21appendix a

8/10/2019 21appendix A

1/91

Appendix A

Worked example 1: Solution based onPDR method

Solution for Case 1

3.4.1 Vertical load capacity

- For the raft: assumed rectangular raft with dimensions of 10m x 6m x 0.52m

1.12 0.1 5.14 0.576u cs c

q F cN MPa= = =

0.576 10 6 34.5ultraft uQ q A MN = = =

- The corresponding factor of safety is

34.5/20 = 1.73, which does not satisfy the design criterion.

- For the pile: assumed mass circle pile with dimensions of 15m long and 0.6m diameter.

( ) (0.6 0.1) 28.26 1.696s s s u s

Q f A S A MN = = = = (compression)

( ) (0.42 0.1) 28.26 1.187s s s u s

Q f A S A MN = = = = (tension)

0(9 ) (9 0.1 0.018 15) 0.2826 0.331e e e u v eQ q A S A MN = = + = + =

0.024 4.239 0.102p c pW V MN = = =

1.696 0.331 0.102 1.925ultpile s e pQ Q Q W MN = + = + = (compression)

1.187 0.102 1.289ult

pile s pQ Q W MN = + = + = (tension)

- If the raft and pile (assumed 9 piles) capacities are added, the total capacity of the

foundation is

34.5 (9 1.925) 51.83MN+ = (3.41)

- The capacity of a block containing the raft and piles + the capacity of the cap outside the

perimeter of block:

2 (8.6 4.6) 0.1 15 8.6 4.6 (9 0.1 0.018 15) (10 6 8.6 4.6) 0.576

39.6 46.29 11.77 97.66MN

+ + + +

= + + =

(3.42)

- Compare between (3.23) and (3.24): the design value of ultimate capacity of the

foundation is

51.83 MN


51.83/20 = 2.59, which satisfies the design criterion.

3.4.2 Moment capacity

- The maximum ultimate moment sustained by the soil below the raft:


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187

2 20.576 6 1043.2

8 8

urm

p BLM MNm

= = =

- The factor of safety for moment loading:

43.2/25 = 1.73, which does not satisfies the design criterion.

- The ultimate moment capacity of the raft:

271

4

ur

m u u

M V V

M V V

=

27 20 2043.2 1 42.6

4 51.83 51.83urM MNm

= =

- The ultimate moment contributed by the piles:

9

1

1.289 (3 4 3 4 3 0) 30.9up uui ii

M P x MNm=

= = + + =

Puui= 1.289 MN = ultimate uplift capacity of typical pile i

- The total moment capacity:

42.6 + 30.9 = 73.5 MNm (3.43)

- The ultimate moment capacity of the block containing the piles and the soil:

+ The average ultimate lateral pressure along the block (conservatively):Figure 3.71

Figure 3.71 Lateral resistance factors at ground surface (0) and at great depth () (after

Poulos and Davis, 1980)

4.5 0.1 0.45u c u

p K S MPa= = =

+ The ultimate moment capacity of the block:

2 20.25 0.45 6 15 151.9uB B u B B

M p B D MNm= = = (3.44)

- Compare between (3.25) and (3.26): the design value of ultimate moment capacity is


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73.5 MNm



3.4.3 Lateral load capacity

- Sum of the ultimate lateral capacity of the raft + all piles:

+ Short pile failure: Equation (7.11) in Poulos and Davis (1980)

( ) ( )9 1.5 9 0.1 0.6 15 1.5 0.6 7.6u uH S d L d MN= = = (per pile)

For 9 piles, the total lateral capacity is: 9 x 7.6 = 68.5 MN (3.45)

+ Long pile failure: Equations (7.9) and (7.14) in Poulos and Davis (1980)

9 9 0.1 0.6

u u

u

H H

f S d= =

2 2 0.45

(1.5 0.5 ) (1.5 0.6 0.5 )

y

u

MH

d f f

= =

+ +

Hu= 0.61 MN (per pile)


Compare (3.27) and (3.28) ==> choose (3.28): 5.49 MN (3.47)

- The ultimate lateral capacity of the block containing piles-raft-soil:

0.45 6 15 40.5u u B B

H p B D MN= = = (3.48)

Compare (3.29) and (3.30) ==> choose (3.29): 5.49 MN

- The factor of safety against lateral failure is:


3.4.4 Load-settlement behavior

- The following calculations will be carried out.

1. A non-linear analysis to estimate the relationship between load and immediatesettlement. From this curve, the immediate settlement is calculated.

2. A linear analysis of both undrained and drained behavior to obtain, by difference,

the consolidation settlement.

3. Long-term settlement = immediate settlement + consolidation settlement.

- Calculation of raft stiffness (elastic or initial)

saE

KI

= (3.49)

a = foundation radius = ( ) / (6 10) / 3.14 4.371B L m = =


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189

4.3710.17

25

a

h= = , 0.5

u = Figure 3.1 ==> 0.98I =

4.3710.17

25

a

h= = , 0.3 = Figure 3.1 ==> 1.22I =

(a)undrained case:3.14 4.371 30

420 / 0.98

riK MN m

= =

(b)drained case:3.14 4.371 15

169 / 1.22

riK MN m

= =

- For long-term settlements (immediate plus consolidation settlements, but excluding

creep), the applied load of 15/(6 x 10) = 0.25 MPa: (raft alone)

+ Average settlement of the raft:

0.25 6 10 0.0888169

applied

r

r

Pw mk

= = = or 89 mm

+ Differential settlements:

1/2 32

2

15.57

1

srri

s r

E B tK

E L L

=

Horikoshi and Randolph (1997) (3.50)

1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10riK

= =

From Figure 3.4 in Chapter 3:

Mid-side and centre: 0.08 x 89 = 7 mm

Corner and centre: 0.2 x 89 = 18 mm

- Calculation of pile stiffness (elastic or initial):

+ Single pile

(a)undrained case

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.561 15

11 1 0.5 1 4.135 0.635 0.310 0.3

1 4 tanh 1 4 1 0.561 151 1

1 3.14 3000 1 0.5 1 0.635 0.3

l

l l

l rk G r

l l

l r

+ +

= =

+ +

(3.51)

1k = 0.217 MN/mm

3010

2(1 ) 2(1 0.5)

EG MPa

= = =

+ +

r0= d/2 = 0.6/2 = 0.3 m

= rb/r0= 0.3/0.3 = 1= 0.5


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190

/ 10 /10 1l bG G = = =

/ 10 /10 1avg l

G G = = =

/ 30000 /10 3000p l

E G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.5 15 / 0.3 4.135l r = =

( ) ( )02 / / 2 / (4.135 3000) 15 / 0.3 0.635l l r = =

( )2 2 0.635

2 2 0.635

1 1tanh 0.561

1 1

l

l

e el

e e

= =

+ +

(b)drained case:

( )

( )

( )

( )

01 0

0

4 2 tanh 4 1 2 3.14 0.434 151

1 1 0.3 1 4.472 0.465 0.35.8 0.31 4 tanh 1 4 1 0.434 15

1 11 3.14 5172.4 1 0.3 1 0.465 0.3

l

l l

l rk Grl l

l r

+ +

= = + +

(3.52)

1k = 0.122 MN/mm

155.8

2(1 ) 2(1 0.3)

EG MPa

= = =

+ +

r0= d/2 = 0.6/2 = 0.3 m

= rb/r0= 0.3/0.3 = 1

= 0.3

/ 5.8 / 5.8 1l b

G G = = =

/ 5.8 / 5.8 1avg l

G G = = =

/ 30000 / 5.8 5172.4p l

E G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.3 15 / 0.3 4.472l r = =

( ) ( )02 / / 2 / (4.472 5172.4) 15 / 0.3 0.465l l r = =

( )2 2 0.465

2 2 0.465

1 1tanh 0.434

1 1

l

l

e el

e e

= =

+ +

+ Piled group:

Assuming that the group factor is approximated as pn (where npis the number of piles),

the following initial piled group stiffness are obtained:

(a) undrained case: 1 217 9 651 / pi pK K n MN m= = =

(b) drained case: 2 122 9 366 / pi pK K n MN m= = =


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+ Piled raft: pri piK X K=

(a)undrained case:

( )

( )

( )

( )

1 0.6 / 1 0.6 420 / 6511.044

1 0.64 420 / 6511 0.64 /

r p

r p

K KX

K K

= =

1.044 651 680 / ue piK X K MN m= = =

(b)drained case:

( )

( )( )

( )

1 0.6 / 1 0.6 169 / 3661.026

1 0.64 169 / 3661 0.64 /

r p

r p

K KX

K K

= =

1.026 366 375 / e piK X K MN m = = =

- Proportion of load carried initially by the piles, p :

(a)undrained case:

( )

0.2 0.2 4200.267

1 0.8( / ) 1 0.8 420 / 651 651

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.267) 0.79p

= + = + =

(b)drained case:

( )0.2 0.2 169 0.146

1 0.8( / ) 1 0.8 169 / 366 366r

r p p

KK K K

= =

1 / (1 ) 1/ (1 0.146) 0.87p = + = + =

- For the undrained case, the non-linear analysis is tabulated in Table 3.1, assuming that

the hyperbolic factors are Rfr= 0.75 and Rfp= 0.5. For each applied load, the values of

p and X from the previous load are used, starting with the initial values for the first load.

( )( )

1 0.6 /

1 0.64 /

r p

r p

K KXK K

0.2

1 0.8( / )

r

r p p

K

K K K

1/ (1 )p = +

p p puV V V= r pV V V= ( )1 /p pi fp p puK K R V V =

( )1 /r ri fr r ruK K R V V = pu

A

p

VV

= Vpu= ultimate capacity of piles

:AV V

1

fp p

pipu

VS

R V

XK V

=


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192

:A

V V> ( ) ( )1

1

A A

pi fp pu

ri fr

ru

V V VS

XK R V VK R

V

= +

puV = ultimate capacity of piles (single pile or block failure, whichever is less).

ruV = ultimate capacity of raft.

Table 3.5 Calculation of load-settlement curve for piled raft foundation in worked

example (undrained case)

V Vp Vr Kr Kp VA S

(MN) X (MN) (MN) (MN/m) (MN/m) (MN) (mm) V>VA

0 1.044 0.790 0.00 0.00 420.0 651.0 21.9 0.0 No

5 1.044 0.789 3.95 1.05 410.4 576.8 21.9 8.3 No10 1.052 0.752 7.52 2.48 397.3 509.8 23.0 18.6 No

15 1.062 0.707 10.61 4.39 379.9 451.7 24.5 31.3 No

20 1.073 0.660 13.21 6.79 358.0 402.8 26.2 46.3 No

25 1.082 0.619 15.48 9.52 333.1 360.2 28.0 64.1 No

30 - - 17.53 12.47 306.2 321.6 28.0 86.1 Yes

35 - - 17.53 17.47 260.5 321.6 28.0 106.6 Yes

40 - - 17.53 22.47 214.9 321.6 28.0 135.9 Yes

45 - - 17.53 27.47 169.2 321.6 28.0 181.1 Yes

50 - - 17.53 32.47 123.6 321.6 28.0 260.4 Yes

52 - - 17.53 34.47 105.3 321.6 28.0 311.7 Yes

+ The computed load-settlement curve is shown in Figure 3.3. At the long-term design

load of 15 MN, the calculated immediate settlement is 31 mm.

Piled raft

Piles

Raft

Figure 3.72 Calculated load-settlement curve for piled raft foundation in worked example

(undrained case).

- It will be assumed that the final consolidation settlement ( )CFS can be computed as the

Verticalappliedload:MN

Settlement: mm

p


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193

difference between the total final and immediate settlements from purely elastic analyses,

so that

1 115 0.0179

375 680CF

e ue

V VS m

K K

= = =

(3.53)

- Thus, the estimated total final settlement is

1 1( ) 0.0313 0.0179 0.0492

TF

u e ue

VS V m

K K K= + = + =

or 49 mm

This satisfies the design criterion of 50 mm maximum long-term settlement.

3.4.5 Differential settlement

- The simplifying assumption is made that the vertical load is uniformly distributed on the

raft. The raft-soil stiffness is defined herein as

1/2 32

2

15.57

1

srrs

s r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10rs

K

= =

- From Figure 3.2: the ratio of the maximum differential settlement to the average

settlement is 0.2 (corner) and 0.08 (mid-side). Assuming that this ratio applies also to the

piled raft, the maximum long-term differential settlement (centre-to-corner) is 0.2 x

0.0492 = 0.00984 m (or 9.8 mm) and (centre-to-midside) is 0.08 x 0.0492 = 0.00394 m

(or 3.9 mm). This satisfies the design criterion of 10 mm maximum long-term differential

settlement.

3.4.6 Pile loads

- At the design ultimate load of 20 MN, the proportion of load carried by the piles (from

Table 3.1) is given by 0.660p =

. Then

max 2

2 2

1 1

20 0.660 25 40

9 6 4p pp y ix i

n n

p

i i

i i

V M yM xP

nx y

= =

= + + = + +

1.47 1.04 2.51MN= + =

min 2

2 2

1 1

20 0.660 25 40

9 6 4p pp y ix i

n n

p

i i

i i

V M yM xP

nx y

= =

= =

1.47 1.04 0.43MN= =


9/91

194

- The maximum axial piled load of 2.51 MN exceeds the ultimate geotechnical piled load

capacity of 1.925 MN, thus implying that the capacity of the outer piles is fully utilized.

3.4.7 Raft bending moments and shears

- Long-term case (purely vertical loading) is considered and the applied loading is

assumed to be uniformly distributed. The average applied pressure is 15/(6 10) = 0.25

MPa and the piles take 87% of the applied load.

+ The average raft contact pressure is

0.25 0.87 x 0.25 = 0.0325 MPa

+ The average load in each pile is

(0.87 x 15)/9 = 1.45 MN

x

y

Figure 3.73 Diving of raft into three strips of equal width (B1 = 2 m)

1.45MN1.45MN1.45MN

0.065MN/m

0.5MN/m

(a) Load diagram

Q (MN)

Shear:MN

Length (m)


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195

(b) Shear diagram

M (MNm)

(c) Moment diagram

Figure 3.74 Load, shear and moment diagrams for strip

- Dividing the raft into three strips of equal width (in each direction) and calculating the

maximum positive (sagging) and the corresponding maximum negative (hogging)

bending moments based on simple statics (Poulos (1991)). Figure 3.4 shows the method

used for dividing of the raft and Figure 3.5a presents the load diagram.

+ In Figure 3.5a:

Applied load on trip (B = 2m): 0.25 x 2 = 0.5 MPa

Pressure under the trip (B = 2m): 0.0325 x 2 = 0.065 MPa

Load of each pile: 1.45 MN

+ Maximum positive bending moments (Figure 3.5c):

In x-direction:

Mx= 0.967/2 = 0.484 MNm/m (at x = 10/3 m) (B = 2 m)

In y-direction:

My= 0+ Maximum negative bending moments (Figure 3.5c):

In x-direction:

Mx= -0.218/2 = - 0.109 MNm/m

In y-direction:

My= -0.218/2 = - 0.109 MNm/m

+ Maximum shear (Figure 3.5b):

Qmax

= +1.015/2 = + 0.508 MN/m

+ Minimum shear (Figure 3.5b):

Bendingmoment:MNm

Length (m)


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196

Qmin= -1.015/2 = - 0.508 MN/m

Solution for Case 2

1. Vertical load capacity- For the raft: assumed rectangular raft with dimensions of 10m x 6m x 0.52m

1.12 0.1 5.14 0.576u cs c

q F cN MPa= = =

0.576 10 6 34.5ultraft uQ q A MN = = =


( ) (0.6 0.1) 33.912 2.035s s s u s


( ) (0.42 0.1) 33.912 1.424s s s u s


0(9 ) (9 0.1 0.018 21.6) 0.19625 0.253

e e e u v eQ q A S A MN = = + = + =

0.024 4.239 0.102p c pW V MN = = =


1.424 0.102 1.526ultpile s pQ Q W MN = + = + = (tension)


foundation is

34.5 (9 2.186) 54.174MN+ = (3.55)


perimeter of block:

2 (8.5 4.5) 0.1 21.6 8.5 4.5 (9 0.1 0.018 21.6) (10 6 8.5 4.5) 0.576

56.16 49.3 12.528 117.988MN

+ + + +

= + + =(3.56)


foundation is

54.174 MN



2. Moment capacity


2 20.576 6 1043.2

8 8

urm

p BLM MNm

= = =



12/91

197

271

4

ur

m u u

M V V

M V V

=

27 20 2043.2 1 42.2

4 54.174 54.174urM MNm

= =


9

1

1.526 (3 4 3 4 3 0) 36.624up uui ii

M P x MNm=

= = + + =



42.2 + 36.624 = 78.8 MNm (3.57)


+ The average ultimate lateral pressure along the block (conservatively): Figure 7.7 in

Poulos and Davis (1980)

4.5 0.1 0.45u c u

p K S MPa= = =


2 20.25 0.45 6 21.6 314.9

uB B u B BM p B D MNm= = = (3.58)


78.8 MNm



3. Lateral load capacity



( ) ( )9 1.5 9 0.1 0.5 21.6 1.5 0.5 9.38u uH S d L d MN= = =



9 9 0.1 0.5

u u

u

H Hf

S d= =

2 2 0.45

(1.5 0.5 ) (1.5 0.5 0.5 )

y

u

MH

d f f

= =

+ +

Hu= 0.62 MNFor 9 piles, the total lateral capacity is: 9 x 0.62 = 5.61 MN (3.60)


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198



0.45 6 21.6 58.3u u B B

H p B D MN= = = (3.62)




4. Load-settlement behavior


4. A non-linear analysis to estimate the relationship between load and immediate

settlement. From this curve, the immediate settlement is calculated.





saE

KI

= (3.63)


4.3710.17

25

a

h= = , 0.5

u = Figure 3.1 ==> 0.98I =

4.3710.17

25

a

h= = , 0.3 = Figure 3.1 ==> 1.22I =

(c)undrained case:3.14 4.371 30

420 / 0.98

riK MN m

= =

(d)drained case:3.14 4.371 15

169 / 1.22

riK MN m

= =




0.25 6 100.0888

169

applied

r

r

Pw m

k

= = = or 89 mm


1/2 32

215.57 1sr

ri

s r

E B tKE L L

= Horikoshi and Randolph (1997) (3.64)


14/91

199

1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10ri

K

= =


Mid-side and centre: 0.08 x 89 = 7 mmCorner and centre: 0.2 x 89 = 18 mm


+ Single pile

(c)undrained case

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.774 21.61

1 1 0.5 1 4.682 1.031 0.2510 0.25

1 4 tanh 1 4 1 0.774 21.61 11 3.14 3000 1 0.5 1 1.031 0.25

l

l l

l rk G r

l l

l r

+ +

= =

+ +

(3.65)

1k = 0.225 MN/mm

3010

2(1 ) 2(1 0.5)

EG MPa

= = =

+ +

r0= d/2 = 0.5/2 = 0.25 m

= rb/r0= 0.25/0.25 = 1

= 0.5

/ 10 /10 1l b

G G = = =

/ 10 /10 1avg l

G G = = =

/ 30000 /10 3000p l

E G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.5 21.6 / 0.25 4.682l r = =

( ) ( )02 / / 2 / (4.682 3000) 21.6 / 0.25 1.031l l r = =

( )2 2 1.031

2 2 1.0311 1tanh 0.7741 1

l

le ele e

= = + +

(d)drained case:

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.64 21.61

1 1 0.3 1 5.019 0.758 0.255.8 0.25

1 4 tanh 1 4 1 0.64 21.61 1

1 3.14 5172.4 1 0.3 1 0.758 0.25

l

l l

l rk Gr

l l

l r

+ +

= = + +

(3.66)

1k = 0.137 MN/mm


15/91


16/91

201

( )

0.2 0.2 4200.248

1 0.8( / ) 1 0.8 420 / 675 675

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.248) 0.80p = + = + =

(d)drained case:

( )

0.2 0.2 1690.123

1 0.8( / ) 1 0.8 169 / 411 411

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.123) 0.89p = + = + =



p

and X from the previous load are used, starting with the initial values for the first load.



V Vp Vr Kr Kp VA S


0 1.041 0.800 0.00 0.00 420.0 675.0 24.6 0.0 No

5 1.041 0.801 4.01 0.99 410.9 606.3 24.5 7.9 No

10 1.048 0.772 7.72 2.28 399.1 542.6 25.5 17.6 No

15 1.056 0.737 11.05 3.95 383.9 485.4 26.7 29.3 No20 1.064 0.699 13.98 6.02 365.0 435.2 28.1 43.2 No

25 1.072 0.662 16.56 8.44 342.9 391.0 29.7 59.6 No

30 1.080 0.630 18.89 11.11 318.6 350.9 31.2 79.2 No

35 - - 21.04 13.96 292.6 314.0 31.2 99.1 Yes

40 - - 21.04 18.96 246.9 314.0 31.2 123.1 Yes

45 - - 21.04 23.96 201.3 314.0 31.2 158.6 Yes

50 - - 21.04 28.96 155.6 314.0 31.2 216.8 Yes

52 - - 21.04 30.96 137.3 314.0 31.2 252.0 Yes

( )( )

1 0.6 /

1 0.64 / r p

r p

K KX

K K

0.21 0.8( / )

r

r p p

KK K K

1/ (1 )p = +



A

p

VV


:AV V

1fp p

pi

pu

VS

R VXK

V

=

p


17/91

202

:A

V V> ( ) ( )1

1

A A

pi fp pu

ri fr

ru

V V VS

XK R V VK R

V

= +





Piled raft

Piles

Raft


(undrained case).



so that

1 115 0.0144

420 703CF

e ue

V VS m

K K

= = =

(3.67)


1 1( ) 0.0293 0.0144 0.0437

TF

u e ue

VS V m

K K K= + = + =

or 44 mm


5. Differential settlement




Settlement: mm


18/91

203

1/2 32

2

15.57

1

srri

s r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10ri

K

= =






settlement.

6. Pile loads


Table 3.1) is given by 0.699p

= . Then

max 2

2 2

1 1

20 0.699 25 40

9 6 4p pp y ix i

n n

p

i i

i i

V M yM xP

nx y

= =

= + + = + +

1.55 1.04 2.59MN= + =

min 2

2 2

1 1

20 0.699 25 40

9 6 4p pp y ix i

n n

p

i i

i i

V M yM xP

nx y

= =

= =

1.55 1.04 0.51MN= =



7. Raft bending moments and shears- Long-term case (purely vertical loading) is considered and the applied loading is




0.25 0.89 x 0.25 = 0.0275 MPa


(0.89 x 15)/9 = 1.48 MN



19/91

204




+ Maximum positive bending moments (Figure 3.5c):In x-direction:

Mx= 0.989/2 = 0.495 MNm/m

x

y


1.48MN1.48MN1.48MN

0.055MN/m

0.5MN/m

(a) Load diagram

Q (MN)

(b) Shear diagram

Sh

ear:MN

Length (m)


20/91

205

M (MNm)

(c) Moment diagram


In y-direction:

My= 0

+ Maximum negative bending moments (Figure 3.5c):

In x-direction:

Mx= -0.223/2 = - 0.112 MNm/m

In y-direction:

My= -0.223/2 = - 0.112 MNm/m


Qmax= +1.038/2 = + 0.519 MN/m


Qmin= -1.038/2 = - 0.519 MN/m

Solution for Case 3

1. Vertical load capacity


1.12 0.1 5.14 0.576u cs c

q F cN MPa= = =

0.576 10 6 34.5ultraft uQ q A MN = = =


( ) (0.6 0.1) 33.912 2.035s s s u s


( ) (0.42 0.1) 33.912 1.424s s s u s


0(9 ) (9 0.1 0.018 21.6) 0.19625 0.253e e e u v eQ q A S A MN = = + = + =

0.024 4.239 0.102p c p

W V MN = = =


Bendingmo

ment:MNm

Length (m)


21/91

206



foundation is

34.5 (9 2.186) 54.174MN+ = (3.69)


perimeter of block:

2 (43 / 6 23 / 6) 0.1 21.6 (43 /6) (23 / 6) (9 0.1 0.018 21.6)

(10 6 (43 / 6) (23 / 6)) 0.576

47.52 35.4062 18.736 101.662MN

+ + +

+

= + + =

(3.70)


foundation is54.174 MN



2. Moment capacity


2 20.576 6 10

43.28 8

ur

m

p BLM MNm

= = =


271

4

ur

m u u

M V V

M V V

=

27 20 2043.2 1 42.2

4 54.174 54.174urM MNm

= =


9

1

1.526 (3 4 3 4 3 0) 36.624up uui ii

M P x MNm=

= = + + =



42.2 + 36.624 = 78.8 MNm (3.71)





22/91

207

4.5 0.1 0.45u c up K S MPa= = =


2 20.25 0.45 6 21.6 314.9uB B u B B

M p B D MNm= = = (3.72)


78.8 MNm






( ) ( )9 1.5 9 0.1 0.5 21.6 1.5 0.5 9.38u uH S d L d MN= = =



9 9 0.1 0.5

u u

u

H Hf

S d= =

2 2 0.45

(1.5 0.5 ) (1.5 0.5 0.5 )

y

u

MH

d f f

= =

+ +

Hu= 0.62 MN




0.45 6 21.6 58.3u u B B

H p B D MN= = = (3.76)


- The factor of safety against lateral failure is:5.61/2 = 2.81, which satisfies the design criterion.



7. A non-linear analysis to estimate the relationship between load and immediate





23/91

208



saE

KI

= (3.77)


4.3710.17

25

a

h= = , 0.5

u = Figure 3.1 ==> 0.98I =

4.3710.17

25

a

h= = , 0.3 = Figure 3.1 ==> 1.22I =

(e)undrained case:3.14 4.371 30

420 / 0.98

riK MN m

= =

(f) drained case:3.14 4.371 15

169 / 1.22

riK MN m

= =




0.25 6 100.0888

169

applied

r

r

Pw m

k

= = = or 89 mm


1/2 32

2

15.57

1

srri

s r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10riK

= =





+ Single pile

(e)undrained case

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.774 21.61

1 1 0.5 1 4.682 1.031 0.2510 0.25

1 4 tanh 1 4 1 0.774 21.61 1

1 3.14 3000 1 0.5 1 1.031 0.25

l

l l

l rk G r

l l

l r

+ +

= = + +

(3.79)


24/91

209

1k = 0.225 MN/mm

3010

2(1 ) 2(1 0.5)

EG MPa

= = =

+ +

r0= d/2 = 0.5/2 = 0.25 m= rb/r0= 0.25/0.25 = 1

= 0.5

/ 10 /10 1l b

G G = = =

/ 10 /10 1avg l

G G = = =

/ 30000 /10 3000p l

E G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.5 21.6 / 0.25 4.682l r = =

( ) ( )02 / / 2 / (4.682 3000) 21.6 / 0.25 1.031l l r = =

( )2 2 1.031

2 2 1.031

1 1tanh 0.774

1 1

l

l

e el

e e

= =

+ +

(f) drained case:

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.64 21.61

1 1 0.3 1 5.019 0.758 0.255.8 0.25

1 4 tanh 1 4 1 0.64 21.61 11 3.14 5172.4 1 0.3 1 0.758 0.25

l

l l

l rk Gr

l l

l r

+ +

= =

+ +

(3.80)

1k = 0.137 MN/mm

155.8

2(1 ) 2(1 0.3)

EG MPa

= = =

+ +

r0= d/2 = 0.5/2 = 0.25 m

= rb/r0= 0.25/0.25 = 1

= 0.3

/ 5.8 / 5.8 1l b

G G = = =

/ 5.8 / 5.8 1avg l

G G = = =

/ 30000 / 5.8 5172.4p lE G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.3 21.6 / 0.25 5.019l r = =

( ) ( )02 / / 2 / (5.019 5172.4) 21.6 / 0.25 0.758l l r = =

( )2 2 0.758

2 2 0.7581 1tanh 0.641 1

l

le ele e

= = + +


25/91

210

+ Piled group:

Assuming that the group factor is approximated asp

n (where npis the number of piles),




+ Piled raft:pri pi

K X K=

(e)undrained case:

( )

( )( )

( )

1 0.6 / 1 0.6 420 / 6751.041

1 0.64 420 / 6751 0.64 /

r p

r p

K KX

K K

= =

1.041 675 703 / ue pi

K X K MN m= = =

(f)drained case:

( )

( )( )

( )

1 0.6 / 1 0.6 169 / 4111.022

1 0.64 169 / 4111 0.64 /

r p

r p

K KX

K K

= =

1.022 411 420 / e pi

K X K MN m = = =

- Proportion of load carried initially by the piles,p

:

(e)undrained case:

( )

0.2 0.2 4200.248

1 0.8( / ) 1 0.8 420 / 675 675

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.248) 0.80p

= + = + =

(f)drained case:

( )

0.2 0.2 1690.123

1 0.8( / ) 1 0.8 169 / 411 411

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.123) 0.89p

= + = + =





example (undrained case)V Vp Vr Kr Kp VA S

p


26/91

211


0 1.041 0.803 0.00 0.00 420.0 675.0 24.5 0.0 No

5 1.041 0.801 4.01 0.99 410.9 606.3 24.5 7.9 No

10 1.048 0.772 7.72 2.28 399.1 542.6 25.5 17.6 No

15 1.056 0.737 11.05 3.95 383.9 485.4 26.7 29.3 No

20 1.064 0.699 13.98 6.02 365.0 435.2 28.1 43.2 No

25 1.072 0.662 16.56 8.44 342.9 391.0 29.7 59.6 No

30 1.080 0.630 18.89 11.11 318.6 350.9 31.2 79.2 No

35 21.04 13.96 292.6 314.0 31.2 99.1 Yes

40 21.04 18.96 246.9 314.0 31.2 123.1 Yes

45 21.04 23.96 201.3 314.0 31.2 158.6 Yes

50 21.04 28.96 155.6 314.0 31.2 216.8 Yes

52 21.04 30.96 137.3 314.0 31.2 252.0 Yes

( )

( )

1 0.6 /

1 0.64 /

r p

r p

K K

X K K

0.2

1 0.8( / )

r

r p p

K

K K K

1/ (1 )p = +

p p puV V V=

r pV V V= ( )1 /p pi fp p puK K R V V =


A

p

VV


:A

V V

1fp p

pi

pu

VS

R VXK

V

=

:A

V V> ( ) ( )1

1

A A

pi fp pu

ri fr

ru

V V VS

XK R V VK R

V

= +



+ The computed load-settlement curve is shown in Figure 3.3. At the long-term designload of 15 MN, the calculated immediate settlement is 29 mm.


27/91

212

Piled raft

Piles

Raft


(undrained case).



so that

1 115 0.0144

420 703CF

e ue

V VS m

K K

= = =

(3.81)


1 1( ) 0.0293 0.0144 0.0437

TF

u e ue

VS V m

K K K

= + = + =

or 44 mm





1/2 32

2

15.57

1

srri

s r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10ri

K

= =






settlement.

Verticalappli

edload:MN

Settlement: mm


28/91

213

6. Pile loads



= . Then

max 2

2 2

1 1

20 0.699 25 40

9 6 4p pp y ix i

n n

p

i i

i i

V M yM xP

nx y

= =

= + + = + +

1.55 1.04 2.59MN= + =

min 2

2 2

1 1

20 0.699 25 40

9 6 4p pp y ix i

n n

p

i i

i i

V M yM xP

nx y

= =

= =

1.55 1.04 0.51MN= =



7. Raft bending moments and shears





0.25 0.89 x 0.25 = 0.0275 MPa


(0.89 x 15)/9 = 1.48 MN






In x-direction:

Mx= 0/2 = 0 MNm/m

In y-direction:

My= 0


In x-direction:Mx= -0.618/2 = -0.309 MNm/m


29/91

214

In y-direction:

My= -0.618/2 = -0.309 MNm/m


Qmax= +0.742/2 = + 0.371 MN/m+ Minimum shear (Figure 3.5b):

Qmin= -0.742/2 = - 0.371 MN/m

x

y


1.48MN1.48MN1.48MN

0.055MN/m

0.5MN/m

(a) Load diagram

Q (MN)

(b) Shear diagram

Shear:MN

Length (m)


30/91

215

M (MNm)

(c) Moment diagram


Solution for Case 4

1. Vertical load capacity- For the raft: assumed rectangular raft with dimensions of 10m x 6m x 0.52m

1.12 0.1 5.14 0.576u cs c

q F cN MPa= = =

0.576 10 6 34.5ultraft uQ q A MN = = =


( ) (0.6 0.1) 16.956 1.017s s s u s


( ) (0.42 0.1) 16.956 0.712s s s u s


0(9 ) (9 0.1 0.018 9) 0.2826 0.3e e e u v eQ q A S A MN = = + = + =

0.024 2.5434 0.061p c pW V MN = = =



- If the raft and pile (15 piles) capacities are added, the total capacity of the foundation is

34.5 (15 1.256) 53.34MN+ = (3.83)


perimeter of block:

2 (8.6 4.6) 0.1 9 8.6 4.6 (9 0.1 0.018 9) (10 6 8.6 4.6) 0.576

23.76 42.01 11.77 77.54MN

+ + + +

= + + = (3.84)


foundation is

53.34 MN


Bendingmoment:MNm

Length (m)


31/91

216


2. Moment capacity


2 20.576 6 1043.2

8 8

urm

p BLM MNm

= = =


271

4

ur

m u u

M V V

M V V

=

27 20 2043.2 1 42.4

4 53.34 53.34urM MNm

= =


9

1

0.773 (3 4 3 2 3 2 3 4 3 0) 27.828up uui ii

M P x MNm=

= = + + + + =



42.4 + 27.828 = 70.2 MNm (3.85)




4.5 0.1 0.45u c u

p K S MPa= = =


2 20.25 0.45 6 9 54.7

uB B u B BM p B D MNm= = = (3.86)


54.7 MNm


54.7/25 = 2.19, which does not satisfy the design criterion. Piled length is increased up to

9.7 m.




( ) ( )9 1.5 9 0.1 0.6 9 1.5 0.6 4.37u uH S d L d MN= = =


32/91

217



9 9 0.1 0.6

u u

u

H Hf

S d= =

2 2 0.45

(1.5 0.5 ) (1.5 0.6 0.5 )

y

u

MH

d f f

= =

+ +

Hu= 0.61 MN


Compare (3.27) and (3.28) ==> choose (3.28):9.15 MN (3.89)


0.45 6 9 24.3u u B B

H p B D MN= = = (3.90)






10.A non-linear analysis to estimate the relationship between load and immediate


11.A linear analysis of both undrained and drained behavior to obtain, by difference,


12.Long-term settlement = immediate settlement + consolidation settlement.


saE

KI

= (3.91)


4.3710.17

25

a

h= = , 0.5

u = Figure 3.1 ==> 0.98I =

4.3710.17

25

a

h= = , 0.3 = Figure 3.1 ==> 1.22I =

(g)undrained case:3.14 4.371 30

420 / 0.98

riK MN m

= =

(h)drained case:3.14 4.371 15

169 / 1.22riK MN m

= =


33/91

218




0.25 6 10 0.0888169

appliedr

r

Pw mk

= = = or 89 mm


1/2 32

2

15.57

1

srri

s r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10ri

K

= =





+ Single pile

(g)undrained case

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.407 91

1 1 0.5 1 3.624 0.386 0.310 0.31 4 tanh 1 4 1 0.407 9

1 11 3.14 3000 1 0.5 1 0.386 0.3

l

l l

l rk G rl l

l r

+ +

= = + +

(3.93)

1k = 0.184 MN/mm

3010

2(1 ) 2(1 0.5)

EG MPa

= = =

+ +

r0= d/2 = 0.6/2 = 0.3 m

= rb/r0= 0.3/0.3 = 1

= 0.5

/ 10 /10 1l b

G G = = =

/ 10 /10 1avg l

G G = = =

/ 30000 /10 3000p l

E G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.5 9 / 0.3 3.624l r = =

( ) ( )02 / / 2 / (3.624 3000) 9 / 0.3 0.407l l r = =


34/91

219

( )2 2 0.407

2 2 0.407

1 1tanh 0.386

1 1

l

l

e el

e e

= =

+ +

(h)drained case:

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.288 911 1 0.3 1 3.961 0.296 0.3

5.8 0.31 4 tanh 1 4 1 0.288 9

1 11 3.14 5172.4 1 0.3 1 0.296 0.3

l

l ll r

k Grl l

l r

+ +

= = + +

(3.94)

1k = 0.09 MN/mm

155.8

2(1 ) 2(1 0.3)

EG MPa

= = =

+ +

r0= d/2 = 0.6/2 = 0.3 m

= rb/r0= 0.3/0.3 = 1

= 0.3

/ 5.8 / 5.8 1l b

G G = = =

/ 5.8 / 5.8 1avg l

G G = = =

/ 30000 / 5.8 5172.4p l

E G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.3 9 / 0.3 3.961l r = =

( ) ( )02 / / 2 / (3.961 5172.4) 9 / 0.3 0.296l l r = =

( )2 2 0.296

2 2 0.296

1 1tanh 0.288

1 1

l

l

e el

e e

= =

+ +

+ Piled group:






+ Piled raft:pri pi

K X K=

(g)undrained case:

( )

( )( )

( )

1 0.6 / 1 0.6 420 / 7131.038

1 0.64 420 / 7131 0.64 /

r p

r p

K KX

K K

= =

1.038 713 740 / ue piK X K MN m= = =


35/91

220

(h)drained case:

( )

( )( )

( )

1 0.6 / 1 0.6 169 / 3491.028

1 0.64 169 / 3491 0.64 /

r p

r p

K KX

K K

= =

1.028 349 359 / e piK X K MN m = = =


:

(g)undrained case:

( )

0.2 0.2 4200.223

1 0.8( / ) 1 0.8 420 / 713 713

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.223) 0.82p

= + = + =

(h)drained case:

( )

0.2 0.2 1690.158

1 0.8( / ) 1 0.8 169 / 349 349

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.158) 0.86p

= + = + =




( )

( )

1 0.6 /

1 0.64 /

r p

r p

K KX

K K

0.2

1 0.8( / )

r

r p p

K

K K K

1/ (1 )p

= +

p p puV V V=



A

p

VV


:AV V

1fp p

pi

pu

VSR V

XKV

=

:A

V V> ( ) ( )1

1

A A

pi fp pu

ri fr

ru

V V VS

XK R V VK R

V

= +




36/91

221



V Vp Vr Kr Kp VA S


0 1.038 0.820 0.00 0.00 420.0 713.0 23.0 0.0 No

5 1.038 0.818 4.09 0.91 411.7 635.6 23.0 7.6 No

10 1.044 0.788 7.88 2.12 400.7 563.9 23.9 17.0 No

15 1.052 0.752 11.28 3.72 386.1 499.5 25.0 28.5 No

20 1.061 0.712 14.23 5.77 367.4 443.7 26.5 42.5 No

25 1.070 0.671 16.77 8.23 344.9 395.6 28.1 59.0 No

30 19.03 10.97 319.9 352.8 28.1 79.6 Yes

35 19.03 15.97 274.2 352.8 28.1 99.0 Yes

40 19.03 20.97 228.6 352.8 28.1 126.1 Yes

45 19.03 25.97 182.9 352.8 28.1 167.0 Yes

50 19.03 30.97 137.3 352.8 28.1 235.3 Yes52 19.03 32.97 119.0 352.8 28.1 277.6 Yes



Piled raft

Piles

Raft


(undrained case).- It will be assumed that the final consolidation settlement ( )

CFS can be computed as the


so that

1 115 0.0215

359 740CF

e ue

V VS m

K K

= = =

(3.95)


1 1( ) 0.0285 0.0215 0.05TFu e ue

VS V mK K K

= + = + =

or 50 mm

p

Verticalappliedlo

ad:MN

Settlement: mm


37/91

222

This equals the design criterion of 50 mm maximum long-term settlement.




1/2 32

2

15.57

1

srri

s r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10ri

K

= =



piled raft, the maximum long-term differential settlement (centre-to-corner) is 0.2 x 0.05

= 0.01 m (or 10 mm) and (centre-to-midside) is 0.08 x 0.05 = 0.004 m (or 4 mm). This

satisfies the design criterion of 10 mm maximum long-term differential settlement.

6. Pile loads



= . Then

max 2 2

2 2

1 1

20 0.712 25 40

15 6 4 6 2p pp y ix i

n n

p

i i

i i

V M yM xP

nx y

= =

= + + = + +

+

0.95 0.83 1.78MN= + =

min 2 2

2 2

1 1

20 0.712 25 40

15 6 4 6 2p pp y ix i

n n

p

i i

i i

V M yM xP

nx y

= =

= =

+

0.95 0.83 0.12MN= =



7. Raft bending moments and shears






38/91

223

0.25 0.86 x 0.25 = 0.035 MPa


(0.86 x 15)/15 = 0.86 MN

- Dividing the raft into three strips of equal width (in each direction) and calculating themaximum positive (sagging) and the corresponding maximum negative (hogging)



x

y


0.86MN0.86MN0.86MN

0.07MN/m

0.5MN/m

(a) Load diagram

Q(MN)

(b) Shear diagram

Shear:MN

Length (m)


39/91

224

M(MNm)

(c) Moment diagram



In x-direction:

Mx= 0.0/2 = 0 MNm/m

In y-direction:

My= 0


In x-direction:

Mx= -0.215/2 = - 0.108 MNm/m

In y-direction:

My= -0.215/2 = - 0.108 MNm/m


Qmax= +0.43/2 = + 0.215 MN/m


Qmin= -0.43/2 = - 0.215 MN/m

Solution for Case 5



1.12 0.1 5.14 0.576u cs c

q F cN MPa= = =

0.576 10 6 34.5ultraft uQ q A MN = = =


34.5/13.5 = 2.56, which satisfies the design criterion.

- For the pile: assumed mass circle pile with dimensions of 15m long and 0.6m diameter.( ) (0.6 0.1) 28.26 1.696

s s s u sQ f A S A MN = = = = (compression)

Bendingmo

ment:MNm

Length (m)


40/91

225

( ) (0.42 0.1) 28.26 1.187s s s u sQ f A S A MN = = = = (tension)

0(9 ) (9 0.1 0.018 15) 0.2826 0.331e e e u v eQ q A S A MN = = + = + =

0.024 4.239 0.102p c p

W V MN = = =




foundation is

34.5 (9 1.925) 51.83MN+ = (3.97)


perimeter of block:

2 (8.6 4.6) 0.1 15 8.6 4.6 (9 0.1 0.018 15) (10 6 8.6 4.6) 0.576

39.6 46.29 11.77 97.66MN

+ + + +

= + + = (3.98)


foundation is

51.83 MN





2 20.576 6 1043.2

8 8

urm

p BLM MNm

= = =




271

4

ur

m u u

M V V

M V V

=

27 13.5 13.543.2 1 37.2

4 51.83 51.83urM MNm

= =


9

1

1.289 (3 4 3 4 3 0) 30.9up uui ii

M P x MNm=

= = + + =


41/91

226



37.2 + 30.9 = 68.1 MNm (3.99)

- The ultimate moment capacity of the block containing the piles and the soil:+ The average ultimate lateral pressure along the block (conservatively):Figure 3.71

4.5 0.1 0.45u c up K S MPa= = =


2 20.25 0.45 6 15 151.9uB B u B B

M p B D MNm= = = (3.100)


68.1 MNm







For 9 piles, the total lateral capacity is: 9 x 7.6 = 68.5 MN (3.101)+ Long pile failure: Equations (7.9) and (7.14) in Poulos and Davis (1980)

9 9 0.1 0.6

u u

u

H Hf

S d= =

2 2 0.45

(1.5 0.5 ) (1.5 0.6 0.5 )

y

u

MH

d f f

= =

+ +





0.45 6 15 40.5u u B B

H p B D MN= = = (3.104)






42/91

227




14.A linear analysis of both undrained and drained behavior to obtain, by difference,the consolidation settlement.



saE

KI

= (3.105)


4.371 0.1725

ah

= = , 0.5u

= Figure 3.1 ==> 0.98I =

4.3710.17

25

a

h= = , 0.3 = Figure 3.1 ==> 1.22I =

(i) undrained case:3.14 4.371 30

420 / 0.98

riK MN m

= =

(j) drained case:3.14 4.371 15

169 / 1.22

riK MN m

= =


creep), the applied load of 10/(6 x 10) = 1/6 MPa: (raft alone)


(1/ 6) 6 100.0592

169

applied

r

r

Pw m

k

= = = or 59 mm


1/2 32

2

15.57

1

srri

s r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10riK

= =





+ Single pile


43/91

228

(i) undrained case

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.561 151

1 1 0.5 1 4.135 0.635 0.310 0.3

1 4 tanh 1 4 1 0.561 151 1

1 3.14 3000 1 0.5 1 0.635 0.3

l

l l

l rk G r

l l

l r

+ +

= = + +

(3.107)

1k = 0.217 MN/mm

3010

2(1 ) 2(1 0.5)

EG MPa

= = =

+ +

r0= d/2 = 0.6/2 = 0.3 m

= rb/r0= 0.3/0.3 = 1

= 0.5

/ 10 /10 1l bG G = = =

/ 10 /10 1avg l

G G = = =

/ 30000 /10 3000p l

E G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.5 15 / 0.3 4.135l r = =

( ) ( )02 / / 2 / (4.135 3000) 15 / 0.3 0.635l l r = =

( )

2 2 0.635

2 2 0.635

1 1

tanh 0.5611 1

l

l

e e

l e e

= = + +

(j) drained case:

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.434 151

1 1 0.3 1 4.472 0.465 0.35.8 0.3

1 4 tanh 1 4 1 0.434 151 1

1 3.14 5172.4 1 0.3 1 0.465 0.3

l

l l

l rk Gr

l l

l r

+ +

= = + +

(3.108)

1k = 0.122 MN/mm

15 5.82(1 ) 2(1 0.3)

EG MPa

= = =+ +

r0= d/2 = 0.6/2 = 0.3 m

= rb/r0= 0.3/0.3 = 1

= 0.3

/ 5.8 / 5.8 1l b

G G = = =

/ 5.8 / 5.8 1avg l

G G = = =

/ 30000 / 5.8 5172.4p lE G= = =


44/91

229

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.3 15 / 0.3 4.472l r = =

( ) ( )02 / / 2 / (4.472 5172.4) 15 / 0.3 0.465l l r = =

( )

2 2 0.465

2 2 0.465

1 1tanh 0.4341 1

l

l

e el e e

= = + +

+ Piled group:






+ Piled raft:pri pi

K X K=

(i)undrained case:

( )

( )( )

( )

1 0.6 / 1 0.6 420 / 6511.044

1 0.64 420 / 6511 0.64 /

r p

r p

K KX

K K

= =

1.044 651 680 / ue pi

K X K MN m= = =

(j)drained case:

( )( )

( )

( )

1 0.6 / 1 0.6 169 / 3661.026

1 0.64 169 / 3661 0.64 /

r p

r p

K KX

K K

= =

1.026 366 375 / e pi

K X K MN m = = =


:

(i)undrained case:

( )

0.2 0.2 4200.267

1 0.8( / ) 1 0.8 420 / 651 651

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.267) 0.79p

= + = + =

(j)drained case:

( )

0.2 0.2 1690.146

1 0.8( / ) 1 0.8 169 / 366 366

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.146) 0.87p

= + = + =




45/91

230




V Vp Vr Kr Kp VA S


0 1.044 0.790 0.00 0.00 420.0 651.0 21.9 0.0 No

5 1.044 0.789 3.95 1.05 410.4 576.8 21.9 8.3 No

10 1.052 0.752 7.52 2.48 397.3 509.8 23.0 18.6 No

15 1.062 0.707 10.61 4.39 379.9 451.7 24.5 31.3 No

20 1.073 0.660 13.21 6.79 358.0 402.8 26.2 46.3 No

25 1.082 0.619 15.48 9.52 333.1 360.2 28.0 64.1 No

30 - - 17.53 12.47 306.2 321.6 28.0 86.1 Yes

35 - - 17.53 17.47 260.5 321.6 28.0 106.6 Yes

40 - - 17.53 22.47 214.9 321.6 28.0 135.9 Yes

45 - - 17.53 27.47 169.2 321.6 28.0 181.1 Yes

50 - - 17.53 32.47 123.6 321.6 28.0 260.4 Yes

52 - - 17.53 34.47 105.3 321.6 28.0 311.7 Yes

( )

( )

1 0.6 /

1 0.64 /

r p

r p

K KX

K K

0.2

1 0.8( / )

r

r p p

K

K K K

1/ (1 )p

= +



A

p

VV


:A

V V

1fp p

pi

pu

VS

R VXK

V

=

:A

V V> ( ) ( )1

1

A A

pi fp pu

ri fr ru

V V VS

XK R V V

K R V

= +





p


46/91

231

Piled raft

Piles

Raft


(undrained case).

- It will be assumed that the final consolidation settlement ( )CF

S can be computed as the


so that

1 110 0.012

375 680CF

e ue

V VS m

K K

= = =

(3.109)


1 1( ) 0.0186 0.012 0.031

TFu e ue

VS V m

K K K= + = + =

or 31 mm





1/2 32

2

15.57

1

srri

s r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10ri

K

= =



piled raft, the maximum long-term differential settlement (centre-to-corner) is 0.2 x 0.031

= 0.0062 m (or 6 mm) and (centre-to-midside) is 0.08 x 0.031 = 0.0025 m (or 3 mm). This

satisfies the design criterion of 10 mm maximum long-term differential settlement.


Settlement: mm


47/91

232

3.4.6 Pile loads

- At the design ultimate load of 13.5 MN, the proportion of load carried by the piles (from


= . Then

max 2

2 2

1 1

13.5 0.721 17 4 09 6 4p p

p y ix i

n n

p

i i

i i

V M yM xPn

x y

= =

= + + = + +

1.08 0.71 1.79MN= + =

min 2

2 2

1 1

13.5 0.721 17 40

9 6 4p pp y ix i

n n

p

i i

i i

V M yM xP

nx y

= =

= =

1.08 0.71 0.37MN= =


capacity of 1.925 MN, thus implying that the capacity of the outer piles utilized is

1.79/1.925 = 93 %.

3.4.7 Raft bending moments and shears


assumed to be uniformly distributed. The average applied pressure is 10/(6 10) = 1/6



(1/6) 0.87 x (1/6) = 0.0217 MPa


(0.87 x 10)/9 = 0.97 MN

x

y



48/91

233

(29/30)MN

(13/300)MN/m

(1/3)MN/m

1m 4m 4m 1m

(29/30)MN (29/30)MN

(a) Load diagram

Q (MN)

(b) Shear diagram

M (MNm)

(c) Moment diagram






+ In Figure 3.5a:

Applied load on trip (B = 2m): (1/6) x 2 = 1/3 MPa

Pressure under the trip (B = 2m): (13/600) x 2 = (13/300) MPa

Shear

:MN

Length (m)

Bendingmoment:MNm

Length (m)


49/91

234

Load of each pile: (29/30) MN


In x-direction:

Mx= 0.644/2 = 0.322 MNm/m (at x = 10/3 m) (B = 2 m)In y-direction:

My= 0


In x-direction:

Mx= -0.145/2 = - 0.0725 MNm/m

In y-direction:

My= -0.145/2 = - 0.0725 MNm/m


Qmax= + 0.677/2 = + 0.3385 MN/m


Qmin= - 0.677/2 = - 0.3385 MN/m

Solution for Case 6



1.12 0.1 5.14 0.576u cs c

q F cN MPa= = =

0.576 10 6 34.5ultraft uQ q A MN = = =




( ) (0.6 0.1) 28.26 1.696s s s u sQ f A S A MN = = = = (compression)

( ) (0.42 0.1) 28.26 1.187s s s u s


0(9 ) (9 0.1 0.018 15) 0.2826 0.331e e e u v eQ q A S A MN = = + = + =

0.024 4.239 0.102p c p

W V MN = = =





50/91

235

foundation is

34.5 (9 1.925) 51.83MN+ = (3.111)


perimeter of block:2 (8.6 4.6) 0.1 15 8.6 4.6 (9 0.1 0.018 15) (10 6 8.6 4.6) 0.576

39.6 46.29 11.77 97.66MN

+ + + +

= + + =(3.112)


foundation is

51.83 MN





2 20.576 6 10

43.28 8

urm

p BLM MNm

= = =




271

4

ur

m u u

M V V

M V V

=

27 35 3543.2 1 35.1

4 51.83 51.83urM MNm

= =


9

1

1.289 (3 4 3 4 3 0) 30.9up uui i

i

M P x MNm=

= = + + =



35.1 + 30.9 = 66 MNm (3.113)


+ The average ultimate lateral pressure along the block (conservatively):Figure 3.71

4.5 0.1 0.45u c u

p K S MPa= = =



51/91

236

2 20.25 0.45 6 15 151.9

uB B u B BM p B D MNm= = = (3.114)


66 MNm


66/17 = 3.88, which satisfies the design criterion.







9 9 0.1 0.6

u u

u

H Hf

S d= =

2 2 0.45

(1.5 0.5 ) (1.5 0.6 0.5 )

y

u

MH

d f f

= =

+ +





0.45 6 15 40.5u u B BH p B D MN= = = (3.119)








17.A linear analysis of both undrained and drained behavior to obtain, by difference,





52/91

237

saE

KI

= (3.120)


4.371 0.1725

a

h= = , 0.5

u = Figure 3.1 ==> 0.98I =

4.3710.17

25

a

h= = , 0.3 = Figure 3.1 ==> 1.22I =

(k)undrained case:3.14 4.371 30

420 / 0.98

riK MN m

= =

(l) drained case:3.14 4.371 15

169 / 1.22

riK MN m

= =


creep), the applied load of 35/(6 x 10) = 7/12 MPa: (raft alone)


(7 /12) 6 100.2071

169

applied

r

r

Pw m

k

= = = or 207 mm


1/2 32

2

15.57

1

sr

ris r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10ri

K

= =




- Calculation of pile stiffness (elastic or initial):+ Single pile

(k)undrained case

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.561 151

1 1 0.5 1 4.135 0.635 0.310 0.3

1 4 tanh 1 4 1 0.561 151 1

1 3.14 3000 1 0.5 1 0.635 0.3

l

l l

l rk G r

l l

l r

+ +

= = + +

(3.122)

1k = 0.217 MN/mm


53/91

238

3010

2(1 ) 2(1 0.5)

EG MPa

= = =

+ +

r0= d/2 = 0.6/2 = 0.3 m

= rb/r

0= 0.3/0.3 = 1

= 0.5

/ 10 /10 1l b

G G = = =

/ 10 /10 1avg l

G G = = =

/ 30000 /10 3000p lE G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.5 15 / 0.3 4.135l r = =

( ) ( )02 / / 2 / (4.135 3000) 15 / 0.3 0.635l l r = =

( )2 2 0.635

2 2 0.635

1 1tanh 0.561

1 1

l

l

e el

e e

= =

+ +

(l) drained case:

( )

( )

( )

( )

0

1 0

0

4 2 tanh 4 1 2 3.14 0.434 151

1 1 0.3 1 4.472 0.465 0.35.8 0.3

1 4 tanh 1 4 1 0.434 151 1

1 3.14 5172.4 1 0.3 1 0.465 0.3

l

l l

l rk Gr

l l

l r

+ +

= = + +

(3.124)

1k = 0.122 MN/mm

155.8

2(1 ) 2(1 0.3)

EG MPa

= = =

+ +

r0= d/2 = 0.6/2 = 0.3 m

= rb/r0= 0.3/0.3 = 1

= 0.3

/ 5.8 / 5.8 1l bG G

= = =

/ 5.8 / 5.8 1avg lG G = = =

/ 30000 / 5.8 5172.4p l

E G= = =

( ) ( )0ln 2.5 1 / ln 2.5 1 1 0.3 15 / 0.3 4.472l r = =

( ) ( )02 / / 2 / (4.472 5172.4) 15 / 0.3 0.465l l r = =

( )2 2 0.465

2 2 0.465

1 1tanh 0.434

1 1

l

l

e el

e e

= =

+ +

+ Piled group:


54/91

239






+ Piled raft:pri piK X K=

(k)undrained case:

( )

( )( )

( )

1 0.6 / 1 0.6 420 / 6511.044

1 0.64 420 / 6511 0.64 /

r p

r p

K KX

K K

= =

1.044 651 680 / ue piK X K MN m= = =

(l)drained case:

( )

( )( )

( )

1 0.6 / 1 0.6 169 / 3661.026

1 0.64 169 / 3661 0.64 /

r p

r p

K KX

K K

= =

1.026 366 375 / e pi

K X K MN m = = =


:

(k)undrained case:

( )

0.2 0.2 4200.267

1 0.8( / ) 1 0.8 420 / 651 651

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.267) 0.79p

= + = + =

(l)drained case:

( )

0.2 0.2 1690.146

1 0.8( / ) 1 0.8 169 / 366 366

r

r p p

K

K K K

= =

1 / (1 ) 1/ (1 0.146) 0.87p = + = + =






V Vp Vr Kr Kp VA S(MN) X (MN) (MN) (MN/m) (MN/m) (MN) (mm) V>VAp


55/91

240

0 1.044 0.790 0.00 0.00 420.0 651.0 21.9 0.0 No

5 1.044 0.789 3.95 1.05 410.4 576.8 21.9 8.3 No

10 1.052 0.752 7.52 2.48 397.3 509.8 23.0 18.6 No

15 1.062 0.707 10.61 4.39 379.9 451.7 24.5 31.3 No

20 1.073 0.660 13.21 6.79 358.0 402.8 26.2 46.3 No

25 1.082 0.619 15.48 9.52 333.1 360.2 28.0 64.1 No

30 - - 17.53 12.47 306.2 321.6 28.0 86.1 Yes

35 - - 17.53 17.47 260.5 321.6 28.0 106.6 Yes

40 - - 17.53 22.47 214.9 321.6 28.0 135.9 Yes

45 - - 17.53 27.47 169.2 321.6 28.0 181.1 Yes

50 - - 17.53 32.47 123.6 321.6 28.0 260.4 Yes

52 - - 17.53 34.47 105.3 321.6 28.0 311.7 Yes

( )

( )

1 0.6 /

1 0.64 /

r p

r p

K KX

K K

0.2

1 0.8( / )

r

r p p

K

K K K

1/ (1 )p

= +

p p puV V V=



A

p

VV


:A

V V

1fp p

pi

pu

VS

R VXK

V

=

:AV V> ( ) ( )1

1

A A

pi fp pu

ri fr

ru

V V VS

XK R V VK R

V

= +






56/91

241

Piled raft

Piles

Raft


(undrained case).

- It will be assumed that the final consolidation settlement ( )CF

S can be computed as the


so that

1 135 0.0419

375 680CF

e ue

V VS m

K K

= = =

(3.125)


1 1( ) 0.1066 0.0419 0.1485

TFu e ue

VS V m

K K K= + = + =

or 149 mm

This exceeds the design criterion of 50 mm maximum long-term settlement.




1/2 32

2

15.57

1

srri

s r

E B tK

E L L

=


1/2 32

2

30000 1 0.3 6 0.525.57 1.15

15 1 0.2 10 10ri

21appendix a

Documents