• Author / Uploaded
• Arnel Suarez

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13.2

A static pile load test was carried on a 0.8 m diameter pile installed 21 m into a loose to medium sandy soil. The pile was driven into the soil. Selected load-displacement data are shown in the table below. (a) Determine the allowable load if the serviceability limit state is 12 mm. (b) Is the maximum load the ultimate load? Justify your answer. (c) Discuss some of the issues you would consider in the interpretation of the data.

Load (kN)

0

800 1100 2250 2800 3200 3500 3600 3620 3618

Displacement (mm)

0

2.5

3.8

7.5

10 12.5

15

20

21

Solution 13.2 4000 3500 3000 2500 Load (kN) 2000 1500 1000 500 0 0

5

10

15

20

Vertical displacement (mm)

(a) Allowable load at displacement of 12 mm = 3100 kN (b) It appears so because the load seems to reach a plateau

25

30

26

13.3

A static pile load test using an O-cell was carried on a 1.8 m diameter, 25 m long (embedded length) drilled shaft. The soil profile is as given in Table P13.3a. Selected load-displacement data are shown in Table P13.3b. (a) Make a neat sketch of the soil profile and the drilled shaft as shown in Example 13.3. (b) Determine the ultimate skin friction and ultimate end bearing capacity. Justify your answer. (c) If an FS of 2 is required, determine the allowable load and settlement. Justify your answers. Table P13.3a Elevation 5 to – 3.4 -3.4 to –17.6 -17.7 to – 38.2 (m) Soil type Sandy fat Silty sand Mudstone or clay (CH) with gravel weak rock (SM) Table P13.3b Load (MN) Displacement up (mm) Displacement down (mm)

0 1 0 0.4 0 -0.5

5 8 10 15 20 25 27 27.2 27.1 0 0.8 1 1.2 1.5 3.4 6.5 8 9.2 10.8 9.0 -6 -10 -11 -16 -21 -29 -43 -40 -41 -37

Solution 13.3 Decide whether the ultimate pile load capacity is well or ill defined. Inspection of the plot shows that the skin friction is fully mobilized at about 9 mm settlement but the end bearing capacity has not been fully mobilized. The initial load-displacement response appears to be from loose material at the bottom the hole 20 10 0 0

Displacement (mm)

5

10

15

20

25

30

35

-10 -20

Possibly loose material

-30 -40 -50

Load (kN)

Determine the ultimate load capacity without consideration of settlement. Qf = 27.2 MN The load to compress the loose material is about 8 MN. The actual base response is Qb > (27.2 MN – 8 MN) > 19.2 MN

Determine the allowable load capacity for FS = 2 and the settlement.

Since the skin friction was fully mobilized, the FS will be applied to it.

The settlement to mobilize a skin friction of 13.6 MN is about 1.3 mm. The end bearing at the same displacement is 2 MN but this is part of the response from the loose material. You can neglect this. The allowable load is 13.6 - 1.53 = 12.1 MN (say 12 MN)

Solution 13.4 Pile : Area = 0.16 m 2 , Perimeter = 1.41m Layer 1 + medium clay Layer 2 = Stiff clay Assume L = 12 m LAYER 1:

5  42.5 kPa 2 f s = 0.5 su 'zo  0.5 30  42.5  17.9kPa

 z  17 

f s = 0.5su0.75  'zo 

0.25

 0.5 300.75  42.5

0.25

 16.4kPa

Use fs = 16.4 kPa β  1  sincs'   OCR  tani  1  sin30 1.6  tan30  0.365 0.5

0.5

LAYER 2: (

( )

)(

(

)

)(

)

f s = 0.5 su 'zo  0.5 60 116.5  41.8kPa f s = 0.5su0.75  'zo 

0.25

 0.5 600.75 116.5

0.25

 35.4kPa

Use fs = 35.4 kPa β  1  sincs'   OCR  tani  1  sin24  4  tan24  0.528 0.5

(

0.5

)

(

Qult  2  275  550kN TSA

Qult = Perimeter



)

f s L  N c  su b Ab

 1.41(16.4  5  35.4  7)  9  60  0.16  551kN  Qult ESA

Qult = perimeter βi  'z   length i  N q  σ'z  Ab j

i =1

i

b

 1.41(0.365  42.5  5  0.528 116.5  7)  12.3 148  0.16  1008kN TSA governs design, L = 12 m is satisfactory

13.5 Determine the allowable load for a steel closed-ended pipe pile, 0.4 m in diameter, driven 20 m into the soil prof le shown in Figure P13.5. Groundwater is at 2 m below the surface, but you can assume it will rise to the surface. A factor of safety of 2 is required. Neglect negative friction.

Solution 13.5 Layer 1: soft clay; Layer 2: Stiff clay Layer 3: Sand Area of pile = 0.13

m 2 , perimeter = 1.26m

TSA (Layers 1 and 2) : fs is lower of

f s = 0.5 su 'zo and f s = 0.5su0.75  'zo 

0.25

Qb = fb Ab = Nc  su b Ab ESA:

β  1  sincs'   OCR  tani 0.5

f s =  'zo Q f    'x  tani'   Perimeter i   Length i j

i

i 1

Qb = fb Ab = N q  σ'z  Ab



b

Janbu: N q = tan '  1  tan 2  '

 exp(2ψ tan ) 2

'

p

Textbook: Nq = 0.6exp(0.126cs' ) (This equationused in the calculation of end bearing capacity) Assuming no negative friction.

Diameter (m)

0.4

Perimeter (m)

1.26

2

Area (m )

0.13

c = clay, s = sand

layer groundwater 1 2 3

Depth m 0 5 10 20

Soil Type

thickness m 0 5 5 10

Clay TSA ESA

c c s

Sand ESA

Unit weight kN/m3 0 18 18.5 17.5

' deg 0 25 23 32

Clay/sand

su kPa 0 15 65 0

O C R 0 1 5 0

FRICTION TSA ESA

center

center

center

base

Total stress kPa 0 45.0 136.3 270.0

Porewater pressure kPa 0.0 24.5 73.5 147.0

Effective stress kPa 0.0 20.5 62.8 123.0

Effective stress kPa 0.0

END BEARING TSA ESA

TSA

Qult ESA

kN

kN

0 51 252 1392

0 35 263 1403

Layer  1 2 3

  0.00 0.27 0.58 0.29

fs kPa 0.0 8.1 31.9 0.0

fs kPa 0.0 5.5 36.3 0.0

TAS control

Q ult = 1392 kN; Q a= 1392/2 =696 kN

fs kPa 0.0

36.1

Nq 0.0

33.8

Qf kN

Qf kN

Qb kN

0 51 252 706

0 35 263 717

0 0 0 686

Qb kN 0 0 0 686

161.5

13.6 A square precast concrete pile of sides 0.4 m is to be driven 12 m into the soil strata shown in Figure P13.6. Estimate the allowable load capacity for a factor of safety of 2. Owing to changes in design requirements, the pile must support 20% more load. Determine the additional embedment depth required.

Solution 13.6 Layer 1: soft clay; Layer 2: Stiff clay Layer 3: Sand TSA (Layers 1 and 2) : fs is lower of

f s = 0.5 su 'zo and f s = 0.5su0.75  'zo 

0.25

Qb = fb Ab = Nc  su b Ab ESA:

β  1  sincs'   OCR  tani 0.5

f s =  'zo Q f    'x  tani'   Perimeter i   Length i j

i

i 1

Qb = fb Ab = N q  σ'z  Ab



b

Janbu: N q = tan '  1  tan 2  '

 exp(2ψ tan ) 2

'

p

Textbook: Nq = 0.6exp(0.126cs' ) (This equationused in the calculation of end bearing capacity) Width (m) Perimeter (m) 2

Area (m ) c = clay

0.4 1.60 0.16

layer groundwater 1 2 3

Depth m 0 4 10 12

TSA

thickness m 0 4 6 2 Clay ESA

Soil Type

Unit weight kN/m3 0 18 18.5 18.5

c c c Clay

' deg 0 24 25 25

Clay/sand

OC R

su kPa 0 16 80 90

0 1.2 9 6

center

center

center

base

Total stress kPa 0 36.0 127.5 201.5

Porewater pressure kPa 0.0 19.6 68.6 107.8

Effective stress kPa 0.0 16.4 58.9 93.7

Effective stress kPa 0.0

FRICTION TSA ESA

END BEARING TSA ESA

102.4

Qult TSA

ESA

layer  1 2 3

  0.00 0.29 0.81 0.66

fs kPa 0.0 8.0 34.3 45.5

fs kPa 0.0 4.7 47.6 61.8

Nc

Qf kN

Nq 0.0

9.0

Qf kN

0 52 381 526

14.0

Qb kN

0 30 487 685

0 0 0 130

Qb kN

kN

0 0 0 229

0 52 381 656

TSA governs design Qult = 656 kN

Qa =

656  328 kN 2

New load = 1.2  328 = 394 kN Qult = 2 x 394 = 788 kN Let L = 14 m (2 m additional length) be the length embedded into the last layer. Clay TSA ESA

Clay

Clay/sand

FRICTION TSA ESA

END BEARING TSA ESA

Qult TSA

ESA

layer fs fs    kPa kPa 0.00 0.0 0.0 1 0.29 8.0 4.7 2 0.81 34.3 47.6 3 0.66 46.5 67.5 Qult = 808 kN > 788 kN Okay Use L = 14 m

Nc

Qf kN

Nq 0.0

9.0

14.0

0 52 381 678

Qf kN 0 30 487 919

Qb kN 0 0 0 130

Qb kN 0 0 0 268

kN 0 52 381 808

kN 0 30 487 1188

kN 0 30 487 914

13.7 Estimate the allowable load capacity of a 0.5-mdiameter steel closed-ended pipe pile embedded 17 m in the soil profile shown in Figure P13.7. The factor of safety required is 2. The N values are blows/ft. Compare the load capacity for a driven pile and a drilled shaft.

Solution 13.7 (a) Driven (displacement) pile

N av =

11  5  20  19 = 13.75 4

 use

N av = 13

Perimeter =  D =  (0.5) = 1.57m, base area = (

fs  1.9Nav  1.9 13  24.7 kPa  100 kPa

 2 D ) = 0.196 m2 4

Use fs  24.7 kPa Qf  24.7 1.57 17  659 kN (f b )=C N60 Ls 17  38  1292  380kPa , Use C  380 D 0.5 N in the vicinity of the base = 19 f b  380 19  7220kPa Qb  7220  0.196  1415kN C  38

Qult  Qf  Qb  659  1415  2074kN Qult 2074   1037 kN FS 2 (b) Drilled shaft Nav < 15

Qa =



N 13 (1.5  0.245 z)  (1.5  0.245 17)  0.42  1.2; use   0.42 15 15

From Table A.11 (Appendix A),   18 kN/m3 Assume groundwater will rise to the surface. Vertical effective stress at center of shaft = 17/2 x (18 – 9.8) = 69.7 kPa Qf = 0.42 x 1.57 x 69.7 x 17 =781 kN Use Quiros and Reese (1977) expression L > 10 m, C = 57.5, fb = CN = 57.5 x 19 = 1092.5 kPa < 2900 kPa; Use fb = 1092.5 kPa Qb = 1092.5 x 0.196 = 214kN Qult = 781 + 214 = 995 kN Q 995 Q a = ult   498kN FS 2 Note: Both of these methods are empirical. The equations are based on field tests on soils that may not be similar to the sand in this problem

13.8 The soil profi le at a site for an offshore structure is shown in Figure P13.8. The height of the pile above the sand surface is 15 m. Determine the allowable load for a driven closed-ended pipe pile with diameter 1.25 m and embedded 10 m into the stiff clay. A factor of safety of 2 is required.

Solution 13.8 Perimeter =  x 1.25 = 3.93 m, Area =

 1.252  1.23m2 4

Layer 1: sand; Layer 2: Stiff clay TSA (Layer 2) : fs is lower of

f s = 0.5 su 'zo and f s = 0.5su0.75  'zo 

Qb = fb Ab = Nc  su b Ab ESA:

β  1  sincs'   OCR  tani 0.5

f s =  'zo Q f    'x  tani'   Perimeter i   Length i j

i

i 1

Qb = fb Ab = N q  σ'z  Ab



b

Janbu: N q = tan '  1  tan 2  '

 exp(2ψ tan ) 2

'

p

0.25

Textbook: Nq = 0.6exp(0.126cs' ) (This equation used in the calculation of end bearing capacity) s =Sand: cs =

32 ,   (1 – sin 32) tan (32) = 0.29

c = Clay: cs = 28.8 ,   (1 – sin 28.8) tan (28.8)61/2 = 0.7; TSA: fs = 51.1 kPa

layer groundwater 1 2

Depth m 0 24 34

thickness m 0 24 10

Clay TSA ESA layer

Soil Type

s c

Sand ESA

Unit weight kN/m3 0 16.8 18.8

Clay

fs fs fs Nc kPa kPa kPa groundwater 0.0 0.0 0.0 1 0.0 0.0 24.7 2 51.1 148.6 9.0 Note: TSA and ESA for the sand are the same TSA governs Q ult = 5216 kN

Qa =

5216  2608 kN 2

' deg 0 32 28.8

Clay/sand

su kPa 0 0 80

OC R 0 0 6

FRICTION TSA ESA Qf kN

Nq 0.0 22.6

Total stress kPa 0 201.6 497.2

0 2325 4332

Qf kN 0 2325 8163

Porewater pressure kPa 0.0 117.6 284.2

Effective stress kPa 0.0 84.0 213.0

END BEARING TSA ESA Qb kN

Qb kN

0 0 884

0 0 7156

TSA

kN 0 2325 5216

Effective stress kPa 0.0 258.0

Qult ESA

kN 0 2325 15319

Solution 13.9 (a) Depth (m) m 3 5 6.5 9 10 14 18 20

'zo kPa 45 65 80 105 115 147 179 195

OCR

su/'zo

2.56 1.77 1.44 1.10 1 1 1 1

0.497 0.371 0.314 0.252 0.235 0.235 0.235 0.235

su kPa 22.4 24.1 25.1 26.5 27.0 34.5 42.0 45.8

Sleeve

Fill

Pile

(b) The fill will cause negative skin friction as it settles. One mitigation method is to put a sleeve over a depth of 3 m (c) Because the undrained shear strength varies with depth, we can integrate it to find the skin friction. In engineering practice, it is best to take average values of undrained shear strengths from 3 m to 10 m and then from 10 m to 18 m 3m to 10 m: su = 25 kPa at an average depth of 6.5 m 10 m to 18 m: su = 35 kPa at an average depth of 14 m Base of shaft: su = 42 kPa Layer 1: Clay from 3 m to 10 m; Layer 2: Clay from 10 m to 18 m TSA: fs is lower of

f s = 0.5 su 'zo and f s = 0.5su0.75  'zo 

0.25

Qb = fb Ab = Nc  su b Ab ESA:

β  1  sincs'   OCR  tani 0.5

f s =  'zo Q f    'x  tani'   Perimeter i   Length i j

i 1

i

Qb = fb Ab = N q  σ'z  Ab b

Textbook: Nq = 0.6exp(0.126cs' ) (This equation used in the calculation of end bearing capacity)

Diameter (m)

0.8

Perimeter (m)

2.51

2

Area (m )

0.50

TSA

layer groundwater 1 2

Dept h m 3 10 18

thickness m 0 7 8

Soil Typ e

c c

Unit weight kN/m3 0 19.8 17.8

' deg 0 28 28

su kPa 0 25 35

Nq = 0.6exp(0.126cs' )  0.6exp(0.126  28)  20.4 TSA : Qult  2.51(13.6  7  22.9  8)  (9 179  0.5)  1504kN ESA : Qult  2.51(11.8  7  28.8  8)  (20.4 179  0.5)  2611kN TSA governs: Qult = 1504 kN

OCR 0 1.44 1

  0.00 0.34 0.28

fs kPa 0.0 13.6 22.9

ESA

fs kPa 0.0 11.8 28.8

Solution 13.10 Single pile 2

Area = 0.13m , Perimeter = 1.26m 1.5m Group pile 2

2

Area = 3.4 = 11.56m , Perimeter = 4  3.4 = 13.6m  z at base = 15(17.5 – 9.8) = 115.5 kPa  z at center of group = 7.5 (17.5 – 9.8) = 57.8 kPa At base

s u = 0.25  115.5 = 28.9 kPa

For friction, take average s u  fs is lower of

28.9 = 14.5 kPa 2

f s = 0.5 su 'zo  0.5 14.5  57.8  14.5kPa and

f s = 0.5su0.75  'zo 

0.25

 0.5 14.50.75  57.80.25  10.2 kPa

Use fs = 10.2 kPa OCR=1.2,  = 0.32, Nq = 26.3 TSA – Block Mode failure

Q f = 1  10.2  13.6  15 = 2080 kN Q b = 9  28.9  11.56 = 3007 kN Q ult = 2080 + 3007 = 5087 kN

1.5m

TSA – Single pile mode failure

Q f = 1  10.2  1.26  15 = 193 kN Q b = 9  28.9  0.13 = 34 kN Q ult = 193 + 34 = 227 kN For 9 piles: Q ult = 9  227 = 2043 kN ESA – Block Mode failure

Q f = 0.32  57.8  13.6  15 = 3773 kN Q b = 26.3  115.5  11.56 = 35115 kN Q ult = 3773+ 35115 = 38888 kN ESA – Single pile mode failure

1.26  350kN 13.6 0.13 Q b = 35115   395 kN 11.56 Q ult = 9(350+395) = 6705 kN

Q f = 3773 

Single pile mode failure governs 2043 TSA governs: Qa   681kN 3

13.11

The soil profile and soil properties at a site are shown in the table below. A group of 12 concrete piles in a 3 × 4 matrix and of length 12 m is used to support a load. The pile diameter is 0.45 m and pile spacing is 1.5 m. Determine the allowable load capacity for a factor of safety of 2. Calculate the total settlement (elastic and consolidation) under the allowable load. Assume Ep = 20 × 106 kPa.

Depth (m) 0 to 3

Type of deposit Sand

 = 17 kN/m3, cs'  28º Eso = 19 MN/m2

Groundwater level at 3 m 3 to 6

Soil test results

3

sat = 17.5 kN/m ,

Sand

cs'  30º

Eso = 18 MN/m2

6 to 15

3

sat = 18.5 kN/m ,

Clay

cs'  27º

su = 30 kPa Cc = 0.4, Cr = 0.06, OCR = 1.5 Eso = 30 MN/m2, v′ = 0.3

15to 17

sat = 18 kN/m3, cs'  24º

Soft clay

su = 20 kPa Cc = 0.8, OCR = 1.0 Eso'  10 MN/m2 , v  0.3

>17

Rock

Solution 13.11 I 6m II

sand 9m clay

III 3m

3m

IV 12 piles, 3x4 matrix, L = 12m D = 0.45m

Soft clay

s =3D = 3(0.45) = 1.35. Try s = 1.5m 2

Single pile: perimeter = D =1.41m, Area = 0.159 m Group: perimeter = 2(2s + D) + 2(3s + D) = 2(2(1.5)+0.45) + 2(3(1.5) + 0.45) = 16.8 m

A b g = (2(1.5) + 0.45)(3(1.5) + 0.45) = 17.08 m 2

TSA – Block failure mode Sand layer I  = 0.28 At center:

z  

L  3 = 17   =25.5 kPa  2 2

Q f I =0.28 (25.5)(16.8)(3) = 360 kN Sand Layer II

 z = 3(17) + (17.5 – 9.8)

3 = 62.6 kPa 2

 = 0.29 Q f II = 0.29 (62.6)(16.8)(3) = 909 kN Clay Layer III fs = 20.3 kPa Q f III = 20.8(16.8)(6) = 2097 kN End Bearing:

Q b = 9(30)(17.08) = 4612 kN

Total skin friction: Qf = 360+909+2097 = 3366 kN Q ult = 3366 + 4612 = 7898 kN TSA – single pile mode Skin friction: Qf = 3366 x 1.41/16.8 = 283 kN End Bearing: Q b == 4612 x 0.159/17.08 = 43 kN

Q ult = 9(283 + 43) = 2934 kN Allowable load capacity

Qa  =

2934 = 1467 kN 2

Elastic Settlement (average from Layers I-III)

Qa =

e

=

283 = 142 kN 2 Qa I E so L

3(19)  3(18)  6(30) = 24.5 MN m 2 , 12 I = 0.5 + log (L/D) = 0.5 + log (26.7) = 1.9 Es 

E p = 20 x 10 Pa ; 3

L 12 = = 26.7 D 0.45

es =

 24.5

142

 103  12

 1.9  0.92 103 m  0.92 mm

0.5

R s = 12 =3.5  es g = 0.92(3.5) = 3 mm Clay Layer III Load is transferred to 2L/3 from surface Depth to center of clay layer from load = 12/3 + 3/2 = 5.5m 1467 At center:  z =  15kPa (3.45  5.5)(4.95  5.5)

zo = 3  17 + 3(17.5 - 9.8) + 7.5(18.5 - 9.8)  139 kPa  final = 139 + 15 = 154 kPa zc = 1.5  154 = 231 kPa >  final

At center:

G e 





2.7  e o 9.8 ; e o =0.91  sat =  s o  w ; 18.5 =    1 eo   1  eo   c =

3000  154   0.06  log 10   = 4 mm 1  0.91  139 

Clay-Layer IV Load is transferred to a depth of 2L/3 Depth of load to center of clay = 12/3 + 3 + 1 = 8 m 1467  10kPa At center:  z = (3.45  8)(4.95  8)

zo = 3  17 + 3(17.5 - 9.8) + 9(18.5 - 9.8) + 1 (18 – 9.8)161 kPa  final = 161 + 10 = 171 kPa

At center:

2.7  e o 18 = 9 .8 1  eo  c =

e o = 1.02

2000  171   0.8  log    21 mm 1  1.02  161 

Total settlement = 3 + 4 + 21 = 28 mm

13.12 The soil at a site consists of a 30 m thick deposit of clay. At a depth 6 m and below it is normally consolidated. A soil sample from this depth was tested in a direct simple shear (DSS) apparatus. The DSS gave a normalized undrained shear strength of

  su  f   '   0.22 where the subscript f denotes   zo  DSS

failure (critical state). The average saturated unit weight is 19.8 kN/m3. Groundwater level is at the surface. From Chapter 11, the normalized undrained shear strength is given by the equation

  su  f  '   zo

 3 sincs'   2  DSS

0.8

 OCR    . (a) Plot the variation of undrained shear strength with depth up to  2 

a depth of 30 m. (b) Estimate the allowable load capacity for a steel cylindrical pile of diameter 1.5 m, length 15 m, wall thickness 65 mm driven with a driving shoe (displacement pile). Assume FS = 2.

Solution 13.12 (a) 0.8   su  f  3 sincs'  OCR   '     2  2    zo  DSS

Solve for cs' 3 sincs' 2

1   2 ' ' sincs  0.442; cs  26.2o

0.22DSS 

Depth (m) m 0 1 2 3 6 8 10.5 12 15 30

'zo kPa 0 10 20 30 60 80 105 120 150 300

0.8

OCR

su/'zo

6.00 3.00 2.00 1.00 1 1 1 1 1

0.000 0.921 0.529 0.382 0.220 0.220 0.220 0.220 0.220 0.220

su kPa 0.0 9.2 10.6 11.5 13.2 17.6 23.1 26.4 32.9 65.9

Pile

(b)

Because the undrained shear strength varies with depth, we can integrate it to find the skin friction. In engineering practice, it is best to take average values of undrained shear strengths from 0 m to 6 m and then from 6 m to 15 m 0 m to 6 m: su = 11.5 kPa at an average depth of 3 m 6 m to 15 m: su = 23.1 kPa at an average depth of 10.5 m Base of shaft: su = 32.9 kPa Layer 1: Clay from 0 m to 6 m; Layer 2: Clay from 6 m to 15 m TSA: fs is lower of f s = 0.5 su 'zo and f s = 0.5su0.75  'zo 

0.25

Qb = fb Ab = Nc  su b Ab

ESA: β  1  sincs'   OCR  tani 0.5

f s =  'zo Q f    'x  tani'   Perimeter i   Length i j

i

i 1

Qb = fb Ab = N q  σ'z  Ab b

Textbook: Nq = 0.6exp(0.126cs' ) (This equation used in the calculation of end bearing capacity) Diameter (m)

1.5

Perimeter (m)

4.71

2

Area (m )

1.77

TSA

ESA

layer groundwater 1 2

Depth m 0 6 15

thickness m 0 6 9

Soil Type

c c

Unit weight kN/m3 0 19.8 19.8

' deg 0 26.2 26.2

su kPa 0 11.5 23.5

OCR

Nq = 0.6exp(0.126cs' )  0.6exp(0.126  26.2)  16.3 TSA : Qult  4.71(7.3  6  17.1 9)  (9  32.9 1.77)  1455kN ESA : Qult  4.71(11.7  6  28.9  9)  (16.3 150 1.77)  5883kN TSA governs: Qult = 1455 kN Qa = 1455/2 = 728 kN

0 2 1

  0.00 0.39 0.27

fs kPa 0.0 7.3 17.1

fs kPa 0.0 11.7 28.9

Solution 13.13 Pile group of 10 piles (drilled shafts) Design load = 2 x 15 = 30 MN Data Straight, prismatic drilled shafts SI Select units Design load 350 kN Shaft diameter 0.5 m Top of base layer 2 m

FS Group Spacing Matrix

Groundwater

2

No. of piles

Use N values

y

m

Group width

Perimeter Area

1.57 0.20

m 2 m

Group length Perimeter

Max end bearing

2900

kPa

Area

layer 0 1 2 3 4 5 6 7 8 9 10

Depth m 0 2 3 5 7 9 10 12 15 20 28

thickness m 0 2 1 2 2 2 1 2 3 5 8

Depth to center m 0 1 2.5 4 6 8 9.5 11 13.5 17.5 24

Soil Type

s s s s s s s s s s

Unit weight kN/m3 0 14 15 17 17.5 18 18 19 20.5 20.5 20.5

N60 0 5 7 12 16 18 19 25 38 38 38

Calculate effective vertical stresses and skin friction factor (Eq. 13.42) center

layer 0 1 2 3 4 5 6 7 8 9 10

Depth m 0 2 3 5 7 9 10 12 15 20 28

Total stress kPa 0 14.0 35.5 60.0 94.5 130.0 157.0 185.0 234.8 316.8 450.0

center

center

base

Porewater pressure kPa 0.0 0.0 4.9 19.6 39.2 58.8 73.5 88.2 112.7 151.9 215.6

Effective stress kPa 0.0 14.0 30.6 40.4 55.3 71.2 83.5 96.8 122.1 164.9 234.4

Effective stress kPa 0.0

Calculate shin friction and end bearing stress (Eq. 13.44) ESA ESA

layer 0 1 2 3 4 5 6 7

Depth m 0 2 3 5 7 9 10 12

fs kPa 0.0 16.8 15.9 32.6 49.8 57.5 62.2 66.5

fb kPa 0.0 0.0 120.8 345.0 920.0 1035.0 1092.5 1437.5

33.2 47.6 63.0 79.4 87.6 106.0 138.1 191.6 277.2

n 0.00 0.33 0.47 0.80 1.00 1.00 1.00 1.00 1.00 1.00 1.00

  0.00 1.20 0.52 0.81 0.90 0.81 0.74 0.69 0.60 0.48 0.30

8 9 10

15 20 28

73.2 78.3 70.3

2185.0 2185.0 2185.0

Calculate load capacity for single pile mode failure and for block mode failure. GROUP FAILURE MODE SINGLE PILE BLOCK FAILURE DIAMETER WIDTH 0.5 10.5 Depth Qult FRICTION END BEARING Qult m ESA ESA ESA ESA 0 Qf Qb MN MN MN MN 0.00 0.00 0.00 0.00 0.53 3.43 0.00 3.43 2 0.89 5.05 51.35 56.40 3 2.22 11.71 146.71 158.42 5 4.78 21.86 391.23 413.09 7 33.58 440.13 473.71 9 6.75 39.92 464.59 504.51 10 7.81 53.50 611.30 664.80 12 10.54 75.90 929.17 1005.07 15 15.40 115.84 929.17 1045.01 20 21.51 173.18 929.17 1102.35 28 30.41 Single pile (shaft) mode governs design. The length for a design load of 30 MN is 28 m.

Solution 13.14 From Example 13.8 3  6.5 = 4.75MPa 2 Averagesleeve resistance  0.075MPa over a depth of 10m.

Xu and Lehane  2005 : q cav =

Calculate load capacity 2   Di 0.2  D*   Di    1  min 1,     2   D    1.5    D  

0.5

0.2 2 D*    0.27    0.27    1  min 1,    2  0.3    1.5    0.3  

0.5

D* 0.5  1  0.575  0.65 0.3 D*  0.65  0.3  0.195m 2   D*   2 Cb = 0.15 1  3     0.15 1  3  0.65   0.34   D  

π  D* 

2

π  (0.1952 ) Ab  = = 0.03m2 4 4 Qb  Cb qcav Ab = 0.34  4.75  0.03 = 0.049MN Qb  f s πDL = 0.075  π  0.3 10 = 0.707 MN

Qult = 0.049 + 0.707 = 0.756 MN = 756 kN If the sleeve friction was not measured, the estimated fs at mid-depth of the pile ( h = 5 m) is calculated as follows. Ars = 0.652 =0.423 m2 [

)]

( [

(

)]

which is approximately 10% of the measure value.

Solution 13.15 Predictions of Program APILES - Version 1.0 (1988). This analysis for APILES was developed by M. Budhu and T. Davies. This PC version of APILES was written by M.Budhu. We are not responsible for any consequences in using this program. This analysis is valid for piles whose lengths are greater than their effective length.

TITLE: problem 13.15 Your input data is as follows PILE DATA Head type: Fixed head Length = 8.00m Diameter = .50m Young's Modulus = .10E+08kPa Allowable stress = .10E+05kPa SOIL DATA Soil type: Stiff clay Unit weight of soil: 10.0000kN/m^3 Earth pressure coefficient: 1.00 Adhesion factor = .50 Young's modulus = .20E+05kPa Undrained shear strength = .40E+02kPa The working load is:

223.61kN

The displacements are computed at the point at which the load is applied. The bending moments and rotations are computed at ground surface

load kN 3.79 20.74 23.40 34.63 44.97 54.23 86.79 91.49 97.05 146.61 151.32 152.25 217.59 218.87 219.65 221.96 226.08 236.99 263.77 275.87 285.89 296.61 296.72 299.01 306.88 325.85 327.31 333.87 357.59 368.64 375.15 376.37 384.51 394.23 422.70 427.94 429.16 436.26 454.17

disp mm .15 .62 .69 1.11 1.65 2.11 4.01 4.28 4.62 9.28 9.72 9.81 19.95 20.24 20.42 20.97 21.94 24.59 31.08 34.01 37.48 41.21 41.25 42.05 44.84 51.72 52.45 55.76 67.70 73.30 76.63 77.25 82.62 89.41 109.43 113.12 114.17 120.51 136.60

moment kNm

rotation radians

.00 .00 .00 .02 .00 .02 .00 .04 .00 .06 .00 .08 .00 .14 .00 .15 .00 .16 .00 .30 .00 .32 .00 .32 .00 .60 .00 .61 .00 .61 .00 .63 .00 .65 .00 .71 .00 .87 .00 .94 .00 1.02 .00 1.10 .00 1.10 .00 1.12 .00 1.18 .00 1.33 .00 1.35 .00 1.41 .00 1.65 .00 1.76 .00 1.82 .00 1.83 .00 1.93 .00 2.05 .00 2.41 .00 2.48 .00 2.49 .00 2.60 .00 2.86

max.B.M. depth to max.B.M. kNm m 1.67 1.18 8.43 1.07 9.57 1.07 16.77 1.10 24.13 1.17 30.93 1.21 53.16 1.37 56.53 1.38 60.17 1.39 108.52 1.51 113.11 1.52 114.00 1.52 202.72 1.67 204.82 1.68 206.08 1.68 209.87 1.69 216.65 1.71 233.26 1.75 279.29 1.86 300.46 1.89 319.79 1.94 340.93 1.99 341.16 1.99 345.72 2.00 361.57 2.03 400.41 2.10 403.48 2.10 417.40 2.14 469.40 2.25 494.39 2.30 509.65 2.33 512.54 2.34 531.72 2.38 555.03 2.42 625.87 2.56 639.33 2.58 642.47 2.59 660.96 2.62 708.62 2.70

500 450 400 350 300 Lateral load 250 (kN) 200 150 100 50 0 0

50

100

Pile head displacement (mm)

(b) The working load is: (c) Max BM = 213 kN.m

223.61kN

(d) Pile head deflection = 22 mm

150