Chapter 14 Foundations for Concrete Structures.pdf

14-1

14Foundations for

Concrete Structures

Manjriker Gunaratne PhD PE

141 Foundation Engineering 14-1Soil Classification bull Strength of Soils bull Compressibility and Settlement bull Groundwater and Seepage bull Dewatering of Excavations bull Environmental Geotechnology bull Design of Landfill Liners

142 Site Exploration14-27Plate Load Tests

143 Shallow Footings 14-32Bearing Capacity of Shallow Footings bull Footings with Eccentricity bull Presumptive Load-Bearing Capacity

144 Mat Footings14-37Design of Rigid Mat Footings bull Design of Flexible Mat Footings

145 Retaining Walls 14-43Determination of Earth Pressures bull Design of Concrete Retaining Walls bull Effect of Water Table bull Reinforced Walls bull Sheet Pile Walls bull Braced Excavations bull Soil Nail Systems bull Drainage Considerations

146 Pile Foundations14-57Advantages of Concrete Piles bull Types of Concrete Piles bull Estimation of Pile Capacitybull Computation of Pile Settlement bull Pile Groups bull Verification of Pile Capacity

147 Caissons and Drilled Piers14-76Estimation of Bearing Capacity

References 14-79

141 Foundation Engineering

Geotechnical engineering is a branch of civil engineering in which technology is applied to the designand construction of structures involving earthen materials and there are many branches of geotechnicalengineering Surficial earthen material consists of soil and rock soil and rock mechanics are fundamentalstudies of the properties and mechanics of soil and rock Foundation engineering is the application of

Professor of Civil Engineering at University of South Florida Tampa expert in various areas of geotechnicalengineering including foundation design numerical modeling and soil stabilization

14-2 Concrete Construction Engineering Handbook

the principles of soil mechanics rock mechanics and structural engineering to the design of structuresassociated with earthen materials It is generally observed that most common foundation types supportedby intact bedrock present no compressibility problems hence when designing common foundationtypes the foundation engineerrsquos primary concerns are the strength and compressibility of the subsurfacesoil and whenever applicable the strength of bedrock

1411 Soil Classification

14111 Mechanical Analysis

According to texture or the feel two different soil types can be identified (1) coarse-grained soil (graveland sand) and (2) fine-grained soil (silt and clay) Whereas the engineering properties (primarily strengthand compressibility) of coarse-grained soils depend on the size of individual soil particles the propertiesof fine-grained soils are mostly governed by moisture content Hence it is important to identify the typeof soil at a given construction site because effective construction procedures invariably depend on thesoil type Soil engineers use a universal format called the unified soil classification system (USCS) to identifyand label soil The system is based on the results of common laboratory tests of mechanical analysis andAtterberg limits (Bowles 1986) To classify a given soil sample mechanical analysis is conducted in twostages (1) sieve analysis for the coarse fraction (gravel and sand) and (2) hydrometer analysis for thefine fraction (silt and clay) Of these sieve analysis is conducted according to ASTM D 421 and D 422procedures using a set of US standard sieves (Figure 141) The most commonly used sieves are numbers20 40 60 80 100 140 and 200 corresponding to sieve openings of 085 0425 025 018 015 0106and 0075 mm respectively

During the test the percentage (by weight) of the soil sample retained on each sieve is recorded fromwhich the percentage (R) passing (or finer than) a given sieve size (D) is determined On the otherhand if a substantial portion of the soil sample consists of fine-grained soils (D lt 0075 mm) then sieveanalysis has to be followed by hydrometer analysis (Figure 142) This is performed by first treating thefine fraction with a deflocculating agent such as sodium hexametaphosphate (Calgonreg) or sodium silicate(water glass) for about half a day and then allowing the suspension to settle in a hydrometer jar kept ata constant temperature As the heavier particles settle followed by the lighter ones a calibrated ASTM152H hydrometer is used to estimate the fraction (R) still settling above the hydrometer bottom at any

FIGURE 141 Equipment for sieve analysis

Foundations for Concrete Structures 14-3

given stage Further the particle size (D) that has settled past the hydrometer bottom at that stage intime can be estimated from Stokesrsquo law It can be seen that R is the weight percentage of soil finer than D

Complete details of the above tests are provided in Bowles (1986) For soil samples that have significantcoarse and fine fractions the sieve and hydrometer analysis results (R and D) can be logically combinedto generate grain (particle) size distribution curves such as those indicated in Figure 143 From Figure143 it can be seen that 30 of soil type A is finer than 0075 mm (US No 200 sieve) with R = 30and D = 0075 mm being the last pair of results obtained from the sieve analysis In combining sieveanalysis data with hydrometer analysis data one has to convert the R (based on the fine fraction only)and D obtained from hydrometer analysis to R based on the weight of the entire sample to ensurecontinuity of the curve

FIGURE 142 Equipment for hydrometer analysis

FIGURE 143 Grain-size distribution curves

0

100

80

60

40

20

19 475 1

084

0

042

0

015

00

10

075

001

000

20

001

Perc

ent F

iner

Grain Diameter (mm)

A

B

As an example let the results from one hydrometer reading of soil sample A be R = 90 and D =005 mm To plot the curve one needs the percentage of the entire sample finer than 005 mm Becausewhat is finer than 005 mm is 90 of the fine fraction (30) used for hydrometer analysis the convertedR for the final plot can be obtained by multiplying 90 by the fine fraction of 30 Hence theconverted data used in Figure 143 are R = 27 and D = 005 mm

14112 Atterberg Limits

As mentioned earlier properties of fine-grained soils are governed by water hence the effect of wateron fine-grained soils has to be considered in soil classification This is achieved by employing the Atterberglimits or consistency limits The physical state of a fine-grained soil changes with increasing water contentas shown in Figure 144 from a brittle to a liquid state Theoretically the plastic limit (PL) is defined asthe water content at which the soil changes from semisolid to plastic (Figure 144) For a given soil samplethis is an inherent property that can be determined by rolling a plastic soil sample into a worm shapeto gradually reduce its water content by exposing more and more of an area until the soil becomessemisolid This change can be detected by the appearance of cracks on the sample According to ASTM4318 the plastic limit is the water content at which cracks develop on a rolled soil sample at a diameterof 3 mm thus the procedure is one of trial and error The apparatus (ground glass plate and moisturecans) used for the test is shown in Figure 145 but the reader is also referred to Bowles (1986) and Wray(1986) for details

On the other hand the liquid limit (LL) which is visualized as the water content at which the stateof a soil changes from plastic to liquid with increasing water content is determined in the laboratoryusing Casagrandersquos liquid limit device (Figure 146) This device is specially designed with a standardbrass cup where a standard-sized soil paste is laid during testing In addition the soil paste is groovedin the middle by a standard grooving tool thereby creating a gap with standard dimensions The brasscup is then made to drop through a distance of 1 cm on a hard rubber base The number of drops (blows)required to close the above gap along a distance of 12 in is counted Details of the test procedure canbe found in Bowles (1986) ASTM 4318 specifies the liquid limit as the water content at which closingof the standard-sized gap is achieved in 25 drops of the cup therefore one has to repeat the experimentfor different trial water contents each time recording the number of blows required to close the above

FIGURE 144 Variation of the fine-grained soil properties with water content

FIGURE 145 Equipment for the plastic limit test

Brittlesolid

Shrinkagelimit

Semi-solid Plastic Liquid

Watercontent

Plasticlimit

Liquidlimit

standard-sized gap Finally the water content corresponding to 25 blows can be interpolated from thedata obtained from all of the trials The plasticity index (PI) is defined as follows

PI = LL ndash PL

14113 Unified Soil Classification System

In the commonly adopted unified soil classification system (USCS) shown in Table 141 the aforemen-tioned soil properties are effectively used to classify soils Example 141 below illustrates the classificationof the two soil samples shown in Figure 143 Defining the following two curve parameters is necessaryto accomplish the classification

Coefficient of uniformity Cu = D60D10

Coefficient of curvature Cc = (D30)2(D60 times D10)

where Di is the diameter corresponding to the ith percentage on the grain-size distribution curve

Example 141

Soil A The percentage of coarse-grained soil is equal to 70 hence soil A is a coarse-grained soil Thepercentage of sand in the coarse fraction is equal to (70 ndash 30)70 times 100 = 57 Thus according to theUSCS (Table 141) soil A is a sand If one assumes clean sand then

Cc = (0075)2(2 times 0013) = 021 does not meet criterion for SWCu = 20013 = 15385 meets criterion for SW

Hence soil A is a poorly graded sand (SP)

Soil B The percentage of coarse-grained soil is equal to 32 hence soil B is a fine-grained soil Assumingthat LL is equal to 45 and PL is equal to 35 (then PI is equal to 10) and using Casagrandersquos plasticitychart (Table 141) it can be concluded that soil B is a silty sand with clay (ML)

1412 Strength of Soils

The two most important properties of a soil that a foundation engineer must be concerned with arestrength and compressibility Because earthen structures are not designed to sustain tensile loads themost common mode of soil failure is shear hence the shear strength of the foundation mediumconstitutes a direct input to the design of concrete structures associated with the ground

FIGURE 146 Equipment for the liquid limit test

TAB

LE

14

1U

nifi

ed S

oil

Cla

ssifi

cati

on S

yste

m

Maj

or D

ivis

ion

s

Gro

up

Sym

bols

(τ

)Ty

pica

l N

ames

Labo

rato

ry C

lass

ifica

tion

Cri

teri

a

12

34

6

Coarse-grained soilsMore than half of the material

is larger than No 200 (75 microm) sieve size

(Particles smaller than the No 200 sieve cannot be seen with the naked eye)

GravelsMore than half of gravel

fraction is larger than No 4 (475 mm) sieve size

(For visual classification 5-mm may be used as equivalent to the No 4 sieve size)

Clean gravels (few or

no fines)

GW

Wel

l-gr

aded

gra

vels

gra

velndash

san

d m

ixtu

res

few

or

no

fin

es

Use grain size curve to identify the fractions as given under field identification

Determine percentage of gravel and sand from grain size curveDepending on percentages of fines (fraction smaller thanNo 200 sieve size) coarse-grained soils are classified as follows

Less than 5 GW GP SW SPMore than 12 GM GC SM SC5ndash12 Borderline cases requiring use of dual symbols

Cu

= D

60D

10 g

reat

er t

han

4C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(

see

Sect

ion

25

)

GP

Poor

ly g

rade

d gr

avel

s g

rave

lndashsa

nd

mix

ture

s f

ew o

r n

o fi

nes

Not

mee

tin

g al

l gr

adat

ion

req

uir

emen

ts f

or G

W

Gravels with fines

(appreciableamount of fines)

GM

Silt

y gr

avel

s g

rave

lndashsa

ndndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Abo

ve A

-lin

e w

ith

PI

valu

es

betw

een

4 a

nd

7 ar

e bo

rder

line

case

s re

quir

ing

use

of

dual

sym

bols

GC

Cla

yey

grav

els

gra

velndash

san

dndashcl

ay m

ixtu

res

Att

erbe

rg li

mit

s ab

ove

A-l

ine

wit

h P

I gr

eate

r th

an 7

SandsMore than half of coarse

fraction is smaller than No 4(475 mm) sieve size

Clean sands (few or

no fines)

SWW

ell-

grad

ed s

ands

gra

velly

san

ds f

ew o

r n

o fi

nes

Cu

= D

60D

10 g

reat

er t

han

6C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(see

Sec

tion

25

)

SPPo

orly

gra

ded

san

ds g

rave

lly s

ands

few

or

no

fin

esN

ot m

eeti

ng

all

grad

atio

n r

equ

irem

ents

for

SW

Sands withs fines

(appreciable amount of fines)

SMSi

lty

san

ds s

andndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Lim

its

plot

tin

g in

hat

ched

zo

ne

wit

h P

I va

lues

bet

wee

n

4 an

d 7

are

bord

erlin

e ca

ses

requ

irin

g u

se o

f du

al

sym

bols

SC

Cla

yey

san

ds s

andndash

clay

mix

ture

sA

tter

berg

lim

its

abov

e A

-lin

e w

ith

PI

grea

ter

than

7

Fine-grained soilsMore than half of material

is smaller than No 200 (75 microm) sieve size

Silt

s an

d cl

ays

liq

uid

lim

it

less

th

an 5

0

ML

Inor

gan

ic s

ilts

and

very

fin

e sa

nds

roc

k fl

our

silt

y or

cl

ayey

fin

e sa

nds

or

clay

ey s

ilts

wit

h s

ligh

t pl

asti

city

Plas

tici

ty C

hart

for

Lab

orat

ory

Cla

ssifi

cati

on o

f Fi

ne-G

rain

ed S

oils

CL

Inor

gan

ic c

lays

of

low

to

med

ium

pla

stic

ity

gra

velly

cl

ays

san

dy c

lays

silt

y cl

ays

lea

n c

lays

OL

Org

anic

silt

s an

d or

gan

ic s

ilty

clay

s of

low

pla

stic

ity

Silt

s an

d cl

ays

liq

uid

lim

it

grea

ter

than

50

MH

Inor

gan

ic s

ilts

mic

aceo

us

or d

iato

mac

eou

s fi

ne

san

dy

or s

ilty

soils

ela

stic

silt

s

CH

Inor

gan

ic c

lays

of

hig

h p

last

icit

y f

at c

lays

OH

Org

anic

cla

ys o

f m

ediu

m t

o h

igh

pla

stic

ity

org

anic

silt

s

Hig

hly

org

anic

soi

lsP

tPe

at a

nd

oth

er h

igh

ly o

rgan

ic s

oils

Sour

ce H

oltz

RD

an

d K

ovac

s W

D

An

Intr

oduc

tion

to

Geo

tech

nica

l En

gine

erin

g P

ren

tice

Hal

l E

ngl

ewoo

d C

liffs

NJ

1981

Wit

h p

erm

issi

on

|

mdash mdash mdash mdash mdash

|

mdash mdash mdash mdash mdashmdash mdash

Liqu

id L

imit

0

10

2

0

30

40

5

0

60

7

0

80

90

10

0

60 50 40 30 20 10 7 4 0

Plasticity Index

Com

parin

g so

ils a

t equ

al li

quid

lim

its

toug

hnes

s and

dry

stre

ngth

incr

ease

with

incr

easi

ng p

last

icity

inde

x

OH or MH

ML or OL

CLndash

ML

A-Lin

e

CH

14121 Effective Stress Concept

Pores (or voids) within the soil skeleton contain fluids such as air water or other contaminants Anyload applied on a soil is partly carried by such pore fluids in addition to being borne by the soil grainstherefore the total stress at any given location within a soil mass can be expressed as the summation ofthe stress contributions from the soil skeleton and the pore fluids as

σ = σprime + up (141)

where

σ = total stress (above atmospheric pressure)σprime = stress in soil skeleton (above atmospheric pressure)up = pore (fluid) pressure (above atmospheric pressure)

The stress in the soil skeleton or the intergranular stress is also known as the effective stress as it indicatesthat proportion of the total stress carried by grain-to-grain contacts

In the case of dry soils in which the pore fluid is primarily air if one assumes that all pores anywherewithin the soil are open to the atmosphere through interporous connectivity then from Equation 141the effective stress would be the same as the total stress

σprime = σ (142)

On the other hand in completely wet (saturated) soils the pore fluid is mostly water and the effectivestress is completely dependent on the pore water pressure (uw) Then from Equation 142

σprime = σ ndash uw (143a)

Using the unit weights of soil (γ) and water (γw) Equation 143a can be modified to a more useful formas shown in Equation 143b

(143b)

where

z = depth of the location from the ground surfacedw = depth of the location from the groundwater table

Finally in partly saturated soils the effective stress is governed by both the pore water and pore airpressures (ua) For unsaturated soils that contain both air and water with a high degree of saturation(85 or above) Bishop and Blight (1963) showed that

σ = σprime + ua ndash χ(ua ndash uw) (144)

where (ua ndash uw) is the soil matrix suction that depends on the surface tension of water and χ is a parameterin the range of 0 to 10 that depends on the degree of saturation One can verify the applicability ofEquation 143a for saturated soils based on Equation 144 as χ = 1 for completely saturated soils

14122 Determination of Shear Strength

The shear strength of soils is assumed to originate from the strength properties of cohesion (c) andinternal friction (φ) Using the basic Coulombrsquos friction principle the shear strength of a soil can beexpressed as

τf = c + σtanφ (145)

However it is also known that the magnitudes of the soil shear strength properties vary with prevailingdrainage conditions and to a minor extent with the stress level hence it is important to characterize thestrength properties in terms of the drainage condition (drained or undrained) employed during testingA wide variety of laboratory and field methods are used to determine the shear strength parameters cand φ of soils The triaxial test the standard penetration test (SPT) and the static cone penetration tests(CPTs) are the most common ones used in foundation engineering

prime = minusσ γ γv w wz d0

14123 Triaxial Tests

In this test a sample of undisturbed soil retrieved from a site is tested under a range of pressures thatencompass the expected field stress conditions due to the building Figure 147 is a schematic diagramof the important elements of a triaxial setup and the actual testing apparatus is shown in Figure 148From the discussion of soil strength it can be seen that the type of soil and the field-loading rate havea bearing on selection of the laboratory drainage conditions and hence the loading rate Accordinglythree types of triaxial tests are commonly conducted (1) consolidated drained (CD) tests (2) consolidatedundrained (CU) tests and (3) unconsolidated undrained (UU) tests In CU and CD tests the pressureexerted on the cell fluid is used to consolidate the soil sample back up to the in situ stress state beforeapplying the axial compression On the other hand in UU tests the cell pressure is applied with noaccompanying consolidation merely to provide a confining pressure Computations involving CU andUU tests are given in Example 142 and Example 143 and the reader is referred to Holtz and Kovacs(1981) for more details regarding the testing procedure

FIGURE 147 Schematic diagram of triaxial cell

FIGURE 148 Triaxial testing apparatus

Axial loading ram

Loading capCell

Cell fluid

Drainage lineBase

Porousstones Sample

Example 142

Assume that one conducts two CU triaxial tests on a sandy clay sample from a tentative site to determinethe strength properties The applied cell pressures deviator stresses and measured pore pressures atfailure are given in Table 142 The strength parameters can be easily estimated using the Mohr circlemethod as follows

bull Total strength parameters The total stresses (σ1 and σ3) acting on both test samples at failure areindicated in Figure 149a Accordingly the Mohr circles for the two stress states can be drawn asin Figure 1410 Then the total strength parameters (sometimes referred to as the undrainedstrength parameters) can be evaluated from the slope of the direct common tangent which is theCoulomb envelope (Equation 145) plotted on the Mohr circle diagram

cu = 40 kPa and φu = 132deg

It is obvious that the generated pore pressure has been ignored in the above solutionbull Effective strength parameters The effective stresses (σ1prime and σ3prime) on both test samples at failure

computed by subtracting the pore pressure from the total stress are indicated in Figure 149bThe Mohr circles corresponding to the two stress states are drawn in Figure 1410 The effective

TABLE 142 Measured CU Triaxial Test Data

TestCell Pressure

(kPa)Deviator Stress at Failure

(kPa)Pore Pressure

(kPa)

1 20 202 52

2 40 304 83

FIGURE 149 Stress states at failure (a) total stresses (kPa) (b) effective stresses (kPa)

FIGURE 1410 Mohr circle diagram for a consolidated undrained (CU) test

(a) (b)

σ3 = 20

σ1 = 402

00

Δu = 52

00

σ1 = 704 σ1 = 350 σ1 = 621

Δu = 83 σ3 = 40 σ3 = 148 = 327ʹ σ3ʹ

10

20

30

10 20 30 40 50 60 70 80 90

τ (k

Pa)

cacute = 55 kPaφacute = 137deg

Effectivestress envelope

Total stress envelopecu = 40 kPaφu = 132deg

σ (kPa)

strength parameters (sometimes referred to as the drained strength parameters) can be found fromthe slope of the Coulomb envelope for effective stresses plotted on the Mohr circle diagram

cprime = 55 kPa and φprime = 137deg

Example 143

Assume that one wishes to determine the strength properties of a medium stiff clayey foundation undershort-term (undrained) conditions An effective method for achieving this is to conduct a UU (quick)test For the results presented in Table 143 the undrained strength parameters have to be estimatedBecause the pore pressure generation is not monitored in these tests only the total stresses can be plottedas in Figure 1411 It can be seen that the deviator stress at failure does not change with the changingcell pressure during this type of test This is because the soil samples are not consolidated to thecorresponding cell pressures during UU (unconsolidated undrained) tests therefore the soil structureis unaffected by the change in cell pressure Hence the following strength parameters can be obtainedfrom Figure 1411

The reader should note that the subscripts u are used to distinguish the UU test parameters

141221 Selection of Triaxial Test Type Based on the Construction SituationThe CD strength is critical for consideration of long-term stability Examples of such situations include

bull Slowly constructed embankment on a soft clay depositbull Earth dam under steady-state seepagebull Excavation of natural slopes in clay

On the other hand CU strength is more relevant for the following construction conditions

bull Raising of an embankment subsequent to consolidation under its original heightbull Rapid drawdown of a reservoir of an earthen dam previously under steady-state seepagebull Rapid construction of an embankment on a natural slope

TABLE 143 Measured UU Triaxial Test Data

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

FIGURE 1411 Mohr circle diagram for an unconsolidated undrained (UU) test

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

Finally the UU strength is applicable under the following conditions

bull Rapid construction of an embankment over a soft claybull Large dam constructed with no change in water content in the clay corebull Footing placed rapidly on a clay deposit

14124 Standard Penetration Test

The standard penetration test (SPT) is the most common field test used to estimate the in situ shearstrength of foundation soil In this test a 140-lb hammer (Figure 1412) that falls 30 in is used to drivea standard split spoon sampler (Figure 1413) 18 in into the ground The number of hammer blowsnecessary to achieve the last 12 in of penetration is recorded as the blow count (N) Although it isrelatively easy to perform SPT suffers because it is crude and not repeatable The basic principleunderlying the SPT test is the relation between the penetration resistance and shear strength of the soilwhich can be visualized as a unique relationship Because the penetration resistance is obviously affectedby the overburden the following correction is applied before determining the soil properties

FIGURE 1412 Standard penetration test hammer

FIGURE 1413 Split spoon sampler

(146)

where σvprime is the effective overburden stress (in tons per square feet) computed as follows

σvprime = γz ndash γωdω (147)

where

γ = unit weight of soilz = depth of test locationγw = unit weight of waterdw = depth of test location from the groundwater table

Once the corrected blow count (N70prime ) is determined one can find the strength parameters based on theempirical correlations shown in Table 144 and Table 145 The subscript 70 indicates 70 efficiency inenergy transfer from the hammer to the sampler This value has been shown to be relevant for the NorthAmerican practice of SPT It should be noted that the undrained strength (cu) of a saturated clay is onehalf the unconfined compression strength (qu)

TABLE 144 Relation between SPT Blow Count and Friction Angle of Granular Soils

Description Very Loose Loose Medium Dense Very Dense

Relative density (Dr) 0 015 035 065 085

SPT N70primeFine 1ndash2 3ndash6 7ndash15 16ndash30

Medium 2ndash3 4ndash7 8ndash20 21ndash40 gt40

Coarse 3ndash6 5ndash9 10ndash25 26ndash45 gt45

φFine 26ndash28 28ndash30 30ndash34 33ndash38 mdash

Medium 27ndash28 30ndash32 32ndash36 36ndash42 lt50

Coarse 28ndash30 30ndash34 33ndash40 40ndash50 mdash

γwet (kNm3) 11ndash16a 14ndash18 17ndash20 17ndash22 20ndash23

a Excavated soil or material dumped from a truck has a unit weight of 11 to 14 kNm3 and must be quitedense to weigh much over 21 kNm3 No existing soil has a Dr = 000 nor a value of 100 Common rangesare from 03 to 07

Note Empirical values for φ and Dr and unit weight of granular soils are based on a normally consolidated(approximately φ = 28deg + 15degDr plusmn 2deg) SPT at about 6-m depth

Source Bowles JE Foundation Analysis and Design McGraw-Hill New York 1995 With permission

TABLE 145 Relation between SPT Blow Count and Unconfined Compression Strength of Clay

Consistency of Saturated Cohesive Soilsa N70prime qu (kPa) Remarks

Very soft 0ndash2 lt25 Squishes between fingers when squeezed

Soft 3ndash5 25ndash50 Very easily deformed by squeezing

Medium 6ndash9 50ndash100

Stiff 10ndash16 100ndash200 Difficult to deform by hand squeezing

Very stiff 17ndash30 200ndash400 Very difficult to deform by hand squeezing

Hard gt30 gt400 Nearly impossible to deform by hand

a Blow counts and OCR division serve as a guide in clay exceptions to the rule are very common

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

14125 Static Cone Penetration Test

The cone penetration test (CPT) has been gaining popularity as a more reliable and repeatable alternativeto SPT In this test a standard cone and a sleeve (Figure 1414) are advanced at a steady rate (1 cmsec)into the ground while the cone resistance (qc) and the sleeve friction (fs) are electronically measured Theentire cone apparatus and the associated computing facilities are usually trunk mounted as shown inFigure 1415 A typical cone profile obtained from a University of South Florida organic soil research siteis shown in Figure 1416 Because it measures the two parameters qc and fs CPT is a useful tool foridentifying soil type as well as for evaluating soil properties A convenient parameter termed the frictionratio (FR) is defined for this purpose as

(148)

Figure 1417 shows a simple chart that can be used for soil classification using CPT data Currentlyit is commonplace to have cone tips fitted with transducers that can produce a continuous record ofthe ground pore pressures at various depths Using CPT data the undrained strength of a clay can beobtained as

(149)

FIGURE 1414 Cone and sleeve

FIGURE 1415 Cone penetration test equipment (From Stinnette P Geotechnical Data Management and AnalysisSystem for Organic Soils PhD dissertation University of South Florida Tampa 1996)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

and po and uc are the effective overburden pressure and the pore pressure respectively measured in thesame units as su and qc a is taken as the approximate diameter ratio (d1D)2 (Figure 1414)

On the other hand the friction angle of a granular soil can be obtained from qc (in megapascals) basedon the following approximate expression

(1412)

For gravel and silty sand corrections of +5deg and ndash5deg respectively have to be made

1413 Compressibility and SettlementSoils like any other material deform under loads hence even if the integrity of a structure is satisfiedsoil supporting the structure can undergo compression leading to structural settlement For most drysoils this settlement will cease almost immediately after the particles readjust to attain an equilibriumwith the structural load This immediate settlement is evaluated using the theory of elasticity howeverif the ground material is wet fine-grained (low permeability) soil then the settlement will continue fora long period of time with slow drainage of water until the excess pore water pressure completelydissipates This is usually evaluated by Terzaghirsquos consolidation theory In some situations involving veryfine clays and organic soils settlement continues to occur even after the pore water pressure in thefoundation vicinity comes to an equilibrium with that of the far field Secondary compression conceptsare required to estimate this secondary settlement

FIGURE 1416 A typical cone profile (From Mullins AG Field Characterization of Dynamic Replacement ofFlorida Soils PhD dissertation University of South Florida Tampa 1996)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

q q u aT c c= + minus( )1

N PImT = +135 5

50

φ = +29 qc

14131 Estimation of Foundation Settlement in Granular Soils

Very often settlement of footings founded on granular soils is determined based on the plate load testsdiscussed in Section 142 The most commonly adopted analytical methods for settlement evaluation ingranular soils are based on the elastic theory however one must realize that reliable estimates of elasticmoduli and Poissonrsquos ratio values for soils are not easily obtained This is mainly because of the samplingdifficulty and particularly the dependency of the elastic modulus on the stress state Reliable fieldmethods for obtaining elastic moduli are also scarce The following expressions can be used to find theimmediate settlement

(1413)

wherese = immediate (elastic) settlementf = 05 or 10 (depending on whether se is at the corner of the foundation)B = width of foundationq0 = contact pressure (PBL where L is the length of the foundation)Es = elastic modulusα = a factor to be determined from Figure 1418

Another widely used method for computing granular soil settlements is the Schmertmann and Hartman(1978) method based on the elastic theory

(1414)

where

C1 = foundation depth correction factor = 1 ndash 05[qq ndash q)]C2 = correction factor for creep of soil = 1 + 02 log(time in years01)

FIGURE 1417 Soil classification using CPT data (From Bowles JE Foundation Analysis and Design McGraw-HillNew York 1995 With permission)

Authorrsquos extension

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silts Silty clayClayey

silt Clay

Peat

0 1 2 3 4 5 6Friction ratio (fR) ()

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

q = stress at foundation levelq = overburden stressIz = strain influence factor in Figure 1419

The elastic properties necessary to manipulate the above expressions are provided in Table 146 andTable 147 Furthermore some useful relationships that can provide the elastic properties from in situtest results are given below

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

A comprehensive example illustrating the use of the above relations is provided in Example 144

14132 Estimation of Foundation Settlement in Saturated Clays

The load applied on a saturated fine-grained soil foundation is immediately acquired by the pore wateras illustrated in Figure 1420a however with the dissipation of pore pressure resulting from drainage ofwater the applied stress (total stress) is gradually transferred to the soil skeleton as an effective stress

FIGURE 1418 Chart for obtaining α factor (From Das BM Principles of Foundation Engineering PWS PublishingBoston MA 1995 With permission)

FIGURE 1419 Strain influence factor (From Schmertmann JH and Hartman JP J Geotech Eng Div Am SocCiv Eng 104(GT8) 1131ndash1135 1978 Reprinted with permission of ASCE)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

For circular foundation

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

(Figure 1420b) The long-term soil skeleton rearrangement taking place during this process is termedthe consolidation settlement The soil properties required for estimation of the magnitude and rate ofconsolidation settlement can be obtained from the laboratory one-dimensional (1-D) consolidation testFigure 1421 shows the consolidometer apparatus where a saturated sample (25-in diameter and 10-inheight) is subjected to a constant load while the deformation and sometimes the pore pressure are

TABLE 146 Elastic Properties of Geomaterials

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Very dense 500ndash1440

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Note Value range for the static stress-strain modulusEs for selected soils (see also Table 56) The valuerange is too large to use an ldquoaveragerdquo value for designField values depend on stress history water contentdensity and age of deposit

Source Bowles JE Foundation Analysis and DesignMcGraw-Hill New York 1995 With permission

TABLE 147 Poisson Ratios for Geomaterials

Type of Soil micro

Clay saturated 04ndash05

Clay unsaturated 01ndash03

Sandy clay 02ndash03

Silt 03ndash035

Sand gravelly sand commonly used ndash01ndash100 03ndash04

Rock 01ndash04(depends somewhat on type of rock)

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

Source Bowles JE Foundation Analysis and Design McGraw-Hill New York1995 With permission

monitored until consolidation is complete A detailed description of this procedure can be found inBowles (1986) The sample is tested in this manner for a wide range of stresses that encompass theexpected foundation pressure Using Terzaghirsquos 1-D consolidation theory the relationship shown in Table148 between the degree of consolidation U (settlement at any time t as a percentage of the ultimatesettlement) and the time factor T can be derived for a clay layer subjected to a constant pressure incrementthroughout its depth

Figure 1422 shows the results of a consolidation test conducted on an organic soil sample Thecoefficient of consolidation (Cv) for the soil can be obtained from these results using Casagrandersquoslogarithm-of-time method (Holtz and Kovacs 1981) Using this method from Figure 1422 one canestimate the time for 90 consolidation as 200 sec Then by using the following expression for the timefactor one can estimate Cv as 25 times 10ndash4 in2sec because U = 90 when t = 200 sec

(1417)

FIGURE 1420 Illustration of consolidation settlement (a) subsurface profile (b) effective stress distribution and(c) pore pressure distribution

FIGURE 1421 Laboratory consolidometer apparatus (Figure courtesy of the University of South Florida Tampa)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

zz InitialEffectiveStress

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

where Hdr is the longest drainage path in the consolidating soil layer It should be noted that the waterin the laboratory soil sample drains through both sides during consolidation so Hdr = 05 in

When the above consolidation test is repeated for several other pressure increments doubling thepressure each time variation of the post-consolidation (equilibrium) void ratio e with pressure p can beobserved using the following relation between e and the sample strain computed from the monitoredsample deformation

(1418)

where e0 and H are the initial void ratio and the sample height and ∆H and ∆e are their respectivechanges A typical laboratory consolidation curve (e vs log p) for a clayey soil sample is shown in Figure1423 The following important parameters can be obtained from Figure 1423

Recompression index (Cr) = (1095 ndash 1045)(log60 ndash log10) = 0064Compression index (Cc) = (1045 ndash 093)(log120 ndash log60) = 0382Preconsolidation pressure (pc) = 60 kPa

TABLE 148 Degree of Consolidation vs Time Factor

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

FIGURE 1422 Settlement vs logarithm-of-time curve (From Stinnette P Engineering Properties of FloridaOrganic Soils Masterrsquos project University of South Florida Tampa 1992)

∆ ∆e

e

H

H1 0+=

DeflectionLog TimeState Road 580 Sample ldquoArdquo 25 T6F

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

All of the above information can be used to estimate the ultimate consolidation settlement of a saturatedclay layer (of thickness H) due to an average pressure increase of ∆p The ultimate consolidation settlement(scon) can be expressed by the following depending on the individual case as illustrated in Figure 1424

Case 1 (σv0prime gt pc)

(1419)

Case 2 (σv0prime + ∆p lt pc)

(1420)

Case 3 (σv0prime + ∆p gt σv0

prime )

(1421)

The average pressure increase in the clay layer can be accurately determined by using Newmarkrsquos chartshown in Figure 1425 When the footing is drawn on the chart to a scale of OQ = dc (the depth of themidplane of the clay layer from the footing bottom) ∆p can be evaluated by

∆p = qIM (1422)

FIGURE 1423 Laboratory consolidation curve (e vs logp)

FIGURE 1424 Illustration of the use of the consolidation equation (a) case 1 (b) case 2 and (c) case 3

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

where q I and M are the contact pressure the influence factor (specific to the diagram) and the numberof elements of the chart covered by the drawn footing respectively

Example 144

Assume that it is necessary to compute the maximum differential settlement of the foundation shown in Figure 1426 which also shows the SPT elastic moduli (using Equation 1415 for sands and 33 of the estimate for clay) and unit weight profiles as well as the strain influence factor plot For the above data

Contact pressure (q) = 200(15)2 kPa = 8889 kPaOverburden pressure at footing depth (q) = 165 times 10 kPa = 165 kPa

Immediate Settlement Areas of the strain-influence diagram covered by different elastic moduli are

FIGURE 1425 Newmarkrsquos chart (From Holtz RD and Kovacs WD An Introduction to Geotechnical EngineeringPrentice Hall Englewood Cliffs NJ 1981 With permission)

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

Then by applying Equation 1414 one obtains the immediate settlement as

From Equation 1413 scorner can be deduced as 05(587) = 294 mm

Consolidation Settlement As for the consolidation settlement the average stress increase in clay can beobtained as

On the other hand the average overburden pressure at the clay layer is found from Equation 143b as

From Figure 1424 one observes that the relevant expression for this situation is Equation 1421 and byusing the above estimates the consolidation settlement is found as

As for the corner the applicable expression from Figure 1424 is Equation 1420 hence

Therefore the total settlement at the center of the footing will be 3039 mm (112 in) while that at thecorner will be 60 mm (024 in)

Total Settlement Check Most building codes stipulate the maximum allowable total settlement to be 10in so the above value is unacceptable

FIGURE 1426 Settlement computation

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

Elev (m)Unit wt (kNm3)

165

175

180

Bedrock

GWT

Scenter = minus minus 1 0 5 16 5 88 89 16 5 1 0 ( ) 888 89 16 5

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

times + ) ( ) 7 10 0 033 2 57 10 5 873 3times + times = mm

∆

pcenter

corner

kPa

p

= times times times =

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

prime = + + minus =σv 0 16 5 2 17 5 1 5 18 0 1 0 9 8 2 75 ( ) ( ) ( ) ( ) 554 8 kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

Differential Settlement Check The differential settlement is equal to (scenter ndash scorner) distance from centerto corner or (3039 ndash 600)106 times 1000 = 0023 According to most building codes the maximumallowable differential settlement to prevent structural cracks in concrete is 0013 hence the above designfails the differential settlement criterion

1414 Groundwater and SeepageStability analysis of water-retaining concrete structures requires that the uplift forces exerted on them beevaluated These structures often exist in groundwater flow regimes caused by differential hydraulic headshence an analysis of groundwater seepage has to be performed invariably when estimating the upliftforces The most common and the simplest means of seepage analysis is the method of flownets In thismethod two orthogonal families of equipotential and flow lines are sketched in the flow domain (Figure1427) using the following basic principles A flow line is an identified or a visualized flow conduitboundary in the flow domain On the other hand an equipotential line is an imaginary line in which thetotal energy head is the same

14141 Rules Governing the Construction of a Flownet

1 Equipotential lines do not intersect each other2 Flow lines do not intersect each other3 Equipotential lines and flow lines form two orthogonal families4 To ensure equal flow in the drawn flow conduits and equal head drop between adjacent equipo-

tential lines individual flow elements formed by adjacent equipotential lines and flow lines bearthe same heightwidth ratio (typically 10)

FIGURE 1427 Seepage under a concrete dam

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

k = 1times10ndash6 cms

Datum (z = 0)

A B

1113D C

Distancedownstream

With seepage velocities being generally very low the pressure (p) exerted by seeping water contributesalong with the potential energy to the total head (energyunit weight) of water as

(1423)

The quantity of groundwater flow at any location in a porous medium such as soil can be expressed byDrsquoArcyrsquos law as

q = kiA (1424)

where k is the coefficient of permeability (or hydraulic conductivity) at that location and i the hydraulicgradient can be expressed by

(1425)

The following example illustrates the flownet method of seepage analysis and evaluation of uplift pres-sures For more accurate and rigorous methods the reader is referred to Harr (1962)

Example 145

Assume that it is necessary to establish the pressure distribution on the bottom of the dam shown in Figure1427 and the seepage under the dam shown in Figure 1427 As the first step in the solution a flownethas been drawn to scale following the rules above Using the bedrock as the datum for the elevation headtotal heads have been assigned using Equation 1423 for all of the equipotential lines as shown It is notedthat the head drop between two adjacent equipotential lines is

(9 m ndash 5 m)12 = 0333 m

Then by applying Equation 1423 to the points where the equipotential lines and the dam bottom (Bi)intersect the following expression can be obtained for the pressure distribution which is plotted inFigure 1427

p = γw(h ndash 30)

Then the total upthrust can computed from the area of the pressure distribution as 34 kPam acting ata distance of 45 m downstream

By applying Equation 1425 to the element ABCD one obtains

i = (5333 ndash 50)11 = 0302

Because k = 1 times 10ndash6 cms one can apply Equation 1424 to obtain the quantity of seepage through ABCDas

q1 = 1s(10ndash9)(0302)(13)(1) m3sm (because AD = 13 m)

Because all of the conduits must carry equal flow (see rule 4 of the flownet construction)

Note the following important assumptions made in the above analysis

1 The subgrade soil is homogeneous2 The bedrock and concrete dam are intact3 There is no free flow under the dam due to piping (or erosion)

Thus the design and installation of an adequate pore-pressure monitoring system that can verify theanalytical results are essential A piezometer with a geomembranesand filter that can be used for mon-itoring pore pressures is shown in Figure 1428

hp

zw

= +γ

idh

dx=

q = times = timesminus minus3 10 0 302 1 3 1 1 18 109 9( )( )( )( ) m sm m3 3sm

1415 Dewatering of Excavations

Construction in areas of shallow groundwater requires dewatering prior to excavation Although con-tractors specialized in such work determine the details of the dewatering program depending on the fieldperformance a preliminary idea of equipment requirements and feasibility can be obtained by a simplifiedanalysis Figure 1429 shows the schematic diagram for such a program and the elevations of the depressedwater table at various distances from the center of the well Observation wells (or bore holes) can beplaced at any location such as those shown at distances of r1 and r2 to monitor the water table depressionWhen analyzing a seepage situation like this Dupuit (Harr 1962) assumed that (1) for a small inclinationof the line of seepage the flow lines are horizontal and (2) the hydraulic gradient is equal to the slopeof the free surface and is invariant with depth For discharge through any general section such as an

FIGURE 1428 Piezometer probes (From Thilakasiri HS Numerical Simulation of Dynamic Replacement ofFlorida Organic Soils PhD dissertation University of South Florida Tampa 1996)

FIGURE 1429 Dewatering of excavations

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

observation well one can write the following expression for the flow by combining Equation 1424 andEquation 1425

(1426)

Noting that q and k are constants throughout the flow regime considered Equation 1426 can beintegrated between distances of r1 and r2 to obtain

(1427)

By defining the extent of dewatering using parameters r1 r2 h1 and h2 one can utilize the aboveexpression to determine the capacity requirement of the pump

1416 Environmental Geotechnology

The amount of solid waste generated in the United States was expected to exceed 510M tons by the year2000 (Koerner 1994) thus the immediate need for construction of adequate landfills cannot be over-emphasized Although the construction of landfills involves political and legal issues properly designedconstructed and maintained landfills have proven to be secure especially if they are provided with linedfacilities These are installed on the bottom or sides of a landfill to control groundwater pollution by theliquid mixture (leachate) formed by the interaction of rainwater or snowmelt with waste material Typesof liners for leachate containment are basically (1) clay liners (2) geomembranes and (3) compositeliners consisting of geomembranes and clay liners Of these until recently the most frequently used linerswere clay liners which minimized leachate migration by achieving permeability values as low as 5 times 10ndash8

to 5 times 10ndash9 cmsec however due to the large thickness requirement (06 to 2 m) and chemical activityin the presence of organic-solvent leachates geomembranes have been increasingly utilized for landfills

1417 Design of Landfill Liners

As shown in Figure 1430 and Figure 1431 the important components of a solid material containmentsystem include (1) a leachate collectionremoval system (2) a primary leachate barrier (3) a leachatedetectionremoval system (4) a secondary leachate barrier and (5) a filter above the collection systemto prevent clogging Some of the design criteria are as follows (Koerner 1994)

bull The leachate collection system should be capable of maintaining a leachate head of less than 30 cmbull Both collection and detection systems should have 30-cm-thick granular drainage layers that are

chemically resistant to waste and leachate and that have a permeability coefficient of not less than1 times 10ndash2 cmsec or an equivalent synthetic drainage material

bull The minimum bottom slope of the facility should be 2

FIGURE 1430 Typical cross-section of a geomembrane-lined landfill (From Koerner RM Designing with Geosyn-thetics 3rd ed Prentice Hall Upper Saddle River NJ 1994 With permission)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

Primary geomembraneSecondary geomembrane

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

4171 Design Considerations for Clay Liners

In the case of clay liners the US Environmental Protection Agency (EPA) requires that the coefficientof permeability be less than 10ndash7 cmsec This can be achieved by meeting the following classificationcriteria

bull The soil should have at least 20 fines (see Section 14111 Mechanical Analysis)bull The plasticity index should be greater than 10 (see Section 14112 Atterberg Limits)bull The soil should not have more than 10 gravel-size (gt475 mm) particlesbull The soil should not contain any particles or chunks of rock larger than 50 mm

It is realized that liner criteria can be satisfied by blending available soils with clay minerals such assodium bentonite

4172 Design Considerations for Geomembrane Liners

Geomembranes are mainly used in geotechnical engineering to perform the functions of (1) separation(2) filtration and (3) stabilization In this application of geotextiles the functions of separation and toa lesser extent filtration are utilized Due to the extreme variation of solid-waste leachate compositionfrom landfill to landfill the candidate liner should be tested for permeability with the actual or synthesizedleachate In addition to the permeability criterion other criteria also play a role in geomembrane materialselection They are as follows

bull Resistance to stress cracking induced by the soilwaste overburdenbull Different thermal expansion properties in relation to subgrade soilbull Coefficient of friction developed with the waste material that governs slope stability criteriabull Axisymmetry in tensile elongation when the material is installed in a landfill that is founded on

compressible subgrade soils

In selecting a geomembrane material for a liner serious consideration should also be given to itsdurability which is determined by the possibility of leachate reaction with the geomembrane and pre-mature degradation of the geomembrane For more details on geomembrane durability and relevanttesting the reader is referred to Koerner (1994) According to US EPA regulations the required minimumthickness of a geomembrane liner for a hazardous waste pond is 075 mm

142 Site Exploration

In addition to screening possible sites a thorough site study can reveal plenty of vital informationregarding the soil and groundwater conditions at a tentative site leading to more efficient selection offoundation depth and type as well as other construction details hence a site investigation that includesa subsurface exploration can certainly aid in economizing the time and cost involved in foundationconstruction projects An exhaustive site study can be separated into two distinct phases (1) preliminaryinvestigation and (2) detailed investigation In the preliminary investigation one would attempt to obtain

FIGURE 1431 Typical cross-section of a claygeomembrane-lined landfill (From Koerner RM Designing withGeosynthetics 3rd ed Prentice Hall Englewood Cliffs NJ 1994 With permission)

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Secondary geomembraneSecondary composite liner

Subsoil

as much valuable information about the site as possible at the least expense Useful information regardingthe site can often be obtained from the following sources

bull Local department of transportation (DOT) soil manualsbull Local US Geological Survey (USGS) soil mapsbull Local US Army Corps of Engineers hydrological databull US Department of Agriculture (USDA) agronomy mapsbull Local university research publications

A preliminary investigation also involves site visits (or reconnaissance surveys) where one can observesuch site details as topography accessibility groundwater conditions and nearby structures (especiallyin the case of expected pile driving or dynamic ground modification) Firsthand inspection of theperformance of existing buildings can also add to this information A preliminary investigation can bean effective tool for screening all alternative sites for a given installation A detailed investigation has tobe conducted at a given site only when that site has been chosen for the construction as the cost of suchan investigation is enormous This stage of the investigation invariably involves heavy equipment forboring therefore at first it is important to set up a definitive plan for the investigation especially interms of the bore hole layout and the depth of boring at each location Generally there are roughguidelines for bore hole spacing as indicated in Table 149

In addition to planning boring locations it is also prudent on the part of the engineer to search forany subsurface anomalies or possible weak layers that can undermine construction As for the depth ofboring one can use the following criteria

1 If bedrock is in the vicinity continue boring until sound bedrock is reached as verified from rockcore samples

2 If bedrock is unreachable one can seek depth guidelines for specific buildings such as those givenby the following expressions (Das 1995)

D = 3S07 (for light steel and narrow concrete buildings)D = 6S07 (for heavy steel and wide concrete buildings)

3 If none of the above conditions is applicable then one can explore up to a depth at which thefoundation stress attenuation reduces the applied stress by 90 (∆pσprimev0 = 01 in Example 144)This generally occurs around a depth of 2B where B is the minimum foundation dimension

Hand augers and continuous flight augers (Figure 1432a) can be used for boring up to a depth of about3 m in loose to moderately dense soil For extreme depths a mechanized auger (Figure 1432b) can beused in loose to medium dense sands or soft clays When the cut soil is brought to the surface a technicallyqualified person should observe the texture color and type of soil found at various depths and preparea bore-hole log identifying the soil types at the different depths This type of boring is called dry sampleboring (DSB) On the other hand if relatively hard strata are encountered investigators have to resortto a technique known as wash boring Wash boring is carried out using a mechanized auger and a water-circulation system that aids in cutting and drawing the cut material to the surface A schematic diagramof the wash-boring apparatus is shown in Figure 1433 and the Florida Department of Transportationdrill rig which utilizes the above technique is shown in Figure 1434

TABLE 149 Approximate Spacing of Boreholes

Type of Project Spacing (m)

Multistory 10ndash30

One-story industrial plants 20ndash60

Highways 250ndash500

Residential subdivisions 250ndash500

Dams and dikes 40ndash80

In addition to visual classification one has to obtain soil type and strength and deformation prop-erties for a foundation design hence the soil at various depths has to be sampled as the bore holesadvance Easily obtained disturbed samples suffice for classification index and compaction propertieswhile triaxial and consolidation tests require carefully obtained undisturbed samples (samples withminimum disturbance) Disturbed granular or clayey samples can be obtained by attaching a standardsplit spoon sampler (Figure 1413) to the drill rods An undisturbed clay sample can be obtained bycarefully advancing and retrieving a Shelby tube (Figure 1435) into a clay layer however if one needsto evaluate a granular material for strength settlement or permeability then in situ tests have to beperformed due to the difficulty in obtaining undisturbed samples in such soils In this regard the readeris referred to the in situ tests shown in Table 1410 A description of the plate load test is presented inSection 1421

FIGURE 1432 Drilling equipment (a) hand-auger and (b) mechanized auger (Figure courtesy of the University ofSouth Florida Tampa)

(a)

(b)

FIGURE 1433 Schematic diagram of wash boring

FIGURE 1434 Florida Department of Transportationrsquos CME-75 drill rig

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

Returningwater withcut soil

To hoist

Cut soilwater

Collection tubor sump

Pressurized water

Drill bit

1421 Plate Load Tests

Plate load apparatus consists of a set of steel plates of standard diameters (12 in 18 in etc) a hydraulicloading and recording mechanism a reaction frame and a deflection gauge (Figure 1436) During thetest the plate is laid at the tentative foundation depth and gradually loaded while the magnitude of theload and plate deflection at different stages is recorded Figure 1437 shows a typical plot of plate loadresults on a sand deposit When one scrutinizes Figure 1437 it can be seen that the ultimate bearingcapacity of the plate can be estimated from the change in gradient of the loadndashdeflection curve hence thebearing capacity and the settlement of a tentative foundation can be predicted in the following mannerbased on the results of a plate load test performed on that location In the following expressions thesubscripts f and p refer to the foundation and the plate respectively

Ultimate bearing capacity in clayey soils

(1428)

Ultimate bearing capacity in sandy soils

(1429)

where Bp and Bf refer to the plate diameter and the minimum foundation dimension respectivelyOne can deduce the above expressions based on the basic expression for the bearing capacity of shallow

footings (Section 143 Equation 1432) when one realizes that predominant contributions for bearingcapacity in clay and sand are made by the terms involving Bc and Nγ terms of Equation 1432 respectively

FIGURE 1435 Shelby tubes

TABLE 1410 Recommended In Situ Tests

Evaluation Parameter Test

Permeability Field pumping testa

Settlement Plate load testb

Shear strength SPT or CPTc

a In Section 1415 Dewatering of Excavationsb In Section 1421 Plate Load Testsc In Section 1412 Strength of Soils

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

Settlement of a footing under a given contact pressure (q) can be estimated by the corresponding platesettlement (sp) (Figure 1437) using the following expressions

Immediate settlement in clayey soils

(1430)

Immediate settlement in sandy soils

(1431)

143 Shallow Footings

A shallow spread footing must be designed for a building column to transmit the column load to theground without exceeding the bearing capacity of the ground and causing an excessive settlement (Figure1438) Plate-load test results clearly demonstrate the existence of a maximum stress (approximately 10psi in Figure 1437) that can be imposed on a plate without causing excessive settlement This maximumstress is termed the bearing capacity of a foundation

1431 Bearing Capacity of Shallow Footings

To avoid catastrophic bearing failures shallow footings are proportioned based on the bearing-capacitycriterion Two expressions extensively used to evaluate the ground bearing capacity are provided below

14311 Terzaghirsquos Expression

(1432)

FIGURE 1436 Plate load test

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

q s cN qN s B Nc c qult = + + ( )γ γγ0 5

14312 Hansenrsquos Expression

(1433)

where

Nc Nq and Nγ are bearing capacity factors (Table 1411)s coefficients are shape factors based on BL (Table 1412)d coefficient are depth factors based on Df B (Table 1412)i coefficients are inclination factors based on load inclination φ (Table 1412)γ is the unit weight of soil in the footing influence zonec and φ are the shear strength parameters of the soil

Thus to avoid bearing-capacity failure

(1434)

where

qnult = net ultimate bearing capacity based on Equation 1435P = structural loadA = footing areaF = safety factor

qnult = qult ndash q (1435)

FIGURE 1437 Typical plate load test results (From AG Mullins AG Field Characterization of Dynamic Replace-ment of Florida Organic Soils PhD dissertation University of South Florida Tampa 1996)

FIGURE 1438 Schematic diagram of a shallow footing

Individual column (1 3)

Plate Deflection (in)0 1 2 3 4

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q s d i cN s d i qN s d i B Nc c c c q q q qult = + + γ γ γ γγ0 5

q

F

P

An ult gt

Example 146

Proportion a suitable footing for a 1000-kN vertical column load on a sandy ground where the SPTresults are as indicated below Assume that the groundwater table is at a depth of 05 m below the groundsurface

An average N value has to be determined from the above data within the influence zone of the footingFor this one has to assume a footing size as the influence zone depends on the size of the footing soassume a circular footing of diameter 15 m placed at a depth of 1 m from the ground surface As indicatedin Figure 1439 the influence zone extends from 05Df above the footing (ie elevation ndash05 m) to 2B

TABLE 1411 Bearing Capacity Factors

Terzaghirsquos Expression Hansenrsquos Expression

φ Nc Nq Nγ Nc Nq Nγ

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

TABLE 1412 Shape Depth and Inclination Factors

Hansenrsquos Expression Terzaghirsquos Expression

Sq = 1 + (BL) tanφ

Sγ = 1 ndash 04(BL) 10 for strip footings06 for circular footings08 for square footings

Depth For Df B lt 1

dc = 1 + 04(Df B)

dq = 1 + 2tanφ(1 ndash sinφ)2(Df B)

dγ = 1

For Df B gt 1

dc = 1 + 04tanndash1(DfB)

dq = 1 + 2tanφ(1 ndash sinφ)2tanndash1(Df B)

dγ = 1

Inclination ic = iq = (1 ndash β90deg)2 a

iγ = (1 ndash βφ)2 a

a Here β is the load inclination to the vertical

Elevation (m) N

10 520 730 1040 1250 12

below the footing (ie ndash40 m) Then by averaging the corrected N values within this range one canobtain the average Nprime as worked out in the table below

Note that the vertical effective stresses (σvprime) are obtained using Equation 143 and assuming unit weightsof 170 kNm3 and 98 kNm3 respectively for sand and water while CN is obtained using Equation 146

Then from Table 144 a φ of 37deg can be found This yields interpolated values Nc = 60 Nq = 49 andNγ = 57 from Table 1411 (Hansenrsquos factors) The following factors can also be evaluated from Table 1412

The following quantities are also needed for Equation 1435

q = σvprime at the foundation level = 121 kPa

γ = γprime (because the foundation is fully submerged) = 170 ndash 98 = 72 kNm3

Finally by substituting the above values in Equation 1433 one obtains the ultimate bearing capacity as

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

Note that the cohesion term is dropped due to negligible cohesion in sandy soils Then using Equation1434 one obtains a safety factor of 13167(10001515) = 296 which provides an adequate designhence a 15 times 15-m footing at a depth of 10 m would suffice Note that if the groundwater table waswell below the footing (usually greater than 2B) then one would revise the following quantities as

FIGURE 1439 Foundation influence zone

Elevation (m) N σvprime (kPa) CN Nprime

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

Note Average Nprime = [14(10) + 150(10) + 19(10) + 203(05)]35 = 17

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

i i ic q (because the loadd is applied vertically)

q = σvprime at foundation level = 17 kPa

γ = kNm3

If the groundwater table was below the footing but still near it (a distance of d below) one can usethe following approximation to evaluate the γ term

(1436)

As an example if the groundwater table was 20 m below the ground one can assume a γdry of 165 kNm3 to modify the above two quantities as

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

1432 Footings with Eccentricity

If a footing has to be designed for a column that carries an axial load (P) as well as a moment (M) oran eccentric axial load the resulting contact pressure distribution is as shown in Figure 1440a Onerealizes however that this is statically equivalent to the uniform distribution shown in Figure 1440bhence a simple method of computing the bearing capacity is to assume that only the portion of thefooting containing the column at its center contributes to bearing capacity When designing such afooting modified dimensions (Bprime and Lprime) have to be used in Equation 1432 or Equation 1433 whereBprime and Lprime are defined as follows

Bprime = B ndash 2ex

Lprime = L ndash 2ey

Example 147

Check the adequacy of the footing shown in Figure 1441 for the soil data obtained from the UU test inExample 143 (cu = 506 kPa) From Figure 1441

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

Then Bprime = 11 and Lprime = 11 Because φ = 0deg one obtains Nc = 514 Nq = 10 and Nγ = 00 from Table1411 hence the only significant term in Equation 1433 is the cohesion term and only the relevantfactors are computed as follows

Finally the safety factor can be computed as

F = (39378)(11)(10)250 = 1733

Because the safety factor has to be more than 25 this is not a safe design This factor can be improvedby increasing the dimensions to about 20 times 20 m depending on the available space

γ γγ γ

= prime +minus prime( )dry d

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

1433 Presumptive Load-Bearing Capacity

The building codes of some cities suggest bearing capacities for a certain building site based on theclassification of the predominant soil type at the site Table 1413 presents a comprehensive list of presump-tive bearing capacities for various soil types however it should be noted that these values do not reflectthe foundation shape depth load inclination location of the water table and settlements that are associatedwith the sites For this reason the use of these bearing capacities is primarily advocated in situations wherea preliminary idea of the potential foundation size is needed for the subsequent site investigation

144 Mat Footings

Because mat footings are larger in dimension than isolated spread footings they are commonly used fortransferring multiple column loads to the ground to prevent bearing-capacity failures Thus an idealapplication of a mat footing would be on relatively weak ground However if the ground has sufficientstrength to produce adequate bearing for isolated spread footings a mat footing will be an economicalalternative only if the combined area of the spread footings is less than 50 of the entire building plan area

1441 Design of Rigid Mat Footings

14411 Bearing Capacity of a Mat Footing

One can use Equation 1432 or Equation 1433 to proportion a mat footing if the strength parametersof the ground are known However because the most easily obtained ground strength parameter is thestandard penetration blow count (N) an expression is available that uses N to obtain the bearing capacityof a mat footing on a granular subgrade This is expressed as follows

FIGURE 1440 Simplistic design of an eccentric footing (a) pressure distribution due to an eccentric load and (b)equivalent pressure distribution

FIGURE 1441 Design of an eccentric footing (a) plan and (b) elevation

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

Mx = 50 kNmMy = 625 kNm

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

qnall = net allowable bearing capacity (kPa)B = width of footings = settlement (mm)Df = depth of footing (mm)

Then the following condition has to be satisfied to avoid bearing failure

(1438)

in which the use of a safety factor is precluded by employing an allowable bearing capacity

Example 148

Figure 1442 shows the plan of a column setup where each column is 05 times 05 m in section Design anadequate footing if the corrected average SPT blow count of the subsurface is 10 and if the allowablesettlement is 254 mm (1 in) Assume a foundation depth of 05 m The bearing capacity can then becomputed from Equation 1437 as

TABLE 1413 Presumptive Bearing Capacities

Presumptive Bearing Capacities from Indicated Building Codes (kPa)

Soil DescriptionChicago (1995)

National Board of Fire Underwriters

(1976)BOCA(1993)a

Uniform Building Code (1991)b

Clay very soft 25 mdash mdash mdash

Clay soft 75 100 100 100

Clay ordinary 125 mdash mdash mdash

Clay medium stiff 175 100 100

Clay stiff 210 mdash 140 mdash

Clay hard 300 mdash mdash mdash

Sand compact and clean 240 mdash 140 200

Sand compact and silty 100 mdash mdash mdash

Inorganic silt compact 125 mdash mdash mdash

Sand loose and fine mdash mdash 140 210

Sand loose and coarse sandndashgravel mixture compact and fine

mdash 140ndash400 240 300

Gravel loose and compact coarse sand 300 mdash 240 300

Sand-gravel compact mdash mdash 240 300

Hardpan cemented sand cemented gravel 600 950 340 mdash

Soft rock mdash mdash mdash mdash

Sedimentary layered rock (hard shale sandstone siltstone)

mdash mdash 6000 1400

Bedrock 9600 9600 6000 9600

a Building Officials and Code Administrators International Incb Bowles (1995) interpretation

Note Values converted from pounds per square foot to kilopascals and rounded Soil descriptions vary widely between codestable represents authorrsquos interpretations

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

By applying Equation 1438

Hence the mat can be designed with a 025-m edge space as shown in Figure 1442For the reinforcement design one can follow the simple procedure of separating the slab into a number

of strips as shown in Figure 1442 Each strip (BCGF in Figure 1442) can be considered as a beam Theuniform soil reaction per unit length (w) can be computed as (4000)(25)[(55)(55)] = 3305 kNmFigure 1443 indicates the free-body diagram of the strip BCGF (Figure 1442) It can be seen from thefree-body diagram that the vertical equilibrium of each strip is not satisfied because the resultant down-ward load is 2000 kN as opposed to the resultant upward load of 1815 kN This discrepancy results fromthe arbitrary separation of strips at the midplane between the loads where nonzero shear forces exist Infact one realizes that the resultant upward shear at the boundaries BF and CG (Figure 1442) accountfor the differencemdashthat is 185 kN To obtain shear and moment diagrams of the strip BCGF one canadd this to modify them as indicated in the figure This was achieved by reducing the loads by a factorof 0954 and increasing the reaction by a factor of 1051 The two factors were determined as follows

For the loads [(2000 + 1815)2]2000 = 0954For the reaction [(2000 + 1815)2]1815 = 1051

FIGURE 1442 Illustration of a mat footing

FIGURE 1443 Free-body diagram for strip BCGF (Figure 1442)

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

The resulting shear and moment diagrams are indicated in Figure 1444 and Figure 1445 Now usingFigure 1444 and Figure 1445 we can determine the steel reinforcements as well as the mat thicknessThis estimation is not repeated here as it can be found in other chapters of this book

14412 Settlement of Mat Footings

The settlement of mat footings can also be found using the methods that were outlined in Section 1413(Compressibility and Settlement) assuming that they impart stresses on the ground in a manner similarto that of spread footings

1442 Design of Flexible Mat FootingsFlexible mat footings are designed based on the principle of slabs on elastic foundations Because of theirfinite size and relatively large thickness one can expect building foundation mats to generally exhibitrigid footing behavior therefore applications of flexible footings are limited to concrete slabs used forhighway or runway construction The most significant parameter associated with the design of beamson elastic foundations is the radius of relative stiffness (1β) given by the following expression

(1439)

where

E = elastic modulus of concretemicro = Poissonrsquos ratio of concretek = coefficient of subgrade reaction of the foundation soil usually determined from the plate load

test (Section 1421) or Equation 1440h = slab thickness

(1440)

FIGURE 1444 Distribution of shear on strip BCGF (Figure 1442)

FIGURE 1445 Distribution of moment on strip BCGF (Figure 1442)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

ndash21954 ndash21954

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

Es = elastic modulus of subgrade soilmicros = Poissonrsquos ratio of subgrade soil

When β has been evaluated for a particular mat the shear moment and reinforcing requirementscan be determined from nondimensional charts that are based on the solution for a concentrated load(P) applied to a slab on an elastic foundation The following expressions can be used along with Figure1446 for the evaluation

FIGURE 1446 Radial and tangential moments and shear coefficients in a slab under point load (From Scott RFFoundation Analysis Prentice Hall Englewood Cliffs NJ 1981 With permission)

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(a) Radial Moment (Mr)

βr

(b) Tangential Moment (Mθ)

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

FIGURE 1447 Influence chart for determining moment at the edge of a slab (From Pickett G and Ray GK AmSoc Civ Eng Trans 116(2425) 49ndash73 1951)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

The moment due to a distributed load can be obtained by drawing the contact area on an influence chartsuch as the one shown in Figure 1447 and then using Equation 1443 It should be noted that the scalefor the drawing should be selected such that 1β is represented by the distance l shown in Figure 1447

(1443)

where

P = distributed loadN = number of elements covered by the loading area drawn

Example 149

Plot the shear and moment distribution along the columns A B and C of the infinite slab of 8-inthickness shown in Figure 1448 Consider it to be a flexible footing Assume a coefficient of subgradereaction of 2600 lbft3 Because Ec = 576 times 108 psf and microc = 015 then one can apply Equation 1439 toobtain β = 01156 ftndash1 Using the above results Figure 1446a and Equation 1441a Table 1414 can bedeveloped for the radial moment (the moment on a cross-section perpendicular to the line ABC in Figure1448) These moment values are plotted in Figure 1449

145 Retaining Walls

When designing a retaining structure one must ascertain that its structural capacity is adequate towithstand any potential instability that can be caused by the lateral earth pressures of the retained backfillhence a major step in the design of a retaining structure is the evaluation of the magnitude directionand the line of action of the lateral force Most of the methods available for analyzing lateral earthpressures assume a yielding soil mass in the vicinity of the retaining structure so the solutions are basedon the limit equilibrium The magnitude of the lateral force depends on the soil failure mechanism Themechanism in which the backfill yields with the outward movement of the wall is known as the active

FIGURE 1448 Illustration of an infinite flexible slab

TABLE 1414 Flexible Footing Moments

Distance from A (ft)

C Coefficient for Load at A

C Coefficient for Load at B

C Coefficient for Load at C

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

250 kips 500 kips 250 kips

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

failure mechanism (Figure 1450a) while the yielding of soil due to inward wall movement is termed thepassive failure mechanism (Figure 1450b) Also indicated on Figure 1450 is the orientation of the failureplanes for each condition in the case of a smooth vertical wall supporting a horizontal backfill The twomost widely used analytical methods are illustrated below

1451 Determination of Earth Pressures

14511 Rankine Method

The Rankine method of analysis can be employed for the relatively simple case of a smooth verticalwall supporting a homogeneous backfill A modified form of Rankinersquos original analysis allows one toobtain the active and passive earth pressure distributions by using the following expressions

(1444)

and

(1445)

where the subscripts a and p stand for active and passive states respectively and the coefficients of earthpressure Ka and Kp can be determined from the following expressions

(1446)

FIGURE 1449 Moment distribution on a flexible slab (in kip-ft)

FIGURE 1450 Illustration of (a) active and (b) passive states

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Retainingwall Backfill

Failureplane

(a)

Active State(b)

Passive State

Failureplane

45deg + φ2 45deg ndash φ2

prime = prime minusσ σa v a aK c K2

prime = prime minusσ σp v p pK c K2

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

β = inclination of the backfillcφ = soil strength parameters

The direction of the resultant lateral pressure and the line of action are indicated on Figure 1451

Example 1410

Determine the lateral pressure on the retaining wall shown in Figure 1451 Assume active conditionsBy using Equations 1446 one obtains Ka = 038 Then by using Equation 1444

At z = 00 m σaprime = ndash2(10)(0616) = ndash1232 kPaAt z =2 0 m σaprime = 06 kPaAt z =8 0 m σaprime = 193 kPa

The line of action of the effective force can be found by discretizing the pressure diagram into fivesegments as shown in Figure 1451

Hence

In obtaining the total force on the wall one should remember to include the water pressure in segment5 in addition to the effective force

FIGURE 1451 Illustration of the Earthrsquos pressure distribution

Segment Area (A)Centroidal

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

14512 Coulomb Method

The Coulomb method of analysis is an attractive alternative to the Rankine method due to its ability tohandle more complex cases involving rough nonvertical retaining walls The relevant equations for lateralearth pressure coefficients are given below

(1448)

(1449)

where α and δ are the inclination of the wall face to the horizontal and the angle of wall frictionrespectively As illustrated in Figure 1452 the direction of the resultant lateral force changes whenconditions change from active to passive states with the line of action remaining on the generators ofthe friction cone The above coefficients can be used in conjunction with Equation 1444 and Equation1445

On the other hand if α = 90deg and β = δ = 0deg it can be easily shown that Equation 1448 and Equation1446 reduce to

(1450)

Hence one can see that Rankinersquos and Coulombrsquos analytical predictions agree only when earth pressuresare predicted on smooth vertical walls supporting horizontal backfills In this case the direction of thelateral force is horizontal

Example 1411

Use the Coulomb method to determine the lateral earth pressure on the wall shown assume passiveconditions Because the angle of wall friction (δ) is not specified in this problem it would be adequate toassume that δ = (23)φ = 1333deg Substituting α = 70deg φ = 20deg δ = 1333deg and β = 0deg in Equation 1449one would obtain Kp = 1968 With c = 00 and a surcharge of q Equation 1445 reduces to the following

FIGURE 1452 Illustration of the application of Coulombrsquos analysis

ndash1333deg+1333deg

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

minus( ) ++( ) minus( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

ssin sinα δ α βminus( ) +( )

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

ssin sinα δ α β+( ) +( )

2

Ka = minus+

1

sin

φφ

(1451)

where σvprime = 120z and q = 610 psf Then one would obtain the passive pressure distribution as

(1452)

and the resultant lateral force will be 44568 kipsft acting at a distance of 60 ft above the base as shownin Figure 1452

14513 Trial-Wedge Method

When the backfill is nonhomogeneous or if there is a finite or a concentrated surcharge on the backfillthe above methods cannot be used to determine the earth pressures In such situations the lateral forceon a retaining wall can be estimated by the graphical construction of the force polygon for a selectedpotential failure wedge The following example is provided to illustrate this technique

Example 1412

Use the trial-wedge solution to estimate the active soil pressure on the retaining wall shown in Figure 1453

Step 1 Select a trial failure surface An appropriate initial estimate would be the one correspondingto a smooth vertical wall with a horizontal backfill indicated in Figure 1450 Hence in this caseselect a surface with an inclination of 45deg + (10deg2) = 50deg to the horizontal (OC)

Step 2 In the case of active conditions estimate the depth up to which tension develops (= 2cγKa)and demarcate the area of tension cracks In this case the vertical depth of tension cracks is 44ft and the tension crack area is indicated by AprimeBprimeCprime Therefore an effective cohesive force isdeveloped on the OC surface only up to Cprime This will have a magnitude of OCprime times 1 times 5 ft times 15 times144 psf = 1526 kips

Step 3 Estimate the total weight of the failure wedge If there is a water table submerged unit weightsmust be used under the water table to account for buoyancy In this case the weight of AOC canbe estimated as (05 times 126 times 31 times 120) + (05 times 126 times 115 times 115) = 10675 kips

Step 4 Draw a force polygon assuming the indicated directions for the two other forces (active forceon the wall and the reaction from the intact soil mass) In this case these are designated as Pa andR on Figure 1454

Step 5 Estimate Pa from the force polygon In this case Pa = 525 kips

FIGURE 1453 Trial wedge method

Trial failuresurfaceCBA

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

Sand fill fordrainage (γ = 120 lbft3)Concrete

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

prime = prime +( )σ σp v pq K

σp z= +236 16 1200

Step 6 Repeat the above procedure for a number of different trial failure planes Estimate the minimumlateral force to be the actual Pa and consider the corresponding trial failure plane to be the actualfailure plane

Step 7 If G is the centroid of the actual failure wedge obtain the line of action of Pa from the pointQ on the wall such that QG is parallel to the failure surface OC

1452 Design of Concrete Retaining WallsThe two basic types of concrete retaining walls are the gravity type and the cantilever type The followingconditions must be satisfied when designing a concrete retaining wall

bull Stability against overturning due to earth pressurebull Stability against base sliding due to earth pressurebull Prevention of any possible tensile stresses in the base soil

In addition to these design criteria the design must also provide for a good drainage system and adequatebearing (see Section 1431 Bearing Capacity of Shallow Footings) and no excessive settlements (seeSection 1413 Compressibility and Settlement)

14521 Design of a Cantilever Retaining Wall

The basic design of a cantilever retaining wall is illustrated by the following example

Example 1413

Design a suitable cantilever retaining wall to support the backfill shown in Figure 1455 Bowles (1995)recommended the tentative dimensions shown in Figure 1456 for a cantilever retaining wall Based onBowlersquos recommendations the dimensions shown in Figure 1455 are assumed for the retaining wall Incomputing the lateral pressure on the wall it is usually assumed that the wall section starts at the cross-section CG shown in Figure 1456 hence the active wall pressure can be evaluated using either of thedescribed methods as 4496 kipsft at a height 5 ft from the base Despite the passive force due to the soilberm on the toe practitioners generally neglect it for a conservative design The weights of the differentretaining wall components are computed as follows

w1 = (50 times 135 times 120)1000 = 81 kipsftw2 = (90 times 15 times 150)1000 = 2025 kipsftw3 = (10 times 135 times 150)1000 = 2025 kipsftw4 = (05 times 05 times 135 times 150)1000 = 0506 kipsft

The locations of the respective centroids from the toe are indicated below

FIGURE 1454 Force polygon

Element Weight (wi) Centroidal Distance (xi) wixi

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

Scale 1 = 5 kipsPA = 105 x 5 = 525 kips

Stability Against Overturning The factor of safety against overturning can be computed as

This is adequate as it is greater than 15

Stability Against Sliding The normal reaction on the base is equal to Σwi = 12656 Thus the maximumfrictional force on the base is equal to 12656(tanφ) = 6449 kipsft Then the safety factor against slidingcan be computed as

This is also acceptable as it is greater than 125

Check for Tension in the Base Soil The eccentricity on the base produced by the above forces is equal to05(81 times 20 ndash 2025 times ndash0506 times 167 ndash 4496 times 5)(12656) = e = ndash0723 m For tension not to developin the base soil e lt B6 Because B60 = 15 the base tension criterion is also satisfied by this design(Figure 1457)

FIGURE 1455 Tentative design dimensions for a cantilever retaining wall (From Bowles JE Foundation Analysisand Design McGraw-Hill New York 1995 With permission)

FIGURE 1456 Cantilever retaining wall design

H48

Minimum batter

200 mm minimum(300 mm preferable)

H12 to H10

B = 04 to 07 H

Below frost depth and seasonal volume change

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

Compacted Granular Fillc = 0φ = 30γ = 120 lbft3

W1

F1

52= =Stabilizing moment

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

4 49= =Stabilizing force

Sliding force

661 43=

Reinforcement Requirements To determine reinforcement requirements the shear and bending momentdiagrams can be drawn separately for the stem and the base (Figures 1458 and 1459) It should be notedthat the maximum shear and moment in the stem occurs at its base and are 3045 kipsft and 16412kip-ftft respectively The reader should note that when obtaining the shear and moment diagrams forthe base the base soil pressure is represented by a uniform distribution within a distance of Bprime (equal toB ndash 2e) as this is statically equivalent to an eccentric force of 12656 kipsft A maximum shear force of3743 kipsft is produced at the base The reader should also note that in the moment diagram withinthe stem area of the base an unbalanced stem moment of 16512 kipsft (Figure 1458) is graduallytransferred to the base Further the heel of the base will have a moment of 42 kip-ftft due to the earthpressure distribution on the base thickness as shown in the load distribution diagram (Figure 1459a)This information can be utilized for reinforcement design

Drainage Design The interested reader is referred to Bowles (1995) for further design details howeverone should remember the following preliminary guidelines

bull Provide a compacted free-draining soil layer adjacent to the retaining wall if the backfill materialis clayey in nature

bull Provide a perforated collector drain with an appropriate soil or geomembrane filter at the base ofthe retaining wall

bull Make provisions for drainage outlets known as weep holes in the stem with a soil or geomembranefilter at the wall face to prevent clogging

1453 Effect of Water TableIn many instances the soil behind an earth-retaining structure is submerged Examples include seawallssheet pile walls in dewatering projects and offshore structures Another reason for saturation of backfillmaterial is poor drainage which leads to an undesirable buildup of water pressure behind the retaining

FIGURE 1457 Design details

FIGURE 1458 Load shear and moment diagrams for the retaining wall stem

5 ft15 ft

Filters

Collector drain with a filter

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

(b) Shear (c) Moment

3645 kipft 1641 kip-ftft

Z Z

(a) Load

wall Drainage failure often results in subsequent failure and collapse of the earth-retaining structure Incases where the design considers the presence of a water table the lateral earth pressure is calculatedfrom the effective soil stress Oddly enough this leads to a reduction in effective horizontal earth pressureas the effective stresses are lower than their total counterpart however the total stresses on the wallincrease due to the presence of the hydrostatic water pressure In other words while the effectivehorizontal stress decreases the total horizontal stress increases

1454 Reinforced Walls

14541 Geogrid-Reinforced Walls

With the increased availability of high-strength geogrid materials in the 1990s geogrid-reinforced wallswere introduced as an alternative to metallic strip reinforcement They provide increased interface area(because the coverage area can be continuous) better interlocking with the backfill (due to the geometryof the openings) resistance to corrosive environments and lower cost The most common type of geogridsused in earth reinforcement is the uniaxial type due to its high strength and stiffness in the main directionFacing panel units may be connected to the geogrid using a steel bar interwoven into the grid (knownas a Bodkin connector) or more recently through special plastic clamps that tie into a geogrid sectionembedded in the concrete panel

Among the concerns associated with the use of geogrids in heavily loaded walls (such as bridgeabutments) are the time-dependent stress relaxation (creep deformation) installation damage andchemical degradation It is therefore crucial to determine the design strength of the geogrid consideringthe various reduction factors due to creep installation damage chemical degradation and biological

FIGURE 1459 (a) Load (b) shear and (c) moment diagrams for the retaining wall base

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(c) Moment (kip-ftft)

degradation (Gunaratne 2006) In addition it is extremely important to ensure that the geogrid is fullystretched during installation and compaction of the subsequent soil layer otherwise significant defor-mation is required before tensile stresses and interface friction are mobilized A closely related problemthat has been identified is the difficulty in keeping the facing elements plumb during installationespecially when close tolerance is required in tall walls

Design and construction procedures for geogrid reinforced walls are almost identical to reinforcedearth walls One distinct exception is that the strength of the geogrid is expressed in terms of force perunit length and the associated horizontal spacing (sh) is taken as a unit length in all calculations Anotherdifference is that because of the effective interlocking of the soil particles within the geogrid openingsthe interface friction angle is usually equal to the internal friction angle of the soil

14542 Geotextile-Reinforced Walls

Unlike metallic and geogrid reinforcement typical geotextile-reinforced wall designs do not require facingelements Instead the geotextile layer is wrapped around the compacted soil at the front to form thefacing (Figure 1460) The finished wall must be covered with shotcrete bitumen or gunite to preventultraviolet radiation from reaching and damaging the geotextile Such walls are usually constructed astemporary structures or where aesthetics are not of prime importance however it is possible to coverthe wall with a permanent faux finish that blends with the surrounding environment The designprocedures for geotextile-reinforced walls are also identical to those described earlier for steel and geogridreinforcement The interface friction angle between the soil and the geotextile sheet is typically equal to12 to 23φ In addition the overlap length (Lo) must be determined from the following equation

(1453)

where sv is the vertical spacing of the geotextile and z is any given depth The minimum acceptable overlaplength is 1 m

1455 Sheet Pile Walls

14551 Cantilever Sheet Piles

A conceptual representation of the lateral earth pressure acting on a cantilever sheet pile is shown inFigure 1461 Only active pressure is present on side A from the ground surface to the depth ofexcavation Below the excavation depth passive conditions are assumed to act on side B of the sheetpile while active conditions persist on side A up to point O where a reversal of conditions occursPoint O can be viewed roughly as the point of rotation of the sheet pile in the ground Such rotationis necessary to achieve static equilibrium of the system Below point O active conditions develop onside B while passive earth pressures are present on side A Cantilever sheet pile design typically involvesthe determination of the embedment depth (D) given other geometric constraints of the problem aswell as soil properties Therefore the first step is to calculate the magnitude of the horizontal stresses

FIGURE 1460 Geotextile-reinforced wall with wrapped-around facing

GeotextileReinforcement

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Chapter 1 Concrete Constituent Materials
Chapter 2 Mineral Admixures
Chapter 3 Chemical Admixtures
Chapter 4 Long-Term Effects and Serviceability
Chapter 5 Properties and Performance of Normal-Strength and High-Strength Concrete
Chapter 6 Design and Placement of Concrete Mixtures
Chapter 7 Design and Construction of Concrete Formwork
Chapter 8 Construction Loading in High-Rise Buildings
Chapter 9 Deflection of Concrete Members
Chapter 10 Structural Concrete Systems
Chapter 11 Construction of Prestressed Concrete
Chapter 12 Unbonded Post-Tensioning System Technology in Building Construction
Chapter 13 Concrete for Offshore Structures
Chapter 14 Foundations for Concrete Structures
Chapter 15 Specialized Construction Applications
Chapter 16 Structural Concrete Repair
Chapter 17 Joints in Concrete Construction
Chapter 18 Automation in Concrete Construction
Chapter 19 Equipment for Concrete Building Construction
Chapter 20 Roller-Compacted Concrete
Chapter 21 Nondestructive Test Methods
Chapter 22 Fiber-Reinforced Composites
Chapter 23 Bonded Concrete Overlays
Chapter 24 Engineered Cementitious Composite (ECC) Material Structural and Durability Performance
Chapter 25 Design of FRP Reinforced and Strengthened Concrete
Chapter 26 Low-Calcium Fly-Ash-Based Geopolymer Concrete
Chapter 27 Performance Evaluation of Structures
Chapter 28 Masonry Design and Construction
Chapter 29 Aesthetics in the Construction and Design of Long-Span Prestressed Concrete Bridges
Chapter 30 Architectural Concrete
Chapter 31 Fire Resistance and Protection of Structures
Chapter 32 Seismic-Resisting Construction
Chapter 33 Prefabricated Bridge Elements and Systems
Chapter 34 Design of Precast Concrete Seismic Bracing Systems
Chapter 35 Cracking Mitigation and Maintenance Considerations
Chapter 36 Proportioning Concrete Structural Elements by the ACI 318-08 Code
Index
Back cover

Page 2: Chapter 14 Foundations for Concrete Structures.pdf

the principles of soil mechanics rock mechanics and structural engineering to the design of structuresassociated with earthen materials It is generally observed that most common foundation types supportedby intact bedrock present no compressibility problems hence when designing common foundationtypes the foundation engineerrsquos primary concerns are the strength and compressibility of the subsurfacesoil and whenever applicable the strength of bedrock

1411 Soil Classification

14111 Mechanical Analysis

According to texture or the feel two different soil types can be identified (1) coarse-grained soil (graveland sand) and (2) fine-grained soil (silt and clay) Whereas the engineering properties (primarily strengthand compressibility) of coarse-grained soils depend on the size of individual soil particles the propertiesof fine-grained soils are mostly governed by moisture content Hence it is important to identify the typeof soil at a given construction site because effective construction procedures invariably depend on thesoil type Soil engineers use a universal format called the unified soil classification system (USCS) to identifyand label soil The system is based on the results of common laboratory tests of mechanical analysis andAtterberg limits (Bowles 1986) To classify a given soil sample mechanical analysis is conducted in twostages (1) sieve analysis for the coarse fraction (gravel and sand) and (2) hydrometer analysis for thefine fraction (silt and clay) Of these sieve analysis is conducted according to ASTM D 421 and D 422procedures using a set of US standard sieves (Figure 141) The most commonly used sieves are numbers20 40 60 80 100 140 and 200 corresponding to sieve openings of 085 0425 025 018 015 0106and 0075 mm respectively

During the test the percentage (by weight) of the soil sample retained on each sieve is recorded fromwhich the percentage (R) passing (or finer than) a given sieve size (D) is determined On the otherhand if a substantial portion of the soil sample consists of fine-grained soils (D lt 0075 mm) then sieveanalysis has to be followed by hydrometer analysis (Figure 142) This is performed by first treating thefine fraction with a deflocculating agent such as sodium hexametaphosphate (Calgonreg) or sodium silicate(water glass) for about half a day and then allowing the suspension to settle in a hydrometer jar kept ata constant temperature As the heavier particles settle followed by the lighter ones a calibrated ASTM152H hydrometer is used to estimate the fraction (R) still settling above the hydrometer bottom at any

FIGURE 141 Equipment for sieve analysis

0

100

80

60

40

20

19 475 1

084

0

042

0

015

00

10

075

001

000

20

001

Perc

ent F

iner

Grain Diameter (mm)

A

B

Brittlesolid

Shrinkagelimit

Watercontent

Plasticlimit

Liquidlimit

PI = LL ndash PL

Example 141

TAB

LE

14

1U

nifi

ed S

oil

Cla

ssifi

cati

on S

yste

m

Maj

or D

ivis

ion

s

Gro

up

Sym

bols

(τ

)Ty

pica

l N

ames

Labo

rato

ry C

lass

ifica

tion

Cri

teri

a

12

34

6

no fines)

GW

Wel

l-gr

aded

gra

vels

gra

velndash

san

d m

ixtu

res

few

or

no

fin

es

Cu

= D

60D

10 g

reat

er t

han

4C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(

see

Sect

ion

25

)

GP

Poor

ly g

rade

d gr

avel

s g

rave

lndashsa

nd

mix

ture

s f

ew o

r n

o fi

nes

Not

mee

tin

g al

l gr

adat

ion

req

uir

emen

ts f

or G

W

Gravels with fines

GM

Silt

y gr

avel

s g

rave

lndashsa

ndndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Abo

ve A

-lin

e w

ith

PI

valu

es

betw

een

4 a

nd

7 ar

e bo

rder

line

case

s re

quir

ing

use

of

dual

sym

bols

GC

Cla

yey

grav

els

gra

velndash

san

dndashcl

ay m

ixtu

res

Att

erbe

rg li

mit

s ab

ove

A-l

ine

wit

h P

I gr

eate

r th

an 7

Clean sands (few or

no fines)

SWW

ell-

grad

ed s

ands

gra

velly

san

ds f

ew o

r n

o fi

nes

Cu

= D

60D

10 g

reat

er t

han

6C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(see

Sec

tion

25

)

SPPo

orly

gra

ded

san

ds g

rave

lly s

ands

few

or

no

fin

esN

ot m

eeti

ng

all

grad

atio

n r

equ

irem

ents

for

SW

Sands withs fines

SMSi

lty

san

ds s

andndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Lim

its

plot

tin

g in

hat

ched

zo

ne

wit

h P

I va

lues

bet

wee

n

4 an

d 7

are

bord

erlin

e ca

ses

requ

irin

g u

se o

f du

al

sym

bols

SC

Cla

yey

san

ds s

andndash

clay

mix

ture

sA

tter

berg

lim

its

abov

e A

-lin

e w

ith

PI

grea

ter

than

7

Silt

s an

d cl

ays

liq

uid

lim

it

less

th

an 5

0

ML

Inor

gan

ic s

ilts

and

very

fin

e sa

nds

roc

k fl

our

silt

y or

cl

ayey

fin

e sa

nds

or

clay

ey s

ilts

wit

h s

ligh

t pl

asti

city

Plas

tici

ty C

hart

for

Lab

orat

ory

Cla

ssifi

cati

on o

f Fi

ne-G

rain

ed S

oils

CL

Inor

gan

ic c

lays

of

low

to

med

ium

pla

stic

ity

gra

velly

cl

ays

san

dy c

lays

silt

y cl

ays

lea

n c

lays

OL

Org

anic

silt

s an

d or

gan

ic s

ilty

clay

s of

low

pla

stic

ity

Silt

s an

d cl

ays

liq

uid

lim

it

grea

ter

than

50

MH

Inor

gan

ic s

ilts

mic

aceo

us

or d

iato

mac

eou

s fi

ne

san

dy

or s

ilty

soils

ela

stic

silt

s

CH

Inor

gan

ic c

lays

of

hig

h p

last

icit

y f

at c

lays

OH

Org

anic

cla

ys o

f m

ediu

m t

o h

igh

pla

stic

ity

org

anic

silt

s

Hig

hly

org

anic

soi

lsP

tPe

at a

nd

oth

er h

igh

ly o

rgan

ic s

oils

Sour

ce H

oltz

RD

an

d K

ovac

s W

D

An

Intr

oduc

tion

to

Geo

tech

nica

l En

gine

erin

g P

ren

tice

Hal

l E

ngl

ewoo

d C

liffs

NJ

1981

Wit

h p

erm

issi

on

|

Liqu

id L

imit

0

10

2

0

30

40

5

0

60

7

0

80

90

10

0

60 50 40 30 20 10 7 4 0

Plasticity Index

Com

parin

g so

ils a

t equ

al li

quid

lim

its

toug

hnes

s and

dry

stre

ngth

incr

ease

with

incr

easi

ng p

last

icity

inde

x

OH or MH

ML or OL

CLndash

ML

A-Lin

e

CH

where

σprime = σ (142)

(143b)

where

Axial loading ram

Loading capCell

Cell fluid

Drainage lineBase

Porousstones Sample

Example 142

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 20 202 52

2 40 304 83

(a) (b)

σ3 = 20

σ1 = 402

00

Δu = 52

00

σ1 = 704 σ1 = 350 σ1 = 621

Δu = 83 σ3 = 40 σ3 = 148 = 327ʹ σ3ʹ

10

20

30

10 20 30 40 50 60 70 80 90

τ (k

Pa)

σ (kPa)

Example 143

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 3: Chapter 14 Foundations for Concrete Structures.pdf

0

100

80

60

40

20

19 475 1

084

0

042

0

015

00

10

075

001

000

20

001

Perc

ent F

iner

Grain Diameter (mm)

A

B

Brittlesolid

Shrinkagelimit

Watercontent

Plasticlimit

Liquidlimit

PI = LL ndash PL

Example 141

TAB

LE

14

1U

nifi

ed S

oil

Cla

ssifi

cati

on S

yste

m

Maj

or D

ivis

ion

s

Gro

up

Sym

bols

(τ

)Ty

pica

l N

ames

Labo

rato

ry C

lass

ifica

tion

Cri

teri

a

12

34

6

no fines)

GW

Wel

l-gr

aded

gra

vels

gra

velndash

san

d m

ixtu

res

few

or

no

fin

es

Cu

= D

60D

10 g

reat

er t

han

4C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(

see

Sect

ion

25

)

GP

Poor

ly g

rade

d gr

avel

s g

rave

lndashsa

nd

mix

ture

s f

ew o

r n

o fi

nes

Not

mee

tin

g al

l gr

adat

ion

req

uir

emen

ts f

or G

W

Gravels with fines

GM

Silt

y gr

avel

s g

rave

lndashsa

ndndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Abo

ve A

-lin

e w

ith

PI

valu

es

betw

een

4 a

nd

7 ar

e bo

rder

line

case

s re

quir

ing

use

of

dual

sym

bols

GC

Cla

yey

grav

els

gra

velndash

san

dndashcl

ay m

ixtu

res

Att

erbe

rg li

mit

s ab

ove

A-l

ine

wit

h P

I gr

eate

r th

an 7

Clean sands (few or

no fines)

SWW

ell-

grad

ed s

ands

gra

velly

san

ds f

ew o

r n

o fi

nes

Cu

= D

60D

10 g

reat

er t

han

6C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(see

Sec

tion

25

)

SPPo

orly

gra

ded

san

ds g

rave

lly s

ands

few

or

no

fin

esN

ot m

eeti

ng

all

grad

atio

n r

equ

irem

ents

for

SW

Sands withs fines

SMSi

lty

san

ds s

andndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Lim

its

plot

tin

g in

hat

ched

zo

ne

wit

h P

I va

lues

bet

wee

n

4 an

d 7

are

bord

erlin

e ca

ses

requ

irin

g u

se o

f du

al

sym

bols

SC

Cla

yey

san

ds s

andndash

clay

mix

ture

sA

tter

berg

lim

its

abov

e A

-lin

e w

ith

PI

grea

ter

than

7

Silt

s an

d cl

ays

liq

uid

lim

it

less

th

an 5

0

ML

Inor

gan

ic s

ilts

and

very

fin

e sa

nds

roc

k fl

our

silt

y or

cl

ayey

fin

e sa

nds

or

clay

ey s

ilts

wit

h s

ligh

t pl

asti

city

Plas

tici

ty C

hart

for

Lab

orat

ory

Cla

ssifi

cati

on o

f Fi

ne-G

rain

ed S

oils

CL

Inor

gan

ic c

lays

of

low

to

med

ium

pla

stic

ity

gra

velly

cl

ays

san

dy c

lays

silt

y cl

ays

lea

n c

lays

OL

Org

anic

silt

s an

d or

gan

ic s

ilty

clay

s of

low

pla

stic

ity

Silt

s an

d cl

ays

liq

uid

lim

it

grea

ter

than

50

MH

Inor

gan

ic s

ilts

mic

aceo

us

or d

iato

mac

eou

s fi

ne

san

dy

or s

ilty

soils

ela

stic

silt

s

CH

Inor

gan

ic c

lays

of

hig

h p

last

icit

y f

at c

lays

OH

Org

anic

cla

ys o

f m

ediu

m t

o h

igh

pla

stic

ity

org

anic

silt

s

Hig

hly

org

anic

soi

lsP

tPe

at a

nd

oth

er h

igh

ly o

rgan

ic s

oils

Sour

ce H

oltz

RD

an

d K

ovac

s W

D

An

Intr

oduc

tion

to

Geo

tech

nica

l En

gine

erin

g P

ren

tice

Hal

l E

ngl

ewoo

d C

liffs

NJ

1981

Wit

h p

erm

issi

on

|

Liqu

id L

imit

0

10

2

0

30

40

5

0

60

7

0

80

90

10

0

60 50 40 30 20 10 7 4 0

Plasticity Index

Com

parin

g so

ils a

t equ

al li

quid

lim

its

toug

hnes

s and

dry

stre

ngth

incr

ease

with

incr

easi

ng p

last

icity

inde

x

OH or MH

ML or OL

CLndash

ML

A-Lin

e

CH

where

σprime = σ (142)

(143b)

where

Axial loading ram

Loading capCell

Cell fluid

Drainage lineBase

Porousstones Sample

Example 142

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 20 202 52

2 40 304 83

(a) (b)

σ3 = 20

σ1 = 402

00

Δu = 52

00

σ1 = 704 σ1 = 350 σ1 = 621

Δu = 83 σ3 = 40 σ3 = 148 = 327ʹ σ3ʹ

10

20

30

10 20 30 40 50 60 70 80 90

τ (k

Pa)

σ (kPa)

Example 143

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 4: Chapter 14 Foundations for Concrete Structures.pdf

Brittlesolid

Shrinkagelimit

Watercontent

Plasticlimit

Liquidlimit

PI = LL ndash PL

Example 141

TAB

LE

14

1U

nifi

ed S

oil

Cla

ssifi

cati

on S

yste

m

Maj

or D

ivis

ion

s

Gro

up

Sym

bols

(τ

)Ty

pica

l N

ames

Labo

rato

ry C

lass

ifica

tion

Cri

teri

a

12

34

6

no fines)

GW

Wel

l-gr

aded

gra

vels

gra

velndash

san

d m

ixtu

res

few

or

no

fin

es

Cu

= D

60D

10 g

reat

er t

han

4C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(

see

Sect

ion

25

)

GP

Poor

ly g

rade

d gr

avel

s g

rave

lndashsa

nd

mix

ture

s f

ew o

r n

o fi

nes

Not

mee

tin

g al

l gr

adat

ion

req

uir

emen

ts f

or G

W

Gravels with fines

GM

Silt

y gr

avel

s g

rave

lndashsa

ndndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Abo

ve A

-lin

e w

ith

PI

valu

es

betw

een

4 a

nd

7 ar

e bo

rder

line

case

s re

quir

ing

use

of

dual

sym

bols

GC

Cla

yey

grav

els

gra

velndash

san

dndashcl

ay m

ixtu

res

Att

erbe

rg li

mit

s ab

ove

A-l

ine

wit

h P

I gr

eate

r th

an 7

Clean sands (few or

no fines)

SWW

ell-

grad

ed s

ands

gra

velly

san

ds f

ew o

r n

o fi

nes

Cu

= D

60D

10 g

reat

er t

han

6C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(see

Sec

tion

25

)

SPPo

orly

gra

ded

san

ds g

rave

lly s

ands

few

or

no

fin

esN

ot m

eeti

ng

all

grad

atio

n r

equ

irem

ents

for

SW

Sands withs fines

SMSi

lty

san

ds s

andndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Lim

its

plot

tin

g in

hat

ched

zo

ne

wit

h P

I va

lues

bet

wee

n

4 an

d 7

are

bord

erlin

e ca

ses

requ

irin

g u

se o

f du

al

sym

bols

SC

Cla

yey

san

ds s

andndash

clay

mix

ture

sA

tter

berg

lim

its

abov

e A

-lin

e w

ith

PI

grea

ter

than

7

Silt

s an

d cl

ays

liq

uid

lim

it

less

th

an 5

0

ML

Inor

gan

ic s

ilts

and

very

fin

e sa

nds

roc

k fl

our

silt

y or

cl

ayey

fin

e sa

nds

or

clay

ey s

ilts

wit

h s

ligh

t pl

asti

city

Plas

tici

ty C

hart

for

Lab

orat

ory

Cla

ssifi

cati

on o

f Fi

ne-G

rain

ed S

oils

CL

Inor

gan

ic c

lays

of

low

to

med

ium

pla

stic

ity

gra

velly

cl

ays

san

dy c

lays

silt

y cl

ays

lea

n c

lays

OL

Org

anic

silt

s an

d or

gan

ic s

ilty

clay

s of

low

pla

stic

ity

Silt

s an

d cl

ays

liq

uid

lim

it

grea

ter

than

50

MH

Inor

gan

ic s

ilts

mic

aceo

us

or d

iato

mac

eou

s fi

ne

san

dy

or s

ilty

soils

ela

stic

silt

s

CH

Inor

gan

ic c

lays

of

hig

h p

last

icit

y f

at c

lays

OH

Org

anic

cla

ys o

f m

ediu

m t

o h

igh

pla

stic

ity

org

anic

silt

s

Hig

hly

org

anic

soi

lsP

tPe

at a

nd

oth

er h

igh

ly o

rgan

ic s

oils

Sour

ce H

oltz

RD

an

d K

ovac

s W

D

An

Intr

oduc

tion

to

Geo

tech

nica

l En

gine

erin

g P

ren

tice

Hal

l E

ngl

ewoo

d C

liffs

NJ

1981

Wit

h p

erm

issi

on

|

Liqu

id L

imit

0

10

2

0

30

40

5

0

60

7

0

80

90

10

0

60 50 40 30 20 10 7 4 0

Plasticity Index

Com

parin

g so

ils a

t equ

al li

quid

lim

its

toug

hnes

s and

dry

stre

ngth

incr

ease

with

incr

easi

ng p

last

icity

inde

x

OH or MH

ML or OL

CLndash

ML

A-Lin

e

CH

where

σprime = σ (142)

(143b)

where

Axial loading ram

Loading capCell

Cell fluid

Drainage lineBase

Porousstones Sample

Example 142

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 20 202 52

2 40 304 83

(a) (b)

σ3 = 20

σ1 = 402

00

Δu = 52

00

σ1 = 704 σ1 = 350 σ1 = 621

Δu = 83 σ3 = 40 σ3 = 148 = 327ʹ σ3ʹ

10

20

30

10 20 30 40 50 60 70 80 90

τ (k

Pa)

σ (kPa)

Example 143

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 5: Chapter 14 Foundations for Concrete Structures.pdf

PI = LL ndash PL

Example 141

TAB

LE

14

1U

nifi

ed S

oil

Cla

ssifi

cati

on S

yste

m

Maj

or D

ivis

ion

s

Gro

up

Sym

bols

(τ

)Ty

pica

l N

ames

Labo

rato

ry C

lass

ifica

tion

Cri

teri

a

12

34

6

no fines)

GW

Wel

l-gr

aded

gra

vels

gra

velndash

san

d m

ixtu

res

few

or

no

fin

es

Cu

= D

60D

10 g

reat

er t

han

4C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(

see

Sect

ion

25

)

GP

Poor

ly g

rade

d gr

avel

s g

rave

lndashsa

nd

mix

ture

s f

ew o

r n

o fi

nes

Not

mee

tin

g al

l gr

adat

ion

req

uir

emen

ts f

or G

W

Gravels with fines

GM

Silt

y gr

avel

s g

rave

lndashsa

ndndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Abo

ve A

-lin

e w

ith

PI

valu

es

betw

een

4 a

nd

7 ar

e bo

rder

line

case

s re

quir

ing

use

of

dual

sym

bols

GC

Cla

yey

grav

els

gra

velndash

san

dndashcl

ay m

ixtu

res

Att

erbe

rg li

mit

s ab

ove

A-l

ine

wit

h P

I gr

eate

r th

an 7

Clean sands (few or

no fines)

SWW

ell-

grad

ed s

ands

gra

velly

san

ds f

ew o

r n

o fi

nes

Cu

= D

60D

10 g

reat

er t

han

6C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(see

Sec

tion

25

)

SPPo

orly

gra

ded

san

ds g

rave

lly s

ands

few

or

no

fin

esN

ot m

eeti

ng

all

grad

atio

n r

equ

irem

ents

for

SW

Sands withs fines

SMSi

lty

san

ds s

andndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Lim

its

plot

tin

g in

hat

ched

zo

ne

wit

h P

I va

lues

bet

wee

n

4 an

d 7

are

bord

erlin

e ca

ses

requ

irin

g u

se o

f du

al

sym

bols

SC

Cla

yey

san

ds s

andndash

clay

mix

ture

sA

tter

berg

lim

its

abov

e A

-lin

e w

ith

PI

grea

ter

than

7

Silt

s an

d cl

ays

liq

uid

lim

it

less

th

an 5

0

ML

Inor

gan

ic s

ilts

and

very

fin

e sa

nds

roc

k fl

our

silt

y or

cl

ayey

fin

e sa

nds

or

clay

ey s

ilts

wit

h s

ligh

t pl

asti

city

Plas

tici

ty C

hart

for

Lab

orat

ory

Cla

ssifi

cati

on o

f Fi

ne-G

rain

ed S

oils

CL

Inor

gan

ic c

lays

of

low

to

med

ium

pla

stic

ity

gra

velly

cl

ays

san

dy c

lays

silt

y cl

ays

lea

n c

lays

OL

Org

anic

silt

s an

d or

gan

ic s

ilty

clay

s of

low

pla

stic

ity

Silt

s an

d cl

ays

liq

uid

lim

it

grea

ter

than

50

MH

Inor

gan

ic s

ilts

mic

aceo

us

or d

iato

mac

eou

s fi

ne

san

dy

or s

ilty

soils

ela

stic

silt

s

CH

Inor

gan

ic c

lays

of

hig

h p

last

icit

y f

at c

lays

OH

Org

anic

cla

ys o

f m

ediu

m t

o h

igh

pla

stic

ity

org

anic

silt

s

Hig

hly

org

anic

soi

lsP

tPe

at a

nd

oth

er h

igh

ly o

rgan

ic s

oils

Sour

ce H

oltz

RD

an

d K

ovac

s W

D

An

Intr

oduc

tion

to

Geo

tech

nica

l En

gine

erin

g P

ren

tice

Hal

l E

ngl

ewoo

d C

liffs

NJ

1981

Wit

h p

erm

issi

on

|

Liqu

id L

imit

0

10

2

0

30

40

5

0

60

7

0

80

90

10

0

60 50 40 30 20 10 7 4 0

Plasticity Index

Com

parin

g so

ils a

t equ

al li

quid

lim

its

toug

hnes

s and

dry

stre

ngth

incr

ease

with

incr

easi

ng p

last

icity

inde

x

OH or MH

ML or OL

CLndash

ML

A-Lin

e

CH

where

σprime = σ (142)

(143b)

where

Axial loading ram

Loading capCell

Cell fluid

Drainage lineBase

Porousstones Sample

Example 142

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 20 202 52

2 40 304 83

(a) (b)

σ3 = 20

σ1 = 402

00

Δu = 52

00

σ1 = 704 σ1 = 350 σ1 = 621

Δu = 83 σ3 = 40 σ3 = 148 = 327ʹ σ3ʹ

10

20

30

10 20 30 40 50 60 70 80 90

τ (k

Pa)

σ (kPa)

Example 143

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 6: Chapter 14 Foundations for Concrete Structures.pdf

TAB

LE

14

1U

nifi

ed S

oil

Cla

ssifi

cati

on S

yste

m

Maj

or D

ivis

ion

s

Gro

up

Sym

bols

(τ

)Ty

pica

l N

ames

Labo

rato

ry C

lass

ifica

tion

Cri

teri

a

12

34

6

no fines)

GW

Wel

l-gr

aded

gra

vels

gra

velndash

san

d m

ixtu

res

few

or

no

fin

es

Cu

= D

60D

10 g

reat

er t

han

4C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(

see

Sect

ion

25

)

GP

Poor

ly g

rade

d gr

avel

s g

rave

lndashsa

nd

mix

ture

s f

ew o

r n

o fi

nes

Not

mee

tin

g al

l gr

adat

ion

req

uir

emen

ts f

or G

W

Gravels with fines

GM

Silt

y gr

avel

s g

rave

lndashsa

ndndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Abo

ve A

-lin

e w

ith

PI

valu

es

betw

een

4 a

nd

7 ar

e bo

rder

line

case

s re

quir

ing

use

of

dual

sym

bols

GC

Cla

yey

grav

els

gra

velndash

san

dndashcl

ay m

ixtu

res

Att

erbe

rg li

mit

s ab

ove

A-l

ine

wit

h P

I gr

eate

r th

an 7

Clean sands (few or

no fines)

SWW

ell-

grad

ed s

ands

gra

velly

san

ds f

ew o

r n

o fi

nes

Cu

= D

60D

10 g

reat

er t

han

6C

c = (

D30

)2 (D

10 times

D60

) be

twee

n 1

an

d 3

(see

Sec

tion

25

)

SPPo

orly

gra

ded

san

ds g

rave

lly s

ands

few

or

no

fin

esN

ot m

eeti

ng

all

grad

atio

n r

equ

irem

ents

for

SW

Sands withs fines

SMSi

lty

san

ds s

andndash

silt

mix

ture

sA

tter

berg

lim

its b

elow

A-l

ine

or

PI

less

th

an 4

Lim

its

plot

tin

g in

hat

ched

zo

ne

wit

h P

I va

lues

bet

wee

n

4 an

d 7

are

bord

erlin

e ca

ses

requ

irin

g u

se o

f du

al

sym

bols

SC

Cla

yey

san

ds s

andndash

clay

mix

ture

sA

tter

berg

lim

its

abov

e A

-lin

e w

ith

PI

grea

ter

than

7

Silt

s an

d cl

ays

liq

uid

lim

it

less

th

an 5

0

ML

Inor

gan

ic s

ilts

and

very

fin

e sa

nds

roc

k fl

our

silt

y or

cl

ayey

fin

e sa

nds

or

clay

ey s

ilts

wit

h s

ligh

t pl

asti

city

Plas

tici

ty C

hart

for

Lab

orat

ory

Cla

ssifi

cati

on o

f Fi

ne-G

rain

ed S

oils

CL

Inor

gan

ic c

lays

of

low

to

med

ium

pla

stic

ity

gra

velly

cl

ays

san

dy c

lays

silt

y cl

ays

lea

n c

lays

OL

Org

anic

silt

s an

d or

gan

ic s

ilty

clay

s of

low

pla

stic

ity

Silt

s an

d cl

ays

liq

uid

lim

it

grea

ter

than

50

MH

Inor

gan

ic s

ilts

mic

aceo

us

or d

iato

mac

eou

s fi

ne

san

dy

or s

ilty

soils

ela

stic

silt

s

CH

Inor

gan

ic c

lays

of

hig

h p

last

icit

y f

at c

lays

OH

Org

anic

cla

ys o

f m

ediu

m t

o h

igh

pla

stic

ity

org

anic

silt

s

Hig

hly

org

anic

soi

lsP

tPe

at a

nd

oth

er h

igh

ly o

rgan

ic s

oils

Sour

ce H

oltz

RD

an

d K

ovac

s W

D

An

Intr

oduc

tion

to

Geo

tech

nica

l En

gine

erin

g P

ren

tice

Hal

l E

ngl

ewoo

d C

liffs

NJ

1981

Wit

h p

erm

issi

on

|

Liqu

id L

imit

0

10

2

0

30

40

5

0

60

7

0

80

90

10

0

60 50 40 30 20 10 7 4 0

Plasticity Index

Com

parin

g so

ils a

t equ

al li

quid

lim

its

toug

hnes

s and

dry

stre

ngth

incr

ease

with

incr

easi

ng p

last

icity

inde

x

OH or MH

ML or OL

CLndash

ML

A-Lin

e

CH

where

σprime = σ (142)

(143b)

where

Axial loading ram

Loading capCell

Cell fluid

Drainage lineBase

Porousstones Sample

Example 142

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 20 202 52

2 40 304 83

(a) (b)

σ3 = 20

σ1 = 402

00

Δu = 52

00

σ1 = 704 σ1 = 350 σ1 = 621

Δu = 83 σ3 = 40 σ3 = 148 = 327ʹ σ3ʹ

10

20

30

10 20 30 40 50 60 70 80 90

τ (k

Pa)

σ (kPa)

Example 143

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 7: Chapter 14 Foundations for Concrete Structures.pdf

where

σprime = σ (142)

(143b)

where

Axial loading ram

Loading capCell

Cell fluid

Drainage lineBase

Porousstones Sample

Example 142

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 20 202 52

2 40 304 83

(a) (b)

σ3 = 20

σ1 = 402

00

Δu = 52

00

σ1 = 704 σ1 = 350 σ1 = 621

Δu = 83 σ3 = 40 σ3 = 148 = 327ʹ σ3ʹ

10

20

30

10 20 30 40 50 60 70 80 90

τ (k

Pa)

σ (kPa)

Example 143

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 8: Chapter 14 Foundations for Concrete Structures.pdf

Axial loading ram

Loading capCell

Cell fluid

Drainage lineBase

Porousstones Sample

Example 142

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 20 202 52

2 40 304 83

(a) (b)

σ3 = 20

σ1 = 402

00

Δu = 52

00

σ1 = 704 σ1 = 350 σ1 = 621

Δu = 83 σ3 = 40 σ3 = 148 = 327ʹ σ3ʹ

10

20

30

10 20 30 40 50 60 70 80 90

τ (k

Pa)

σ (kPa)

Example 143

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 9: Chapter 14 Foundations for Concrete Structures.pdf

Example 142

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 20 202 52

2 40 304 83

(a) (b)

σ3 = 20

σ1 = 402

00

Δu = 52

00

σ1 = 704 σ1 = 350 σ1 = 621

Δu = 83 σ3 = 40 σ3 = 148 = 327ʹ σ3ʹ

10

20

30

10 20 30 40 50 60 70 80 90

τ (k

Pa)

σ (kPa)

Example 143

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 10: Chapter 14 Foundations for Concrete Structures.pdf

Example 143

TestCell Pressure

(kPa)Pore Pressure

(kPa)

1 40 1022 NA

2 60 1014 NA

20

40

60

20 40 60 80 100

τ (k

Pa)

φu = 0

σ (kPa)

Cu = 506 kPa

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 11: Chapter 14 Foundations for Concrete Structures.pdf

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 12: Chapter 14 Foundations for Concrete Structures.pdf

(146)

where

Incr

easi

ngO

CR

Age

dce

men

ted

NC

You

ngcl

ay

prime =prime

N Nv

1

σ

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 13: Chapter 14 Foundations for Concrete Structures.pdf

(148)

(149)

60

356 cm

Sleeve

Cone

d1

D

Ff

qR

s

c

=

sqt p

Nu

kT

= minus 0

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 14: Chapter 14 Foundations for Concrete Structures.pdf

where

(1410)

(1411)

(1412)

Frict

ion

Ratio

0

100

200

300

400

Tip

Resis

tanc

e (ts

f)

01234

Loca

l Fric

tion

(tsf)

0 2 4 6 8Depth (m)

20 4 6 8

20 4 6 802468

N PImT = +135 5

50

φ = +29 qc

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 15: Chapter 14 Foundations for Concrete Structures.pdf

(1413)

(1414)

where

Siltysand(SMSC)

Sand(SW SP)

400

200

100806040

20

1086

4

2

1

Cone

bea

ring

(qc)

(kPa

) (times1

00)

Sandysilts and

silt Clay

Peat

s fB

Ec

ss

q= minus( )0 12

2micro α

s C C q qI

Eze

z

s

z

= minus sum1 2

0

( ) ∆

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 16: Chapter 14 Foundations for Concrete Structures.pdf

Es (tsf) = 8N (1415)

andEs = 2qc (1416)

1 2 3 4 5 6 7 8 9 10

30

25

20

15

10

05

αα

αav

αr

α = 1αav = 085αr = 088

αav αr

LB

+ + + + + + + + + + + + +

B Iz

z = B2

z = 0

z = 2B

06

z

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 17: Chapter 14 Foundations for Concrete Structures.pdf

Soil Es (MPa)

Clay

Very soft 2ndash15

Soft 5ndash25

Medium 15ndash50

Hard 50ndash100

Sandy 25ndash250

Glacial till

Loose 10ndash150

Dense 150ndash720

Loess 15ndash60

Sand

Silty 5ndash20

Loose 10ndash25

Dense 50ndash81

Sand and gravel

Loose 50ndash150

Dense 100ndash200

Shale 150ndash5000

Silt 2ndash20

Type of Soil micro

Silt 03ndash035

Loess 01ndash03

Ice 036

Concrete 015

Steel 033

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 18: Chapter 14 Foundations for Concrete Structures.pdf

(1417)

+ + + + + + + ++ + + + + + + +

Surcharge ∆σ

+ + + + + + ++ + + + + +Sandy soil

GWT

Clay layer

Bedrock

∆σ ∆σ uu0

u0 u

FinalEffective

Stress

InitialPore

Pressure

FinalPore

Pressure(a) (b) (c)

∆

σvʹ σvʹ

σv0ʹ σv0ʹ

TC t

Hv

dr

=2

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 19: Chapter 14 Foundations for Concrete Structures.pdf

(1418)

Uavg T

01 0008

02 0031

03 0071

04 0126

05 0197

06 0287

07 0403

08 0567

09 0848

095 1163

10 infin

∆ ∆e

e

H

H1 0+=

01

0095

009

0085

008

0075

007

0065

006

U = 0

u = 50

u = 100

01 1 10 100 1000 10000 100000t50Time (sec)4t1t1Cv = 0196^H^2t50

00806

Defl

ectio

n (in

)

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 20: Chapter 14 Foundations for Concrete Structures.pdf

(1419)

(1420)

prime )

(1421)

∆p = qIM (1422)

e

10

11

10

09cc

pc = 60 kPa 100 120p (kPa)

cr

093

1045

1085

log p

pc

cc

∆p

e

pc

log p ∆p

cr

e

pc

∆p

log p

e

(a) (b) (c)

σv0ʹ

σv0ʹ σv0ʹ

sC H

e

pc v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

pr v

vcon =

+prime +

prime1 0

0

logσ

σ∆

sC H

e

p C H

e

p

pr c

v

c v

ccon =

+ prime+

+prime +

1 10 00

0log logσ

σ ∆

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 21: Chapter 14 Foundations for Concrete Structures.pdf

Example 144

dc

O Q

I = 0001

A1 0 5 0 75 0 6 0 5 0 25 0 533 0 6 0 367= times + + = ( ) ( )( ) mm

mA

A

2

3

0 5 1 5 0 533 0 133 0 5

0 5 0 5

= + =

=

( )( )

( )(00 133 0 033 ) = m

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 22: Chapter 14 Foundations for Concrete Structures.pdf

ndash71 m

200 kN

ndash35

ndash10ndash175

ndash40

ndash20

15 m times 15 m

SPT (N) E (MPa)

15 115

14 107

10 256

SandySoil

ClayeySoil

0133

5330 6

165

175

180

Bedrock

GWT

0 367 1 0 11 5 10 0 5 103

( ) ( ) (

minus

∆

pcenter

corner

kPa

p

=

4 19 88 89 0 001 6 75

5

88 88 89 0 001 5 2times times = kPa

scenter = + ( )+

0 064 1 1 06 2 5 60 54 8

0

( ) log( )

( ) ( )log ( ) 382 1 1 06 2 5 54 6 75 60 0+ + = 00819 8 19m mm=

scorner = + +0 064 1 1 06 2 5 54 8 5 2 54 ( ) ( )log 8 3 06( ) = mm

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 23: Chapter 14 Foundations for Concrete Structures.pdf

533 50 4667 4333 40 3667 3333 30 267

9 13 13 15 13 14 15 08

445 m

h = 7

h = 9 m h = 5 m

h = 6 h = 6667

h = 80 h = 7667 h = 7333

h = 6333 h = 5667

h = 5333h = 8333

h = 8667

34 5 6 7 8 9 10

11

12

13

2

1

ν = 391 kPam

(pγ

w)

20

40

60

m

5 m

11 m1 m

10 m1 m z = 3 m

4 m

Datum (z = 0)

A B

1113D C

Distancedownstream

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 24: Chapter 14 Foundations for Concrete Structures.pdf

(1423)

q = kiA (1424)

(1425)

Example 145

(9 m ndash 5 m)12 = 0333 m

p = γw(h ndash 30)

i = (5333 ndash 50)11 = 0302

hp

zw

= +γ

idh

dx=

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 25: Chapter 14 Foundations for Concrete Structures.pdf

Observationwell

h2

r2

DepressedGWT

Original GWT

Ground surface

Well point

Riser

rImpervious

layer

+ + + + + + + +

h1

h

Intendedexcavation

r1

Header

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 26: Chapter 14 Foundations for Concrete Structures.pdf

(1426)

(1427)

Perforated pipes

Gravel

Subsoil

Waste

Filter soil

q kdh

dxh= minus

qk h h

r r=

minus( )( )

π 12

22

1 2ln

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 27: Chapter 14 Foundations for Concrete Structures.pdf

Gravelwith

perforated pipeClay

Waste

Filter soil

Primary geomembrane

Subsoil

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 28: Chapter 14 Foundations for Concrete Structures.pdf

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 29: Chapter 14 Foundations for Concrete Structures.pdf

(a)

(b)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 30: Chapter 14 Foundations for Concrete Structures.pdf

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

To pump

Drill rig

Casing

To hoist

Cut soilwater

Pressurized water

Drill bit

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 31: Chapter 14 Foundations for Concrete Structures.pdf

(1428)

(1429)

q qu f u p( ) ( )=

q qB

Bu f u p

f

p( ) ( )=

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 32: Chapter 14 Foundations for Concrete Structures.pdf

(1430)

(1431)

(1432)

s sB

Bf p

f

p

=

s sB

B Bf p

f

f p

=+

2

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 33: Chapter 14 Foundations for Concrete Structures.pdf

(1433)

where

(1434)

where

15

10

5

0

ndash5

q

Sρ

Bear

ing

Stre

ss (p

sf)

(Thou

sand

s)

B

P

Df

q

F

P

An ult gt

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 34: Chapter 14 Foundations for Concrete Structures.pdf

Example 146

0 57 10 00 514 10 00

5 73 16 05 649 16 01

10 96 27 12 834 25 04

15 129 44 25 110 39 12

20 177 74 50 148 64 29

25 251 127 97 207 107 68

30 372 225 197 301 184 151

35 578 414 424 464 335 344

40 957 813 1004 7525 641 794

45 1723 1733 2975 1335 1347 2005

Sq = 1 + (BL) tanφ

Depth For Df B lt 1

dc = 1 + 04(Df B)

dγ = 1

For Df B gt 1

dγ = 1

Elevation (m) N

10 520 730 1040 1250 12

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 35: Chapter 14 Foundations for Concrete Structures.pdf

qnult = (1657)(1176)(121)(49 ndash 1) + (05)(06)(10)(15)(72)(57) = 13167 kPa

10 5 121 281 1420 7 193 223 15630 10 265 19 1940 12 337 169 203

15 m

1 m N = 5

N = 7

N = 10

N = 12

ndash05 m

ndash1 m

ndash20 m

ndash30 m

ndash40 m

Elev

1000 kN

0

Influencezone

GWT

s s s

d d d

c q

= = =

= =

1 816 1 657 0 6

1 266 1 176

γ

γγ

γ

=

= = =

1 0

1 0 1 0 1 0

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 36: Chapter 14 Foundations for Concrete Structures.pdf

γ = kNm3

(1436)

γ = 72 + (165 ndash 72)(10)2(15) = 103 kNm3

Example 147

ex = MxP = 500 kNm2500 kN = 02 m

ey = MxP = 625 kNm2500 kN = 025 m

F = (39378)(11)(10)250 = 1733

γ γγ γ

B2

S

d

i

c

= + =

1 1 0 5 14 1 195

1 0 4 1 0 1 5 1 267

( )

cc

ultq

=

= =

1 0

1 195 1 267 50 6 5 14 393 78

( )( )( ) kPPa

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 37: Chapter 14 Foundations for Concrete Structures.pdf

144 Mat Footings

ndashPe (B2) I

P

ex

PBL

+ Pe (B2)I

(a) (b)

P

ex

B

2ex

PBL

15 m

P = 250 kN

GWT

250 kN

02 m 1 m

Saturated clay

(a) (b)

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 38: Chapter 14 Foundations for Concrete Structures.pdf

(1437)

where

(1438)

Example 148

(1976)BOCA(1993)a

Clay soft 75 100 100 100

Bedrock 9600 9600 6000 9600

qN

B

D

B

sn

f

all = +

+

0 08

11

3 281

0 332

225 4

qP

An all gt

qn ( ) ( all = + + 10 1 0 33 0 5 5 0 1 0 3 5 0 0 088 136 87) = kPa

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 39: Chapter 14 Foundations for Concrete Structures.pdf

25 m

25 m 2 5 m

500 kN 1000 kN 500 kN

250 kN

500 kN

250 kN 250 kN

250 kNA

B

C

D

E

F

G

H

e

500 1000 500 kN

477 477954

025 m25 m 025 m 25 m

Adjustedloads

Adjustedreaction

330 kNm34683

F GBC

4000 5 0 2 136 87

0 2029

2

+( ) lt

gt

e

e m

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 40: Chapter 14 Foundations for Concrete Structures.pdf

(1439)

where

(1440)

8675

47629

39029

ndash8675

13751375

ndash39029

kN

1084 108410825

kNm

14

12 1

3

2β micro=

minus( )Eh

k

kE

Bs

s

=minus( )1 2micro

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 41: Chapter 14 Foundations for Concrete Structures.pdf

where

0 04 08 12 16 20 24 28 32 36 40 44 48 52 56 60

βr

(c) Shear

C

+04

+08

+12

+16

+20

C

+04

+08

+12

+16

+20

C

ndash20

ndash40

ndash60

ndash80

ndash100

E

D

C

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 42: Chapter 14 Foundations for Concrete Structures.pdf

(1441a)

(1441b)

(1442)

Positive Blocks

n

0416

0337

09552

01889

01412

0990

0757

01412

0930

01889

09552

0745

0337

0416

0352

0251

0199

0243

Negative Blocks

1

MP

Cr = minus4

MP

Dθ =4

VP

E= minus β4

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 43: Chapter 14 Foundations for Concrete Structures.pdf

(1443)

where

Example 149

145 Retaining Walls

Moment(kip-ft)

00 16 018 00 1225

10 08 025 002 3225

20 05 04 005 8438

30 04 05 008 925

40 025 08 01 12188

50 018 16 018 2225

60 01 08 025 12188

70 008 05 04 925

80 005 04 05 8438

90 002 025 08 3225

100 00 018 16 1225

A B C

5 ft5 ft

MP N

= ( )1

10 000

2

β

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 44: Chapter 14 Foundations for Concrete Structures.pdf

(1444)

and

(1445)

(1446)

1225 1225

32253225

844 844925 925

1219 1219

2225

A B C

Failureplane

(a)

Active State(b)

Passive State

Failureplane

Ka =minus minus

+ minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 45: Chapter 14 Foundations for Concrete Structures.pdf

and

(1447)

where

Example 1410

Hence

Distance (z) Az

1 ndash1177 064 ndash7492 0027 197 00533 1122 60 67324 36 50 180

657 m

2 m

5deg

GWT

γ = 17 kNm3

Pa

φ = 30degC = 10 kPaγ = 18 kNm36 m

588 kPau

193 kPaσa

ndash1232 kPa2

4

5

3

106 kPa

ʹ

K p =+ minus

minus minuscos

cos cos cos

cos cos cosβ

β β φ

2 2

z = minus + + + minus + +( ) ( 7 49 0 053 673 2 18 0 11 77 0 027 112 ) 2 3 6 6 57+ = m

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 46: Chapter 14 Foundations for Concrete Structures.pdf

(1448)

(1449)

(1450)

Example 1411

Active

Passive

15 ft

5

75

18 kipft

610 psf

Pp

3540 psf 1200 psf

2656 kipft

Ka =+( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

K p =minus( )

+( ) ++( ) +( )

sin

sin sinsin sin

2

2 1

α φ

α α δφ δ φ β

2

Ka = minus+

1

sin

φφ

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 47: Chapter 14 Foundations for Concrete Structures.pdf

(1451)

(1452)

Example 1412

10deg15deg

5o

Pa

Q

50deg75deg

E

D A BC

R G

wall

12 6Clayey backfill

c = 10 psi φ = 10degγ = 115 lbft3

δ = 15deg

Scale 1 = 5

O

σp z= +236 16 1200

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 48: Chapter 14 Foundations for Concrete Structures.pdf

Example 1413

1 81 65 52652 2025 45 9113 2025 35 7094 0506 283 143

RW = 10675

PA

C = 1526

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 49: Chapter 14 Foundations for Concrete Structures.pdf

H48

Minimum batter

H12 to H10

B = 04 to 07 H

B3

1

135

05 1

15

525

W4

W3

W2

Pa = 449 kipsft

C

D

G1 ft

W1

F1

Overturning moment

665 9 11 7 09 1 43

4 496 53 13

+ + +( )times

=

F2

6 449

Sliding force

661 43=

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 50: Chapter 14 Foundations for Concrete Structures.pdf

5 ft15 ft

Filters

1ftBerm

25 ft

Weepholes(5)

15 ft

1 ft

Z

W = 40 Z V = 20 Z 2 M = 667 Z3

05395 kipft2

Z Z

(a) Load

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 51: Chapter 14 Foundations for Concrete Structures.pdf

25

05

10

50

225

0225

16411286

1845

404615

4496

(a) Load (kipft2)

2373354

3743 3134

(b) Shear (kipft)

44253195

472

733

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

Page 52: Chapter 14 Foundations for Concrete Structures.pdf

(1453)

Ls

FSov h z

v z i

= timestimes

timesσσ ϕ

tan2overlap

Front cover
Contents
Preface
Acknowledgments
Editor-in-Chief
Contributors
Index
Back cover

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