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 sincs' OCR tani 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 sincs' OCR tani 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 sincs' OCR tani 0.5
f s = 'zo Q f 'x tani' 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(0.126cs' ) (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 sincs' OCR tani 0.5
f s = 'zo Q f 'x tani' 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(0.126cs' ) (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 1.9N av 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 sincs' OCR tani 0.5
f s = 'zo Q f 'x tani' 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(0.126cs' ) (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 sincs' OCR tani 0.5
f s = 'zo Q f 'x tani' Perimeter i Length i j
i 1
i
Qb = fb Ab = N q σ'z Ab b
Textbook: Nq = 0.6exp(0.126cs' ) (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(0.126cs' ) 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
sat = 17.5 kN/m3,
Sand
cs' 30º
Eso = 18 MN/m2
6 to 15
sat = 18.5 kN/m3,
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/m , cs' 24º 3
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 0.92 103 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 sincs' 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 sincs' OCR ' 2 2 zo DSS
Solve for cs' 3 sincs' 2
1 2 ' ' sincs 0.442; cs 26.2o
0.22DSS
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 sincs' OCR tani 0.5
f s = 'zo Q f 'x tani' Perimeter i Length i j
i
i 1
Qb = fb Ab = N q σ'z Ab b
Textbook: Nq = 0.6exp(0.126cs' ) (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(0.126cs' ) 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 2 0.53 3.43 0.00 3.43 3 0.89 5.05 51.35 56.40 5 2.22 11.71 146.71 158.42 7 4.78 21.86 391.23 413.09 9 6.75 33.58 440.13 473.71 10 7.81 39.92 464.59 504.51 12 10.54 53.50 611.30 664.80 15 15.40 75.90 929.17 1005.07 20 21.51 115.84 929.17 1045.01 28 30.41 173.18 929.17 1102.35 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 cav =
Calculate load capacity 2 Di 0.2 D* Di 1 min 1, 2 D 1.5 D
0.5
0.27 0.2 0.27 2 D* 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 qcav 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