HS Model Presentation

Hardening Soil model with small strain stiffness

Andrzej TrutyZACE Services

25.08.2009

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 pressuremodified form of the cap yield surface (2009)

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


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


Influence of void ratio and confining stress p′

onG/Go (γ))

Cohesionless soils

Wichtmann and Triantafyllidis (after Benz)


Influence of plasticity index PI on G/Go (γ))

Cohesive soils

(Vucetic and Dobry (after Benz (PhD thesis))

Remarks

1 Results for PI < 30 are confirmed by other researchers whilethese for PI > 30 should be used with a special care (Benz)

2 Stokoe proposed linear interpolation for γ0.7

γ0.7 = 10−4 for PI = 0 to γ0.7 = 6× 10−4 for PI = 100


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


Shear modulus for very small strains

Go = A f (e)OCRk

(p′

pref

)m

Hardin and Black (1978)

Soil emin emax A [kPa] f (e) m Ref.

Clean sands 0.5 1.1 57(2.17− e)2

1 + e0.4 Iwasaki

Undisturbedclayey soilsand crushedsand

0.6 1.5 33(2.97− e)2

1 + e0.5 Hardin

andBlack

Undisturbedcohesive soils

0.6 1.5 16(2.97− e)2

1 + e0.5 Kim

Loess 1.4 4.0 1.4(7.32− e)2

1 + e0.6 Kim


Poisson coefficient for very small strains

Poisson ratio varies in the range ν = 0.1..0.3 in small straindomain

Its value in further derivations will be kept constant (bydefault ν = 0.25)


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 and cap yield surfaces

Graphical representation of shear mechanism and cap surface


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


HS model: shear mechanism

0

100

200

300

400

500

600

700

0 100 200 300 400 500 600

p [kPa]

q [k

Pa]

M-C

γ=0.1=const.

γ=0.01=const.

γ=0.001=const.γ=0.0001=const.

f1 =qa

E50

q

qa − q− 2

q

Eur− γPS

qf =2 sin(φ)

1− sin(φ)(σ3 + c ccotφ)

qa =qf

Rf


Flow rule for shear mechanism, dilatancy and hardening

g1 =σ1 − σ3

2− σ1 + σ3

2sinψm

sinψm =sinφm − sinφcs

1− sinφm sinφcs

sinφm =σ1 − σ3

σ1 + σ3 + 2c cotφ

-0.5-0.4-0.3-0.2-0.1

00.10.20.30.40.50.6

0 10 20 30 40 50 60phi_m [deg]

sin

psi_

m

Domain ofcontractancy

Domain ofdilatancy

Contractancy cut-off

Rowe’s dilatancy

dγPS = dλ1

(∂g1

∂σ1− ∂g1

∂σ2− ∂g1

∂σ3

)= dλ1


Cap mechanism

Yield condition: f2 =q2

M2 r2(θ)+ p2 − p2

c

r(θ) is defined via van Ekelen’s formula (like in Cam-Claymodel

Plastic potential: g2 =q2

M2+ p2

Hardening law: d pc = dλ2 2H

(pc + c cotφ

σref + c cotφ

)m

p

Remarks:

1 M and H parameters can be estimated for assumed KNCo and

tangent Eoed modulus set up at a given vertical stress


Additional strength criteria

Mohr-Coulomb yield condition

f ∗1 = σ1 −1− sinφ

1 + sinφσ3 −

2c cosφ

1 + sinφ= 0

Mohr-Coulomb plastic flow rule

g∗1 = g1

NB. Here same plastic flow rule is used as for the shearmechanism f1Rankine yield condition (tensile cut-off)

f3 = σ1 − ft = 0

where: ft is the assumed tensile strength (default is ft = 0)

Rankine plastic flow rule(associated flow rule is used)

g3 = f3


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

4 Hardening laws for γPS and pc are modified by introducing hi

factor; this factor for very small strains is much larger than1.0 and decreases to 1.0 once the shear strain γhist exceedscertains strain amplitude γc

5 Certain constractancy is allowed in the plastic flow rule forshear mechanism


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

Modifications: Dilatancy

PHI = 40, PSI=10

-35

-30

-25

-20

-15

-10

-5

0

5

10

0 10 20 30 40

PHI_m [deg]

PSI_

m [d

eg]

Dafalias,Li(after Benz)

Rowe’s dilatancy

Scaled Rowe’s dilatancyD = 0.25

PHI = 30, PSI=5

-30

-25

-20

-15

-10

-5

0

5

10

15

20

0 10 20 30 40

PHI_m [deg]PS

I_m

[deg

]

Dafalias,Li(after Benz)

Rowe’s dilatancy

Scaled Rowe’s dilatancyD = 0.25


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


Setting M and H parameters based on oedometric test

Assumptions:

1 At a given σrefoed vertical stress both shear and volumetric

mechanisms are active

2 p∗ =1 + 2KNC

o

3σref

oed while q∗ = (1− KNCo )σref

oed

3 A strain driven program is applied with vertical strainamplitude ∆ε = 10−5 and resulting tangent oedometric

modulus is computed as Eoed =∆σ

∆ε4 The two conditions must be fulfiled: Ko coefficient generated

by the model must be equal to the one set by the user (usingJaky’s formula for instance Ko = 1− sinφ) and tangentoedometric modulus generated by the model must be equal tothe value given by the user

5 If we take the data from the experiment we must be sure thatthe given oedometric modulus corresponds to the primaryloading branch of σ − ε curve


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


User interface

Remark1 HS/HS-small model can be actived only in the � Advanced

modeAndrzej Truty ZACE Services Hardening Soil model with small strain stiffness

User interface: Elastic properties


User interface: Elastic properties (HS)

Remarks

1 Standard HS model is activated if � Advanced checkbox isset OFF

2 E refur is the unloading/reloading Young modulus given at the

reference stress σref

3 νur is the unloading/reloading Poisson coefficient; it variesfrom 0.15 to 0.3, hence for sands it is recommended toassume νur = 0.2..0.25 and for clays νur = 0.25..0.3

4 m is the exponent in stress dependency power law; it variesfrom m = 0.4 to m = 0..6; it is smaller for dense sands andlarger for clays

5 σL is the minimum allowed reference stress value used forevaluation of stiffness moduli


User interface: Elastic properties (HS-small)

Remarks

1 HS-small model is activated if � Advanced checkbox is setON

2 The HS-small model requires two additional parameters:Young modulus at very low strains E ref

o at the reference stressσref and threshold shear strain γ0.7

3 In case of lack of information on E refo one may try to estimate

E refo based on Alpan’s diagram assuming Es = Eur

4 In the current implementation γ0.7 is assumed to be constant

5 In case of lack of information on γ0.7 the diagram by Vuceticand Dobry can be used for cohesive soils and diagram byWichtmann and Triantafyllidis for cohesionless ones


User interface: Plastic properties (HS/HS-small)



Remarks1 All material properties collected in group Nonlinear are

common for HS and HS-small models2 In the advanced mode one may activate tensile and dilatancy

cut-off conditions, set up the multiplier D for Rowe’sdilatancy law in the contractant domain (for HS model thedefault value is D = 0.0 and for HS-small D = 0.25),

3 E ref50 is the secant Young modulus at 50 % of failure deviatoric

stress qf derived from the q − ε1 curve in drained triaxial test4 φ is the friction angle5 ψ is the dilatancy angle6 c ′ is the effective cohesion7 Rf is the failure ratio (default Rf = 0.9)8 ft is the tensile strength (default ft = 0)9 emax is the maximum allowed void ratio; if current void ratio

exceeds the emax dilatancy angle is switched to ψ = 0



Remarks

1 Cap surface parameter M and hardening parameter H arederived by using a simple calculator which simulates anoedometric test; for given tangent oedometric modulus Eoed

at a given reference vertical stress σrefoed and for assumed KNC

o

parameter (here Jaky’s formula can be used) values of H and

M are evaluated (press button Evaluate M,H ); one may

assume Eoed = E ref50

(σref

oed + c cotφ

σref + c cotφ

)m

as a default value

2 Setting the initial state variables γPSo and pco can be carried

out by means of assumed OCR or preoverburden pressureqPOP

3 To compute KNCo from Jaky formula press button

Use Jaky’s formula for KoNC



Remarks

1 Pairs KSRo and OCR (OCR ≥ 1.0) or KSR

o and qPOP areneeded to setup the initial position of the cap surface and theinitial value of the hardening parameter γPS

2 pminco is the minimum allowed value for the initial

preconsolidation stress


Converting MC to HS model: general idea

Question: Having calibrated standard MC can we convert it to HSmodel ?

Stiffness modulus E refur and cap surface parameters H and M

can be estimated by running an inverse analysis of a planestrain problem of a soil layer loaded by a strip loading q

q = 0.5 qult with qult being the approximate ultimate limitload density

The template data files for MC and HS model can be found inthe CFG directory under names: template-foot-MC andtemplate-foot-HS


Converting MC to HS model: indentation problem

10m

10m

q = 0.5 qult

1m

A


User interface: Converting MC to HS model

Given: γdry , K insituo ,νur , σref , σL, m, φ, ψ, c ′, OCR, KSR

o ,

E refur

E ref50

= .... andE ref

50

E refoed

= .... and Young modulus that user

would assume in the simulation with a standard MC model

Find: E refur


Convert MC to HS model: algorithm

The estimation idea is as follows:

1 We know parameters to be used in the simulation with aid ofa standard MC model: E , γdry , K insitu

o ,νur , φ, ψ, c ′

2 Now we want to use HS/HS-small model but we do not knowon how to estimate E ref

ur , H and M parameters

3 We select a plane-strain problem of a strip loading q appliedto a uniform layer of soil as a template problem

4 We assume the additional parameters for HS model: σL, m,

OCR, KSRo and the two coefficients

E refur

E ref50

= .... (default is 3)

andE ref

50

E refoed

= .... (default is 1.0)

5 We run the optimization procedure which yields the E refur , M

and H such that the settlement at point A obtained from MCand standard (!!!) HS model are the same


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-qplane

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-qplane

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



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



Plot Eo vs σ3

0

100000200000

300000

400000

500000600000

700000

800000

0 100 200 300 400 500 600 700

E

o[k

Pa]

3σ [kPa]



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

Note: this is very rough estimation

Best solution: Perform triaxial testing and project on CPTprofile to adjust empirical coefficient (49.4) for a given site


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


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


Documents

HS Model Presentation