# Lecture 02 ConstitutiveModel Lecture Handout

#### Short Description

geotechnical finite element lecture note...

#### Description

For finite element analysis

1

Q: What are constitutive models and why are they needed?* A: “They form a central component of most, if not all, the predictive models we develop of ground response. From the simplest conceptual models, to the most sophisticated mathematical model, we need to idealise the behaviour of small elements of soil.” A constitutive model can be defined as a set of mathematical relationships between for example, components of stress and components of strain. *Carter (2006)

2

Soil compression p s h

Odemeter Test

3

Mean stress : p'

(

Deviator stress : q

1

2 3 1

3

)

3

Shearing of soil - Contractant

3

triaxial

4

Shearing of soil - Dilatant

V

5

Work hardening Stress Peak Perfectly plastic

Work softening Yield point

Strain

6

Stress

Stress

Strain

Strain

(b) Rigid plastic

(a) Elastic Stress

Stress

Strain (c) Elastic-Plastic

Strain (d) Elastic-plastic softening 7

There is a one-to-one relationship between stress and strain. i.e. = E

Stress

It can be linear or non linear Strain

Elastic

The stress-strain relationship is not unique. It is non linear.

Stress

Strain Elastic-Plastic

9

Conditions of application of elastic model:

• For problems where loads are at the working load limit (i.e. much less than the ultimate load limit) • For initial displacement estimation of a structure under loading.

• No-one can pretend that soil behaves as an elastic material except under strictest conditions. • Choice of elastic modulus and Poisson’s ratio are critical !!

10

Yield surface is a boundary in soil element stress field. It defines the state of stress at which soil response changes from elastic to plastic. Stress Yield stress- y Strain

f(

ij

) 0

Elastic behaviour

f(

ij

) 0

Onset of inelastic behaviour

f(

ij

) 0

Not allowed 11

Stress tensor :

ij

xx

xy

xz

xy

yy

yzx

xz

yz

zz

y zz

xy yz

zz yz

xy

xx

xz xz

xz xz

xx

x

yz

xy xy

zz yz

z

zz

12

y v h

h

q

h

Yield point

x

h

Yield Surface z

v

Mean stress : p'

(

Deviator stress : q

v

2 3 v

h)

h

p’

13

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 a stress increment Yield function f( ) tells us whether plastic strain is occurring or not, however, we would like to know direction and magnitude of plastic strain For that we need flow rule

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q

g

p

q

p Critical State Line (CSL) p

q

p

Potential Surface [g( )]

v

p’

p

v

Yield Surface [f( )]

15

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 response Advantages: 1) The solutions of the equations that emerge in the analyses is faster 2) 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 actually observed.

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(stress)

Real soil response

Idealised soil model – MC model

strain or displacement)

17

Basic Law:

i i i

i

p=

=

i

e

+

i

p

e

= reversible (elastic) strain

p

= irreversible (plastic) strain

0 for f < 0

f = f(

xx,

yy,

zz,

xy,

xz,

yz)

Yield function p i

e i i

First order approximation 18

c cos

z xz

c

s sin

xz x

s

Failure criterion:

r f

= c’ + ’atan ’ f

Or r c cos + s sin s : centre of Mohr’s stress circle r: radius of Mohr’s stress circle 19

Hexagonal shape

20

Flow rule for plastic strains :

g

p i

i

This means :

p x

g x

p y

g

, etc

y

a multiplier that determines the magnitude of plastic strains g

determines the direction of plastic strains

i

21

p

s

’n

f=0 g=0

’n

p

v

The assumption of normality of plastic strain increment vectors to the yield laws would result in much greater plastic volumetric dilation than actually observed. f M-C model: f r – s sin – c cos g r – s sin - c cos

g

: dilatancy angle 22

Dense sand

Loose sand

v

=

x

+

y

+

z

=2

x

+

y

Dense sand

v

Loose sand 23

i

V

Dense packing of grains

• Sliding takes place NOT on horizontal planes, but on planes inclined at an angle to the horizontal. i

(strength = dilatancy + friction)

i

• The apparent externally mobilised angle of friction on the horizontal planes ( ) is larger than the angle of friction resisting sliding of the inclined planes ( i). 24

Only plastic strains f=0

xy

yy xy yy

f 0.2 kPa to avoid numerical difficulties. In advanced MC model, c can increase with soil depth. 30

Sand Undrained Clay c

u

c=0

Effective stress = Total stress – pore pressure ’= u = 0 can be used for undrained analysis. A high value is for dense sand and can make analysis take long time. A not too high value is recommended for initial preliminary analysis. 31

Except for very over-consolidated clays, we can use = 0 for clay layers. The dilatancy of sand depends on the sand density and its friction angle . For quartz sand, For sand with

< 30,

– 30 . = 0.

32