SSIsokocalo.engr.ucdavis.edu/~jeremic/LectureNotes/Lecture...The arti cial oscillation are diminished by introducing some numerical damping into the analysis through = 0.6 and = 0.3025

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Real-ESSI Lecture Notes 307.6. VERIFICATION OF 4NODEANDES ELEM . . . page: 1601 of 3100

Figure 307.96: 4NodeANDES edge clamped circular plate with element side length 5m.


Jeremic et al. University of California, Davis version: 7. May, 2021, 13:33

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Figure 307.99: 4NodeANDES edge clamped circular plate with element side length 0.5m.


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Figure 307.100: 4NodeANDES edge clamped circular plate with element side length 0.25m.

Table 307.28: Results for 4NodeANDES circular plate with four edges clamped.

Element type 4NodeANDES Theoretical

displacementElement side length Height:1.00m

10m 1.69E-003 m 1.706E-03 m

5m 1.97E-003 m 1.706E-03 m

2m 1.97E-003 m 1.706E-03 m

1m 1.96E-003 m 1.706E-03 m

0.5m 1.96E-003 m 1.706E-03 m

0.25m 1.96E-003 m 1.706E-03 m

Table 307.29: Errors for 4NodeANDES circular plate with four edges clamped.

Element type 4NodeANDES

Element side length Height:1.00m

10m 0.71%

5m 15.43%

2m 15.31%

1m 15.16%

0.5m 15.13%

0.25m 15.12%


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7cm

Figure 307.101: Error scale 0% - 20%.

7cm


Figure 307.103: 4NodeANDES circular plate with edge clamped˙ Displacement error versus Number of

side division


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307.6.7 Verification of 4NodeANDES circular plate with all edges simply supported

Problem description: Diameter=20m, Height=1m, Force=100N, E=1E8Pa, ν = 0.3.

The four edges are simply supported.

The load is the uniform normal pressure on the whole plate.

The plate flexural rigidity is

D =Eh3

12(1− ν2)=

108N/m2 × 13m3

12× (1− 0.32)= 9.1575× 106 N ·m (307.11)

The theoretical solution7 is

d =(5 + ν)qa4

64(1 + ν)D=

(5 + 0.3)× 100N/m2 × 104m4

64× (1 + 0.3)× 9.1575× 106 N ·m = 6.956× 10−3m (307.12)

The 4NodeANDES were shown in Figure (307.104) - (307.109).

Figure 307.104: 4NodeANDES edge simply supported circular plate with element side length 10m.

The results were listed in Table (307.30).

The errors were listed in Table (307.31).

The errors were plotted in Figure (307.112).

The Real-ESSI model fei/DSL files for the table above are HERE.7Stephen Timoshenko, Theory of plates and shells (2nd edition). MrGRAW-Hill Inc, page55, 1959.


http://sokocalo.engr.ucdavis.edu/~jeremic/lecture_notes_online_material/Real-ESSI-Verification/4NodeANDES/circular_plate_simply_support/circular_plate_simply_support.tar.gz

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Figure 307.108: 4NodeANDES edge simply supported circular plate with element side length 0.5m.


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Figure 307.109: 4NodeANDES edge simply supported circular plate with element side length 0.25m.

Table 307.30: Results for 4NodeANDES cicular plate with four edges simply supported.

Element type 4NodeANDES Theoretical

displacementElement side length Height:1.00m

10m 7.50E-003 m 6.956E-03 m

5m 7.29E-003 m 6.956E-03 m

2m 7.25E-003 m 6.956E-03 m

1m 7.23E-003 m 6.956E-03 m

0.5m 7.22E-003 m 6.956E-03 m

0.25m 7.22E-003 m 6.956E-03 m


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Table 307.31: Errors for 4NodeANDES cicular plate with four edges simply supported.

Element type 4NodeANDES

Element side length Height:1.00m

10m 7.75%

5m 4.73%

2m 4.15%

1m 3.89%

0.5m 3.84%

0.25m 3.82%

7cm


7cm


Figure 307.112: 4NodeANDES circular plate with edge simply supported˙ Displacement error versus

Number of side division


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Chapter 308

Verification and Validation for Static and

Dynamic Behavior of Special Elements

(Contacts/Interfaces/Joints,

Gap/Frictional, Isolators) (2010-2011-2016-2017-2019-2021-)

(In collaboration with Mr. Sumeet Kumar Sinha and Dr. Yuan Feng, and Dr. Han Yang)

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Real-ESSI Lecture Notes 308.1. CHAPTER SUMMARY AND HIGHLIGHTS page: 1611 of 3100

308.1 Chapter Summary and Highlights

308.2 Verification of Static Penalty Contact/Interface/Joint Element

Modeling and Simulation

References for interface elements: Hird and Russell (1990), AG (2020).

This section presents the verification of Penalty Stiffness based Frictional Contact/Interface/Joint

Element using analytical simple solutions to verify the numerical solutions obtained by the application of

the developed model. The examples show the response of element for different cases. Solution sensitivity

on penalty stiffness is also discussed in details for those examples.

Theoretically, the penalty stiffness should be infinite, but for numerical stability of the solution, it

can go upto 1016. This is because for a double precision computer, machine epsilon ε ≈ 10−16 and

thus the corresponding displacement for the penalty springs can go low only till 10−16. For all the cases

considred below in this cases, the convergence criteria was as ||δU || <= 10−12.

308.2.1 Static Normal Contact/Interface/Joint Verification

A Two-bar truss example is considered here to verify the normal contact/interface/joint for different

normal loading conditions and different penalty stiffness Kn. This is an example of normal loading on a

1-D contact/interface/joint between two bars separated by an initial gap of δin = 0.1m. An illustrative

diagram of the problem statement is shown below.

Figure 308.1: Illustration of two bar normal Contact/Interface/Joint problem under monotonic loading

with inital gap.

A snapshot of the code for the contact/interface/joint element is shown below. The Real-ESSI model

fei/DSL files for this example can be downloaded HERE.

1 add element #3 type FrictionalPenaltyContact with nodes (2,3)2 normal_stiffness =1e10*N/m3 tangential_stiffness = 1e10*Pa*m4 normal_damping = 0*kN/m*s5 tangential_damping = 0*kN/m*s6 friction_ratio = 0.37 contact_plane_vector = (1,0,0);


http://sokocalo.engr.ucdavis.edu/~jeremic/lecture_notes_online_material/Real-ESSI-Verification/contact/Two_Bar_Contact_Under_Normal_Monotonic_Loading.tgz

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Real-ESSI Lecture Notes 308.2. VERIFICATION OF STATIC PENALTY C . . . page: 1612 of 3100

308.2.1.1 Case 1: Monotonic Loading with initial gap δin = 0.1m

In this case a force of 0.3N is applied to Node 2. From Figure 308.2, the solution converges to the

analytical result for Kn = 100N/m i.e 100 times the stiffness of bar element. Please note that, the for

penalty stiffness > 10e15, the convergence fails (when the bars contact/interface/joint mode chnges),

as the global stiffness matrix becomes ill conditions. Thus, the penalty stiffness cannot be too large.

308.2.1.2 Case 2: Monotonic Loading with no initial gap δin = 0m

In this case a force of 0.3N is applied to Node 2. From Figure 308.3, the solution again converges to

the analytical result for Kn = 100N/m i.e 100 times the stiffness of bar element.

308.2.1.3 Case 3: Cyclic Loading with initial gap δin = 0.1m

For cyclic loading cases considered below, the loading force Fn applied is shown in Figure 308.4. From

Figure 308.5, the solution again converges to the analytical result for Kn = 100N/m i.e 100 times the

stiffness of bar element.

308.2.1.4 Case 4: Cyclic Loading with no initial gap δin = 0m

The same cyclic load shown in Figure 308.4 is again applied fpor this case. From Figure ??, the solution

again converges to the analytical result for Kn = 100N/m i.e 100 times the stiffness of bar element.


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Force [N]0 0.05 0.1 0.15 0.2 0.25 0.3

Nor

mal

Dis

plac

men

t [m

]

0

0.05

0.1

0.15

0.2

0.25

Node 1Node 3Anaytical Solution

Force [N]

0

0.1

0.2

0.30.25

Normal Displacment [m]

0.20.150.10.050

10 15

10 10

10 5

10 0

Nor

mal

Pen

alty

Stif

fnes

s N

/m


Figure 308.2: Displacements of Node 2 and Node 3 with change in normal penalty stiffness for

δin = 0.1m


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Force [N]0 0.05 0.1 0.15 0.2 0.25 0.3

Nor

mal

Dis

plac

men

t [m

]

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2Node 1Node 3Anaytical Solution

00.1

Force [N]

0.20.3


0.20.15

0.10.05

10 0

10 5

10 10

10 15

0

Nor

mal

Pen

alty

Stif

fnes

s N

/m


Figure 308.3: Displacements of Node 2 and Node 3 with change in normal penalty stiffness for δin = 0m


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Time Step0 5 10 15 20 25 30

For

ce F

n [N

]

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Figure 308.4: Cyclic normal load applied on two bar contact/interface/joint problem.


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Time Step0 5 10 15 20 25 30

Nor

mal

Dis

plac

men

t [m

]

-0.1

0

0.1

0.2

0.3

0.4

0.5


0

Time Step

1020

300.6


0.40.2

0-0.2

10 10

10 5

10 0

10 15

Nor

mal

Pen

alty

Stif

fnes

s N

/m




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Time Step0 5 10 15 20 25 30

Nor

mal

Dis

plac

men

t [m

]

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45


0

Time Step

10

20

300.5


0.40.3

0.20.1

010 0

10 5

10 10

10 15

Nor

mal

Pen

alty

Stif

fnes

s N

/m




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308.2.2 Static Frictional Tangential Contact/Interface/Joint Verification

A simple 3-D truss example with Normal confinement in z-direction of FN = 0.5N , friction coefficient

µ = 0.2 and shear loading of magnitude Fs = 0.5N is considered to verify the tangential behaviour of

contact/interface/joint element. Different cases as dicussed below are considered.

Figure 308.7: Illustration of 3-D three bar contact/interface/joint problem.

A snapshot of the properties of contact/interface/joint element os shown below. The Real-ESSI

model fei/DSL files for this example can be downloaded HERE.

1 add element #4 type FrictionalPenaltyContact with nodes (1,2)2 normal_stiffness =1e10*N/m3 tangential_stiffness = 1e10*Pa*m4 normal_damping = 0*kN/m*s5 tangential_damping = 0*kN/m*s6 friction_ratio = 0.27 contact_plane_vector = (0,0,1);


http://sokocalo.engr.ucdavis.edu/~jeremic/lecture_notes_online_material/Real-ESSI-Verification/contact/Two_Bar_Contact_Under_Normal_Monotonic_Loading.tgz

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308.2.2.1 Case 1: Verfication of the yield surface for different loading angles with fixed normal

confinement.

A Shear force of magnitude of Fs = 0.5N was applied in 20 steps in different loading directions. The

response of the contact/interface/joint element and the displacement of node 2 is shown in Figure

308.8 and Figure 308.9 respectively . It can be bserved that the contact/interface/joint element slips

at magnitude of force ||F ||− > (Fn = 0.5) ∗ (µ = 0.2)− > 0.1N for all loading angles.

In Figure 308.9, it can be observed that for the first 4 steps, there is no (zero) displacement for

node 2 because of the stick case. When the load exceeds 0.1N , slip occurs and node 2 starts to undergo

deformation.


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Force in x direction Fx [N]

-0.1 -0.05 0 0.05 0.1

For

ce in

y d

irect

ion

Fy [N

]

-0.1

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

Anaytical Solutionloading direction

Force in y direction Fy [N]

0.1-0.100.10.05Force in x direction F

x [N]

0-0.05-0.1

14

12

10

8

6

4

2

0

16

18

20

Tim

e S

teps

Anaytical Solutionloading direction

Figure 308.8: Response of contact/interface/joint element for different loading angles for confinement

of 0.5N and coefficient of friction as 0.2

.


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Displacment in x direction Ux [N]

-0.5 0 0.5

Dis

plac

men

t in

y di

rect

ion

Uy [N

]

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

Tim

e S

teps

0

2

4

6

8

10

12

14

16

18

20

Displacment in x direction Ux [N]

-0.5 0 0.5

Figure 308.9: Displacement of Node No. 2 in x and y direction for different loading angles for

confinement of 0.5N and coefficient of friction as 0.2

.


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Real-ESSI Lecture Notes 308.3. VERIFICATION OF STATIC AND DYNA . . . page: 1622 of 3100

308.3 Verification of Static and Dynamic Contact/interface/Joint Ele-

ment Modeling and Simulation

Solution verification of the contact/interface/joint element formulation and its implementation is pre-

sented in what follows. Analytical simple solutions for the frictional contact/interface/joint element

are used to verify numerical solutions obtained by application of the developed model. The examples

provided show the response of the contact element in several situations. Initially, the element is tested by

connecting two nodes that each have 3dof, subsequently, the element is even implemented to simulate a

contact/interface/joint between two nodes that each have 7dof and two nodes with different dofs: 3dof

for the first one and 7dof for the second one.

The parameters used for the contact/interface/joint element are listed in Table 308.1.

Parameter Value

CN [kN/m] 10420

vmax [m] 0.001

KT [kN/m] 1e7

µ [-] 0.6

Table 308.1: Contact/interface/joint element parameters.

Figure 308.10: Input signal: time history of displacement, acceleration and velocity.


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308.3.1 Truss Examples

The first example (Figure 308.11) represents five nodes: 1,2,3,4,5. Nodes 2,3,4,5 are connected

by three truss elements and a contact/interface/joint element links node 1 and 2. All degrees of

freedom of nodes 1,3,4,5 are fixed, whereas a sine wave displacement time-history (Figure 308.10)

is applied to node 2 along x direction and the normal force acting within the contact/interface/joint

element is recorded. The results, represented in Figure 308.12, show the normal response of this new

contact/interface/joint element. As the timestep decreases, the force-displacement curves tends to be

similar to the one represented by (??).

Figure 308.11: System composed of one contact/interface/joint element and three truss elements. A

sine wave displacement time-history is applied to node (2).

Figure 308.12: Normal force vs normal-relative displacement in contact/interface/joint element.


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The second example (Figure 308.13) shows the tangential response of this new contact/inter-

face/joint element. The geometry of the problem is the same as Figure 308.11, but an axial force

(p = 1140 kN), constant in time, and a sine wave time-history displacement are applied to node 2. The

results, represented in Figure 308.14, show that the response is not dependent on the timestep used for

the analysis. Due to the elastic-perfectly-plastic behavior, associated with Mohr-Coulomb yield criteria,

the maximum shear force (tmax) that the contact/interface/joint element can sustain is 684 kN , equal

to tmax = µ · p = 0.6 · 1140kN .

Figure 308.13: Same system used in Figure 308.11. A normal force constant in time (F = 1140 kN)

and a sine wave displacement time history are applied to node (2).

Figure 308.14: p = 1140 kN. Transversal force in the contact/interface/joint element vs transversal-

relative displacement


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308.3.2 Single Brick Element Examples

The contact/interface/joint element is now used to connect each node (5,6,7,8) of the bottom face of

an eight node hexahedral element (hex8) to the corresponding node attached to the ground (1,2,3,4).

A linear elastic constitutive model is used for the eight node brick and the parameters are listed in

Table 308.2.

Parameter Value

E [kPa] 1.5e10

ν [-] 0.0

Table 308.2: Brick Element Parameters

In the first example (Figure 308.15) the time-history displacement, shown in Figure 308.10, is applied

in the vertical direction to each node of the top surface of the brick element (9,10,11,12) and the

time-history normal force induced in each contact element is represented in Figure 308.16. It is worth

noting that the normal force is positive if the displacement of the top surface is downward, whereas it is

zero if the detachement occurs caused by the upward movement.

Through the second example, shown in Figure 308.17, the transversal response of the contact/in-

terface/joint elements is highlighted. The vertical normal force (Fv = 50kN), constant in time, and

a horizontal time-history displacment are applied to each node of the top surface and the transversal

response is shown in Figure 308.18. As stated in section 308.3.1, the maximum shear force that the

contact element can sustain is is 30 kN , equal to tmax = µ · p = 0.6 · 50kN .

The third example is focused on the transversal response of the contact/interface/joint element under

variable normal forces. In fact, a sine wave time-history horizontal displacement and a vertical force are

applied to each node of the top surface of the brick. The normal force is variable in time according to the

factor (Fact(t)) shown in Figure 308.20 and the vertical force is computed as Fv(t) = Fact(t) · Fv,max,

and Fv,max equal to 50 kN. The response of the contact element, shown in Figure 308.21, is independent

of the timestep used for the analysis enphasizing the correct numerical implementation.

308.3.3 Double Brick Element Examples

Few other examples are produced taking in consideration two brick elements. The constitutive model used

for these two brick elements is linear elastic with the same parameters listed in Table 308.2. Vertical and

horizontal time-history displacement are applied to the top surface, shown in Figure 308.22 and Figure

308.23, and variable vertical forces are considered in the example represented in Figure 308.24. The

results are the same shown in Figure 308.16, Figure 308.18 and Figure 308.19.


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Figure 308.15: Eight-node brick element over four contact/interface/joint elements. A sine wave time-

history vertical displacement applied to each node of the top surface.

Figure 308.16: Normal force vs time in each contact element.


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Figure 308.17: Eight-node brick element over four contact/interface/joint elements. Fv= 50 kN and a

sine wave time-history horizontal displacement applied to each node of the top surface.

Figure 308.18: Transversal force vs transversal relative-displacement in the contact element with normal

force F equal to 50 kN.


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Figure 308.19: Eight-node brick element over four contact/interface/joint elements. Variable vertical

force (Fv(t)) and a sine wave time-history horizontal displacement applied to each node of the top

surface.

Figure 308.20: Time-history of the normal force factor Fact(t).


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Figure 308.21: Variable normal force. Transversal force vs transversal relative-displacement in each

contact element.

Figure 308.22: Two eight-node brick elements connected by four contact/interface/joint elements. Ver-

tical time-history displacement applied to the nodes of the top surface.


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tical force, equal to 50 kN and constant in time, and horizontal time-history displacement applied to the

nodes of the top surface.


tical force, variable in time, and horizontal time-history displacement applied to the nodes of the top

surface.


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308.4 Verification of Static and Dynamic Coupled (Saturated) Contac-

t/Interface/Joint Element Modeling and Simulation

308.4.1 Dry u-p-U Contact/Interface/Joint to the Ground

A single brick u-p-U finite element is used to model an oedometric compression shown in Figure 308.25.

Horizontal displacements and pore pressure are fixed in each node in order to guarantee the one-

dimensional and dry conditions. As the ground is modeled as an undeformable and impermeable layer

(1,2,3,4), vertical soil and fluid displacements are fixed. The time-history displacement, shown in

Figure 308.10, is applied in vertical direction to each node of the top surface (9,10,11,12).

The time-history normal force induced in each contact/interface/joint element is represented in Figure

308.16 and compared with the one obtained with the dry brick element with u formulation, shown in

Figure 308.15. It is worth noting that the normal force patterns are perfectly overlapped: this is due to

the fact that excess pore pressure is fixed to zero and the oedometric stiffness are the same in the two

cases.

Parameter Symbol Value

Young’s Modulus E [kPa] 1.5 · 1010

Poisson ratio ν [-] 0.0

Solid particle bulk modulus Ks [kPa] 3.6 · 107

Fluid bulk modulus Kf [kPa] 2.17 · 106

Solid density ρs [Mg/m3] 2.7

Fluid density ρf [Mg/m3] 1.0

Porosity n [-] 1.0 · 10−8

Darcy permeability K [m/s] 1.0 · 103

Table 308.3: Soil parameters.


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Figure 308.25: Single eight-node brick element. The nodes of the bottom surface are connected to the

ground floor through contact/interface/joint elements. Vertical time-history displacement applied to the


Figure 308.26: Normal force vs time in each contact/interface/joint element. u-formulation represents

the results shown in Figure 308.16 whereas upU-formulation represents the results obtained through the

model shown in Figure 308.25.


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308.4.2 u-p-U Contact/Interface/Joint to the ground

A column of four brick u-p-U finite elements is used to model the horizontal layer. The height of the

soil column is 1 m and the height of each element has dimensions 1m · 1m · 1m, illustrated in Figure

308.27.

The following boundary conditions are applied to the model. As the bottom of the column is modeled

as an undeformable and impermeable layer, both the solid and fluid displacements are fixed. The pore

pressure is kept constant as zero at the top surface of the soil column because of the perfectly drained

condition. In order to simulate the 1D compression problem, all the lateral movement of the solid and

fluid phase are constrained so that the vertical displacement is the only non-zero displacement. A vertical

time-history displacement, shown in Figure 308.10, is applied to the solid dof of the top surface nodes.

In order to simulate the one dimensional compression, all the degrees of freedom at the same level are

connected in a masterslave fashion.

The nodes of the bottom surface (5,6,7,8) are connected to the ground (1,2,3,4) through con-

tact/interface/joint elements. Under the hypothesis of laminar flow, no cavitation and one-dimensional

type of problem, the water has to fill up all the voids generated into the media while the displacement

filed is acting on the top surface. This means that if the soil and the ground are separated because of the

detachment of the contact/interface/joint element, the water has to fill the gap. Even the pore pressure

at both sides of the gap has to assume the same value. Such boundary conditions can be introduced

by adding masterslave between each node of the bottom surface (5,6,7,8) and the corresponding one

belonging to the ground (1,2,3,4).

In this paragraph a parametric study is performed in order to examine the performance of this u-p-U

contact/interface/joint element. The analyses consider several configuration of permeability and soil

stiffness. In fact, three values of elastic modulus (E) are taken into consideration referring to a Stiff,

Medium and Soft soil (called respectively StS, MS and SoS) and three values of the darcy permeability

are similarly defined. All soil and contact/interface/joint parameters are listed respectively in Table 308.4

and Table 308.1.

This kind of excitation (a sine wave) applied at the top of the model is clearly composed of waves

of all kinds of frequency, expetially at the initial strong change in acceleration and velocity for t = 0s.

Due to this fact, a fairly dense mesh of 100 u-p-U brick finite elements of dimensions 1m · 1m · 1cm was

chosen. Therefore, the time step ∆t needs to be limited to ∆t = ∆h/v , where v is the highest wave

velocity. In our case, the temporal integration involves 4000 steps of , which allows a maximum wave

velocity of . The propagation velocity can be calculated by the following equation given by de Boer et


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al. (1993) and is equal to

v =

√n2(1− ν)E

(1 + ν)(1− 2ν)[n2(1− n)ρs + (1− n)2nρf ](308.1)

The artificial oscillation are diminished by introducing some numerical damping into the analysis

through α = 0.6 and β = 0.3025 for the Newmark integrator.

Figure 308.27: Four eight-node brick elements. The nodes of the bottom surface are connected to the

ground floor through contact/interface/joint elements. Vertical time-history displacement applied to the


The specific permeability, k and the time needed for completion of the 1D consolidation process, t,

can be estimated using the darcy permeability K through (308.2) and (308.3) respectively, where ρw

is the mass density of the fluid (water), g is the acceleration of gravity equal to 9.81m/s2, H is the

thickness of the soil layer and Eoed is the one dimensional soil stiffness:

k =K

ρw · g(308.2)

t =H2

cv=H2 · ρw · gK · Eoed

(308.3)

The time t is computed and listed in Table 308.4.2 for each soil type condition. As can be seen,

for the stiff soil (StS), the consolidation time is much lower than the loading acting on the top surface,

therefore no excess pore pressure is developped into the soil and drained response is occurred.


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STIFF SOIL - (StS)


Young’s Modulus E [kPa] 1.0 · 1010 1.0 · 1010 1.0 · 1010

Poisson ratio ν [-] 0.0 0.0 0.0

Solid particle bulk modulus Ks [kPa] 3.6 · 107 3.6 · 107 3.6 · 107

Fluid bulk modulus Kf [kPa] 2.17 · 106 2.17 · 106 2.17 · 106

Solid density ρs [Mg/m3] 2.7 2.7 2.7

Fluid density ρf [Mg/m3] 1.0 1.0 1.0

Porosity n [-] 0.46 0.46 0.46

Darcy permeability K [m/s] 1.0 · 10−3 1.0 · 10−5 1.0 · 10−7

MEDIUM SOIL - (MS)







Porosity n [-] 0.46 0.46 0.46


SOFT SOIL - (SoS)







Porosity n [-] 0.46 0.46 0.46


Table 308.4: Stiff, Medium and Soft soil parameters (StS, MS and SoS).


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Soil type Symbol Value

K [m/s] 1.0 · 10−3 1.0 · 10−5 1.0 · 10−7

Stiff soil (StS) t [s] 1.0 · 10−6 1.0 · 10−4 1.0 · 10−2

Medium-Stiff soil (MS) t [s] 1.0 · 10−3 1.0 · 10−1 1.0 · 101

Soft soil (SoS) t [s] 1.0 · 100 1.0 · 102 1.0 · 104

Table 308.5: The time needed for completion of the 1D consolidation process, t, estimated thrugh the

darcy permeability K for each soil type condition.


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Real-ESSI Lecture Notes 308.5. VERIFICATION OF STATIC, ISOLATOR . . . page: 1637 of 3100

308.5 Verification of Static, Isolator Element Modeling and Simulation


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Chapter 309

Verification and Validation for Coupled,

Porous Solid – Pore Fluid Problems

(2000-2003-2007-2009-2010-2016-2017-2020-2021-)

(In collaboration with Prof. Zhao Cheng, Dr. Panagiota Tasiopoulou, Ms. Fatemah Behbehani and Mr Yusheng

Yang)

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309.2 Introduction

presented here are verification examples for u-p-U formulation. Examples include:

1. Drilling of a borehole

2. The case of a spherical cavity

3. Consolidation of a soil layer

4. Line injection of a fluid in a reservoir

5. Shock wave propagation

6. Vertical Consolidation of a soil layer by Coussy (2004)

7. One dimensional shock wave propagation with a step displacement boundary condition by Gajo

and Mongiovi (1995)

8. One dimensional shock wave propagation with step loading at the surface by de Boer et al. (1993)

9. One dimensional shock wave propagation with a step velocity boundary condition by Hiremath

et al. (1988)

309.3 Drilling of a well

309.3.0.0.1 The Problem Let us consider an infinite half space domain composed of an isotropic,

homogeneous and saturated thermoporoelastic material. At its reference state, it is assumed that the

temperature, fluid pressure and stress fields are uniform and equal respectively, to T0, p0 and σ0 =

σ01(with σ0 < 0). At time = 0, an infinite cylinder of radius r0 is instantaneously drilled parallel to the

vertical axis Oz. It is filled with a fluid of the same nature as that saturating the porous medium but

at a different pressure and temperature at the values of p1 and T1 respectively. The interface r = r0

between the well and the porous medium is assumed to be in thermodynamic equilibrium.

In cylindrical coordinates (r, θ, z), the boundary conditions can be summarized as follows (see


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Real-ESSI Lecture Notes 309.3. DRILLING OF A WELL page: 1640 of 3100

Fig.309.1):

t ≤ 0→ σ0 = σ01 p(r) = p0 T (r) = T0 (309.1)

t > 0→ σrr(r0) = −p1 σrθ(r0) = σrz(r0) = 0

σrr(r →∞)→ σ0 σrθ(r →∞) = σrz(r →∞)→ 0

p(r0, t) = p1 p(r →∞)→ p0

T (r0) = T1 T (r →∞)→ T0 (309.2)

Figure 309.1: Boundary Conditions for Drilling of a Borehole

309.3.0.0.2 Analytical Solution Since the well is assumed to be infinite long in its vertical axis Oz,

the analysis is performed under plane strain hypothesis(εzz = 0). Therefore,

ξ = ξr(r)er p = p(r) T = T (r) (309.3)

in which ξr is the radial displacement. In cylindrical coordinates, Eqn. 309.48 yields

εrr =∂ξr

∂rεθθ =

ξr

rother εij = 0 (309.4)

Based on the constitutive equations from Coussy (1995), it follows that

σrr = σ0 + λ0(∂ξr

∂r+ξr

r) + 2µ

∂ξr

∂r− b(p− p0)− 3αK0(T − T0) (309.5)

σθθ = σ0 + λ0(∂ξr

∂r+ξr

r) + 2µ

ξr

∂r− b(p− p0)− 3αK0(T − T0) (309.6)


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σzz = σ0 + λ0(∂ξr

∂r+ξr

r)− b(p− p0)− 3αK0(T − T0) (309.7)

other σij = 0 (309.8)

Finally combined with the Eqns. 309.5-309.8, it yields the near field or long-term solution (Coussy,

1995)

ξr =σ0 + p1

2µ

r02

r+r0[b(p1 − p0) + 3α0K0(T1 − T0)]

2(λ0 + 2µ)(r

r0− r0

r) (309.9)

σrr = −p1r0

2

r2− σ0 −

µ[b(p1 − p0) + 3αK0(T1 − T0)]

λ0 + 2µ(1− r0

2

r2) (309.10)

σθθ = (2σ0 + p1)r0

2

r2− µ[b(p1 − p0) + 3αK0(T1 − T0)]

λ0 + 2µ(1 +

r02

r2) (309.11)

σzz = σ0 − 2µ

λ0 + 2µ[b(p1 − p0) + 3αK0(T1 − T0)] (309.12)

And the diffusion process can be achieved if that the time are large enough with respect to the charac-

teristics diffusion time relative to point r. When the boundary conditions for r = r0 in fluid pressure and

temperature which are p = p1 and T = T1 apply for the whole model, the following equations correspond

to the undrained solution of the instantaneous drilling of a borehole in an infinite elastic medium.

ξr =σ0 + p1

2µ

r02

rσrr = σ0 − (σ0 + p1)

r02

r2

σθθ = σ0 + (σ0 + p1)r0

2

r2σzz = σ0 (309.13)

309.3.0.0.3 Discussion of the Results As the problem is Axisymmetric, we construct the model

as a quarter of a donut. The inside diameter of the donut is 10 cm and the outside diameter is 1 m.

To accommodate both the plain strain hypothesis and the geometry of the element for finite element,

the thickness of the model is chosen to be 5 cm. The final mesh is generated as Fig.309.2. And the

boundary conditions is as follows: As a consequence of plain strain problem, all the movements for solid

and fluid in vertical direction Oz are suppressed; the solid and fluid displacement for the nodes along

the X axis and Y axis are fixed in Y and X direction respectively for the reason of axisymmetry; the

nodes along the outside perimeter are fixed in the solid and fluid displacement with the assumption

of infinite medium. the pressure is translated into nodal forces and applied on the nodes along the

inside perimeter. For simplicity, the hydrostatic stress σ0 is equal to zero and with the assumption of

thermodynamic equilibrium through the process, the temperature factor can be neglected. Also the


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Figure 309.2: The mesh generation for the study of borehole problem

Parameter Symbol Value Units

Poisson Ratio ν 0.2 -

Young’s Modulus E 1.2E+6 kN/m2

Solid Bulk Modulus Ks 3.6E+7 kN/m2

Fluid Bulk Modulus Kf 1.0E+17 kN/m2

Solid Density ρs 2.7 ton/m3

Fluid Density ρf 1.0 ton/m3

Porosity n 0.4 -

Table 309.1: Material Properties used to study borehole problem

initial fluid pressure p0 is set to be 0 kPa. The analytical solution is studied below using the following

set of parameters shown in Table 309.1.

In the analysis, ten loading cases for final fluid pressure from 10 kPa to 100 kPa are studied. And

by manipulating the permeability, it is possible to investigate both the drained behavior and undrained

behavior. For the drained behavior, we choose the permeability as k = 3.6×10−4m/s, which is a typical

value for sand, the comparison between the close solution and experimental result is shown in Fig.309.3.

From the results, we can see that along the inside perimeter, the close solution and experimental result

provide very good agreement to each other. But as the increase of the radius, we can see the analytical

solution is getting more and more distant from the experimental results. In another word, the analytical


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solution can be interpreted as that with the increase from the loading surface, the radial displacement is

larger. This is unreasonable in the point of view in soil mechanics. While the experimental result show

the effect that with the increase of the radial distance, the radial displacement is decreasing.

0 10 20 30 40 50 60 70 80 90 1000

0.5

1

1.5

2

2.5

3

3.5x 10

−3

Pressure (kPa)

Rad

ial D

ispl

acem

ent (

cm)

r=0.1m, close solutionr=0.1m, experimentalr=0.5m, close solutionr=0.5m, experimentalr=0.9m, close solutionr=0.9m, experimental

Figure 309.3: The comparison of radial solid displacement between analytical solution and experimental

result for drained behavior

For the undrained behavior, the permeability of k = 3.6 × 10−8m/s is selected as a representative

value for typical clayey soil. The comparison between the close solution and experimental result is

provided as well. From the Fig.309.4 we can see that, the analytical solution is linearly away from the

experimental result by a ratio of approximately 1.6. It should also be noticed that the close solution

of the drained and undrained behavior for the nodes along the inside perimeter are exactly the same,

which is contradictory to the definition of drained and undrained behavior. For the drained behavior,

as the water easily dissipate from the soil body, the problem can be treated with the knowledge of

continuum mechanics using the parameters of the solid skeleton. While for the undrained behavior, with

the involvement of the pore water, the elastic parameters for the mixture should be different, so the

response will not be the same as well. As a result of this, the experimental results give a more reasonable

conclusion.

As for the drained behavior, the fluid totally flows out of the soil body and all excessive pore pressure

dissipates, there is small coupling between the solid and fluid phase. We can use the continuum mechanics


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0 10 20 30 40 50 60 70 80 90 1000

0.2

0.4

0.6

0.8

1x 10

−3

Pressure (kPa)

Rad

ial D

ispl

acem

ent (

cm)



result for undrained behavior

to treat this problem. Here introduces a problem of an infinite cylindrical tube, with the inner radius R1

and outer radius R0, subjected to an internal pressure P1 and an external pressure P2. The displacement

field as follows (S.Timoshenko and D.H.Young, 1940):

ξr =R1

2P1

2(R02 −R1

2)(

r

λ+ µ+R0

2

µr) (309.14)

With P0 = 0 and take the limit of R0 →∞, we can obtain the following equation:

ξr =P1

2µ

R12

r(309.15)

which is identical to Eq.309.10. Also to minimize the effect of infinite boundary, we introduce the result

from another model which is exactly the same as the previous one besides the expansion of the outer

radius to 30m. At the final fluid pressure of 50 kPa, the results are shown in Fig.309.5. From the plot we

can make a conclusion that the undrained analytical solution from Coussy (1995) is actually the drained

solution and the undrained solution still needs to be investigated.


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Real-ESSI Lecture Notes 309.4. THE CASE OF A SPHERICAL CAVITY page: 1645 of 3100

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

3

4

5

6x 10

−4

Radius (m)

Rad

ial D

ispl

acem

ent (

cm)

Coussy Undrained SolutionTimoshenko SolutionExperimental(1m Boundary)Experimental(30m Boundary)

Figure 309.5: The comparison of radial solid displacement between two analytical solutions and ex-

panded boundary

309.4 The Case of a Spherical Cavity

309.4.0.0.1 The Problem Considering a medium composed of an isotropic, homogeneous, saturated

thermoporoelastic material. In its initial state, it is assumed that the temperature, fluid pressure and

stress fields are uniform and equal respectively, to T0, p0 and σ0 = σ01(with σ0 < 0). At time t=

0, a spherical cavity of radius r0 is immediately drilled and filled with the same saturating fluid in the

medium. For t > 0, the temperature and the pressure of the fluid are kept constant with the value of

T1 and p1 respectively. The interface r = r0 between the well and the porous medium is assumed to be

in the thermodynamic equilibrium.

In spherical coordinates (r, θ, ϕ), the boundary conditions can be summarized as follows:


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t ≤ 0→ σ0 = σ01 p(r) = p0 T (r) = T0 (309.16)

t > 0→ σrr(r0) = −p1 σrθ(r0) = σrϕ(r0) = 0

σrr(r →∞)→ σ0 σrθ(r →∞) = σrϕ(r →∞)→ 0

p(r0, t) = p1 p(r →∞)→ p0

T (r0) = T1 T (r →∞)→ T0 (309.17)

Strictly speaking, the expressions for r →∞ are not boundary conditions. They are complementary

conditions to be satisfied by the solution. It is used to model that at the point far from the disturbed

area, the state of the medium are held as its initial state.

309.4.0.0.2 Analytical Solution This is a problem of spherical symmetry. The radial displacement

is the only non-zero displacement and all the fields are r and t dependent. Therefore,

ξ = ξr(r)er p = p(r) T = T (r) (309.18)

in which ξr is the radial displacement. In spherical coordinates, Eqn.309.18 yields

εrr =∂ξr

∂rεθθ =

ξr

rother εij = 0 (309.19)


σrr = σ0 + λ0(∂ξr

∂r+ξr

r) + 2µ

∂ξr

∂r− b(p− p0)− 3αK0(T − T0) (309.20)

σθθ == σϕϕ = σ0 + λ0(∂ξr

∂r+ξr

r) + 2µ

ξr

∂r− b(p− p0)− 3αK0(T − T0) (309.21)

other σij = 0 (309.22)

Finally combined with the Eqns. 309.19-309.22, it yields the near field or long-term solution (Coussy,

1995)

ξr =σ0 + p1

4µ

r03

r2+r0[b(p1 − p0) + 3αK0(T1 − T0)]

2(λ0 + 2µ)(1− r0

2

r2) (309.23)

σrr = −p1r0

3

r3+ σ0(1− r0

3

r3)− 2µ[b(p1 − p0) + 3αK0(T1 − T0)]

λ0 + 2µ(r0

r− r0

3

r3) (309.24)

σθθ = σϕϕ = p1r0

3

2r3− σ0(1 +

r03

2r3)− µ[b(p1 − p0) + 3αK0(T1 − T0)]

λ0 + 2µ[r0

r(1 +

r02

r2)]

(309.25)


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And the diffusion process can be achieved if that the time are large enough with respect to the charac-

teristics diffusion time relative to point r. When the boundary conditions for r = r0 in fluid pressure and

temperature which are p = p1 and T = T1 apply for the whole model, the following equations correspond

to the undrained solution of the instantaneous drilling of a borehole in an infinite elastic medium.

ξr =σ0 + p1

4µ

r03

r2σrr = −p1

r03

r3+ σ0(1 +

r03

r3)

σθθ = σϕϕ = p1r0

3

2r3+ σ0(1 +

r03

2r3) (309.26)

309.4.0.0.3 Discussion of the Results The model is constructed as a quarter of a half ball. The

cavity radius is 10cm. As the outside boundary is fixed, to minimize the possibility of the sudden increase

of the fluid bulk modulus, the outside radius of the sphere is set to be 2 m. The final mesh is generated as

Fig.309.6. And the following boundary conditions apply: The nodes on XZ and Y Z plane are fixed for

solid and fluid displacement in Y and X direction respectively; the vertical solid and fluid displacement

for the nodes on the XY plane are suppressed; for the nodes along the outside surface, to satisfy the

complementary conditions, all the solid and fluid displacements are set to be zero as well. The pressure

is translated in to nodal forces and applied in the radial direction. For simplicity, the hydrostatic stress

σ0 is equal to zero and with the assumption of thermodynamic equilibrium through the process, the

temperature factor can be neglected. Also the initial fluid pressure p0 is set to be 0 kPa. The analytical

solution is studied below using the following set of parameters shown in Table 309.2.

Figure 309.6: The mesh generation for the study of spherical cavity

As the same procedure in the previous drilling of borehole problem, we compared both the drained

and undrained behavior. The drained and undrained behavior are tested by the permeability of k =


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Real-ESSI Lecture Notes 309.5. LINE INJECTION OF A FLUID IN A RES . . . page: 1648 of 3100








Porosity n 0.4 -

Table 309.2: Material Properties used to study spherical cavity problem

3.6 × 10−4m/s and k = 3.6 × 10−8m/s respectively. In drained behavior, we can see along the cavity

surface, the experimental result of the radial displacement match the analytical solution very well. While

with the increase of the radius, the decrease of the radial displacement for close solution is much smaller

that of the experimental results. For the undrained behavior, we can see the radial displacement of the

experimental results are always smaller than the close solution. Again it should be noted that the close

solutions for the drained and undrained behavior along the cavity surface are exactly the same. This can

be explained in the same way as the previous drilling of the borehole problem. When the experimental

results from drained behavior are compared with the analytical undrained solution, it is observed they

provide good agreement to each other as well.

309.5 Line Injection of a fluid in a Reservoir

309.5.0.0.1 The Problem Liquid water is usually injected into a reservoir from a primary well in

order to recover the oil from a secondary well in petroleum engineering. This induces a problem of

injecting a fluid into a cylindrical well of negligible dimensions.

Consider a reservoir of infinite extent composed of an isotropic, homogeneous and saturated poroe-

lastic material. Through a cylindrical well of negligible dimensions, the injection of the same fluid is

performed in all directions orthogonal to the well axis forming the Oz axis of coordinates. As a result of

the axisymmetry and cylindrically infinite, all quantities spatially depends on r only. The injection starts

at time t = Γ and stops at time t = Γ. The flow rate of fluid mass injection is constant and equal to

q.As a finite amount of Ω of fluid mass is injected instantaneously(i.e. Γq → ΩasΓ→ 0).


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0 10 20 30 40 50 60 70 80 90 1000

1

2

3

4

5

6x 10

−4

Pressure (kPa)

Rad

ial D

ispl

acem

ent (

cm)



result for drained behavior

309.5.0.0.2 Analytical Solution This is a problem of cylindrically symmetry. Consequently the

cylindrical coordinates(r,θ,z)is adopted. The vector of relative flow of fluid mass w reads

w = w(r, t)er (309.27)

where er is the unit vector along the radius. Using the fluid mass balance relationship, it yields

∫ r

0

∂m

∂t(r, t)2πrdr = q − 2πrw(r, t) ∀r, t (309.28)

In addition, we require the fluid flow to reduce to zero infinitely far from the well

rw → 0 r →∞ t <∞∫ ∞0

∂m

∂t(r, t)rdr =

q

2r∀0 < t <∞ (309.29)


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0 1 2 3 4 5 6 7 8 9 10

x 104

0

1

2

3

4

5

6x 10

−4

Pressure (kPa)

Rad

ial D

ispl

acem

ent (

cm)



result for undrained behavior

Based on above Eqs.309.28-309.29, the radial displacement is derived in the form

p =Ω

4πρfl0 ktexp(− r2

4cmt)

ξr =bMΩ

2πρfl0 (λ+ 2µ)r[1− exp(− r2

4cmt)] (309.30)

Using the constitutive equation, the stress field can be derived as follows:

σrr = −2µξrr

σθθ = 2µξr

r− 2µb

λ0 + 2µp

σzz = − 2µb

λ0 + 2mup (309.31)


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0 1 2 3 4 5 6 7 8 9 10

x 104

0

1

2

3

4

5

6x 10

−4

Pressure (kPa)

Rad

ial D

ispl

acem

ent (

cm)

r=0.1m, undrained(c)r=0.1m, drained(e)r=0.15m, undrained(c)r=0.15m, drained(e)r=0.25m, undrained(c)r=0.25m, drained(e)

Figure 309.9: The comparison of radial solid displacement between analytical solution for undrained

behavior and experimental result for drained behavior

309.5.0.0.3 Discussion of the Results As a result of axisymmetry, the model can be constructed

as a quarter of a pie. The radius of the pie is 1 m and the thickness of the pie is 5 cm. A cylindrical

well is drilled at the center of the pie, and its radius is 1 cm, which can be neglected in dimension when

compared with the whole pie. The final mesh is shown as Fig.309.10. And the boundary conditions is

as follows: As a consequence of plain strain problem, all the movements for solid and fluid in vertical

direction Oz are suppressed; the solid and fluid displacement for the nodes along the X axis and Y

axis are fixed in Y and X direction respectively for the reason of axisymmetry; the nodes along the

outside perimeter are fixed in the solid and fluid displacement with the assumption of infinite medium.

To the difference with the previous problems, the traction boundary conditions are applied on the fluid

displacement. It should be noted that the Ω mention in the above equations is the volume of the fluid

injected per unit of vertical well length and has a unit of m3/m. In order to generate the volume of

1 cm3/m, the corresponding fluid displacement of the nodes along the well has been calculated and

applied as a step function at the time of 0 sec. For simplicity, the initial fluid pressure p0 is set to be

0 kPa. The analytical solution is studied below using the following set of parameters shown in Table

309.3.


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Figure 309.10: The mesh generation for the study of line injection problem






Undrained Bulk Modulus Ku 6.0E+7 kN/m2

Bulk Modulus K 6.7E+5 kN/m2



Fluid Diffusivity coefficient cf 0.4973 m2/s

Porosity n 0.4 -

Permeability k 3.6E-6 m/s

Table 309.3: Material Properties used to study the line injection problem

In the analysis, the pore pressure and the radial displacement are studied. The results are recorded

from three points at the radius of 10 cm, 50 cm and 85 cm. The close solution and experimental results

are shown in Fig.?? and Fig.??. As the time step is set to be 1 sec, the first data point starts at the time

of 1 sec. From the pore pressure plot we can see that the build-up of the pore pressure reach the peak

value of 34 kPa at the radius of 85 cm. With the decrease of the radius, the pore pressure decreases

as well. This can be explained by the fact that the closer the point to the injection location, the earlier

and the larger load is applied, so the pore pressure dissipates faster. And as time passes by, we can see


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the pore pressure progressively dissipates and finally almost reaches the same value within the model.

The same phenomena can be been observed from the radial solid displacement. The maximum solid

displacement occurs at the radius of 85 cm, which means more coupling between the solid and fluid

phase, as consequence, the pore pressure should have the largest value. This corresponds to the previous

result. With the increase of the time, the radial solid displacement get closer to zero, which means the

fluid moves out the solid skeleton.

0 5 10 15 20 25 300

5

10

15

20

25

30

35

Time (sec)

Por

e P

ress

ure

(kP

a)

Radius = 10cm(c)Radius = 10cm(e)Radius = 50cm(c)Radius = 50cm(e)Radius = 85cm(c)Radius = 85cm(e)

Figure 309.11: The comparison between analytical solution and experimental result for pore pressure


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0 5 10 15 20 25 300

1

2

3

4

5

6

7

8x 10

−4

Time (sec)

Rad

ial D

ispl

acem

ent (

cm)


Figure 309.12: The comparison between analytical solution and experimental result for radial displace-

ment


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0 5 10 15 20 25 300

1

2

3

4

5

6

7

8x 10

−4

Time (sec)

Rad

ial D

ispl

acem

ent (

cm)


Figure 309.13: The comparison between analytical solution and experimental result for radial displace-

ment


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Real-ESSI Lecture Notes 309.6. SHOCK WAVE PROPAGATION IN SATU . . . page: 1656 of 3100

Table 309.4: Simulation parameters used for the shock wave propagation verification problem.


Poisson ratio ν 0.3

Young’s modulus E 1.2× 106 kN/m2

Solid particle bulk modulus Ks 3.6× 107 kN/m2

Fluid bulk modulus Kf 2.17× 106 kN/m2

Solid density ρs 2700 kg/m3

Fluid density ρf 1000 kg/m3

Porosity n 0.4

Newmark parameter γ 0.6

309.6 Shock Wave Propagation in Saturated Porous Medium

In order to verify the dynamic behavior of the system, an analytic solution developed by Gajo (1995) and

Gajo and Mongiovi (1995) for 1C shock wave propagation in elastic porous medium was used. A model

was developed consisting of 1000 eight node brick elements, with boundary conditions that mimic 1D

behavior. In particular, no displacement of solid (ux = 0, uy = 0) and fluid (Ux = 0, Uy = 0) in x and y

directions is allowed along the height of the model. Bottom nodes have full fixity for solid (ui = 0) and

fluid (Ui = 0) displacements while all the nodes above base are free to move in z direction for both solid

and fluid. Pore fluid pressures are free to develop along the model. Loads to the model consist of a unit

step function (Heaviside) applied as (compressive) displacements to both solid and fluid phases of the

model, with an amplitude of 0.001 cm. The u–p–U model dynamic system of equations was integrated

using Newmark algorithm (see section 108.3). Table 309.4 gives relevant parameters for this verification.

Two set of permeability of material were used in our verification. The first model had permeability

set k = 10−6 cm/s which creates very high coupling between porous solid and pore fluid. The second

model had permeability set to k = 10−2 cm/s which, on the other hand creates a low coupling between

porous solid and pore fluid. Comparison of simulations and the analytical solution are presented in

Figure 309.14.

Before proceeding to the analysis, the following assumptions are made: For high-frequency compo-

nents, the permeability remains constant; thus, the dependency of the permeability on the frequency is

neglected. Unless specified, all the models in this report are elastic isotropic.


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Real-ESSI Lecture Notes 309.7. VERTICAL CONSOLIDATION OF A SOI . . . page: 1657 of 3100

4 6 8 10 12 140

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

time (µsec)

Sol

id D

ispl

. (x1

0−3 cm

)

4 6 8 10 12 140

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

time (µsec)

Flu

id D

ispl

. (x1

0−3 cm

)

K=10−6cm/s, FEM

K=10−6cm/s, Closed Form

K=10−2cm/s, FEM

K=10−2cm/s, Closed Form

Figure 309.14: Compressional wave in both solid and fluid, comparison with closed form solution.

309.7 Vertical Consolidation of a soil layer by Coussy (2004)

309.7.1 Brief review of Analytical Solution for Consolidation by Coussy (2004)

The consolidation process can be defined as follows: When an elastic soil layer is subjected to an external

change in mean normal stress, immediately the water will alone sustain this increment of mean normal

stress and cause the build-up the excessive pore water pressure. In the progress of the flow of the water

to the surface, the load is gradually transferred to the soil skeleton and the excessive pore water pressure

will dissipate. At the same time, the settlement of the soil layer occurs. As settlement is usually a major

concern in geotechnical engineering, this is a key problem in soil mechanics.

Consider a soil layer composed of an isotropic, homogeneous and saturated thermoporoelastic mate-

rial. The layer has a thickness of h in the Oy direction and of infinite extent in the two other directions

Ox and Oy. The layer is underlain by a rigid and impervious base at y = 0. And the top surface at

y = h is so perfectly drained that the pore pressure is held constant as zero.

At the initial state of the soil layer, the thermal effects are neglected so that the boundary conditions

follow that:

t ≤ 0→ y = h p = 0 (309.32)

y = 0∂p

∂z= 0 (309.33)

At time t = 0, tan instantaneous vertical load −$ey is suddenly applied on the top surface y = h,


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the induced boundary conditions require that

t > 0→ y = h σey = −$ey (309.34)

The undeformability of the substratum reads

y = 0 ξ = 0 (309.35)

The impermeability implies

y = 0 − w · ey = −wy = 0 (309.36)

The problem is then to determine the new fields of fluid pressure, stress and displacement induced

by the external loading.

Since this is a one-dimensional problem, the only non-zero displacement is the vertical displacement

ξy. But in particular the fluid pressure depends only on y and t.

ξ = ξy(y, t)ey p = p(y, t) (309.37)


σyy = (λ0 + 2µ)∂ξy

∂y+ bp (309.38)

σxx = σzz =λ0

(λ+ 2µ)σyy −

2µb

λ0 + 2µp (309.39)

And because the fluid pressure p must be an ordinary function of time t, although the derivative of

p is infinite at time t = 0 according to the consolidation equation (Coussy, 2004), the discontinuity of

the fluid pressure p at time t = 0 must satisfy


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p(y, t = 0+) = η$ $ =ν − ν0

(1− ν)(1− 2ν0)b(309.40)

where ν and ν0 are the drained and undrained Poisson ratio, respectively. For time t > 0, the vertical

stress σyy = −$ is constant in time and space, therefore the diffusion equation reads

t > 0 cm∂2

∂y2p =

∂

∂tp (309.41)

Collecting the above results, finally the fluid pressure reads

p(y, t) = η$∞∑n=0

4(−1)n

π(2n+ 1)cos[

(2n+ 1)π

2

y

h]exp[−(2n+ 1)2π2

4

t

τ] (309.42)

Each term of the series decreases exponentially with respect to the ratio tτ , in which τ is a charac-

teristics consolidation time

τ =h2

cmcm = kM

λ0 + 2µ

λ+ 2µ(309.43)

where λ and λ0 are the drained and undrained Lame coefficient, respectively. Given by the Eqn.309.39,

the only non-zero displacement ξy satisfies

∂ξy

∂y=

1

λ0 + 2µ(σyy + bp) (309.44)

By substituting the value of −$ of the vertical stress and expression of (309.44), the series converges

and it can integrated term by term yielding

ξy(y, t) =$

λ0 + 2µ(y

h+

8ηb

π2)

∞∑n=0

(−1)n

(2n+ 1)2sin[

(2n+ 1)π

2

y

h]exp(−(2n+ 1)2π2

4

t

τ] (309.45)

(309.46)

Using Eqn.309.46 and substitute y = h, the settlement can be expressed as

s(t) = s∞ + (s0+ − s∞)∞∑n=0

8

π2(2n+ 1)2exp−(2n+ 1)2π2

4

t

τ (309.47)

s0+ =h$

λ+ 2µs∞ =

h$

λ0 + 2µ(309.48)


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309.7.2 Numerical Analysis

A soil column of ten brick u − p − U finite elements is used to model the horizontal layer. The

height of the soil column is 10 m and the height of each element has dimensions 1m × 1m × 1m,

illustrated in Fig. 309.15. The material properties, shown in Table 309.5, are chosen as representative

values for the natural soil deposit. Developed u − p − U finite element model, can simulate realistic

compressibility of the pore fluid. However, it is important to note that the analytical solution for the

vertical consolidation is based on the assumption that both the soil particles and the pore fluid (water)

are completely incompressible. A uniform vertical pressure of 400 kPa is applied on the top surface of

the soil column. The numerical analysis was performed in two stages:

(I) Self-Weight application (see Figures 309.15 to 309.19).

(II) Consolidation with drainage at top due to uniform vertical pressure of 400 kPa at the surface

(see Figures 309.20 to 309.26).

Table 309.5: Material Properties used to study consolidation of a soil layer.


gravity acceleration g 9.81 m/s2

soil matrix Young’s Modulus E 10× 103 kN/m2

soil matrix Poisson’s ratio v 0.25

solid particle density ρs 2.65× 103 kg/m3

water density ρf 1.0× 103 kg/m3

solid particle bulk modulus Ks 37.0× 106 kN/m2

fluid bulk modulus Kf 2.2× 106 kN/m2

porosity n 0.46

Biot coefficient α 1.0

The following boundary conditions are applied to the model (Fig 309.27): As the bottom of the soil

column is modeled as an undeformable and impermeable layer, both the solid and fluid displacements

are fixed. The pore pressure is kept constant as zero at the top surface of the soil column because

of the perfectly drained condition. In order to simulate the 1D consolidation problem, all the lateral

movement of the solid and fluid phase are suppressed so that the vertical displacement is the only non-

zero displacement for the intermediate nodes. To capture both the long term (t >0.1 sec) and short term

(t <0.1 sec) response of the soil column, two different time steps are adopted: 0.1 sec and 0.005 sec,

respectively. In order to observe the dissipation of the excessive pore water pressure in a reasonable and


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NNUMERICAL UMERICAL MMODELODEL

u u –– p p –– U ElementsU Elements

u: soil skeleton displacements

p: pore water pressureU: pore water displacements

x

y

z

7 DOFs at each node(3u, p, 3U)

u = U

p = f (U, Kw)

1 m

10 m

Figure 309.15: Numerical Simulation of self weigh application and consolidation of a soil layer using

u-p-U brick elements.

convenient period, we select k = 1.0× 10−4m/s as the value for the permeability. To cure the artificial

oscillation, some numerical damping is introduced into the analysis by using γ = 0.6 and β = 0.3025

for the Newmark integrator. Fig. 309.20 illustrates the physical geometry of the problem whereas Fig.

309.21 shows the numerical modeling.

Based on the above parameters, the other relative parameters can be calculated as follows:

The bulk modulus of the mixture:

K =E

3(1− 2ν)= 6.67× 103kPa (309.49)

λ =Eν

(1 + ν)(1− 2ν)= 6.67× 103kPa µ =

E

2(1 + ν)= 4× 103kPa (309.50)

The Biot coefficient:

b = 1− K

Ks= 0.9998 (309.51)

The undrained bulk modulus of the mixture:

N =Ks

b− n = 6.85× 107kPa M =KfN

Kf +Nn= 4.47× 106kPa (309.52)


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10 m

0

1

2

3

4

5

6

7

8

9

10

0 20 40 60 80 100

Effective Stresses at Gauss points (kPa)

depth (m

)Gauss Pointsγeff = 8.74 kN/m3

Figure 309.16: The distribution of the vertical effective stresses with depth due to self-weight. The

stresses are calculated at the Gauss points within each brick element. The stresses obtained from the

numerical analysis are equal for all the Gauss points within the element, due to the compatibility of

deformations at the element interfaces.

Ku = K + b2M = 4.4× 106kPa (309.53)

The diffusion coefficient and characteristic time of consolidation:

cf =kM

γw

K + 4µ3

Ku + 4µ3

= 1.2m2/s t =h2

cf= 83.33s (309.54)

309.7.3 Discussion of Numerical Results - Conclusions

The stage of self weight application shows in Fig. 309.19 that the expected estimated settlement is

quite close to the one obtained from the analysis. The difference in the two values is due to the stress

distribution coming out of the analysis which is slightly different than the one considered in theory (see

Fig. 309.16) and possibly due to the compressibility of soil particles in the numerical analysis.

In Fig.309.24, the normalized fluid pressure is plotted against the location for various normalized

times. For normalized time Tv = 0.1, only the nodes close to the top free flow surface display the

dissipation of the pore pressure. The experimental result provide good agreement with the analytical


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10 m

0

1

2

3

4

5

6

7

8

9

10

0 50 100 150 200

Stresses (kPa)

depth (m

)Effective Stresses

Hydrostatic Pressures

Total Stresses

γ = 18.55 kN/m3

Figure 309.17: The distribution of the hydrostatic pore pressures, the effective and total stresses with

depth, obtained from the numerical analysis after self-weight application.

solution. With the increase of the normalized time, we can clearly see the tendency of the dissipation

of the water. At normalized time Tv = 1.0 (natural time t = 83 sec), the maximum normalized pore

pressure is only about 0.11. It can be concluded that the numerical analysis can effectively demonstrate

the process of the dissipation of the pore pressure.

In Fig. 309.26, the change of the porosity is predicted due to the consolidation of the soil layer.

Both the change of volume of the soil (0.1782m3) and the fluid (0.1794m3) have been calculated.

Theoretically, these two values should be the same, according to the fact that the settlement of the soil

layer is due the fluid which is squeezed out, assuming that the soil grains are incompressible. However,

this difference in the values is due to the compressibility of the soil grains, which was not considered

infinite in the numerical analysis.


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Real-ESSI Lecture Notes 309.8. ONE DIMENSIONAL WAVE PROPAGATI . . . page: 1664 of 3100

10 m

Porosity, n = 0.46

Vf = 4.6 m3 Vs = 5.2 m3

ρf g Vf = 46 kPa Gs gVs = 138 kPa

Total self load = 184 kPa

Effective Stress at 5 m = 42 Effective Stress at 5 m = 42 kPakPa

TTOTAL OTAL SSETTLEMENT ETTLEMENT DDUE UE TTO O SSELF ELF WWEIGHTEIGHT

Figure 309.18: Estimation of the total self load, as it is obtained from the numerical analysis, given the

porosity and the densities of the fluid (ρf ) and the grains (ρs or Gs).

309.8 One dimensional wave propagation in elastic porous media sub-

jected to step displacement boundary condition

309.8.1 Brief review of Analytical Solution by Gajo and Mongiovi (1995)

An one-dimensional exact analytical solution of the Biot’s equations is provided by Gajo and Mongiovi

(1995) for the completely general solution of the transient problem in saturated, linear, elastic, porous

media. The analytical solution was obtained was obtained by Fourier series. This solution is considered

to be completely general because it is not based on any assumptions with respect to the inertial, viscous

or mechanical coupling. Furthermore, it can be applied to any type of boundary-initial value problem.

The advantage of this analytical solution consists of showing the mechanics of dispersive wave

propagation in saturated elastic solids. This is achieved by allowing the detailed analysis of wave fronts

of the first and second kind of longitudinal waves and by analyzing accurately the effects of each term

of coupling on the transient behavior of saturated porous media. In particular, since each term of the

Fourier series represents a frequency component of the excitation signal, the analytical solution can


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10 m

Constraint Modulus, D =E (1‐v)

(1+v) (1‐2v)

D = 12 MPa

S = σ’H/D = 42x10/12000 = 0.035 m

The settlement from the analysis ⇒ 0.0363 m

TTOTAL OTAL SSETTLEMENT ETTLEMENT DDUE UE TTO O SSELF ELF WWEIGHTEIGHT

Figure 309.19: Prediction of the total settlement due to self-weight and comparison with the numerical

result.

describe the behavior of each frequency component. Thus, it can illustrate the mechanics of dispersive

wave propagation in which higher frequencies propagate with two waves and lower frequencies with only

one wave, as a function of permeability and travel length.

Considering the above mentioned arguments, the analytical solution can provide a useful comparative

term towards the verification and the validation of the existing numerical solutions based on the finite

element method. Such a study was conducted by Gajo et al. (1994), by comparing analytical results

with numerical ones obtained by a u− p− U numerical formulation.

In the paper by Gajo and Mongiovi (1995), the transient response of porous media is shown for

typical material properties of a natural granular deposit and for different degrees of viscous coupling.

Specifically, analytical results are given from the solution of the following one-dimensional boundary value

problem: at the top and bottom surfaces of a soil layer of finite thickness L, the excitation consisting of a

step displacement boundary condition (Heaviside function) is applied to both solid and fluid phases. This

problem can demonstrate better the mechanics of dispersive wave propagation, since the step excitation

contains waves of all kind of frequencies. The analytical solution is relative only to the first arrival of

the waves of the first and second kind.


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CCONSOLIDATION WITH ONSOLIDATION WITH SSINGLE INGLE DDRAINAGE AT RAINAGE AT TTOPOP

HE, v, k

poTime Factor:Time Factor: TTvv = = CCvv t / Ht / H22

Coefficient of Consolidation:Coefficient of Consolidation:

CCvv = k D / = k D / γγww

CCvv = 10= 10‐‐33 12000/10 = 1.2 12000/10 = 1.2

In our case:In our case:

tttotaltotal = H= H22/C/Cvv = 100/1.2 = 83.33 sec= 100/1.2 = 83.33 sec

z

Figure 309.20: The physical geometry of the problem.


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x

y

z

10 m

‐100 kN

Boundary Conditions as beforeBoundary Conditions as before……

‐100 kN

Load

time

Nodal load at top Nodal load at top ⇒⇒ total = 400kNtotal = 400kN

NewmarkNewmark Integration MethodIntegration Method((ββ = 0.3025; = 0.3025; γγ = 0.6)= 0.6)

Dissipation of vibrationsDissipation of vibrations……

Figure 309.21: Numerical simulation of 1D consolidation of a soil layer.


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0

0.2

0.4

0.6

0.8

1

1.2

0 20 40 60 80 100

time (sec)

Normalized

Excess Po

re

Pressures

z/H = 0.2

z/H = 0.5

z/H = 1

Quicker dissipation towards the surfaceQuicker dissipation towards the surface

Numerical Analysis

Figure 309.22: Time history of the normalized excess pore pressure for three different normalized depths

indicating faster dissipation close to the surface. The dissipation has practically been completed at

t = 83.33sec, as predicted.


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0

0.2

0.4

0.6

0.8

1

1.2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

time (sec)

Normalized

Excess Po

re

Pressures

Analytical Solution

Numerical Analysis

z/H = 0.2

z/H = 0.5

z/H = 1

Time factor Tv

Figure 309.23: Normalized excess pore pressures versus normalized time for three different normalized

depths. Comparison of numerical results with the analytical ones.


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0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Normalized Excess Pore Pressure

Normalized

dep

th

10 m

Tv = 0.1

Tv = 0.25

Tv = 0.5

Tv = 1

Analytical Solution

Numerical Analysis

Figure 309.24: Distribution of normalized excess pore pressures with normalized depth for four different

time factors. Comparison of numerical results with the analytical ones.


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‐0.40

‐0.30

‐0.20

‐0.10

0.00

0 20 40 60 80 100 120

time (sec)

Settlemen

t of soil ske

leton (m

)

Numerical Analysis

z = 0

z = 2 m

z = 5 m

z = 9 m

Figure 309.25: Time history of the settlements for four different depths.


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CCHANGE OF HANGE OF PPOROSITYOROSITY

The vertical displacement of the fluid (upwards)The vertical displacement of the fluid (upwards)⇒⇒ 0.39 m0.39 m

Fluid

46% 54%

Solid Part

Initial Porosity, n = 0.46Volume of fluid that escaped = 46% 0.39mVolume of fluid that escaped = 46% 0.39m

⇒⇒ 0.1794 m0.1794 m33

The settlement of the solid part from the analysis ⇒ 0.33 m

Change of volume of soil = 54% 0.33 mChange of volume of soil = 54% 0.33 m3 3

0.1782 m0.1782 m33

New Porosity, n = (4.6‐0.1794)/(10‐0.1782)

New Porosity, n = 0.45

Figure 309.26: Prediction of the change in porosity of the soil layer due to consolidation.


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x

y

z

10 m

SSELF ELF WWEIGHT EIGHT AAPPLICATIONPPLICATION

Boundary ConditionsBoundary Conditions

Base: all uuii, , UUii DOFs fixed, pp free

uuyy,u,uxx, , UUxx, , UUyy fixed at every node

Top: pp fixed (p = 0)

Self Weight is applied Self Weight is applied ……..

Figure 309.27: Boundary Conditions applied to the numerical model throughout the self-weight applica-

tion and the process of consolidation due to extra load applied to the surface of the soil column.


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Numerical examples for three different values of viscous coupling (k = 10−8cm3s/g, k = 10−6cm3s/g,

k = 10−5cm3s/g) were solved in order to verify the previously mentioned u−p−U formulation in a wide

range of drag. The numerical model used for the simulation of the 1C shock wave propagation consists

of 400 u-p-U brick finite elements of dimensions 0.01cm × 0.01cm × 0.01cm creating a soil column

4cm thick. Figure 309.28 illustrates the transition from the physical configuration of the problem to its

numerical simulation. Table 309.6 shows the soil properties of the numerical model.

0.01 cm

Figure 309.28: The numerical model used for the verification of the finite element implementation

through comparison with the analytical results provided by Gajo and Mongiovi (1995).

At the top surface of the soil column, a step displacement of 1.0 × 10−3cm is applied both to the

solid and the fluid phase. Only the vertical displacement is free. There is no lateral flow or displacement.

The degree of freedom related to the pore pressures is constrained at the top surface to be equal to the

atmospheric pressure and is free at the rest of the nodes. The base of the model is rigid and impervious.

This kind of excitation (Heaviside function) applied at the top of the model, results clearly in waves

of all kinds of frequency, first due to its nature and secondarily due to the way of its application. This

fact together with the great stiffness of the solid skeleton (see Table 309.6 require a very dense spatial


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Table 309.6: Soil Properties for 1C shock wave propagation for the problem by Gajo and Mongiovi

(1995).



soil matrix Young’s Modulus E 1.2× 106 kN/m2






porosity n 0.4


discretization. Here, 400 u-p-U brick finite elements of dimensions 0.01cm×0.01cm×0.01cm, following

similar discretization with Gajo et al. (1994). The time step, δt required needs to be limited to

δt <δh

v(309.55)

(309.56)

where v is the highest wave velocity. In our case, the temporal integration involves 800 steps of

2.0× 10−8sec, which allows a maximum wave velocity of 5.0× 105m/s.

Two different time integration methods where used: i) The Newmark integrator and ii) The Hilber-

Hughes-Taylor (HHT) Integrator. Sets of parameters, assuring unconditionally numerical stability, were

chosen for both integrators. For the case of Newmark integrator (see Figures 309.29 - 309.44), the

following sets of parameters were used:

a) γ = 0.5 and β = 0.25,

b) γ = 0.6 and β = 0.3025,

c) γ = 0.7 and β = 0.4.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

solid

displacem

ent (cm

)

FEM (Newmark, gamma=0.5; beta=0.25)FEM (Newmark, gamma=0.6; beta=0.3025)Analytical Solution

Figure 309.29: Time history of solid displacements of longitudinal waves at 1 cm below the surface.

Comparison of numerical results (FEM) with the analytical solution by Gajo and Mongiovi (1995) for

the case of viscous coupling (k = 10−8cm3s/g). Two different sets of Newmark parameters were used

for the numerical analysis.


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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

1.2

1.4x 10

‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

solid

displacem

ent (cm

)


Figure 309.30: A magnified view of Figure 309.29 illustrating the details of wave front of the longitudinal

wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)


Figure 309.31: Time history of fluid displacements of longitudinal waves at 1 cm below the surface.





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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

1.2

1.4x 10

‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)



wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

1.2

1.4x 10

‐3 k = 10‐6 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)







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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐6 cm3s/g

t [10‐6 sec]

solid

displacem

ent (cm)

FEM (Newmark, gamma=0.6; beta=0.3025)

FEM (Newmark, gamma=0.7; beta=0.4)

Analytical Solution


wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

1.2

1.4x 10

‐3 k = 10‐6 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)







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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

1.2

1.4x 10

‐3 k = 10‐6 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)



wave of first kind.

For the HHT integrator (see Figures 309.45 - 309.56), the following sets of parameters were used:

a) α = −0.1, γ = 0.6 and β = 0.3025,

b) α = −0.3, γ = 0.8 and β = 0.4225.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐5 cm3s/g

t [10‐6 sec]

solid

displacem

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)







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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1x 10

‐3 k = 10‐5 cm3s/g

t [10‐6 sec]

solid

displacem

ent (cm

)



wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6x 10

‐3 k = 10‐5 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)







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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6x 10

‐3 k = 10‐5 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)



wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

x 10‐3 FEM Vs Analytical Solution

t [10‐6 sec]

solid

displacem

ent (cm

)

FEM (k = 10‐5 cm3s/g)

FEM (k = 10‐6 cm3s/g)

FEM (k = 10‐8 cm3s/g)

Analytical Solution (k = 10‐5 cm3s/g)





three different values of viscous coupling. The Newmark set of parameters used for the numerical solution

was: γ = 0.6 and β = 0.3025.


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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

x 10‐3 FEM Vs Analytical Solution

t [10‐6 sec]

solid

displacem

ent (cm

)

FEM (k = 10‐5 cm3s/g)

FEM (k = 10‐6 cm3s/g)

FEM (k = 10‐8 cm3s/g)





wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6x 10

‐3 FEM Vs Analytical Solution

t [10‐6 sec]

fluid displacemen

t (cm

)

FEM (k = 10‐5 cm3s/g)

FEM (k = 10‐6 cm3s/g)

FEM (k = 10‐8 cm3s/g)






three different values of viscous coupling. The Newmark set of parameters used for the numerical solution

was: γ = 0.6 and β = 0.3025.


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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6x 10‐3 FEM Vs Analytical Solution

t [10‐6 sec]

fluid displacemen

t (cm

)

FEM (k = 10‐5 cm3s/g)

FEM (k = 10‐6 cm3s/g)

FEM (k = 10‐8 cm3s/g)





wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

solid

displacem

ent (cm

)

FEM (HHT, alpha=‐0.1)FEM (HHT, alpha=‐0.3)Analytical Solution


Comparison of numerical results (FEM) with the analytical solution by Gajo and Mongiovi (1995)

for the case of viscous coupling(k = 10−8cm3s/g). Two different sets of unconditional stable HHT

parameters were used for the numerical analysis.


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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

solid

displacem

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)



wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)



Comparison of numerical results (FEM) with the analytical solution by Gajo and Mongiovi (1995) for the

case of viscous coupling(k = 10−8cm3s/g). Two different sets of unconditional stable HHT parameters

were used for the numerical analysis.


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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)



wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐6 cm3s/g

t [10‐6 sec]

solid

displacem

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)







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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

solid

displacem

ent (cm

)



wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

x 10‐3 k = 10‐6 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)







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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐8 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)



wave of first kind.


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4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐5 cm3s/g

t [10‐6 sec]

solid

displacem

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)







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5 5.5 6 6.50

0.2

0.4

0.6

0.8

1

x 10‐3 k = 10‐5 cm3s/g

t [10‐6 sec]

solid

displacem

ent (cm

)



wave of first kind.


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4 6 8 10 12 14 160

0.5

1

1.5

2x 10

‐3 k = 10‐5 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)







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5 5.5 6 6.50

0.5

1

1.5

2x 10

‐3 k = 10‐5 cm3s/g

t [10‐6 sec]

fluid displacemen

t (cm

)



wave of first kind.


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Biot has shown than when dissipation is present, each frequency component propagated with its own

velocity. Thus, especially in the case of numerical solutions using a finite element procedure, the response

is very sensitive to the numerical damping introduced to the system. Generally, a drawback of all types

types of numerical solutions is the distortion and the smearing of the wave fronts, which are linked to the

highest frequency that is allowed by the computational grid and the numerical damping due to the time

integration method. The numerical results presented here, show a larger rise time than the analytical

solution, that it could potentially be improved by using a finer spatial and temporal discretization.

In particular, the dissipation of high frequency oscillations is achieved more efficiently by the Newmark

integrator that the HHT one. Due to the fact that the filtering of high frequencies is less in case of HHT

integrator, the smearing of the wave front of the first kind of longitudinal waves is not so extensive as in

the Newmark case. Obviously, for both cases, as the numerical damping increases by changing the sets

of parameters, the rise time of the water fronts increases too. It is also worth mentioning that the rise

time of the wave front of the second kind low-frequency longitudinal wave is even longer than that of

the first kind.

Figures 309.41 and 309.44 illustrate the comparative results for all the three different values of viscous

coupling using the Newmark integration method. In general, it is worth noting that the numerical results

are in good agreement with the main characteristics of the mechanics of dispersive wave propagation in

fully saturated, porous media, as indicated by the analytical results. For example, numerical results well

demonstrate that during the propagation of the first wave, the solid and fluid displacement are in phase

with each other, whereas during the propagation of second wave, the displacements of the two phases

are in opposition. Overall, the finite element solutions reproduce correctly the forms of wave propagation

for a wide range of permeability.


jected to step loading at the surface

309.9.1 Brief review of Analytical Solution by de Boer et al. (1993)

An analytical solution for the one-dimensional transient wave propagation fluid-saturated elastic porous

media is provided by de Boer et al. 1993. The fluid-saturated porous material is modeled as a two phase

system composed of an incompressible solid phase and an incompressible fluid phase. An exact analytical

solution is obtained via Laplace transform technique considering initial and boundary conditions, which

exhibits only one independent compressive wave in both the solid and fluid phases, as a result of the


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incompressibility constraint.

The problem configuration, which the analytical solution is addressed to, consists of an one-dimensional

infinitely long column, separated from the half-space of a fluid-saturated porous elastic skeleton material.

The motion of both the solid and the fluid materials is constrained to occur in the vertical direction.

Loading as a function of time, σ(z = 0, t) = f(t), is applied to the half space surface boundary by a

permeable punch with ideal permeability. Homogeneous pore distribution and free pore fluid surface are

assumed. The wave motion in the porous medium is expressed by the solid and fluid displacements or

the solid extra stresses, respectively, but it cannot be expressed by the pore pressure which is just the

Lagrangian multiplier corresponding to the incompressibility constraint of the medium.

In particular, in the paper by de Boer et al. (1993), the solid and the fluid displacements, the solid

skeleton extra stresses and the pore pressure are given with respect to time and with respect to different

depths in the soil column within the framework of three loading forms: i) sinusoidal, ii) step loading and

iii) impulsive loading. These results can be taken for a quantitative comparison to various numerical

solutions.


Numerical example for the step loading case was solved in order to verify the previously mentioned

u− p−U formulation. The numerical model used for the simulation of the 1C shock wave propagation

consists of 1000 u-p-U brick finite elements of dimensions 1cm × 1cm × 1cm creating a soil column

10m thick. Obviously, the numerical simulation of a semi infinite soil column is not possible; thus,

a soil column of thickness of 10 cm was considered adequate for the current problem configuration.

Figure 309.57 illustrates the transition from the physical configuration of the problem to its numerical

simulation. Table 309.7 shows the soil properties of the numerical model, which are the same with

those used for the analytical results presented in the paper by de Boer et al. (1993).

The only difference is noted on the elastic modulus, which was selected to be 20MN/m2 for the

numerical solution (FEM) instead of 30MN/m2, as it is mentioned in the previously mentioned paper.

This is due to the fact that the numerical results indicated that the results given in the paper correspond

to a soil column with elastic modulus equal to 20MN/m2 instead of 30MN/m2. Moreover, it should

be mentioned that the solid and fluid compressibility were given realistic values (see Table 309.7), which

practically means that the two constituents are incompressible.

At the top surface of the soil column, a step loading of σ(z = 0, t) = 3kN/m2 is applied to the solid

part, as a nodal load equally distributed to the four top nodes. The nodal load is expressed as:


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z

z = 0

σ(0,t) = 3 kPap(0,t) = 0

1000 brickelements

1 cm

FN = 7.5 x 10‐5 kNFN

FNFN

Numerical model

Excitation at top:Heaviside function


through comparison with the analytical solution provided by de Boer et al. (1993).

FN (z = 0, t) =σ(z = 0, t)×A

4=

3kN/m2 × 0.01m× 0.01m

4= 7.5× 10−5kN (309.57)

Only the vertical displacement is free. There is no lateral flow or displacement. The degree of

freedom related to the pore pressure is constrained at the top surface to be equal to the atmospheric

pressure, while it is free at the rest of the nodes. The base of the model is rigid and impervious.


of all kinds of frequency, first due to its nature and secondarily due to the way of its application. Due to

this fact, a fairly dense mesh of 1000 u-p-U brick finite elements of dimensions 1cm× 1cm× 1cm was

chosen. The time step, δt required needs to be limited to

δt <δh

v(309.58)

(309.59)


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Table 309.7: Soil Properties for 1C shock wave propagation for the problem by de Boer et al. (1993)



soil matrix Young’s Modulus E 20× 103 kN/m2


soil matrix Lame’s constant λ 5.55× 103 kN/m2

soil matrix shear modulus µ 8.33× 103 kN/m2





porosity n 0.33

Darcy’s permeability kD 0.01 m/s


1.0 × 10−4sec, which allows a maximum wave velocity of 100m/s. The propagation velocity can be

calculated by the following equation given by de Boer et al. (1993) and is equal to 90.7m/s.

v =

√n2(λ + 2µ)

n2(1− n)ρs + (1− n)2(nρf )= 90.7m/s (309.60)

The Newmark time integration method was used, which dissipates more efficiently the high fre-

quencies introduced in the system due to numerics than HHT integrator, as shown in section 4.3.

The following set of parameters was chosen, assuring unconditionally numerical stability: γ = 0.7 and

β = 0.4.


Figures 309.58 to 309.67 illustrate the comparative results between analytical and numerical solution.

In general, it is worth noting that the numerical results are in good agreement, with respect to time and

with respect to depth, with those obtained by the analytical solution. The responses of the medium due

to step loading are indicative of the consolidation process in case of a free pore water surface. The solid

moves downwards, indicating that settlement occurs and the fluid is squeezed out from the pore volume

creating an upward flow. During the consolidation process, the extra solid skeleton stresses increase with


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time at a certain depth. However, they decrease with the distance from the loading surface at a certain

time. In opposition to the extra solid skeleton stresses, the pore pressure decreases with time tending to

zero, while it increases with depth.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

0.01

0.02

0.03

0.04

0.05

t (sec)

solid

displacem

ent (cm

)

FEMAnalytical Solution

z = 1 mz = 0.4 m

z = 0 m

Figure 309.58: Time history of solid displacements at different depths due to step loading. Comparison

of numerical results (FEM) with the analytical solution by de Boer et al. (1993).

Overall, it is worth mentioning that the results obtained from the finite element procedure practically

coincide with the ones given by the analytical solution. The only difference is located to the pore pressure

(see Figures 309.64 to 309.67), where numerical response is oscillatory in contrast to the analytical

solution. This may be due to the high frequencies introduced to the system by the temporal and spatial

discretization and/or the compressibility of the solid and fluid phases. It should be mentioned again,

that in the analytical solution, the two constituents are assumed incompressible whereas in the numerical

model, the solid and fluid bulk moduli have realistic values (see Table 309.7. That is why in Figure

309.68, numerical examples with different values of fluid compressibility were solved. It is obvious that

the oscillations decrease as the the fluid becomes more and more compressible because the stiffness of

the system decreases and the high frequencies are dissipated faster. Moreover, Figure 309.68 indicated

that better quantitative agreement between the numerical and analytical solution is achieved when the

bulk modulus of the fluid is 2.2 × 106 kPa - realisic value - instead of 2.2 × 109 kPa, which would be

expected since the pore fluid is assumed to be incompressible in the framework of the analytical solution.


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Real-ESSI Lecture Notes 309.10. ONE DIMENSIONAL WAVE PROPAGAT . . . page: 1709 of 3100

0 1 2 3 4 5 60

0.005

0.01

0.015

0.02

0.025

depth (m)

solid

displacem

ent (cm

)FEMAnalytical Solution

t = 0.15 s

t = 0.1 s

Figure 309.59: Response of solid displacements versus depth at different time moments due to step

loading. Comparison of numerical results (FEM) with the analytical solution by de Boer et al. (1993).

Comparing the pore pressures obtained from these two cases, it can be observed that the pore pressure

generation in the more compressible fluid is slightly higher than that of the almost incompressible fluid

because of the existence of the oscillatory waves, as mentioned by (Zienkiewicz and Shiomi, 1984).


jected to step velocity boundary condition

309.10.1 Brief review of Analytical Solution by Hiremath et al. (1988)

Hiremath et al. (1988) present a solution of Biot’s dynamic equation of motion for one-dimensional

wave propagation in a fluid-saturated linear elastic isotropic soil using Laplace transformation followed

by numerical inversion. This study is considered to be an extension of the exact transient solution

presented by Garg et al. (1974) for two limiting cases of infinitely small and infinitely large viscous

coupling. In both cases, a soil column of finite dimension subjected to velocity boundary conditions was

analyzed, allowing for reflection of waves at the boundaries.

In particular, Hiremath et al. (1988) examines two cases allowing for weak and strong viscous coupling,


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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4‐0.1

‐0.08

‐0.06

‐0.04

‐0.02

0

t (sec)

fluid displacemen

t (cm


z = 0 m

z = 0.4 m

z = 1 m

Figure 309.60: Time history of fluid displacements at different depths due to step loading. Comparison


or else, as it is referred in the related paper, low and high drag, respectively. Moreover, two different

types of excitations were applied at the boundary surface in terms of solid and fluid velocity. In the

first case, a unit step boundary condition boundary condition was applied at the top surface for both

solid and fluid phases. In the second case, the fluid velocity specified at the boundary is different from

the specified solid velocity increasing gradually to unity over the time scale. The results obtained from

the numerical inversion allowed for six reflections of the fast compressional wave of first kind and two

reflections of the secondary slow longitudinal wave.

One of the most important observations which both Garg et al. (1974) and Hiremath et al. (1988)

concluded to, is that in case of strong viscous coupling (high drag), the material behaves as a single

continuum with internal dissipation and the two wave fronts tend to become a single one.

Hiremath et al. (1988) presented a comparison of finite element solution of Biot’s equations of motion

with one based on numerical inversion of the Laplace transform solution. It is explained in the paper that

a proper choice of element type and time domain integration is essential for capturing the results coming

from the semi-analytical solution. Moreover, the spatial discretization needs to be combined with an

appropriate temporal one, so that the wave does not traverse more than one element length during a

single time step. In detail, Hiremath et al. (1988) suggests 100 linear elements of 0.005 m length and


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0 1 2 3 4 5 6‐0.06

‐0.05

‐0.04

‐0.03

‐0.02

‐0.01

0

depth (m)

fluid displacemen

t (cm

)


t = 0.15 s

t = 0.1 s

Figure 309.61: Response of fluid displacements versus depth at different time moments due to step


986 time steps of size 10−6 sec.


Numerical examples for two extreme values of viscous coupling: a) high drag (k = 0.148×10−8cm3s/g)

and b)low drag (k = 0.148 × 10−2cm3s/g), were solved in order to verify the previously mentioned

u− p− U formulation by comparing the results with the semi-analytical solution provided by Hiremath

et al. (1988). The numerical model used for the simulation of the 1C shock wave propagation consists

of 100 u-p-U brick finite elements of dimensions 0.005m × 0.005m × 0.005m creating a soil column

50cm thick. Figure 309.69 illustrates the transition from the physical configuration of the problem to

its numerical simulation. Table 309.8 shows the soil properties of the numerical model.

At the top surface of the soil column, a step velocity of 1.0 × 10−2m/sec is applied both to the

solid and the fluid phase. Only the vertical translational degrees of freedom are free. The horizontal

translational degrees of freedom are constrained so that there is no lateral flow or displacement. The

base of the model is rigid and impervious.


of all kinds of frequency, first due to its nature and secondarily due to the way of its application. This


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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

‐2.5

‐2

‐1.5

‐1

‐0.5

0

t (sec)

solid

skeleton stress (k

Pa)


z = 1.2 m

z = 0.4 m

Figure 309.62: Time history of solid skeleton stresses at different depths due to step loading. Comparison


Table 309.8: Soil Properties for 1C shock wave propagation for example by Hiremath et al. (1988).



soil matrix Young’s Modulus E 23.21× 106 kN/m2






porosity n 0.18



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0 1 2 3 4 5 6 7 8 9 10‐3

‐2.5

‐2

‐1.5

‐1

‐0.5

0

depth (m)

solid

skeleton stress (k

Pa)


t = 0.35 s

t = 0.25 s

t = 0.15 s

Figure 309.63: Response of solid skeleton stresses versus depth at different time moments due to step


fact together with the great stiffness of the solid skeleton (see Table 309.8 require a very dense spatial

discretization. Here, 100 u-p-U brick finite elements of dimensions 0.005m × 0.005m × 0.005m were

chosen, following similar discretization with Hiremath et al. (1988). The time step, δt required needs to

be limited to

δt <δh

v(309.61)

(309.62)


5.0× 10−7sec, in comparison with 986 time steps of size 10−6sec, used by Hiremath et al. (1988). The

time integration method used was the Newmark integrator with parameters: γ = 0.6 and β = 0.3025.


Biot has shown than when dissipation is present, each frequency component propagates with its own

velocity. Thus, especially in the case of numerical solutions using a finite element procedure, the response


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0 0.05 0.1 0.15 0.2 0.25 0.30

0.5

1

1.5

2

2.5

3

3.5

t (sec)

pore pressure (kPa


z = 0.2 m

z = 0.4 m

z = 1 m

Figure 309.64: Time history of pore pressure at different depths due to step loading. Comparison of

numerical results (FEM) with the analytical solution by de Boer et al. (1993).

is very sensitive to the numerical damping introduced to the system. Generally, a drawback of all types

types of numerical solutions is the distortion and the smearing of the wave fronts, which are linked to the

highest frequency that is allowed by the computational grid and the numerical damping due to the time

integration method. The numerical results presented here, show some oscillations at the rough changes

in velocity due to reflection of wave fronts that could be possibly diminished by using a finer spatial and

temporal discretization.

Figures 309.70 to 309.77 illustrate the comparative results for both extreme cases of viscous cou-

pling. In general, it is worth noting that the numerical results are in good agreement with the main

characteristics of the mechanics of dispersive wave propagation in fully saturated, porous media, as in-

dicated by the semi-analytical results. For example, numerical results well demonstrate that for the case

of strong viscous coupling (high drag), the solid and fluid are in phase with each other, implying that

the two-phase material behaves as a single continuum. Overall, the finite element solutions reproduce

correctly the trends of wave propagation in both limiting cases of viscous coupling.


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0 1 2 3 4 5 6 7 80

0.5

1

1.5

2

2.5

3

3.5t = 0.01 sec

depth (m)

pore pressure (kPa

)


Figure 309.65: Response of pore pressure versus depth at t = 0.01sec due to step loading. Comparison



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0 1 2 3 4 5 6 7 80

0.5

1

1.5

2

2.5

3

3.5t = 0.05 sec

depth (m)

pore pressure (kPa

)





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0 1 2 3 4 5 6 7 80

0.5

1

1.5

2

2.5

3

3.5t = 0.1 sec

depth (m)

pore pressure (kPa

)





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0 0.05 0.1 0.15 0.2 0.25 0.3‐0.5

0

0.5

1

1.5

2

2.5

3

3.5Different fluid compressibilities

t (sec)

pore pressure (kPa

)

FEM (Kf = 2.2 104 kPa)

FEM (Kf = 2.2 106 kPa)

FEM (Kf = 2.2 109 kPa)

Analytical Solution (Kf = 2.2 106 kPa)

Figure 309.68: Time history of pore pressure at 1 m below the ground surface due to step loading for

different fluid compressibility. Comparison of numerical results (FEM) with the analytical solution by

de Boer et al. (1993).


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z

z = 0

vs(0,t) =vf(0,t) = 1 cm/s

100 brickelements

5 cm

vs = vf = 1 cm/s

Numerical model

Excitation at top:Heaviside function

z = 50 cm


through comparison with the semi-analytical results provided by Hiremath et al. (1988).


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10‐3

0

2

4

6

8

10

12x 10

‐3

t (sec)

solid

velocity (m

/s)

FEM

Analytical Solution

Figure 309.70: Time history of solid velocity at 10 cm below the surface. Comparison of numerical

results (FEM) with the semi-analytical solution by Hiremath et al. (1988) for the case of high drag

(k = 0.148× 10−8cm3s/g).


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10‐3

0

2

4

6

8

10

12x 10

‐3

t (sec)

solid

velocity (m

/s)

FEM

Analytical Solution



(k = 0.148× 10−8cm3s/g).


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10‐3

0

2

4

6

8

10

12x 10

‐3

t (sec)

fluid velocity (m

/s)

FEM

Analytical Solution

Figure 309.72: Time history of fluid velocity at 10 cm below the surface. Comparison of numerical


(k = 0.148× 10−8cm3s/g).


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10‐3

0

2

4

6

8

10

12x 10

‐3

t (sec)

fluid velocity (m

/s)

FEM

Analytical Solution



(k = 0.148× 10−8cm3s/g).


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10‐3

0

2

4

6

8

10

12x 10

‐3

t (sec)

solid

velocity (m

/s)

FEM

Analytical Solution



(k = 0.148× 10−2cm3s/g).


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10‐3

‐2

0

2

4

6

8

10

12x 10

‐3

t (sec)

solid

velocity (m

/s)

FEM

Analytical Solution



(k = 0.148× 10−2cm3s/g).


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10‐3

0

0.002

0.004

0.006

0.008

0.01

0.012

t (sec)

fluid velocity (m

/s)

FEM

Analytical Solution



(k = 0.148× 10−2cm3s/g).


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10‐3

‐2

0

2

4

6

8

10

12x 10

‐3

t (sec)

fluid velocity (m

/s)

FEM

Analytical Solution



(k = 0.148× 10−2cm3s/g).


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Chapter 310

Verification and Validation for Seismic

Wave Propagation Problems

(1989-2000-2004-2005-2008-2009-2010-2011-2017-2018-2019-2021-)

(In collaboration with Dr. Nima Tafazzoli, Dr. Matthias Preisig, Dr. Federico Pisano, Mr. Kohei Watanabe, and

Dr. Hexiang Wang)

1728

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310.2 Wavelet Seismic Signals

A wavelet is a wave with specific definitions and parameters. The amplitude of a wavelet usually starts

at zero, increases by time and ended up at zero again. Typically a wavelet can be plotted as a brief

oscillation such as a the small oscillation recorded by seismogram. There are different types of wavelets

each with their own properties used for specific purpose in signal processing. For specific purposes

different wavelets might be summed up to come up with new type of wave. A recently developed

wavelet analysis has become a powerful tool to analyze the soil-structure systems for transient loads

providing information both in time and frequency domains. In wavelet representation the basis functions

are localized and contained in finite time domains (Sarica and Rahman (2003)).

310.2.1 Ricker Wavelet

One type of wavelet motions is the Ricker wave (Ryan (1994), Mavroeidis and Papageorgiou (2003)).

The formulation of Ricker wavelet is shown in Equations (310.1):

R(t) = A (1− 2π2f2 t2) exp(−π2f2t2) (310.1)

where R(t) is the amplitude of the function in time, A if the maximum amplitude, and f is the peak

frequency on the wavelet’s frequency spectrum. Figure (310.1) shows the actual time history and fast

Fourier transform of Ricker wavelet, where A is taken as 1 and f is taken as 5Hz. As it is shown, the

frequency range of the motion is narrower compared to the real earthquake motion.

−0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Time [s]

Ric

ker

wav

elt a

mpl

itude

0 5 10 150

0.05

0.1

0.15

0.2

Frequency [Hz]

Ric

ker

wav

elet

FF

T a

mpl

itude

Figure 310.1: Frequency content and a time domain representation of Ricker wavelet


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Real-ESSI Lecture Notes 310.2. WAVELET SEISMIC SIGNALS page: 1730 of 3100

310.2.2 Ormsby Wavelet

Another example of interesting wavelet is called Ormsby wavelet (Ryan (1994)) which features a con-

trollable flat frequency content with formulation shown in Equation (310.2).

f(t) = A((πf4

2

f4 − f3sinc(πf4(t− ts))2 − πf3

2

f4 − f3sinc(πf3(t− ts))2)

− (πf2

2


2

f2 − f1sinc(πf1(t− ts))2)) (310.2)

where f1 and f2 define the lower range frequency band, f3 and f4 define the higher range frequency

band, A is the amplitude of the function, and ts is the time that maximum amplitude is happening, and

sinc(x) = sin(x)/x.

Figure (310.2) shows an example of Ormsby wavelet in time domain and frequency domain. In this

case, wave has a flat frequency range of 5Hz to 20Hz. Shown in Figure (310.3) is half of the Ormsby

wavelet in frequency domain which the frequency range starts from 0 and remains constant up to 20Hz.

This type of motion could be useful when low frequency range of motions are required for dynamic

analysis of the systems.

Such broad band signals could be used to assess different aspects of soil-structure systems and with

different incoming wave inclinations. While wavelet time domain motions are not the same as actual

earthquakes, the idea is to use them for dynamic analysis of soil-structure systems for possible problems

coming out of dynamic behavior, at different frequencies and for different energy input levels.

When used with the DRM, motions developed from different directions, different incident angles

and different energies, will create a full envelope of these motions, which then can be used to evaluate

performance based response of the soil-structure systems.


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Real-ESSI Lecture Notes 310.2. WAVELET SEISMIC SIGNALS page: 1731 of 3100

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0 1 2 3 4 5 6

Orm

sby

Fun

ctio

n A

mpl

itude

Time [s]

0

5e-05

0.0001

0.00015

0.0002

0.00025

0.0003

0.00035

0.0004

0.00045

0.0005

0 5 10 15 20 25 30

Orm

sby

FF

T A

mpl

itude

Frequency [Hz]

Figure 310.2: Frequency content and a time domain representation of an Ormsby wavelet, with constant

frequency content between 5Hz and 20Hz

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0 0.5 1 1.5 2 2.5 3

Orm

sby

Fun

ctio

n A

mpl

itude

Time [s]

0

0.0001

0.0002

0.0003

0.0004

0.0005

0.0006

0.0007

0.0008

0.0009

0.001

0 5 10 15 20 25 30Orm

sby

FF

T F

unct

ion

Am

plitu

de

Frequency [Hz]

Figure 310.3: Frequency content and a time domain representation of half of Ormsby wavelet formulation,

with minimum frequency of zero and maximum of 20Hz


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Real-ESSI Lecture Notes 310.3. FINITE ELEMENT MESH SIZE EFFECTS . . . page: 1732 of 3100

310.3 Finite Element Mesh Size Effects on Seismic Wave Propagation

Modeling and Simulation

310.3.1 Analysis Cases

Summary of the cases is shown in Table below. the input motion used is Ormsby wavelet which the

corner cutoff frequency is shown in the table.

Case NumberModel Height

(m)

Shear Wave

Velocity

(m/s)

Element Size

(m)

Frequncy

Cutoff (Hz)

Maximum Propagation

Frequency (Hz)

1 1000 1000 10 3 10

2 1000 1000 20 3 5

3 1000 1000 10 8 10

4 1000 1000 20 8 5

5 1000 1000 50 3 2

6 1000 1000 50 8 2

7 100 100 1 3 10

8 100 100 2 3 5

9 100 100 1 8 10

10 100 100 2 8 5

11 100 100 10 8 1

12 1000 100 10 8 1

13 1000 100 20 8 0.5

14 1000 100 50 8 0.2

310.3.2 Comparison of Case 1 and 2


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-0.001

-0.0005

0

0.0005

0.001

0.0015

0.002

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]

Figure 310.4: Displacement time history of input motion (Ormsby Wavelet)

-0.0015

-0.001

-0.0005

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0.004

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]

Element Size: 10m

Element Size: 20m

Figure 310.5: Comparison of displacement time histories of case 1 and 2 at top of the model


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0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0.00014

0.00016

0 2 4 6 8 10

FF

T A

mpl

itud

e

Frequency [Hz]

Input MotionElement Size: 10mElement Size: 20m

Figure 310.6: Comparison of FFT of case 1 and 2 at top of the model and input motion at the bottom

of model


310.3.4 Comparison of Cases 3, 4, and 6



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-0.002

-0.001

0

0.001

0.002

0.003

0.004

0.005

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]


-0.004

-0.002

0

0.002

0.004

0.006

0.008

0.01

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]

Element Size: 10m

Element Size: 20m



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0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0.00014

0.00016

0 2 4 6 8 10

FF

T A

mpl

itud

e

Frequency [Hz]



of model

-0.002

-0.001

0

0.001

0.002

0.003

0.004

0.005

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]



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-0.006

-0.004

-0.002

0

0.002

0.004

0.006

0.008

0.01

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]

Element Size: 10m

Element Size: 20m

Element Size: 50m

Figure 310.11: Comparison of displacement time histories of case 3, 4, and 6 at top of the model

0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0.00014

0.00016

0 2 4 6 8 10

FF

T A

mpl

itud

e

Frequency [Hz]

Input MotionElement Size: 10mElement Size: 20mElement Size: 50m

Figure 310.12: Comparison of FFT of case 3, 4, and 6 at top of the model and input motion at the

bottom of model


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-0.001

-0.0005

0

0.0005

0.001

0.0015

0.002

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]


-0.0015

-0.001

-0.0005

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0.004

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]

Element Size: 1m

Element Size: 2m



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Real-ESSI Lecture Notes 310.4. DAMPING OF THE OUTGOING WAVES page: 1739 of 3100

0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0.00014

0.00016

0 2 4 6 8 10

FF

T A

mpl

itud

e

Frequency [Hz]



of model

310.3.6 Comparison of Case 9, 10, and 11

310.3.7 Comparison of Case 12, 13, and 14

310.4 Damping of the Outgoing Waves

310.4.1 Comparison of Rayleigh Damping and Caughey 4th Order Damping

As mentioned before, Caughey damping in general will damp out the motions at specified modes (fre-

quencies) to be specified which could also be the natural frequencies of the system. Depending on the

type of damping to be used, the response of those modes would be affected. In order to observe the

damping effect on certain modes, a soil profile is made with thickness of 50m and shear wave velocity

of 100m/s. For input motion, an Ormsby wavelet with frequency range of 0 to 7 Hz is considered at

the base of model. The wave is propagated through the soil layer using the elastic transfer functions

and comparison is made between the case which Rayleigh damping is used versus the case which the

frequency independent damping is used in the model. The same procedure is done by using Caughey

damping.

Figures (310.22) and (310.23) show the base motion (Ormsby wavelet), motion at the surface


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-0.002

-0.001

0

0.001

0.002

0.003

0.004

0.005

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]


-0.004

-0.002

0

0.002

0.004

0.006

0.008

0.01

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time [s]

Element Size: 1m

Element Size: 2m

Element Size: 10m



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0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0.00014

0.00016

0 2 4 6 8 10

FF

T A

mpl

itud

e

Frequency [Hz]


Element Size: 10m


bottom of model

-0.002

-0.001

0

0.001

0.002

0.003

0.004

0.005

0 2 4 6 8 10 12 14 16 18 20

Dis

plac

emen

t (m

)

Time [s]



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-0.0015

-0.001

-0.0005

0

0.0005

0.001

0 2 4 6 8 10 12 14 16 18 20

Dis

plac

emen

t (m

)

Time [s]

Element Size: 10m

Element Size: 20m

Element Size: 50m


0

5e-06

1e-05

1.5e-05

2e-05

2.5e-05

3e-05

3.5e-05

0 2 4 6 8 10

FF

T A

mpl

itud

e

Frequency [Hz]

Input MotionElement Size: 10mElement Size: 20mElement Size: 50m


bottom of model


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considering the frequency independent damping, motion at the surface using Rayleigh wave (frequency

dependent), motion at the surface using Caughey damping of 4th order (frequency dependent) as well as

how Rayleigh and Caughey damping ratio change with frequency. It can be observed how the response

is affected at different modes using Rayleigh damping versus using Caughey damping.

0

0.05

0.1

0.15

0.2

0.25

0.3

0 1 2 3 4 5 6 7 8

Acc

eler

atio

n (g

)

Frequency (Hz)

Motion at the Base Motion at the surface (Damping = f(ω)) Motion at the surface (Damping = fixed)

Figure 310.22: Comparison of obtained motion at the surface using frequency independent damping and

frequency dependent Rayleigh damping.

310.4.2 Parametric Study on Effect of Rayleigh Damping on Reflected Waves

As mentioned in previous chapter, one of the issues of the modeling in dynamic analysis is reflecting

of the motions from the boundaries since there are limitations regarding the size of the problems we

can model. In order to reduce the computational cost of the problems, the size of the mesh has to be

reduced. By reducing the size of the model the chance of reflecting the motions from the boundaries

gets higher since there is less volume for the waves to get dissipated.

There are different ways to reduce reflection of the waves from the numerical boundaries such as PML,

viscous dampers, infinite elements, or considering Rayleigh damping for specific elements. Presented here

show the results of wave propagation models considering Rayleigh damping. There are different damping

patterns used here such as constant damping ratio for all the elements in the damping zone or linear

pattern of increasing the damping ratio.

In order to find the Rayleigh damping coefficients, two frequencies have to be considered. In these

examples both cases of using the natural frequencies of the soil column and also using the dominant


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0

0.05

0.1

0.15

0.2

0.25

0.3

0 1 2 3 4 5 6 7 8

Acc

eler

atio

n (

g)

Frequency (Hz)

Motion @ Base Motion @ Surface (Damping = f(ω)) Motion @ Surface (Damping = fixed)

Figure 310.23: Comparison of obtained motion at the surface using frequency independent and frequency

dependent Caughey 4th order damping.

periods of the motions are used and comparison is made. Different shear wave velocities and input motion

frequencies are used which is mentioned for each case. The input motion considered for simulation is

Ricker wavelet considering different dominant frequencies. Vs is the soil profile shear wave velocity and

fr is the frequency of the Ricker wavelet, and xi is the Rayleigh damping ratio at considered frequencies.

The height of the finite element model is 60m and boundary conditions are introduced in order to

model 1C wave propagation. The motion is imposed at one side of the model and Rayleigh damping is

applied to couple of the elements on the other side of the model in order to damp out the waves. Results

are recorded at the boundary of damped and undamped zones.

Figure (310.24) shows the comparison of time histories for the soil column with shear wave velocity of

100m/s and input motion frequency of 8Hz. Frequencies used to calculate Rayleigh damping coefficients

in this case are natural frequencies of the soil column. The same damping ratio is used for all the damping

zone elements. It can be observed that the one with constant damping ratio of 0.5 has done better job

in terms of damping out the reflected motions.

Same analysis is done by using frequencies of 6Hz and 12Hz for the Rayleigh damping. As shown in

Figure (310.25), in this case the reflected waves are damped out more comparing to previous case where

natural frequencies of the soil were used for Rayleigh damping. This fact shows that the frequencies to

be used for calculating the Rayleigh damping coefficients, do not have to be the natural frequencies of

the soil which sometimes used in practice and depends on the frequency range of the input motion as


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well. Since the Ricker wavelet used here has a peak frequency of 8Hz, the higher values of frequencies

should be used for Rayleigh damping coefficients. Figure (310.26) shows the results of the same case

except that linear increasing pattern is used for damping ratio of the 5 elements in damping zone. It

seems that using the linear pattern starting from 0.3 to 1.1 results in less reflected motions.

The reason could be because of the nature of Rayleigh damping which is frequency dependent. So

different damping ratios are observed at different frequencies. In deed by changing the damping ratio at

each element, five different patterns of Rayleigh damping are being used which has more capability of

absorbing motions with different frequencies and amplitudes.

-0.05-0.04-0.03-0.02-0.01

0 0.01 0.02 0.03

0 0.5 1 1.5 2 2.5 3Dis

plac

emen

t (m

)

Time (s)

xi=0.5xi=0.3xi=0.1

Figure 310.24: Displacement time history considering Rayleigh damping using natural frequencies of the

soil, Vs=100m/s , fr=8Hz

Figures (310.27) and (310.28) show the displacement time histories for same patterns of constant

damping ratio for all elements in the damping zone and linearly increasing damping ratios respectively

but for shear wave velocity of Vs = 300m/s and input motion frequency of 5Hz. Same conclusion can

be made here as previous case regarding the pattern of damping ratios and frequencies to be used for

Rayleigh damping.

Figures (310.29) and (310.30) are comparisons of recorded displacement time histories between

patterns of same damping ratio for damping zone elements, damping ratio changes along the length of

damping zone, and case of with out damping. Figure (310.29) is the case which shear wave velocity of

the soil column is 100m/s and frequency of input Ricker motion is 1Hz while soil profile used in Figure

(310.30) has shear wave velocity of 300m/s with input motion frequency of 8Hz.

It can be observed that in case of having no physical damping, waves are getting trapped in the model

and are reflecting back from the boundaries. Displacement time histories obtained from mentioned


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-0.05-0.04-0.03-0.02-0.01

0 0.01 0.02 0.03

0 0.5 1 1.5 2 2.5 3Dis

plac

emen

t (m

)

Time (s)

xi=0.5xi=0.3xi=0.1

Figure 310.25: Displacement time history considering Rayleigh damping using f1,f2 = 6,12 Hz ,

Vs=100m/s , fr=8Hz

-0.05-0.04-0.03-0.02-0.01

0 0.01 0.02 0.03

0 0.5 1 1.5 2 2.5 3Dis

plac

emen

t (m

)

Time (s)

xi=0.1, 0.2, 0.3, 0.4, 0.5xi=0.1, 0.3, 0.5, 0.7, 0.9xi=0.2, 0.4, 0.6, 0.8, 1.0xi=0.3, 0.5, 0.7, 0.9, 1.1


Vs=100m/s , fr=8Hz

patterns of damping ratios have minor differences which does not mean always will be this close but

still the pattern of using linearly increasing of damping ratio seems to do a better job for damping the

reflecting waves.

In order to have a better understanding of these patterns of damping, wave propagation through the

depth of model is recorded for case of shear wave velocity of 100m/s and input motion frequency of 8Hz.

Displacement time histories in Figures (310.31) to (310.33) show wave propagation through the model

for cases of using uniform damping ratios, linearly increasing damping ratios, and with out damping


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-0.06-0.05-0.04-0.03-0.02-0.01

0 0.01 0.02 0.03

0 0.5 1 1.5 2 2.5 3Dis

plac

emen

t (m

)

Time (s)

xi=0.5xi=0.3xi=0.1


Vs=300m/s , fr=5Hz

-0.06-0.05-0.04-0.03-0.02-0.01

0 0.01 0.02 0.03

0 0.5 1 1.5 2 2.5 3Dis

plac

emen

t (m

)

Time (s)

xi=0.1, 0.2, 0.3, 0.4, 0.5xi=0.1, 0.3, 0.5, 0.7, 0.9xi=0.2, 0.4, 0.6, 0.8, 1.0xi=0.3, 0.5, 0.7, 0.9, 1.1


Vs=300m/s , fr=5Hz

respectively. By looking at the wave propagation through the whole soil profile it can be concluded that

for this soil profile using the linearly increasing of damping ratios does a better job for damping the

reflected motions at different depths.

Figure (310.34) shows the comparison of cumulative total energy time histories for the soil profile

with shear wave velocity of 100m/s and input motion frequency of 8Hz for different Rayleigh damping

patterns of uniform, increasing linearly, and case of no damping. What is expected to be observed is that

total energy keeps increasing until the input motion gets to zero in time which energy should remain


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-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

plac

emen

t (m

)

Time (s)

xi=0.5xi=0.3, 0.5, 0.7, 0.9, 1.1

No Damping

Figure 310.29: Displacement time history considering Rayleigh damping using f1,f2 = 0.5,2 Hz ,

Vs=100m/s , fr=1Hz

-0.05

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 0.5 1 1.5 2 2.5 3

Dis

plac

emen

t (m

)

Time (s)

xi=0.5xi=0.3, 0.5, 0.7, 0.9, 1.1

No Damping


Vs=300m/s , fr=8Hz

constant unless there are waves reflecting back from boundaries. As it is shown, total energy slightly

increase by time due to the reflected motions. This difference is much higher for case of no physical

damping used since higher portion of the motions will get trapped in the model.

In order to be able to see the effect of size of damping zone on reflected motions, analysis is done on

the soil profile with shear wave velocity of 100m/s and frequency of 8Hz for input motion. Comparison

of displacement time histories for different size of the damping zones is shown in Figure (310.35). As

expected, by reducing the size of damping zone, more waves are reflecting back from model boundaries.

The effect of number of elements to be used in damping zone is also studied here. Comparison is


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Figure 310.31: Displacement time history in depth considering homogeneous damping (15m), xi=0.5,

Vs=100m/s, fr=8Hz

made for cases which the size of the damping zone is the same but the size of the elements (and therefore

number of the elements) in that zone is changed. The size of the damping zone assumed to be 15m

while the number of the elements used in that zone is considered to be 3, 5, and 15. The comparison

for this change of number of the elements is shown in Figure (310.36). Rayleigh damping ratio with

pattern of increasing linearly from 0.3 to 1.1 is used. As it is observed, by reducing number of elements

in the damping zone, the amount of reflected waves are getting higher.


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Figure 310.32: Displacement time history in depth considering linear increasing of xi in Rayleigh damping

(every 3m), xi=0.3, 0.5, 0.7, 0.9, 1.1, Vs=100m/s, fr=8Hz


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Figure 310.33: Displacement time history in depth with out considering damping, Vs=100m/s, fr=8Hz

0 5

10 15 20 25 30 35 40 45

0 0.5 1 1.5 2 2.5 3Ene

rgy

(kJ/

m/m

)

Time (s)

xi=0.5xi=0.3, 0.5, 0.7, 0.9, 1.1

No Damping

Figure 310.34: Comparison of energy time history by considering different Rayleigh damping patterns

(in the non-damping zone at the middle of model), Vs=100m/s, fr=8Hz


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-0.05-0.04-0.03-0.02-0.01

0 0.01 0.02 0.03 0.04

0 0.5 1 1.5 2 2.5 3Dis

plac

emen

t (m

)

Time (s)

15m of damping layer10m of damping layer

5m of damping layer2m of damping layer

Figure 310.35: Comparison of displacement time histories for different size of damping zones

-0.05-0.04-0.03-0.02-0.01

0 0.01 0.02 0.03

0 0.5 1 1.5 2 2.5 3Dis

plac

emen

t (m

)

Time (s)

15m (15 Elements)15m (5 Elements)15m (3 Elements)

Figure 310.36: Displacement time history at a point in the non-damping zone close to the boundary of

imposing motion


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Real-ESSI Lecture Notes 310.5. MESH SIZE EFFECTS FOR LINEAR (8 . . . page: 1753 of 3100

310.5 Mesh Size Effects for Linear (8 Node Brick) and Quadratic (27

Node Brick) Finite Elements on Wave Propagation

Generally, the results of numerical analysis using finite element method technique for the dynamic problem

are affected by size of mesh (grid spacing). According to Argyris and Mlejnek (1991), about 10 nodes

per wavelengths are required to simulate accurately for the given frequency and fewer than 10 nodes

may induce an artificial damping due to the numerical reason.

Figure 310.37: One dimensional column test model to inspect the mesh size effect

In this section, mesh size effect is inspected to decide an appropriate size of the mesh to build finite

element model for verification. One dimensional column model is built as shown in figure 310.37. Total

height of the model is 1000 m. Two models are built with element height of 20 m and 50 m, and each

model have two different shear wave velocities (100 m/s and 1000 m/s). Density is set as 2000 kg/m3,

and Poisson’s ratio is set as 0.3, for all test models. Various cases are set and tested as shown in table

310.1. Both 8 node and 27 node brick elements are used for all models. Thus, total 24 parametric study

cases are inspected. Linear elastic elements are used for all analyses. All analyses are performed in time

domain with Newmark dynamic integrator without any numerical damping (γ = 0.5, and β = 0.25, no

numerical damping, unconditionally stable).

Ormsby wavelet (Ryan, 1994) is used as an input motion and imposed at the bottom of the model.


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Table 310.1: Analysis cases to determine a mesh size

Case Vs Cutoff Element Max. propagation

number (m/s) freq. (Hz) height (m) freq. (Hz)

1 1000 3 10 10

2 1000 8 10 10

3 1000 15 10 10

4 100 3 10 1

5 100 8 10 1

6 100 15 10 1

7 1000 3 20 5

8 1000 8 20 5

9 1000 15 20 5

10 100 3 20 0.5

11 100 8 20 0.5

12 100 15 20 0.5

Figure 310.38: Ormsby wavelet in time and frequency domain with flat frequency content from 5 Hz to

20 Hz

Ormsby wavele features a controllable flat frequency content with formulation shown in equation 310.3.

f(t) = A((πf4

2


2

f4 − f3sinc(πf3(t− ts))2)

− (πf2

2


2

f2 − f1sinc(πf1(t− ts))2) (310.3)

where f1 and f2 define the lower range frequency band, f3 and f4 define the higher range frequency

band, A is the amplitude of the function, and ts is the time that maximum amplitude is happening, and


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sinc(x) = sin(x)/x. Figure 310.38 shows an example of Ormsby wavelet with flat frequency content

from 5 Hz to 20 Hz.

Figure 310.39: Number of nodes per wavelength along frequencies, and element sizes (a) Vs = 1000

m/s (b) Vs = 100 m/s

For this example, cutoff frequencies of Ormsby wavelets are set as 3, 8, and 15 Hz (table 310.1).

Figure 310.39 shows number of nodes per wavelength along frequencies and figure 310.40 – 310.46

show comparison of analysis results. As shown in figure 310.40, case 1 and 7 (analysis using Ormsby

wavelet with 3 Hz cutoff frequency) predict exactly identical results to the analytic solution in both time

and frequency domain. Since, number of nodes per wavelength for both cases are over 10 (see figure

310.39(a) and table 310.1, all cases under 3 Hz shows more than 10 nodes per wavelength), those exact

results are expected.

Increasing cutoff frequency from 3 Hz to 8 Hz induces numerical errors as shown in figure 310.41.

In frequency domain, both 10 m and 20 m element height model with 27 node brick element predict

exactly same results with the analytic one. However, in time domain, asymmetric shape of time history


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0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0Time (s)

−0.002

−0.001

0.000

0.001

0.002

0.003

0.004

Dis

pla

cem

ent

(m) Input

Analytic solution at top

Observed at top (8 node)


0 1 2 3 4 5 6Frequency (Hz)

0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input




0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0Time (s)

−0.002

−0.001

0.000

0.001

0.002

0.003

0.004

Dis

pla

cem

ent

(m) Input





0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input




Figure 310.40: Comparison between (a) case 1 (top, Vs = 1000 m/s, 3 Hz, element size = 10m) and

(b) case 7 (bottom, Vs = 1000 m/s, 3 Hz, element size = 20m)


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0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0Time (s)

−0.004

−0.002

0.000

0.002

0.004

0.006

0.008

0.010

Dis

pla

cem

ent

(m) Input




0 2 4 6 8 10 12 14 16Frequency (Hz)

0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input




0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0Time (s)

−0.004

−0.002

0.000

0.002

0.004

0.006

0.008

0.010

Dis

pla

cem

ent

(m) Input




0 2 4 6 8 10 12 14 16Frequency (Hz)

0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input







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0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0Time (s)

−0.010

−0.005

0.000

0.005

0.010

0.015

0.020

Dis

pla

cem

ent

(m) Input




0 5 10 15 20 25 30Frequency (Hz)

0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input




0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0Time (s)

−0.010

−0.005

0.000

0.005

0.010

0.015

0.020

Dis

pla

cem

ent

(m) Input




0 5 10 15 20 25 30Frequency (Hz)

0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input







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Figure 310.43: Schematic cartoon to show that nodes can behave like frequency contents filter

displacements are observed. Observations from top of 8 node brick element models show more numerical

error in both time and frequency domain due to the decreasing number of nodes per wavelength (figure

310.39). Figure 310.42 shows analysis results with 15 Hz cutoff frequency. Results from 27 node brick

elements are almost same in frequency domain but asymmetric shapes are also observed in time domain.

Decreasing amplitudes in frequency domain along increasing frequencies are observed from 8 node brick

element cases.

Figure 310.44 – 310.46 show results predicted from Vs = 100 m/s cases. Similar as Vs = 1000

m/s cases, decreasing amplitude along increasing frequencies are observed in all cases. One interesting

observation is bumps in frequency domain which can be seen at natural frequencies (natural modes) of

the elements (n th mode of elements, f = (2n−1)V s/4H, 2.5 Hz, 5.0 Hz, and so on). This observation

may mean that if certain condition is satisfied between modes and size of the element, it will behave like

frequency contents filter. Figure 310.43 shows possible explanation of this observation. As in the case

of figure 310.43, nodes (circle in the figure) cannot capture harmonic oscillation of the frequency since

amplitude of the oscillation is always zero. As a result, the frequency contents at the frequency cannot

be predicted by the analysis.

The results shown here are used as a reference to determine mesh size and frequency range of input

motions for the verification.


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0 5 10 15 20Time (s)

−0.002

−0.001

0.000

0.001

0.002

0.003

0.004

Dis

pla

cem

ent

(m) Input





0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input




0 5 10 15 20Time (s)

−0.002

−0.001

0.000

0.001

0.002

0.003

0.004

Dis

pla

cem

ent

(m) Input





0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input




Figure 310.44: Comparison between (a) case 4 (top, Vs = 100 m/s, 3 Hz, element size = 10m) and (b)

case 10 (bottom, Vs = 100 m/s, 3 Hz, element size = 20m)Jeremic et al. University of California, Davis version: 7. May, 2021, 13:33

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0 5 10 15 20Time (s)

−0.004

−0.002

0.000

0.002

0.004

0.006

0.008

0.010

Dis

pla

cem

ent

(m) Input




0 2 4 6 8 10 12 14 16Frequency (Hz)

0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input




0 5 10 15 20Time (s)

−0.004

−0.002

0.000

0.002

0.004

0.006

0.008

0.010

Dis

pla

cem

ent

(m) Input




0 2 4 6 8 10 12 14 16Frequency (Hz)

0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input




Figure 310.45: Comparison between (a) case 5 (top, Vs = 100 m/s, 8 Hz, element size = 10m) and (b)

case 11 (bottom, Vs = 100 m/s, 8 Hz, element size = 20m)Jeremic et al. University of California, Davis version: 7. May, 2021, 13:33

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0 5 10 15 20Time (s)

−0.005

0.000

0.005

0.010

0.015

0.020

Dis

pla

cem

ent

(m) Input




0 5 10 15 20 25 30Frequency (Hz)

0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input




0 5 10 15 20Time (s)

−0.005

0.000

0.005

0.010

0.015

0.020

Dis

pla

cem

ent

(m) Input




0 5 10 15 20 25 30Frequency (Hz)

0.00

0.01

0.02

0.03

0.04

0.05

0.06

FFT a

mplit

ude

Input





(b) case 12 (bottom, Vs = 100 m/s, 15 Hz, element size = 20m)Jeremic et al. University of California, Davis version: 7. May, 2021, 13:33

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Real-ESSI Lecture Notes 310.6. VERIFICATION OF THE SEISMIC INPUT . . . page: 1763 of 3100

310.6 Verification of the Seismic Input (Domain Reduction Method)

for 3C, Inclined Seismic Wave Fields

310.6.1 Inclined, 3C Seismic Waves in a Free Field

In this section verification of the 3C wave propagation problem using Domain Reduction Method will

be studied. In order to do so, a finite element model with dimensions of 10000m × 50m × 5000m

is considered. Two cases are studied here with the source of motion (fault) to be located at (x =

3000m, y = 0, z = 3000m) and (x = 3000m, y = 0, z = 3000m). Figures (310.47) and (310.48) show

these two models respectively.

0m 5000m 10000m

5000m

0m

2000m

2000

m

Fault

Figure 310.47: Domain to be analyzed for the 1st stage of DRM with fault located at an angle of 45

with respect to the top middle point of the model

0m 5000m 10000m

5000m

0m

2000

m

3000mFault

Figure 310.48: Domain to be analyzed for the 1st stage of DRM with fault located at an angle of 34

with respect to the top middle point of the model


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The size of the elements is chosen to be 50m in all directions for both cases in order to reduce the

computational time. The soil parameters are: shear wave velocity of 700m/s, density of 1800kg/m3,

and Poisson’s ration of 0.1. Analyses for the fault slip model are done by applying the motion at the

nodes of one element. This is done in order to represent the the wave propagation starting from the fault

using Multiple Support Excitation. This is representing the first stage of analysis of DRM in which a big

model including the fault is considered for free field case in order to obtain the required motions for DRM

layer. For simulating the second stage of DRM, a smaller model with dimensions of 240m× 5m× 70m

is considered as shown in Figure (310.49). The size of the plastic bowl is 200m × 5m × 50m. Size of

the elements for this model is chosen to be 5m.

0m

0m

70m

240m

DRM Layer

Figure 310.49: Domain to be analyzed for the 2nd analysis stage of DRM with smaller size comparing

to the original model

Displacement and acceleration time histories of corresponding nodes of DRM layer are obtained by

interpolating between the the results obtained from the first model. These displacement and accelerations

are used to calculate the effective forces as an input for DRM analysis. Input motions to be used here are

Ricker wave, Morgan Hill, and Kocaeli earthquakes. The maximum allowable frequency to be propagated

through this model can be calculated based on Equation (310.4):

∆h ≤ λ/10 = Vs/(10 fmax) (310.4)

Based on the shear wave velocity of 700m/s and element size of 50m, maximum allowable frequency

to be propagated through this model would be 1.4Hz for the original model and based on element size

of 5m would be 14Hz for the DRM model.


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310.6.1.1 Ricker Wavelets

Figure (310.50) show the displacement time history and FFT of Ricker wave of 2nd order with dominant

frequency of 1Hz and maximum amplitude occurring at 1 second.

-0.005

-0.004

-0.003

-0.002

-0.001

0

0.001

0.002

0.003

0 1 2 3 4 5 6

Dis

plac

emen

t (m

)

Time (s)

0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0.00014

0.00016

0 2 4 6 8 10

Ric

ker2

nd F

FT

Am

plitu

de

Frequency (Hz)

Figure 310.50: Displacement time history and FFT of Ricker wave with dominant frequency of 1Hz

The first case to be studied here is the one with the fault source located at (x = 3000m, y = 0, z =

3000m) which has the angle of 45 with respect to the top middle point of the model. Results to be

discussed here are comparison of displacement and acceleration time histories at the top middle point

of the model (x = 5000m, y = 0, z = 5000m) between the fault slip and DRM models. Comparison of

displacement time histories in X and Z directions are shown in Figure (310.51). As it can be observed,

the results of DRM model matches perfectly with the ones obtained from the fault slip model.

Figure (310.53) is the displacement and acceleration time history of a point located outside of DRM

layer in X direction (x = 10m, y = 0, z = 40m). As mentioned before in definition of DRM, no motion

should come out of the DRM layer in case of free field. As shown in these figures, displacement and

acceleration time histories at this point are zero which verifies this fact.

The same motion is applied to the model with fault source located at (x = 2000m, y = 0, z = 3000m)

which has the angle of 34 with respect to the top middle point of the model. Displacement time histories

of the top middle point show the perfect match between results obtained from fault slip model with the

ones obtained from DRM mode.

As shown in Figure (310.55), the second motion to be used for analysis is Ricker wave with frequency

of 0.5Hz and maximum amplitude occurring at 3 seconds. Figure (310.56) shows the displacement time

histories of X and Z directions for the same point as before (x = 5000m, y = 0, z = 5000m). As it

is shown, results of the fault slip and DRM model are the same which verifies the solution from DRM

formulation for this motion as well.


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(X) (Z)

-0.0004

-0.0003

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0.0004

0 2 4 6 8 10 12

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

-0.0004

-0.0003

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0.0004

0 2 4 6 8 10 12

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

Figure 310.51: Comparison of displacements for top middle point using Ricker wave (f = 1Hz) as an

input motion

(X) (Z)

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0 2 4 6 8 10 12

Acc

eler

atio

n (m

/s2 )

Time (s)

DRMFault Slip Model

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0 2 4 6 8 10 12

Acc

eler

atio

n (m

/s2 )

Time (s)

DRMFault Slip Model

Figure 310.52: Comparison of accelerations for top middle point using Ricker wave (f = 1Hz) as an

input motion

The third motion to be used is Ricker wave with frequency of 2Hz and maximum amplitude happening

at 1 second as shown in Figure (310.57). Comparison of displacement time histories between the fault

slip and DRM model has been done and shown in Figure (310.58) along X and Z directions respectively.

In this case, results do not match for the top middle point of the model. The main reason is due to

the frequency of the motion. The maximum allowable frequency to be propagated in the fault slip model

is 1.4Hz while it is 14Hz in DRM model. Dominant frequency of the Ricker wave as input motion


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Displacement Acceleration

-0.0004

-0.0003

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0.0004

0 2 4 6 8 10 12

Dis

plac

emen

t (m

)

Time (s)

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0 2 4 6 8 10 12

Acc

eler

atio

n (m

/s2 )

Time (s)

Figure 310.53: Displacement and acceleration time history for a point outside of DRM layer in (x)

direction

(X) (Z)

-0.0004

-0.0003

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0.0004

0 2 4 6 8 10 12

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

-0.0004

-0.0003

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0.0004

0 2 4 6 8 10 12

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model


input motion

is 2Hz. Frequencies above the 1.4Hz can not be propagated in the fault slip model while they will

propagate in the DRM model. this can change the characteristics of the motion propagating through

the model and is the main reason of differences between the obtained results.


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-0.005

-0.004

-0.003

-0.002

-0.001

0

0.001

0.002

0.003

0 1 2 3 4 5 6

Dis

plac

emen

t (m

)

Time (s)

0

5e-05

0.0001

0.00015

0.0002

0.00025

0.0003

0 2 4 6 8 10

Ric

ker2

nd F

FT

Am

plitu

de

Frequency (Hz)

Figure 310.55: Displacement time history and FFT of Ricker wave with dominant frequency of 0.5Hz

(X) (Z)

-0.0003

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0 1 2 3 4 5 6 7 8 9 10

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

-0.0003

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0 1 2 3 4 5 6 7 8 9 10

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

Figure 310.56: Comparison of displacements for top middle point using Ricker wave (f = 0.5Hz) as an

input motion


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-0.005

-0.004

-0.003

-0.002

-0.001

0

0.001

0.002

0.003

0 1 2 3 4 5 6

Dis

plac

emen

t (m

)

Time (s)

0

1e-05

2e-05

3e-05

4e-05

5e-05

6e-05

7e-05

8e-05

0 2 4 6 8 10

Ric

ker2

nd F

FT

Am

plitu

de

Frequency (Hz)

Figure 310.57: Displacement time history and FFT of Ricker wave with dominant frequency of 2Hz

(X) (Z)

-0.0004

-0.0003

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0.0004

0 1 2 3 4 5 6 7 8 9 10

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

-0.0004

-0.0003

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0.0004

0 1 2 3 4 5 6 7 8 9 10

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model


input motion


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310.6.2 Vertical (1C) Seismic Waves in a Free Field

310.6.2.1 Morgan Hill and Kocaeli Earthquakes

In order to investigate more, Morgan Hill and Kocaeli earthquakes are used as an input motions for the

same models as before. These earthquakes were recorded during the ground shaking and obtained from

PEER motion database. Figure (310.59) shows the acceleration time history and FFT of Morgan Hill

earthquake with major frequency range of up to 4Hz. Acceleration time history and FFT of Kocaeli

earthquake are shown in Figure (310.60). Major part of the frequency range for Kocaeli earthquake is

up to frequency of 4Hz.

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0 1 2 3 4 5

Acc

eler

atio

n (g

)

Time (s)

0

0.002

0.004

0.006

0.008

0.01

0.012

0 2 4 6 8 10 12 14

FF

T A

mpl

itude

(gs

)

Frequency (Hz)

Figure 310.59: Acceleration time history and FFT of Morgan Hill earthquake

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0 5 10 15 20 25

Acc

eler

atio

n (g

)

Time (s)

0

0.002

0.004

0.006

0.008

0.01

0.012

0 2 4 6 8 10 12 14

FF

T A

mpl

itude

(gs

)

Frequency (Hz)

Figure 310.60: Acceleration time history and FFT of Kocaeli earthquake

Figure (310.61) shows the displacement time histories of the top middle point of the model for


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Morgan Hill earthquake while the ones from Kocaeli earthquake are shown in Figure (310.62). As it is

observed, results of fault slip model and DRM model do not match since the majority of the energy in

the earthquake is in the range of up to 4Hz which is higher than the maximum allowable frequency to

be propagated in the original model (1.4Hz).

(X) (Z)

-0.00012

-0.0001

-8e-05

-6e-05

-4e-05

-2e-05

0

2e-05

4e-05

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

-7e-05

-6e-05

-5e-05

-4e-05

-3e-05

-2e-05

-1e-05

0

1e-05

2e-05

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5D

ispl

acem

ent (

m)

Time (s)

DRMFault Slip Model

Figure 310.61: Comparison of displacements for top middle point using Morgan Hill earthquake as an

input motion

(X) (Z)

-0.05

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0 5 10 15 20 25 30

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 5 10 15 20 25 30

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

Figure 310.62: Comparison of displacements for top middle point using Kocaeli earthquake as an input

motion

In order to investigate more regarding the frequency content issue, Kocaeli acceleration time history

is considered and frequencies above 1.4Hz are filtered out of the record. Acceleration time history and


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FFT of the filtered record are shown in Figure (310.63). The majority of the energy is in the frequency

range of below 1.4Hz while still there are frequencies up to 2Hz in the motion as can be observed in

FFT of the filtered motion.

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 5 10 15 20 25

Acc

eler

atio

n (g

)

Time (s)

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0.004

0.0045

0.005

0 2 4 6 8 10 12 14

FF

T A

mpl

itude

(gs

)

Frequency (Hz)

Figure 310.63: Acceleration time history and FFT of filtered Kocaeli earthquake

Figure (310.64) shows the displacement time histories for the same point as the one studied for the

actual record. As it is observed, the obtained time histories match perfectly between the case of fault slip

and DRM models. Figure (310.65) shows the acceleration time histories. Comparing the time histories

shows an acceptable match between the results. There are tiny differences in acceleration time histories

(specially at the peaks) which can be due to the fact that there are still frequencies above 1.4Hz in the

input motion but with much less impact in terms of amplitude.


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(X) (Z)

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 5 10 15 20 25 30

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 5 10 15 20 25 30

Dis

plac

emen

t (m

)

Time (s)

DRMFault Slip Model

Figure 310.64: Comparison of displacements for top middle point using filtered Kocaeli earthquake as

an input motion

(X) (Z)

-0.2

-0.1

0

0.1

0.2

0 5 10 15 20 25 30

Acc

eler

atio

n (m

/s2 )

Time (s)

DRMFault Slip Model

-0.2

-0.1

0

0.1

0.2

0 5 10 15 20 25 30

Acc

eler

atio

n (m

/s2 )

Time (s)

DRMFault Slip Model

Figure 310.65: Comparison of accelerations for top middle point using filtered Kocaeli earthquake as an

input motion


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310.6.3 Earthquake-Soil-Structure Interaction Verification for Simulated Northridge

Seismic Motions

Figure 310.66: Finite element model to be used on analyses with input motions computed by integration

equation (x-z plane view)

More realistic example is shown here. Seismic wave fields of Northridge earthquake simulated by

program fk are applied as an input motion for this example. Figure 310.66 shows x-z plane view of three

dimensional model. Similar as analytic case, using fk program, acceleration and displacement fields are

generated at all nodes in DRM layer.

Figure 310.67 shows analysis results observed at the top-midpoint of the finite element model. As

shown in figure 310.67, both results show perfect match.

310.6.4 Curious Case of 1C versus 3C modeling

To inspect more, artificial downhole array is prepared as shown in fingure 310.68. Total 2 observation

points are set on 0 m, and 50 m depth from the ground surface. one dimensional site response analyses

are performed along artificial downhole array usind DEEPSOIL v5.0 (Hashash and Park, 2002). 1D soil

column model is built to run DEEPSOIL with identical soil properties to finite element model. Linear

time domain site response analyses are performed. Displacements recorded at 200 m depth are used as

an input motion. Site response analyses results on the observation points are compared with fk, and


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0 5 10 15 20 25Time (s)

−0.4

−0.3

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

Dis

pla

cem

ent

(m) fk results at top

Observed at top, FEM

pdf

0 5 10 15 20 25Time (s)

−0.4

−0.3

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

Dis

pla

cem

ent



pdf

0 5 10 15 20 25Time (s)

−0.4

−0.3

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

Dis

pla

cem

ent



.pdf

Figure 310.67: Comparison between results computed from program fk and finite element analysis,

observed at the top middle point of the finite element model


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finite element analyses results.

Figure 310.68: Comparison between analytic solution and FEM analysis result observed at top, middle

point of the model (SV (imposed on x direction) Ricker wave input with 0, x component)

Figure 310.69 – 310.74 are analyses results. Figure 310.69, 310.70, and 310.71 show comparison of

results observed at the ground surface, EW, NS, and UD components, respectively. For the case of EW

and NS components, 1C site response analyses results predict similar response as fk and FEM results

compared to UD case. For all cases, 1C analyses results shows larger amplitude especially on UD case,

1Hz frequency contents show unrealistic response amplification. The same trend can be observed at 50

m depth cases (figure 310.72 – 310.74).

Possible explanation are as follows. fk results includes all components of waves (body and surface)

and interaction between them. However, 1C wave propagation analyses cannot incorporate such effect.

Also, 1C analyses is very sensitive to material properties (stiffness, damping ratio, and so on) and

frequency contents of input waves.


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0 5 10 15 20 25Time (s)

−0.4

−0.3

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

Dis

pla

cem

ent

(m)

fk results at top

1D analysis


0 2 4 6 8 10Frequency (Hz)

0

10

20

30

40

50

60

70

FFT a

mplit

ude

fk results at top

1D analysis


Figure 310.69: Comparison between results computed from program fk, finite element analysis, and 1C

analysis, observed at the top middle point, EW component


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0 5 10 15 20 25Time (s)

−0.4

−0.3

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

Dis

pla

cem

ent

(m)

fk results at top

1D analysis



0

10

20

30

40

50

60

70

FFT a

mplit

ude

fk results at top

1D analysis



analysis, observed at the top middle point, NS component


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0 5 10 15 20 25Time (s)

−0.4

−0.3

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

Dis

pla

cem

ent

(m)

fk results at top

1D analysis



0

10

20

30

40

50

60

70

FFT a

mplit

ude

fk results at top

1D analysis



analysis, observed at the top middle point, UD component


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0 5 10 15 20 25Time (s)

−0.4

−0.3

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

Dis

pla

cem

ent

(m) fk results at depth = 50m

1D analysis

Observed at depth = 50m, FEM


0

10

20

30

40

50

60

70

FFT a

mplit

ude

fk results at depth = 50m

1D analysis



analysis, observed at the depth = 50m, EW component


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0 5 10 15 20 25Time (s)

−0.4

−0.3

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

Dis

pla

cem

ent


1D analysis



0

10

20

30

40

50

60

70

FFT a

mplit

ude


1D analysis



analysis, observed at the depth = 50m, NS component


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0 5 10 15 20 25Time (s)

−0.4

−0.3

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

Dis

pla

cem

ent


1D analysis



0

10

20

30

40

50

60

70

FFT a

mplit

ude


1D analysis



analysis, observed at the depth = 50m, UD component


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310.6.5 Earthquake-Soil-Structure Interaction for Surface and Embedded Structures

Luco (1974) Gazetas and Roesset (1979) Wong and Luco (1978) Iguchi and Luco (1981) Kausel and

Roesset (1975) Wong and Luco (1985) Day (1977) Zhang and Chopra (1991) Papageorgiou and Pei

(1998) Luco et al. (1990) Pais and Kausel (1989) Apsel and Luco (1987) Gazetas and Roesset (1979)

Kausel and Roesset (1975)

310.6.5.1 Uniform half-space

310.6.5.2 Layered half-space

310.6.5.3 Layered Layered over rigid lower boundary


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Real-ESSI Lecture Notes 310.7. CASE HISTORY: SIMPLE STRUCTURE . . . page: 1784 of 3100

310.7 Case History: Simple Structure on Nonlinear Soil

310.7.1 Simplified Models for Verification

Due to the complexity of full scale finite element models it is helpful to perform preliminary tests on

simplified models in order to verify the adequacy of the time and mesh discretization with respect to

the input motion. It also provides good insight in the performance of the nonlinear material model. To

achieve this a series of tests on a one-dimensional soil column have been proposed:

• Static pushover test on nonlinear soil column

Through the static pushover test the behavior of the nonlinear material model can be verified.

• Dynamic test of elastic soil column

By applying an earthquake motion to the elastic soil column it can be tested whether the selected

grid spacing is capable of representing the motion correctly without filtering out any relevant

frequencies. This test also allows to choose appropriate damping parameters. It should be noted

that this is additional (small) damping that is used for stability of the numerical scheme and should

not be relied upon to provide major energy dissipation. Major energy dissipation should be coming

from inelastic deformations of the SFS system.

• Dynamic test of nonlinear soil column

Finally the stability and the accuracy of the numerical method can be examined by applying the

earthquake motion to the nonlinear column of soil. A second analysis with a time step reduced by

50% should not give a significantly different result.

Furthermore it will be examined how propagation through an elastic-plastic material will change

the frequency content of the motion.

310.7.1.1 Model Description

The one-dimensional soil column used for verification has the same depth and element sizes as the 2d

and 3d models that will be addressed later. Its total depth is 10.5 meters and it consists of a single stack

of 8-node brick elements of 1.5 meters side length. In order to achieve one-dimensional wave propagation

in vertical direction the movement of four nodes at each level of depth is constrained to be equal. The

input motion is applied to the four nodes at the base of the model. As input motions four time histories

from the Northridge Earthquake are selected (Figure 310.75).

The material properties of the soil are given in Table 310.7.1.1.


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0 10 20 30 40

−1

0

1

LA University Hospital

Acc

eler

atio

n [g

]

0 5 10 15 20 250

0.01

0.02

0.03

Fou

rier

Am

plitu

de [g

]

0 10 20 30 40

−1

0

1

Lake Hughes

Acc

eler

atio

n [g

]

0 5 10 15 20 250

0.01

0.02

0.03

Fou

rier

Am

plitu

de [g

]

0 10 20 30 40

−1

0

1

2

Century City 090

Acc

eler

atio

n [g

]

0 5 10 15 20 250

0.01

0.02

0.03

0.04

Fou

rier

Am

plitu

de [g

]

0 10 20 30 40

−1

0

1

Century City 360

Acc

eler

atio

n [g

]

Time [s]0 5 10 15 20 25

0

0.02

0.04

0.06

Fou

rier

Am

plitu

de [g

]

Frequency [Hz]

Figure 310.75: Acceleration time histories and Fourier amplitude spectra’s of the selected ground motions

Friction angle φ′ 37

Undrained shear strength cu 10 kPa

Mass density ρ 1800 kg/m3

Shear wave velocity vs 200 m/s

The discretization parameters, the time step ∆t and the maximum grid spacing ∆h, are determined

following the guidelines outlined in Section 502.3.1. This yields a maximum grid spacing of

∆h ≤ vs10 fmax

=200

10 10= 2 m (310.5)

For the following analysis ∆h = 1.5 m is selected. The maximum time step is

∆t ≤ ∆h

vs=

1.5

200= 0.0075 s (310.6)

Taking into account a further reduction of the time step by about 60% due to the use of nonlinear

material models ∆t = 0.002 s is chosen.


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310.7.1.2 Static Pushover Test on Elastic-Plastic Soil Column

For the static pushover test, an elastic perfectly plastic Drucker-Prager material model as specified in

Table 310.7.1.1 is used.

After applying self weight a horizontal load of 100 kN is applied to a surface node in increments of

0.1 kN. The system of equations is solved using a full Newton-Raphson algorithm. The predicted shear

strength of the first element that is expected to fail, the one at the surface, is:

τf = cu + z ρ g tanφ′

= 10 + 0.75x 1.8x 9.81 tan 37

= 19.98 kPa (310.7)

where z is the depth of the center of the first element.

Self weight produces the following stresses in the element at the surface:

σx = σy = 8.83 kPa

σz = 13.24 kPa

The maximum shear stress is

τmax =

√(σz − σx

2

)2

+ τ2xz (310.8)

The theoretical failure load can be obtained as follows:

Pf = τxzA

=

√τ2f −

(σz − σx

2

)2

A

= 44.7 kN (310.9)

The static failure load is underestimated by about 6%. This accuracy is acceptable for the given

model because the boundary conditions cannot assure constant stresses at a given depth (no shear stress

is applied to the lateral surfaces).

310.7.1.3 Dynamic Test on Elastic Soil Column

In order to test the spatial discretization of the model an earthquake motion is propagated through an

elastic soil column. The grid spacing of the finite element mesh can be considered sufficiently fine if

frequencies up to fmax = 10 Hz are represented accurately in the numerical analysis. A good way

to verify this is to calculate transfer functions between the base and the surface of the soil column.


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Because transfer functions don’t depend on the input motion they can easily be compared with closed

form solutions.

The transfer function of a soil deposit describes the amplification between the frequencies of the

motion at the base and at the soil surface:

TF (ω) =u(z = 0, ω)

u(z = H,ω)(310.10)

where z is the depth measured from the surface and H is the thickness of the soil deposit above the

bedrock. ω = 2π f is the circular frequency.

For elastic soil with viscous damping the wave equation can be written as (Kramer, 1996a)

ρ∂2u

∂t2= G

∂2u

∂z2+ η

∂3u

∂z2∂t(310.11)

η is the damping coefficient, defined as

η =2G

ωξ (310.12)

where ξ is the frequency independent hysteretic material damping.

After solving the wave equation the transfer function can be written as

TF (ω) =1

cosωH/v∗s(310.13)

where v∗s is the complex shear wave velocity

v∗s =

√G∗

ρ=

√G(1 + i 2 ξ)

ρ(310.14)

In a finite element model with mass- and stiffness proportional Rayleigh damping the damping

coefficient η is constant. Therefore the hysteretic material damping ratio ξ needs to be frequency

dependent in order to satisfy Equation 310.12. Solving Equation 310.12 for ξ and substituting it into

Equation 310.14 and then into Equation 310.13 yields a new transfer function:

TF (ω) =1

cos

(ωH

√ρ

G + iωη

) (310.15)

Figure 310.76 shows a comparison between the closed form solution and the numerical transfer

functions obtained from the finite element analysis. Rayleigh damping is used to obtain the damping

matrix C:

C = αM + βK (310.16)

The analysis are performed using stiffness proportional Rayleigh damping of β = 0.001 and β = 0.01.

No mass proportional damping is applied (α = 0). The damping coefficients of the closed form solution

are chosen to be η = β G .


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0 5 10 150

5

10

15Solution obtained from FE analysis

Frequency [Hz]

Stiffness proportional Rayleigh Damping: β = 0.001Stiffness proportional Rayleigh Damping: β = 0.01

0 5 10 150

5

10

15Closed Form Solution

Frequency [Hz]

Frequency dependent Hysteretic Material Damping η = 0.001*GFrequency dependent Hysteretic Material Damping η = 0.01*GNo Damping

Figure 310.76: Transfer Function between Surface and Base of Soil Layer

It can be seen that the numerical transfer functions are very close to the closed form solutions for

η = β G. The peak corresponding to the second natural frequency of the soil layer is slightly shifted to

the right in the result of the FE analysis. For the FE analysis the Rayleigh damping cannot be reduced

any further as the solution would become unstable. This result proves that a FE analysis involving

Rayleigh damping with α = 0 and β = η /G is equivalent to the closed form solution of the wave

equation with frequency-dependent hysteretic material damping.

Based on the above observations a stiffness proportional Rayleigh damping of β = 0.01 is selected

for the finite element analysis. This choice damps frequencies above 10 Hz appropriately.

310.7.1.4 Dynamic Test on Elastic-Plastic Soil Column

As the next step an elastic-plastic material model of Drucker-Prager type with kinematic strain hardening

has been selected. Previous analysis involving material with isotropic hardening have proved to be

unsuitable because energy can only be dissipated as the yield surface expands. For dynamic problems

this can lead to an unreasonably large extension of the yield surface, especially if resonance frequencies

are present. Therefore only kinematic hardening has been selected in this analysis.

The analysis were performed with four different ground motions using time steps of ∆t = 0.002s and


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0 5 10 15 20 25 30

−0.1

0

0.1

Time [s]

Dis

plac

emen

t [m

] ElasticElastic−Plastic


0 5 10 15 20

−0.1

0

0.1

0.2

Time [s]

Dis

plac

emen

t [m

]

Lake Hughes

0 5 10 15 20 25

−0.2

0

0.2

Time [s]

Dis

plac

emen

t [m

]

Century City 090

0 5 10 15 20 25

−0.2

0

0.2

Time [s]

Dis

plac

emen

t [m

]

Century City 360

Figure 310.77: Displacement Time-Histories at surface of 1d Soil Column, elastic and elastic-plastic

material

∆t = 0.001s. A linear integrator without iterations within a time step was used. All ground motions

were scaled to a maximum acceleration of 1g. For comparison the analysis were also performed on elastic

material. Figure 310.77 shows the displacement time histories at the surface for all four ground motions.

While the overall shapes of the displacements are the same as for the elastic case there is some residual

plastic displacement resulting in the time histories of the Century City motions.

The Fourier amplitude spectra’s of the acceleration recorded at the surface (Figure 310.78) have the

same general shape for the case of elastic and elastic-plastic material. The amplification at the first

resonance frequency (f = 4.75Hz) is bigger in the elastic analysis. Higher frequencies resulting from

plastic slip are damped out effectively in the nonlinear analysis.

Figure 310.79 shows the acceleration time history at the upper node of the lowest element, that is

the first free node above the base. The record shows large peaks of the order of about 6 g. These

peaks are caused by plastic slip and counter balancing of the resulting plastic deformation. The periods

of the peaks are of the order of a few time steps, they add a very high frequency component to the


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0 5 10 150

0.2

0.4

0.6

Frequency [Hz]

Fou

rier

Am

plitu

de [m

/s2 ]


0 5 10 150

0.2

0.4

0.6

Frequency [Hz]

Fou

rier

Am

plitu

de [m

/s2 ]

Lake Hughes

0 5 10 150

0.5

1

Frequency [Hz]

Fou

rier

Am

plitu

de [m

/s2 ]

Century City 090

0 5 10 150

0.2

0.4

0.6

Frequency [Hz]

Fou

rier

Am

plitu

de [m

/s2 ]

ElasticElastic−Plastic

Century City 360

Figure 310.78: Fourier Amplitudes at surface of 1d Soil Column

acceleration. Because these frequencies are due to a purely numerical phenomenon, they should not

be allowed to propagate through the model. This can be achieved easily by specifying an appropriate

numerical procedure (Newmark with appropriate combination of γ and β) or with Rayleigh Damping.

As for the elastic model transfer functions were also computed for the nonlinear model. In Figure

310.80 the transfer functions between the acceleration at the soil surface and the base are compared.

The functions for the nonlinear model are not smooth anymore but the general shape is the same as for

the linear elastic model, i.e. the first natural frequency of the layer is clearly visible. The peaks that are

present in the range of 25 Hz are purely numerical as they appear due to the division by a very small

value.

A second set of analysis performed with half the time step of the previous analysis gives an idea of

the accuracy of the numerical method. In Figure 310.81 the difference between the displacement (or

acceleration) of the analysis with ∆t = 0.002s and ∆t = 0.001s, divided by the corresponding maximum

value is given for the entire time history:

∆d =d0.002(t)− d0.001(t)

max d0.001and ∆a =

a0.002(t)− a0.001(t)

max a0.001(310.17)


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12.6 12.65 12.7 12.75−40

−20

0

20

40

60

Time [s]

Acc

eler

atio

n [m

/s2 ]

0 5 10 15 20 25−60

−40

−20

0

20

40

60

Time [s]

Acc

eler

atio

n [m

/s2 ]

∆ t = 0.002 s∆ t = 0.001 s

Figure 310.79: Acceleration time history at lowest free node

In Figure 310.82 an integral measure for the difference in displacements and accelerations between the

two analysis is given for all depths. The integral measures are defined as

diffd =1

max |d|1

T

T∑0

|d0.002(t)− d0.001(t)| dt (310.18)

diffa =1

max |a|1

T

T∑0

|a0.002(t)− a0.001(t)| dt (310.19)

(310.20)

The integral differences in accelerations are quite large in the elements that are close to the base,

that is where the motion is applied. Toward the surface the difference becomes smaller than 1%. This

is a result of the fact that most of the plastic deformation occurs near the base which represents an

undesired boundary effect. Again this result underlines the importance of an appropriate choice of the

size of the computational domain.

With a point wise difference not exceeding 5% for accelerations and 2% for displacements the time

step ∆t = 0.002s is sufficiently small to ensure stable and accurate results.


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0 5 10 15 20 250

1

2

3

4

5

Frequency [Hz]

TF

[−]

ElasticElastic−Plastic


0 5 10 15 20 250

1

2

3

4

5

Frequency [Hz]

TF

[−]

Lake Hughes

0 5 10 15 20 250

1

2

3

4

5

Frequency [Hz]

TF

[−]

Century City 090

0 5 10 15 20 250

1

2

3

4

5

Frequency [Hz]

TF

[−]

Century City 360

Figure 310.80: Transfer functions between acceleration at the soil surface and the base

310.7.1.5 2d Model

A 2d-model is proposed as a simplification of the full 3d-model. Representing a cross section of the

full model it is expected to provide insight into its dynamic behavior while requiring considerably less

computational resources. The 2d-model consists of one slice of eight-node brick elements as shown in

Figure 310.83. The nodes of the two lateral faces are constrained to move together in x- and z-direction,

the out-of-plane displacement in y-direction is fixed. The model approximates a plane strain situation.

The earthquake motion is applied to the model by the DRM method.

310.7.1.6 Input Motions

As input for the 2d model the motion from the Northridge earthquake recorded at LA University Hotel

(Figure 310.75) is used. The acceleration time history is scaled to a peak ground acceleration of 1 g.

Motion is applied in x-direction only, that is, this is a 1–D wave propagation.

Acceleration time histories at all the nodes of the boundary layer are obtained by vertically propagating


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0 5 10 15 20 25 30−5

0

5

Time [s]D

iffer

ence

[%] Displacement

Acceleration


0 5 10 15 20−5

0

5

Time [s]

Diff

eren

ce [%

]

Lake Hughes

0 5 10 15 20 25−5

0

5

Time [s]

Diff

eren

ce [%

]

Century City 090

0 5 10 15 20 25−5

0

5

Time [s]

Diff

eren

ce [%

]

Century City 360

Figure 310.81: Difference between results of analysis with different time steps, in percent of the maximum

value

a plane wave using the program SHAKE91 (Idriss and Sun, 1992). Because the free-field model has to

match the properties of the free field as represented by the finite element model for the reduced domain,

only linear elastic material without strain dependent reduction of shear modulus and a constant amount

of hysteretic material damping is used in the SHAKE91-analysis. The earthquake motion obtained in

this way corresponds to a shear wave propagating upward through a homogeneous linearly elastic half

space.

The acceleration time histories from the SHAKE91-analysis are then integrated twice to obtain

displacements. Before integration the acceleration and velocity time histories are transformed into Fourier

space, multiplied with a high pass filter and transformed back into time domain. Then a simple parabolic

baseline correction is performed in order to obtain zero initial, final and mean values.


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LA University HospitalCentury City 090

Lake HughesCentury City 360

−9−7.5

−6−4.5

−3−1.5

0

0

0.5

1

Displacements

Depth [m]

Diff

eren

ce [%

]

LA University HospitalLake Hughes

Century City 360Century City 090

−9−7.5

−6−4.5

−3−1.5

0

0

2

4

6

8

Accelerations

Depth [m]

Diff

eren

ce [%

]

Figure 310.82: Averaged differences between results of analysis with different time steps

14 elements

y x

10.5 m

21 m

0

1.5 m

z

outside layer

boundary layer

plastic bowl

7 elements

Figure 310.83: Two-dimensional quasi-plane-strain model


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310.7.1.7 Boundary Conditions

Different boundary conditions are tested on the free-field model. First all outside boundary nodes are

fully fixed as shown in Figure 310.84 a). Then they are released and attached to dash pots that are

b)a)

Figure 310.84: The boundary conditions of the 2d model

both perpendicular and tangential to the boundary (schematically shown in Figure 310.84 b)). The dash

pots perpendicular to the boundary are specified to absorb p-waves, those tangential to the boundary

to absorb s-waves. Because in this configuration no displacement constraint is imposed to the model on

the faces at x = ± 10.5 meter the horizontal at-rest soil pressure has to be applied to the corresponding

nodes manually. This is done by recording the reaction forces in the model with fixed boundaries and

applying them with opposite sign to the model with absorbing boundaries. The horizontal displacements

after applying self weight should be very small.

This configuration of boundary conditions has no fixed point in x-direction. Because the dash pots

only provide resistance to high velocity motions the model is very sensitive to low frequency components

of the motion. The slightest imbalance in acceleration causes the entire model to move as a rigid body

in x-direction. To avoid this to happen the node at the center of the base (x = 0.0 m, z = -10.5 m) is

fully fixed in the following analysis.

Figure 310.85 shows results from a free-field analysis on a homogeneous elastic model. Displacements

on an exterior boundary node as well as transfer functions between a point at the surface and a point on

the exterior boundary of the plastic bowl are presented for the two configurations of boundary conditions

shown in Figure 310.84. It can be seen that the displacements outside the plastic bowl in the model

using absorbing boundary conditions are much larger compared to the model with fixed boundaries. This

result is as expected considering the immediate proximity of the boundary. It also gives an idea about

the constraints the fixed boundary imposes on the motions. The transfer function in Figure 310.85 b) is

defined as the ratio between the Fourier amplitude spectra of a point at the surface and a point on the

exterior boundary:

TF (ω) =A1(ω)

A2(ω)(310.21)


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0 5 10 15 20 25 30−5

0

5

10x 10

−3

Time [s]

Dis

plac

emen

t [m

]

Fixed boundaryAbsorbing boundary

0 2 4 6 8 10 12 14 16 18 200

5

10

15

Frequency [Hz]

[−]

maximum value = 5

maximum value = 60

Fixed boundaryAbsorbing boundary

a)

b)

Figure 310.85: Elastic homogeneous free-field model: a) Displacements of an exterior boundary node

(x,z) = (9.0,0.0) m, b) Acceleration transfer function between surface and depth A(ω)1

A(ω)2

where A1(ω) is the Fourier amplitude spectrum of the acceleration time history at the point (x,z) =

(0,0)m and A2(ω) the corresponding spectrum at the point (x,z) = (9.0,-7.5)m. The figure shows that

the large peak representing the first natural frequency of the system, corresponding to a standing shear

wave in a soil layer of 10.5 meter depth, gets reduced considerably by the absorbing boundary. An energy

build-up in the model due to reflection of waves on the model boundaries can be reduced effectively with

the configuration of boundary conditions shown in Figure 310.84 b). By releasing the fixed node at (x,z)

= (0,-10.5)m the resonance peak could be reduced by another 10% approximately, however at the cost

of remaining permanent displacements at the end of the analysis.

Alternatively to imposing a rigid constraint to a single node at the base the model can be prevented

to move horizontally as a rigid body through uniaxial springs. This gives the possibility to adjust

the frequency of the eigenmode that corresponds to a vertically propagating plane shear wave. By

appropriately choosing the spring constants the model can therefore be adjusted in such a way that it

represents the natural frequency of a soil deposit on bedrock.

310.7.1.8 Structure

Four very simple structures are chosen to illustrate the effects of dynamic SFSI. A beam-column element

of length L and moment of inertia Iy is fixed to a footing. A lumped mass of M = 100, 000 kg is


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mass

Figure 310.86: The 2d SFSI-model

b)a)

Figure 310.87: a) First eigenmode, b) second eigenmode of SFSI-system

added to the translational degrees of freedom of the top of the structure. The footing is 0.5 m deep,

spans over four soil elements and is rigidly connected to the adjacent soil nodes. Its Young’s modulus

is chosen large enough so that the footing can be considered rigid. The mass density of the footing is

ρ = 2400 kg/m3, the column is considered massless. The moment connection between the nodes of the

footing, having 3 (translational) degrees of freedom, and the 6 degrees of freedom of the nodes of the

column is assured by a very stiff beam element that is connected to a node at the bottom and a node at

the top of the footing. The column is then simply connected to the upper node of this auxiliary beam

element.

The parameters of the four columns are chosen such that the second natural frequency, that is the

natural frequency attributed to bending of the column (Figure 310.87 b)), is evenly distributed over

the frequency range of the input motion (Figure 310.88). Structure 4 is designed such that it’s second

natural frequency matches the largest spike in the input motion. Table 310.2 lists the properties of


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0 1 2 3 4 5 60

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

Fou

rier

Am

plitu

de [g

]

Frequency [Hz]

fn 1 f

n 2 f

n 3f

n 4

Figure 310.88: Fourier amplitude spectrum of input motion with second natural frequencies of the 4

SFSI-systems

the structures used in the analysis. For the nonlinear columns a strain hardening material is chosen

Structure Length Stiffness E Iy Mass Yield Moment

[m] [MN m2] [kg] [kNm]

1 5.5 1680 100,000 800

2 3.5 5670 100,000 1,800

3 2.5 13440 100,000 320

4 5.0 5670 100,000 1,800

Table 310.2: Properties of the analyzed structures

that consists of an initial elastic branch with tangent modulus E and a post-yield branch with tangent

modulus 0.2E. The Young’s modulus for all four structures is E = 210 GPa. The yield stress fy for

structures 1, 2 and 4 is 20 MPa and for structure 3 it is 2 MPa.

310.7.1.9 Structure with Fixed Base

To begin with a parametric study of a series of structures with varying stiffnesses is analyzed. The

stiffness is varied by changing the width of the column section. The different structures are expected

to respond specifically to the frequency range of the input motion that is in the neighborhood of the

natural frequency of the column. The input motion that is applied at the base of the structure has been

recorded in a previous free-field analysis of the 2d-model.

The results of this parametric study are shown in Figures 310.89 and 310.90 for linear and nonlinear

structures, respectively. The Fourier amplitudes spectra’s of the acceleration at the top of the structure


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01

23

45

6 2.5

3

3.5

4

4.50

2

4

6

f0 [Hz]

Linear structure

Input motions

Frequency [Hz]

Fou

rier

Acc

eler

atio

n A

mpl

itude

Figure 310.89: Parametric study of 15 linear structures with varying natural frequency.


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01

23

45

6 2.5

3

3.5

4

4.50

0.5

1

1.5

f0 [Hz]

Nonlinear structure

Input motions

Frequency [Hz]

Fou

rier

Acc

eler

atio

n A

mpl

itude

Figure 310.90: Parametric study of 15 nonlinear structures with varying natural frequency


Documents

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