[] Practical Problems in Soil Mechanics and Foundation

8/18/2019 [] Practical Problems in Soil Mechanics and Foundation

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Further t i t les in this series:

1. G . S A N G L E R A T - T H E P EN ^ T R O M E T E R A N D S O IL E X P L O R A T IO N

2 . Q . Z A R U B A A N D V . M E N C L - L A N D S L I D E S A N D T H E I R C O N T R O L

3 . E .E . W A H L S T R O M - T U N N E L I N G I N R O C K

4 .

R . S I L V E S T E R - C O A S T A L E N G I N E E R I N G , 1 a nd 2

5 . R . N . Y O N G A N D B.P . W A R K E N T I N - S O I L P R O P E R T I E S A N D B E H A V I O U R

6 . E .E . W A H L S T R O M - D A M S , D A M F O U N D A T I O N S , A N D R E S E R V O I R S I T ES

7 . W . F . C H E N - L I M I T A N A L Y S I S A N D S O I L P L A S T I C I T Y

8 . L . N . P E R SE N - R O C K D Y N A M I C S A N D G E O P H Y S I C A L E X P L O R A T I O N

Introduct ion to Stress Waves in Rocks

9 . M . D . G I D I G A S U - L A T E R I T E S O I L E N G I N E E R I N G

10. Q . Z A R U B A A N D V . M E N C L - E N G I N E E R I N G G E O L O G Y

1 1 . H . K. G U P T A A N D B .K . R A S T O G I - D A M S A N D E A R T H Q U A K E S

12. F . H . C H E N - F O U N D A T I O N S O N E X P A N S I V E S O I L S

13 .

L . H OB S T A N D J . ZA J IC - A N C H O R IN G IN R OC K

14. B. V O I G H T E d i to r ) - R O C K S L I D E S A N D A V A L A N C H E S , 1 a n d 2

15. C . L O M N I T Z A N D E . R O S E N B L U E T H E d it o rs ) - S E I S M I C R I S K A N D E N G I N E E R I N G D E C I S I O N S

16.

C . A. B A A R - A P P L I E D S A L T - R O C K M E C H A N I C S , 1

The In-Si tu Behavior of Sal t Rocks

17. A .P .S . S E L V A D U R A I - E L A S T I C A N A L Y S I S O F S O I L - F O U N D A T I O N I N T E R A C T I O N

18.

J . F E D A - S T R E S S IN S U B S O I L A N D M E T H O D S O F F I N A L S E T T L E M E N T C A L C U L A T I O N

19.

A . K E Z D I - S T A B I L I Z E D EA R T H RO A D S

20 . E .W . B R A N D A N D R .P . B R E N N E R E d i to r s ) - S OF T-C LA Y E N GIN E E R IN G

2 1 . A . M Y S L I V E C A N D Z . K Y S E L A - T H E B E A R I N G C A P A C I T Y O F B U I L D I N G F O U N D A T I O N S

2 2 . R . N . C H O W D H U R Y - S LO P E A N A L Y S I S

23 . P. B R U U N - S T A B I L I T Y O F T I D A L I N L E T S

Theory and Eng ineer ing

24 . Z . B A Z A N T - M E T H O D S O F F O U N D A T I O N E N G I N E E R I N G

25 .

A . K E Z D I - S O I L P HY S IC S

Selected Topics

2 6 .

H . L . J E S SB E R G E R E d i to r ) - G R O U N D F R E E Z I N G

27 .

D . S T E P H EN S O N - R O C K F I L L I N H Y D R A U L I C E N G I N E E R I N G

28 . P .E . F R I V I K , N . J A N B U , R . S A E T E R S D A L A N D L . I . F I N B O R U D E d it o rs ) - G R O U N D F R E E Z I N G

1980

2 9 . P. P E T E R - C A N A L A N D R I V ER L E VE E S

3 0 .

J . F E D A - M E C H A N I C S O F P A R T I C U L A T E M A T E R I A L S

The Pr inc ip les

3 1 . Q . Z A R U B A A N D V . M E N C L - L A N D S L I D E S A N D T H E I R C O N T R O L

Second com ple te ly revised ed i t ion

32 . I.W . FA R M E R E d i t o r ) - S T R A TA ME C H A N IC S

33 . L . H O B S T A N D J . Z A J I C - A N C H O R I N G I N R O C K A N D S O I L

Second com ple te ly revised ed i t io n

3 5 .

L . R E T H A T I - G R O U N D W A T E R I N C I V I L E N G I N E E R I N G


2/262

DEVELOPMENTS

IN

GEOTECH NICAL ENGINEERING

4B

PR CTIC L PROB LEMS

IN

SOIL MECHANICS

AND

FOUNDATION

ENGINEERING

WALL

AND

FOUNDATION CALCULATIONS

SLOPE STABILITY

GUYSANGLERAT

GILBERTOLIVARI

BERNARD CAM BO U

Translated

by

G. GENDARME

ELSEVIER

Amsterdam

Oxford

New York

Tokyo 985


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ELSEVIER SCIENCE PUBLISHERS B.V.

Molenwerf 1

P.O.

Box 211, 1000 AE Amsterdam, The Nether lands

Distributors for the United States and Cana da

ELSEVIER SCIENCE PUBLISHING COMPANY INC.

52, Vanderbi l t Avenue

New York , N.Y. 10017

ISBN 0-444-42133-8 (Vol . 34B)

ISBN 0-444-41662-5 (Series)

ISBN 0-444-42109-2 (Set)

© Elsevier Science Publishers B.V., 1985

All rights reserved. No part of this publication may be reproduced, stored in a retrieval

system or t ransm it ted in any form or by any mea ns, e lectronic, mech anical , ph oto

copying, recording or otherwise, wi thout the pr ior wri t ten permission of the publ isher ,

Elsevier Science Publ ishers B.V. /Science Tech nology Division, P.O. Box 330, 1000 AH

Amsterdam, The Nether lands .

Special regulations for readers in the USA This publication has been registered with the

Copy right Clearance Center Inc. (CCC ), Salem, Massachu set ts . Inform at ion can be ob

tained from the CCC about condi t ions under which photocopies of par ts of this publ ica

t ion may be made in the USA. All other copyright quest ions, including photocopying

outside of the USA, should be referred to the publishers.


4/262

I N T R O D U C T I O N

Gu y Sangle ra t has tau gh t geo techn ica l engineer ing a t th e Ec ole Cent ra le

de L y o n s ince 19 67 . Th i s d i sc ip l i ne w a s i n t rod uc e d th e re by Je a n Cos t e t .

S ince 1968 and 1970, respec t ive ly , Gi lber t Ol ivar i and Bernard Cambou

ac t ive ly ass i sted in th i s respo ns ib i l i ty . Th ey di rec ted labo ra to ry wo rk,

outs ide s tudies and led spec ia l s tudy groups .

In o rd er to ma ster an y scient if ic discipl in e , i t is necess ary to ap ply i ts

theore t ica l pr inc ip les to prac t ice and to readi ly so lve i t s problems. This holds

t rue a l so for theore t ica l so i l mechanics when appl ied to geotechnica l engin

eering.

From Coste t ' s and Sangle ra t ' s exper iences wi th the i r previous ly publ i shed

te x tbooks i n ge o te c hn ic a l e ng ine e r ing , w h ic h c on ta in e xa mple -p rob le ms a nd

answers , i t became evident tha t one e lement was s t i l l miss ing in conveying

the und e rs t a n d ing of t h e sub je c t m a t t e r t o t he so lu t ion o f p ra c t i c a l p ro b le m s :

p rob le ms a ppa re n t ly ne e de d de t a i l e d , s t e p -by-s t e p so lu t ions .

For th i s reason and a t the reques t of many of the i r s tudents , Sangle ra t ,

Ol ivar i and Ca m bo u de c ided to pu bl i sh pr ob lem s. Over th e years s ince 1967

the pro ble m s in th i s te xt have been g iven to s tu den ts of th e Ec ole C ent ra le

de L y o n and since 19 76 to spec ial geote chnica l engineer ing s tu dy grou ps of

the Publ ic Works Depar tment of the Nat iona l School a t Vaulx-en-Vel in ,

where Gilbert Olivari was assigned to teach soi l mechanics.

In order to ass i s t the reader of these volumes , i t was dec ided to ca tegor ize

problems by degrees of so lu t ion d i f f icul ty . There fore , easy problems a re

prec ede d by on e s ta r (*) , tho se cons id ered m ost d i f f icul t by 4 s ta rs (* ** *) .

Depending on h is degree of in te res t , the reader may choose the types of

problems he wishes to so lve .

Th e a u th ors d i re c t t he p rob le m s n o t on ly t o s tud e n t s bu t al so t o t he

prac t ic ing Civil Eng ineer an d to o th ers w ho , on occa s ion, need to so lve geo

techn ica l eng ineer ing pro ble m s. To a l l, th i s work offers an easy re fe r ence ,

provided tha t s imi la r i t i e s of ac tua l condi t ions can be found in one or more

of the solu t ions presc r ibed here in .

Mainly , the S . I . (Sys teme In te rna t iona l ) uni t s have been used. But , s ince

pra c t i c e c a nno t be i gnore d , i t w a s de e me d ne c e ssa ry t o i nc orpora t e o the r

wide ly accepted uni t s . Thus the C.G.S . and Engl i sh uni t s ( inch, foot , pounds

per cub ic foo t , e tc . ) have been inc lu ded b ecause a la rge qu an t i ty of l i t e r a tur e

i s based on these uni t s .

The authors a re gra te ful to Mr. Jean Kerise l , pas t pres ident of the In te r

na t io na l Soc ie ty for Soi l Me chanics and Fo un da t io n Eng ineer ing , for having


5/262

VI

I NTRODUCTI ON

written the Preface to the French edition and allowing the authors to include

one of the problems given his students while Professor of Soil Mechanics at

the Eco le Nationale de Ponts et Chaussees in Paris. Their gratitud e also

goes to Victor F.B. de Mello, President of the International Society for

Soil Mechanics, who had the kindness to preface the English edition.

The first problems were originally prepared by Jean Costet for the course

in soil mechanics which he introduced in Lyon.

Thank s are also due to Jean-Claude Ro uault of Air Liqu ide and Henri

Vidal of Re info rced E ar th and also to our Brazilian friend Lucien De cou rt

for contributing problems, and to Thierry Sanglerat for proofreading manu

scripts and printed proofs.


6/262

IX

N O T A T I O N S

The fo l low ing ge ne ra l no t a t i ons a ppe a r i n t he p rob le ms :

B

c

c

n

C

C

u

c

c

d

D

E

FR

G

h

H

i

IP

k

Sk e m pto n s s e c ond c oe f f ic i e n t ( so me t im e s

A

refers also to

c ross-sec t iona l a rea ) .

value of

A

a t failure

foo t ing w id th ( some t ime s

B

re fe rs a l so to Sk em pt on s f i rs t

coeff ic ient ) .

so il coh es ion (un di f fe rent ia ted)

effect ive cohesion

reduced cohes ion (s lope s tabi l i ty)

undra ine d c ohe s ion

c onso l ida t e d -undra ine d c ohe s ion

c ompre ss ion inde x

uni formi ty coeff ic ient , de f ined as d

6 0

d

1 0

coeff ic ient of consol ida t ion

so i l pa r t i c l e d i a me te r ( some t ime s : ho r i z on ta l d i s t a nc e

be tween adjacent , s imi la r s t ruc tures , a s in the case of sub

surface drains)

equ ival ent d iam ete r of s ieve op eni ng s in grain-size distr i

b u t i o n

de p th t o bo t tom of foo t ings ( some t ime s

D

re fe rs to depth

to hard layer under the toe of a s lope) .

vo id ra t i o ( some t ime s :

e

re fe rs to eccent r ic i ty of a concen

t ra ted force ac t ing on a foot ing)

m a x i m u m a n d m i n i m u m v o i d r a t i o s

Y o u n g s m o d u l u s

p r e s s u r e m e t e r m o d u l u s

f r i c t i on ra t i o ( s t a t i c pe ne t rome te r t e s t )

accelerat ion due to gravi ty (gravie)

she a r modu lus

hydra u l i c he a d

soi l l ayer th ickness (or normal cohes ion: H — c c o t


7/262

X

NOTATIONS

^ P 7 > ^ p q ^ p c

Kpy> *^Pq

K

v

K

K

0

I

L

m

v

M

m

M

R

M

N N N

Pi

Pi

Q

Q

Q

f

Q P

9

d


8/262

NOTATI ONS

XI

W

w

u

w

p

x,y,z

7

7s

7sat

7h

7w

Td

7

x y » yz ? z:

^ x > ^ y ? ^ z

V

i

O

o

o*

o

m

r

x y

?

y z

?

z

ti

: w a te r c o n te n t o r s e t t l e m e n t

: l iquid l imit , plast ic l imit

: Car tes ian coordina tes , wi th

Oz

usua l ly cons idered the ver t i

c a l, do w n w a rd a x i s

: angle be tween or ienta t ions , usua l ly rese rved for the angle

be tw ee n tw o soi l faces. Also used t o c lassify soi ls for th e

purpose o f t he i r c ompre ss ib i l i t y f rom s t a t i c c one pe ne t ro -

mete r tes t da ta C.P .T.

: s lop e of th e surface of backfi l l beh ind a re ta inin g wall

(angle of s lope)

unit weight of soi l (unspecified)

soi l part ic les uni t weight (specific gravi ty)

sa tura ted uni t we ight of so i l

wet uni t weight of soi l

un i t w e igh t o f w a te r = 9 . 81 kN /m

3

.

dry uni t weight of soi l

effect ive uni t weight of soi l

shear s t ra in , twice the angula r de format ion in a rec tangula r ,

3-dimensiona l sys tem

angle of fr ic t ion between soi l and re ta ining wall surface in

pass ive or ac t ive ea r th pressure problems, or the angle of

inc l ina t ion of a po in t load ac t ing on a f oot in g

dynamic v iscos i ty of wate r

axia l s t ra ins in a rec tangula r , 3-dimensiona l sys tem

principal s t ress

volumetr ic s t ra in

angle of radius in pola r coordina tes sys tem (somet imes:

t e m p e r a t u r e )

: Poisso n s ra t io

: effect ive normal s t ress

: to ta l normal s t ress

: no rm al s tresses in a rec tan gula r , 3-dimension a l sys tem

: major principal s t resses

: average stress

: shear stress

: average shear s t ress

: shear s tresses in a rec tan gula r , 3-dim ensiona l sys tem

: angle of in te rn a l f r ic t ion (un def in ed)

: effect ive angle of int ern al fr ic t ion

: re du ce d, effect ive angle of inte rna l fr ic t ion (slop e-stab i l i ty

ana lyses)

: angle of in te rna l f r ic t ion , consol ida ted , undra ined

: s lope of a wall from the vert ical

: auxi l iary angles defined by sin top = sin j3/sin y and

sin co

6

= sin 8 / s in


9/262

XII

NOTATI ONS

:

3 1416

p : d i s tan ce f rom or ig in to a po int in pola r co ord ina te sys tem

\p : angle of ma jor prin cipa l s t ress wi th rad ius ve cto r (pla st ic i ty

p rob le ms)


10/262

XIII

NGIN RING UNITS

It is presently required that all scientific and technical publications resort

to the S.I. units (Syst&me International) and their multipliers (deca, hecta,

kilo, Mega, Giga). Geotechnical engineering units follow this requirement

and most of the problems treated here are in the S.I. system.

Fundamental S.I. units:

length

mass

time

meter (m)

kilogram (kg)

second (s)

S.I. Units derived from the above

surface

volume

specific mass

velocity (permeability)

acceleration

discharge

force (weight)

unit weight

pressure, stress

work (energy)

viscosity

square meter (m

2

)

cubic meter (m

3

)

kilogram per cubic meter (kg/m

3

)

meter per second (m/s)

meter per second per second (m/s

2

)

cubic meter per second (m

3

/s)

Newton (N)

Newton per cubic meter (N/m

3

)

Pascal (Pa) 1 Pa = 1 N /m

2

Joule (J) 1 J = 1 N x m

Pascal-second* Pa x s

How ever, in practice, oth er units axe enc oun tered frequen tly. Table A

presents correlations between the S.I. and two other unit systems encoun

tered worldwide. This is to familiarize the readers of any publication with

the units used therein. For that purpose also, British units have been adopted

for some of the presented problems.

Force pressure) conversions

Force units

Pressure units

Weight unit

see Table B

see Table C

l k N / m

3

= 0.102 tf/m

3

*T his un it used to be called the po ise uill e , bu t i t has no t been officially ad op ted .


11/262

XIV

ENGI NEERI NG UNI TS

T A B L E

A

Corre la t ions be tween most common uni t sys tems

Length

Mass

Time

Force

Pressure

(stress)

Work

(energy)

Sys teme In te rna t iona l

(S.I.)

uni ts

m e t e r

(m)

kilogram (kg)

second

(s)

N e w t o n (N)

Pascal

(Pa)

Joule

(J)

c o m m o n

mult iples

km

t onne (t)

—

kN

kPa

MPa

k J

Meter-Kilogram

(M.K.)

uni ts

m e t e r (m)

gravie*

second (s)

kilogram force

(kgf)

kilogram force

per square

meter (kgf /m

2

)

ki logram meter

(kgm)

system

c o m m o n

mult iples

km

—

—

tf

( t / m

2

1 kg/cm

2

t f . m

Cent imeter -Gram-

Second system

(C.G.S

uni ts

cm

g

s

dyne

barye

erg

:.)

c o m m o n

mult iples

m

—

—

bar

( 1 0

6

baryes)

Joule

( 1 0

7

ergs)

*Note tha t 1 gravie = 9 .81 kg (in most problem s rounde d off to 10).

Th e unit weight of wa ter is: 7

W

= 9.81 kN /m

3

bu t it is ofte n ro un de d off

t o :

7

W

= 10 kN/m

3

.

Energy units

1 Jou le = 0.10 2 kg .m = 1.02 x 10~

4

t .m

1 k g f . m = 9 .81 Joules

1 tf .m = 9.81 x 10

3

Joules

Dynamic viscosity units

1 Pascal-second (Pa.s) = 10 poises (Po).

British units

1 inch

1 foot

1 square inch

1 square foot

l m

2

1 cubic inch

1 cubic foot

l m

3

1 pound (lb)

1 Newton

1 lb/cu. in.

l m = 39.37 0 in.

l m = 3 . 280 8 foot

1 cm

2

= 0. 15 5 sq. in.

l c m

3

= 0.06 1 Ocu. in.

= 0.0 25 4 m

= 0.30 4 8 m

-

6 .45 16 cm

2

= 14 4 sq. in. = 0 .0 92 9 m

2

= 10.764 sq.f t .

= 1 6.38 7 cm

3

= 17 28 cu. in. = 0 .0 28 31 7 m

3

= 35 .3 14 cu. ft.

= 4.4 49 7 Ne w to n = 0 .4 53 59 kgf

= 0.2 25 lb = 0.1 12 4 x 10

3

sh. to n. (1 sh. to n. = 2 kip)

= 1.003 xl O

- 4

ton.

= 270.27 kN/m

3


12/262

NGIN RING UNITS

1 lb/cu . ft. = 0.156 99 kN /m

3

1 kN/m

3

= 3.7 x 10

3

lb/cu. in. = 6.37 Ib/cu. ft.

1 lb/sq. in. (p.s.i.) = 6.896 55 x 10

3

Pa

1 Pascal = 14 .50 x 10

5

p.s.i.

100 kPa = 1 bar = 14.50 p.s.i.


13/262

T

A

B

L

B

F

c

u

s

c

o

V

a

u

y

/

o

/

i

/

e

e

i

n

-

N

e

w

t

o

D

e

w

t

o

K

i

o

w

t

o

K

i

o

a

m

f

o

c

T

f

o

c

D

y

N

e

w

t

o

1 1 1

9

8

9

8

1

1

5

D

e

w

t

o

1

1

1

9

8

x

1

1

9

8

1

1

K

i

o

w

t

o

1 1

1 9

8

1

3

9

8

1

8

K

i

o

a

m

f

o

c

1

0

1

1

0

1

0

1

1

1

1

0

1

6

T

f

o

c

1

0

1

1

0

1

3

1

0

1

1

1 1

1

0

1

9

D

y

1

1 1

9

8

1

9

8

1

1

T

A

B

L

E

C

P

e

u

e

u

s

V

a

u

/

o

^

o

y

.

e

e

y

i

n

*

P

K

io

B

a

H

e

o

B

a

y

k

c

m

2

k

m

m

2

t

m

2

c

m

o

f

w

a

e

A

tm

o

p

e

P

1 1 1 1

0

1

9

8

x

1

9

8

x

1

9

8

x

1

9

8

x

1

1

0

x

1

k

i

o

3

1

1

1 I

O

9

8

x

9

8

x

9

8

9

8

x

1

1

1

0

x

-

3

-

2

1

b

i

o

-

5

I

O

1 1 1

6

0

9

9

8

x

9

8

x

9

8

x

I

O

I

O

1

0

h

I

O

1 I

O

1

1

8

9

8

x

0

9

9

8

x

9

8

x

1

1

1

0

x

3

-

4

-

6

I

O

b

y

1

I

O

4

I

O

6

I

O

8

1

9

8

x

9

8

x

9

8

x

9

8

x

I

O

7

I

O

4

I

O

2

1

0

x

O

6

k

c

m

2

1

0

x

O

1

0

x

O

-

2

1

0

1

0

x

O

2

1

0

x

O

-

6

1 I

O

2

0

1

I

O

1

0

k

m

m

2

1

0

O

-

7

1

0

x

1

4

1

0

x

O

1

0

1

0

x

1

8

I

O

1

I

O

I

O

1

0

x

1

2

t

m

2

1

0

x

O

-

4

1

0

x

O

1

2

1

0

x

O

3

1

0

x

1

5

1

I

O

3

1

I

O

'

1

0

x

O

1

c

m

o

w

a

e

1

0

x

O

1

2

1

0

x

O

3

1

0

x

1

1

0

x

1

3

I

O

3

1

I

O

2

1

1

0

x

O

3

a

m

.

9

8

x

O

9

8

x

O

0

9

9

9

8

x

O

1

.

9

8

x

O

0

9

1

9

6

x

O

1

9

6

x

O

9

6

x

O

1


14/262

Chapter 7

1

RETAINING WALLS

^Problem 7.1

Earth pressures on a vertical wall, horizonta l backfill, above

the water table

A 4m high wall serves as a retaining-wall for a mass of horizon tal flattened

dry sand (Fig. 7.1). The dry sand's unit weight is 18.3 kN/m

3

and its internal

angle of friction is 36° .

What is the magnitude of the earth force P on a 1 m wide wall slice, as

suming that the wall does not deflect? Calculate also the earth force P

x

if the

wall deflects sufficiently to generate active (Rankine) pressure conditions in

the backfill. Assume that the back face of the w all is frictionless.

Fig. 7 .1 .

Solution

If no wall deflection occurs, the earth pressure at rest condition prevails,

i.e. that pressure P

0

, then acting on the wall, may be represented by the

Mohr's circle equilibrium condition comprised between the Coulomb's

envelopes (Fig. 7.2 ). In general, for a sand: 0.3 3 < K

0

< 0.7. (cf. 6.1.4 in

Costet-Sanglerat, where the values of K

0

are calculated from empirical

formulas.) The pressure distribution on the inner wall face is triangular and

because it is assumed that the face is frictionless, the pressures act perpen

dicular to the wall.

D r y

u-

if--

s a n d

: 1 8 . 3 k N /

m

3

3 6


15/262

2

RETAINING WALLS

Fig. 7.2.

So, for a i m wide wall slice, we have:

A> = ?K

0

y

d

H

2

b

where

b =

1.00 m and 7

d

= 18.3 kN/m

3

.

K

0

calculated by the formula of Jaky gives: K

0

= 1

—

sin y .

F or */ /= 36 °: sin


16/262

PROBLEM 7.2

3

irkProblem 7.2 Earth pressure con sider ing th e wa ter table on a vertical wall

Assuming the givens of the preceding problem, what is the total resultant

earth pressure acting on the wall and its location with respect to the base of

the w all, if there is a water table at 1 m below the back fill grade (assume a

sand porosity of 0.31) (see Fig. 7.3).

'. h

=

1.00 m'.' :':'•

• ' : '•;•: '.•-.-Water t ab le

H - h = 3 . 0 0 m

Fig. 7.3.

So l ut i o n

F r o m t h e p r e c e d i n g p r o b l e m , w e h a v e :

fe

a7

= t a n

2

( 2 7 ° ) = 0 . 2 5 9 6 , s a y 0 . 2 6 .

The buoyan t we i gh t o f t he s and i s :

1 = Tsat - 7w = 7d + «7w - 7w = 7d — (1 — rc)7v

7 '

= 1 8 . 3 - ( 1 - 0 . 3 1 ) x 1 0.0 = 1 1 .4 k N / m

3

.

o r :

The d i s t r ibu t ion of the s t resses beh ind the wal l i s ( see Figs . 7 .3 and 7 .5) :

O n AB: t he d i s t r i bu t i on i s t r i angu l a r and we have :

= 0;

= k

ay

xy

d

xh = 0 . 2 6 x 1 8 . 3 x 1 . 0 0 = 4 .7 6 k N /m

2

O n

BC:

the d i s t r i bu t ion is s t il l t r i an gu lar , b u t a t

B

the s lope of the hy

p o t e n u s e c h a n g e s : h e r e t h e b u o y a n t w e i g h t a n d t h e h y d r o s t a t i c w a t e r

pressu re m us t be t a ke n i n t o ac co un t , as wel l as th e we igh t o f d ry sand , to b e

cons i de red a s a un i fo r m s u rcha rg e . T he re fo re :

— pres s u re due t o t he buo ya n t we i gh t of t he s a nd :

fflB

0;

nc

fc

a7

xy'x(H-h) =

0 . 2 6 x 1 1 . 4 x 3 . 0 0 = 8.8 9 k N /m

2

— pressu re du e to the un i for m d i scharge of th e sand ( rec tang ular d i s t r i

b u t i o n ) :

° 2B

o

2

c kqxq = k^xh xy

d

w h e r e :

k

c

w

a 7

cos(/3 —X)

(Fig. 7 .4)


17/262

RETAINING WALLS

Fig. 7.5.

4 . 7 6 k N / m

2

43.65 kN/m*

This equation derived from Coulomb's hypothesis is also valid for Rankine-

conditions (6.24 in Costet-Sanglerat); but:

j3 = X = 0, fe

q

=

fe

a7

,

so:

a

2B

=

o

2C

=

/e

a7

x

h

x 7

d

= a

B

computed previously as = 4.76 kN/m

2

—

hydrostatic pressure (triangular distribution):


18/262

PROBLEM 7.3

5

^3B = 0; a

3 C

=

(H-h)y

w

=

3 0 k N / m

2

.

So we end up with the diagram shown in Fig. 7.5: the total force acting on

the wall is the resultant of the forces

R

x

, R

2

and

R

3

of th at Figure, and we

have:

R

x

= (1/2 ) x 4.76 x 1.00 x 1.00 = 2. 40 kN per m eter of wall loca ted at

3.00 + 0.33 = 3.33 m from C (1/3 of AB).

R

2

= 3 .0 0 x 4 .7 6 x 1 .0 0 = 14.3 kN (per meter of wall ) acting a t 1 .5 m dis

tance of

C

(middle of

BC).

R

3

= (1/2 ) x 38.9 x 3.00 x 1.00 = 58.3 kN (per m eter of w all) acting a t

1.00 m distance of C (lower 1/3 of BC).

The resultant force thus is: P = R

}

+ R

2

+-R3 — 75 kN and this force acts

at such a distance d from C that :

Pd =

R

x

d

x

+

R

2

d

2

+

R

3

d

3

,

2.4

3.33

+

14.3

1.50

4- 58.3

1

.00

d

= - 1.17m

75.0

Summary of answers

P = 75 kN per meter of wall, d = 1.17 m.

++Problem

7.3

Retaining wall with horizontal backfill; overturning stability

and sliding stability

Suppose you are asked to determine the stability of the quay wall shown

on F ig. 7.6. (It is assum ed that the steps of the wall are comparable to a

straight line AB because the weight of the soil is not significantly different

from that of the concrete in the small triangular areas.)

The ba se of the founda tion's upper part is at the level of the water table

and that of the natural soil, in which the footing, completely subm erged, is

embedded. The retaining-wall supports the soil above the water table.

Assume the following values:

Concrete : unit weight 23 kN/m

3

Fill : unit weigh 118 kN/m

3

internal angle of friction $

x

= 30°

cohesion c = 0

earth pressure coefficients on AB (8 =


19/262

1.00

RET AINING W ALLS

q =10 kPa

Water tab le

Fig. 7.6

Wall:

h

x

=

6.50m

h

2

= 2.50m

FA = lm, KB = 4m, DC = 5m.

As a security precaution, ignore the passive earth pressure on plane ED of

the foundation.

Find:

(1) The eccentricity of the resultant force acting on base CD . Is there tension?

(2) The maximum bearing pressure on the foundation soil.

(3) The safety factor against overturning.

(4) The safety factor against lateral sliding (assume the friction coefficient

between the bottom of the foundation and the soil is tan


20/262

PROBLEM 7.3

7

—

the active earth force P increased by the value of the lateral force Q due to

the surcharge imposed by the fill;

—

the passive earth force B acting on p lane ED of the foundation;

—

the foundation soil reaction R.

For the wall to be in equilibrium, the resultant of all these forces must be

zero which allows the calculation of the value of the reaction R.

For the sake of safety, it is general practice to ignore the passive force B

acting on the side of the footing. There are two reasons for this. Firstly, the

wall displacement is generally not sufficiently large to actually mobilize the

passive condition: a displacement of about 0.05 to 0.10 ft (ft being the height

of plane ED) would be need ed. In our case, this would mean a displacement

of 12—25 cm , considerably m uch m ore than w all mo vem ents associated w ith

the development of active conditions. Secondly, in practice, the possibility

of an excavation being made along

ED

after construction, always must be

taken into account.

(a) Wall weight and hydrostatic pressure

As indicated above, we assume the back of the wall, AB, to be a straight

line. Then (Fig. 7.7) we have:

wall: rectangular section AHKF

W

x

= 1.00 x 6.50 x 23 = 14 9.5 kN (per m eter length of wall)

triangular section AHB:

W

2

= \

x 3.00 x 6.50 x 23 = 22 4.3 kN (per meter length of wall)

q = l O k P a

F A

j

E

. 6 . 5 0 m

XSX/y^^A^X^/

Fi l l

= 30°

: 2 . 5 0 m

j^*28^^

6

7Tv

2

W

3

-7T

t^ '

2

N a t u r a l s o i l

< >

2

=

2 5 °

W a t e r

t a b l e

Fig. 7.7.


21/262

8

RET AINING W ALLS

footing: BCDE (taking into account the uplift pressure due to hydrostatic

pressure and using the buoyant unit weight of concrete:

7

;

e

t on = 13kN /m

3

) .

W

3

= W

3

-

n = 2.50 x 5.00 x 13 = 162 .5 kN.

In the axes-system (Cx, Cy) (Fig. 7.7) these forces have following action

points:

Vf

t

:(x = 3.50; y = 5.75)

W

2

: (x = 2.00; y = 4.67)

W

3

: (x = 2.50; y =

1.25).

(b) Forces on the plane

AB

(b

x

)

Earth pressure for ce

P

x

^ i , o , , , ^ . « . hy 6.50

^ i = hdl

2

k

R1

where

fe

a7

= 0.474 , / = —*— = = 7 .17m.

cos

A

cos 25

So, P

x

= \ x 1 8 x 7 1 7

2

x 0.474 = 21 9.3 kN (per m length of wall).

Horizontal component : P

1 H

=

P

x

cos(5

+ \) = P

x

cos 55° = 125 .8 kN (per

m length of wall).

Vertical component: P

l v

= P

x

sin 55° = 179.6 kN (per m length of wall).

Remark

Angle 8 = (f has been chosen because when the state of plasticity is

developed,

AB

is a line of failure. The portions of soil located to the left

of this line and above the steps are not in a plastic equilibrium state. The

shear will be that of soil along

A B

and therefore 5 =


22/262

PROBLEM 7.3

9

Remark

Angle 5' = \ $ is the usually assumed value in the case of friction between

soil and concrete. The footing of the wall is below the water table. Since

the hydrostatic pressure acts on both vertical faces of the footing, but

in opposite directions, it does not have to be accounted for.

( c j Earth pressures: tr iangular distribution:

Pi = Wh

2

2

k'

ai

P

2

= 0.5 x 11 x 2750

2

x 0.364 = 12.5 kN (per m length of wall)

Horizontal component :

P

2H

=

12.5 x 0.958

—

12 kN (per m leng th of wa ll).

Vertical component: P

2 V

= 12.5 x 0.287 — 3.6 kN (per m length of wall).

The point through which the force acts is at 1/3 up from C on BC, or, in

our coordinate system, at

x =

0.0 and

y =

0.83 m.

(c

2

) Earth pressure due to the surcharge fill and to the mass of earth above

the water table: Q

2

The su rcharge fill is 10 kPa. T he w eight of th e soil above th e wa ter table is:

6.50 x 18 = 117 kPa, and the total is: q = 127 kPa.

Therefore, we have:

Q

2

= q'-h

2

-

fc^ = 127 x 2.50 x 0.364 = 11 5.6 kN (per m length of wall).

Horizontal component : Q

2H

= 115 .6 x 0.95 8 = 110.7 kN (per m length of

wall).

Vertical component: Q

2 V

= 115.6

x

0.287 = 33.2 kN (per m length of wall).

Since the pressure distribution is rectangular, the point of application of

the force is half-way up BC, or x = 0, y = 1.25 m. The resultant of all the

forces acting on the wall (with the exception of the soil reaction on the

footing) is F, and its line of action through plane DC (Fig. 7.8) can be de

termined. At P, the equivalent force F' gives:

M

c

= moment of F with respect to C = moment of F ' a t C = F

v

x d.

Therefore, point P is defined by: d =

M

c

/F

v

, whe re:

M

c

= S moments of exterior forces with respect to C

F

v

= 2 vertical components of exterior vertical forces.

The eccentricity of P with respect to the axis of symmetry of the footing

is

e

= |d— D C /2 | and the resultant F goes through the middle third if:

e


23/262

10

RETAINING WALLS

Fig. 7.8.

TABLE 7A

Forces

^1 H

^ 1 V

Qm

Qiv

Pm

PTV

Q2H

Qrv

W

x

W

2

w

3

= w

3

- n

Forces

(kN)

vertical

179.6

30.6

3.6

33.2

149.5

224.3

162 .5

horizontal

125.8

21.5

12.0

110.7

Lever

arm by C

(m)

4.67

1.00

5.75

1.50

0.83

0

1.25

0

3.50

2.00

2.50

Moment

by C

kN-m)

587.5

179.6

123 .6

45.9

10 .0

0

138.4

0

523.3

448.6

406.3


24/262

PROBLEM 7 .3

11

(2) Calculation of the maximal stress in the bottom of the footing

The stress distribution results in a force R which must be in equilibrium

with F \ The usual calculation is to resolve this force in horiz ontal and

vertical components.

For the vertical components, the trapezoidal distribution resultant must

equal F

v

.

Referring to Fig. 7.9, we have:

' m a x ' ^ m in

1

2 \

w

m a x

o,

a„

xB = Fv

2B B\

)xBx\— - - ) = F

v

xe

(1)

(2)

from which: a

n

783.3

(F

v

/B)(l

+

6e/B),

and therefore:

6 x 0.64 .

= 277 kPa, or a

r

2.8daN/cm

2

'max

Fig. 7.9.

m m

Remark

For calculating the allowable bearing capacity for an eccentric, inclined

load, Meyerhof proposes the following formula for the vertical component

of the allowable stress (for sands):


25/262

1 2

RET AINING W ALLS

(3 )

Calculation of the safety factor against overturning of the wall

To e s t im ate th e safe ty fac to r aga ins t over tu rn ing of th e wal l , i t is necess

ary to k no w th e loca t ion of the axi s o f ro ta t io n of th e wal l . If th e foun

da t i o n s o il were non de fo rm a b l e , t h i s ax i s w ou l d be t h ro u g h ! ) (F i g . 7 .8 ) a t

the toe of the foo t ing . Since the so i l deforms , the loca t ion of the ro ta t ion

ax i s i s no t known and m ay we l l va ry du r i ng t he ove r t u rn i ng p roces s . The re

fore th e safe ty fac tor var ies du r ing th e cou rse of the m ov em en t .

I f i t is assu me d tha t the axi s o f ro ta t io n i s th ro ug h p o in t D, w e can w r i t e :

M om en t s o f s t ab i l i z i ng fo rces t h rough

D:

w

l

w

2

w^

Pxv

Qiv

?2V

QlV

1 4 9 . 5 x 1 . 5 0 =

224 .3 x 3.00 =

1 6 2 . 5 x 2 . 5 0 =

179.6 x 4.00 =

30.6 x 3.50 =

3.6 x 5.00 =

33.2 x 5.00 =

224.3

672.9

406.3

718.4

107.1

18.0

166.0

2 M i = 2 3 1 3 m - k N

M o m e n t s o f o v e r t u r n i n g f o r c e s t h r o u g h D:

Pm

Q\H

^ 2H

© 2H

125.8 x 4.67 =

2 1 . 5 x 5 . 7 5 =

12.0 x 0.83 =

1 1 0 . 7 x 1 . 2 5

=

2 M

2

=

=

587.5

=

123.6

= 10.0

=

138.4

=

859.5 m

•kN

The s a fe t y f ac t o r aga i n s t ove r t u rn i ng fo r t he cond i t i on o f an unde fo rm -

able foundat ion so i l then i s :

S M

2

2 31 3

F

r

= = = 2 . 6 9 - 2 . 7 > 1 .5 .

2 M

2

8 5 9 . 5

F

r

is qu i t e a b i t m or e than 1 .5 w hich is th e usual ly accep tab le va lue of th e

safe ty fac tor . In p rac t i ce , i t is n o t necessary t o con t ro l the o ver turn ing

s tab i l i ty safe ty fac to r i f th e res u l t a n t o f a l l fo rces a c t ing on the wal l , passes

t h ro ugh t he m i dd l e t h i rd o f t he fou nd a t i o n . Th i s r e s u l t an t s hou l d , how

ever , be as c lose as poss ib le to the foo t ing cen ter , when the sof tness o f the

foundat ion so i l increases .

(4 ) Safety factor against sliding

O f i n t e re s t now a re t he ho r i z on t a l fo rces . The ho r i z on t a l com ponen t F

H

of F ' mu s t be in equ i l ib r ium wi th th e f r i c t ion force ac t ing agains t th e

bo t t om o f t he foo t i ng .


26/262

PROBLEM 7.4

13

The general equation for the safety factor against sliding is:

aB + F

v

tan

5

F

s

= ~ ,

where

a =

adherence between soil and footing (|a|


27/262

14

RET AINING W ALLS

S o l u t i o n

We fi rs t m us t f ind th e earth pre ssu re coeff icie nt of th e fi ll w i th 5 / ^ = 1,

fJl


28/262

PROBLEM 7.5

15

When,

as a

first approximation,

the

soil

is

assumed

to be

homogeneous

and

of

unit w eight j

; one

finds:

?2 = WhlKy

= i x 11 x

2^50

2

x 0.546

=

18.8 kN

per m

length

of

w all).

Q

2

=

q'-h

2

- fc

aq

= 127 x 2.50 x

0.546

=

173.4 kN

per m

length

of

wall).

Conclusion

In

the

case

of an

inclined backfill

at j3 = 20° the

lateral forces increase

by

over 50%

in

comparison

to the j3 = 0°

condition.

irkProblem

7.5

Comparison

of

lateral forces

on a

vertical wall with horizon

tal backfill and different assum ptions (Boussinesq eq uilib

rium and graphical method

of Culmann)

Referring to the giuens of problem 7.1 (wall 4 m high), a vertical-face

(X = 0) dry

sand,

an horizontal backfill

fj3 =

0),


29/262

1 6 RET A INING W ALLS

Fig. 7 .10.

4 a x = 0.5 x 18.3 x 4.00 x 0.809 x 1.175 = 34.8 kN (per m length of wall) .

Then we have:

k

ay

Caquot-Kerisel

—

1 in this par ticula r case.

fe

a7

Culmann

**Problem 7.6

Detect ing errors made in the design of failing retaining

structures (ruptures, collapses, etc.) of reinforced concrete or

masonry

The five walls of Fig. 7.11 all failed. Can you identify the causes of these

failures ?

Solution

—

Wall 1. No calculation made. Footing width obviously too narrow. Failure

plane at contact face between sand and rock.

—

Wall 2. Insufficient drainage of t he fill mass and n o 'we ep ho les '. An angle

of inter nal friction of 20° indica tes a clayey soil, ther efor e on e which w ould

not easily drain.


30/262

PROBLEM 7.6

17

7ZW777777ZK

Li^J

Fig. 7.11.

1.20

;/;////////////////;////;;/,

v

S o M c l a y

©

—

Wall 3. Although 'weep ho les' are indicated , there is no indication of a

drainage blanket in the clay-fill behind the wall.

—

Wall 4. Steel reinf orce m ent placed on th e com pression side of the wall

stem, but no steel on the tension side, leading to ruptures in the wall.

—

Wall 5 . Failure due to dee p slip surface. The overall stability w as not

properly evaluated.


31/262

1 8

RET AINING W ALLS

**Problem 7.7

Diagram of stresses behind a gravity wall. Stratified soil and

water table. Uniformly loaded backfill

It is required to draw the distribution of horizontal stress comp onents

acting on the gravity wall of Fig. 7.12(a), knowing that:

- the inner wall face is straight and inclined 10° with the vertical.

- the backfill of the wall is horizontal a nd uniformly loaded by 20kPa

- the soil behind the wall consists of 4 distinctly horizontal layers having

the properties indicated on Fig. 7.12(a). The low ermost layer is partly sub

merged to ground water table.

q = 2 0 kPa

1 1

1 , 1 I I i *

f W M ^

(7) ^1=35°

v

-

y

1=

1&k

© „ .

I

A\

/ / ^ \ //AW/AW //'WN

N / n

f? \ Y>

2

=35°

W y ? = l 6 k N / m

3

y? = 1 6 kN /

^ 3 = 2 0 °

18 k N / m

3

A) ^ 4 = 3 5 °

^ 4 =1 6 k N / m

3

W a t e r

5

= 35°

5

= 11 k N / m

3

f h ( m )

1

r ( m )

( a ) G i v e n s o f t h e p r o b l e m

(b ) D i r e c t i o n o f l a t e ra l S t resses

Fig. 7 .12.

Solution

The required diagram is shown on Fig. 7.13. The details of the compu

tat io n are given in C ostet-Sanglerat vol. 1, sect. 6.2. 5. Fro m Fig 7 13 it is

possible to calculate th e safety factors against overturning an d' against

sliding, as described in problem 7.3, provided that the dimensions of the wall

are known.


32/262

PROBLEM 7.8

19

Fig. 7.13. Lateral pressure distribution.

+rk+Problem

7.8 The influence of drainage con dition s on the earth pressures

acting on a retaining wall

A retaining-wall, 5 m high, supports a horizontal backfill of cohesionless

sand (Fig. 7.14). The inner w all face is rough, so assume 5 =

\p

(assume

k

ay

= 0.308). The internal angle of friction of the sand

( . 4

S a t u r a t e d s a nd

I m p e r v i o u s s o i l

Fig. 7.14. Wall with submerged, undrained backfill.


33/262

20

RETAINING WALLS

Dra i n a ge b l a n ke t

I m p e r v i o u s n a t u r a l s oi L

Fig. 7.15. Wall with backfill over the drainage blanket.

royif^»^^finff^ fi»i»iff^^~

Dra inage b lanket

Satura ted sand

I m p e r v i o u s s o i l

Fig. 7.16. Wall with backfill against vertical drainage blanket.

Solution

(1) Dry-sand backfill

Js

27

1 +e

1 + 0.53

17.6 kN/m

3

The unit-weight of sand is: y

d

The lateral earth pressure is:

P =

\k

a

-f

d

H

2

= \ x 0.308 x 17.6 x 5

2

= 67.8 kN:

^hor. = 67 .8 x cos 30° = 58.7 kN; P

v e r t

. = 67 .8 x sin 30° = 33.9 kN.

(2) Both wall and backfill are completely submerged

Here, the submerged or buoyant soil unit-weight must be used:

7h = Id +

eju

1 7 . 6 +

5.3

= 21.1 kN/m

2

1 + e 1.53

7 = Ih-Jw =

2 1 . 1 - 1 0 = 1 1 .1 k N /m

3

.

The lateral earth pressure is:

P =

hk

a

y H

2

= \ x 0.308 x 11.1 x 5

2

= 42.7 kN:

P

h o r

. = 42.7 x cos 30° = 37 kN; P

v

42.7 x sin 30° = 21.4 kN.

In this case, the hydrostatic pressures act on both sides of the wall, but in

opposite directions and therefore cancel themselves.


34/262

PROBLEM 7 .8

2 1

(3) Backfill alone is submerged

To t he ca l cu l a t ed bu oy an t s o il p re s s u re m u s t now be add ed t he hyd r o

s ta t i c p ressure : P

w a t e

r = y

w

H

2

/2 = 10 x 5

2

/ 2 = 1 25 kN . T hus :

hor .

= 37 + 12 5 - 16 2 kN ; P

v

21 .4 kN .

(4) Backfill saturated and drained through a sloping drainage blanket (Figs.

7.15, 7.17)

In th i s ca se , as m ay occ ur w he n a heav y ra in fa ll s do w n, in th e backf i l l

the re co me s to e xi s ten ce a f low net as show n on Fig . 7 .1 7 , wh ere f low

l ines a re ver t i ca l and equipo ten t i a l l ines a re hor izonta l . Assuming tha t

th e dra ina ge b la nk et is n o t ' l o ad ed ' , th e por e-w ater p ressures in i t a re zero

as on th e f ree ho r izo nta l sur face . Th eref ore th e por e pressure is zero

th ro ug ho ut th e backf i l l . T he ca lcu la t ion is th e same as fo r th e case of the

dry-san d backf i ll if we rep la ce th e dry un i t -w eigh t by th e sa tu ra te d un i t -

weigh t .

Flow L ines

h : 5 m

4 m

Equipotent ia ls \

3 m

2 m

1 m

Ze r o p o r e p r e s s i o n

in the dr a in ag e b lanke t

Fig. 7 .17. Flow-net due to heavy rainfall over the backfil l with a sloping drainage blanket.

Thus :

P =

?k

a

y

h

H

2

= \

x 0 .3 08 x 21 .1 x 5

2

81 .2 kN :

P

h o r

. - 81 .2 x cos 30° = 70 .3 kN ; P

v e r t

. = 81 .2 x s in 30° = 40 .6 kN .

(5) The backfill is saturated, but now d rained through a vertical drainage

blanket

Fo r th i s s i tua t ion , the f low n et i s sho wn on Fig . 7 .18 . I t is impo ss ib le to

g ive a s i m p l e m a t hem at i ca l s o l u t i on . Fo l l ow i ng t he m e t ho d o f Cou l om b

severa l so i l wed ges are t es ted in o rder t o f ind on e wh ich y ie lds the m axim al

la tera l ear th p ressure . In each case , pore-water p ressure mus t be evalua ted

along the fa i lu re p l an e , and the res u l t a n t p ressu re m us t be ca lcu la ted by

g raph i ca l s o l u t i on , fo r e xam pl e . Th i s po re -wa t e r p re s s u re m us t be t ake n i n t o


35/262

22

RET AINING W ALLS

account in the equilibrium of Coulomb's wedge to calculate the lateral earth

pressure.

Because there exist pore pressures all along the boundary, the lateral

pressure with a vertical drainage blanket will be larger than that of a sloping

drainage blanket.

Fig. 7 .18 . Flow -net for rainwa ter draining with a vert ical bl ank et.

Fig. 7.19 shows a graphical method to determine pore pressures for a soil

wedge whose boundary conditions correspond to an angle of 45° with the

horizontal .

Consider an equipotential line, such as NM

9

where the loads at N and M

are equal (h

N

= h

M

). On the other hand, the pore pressure at N is zero (no

hydro static head in th e drainage blanke t). We then have:

h

M

— u

M

/y

w

4- z

M

, h

N

— z

N

where z *

u

M

hu

from which each point in the diagram can be analyzed. The resultant of

the pore pressure is U = 60.7 kN. T he equilibrium state of th e soil wedge

(Fig. 7.20) is then calculated as follows:

W = 4 x 5

2

x 21.1

Fur thermore , P =

= 263.8 kN.

W - E 7 c o s f l ) t a n ( f l - i ) +

U

sin

6

sin 5 tan

(9

—


36/262

PROBLEM 7.8

23

P o r e w a t e r

p r e s s u r e d i a g r a m

P o i n t s u A L u

m

A L

10 kPa

m

0 0 0 . 3 0 0 . 0 6

1 0.4 0 .4 0 0.2 4

2 0 . 8 0 . 55 0 . 50

3 1 0 . 6 0 0 . 66

4 1.2 0 .8 0 1.00

5 1.3 0. 9 5 1.19

6 1.2 1.4 0 1.47

7 0.9 2 .10 0. 95

8 0 — —

U = 6 .07-10

KN

M

c alculate d a t the center of each segm ent of length A L )

Fig. 7 .19. Resultant of forces due to pore pressure for 6 — 45°

P

A

W - 2 6 3 . 8 k N

U = 60 .7 k N

Fig. 7.20. Graphical determination of lateral earth pressure.

w he re: 5 =


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24

RET AINING W ALLS

The p roc edu re is r epe a t ed fo r o t he r va l ues o f 0 and wil l resul t in the curve

of Fig . 7 .2 1 , wh ich g ives P as a fun ct ion of 0 , reach in g a m axim al va lue for

6 = 45° . The va lue ca lcu la ted above i s the one for which the wal l should be

des igned .

Conclusion

Thi s p rob l em i l l u s t r a t e s c l ea r l y on t he one hand , t he i m por t ance o f

prov id ing a d ra inage for a backf i l l sub jec t to sa tura t ion and , on the o ther ,

the in f luence of the type of the dra inage . The va lue of the pressures increases

as fol lows:

— wa l l and backfi ll co m p l e t e l y s ubm erged P — 42 .7 kN

— dry backf i l l P= 6 7 . 8 k N

— satu ra te d backf i l l wi th s lop ing b la nk et

P =

81 .2 kN

— satu ra te d backf i l l wi th ver t i ca l b lan ke t

P =

1 0 2 . 1 k N

— s a t u ra t ed backf il l w i t h ou t a b l an ke t P = 1 6 2 . 0 kN

102.7 kN

Fig. 7 .21. Variation of P as a function of 6.

ick+Problem 7.9

Analys i s o f the failure of a reinfo rced con cre te retaining-wall

Corrective measure by using rock anchors

A reinforced concrete retaining-wall along a motorway consisted of 21

elements each 6 m in length. Shortly after construction, the wall failed:

several elements were pushed over and in others h ad developed large d iagonal

cracks. It was observed that most of the drain holes in the wall were plugged

up. The wall dimensions are shown on Fig. 7.22.

A review of the construction procedures showed that the excavations for

the wall had been done under adverse conditions:


38/262

PROBLEM 7 .9

25

—

already during the excavations numerous seepages had been observed in

the cuts;

— the graded filter material specified for the drainage blanket had not been

used,

but was replaced by excavated material;

— the wall footings were not bearing on solid rock, in particular not at the

toe.

(1) Analyse the wall stability a nd explain the observed failures.

(2) Recomm end a repair method by tie rods (2 rows) anchored in rock.

The backfill material properties were: ^=34° , y

h

= 19kN/m

3

, y' =

llkN/m

3

. The backfill behind the wall was replaced at an angle

jS

= 34 °

with the horizontal. A ssume that the reinforced concrete unit-weight was

23kN/m

3

and the angle of friction between concrete and rock was 8 — 30° .

Solution

(1) Analysis of wall stability

Assum ptions used for calculation

Because of the poor quality of the drainage material, the calculation must

consider th e h ydr ostatic pressure (assuming tha t the water level is at th e to p

of the wall).

The earth pressure at th e heel of the wall is considered as non-ex istent

since it is encompassed in the rock. However, the hydrostatic pressure acts

3.70 m

Fig. 7 .22. Failed wall .


39/262

26

RET AINING W ALLS

along BC because the rock is fractured. The passive pressure at the toe of the

wall also may be overlooked (poor-quality rock).

Assume that the backfill volume EDCF is part of the wall weight (Fig.

7.22). Thus the lateral and water pressures on the fictive surface

BCF

which

act on the wall and volume of the soil EDCF must be calculated.

To determine the overturning stability, bending moments can be calcu

lated with respect to A (at the toe), because the foundation soil can be

assumed to be rigid (rock) and the center of rotation will be at point A. It

also can be assumed that a limit Rankine-equilibrium condition exists on

plane CF.

The stress tensor at depth h may be represented by a Mohr's circle as

shown on Fig. 7.23. The pole of this circle can be easily constructed and

then the failure lines of Rankine equilibrium can be drawn (Fig. 7.24).

Fig. 7 .23.

Stability calculation

The first failure line which intersects the wall is the line

CF.

Thus the

Rankine equilibrium condition will be modified only between the back face

of the wall and

CF

plane. It is therefore justified to calculate the Rankine

earth pressure acting on plane CF. In practice, the lateral earth pressure

calculated as above, is slightly overestimated because the critical failure

wedge intersects the rock zone which cannot slip.

The stress acting on a vertical face along CF is jh cos

]3

(see Fig. 7.2 3).

It is inclined up w ards a t an angle |3 = 34° with th e norm al to CF. Thus we

get the earth pressure coefficients:

hor izontal :

fc

ah

= cos

2

/3 = cos

2

(34°) = 0.687


40/262

PROBLEM 7 .9

/? = = 3 4

Fa i Lu re L ines o f

Rank ine equ i l i b r iu r r

Fig. 7 .24.

vert ical :

k

av

= co s j3s in j3 = cos (34 °) s in (34 °) = 0 .4 63

Over turn ing s tab i l i ty and s l id ing s t ab i l i ty a re s tud ied us ing resu l t s o f

Table 7B (ca lcu la t ions are ma de pe r one m ete r o f wall l eng th) . The s t ab i l i z ing

m om en t s a r e a s s um ed t o be pos i t i ve and t he ove r t u rn i ng m om en t nega t i ve .

Upl i f t pore p ressures a re d i s regarded .

TABLE 7B

Forces (kN) (per m leng th)

Lever arm

a b o u t

A

(m)

3.20

3.20

2.45

1.1

3.7

2.77

M o m e n t

A

(k N • m )

(per m length)

+ 403

+ 36.8

+ 183

+ 27.8

+ 419

- 4 6 5

Weight of concrete and soil

Pi

= 6 X 1 X 2 1 = 1 2 6 kN

p

= 0 . 5 X 1 X 23 = 11.5 kN

p

- 0 . 5 X 6.5 X 23 = 74.7 kN

p

4

= 0.5

X

2.2 X 23 = 25 .3k N

Earth p ressure on

CF

p

v

= | x l l x (6 .6 7 )

2

X 0 .4 6 3 = 1 1 3 k N

p

H

= h

X 11 X (6.67 )

2

x 0.687 = 16 8 kN

Earth p ressure on

BF

Pw a te r = 2 X 10 X ( 7 . 1 7 )

2

= 2 5 7 k N

2 . 3 9 - 6 1 4 . 2

Overturning stability:

2(M/A>0)

F

R

=

1 0 7 0

X(M/A < 0 ) 1 0 7 9

= 0 .99 .


41/262

2 8

RET AINING W ALLS

Th e safe ty fac tor aga ins t ov er tu rn in g is l ess th an 1 and thu s ov er tu rn in g is

a cer ta in ty . In add i t ion , because the foo t ing does no t bear en t i re ly on so l id

rock , t he cen t r e o f ro t a t i on w i l l s hove back t o a po i n t unde r t he foo t i ng

ins tead of to p o in t A , and th i s in tu rn w i ll s t il l decre ase th e safe ty fac to r .

Sliding stability

Th e sum of th e ver t i ca l fo rces is equ al to 3 50 .5 kN (per m of wal l l eng th) ,

th e sum of the hor iz on ta l fo rces is 42 5 kN . Th e angle o f f r i c t ion be tw ee n th e

conc re t e and t he rock i s 30° , t he re fo re : F

G

= 3 5 0 . 5 x t a n ( 3 0 ° ) / 4 2 5 = 0 . 4 7 :

the wal l would also fai l in s l iding.

To c on clu de , the l ack of d ra inag e b eh i nd the wal l causes i t to be un s ta b le

and crea tes two modes of fa i lu re , namely by over turn ing and s l id ing .

Th e va r ious wa ll pane l s un de rw en t i m p or t an t d i s p l ac em en t s o f va ry i ng

m a gn i t ud es a s a con s equ enc e o f t he b ed ro ck qua l i t y . Th i s caus ed t he pane l s

t o i n t e rac t w i t h each o t he r wh i l e t hey , t heo re t i ca l l y , were s uppos ed t o ac t

ind ep en de nt ly o f each o the r . S ince no re info rc ing w as des igned to res i s t th e

bend i ng m om en t s , c r acks deve l oped i n t h e ou t e r face of t he pane l s .

Remark

A s s um i ng t h a t a p rop e r d ra i nage had been i n s ta l l ed and t ha t t he roc k was

sound , the fo rces ac t ing on the wal l would have been (per meter o f wal l

l eng t h ) :

p

v

= \ x 19 x (6 .6 7 )

2

x 0 . 4 6 3 = 1 9 6 k N

p

H

= $ x 19 x (6 .6 7)

2

x 0 .68 7 - 29 0 kN

p

x

= 19 x 6 x 1 = 1 1 4 k N

M o m e n t w i t h r e s p e c t t o A due t o p

v

= 7 22 kN • m

M om en t w i t h r e s pec t t o A due t o p

H

= 80 3 kN • m

M o m e n t w i t h r e s p e c t t o A due t o p

x

= 36 5 kN • m .

The safe ty fac tor aga ins t over tu rn ing

is: F

R

=

13 35 /8 03 = 1.7 ( accep t ab l e )

and the coef f i c ien t aga ins t s l id ing would have been :

F

G

=

4 2 0 t a n 3 0 ° / 2 9 0 =

0 .8 4 , which i s to o low.

Th e des igner p rob ab ly assum ed the pres enc e of a pass ive pres sure a t th e

t o e .

If the ro ck the re ha d b een so un d , s l id ing cou ld no t oc cu r . Th e er ror s in

th e des ign cons i s t ed of : (1 ) unrea l i s t i c appra i sa l o f the ro ck qu al i ty ; (2 ) a

poo r cons t ruc t i on p rac t i ce ( f au l t y d ra i nage b l anke t ) .

(2) Corrective measures

Th e f i rs t s t ep to repa i r wou ld be , as fa r as poss ib le , to imp rove t h e dra inage

of the backf i l l by c lear ing ou t the p lugged up dra in ho les and by adding a

dra inage b lanket . I f th i s would no t be poss ib le , i t would be requ i red to se t

up fo r t h e fo r t i f ica t i on a ca l cu l a t i on , t ak i ng i n t o acco un t t he wa t e r p re s s u re

ac t ing on the wal l .


42/262

PROBLEM 7.9

29

For instance, two rows of rock anchors may be placed, one located just

above the footing, to prevent sliding, and the other at height z above the

base.

Each row of anchors is assumed to equal a tension T (per m length of

wall).

To realise a safety factor of 1.5 against sliding and overturning, we

would have:

F

G

= (201 + 2T )/ 42 5 = 1.5,

or T = \ [ (1 .5 x 425) - 2 0 1 ] = 218 kN

and F

R

= (107 0 + T x 0.5 + T

x

*)/1 07 9 = 1.5

from which:

z= [ (1 .5 x 1 0 7 9 ) - 1 0 7 0 - ( 2 1 8 x 0 .5 )] / 2 1 8 : assu me z = 2.

In this calculation, it is assumed that the placement (thus: the tension) of

the anchors did not alter the magnitude of the earth pressures. This corrective

method only seeks to avoid further failures and not to replace the wall to

its original design position. (This would engender passive pressures.)

The calculation neither did account for the poor rock quality at the toe of

the wall. It is therefore not possible to determine the point of rotation. This

unknown is partly taken care of by seeking a design yielding a safety factor

of 1.5 which can be dangerous. It could also be taken care of by increasing

R e i n f o r c e d

p a n e l

3.00 m

Fig. 7.25. Remedial methods of support.


43/262

30

RET AINING W ALLS

the load which the upper anchor is designed to take or by increasing the

height of the row.

To conclude, the remedial measure for the wall (Fig. 7.25) would be:

—

placing a reinforced conc rete panel against th e cen ter wall face;

—

installing tw o whalers located just above the footing and 3 m above the

base,

respectively;

—

installing two anchor lines deriving their tension in the bedrock and on

the whalers, designed to withs tand a tension of 2 18 kN per m of length

of wall.

+++Problem

7.10 Design of a reinforced-earth retaining-wall with horiz onta l

backfill

A motorway is planned to cross an unstable slope as shown on Fig. 7.26.

It is proposed to construct the pavem ents on engineered fill placed over the

unstable a reas and to support the fill by a retaining-wall.

Two solutions are being considered, one with a conven tional reinforced-

concrete wall, the other w ith a reinforced-earth structure.

(l)List the conditions favorable for the choice of a reinforced-earth

design.

(2) In a general manner, wha t are the problem s that could affect the

performance of such a structure?

(3) The height of the reinforced-earth wall must be H = 20 m. Assume the

wall thickness to be L = 0.8 H (generally accepted value).

Backfill and fill of the wall consist of the same material whose unit-weight

is 18kN/m

3

and angle of internal friction


44/262

PROBLEM 7.10

31

The technology of the wa ll surface elements imposes the following added

restrictions: The strip layers are laid 0.25 m apart. The reinforcement can

only be attached every 50 cm to the w all panels.

Design the wall to meet the safety-factor criteria. Assum e all backfill and

fill to be

sand.

The following assumptions are necessary to determine the internal stability

of the wall:

— the principal stresses near the wall skin are horizontal and vertical;

—

the vertical stresses in the wall mass along any line of elevation h is uniform

over a width L

—

2e, where e is the eccentricity of the resultant of the forces

acting at that elevation. The coefficient of friction between the soil and the

reinforcement is 0.2.

Because of the spacing of the tie points between reinforcing and wall skin

(every 0.50 m), it is necessary to attach a larger number of strips than

strictly required. Calculate the corresponding safety factor wh ich, in any

event, cannot be less than 1.5.

Solution

(1) The stability of reinforced-concrete walls would have been very diffi

cult to guarantee because they would have imposed heavy, concentrated

loads on the foundation soils. It would have been necessary to anchor the

foundation into the underlying bedrock. Small movements in the unstable

soils above would have sheared the anchors. Retaining-structures of rein

forced earth, however, can be supported directly by unstable masses because

they can withstand small deflections.

(2) There are basically 3 typ es of prob lem s related to the stability of a

reinforced-earth wall:

(a) The overall wall-mass stability of the slope. This problem is the same

as that encountered with reinforced-concrete walls. It can be analyzed by

the 'circular slide' method. For the present problem, it is assumed that this

overall stability has already been assessed.

(b) The wall stability under the lateral pressure of the fill. This is similar

to the classical retaining-wall problem (external stability).

(c) The prob lem of internal stability of the wall th at de termin es th e

dimensions and the spacing of the reinforcements.

(3 ) External stability. Since we have assumed th at th e wall has an overall

stability, let us look at its external stability.

Assuming a Ran kine eq uilibrium state behind th e wall, th e earth pressure

on the vertical face is horizontal. We then have:

Active pressure: P = k

a

•

y(H

2

/2)

V =

35° , 5 = 0,

k

a

=

0.27

P =

0.27 x 18 (20

2

/2 ) - 970 kN .


45/262

32

RET AINING W ALLS

The resultant of the forces applied to the foundation of the wall will have

the following components:

horizo ntal = 970 kN, vertical = y-H-L = 5760 kN.

The resultant will act at a distance e from the center of the footing so that

e = (970 x 6 .6)/5750 = l . l m .

Since the resultant falls within the middle third of the wall footing, the

wall is safe against overturning.

If we assume a coefficient of friction of 0.3 between the wall and the

foundation soil, the safety factor against sliding will be:

(5750 x 0.3)/970 = 1.8, which is satisfactory.

A failure through punch is not likely because of the relatively high angle

of internal friction of the foundation soil, 0 = 35°. The external stability of

the wall is satisfactory.

(4 ) Internal stability

(a) Tension in the reinforcem ent. For de termining the internal stability,

we have to consider the tension stresses in the reinforcements and the length

of the reinforcing elements. As for the tension stresses, we must first evaluate

the vertical stresses acting at a depth of h from the top of the wall (Fig.

7.27). The vertical stress is due to the overburden above h and to the earth

pressure of the fill being retained. The resultant of the forces applied at this

level has the following components (per m length of wall):

R

v

= W = yhL along th e vertical,

R

h

= P = k

a

y(h

2

/2) along the horiz onta l,

the eccentricity of the resultant is:

_

k

a

h

2

yh/3 _ k

a

h

2

2yhL 3 x 2 x L

L =16m


46/262

PROBLEM 7.10

33

In accordance with Meyerhof's hypothesis, we assume that the stress

distribu tion is uniform over a wid th L — 2 e. The magnitude of the vertical

stress o

v

is:

_

yhL _ yhL

°

v

~ L-2e ~ L-k

a

(h

2

/3L)

If we assume that the soil at the contact with the wall skin is in a state of

active pressure, then the horizontal stress is: o

h

= k

a

o

v

. If we now assume

that the l ine of reinforcement at depth

h

is designed to withstand all the

horizontal stresses at that level and above height

Ah,

the tension in the rein

forcing elements must be equal to:

Ah

T = o

h

x Ah = k

a

o

v

Ah = k

a

yh :

l-(l/3)k

a

(h/L)

2

for a wall length of 1 m.

The maximal reinforcing tensions will occur at the bottom of the wall.

The tensions there are:

h = H = 20 m, L = 16 m, y = 1 8 k N / m

3

, k

a

= 0.27, Ah = 0.25 m.

Thus T - 0 .27 x 18 x 20 x 0. 25 /[ l - (0 .27/3 )(20/1 6)

2

] = 28.3 kN

Fo r a length of 1 m of w all, it will be necessary to design the reinf orce m ent

to withstand a tensile stress of 28.3 kN.

The cross-sectional area of each reinforcing element, assuming each to be

6 cm wide an d 2 mm thic k, and taking into acc ou nt the safety factor, will

be: ^(6 x 10~

2

x 2 x 10~

3

) = 6 x 10~

5

m

2

. Each element can resist a tension

of: 6 x 10

- 5

x 250 x 10

6

= 15 kN.

At the bottom of the wall, one element will have to be placed every 50 cm.

Let us now compute the height h, from which o nly o ne elem ent per m will

be required.

We have: 15 = 0.27 xl8xh

x

x 0 .25 / [ l - (0.27/3)(h

1

/16)

2

] .

For h less than 16 m, the te rm (0.27/3)(/z

1

/16)

2

may be neglected. A

simple calculation leads to: h

x

= 12.0 m.

One reinforcing elem ent per m will suffice for a height of 8 m upw ard and

one elem ent every 2 m from 14 m and up .

The tens ion diagram for the entir e wall, 1 m in length is show n on

Fig. 7.28.

The safety factor obtained with this design is:

F = SLYeaABCDEFGO /aieaACEHO:

Area ABCDEFGO = 30 x 8.5 4- 15 x 5.5 + 7.5 x 6 = 255 + 83 + 45 = 383,

the area ACEHO varies little from the area of triangle AHO or: 30 x (20/2) =

300.

Therefore the safety factor is:

F =

38 3/3 00 = 1.28.


47/262

34

RET AINING W ALLS

1 E l em en t ev e r y 2 m e t e r s

1 E l em en t ev e r y m e t e r

Tens ion fo rces ins ide the wa l l

„ C a pa c i t y o f t en s i on

f o r c es i n e l em en t s

H 30

kN/mL

Fig. 7 .28.

This coefficient is to o low. To increase it, it is for instance possible to

place strips at 0.50 m intervals half-way up the wall and at meter intervals in

the upper part of the wall.

The safety factor becomes:

F =

AIJKG

_ 30

x

10 + 15

x

10 _ 450

AOH

300 ~ 300

= 1.5,

which is acceptable.

(b) Length of reinforcing elements. The vertical stress o is very close

to the value of yh. Iff is th e coefficient of friction be tw een th e soil and the

reinforcing element and b is the element w idth, the adhesion requirem ent is:

T

k

n

Ah

2bfyh 2bfn

where n is the number of elements per meter.

Usually the value of 0.2 is assumed for the friction coefficient between

soil and strip, and thus:

L

n

>

0.27 x 0.25

2 x 6 x 10"

2

x 0.2 x 1

= 2.8 m

This length should be added to the width of Coulomb's edge, at the elevation

h,

i.e.:


48/262

PROBLEM 7.10

35

Fig. 7.29.

L

c

= ( t f - ^ t a n ? - |

fo r

h =

0:

(L

c

)»

=

10.4 m.

Therefore, the maximum length of the reinforcing elements is : 2.80 + 10.40 =

13.20 m. This condition is fulfilled since the wall's thickness is 16 m.

Remark

The assumptions made above to calculate the tension in the reinforcing

strips are somewhat arbitrary and other assumptions could be made, in

particular for the calculation of the tensions in the strips (see Fig. 7.29),

where a trapezoidal vertical stress distribution is assumed over the width of

the wall. Force F is the vertical component of the resultant force applied at

the level h, and is in equilibrium with the stresses of trapezoidal distribution:

F = ~ T ;>

F

*

e

= h(o

a

-o

h

)Lx{\L-\L)

2(o

1

+ o

2

)

or:

F-e = (o

a

-o

b

)L

2

/12

which gives: o

a

4- o

b

= 2F/L, o

a

— o

b

- 12F-e/L

2

from which: o

a

= (F/L)(l + 6e/L), o

b

- (F/L)(l - 6e/L).

The vertical component of the resultant, F, is yhL, and its eccentricity:

e = k

a

h

2

/6L, from wh ich:

o

a

= yh[l+k

a

(h/L

2

] and o

b

= j

•

h[l - k

a

(h/L)

2

].

If w e assum e, as we did abov e, th at th e soil pressure against th e skin is active,

then the horizontal pressure:


49/262

36 RETAINING WALLS

OH =k

a

O

v

=k

a

O

a

=k

a

j - f t [ l + k

a

(h/L)

2

],

whi le the maximum s t ress a t the base of the wal l i s :

k

a

-yH[l + k

a

(H/L)

2

]

The tens ion in a row of s t r ips a t the wal l base , per meter l eng th of wal l i s :

T _ =

k

a

y-H-AH[l + k

a

(H/L)

2

]

m a x

= 0 . 2 7 x 1 8 x 2 0 x 0 . 2 5 [ 1 + 0 . 2 7 ( 2 0 / 1 6 )

2

] = 3 4 . 5 5 k N .

The m ag n i t ud e found on t he bas is o f t h i s a s s u m pt i on , t he re fo re is ab ou t 20%

hi ghe r t ha n i n t h e p rev i ous a s s u m p t i on o f un i fo rm s t r e ss ove r w i d t h L — 2e:

T

m a x

= 2 8 . 3 k N .

Calculation of the tension force in the reinforced strips by the method of

'Coulomb's wedge'

This method cons i s t s in cons ider ing the t r i ang le o f re in forced ear th

b o u n d e d b y t h e p o t e n t i a l r u p t u r e p l a n e s AC pass ing th ro ug h the basi s o f th e

wal l (Fig . 7 .30) .

I t is assum ed in th e m et ho d th a t th e so i l be tw ee n th e s t r ips i s in a p las t i c

equ i l i b r i um a l ong t he po t en t i a l f a i l u re p l ane .

The forces ac t ing on the pr i sm are :

— t he we i gh t o f t h e s o i l we dge : W = \yH

2

co t 0 ;

— r e a c t i o n

R

of the soi l on plane

AC

(th i s rea c ta n t is inc l ined by an angle

y

wi t h r e s pec t t o t he no rm a l

AC);

— t h e t o t a l t en s i o n fo rce T

t

i n t h e s t r i p s a t t he d i f f e ren t i n t e r s ec t i on p o i n t s

w i t h AC (th i s fo rce is ho r iz on ta l ) .

The equ i l i b r i um o f t he t h ree fo rces r equ i r e s t ha t :

T

t

= \yH

2

c o t 0 • t a n ( 0 - ^ )

T

t

is a fun ct ion of ang le 0 . Thi s fun ct io n has a m ax im um for dT

t

/dd = 0 ,

w hi ch g ives 0 = - 4- - ,

B

m:mm

: 4 - : • / ' i ' i i • '

(*;•. w

T t

'./ " ur

A

9

Fig. 7.30.


50/262

P R 0 B L E M 7 . i l

37

from which:

T

u

= - t a n

2

U 2/

yH

2

= -KlH

2

One more assumption can be made regarding the distribution of tension

forces in the reinforcing strip, nam ely tha t it is triangular. The forces th at

undergo the greatest tension are the ones located near the base of the wall.

For a wall length of l m , th e tension in the bo tto m row of strip is:

T = k

a

yHAH = 0.27 x 18 x 20 x 0.25 = 24.3 kN.

This tension is abou t 15% less than th at calculated with the assum ption

of a uniform stress distribution over the width L

—

2e: T = 2 8.3 kN.

+++Problem

7.11

Design of a reinforc ed ea rth retaining-wall with a reinforced

con crete skin and a sloped, surcharged backfill

The dimensions of the reinforced earth structure are shown on Fig. 7.31,

and the skin consists of reinforced concrete slabs of the type shown in

Fig. 7.32. The internal angle of friction of the fill is 35° , its unit-weight is

7 !

=

20kN/m

3

. The horizontal portion of the backfill supports a uniformly

distributed load of 10 kPa.

H' (m)

H ( m )

m f n T n U m U

m t m m

d 0.75

C o n c r e t e ,

sk in

Z (m)

- ' r

7

2

(p2

B (m)

Fig . 7 .31 .

http://pr0blem7.il/http://pr0blem7.il/


51/262

38

RETAINING WALLS

1.50

1.50 m

LA

Fig. 7.32.

i i

i

3 7 - cm

Minimum s lab dimensions

1.50 x 1.50 m (2.25 m

2

Thickness E= 1 8 , 22 or 2 6cm

The reinforcing strips are smoo th and mad e of galvanized steel, 60 x 3 and

80 x 3 mm. For considering corrosion loss, assum e thickness of the strips to

be 2 mm . The allowable tensile stress o'

a

is less or equal to 2/3 of the elastic

limit

or 1.6 x 10

s

kPa.

To calculate the strip spacings, assum e the gross cross-section, taking into

accoun t that experimental results on models and full-size test wa lls indicate

that the

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[] Practical Problems in Soil Mechanics and Foundation