SW1451_c001

7/27/2019 SW1451_c001

http://slidepdf.com/reader/full/sw1451c001 1/81

Chapter 1

Dynamic soil structure interaction

1.1 INTRODUCTION

This chapter deals with some of the basic concepts of dynamic soil-structure interactionanalysis. At the advent of this chapter we expect you to have some background on

• Static soil structure interaction• Theory of Vibration/structural dynamics• Basic theory of soil dynamics

Based on the above∗∗

, we build herein the basic concepts of dynamic soil structureinteraction, which is slowly and surely gaining its importance in analytical procedurefor important structures.

1.1.1 The marriage of soil and structure

As was stated earlier in Chapter 4 (Vol. 1) even twenty years ago struc-tures and foundations were dealt in complete isolation where the structural andgeo-technical/foundation engineers hardly interacted1.

While the structural engineer was only bothered about the structural configuration

of the system in hand he hardly cared to know anything more about soil other than theallowable bearing capacity and its generic nature, provided of course the foundationdesign is within his scope of work. On the other hand the geotechnical engineer onlyremained focused on the inherent soil characteristics like (c, φ, N c, N q, N γ , eo, Cc, Getc.) and recommending the type of foundation (like isolated footing, raft, pile etc.)or at best sizing and designing the same.

The crux of this scenario was that nobody got the overall picture, while in realityunder static or dynamic loading the foundation and the structure do behave in tandem.

∗∗ For theoretical background on these topics please consider Volume 1.1 Even today there are companies which has divisions like structural and civil engineering!! Where the

responsibility of the structural division is to design the superstructure considering it as fixed base frame,furnish the results (Axial load, Moments and Shear) and the column layout drawing to the civil divisionwho releases the foundation drawing based on this input data.

© 2009 Taylor & Francis Group, London, UK

7/27/2019 SW1451_c001


2 Dynamics of Structure and Foundation: 2. Applications

In chapter 4 (Vol. 1), in the problem Example 1.3.1, we have shown how thesoil stiffness can affect the bending moment and shear forces of a bridge girder andignoring the same how we can arrive at a result which can be in significant variationto the reality.

Drawing a similar analogy one can infer that ignoring the soil stiffness in the overall response (and treating it as a fixed base problem) the dynamic response of structure(the natural frequencies, amplitude etc.) can be in significant variation to the realityin certain cases.

This aspect came to the attention of engineers while designing the reactor buildingof nuclear power plant for earthquake. Considering its huge mass and stiffness, thefundamental time period for the fixed base structure came around 0.15 sec whileconsidering the soil effect the time period increased to 0.5 second giving a completelydifferent response than the fixed base case.

With the above understanding – that underlying soil significantly affects the response

of a structure, research was focused on this topic way back in 1970, and under thepioneering effort of academicians and engineers, the two diverging domain of technol-ogy was brought under a nuptial bond of “Dynamic soil structure interaction”, wheresoil and structure where married off to a unified integrated domain. To our knowl-edge the first significant structure where the dynamic effect of soil was considered inthe analysis in Industry in India was the 500 MW turbine foundations for Singrauliwhere the underlying soil was modeled as a frequency independent linear spring andthe whole system was analyzed in SAP IV (Ghosh et al. 1984).

1.1.2 What does the interaction mean?

We have seen earlier that considering the soil as a deformable elastic medium thestiffness of soil gets coupled to the stiffness of the structure and changes it elasticproperty. Based on this the characteristic response of the system also gets modified.This we can consider as the local effect of soil.

On the other hand consider a case of a structure resting on a deep layer of softsoil underlain by rock. It will be observed that its response is completely differ-ent than the same system when it is located on soft soil which is of much shallowdepth or resting directly on rock2. Moreover the nature of foundation, (isolated

pad, raft, pile), if the foundation is resting or embedded in soil, layering of soil,type of structure etc. has profound influence on the over all dynamic response of thesystem.

We had shown for static soil-structure interaction (Chapter 4 (Vol. 1)) case that thesoil can be modeled as equivalent springs or as finite elements and are coupled with thesuperstructure.

Thus for a simple beam resting on an elastic support can be modeled as shownin Figure 1.1.1 and an equivalent mathematical model for the same is shown inFigure 1.1.2.

Based on matrix analysis of structure the element stiffness for this element may bewritten as

2 The reason for these effects we will discuss subsequently.


7/27/2019 SW1451_c001


Dynamic soil structure interaction 3

Node iiNode i Node jNode j

Soil Spring K iSoil Spring K i Soil Spring K jSoil Spring K j

Figure 1.1.1 Equivalent beam element connected to soil springs.

2 4

1 1 3 2

Figure 1.1.2 Mathematical model of the equivalent beam element.

[Kbeam] = EIz

L3

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

12 6L 0 −12 6L 0

6L 4L2 0 −6L 2L2 0

0 0IxL2

2Iz(1 + ν)0 0

−IxL2

2Iz(1 + ν)0−12 6L 0 12 6L 0

6L 2L2 0 6L 4L2 0

0 0 −IxL2

2Iz(1 + ν)0 0 IxL2

2Iz(1 + ν)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(1.1.1)

and the displacement vector is given by

{δ} = <δ1 θ 1 θ 2 δ2 θ 3 θ 4>T (1.1.2)

When the soil springs are added to the nodes, the overall stiffness becomes

[Kbeam] = EIz

L3

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

12 + L3Kii

EIz

6L 0 −12 6L 0

6L 4L2 0 −6L 2L2 0

0 0IxL2

2Iz(1 + ν)0 0

−IxL2

2Iz(1 + ν)

−12 6L 0

12 + L3Kjj

EIz

6L 0

6L 2L2 0 6L 4L2 0

0 0−IxL2

2Iz(1 + ν)0 0

IxL2

2Iz(1 + ν)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(1.1.3)


7/27/2019 SW1451_c001



where, [Kbeam

] = combined stiffness matrix for the beam and the spring; Kii = K jj =spring values of soil at node i and node j of the beam respectively.

The above is a very convenient way of representing the elastic interaction behaviorof the underlying soil and can be very easily adapted in a commercially available finiteelement or structural analysis package.

1.1.3 It is an expensive analysis do we need to do it?

This is a common query comes to the mind of an engineer before starting of an analysis.Based on this fact an engineer do become apprehensive if his/her analysis would sufferfrom a cost over run or whether he/she will be able to finish the design within theallocated time frame.

If he is convinced that soil structure interaction do takes place and the structure isa crucial one3 our recommendation would be ‘its worth the effort rather than to be

sorry later’. The additional engineering cost incurred is trivial compared to the riskand cost involved in case of a damage under an earthquake or a machine induced load.

Now the first question is for what soil condition does dynamic soil structureinteraction takes place?

Veletsos and Meek (1974) suggest that chances of dynamic soil structure interactioncan be significant for the expression

V s

fh≤ 20 (1.1.4)

where V s = shear wave velocity of the soil; f = fundamental frequency of the fixedbase structure; h = height of the structure.

Let us now examine what does Equation (1.1.4) signifies?Knowing the time period T = 1/ f , the above expression can be rewritten as

V sT

h≤ 20 (1.1.5)

For a normal framed building considering the fixed base time period as (0.1n), wheren is the number of stories and thus, we have

V sn

h≤ 200 (1.1.6)

For a normal building the average ratio of h/n (height : storey ratio) is about 3 to3.3 meter. Thus considering h/n = 3, we have

V s

≤600 m/sec. (1.1.7)

3 Like Power House, Turbine foundations, Nuclear reactor Building, Main process piper rack, distillationcolumns, bridges, high rise building catering to large number of people etc.


7/27/2019 SW1451_c001



From which we conclude that for ordinary framed structure, when shear wavevelocity is less or equal to 600 meter/sec we can expect dynamic soil structureinteraction between the frame and the soil.

Incidentally, V s = 600 m/sec is the shear wave velocity which is associated with rock.Thus it can be concluded that for all other type of soil, framed structures will behavedifferently than a fixed base problem-unless and until it rests on rock. For Cantileverstructures like tall vessels, chimneys etc of uniform cross section fundamental timeperiod T is given by

T = 1.779

mh4

EI (1.1.8)

where, m = mass per unit length of the system; h = height of the structure; EI =flexural stiffness of the system.

Substituting the above value in Equation (1.1.5) we have

V sT

h≤ 20; or

V s1.779

mh4

EI

h≤ 20; or, V s ≤ 11.24

h

EI

m(1.1.9)

Considering, I = Ar2 and m = ρ · A, where A = area of cross section; r = radius of gyration; ρ = Mass density of the material, we have

V s ≤ 11.24r

h

E

ρ(1.1.10)

Shear Wave Velocity for Soil-Structure interaction for

Chimneys

0.00

200.00

400.00

600.00

800.00

1000.00

1200.001400.00

1 0 0

1 2 5

1 5 0

1 7 5

2 0 0

2 2 5

2 5 0

2 7 5

3 0 0

Slenderness Ratio

S h e a r W a v e

v e l o c i t y ( m / s e c )

Shear Wave velocity

steel chimney

Shear Wave velocity

concrete chimney

Figure 1.1.3 Chart to assess soil-structure interaction for steel and concrete chimney.


7/27/2019 SW1451_c001



For steel structure the above can be taken as, V s ≤ 57580/λ where λ = h/r, theslenderness ratio of the structure.

For concrete structure we have

V s ≤123970

λ (1.1.11)

Based on the above expressions one can very easily infer if soil structure interactionis significant or not.

The chart in Figure 1.1.3 shows limiting shear wave velocity below which soil-structure interaction could be significant for a steel and concrete chimney.

1.1.4 Different soil models and their coupling

to superstructure

The various types of soil model that are used for comprehensive dynamic analysis areas follows:

1 Equivalent soil springs connected to foundations modeled as beams, plates, shelletc.,

2 Finite element models (mostly used in 2D problems),3 Mixed Finite element and Boundary element a concept which is slowly gaining

popularity.

Of all the options, spring elements connected to superstructure still remain the mostpopular model in design practices due to its simplicity and economy in terms of analysisespecially when the superstructure is modeled in 3-dimensions.

It is only in exceptional or very important cases that the Finite elements and Bound-ary elements are put in to use and that too is mostly restricted to 2 dimensional cases.

1.2 MATHEMATICAL MODELING OF SOIL & STRUCTURE

We present hereafter some techniques that are commonly adopted for coupling thesoil to a structural system.

1.2.1 Lagrangian formulation for 2D frames or stick-models

This formulation is one of the most powerful tool to couple the stiffness of soil to thesuperstructure-specially when one is using a stick model or a 2D model.

For the frame shown hereafter we formulate the coupled stiffness and mass matrix

for the soil structure system which can be effectively used for dynamic analysis.In the system shown in Figure 1.2.1, mf , J θ = mass and mass moment of inertia

of the foundation; m1, J 1 = mass and mass moment of inertia of the 1st story; m2, J 2 = mass and mass moment of inertia of the top story; Kx, Kθ = translational androtational stiffness of the soil; and k1, k2 = stiffness of the columns.


7/27/2019 SW1451_c001



y2

m2J2

k 2

y1

h2

m1, J

1

k 1 h

1

K x

K mf , J

Figure 1.2.1 2D Mathematical model for soil structure interaction.

The equation for kinetic energy of the system may be written as

T = 1

2mf u

2 + 1

2 J θ θ

2 + 1

2m1(u + h1θ + y1)

2 + 1

2 J 1θ

2

+1

2m2(u + (h1 + h2)θ + y2)

2 + 1

2 J 2θ

2

(1.2.1)

U = 1

2Kxu2 + 1

2Kθ θ

2 + 1

2k1y2

1 + 1

2k2 (y2 − y1)

2 (1.2.2)

Considering the expression4, d dt

∂T ∂qi

+ ∂U

∂qi= 0, we have the free vibration

equation as

⎡⎢⎢⎢⎢⎣mf + m1 + m2 m1h1 + m2H m1 m2

m1h1

J + m1h21 + m2H

2

m1h1 m2H m1 m1h1 m1 0

m2 m2H 0 m2

⎤⎥⎥⎥⎥⎦⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

u

θ y1

y2

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭

+

⎡⎢⎢⎢⎢⎣

Kx 0 0 0

0 Kθ 0 0

0 0 k1 + k2 −k2

0 0 −k2 k2

⎤⎥⎥⎥⎥⎦

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

u

θ

y1

y2

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭

= 0 (1.2.3)

4 Refer Chapter 2 (Vol. 2) for further application of this formulation where we have derived a 2D soil-structure interaction model for a Turbine framed foundation.


7/27/2019 SW1451_c001



Figure 1.2.2 Typical finite element mesh with soil springs, for a flexible raft.

where J = J θ + J 1 + J 2 sum of all mass moment of inertia;

H = h1 + h2 = the total height of the structure.

Above formulation can very well be used in cases the foundation is significantly rigidand can be modeled as rigid lumped mass having negligible internal deformation5.

However for cases where the foundation is more flexible one usually resorts to finiteelement modeling of the base raft which is connected to the soil springs as shown inFigure 1.2.2.

For the problem as shown above irrespective of the raft being modeled as a beamor a plate the soil stiffness is directly added to the diagonal element Kii of the globalstiffness matrix to arrive at the over all stiffness matrix of the system.

Before we proceed further we explain the above assembly by a conceptual problemhereafter.

Example 1.2.1

For the beam as shown in Figure 1.2.3, compute the global stiffness matrix whensupported on a spring at its mid span. Take EI as the flexural stiffness of the beam. The spring support has stiffness @ K kN/m.

Solution:

For a beam having two degrees of freedom per node as shown in Figure 1.2.4,the element stiffness matrix is expressed as follows.

5 A classic example is a turbine frame foundation resting on a bottom raft whose thickness is usuallygreater than 2.0 meter.


7/27/2019 SW1451_c001



L L

K

Figure 1.2.3 Spring supported beam.

2

1 43

Figure 1.2.4 Two degrees of freedom of a beam element.

The element matrix for such case is given by

1 2 3 4

Kij =

⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

12EI

L3

6EI

L2

−12EI

L3

6EI

L2

6EI

L2

4EI

L −6EI

L2

2EI

L−12EI

L3

−6EI

L2

12EI

L3

−6EI

L2

6EI

L2

2EI

L

−6EI

L2

4EI

L

⎤

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Assembling the element matrix for the two beams we have

[K] g =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

12EI

L3

6EI

L2 −12EI

L3

6EI

L2 0 0 0 0

6EI

L2

4EI

L

−6EI

L2

2EI

L0 0 0 0

−12EI

L3

−6EI

L2

12EI

L3+ 12EI

L3

−6EI

L2+ 6EI

L2

−12EI

L3

6EI

L20 0

6EI

L2

2EI

L

−6EI

L2+ 6EI

L2

4EI

L+ 4EI

L

−6EI

L2

2EI

L0 0

0 0−12EI

L3

−6EI

L212EI

L3−6EI

L2 0 0

0 06EI

L2

2EI

L−6EI

L24EI

L 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦


7/27/2019 SW1451_c001



As Left hand support is fixed hence we have to eliminate row and column 1and 2.

Similarly, as right hand support is hinged we have to eliminate row and column5 from the above when we have

[K] g =

⎡⎢⎢⎢⎢⎢⎢⎣

24EI

L30

6EI

L2

08EI

L

2EI

L

6EI

L2

2EI

L

4EI

L

⎤⎥⎥⎥⎥⎥⎥⎦

with appropriate boundary conditions.

To use the spring support, the spring is now directly added to the diagonal

element of the global matrix.Thus the combined stiffness matrix is given by

[K] g =

⎡⎢⎢⎢⎢⎢⎢⎣

24EI

L3+ Ks 0

6EI

L2

08EI

L

2EI

L

6EI

L2

2EI

L

4EI

L

⎤⎥⎥⎥⎥⎥⎥⎦

The above is the normal practice adapted in global assemblage of soil springin a finite element assembly.

We further elaborate the phenomenon with a suitable practical numericalexample.

Example 1.2.2

Shown in Figure 1.2.5 is a bridge girder across a river is resting at points A and Bon rock abutments at ends, and resting on a pier at center of the girder (point C)

A 5.0 m C 5.0 m B

Water Level

Figure 1.2.5 Bridge girder across abutments.


7/27/2019 SW1451_c001



A C B

1 12 2 3 3

44

5

Figure 1.2.6 Idealisation of the bridge girder ignoring soil effect.

which is resting on the soil bed of the river. The flexural stiffness of the girder is EI = 100,000 kN · m2. Area of girder is 5.0 m2. The dynamic shear modulusof soil is G = 2500 kN/m2. The bridge pier foundation has plan dimension of 6 m × 6 m. Determine the natural frequencies of vibration of the girder consid-ering with and without soil effect. Unit weight of concrete = 25 kN/m3. Mass

moment of inertia per meter run=

30 kN·

sec2·

m.

Solution:

The bridge girder can be mathematically represented by a continuous beam asshown in Figure 1.2.6. Here node 2 and 4 are at the center of beam.

Thus, for beam element 1, 2, 3, and 4, we have element stiffness matrix as

[Kij] = EI L3

⎡⎢⎢⎢⎣12 6L

−12 6L

6L 4L2 −6L 2L2

−12 −6L 12 −6L

6L 2L2 −6L 4L2

⎤⎥⎥⎥⎦

The unconstrained combined stiffness matrix as

[Kij

]

= EI

L3

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

12 6L −12 6L 0 0 0 0 0 0

6L 4L2 −6L 2L2 0 0 0 0 0 0

−12 −6L 24 0 −12 6L 0 0 0 0

6L 2L2 0 8L2 −6L 2L2 0 0 0 0

0 0 −12 −6L 24 0 −12 6L 0 0

0 0 6L 2L2 0 8L2 −6L 2L2 0 0

0 0 0 0 −12 −6L 24 0 −12 6L

0 0 0 0 6L 2L2 0 8L2 −6L 2L2

0 0 0 0 0 0 −12 −6L 12 −6L

0 0 0 0 0 0 6L 2L2 −6L 4L2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Substituting the values we have


7/27/2019 SW1451_c001



[K] =

76800 96000 −76800 96000 0 0 0 0 0 096000 160000 −96000 80000 0 0 0 0 0 0

−76800 −96000 153600 0 −76800 96000 0 0 0 0

96000 80000 0 320000−

96000 80000 0 0 0 00 0 −76800 −96000 153600 0 −76800 −96000 0 00 0 96000 80000 0 320000 −96000 80000 0 00 0 0 0 −76800 −96000 153600 0 −76800 960000 0 0 0 96000 80000 0 320000 −96000 800000 0 0 0 0 0 −76800 −96000 76800 960000 0 0 0 0 0 96000 80000 −96000 160000

Now imposing the boundary condition that vertical displacement are zero at1, 3, 5,6 we have

[K] =

160000 −96000 80000 0 0 0 0−96000 153600 0 96000 0 0 0

80000 0 320000 80000 0 0 00 96000 80000 320000 −96000 80000 00 0 0 −96000 153600 0 960000 0 0 80000 0 320000 800000 0 0 0 96000 80000 160000

Lumped mass at each node is given by → Mii = 25 × 5 × 2.5/9.81 = 31.85kN · sec2 /m.

Mass moment of inertia at each node is given by → J ii = 30 × 1.25 = 37.5.Thus combined mass matrix is given by

[M] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

37.5 0 0 0 0 0 0 0

0 31.85 0 0 0 0 0 00 0 65 0 0 0 0 0

0 0 0 31.85 0 0 0 0

0 0 0 0 37.5 0 0 0

0 0 0 0 0 31.85 0 0

0 0 0 0 0 0 65 0

0 0 0 0 0 0 0 37.5

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

6 We assume that since the bridge is supported on hard rock at ends, displacement at node 1 and 5are zero.


7/27/2019 SW1451_c001



A C B

1 23 4 5

Kz

Figure 1.2.7 Idealisation of the bridge girder considering soil effect.

Considering the equation[K] − [M]ω2 = 0 we have

MODE 1 2 3 4 5 6 7

Eigen value 692 1328 2684 4897 7448 7787 11722Natural 26.30 36.44 51.80 69.97 86.59926 88.24996 108.26855

frequency(rad/sec)

Considering the effect of soil we can construct the model as in Figure 1.2.7.

Here Kz = 4Gr0

1 − νwhere r0 =

LxB

π, Here L = B = 6.0 m

Here r0

=3.38 m and for G

=2500 kN/m2 and ν

=0.3 Kz

=48285.71

kN/m.Now imposing the boundary condition that vertical amplitude at node 1 and

5 are zero (node 3 is not zero) we have[K] =

160000 −96000 80000 0 0 0 0 0−96000 153600 0 −76800 96000 0 0 0

80000 0 320000 −96000 80000 0 0 00 −76800 −96000 201959.1 0 −76800 −96000 0

0 96000 80000 0 320000 −96000 80000 00 0 0 −76800 −96000 153600 0 960000 0 0 96000 80000 0 320000 800000 0 0 0 0 96000 80000 160000

The Mass matrix remains same as derived earlier.Performing the eigen value solution we have

Modes 1 2 3 4 5 6 7 8

Eigen-values 75 692 2684 3045 7067 7448 9489 11722Natural 8.660 26.30 51.80 56.18 84.06 86.30 97.41 108.27

frequency(rad/sec)


7/27/2019 SW1451_c001



Having established the fact as to how soil affects the dynamic response let ussee further what different type of soil model is possible. For design office practicesspring values considered are usually based on Richart/Wolf’s model which are effec-tively combined with structure as shown above to find out the overall response of asystem.

The example above, though it has been worked out based on beam the theory, it iseffective for any kind of structural elements like plates, shells, 8-nodded brick elementetc. Thus implementing the above in a general purpose Finite element package isquite straight forward. For raft modeled as beam with underlain spring, the essenceof arriving at individual springs at each node is same as shown in the case of staticanalysis based on influence zone7.

The only difference being that the nodal influence area is to be converted into anequivalent circular area to arrive at vertical spring values. The horizontal springs arebased on the full area and are divided equally at the end.

1.2.2 What happens if the raft is f lexible?

Methodology described in previous section is usually adapted when the raft is uncon-ditionally rigid. However there could be cases where the raft could be perfectly flexibleor intermediate (i.e. somewhere between perfectly rigid and perfectly flexible) whenthe calculation of spring values is different than what has been mentioned in thepreceding.

Before we get into this issue the obvious query would be what is the boundary

condition for raft rigidity in terms of dynamic loading?Unfortunately there is none, and the condition pertaining to static load still applies8.Thus as explained in Chapter 4 (Vol. 1), if L is the c / c distance between the columns,

then for

• λL ≤ π4 the raft will behave as rigid raft

• For λL ≥ π the raft will behave as flexible raft • For all values between π/4 ≤ λL ≤ π , the slab behave in between rigid/flexible

in which λ = 4

kB

4EcI , k = modulus of sub-grade reaction, (in kN/m3); B = width

of raft in meter; Ec = modulus of elasticity of concrete (in kN/m2); I = moment of inertia of the raft (in m4).

1.2.2.1 Calculation of spring constant for rigid raft

The rigidity of raft plays a significant role in the soil spring values connected to theplate elements as mentioned above.

7 Refer Example 4.6.1 in Chapter 4 (Vol. 1) for further details.8 This is not illogical for dynamic load can be conceived as a system under static equilibrium at a time t .

Thus condition of rigidity as explained in Chapter 4 (Vol. 1) should hold good.


7/27/2019 SW1451_c001



When the raft is rigid the gross spring value is obtained based on the full raftdimension and then are broken up into discrete values

kz =

Kz A p

AG (1.2.4)

where, kz = value of discrete spring for the rigid finite element; Kz = value of gross

spring considering the overall dimension of the raft; Ap = area of the finite elementplate, and AG = gross area of the raft.

1.2.2.2 Calculation of spring constant for flexible raft

When the raft is flexible an equivalent radius within which the load gets dispersed is

first obtained from the formula

r0 = 0.8t s

Ec

Gs

1 − ν

1 − ν2c

13

(1.2.5)

The gross spring value is then obtained based on this equation. Finally the discretespring for the finite element is obtained as

kz = Kz

A p

πr20

(1.2.6)

where, r0 = equivalent radius within which the load gets dispersed; Ec = dynamicmodulus of the concrete raft; Gs = dynamic shear modulus of the soil; ν = Poisson’sratio of soil; νc = Poisson’s ratio of the raft, and t s = thickness of the raft.

A suitable problem cited hereafter elaborates the above more clearly.

Example 1.2.3

A raft of dimension 30 m × 15 m is resting on a soil having dynamic shearmodulus of 35000 kN/m2 and Poisson’s ratio of soil = 0.4. Determine the soilsprings for plate elements of size 2.0 m × 2.0 m for finite element analysisconsidering,

• The raft as rigid• Considering the raft as flexible.

The thickness of the raft is 1.8 m.


7/27/2019 SW1451_c001



Solution:

Considering the raft as rigid:

r0 = 30

×15

π = 11.968 meter;

Kz = 4Gr0

1 − ν= 4 × 35000 × 11.96

0.6= 2790666.67 kN/m

For finite element of size 2 m × 2 m discrete spring value will be

kz = Kz

A p

AG

➔ k

z = 2790666.67

2 × 2

30 × 15

= 24806 kN/m

Thus spring values at four nodes are 6201 kN/m i.e 1/4th of the abovecalculated value. When the raft is considered flexible, we have:

r0 = 0.8t s

Ec

Gs

1 − ν

1 − ν2c

1/3

Here Ec = 3 × 108 kN/m2; νc = 0.25(say),

then r0 = 0.8 × 1.8

3 × 108

35000

1 − 0.4

1 − 0.252

1/3

= 25.39 m

Thus Kz = 4Gr0

1 − ν= 4 × 35000 × 25.39

0.6= 5924333.333 kN/m

Thus for finite element of size 2 m × 2 m the discrete spring value is

kz = 5924333.333

2 × 2

π × 25.392

= 11701 kN/m

Thus spring values at four nodes are 2925 kN/mIt will be observed that the spring values vary considerably for the two different

approach.

1.2.2.3 What sin thou make in treating foundation

& the structure separately?

Difficult to pass a sweeping judgment for depending on the situation, the sin could becardinal or even trivial.


7/27/2019 SW1451_c001



Based on a number of analysis carried out it can be stated that treating themin isolation can result in conservative design9 or dangerously un-conservative, thusresulting in an unsafe structure which could be a danger to human life and property.

Having made the above statement a number of questions obviously come tomind10 like

1 How conservative or how susceptible the system can be ignoring the soil effect?2 Considering soil effect (specially for FEM analysis) makes the analysis more

laborious and time consuming – thus more costly – is it worth?3 My boss is a traditionalist and under project time pressure – can I convince him

it is worth the effort.4 Before doing the detailed analysis itself can I come up with a quantitative value

based on which I can assess how far this effect will be (for good or worse) andthus convince my boss on the value addition to this effort?

5 What is the risk in terms of cost and safety if I do not do this analysis?The questions are surely pertinent and not always very easy to answer. However with

a little bit of intelligent analysis it is not difficult to come up with a logical conclusionon this issue.

We try to explain. . .The obvious answer is ‘it essentially could modify the natural frequency/time period

of the system’11.What needs to be evaluated is – what is the effect of this modified time period

on the system compared to, if the soil is ignored (i.e. it is considered a fixed base

problem).The two classes of problems under which dynamic soil structure interaction plays asignificant role are

• Systems subjected to vibration from machines like block foundations (machinefoundations for pumps, compressors, gas turbines etc), frame foundations (turbinefoundations, compressor foundations, boiler feed pump foundations)

• Structures subjected to earthquake.

For the machine foundation source of disturbance is the machine mounted on the

system the dynamic waves generated are transferred from the machine – via structureto the surrounding soil-which is an infinite elastic half space.

While for earthquake the source of disturbance is the ground itself where elasticwaves generate within the soil mass due to the tectonic movement/rupture of the rockmass (geologically known as faults).

It is obvious that soil will affect these two classes of problem in different ways.For instance a machine supported on a frame- the frame is usually made signifi-

cantly stiff to ensure stress induced in it are not significant and are generally made

9 For big projects which could mean a cost over run.10 SpeciallyforfreshmannewtothetopicwhohasgotaleadengineerandadepartmentalHEADtoanswerto.11 We say the word “could” as because the extent of modification will depend upon the shear wave velocity

of the soil. We had shown previously the boundary limits within which it can have a significant effect.


7/27/2019 SW1451_c001



over tuned for medium or low frequency machine when considered as a fixed basedproblem.

But in reality considering the soil effect, the foundation may actually be under tunedor even hover near the resonance zone when the underlying soil participates in thevibration process. Thus the amplitude of vibration could significantly vary than thecalculated one.

Generically, considering the soil stiffness will make the system more flexible thena fixed base problem and it can be intuitively deduced that though the stress mightremain within the acceptable level the amplitude of vibration will be more and couldwell exceed the acceptable limit which might have secondary damaging effect to themachine and its appurtenances.

For earthquake the effect is quite different. In this case the structure resting on thesite can be visualized as a body resting on an infinite elastic space (similar to a shipfloating in sea). Due to rupture in the fault as waves dissipate in all direction the soil

mass starts vibrating at its own fundamental frequency known as the free field timeperiod of the site.

In such case the earthquake acts as an electronic filter and tries to excite the super-structure resting on it to its own fundamental frequency and suppressing or eveneliminating other modal frequencies12. Thus if the fixed base frequency of the struc-ture matches the fundamental frequency of the soil strata on which it is resting, theyare in resonance and catastrophe could well be a reality.

Before dwelling into the mathematical aspect of it we further substantiate the abovestatement by some real life facts and observations.

Dowrick (2003) reports that in the Mexico earthquake in 1957 extensive damageoccurred to the buildings that were tall and were found to be resting on alluvium soilof depth >1000 m. In 1967, the Caracas earthquake showed identical result where thetall structures underwent extensive damage and those were resting on deep alluviumsoil overlying bedrock. In 1970 earthquake at Gediz in Turkey a part of a factorywas demolished in a town about 140 Km from the epicenter while no other build-ings in the town underwent any damage! Subsequent investigation revealed that thefundamental period of the building matched the free field time period of the site. TheCaracas earthquake as cited earlier also showed a distinctive pattern where mediumrise buildings (5–9 storeys) underwent extensive damage where depth to bedrock wasless than 100 m, while buildings over 14 stories were damaged where the depth tobedrock was greater than 150 meters.

Let us see why such thing happened and how does it substantiate the free field timeperiod phenomenon as stated earlier.

The free field time period of a site is given by the equation

T n = 4H

(2n − 1)V s(1.2.7)

12 It can be visualized as a giant hand trying to shake a small body resting on it. Since the body is muchweaker to the giant it tries to follow the same phase of vibration as the soil medium.


7/27/2019 SW1451_c001



0

20

40

60

80

100

120

0 0 . 1 5 0 .

3 0 .

4 5 0 . 6

0 . 7 5 0 .

9 1 . 0 5 1 .

2 1 . 3 5 1 .

5

Depth of soil/Shear wave Velocity

N u m b e r o f S t o r i e s

n for RCC frame

n for steel frame

Figure 1.2.8 Limiting value of storeys for frames.

where, T = time period of the free field soil (i.e. without the structure); H = depth of soil over bedrock13; n = number of mode; and V s = shear wave velocity of the soil.

Thus based on the explanation above it can be argued that if the fixed base frequencyof structure is in the close proximity of the free field time period of the site the structuremay be subjected to significant excitation.

The above statement can be extended to a very interesting hypothesis.If we equate the free field time period of the site to the fixed base time period of

the structure we can arrive at some limiting design parameters which can result insignificant dynamic amplification and which should be avoided at the very out set of planning of the structure.

For instance as per IS-1893 RCC moment resisting frames with no infill brick work,the fundamental time period is given by

T = 0.075h0.75 (1.2.8)

Thus equating it to fundamental free field time period of the site we have

0.075h0.75 = 4H

V s, which gives h =

160H

3V s

4/3

(1.2.9)

Considering 1 floor is of height 3.3 m, we can further simplify the equation to

n = 0.303160H

3V s4/3

(1.2.10)

13 Here bedrock is perceived as that level where the shear wave velocity of soil is greater or equal to600m/sec.


7/27/2019 SW1451_c001



The curves shown in Figure 1.2.8 give limiting stories for RCC and steel frames forwhich resonance can occur in a structure during an earthquake as per IS-189314 forvarious values of H / V s.

Let us now probe the problem a bit more based on a suitable numerical problem.

Example 1.2.4

A particular site has been found to consist of 100 m soil overlying bedrockwhen the shear wave velocity of the soil is 222.22 m/sec. Find the limitingnumber of stories of height 3.3 meter for an RCC frame for which resonancecan occur. What would be resonance story if the depth of the overlying soft soil isonly 30 m.

Solution:Based on above data H /V s = 100/222.2 = 0.45 when H = 100 m.

As per the chart as shown above the limiting story for which resonance canoccur is 18.

Thus for a 18 storied building resonance can very well occur and the strategywould be to build the building at least (±)25% away i.e. either it should be 23storied or more or 14 storied or less.

When the depth of soil is only 30 m, H /V s = 30222.2 = 0.135.

Based on the above chart the limiting story height is roughly 4-storey only.

Thus to avoid resonance the building should be either more than 5-storey or lessthan 3-storey.

The above problem well explains the phenomenon as to what happened in theMexico and Turkey earthquakes and perhaps challenges the myth quite prevalent inmany design offices – that for one or two storied building earthquake is not important and can well be ignored.

It is evident from the above problem that the response depends on the depth of soil

on which it is resting and depending on the free field time period the response caneither amplify or attenuate. It can well affect even a one storied building.

The chart in Figure 1.2.9 shows limiting story height of buildings with infill brickpanels and all other type of frames as per IS 1893 for different width of buildingvarying from 10 meter to 50 m15.

The above theory is though explained in terms of building, can very well be adaptedfor any class of structure for which it is possible to establish the fundamental timeperiod expression.

14 In this case time period for steel frame is considered as T = 0.085(h)0.75 as per IS-1893.15 Time period of the fixed base structure considered as T = 0.09h /(d )0.5 as per, Indian Standards Institution

(1984, 2002). “Indian Standard Criteria for Earthquake Resistant Design of Structures”, IS: 1893(Part 1), ISI, New Delhi, India.


7/27/2019 SW1451_c001



0

5

10

15

20

25

30

35

40

0 0 . 0 8

0 . 1 5

0 . 2 3 0 .

3 0 . 3 8

0 . 4 5

0 . 5 3 0 .

6 0 .

6 8 0 . 7 5

Depth of soil/Shear wave velocity

N u m b e r o f s t o r y

n for d=10m

n for d=15m

n for d=20m

n for d=25m

n for d=30m

n for d=35m

n for d=40m

n for d=45m

n for d=50m

Figure 1.2.9 Limiting story for building with infill brick panel.

Having assessed the resonance criteria and making sure at planning stage that thetwo periods do not match one would still like to quantify the combined time periodof the overall soil structure system and assess whether there is any amplification or

attenuation of the earthquake force.Before plunging into detailed analysis based on FEM or otherwise it would be useful

to have a rough estimate as to how much the underlying soil affects the overall response.Veletsos and Meek (1974) has given a very useful expression based on which it is

possible to estimate the modified time period of a structure, and is given by

T = T

1 + k

Kx

1 + Kxh2

Kθ

(1.2.11)

where T = modified time period of the structure due to the soil stiffness, T = time

period of the fixed base structure, k = stiffnessof the fixed base structure @ 4π2W gT 2

,

Kx, Kθ = horizontal and rotational spring constant of the soil (IS-1893), h = effectiveheight or inertial centroid of the system, and, W = total weight of the structure.

Based on the above expression one can immediately arrive at a rough estimate asto how strong could be soil response at the very outset of a design. We elaborate theabove based on two suitable problems hereafter.

Example 1.2.5

An RCC Chimney 150 meter in height has a uniform cross section area of Ac =8.5 m2 and moment of inertia I = 92.5m4. Evaluate the base moment and


7/27/2019 SW1451_c001



shear under earthquake considering the problem as fixed base as well as thesoil effect. The structure is located in zone IV as per IS 1893. The structureis supported on raft of diameter 18 meter. The soil has a dynamic shear wavevelocity of 120 m/sec and unit weight of 19 kN/m3. Consider 5% dampingfor the analysis.16 The grade of concrete used is M30 having dynamic Econc =

3.12 × 108 kN/m2.

Solution:

Height of the structure = 150 m; Area of shell = 8.5 m2

Weight of chimney = 150 × 8.5 × 25 = 31875 kN (unit weight of conc. =25 kN/m3)

Radius of gyration of the chimney = I /A = 92.5/8.5 = 3.298 m

Thus slenderness ratio H / r = 1503.298

= 45.4.

As per IS 1893 CT = 82.8.

As per IS 1893 time period of a fixed base chimney is given by, T =CT

3.13

WH

EcAc.

where, W = weight of chimney in N; Ec = Dynamic Young’s modulus of conc.@ 3

×108 kN/m2

Thus, T = 82.8

3.13

31875 × 150

3.12 × 108 × 8.5= 1.13 sec

For 5% damping referring to chart in IS-1893 we have Sa / g = 0.10.

Thus the horizontal seismic coefficient is given by, αh = βIF oSa

g .

Here β (Soil foundation factor) = 1.0 for chimney resting on raft, I = 1.5Importance factor, F o = Zone factor @ 0.25 for zone IV.

This gives

αh = 1.0 × 1.5 × 0.25 × 0.10 = 0.0375

The Bending moment and shear force are given by,

M

=αhW H 0.6 x

H

12

+0.4 x

H

4

and V

=CvαhW 5x

3H −

2

3 x

H

2

16 In this case it is presumed that reader has some idea of how to use the code IS-1893 or is at least

familiar with it.


7/27/2019 SW1451_c001



Here, Cv = a coefficient depends on the slenderness ratio and as per the presentproblem is 1.47 as per IS 1893; H = height of c.g. of the structure above base@ 75 meter for the problem; x = distance from the top.Substituting the appropriate values, we have

M = 0.0375 × 31875 × 75 [0.6(1.0)12 + 0.4(1.0)4] = 89648 kN/m

V = 1.47 × 0.0375 × 31875 [(5/3)− (2/3)] = 1757 kN.

Considering the soil effect we have the dynamic shear modulus of soil, G = ρv2s

Or G = (19/9.81) × 120 × 120 = 27890 kN/m2.

With radius of raft = 9.0 m, Kx = 8GR

2 − ν, ν = Poisson’s ratio of the soil

considered as 0.35,

Kx = 8 × 27890 × 9

2 − 0.35= 1217018.2 kN/m

And Kθ =8GR3

3(1 − ν)which gives, Kθ =

8 × 27890 × 93

3(1 − 0.35)= 83412554 kN/m

The fixed base stiffness of chimney is given by

k = 4π2W

gT 2= 4 × π2 × 31875

9.81 × 1.132= 100458 kN/m

Substituting the above numerical values in Veletsos’ equation we have

T = T

1 + k

Kx

1 + Kxh2

Kθ

➔ T = 1.13

1 + 100458

1217018.2

1 + 1217018.2 × 752

83412554

= 3.2 sec

As per IS 1893 for T = 3.2 sec, Sa / g = 0.05, which gives

αh = 0.0375

0.10× 0.05 = 0.01875 (By proportion)

The base moment and shear are given by

M = 89648

0.0375× 0.01875 = 44824 kN · m;


7/27/2019 SW1451_c001



and V = 1757

0.0375× 0.01875 = 878.5 kN

The results are compared hereafter17

Case Moment Shear Remarks

Without soil 89648 1757 Reduction in moment and shear by 34%With soil 44824 878.5

The problem shows a clear attenuation of the response.

We show another example hereafter.

Example 1.2.6

Shown in Figure 1.2.10 is a horizontal vessel having empty weight of 340 kNand operating weight of 850 kN is placed on two isolated footing of dimension8.5m × 3 m. The center to center distance between the two foundations is 5.5meter. The center line of vessel is at height ( H f ) of 4.5 meters from the bottom of the foundation. Thickness of the foundation slab is 0.3 meter. The RCC pedestal

is of width 1.0 meter, length 6 meter having height of 3.45 meter. The shear wavevelocity of the soil is 200 m/sec having Poisson’s ratio of 0.3. Allowable bearingcapacity of the foundation is 150 kN/m2. Calculate the design seismic momentconsidering the effect of soil and without it, if the site is in zone III as per IS-1893.Consider soil density @ 18 kN/m3 and unit weight of concrete as 25 kN/m3?

Solution:

Plan are of footing = 8.5 × 3 = 25.5 m2

Equivalent circular radius =

Af π

=

25.5π

= 2.849 m

Moment of inertia of the foundation about X -axis

1

12BL3

= 1

123×8.53 =

153.5313 m4

Moment of inertia of the foundation about Y -axis

1

12LB3

= 1

128.5×33 =

19.125 m4

17 Without an elaborate analysis it could be an effective calculation to convince the boss thatyou can save some money and the worth of a dynamic soil-structure interaction analysis.


7/27/2019 SW1451_c001



yp

y p

Hf

Hp

Ds

Wp

Lp

Y

Lf

X

Bf

Bf

Ls

Figure 1.2.10 A horizontal vessel.

Equivalent circular radius about X axis = 1

2

64I xx

π

0.25

= 3.739183 m

Equivalent circular radius about Y axis = 1

2

64I yy

π

0.25

= 2.221 m

Mass density of soil (ρ) = 18

9.81= 1.835 kN/m3

Dynamic shear modulus = ρv2s = 1.835 × 2002 = 73400 kN/m2


7/27/2019 SW1451_c001



Lateral spring in X and Y direction = 32Gr0(1 − ν)

7 − 8ν= 1018306 kN/m

Rocking spring about X axis

=

8Gr3x

3(1 − ν) =14627886 kN/m

Rocking spring about Y axis =8Gr3

y

3(1 − ν)= 3063462 kN/m

Moment of Inertia of the pedestal about X axis = 1

12× 1 × 63 = 18 m4

Moment of Inertia of the pedestal about X axis

=1

12 ×6

×13

=0.5 m4

Structural stiffness of pedestal about X axis = 3EI x

L3= 3 × 3.2 × 108 × 18

3.453=

3.95 × 108 kN/m

Structural stiffness of pedestal about Y axis = 3EI y

L3= 3 × 3.2 × 108 × 0.5

3.453=

1.10 × 107 kN/m

Contributing mass for the vessel empty case = 3402 × 9.81

= 17.33 kN-sec2 /m

Contributing mass for the vessel operating case = 850

2 × 9.81=

43.323 kN-sec2 /m

Contributing uniformly distribute load for the pedestal = 25.5 × 25/9.81 =64.98 kN/m

The mathematical model for the pedestal thus constitute of a beam element(pedestal) having a mass lumped at its tip (mass contribution from the vessel) isshown in Figure 1.2.11.

The time period of such fixed base model is given by (Paz 1991)

T = 2π

(M + 0.25mb)

K

mb

M

Figure 1.2.11 Mathematical model for the pedestal.


7/27/2019 SW1451_c001



and the modified time period considering soil effect is given by

T = T

1 + k

Kx

1 + Kxh

2

Kθ

The time periods and the corresponding Sa / g values as per IS-1893 for 5%damping are as show hereafter.

Time period Time period Time period Time period (vessel empty ) (vessel empty ) (operating ) (operating )about X about Y about X about

Sl no Case direction direction direction Y direction

1 Without soil 0.0018 0.011 0.0024 0.0152 With soil effect 0.0579 0.1057 0.0771 0.1408

Corresponding Sa / g value is given by

Sa/g Sa/g Sa/g Sa/g (vessel empty ) (vessel empty ) (operating ) (operating )about about about about

Sl no Case X direction Y direction X direction Y direction


Base shear as per IS 1893 considering Importance factor as 1.0 for vessel emptycase and 1.25 for vessel in operation case we have

Shear Shear Shear Shear (vessel empty ) (vessel empty ) (operating ) (operating )about about about about


1 Without soil 21.25 25.5 26.5625 34.53125

2 With soil effect 42.5 42.5 53.125 53.125

The moment at the foundation level is given by

Moment Moment(vessel (vessel Moment Momentempty ) empty ) (operating ) (operating )about about about about



This case clearly shows an amplification of force considering the soil effect.


7/27/2019 SW1451_c001



Having established a basis of how to evaluate the coupled soil-structure interactionunder dynamic loading we now extend the above theory to system with multi degreeof freedom where the theory can be very well be adapted as a powerful tool for adetailed yet economic dynamic analysis.

1.3 A GENERALISED MODEL FOR DYNAMIC SOIL STRUCTURE

INTERACTION

In this section we present a generalised model for dynamic soil-structure interaction.Though the model is developed based on 3D frames can also be adapted for a threedimensional Finite Element analysis.

1.3.1 Dynamic response of a structure with multi degree

of freedom considering the underlying soil sti ffness

We had shown earlier that for a single degree of freedom system the modified timeperiod of a structure considering the soil effect is given by

T = T

1 + k

Kx

1 + Kxh2

Kθ

(1.3.1)

Both ATC (1982) and FEMA has adapted this formula for practical design office usage

(Veletsos & Meek 1974, Jennings & Bielek 1973). The nomenclatures of the formulaare as explained earlier. Now squaring both sides of the above equation we have

T 2 = T 2

1 + k

Kx+ kh

2

Kθ

(1.3.2)

Considering the expression T = 2πω

we have

4π2

ω2= 4π

2

ω2

1 + 4π

2

mT 2Kx

+ 4π2

mh2

T 2Kθ

or 4π

2

ω2= 4π

2

ω2

1 + mω

2

Kx+ ω

2

mh2

Kθ

(1.3.3)

Simplifying and expanding the above we have

1

ω2= 1

ω2+ m

Kx+ mh2

Kθ

which can be further modified to

1ω2

= 1ω2

+ 1ω2

x

+ 1ω2θ

(1.3.4)

which gives the modified natural frequency relation for a system with single degree of freedom. This formulation has also been shown in, Kramer, S. (2004).


7/27/2019 SW1451_c001



Now considering generically ω =

k/m we have

m

ke= m

k+ m

Kx+ mh2

Kθ

, or,1

ke= 1

k+ 1

Kx+ h2

Kθ

(1.3.5)

where ke = equivalent stiffness of the soil structure system having single degree of freedom.

We shall extend the above basis to multi degree of freedom hereafter (Chowdhuryand Dasgupta 2002).

1.3.2 Extension of the above theory to system with multi degree of freedom

A 3-D frame shown in Figure 1.3.1, is considered for the presentation of the proposed

method. The frame structure has n degrees-of freedom and subjected to soil reactionsin the form of translational and rotational springs.

For a system having n degrees of freedom the above equation can be written in theform

[M]n×n

[Ke]n×n

= [M]n×n

[K]n×n

+ [M]n×n

Kx+

[M]n×n

h2

n×n

Kθ

(1.3.6)

Y

XO

Z

Kx

Kθ -- mass points

Figure 1.3.1 A 3-D Frame having multi-degree-of freedom with representative foundation spring.


7/27/2019 SW1451_c001



Here, [Ke] = equivalent stiffness matrix of the soil structure system of order n, [M] =a diagonal mass matrix of order n having masses lumped at the element diagonals,

[h2] = radius vectors of the lumped masses to the center of the foundation springsof order n, Kx, Kθ = translation and rotation spring stiffness of the total foundationsystem represented by a unique value.

Taking out the common factor [M], we have

[I ]

[Ke]= [I ]

[K]+ [I ]

Kx+h2

Kθ

(1.3.7)

where, [I ] = identity matrix of order n having its diagonal element as 1.or [I ][Ke]−1 = [I ][K]−1 + [I /Kx] + [h2/Kθ ]

⇒ [Fe] = [F ] + [F x] + [F θ ] (1.3.8)

where [F ] = Flexibility matrix of the system with suffixes as mentioned earlier forstiffness matrices.

Once the flexibility matrix of the equivalent soil structure system is known thestiffness matrix may be obtained from the expression

[Ke] = [Fe]−1 (1.3.9)

Now knowing the modified stiffness matrix the eigen solution may be done basedon the usual procedure of

[Ke] [ϕ] = [λe] [M] [ϕ]. (1.3.10)

1.3.3 Estimation of damping ratio for the soilstructure system

While calculating the damping ratio, the normal process is to guess a damping ratiofor the structure like 2–5%, and consider the same damping ratio for all the mode andobtain the value of Sa / g value for the particular structure per mode corresponding tothe time period based on the curves given in IS-1893.

The basis of assuming this damping ratio is purely judgmental and is dependenton either the experience of the engineer, recommendation of codes, or based on fieldobservations on the performance of similar structure under previous earthquakes.

When the effect of soil is neglected it is possible to obtain the material damping ratioof the structure depending on what constitute the material like steel, RCC etc.

However when the whole system is resting on soil an analyst is usually faced withthe following stumbling blocks for which clear solution is still eluding us specially formodal analysis in time domain.


7/27/2019 SW1451_c001



The difficulties encountered can be summarised as follows

• The damping matrix of the coupled soil-structure system becomes non-proportional for which the damping matrix does not de-couple based onorthogonal transformation.

• As the damping ratio of the structure and the soil could be widely varying itbecomes difficult to assess a common damping ratio which would affect the soilas well as structure.

• Even after elaborate FEM modelling of the soil, the damping ratio contributionper mode still remains guess estimation at the best.

We present hereafter a method by which one can estimate approximately the contri-bution of combined soil structure system under earthquake for various modes, withoutresorting to an elaborate modelling of the soil itself.

We only estimate the contribution of the soil damping to the structural system whoseresponse we are interested in. The estimation is surely approximate but at least givesa rational mathematical basis to arrive at some realistic damping value rather thanguessing a damping value at the outset and presuming that it remain same for eachmode, specially for coupled soil structure system where widely varying damping for thefoundation and structure makes it difficult for the analyst to arrive at unified rationalvalue applicable to the system.

1.3.4 Formulation of damping ratio for single degree

of freedom

Neglecting the higher order, the material damping ratio for a soil structure systemhaving single degree of freedom is given (Kramer 2004) by

ζ

ω2= ζ

ω2+ ζ x

ω2x

+ ζ θ

ω2θ

(1.3.11)

where, ζ

=damping ratio of the equivalent soil structure system; ζ

=damping ratio

of the fixed base structure; ζ x = horizontal damping ratio of the soil, where ζ x = 0.288√ Bx

and Bx = (7−8ν)mg 32(1−ν)ρsr3

x, where m = total mass of the structure and foundation; g =

acceleration due to gravity; ν = Poisson’s ratio of the soil; ρs = mass density of thesoil; rx = Equivalent circular radius in horizontal mode; ζ θ = damping ratio of the

soil in rocking mode ζ φx = 0.15(1+Bθ )

√ Bθ

and Bθ = 0.375(1−ν) J θ g ρsr5

θ

; and J θ = mass moment

of inertia of the foundation and the structure.Converting the damping ratio equation to stiffness-mass basis we have

mζ ke

= mζ k

+ mζ xKx

+ mh2ζ θ

Kθ

or ζ ke

= ζ k

+ ζ xKx

+ h2ζ θ

Kθ

;

➔ ζ = ke

ζ

k+ ζ x

Kx+ ζ θ

Kθ

(1.3.12)


7/27/2019 SW1451_c001



For very high value of Kx and Kθ ke → k when ζ → ζ .

1.3.5 Extension of the above theory to systems with multi-degree freedom

On extending the above to multi degree of freedom of order n, we have

[ζ ][M]n×n

[Ke]n×n= [ζ ][M]n×n

[K]n×n+ [ζ x][M]n×n

Kx+ [ζ θ ][M]n×n[h2]n×n

Kθ

➔ [ζ ] = [Ke]{[ζ ][F ] + [ζ x][F x] + [ζ θ ][F θ ]} (1.3.13)

[ζ ] = Damping ratio matrix of the combined soil structure system having n number

of modes.It is to be noted that [ζ ] is non-proportional and not a diagonal matrix, and based on the matrix operation as shown above has off-diagonal terms.

A study on the parametric effect shows that [ζ ] becomes nearly a diagonal matrix(i.e. the off diagonal terms vanishes or approaches zero) when damping ratio of thestructure and the soil foundation system are nearly equal.

However, when the damping ratio are widely varying the off diagonal terms do notvanish however there magnitudes are relatively smaller than the diagonal terms (ζ ii)which has the most dominant effect on the system.

Thus if it is possible to arrive at a foundation layout where the damping ratio of the

structure and foundation are closely spaced considering the diagonal terms as modaldamping ratio per mode is quite correct.

Even when the off diagonal term exists due to widely varying values for practicaldesign engineering purpose considering the ζ ii term of damping ratio matrix is realisticfor it gives a reasonably rational basis of estimation of the damping ratio per moderather than guessing a value based on gut feeling.

We explain the above theory based on suitable example hereafter

Example 1.3.1

Shown in Figure 1.3.1 is a three storied steel frame subjected to dynamic forces.The damping ratio for steel is found to vary between 2 to 5%. Determine

• The fixed base natural frequencies of the structure.• The fixed base eigen-vectors.• Modified natural frequency with foundation stiffness.

•Modified eigen.

• Take K x = 35000 kN/m and K θ = 50000 kN/m for the soil-foundation.• Analyse the floor shears for earthquake based on IS-1893 Zone III for

◦ Fixed base.◦ Considering the soil effect.


7/27/2019 SW1451_c001



G H X3

3000

E F X2

3000

C D X1

3000

A B

Figure 1.3.2

Here,

1 K AC = K DB = 1.5× 103 kN/m M GH = 200 kN sec2 /m2 K CE = K DF = 1.0 × 103 kN/m M EF = 400 kN sec2 /m3 K EG = K FH = 0.75 × 103 kN/m M CD = 400 kN sec2 /m

Solution:The stiffness and mass matrix is given by

[K] =⎡⎣ 5000 −2000 0

−2000 3500 −15000 −1500 1500

⎤⎦ and [M] =

⎡⎣400

400200

⎤⎦

• Based on Figure 1.3.2, we have found earlier that

ω1 =√

1.6426 = 1.281 rad/sec; ω2 =√

10.00 = 3.162 rad/sec; ω3 =√ 17.104 = 4.135 rad/sec.


7/27/2019 SW1451_c001



Thus the time periods for the fixed base structure is given by18

T 1 = 4.97 sec, T 2 = 1.987 sec, T 3 = 1.52 sec

• The mode shapes or the eigen-vectors are

[φ] =⎡⎣ 1.00 1.0 1.0

2.1715 0.5 −0.92082.7816 −1.50 0.719

⎤⎦

• Normalised eigen vectors

[ϕi] = ⎡⎣ 0.01615 0.03244 0.0344512

0.0350718 0.01622 −0.031720.04493 −0.02433 0.02477

⎤⎦

Calculation for the combined soil-structure system

Here stiffness matrix of the fixed base structure

[K] =⎡⎣

5000 −2000 0−2000 3500 −1500

0 −1500 1500

⎤⎦which on inversion gives

[F ] =⎡⎣0.000333 0.000333 0.000333

0.000333 0.000833 0.0008330.000333 0.000833 0.003145

⎤⎦

[F x]

= ⎡⎣1/35000 0 0

0 1/35000 0

0 0 1/3500⎤⎦

=⎡⎣2.85714 0 0

0 2.85714 00 0 2.85714

⎤⎦× 10−5

[h2] =⎡

⎣

9 0 00 36 00 0 81

⎤

⎦18 You can check the value by any of the method as explained in Chapter 5 (Vol. 1) for eigen value

analysis.


7/27/2019 SW1451_c001



Thus

[F θ ] =⎡⎣

9/50000 0 00 36/50000 0

0 0 81/50000

⎤⎦

=⎡⎣0.00018 0 0

0 0.00072 00 0 0.00162

⎤⎦

As [Fe] = [F ] + [F x] + [F θ ] we have

[Fe] = ⎡⎣0.000542 0.000333 0.000333

0.000333 0.001581905 0.0008330.000333 0.0008333 0.001982

⎤⎦

which is combined flexibility matrix of the soil structure system.Inversion of the above flexibility matrix gives

[Ke] =⎡⎣

2195.19 −344.34 −224.4212−344.34 804.662 −306.223

−224.42

−306.223 671.0682

⎤⎦

The above gives the combined stiffness matrix for structural system consider-ing the soil compliance19.

Thus based on the above modified stiffness matrix and mass matrix as

[M] =⎡

⎣

400400

200

⎤

⎦We have, based on eigen solution, ω1 = √

1.2163 = 1.10286 rad/sec; ω2 =√ 3.9666 = 1.9916 rad/sec; ω3 =

√ 5.8255 = 2.4136 rad/sec.

Thus the time periods for the combined soil-structure system is given by

T 1 = 5.697 sec, T 2 = 3.154 sec, T 3 = 2.603 sec

19 Watch the numbers. . . .. it is symmetric and is completely different than when you add thesprings directly to the diagonal. This matrix has no rigid body mode and can be used directlyfor static analysis too.

Moreover if we take Kx and Kθ very high the Ke converges to the fixed base matrix K.


7/27/2019 SW1451_c001



Normalised modified eigen vectors considering soil stiffness is given by

[ϕi] =⎡⎣

−0.013 0.0479 0.00589−0.0409 −0.00772 −0.0276

−0.0360 −0.0169 0.05835

⎤⎦

Calculation of modal damping

Considering, ζ = 5% for the structure, ζ x = 10% for the soil in translationmode, ζ θ = 15% for the soil in rocking mode

We have, [ζ ] = [Ke]{[ζ ][F ] + [ζ x][F x] + [ζ θ ][F θ ]}

Substituting the values as mentioned and calculated above we have

[ζ ] =⎡⎣ 0.092 −0.028 −0.020

−0.007 0.10908 −0.028−0.002 −0.0126 0.1115

⎤⎦

It will be seen that that the main diagonal terms are dominant andcan be considered as the modal damping ratio contribution for eachmode.

Suppose we had closely spaced damping data like ζ = 5% for the structure;ζ x = 6% for the soil in translation mode; ζ θ = 5.5% for the soil in rockingmode, the modal damping matrix reduces to

[ζ ] =⎡⎣ 0.0525 −0.0015 −0.001016

−0.0004 0.05312 −0.00144−0.00014 −0.00066 0.05315

⎤⎦

When the matrix become practically diagonal dominant with off diagonalterms having very low values.

Thus for the present problem ζ may be considered as ζ 1 = 9.2% forfirst mode, ζ 2 = 10.9% for second mode; and ζ 3 = 11.1% for the thirdmode.

Calculation of earthquake force fixed base structure

m φ1 mφ1 mφ21 φ2 mφ2 mφ2

2 φ3 mφ3 mφ23

400 0.01615 6.46 0.104329 0.03244 12.976 0.420941 0.03445 13.7804 0.47475407

400 0.03507 14.028 0.491962 0.01622 6.488 0.105235 −0.01372 −5.488 0.07529536

200 0.04493 8.986 0.403741 −0.02433 −4.866 0.118389 0.02477 4.954 0.12271058

29.474 1.000032 14.598 0.644565 13.2464 0.67276001


7/27/2019 SW1451_c001



Modal mass participation factor

κ1 = 29.474

1.000032= 29.47306 for the first mode,

κ2 = 14.598

0.644565= 22.64777 for the second mode,

κ3 = 13.2464

0.67276= 19.689 for the third mode.

Assuming 5% damping for the structure we have,

Mode Time period (secs) Sa (m/sec 2) Remarks

1 4.9 0.4905 Sa value obtained from the chartgiven in IS-1893 for 5% damping

2 1.98 0.6867 Do3 1.52 0.7848 Do

For zone III:K = 1.0, β = 1.0, I = 1.2F 0 = 0.2 as per the code

Thus base shear is given by; V =3i=1 K · I , β · F 0 · κi · Samiφi

Substituting data on the above formula we have

Mode Base shear V Remarks

1 102 Fixed base case2 5.453 4.91

Calculation for coupled soil-structure interaction.

m φ1 mφ1 mφ21 φ2 mφ2 mφ2

2 φ3 mφ3 mφ23

400 −0.013 −5.2 0.0676 0.0479 19.16 0.917764 0.006 2.4 0.0144400 −0.041 −16.4 0.6724 −0.0077 −3.08 0.023716 −0.0276 −11.04 0.304704200 −0.036 −7.2 0.2592 −0.0169 −3.38 0.057122 0.0583 11.66 0.679778 −28.8 0.9992 12.7 0.998602 3.02 0.998882

Modal mass participation factor

κ1 = −28.8

0.9992= −28.8231 for the first mode,

κ2 = 12.70.998602

= 12.717 for the second mode, and

κ3 = 3.02

0.9988= 3.0233 for the third mode.


7/27/2019 SW1451_c001



Modal damping for each mode, as calculated earlier.

Mode Damping Time (sec ) Sa (m/sec 2) Remarks

1 9.2% 5.7 0.343 Calculated from curve based on interpo-

lation corresponding to 9.2% damping2 10.9% 3.2 0.294 Do- with 10.9% damping3 11.15% 2.6 0.245 Do- with 11.15% damping

Calculation for Base shear

Base shear for the frame with coupled soil-structure interaction is given by

Mode Base shear V Remarks

1 68.4 Couple soil-foundation system2 11.43 0.537

Calculation of storey forces

The storey forces for the two cases are calculated hereafter

Coupled soil structure system Fixed base

Base Base Base Base Base Baseshear shear shear shear shear shear

Storey m h mh2 mh23i =1 mh2

mode 1 mode 2 mode 3 mode 1 mode 2 mode 3

1st 400 3 3600 0.10526 7.20 ×10+00 1.20 ×10+00 5.66 ×10−02 1.08 ×10+01 5.74 ×10+00 5.17 ×10+00

2nd 400 6 14400 0.42105 2.88×10+01 4.80 ×10+00 2.26 ×10−01 4.31 ×10+01 2.29 ×10+01 2.07 ×10+01

Top 200 9 16200 0.47368 3.24×10+01 5.40 ×10+00 2.55 ×10−01 4.84 ×10+01 2.58 ×10+01 2.33 ×10+01

Comparison of results

Time period

Structure type T1 T2 T3

Fixed base structure 4.9 1.987 1.52Soil-structure interaction 5.697 3.154 2.603

The time periods are increasing with introduction of soil springs as predictedat the outset.

Acceleration

Structure type Mode 1 Mode 2 Mode 3

Fixed base structure 0.4905 0.6867 0.7848Soil-structure interaction 0.34335 0.2943 0.245


7/27/2019 SW1451_c001



The acceleration decreases with soil-structure effect in this case

Damping


Fixed base structure 5% 5% 5%Soil-structure interaction 9.2% 10.9% 11.15%

Damping constant for all mode for fixed base case varies with mode forcoupled analysis but is neither 5% min. nor 15% maximum but somewherein-between which is quite logical.

Base Shear (kN)


Fixed base structure 102 54.5 49.1Soil-structure interaction 68.4 11.4 0.537

➔ A significant reduction in base shear, considering the soil effect, thoughconceptually it can be predicted that amplitude of vibration will increase.

Shear Force per floor

Fixed Coupled Fixed Coupled Fixed Coupled base with soil base with soil base with soil

Modes ➔ 1 1 2 2 3 3Storey1 10.8 7.2 5.74 1.2 5.17 0.00562 43.1 28.8 22.9 4.8 20.7 0.226Top 48.4 32.4 25.8 5.4 23.3 0.255

➔ Significant variation in floor shears per mode.

Based on the above example it can be concluded that

• The major advantage with this technique is the calculation of the time periodwithout resorting to an elaborate modelling of the soil. Two representative springvalue for the foundation is capable of modifying the stiffness of the super-structurehaving any conceivable degree of freedom.

• This cuts down significantly the modelling as well as the cost of computation.• No rigid body motion exists.• Stiffness matrix of the soil structure system is symmetric and real.• The structure can be discretized to as many degrees of freedoms one choose to

select.


7/27/2019 SW1451_c001



• Beam, plates, shell, bricks anything can be used to model the super structuresystem thus do not generically violate the procedures followed for FEM analysisof the superstructure.

• Since the matrix has no rigid body mode may be also be used directly for calculatingthe static response too. No additional computational effort is required.

• Though approximate, furnishes a rational basis of estimating the modal dampingratio per mode for the coupled soil structure-system.

• The results are logical and in general satisfies the trend as observed based on morerigorous analysis based on complex damping and eigen value problem (where amatrix of order n × n gets inflated to the order 2n × 2n thus adding to the cost of computation).

1.3.6 Some fallacies in coupling of soil and structure(Chowdhury 2008)

You will observe here that we had advocated two types of coupling of soil spring, onevide Equation (1.1.3) where the soil spring is directly added to the diagonal stiffnesselement of the structural matrix, meaning thereby that it is a parallel connection andthe other by Equation (1.3.7) which shows that the spring are in series.

The first method has developed from the theory of nodal compatibility and is a verypopular technique in practice for the root of its development is in the realms of matrixanalysis of structure and can very well be adapted in commercially available software.

While the second formulation is developed in the frequency domain analysis as

suggested by Veletsos for a harmonic oscillator having single degree of freedom coupledto a translational and rocking spring.The question that remains as to which one is more realistic and gives the true

interaction of the soil with structure especially when we model the soil as boundarysprings.

One of the major flaws in parallel spring model is, as the boundary elements arediscrete and not a continuum it only gives a local effect and also affects the structuralnode only locally.

The intention here is not to challenge or shock the structural engineers who havebeen doing this for ages. But putting on the hat of a theoretical physicist and probing

this formulation a bit more – it comes up with some very interesting result.Let us imagine that the beam in Figure 1.1.1 is made of RCC of say dimension

450× 900 supported on a compliant foundation where the soil is modeled as a spring.Now we put a motor on the beam which gives some dynamic force Psin wmt -say. Wewant to find out the dynamic response of the beam. The problem shows no ambiguityfor the beam along with the soil spring vibrates with natural frequencies that can beobtained based on the lumped mass matrix at node i & j and the stiffness matrix derivedvide Equation (1.1.3) and then subsequent amplitude and stresses can be calculatedby the usual procedure.

Now let us presume 100 years down the road scientists have developed a materialwhose Young’s Modulus is say 9 × 1020 kN/m2 and we build this beam (of samedimension) with this material (not the spring which represents the soil) and pose thequestion as to does the system vibrate? Looking at Figure 1.1.1 one can intuitively saythat yes it does vibrate, but the beam here being very stiff (lim k → ∞) undergoes rigid


7/27/2019 SW1451_c001



body mode and the vibration is now guided by the stiffness of the spring only. Nowthe question is – does Equation (1.1.3) reflects this phenomenon – amazingly not! For

putting this value of E = 9 × 1020 kN/m2 we find that [K] g becomes an infinitely stiff

matrix where the poor Kii and K jj (whose order would be of 105 to 106) is completely

gobbled up by the stiffness values of the beam that are exponentially higher and wouldstart giving time periods that are zero.Like patch test in FEM, it is a test we can use to check the sanctity of a stiffness

formulation. We call this an RB (short of Rigid Body) test and we see it fails thistest with parallel spring connection, especially when the structure has got significantstiffness compared to soil.

Now if we put Equation (1.3.7) which is the series connection, to RB test, we findthat it passes the test with flying colors for as Limit of K → ∞ the first term in the righthand side of Equation (1.3.7) approaches zero and we are left with the soil springsvalues only based on which the body vibrates and satisfies RB test conditions posed

earlier.In Equations (1.3.8) and (1.3.9) it is clearly seen that the soil flexibility gets directly

added to the diagonal and then on inversion affects all the terms of the [Ke] and gives thetrue interaction unlike parallel spring which affects only locally the interaction effectsand does not possibly gives a true picture when the stiffness of the superstructurebecomes quit high compared to that of the soil.

1.3.7 What makes the structural response attenuateor amplify?

In Example 1.2.5 and 1.2.6 we had shown two opposite cases of dynamic soil structureresponse. While in the case of the chimney the response is attenuated, in case of thehorizontal vessel the response is however significantly amplified. One would obviouslybe curious and wonder why does it happen? The riddle is surely not difficult to answer.Shown in Figure 1.3.3 is the generic nature of the acceleration curve used for designof structures under earthquake. The nature of the curve is almost common/similar forall the earthquake codes around the world.

Based on the curve (Figure 1.3.3) it is evident that when the structure is very stiff or massive and its fixed base time period hovers around the vicinity of point A, the

dynamic soil structure effect can show significant amplification so long as the cou-pled time period of the soil-structure system is within the zone C. Thus structureslike massive gravity dams, nuclear reactor buildings, Massive turbine foundations20,large vessels supported on short pedestals (which are stiff) could show significantamplification in response when the effect of soil is considered in the analysis.

While for any structure whose fixed base time period is somewhere between pointB and C, if exceeds the point C consideration of the soil effect can undergo a majorattenuation. Normal buildings, RCC, Steel Chimneys, elevated water tanks etc wouldpossibly fall in this category. Thus depending on the stiffness of the structure, its mass

distribution, dynamic property of the soil one can either save some money (if there is

20 For instance the structural configurations used for old LMW type Russian turbo-generators usedcommonly in India for 210 MW plant.


7/27/2019 SW1451_c001



B C

g Sa

A

D

Time period (sec)

Figure 1.3.3 Generic response spectra curve for earthquake.

attenuation) or could result in more costly design (for amplified response) which mayvary from case to case.

1.4 THE ART OF MODELLING

Computer Modelling of soil & structure optimally to arrive at a meaningful solutionis an art by itself, and can well be a topic of a complete book. We present hereaftersome major techniques that has been found to effective & reasonable.

1.4.1 Some modelling techniques

Experience shows that in many cases young engineers eager on get-going mode wouldstart from the very outset with an elaborate model of the whole soil-structure system21.

They spend significant amount of time on data input and checking of such massive

model and come up with a result whose qualitative difference with a much simplermodel is only marginal. Moreover trying to handle a big data-base, an inadvertentmodeling or input data error passing the scrutiny is not at all uncommon.

So at the very outset our suggestion would be, start with a simple model withouttrying to over sophisticate the issue from the very out set22.

Start with a test case or a simplified model to check the results. For instance a simplemodel given in a book or the user manual of the software in use is a very good startingto have some idea what types of element to choose, what order of refinement sufficeand what type of simplified idealization is acceptable.

21 Problem modeled with a minimum 1000 degrees of freedom!22 Mentioning the fact that you have used eight nodded brick elements, or 9-noded plate elements based

on iso-parametric formulation may look impressive as technical jargons in a design basis report but maynot always be cost-effective solution.


7/27/2019 SW1451_c001



Super structures above ground

Ground level

Railway Carriage

Underground Tunnel

Figure 1.4.1 An underground tunnel for movement of train in a metro city.

You will be amazed to find that in most of the cases, modeling the soil intelligentlyas linear springs (whose values are judiciously chosen) can be good enough for manymajor soil structure interaction analyses. Specially, when the structure is modeled in3D, avoid using Finite elements to model soil and coupling it to the structure.

Firstly, the model becomes huge resulting in more engineering time plus gives resultswhich become difficult to decipher and does not necessarily always gives a moreaccurate or better result compared to a relatively simplified model.

Start with a simple model (preferably a stick model) and add the soil spring to geta first order feel of how much the soil affects its response 23.

Get a basic feel as to how much the results vary in terms of fixed base problem-if found significant one should then and only then resort to a much more detailedanalysis.

If the variation is say within 15%, one can well ignore the soil effect and considerthe problem as a standard fixed base problem and proceed with the analysis.

Keep your eyes open but do not be biased on the issue. Optimize your engineeringeffort to the best possible way.

There are certain types of problem where resorting to FEM however would becomealmost essential. For instance for the problem considered in Figure 1.4.1, it would beimpossible to arrive at reasonable solution without an application of FEM.

Shown in Figure 1.4.1 is a sketch of an underground tunnel catering to movementof high speed trains. The movement of train generates dynamic forces which travelsthrough the soil to the surface and could adversely affect the structures built on thesurface like buildings, water tanks etc and becomes an important study for engineers

undertaking such kind of projects.

23 A computer analysis is not mandatory at this stage, a simple hand calculation or an analysis in spreadsheet or MATHCAD would suffice.


7/27/2019 SW1451_c001



It is but evident that for these cases of modeling, the soil as spring element will notwork and a comprehensive finite element modeling of the soil based on plane strainelement is required. Here also, while doing the modeling, our suggestion would bestart with a crude model (say 20 to 30 elements) to get a fill of the first order effectsand then progressively refine the model to get a more accurate result.

In static loading case in Chapter 4 (Vol. 1) we had explained the principles of meshing of such plane strain problem. Under dynamic loading the principles meshingare generally done based on the following

1 Find the time period of the exciting frequency (T s) of the soil medium as 4H / vs.2 If vs is the shear wave velocity of the soil medium then for λ being the wavelength of

the propagating waves they are related by vs = f λ. Here f is the natural frequencyof the medium and f = 1/T s.

3 Thus obtain λ

=vs

·T s.

4 The mesh size should preferably be λ/10 to λ/4 for linear or bilinear/quadraticelements chosen.

One of the major limitations in FEM for wave propagation problem is that theboundary has to be taken to a significant distance away from the source to ensureno waves are reflected back which would otherwise generate spurious modes. Thisoften makes the problem expensive in terms of data input, checking and run time.Moreover, it is difficult to gauge at the outset as to where can the boundary beterminated.

Infinite finite element as discussed in Chapter 4 (Vol. 1) is one alternative which hasbeen found to have a strong potential for catering to such problem.Other than this, paraxial boundaries or providing viscous dampers at the boundary

of soil domain capable of absorbing the propagating waves are often used for this typeof problems24.

Else boundary elements have also been used to model such infinite domains and arecoupled to the superstructure (modeled by FEM) and an effective solution has beensought.

Unfortunately most of the commercially available software do not have the provi-sion of adding matrix which can be assembled to the FEM matrix and an engineer hasto write his own special purpose software to cater to such problems.

Finally a word on the soil. . . . . .

Irrespective of whether we use springs or finite element to model the soil, the fun-damental property on which the stiffness depends, are the value G (Dynamic shearmodulus) and ν, the Poisson’s ratio. We had discussed in detail as to how to arrive atthe appropriate design values of these two parameters in the next section.

In spite of all the techniques used it should be clearly mentioned that the parameters

are still marred by uncertainties and the results thus obtained should be mellowed withsome judgment which comes out only of experience and sustained practice.

24 Refer Chapter 5 (Vol. 1) for detailed discussion on this issue.


7/27/2019 SW1451_c001



It is always preferable to do some parametric study by varying the design soil valuesby (±)15 to 20% (depending on how reliable and exhaustive has been the geotechnicalinvestigation) and check how much these results affect the design values and preferablya conservative and safe value should be chosen (based on this variance).

We mention in Table 1.4.1, some suggestive models for different classes of structureswhere we start with a primary model (i.e. to get a basic feel of the response) and asecondary model which is a further improvement to the primary model.

Table 1.4.1 Some suggestive models.

Sl. StructureNo. type Primary model Secondary model Remarks

1 Framed Stick model with soil 2D or 3D frame system

building considered as two uni- with masses lumped atque spring (rotational nodes. Soil modeled asand translational) consti- springs under eachtuting all the foundations individual foundation

2 High-rise 2D frame for the beam 3D frame for the beam The horizontal slabbuilding column system while the column system with need not be toowith shear shear wall modeled as shear wall modeled as refined and shouldwalls an equivalent cantile- plain stress elements. be good enough to

ver with soil springs The horizontal slabs generate requisiteunder each column and modeled as plane stress stiffness in its ownthe shear wall elements. Soil modeled plane

as springs below eachfoundation3 Chimneys 2D stick model with soil No further refinement is For local effect model

and eleva- idealized as springs usually warranted unless the shell or super-ted water some local effect of soil structure as a stick tanks is required to be studied and the soil a axis

on surrounding structure. symmetric plainstrain element

4 Frames 2D frames with soil A detailed 3D model Refer Chapter 2 (Vol. 2)support- modeled as springs. constituting of beam on detailed modelinging rotary Bottom raft considered elements with master and technique for thesemachines infinitely stiff hence only slave node option. The type of foundation.

lumped mass contribu- bottom raft discretisedtion is taken. Soil mode- into beam or plate elem-led as springs. 3 to 4 ents with soil modeleddegrees of freedom as springs and connectedusually suffice. at each node of the raft

elements.5 Dams and A simple stick model with A comprehensive 2D

embank- soil modeled as springs model with the damments else time period may be broken up into plain

found from formula strain element and soilsuggested in code and modeled as springs or

modified by Veletsos’s further refined into 2Dformula. plain strain element

depending on the comp-lexity of the soil or theimportance of the dam


7/27/2019 SW1451_c001



1.4.2 To sum it up

Dynamic soil structure interaction is still in its early days and investigators are stilllooking for answers to many problems which are encountered in practice.

For instance soil are modeled as linear springs based on elastic half space theory,

considering it as a linear isotropic medium, but in reality it is not so. Layered soilphenomenon, pore pressure dissipation under dynamic loading, liquefaction potentialand its effect, infinite domain problem, non linear and inelastic behaviour, radiationand geometric damping are some of the important factors on which research is still inprogress to arrive at a more realistic model amenable to design office practice.

What has been presented in this chapter is only an introductory concept and whatis in vogue in practice at the present.

Hopefully in days to come our understanding in some of the issues mentioned abovewill be more profound and engineers and researchers would come up with results which

would be more realistic and reliable.However a word of caution should be pertinent at this juncture.As stated earlier as the uncertainty plaguing the problem is many, one should not

loose the final outcome of what we are trying to achieve i.e. a safe and sound structurewhich can stand the vagaries of nature.

So one should not get lost in the maze of sophisticated mathematics and try to alwayseconomize on the structure based on what the computer out put reflects25.

For facilities important to society the results should always be mellowed with soundengineering practice like good detailing, robust geometric configuration, and goodquality of time tested construction practice.

All these aspects are equally important for a structure to survive the wrath of MotherNature whose ways are still not very clearly known to us.

1.5 GEOTECHNICAL CONSIDERATIONS FOR DYNAMIC

SOIL STRUCTURE INTERACTION

In this section we deal with the geotechnical considerations which go into the processof a successful dynamic soil-structure interaction analysis.

At the very outset we would request readers specially with a strong structural leaning not to ignore this section. For our experience shows that nemesis of many mistakeslies in misinterpretation of this particular topic. As such before launching yourself into linear or non-linear finite element analysis of soil-structure system, the conceptualaspect of the influencing soil parameters, its limitations and its effects should be clearlyunderstood.

As a pre-requisite, we expect that you have some background on . . .

• Some fundamental concepts of Soil Mechanics

•Basic concepts in Soil Dynamics

25 The output is nothing but a reflection of man’s limited knowledge of nature and only an approximatequantification of an idealized mathematical model which could be in significant variance to reality inspite of our best effort.


7/27/2019 SW1451_c001



1.5.1 What parameters do I look for in the soilreport?

To start with we pose the above fundamental question. To readers having some back-ground on this issue may find it intriguing for his obvious answer would be the dynamic

shear modulus (G) and Poisson’s ratio (ν).The obvious query that subsequently comes to mind is, does it require a full section

to be devoted to this issue?The answer would surely be an emphatic yes, for in our opinion the values

adapted are often misunderstood/abused in many a case, and often makes the analysisquestionable or unrealistic.

The reasons that could be attributed to it are as follows:

•Geotechnical test (lab or field) based on which data evaluated are not understood

properly. As the limitations of such data are not clearly made; often results inincorrect interpretation.

• Data considered are often not relevant or correct in terms of real situation in thefield, specially for layered soil.

• Insufficient data and or lack of knowledge on the strain level to which thefoundation-structure system will be subjected to – specially during earthquake.

• Lack of dynamic test data and improperly co-related value from static soilparameter which could be widely varying with the reality.

• Finally, often forgetting the bottom line that unlike man made material like con-

crete and steel, soil is far more heterogeneous and unpredictable; thus for a realsoil structure interaction it is unfair to have an analysis on an absolute scale. Itshould preferably be done for a particular range of values and the best estimate isto be made out of it – and this is where engineering judgment would count to alarge extent.

Having made the above statements, let us evaluate various aspects of dynamicproperty of soil which are important for an integrated soil-structure interactionanalysis.

Before even looking at soil report the analyst should be clear with himself on

• The type of structure he is dealing with• Type of foundation that is anticipated like shallow foundation (could be isolated

or combined footing), raft or piles etc.• What analysis he is looking for like is it an analysis for machine induced load,

earthquake, blast force etc.

Understanding of the above criteria will not only help him in understanding the

data obtained from different tests but could also possibly make him realize theirinterpretation in a more realistic perspective.

The engineering parameters we look for in the soil report for develop-ing the soil model either for finite element or linear/non-linear spring dashpotmodel are


7/27/2019 SW1451_c001



(shear stress)

2

1

1 2

G1

G2

Figure 1.5.1 Shear stress-strain curve of soil under cyclic loading.

1 Dynamic shear modulus (G) or shear wave velocity (vs)26

2 Poisson’s ratio (v)3 Damping value of soil both radiative and material.

The values are usually obtained either from field test, laboratory test or from theo-retical co-relation with other engineering soil parameters. Before we step further intothe topic it would possibly be worthwhile to understand how soil behaves under cyclicloading and what its characteristics are.

It should be remembered that even under low strain, soil behavior is essentiallynon-linear though at low strain it does show some kind of linearity.

Shown in Figure 1.5.1, is the shear stress-strain curve of soil under cyclic loading.It is evident from the above figure that shear strain varies with stress, and goes onincreasing with number of cycles of loading.

Thus before an analysis is being carried out one has to have an idea about the averagestrain range to which the soil will be subjected to under the induced dynamic loading.

The characteristic curve which shows the variation of shear modulus with respectto shear strain is shown in Figure 1.5.1a.

The curve shown above is otherwise known as Seed and Idriss’s (1970) curve whichshows the variation of dynamic shear modulus of soil with shear strain.

26 Relationship being G = ρv2s .


7/27/2019 SW1451_c001



0

0.2

0.4

0.6

0.8

1

1.2

0 0.01 0.1 1 10 100 1000Strain Ratio

G / G 0

Figure 1.5.1a Variation of Shear Modulus with strain under cyclic loading. (Seed & Idriss 1970).

Soil subjected to stress by machine foundation are usually low strain and variesbetween to 10−4 to 10−3%.

However for an earthquake of even moderate magnitude this will be much higher-having strain range varying to 10−2 to even 10−1% for very severe earthquake.

Since it is difficult to gauge at the outset of an analysis how much strain the soil willbe subjected to, the correction factor to be used to modify the data as obtained in thesoil report becomes difficult to quantify.

On the contrary rendering no correction would result in assuming a more stiff soil

and the result obtained based on this could be significantly varying from the reality.Fortunately or unfortunately most of the tests carried out in the field or in the

laboratory for determination of the dynamic shear modulus is based on low strainrange having values restricted to 10−4%.

Thus it should be clearly understood that the dynamic shear modulus data furnishedin the soil report is only valid for LOW strain range and can be only used directlyfor analysis where the strain induced in the soil is significantly low like in design of machine foundations only. For earthquake analysis where the site is situated in anarea of moderate to severe earthquake zone, direct use of such soil dynamic data may

not be valid for design of normal structures, for the strain induced in soil is muchhigher.

1.6 FIELD TESTS

The most common field tests that are carried out at site for evaluation of dynamicshear modulus or shear wave velocity are

1 Block Vibration Test

2 Seismic cross hole

1.6.1 Block vibration test

In block vibration test as shown in Figure 1.6.1, an oscillator is placed on a concreteblock of size 1.5 m × 0.75 m × 0.7 m resting at foundation level and induces dynamic


7/27/2019 SW1451_c001



Oscillator

Lx

Fdn Level

Propagating waves

H=0.6 to 1.2 m

Figure 1.6.1 Schematic diagram for block vibration test.

loading on the soil. Two geo-phones are placed at a distance to pick up the signal fromthe oscillator.

Once the oscillator induces dynamic force on the soil the geo-phones pick up thissignal and transfer them to an oscilloscope which shows an elliptical figure of Lissajous.The operating speed of the oscillator is varied till the time the natural frequency of thesoil and the operating frequency of the oscillator matches (the Lissajous’ figure in theoscilloscope becomes a perfect circle).

The shear wave velocity of the site is then given by

vs = 4fLx (1.6.1)

where vs = shear wave velocity of the soil; f = operating frequency of the oscillatorin cps; Lx = distance between the two geo-phones.

For arriving at meaningful results usually high frequency oscillators (>100 cps) areput to use for which the waves generated are of the order of 0.6 to 1.2 m.

Thus results obtained from this test only influence soil of depth 0.6 to 1.2 m belowthe depth of foundation and should not be used where piles or other types of deepfoundations having influence area propagating much deeper is used.

Trying to induce lower frequency calls for much heavier oscillators which make thetest uneconomical compared to other types of tests.

1.6.2 Seismic cross hole test

As shown in the schematic sketch in Figure 1.6.2, a probe is placed in a bore hole

to the desired depth and shear wave is generated in the soil by hitting it hard with ahammer.

The waves are picked up by a geo-phone entrenched firmly to the casing of anotherbore hole located at a known distance (Lx) from the first hole.

The time taken to pick up the signal is measured by the oscilloscope.


7/27/2019 SW1451_c001


7/27/2019 SW1451_c001



project. Though not unusual, but should not happen as a rule, for this shows the lackof foresight on the part of the engineer while submitting the technical and commercialproposal for a project. Even at the proposal stage the process involved in a plant is wellknown to the bidder and all the concerned civil engineer has to do is to check with hisprocess department and find out if rotating machines are part of the process or not.

On the other hand knowing the location of a particular site one can easily find outfrom the codes how active this zone is seismically and if felt reasonable all he hasto do is to include this additional cost of dynamic geotechnical investigation in hiscommercial bid. People suffer from misnomer that dynamic tests are expensive-whichis actually not true, for an average dynamic test in international market takes roughlyUS$ 20,000–25,000 which would however be 0.25% of a small petrochemical refineryand possibly 0.1% of a combined cycle 350 MW power plant.

Lack of these tests can land up some of the equipments operating in such projectsinto serious problem whose cost itself would constitute 30–40% of the whole

project cost!So one has to decide on the risk involved – and come to a conclusion of its worth.

Though theoretical co-relation exist for evaluation of dynamic shear modulus of soilfrom static soil test (which has been successfully used in project works), it is alwayspreferable to have these dynamic tests carried out at site, for not only does it imbibemore confidence in the design process but engineer should also be aware that “theo-retically co-related values have also varied widely with respect to actual field data, andshould be mellowed with judgment.” Considering the uncertainty prevalent in soil, issurely not an easy task to accomplish.

1.7 THEORETICAL CO-RELATION FROM OTHER SOIL

PARAMETERS

The most outstanding work in establishing theoretical co-relation for evaluating thedynamic property of soil has been done by Hardin, Drnevich, Richart, Seed, Idrissto name a few27. The expressions suggested by them have been successfully used formany real projects by the engineers in the past. We are going to have a look at some

of them hereafter and understand their limitations if any.

1.7.1 Co-relation for sandy and gravelly soil

1.7.1.1 Hardin and Richart’s Formula

For rounded grained soil having void ratio less or equal to 0.8 the dynamic shearmodulus is given by (Hardin and Richart 1963)

G=

2630(2.17 − e)2

1 + e

√ σ

0in psi (1.7.1)

27 This is by no mean to ignore other researchers who have contributed significantly to this difficult study.We only name a few, which are popular in practice.


7/27/2019 SW1451_c001



For angular grained soil the dynamic shear modulus is given by

G = 1230(2.97 − e)2

1 + e

√ σ 0 in psi. (1.7.2)

where, G = dynamic shear modulus of the soil in psi, e = in-situ void ratio of the soilsample, σ 0 = mean effective stress in psi = 0.333σ v(1 + 2K0), σ v = vertical effectivestress in psi, σ h = horizontal effective in psi = K0σ v, K0 = earth pressure at rest, andis a function of the plasticity index and the over-consolidation ratio.

The relationship between plasticity index, over-consolidation ratio and K0 is asshown in the following figure.

Figure 1.7.1 Value of the K0 after Brooker & Ireland (1965) Reproduced by permission of the NationalResearch Council of Canada from the candian geotechnical Journal Vol-2 (1965).

1.7.1.2 Seed and Idriss Formula

The formula for dynamic modulus in this case, Seed and Idriss (1970) have been relatedto relative density of sand which can usually be quantified from SPT test and is given by

G = 83.3K2√ σ 0 in psi (1.7.3)

Here K2 is a function of the relative density of the sand which can again be estimatedfrom the SPT value.

The relationship between SPT value and the relative density is as given Table 1.7.1.


7/27/2019 SW1451_c001



Table 1.7.1 Soil properties with SPT values.

SPT value Compactness Relative density Angle of friction

0–4 Very loose 0–15 <28

4–10 Loose 15–35 28−3010–30 Medium 35–65 30−3630–50 Dense 65–80 36−41>50 Very dense >85 >41

Table 1.7.2 Values of K 2 versus relative density atstrainof 10−3% (Seed and Idriss 1970).

Relative density (%) K 2

90 7075 6160 5245 4340 4030 34

For case of computer programming K2 can also be represented by the expression

K2 = 0.6Dr + 16 (1.7.3a)It is to be noted that in this case to determine the relative density, the observed SPT

value has to be corrected for the overburden pressure and dilatancy to arrive at thedesign SPT value before it is co-related with the above table.

1.7.1.3 Corrections to SPT value

Though available in standard textbooks of Soil Mechanics and Foundation Engineer-ing, for brevity we present the correction expressions as mentioned hereafter.

For dilatancy correction if the observed SPT value (N 0) is greater than 15 then the

corrected SPT value N is given by (Terzaghi and Peck 1967).

N = 15 + 1

2(N 0 − 15) (1.7.4)

The overburden correction as per Peck et al. 1980 is given by

N = 0.77N log10

2000

p for p ≥ 25 kPa (1.7.5)

For p ≤ 25 kPa, as per Murthy (1991)

N = 4N

2 + 0.034 p (1.7.6)


7/27/2019 SW1451_c001



in which, N = corrected SPT value for overburden, N = corrected SPT value fordilatancy, p = gross overburden pressure in kN/m2.

1.7.1.4 Ohsaki and Iwasaki’s formula

Ohsaki and Iwasaki (1973) have given co-relation for dynamic shear modulus directlyco-related to SPT value and is expressed as

G = 12000 N 0.8 in kPa (1.7.7)

Here N = design SPT value at the site after relevant corrections.

Example 1.7.1

As shown in Figure 1.7.2 is a small site having dimensions 18 m × 6 m whichwould be supporting a Compressor unit and a few pumps, for which four boreholes were dug at four corners as shown. The soil was found to be cohe-sionless in nature and SPT values observed at the four bore holes are as tabledhereafter

Depth BH1 BH2 BH3 BH4(meter) (SPT value) (SPT value) (SPT value) (SPT value)

2 4 6 4 34 8 6 6 56 12 9 11 88 15 12 16 1110 20 18 24 1614 22 24 28 20

18.0

BH2

6.0

BH4

BH1

BH3

Figure 1.7.2


7/27/2019 SW1451_c001



Based on Laboratory and field analyses following parameters were furtherestablished:

Ground water table = 1.6 m, below grade level

Saturated density of soil = 22 kN/m3

Void ratio e0 = 0.58; Plasticity Index = 0.0; Poisson’s ratio = 0.32Determine the best estimate of dynamic shear modulus (G) of soil at 10.6meter below ground level presuming no dynamic soil test was done duringgeo-technical investigation.

Solution:

Average observed SPT value at a depth of 10.0 meter

=20 + 18 + 24 + 16

4 =19.5

=20 (say)

Average observed SPT value at a depth of 14.0 meter

= 22 + 24 + 28 + 20

4= 23.5 = 24 (say)

At a depth of 10.6 meter below ground level based on linear inter-polationaverage observed SPT Value

=24

−20

4 × 0.6 + 20 = 20.6 ∼= 21 (say)

The above observed SPT value has now to be corrected for dilatancy andoverburden pressure

1 Correction for dilatancy

As per Terzaghi, corrected SPT (N ) value is given by

N = 15 +1

2 (N 0 − 15) for N > 15; (1.7.8)

or N = 15 + 1

2(21 − 15) = 18 (1.7.9)

2 Correction for overburden pressure

As per Peck

N = 0.77N log10 2000 p for p ≥ 25 kPa (1.7.10)

N = Corrected SPT value for overburden; N = Corrected SPT value fordilatancy; p = Gross overburden pressure in kN/m2.


7/27/2019 SW1451_c001



Here p = 22 × 10.6 = 233.2 kN/m2

Substituting above in Peck’s formula we have, N = 0.77 N log10

2000

233.2= 13

Thus, corrected design SPT value = 13Referring to table 1.7.1 for N = 13, Dr (Relative density) = 39.5%Net overburden pressure at 10.6 meter level is expressed as

σ v = (22 − 10) × 9 + 22 × 1.6 = 131.6 kN/m2(18.718 p.s.i)

As there is no previous history of loading on the site O.C.R. = 1.Thus for P.I.

=0.0 and O.C.R

=1 as per Brooker and Ireland’s curve we

have K0 = 0.48

Considering confining pressure

σ o = 0.333σ v(1 + 2K0); we have σ 0 = 18.718

3(1 + 2 × 0.48) = 12.22 p.s.i.

As per Hardin and Richart’s formula

G = 2630(2.17 − e)2

1 + e

√ σ 0

G = 2630(2.17 − 0.58)2

1 + 0.58

√ 12.22 = 14710.5 p.s.i. (101426 kN/m2

)

(1.7.11)

As per Seed and Idriss formula

Referring to the chart given above for Dr = 39.5% and strain in the range of 10−3% (usually valid for machine foundation) K2 = 40.And as G = 83.3 K2

√ σ 0 we have

G = 83.3 × 40 ×√

12.22 = 11647 p.s.i. (80308 kN/m2)

Thus taking average value of G based on Hardin and Seed’s method

Average G = 101426 + 80308

2= 90867 kN/m2

As per Ohsaka and Iwaski’s formula


7/27/2019 SW1451_c001



G = 12000 N 0.8 in kPa➔ G = 12000 × (13)0.8

= 93397.6 kpa (93398 kN/m2)

Thus it will be observed that variation with average G obtained based onHardin, Seed’s and Ohsaka’s formula is not significant and is of the order of 2.7%28.

1.7.2 Co-relation for saturated clay

1.7.2.1 Hardin and Drnevich formula

Hardin and Drnevich (1973) have given the following formula applicable to clayey

soil as

Gmax = 1230(2.973 − e)2

(1 + e)(OCR)k(σ 0)

0.5 in psi (1.7.12)

where, e = void ratio; OCR = over consolidation ratio; σ 0 = mean effective stress inpsi = 0.333 (σ v + 2σ h); σ v = vertical effective stress in psi; σ h = horizontal effectivestress in psi = K0σ v, K0= earth pressure at rest, and is a function of the plasticity indexand the over-consolidation ratio.

k=

is a function of the plasticity index (PI) of the soil and is given as

k = −5 × 10−8(PI )3−4 × 10−5(PI )2 + 0.0092(PI ) + 0.0025 (1.7.12a)

It is to be noted that Gmax as obtained above corresponds to a shear strain range of 0.25×10−4% and needs to be modified for the appropriate strain range as appropriatefor a problem in hand based on the expression

G=

Gmax

(1 + ψ/ψr)(1.7.13)

Here ψ = desired strain range; ψr = reference strain range and is expressed as

ψr = τ max

Gmax× 100 and

28 The point we are trying to make here is not to go by one formula, but check with possibly all of themand comparing them to arrive at result which would possibly be best fit and hopefully be most realistic.Here again it is to be noted that we had not used the angular sand formula of Hardin, if the soildescription does not reflect it or the soil has both rounded and angular grains an intermediate value hastobe chosen judiciously.


7/27/2019 SW1451_c001



τ max =

1 + K0

2(σ v − u) sinφ + c cosφ

2

−

1 − K0

2(σ v − u)

20.5

(1.7.14)

in which, σ v = total vertical stressing in soil; u = pore pressure; c = cohesion of soil;φ = angle of friction of soil, and K0 = coefficient of earth pressure at rest.

Example 1.7.2

It has been decided to place foundation of an industrial structure at 4.0 meter below the existing ground level.

Based on laboratory and field tests it has been found that the Ground watertable is at a depth of 1.0 meter below GL.

Unconsolidated undrained triaxial tests reveal the sample to have the followingvalues:

• Cohesion value c = 0.21 kg/cm2

• Angle of resistance = 18 degrees• Pore pressure = 0.0 kg/cm2

Consolidation tests reveal that it had a history of pre-consolidation pressureof 200 kN/m2:

• Initial void ratio = 0.61• Plasticity limit PI = 35• Saturated unit weight of soil = 19 kN/m3

The site has a history of moderate to severe earthquake when from previous

record it is observed to generate a strain range up to 0.1%.Calculate the dynamic shear modulus of soil for this predicted strain

range.

Solution:

For foundation located at 4.0 meter below the ground level net vertical pressure

σ v = 19 × 1.0 + (19 − 10) × 3.0 = 46 kN/m2(6.54 psi)

OCR = 200

46= 4.34, for plasticity index of 35 from Brooker and Ireland’s,

chart K0 = 1.1


7/27/2019 SW1451_c001



Thus considering σ o = 0.333σ v(1 + 2K0), we have

σ o = 0.333 × 6.54(1 + 2 × 1.1) = 6.976 psi

Gmax = 1230 (2.973 − e)2

(1 + e)(OCR)k(σ 0)0.5 in psi

Here k = 0.27 for PI = 35 as per Equation 1.7.12aThus substituting the values we have

Gmax = 1230(2.973 − 0.61)2

(1 + 0.61)(4.34)0.27(6.976)0.5

= 16746 psi (115465 kN/m2

)

Calculation for Shear stress

τ max =

1 + K0


2

−

1 − K0

2(σ v − u)

20.5

or, τ max =

1 + 1.12

(6.54) sin18 + 3.0457 cos 182

−

1 − 1.12

(6.54)20.5

= 5.00 psi.

ψr = Reference strain range and is expressed as

ψr = τ max

Gmax× 100 = 5.00

16746× 100 = 0.0299%

Thus for 0.1% strain. . .

G = Gmax1 + ψ

ψr

➔ G = 1154651 + 0.1

0.0299

= 26577 kN/m2

It is thus observed that dynamic shear modulus is 23% of the theoretically calc-ulated data.

Based on the above example it would perhaps be not difficult to realize that howimportant role does the strain range plays on the design value of dynamic shearmodulus of soil.


7/27/2019 SW1451_c001



1.8 ESTIMATION OF MATERIAL DAMPING OF SOIL

Damping plays a significant part in the overall response of soil structure system. Whilefor structural members material damping plays a significant part (mostly consideredas Rayleigh damping), for soil, two types of damping are basically involved.

• Radiation damping• Material damping

Radiation or geometric damping of a soil foundation system is a mean by whichthe energy is dissipated by means of radiation from the source and is a function of mass and inertia of the system29. Material damping of the soil foundation system isa mean by which the energy is dissipated by hysteresis and is an inherent property of

the constituting material of the soil.This can very well be found from resonant column test in the laboratory when

after the soil has been vibrated the exciter is stopped and successive amplitudes aremeasured. If a1 and a2 are two successive amplitudes then

Dm =

lna2

a1

4π2 +

ln

a2

a1

2

(1.8.1)

The total damping ratio of a soil foundation system is sum of radiation and materialdamping. It is generally observed that material damping has a significant magnituderelative to radiation damping specially in rotational modes. In such cases total dampingrather than geometric damping should be used to obtain the response of the structurefoundation system.

For translatory mode, on the contrary material damping plays an insignificant roleand may be neglected in the analysis. Thus for tall narrow structures like chimney,Boiler structures, tall buildings where the coupled horizontal and rocking mode couldplay significant role it would perhaps be realistic to also consider the material damping

of soil in order to have a meaningful response.

1.8.1 Whitman’s formula

Whitman (1973) has suggested that total damping (geometric + material) for amachine foundation can be obtained from the expressions

Horizontal Mode

Dh =

0.31 Mρr3

0

(1.8.2)

29 We have dealt this detail in Chapter 2 (Vol. 2) – “Design of Machine foundations”.


7/27/2019 SW1451_c001



For vertical mode

Dv = 0.49

Mρr3

0

(1.8.3)

For rocking mode

Dθ = 0.05 + 0.1

⎡⎣ I θ

ρr50

0.5 1 +

I θ

4ρr50

⎤⎦−1

(1.8.4)

Here M = mass of foundation plus structure or machine vibrating; I θ = massmoment of inertia of foundation plus machine/structure about a horizontal axis

through the base of the foundation perpendicular to the plane of rocking; r0 =equivalent radius of footing, and ρ = mass density of soil.

1.8.2 Hardin’ formula

Hardin (1965) has expressed material damping of sandy soil by the expression

Dm = 0.985ψ0.2r√

σ 0(1.8.5)

Here notations are same as expressed earlier except the fact that the confining pres-sure σ 0 is expressed in kPa. The equation is valid for shear strain amplitude of 10−6

to 10−4 with a confining pressure of 24 kPa to 144 kPa.For a particular strain range the value obtained above can be corrected based on the

expression

Dc

Dm= ψ/ψr

1 + ψ/ψr(1.8.6)

Example 1.8.1

For the example as shown in Example 1.7.2, estimate the damping ratio of thesoil as per Hardin’s formula. The soil properties remain same as given in Example1.7.2.

Solution:

Based on the solution furnished in Example 1.7.1 value of dynamic shearmodulus is given by

G = 90867 kN/m2; σ 0 = 12.22 p.s.i. (85.9 kN/m2); K0 = 0.48


7/27/2019 SW1451_c001



For design SPT value N = 13, φ = 31◦, and σ v = 131.6 kN/m2

Considering, τ max =

1 + K0


2

−

1 − K0

2(σ v − u)

2

0.5

, we have

τ max =

1 + 0.48

2131.6sin31

2

−

1 − 0.48

2131.6

20.5

= 36.67 kN/m2

As ψ = τ G

× 100, we have ψr = 36.6790867

× 100 = 0.0404%

Considering Dm = 0.985ψ0.2r√

σ 0, we have

Dm = 0.985(0.0404)0.2

√ 85.9

= 0.056

Thus material damping ratio is estimated as 5.6%.

1.8.3 Ishibashi and Zhang’s formula

Ishibashi and Zhang (1993) has proposed an expression for the damping ratio of

plastic and non-plastic soil and is given by

ζ = 0.3331 + exp(−0.0145PI1.3)

2

0.586

G

Gmax

2

− 1.547G

Gmax+ 1

(1.8.7)

The notations for the above expression are already explained in earlier formulas.We show below variation of damping ratio with G/Gmax for different Plasticity

Index based on the above formula.It will observed (Figure 1.8.1) that as G/Gmax reduces, as damping ratio goes

on increasing meaning thereby that as strain increases damping ratio goes onincreasing. Variation of Damping with strain vide Equation (1.8.6) is shown inFigure 1.8.1a.


7/27/2019 SW1451_c001



Variation of damping ratio

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

1 0 . 9

0 . 8 0 .

7 0 .

6 0 . 5

0 . 4

0 . 3

0 . 2

0 . 1 0

G/Gmax

D a m p i n g r a t i o

PI = 10

PI = 20

PI = 30

PI = 40

PI = 50

PI = 60

PI = 70

PI = 80

PI = 90

PI = 100

Figure 1.8.1 Variation of damping with plasticity index as per Ishibashi and Zhang (1993).

0

0.2

0.4

0.6

0.8

1

1.2

0 0.01 0.1 1 10 100 1000

Strain Ratio

D / D r

Figure 1.8.1a Variation of damping ratio with strain under cyclic loading.

Example 1.8.2

For the clayey soil sample as shown in Example 1.7.2, determine the dampingratio for the strain range level of 0.1% based on Zhang’s formula. Consider allsoil properties same as Example 1.7.2?

Solution:

Based on earlier example we have seen that plasticity index PI = 35.


7/27/2019 SW1451_c001



In Example 1.7.2 we have already calculated that for 0.1% strain G/Gmax =0.230 Substituting the above in Ishibashi and Zhang’s formula we have

ζ = 0.3331 + exp(−1.47)

2[0.586(0.23)2 − 1.547 × 0.23 + 1] = 0.1382

Thus estimated damping ratio is 13.82%.

1.9 ALL THINGS SAID AND DONE HOW DO WE ESTIMATE

THE STRAIN IN SOIL, SPECIALLY IF THE STRAIN IS LARGE?

We acknowledge at the very outset that posing the question, though easy, is not veryeasy to answer. The uncertainties involved are so widely varying that it would bedifficult to give a precise answer to this issue. To the best of our knowledge there isno straight forward answer to this problem and the best one can achieve is a reason-able estimate or can possibly study a range of values and try to predict the overallbehavior.

For high speed centrifugal machine foundation it does not pose a serious problemfor at the low strain range a few percent here and there does not contribute a significant

variation to these values.But for impact type of machines (hammer foundations) and slow speed machines

(coal mill foundations, reciprocating compressors) induced strain could be larger thanstrain developed during field test, for which the correct estimation of Gdyn and dampingbecomes important.

For earthquake of course the strain would invariably be larger than measured duringtest, even for a moderate earthquake when as the strain range increases, degrada-tion in soil stiffness becomes significant and has a major contribution to the overallresponse.

It is obvious that strain induced in soil will depend upon the strength of dynamicloading, the geological condition of the site, stress history of soil and a number of other factors. So the point remains that if there exists no previous records of strainfrom similar machine in same site or from previously occurring earthquake data howdoes one rationalize the strain?

We discuss below some of the techniques which could be used for evaluation of thestrain induced in the soil.

1.9.1 Estimation of strain in soil for machinefoundation

For machine foundations the present practice of arriving at strain dependent dynamicshear modulus and damping can be structured as follows:


7/27/2019 SW1451_c001


7/27/2019 SW1451_c001



on the shaft is 3.5 kN · sec2 /m rotating at an eccentricity of 0.4 mm havingoperating frequency of 1800 rpm.

The soil investigation has revealed the soil data as follows

• SPT=

13 (After dilatancy and overburden correction)• Plasticity Index (PI) = 0• Poissons Ratio = 0.3

Calculate the correct value of dynamic shear modulus and damping.

Solution:

Based on Ohsaka’s formula G = 12000 × (13)0.8 = 93397.65 kN/m2

Assumed strain level=

1×

10−

4%Weight of foundation = 6 × 3.2 × 2.5 × 25 = 1200 kN; Weight of machine =

300 kN, Total weight = 1500 kNMass of foundation + machine = 1500/9.81 = 152.9052 kN · sec2/m

Equivalent radius of the foundation (r0) =

6 × 3.2

π= 2.472155 m

Equivalent vertical spring stiffness of soil30.

Kz = 4Gr0

1 − υ= 4 × 93397.65 × 2.472

0.7= 1.32 × 106 kN/m

ωn =

Kz

m=

1.32 × 106

152.9= 92.89 rad/sec;

ωm =1800

×2

×π

60 = 188 radian/sec r = ωm/ωn = 2.029.

Pdyn = m · e · ω2m = 3.5 × 0.4

1000× (188)2 = 49.7428 kN.

Considering transmissibility as

T r =

r2 1 +(2ζ r)2

(1 − r2)2 + (2ζ r)2we have, T

r =0.65078.

30 Refer to Chapter 5 (Vol. 1) – Basic Concepts of Soil Dynamics, for details of this formula.


7/27/2019 SW1451_c001



Equivalent static force on foundation = 0.65078 × 49.7428

6 × 3.2= 1.686 kN/m2

Considering ψ(%)

=

12qdyn

G =

12 × 1.686

93397.65 =2.17

×10−4%

Considering G = Gmax

(1 + ψψr)

we have, New G = 93397.651 + 2.17×10−4

1×10−4

= 29497.88 kN/m.

We proceed with second cycle of iteration with this new value of G.

Shown below is such iteration for 14 cycles

Cycles 1 2 3 4 5 6 7

G dyn 93397.65 29497.88 22553.77 20646.63 20025.37 19811.83 19737.08Damping 0.012987 0.189764 0.21998 0.228656 0.231517 0.232505 0.232851K z 1.32 × 4.17 × 3.19 × 2.92 × 2.83 × 2.80 × 2.79 ×

10+06

10+05

10+05

10+05

10+05

10+05

10+05

ωn 92.89143 52.20392 45.64753 43.67493 43.01282 42.78288 42.7021r (w m/w n) 2.029203 3.610755 4.129371 4.315875 4.382311 4.405864 4.414199T r 0.65078 0.298033 0.255622 0.243327 0.239243 0.23783 0.237334P dyn 49.74281 49.74281 49.74281 49.74281 49.74281 49.74281 49.74281P eq 32.37161 14.82497 12.71537 12.10377 11.90063 11.83032 11.80565qdyn 1.686022 0.772134 0.662259 0.630405 0.619824 0.616163 0.614878Strain 2.17 × 3.14 × 3.52 × 3.66 × 3.71 × 3.73 × 3.74 ×

10−04 10−04 10−04 10−04 10−04 10−04 10−04

Cycles 8 9 10 11 12 13 14

G dyn 19710.75 19701.45 19698.17 19697.01 19696.60 19696.45 19696.40Damping 0.232973 0.233016 0.233031 0.233037 0.233039 0.233039 0.23304K z 2.78 × 2.78 × 2.78 × 2.78 × 2.78 × 2.78 × 2.78 ×

10+05 10+05 10+05 10+05 10+05 10+05 10+05

ωn 42.6736 42.66353 42.65998 42.65872 42.65827 42.65812 42.65806r (w m/w n) 4.417147 4.418189 4.418558 4.418688 4.418734 4.41875 4.418756T r 0.237159 0.237097 0.237075 0.237068 0.237065 0.237064 0.237064P dyn 49.74281 49.74281 49.74281 49.74281 49.74281 49.74281 49.74281P eq 11.79696 11.79388 11.7928 11.79241 11.79228 11.79223 11.79221qdyn 0.614425 0.614265 0.614208 0.614188 0.614181 0.614179 0.614178Strain 3.74 × 3.74 × 3.74 × 3.74 × 3.74× 3.74 × 3.74 ×

10−04 10−04 10−04 10−04 10−04 10−04 10−04

We show hereafter the variation of strain, damping and G dyn per cycle in Figs. 1.9.1 to 1.9.3.


7/27/2019 SW1451_c001



Variation of strain per cycle

0.00E+00

5.00E-05

1.00E-04

1.50E-04

2.00E-04

2.50E-04

3.00E-04

3.50E-04

4.00E-04

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Number of cycles

S t r a i n ( % )

Figure 1.9.1 Variation of strain (%) per cycle.

Variation of damping ratio with strain

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Number of cycles

D

a m p i n g r a t i o

Figure 1.9.2 Variation of material damping per cycle.


7/27/2019 SW1451_c001



Variation of Gdyn with strain

0

20000

40000

60000

80000

100000

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Number of cycles

G d y n ( k N / m 2 )

Figure 1.9.3 Variation of dynamic shear modulus per cycle.

From the tables and the above plots it is observed that the value becomesconstant after 7th cycle of iteration based on which we conclude that designvalues are as follows

Gdyn

=19700 kN/m; Material Damping ratio

=0.23, and Estimated strain

range = 3.74 × 10−4%.

Thus actual design of the foundation shall be carried out based on this correctedvalue instead of the initial values as mentioned in the soil report.

1.9.2 Estimation of soil strain for earthquakeanalysis

For earthquake analysis things are surely more complicated for not only the forcesinduced in the soil is much more complex, the behavior itself is different from machinefoundations.

While in machine foundation the force is induced in the soil from the structurein earthquake the force is induced within the soil where the soil first start vibrat-ing based on the waves propagating through it. Thus acceleration, it is excited to,depends on the free field vibration of the site. This acceleration induced in the soil

generates shear strain on which the stiffness degradation and damping ratio woulddepend.

Though the analysis shown hereafter is based for isotropic homogenous mediumit can well be extended to layered soil having variable property based on weightedaverage.


7/27/2019 SW1451_c001



Propagation of Earthquake

H

Figure 1.9.4 Schematic diagram of an industrial site with propagating earthquake waves.

Shown in Figure 1.9.4 is a schematic diagram of an industrial site with propagatingearthquake waves. The depth of the site (H ) is considered to the bedrock level fromwhere the waves are presumed to be propagating31.

The waves propagating at bedrock level travels upward and hits the surface (z = 0)when the site surface undergoes a motion. However as surface is free, it is free to shakeas such no strain energy develops at the surface.

The motion of such elastic waves propagating through an elastic medium can bedefined by the partial differential equation

∂2u∂t 2

= v2s∂2u∂z2

(1.9.4)

Here u = displacement of the soil and is a function of time t and depth z, vs = Shearwave velocity of the soil.

Considering u(z, t ) = φ(z)ψ(t ), (1.9.5)

31 For site having no bedrock this level is usually considered at the depth where shear wave velocity of the site is greater or equal to 600 m/sec. Based on SPT value this can be considered as the depth wheredesign SPT value is greater than 50.


7/27/2019 SW1451_c001



we have substituting in the equation above

φ(z)ψ(t ) = v2s φ(z)ψ(t )

or ¨ψ(t )

v2s ψ(t ) = ¨

φ(z)

φ(z) = − p2(say)

The above on separation gives two homogenous equations

φ(z) + p2φ(z) = 0 and ψ(t ) + p2v2s ψ(t ) = 0 (1.9.6)

The above gives solution

φ(z) = A cos pz + B sin pz (1.9.7)

at z = 0 as there will be no shear strain hence du/dz = 0

φ(z) = −Ap sin pz + Bp cos pz = 0, at z = 0

The above gives the constant B = 0, from which we deduce, φ(z) = A cos pzAt z = H as the soil is confined, hence we have, u(z, t ) = 0 → A cos pH = 0

p = (2n − 1) π

2H (1.9.8)

Considering, ψ(t ) = C cos λt + D sin λt (1.9.9)

we have, λ = p2

v2s , where λ is the eigen-value of the problem.

Knowing from our fundamental knowledge of vibration that

ω2n = λ = (2n − 1)2 π2

4H 2v2

s ➔ or ωn = (2n − 1) π

2H vs rad/sec (1.9.10)

Considering T = 2πωn

, we have

T n = 4H

(2n − 1) vssecs (1.9.11)

Here T n is known as the free field time period of the site for n numbers of mode.


7/27/2019 SW1451_c001


7/27/2019 SW1451_c001



Considering shear strain γ z = ∂u∂z we have

γ z = − 16βSaH

(2n

−1)π(π

+2)v2

s

sin(2n − 1)πz

2H (1.9.19)

Considering G = ρv2s we have

γ z = − 16βSaH ρ

(2n − 1)(π + 2)πGsin

(2n − 1)πz

2H (1.9.20)

Here G = dynamic shear modulus of the soil, ρ = mass density of the soil.For foundation at a particular depth below the free surface for which we have

obtained the dynamic shear modulus based on field or lab test34. We start initially tofind out the shear strain in the soil based on this value considering a strain range of 10−3/10−4%.

The steps that are followed subsequently to arrive at the corrected G and dampingvalue are furnished hereafter (Chowdhury 2008).

1 Identify the bedrock level (H ) of the site2 Find out the shear wave velocity from the expression G = ρv2

s3 Find out the free field time period of the site from the expression T n

=4H

(2n−

1)vs4 Based on the site response spectra/spectra given in code and damping value as

obtained in soil report obtain the acceleration Sa5 Obtain shear strain for the soil profile based on the expression γ z =

− 16βSaH ρ(2n−1)π(π+2)G sin (2n−1)πz

2H

6 Check if this strain is near or equal to the initial strain(10−3 to 10−4)%35.7 If there exists a significant variation correct G based on the equation G = Gmax

(1+ψ/ψr)

8 Find out the ratio G / Gmax

9 Obtain new damping ratio based on Zhang’s expression

ζ = 0.3331 + exp(−0.0145PI1.3)

2

0.586

G

Gmax

2

− 1.547G

Gmax+ 1

10 Repeat the steps as mentioned from 2 to 7 till the strain is same as previous cycle.

The value for which the strain becomes constant is the corrected Dynamic shearmodulus of the soil.

The above steps will now be further elaborated by a suitable problem.

34 Or from theoretical co-relation.35 In absence of this input in soil report consider 10−3% for soft soil and 10−4% for medium stiff soil.


7/27/2019 SW1451_c001



Example 1.9.2

For a particular site susceptible to earthquake it was observed based on soilinvestigation that bed rock exists at 20 meters below ground level. Seismic cross-

hole test reveals average dynamic shear modulus of the soil to be 154897 kN/m2

at a reference strain of 1× 10−5.Considering density of soil as 19 kN/m3, and plasticity index as 35. Calculate

the corrected dynamic shear modulus of soil and damping at 2.5 meter belowGL where foundation of a particular structure will be placed. Consider IS 1893curves to evaluate the acceleration pertaining to a particular time period.

Solution:

Depth of soil over bedrock

=20 m, Density of soil

=19 kN/m3, Thus mass

density of soil (ρ) = 199.81 = 1.936 kN · sec2 /m, Dynamic shear Modulus of soil(G) = 154897 kN/m2.

Shear wave velocity of soil (vs) =

G

ρ= 282.8 m/sec.

Considering T n = 4H

(2n

−1)vs

we have for fundamental mode

T 1 = 4 × 20

282.8= 0.283 sec

Considering Zhang’s formula

ζ = 0.3331 + exp(−0.0145PI1.3)

2

0.586

G

Gmax

2

− 1.547G

Gmax+ 1

Taking G/Gmax = 1 for first cycle and PI = 35 we have, ζ = 0.798%For damping @ 0.798% and time period of 0.283 sec Sa = 5.75 m/sec2 from

IS code.For this case the code factor Z is considered as 0.24, I = 1.2, R = 3. The value

of R is chosen as 3 in this case because for PI = 35 it is assumed that the soil hashigh plasticity and thus has reasonable ductility. As a matter of fact there is noguideline at moment prevalent in the code and an engineer has to use his ownjudgment here.

Substituting the above data in the expression

γ z = − 2SaH ρ

(2n − 1)πGsin

(2n − 1)πz

2H


7/27/2019 SW1451_c001



We have at depth of 2.5 meter below ground level γ z = 2.5145 × 10−5

Considering G = Gmax

(1

+

ψψr)

we have G = 44073.2 kN/m2

We proceed with next cycle of iteration with this new value of G.In table below we show how the data converges for 10 successive cycles

Cycles 1 2 3 4 5

G kN/m2 154897 44073.21001 39512.54 37131.85 35405.98V sm/sec 282.7998707 150.8499063 142.8319 138.4621 135.206T sec 0.282885561 0.53032847 0.560099 0.577775 0.59169

Sa m/sec2

5.75 1.9 1.85 1.85 1.8Strain 2.51454 × 10−05 2.9202 × 10−05 3.17 × 10−05 3.37 × 10−05 3.44 × 10−05

G /G max 1 0.284532367 0.255089 0.23972 0.228578Damping

ratio(%) 0.798 12.425 13.167 13.562 13.852

Cycles 6 7 8 9 10

G kN/m2 34857.41 34437.41 34113.86 33863.42 33668.85V sm/sec6 134.1545 133.3438 132.716 132.2279 131.8475

T sec 0.596327 0.599953 0.602791 0.605016 0.606762Sa m/sec2 1.8 1.8 1.8 1.8 1.8

Strain 3.5 × 10−05 3.54 × 10−05 3.57 × 10−05 3.6 × 10−05 3.62E−05

G /G max 0.225036 0.222325 0.220236 0.218619 0.217363Damping

ratio(%) 13.945 14.016 14.071 14.114 14.147

Thus based on above calculation we may take, corrected G value =33600 kN/m2; Damping ratio = 0.14. Variation of shear modulus and dampingratio (%) with number of cycles are shown in Figs. 1.9.5 and 1.9.6.

Varaition of Gdyn at foundation level

0

50000

100000

150000

200000

1 2 3 4 5 6 7 8 9 10

Iteration number

G

Figure 1.9.5 Variation of shear modulus with number of cycles.


7/27/2019 SW1451_c001



Varaition Damping ratio(%) at foundation level

0.000

5.000

10.000

15.000

1 2 3 4 5 6 7 8 9 10

Number of iterations D a m p i n g

R a t i o ( % )

Figure 1.9.6 Variation of damping ratio (%) with number of cycles.

1.9.3 What do we do if the soil is layered with varyingsoil property?

Till now the theories we have presented assumes soil as a homogenous isotropicmedium but in reality in all possibility the soil encountered at a particular site willbe layered in nature.

Shown in Figure 1.9.7, is a typical stratified soil profile where the shear modulus,density of soil and Poisson’s ratio vary with depth.

For most of the cases taking a weighted average is the normal practice where the

average dynamic property may be taken as

Gav = G1 · H 1 + G2 · H 2 + G3 · H 3 + G4 · H 4

H 1 + H 2 + H 3 + H 4(1.9.21)

and same principle be applied for mass density and Poisson’s ratio.However for very important structures or site susceptible to major earthquakes

methods based on finite element analysis may be applied to arrive at a design dynamic

modulus and damping value36.Shown in Figure 1.9.8, is the finite element model of a site having layered soil

property. In this case the soil is modelled as plane strain element to the bedrockboundary and each individual layers having different properties can very easily becatered to.

To start with we assume G value as obtained from soil report and consider thedamping ratio based on Zhang’s formula considering G/Gmax = 1 at the strain levelof 10−3/10−4% say.

Suppose the previous earthquake history shows that shaking has taken place for

duration of 3 sec maximum, we select duration of 6 sec for analysis.

36 In such cases preferably site response spectra of the particular should be used. Moreover some previoushistory of shaking and its duration should be available for analysis.


7/27/2019 SW1451_c001


7/27/2019 SW1451_c001



Calculation of shear modulus of soil based on the free field time period is an effectivetool for assessing the dynamic shear modulus of soil. However, there is a possibilitythat the time period obtained by this method could be higher than the reality unlessproper consideration are given for the confining effect of the surrounding soil andproper judgement of the depth is made. ATC (1982) has defined H max as the depthlimited to 183 m having low strain shear wave velocity of 760 m/sec.

1.9.4 Checklist of parameters to be looked in thesoil report

Based on above discussion, the parameters which require particular attention in a soilreport from the engineer are summarised as follows:

1.9.4.1 Field test

• Has SPT test been carried out?The obvious intention is to find out N on which G value depends. This can also be

utilised to cross check the field observed dynamic data.

• If SPT values are furnished are the observed data or corrections need to be done?A point to be checked for field observed data as shown earlier needs to be corrected.While the soil consultant will do this correction during his own calculation of bearing

capacity of soil for foundation recommendations, usually furnishes observed field data

while furnishing the bore log detail in the report. So for your calculation this data needsto be corrected. If you are not too sure you can back calculate it from recommendedφ value.

• Has Ground Water Table been established during boring?Usually provided in a soil report but better to check for this has significant effect on

the net vertical stress.

• Has any dynamic field test carried out?

◦ Block Vibration test◦ Seismic cross hole test

One of the tests should be a part of the soil report. But do not take the valuesfurnished sacrosanct. Back check with theoretical co-relation to establish if the orderis close, if not you do have the right to ask your soil consultant why there is thisdiscrepancy. There could be special geological condition which could result in suchdiscrepancy and you should be clear about it.

•If the above tests are carried out, what is the strain range induced in the soil during

the test?This is something usually not supplied by the soil consultant who usually would

recommend a unique G value. This should not be acceptable to you.You should clear it at the very outset when providing him the specification for

Geotechnical investigation that this is an input you are looking for and it should be


7/27/2019 SW1451_c001



a part of his report. It is more realistic to start with this value rather than guessing atheoretical value of 10−3 /10−4%.

1.9.4.2 Laboratory tests

• Check Atterberg’s limit – gives values of liquid limit, plastic limit, plasticityindex etc.

Generally speaking37 as a ritual, structural engineers/analysts ignore this topic. Oursuggestion would be, do not disregard this for this is the basic data which gives youthe first insight into the fact as to how the soil behaves. Moreover plasticity indexbeing an important property it is all the more important that you should pay attentionto this.

•Triaxial test gives values of c, φ and pre-consolidation history

Again given a back handed treatment by the analyst who without going throughthe test data would prefer to pick up the numbers which concerns him (ca and φ). Wesuggest go through the test and develop enough skill to interpret the pre-consolidationstress. Make sure to ask during enquiry to the consultant to supply this data. For OverConsolidation Ratio (OCR) plays a very important role in arriving at the correct valueof Gdyn.

During interpretation if need be, seek help of a geotechnical specialist to make surewhat has been understood is correct – this will save a lot of headache in the long run.

• Bulk density and void ratio of soil• Grain distribution to check if the soil is gap graded, uniformly graded, or poorly

graded. Relative density of cohesion less soil is highly dependent on this.

1.10 EPILOGUE

The technology described in this chapter to our perception is still in its infancy and weare optimistic that with time and research that is being carried out all over the world,

we shall be in a better position in future to predict more realistically the dynamicproperties of soil which affect the response of structure.

Whatever we have presented here is what we believe is simple to apply, providesreasonably realistic results and practical for day to day design office practice.

There is hardly any comprehensive text which gives a defined picture on thisissue. Most of the techniques developed herein are based on research papers(names furnished in the reference) and typical practices followed in some designoffices38.

We urge the readers to go through these papers which we believe will give them

further insight to the problem.

37 Exceptions are always there. . ..38 Even consultants who require to use these type of technology is very limited.


7/27/2019 SW1451_c001



The ideas presented in this chapter is to make the reader aware of the limitationsprevalent with soil and also to caution him on the fact that without these valuesrealistically estimated, the whole analysis related to dynamic soil structure interactioncould become a questionable exercise.

So be aware and use your judicious best to furnish a meaningful design.

SUGGESTED READING39

1 Cohen, M. & Jennings, P., ‘Silent Boundary Methods For Transient Analysis’, Computa-tional Method in Transient Analysis – Computational Method in Mechanics, Vol. 1, NorthHolland.

2 Dasgupta, S.P. & Kameswara Rao, N.S.V.K. 1976, ‘Some finite element solutions in thedynamics of circular footings’, Proc 2nd International Conference on Numerical Methodsin Geomechanics, Blacksburg USA.

3 Dasgupta S.P. & Kameswara Rao, N.S.V.K. 1978, Dynamics of rectangular footings byFinite elements, Journal of GT Division ASCE, Vol. 104, No. 5.4 Gazetas, G & Tassoulas, A.L. 1987, ‘Horizontal Stiffness of Arbitrarily shaped embedded

foundation’, Journal of GT Division, ASCE, Vol. 113, No. 5.5 Kameswar Rao, N.S.V. 1977, ‘Dynamic soil structure system – A Brief Review’, J. Struct.

Engg ., India, Vol. 4.6 Lysmer, J. & Kuhlemeyer, R.L. 1969, ‘Finite dynamic model of infinite media’, J.EM.Divn,

ASCE, EM4.7 Segol, G., Abel, J.F. & Lee, P.C.Y. 1975, ‘Finite element Mesh Gradation of surface waves’,

J. GT Division, ASCE, Vol. 101, GT 11.8 Wolf, J.P. 1985, Dynamic Soil Structure Interaction, Prenctice-Hall Englewood Cliffs, NJ.9 Wolf, J.P. 1988, Dynamic Soil Structure Interaction in Time Domain, Prenctice-Hall,

Englewood Cliffs, NJ.10 Wolf, J.P. 1994, Foundation Vibration Analysis: Using Simple Physical Model , Prentice-

Hall, Englewood-Cliffs, NJ.11 Whitman, R.V. 1970, Soil Structure Interaction – Seismic design for Nuclear power plants,

The MIT press, Cambridge, Massachusets.

Documents

SW1451_c001