foundations analysis and design

ENCE 461Foundation Anal ysis and

Design

Retaining WallsLateral Earth Pressure Theory

Retainin g Walls

� Necessary in situations where gradual transitions either take up too much space or are impractical for other reasons

� Retaining walls are analysed for both resistance to overturning and structural integrity

� Two categories of retaining walls

� Gravity Walls (Masonry, Stone, Gabion, etc.)

� In-Situ Walls (Sheet Piling, cast in-situ, etc.)

Lateral Earth Pressure Coefficient

� K = lateral earth pressure coefficient

� �x’ = horizontal effective stress

� ��

� ��

� ��

K��x '

�z'

Mohr’s Circle and Lateral

Earth Pressures

=�z'� x ' =

Development of Lateral Earth Pressure

Po

b��1 z1

2 K o

2

Note Pore Water Effect! subtract vertically add horizontally

Groundwater Effects

� Steps to properly compute horizontal stresses including groundwater effects:

� Compute total vertical stress

� Compute effective vertical stress by removing groundwater effect through submerged unit weight; plot on P

o diagram

� Compute effective horizontal stress by multiplying effective vertical stress by K

� Compute total horizontal stress by directly adding effect of groundwater unit weight to effective horizontal stress

Groundwater Effects

Conditions of Lateral Earth Pressure Coefficient

� At-Rest Condition

� Condition where wall movement is zero or “minimal”

� Ideal condition of wall, but seldom achieved in reality

� Active Condition

� Condition where wall moves away from the backfill

� The lower state of lateral earth pressure

� Passive Condition

� Condition where wall moves toward the backfill

� The higher state of lateral earth pressure

Effect of Wall Movement

Wall Movements Necessar y to Achieve Active or Passive

States

Estimates of At Rest Lateral Earth Pressure Coefficient

� Jaky’s Equation

� Modified for Overconsolidated Soils

� Applicable only when ground surface is level

� In spite of theoretical weaknesses, Jaky’s equation is as good an estimate of the coefficient of lateral earth pressure as we have

K o�1�sin� '

K o��1�sin� ' �OCRsin� '

Relationship of Poisson’s Ratio with Lateral Earth Pressure

Coefficient

K o��

1��

��2��1��1

��tan��1tan��2

(Normally Consolidated Soils)

Example of At Rest Wall Pressure

� Given

� Retaining Wall as Shown

� Find

� PA,

from At Rest Conditions

At Rest Pressure Example

� Compute at rest earth pressure coefficient

� Compute Effective Wall Force

K o�1�sin� 'K o�1�sin30º�0.5

Po

b��1 z1

2 K o

2Po

b�

1202020.52

Po

b�12000

lbsft�12

kipsft

hPA�203�6.67 ft.

(valid for all theories)

Development of Active Earth Pressure

Development of Passive Earth Pressure

Earth Pressure Theories

Rankine Earth Pressure EquationsLevel Backfills

Rankine Theor y with Inclined Backfills

Rankine Coefficients with Inclined Backfills

Inclined and level backfill equations are identical when � = 0

Example of Rankine Active Wall Pressure

� Given

� Retaining Wall as Shown

� Find

� PA,

from At Rest Conditions

Rankine Active Pressure Example



Po

b��1 z1

2 K a

2Po

b�

1202020.3332

Po

b�8000

lbsft�8

kipsft

K A�tan2�45º��

2�

K A�tan2�45�15��13

Rankine Passive Pressure Example

Rankine Passive Pressure Example



Po

b��1 z1

2 K p

2Po

b�

12020232

Po

b�72000

lbsft�72

kipsft

K P�tan2�45º�

2�

K P�tan2�4515��3

Summar y of Rankine and At Rest Wall Pressures

72,000 lbs.

12,000 lbs. 8000 lbs.

Coulomb Theor y

K p�cos2��

cos2�cos��1�sin��sin��

cos��cos� ��

2

K a�cos2��

cos2�cos��1sin��sin��

cos��cos��

2�

Typical Values of Wall Friction

Example of Coulomb Theor y

� Given

� Wall as shown above

� Find

� KA, K

P, P

A

Solution for Coulomb Active Pressures

� Compute Coulomb Active Pressure

� KA = 0.3465

� Compute Total Wall Force

� PA = 8316 lb/ft of wall

Solution for Coulomb Passive Pressures

� Compute Coulomb Passive Pressure

� KP = 4.0196

� Compute Total Wall Force

� PA = 96,470 lb/ft of wall

Walls with Cohesive Backfill

� Retaining walls should generally have cohesionless backfill, but in some cases cohesive backfill is unavoidable

� Cohesive soils present the following weaknesses as backfill:

� Poor drainage

� Creep

� Expansiveness

� Most lateral earth pressure theory was first developed for purely cohesionless soils (c = 0) and has been extended to cohesive soils afterward

Theor y of Cohesive

Soils

Active Case(Overburdendriving)

Passive Case(Wall Driving)

1sin�1�sin�

� tan2��

4�

2�

Rankine Pressures with Cohesion (Level Backfill)

�3��1 tan2��

4��

2��2c tan�

�

4��

2�

�1�� H

K A��3

�1

�tan2��

4��

2��

2c� H

tan��

4��

2�

� Active

� Passive

�1��3 tan2��

4�

2�2c tan�

�

4�

2�

�3�� H

K P��1

�3

�tan2��

4�

2�

2c� H

tan��

4�

2�

Overburden Driving

Wall Driving

Comments on Rankine

Equations

� Valid if wall-soil friction is not taken in to account

� Do not take into consideration soil above critical height

� Do not take into consideration sloping walls

� For practical problems, should use equations as they appear in the book

H c�2c

� K a

Equivalent Fluid Method

� Simplification used to guide the calculations of lateral earth pressures on retaining walls

� Can be used for Rankine and Coulomb lateral earth pressures

� Can be used for at rest, active and passive earth pressures

� Transforms the soil acting on the retaining wall into an equivalent fluid

Example of Equivalent Fluid Method

� Given

� Wall as shown above

� KA = 0.3465

� KP = 4.0196

� �w

= 3 degrees

� Find

� Forces acting on the wall (both horizontal and vertical)

Example of Equivalent Fluid� Compute Equivalent Fluid Unit Weights (Active

Case)Gh��K acos�w

Gh�1200.3465cos3ºGh�41.52pcfGv��K asin�w

Gv�1200.3465sin3ºGv�2.18pcf

Example of Equivalent Fluid� Compute Wall Load (Active Case)

Pa

b�

Gh H 2

2Pa

b�

41.52202

2�8304 lb/ft

V a

b�

Gv H 2

2V a

b�

2.18202

2�436 lb/ft

Example of Equivalent Fluid� Compute Equivalent Fluid Unit Weights (Passive

Case)Gh��K pcos�w

Gh�1204.0196cos3ºGh�481.69pcfGv��K psin�w

Gv�1204.0196sin3ºGv�25.24pcf

Example of Equivalent Fluid� Compute Wall Load (Passive Case)

Pp

b�

Gh H 2

2Pp

b�

481.69202

2�96338 lb/ft

V p

b�

Gv H 2

2V p

b�

25.24202

2�5048 lb/ft

Terzaghi Model

� Assumes log spiral failure surface behind wall

� Requires use of suitable chart for K

A

and KP

� Not directly used in this course, but option in SPW 911

Presumptive Lateral Earth Pressures

� Based on Terzaghi theory

� Suitable for relatively simple retaining walls in homogeneous soils

� Classifies soils into five types:

1. “Clean” coarse grained soils

2. Coarse grained soils of low permeability; mixed with fine grained soils

3. Residual soils with granular materials and clay content

4. Very soft clay, organic silts, or silty clays

5. Medium or stiff clay, very low permeability

Presumptive Lateral Earth Pressures

Presumptive Lateral Earth

Pressures

Effects of Surface Loadin g

Surchar ge and Groundwater Loads

Homework Set 5

� Reading

� McCarthy: Chapter 16

� Coduto: Chapters 22, 23, 24 & 25

� Homework Problems

� McCarthy: 16-1, 16-8, 16-12a, 16-17

� Coduto: 25.3 (Hand and Chart Solutions); 25.5 (SPW 911)

� Due Date: 17 April 2002

Questions

ENCE 461Foundation Anal ysis and

Design

Mat Foundations (Part II)

Nonri gid Methods

� Nonrigid methods consider the deformation of the mat and their influence of bearing pressure distribution.

� These methods produce more accurate values of mat deformations and stresses

� These methods are more difficult to implement than rigid methods because of soil-structure interaction

Nonri gid Methods

� Coefficient of Subgrade Reaction

� Winkler Methods

� Coupled Method

� Pseudo-Coupled Method

� Multiple-Parameter Method

� Finite Element Method

Coefficient of Sub grade Reaction

� Nonrigid methods must take into account that both the soil and the foundation have deformation characteristics.

� These deformation characteristics can be either linear or non-linear (especially in the case of the soils)

� The deformation characteristics of the soil are quantified in the coefficient of subgrade reaction, or subgrade modulus, which is similar to the modulus of elasticity for unidirectional deformation


� Definition of Coefficient of Subgrade Reaction

� ks = coefficient of subgrade reaction, units of

force/length3 (not the same as unit weight!)

� q = bearing pressure

� � = settlement

ks�q�


� Plate load test for coefficient of subgrade reaction


� Application of coefficient of subgrade reaction to larger mats

Coefficient of Subgrade Reaction

� Portions of the mat that experience more settlement produce more compression in the springs

� Sum of these springs must equal the applied structural loads plus the weight of the mat

�P�W f�uD��qdA��ksdA

Winkler Methods

� The earliest use of these "springs" to represent the interaction between soil and foundation was done by Winkler in 1867; the model is thus referred to as the Winkler method

� The one-dimensional representation of this is a "beam on elastic foundation," thus sometimes it is called the "beam on elastic foundation" method

� Mat foundations represent a two-dimensional application of the Winkler method

Beams on Elastic Foundations

Application to Spread Footin gs

Note non-linear behaviour

Non-Linear Characteristics of Soil Deformation

Limitations of Winkler Method

� Load-settlement curves are not really linear; we must make a linear approximation to use the Winkler model

� Winkler model assumes that a uniformly loaded mat underlain by a perfectly uniform soil will uniformly settle into the soil.

� Actual data show that such a mat-soil interaction will deflect in the centre more than the edges

� This is one reason why we use other methods (such as Schmertmann's) to determine settlement


� Soil springs do not act independently. Bearing pressure on one part of the mat influences both the "spring" under it and those surrounding it (due to lateral earth pressure)

� No single value of ks truly represents the

interaction between the soil and the mat

� The independent spring problem is in reality the largest problem with the Winkler model

Coupled Method

� Ideally the coupled method, which uses additional springs as shown below, is more accurate than the Winkler method

� The problem with the coupled method comes in selecting the values of k

s for the coupling springs

Pseudo-Coupled Method

� An attempt to overcome both the lack of coupling in the Winkler method and the difficulties of the coupling springs

� Does so by using springs that act independently (like Winkler springs), but have different k

s

values depending upon their location on the mat

� Most commercial mat design software uses the Winkler method; thus, pseudo-coupled methods can be used with these packages for more conservative and accurate results

Pseudo-Coupled Method� Implementation

� Divide the mat into two or more concentric zones

� The innermost zone should be about half as wide and half as long as the mat

� Assign a ks value to each zone

� These should progressively increase from the centre

� The outermost zone ks should be about twice as large as the

innermost zone

� Evaluate the shears, moments and deformations using the Winkler method

� Adjust mat thickness and reinforcement to satisfy strength and serviceability requirements

Pseudo-Coupled Method

Multiple-Parameter Method

� This method replaces the independently-acting linear springs of the Winkler method with springs and other mechanical elements

� The additional elements define the coupling effects

� Method bypasses the guesswork involved in distributing the k

s values in the pseudo-coupled

method; should be more accurate

� Method has not been implemented into software packages and thus is not routinely used on design projects

Finite Element Method

� Models the entire soil-mat system in a three-dimensional way

� In theory, should be the most accurate method

� Method is not yet practical because

� Requires large amount of computing power to perform

� Difficult to determine soil properties in such a way as to justify the precision of the analysis, especially when soil parameters are highly variable

� Will become more in use as these problems are addressed


� Finite element method is used for structural analysis

� Mat is modelled in a similar way to other plate structures with springs connected at the nodes of the elements

� Mat is loaded with column loads, applied line loads, applied area loads, and mat weight

� Usually superstructure stiffness is not considered (conservative)

� Can be done but is rarely performed in practice

Determinin g the Coefficient of Subgrade Reaction

� Not a straightforward process due to:

� Width of the loaded area; wide mat will settle more than a narrow one because more soil is mobilised by a wide mat



� Shape of the loaded area: stresses beneath long, narrow loaded area is different from those below square loaded areas



� Depth of the loaded area below the ground surface

� Change in stress in the soil due to q is a smaller percentage of the initial stress at greater depths



� The position of the mat

� To model the soil accurately, ks needs to be larger near the

edges of the mat and smaller near the centre

� Time

� With compressible (and especially cohesive compressible soils) mat settlement is a process which may take several years

� May be necessary to consider both short and long term cases

� Non-linear nature of soil deformation makes unique value of k

s non-existent


� Methods used to determine coefficient

� Plate load tests

� Test results must be adjusted between the shape of the loading plate and the actual shape of the foundation

� Adjustment must also be made for the size of the plate vs. the size of the foundation, and the influence of size on the depth of soil stress

� Attempts to make accurate adjustments have not been very successful to date

� Derived relationships between ks and E

s

� Relationships developed are too limited in their application possibilities


� Methods used to determine coefficient

� Use settlement techniques such as Terzaghi's consolidation theory, Schmertmann's method, etc., and express the results in a k

s value

� If using a pseudo-coupled value, use values of ks in the centre

of the mat which are half those along the perimeter

� This methodology has the potential of eliminating the problems described earlier while at the same time yielding values of k

s which then can be used in a structural analysis of

the mat with some degree of confidence

Example of Determinin g Coefficient of Sub grade

Reaction� Given

� Structure to be supported on a 30 m wide by 50 m long mat foundation

� Average bearing pressure is 120 kPa

� Average settlement determined � = 30 mm using settlement analysis method

� Find

� Design values of ks used in a pseudo-coupled analysis


Reaction� Solution

� Compute average ks for entire mat

ks�q�

ks�120 kPa0.030 m

�4000kN �m2



� Divide mat into three zones as shown

½ L

½ W

(ks)

C = 2 (k

s)

A

(ks)

B = 1.5 (k

s)

A

(ks)

A



� Compute the area of each zone

AA = (25)(15) = 375 m2

AB = (37.5)(22.5) – 375 = 469 m2

AC = (50)(30) – 469 = 656 m2



� Compute the design ks values

� ACI suggests varying ks from ½ its computed value to

5 or 10 times the computed value, then base the structural design on the worst condition

AA�k sA�AB�k sB�AC�k sC��AA�AB�AC�k savg

375�k sA��469�1.5�k sA��656�2�k sA�1500�k savg

2390�k sA�1500�k savg

�k sA�0.627�k savg�k sA��0.627�4000�2510kN �m2

�k sB��0.627�1.5�4000�3765kN �m3

�k sC��0.627�2�4000�5020kN �m3

Structural Desi gn of Mats� Structural design requires two analyses

� Strength

� Evaluate these requirements using factored loads and LRFD design methods

� Mat must have sufficient thickness T and reinforcement to safety resist these loads

� T should be large enough so that no shear reinforcement is required

� Servicability

� Evaluate using unfactored loads for excessive deformation at places of concentrated loads, such as columns, soil non-uniformities, mat non-uniformities, etc.

� This is the equivalent of a differential settlement analysis

� Mat must be made thicker if this is a problem

Structural Desi gn of Mats

Structural Desi gn of Mats

� Closed form solutions

� Once popular; however, with the advent of computers, have fallen out of favour� For example see http://www.vulcanhammer.net/download/piletoe.pdf

� Finite difference methods

� Finite element methods

� Spring values as computed in the example can then be used in finite element analysis

� The stiffer springs at the edges will encourage the foundation to sag in the centre, which is what we actually see in foundations

Other Considerations in Mat Foundations

� Total settlement

� "Bed of springs" solution should not be used to compute total settlement; this should be done using other methods

� Bearing capacity

� Mat foundations generally do not have bearing capacity problems

� With undrained silts and clays, bearing capacity needs to be watched

� Methods for spread footings can be used with mat foundations, including presumptive bearing capacities

Questions

foundations analysis and design

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