2. Example 1:
A 400 mm square prestressed concrete pile is to be
driven 19 m into the soil profile shown in the figure.
a) Compute the net load transferred to the ground
through toe bearing
b) Compute the nominal side friction capacity
c) Compute the ASD downward load capacity using a
factor of safety of 2.8.
6. Example 2:
A straight drilled shaft of diameter 1 m is
installed in a soil profile, as shown in the figure.
SPT were performed at intervals of
approximately 1 m below the base. Determine
the toe bearing capacity.
7. Solution 2:
Use average N60 value over a depth of 2D from the base or tip.
OβNeill and Reese (1999)
8. Example 3:
The drilled shaft shown is to be designed
without the benefit of any on-site static load
tests. The soil conditions are uniform and the
site characterization program was average.
Unit weights of soil layers above the water
table are 17kN/m3 and below are 20kN/m3.
Compute the allowable downward load
capacity.
11. Example 4:
A concrete pile 40 cm in diameter and 10 m long is driven into a homogenous mass of clayey soil of medium
consistency. The water table is at the ground surface.The undrained shear strength is 80 kPa and the
adhesion factor Ξ± = 0.75. Compute the ultimate load bearing capacity.
Solution 4:
12. Example 5:
It is designed that the steel pile having a
diameter 40cm given in the figure below has
to have an allowable bearing capacity which
should be maximum 490 kN. Calculate the
length of the pile by considering the factor of
safety as 3.
Solution 5:
ππ’ππ‘ = ππ + ππ
ππ = 9 β ππ’ β π΄π = 9 β 125 β
0.42 β π
4
= 141.4 ππ
ππ = ΰ· ππ’ β πΌ β π β βπ
According to Randolph and Murphy (1985),
πβ²π§π· 1 = 3 β 17 = 51πππ β
ππ’
πβ²
π§π· 1
=
50
51
β 1 β πΌ1 β 0.5 (πππ’π)
πβ²π§π· 2 = 6 β 17 + 2 β 19 β 9.8 = 120.4πππ β
ππ’
πβ²
π§π· 2
=
50
120.4
β 0.41 β πΌ2 β 0.78 (πππ)
14. Example 6:
The load-settlement data shown in the figure were
obtained from a full-scale static load test on a 400mm
square, 17m long concrete pile (fΒ΄c=40MPa). Use
Davissonβs method to compute the ultimate
downward load capacity.
17. Example 7:
Given that n1 = 4, n2 = 3, D = 305 mm, d = 1220 mm,
and L = 15 m. The piles are square in cross section
and embedded in a homogenous clay with Ι£ = 15.5
kN/m3 and cu = 70 kN/m2. Using a factor of safety
equal to 4, determine the allowable load bearing
capacity of the group pile.
21. Example 8:
A group pile in clay is shown in the figure on
the right. Determine the consolidation
settlement of the pile groups.
Assume that all the clay layers are normally
consolidated.
30. Deep Foundations-Pile Axial Load Capacity
-Shaft resistance (Undrained condition, clayey soil, non-free draining)
οΆ Ξ±-method (ππππ£ππ πππππ )
ππ = Ξ±(ππ’)π΄π
As= shaft area of the of the pile
Ξ±= adhesion factor and f(slenderness, shear strength)
πΉππ
ππ’
Οβ²π£0
β€ 1; Ξ± = 0.5Fp
ππ’
Οβ²π£0
β0.5
β€ 1
πΉππ
ππ’
Οβ²π£0
β₯ 1; Ξ± = 0.5Fp
ππ’
Οβ²π£0
β0.25
Fp = 1.0 if
πΏπ
π΅
β€ 50
= 0.7if
πΏπ
π΅
β₯ 120
31. Deep Foundations-Pile Axial Load Capacity
-Shaft resistance (Undrained condition, clayey soil, non-free draining)
οΆ Ξ±-method (πππππ πππππ )
ππ = Ξ±(ππ’)π΄π
As= shaft area of the of the pile
Ξ±= adhesion factor and f(shear strength)
Ξ± = 1.0 πππ ππ’ β€ 30
Ξ± = 1. 16 β
ππ’
185
πππ 30 < ππ’ β€ 150
Ξ± = 0.35 πππ ππ’ > 150
32. Deep Foundations-Pile Axial Load Capacity
-Shaft resistance (Drained condition, granular soil, free draining)
ππ = πΎπ Οβ²π£0 tan Ξ΄β² π΄π (πππππ πππ ππππ£ππ πππππ )
As= shaft area of the of the pile
Ks= coefficient of horizontal soil stress
Ξ΄'= the angle of friction b/w pile and soil
Ξ΄β² = Ο = 0.75Οβ²
πΎ0 = 1 β π ππΟβ²
33. Deep Foundations-Pile Axial Load Capacity
-Shaft resistance (Drained condition, granular soil, free draining)
ππ = Ξ²Οβ²π£0π΄π
Ξ²=0.18+0.65Dr
Dr=relative density in decimal form
οΆ Ξ²-method (ππππ£ππ πππππ )
ππ = Ξ²Οβ²π£0π΄π
Ξ²=1.5-0.245 π§
z= depth to the mid-layer (m)
οΆ Ξ²-method (πππππ πππππ )
34. Ex1: A bridge pier is to be supported by 1.0 π diameter driven-cast in place concrete
piles. The design load per pile is determined as 700 ππ by the structural engineer.
Calculate the total pile length using a factor of safety of 3.0 against failure.
Deep Foundations- Single Pile Capacity
6 m Sand Ξ³=17 kN/m3 Ο'=36Λ
Clay Ξ³=20 kN/m3 cu=100 kPa
Οπ£0 (πππ) π’ (πππ) Οβ²π£0 (πππ)
51
102
102 + 20βππππ¦ 10βππππ¦
51
102
102 + 10βππππ¦
35. Ex1: A bridge pier is to be supported by 1.0 π diameter driven-cast in place concrete
piles. The design load per pile is determined as 700 ππ by the structural engineer.
Calculate the total pile length using a factor of safety of 3.0 against failure.
Deep Foundations- Single Pile Capacity
6 m Sand Ξ³=17 kN/m3 Ο'=36Λ
Clay Ξ³=20 kN/m3 cu=100 kPa
β’
πΏπ
π΅
β₯ 12, π‘βπ’π πΏπ,πππ β₯ 12 π
π»ππππ ππππππππ ππ’ππ‘ = βππ + ππ
β’ πΉππ π πππ πππ¦ππ
ππ = πΎπ Οβ²π£0 tan Ξ΄β² π΄π
πΎπ = 0.41 1.5 = 0.62 Ξ΄β² = Ο = 0.75 Οβ² = 27Λ
πΎ0 = 1 β π ππΟβ² = 0.41
38. Ex2: A bridge pier is to be supported by bored-cast in place concrete piles (C30). The
design load per pile is determined as 2000 ππ by the structural engineer. Calculate the
total pile length using a factor of safety of 3.0 against failure.
Deep Foundations- Single Pile Capacity
6 m Sand Ξ³=17 kN/m3 Ο'=28Λ
Clay Ξ³=18 kN/m3 cu=30 kPa
Οπ£0 (πππ) π’ (πππ) Οβ²π£0 (πππ)
51
102
210 60
Sand Ξ³=19 kN/m3 Ο' =38Λ
Clay Ξ³=18 kN/m3 cu=150 kPa
6 m
6 m
120
324
51
102
150
204
39. Ex2: A bridge pier is to be supported by bored-cast in place concrete piles (C30). The
design load per pile is determined as 2000 ππ by the structural engineer. Calculate the
total pile length using a factor of safety of 3.0 against failure.
Deep Foundations- Single Pile Capacity
6 m Sand Ξ³=17 kN/m3 Ο'=28Λ
Clay Ξ³=18 kN/m3 cu=30 kPa
Οπ£0 (πππ) π’ (πππ) Οβ²π£0 (πππ)
51
102
210 60
Sand Ξ³=19 kN/m3 Ο' =38Λ
Clay Ξ³=18 kN/m3 cu=150 kPa
6 m
6 m
120
324
51
102
150
204
β’ πΈπΆ7 Οππππ ππππ =
πππππ
π΄π
β€
Οπ,π
4
Οππππ ππππ β€
30
4
= 7.5 πππ
πππππ
π΄π
=
2000000
π΄π
= 7.5 ππππ ππππππ‘ππ β π΅ = 600ππ
42. -Load vs settlement: Load controlled test, to obtain ultimate load
οΆ Load increment= 25% of working load
οΆ Ultimate load= 200% of working load
Deep Foundations- Pile Load Test (Davisson Approach)
1) Clayey soil
2) Granular soil
Total
settlement
Elastic
settlement
Net settlement=Total settlement- pile elastic settlement
Ξ΄ππππ ππππ π‘ππ =
(ππ€π+0.6ππ€π )πΏπ
π΄ππΈππππ
ππ€π = π€ππππππ ππππ
ππ€π = π βπππ‘ πππ ππ π‘πππ ππππ @π€ππππππ
43. -Load vs settlement
οΆ Use net settlement values
Deep Foundations- Pile Axial Load Test
Ξ΄π’ππ‘ = 0.012 π΅πππ + 0.1
π΅
π΅πππ
+
ππ’πΏπ
π΄ππΈπ
K M
K
M
Net settlement (mm)
ππ’ππ‘ = πππ‘ππππ‘π ππππ (ππ)
π΅ = ππππ ππππππ‘ππ (ππ)
π΅πππ = π ππππππππ ππππ π€πππ‘β ππ
= 300 mm
πΏπ = ππππ πππππ‘β (ππ)
π΄π = πΈππ πππππππ ππππ ππ2
πΈπ = ππππ ππππ π‘ππ ππππ’ππ’π (πΊππ)
Ξ΄π’ππ‘
ππ’ππ‘
44. -Ex3: Lp= 20 m, BxB=406x406 mm, Ep=30 GPa (C30 concrete), Qult ?
Deep Foundations- Pile Axial Load Test
Ξ΄π’ππ‘ = 0.012 π΅πππ + 0.1
π΅
π΅πππ
+
ππ’πΏπ
π΄ππΈπ
K=3.735
Net settlement (mm)
Ξ΄π’ππ‘ = 0.012 300 + 0.1
406
300
+
ππ’20000
(406)(406)(30)
ππ’ππ‘ = 1460 ππ
45. -Downdrag force acting on pile: force exerted by settling soil on pile (Ξ΄soil> Ξ΄pile)
Deep Foundations- Negative Skin Friction
οΆ Placement of fill (granular) settlement of the fill by selfweight
additional consolidation of soft clayey layers beneath fill
οΆ Lowering of GWT Ο' and settlement increases
additional
settlement
for the pile
46. -Downdrag force acting on pile
Deep Foundations- Negative Skin Friction
β’ ππ = π0 + ππ πππ π€πππβπ‘ + ππππ€πππππ
ππππ€πππππ = 0.25 Οβ²π£(Οπ΅πππππ¦ πππ¦ππ)
β’ For clay layer upon consolidation
β’ For sand layer upon settling
ππππ€πππππ = πΎ0Οβ²π£ tan Ξ΄β² (Οπ΅ππ πππ)
Ξ΄β² = (0.6)Οβ²
Resistance
Load
Force
P0
Qb
πΉπ =
ππ + ππ
π0 + ππ πππ π€πππβπ‘ + ππππ€πππππ
β₯ 2.0
ππ + ππ
47. Ex2 (cont.): 2 m height of a very large fill (Ξ³=20 kN/m3) will be placed next to the pile
on the behalf of approach fill. Check whether adequate FS exists or not?
Deep Foundations- Negative Skin Friction
6 m Sand Ξ³=17 kN/m3 Ο'=28Λ
6 m
6 m
Clay Ξ³=18 kN/m3 cu=30 kPa
Sand Ξ³=19 kN/m3 Ο' =38Λ
Clay Ξ³=18 kN/m3 cu=150 kPa
β’ πΉπ =
ππ +ππ
π0+ππ πππ π€πππβπ‘+ππππ€πππππ
β₯ 2.0
πΉπ =
6614.3
2000 + 80.6 + ππππ€πππππ
ππππ€πππππ = 0.25 Οβ²π£(Οπ·πππππ¦ πππ¦ππ)
Οβ²π£0 =
102 + 150
2
+ (2)(20) = 166 πππ
Fill
ππππ€πππππ = 0.25 166 Ο 0.6 6 = 469 ππ
πΉπ =
6614.3
2000 + 80.6 + 469
= 2.59
48. Piles working under tension:
Deep Foundations- Uplift Capacity
β’ πΉπ =
ππ +ππ πππ π€πππβπ‘
ππ’ππ€πππ
β₯ 3.0~6.0
οΆ Ship building docks οΆ Wind turbines, chimneys or any structures
exposed to lateral loads
49. -Group action: zone of soil or rock which is stressed by the entire group extends to a
much greater width and depth than the zone beneath the single pile. Spacing (s) between
adjacent piles should be increased (>5B)
Deep Foundations- Piles in Group
οΆ (# of piles) x Qult (single pile) > Qult (group) under same displacement
οΆ Even though loading tests made on a single pile have indicated satisfactory performance, failure or
excessive settlement may take place
Friction pile
(clayey soil, adhesion)
End bearing pile
(granular soils)
50. -Group capacity:
Deep Foundations- Piles in Group
Group Efficiency Approach
Qult, group = (Ξ·) x (# of piles) x Qult (single pile)
Ξ· (efficiency)=1βΟ
[ π β 1 π + π β 1 π]
90(π)(π)
Ο in degrees = tanβ1
π΅
π
B
B
B
B B
π = ππ ππ πππ€π
π = ππ ππ ππππ’πππ
s = spacing b/w piles measured from center to center
51. -Group capacity:
Deep Foundations- Piles in Group
Terzaghi-Peck Block Approach
οΆ shallow spread foundation
οΆ Crucial for soil profiles consisting of soft clay layers (low cu values)
οΆ ππ’ππ‘ = βππ + ππ
ππ = 2 πΏπππππ + π΅πππππ βπ·πΞ±(ππ’) or ππ = 2 πΏπππππ + π΅πππππ βπ·ππΎπ Οβ²π£0 tan Ξ΄β²
clayey soil granular soil
ππ = (πΏππππππ΅πππππ) π πππππ’ ππ = (πΏππππππ΅πππππ) ππΟβ²ππ π
clayey soil granular soil
π· = πΏπ
52. -Ex4: A group of bored-cast in place piles are constructed in a deep layer of clay. Find
Qult, group? B=0.6 m
Deep Foundations- Piles in Group
πΊπππ’π ππππππππππ¦ πππππππβ
s=1.8 m
β’ ππ’ππ‘,π πππππ ππππ = ππ + ππ
Οβ²π£0 (πππ)
114
228
ππ = π΄π π πππππ’ = Ο 0.3 2 9 60 = 152.7 ππ
ππ = Ξ±(ππ’)π΄π = 1.16 β
60
185
60 Ο 0.6 12 = 1134.2 ππ Ξ± = 1. 16 β
ππ’
185
πππ 30 < ππ’ β€ 150
Pile layout
s=1.8 m
53. -Ex4: A group of bored-cast in place piles are constructed in a deep layer of clay. Find
Qult, group? B=0.6 m
Deep Foundations- Piles in Group
Pile layout
s=1.8 m
β’ Qπ’ππ‘,π πππππ ππππ = ππ + ππ = 152.7 + 1134.2 = 1286.9 ππ
β’ Qult, group = (Ξ·) x (# of piles) x Qult (single pile)
Ξ· (efficiency)=1βΟ
[ π β 1 π + π β 1 π]
90(π)(π)
Ο in degrees = tanβ1
π΅
π
= tanβ1
0.6
1.8
= 18.4Λ
Ξ· =1β18.4
[ 3 β 1 4 + 4 β 1 3]
90(4)(3)
= 0.71
π = 3
π = 4
β’ Qult, group = (0.71)(12)(1286.9)=10964.4 kN
s=1.8 m
54. -Ex4: A group of bored-cast in place piles are constructed in a deep layer of clay. Find
Qult, group? B=0.6 m
Deep Foundations- Piles in Group
Pile layout
s=1.8 m
s=1.8 m
TerzaghiβPeck Block πππππππβ
ππ = 2 πΏπππππ + π΅πππππ Ξ±(ππ’)
οΆ ππ’ππ‘,πππππ = ππ + ππ
ππ = 2[ 1.8 3 + 1.8 2 12 0.84 60 = 10886.4 kN
Ξ± = 1. 16 β
ππ’
185
πππ 30 < ππ’ β€ 150
ππ = πΏππππππ΅πππππ π πππππ’ = πΏππππππ΅πππππ 9ππ’
ππ = 5.4 3.6 9 60 = 10497.6ππ
οΆ ππ’ππ‘,πππππ = ππ + ππ = 10886.4 + 10497.6 = 21384 ππ
οΆ ππ’ππ‘,πππππ = min πΊπππ’π ππππππππππ¦ πππππππβ, ππππ§ππβπ β ππππ πππππ πππππππβ = 10694.4 ππ