload.

If the specimen were reload to a vertical stress of 500 kPa what is the void ratio attained?

e

Log (kPa)200 350 500

1.52

1.45

1.43

e500= ?

Cc

Cr

A

BC

D

37.0200/350log

52.143.1

200log350log

52.143.1

cCSlope of AB,

Slope of CB,

08.0200/350log

45.143.1

200log350log

45.143.1

rC

'

'

.

max

zstresseffcurrent

stresseffimumpastOCR zc

75.1200

350'

'

z

zcOCR

37.143.1log37.043.1350

500log37.043.1500

BDB eee

E

BE

350log500log37.0 EDCBE c

ncl

He

eS

01

To determine

settlement, S

from void

ratio change

url

url

?

?

Stress path along the state boundary I t i lid ti t O C


35/52

35

q

p

v

NCL

CSL

CSL in

elevation

O

CSL in plan view

Stress path along the state boundary

surface i.e. Roscoe surface for

normally consolidated clay

C

U

D

C

U

D

Isotropic consolidation stage: O C

Undrained shearing : C U

Drained shearing : C D

During consolidation under isotropic

stress p the stress path is O-C and the

volume change path moves along the

normal compression line (NCL).

Definition of CSL

Critical state line (CSL) is a curve

drawn on a three-dimensional state

boundary surface in q-p-v space.

Six triaxial compression tests


36/52

36

Critical state line (CSL) intriaxial compression

p

q

p

v

OC1 C2 C3

U1

U2

U3

D1

D2

D3

U1

U2

U3

D1

D2D3

NCL

CSL

CSL

ln p

v

U1

U2

U3

D1

D2

D3

NCLCSL

C1

C2

C3

p01 p02 p03

Consolidation stages: O C1, O C2, O C3

Undrained shearing : C1 U1, C2 U2, C3 U3

Drained shearing : C1 D1, C2 D2, C3 D3

Six triaxial compression tests

During (uniaxial) shearing stages

Undrained : Specimen volume remain constant

Drained : Change in volume takes place

During isotropic consolidation, p; the

volum e change path wil l move along the

Normal Comp ression Lin e (NCL)

ln p = ln 1 = 0

1

N

Formed

Roscoe surface for

normally

consolidated clay

q = ( - ) = ( - )


37/52

37

p

q

p

v

OC1 C2 C3

U1

U2

U3

D1

D2

D3

U1

U2

U3

D1

D2D3

NCL

CSL

CSL

ln p

v

U1

U2

U3

D1

D2

D3

NCLCSL

C1

C2

C3

p01 p02 p03 ln p = ln 1 = 0

1

Defining equations for CSLq = (1 - 3) = ( 1 - 3)p = [(1 + 23)/3] - u

Defining equations for CSL

'

Mpq

'ln pv

M = slope of the CSL in q-p plane.

= the specific (v) at p = 1.0 kPa

= slope of CSL in the v-ln p plane

'

Mpq

'ln pv

vp exp'

vMMpq exp'

or

NFormed

Roscoe surface for normally

consolidated clay

( )


38/52

38

What is governing the soil shear

strength and volumetric behaviour ?

There used to be a strong believe that effective stress controlcompletely both the shear strength and volumetric behaviour of soils.

Terzaghi (1936) described the controlling factor for thebehaviour ofsaturated soilsas follows :

All the measurable effects of a change in stress,such as compression, distortion, and a change in shearingresistance, are exclusively due to changes in effective

stress1, 2,3,.Shear strength Volumetric behaviour

Net / effective

stress

SuctionYield stress

Mobilised shear

strength

Applied stress

Mean normal

effective stress, p ?

q = (1 - 3)p = [(1 + 23)/3] - u


39/52

39

Roscoe surface and stress path for

normally consolidated clay

p

q

p

v

OC2

U2

D2

U2

D2

NCL

CSL

CSL

p01

'

Mpq

C2

M

1

u

ln p

v

U2

D2

NCL

CSL

ln p01

C2

ln p = ln 1.0 = 0

N

11

Roscoe surface is applicable for

normally consolidated clay.

For normally consolidated clay the

stress path commences on the NCL.

Stress path along the state boundary surface i e


40/52

40

q

p

v

NCL

CSL

CSL in

elevation

OCSL in plan view

Roscoe state boundary surface after

the late Professor K. H. Roscoe

Applicable to normally consolidatedor lightly overconsolidated soil.

Stress path along the state boundary surface i.e.

Roscoe surface for normally consolidated clay

Note:

Notice the projection of the of the stress paths

on the Roscoe surface on to the q-p plane.No volume change on undrained path.

Fundamental concept in

Critical state model is

that

a unique failure surface

exist in (q, p, v) space

irrespective of the

history of loading or thestress paths followed


41/52

41

Roscoe surface and stress path for

lightly over-consolidated clay

p

q

p

v

OL

U

D

U

D

NCL

CSL

CSL

p0

'Mpq M

1

u

L q/p = 0

q/p = M

pm

SRL

vL

vcrit

Forlightly overconsolidated soil the

stress path commences on the SRL at

point L, which is located between the

NCL and the CSL.

The specific volume at L is greater

than at critical state, and water

content wetter.

Hvorslev surface (T-S) and The past maximum pressure is represented by


42/52

42

Hvorslev surface (T-S) and

stress path for heavily over-

consolidated clay

p

q

p

v

O

T

S

H

p0

'Mpq

M

1

DH

pC

UH

H

UH

DH

Hvorslev surface

No-tension

cut-off(3=0)

L A

p p p y

point A (preconsolidation pressure).

For lightly overconsolidated soil the current stress

condition is like at L.

For heavily overconsolidated soil the net mean

stress is much further away from point A along the

SRL or URL such as point H, below the CSL on v: p

plane.

Under undrained loading (i.e. volume constant) the

stress path is H UH, where UH is above CSL inq: p plane. After yielding the stress path will

continue with further straining along a straight line

TS to meet the CSL at S. The greater the degree of

overconsolidation, the greater is the strain required

to bring the soil to its critical state.

Under drained loading of heavily overconsolidated

soil, HDH, the soil will expand and the volumewill continue to increase after yielding. DH is a

failure point located on the line TS.

Therefore TS represents that part of the stateboundary surface which governs the yielding of

heavily overconsolidated soils and is called the

Hvorslev surface.

The third part of the state boundary surface is OT

which is called no-tension cut-off, which

represents the condition of zero tensile stress (i.e.

3= 0, then p = (q+0+0)/3 = q/3 or q = 3p).


43/52

43

Roscoe surface

Hvorslev

surface

No-tension

cut-off (3=0)p

q

O

T

S

No-tension

cut-off(3=0)

3-D

Question 1


44/52

44

Question 1

a) Differentiate between yield and failure in soil mechanics.

b) Show that in triaxial test where the cell pressure is kept constant during the shearing

stage the relationship between the small increase in net mean stress, 'p and the small

increase in the deviator stress, q is as follows irrespective of the drainage conditions i.e.

whether drained or undrained.

3/' qp

c) In drained triaxial test the stress path OBF would stop at critical state strength at point

F cscs qp ,' on the critical state line as shown in Figure 1. The location of point F in the q p space is dependent on the value of M and the initial net mean stress,

'

o

p . The

gradient of the drained stress path is fixed as 3 vertical: 1 horizontal and the final failure

point will stop at the critical state line. The undrained stress path will also stop at the

critical state line but at a lower strength. The expressions for the possible values of'

csp

and csq are as follows;

M

pp ocs

3

3 ''

M

MpMpq ocscs

3

3 ''

Briefly discuss on the impossible stress states where soil cannot have infinite strength and

soil cannot sustain tension. Sketch the zone in the q p space. The zone is also known

as no tension cut off zone.

3

1

A

B

F

po

q

pO

CSL

A Q ti 1( )


45/52

45

Answer Question 1(a)Yields:


purely elastic to partly plastic OR when the deformation stops being

recoverable upon unloading. This is often marked by an abrupt change in

the slope (i.e. stiffness) of stress-strain curve. Yielding does notnecessarily mean failure.

Failure:






Yields:


purely elastic to partly plastic OR when the deformation stops being

recoverable upon unloading. This is often marked by an abrupt change in

the slope (i.e. stiffness) of stress-strain curve. Yielding does notnecessarily mean failure.

Failure:






Answer Question 1(b)


46/52

46

SHOW THATTHE VARIATION OF q and p

IN TRIAXIAL TEST IS

q=3p irrespective of drained orundrained condition

In triaxial test 3 is kept constanttherefore 3 = 0

'

3

'

1 q 3 = 0

'

1 q

'3'1'3'2'1' 23

1

3

1 p

'

13

1' p '

1

3' p

'

13 pq

q

p

3

1

Drained stress

path since no pwp

eveloped

3/' qp Undrained stresspath since pwp

eveloped

SHOW THATTHE VARIATION OF q and p

IN TRIAXIAL TEST IS

q=3p irrespective of drained orundrained condition

In triaxial test 3 is kept constanttherefore 3 = 0

'

3

'

1 q 3 = 0

'

1 q

'3'1'3'2'1' 23

1

3

1 p

'

13

1' p '

1

3' p

'

13 pq

q

p

q

p

3

1

3

1

Drained stress

path since no pwp

eveloped

3/' qp Undrained stresspath since pwp

eveloped

Answer Question 1(b)

Answer Question 1(c)


47/52

47

Answer Question 1(c)

M

pp ocs

3

3'

'

M

MpMpq ocscs

3

3 ''

When M =3 then net mean stress and deviator stress is infinity and this is not possible.

When M > 3 then net mean stress and deviator stress is negative and this is also not

possible.

3

1

B

F

po

q

pO

CSL

3

1

Impossiblestress

state

Question 1


48/52

48

Question 1

A clay specimen was normally consolidated isotropically to'

ip kPa. Then the specimen

was drained shear until critical state. Assuming the clay response to stress is in

accordance to critical state Cam clay model, show that the increase in net mean stress,

p to reach failure at critical state is as follows.

M

Mpp i

3'

'

(10 marks)

Where'

ip is the net mean stress at the end of the isotropic compression

M is the stress ratio at critical state where 'Mpq

Apply the sketch in Figure 1 to assist your derivation.

If kPapi 100' , initial specific volume at the end of isotropic compression, vi = 1.849,

M = 0.9, = 0.25 and 0.3 , determine the final stress state at failure. Illustrate yourcalculation by sketching the stress plane and the compression plane. (10 marks)

Answer Question 1


49/52

49

3

1

Drained compression

stress path

CSL

M

1

A(pi, 0)p

q

p

k

B

O

Q

To determine the coordinate at the intersect k.

Equation of line kAB is,

kpq '3

Since the line passes through A then, when q = 0,''i

pp

Then,

kpi '30

Therefore,'3 ipk

To solve for coordinate of point B

'Mpq (1)

'3'3 ippq (2)Solve Equation 1 and 2,

'3'3' ippMp

'33' ipMp

3

3'

'

M

pp i or

M

pp i

3

3'

'

Note that this is the value of p at point B

Then 'p is given by;

''

3

3' i

i pM

pp

M

Mp

M

Mppp

M

pM

M

pp iiiiii

33

33

3

3

3

3'

''''''

Proved.

f kPap 100' initial specific volume at the end of isotropic compression v = 1 849


50/52

50

f kPapi 100 , initial specific volume at the end of isotropic compression, vi = 1.849,

= 0.9, = 0.25 and 0.3 , determine the final stress state at failure. Illustrate yourcalculation by sketching the stress plane and the compression plane. (10 marks)

3

1

Drained compression

stress path

CSL

M

1

A(pi, 0)p

q

p

k

B

O

ln p

csl

1'ln cscs pv

p=1

?csv

Failure stress state is the stress state at critical state.

Value of p at B is kPaM

ppp ics 86.142

9.03

1003

3

3'

''

Then the corresponding value of deviator stress, qcs is,

'Mpq 'cscs Mpq kPaqcs 57.12886.1429.0

76.124.10.396.425.00.386.142ln25.00.3 csv

'ln pv


51/52

51

The end

Thank you

What is meant by stress path in critical state ?


52/52

52

What is meant by stress path in critical state ?

In t/ s space:

s = 0.5(1 + 3)t = 0.5(1 - 3)Stress states can be conveniently represented by Mohr circle.

Stress path is the locus of the point of maximum shear stress on a Mohr

circle during shearing stage.

In q / p space:

q = (a - c)p = [(a + 2c)/3] uStress path is the locus of point (p , q) during shearing.

Half sum and half difference

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