1

Soil Structure Interaction

Continuous Footings & Rafts:

K. N. Sheth Civil Engg. Dept

DDU- Nadiad

Short term training program on

“Ground Improvement Techniques and

Soil Structure Interaction” at

BIRLA VISHVAKARMA MAHAVIDYALAYA Vallabh Vidyanagar

2

INTRODUCTION :

A large no. of problems in Soil Engineering are attempted using Theory of Plasticity (Limit Equilibrium Analysis) disregarding the Load – Deformation behavior. An appropriate FoS to the ultimate load serves the design purpose. Bearing Capacity of Footing – Footing designed for Uniform Pressure Earth Pressure Theories – Wall designed for active/passive pressure Slope Stability Analysis Vertical and lateral load carrying capacity of Pile Pile groups

3

The accuracy in the analysis when generalized and put to practice can attain a level of crude estimate only, due to

1. Representative parameter varies with time & inevitable changes in moisture content

2. Inability to collect representative samples – cohesionless soil and to test strength parameter especially for c-phi soils

3. Strength parameters obtained by sounding are highly empirical in nature

With all the limitation an experienced Geotechnical Engineer with the perception of theory and soil behavior provides safe and viable solutions.

4

There are some critical problems where Load-Deformation analysis at

working load is essential for the design of structural elements.

These problems are grouped as Soil Structure Interaction problems

Analysis of Continuous Footings, Rafts, Piles, Piled Rafts

Analysis Rigid Retaining walls, Braced Excavations, Sequencing

of Excavation

5

It is evident that the level of accuracy attained for such problems when generalized will be very much limited. However it provides the perception of the behavior to a Geotechnical Engineer, with the adjustment of the parameters and advanced methods of analysis can allocate appropriate safety margins or over design.

6

7

Soil Structure Interaction is essentially defining pressure –

deformation relationship at the interface of the Structural/

Foundation Element with the soil.

Even Nonlinear Elasto-Plastic Analysis is a crude tool to handle

different soil type, in its present state.

Measure it with a Micrometer

Mark it with a Chalk

Cut it With an Axe

Hence apart from the software and methods available, the solution

of SSI problem demands very good insight and perception of the

problem

8

Contact Pressure Distribution and Settlement profile depends upon

flexural rigidity of the foundation and stiffness of the soil support.

Typical cases:

Rigid footing on ideal sand, Rigid footing on ideal clay, flexible

footing on ideal sand and flexible footing on ideal clay

9

Soil Models for load–deformation at the Interface: 1. Elastic Half Space : The soil mass is semi-infinite, Elastic, Isotropic and

Homogeneous. (Material Const. E and µ)

2. Winkler’s Spring Model : The soil is replaced by a bed of very closely

spaced isolated elastic springs.

Material Const = Modulus of Subgrade Reaction, ko = p / δ (kN/m3)

3. Models based on laboratory tests - e.g Consolidation parameters etc.

10

Cohesive soils are mainly OC Clays modelled as Elastic Half Space

In Cohesionless Soils, Es increases with depth/confinement.

Hence, it is approximately idealized with Winkler’s Spring Model.

11

Beam/Continuous Footings on soil as Winkler’s Spring Bed.

EI . d4y/dx4 + k . y = q

12

The close form solution of the equation is CF + PI.

It is used for the analysis as:

y, dy/dx, EI.d2y/dx2 (BM) and EI. d3y/dx3 (SF)

Long beams and short beams for different load types and positions :

M. Hetenyi

13

Finite Element Formulation : soil modelled as Winkler’s Spring Bed.

- Finite Element Method Converts a Continuum Problem (Infinite

D.O.F.) to a finite D.O.F. problem.

- It reduces the problem to solution of Simultaneous Equations from

Solution of differential equations.

- It requires a minimizing Function for the differential equation to

obtain the solution. There can be a Function like Minimum Potential

energy as in solid Mechanics.

- FEM is the Tool evolved by Engineers and its Mathematical Ground

is proved after real life applications done.

14

15

Find { } in terms of Nodal Displacements by substituting nodal co-ordinates

16

17

18

19

2 2

2

156 22 . 54 13 .4 . 13 . 3 .

156 22 .4 .

e s

B B L B B LB L B L B L

kB B L

B L

20

Determination of Modulus of Subgrade Reaction ‘k’ :

1. Based on the Soil Investigation Report of SBC k = Allowable Bearing Pressure / Settlement at this pressure For the given size of footing

2. Based on Plate Load Test as per IS 9214 The basic value of the ‘k’ is adopted for 30x30cm plate. The slope of p v/s δ curve is taken either at 0.125mm settlement or at 70 kPa pressure. The va;lue obtained is extrapolated for the size of foundation and its embedment depth.

3. From Lab Test Results : Vague guidelines are given in IS 2950

k = 0.85 *[Es.B4.Es / { E.I. (1-µ2).B}] (1/12)

21

4. Empirical Values as per IS 2950 The basic value of the ‘k’ is adopted for 30x30cm plate and extrapolated for the size of foundation and its embedment depth..

Table : Modulus of Subgrade Reaction (k) for Cohesionless soils (30x30cm plate/30cm wide Beam)

Soil Characteristics Modulus of Subgrade Reaction : k (kg/cm3)

Relative Desnsity

SPT N-Value Dry & Moist State

Submerged State

Loose

<10 1.5 0.9

Medium

10 to 30 1.5 to 4.7 0.9 to 2.9

Dense

>30 4.7 to 18 2.9 to 10.8

22

Table : Modulus of Subgrade Reaction (k) for Cohesive soils (30x30cm plate)

Soil Characteristics Modulus of Subgrade

Reaction : k (kg/cm3)

Consistency UCS (kg/cm2)

Stiff

1 to 2 2.7

Very stiff

2 to 4 2.7 to 5.4

Hard

>4 5.4 to 10.8

23

Corrections : 1. Size of Footings :

(a) Clays : kf = kp * Bf / Bp

(b) Sand : kf = kp [Bp.(Bf+0.3)/{Bf*(Bp+0.3)}]2 Bf & Bp in ‘m’.

2. For Shape :

kf = ksq. * (1 + 0.5*B/L)/1.5

3. For Placement Depth ;

kf = kf (1+2*Df/B)

24

Problem1: Continuous Footing

Footing width = 2m SBC = p = 250 kN/m2

As beam & Slab type Combined footing Beam – 350 x 1000 Slab – 500 thk Approximate Max. B.M. = 1000 kN.m

25

1. Direct Solution 2. Moment Distn. (Reaction changed – 1371,2166,1927,2166,1371) 3. Use Beam on Elastic Foundation (Nodal Springs)

Comparison of BM at Critical Sections

Case 1 Case 2 Adjust Reaction

Case2

Case 3 k = 8000 kN/m3

Case 4 k = 12500

kN/m3

Case 5 k = 15000

kN/m3

A 250 250 250 270 280 280 B 250 770 740 510 540 560 C 250 620 650 630 660 670 1 -750 -505 -540 -610 -590 -580 2 -750 -305 -305 -400 -360 -350

1

FINITE ELEMENT ANALYSIS OF RAFT FOUNDATION

SUPPORTED ON SOIL SPRING

7

Winkler’s spring model is characterized by the equation,p = Ks∗w

where,p = pressure,Ks = modulus of sub grade reactions bedw = deflection.For this case the differential equation for the plate becomes,

Where, Ks and w are as defined above, q is intensity of transverse load and D is flexural rigidity of the plate.

4 4 4

4 2 2 42 sK ww w w qx x y y D D

∂ ∂ ∂+ + = −

∂ ∂ ∂ ∂

5

FINITE ELEMENT METHOD

For the analysis of the raft following two type ofidealization is made,

(i) Raft is idealized as a Thin Plate.

(ii) Soil is idealized as Winkler's Foundation or Elastic Half Space Model

10

A four nodded rectangular plate bending element with 3 D.O.F. per node is to be used for Finite element analysis of Raft Slab and the extension is to be made for analysis of the Beam and Slab Raft.For analysis, the contribution of the soil is to be considered on the of concept of the Winkler’s Foundation.

12

Mx

My

My+ dyMyy

Mx + MxxX

Y

ydy+

Mx

Y

X +x

y

Mxy Mxy

Myx

xMy

dx

dx

xMy

X

Y

AJ(3K) AJ(3K-2)

AJ(3K-1)

x

y w

i

ii

ZX

YZ

t

SIGN CONVENTION FOR INTERNAL FORCES IN PLATE

RECTANGULAR PLATE BENDING ELEMENTJOINT LOADS FOR A PLATE

(a) (b)2 a

2 b

k l

i j

DERIVATION OF STIFFNESS MATRIX FOR PLATE BENDING ELEMENT SUPPORTED ON WINKLER’S SPRING BED:Step I: CHOOSE SUITABLE COORDINATE SYSTEM, DIMENSIONS AND NODE NUMBERING

26

2. Raft supported on soil as Winkler’s Spring Bed.

Loads : * Corner Cols. : 1800 x 4 nos = 7,200 kN * Edge Col.s : 2400 x 12 nos = 28,800 kN * Inner Col.s 3200 x 9 nos = 28,800 kN Total Load = 64,800 kN

SBC=200 kN/m2 Raft area 18m x 18m Raft Thk.500mm

27

SOLUTION 1. Conventional Plate analysis with Uniform Pressure 2. Use Beam on Elastic Foundation (Nodal Springs)

Critical Location

Mx My

Case1 Case 2 k = 8000 kN/m3

Case 3 k = 15000

kN/m3

Case1 Case 2 k = 8000 kN/m3

Case 3 k = 15000

kN/m3 1 200 240 245 200 245 245 2 445 400 410 320 320 325 3 370 440 440 280 325 325 4 645 510 515 645 510 515 5 565 546 550 590 510 520 6 505 555 555 505 555 555 7 -175 -200 -180 -175 -200 -180 8 -95 -110 -100 -175 -195 -1809 -80 -100 -90 -80 -100 -90