6 design for deep excavation civil engineering

8/13/2019 6 Design for deep excavation civil engineering

1/121

10.1 Introduction

10.2 Design Methods and Factors of Safety

10.3 Retaining Walls

10.3.1 Soldier Piles

10.3.2 Sheet Piles

10.3.3 Column Piles10.3.4 Diaphragm Walls

Ch10 Design of Retaining Structural Components


2/121

10.4 Structural components in Braced Excavations10.5 Strut Systems

10.5.1 Horizontal struts

10.5.2 End Braces and Corner Braces10.5.3 Wales

10.5.4 Center Posts

10.6 Structural components in Anchored Excavations


3/121

10.7 Anchor Systems

10.7.1 Components of Anchors

10.7.2 Analysis of Anchor Load

10.7.3 Arrangement of Anchors

10.7.4 Design of Anchor Heads, Anchor Stands, and Wales

10.7.5 Design of the Free Section

10.7.6 Design of the Fixed Section10.7.7 Preloading

10.7.8 Design of Retaining Walls

10.8 Tests of Anchors

10.8.1 Proving Test

10.8.2 Suitability Test

10.8.3 Acceptance Test

10.9 Summery and General Comments


4/121

10.2 Design Methods and Factors of Safety

The design methods for reinforced concreteworking stress method and

strength design method

The design methods for steel structuresThe allowable stress method (abbreviated as the ASD method)

and the ultimate strength design method.

For diaphragm walls served as permanent structure, the

allowable stress would be magnified by a factor withoutthe earthquake force considered. Some country building

codes suggest =1.25.


5/121

To design the sections and dimensions of a retaining

wall, the first thing is to carry out the stress analysis.Three methods can be adopted for the stress analysis of

a retaining wall

Assumed hinge method too simple, only suitable on

shallow excavation

Finite element method too complicated for input and

output, need professional geotechnicalengineering training

Beam on elastic foundation method suitable for any

excavation, simple to input and out put, suitable

for common engineer

10.3 Retaining Walls


6/121

FIGURE 10.1 Typical bending moment and shear diagrams of a retaining wall by

stress analysis

-60 -30 0 30 60

28

24

20

16

12

8

4

-90 -60 -30 0 30 60 900

Shear (t)Bending moment (t-m)

D

epth

(m)


7/121

The commonly used types of soldier piles in

excavations are the H steel, I steel and rail piles.

10.3.1 Soldier Piles


8/121

Take the maximum bending moment (M max ) from the

typical bending moment envelope (Figure 10.1).According to the allowable stress method (ASD method),

we can obtain the section modulus of the soldier piles as

a = allowable stress of the steel

l = short-term magnified factor of the allowable stress, which

can be found from the country building codes

a

MSl

max= (10.1)


9/121

The sections of sheet piles are various. U-shaped, Z-shaped, and line-shaped

sheet piles are frequently used in some countries, as shown in Figure 3.19.

10.3.2 Sheet Piles

( a )

( b )

( c )

FIGURE 3.19 Sections of steel sheet piles (a) U pile (b) Z pile (c) straight pile


10/121

The dimensions of a sheet pile are determined on the basis

of the results of the stress analysis. According to the

envelope of bending moments (Figure 10.1), take the

maximum bending momentM max and compute the section

modulus using Eq.10.1, which is then used to find the

dimension of the sheet pile

a

MSl

max= (10.1)


11/121

Column piles used in excavations include the PIP pile,reinforced concrete pile, and the mixed pile.

Column piles bear the axial load and flexural load

simultaneously. Therefore, their behavior is similar to thatof the reinforced concrete columns.

The thus obtained bending moment and shear envelopes

are then used for the design of reinforced concretecolumns. Please refer to the design chart of reinforced

concrete columns or the ACI code.

10.3.3 Column Piles


12/121

The design of a diaphragm wall includes specifying the wallthickness and the reinforcements.

The thickness is usually determined according to the results

of the stress analysis, the deformation analysis and the

feasibility of detailing of concrete reinforcements.

According to the experience of excavations, the thickness ofa diaphragm wall can be assumed to be 5 %He (He is the

excavation depth) in the preliminary design.

10.3.4 Diaphragm Walls


13/121

The deformation of the retaining wall in the central section of

the site is usually assumed to be in the plane strain condition

during analysis. Therefore, the unit width (b =1 m) of the

diaphragm wall is usually used for flexural stress analysis.

The definitions of the designed and nominal bending moments

and shears are as follows

l=

MM

u

LF

u

n

MM=

un

VV=

(10.3)

(10.2)

(10.4)

(10.5)

l=

VVu

LF


14/121

Mu = bending moment for design

Mn = nominal bending moment (capacity of bending

moment)Vu = shear for design

Vn = nominal shear, also called the capacity of shear

M = bending moment obtained from stress analysis

V = shear obtained from stress analysis

LF= load resistance factor; ACI (2005),LF=1.6; ACI

(2002),LF=1.7 = strength reduction factor; ACI (2005),= 0.9 (bending

moment ), = 0.75 (shear); ACI (2002), = 0.9

(bending moment ), = 0.85 (shear)


15/121

FIGURE 10.3 Stress in the ultimate state on a section of the reinforced concrete

beam (reinforcements are smaller than those in the balanced state)

d

b

sA

0.003

c

c

a

ysfAT=yfys

nM

f 85.0 abfC c= 85.0

2

ad

2a


16/121

Suppose the thickness (t) of a reinforced concrete beam is

given. The nominal resistant bending moment of concrete ofthe width of b is

d = distance from extreme compression fiber to centroid of

tension reinforcement

= maximum reinforcement ratio that does not contain

compression reinforcement, max = 0.75 b= compressive strength of concrete

= yielding strength of reinforcements

b = reinforcement ratio producing balanced strain conditions

(1) Vertical main reinforcement

max

yfc

f

2max

max 59.011

bdf

ffM

c

y

yR

= (10.6)


17/121

The balanced reinforcement ratio can be computed as follows

where the measurement unit of and is ; b1 relates to

the strength of concrete, which is generally under 280 .

High performance concrete has used to construct diaphragm

walls recently. Thus, b1 can also be computed by the followingequation

where the measurement unit of and is ; b1 relates to

the strength of concrete.

2kg/cm2kg/cm

2kg/cm0.85-0.05( )0.65

0.85

70

280 kg/cm2

280 kg/cm

2b1= - 280cf

cf

cf

cf

yf

cf

(10.7)b=

y

b

f 6120 +fy

61200.851

cf

(10.8)

yf


18/121

When Mu MR , the only item to be designed is tensionreinforcements, which can be computed as follows

Let the strength ratio of the material be

Since

m=0.85 cf

yf

259.0

1 bd

f

ffM

M

c

y

ynu


19/121

We can then have the reinforcement ratio,

deriving from the above equation

The reinforcement ratio, deriving from the

above equation

As = bd

=

2

211

1

bdf

mM

m y

n (10.9)

(10.10)


20/121

IfMu MR , it follows that the maximum resistance of

concrete (under the condition that the tension

reinforcement has achieved its yielding strength) is still

smaller than the designed bending moment.

That is to say, the diaphragm wall is to be thickened or

compression reinforcements have to be designed.


21/121

The design of compression reinforcements can follow the

method of the doubly reinforced beam, which is as follows

Let the ratio of the reinforcement be 1= 0.75 b

As1 = 1bd

T1 = As1fy

Cc = T1 = 0.85 ba

From above, we obtain

a =0.85 b

T1

cf

cf

(10.11)

(10.12)

(10.13)

(10.14)


22/121

The bending moment provided by the tension

reinforcement is

M2 = Mn - M1

Thus, the area of tension reinforcements corresponding to

M2 is

= distance from extreme compression fiber of the wall to

the centroid of compression reinforcements

M1= T1 ( d )2

a

d

As2

=fy ( d )

M2

d (10.17)

(10.16)

(10.15)


23/121

The required area of tension reinforcements is

As =As1 + As2

The required area of compressive reinforcements is

=Es fy

Es = Young's modulus of reinforcements

= strain of the compression reinforcements, which

can be computed as follows

c = a / b1 (see Figure 10.3)

=As2fy

= ( ) 0.003c c

d

s

s

s

sf

sA

sf

(10.21)

(10.20)

(10.19)

(10.18)


24/121

The retaining wall with one dimensional deformation does not need to

be reinforced horizontally. If shrinkage and temperature are to be

considered, horizontal reinforcements will be needed. According to

the ACI code, the reinforcement with shrinkage and temperature

effects is

(2) Horizontal main reinforcement

As = 0.0020 Ag 4200 kg/cm2 0.0018 Ag = 4200 kg/cm2 0.00184200/ Ag 0.0014Ag > 4200 kg/cm2 Ag= thickness of the retaining wall unit width;

The reinforcement for shrinkage and temperature effects (As) has

to be placed evenly on both sides of the wall

yf

yf

yf

yf

(10.22)


25/121

According to the ACI code, the nominal shear ofconcrete is

Vc is the nominal shear of concrete and is measured by

kg. The unit of is kg/cm2

. b is the unit width and isusually taken to be 100 cm.

When the designed shear (Vu) is smaller thanfVc, it is

theoretically unnecessary to design reinforcement. Inpractice, the shear reinforcement still has to be

designed in order to be able to hang the steel cage into

the trench.

(3) Shear reinforcement

Vc = 0.53 bdcf

cf

(10.23)


26/121

FIGURE 10.2 Plan, 3D view, side view of a steel cage of diaphragm wall

(1, 2, and 3 represent shear reinforcements)

b

b

A

C

E

F

B

D

A

C

E

F

A C

B D

1

1

1

2 2

2 2

2

3 3

3

3

3

Vertical main renforcement

Plan

Horizontal

reinforcement

Vertical main

reinforcement

Side view3-D

view

Horizontal

reinforcement

Sv

Sv

Sv

Sv

Sv

Sv

Sv

Sv

Sv

Sh

Sh

Sh Sh Sh Sh

Sh Sh Sh


27/121

Three types of shear reinforcements are used in the steelcage of a retaining wall, main shear reinforcement-type 1,

and two small slant reinforcements-type 2, and 3. As a

result, the nominal shear (capacity of shear) of the

retaining wall per unit width is

Vn = Vc + Vs

where Vs is nominal shear of shear reinforcements

(10.24)


28/121

Since the horizontal distance between any two shear

reinforcements is identical, the sectional area of the shearreinforcement per unit width (b = 100 cm) is

Av = total sectional area of all shear reinforcements on the

horizontal section per unit width (cm2)

Ab = sectional area of a single shear reinforcement (cm2)

Sh = horizontal distance between shear reinforcements (cm)

Av=100Ab

Sh(10.25)

Th i l h f th t 1


29/121

The nominal shear of the type 1

Sv =vertical distance between the main shear reinforcements


= angle between the small slant reinforcement and thehorizontal reinforcement


b = angle between the small slant reinforcement and the vertical

reinforcement

The nominal shear offered by all shear reinforcements per unit

width is

Vs1 = Sv

Avfy d

Vs2 = SinSv

Avfy d

Vs3 = Sin bSv

Avfy d

Vs = Vs1 + Vs2 + Vs3

(10.28)

(10.27)

(10.29)

(10.26)


30/121

Minimum lap splice length and development length of reinforcements

of a diaphragm wall can be designed according to the ACI code, or

they can be determined using Eqs.10.30 and 10.31. The coefficient1.25 is to magnify the lap splice length and development length,

considering the effects of concrete casting in bentonite, which leads to

a smaller bond stress

Development length

Lap splice length

Ab = the sectional area of a single reinforcement; if applied a vertical

reinforcement, c = 1.0to horizontal reinforcement, c = 1.4

(4) Lap splice length and development length

Ld = 1.25cld = 0.075c Abfy

Ld = 1.25cld =0.091c Abfy

cf

cf

(10.30)

(10.31)


31/121

FIGURE 10.5 Design of reinforcements in a steel cage of the diaphragm wall

(a) profile (b) side view

(a) (b)

Floor slab

Raft foundation

Inside the

excavation zone

Outside the

excavation zone


32/121

FIGURE 10.6 Components of a strutting retaining system

Soldier pile

Lagging

Horizontal

struts

Wale

Center post

Sheetpile

Jack

Jack

End brace

Corner brace

10.4 Structural components in Braced Excavations


33/121

FIGURE 10.7 Single strutting system

End brace

waleRetaining wall

Horizontal struts


34/121

X

Horizontal strutof the higher level

Horizontal strut

of the lower level

Y

U-clip

Center

post

(b)(a)

FIGURE 10.8 Joint of a single strut and a center post (a) 3D view (b) plan view

U clip

Bracket

Horizontalstrut of the

higher level

Horizontal strut of the

lower level

Center post

xe


35/121

FIGURE 10.8 Joint of a single strut and a center post (c) photo


36/121

FIGURE 10.9 Double strutting system

End braces

Retaining wall

Horizontal struts


37/121

Horizontal strut of the higher level

U-clip

Horizontal strut

of the lower

level Center postHorizontal

strut of the

higher level

U-clip

Horizontal strut of the lower level

Center post

(a) (b)

FIGURE 10.10 Joint of a double strut and a center post (a) 3D view (b) plan view


38/121

FIGURE 10.10 Joint of a double strut and a center post (c) photo

10 5 Strut Systems 10 5 1 H i t l t t


39/121

10.5 Strut Systems 10.5.1 Horizontal struts

(1) Stress computation

A strut is usually subjected to the axial compressive load as well as

the flexural load. The axial compressive stress can be computed as:

N

Afa =

A = sectional area of the strutN = axial load =N 1 +N2

N1 = strut load induced by excavation, which can be computed

using the beam on elastic foundation method, the finite

element method, or the apparent earth pressure methodN2 = strut load induced by the temperature change = DtEA

= coefficient of thermal expansion of struts; for the steel strut,

= 1.32 10 5 /

Dt = temperature change of struts ()

E = Young's modulus

(10.32)


40/121

Because the result using DtEA to compute the effect of

temperature change of the strut load,N2, usually comes

out too large, the empirical formula is often used

instead.

The JSA (1988) suggests

N2 = 10 t ~ 15 t orN2 = ( 1.0 t ~ 4.0 t ) Dt, where

Dtis the temperature change () in the air (not thetemperature change of the steel).


41/121

The flexural stress can be computed as follows

M1 = bending moment produced by the strut weight and the live load;

taking the center post as the simply supported hinge, thenM1 =

wL2

/8w = strut weight + live load 0.5 t/m

L = distance between two adjacent center posts

M2 = bending moment caused by the uplift of the center post; since

struts are constructed level by level during excavation, the

influence ofM2 on the top level would be largest while that on

the lowest level would be largest while that on the lowest level

would be the smallest

S= section modulus

M1 +M2S

fb = (10.33)

(2) Allowable stress


42/121

The allowable axial compressive stress of a strut can be selected from the tables

and figures offered by the AISC Specification or using the following equation

KL / ry = effective slenderness ratio of the strut on the flexural plane where K can

be taken as 1.0

L = unsupported length of the strut, usually distance between the two adjacent

center posts

ry = radius of gyration of the cross section of the strut in the direction of theweak axis

Cc = critical slenderness ratio =

E= Young's modulus of struts

Fy = yielding stress of struts

l = short-term magnified factor of the allowable stress, which can be found incountr buildin code, l = 1.25.

( ) owab e s ess

KLry

> Cc

KLry

Cc

l

=3

/

8

1/

8

3

3

5

/

2

1

1

c

y

c

y

c

y

a

C

rKL

C

rKL

FC

rKL

Fy

l

= 2

2

23

12

y

a

r

KL

EF

yFE/22

(10.34)

(10.35)


43/121

The allowable flexural stress (Fb) of a strut can be

derived from the tables and figures offered by the

AISC Specification.

However, the flexural stress (fb) of a strut, during

normal excavation (that is, the uplift of the center post

is not much), is not large. To simplify the design, wecan assumeFb = 0.6Fyl.

(3) E i ti f bi d t


44/121

According to the AISC Specification, the stress on each section of a strut

should satisfy the following equation

Cm = coefficient of modification = 0.85

1 /(1 fa / = amplification factor= allowable Euler stress = l122E / [23KL / rx2]

KL / rx = effective slenderness ratio on the flexural plane, we can

assumeK= 1.0

rx = ratio of gyration of the strut in the direction of the strong axis

E= Young's modulus of struts

l = short-term magnified factor of the allowable stress, which can be

found in country building codes, we can assume l = 1.25

(3) Examination of combined stresses

+ 1.0fa

Fa 15

faFa

fbFb

faFa

> 15

eF

0.1

1

be

a

bm

a

a

FF

f

fC

F

f

(10.36)

(10.37)


45/121

10.5.2 End Braces and Corner Braces

FIGURE 10.11 Distance and angle between struts and end braces or corner braces

2

211 1

3

1 2 3 2 4

6

5

p

p

L1 L2


46/121

The axial force on the end brace is

p = apparent earth pressure or the strut load per unit width

1 2 = spans (see Figure 10.11)

1 = angle between the end brace and the wale

(usually 45)

N =p 12 1 + 2 Sin1 (10.38)


47/121

The axial force on an corner brace is

3 4 5 6 = the spans

23 = angles between the corner brace and the wale,

45 in most cases

To be conservative, the design load can be assumed to be

the maximum value betweenN1 andN2 .

N1 =p 12 3 + 4 Sin2N2 =p 12

5 + 6

Sin3 (10.40)

(10.39)


48/121

The function of wales is to transfer the earth pressure

on the retaining wall to the struts.

For analysis, the earth can therefore be assumed toact on the wale directly.

The earth pressure can be obtained from the apparent

earth pressure method or by transforming the strut

load, computed using the finite element method or

beam on elastic foundation method.

10.5.3 Wales


49/121

The wale is usually acted on by the earth pressure as

well as the axial force from the end brace or corner

brace. That is to say, the wale bears simultaneously

the moment and the axial force and its design falls in

the domain of the beam-column system. With ample

lateral support, the analyses of secondary moment and

buckling for wales can be saved.


50/121

FIGURE 10.12 Computation of the bending moment of a wale

L

Joint

p

WaleStrut

Earth pressure

Fixed end beam mode

Simplly supported beam modeAnalysis method

considering weak joints

Mmax = pL2

8

1 Mmax pL210

14

L

Mmax = pL212

1

To compute the maximum bending moment and shear of wales the wales can be


51/121

To compute the maximum bending moment and shear of wales, the wales can be

viewed as simply supported beams with struts as supporting hinges, or viewed as

fixed end beams.

Simply supported beam

Fixed end beam

Mmax = maximum bending moment of the wale

Qmax = maximum shear of the wale

L = distance between struts

p = earth pressure

Mmax = pL281

Qmax = pL2

1

Mmax = pL212

1

Qmax = pL2

1

(10.41a)

(10.41)

(10.42a)

(10.42)


52/121

Mmax pL210

1

Because the length of a wale is limited, the wales

have to be joined in the field. The strength of the

joint is not fully rigid and the joint could easily

become the weakest part of the structure. Thus,

the joints had better be located at the places

where the stress is smaller.According to the bending moment distribution

diagram of a continuous or a simply supported

beam, if the joint is located at theplaces 1/4 of

the span from the support, the maximum bendingmoment of the wale would be

(10.43)


53/121

In design practice, if the wale is assumed to be a

simply supported beam, it may not work out as

economical.

If it is designed to be a fixed end beam, it may tend

to be insecure. The more reliable method is to

locate the joint at theplace 1/4 of the span from the

support and compute the maximum bending

moment and shear as follows

Mmax = pL2101

Qmax = pL2

1

(10.44)

(10.44a)


54/121

When there are no end braces,L is the horizontaldistance between struts.

When there are end braces, the distance between struts

can be reduced properly.

Generally speaking, when 60 ,L = 1 + 2. When

> 60 , less conservative design can be madeL = 1


55/121

Viewed from the view point of mechanics, the wale is also

subjected to the axial stress.

If sheet piles or soldier piles with end braces are used, the

axial force of the wale can be designed by choosing the

larger one between the following two computing results

or

N = p 12 1 + 2 tan1N = p 5 + 2 6 (10.45)

(10.46)


56/121

Since the corners of a diaphragm wall have an arching

effect, to determine the axial load of a wale around a

corner of an excavation with diaphragm walls, it is

necessary to use Eq. 10.46.

If sheet piles or soldier piles are used, take the

maximum value among the computed results of Eqs.

10.45 and 10.46.


57/121

10.5.4 Center Posts

Center post are usually set to

bear the weight of struts, the materials on the struts,

and other extra loading out of the movement of the

retaining system.

Center posts, usually H steels,

(1) installed by striking the piles into soils directly.

(2) embedment into soils by way of pre-boring or

inserted into a cast-in place pile.


58/121

FIGURE 10.13 Installation of center posts onto cast in-situ piles

Cast-in-place pile

Retaining

wall

Central post

Horizontal strut


59/121

The design of a center post includes the design of the section and

the embedment depth.

The possible axial loads on each center post are

(1) Weight of the horizontal strut and the live load,P1

wi = strut weight of each level and its live loadL1,L2 = distance between struts

n = number of the levels

p1= wi (L1 +L 2 )n

i=1

FIGURE 10.14 Distribution of strut weight on a center post

L1

L1

L2L2

(10.47)


60/121

(2) Weight of the center post above the excavation

surface,P2


61/121

(3) Slant compressive force on the horizontal strut,P3

Suppose the tilt angle of a horizontal strut is , the downward or

upward force on the center post would be

sin)(2 ,,1

3 iyix

n

i

NNP ==

(10.48)

= load on the struts of each level in thex direction

= load on the struts of each level in they direction= tilt angle of the horizontal strut

ixN ,

iy

N,

FIG. 10.15 Action of the axial force of struts on a center post

(a) when the center post settles (b) when the center post heaves

where

1,yN

1,yN

1,yN

1,yN1,xN

1,xN

1,xN

1,xN

2,yN

2,yN

2,yN

2,yN2,xN

2,xN

2,xN

2,xN

(a) (b)

The first level

of struts

The second level

of struts


62/121

In analysis, the tilt angle of the horizontal strut is

difficult to estimate. According to the data of field

observations of excavations, the tilt angle can beassumed to be sin1/50.

To be conservative in analysis, the center post can be

assumed to be subject to the axial force of the strut and

downward force. Thus, the total load on the center post

would be

(10.49)321 PPPP =

In the double strutting system the moments generated by the weight of


63/121

In the double strutting system, the moments generated by the weight oftwo struts, which eccentrically act of the center post, can be assumed to

be mutually offset.

Horizontal strut of the higher level

U-clip

Horizontal strut

of the lower

level

Center post Horizontalstrut of the

higher level

U-clip

Horizontal strut of the lower levelCenter post

(a) (b)

FIGURE 10.10 Joint of a double strut and a center post (a) 3D view (b) plan view


64/121

In the single strutting system, the moment caused by the strut

weight eccentrically acting on the center post can be computed

by the following equation

(10.50)= eccentricity distance

xePPM )( 32xe

FIGURE 10.8 Joint of a single strut and a center post (a) 3D view (b) plan view

X

Horizontal strut

of the higher level

Horizontal strut

of the lower level

Y

U-clip

Center

post

(b)(a)

U clip

Bracket

Horizontal

strut of the

higher level

Horizontal strut of the

lower level

Center post

xe

Buckling length


65/121

The buckling length of a center post should take the maximum

unsupported length during the process of excavation, floor

construction and dismantling. the buckling length of the center

postL is(10.51)),,,,max( 54321 LLLLLL=

Strut Center post

etaining wall

Excavation bottom

Raft foundationFloor slabs

213,usually LLL

FIGURE 10.16 Unsupported length of center posts

1L

2L

3L

4L5L


66/121

Because the center post bears, simultaneously, the

axial force and bending moment, it is the beam-

column system that is to be adopted in analysis.Choose a proper section and then use Eq.

10.47~10.51 to examine it.


67/121

The center post may bear

(1) vertical loads

(2) uplift forces caused by the tilt of the horizontal

strut.

The analysis method is identical with that

for piles though a pile is of circular section.


68/121

FIGURE 10.17 Area to be adopted for the computation the vertical

bearing capacity of center posts of the H pile

columntheofdepthembedded)(2 =

=

HBA

HBA

s

p

B

H

1t

2t

(1) Vertical bearing capacity


69/121

(1) Vertical bearing capacity

The ultimate vertical bearing capacity can be expressed as :

(10.52)

= ultimate vertical bearing capacity of the center post

= point load resistance of the center post

= skin frictional resistance of the center post

= unit frictional resistance of the center post

= surface area of the center post

= the unit point resistance per unit area of the center post

= sectional area of the center post

The allowable vertical bearing capacity of the center post is

(10.53)

= allowable vertical bearing capacity

= factor of safety

ppssspu AqAfQQQ ==uQ

pQ

sQ

s

f

sApq

pA

FSQQ

FSQQ psua ==

aQ

FS


70/121

Under general conditions (low possibility of

earthquake, general excavation, good

geological conditions), the above factor ofsafety can be taken to beFS=2.0.

If considering the possibility of earthquakeor the excavation is of high risk (in soft soils,

for example), the above factor of safety can

be taken to beFS=3.0.


71/121

In sandy soils, the point resistance of the center

post can be computed as

Driven pile (10.54)

Drilling pile (10.55)

N= the average standard penetration number within the influence

range of the bottom end of the center post.

Generally speaking,Nvalue can be taken as the average value

within the range four times of the center post diameter above the

bottom end of the center post diameter below it.

Nfp

40=

2/mtNfp 15=

2/mt

The frictional resistance of the center post for both the driven and


72/121

drilling piles is

N= the average standard penetration number within the depth of the center postembedded in the sandy soils

In clayey soils, to be conservative, the value of the center post is

usually assumed to be 0 and the can be computed as follows

= adhesion between the surface of the center post and the surrounding soils

= undrained shear strength of clay

= reduction value of undrained shear strength.

-values relates to the undrained shear strength of the clay, the installation method of

the center post, and its embedment depth. Besides, the cast-in-situ piles embedded in

clayey soils have lower -values because they do not compress the soil during the

construction and are usually installed in boreholes filled with bentonite. Skempton

suggests the -value be between 0.3 and 0.6, 0.45 in most cases.

Nfs 2.0=2/mt

pf

uws scf ==wcus

sf

(10.56)

(10.57)


73/121

API

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

1.2

vu/s

Cast-in-

place pile

FIGURE 4.12 Relation between adhesion and undrained shear strength of clay

(2) Pullout resistance


74/121

(2) Pullout resistanceThe allowable pulling out resistance of a center post is

(10.58)

= weight of the center post, the influence of groundwater

considered= frictional resistance per unit area of the center post,

which can be estimated using Eq. 10.57 or Eq. 10.58

= surface area of the center post

= factor of safety. As far as the sort-term behavior of a

structure (such as a center post) is concerned,FScan be

assumed to be 3.0

sspa AfFS

WR 1=

pW

sf

sA

FS

10 6 Structural components in Anchored Excavations


75/121

10.6 Structural components in Anchored Excavations

FIGURE 10.18 Anchored excavation

m2

Anchor head

Anchora

gesection

Potentialfailu

resurfa

ce

Excavation surface

Freesect

ion

10 7 Anchor Systems


76/121

Anchors are categorized into permanent anchors andtemporary ones. They are applied extensively. The design

of an anchor involves the soil properties, the materials of

the anchor, grouting, and the construction details. The last

is especially crucial in determining the quality of theanchor. Concerning the general rules of the design and

construction of an anchor, please refer to related literature

and specifications referred to in this section.

Anchors in excavations belong to the category of

temporary anchors. This section will only introduce the

anchor system of the retaining wall in excavations.

10.7 Anchor Systems

General rules of anchor


77/121

General rules of anchor

American Association of Sate Highway and Transportation Officials

(AASHTO) (1992)British Standard Institute (BSI DD81)(1989)

Deutsche Industrie Norm (DIN) (1988)

Federation Internationale de la Precontrainte (FIP) (1982);

Post-Tensioning Institute (PTI) (1980)

Geotechnical Control Office (GCO)(1989)

JSF(1990)

A h i b i ll d f th h


78/121

An anchor is basically composed of the anchor

head, the free section, and the fixed section.

FIGURE 10.19 Configuration of an anchor

Anchor head

Anchor stand

Retaining wall

Casing

Cement mortar Tendon

)(anchortheofLength L

)(sectionfreef

L

)(sectionfixeda

L

Types of anchor heads


79/121

FIG. 10.20 Types of anchor heads (a) locked by wedges (b) locked by screw nuts

(c) composite lock (CICHE, 1998)

clips Male awl

Female awl

Screw nut

PC steel bar

Screw nut

Steel strandsteel bar connector

Multiple PC steel strand

(a)

(b) (c)

Anchors can be categorized into the resistance type, the bearing


80/121

resistance type, and the composite type according to the characteristics of

the bearing force provided by the fixed section.

FIG. 10.21 Types of anchorage body (a) frictional resistance type (b) bearing resistance type

(c) composite resistance type

Tension type

Compression type

Bearing body

Bearing resistance type

Composite (friction + bearing)

resistance type

Frictional resistance type{


81/121

The horizontal component of the anchor

load can be computed using the half

method or underneath pressure methodof the apparent earth pressure method.

The related analytical method is identical

to that for the braced excavation.

10.7.2 Analysis of Anchor Load


82/121

FIGURE 10.23 Anchorage force and its horizontal component

(Horizontal component )

(Anchorage force )


83/121

The anchor load is

(10.59)

where = designed load of the anchor= horizontal component obtained from analysis

= angle between the axis of the anchor and the

horizontal plane

cosPT

w=

wTP


84/121

In literature and specifications on anchors, group anchor

effects are seldom discussed. The design of anchor is

usually based on a single anchor pattern. Under suchconditions, the minimum distance between anchors has to be

limited so that the group anchor effect will not be produced

or the installation of an anchor may cast a bad influence on

the adjacent anchors. Table 10.3 lists the related

specifications on the minimum distance between anchors.

10.7.3 Arrangement of Anchors


85/121

TABLE 10.3 Specifications for the minimum distance between the fixed sections

bdNote : refers to the diameter of the anchorage section

or 1.2 m

Larger than 4

Larger than 4

Larger than 6

Larger than 4

For anchors with working load large than 70 tons, the

minimum center-to-center distance of the anchorage

section is 1.0 mFor anchors with working load largerthan 130 tons, the minimum center-to-center distance of

the anchorage section is 1.5 m

DIN(1988)

AASHTO(1992)

or 1.2 mPTI(1989)

, generally 1.5 m to 2.0 m in applicationBSI(1989)

, or not shorter than 1.5 mFIP(1982)

Suggested minimum distanceRule

bd

bd

bd

bd


86/121

The determination of the vertical distance between

anchors depends on the analytical results of theanchor load. The vertical distance between anchors

is about 2.5~4.5 m and at least 1.0~1.5 m above the

floor slabs. Besides, both FIP (1982)and BSI (1989)require the vertical distance between the fixed

section and underground structure be longer than 3

m. The fixed section has to be 5 m away from theground surface (see Figure 10.24).

The anchorage length must be embedded within the shadow area and

k di h 1 8 f h d f il f


87/121

FIGURE 10.24 Locations of the anchorage sections in soil (redraw after BSI, 1989)

keep a distance more than 1.8 m from the assumed failure surface

Assumed failure surface

Assumed support

Depth

C B

D

A E

F

G

2/45o

2/45

o

2/45 o

Locations of the anchorage sections


88/121

FIGURE 10.25 Distance between an anchorage section and the potential failure surface

Assumed failure surface

(potential failure surface)

2 m

b d f


89/121

FIGURE 10.26 Potential failure surfaces in excavations

Excavation surface

Assumed supports

Wall bottom 2

45

C

a

b d f

e

The installation angle of anchors


90/121

g

Theoretically, if an anchor is installed with the same direction of

loading, it is able to develop the maximum capacity.

Considering the installation quality, to clear the dregs in the drillingbore of the anchor, the installation angle should not be smaller than 10.

The fixed section should be placed in a bearing layer (such as a

sandy layer, gravelly layer, or rocks) or soils with high strength.

Generally speaking, the deeper the soils, the higher the strength. It ismore suitable to place the fixed section in deeper soil. As a result, an

anchor is usually installed with a certain slope. The steeper the slope,

however, the larger the dragging down force on the retaining wall.

When the installation angle exceeds 45, it becomes dangerous.

That is to say, without going beyond 45, the installation angle

should fall within the range .

4510

10 7 4 Design of Anchor Heads Anchor Stands and


91/121

The locking device should be tested before using and its

locking capacity should be large enough to make the

tendon bear 80% of the ultimate loading without being

damaged. If the locking device is used with a wedge clip,the locked tendon should not slide too much. The bearing

strength of the bearing plate should be large enough to

resist the maximum pulling force during the process of

preloading and excavation.

10.7.4 Design of Anchor Heads, Anchor Stands, and

Wales


92/121

The function of an anchor stand is to transfer the load on the wale

or the retaining wall to the anchored soil layer. Anchor stands

can be categorized into reinforced concrete anchor stands and

steel anchor stands. The detailing of reinforcements of the

concrete anchor stand should be carefully designed to meet safety

requirements. Under the working load, the allowable

compressive stress of the concrete anchor stand should be smaller

than 30% of the 28th-day strength of concrete.


93/121

FIGURE 10.27 Commonly used anchor stands

Steel bond

Steel anchorstandWale

Retaining wall

Anti-erosion

materialsConcrete anchor

stand

Bearing

plate

Steel anchor stand

L1


94/121

Steel bond

Wale

hor stand

Lower/ upper brackets

Upper wale

Anchor stand

Lower wale

Soldier pile

FIGURE 10.28 Configuration of wales in anchored system

PL1

Pv

Anchor force

L1

L2

The stress of a wale can be examined as follows


95/121

(1) Stress computation in the direction of strong axis

Viewing the wale between two anchors as a simplysupported beam structure, the maximum bending

moment and shear are(10.60)

(10.61)

where = maximum bending moment in the direction ofstrong axis

= maximum shear in the direction of strong axis

= lateral earth pressure on the wale

= horizontal distance between anchors

2

1max 8

1

pLM

1max2

1pLQ

maxM

maxQ

p

1L

A i h d i f l i h b d i


96/121

As in the design of a wale in the braced excavation

introduced in Section 10.5.3, if the joint is placed at

the distance of 1/4 of the span from the support, themaximum bending moment can be computed by the

following equation

(10.62)2max

10

1pLM =

The bending stress and shear stress of the wale in the direction of


97/121

The bending stress and shear stress of the wale in the direction of

the strong axis should satisfy the following criteria:

(10.63)

(10.64)

= section modulus of the member in the direction of thestrong axis

= area of the web of the member

= allowable bending stress of the member=(0.6~0.66)

= allowable shear stress of the member=0.4

= yielding stress of the member

=short-term magnified factor of the allowable stress

axS

M

2/max

awA

Q 2/max

xS

wA

a

ly

Fa

yF

l

lyF

(2) Stress computation in the direction of the weak


98/121

axis

Viewing the wale between the brackets supporting the anchor as a

beam, the maximum bending moment and shear of the wale are

(10.65)

(10.66)

(10.67)

= maximum bending moment in the direction of weak axis

= maximum shear in the direction of weak axis

= vertical component of the anchorage force at the support

= distance between the brackets supporting the anchor

tan1pLPv=tan4

1

4

1212max LpLLPM v

tan2

1

2

11max pLPQ v=

maxM

maxQ

vP

2L

The bending stress and shear stress of the wale in the direction of the

k i h ld ti f th f ll i diti


99/121

weak axis should satisfy the following conditions

= section modulus of the member in the direction of weak axis

= area of a single flange of the member

= allowable bending stress of the member=0.75

= allowable shear stress of the member=0.4

= yielding stress of the member= short-term magnified factor of the allowable stress

ayS

M

2/max

afA

Q 2/max

yS

fA

a lyF

a

yFl

lyF

(10.68)

(10.69)

10 7 5 Design of the Free Section


100/121

The free section is composed of a tendon and a plastic

casing. Three types of tendons are availablesteel bars,steel wires, and strands. The 7-wire strand is common in

many countries, usually available in size of 13 mm(0.5 1n),

15mm (0.6 in) and 18 mm (0.7 in). The ultimate tensile

strength varies from 1570 to 1765 N/mm2

. The allowabletensile force can be computed as follows

(10.70)

where = designed safety factor for the tendon. Table 10.4

lists the commonly used factors of safety.

tuw FPP /=

tF

10.7.5 Design of the Free Section

TABLE 10.4 Factor of safety for single anchor

il f h i


101/121

ClassificationTensile force

of tendons

( )

Anchoring

force

( )

Bond force of

tendons ( )

Temporary anchors whose working

period is not longer than 6 months and

which dont affect public safety when

failing(2)

1.4 2.0 2.0

Temporary anchors whose workingperiod is not longer than 2 years and

which dont affect public safety,

though having certain influence, when

failing without alert

1.6 2.5 (3) 2.5 (3)

Permanent or temporary anchors which

are highly risky in rusting or which

affect public safety seriously due to

failure(2)

2.0 3.0 (4) 3.0 (3)

tF

aF b

F


102/121

NOTE:

(1) The table is from CICHE (1998), incorporating specifications

made by BSI (1989) and FIP (1982)

(2) The temporary anchors in this table refer to those whoseworking periods are not longer than 2 years. Otherwise, they

are classified as permanent anchors.

(3) With complete proving test results, the minimum factor of

safety can be 2.0.

(4) If creep of soil is to be encountered, the factor of safety can be

increased to 4.0.

Th b f t d i


103/121

The number of tendons is

(10.71)

As discussed above, is computed usingEq. 10.59, transforming the horizontal

component, as analyzed using the apparent

earth pressure method, the beam on elastic

foundation method, or the finite elementmethod.

w

w

P

Tn =

wT


104/121

As Figures 10.24 and 10.25 illustrate, the free section

has to extend beyond the potential failure zone by at

least 2.0 m. If the free section is too short, the

stresses in soils caused by the fixed section will

easily affect the retaining wall. In addition, theanchor behavior will tend to be fragile, any small

displacement possibly bring about large stress, and

threatening the security of the anchor. Therefore, thefree section has to be at least 4.0 m long.

10.7.6 Design of the Fixed Section


105/121

The fixed section of an anchor should be placed 2.0 m

away from the potential failure zone. The design of a fixed

section includes the bond forces between tendons and

grouts and the anchorage force between the fixed section

and soils.The bond forces between tendons and grout have to be

large enough so that the designed strength of the anchor can

fully develop. Table 10.4 suggests the safety factors

between tendons and grout. As for the allowable bondforces between tendons and cement grout, we can use the

values as suggested by JSF (1990), as shown in Table 10.5.

10.7.6 Design of the Fixed Section


106/121

TABLE 10.4 Factor of safety for single anchor (CICHE, 1998)

ClassificationTensile forceof tendons

( )

Anchoring force

( )

Bond force of

tendons ( )

Temporary anchors whose working period

is not longer than 6 months and which

dont affect public safety when failing(2)1.4 2.0 2.0

Temporary anchors whose working period

is not longer than 2 years and which dont

affect public safety, though having certain

influence, when failing without alert

1.6 2.5 (3) 2.5 (3)

Permanent or temporary anchors which arehighly risky in rusting or which affect

public safety seriously due to failure(2)2.0 3.0 (4) 3.0 (3)

tF aF bF


107/121

NOTE:

(1) The table is from CICHE (1998), incorporating specifications

made by BSI (1989) and FIP (1982)

(2) The temporary anchors in this table refer to those whose

working periods are not longer than 2 years. Otherwise, theyare classified as permanent anchors.

(3) With complete proving test results, the minimum factor of

safety can be 2.0.

(4) If creep of soil is to be encountered, the factor of safety can

be increased to 4.0.

TABLE 10 5 All bl b d f f t d i t t (JSF 1990)


108/121

TABLE 10.5 Allowable bond force of tendons in cement mortar (JSF, 1990)

Designed compressive strength of

cement mortar ( )2cm/kgTypes of tendons

150 180 240 300 >400

Temporary

anchors

Steel wireRound steel bar

Steel strand

Multiple steel

strand

8

8

8

8

10

10

10

10

12

12

12

12

13.5

13.5

13.5

13.5

15

15

15

15

Permanent

anchors

Steel wireRound steel bar

Steel strand

Multiple steel

strand

8

8

8

8

9

9

9

9

10

10

10

10

The anchorage force between the fixed section and soils


109/121

(1) Friction type of anchor

The ultimate anchorage force, Tu, for a friction type ofanchor can be computed by the following equation

= ultimate anchorage force

= diameter of the fixed section

= length of the fixed section

= average ultimate shear resistance strength per unit area

(also called the frictional strength) between the fixed

section and soils

uT

ultabu LdT

bd

aL

ult

(10.72)

h d i d l d h h i


110/121

The designed load on the anchor is

(10.73)

where = safety factor of the designed anchorage force; the

commonly used factors of safety are listed in Table 10.4.

a

u

w

F

TT =

aF


111/121

The strength between the fixed sectionand the surrounding soil, , changes with

the types of soils where the fixed section is

placed, the failure mode of the fixed section,

the grouting pressure, and the installation

method. Many investigators have proposed

some empirical formulas to estimate .

ult

ult

(a) Anchorage in rocks


112/121

Null pressurized anchors are often used for rocks (see

Figure 10.22). Little john's suggestion (1970) is usually

taken for the ultimate shear resistance strength between

the fixed section and rocks. That is, let be 0.1 qu (forblock rocks) or 0.25 qu (for weathered rocks) where qu is

the axial compressive strength of rock. Anchor choice is

to adopt the suggestions by JSF (1990), as sown in Table10.6.

ult

TABLE 10.6 Ultimate frictional strength of an anchorage body (JSF, 1990)


113/121

ult )cm/kg( 2

N

Type of soil

Rock

Hard rockSoft rock

Weathered rock

Mudstone

15 ~ 25

10 ~ 15

6 ~ 10

6 ~ 12

Gravel

=10

=20

=30=40

=50

1.0 ~ 2.0

1.7 ~ 2.5

2.5 ~ 3.53.5 ~ 4.5

4.5 ~ 7.0

Gravel

=10

=20

=30

=40

=50

1.0 ~ 1.4

1.8 ~ 2.2

2.3 ~ 2.7

2.9 ~ 3.5

3.0 ~ 4.0

Clay 1.0 c

N

NNNNNNNN

Note :Nis standard penetration test number, c is cohesion, and

is ultimate frictional strength after pressure groutingult

(b) Anchorage in sandy soils


114/121

The -value of an anchor in sandy soils can be computed as

follows

(10.74)

= the average effective overage effective overburden pressure

above the fixed section= the angle of friction between the fixed section and soils

Because anchors in sandy soils are usually the low

pressure type, the average ultimate shear strength of sandis usually determined on the basis of experience or field

tests.

ult

tanvult v

(c) Anchorage in clayey soils


115/121

Anchors in clays are usually installed following the null

pressurized grouting method (see Figure 10.22). The ultimate

shear resistance strength of the fixed section in clayey soils can be

expressed as follows

(10.75)

where = undrained shear strength of clay

= reduction factor for undrained shear strength

uult

s =u

s

1.2


116/121

API

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

vu/s

Cast-in-

place pile

FIGURE 4.12 Relation between adhesion and undrained shear strength of clay


117/121

As Eq. 10.72 shows, the longer the fixed section, the

larger the anchorage force. However, many investigatorshave found that when the length of the fixed section

exceeds some critical value, more lengthening of the

fixed section can hardly increase its anchorage force.

Thus, the length of the fixed section has to be limited. In

principle, the length of the fixed section is best when

between 3.0 and 10.0 m. If out of this range, it is better

to carry out field tests to examine the ultimate anchorageforce. Table 10.7 lists specifications on the minimum and

maximum length of the fixed section.

TABLE 10.7 Specifications for the length of the fixed section (CICHE, 1998)


118/121

Specifications Minimum suggested distance

FIP (1982) and

BSI (1989)

No shorter than 3 m, except if the proving test

result is satisfactory in fulfilling the design, and

not longer than 10 m, under normal condition

PTI (1980) No shorter than 3 m

JSF (1990) Between 3 ~10 m

GCO (1989) No shorter than 3 m

AASHTO

(1992)In soil, no shorter than 4.6 m; in rock, 3 m

10.7.7 Preloading


119/121

Locked Preloading or the Locked Load

The anchor installed, it usually needs to be

preloaded and locked, which is called locked

preloading or the locked load.

The locked load is usually a little larger than

the designed load of the anchor because the

slipping of the wedge clips during the process of

locking will cause some loss of the anchorpreload. After locking, the remaining force is

called the effective load.

10.7.7 Preloading


120/121

10.8 Tests of Anchors


121/121

Influenced by the process of fabrication, grouting, and

geological conditions, the actual ultimate load of an anchorhas to be examined through field tests. The tests are

classified into many methods in terms of their different

aims, though their common goal is to understand the

deformation behaviors and load capacities of the anchor.

Anchor tests can be distinguished into theproving test, the

suitability test, and the acceptance test.

6 design for deep excavation civil engineering

Documents