lecture 02 constitutivemodel lecture handout

7/31/2019 Lecture 02 ConstitutiveModel Lecture Handout

1/32

For f inite element analysis

1


2/32

Q: What are const itutive models and why are theyneeded?*

A: They form a central component of most, if not all,

the predic tive models we develop of ground

response. From the simplest conceptual models, tothe most sophisticated mathematical model, we

need to idealise the behaviour of small elementsof soil.

A const itutive model can be defined as a set ofmathematical relationships between for example,

components of stress and components of strain.

2*Carter (2006)


3/32

Soil compression

Odemeter Test

3

p

s

h


4/32

Shear ing of soil - Contractant

triaxial

4

31

31

q:stressDeviator

3

)2(p':stressMean

3


5/32

Shearing of soil - Dilatant

5

V


6/32

6

Strain

StressWork hardening

Peak

Work softening

Yield point

Perfectly plastic


7/32

7

Strain

Stress

Strain

Stress

Strain

Stress

Strain

Stress

(a) Elastic

(d) Elastic-plastic softening

(b) Rigid plastic

(c) Elastic-Plastic


8/32

8

Elastic

Ther e is a one- t o- one

r elat ionship bet ween st r essand st r ain.

i.e. = E

I t can be linear or non linear

Af t er loading and unloading,mat er ial r et urn t o it s or iginal

condit ion.

Soil f low (def or mat ion)mechanism depends on t hest ress increment .

Strain

Stress


9/32

9

Strain

Stress

Elastic-Plastic

The st r ess-st r ain r elat ionshipis not unique.

I t is non linear .

Af t er loading and unloading,mat er ial doesnt r et ur n t o i t sor iginal condit ion.

Soil f low (def or mat ion)

mechanism depends on t hest r ess level.


10/32

Conditions of application of elastic model:

For problems where loads are at the working loadlimit (i.e. much less than the ultimate load limit)

For initial displacement estimation of a structureunder loading.

Implications of adopting elastic model:

No-one can pretend that soil behaves as an elasticmater ial except under str ictest conditions.

Choice of elastic modulus and Poissons ratio arecr itical !!

10


11/32

Yield surface is a boundary in soil elementstress field.

It defines the state of stress at which soilresponse changes from elastic to plastic.

11

allowedNot0)f(

behaviourinelasticofOnset0)f(behaviourElastic0)f(

ij

ij

ij

Strain

StressYield stress- y


12/32

xyyz

yzxz

zz

xz

xy

12

x

z

y

xx

xy

xzxx

yz

xz

zz

xy

zz

yz

zz

zzyzxz

yzxyyxy

xzxyxx

ij:tensorStress


13/32

h

13

q

hv

h

q:stressDeviator

(p':stressMean v

3

)2

p

Yield Surface

Yield point x

z

y

hh

h

v

v


14/32

14

Plastic strain increment arise if:1) The stress state is located on the yield surface, AND

2) The stress state remains on the yield surface after astress increment

Yield function f() tells us whether plastic strain isoccurring or not, however, we would like to know directionand magnitude of plastic strainFor that we need flow rule


15/32

15

q

p

Yield Surface [f()]

Potential Surface [g()]

Critical State Line (CSL)

vp

qp

qp

vp

g

p


16/32

16

Associated flow rule f() =g()Non associated flow rule f() g()

It would be great advantage that f() =g() , only 1 function is to be

generated to describe plastic responseAdvantages:

1) The solutions of the equations that emerge in the analyses is faster2) The validity of the numerical predictions can be more easily

guaranteed

For metals f() =g() For soil f() g()

The assumption of normality of plastic strain vectors to the yield locus

would result in much greater plastic volumetric dilation than actuallyobserved.


17/32

17

st r ain or displacement )

(st r ess) Real soil response

Idealised soil model MC model


18/32

18

First order approximation

Basic Law: i = ie + ip

ie = r ever sible (elast ic) st r ain

ip = ir r ever sible (plast ic) st r ain

ip

= 0 f or f < 0f = f (xx, yy, zz, xy, xz, yz)

Yield f unct ion

ip ie

i


19/32

19

z

x

xz

xz

c

s r

Failure criterion: f = c + at an f

Or r c cos + s sins : cent r e of Mohr s st r ess circler : r adius of Mohr s st r ess cir cle

s sin

c cos


20/32

Hexagonal shape

20


21/32

21

strainsplasticofdirectionthedetermines

strainsplasticofmagnitudethedeterminesthatmultipliera

etc,:meansThis

:strainsplasticforruleFlow

py

px

pi

i

yx

i

g

gg

g


22/32

22

g = 0

f = 0

n

nsp

vp

The assumption of normality of plastic strain increment vectors to

the yield laws would result in much greater plastic volumetricdilation than actually observed.

f g

M-C model: f r s sin c cosg r s sin - c cos : dilatancy angle


23/32

23

Dense sand

Loose sand

vDense sand

Loose sand

v = x + y + z = 2x + y


24/32

24

Dense packing of grains

V

Interlocking saw blades

i

Sliding takes place NOT on horizontal planes, but on planes inclined at anangle to the horizontal.

i (strength = dilatancy + friction)

The apparent externally mobilised angle of friction on the horizontal

planes () is larger than the angle of friction resisting sliding of the inclinedplanes (i).

i


25/32

25

yyxy

xy 1

yy

xy

xy

G

Only plastic strainsf = 0

f < 0

tan

)1(2

xy

yy

EG

xy

yy

tan

xyyy


26/32

26

y

x= z

z

|y - x|

y

E = 2G(1+)

y

v

v = x + y + z = 2x + y

sin1

sin2tan

11-2


27/32

27


28/32

28

E0 E50y

2

y

Select E0 when soil behaveslinear elast ic f or a lar ger ange.

Select E50 f or gener al soils.

Select Eur f or unloadingproblems (t unnelling,

excavat ion)

However, E is relat ed t o

conf ining pr essur e,

loading/ unloading/ r eloading

Eur


29/32

29

For one-dimensional compression:

1

0

v

hK

Select to match K0.

Under loading: = 0.3 ~ 0.4

Under unloading: = 0.15 ~ 0.25


30/32

For undrained clay analysis, c is theundrained shear strength of the soil.

For drained sand analysis, c is normally zero.

f = c + tan

However, in FE analysis, a small valuec > 0.2 kPa to avoid numerical dif ficulties.

In advanced MC model, c can increase withsoil depth.

30


31/32

= 0 can be used f or undrained analysis.

A high value is f or dense sand and can make

analysis t ake long t ime. A not t oo high value is r ecommended f or init ial

preliminar y analysis.

31

c

u

Ef f ect ive st r ess = Tot al st r ess por e pr essur e = u

c = 0

Undrained ClaySand


32/32

Except for very over-consolidated clays, wecan use = 0 for clay layers.

The dilatancy of sand depends on the sanddensity and its fr ict ion angle .

For quar tz sand, 30.

For sand with < 30, = 0.

32

lecture 02 constitutivemodel lecture handout

Documents