6 Design for deep excavation civil engineering

Ch10 Design of Retaining Structural Components 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 Piles

10.3.4 Diaphragm Walls

10.4 Structural components in Braced Excavations 10.5 Strut Systems 10.5.1 Horizontal struts 10.5.2 End Braces and Corner Braces 10.5.3 Wales 10.5.4 Center Posts 10.6 Structural components in Anchored Excavations

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 Section 10.7.7 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

10.2 Design Methods and Factors of Safety The design methods for reinforced concrete： working stress method and strength design method

The design methods for steel structures： The 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 λ without the earthquake force considered. Some country building codes suggest λ =1.25.

10.3 Retaining Walls 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 geotechnical engineering training Beam on elastic foundation method─ suitable for any excavation, simple to input and out put, suitable for common engineer

Bending moment (t-m) -90 -60 -30 0 30 60 90 0

Shear (t) -60 -30 0 30

60

Depth (m)

4 8 12 16 20 24 28

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

10.3.1 Soldier Piles The commonly used types of soldier piles in excavations are the H steel, I steel and rail piles.

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 S=

Mmax

l a

(10.1)

a = allowable stress of the steel l = short-term magnified factor of the allowable stress, which can be found from the country building codes

10.3.2 Sheet Piles 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.

(a)

(b)

(c)

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

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 moment M max and compute the section modulus using Eq.10.1, which is then used to find the dimension of the sheet pile

S =

M

l

max a

(10.1)

10.3.3 Column Piles 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 that of the reinforced concrete columns. The thus obtained bending moment and shear envelopes are then used for the design of reinforced concrete columns. Please refer to the design chart of reinforced concrete columns or the ACI code.

10.3.4 Diaphragm Walls The design of a diaphragm wall includes specifying the wall thickness 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 of a diaphragm wall can be assumed to be 5 % He (He is the excavation depth) in the preliminary design.

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： Mu = LF M l Mu = M n



LFV l V Vn = u Vu =



(10.2)

(10.3) (10.4) (10.5)

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)

b

0.85 fc

0.003 c

C = 0.85 fcab

a d

d

s y As

a2

fy

a 2

Mn

T = As fy

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)

(1) Vertical main reinforcement Suppose the thickness (t) of a reinforced concrete beam is given. The nominal resistant bending moment of concrete of the width of b is  max f y  1 M R =  max f y 1  0.59   f c 

 2 bd 

(10.6)

d = distance from extreme compression fiber to centroid of tension reinforcement max = maximum reinforcement ratio that does not contain compression reinforcement, max = 0.75 b f c = compressive strength of concrete f y = yielding strength of reinforcements b = reinforcement ratio producing balanced strain conditions

The balanced reinforcement ratio can be computed as follows

f c 6120 b1 (10.7) 6120 + fy where the measurement unit of f c and f y is kg/cm2 ; b1 relates to the strength of concrete, which is generally under 280 kg/cm2. High performance concrete has used to construct diaphragm walls recently. Thus, b1 can also be computed by the following equation：  = 0.85 b fy

﹛ b1 =

f c ≦ 280 kg/cm2

0.85

0.85-0.05(

f c- 280 70

) ≧0.65

f c  280 kg/cm2

(10.8)

where the measurement unit of f c and f y is kg/cm2 ; b1 relates to the strength of concrete.

When Mu ≦  MR , the only item to be designed is tension reinforcements, which can be computed as follows：

Let the strength ratio of the material be

Since

m=

 0.59 f y = M n = f y 1   f c 

Mu

fy 0.85 f c

 2 bd 

We can then have the reinforcement ratio, deriving from the above equation： 2mM n 1  = 1 1 m  f y bd 2

   

(10.9)

The reinforcement ratio, deriving from the above equation：

As = bd

(10.10)

If Mu   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.

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 (10.11)

T1 = As1 fy Cc = T1 = 0.85

ba

(10.12)

f c (10.13)

From above, we obtain T1 a= 0.85 f c b

(10.14)

The bending moment provided by the tension reinforcement is a M1= T1 ( d  ) 2 M2 = M n - M1

(10.15) (10.16)

Thus, the area of tension reinforcements corresponding to M2 is As2=

M2 fy ( d  d  )

(10.17)

d  = distance from extreme compression fiber of the wall to the centroid of compression reinforcements

The required area of tension reinforcements is As = As1 + As2

(10.18)

The required area of compressive reinforcements is As = As2

fy f s

f s = Es  s ≦ fy

(10.19) (10.20)

Es = Young's modulus of reinforcements  s = strain of the compression reinforcements, which can be computed as follows： c  d  s = ( ) × 0.003 (10.21) c c = a / b1 (see Figure 10.3)

(2) Horizontal main reinforcement 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 0.0020 Ag（ f y ＜ 4200 kg/cm2 ） 2 (10.22) f 0.0018 A （ = 4200 kg/cm ） g y As = 0.0018（4200/ f y ）Ag ≧0.0014Ag（ f y > 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

(3) Shear reinforcement According to the ACI code, the nominal shear of concrete is Vc = 0.53 f c bd (10.23) Vc is the nominal shear of concrete and is measured by kg. The unit of f c is kg/cm2. b is the unit width and is usually taken to be 100 cm. When the designed shear (Vu) is smaller than fVc, it is theoretically unnecessary to design reinforcement. In practice, the shear reinforcement still has to be designed in order to be able to hang the steel cage into the trench.

Sh Sh Sh Sh Sh 1 3

ew vi

D E 3- Side view Horizontal reinforcement

Sh Sh Sh Sh F

3

3

A

2

~

2 1

A

E

Sv Sv

b

C

~

C

2



~

~



~

2

Plan ~

~

A

F

~

~

C

Vertical main renforcement

3 1 2

Sv

b 3

Sv

Sv

Sv

Sv

Sv

Sv ~

~ B

~

D

Vertical main reinforcement

Horizontal reinforcement

D

B

FIGURE 10.2 Plan, 3D view, side view of a steel cage of diaphragm wall (1, 2, and 3 represent shear reinforcements)

Three types of shear reinforcements are used in the steel cage 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

(10.24)

where Vs is nominal shear of shear reinforcements

Since the horizontal distance between any two shear reinforcements is identical, the sectional area of the shear reinforcement per unit width (b = 100 cm) is 100Ab Av = Sh

(10.25)

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)

The nominal shear of the type 1 Av fy d (10.26) Vs1 = Sv Sv =vertical distance between the main shear reinforcements The nominal shear of the type 2 Av fy d Vs2 = Sinα Sv α= angle between the small slant reinforcement and the horizontal reinforcement The nominal shear of the type 3

(10.27)

Av fy d (10.28) Vs3 = Sin b Sv b = angle between the small slant reinforcement and the vertical reinforcement The nominal shear offered by all shear reinforcements per unit width is Vs = Vs1 + Vs2 + Vs3 (10.29)

(4) Lap splice length and development length 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 coefficient 1.25 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

Ld = 1.25﹙cld ﹚=

Ld = 1.25﹙cld ﹚=

0.075c Ab fy

f c 0.091c Ab fy

f c

(10.30)

(10.31)

Ab = the sectional area of a single reinforcement; if applied a vertical reinforcement, c = 1.0；to horizontal reinforcement, c = 1.4。

Inside the excavation zone

Outside the excavation zone

Floor slab

Raft foundation

(a)

(b)

FIGURE 10.5 Design of reinforcements in a steel cage of the diaphragm wall (a) profile (b) side view

10.4 Structural components in Braced Excavations Soldier pile Sheetpile

Wale

Lagging

Corner brace

End brace

Horizontal struts

Jack

Center post Jack

FIGURE 10.6 Components of a strutting retaining system

End brace wale

Retaining wall

Horizontal struts

FIGURE 10.7 Single strutting system

Y

Center post

ex Horizontal strut of the lower level

Horizontal strut of the lower level Horizontal strut of the higher level

U clip

Horizontal strut of the higher level

Center post

Bracket

X

U-clip

(a)

(b)

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

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

Retaining wall Horizontal struts End braces

FIGURE 10.9 Double strutting system

Horizontal strut of the higher level

Horizontal strut of the lower level Center post

U-clip

U-clip

Horizontal strut of the lower level

Center post

(a)

Horizontal strut of the higher level (b)

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

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

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:

fa = N A

(10.32)

A = sectional area of the strut N = 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 method N2 = 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

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 or N2 = ( 1.0 t ~ 4.0 t ) × Dt, where Dt is the temperature change (℃) in the air (not the temperature change of the steel).

The flexural stress can be computed as follows： M1 + M2 fb = S

(10.33)

M1 = bending moment produced by the strut weight and the live load; taking the center post as the simply supported hinge, then M1 = wL2/8 w = 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 of M2 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

(2) Allowable stress 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：  1  KL / ry   F 1   2 C  c   Fa = 3 l 5 3  KL / ry  1  KL / ry       3 8  Cc  8  Cc  y

KL ry > Cc KL ry ＜ Cc

12  2 E Fa = l 2 23    KL   ry   

(10.34)

(10.35)

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 the weak axis 2 2 E / Fy 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 in country building code, l = 1.25.

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, we can assume Fb = 0.6Fy．l.

(3) Examination of combined stresses According to the AISC Specification, the stress on each section of a strut should satisfy the following equation： fa ≦ 15 ％ Fa

fa Fa

fb + ≦ 1.0 Fb

(10.36)

fa Cm f b   1.0 (10.37) Fa  f 1  a  Fb  Fe  Cm = coefficient of modification = 0.85 1 / (1  fa / Fe ) = amplification factor = allowable Euler stress = l．12π2 E / [23（KL / rx）2] KL / rx = effective slenderness ratio on the flexural plane, we can assume K = 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。 fa > 15 ％ Fa

10.5.2 End Braces and Corner Braces 4

3

θ2

5

2

2

1

2

p

θ1

θ1 θ1

θ3 6

L1

L2

p

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

The axial force on the end brace is

﹙ ﹚

N =p

 1+  2 2

1 Sinθ1

(10.38)

p = apparent earth pressure or the strut load per unit width  1、 2 = spans (see Figure 10.11) θ1 = angle between the end brace and the wale (usually 45°)

The axial force on an corner brace is

﹙ ﹚ ﹙ ﹚

N1 = p

 3+  4 2

1 Sinθ2

(10.39)

N2 = p

 5+  6 2

1 Sinθ3

(10.40)

 3 、 4、 5 、 6 = the spans θ2、θ3 = 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 between N1 and N2 .

10.5.3 Wales 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 to act 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.

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.

1 Mmax = pL2 12

Earth pressure p

Strut L

Wale Fixed end beam mode

Joint

L 4 1 Mmax = pL2 8 Simplly supported beam mode

Mmax ≦

1 pL2 10

Analysis method considering weak joints

FIGURE 10.12 Computation of the bending moment of a wale

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：

1 Mmax = pL2 8 1 Qmax = pL 2 Fixed end beam：

1 Mmax = pL2 12 1 Qmax = pL 2

Mmax = maximum bending moment of the wale Qmax = maximum shear of the wale L = distance between struts p = earth pressure

(10.41) (10.41a)

(10.42) (10.42a)

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 the places 1/4 of the span from the support, the maximum bending moment of the wale would be 1 Mmax ≦ pL2 10

(10.43)

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 the place 1/4 of the span from the support and compute the maximum bending moment and shear as follows： 1 Mmax = pL2 10 1 Qmax = pL 2

(10.44) (10.44a)

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

can be reduced properly. Generally speaking, when θ ≦ 60° , L =  1 +  2. When θ > 60° , less conservative design can be made： L =  1 。

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：

﹙ ﹚

(10.45)

﹙ ﹚

(10.46)

6 N= p 5 + 2

or N= p

 1+  2 2

1 tanθ1

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.

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.

Horizontal strut

Retaining wall Central post

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

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 n

p1 =Σ wi ( L1 + L 2 )

(10.47)

i=1 wi = strut weight of each level and its live load L1, L2 = distance between struts n = number of the levels L2

L2

L1

L1

FIGURE 10.14 Distribution of strut weight on a center post

(2) Weight of the center post above the excavation surface, P2

(3) Slant compressive force on the horizontal strut, P3 Suppose the tilt angle of a horizontal strut is θ, the downward or upward force on the center post would be n

P3 =  2( N x ,i  N y ,i ) sin 

(10.48)

i =1

where N x ,i = load on the struts of each level in the x direction N y ,i = load on the struts of each level in the y direction  = tilt angle of the horizontal strut N y ,1

N x ,1

N x ,1

N y ,1

N y ,2

N x ,2

The first level of struts

N y ,1

N x ,1

N x ,1

N y ,1

N y ,2

N x ,2

The second level N x ,2 of struts

N x ,2

N y ,2

(a)

N y ,2

(b)

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

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 θ can be assumed to be sinθ≈1/50.

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

P = P1  P2  P3

(10.49)

In the double strutting system, the moments generated by the weight of two struts, which eccentrically act of the center post, can be assumed to be mutually offset. Horizontal strut of the lower level Horizontal strut of the higher level Center post U-clip

U-clip

Horizontal strut of the lower level

Center post

(a)

Horizontal strut of the higher level (b)

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

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 e x = eccentricity distance

M = ( P2  P3 )ex

Center post

Bracket

Y

ex Horizontal strut of the lower level

U clip

(10.50)

Horizontal strut of the higher level

Center post

Horizontal strut of the lower level Horizontal strut of the higher level X

U-clip

(a)

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

Buckling length： 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 post L is

L = max( L1 , L2 , L3 , L4 , L5 )

Strut

(10.51)

Center post

Retaining wall

L1 L2

L5 Raft foundation

L4

Floor slabs

L3

Excavation bottom

usually L3  L1 , L2 FIGURE 10.16 Unsupported length of center posts

Because the center post bears, simultaneously, the axial force and bending moment, it is the beamcolumn system that is to be adopted in analysis. Choose a proper section and then use Eq. 10.47~10.51 to examine it.

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.

t2 B

t1 H Ap = B  H As = 2( B  H )  embedded depth of the column

FIGURE 10.17 Area to be adopted for the computation the vertical bearing capacity of center posts of the H pile

(1) Vertical bearing capacity The ultimate vertical bearing capacity can be expressed as : (10.52) Qu = Q p  Qs = f s As  q p Ap Qu = ultimate vertical bearing capacity of the center post Q p = point load resistance of the center post Qs = skin frictional resistance of the center post f s = unit frictional resistance of the center post q p = surface area of the center post As = the unit point resistance per unit area of the center post A p = sectional area of the center post

The allowable vertical bearing capacity of the center post is Qs  Q p Qu Qa = = FS FS Qa = allowable vertical bearing capacity FS = factor of safety

(10.53)

Under general conditions (low possibility of earthquake, general excavation, good geological conditions), the above factor of safety can be taken to be FS=2.0. If considering the possibility of earthquake or the excavation is of high risk (in soft soils, for example), the above factor of safety can be taken to be FS=3.0.

In sandy soils, the point resistance of the center post can be computed as Driven pile：

f p = 40 N

t / m2

(10.54)

Drilling pile：

f p = 15 N

t/m

(10.55)

2

N= the average standard penetration number within the influence range of the bottom end of the center post. Generally speaking, N value 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.

The frictional resistance of the center post for both the driven and drilling piles is (10.56) f s = 0.2 N t / m 2 N = the average standard penetration number within the depth of the center post embedded in the sandy soils

In clayey soils, to be conservative, the f p value of the center post is usually assumed to be 0 and the f s can be computed as follows：

f s = cw = su

(10.57)

cw= adhesion between the surface of the center post and the surrounding soils su = undrained shear strength of clay  = reduction value of undrained shear strength. α-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 α-values because they do not compress the soil during the construction and are usually installed in boreholes filled with bentonite. Skempton suggests the α-value be between 0.3 and 0.6, 0.45 in most cases.

1.2 Cast-inplace pile

1.0 API



0.8 0.6 0.4 0.2 0.0 0.0

0.2

0.4

0.6

0.8

1.0

su /  v

FIGURE 4.12 Relation between adhesion and undrained shear strength of clay

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

1 Ra = W p  f s As FS

(10.58)

W p= weight of the center post, the influence of groundwater considered f s = frictional resistance per unit area of the center post, which can be estimated using Eq. 10.57 or Eq. 10.58 As = surface area of the center post FS = factor of safety. As far as the sort-term behavior of a structure (such as a center post) is concerned, FS can be assumed to be 3.0

10.6 Structural components in Anchored Excavations te Po a nt i i lu l fa e fa c sur re

Anchor head

 2m

t ion c e s rage o h c An

FIGURE 10.18 Anchored excavation

ct ion e s Free

Excavation surface

10.7 Anchor Systems Anchors are categorized into permanent anchors and temporary 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 the anchor. 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.

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)

An anchor is basically composed of the anchor head, the free section, and the fixed section.

Length of the anchor ( L)

Anchor head

free section( L f )

Anchor stand fixed section ( L a )

Casing Retaining wall Cement mortar

FIGURE 10.19 Configuration of an anchor

Tendon

Types of anchor heads： Male awl Female awl

clips

(a)

Screw nut

PC steel bar

Screw nut

Steel strand─steel bar connector

Multiple PC steel strand

(b)

(c)

FIG. 10.20 Types of anchor heads (a) locked by wedges (b) locked by screw nuts (c) composite lock (CICHE, 1998)

Anchors can be categorized into the resistance type, the bearing resistance type, and the composite type according to the characteristics of the bearing force provided by the fixed section. 。 Tension type

{

Frictional resistance type

Compression type Bearing body

Bearing resistance type

Composite (friction + bearing) resistance type FIG. 10.21 Types of anchorage body (a) frictional resistance type (b) bearing resistance type (c) composite resistance type

10.7.2 Analysis of Anchor Load The horizontal component of the anchor load can be computed using the half method or underneath pressure method of the apparent earth pressure method. The related analytical method is identical to that for the braced excavation.



(Horizontal component )

(Anchorage force )

FIGURE 10.23 Anchorage force and its horizontal component

The anchor load is

P Tw = cos where

Tw = designed load of the anchor

P 

= horizontal component obtained from analysis = angle between the axis of the anchor and the horizontal plane

(10.59)

10.7.3 Arrangement of Anchors 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 such conditions, 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.

TABLE 10.3 Specifications for the minimum distance between the fixed sections

Rule

Suggested minimum distance

FIP(1982)

Larger than 4 d b, or not shorter than 1.5 m

BSI(1989)

Larger than 4 d b, generally 1.5 m to 2.0 m in application

PTI(1989)

Larger than 6 d b or 1.2 m

AASHTO(1992)

Larger than 4 d b or 1.2 m

DIN(1988)

For anchors with working load large than 70 tons, the minimum center-to-center distance of the anchorage section is 1.0 m；For anchors with working load larger than 130 tons, the minimum center-to-center distance of the anchorage section is 1.5 m

Note : d b refers to the diameter of the anchorage section

The determination of the vertical distance between anchors depends on the analytical results of the anchor 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 the ground surface (see Figure 10.24).

The anchorage length must be embedded within the shadow area and keep a distance more than 1.8 m from the assumed failure surface

Assumed failure surface A E 45o    / 2

Depth

F

B

C

G

45    / 2 o

Assumed support

45    / 2 o

D

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

Locations of the anchorage sections：

2m

Assumed failure surface (potential failure surface)

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

b d

Excavation surface

Assumed supports

a

C

45  

Wall bottom

f

 2

e

FIGURE 10.26 Potential failure surfaces in excavations

The installation angle of anchors： 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 drilling bore 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 is more 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 .

10    45

10.7.4 Design of Anchor Heads, Anchor Stands, and Wales 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.

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.

Bearing Retaining wall plate

Steel bond

Steel anchor stand Wale

Anti-erosion materials Concrete anchor stand

Steel anchor stand

FIGURE 10.27 Commonly used anchor stands

L1 Upper wale Lower wale Steel bond

el anchor stand

Soldier pile Wale

L2

Lower/ upper brackets

Anchor stand

PL1 Anchor force

Pv

FIGURE 10.28 Configuration of wales in anchored system

The stress of a wale can be examined as follows： (1) Stress computation in the direction of strong axis Viewing the wale between two anchors as a simply supported beam structure, the maximum bending moment and shear are 1 2 (10.60) M max = pL1 8

Qmax

1 = pL1 2

(10.61)

where M max = maximum bending moment in the direction of strong axis Qmax = maximum shear in the direction of strong axis p = lateral earth pressure on the wale L1 = horizontal distance between anchors

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, the maximum bending moment can be computed by the following equation：

M max

1 2 = pL 10

(10.62)

The bending stress and shear stress of the wale in the direction of the strong axis should satisfy the following criteria: M max / 2 =  a Sx

Qmax / 2 =  a Aw

(10.63)

(10.64)

S x = section modulus of the member in the direction of the strong axis Aw = area of the web of the member  a = allowable bending stress of the member=(0.6~0.66) Fy  l  a = allowable shear stress of the member=0.4 Fy  l Fy = yielding stress of the member l =short-term magnified factor of the allowable stress

(2) Stress computation in the direction of the weak axis Viewing the wale between the brackets supporting the anchor as a beam, the maximum bending moment and shear of the wale are

Pv = pL1 tan 

(10.65)

M max

1 1 = Pv L2 = pL1L2 tan  4 4

(10.66)

Qmax

1 1 = Pv = pL1 tan  2 2

(10.67)

M max = maximum bending moment in the direction of weak axis Qmax = maximum shear in the direction of weak axis

Pv

L2

= vertical component of the anchorage force at the support = distance between the brackets supporting the anchor

The bending stress and shear stress of the wale in the direction of the weak axis should satisfy the following conditions：

M max / 2 =  a Sy

Qmax / 2 =  a Af

(10.68) (10.69)

S y = section modulus of the member in the direction of weak axis A f = area of a single flange of the member  a = allowable bending stress of the member=0.75 Fy  l  a = allowable shear stress of the member=0.4 Fy  l Fy = yielding stress of the member l = short-term magnified factor of the allowable stress

10.7.5 Design of the Free Section The free section is composed of a tendon and a plastic casing. Three types of tendons are available：steel 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 allowable tensile force can be computed as follows：

Pw = Pu / Ft

(10.70)

where Ft = designed safety factor for the tendon. Table 10.4 lists the commonly used factors of safety.

TABLE 10.4 Factor of safety for single anchor

Classification

Tensile force of tendons ( Ft )

Anchoring Bond force of force tendons ( F ) b ( Fa )

Temporary anchors whose working period is not longer than 6 months and which don’t 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 don’t 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)

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, 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.

The number of tendons is Tw n = Pw

(10.71)

As discussed above, Tw is computed using Eq. 10.59, transforming the horizontal component, as analyzed using the apparent earth pressure method, the beam on elastic foundation method, or the finite element method.

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, the anchor behavior will tend to be fragile, any small displacement possibly bring about large stress, and threatening the security of the anchor. Therefore, the free section has to be at least 4.0 m long.

10.7.6 Design of the Fixed Section 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 bond forces between tendons and cement grout, we can use the values as suggested by JSF (1990), as shown in Table 10.5.

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

Tensile force of tendons ( Ft )

Anchorin g force ( Fa)

Bond force of tendons ( Fb )

Temporary anchors whose working period is not longer than 6 months and which don’t 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 don’t 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)

Classification

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, 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.

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

Designed compressive strength of cement mortar ( kg / cm2 )

Types of tendons

150

180

240

300

>400

Temporary anchors

Steel wire Round 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 wire Round 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： (1) Friction type of anchor The ultimate anchorage force, Tu, for a friction type of anchor can be computed by the following equation：

Tu = d b La ult

(10.72)

Tu = ultimate anchorage force d b = diameter of the fixed section La = length of the fixed section

 ult = average ultimate shear resistance strength per unit area (also called the frictional strength) between the fixed section and soils

The designed load on the anchor is Tw

Tu = Fa

(10.73)

where Fa = safety factor of the designed anchorage force; the commonly used factors of safety are listed in Table 10.4.

The strength between the fixed section and the surrounding soil,  ult , 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 .

(a) Anchorage in rocks 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  ult be 0.1 qu (for block 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 Table 10.6.

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

Type of soil Rock

Gravel

Gravel Clay

Hard rock Soft rock Weathered rock Mudstone

N =10 N =20 N =30 N =40 N =50 N =10 N =20 N =30 N =40 N =50

 ult(kg / cm2 ) 15 ~ 25 10 ~ 15 6 ~ 10 6 ~ 12 1.0 ~ 2.0 1.7 ~ 2.5 2.5 ~ 3.5 3.5 ~ 4.5 4.5 ~ 7.0 1.0 ~ 1.4 1.8 ~ 2.2 2.3 ~ 2.7 2.9 ~ 3.5 3.0 ~ 4.0 1.0 c

Note : N is standard penetration test number, c is cohesion, and  ult is ultimate frictional strength after pressure grouting

(b) Anchorage in sandy soils The  ult -value of an anchor in sandy soils can be computed as

follows：

 ult =  v tan   v



(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 sand is usually determined on the basis of experience or field tests.

(c) Anchorage in clayey soils 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：

 ult = su where s u = undrained shear strength of clay  = reduction factor for undrained shear strength

(10.75)

1.2 Cast-inplace pile

1.0 API



0.8 0.6 0.4 0.2 0.0 0.0

0.2

0.4

0.6

0.8

1.0

su /  v

FIGURE 4.12 Relation between adhesion and undrained shear strength of clay

As Eq. 10.72 shows, the longer the fixed section, the larger the anchorage force. However, many investigators have 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 anchorage force. 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)

Specifications

Minimum suggested distance

No shorter than 3 m, except if the proving test FIP (1982) and result is satisfactory in fulfilling the design, and BSI (1989) 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 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 anchor preload. After locking, the remaining force is called the effective load.

10.7.8 Design of Retaining Walls The stress analysis of an anchored wall can be use the assumed support method, the beam on elastic foundation method, or the finite element method. Because of the installation angle of the anchor, however, a downward dragging force will be produced on the retaining wall. Thus, we have to examine whether the vertical bearing capacity of the wall is larger than the total downward dragging force of each of the levels of anchors. The vertical bearing capacity of the wall equals the sum of point bearing capacity of the wall bottom and the frictional resistance of the wall surface. The computing is similar to that of foundation piles. We can also use Eq. 10.52~10.57 directly.

10.8 Tests of Anchors Influenced by the process of fabrication, grouting, and geological conditions, the actual ultimate load of an anchor has 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 the proving test, the suitability test, and the acceptance test. This section will explicate the objectives and procedures according to the specification of FIP.