Retaining Wall Note

III

8.14 Retaining Walls with Metallic Strip Reinforcement

413

Calculation of Active Horizontal an d Vertical Pressure

Figure 8.29 shows a retaining wall with a granular backfill having a unit weight of y , and a friction angle of 4J;. Below the base of the retaining wall, the ill situ soil has been excavated and recompacted, with granular soil used as backfill. Below the backfill, the in situ soil has a unit weight of Y2' friction angle of 4J;, and cohesion of A surcharge having an intensity of q per unit area lies atop the retaining wall, which has reinforcement ties at depths z = 0, S v, 2S v , . . . , N S v. The height of the wall is N S v = H . According to the Rankine active pressure theor y (Section 7.3)

c;.

a'.a = a'.0 K a - 2 C t-vV ~ K Ka u~ = Rankine active pressure at any depth z. For dry gran ular soils with no surcharge at the top, c' = 0, u~ = y,z, and K; = tarr' (45 - 4JU2) . Th us,

where

u~(I ) =

y ,zKa

When a surcharge is added at the top , as shown in Figure 8.29,

I+- b' -.j+ a' -.J

qlunit area y

A

1

of Braja M. Das.

/

Sand

/

SV

/

Z

/

r

(a)

Sv

/

l,

) /(.

Ie

/

'YI H

Sv

/ /

/

Sv

/

1

Sv

(b)

iT~(I )

-

G

Sv-.J

8

, iT"

K,,'YIZ

re skin (Courtesy of Figure 8 .29 Analysis of a reinforce d-eart h retaining wall

(8.3 1)

III II

41 4

Chapter 8: Retaining Walls

(8.32) Due to the surchar ge

= "Y IZ

Due to soil only

The magnitude of (T~(Z ) can be calculated by using the 2:1 method of stress distribution describ ed in Eq . (5.14) and Figure 5.5. The 2:1 method of stress distr ibution is shown in Figure 8.30a. According to Laba and Kennedy (1986),

(for Z :::; 2b')

(8.33)

and

qa'

(T;J(Z) = _ _CO-_ _

(fo r Z

a' + ~ + b'

> 2b')

(8.34)

2

Also, when a surcharge is added at the top, the lateral pressure at any depth is

Ka"YIZ

=

Due to soil only

(8.35)

Due to the surcharge

I+-b'~a'~

tt

t t t t t q/unit area

q/unit area

\

\

11 H

\

L

Sand

H

\

Yl;4>;

\

\

f - - - - - - -\Reinforcement ---r---

strip

(a)

- ' -_

_

c

Reinforcement strip ----'-

_

(b)

Figure 8.30 (a) Notation for the relationship of (T~(2) in Eqs . (8.33) and (8.34); (b) notation for the relationship of (T~(2) in Eqs. (8.36) and (8.37)

415

8.14 Retaining Walls with Metallic Strip Reinforcement

(8.32

According to Laba and Kennedy ( 1986),

U~(2) =

U~ (2 )

M[;

may be expressed (see Figure 8.30b) as

(f3 - sinf3 cos 20') ]

1"

of stress distribustribution is shown j

(8.36)

(in radians)

where (8.33

M

(8.34 1

=

OAb' > I 104 - O.14H -

(8.37)

The net active (latera l) pressure distribution on the retaining wall calculated by using Eqs. (8.35), (8.36), and (8.37) is shown in Figur e 8.29b.

Tie Force The tie for ce per unit length of the wall developed at any depth z (see Figure 8.29) is

pth is

T

= active earth pre ssure at depth z X area of the wall to be suppo rted by the tie = (u:/) (5 v5 H )

(8.35 )

(8.38)

Factor of Safety against Tie Failure The reinforc ement ties at each level, and thus the walls, could fail by either (a) tie breaking or (b) tie pullout. The factor of safety against tie breaking may be determined as

area FS (B) =

yield or bre aking strength of each tie . f . . maximum orce III any tie

ux];

(8.39)

u~5v5H

where

w

= width of each tie t = thickn ess of each tie f y = yie ld or breaking strength of the tie material

(b )

notation for

A factor of safe ty of about 2.5 to 3 is generally recommended for ties at all levels. Reinforcin g ties at any de pth z will fail by pullout if the frictional resistance developed along the surfaces of the ties is less than the force to wh ich the tie s are being subj ected. Th e effective length of the ties along which frictional resistance is developed

416

Chapter 8: Retaining Walls

may be conservatively taken as the length that extends beyond the limits of the Rankin e active f ailure zone, whic h is the zone ABC in Figure 8.29. Lin e BC make s an angle of 45 + 4>;;2 with the horizontal. Now, the maximum friction force that can be realize d for a tie at depth z is (8.40) where

l, 0":,

= effective length = effecti ve vertical pressure at a depth z

4>~

= soil-tie friction angle

Thus, the factor of safety again st tie pullout at any depth z is

F

TR

FS ( p) =

(8.4 1)

Substituting Eqs. (8.38) and (8.40) into Eq. (8.41) yields

(8.42)

Total Length of Tie The total length of ties at any depth is L = I,

+ l,

(8.43)

where

I, l,

= length within the Rankine =

failure zone

effective length For a given

FS (p )

from Eq. (8.42), FS (p)O"~SvS H

1 = - - -- e

2'W0"~ tan4>~

(8.44)

Agai n, at any depth z.

I I

=

(H - z) tan ( 45

+

~;)

(8.45)

So, combining Eqs. (8.43), (8.44), and (8.45) gives

(H - z) L = ---'-- - -'-----4>') tan ( 45 +

--f

FS (P)O"~SVSH

+ - -'----- - -

2'W0"~ tan 4>~

(8.46)

417

8.15 Step-b y-Step-Design Procedure Using Metallic St rip Reinforcement

its of the Rank ine lakes an ang le of It ca n be reali zed

(8.40 )

_ _ Step-by-Step-Design Procedure Using Metallic Strip Reinforcement Following is a step-by-step procedure for the design of reinforced-earth retain ing walls.

General Step 1. De te rmi ne the height of the wall , H, and the properties of the granul ar backfill ma teri al, such as the unit wei gh t (-y ,) and the angl e of frict ion

(

and the req uired valu e of FS(B) and

FS(p)(8.41 )

Interna l Stability Step 3. Assume values for horizontal and vertica l tie spacing. Also, ass ume the wid th of reinforcing strip, w, to be used .

Step 4. Calcula te CT~ from Eqs. (8.35) , (8.36) , and (8.37). Step 5. Calculate the tie for ces at vario us level s from Eq . (8.38) . Step 6. For the know n values of FS(B)' calculate the thickness of ties, t, required to (8.42 )

resist the tie breakout:

T

=

CT~SVSl/

un]; =

FS(B)

or (8.43 )

t

(8.44)

(8 .45)

(CT~SVSH )[FS(B )] =

ui];

The convention is to keep the magni tude of t the same at all level s, so CT;, in Eq. (8.47 ) should eq ual CT;,(max) ' Step 7. For the known value s of
External Stability Step 9. Check for overturning, using Figure 8.31 as a guide. Taki ng the moment about B yields the overturning mom ent for the uni t length of the wall :

M;

=

Paz'

(8.48 )

Here, (8 .46 )

(8 .47)

ll

P; = acti ve for ce =

0 CT~dz

J

The resisting moment per unit length of the wa ll is

.. 418

Chapter 8: Retaining Walls ~ b' ~

a ' -----.j

q/unit area

t . . . . ... t

A

1_ Z

-

- L = L,- -

-

-+

- Xl-~ Wl E

H

D

Figure 8.31 Stab ility check for the retaini ng wall

III situ soil Y2; Q;1; cl

(8.4 9 1

where WI = (area A F EGI) (1 ) (')II) WZ = (area FBD E ) (1) ('y,) So,

(8.50

Step 10. The check for sliding can be done by using Eq . (8 .11) , or

FS (sliding)

where k =~ .

(W, + Wz + ... + qa')[ tan ( kepi) ]

= -

-

-

-

- --

-

-----'- -=----------'-----'--.::

(8 .5 :

III II

~"

~I

8.15 Step-by-Step-Design Procedure Using Metallic Strip Reinforcement

419

Step 11. Check for ultimate bearing capaci ty failure, which can be given as qll

= C2 Nc + h2 L 2N y

(8.52)

The bearing capacity factors N; and N; corr espond to the soil frictio n angle ~2 ' (See Table 3.3.) Fro m Eq. 8.32, the vertica l stress at Z = H is (]"~(H ) -

y ,H

+

(]"~{2)

(8.53)

So the factor of safe ty against beari ng capacity fail ure is

quIt

FS(bearingcapacity) - (]"~(H)

(8.54)

Generally, minimum value s of FS(ovenurning) = 3, FS(sliding) - 3, and FS (bearingcapacityfailure) = 3 to 5 are rec ommended. 'o r the

Example 8.5 (8.49,

A 10 m high retaining wall with galv anized steel-strip reinforceme nt in a gra nular backfill has to be co nstruc ted. Referring to Fi gure 8.29, given :

Granular backfill :

~; =

Foundation soil:

~2

36° Y, = 16.5 kN/m 3

= 28°

17.3 kN/m3 C2 = 50 kN/ m2

Y2 =

Galvanized steel reinf orcement: Width of strip, (8.50 )

= 75 mm = 0.6 m ce nter -to-ce nter = 1 m cen ter-to-center h = 240 ,00 kN/m2 ~~ = 20°

w Sv SH

Req uired

FS(B) =

3

Requ ired

FS( p ) =

3

Check for the exte rna l and intern al stab ility. Assu me the corrosion rate of the galvanized steel to be 0.025 mm/year and the life span of the structure to be 50 years.

(8.51)

Solution Internal Stability Check Tie thickne ss: Maximum tie force , Tmax =

(]"~(max)

SvS H

420

Chapter 8: Retaining Walls

2(

(T a(max) --

'YI HK a -- 'YI H tan 45 -

2cPi)

so

Tmax

-

'Y I H

2(

2cP;)svsH

tan 45 -

From Eq. (8.47), for fi e break, ( T~SvS H) [FS(B)] f =

wf

2(4

[ 'Y IHtan

~)SvSH]FS(B)

5 -

wf y

y

or [ (16.5)( 10) tan

t =

2(4

5 - ¥ )(0.6)(1) ] ( 3)

(0.075 m ) ( 240,000 kN/ m2)

= 0.00428m = 4.28 mm

If the rate of corrosion is 0.025 mm/yr and the life span of the structure is 50 yr, then the actual thickness, t, of the ties will be t = 4.28

+ (0.025) (50)

=

5.53 mm

So a tie thickness of 6 mm would be enough . Tie length: Refer to Eq. (8.46). For this case, (T~ =

'Yl zKa

and (T~ =

'YI Z ,

so

Now the following table can be prepared. (Note: FS(p) = 3, H = 10 m, w = 75 mm, and cP~ = 20°.) z{m)

Tie length L (m) [Eq . (8.46)]

2 4 6 8

12.65 11.63 10.61 9.59

IO

8.57

= 13 m.

So use a tie length of L

Externa l Stability Check Check for overturning: Refer to Figure 8.32. For this case, using Eq. (8.50) FS (oven uming)

=

[i

W1x j H

o

]

(J"~ d z z'

,I

""1_

1111 1

8.15 Step-by-Step-Design procedure Using Metallic Strip Reinforcement

1

10m 11-

421

l. WI ~/L= 13 m

'12

1FS

= 17.3 kN/m 3


(B)

2

WI n

= y 1H L =

Xl

= 4.28 mm

Figure 8.32 Retaining wall with galvanized steel-strip reinforcement in the backfill

r

= 6.5 m

P, = ~ is 50 yr, then the

(J~ dz = h

10

z/ = -3

YIZ,

so

= 2 145 kN/m

( 16.5 )( 10)( 13)

FS(overturning)

=

lKaH2 = 0)( 16.5)(0.26)( 10 )2 = 214. 5 kN /m

= 333 . m = 19.52> 3-0K

(2 145) (6.5) (2 14 .5) ( 3.33 )

Check for sliding: From Eq. (8.5 1)

m, W

= 75 rnrn,

FS(

lidi 5 I

-

ing} -

2 145 tanl

_W_I_ta_n~(_ k o/~;')J

r,

(~}36)1= 4.45> 3-0K

214.5

0/2 = 28 N; = 25 .8, N; = 0

Check for bearing capacity: For

,

16.78 (Tab1e 3.3). From

Eq . (8.52), quIt

= c2Nc + h2 L N;

quit

=

(50)(25.8)

+

(~)( 1 7.3) ( 13)( 16.72) = 3170. 16kN/ m

From Eq. (8.53) ,

(J~(H) =

(8.50) FS

(bearingcapacity)

yl H = ( 16.5 ) ( 10) quit

= - ,- = (J o(H )

3170 .16 165

=

2

165 kN/ m

= 19.2 >

5-0 K

2

-

422

-

-

- - - - - - - -- - - - - - - - - -

Chapter 8: Retaining Walls

_ _ Retaining Walls with Geotextile Reinforcement Figure 8.33 shows a retaining wall in which layers of geotextile have been used as reinforcement. As in Figure 8.31, the backfill is a granular soil. In this type of retaining wall, the facing of the wall is formed by lapping the sheets as shown with a lap length of I,. When construction is finished, the exposed face of the wall must be covered ; otherwise , the geotextile will deteriorate from exposure to ultraviolet light. Bitumen emulsion or Gunite is sprayed on the wall face. A wire mesh anchored to the geotextile facing may be necessary to keep the coating on. Figure 8.34 shows the construction of a geotextile-reinforced retaining wall. Figure 8.35 shows a completed geosynthetic-reinforced soil wall. The wall is in DeBeque Canyon , Colorado . Note the versatility of the facing type. In this case, single-tier concrete block facing is integrated with a three-tier facing via rock facing. /

s/ Geotextile /

H

III situ soil 'Y2; 4>2;c2

Sv

Geotextile )1 Geotextile Sv Sand.'YI ; ; Geotextile Sv

Sv

Ie

Figure 8 .33 Retaining wall with geotextile reinforcement

Figure 8.34 Construct ion of a geotextile-reinforced retaining wall (Courtesy of Jonathan T. H. Wu, University of Colorado at Denver; Denver, Colorado)

I I

~------_.

III II

111

111'

423

8.16 Retaining Walls with Geotexti/e Reinforcement

nent been used as rein-

ie of retaining wall.

p length of fl' When otherwise, the geotulsion or Gunit e is g may be necessary Ie-reinforced retainwall. The wall is in this case, single-tier ng.

Figure 8.35 A completed geotextile-reinforced retaining wall in DeBeque Canyon, Colorado (Co urtesy of Jo na tha n T. H. Wu, University of Colorado at De nver, Denver, Colorado )

aining wall with

.cm cnt

The design of this type of retaining wall is similar to that presented in Section 8.15 . Following is a step-by-step procedure for design based on the recom mendation s of Bell et a1. (1975) and Koerner (2005):

Internal Stability Step 1.

Determine the active pressure distribution on the wall from the formula CT;/

KaCT;,

(8.55)

KIlYIZ

where

K; = Rankine active pressure coefficient = tan 2 ( 45 - cjJ;j2) Y1 = unit weight of the granular backfill = friction angle of the granular backfill

cjJ; Step 2.

Select a geotextile fabr ic with an allowable tensile strength,

Tall

(lb/ft or

kN/ m ).

The allowable tensile strength for retaining wall construction may be expressed as (Koerner, 2005) T all

-

T ult RF id X RF cr X RF cbd

where

of Jo nathan

Tu1t = ultimate tensile stren gth RF id = reduction factor for installation damage RF cr = reduction factor for creep RF cbd = reduction factor for chemical and biological degradation

(8.56)

I!

424

Chapter 8: Retaining Walls

The recommended values of the reduction factor are as follows (Koerner, 2005) 1.1- 2.0

2-4 1-1 .5 Step 3. Determine the vertical spa cing of the layers at any depth z from the form ula (8.57 ) Note that Eg. (8.57) is similar to Eg . (8.39). Th e magnitude of FS(B) is generally 1.3 to 1.5. Step 4. Determ ine the length of each layer of geotextile from the formula

L

=

l, + t ,

(8.58)

where

H -

z

(8.59 )

and (8.60 ) in which

F S (p ) =

4>F

1.3 to 1.5

= friction angle at geotextile-soil interface

= ~4>; Note th at Egs . (8.58) , (8.59), and (8.60 ) are simi lar to Egs . (8.43), (8.45). and (8.44) , respec tively . Based on the published results, the assumption of 4>'FI 4>: = ~ is reasona ble and appears to be conservat ive. Martin et al. (1984) presente d the following laboratory test results for 4>,,14>; between various types of geotextiles and sand. Type

Woven-monofilament/concrete sand Woven-silt film/concrete sand Woven-silt film/rounded sand Woven- silt film/ silty sand Nonwoven-melt-bonded/conc rete sand Nonwoven-needle-punched/concrete sand Nonwoven-needle-punched/rounded sand Nonwoven -needle-punched/silty sand

0.87 0.8 0.86 0.92 0.87 1.0

0.93 0.91

8.16 Retaining Walls with Geotextile Reinfo rce ment

lows (Koerner, 2005)

Step 5. Determine the lap length, I{ , from II

J Z

425

Svo-:,FS(p)

- - --

= ~ '

4o-~ tan 4>F

(8.6 1)

The minimum lap length should be I m.

from the formul a

External Stability (8.57 )

Step 6. Check the factors of safety against overtu rning, sliding, and bearing capacity failure as described in Section 8.15 (Steps 9, 10, and 11).

ude of FS(B) is gen-

Example 8.6

Ie formu la (8.58)

(8.5 9 )

A geotextile-reinfor ced retaining wall 5 m high is shown in Figure 8.36. For the gran ular backfill , 'Y I = 15.7 kN/m 3 and 4>; = 36°. For the geotextile, TUll = 52.5 kN/m. For the design of the wall , determine Sv, L, and II' Use RF id = 1.2, RF cr = 2.5, and RF cbd = 1.25.

Solution We have

4>;) 2

K a = tan 2 ( 45 (8.60)

= 0 ' 26

Dete rmination of S v To find Sv, we make a few trials. From Eq. (8.57), Sv =

-

-

T all -='----------,:

('YlzK a ) [FS(B)]

ace

f.

.qs. (8.43), (8.45). ; = ~ is reasonab le the following labexti les and sand.

=:,.

2,5 m------"I

Sv =O.5 m

5m

j

'Yl = 15.7 kN/m 3

cP' = 36°

.w.; Q87 Q8 Q~ Q~ Q~

1.0 Q~

0.9 1

'Y2 = 18 kN/m 3

cP; =22° e'2 =28 kN/m 2

Figure 8.36 Geotextile-reinforced retaining wall

426

Chapter 8: Retaining Walls Fro m Eg. (8.56), Tall

=

i.;

- - ----=..::.:.--

-

52 .5 1.2 X 2.5 X 1.25

-

------=

RF id X RF cr X RF cbd

14kN/m

With FS(B) = 1.5 at Z = 2 m,

S

=

v

14 = 114 m ( 15.7) (2) (0 .26 ) ( 1.5 ) .

At z = 4 m,

_

Sy At

-

14 (15 .7 ) (4) (0.26) ( 1.5) = 0.57 m

z = 5 m, 14

So, use Sv

= 0.5 m for z = 0 to z = 5 m (See Fig ure 8.36 .)

Determination of L From Egs. (8.58) , (8.59) , and (8.60) , (H - z)

L =

tan ( 45 For FS(p) = 1.5,

tancPF =

SyKa[FS(d

+ - - --'cP; ) 2 tan cPF +2

tan [ (~) (36)] = 0.445, and it follows that

L

= (0.51)

(H - z)

+ 0.438Sy

H = 5 m, Sy = 0.5 m

At At

z = 0.5 m: L = (0.51)(5 - 0.5) + (0.438)(0.5) = 2.514 m z = 2.5 m: L = (0.51)(5 - 2.5) + (0.438)(0.5) = 1.494 m

So, use L

= 2.5 m throughout.

Determination of II From Eg . (8.61) ,

So, use I,

= 1 m.

SyKa[FS(p)]

=

l,

= 0.219Sy = (0 .219)(0.5) = 0.11 m ::; 1 m

4 tan

cPF

=

Sy(0 .26) (1.5)

Ii

.

4tan[(D (36)J

= 0.219Sy

•

I

III

8.16 Retaining Walls with Geotextile Reinforcement

427

Example 8.7 Consider the results of the internal stability check given in Example 8.6. For the geotextile-reinforced retaining wall, calculate the factor of safety against overturning, sliding, and bearing cap acity failure .

kN/m

Solution Refer to Figure 8.37. Factor of Safety Against Overturning

(H)

W,XI From

Eg. (8.50),

FS (overtuming) =

(Pa )

3"

WI = (5)(2.5)(15.7) = 196.25 kN/m XI

2.5 2

=-

=

1.25m

r, = ~'YH2Ka = (~)(l5.7)(5)2(0.26) = 51.03 kN/m Hence,

FS(overtuming)

=

(1 96.25)( 1.25) 51.03(5/3) = 2.88 < 3 (increase length of geotextile layers to 3 m)

hat

r.

" 2.5 m- - .H

sv = 0.5 m 5m

WI

Y1 = 15.7 kN /m 3


=36°

Y2 = 18 kN/m 3 LJ.


•

C2 =28 kN/m 2

Figure 8.37 Stability check

~

428

Chapter 8: Retaining Walls

Factor of Safety Against Sliding From Eg. (8.51),

Wltan(~
(196.25)

=

p a

[tan(~ X 36)] 51.03

= 1.71 > 1.5 - O.K.

Factor of Safety Against Bearing Capacity Failure From Eg . (8.52), qu = c;Nc +

~Y2 L 2 Ny

L

Given: Y2 = 18 kN/m 3, 2 = 2.5 m, c~ = 28 kN/m2, and
q"

=

(28)(16.88)

+

(~)(l8)(2.5)(7.13)

= 633 kN/m

2

From Eg. (8.54),

q"

FS(bearingcapacity)

= - ,(T o(H)

I\IIiJEIII

633 H YI

633 (15 .7) (5)

8.06

>3-

•

O.K.

Retaining Walls with Geogrid Reinforcement-General Geogrids can also be used as reinforcement in granular backfill for the construction of retaining walls. Figure 8.38 shows typical schematic diagrams of retaining walls with geogrid reinforcement. Figure 8.39 shows some photographs of geogrid-reinforced retaining walls in the field. Relatively few field measurements are available for lateral earth pressure on retaining walls constructed with geogrid reinforcement. Figure 8.40 shows a comparison of measured and design lateral pressures (Berg et aI., 1986) for two retaining walls constructed with precast panel facing. The figure indicates that the measured earth pressures were substantially smaller than those calculated for the Rankine active case.

Design Procedure for Geogrid-Reinforced Retaining Wall Figure 8.41 shows a schematic diagram of a concrete panel-faced wall with a granular backfill reinforced with layers of geogrid. The design process of the wall is essentiall y Si IL i1ar to that with geotexti1e reinforcement of the backfill given in Section 8.16. The following is a brief step-by-step procedure. Internal Stability Step 1. Determine the active pressure at any depth z as [similar to Eq. (8.55)] : (T~ = K a Y1Z

(8 .6=

where

x, =

Rankine active pressure coefficient = tan 2 ( 45 _

~l

)

8. 78 Design Procedure fore Geogrid-Reinforced Retaining Wall

429

Geogrids - biaxial

1.5 - O.K.

>

L

'·c'"

Geogrids - uniaxial

~.~~~~;~~:~;;~~~~-i:'b;~%;~tt~,~~ ~~~~~~,~~5s~:~~~~~~,i)~¢~~{~~i2~ (a)

Fro m Tab le 3.3.

Gabion facing ~, I

/ m2

."-{

,!'I -i '.

. . . . !<-

•

1 - O.K.

, !~ _

it-s-Gener. t ruction of reta ining h geogrid reinforce~ walls in the field. 'ess ure o n retaining .ariso n of measured mstructed wi th pre, were substantiall x

:1: [ I

.

I -I

•

Geogrids I

~;::~{~~:~~;:;:~~~~;~~~~~~{~~j:r~t~~ (b) I I I I

Precast concrete ~

panel

/ Pinned connection

I I I I

/ Geogrids

I I I I I I

ill with a gra nular , is essentia lly simI 8. 16. Th e fo llow -

I I

~~~tW~?gl@1,f~~~ ).:;;2,.,~: ~~; Leveling pad (c)

o Eq. (8.55) ]: (8.62 )

~'l

)

Figure 8.38 Typic al schematic diagrams of retaining walls with geogr id reinforcement: (a) geogrid wraparound wall; (b) wall with gabion facing; (c) concrete panel-faced wall (After The Tensar Corporation , 1986)

430

Chapter 8: Retaining Walls

(a)

(c)

Lateral pressure, (T~ 10 20 30

Figure 8.39 (a) HDPE geogridreinforced wall with precast concrete panel facing under construction ; (b) Mechanical splice between two pieces of geogrid in the working direction ; (c) Segmented concre te-block faced wall reinforced with uniaxial geogrid (Courtesy of Tensar International Corporation, Atlanta, Georgia )

(kN/m2 )

o

40

0 -+-- - --'-- - - --'----- - --'-- - - -'----. ---Wall at Tuscon, Arizona, H

=

4.6 m

- - - - Wall at Lithonia, Georgia, H = 6 m

2

3

4

5

Height of fill above load cell (m)

Figure 8.40 Comparison of theoretical and meas ured lateral pressures in geogri d reinforce d retaining walls (Based on Berg et a!., 1986)

I

8.18 Design Procedure fore Geogrid-Reinforced Retaining Wall

T

I

431

w,

z

Sv

Granular backfill

L,

y,


H

Wz

Lz

(b)

(a) HOPE geogridall with precast conicing under , (b) Mechanical en two pieces of Ie working direction: ed concrete-block .inforced with uniaxCourtesy of Tensa r I Corporation, lrgia)

Leve ling pad

Found ation soil Yz.
Figure 8.41 Desi gn of geogrid-reinforced reta ining wall

Step 2.

Select a geogrid with allowable tensile streng th, Tall [similar to Eq. (8.56)] (Koerner, 2005) :

r:

Tall = RF id X RFcr X RF cbd

(8.63)

where RFid = reducti on factor for installation dam age ( l.l to 1.4) RFcr = redu ction factor for cree p (2.0 to 3.0) RFcbd = reduction factor for chemical and biological degradation (1.1 to 1.5). Step 3. Obtain the vertical spacing of the geogrid layers, Sv, as Sv

T allCr

=----:--""--

(J~

(8.64)

FS (B )

where C, = coverage ratio for geogrid. The coverage ratio is the fractional plan area at any particular elevation that is actually occupied by geogrid. For examp le, if there is a 0.3 m ( 1 ft) wide space between each 1.2 m (4 ft) wide piece of geogrid, the coverage ratio is C = r

1.2m = 08 1.2 m + 0.3 m .

Step 4. Calculate the length of each layer of geogrid at a depth z as [Eq. (8.58)] L rison of theoretical press ures in geog rid valls (Based on Berg

=

l,

+ l,

1r = tan

_~.L

H - Z

2(45 -""2cP;) ~~=~~

(8.65)

_

432

Chapter 8: Retaining Walls

For determination of

t, [similar to Eq. (8.60)],

resistance to pullout at a given normal effective stress FS(p) = - - - - --'-- - - ---=-- - - - -- -- pullout force

(2) (ie) (CilTbtan c/J;) (C r) SvlT~

(2) (ie) (Ci tan c/Ji) ( Cr)

(8.66)

SVKa where C, = intera ction coefficient or

SVKa FS(p) t,

(8.67)

= 2Cr C.tan o -I.' .

Thu s, at a given depth z, the total length, L, of the geogrid layer is

H -

L = I + I = r

e

(

tan 45

z

SVK a FS(p)

c/J ;)

+ - - -2C rCitan

+-

(8.68)

c/J;

2

The interacti on coefficient, C;, can be determined experimentally in the laboratory. The followi ng is an approximate range for C, for various types of backfill. Gravel, sandy gravel Well graded sand, gravelly sand Fine sand , silty sand

0.75-0.8 0.7-0.75 0.55-0.6

External Stability Check the factors of safety against overturning, sliding, and bearing capacity failure as described in Section 8.15 (Steps 9, 10, and II) .

Example 8.8 Consider a geogrid-reinforced retaining wall. Referring to Figure 8.4 1, given : H = 6 m, Yl = 16.5 kN/m 3, c/Jr = 35°, Tall = 45 kN/m , FS(B ) = 1.5, FS(?) = 1.5, C, = 0.8, and C, = 0.75. For the design of the wall, determine Sv and L.

Solution

tc,

=

tan 2 ( 45 -

~;) =

tan 2 ( 45 -

3;)

=

0.27

Determi nat ion of S v From Eq. (8.64),

I

II

III I

(45)( 0.8)

5.39

(1 6.5)(z )(0.27 )( 1.5)

z

II

I'll

Problems

: stress

(8.66

At z

= 2m:

S;

=

5.39 2

= 2.7 m

At z

= 4m:

Sv

=

5.39 4

= 1.35 m

At z

= Sm:

S;

=

5.39 = 1.08 m 5

433

Use Sv > 1m Deter mination of L From Eq. (8.68),

(8.67

L =

H -

(8.6

At z At z At z

= 1 m: = 3m: = S m:

SVK a FS(p)

(M )

2CrCi tancP{

2

=

L 0.52(6 - 1) L = 0.52(6 - 3) L 0.52(6 - 5)

=

So, use L = 3 m for

in the laborator~ . ill.

+ ------'---'--

+-

(

tan 45

s

z

z=

6 -z

----,-------'-----

tan( 45

+

+-

(1 m )( 0.27 )( 1.5)

(2) (0.8) (0.75) ( tan 35°) 3;)

+ 0.482 = 3.08 m = 3. 1 m + 0.48 2 = 2.04 m = 2. 1 m + 0.48 2 = 1.0 m

•

0 to 6 m.

Problems In Problems 8.1 through 8.4, use "Yconcrete = 23.58 kN/m 3 • Also, in Eq. (8.1 1), use k, = k2 = 2/3 and Pp = O.

8.1

For the cantilever retaining wall shown in Figure P8.1, let the following data be given: Wall dimensions:

ipacity failure as

H = 8 m, X I = 0.4 m, X 2 = 0.6 m, X3 = 1.5 m, X4 = 3.5 m, = 0.96 rn, D = 1.75 m, (l' = 10°

Xs

Soil properties:

8.2 iven: H = 6 m,

cPl

= 32°'"Y2 = 17.6 kN/m3,

cP2

= 28°, c; =

Calculate the factor of safety with respect to overturning , sliding, and bearing capacity. Repeat Problem 8.1 with the following : Wall dimensions :

', = 0.8, and C,

"YI = 16.5 kN/m 3 , 30kN/m 2

H = 6.5 m, XI = 0.3 m, X2 = 0.6 m, X 3 = 0.8 m, X4 = 2 m, = 0.8 m, D = 1.5 m, (l' = 0°

Xs

"Y, = 18.08 kN/m 3, cPi = 36°'"Y2 = 19.65 kN/m 3, cP2= 15°,c; = 30 kN/m 2 A grav ity retaining wall is shown in Figu re P8 .3. Calculate the factor of safety with respect to overturning and sliding , given the following data: Soil properties:

8.3

.39

z

Wall dimensions:

H = 6 m, X I = 0.6 m, X 2 = 2 m, X 3 = 2 m, X 4 = 0.5 m, X s = 0.75 m.x, = 0.8 m, D = 1.5 m

Soil properties:

"YI = 16.5 kN/m 3 , 40 kN/m 2

cPi =

32°' "Y2 = 18 kN/m 3 ,

Use the Rankine active ear th pressure in your calculation.

cP2=

22°. c; =