Andrzej Truty ZACE Services - Z_SOIL · Hardening Soil model with small strain sti ness Andrzej...

Hardening Soil model with small strain stiffness

Andrzej TrutyZACE Services

1.09.2008

Andrzej Truty ZACE Services Hardening Soil model with small strain stiffness

Introduction

Hardening Soil (HS) and Hardening Soil-small (HS-small)models are designed to reproduce basic phenomena exhibitedby soils:

densificationstiffness stress dependencyplastic yieldingdilatancystrong stiffness variation with growing shear strain amplitudein the regime of small strains (γ = 10−6 to γ = 10−3)this phenomenon plays a crucial role for modeling deepexcavations and soil-structure interaction problems

NB. This model is limited to monotonic loads


Introduction

HS model was initially formulated by Schanz, Vermeer andBonnier (1998, 1999) and then enhanced by Benz (2006)

Current implementation is slightly modified with respect tothe theory given by Benz:

simplified treatment of dilatancy for the small strain version(HS-small)modified hardening law for preconsolidation pressure

This model seems to be one of the simplest in the class ofmodels designed to handle small strain stiffness

It consists of the two plastic mechanisms, shear and volumetric

Small strain stiffness is incorporated by means of nonlinearelasticity which includes hysteretic effects


Triaxial test: illustration

Undrained triaxial test: video

Drained triaxial test : video

Annimations by P.Baran (University of Agriculture, Krakow,Poland)


Notion of tangent and secant stiffness moduli

Initial stiffness modulus Eo

Unloading-reloading modulus Eur

Secant stiffness modulus at 50 % of the ultimate deviatoricstress qf

0

50

100

150

200

250

0 0.05 0.1 0.15 0.2 0.25

EPS-1 [-]

q [k

pa] 1

Eo

1E50

1Eur

qf

0.5 qf

σ3=const

q50

qun

Remark: All classical soil models require specification of Eur

modulus (Cam-Clay, Cap etc..)Andrzej Truty ZACE Services Hardening Soil model with small strain stiffness

Stiffness-strain relation for soils (G/Go (γ))

G - current secant shear modulus

Go - shear modulus for very small strains

Atkinson 1991


Notion of treshold shear strain γ07

To describe the shape ofG

Go(γ) curve an additional

characteristic point is needed

It is common to specify the shear strain γ0.7 at which ratioG

Go= 0.7

0.7

γ07


Dynamic vs static modulus

Relation between ”static” Young modulus Es , obtained fromstandard triaxial test at axial strain ε1 ≈ 10−3, and ”dynamic”Young modulus (the one at very small strains) Ed = Eo isshown in diagram published by Alpan (1970) (after Benz)

1

10

100

1000 10000 100000 1000000

cohesive soils

granular soils

Rockss

d

EE

[kPa]sE


HS model: general concept

Double hardening elasto-plastic model (Schanz, Vermeer,Benz)Nonlinear elasticity for stress paths penetrating the interior ofthe elastic domain

0

100

200

300

400

500

600

0 100 200 300 400 500

p [kPa]

q [k

Pa]

Cap surface

Graphical representation of shear mechanism and cap surfaceAndrzej Truty ZACE Services Hardening Soil model with small strain stiffness

HS model: shear mechanism

Duncan-Chang model as the origin for shear mechanism

0

50

100

150

200

250

0 0.01 0.02 0.03 0.04 0.05

eps-1

q [k

Pa]

1E50

qfM-C limit

1

Eur½ qf


Stiffness stress dependency

Eur = E refur

(σ∗3 + c cotφ

σref + c cotφ

)m

E50 = E ref50

(σ∗3 + c cotφ

σref + c cotφ

)m

Remarks

1 Stiffness degrades with decreasing σ3 up to σ3 = σL (bydefault we assume σL=10 kPa)


Extension to small strain: new ingredients

To extend standard HS model to the range of small strain Benzintroduced few modifications:

1 Strain dependency is added to the stress-strain relation, forstress paths penetrating the elastic domain

2 The modified Hardin-Drnevich relationship is used to relatecurrent secant shear modulus G and equivalent monotonicshear strain γhist

3 Reversal points are detected with aid of deviatoric strainhistory second order tensor Hij ; in addition the currentequivalent shear strain γhist is computed by using this tensor


How does it work ?

N

N+1N-1

plot from paper by Ishihara 1986

At step N : γhistN−1= 8× 10−5 γhistN = 10−4

At step N + 1 : γhistN = 0 γhistN+1= 2× 10−5

Primary loading: γhistN+1> γmax

hist

Unloading/reloading: γhistN+1≤ γmax

hist

Hardin-Drnevich law: G =Go

1 + aγhist

γ0.7

(secant modulus)


Shear tangent modulus cut-off

γc

G

γ

Gur

γc =γ0.7

a

(√Go

Gur− 1

)Andrzej Truty ZACE Services Hardening Soil model with small strain stiffness

Setting initial state variables: γPSo and pco

Given: σo , OCRFind: γPS

o and pco

0

100

200

300

400

500

600

0 100 200 300 400 500

p [kPa]

q [k

Pa]Cap surface

Shear mechanism

σSR

σο

Procedure:

Set effective stress state at the SR pointσSR

y = σyo OCR

σSRx = σSR

z = σy KSRo



0

100

200

300

400

500

600

0 100 200 300 400 500

p [kPa]

q [k

Pa]

Cap surface

Shear mechanism

σSR

σο

Procedure:

For given σSR state compute γPSo from plastic condition

f1 = 0

For given σSR state compute pco from plastic condition f2 = 0



Remarks

1 KSRo = KNC

o ≈ 1− sin(φ) in the standard applications(approximate Jaky’s formula)

2 KSRo = 1 for case of isotropic consolidation (used in triaxial

testing for instance)

3 For sands notion of preconsolidation pressure is not asmeaningful as for cohesive soils hence one may assumeOCR=1 and effect of density will be embedded in H and Mparameters


Setting M and H parameters based on oedometric test

0

100

200

300

400

500

600

0 100 200 300 400 500

p [kPa]

q [k

Pa]

p*

q*

σ

εσref

1

Eoed

oed


Material properties

Parameter Unit HS-standard HS-smallE ref

ur [kPa] yes yesE ref

50 [kPa] yes yesσref [kPa] yes yesm [—] yes yesνur [—] yes yesRf [—] yes yesc [kPa] yes yesφ [o ] yes yesψ [o ] yes yesemax [—] yes yesft [kPa] yes yesD [—] yes yesM [—] yes yesH [kPa] yes yesOCR/qPOP [—/kPa] yes yesE ref

o [kPa] no yesγ0.7 [—] no yes


Converting MC to HS model: indentation problem

Assumption: q = 0.5 qult

Given: E for MC model andEur

E50= ...,

E50

Eoed= ...

10m

10m

q = 0.5 qult

1m

A

Find: E refur , M and H for standard HS model


Example: triaxial test on dense Hostun sand

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

0 0.02 0.04 0.06 0.08 0.1-EPS-Y [-]

SIG

-1 /

SIG

-3 [k

Pa]

HS-stdHS-small

0

20000

40000

60000

80000

100000

120000

0.00001 0.0001 0.001 0.01 0.1 1EPS-X - EPS-Y [-]

G [k

Pa]

HS-stdHS-small

(a)σ1

σ3

(ε1) (Z Soil) (b) G (γ) (Z Soil)

1

1.5

2

2.5

3

3.5

4

0 0.002 0.004 0.006 0.008 0.01-EPS-Y [-]

SIG

-1 /

SIG

-3 [k

Pa]

HS-stdHS-small

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.080 0.02 0.04 0.06 0.08 0.1

-EPS-Y [-]

-EPS

-V [-

]

HS-stdHS-small

(c)σ1

σ3

(ε1) (zoom) (Z Soil) (d) εv (ε1) (Z Soil)

(e) Solution by Benz [?]



1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

0 0.02 0.04 0.06 0.08 0.1EPS-1 [-]

SIG

-1 /

SIG

-3 [k

Pa]

HS-stdHS-small

0

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

0.00001 0.0001 0.001 0.01 0.1 1EPS-1 - EPS-3 [-]

G [k

Pa]

HS-stdHS-small

(a)σ1

σ3


1

1.5

2

2.5

3

3.5

4

0 0.002 0.004 0.006 0.008 0.01EPS-1 [-]

SIG

-1 /

SIG

-3 [k

Pa]

HS-stdHS-small

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.080 0.02 0.04 0.06 0.08 0.1

EPS-1 [-]

EPS-

V [-]

HS-stdHS-small

(c)σ1

σ3


(e) Solution by Benz [?]



1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

0 0.02 0.04 0.06 0.08 0.1

EPS-1 [-]SI

G-1

/ SI

G-3

[kPa

]

HS-stdHS-small

0

50000

100000

150000

200000

250000

300000

0.00001 0.0001 0.001 0.01 0.1 1EPS-1-EPS-3 [-]

G [k

Pa]

HS-stdHS-small

(a)σ1

σ3


1

1.5

2

2.5

3

3.5

4

0 0.002 0.004 0.006 0.008 0.01

EPS-1 [-]

SIG

-1 /

SIG

-3 [k

Pa]

HS-stdHS-small

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.080 0.02 0.04 0.06 0.08 0.1

EPS-1 [-]

EPS-

V [-]

HS-stdHS-small

(c)σ1

σ3


(e) Solution by Benz [?]Andrzej Truty ZACE Services Hardening Soil model with small strain stiffness

Estimation of material properties: input data

Given 3 drained triaxial test results for 3 confining pressures:σ3 = 100 kPaσ3 = 300 kPaσ3 = 600 kPa

Shear characteristics q − ε1

Dilatancy characteristics εv − ε1

Stress paths in p − q planeMeasurements of small strain stiffness moduli Eo (σ3) for theassumed confining pressures (through direct measurement ofshear wave velocity in the sample)


Estimation of material properties: stress paths in p-q plane

Estimation of friction angle φ = φcs and cohesion c

p

q

φφ

sin3cos6*−

=cc

1

φφ

sin3sin6*−

=MResidual M-C envelope

If we know M∗ and c∗ then we can compute φ and c :

φ = arcsin3 M∗

6 + M∗ c = c∗3− sinφ

6 cosφ


Estimation of material properties: stress paths in p-q plane

Estimation of friction angle φ = φcs and cohesion c

0

500

1000

1500

2000

2500

3000

0 300 600 900 1200 1500 1800

p [kPa]

q [k

Pa]

1386

2358 12358/1386=1.7

Here: φ = arcsin3 ∗ 1.7

6 + 1.7≈ 42o c = 0


Estimation of material properties: dilatancy angle

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0 0.02 0.04 0.06 0.08 0.1

EPS-1 = - EPS-3 [-]

EPS-

V [-]

1

d Dilatancy cut-off

ψ = arcsin

(d

2 + d

)


Estimation of material properties: dilatancy angle

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

1

d=0.75Vε

1ε

ψ = arcsin

(0.75

2 + 0.75

)≈ 16o


Estimation of material properties: E refo and m

Analytical formula: Eo = E refo

(σ∗3 + c cotφ

σref + c cotφ

)m

Measured: shear wave velocity vs at ε1 = 10−6 and at givenconfining stress σ3

Compute : shear modulus Go = ρv2s

Compute : Young modulus Eo = 2 (1 + ν) Go

σ3 [kPa] Eo [kPa]

100 250000

300 460000

600 675000


Estimation of material properties: E refo and m

Reanalyze Eo vs σ3 in logarithmic scales

Averaged slope yields m; here m =13.1− 12.55

1.0= 0.55

Find intersection of the line with axis ln Eo at

ln

(σ∗3 + c cotφ

σref + c cotφ

)= 0

Here the intersection is at 12.43 henceE ref

o = e12.43 ≈ 2.71812.43 = 250000 kPa

12.2

12.4

12.6

12.8

13

13.2

13.4

13.6

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

⎟⎟⎠

⎞⎜⎜⎝

⎛++

φσφσ

cotcotln 3

cc

ref

oEln

1

m

12.43


Estimation of E refo from CPT testing

To estimate small strain modulus Go at a certain depth onemay use empirical formula by Mayne and Rix:

Go = 49.4q0.695t

e1.13[MPa]

qt is a corrected tip resistance expressed in MPa

e is the void ratio


Estimation of material properties: E ref50

Lets us find E50 for each confining stress

0

500

1000

1500

2000

2500

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

1)100( 350 kPaE =σ

)300( 350 kPaE =σ1

)600( 350 kPaE =σ1

)100( 350 =σfq

)100( 350 =σfq

)100( 350 =σfq


Estimation of material properties: E ref50

Reanalyze E50 vs σ3 in logarithmic scalesHere we can fix m to the one obtained for small strain moduliFind intersection of the line with axis ln E50 at

ln

(σ∗3 + c cotφ

σref + c cotφ

)= 0

Here the intersection is at ≈ 10.30 henceE ref

50 ≈ e10.30 ≈ 2.71810.30 ≈ 30000 kPa

10.2

10.4

10.6

10.8

11

11.2

11.4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

50ln E

⎟⎟⎠

⎞⎜⎜⎝

⎛++

φσφσ

cotcotln 3

cc

ref10.30


Estimation of material properties: E refur

The unloading reloading modulus as well as oedometricmoduli are usually not accessible

We can use Alpans diagram to deduce E refur once we know

E refo (default is

E refur

E refo

= 3); for cohesive soils like tertiary clays

this value is larger

For oedometric modulus at the reference stress σref = 100kPa we can assume E ref

oed = E ref50

γ0.7 = 0.0001...0.0002 for sands and γ0.7 = 0.00005...0.0001for clays

Smaller γ0.7 values yield softer soil behavior


Excavation in Berlin Sand: engineering draft


Excavation in Berlin Sand: FE discretization


Excavation in Berlin Sand: Bending moments

-35

-30

-25

-20

-15

-10

-5

0-500 -400 -300 -200 -100 0 100 200 300 400 500

M [kNm/m]

Y [m

[] HSHS-smallMC


Excavation in Berlin Sand: Wall deflections

-35

-30

-25

-20

-15

-10

-5

0-0.015 -0.01 -0.005 0 0.005 0.01

Ux [m]

Y [m

] HS-smallHSMC


Excavation in Berlin Sand: Soil deformation in crosssection x =20m

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10

00 0.01 0.02 0.03 0.04

Uy [m]

Y [m

] HSHS-smallMC

Vertical heaving of subsoil at last stage of excavation, relative tothe step when dewatering was finished (t = 2)


Excavation in Berlin Sand: Soil deformation in crosssection x =20m

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10

0-0.004 -0.003 -0.002 -0.001 0

Ux [m]

Y [m

] HSHS-smallMC

Horizontal movement in cross section x=20m at last stage ofexcavation, relative to the step when dewatering was finished

(t = 2)


Conclusions

Model properly reproduces strong stiffness variation with shearstrain

It can be used in simulations of soil-structure interactionproblems

Implementation is ”rather heavy”

It should properly predict deformations near the excavations

Model reduces excessive heavings at the bottom of theexcavation


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