Bogazici University Foundation

1. CE 336- FOUNDATION ENGINEERING
SPRING 2021
28.04.21
Deep Foundations

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.

3. Solution 1:
𝐸 = 𝛽0 𝑂𝐶𝑅 + 𝛽1𝑁60
𝐸 = 5000 1 + 1200 ∗ 25 = 35000𝑘𝑃𝑎
𝐼𝑟 =
35000
2 1 + 0.3 ∗ 188 ∗ 𝑡𝑎𝑛36°
= 99
𝜎′𝑧𝐷 = 17.8 ∗ 3 + 18.2 − 9.81 ∗ 16 =188kPa
According to the figures,
𝑁𝛾
∗ = 15; 𝑁𝑞
∗ = 76
(a)
* *
' '
t ZD
q B N N
 
 
    
   
35000
114
2 1 ' tan ' 2 1 0.30 163 tan36
s
r
zD
E
I
v  
  
        

4. Solution 1:
𝑞′𝑛 = 𝐵𝛾𝑁𝛾
∗ + 𝜎′𝑧𝐷𝑁𝑞
∗
𝑞′𝑛 = 0.4 ∗ 18.2 − 9.8 ∗ 15 + 188 ∗ 76 = 14338𝑘𝑃𝑎
𝑞′𝑛𝐴𝑡 = 14338 ∗ 0.42 = 2294𝑘𝑁
(b)

5. Solution 1:
𝐾0 = 1 − 𝑠𝑖𝑛𝜑′ 𝑂𝐶𝑅𝑠𝑖𝑛𝜑′
𝐾0 = 1 − 𝑠𝑖𝑛36 (1) 𝑠𝑖𝑛36
𝑡𝑜 1 − 𝑠𝑖𝑛36 (4) 𝑠𝑖𝑛36
𝐾0 = 0.41 𝑡𝑜 0.93
𝛽 = 𝐾0
𝐾
𝐾0
𝑡𝑎𝑛 𝜑′
𝜑𝑓
𝜑′
𝛽 = 0.41 ∗ 1.2 ∗ tan 36 ∗ 0.8 𝑡𝑜 0.93 ∗ 1.2 ∗ tan 36 ∗ 0.8 = 0.27 𝑡𝑜 0.61
(c) 𝑃𝑎 =
𝑞′𝑛𝐴𝑡 + σ 𝑓𝑛𝐴𝑠
𝐹
=
2294 + 1272
2.8
= 1274 𝑘𝑁

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.
O’Neill 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.

9. Solution 3:
We assume that;
Silt sand above groundwater table ɣ = 17 kN/m3
Silt sand below groundwater table ɣ = 20 kN/m3
Sand above groundwater table ɣ = 20 kN/m3
For Drilled Shaft
𝛽 = 1,5 − 0,245 𝑧
𝑓𝑠 = 𝛽 ∗ 𝜎′𝑧

10. Solution 3:
Layer z (m) β σ'z (kPa) fs (kPa) As (m2
) Ps (kN)
Silty sand above GWT 1 1.26 17.0 21.34 3.77 80.43
Silty sand below GWT 2.75 1.09 41.6 45.54 2.83 128.78
Sand below GWT (3.5m-9m) 6.25 0.89 77.3 68.61 10.37 711.30
Sand below GWT (9m-14m) 11.5 0.67 130.8 87.53 9.42 824.95
Total 1745.46
60
' 57.5* 57.5 22 1265
t
q N kPa
   
 
2
2
0.6
0.283
4
t
m
A m
 
 
  2
' 1265 0.283 1745
701
3
t t s s
a
q A f A kPa m kN
P kN
FS
    
  

(For drilled shaft)

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 α = 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:
𝑄𝑢𝑙𝑡 = 𝑄𝑝 + 𝑄𝑠
𝑄𝑝 = 9 ∗ 𝑐𝑢 ∗ 𝐴𝑝 = 9 ∗ 125 ∗
0.42 ∗ 𝜋
4
= 141.4 𝑘𝑁
𝑄𝑠 = ෍ 𝑐𝑢 ∗ 𝛼 ∗ 𝑝 ∗ ∆𝑙
According to Randolph and Murphy (1985),
𝜎′𝑧𝐷 1 = 3 ∗ 17 = 51𝑘𝑃𝑎 →
𝑐𝑢
𝜎′
𝑧𝐷 1
=
50
51
≈ 1 → 𝛼1 ≈ 0.5 (𝑏𝑙𝑢𝑒)
𝜎′𝑧𝐷 2 = 6 ∗ 17 + 2 ∗ 19 − 9.8 = 120.4𝑘𝑃𝑎 →
𝑐𝑢
𝜎′
𝑧𝐷 2
=
50
120.4
≈ 0.41 → 𝛼2 ≈ 0.78 (𝑟𝑒𝑑)

13. Solution 5:
For safety, 𝐿 can be assumed zero while determining the lowest possible value of 𝛼3
𝑝 = 𝜋 ∗ 0.4 = 1.26 𝑚
𝑄𝑠 = 50 ∗ 0.5 ∗ 6 + 50 ∗ 0.78 ∗ 4 + 125 ∗ 0.52 ∗ L ∗ 1.26 = 385.56 + 81.9𝐿
𝜎′𝑧𝐷 3 = 6 ∗ 17 + 4 ∗ 19 − 9.8 +
𝐿
2
∗ 19.5 − 9.8 = 138.8 + 4.85𝐿
→
𝑐𝑢
𝜎′
𝑧𝐷 2
=
125
138.8
≈ 0.9 → 𝛼3 ≈ 0.52 (𝑔𝑟𝑒𝑒𝑛)
𝑄𝑎𝑙𝑙 = 500𝑘𝑁 → 𝑄𝑢𝑙𝑡 = 1500𝑘𝑁 → 141.4 + 385.56 + 81.9𝐿 = 1500 → 𝐿 = 11.2 𝑚
𝑃𝑖𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ = 6 + 4 + 11.2 = 21.2 ≈ 22𝑚

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 (f´c=40MPa). Use
Davisson’s method to compute the ultimate
downward load capacity.

15. Solution 6:
Interpratation of static load tests

16. Solution 6:
𝐸 = 4700 𝑓′𝑐 = 4700 40 𝑀𝑃𝑎 = 30000 𝑀𝑃𝑎
4𝑚𝑚 +
𝐵
120
+
𝑃𝐷
𝐴𝐸
= 4𝑚𝑚 +
40𝑚𝑚
120
+
𝑃 17000𝑚𝑚
0.42 ∗ 30000000𝑘𝑃𝑎
= 7.3𝑚𝑚 + 0.0035𝑃
𝑃𝑢𝑙𝑡 = 1650𝑘𝑁

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 ɣ = 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.

18. Solution 7:
Determine the ultimate capacity assuming that the piles in the group act as individual piles.
σ′𝑧 = ɣ × ൗ
𝐿
2 = 15.5 × ൗ
15
2 = 116.25𝑘𝑃𝑎
The average value for Τ
𝑐𝑢
σ′𝑧 = Τ
70
116.25 = 0.6
 
 
 
1 2
1 2
9
9
u p s
p p u
s u
u p u u
Q n n Q Q
Q A c
Q pc L
Q n n A c pc L


 

 
  


 
  
  
2
0.305 0.305 0.093 m
4 0.305 1.22 m
p
A
p
 
 
from figure for 70 kPa, 0.63
u
c 
 
           
 
4 3 9 0.093 70 0.63 1.22 70 15
12 58.59 807.03 10387 kN
ult
Q  
 
 
  


19. Solution 7:
Determine the ultimate capacity assuming that the piles in the group act as a block with dimensions
Lg x Bg x L.
 
 
   
* *
*
Skin resistance of the block:
2
Point bearing capacity :
Thus the ultimate load is equal to:
2
g u g g u
p p p u c g g u c
ult g g u c g g u
p c L L B c L
A Q A c N L B c N
Q L B c N L B c L
   
 
   
 
 
      
      
1
2
*
1 2 4 1 1.22 0.305 3.965 m
2
1 2 3 1 1.22 0.305 2.745 m
2
15 3.965
5.46 and 1.44
2.745 2.745
From figure, 8.6
g
g
g
g g
c
D
L n d
D
B n d
L
L
B B
N
 
      
 
 
 
      
 
 
   


20. Solution 7:
        
Block capacity 3.965 2.745 70 8.6 2 3.965 2.745 70 15
6552 14091 20643 kN
  
  
 
 
 
10387 kN 20643 kN
10387
2597 kN
4
g ult
g ult
g all
Q
Q
Q
FS
 
  

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.

22. Solution 8:
Because the lengths of the piles are 15 m each, the stress distribution starts at a depth of 10 m below
the top of the pile.
Settlement of Layer I
0(1) (1)
1
(1)
0(1) 0(1)
' '
log
1 '
c
c
C H
e
 


   
 
    
   

   
     
(1)
1 1
2000
' 51.6
3.3 3.5 2.2 3.5
g g
Q
kPa
L z B z

   
 
 
 
0(1)
' 2 16.2 12.5 18 9.81 134.8kPa
      
(1)
0.3 7 134.8 51.6
log 162.4
1 0.82 134.8
c mm

 
   
 
   

   

23. Solution 8:
Settlement of Layer II
0(2) (2)
2
(2)
0(2) 0(2)
' '
log
1 '
c
c
C H
e
 


   
 
    
   

   
     
(2)
2 2
2000
' 14.52
3.3 9 2.2 9
g g
Q
kPa
L z B z

   
 
 
 
0(2)
' 2 16.2 16 18 9.81 181.62kPa
      
(2)
0.2 4 181.62 14.52
log 15.7
1 0.7 181.62
c mm

 
   
 
   

   

24. Solution 8:
Settlement of Layer III
0(3) (3)
3
(3)
0(3) 0(3)
' '
log
1 '
c
c
C H
e
 


   
 
    
   

   
     
(3)
3 3
2000
' 9.2
3.3 12 2.2 12
g g
Q
kPa
L z B z

   
 
 
 
0(3)
' 2 16.2 19 18 9.81 208.99kPa
      
(3)
0.25 2 208.99 9.2
log 5.4
1 0.75 208.99
c mm

 
   
 
   

   
Total settlement
( ) (1) (2) (3) 162.4 15.7 5.4 183.5
c T c c c mm
   
      

25. -Ultimate (fully mobilized) load capacity
Deep Foundations-Pile Load Transfer Mechanism
𝑻𝒐𝒕𝒂𝒍 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 = 𝑠ℎ𝑎𝑓𝑡 𝑠𝑘𝑖𝑛 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 + 𝑡𝑖𝑝 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 − 𝑊𝑝𝑖𝑙𝑒
𝑅 = 𝑄𝑢𝑙𝑡 = 𝑄𝑠 + 𝑄𝑏 − 𝑊
𝑝
Buoyancy effect
should be taken
into account

26. -Ultimate (fully mobilized) load capacity
Deep Foundations-Pile Load Transfer Mechanism
𝑻𝒐𝒕𝒂𝒍 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚
=
𝑠𝑘𝑖𝑛 𝑠ℎ𝑎𝑓𝑡 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
+
𝑡𝑖𝑝 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
Friction pile
(clayey soil, adhesion)
End bearing pile
(rock formation)
𝑅 = 𝑄𝑢𝑙𝑡 = 𝑄𝑠 + 𝑄𝑏 − 𝑊
𝑝
End bearing pile
(granular soils)
>3 m
or
5B

27. -Ultimate (fully mobilized) load capacity
Deep Foundations-Pile Load Transfer Mechanism
𝑻𝒐𝒕𝒂𝒍 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 = 𝑠𝑘𝑖𝑛 𝑠ℎ𝑎𝑓𝑡 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 + 𝑡𝑖𝑝 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛
to attain fully mobilised capacity
10 to 25% of the pile width
0.5 to 1% of the pile width

28. -Bearing resistance (Undrained condition, clayey soil, non-free draining)
Deep Foundations-Pile Axial Load Capacity
𝑄𝑏 = 𝐴𝑏 𝑠𝑐𝑁𝑐𝑐𝑢 + 𝑁𝑞σ′𝑞 (𝑏𝑜𝑟𝑒𝑑 𝑎𝑛𝑑 𝑑𝑟𝑖𝑣𝑒𝑛 𝑝𝑖𝑙𝑒𝑠)
Ab=cross-sectional area of the base of the pile
-Bearing resistance (Drained condition, granular soil, free draining)
𝑄𝑏 = 𝐴𝑏 𝑁𝑞σ′𝑞 (𝑏𝑜𝑟𝑒𝑑 𝑎𝑛𝑑 𝑑𝑟𝑖𝑣𝑒𝑛 𝑝𝑖𝑙𝑒𝑠)
𝑠𝑐𝑁𝑐 = 9.0 𝑓𝑜𝑟 𝑠𝑞𝑢𝑎𝑟𝑒 𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑝𝑖𝑙𝑒
may be neglected (Nq=1)

29. Deep Foundations-Pile Axial Load Capacity
-Bearing resistance (Drained condition, granular soil, free draining)
𝑄𝑏 = 𝐴𝑏 𝑁𝑞σ′𝑞
φ Nq
20.00 12.40
21.00 13.80
22.00 15.50
23.00 17.90
24.00 21.40
25.00 26.00
26.00 29.50
27.00 34.00
28.00 39.70
29.00 46.50
30.00 56.70
31.00 68.20
32.00 81.00
33.00 96.00
34.00 115.00
35.00 143.00
36.00 168.00
37.00 194.00
38.00 231.00
39.00 276.00
40.00 346.00
41.00 420.00
42.00 525.00
43.00 650.00
44.00 780.00
45.00 930.00
𝐷
𝐵
𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠
𝐿𝑝𝑖𝑙𝑒
𝐵
𝑀𝑒𝑦𝑒𝑟ℎ𝑜𝑓

30. Deep Foundations-Pile Axial Load Capacity
-Shaft resistance (Undrained condition, clayey soil, non-free draining)
 α-method (𝑑𝑟𝑖𝑣𝑒𝑛 𝑝𝑖𝑙𝑒𝑠)
𝑄𝑠 = α(𝑐𝑢)𝐴𝑠
As= shaft area of the of the pile
α= adhesion factor and f(slenderness, shear strength)
𝐹𝑜𝑟
𝑐𝑢
σ′𝑣0
≤ 1; α = 0.5Fp
𝑐𝑢
σ′𝑣0
−0.5
≤ 1
𝐹𝑜𝑟
𝑐𝑢
σ′𝑣0
≥ 1; α = 0.5Fp
𝑐𝑢
σ′𝑣0
−0.25
Fp = 1.0 if
𝐿𝑝
𝐵
≤ 50
= 0.7if
𝐿𝑝
𝐵
≥ 120

31. Deep Foundations-Pile Axial Load Capacity
-Shaft resistance (Undrained condition, clayey soil, non-free draining)
 α-method (𝑏𝑜𝑟𝑒𝑑 𝑝𝑖𝑙𝑒𝑠)
𝑄𝑠 = α(𝑐𝑢)𝐴𝑠
As= shaft area of the of the pile
α= adhesion factor and f(shear strength)
α = 1.0 𝑓𝑜𝑟 𝑐𝑢 ≤ 30
α = 1. 16 −
𝑐𝑢
185
𝑓𝑜𝑟 30 < 𝑐𝑢 ≤ 150
α = 0.35 𝑓𝑜𝑟 𝑐𝑢 > 150

32. Deep Foundations-Pile Axial Load Capacity
-Shaft resistance (Drained condition, granular soil, free draining)
𝑄𝑠 = 𝐾𝑠σ′𝑣0 tan δ′ 𝐴𝑠 (𝑏𝑜𝑟𝑒𝑑 𝑎𝑛𝑑 𝑑𝑟𝑖𝑣𝑒𝑛 𝑝𝑖𝑙𝑒𝑠)
As= shaft area of the of the pile
Ks= coefficient of horizontal soil stress
δ'= the angle of friction b/w pile and soil
δ′ = ϕ = 0.75ϕ′
𝐾0 = 1 − 𝑠𝑖𝑛ϕ′

33. Deep Foundations-Pile Axial Load Capacity
-Shaft resistance (Drained condition, granular soil, free draining)
𝑄𝑠 = βσ′𝑣0𝐴𝑠
β=0.18+0.65Dr
Dr=relative density in decimal form
 β-method (𝑑𝑟𝑖𝑣𝑒𝑛 𝑝𝑖𝑙𝑒𝑠)
𝑄𝑠 = βσ′𝑣0𝐴𝑠
β=1.5-0.245 𝑧
z= depth to the mid-layer (m)
 β-method (𝑏𝑜𝑟𝑒𝑑 𝑝𝑖𝑙𝑒𝑠)

34. Ex1: A bridge pier is to be supported by 1.0 𝑚 diameter driven-cast in place concrete
piles. The design load per pile is determined as 700 𝑘𝑁 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 γ=17 kN/m3 ϕ'=36˚
Clay γ=20 kN/m3 cu=100 kPa
σ𝑣0 (𝑘𝑃𝑎) 𝑢 (𝑘𝑃𝑎) σ′𝑣0 (𝑘𝑃𝑎)
51
102
102 + 20ℎ𝑐𝑙𝑎𝑦 10ℎ𝑐𝑙𝑎𝑦
51
102
102 + 10ℎ𝑐𝑙𝑎𝑦

35. Ex1: A bridge pier is to be supported by 1.0 𝑚 diameter driven-cast in place concrete
piles. The design load per pile is determined as 700 𝑘𝑁 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 γ=17 kN/m3 ϕ'=36˚
Clay γ=20 kN/m3 cu=100 kPa
•
𝐿𝑝
𝐵
≥ 12, 𝑡ℎ𝑢𝑠 𝐿𝑝,𝑚𝑖𝑛 ≥ 12 𝑚
𝑻𝒐𝒕𝒂𝒍 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 𝑄𝑢𝑙𝑡 = ∑𝑄𝑠 + 𝑄𝑏
• 𝐹𝑜𝑟 𝑠𝑎𝑛𝑑 𝑙𝑎𝑦𝑒𝑟
𝑄𝑠 = 𝐾𝑠σ′𝑣0 tan δ′ 𝐴𝑠
𝐾𝑠 = 0.41 1.5 = 0.62 δ′ = ϕ = 0.75 ϕ′ = 27˚
𝐾0 = 1 − 𝑠𝑖𝑛ϕ′ = 0.41

36. Ex1 (cont.)
Deep Foundations- Single Pile Capacity
𝑄𝑠 = 𝐾𝑠σ′𝑣0 tan δ′ (π𝐷𝑙𝑠𝑎𝑛𝑑)
σ′𝑣0 @𝑧=3𝑚 = 3 17 = 51 kPa 𝑄𝑠,1 = 0.62 54 tan 27 π 1 6 = 303.7 𝑘𝑁
• 𝑭𝒐𝒓 𝒄𝒍𝒂𝒚 𝒍𝒂𝒚𝒆𝒓
𝑄𝑠 = α(𝑐𝑢)𝐴𝑠 𝐹𝑜𝑟
𝑐𝑢
σ′𝑣0
≤ 1; α = 0.5Fp
𝑐𝑢
σ′𝑣0
−0.5
≤ 1 Fp = 1.0 if
𝐿𝑝
𝐵
≤ 50
= 0.7if
𝐿𝑝
𝐵
≥ 120
σ′𝑣0 @𝑧=6𝑚 = 6 17 = 102 kPa
𝑐𝑢
σ′𝑣0
=
100
102
< 1 α = 0.5Fp
𝑐𝑢
σ′𝑣0
−0.5
= 0.5
100
102 + 20 − 10 𝑙𝑐𝑙𝑎𝑦
−0.5
𝑄𝑏 = 𝐴𝑏 𝑠𝑐𝑁𝑐𝑐𝑢 = π 0.5 2 9 100 = 706.9 𝑘𝑁
𝐴𝑠 = π𝐵(2𝑙𝑐𝑙𝑎𝑦)
𝑄𝑠,2 = 0.5
100
102 + 10 𝑙𝑐𝑙𝑎𝑦
−0.5
100 π(1)(2𝑙𝑐𝑙𝑎𝑦)
• 𝑭𝒐𝒓 𝒔𝒂𝒏𝒅 𝒍𝒂𝒚𝒆𝒓
Sand γ=17 kN/m3 ϕ'=36˚
Clay γ=20 kN/m3 cu=100 kPa 2𝑙𝑐𝑙𝑎𝑦

37. Ex1 (cont.)
Deep Foundations- Single Pile Capacity
• 𝑄𝑢𝑙𝑡 = 𝑄𝑠,1 + 𝑄𝑠,2 + 𝑄𝑏 − 𝑊
𝑝
303.7 + 𝑄𝑠,2 + 706.9 = π 0.5 2
25 ∗ 6 + 15 ∗ 2𝑙𝑐𝑙𝑎𝑦 + 700(3) 𝑇𝑟𝑖𝑎𝑙 𝑎𝑛𝑑 𝑒𝑟𝑟𝑜𝑟
𝑜𝑟 𝑒𝑞𝑛 𝑠𝑜𝑙𝑣𝑖𝑛𝑔
• 𝑄𝑎𝑙𝑙𝑜𝑤 =
𝑄𝑠,1+𝑄𝑠,2+𝑄𝑏
𝐹𝑆
− 𝑊
𝑝 = 700
1010.6 + 314.2
100
102 + 10 𝑙𝑐𝑙𝑎𝑦
−0.5
𝑙𝑐𝑙𝑎𝑦 = 117.8 + 23.6 𝑙𝑐𝑙𝑎𝑦 + 700(3)
1010.6 + 314.2
100
102 + 10 𝑙𝑐𝑙𝑎𝑦
−0.5
𝑙𝑐𝑙𝑎𝑦 = 2217.8 + 70.8𝑙𝑐𝑙𝑎𝑦
𝑙𝑚𝑖𝑑 𝑐𝑙𝑎𝑦 = 4 𝑚 𝐿𝑝 = 6 + 2 ∗ 4 = 𝟏𝟔 𝒎
Eqn is written
wrt. Midpoint
of clay layer
γ𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒′ = 25 − 10
γ𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 25 𝑘𝑁/𝑚3
𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦

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 𝑘𝑁 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 γ=17 kN/m3 ϕ'=28˚
Clay γ=18 kN/m3 cu=30 kPa
σ𝑣0 (𝑘𝑃𝑎) 𝑢 (𝑘𝑃𝑎) σ′𝑣0 (𝑘𝑃𝑎)
51
102
210 60
Sand γ=19 kN/m3 ϕ' =38˚
Clay γ=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 𝑘𝑁 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 γ=17 kN/m3 ϕ'=28˚
Clay γ=18 kN/m3 cu=30 kPa
σ𝑣0 (𝑘𝑃𝑎) 𝑢 (𝑘𝑃𝑎) σ′𝑣0 (𝑘𝑃𝑎)
51
102
210 60
Sand γ=19 kN/m3 ϕ' =38˚
Clay γ=18 kN/m3 cu=150 kPa
6 m
6 m
120
324
51
102
150
204
• 𝐸𝐶7 σ𝑝𝑖𝑙𝑒 𝑐𝑜𝑚𝑝 =
𝑄𝑝𝑖𝑙𝑒
𝐴𝑏
≤
σ𝑐,𝑘
4
σ𝑝𝑖𝑙𝑒 𝑐𝑜𝑚𝑝 ≤
30
4
= 7.5 𝑀𝑃𝑎
𝑄𝑝𝑖𝑙𝑒
𝐴𝑏
=
2000000
𝐴𝑏
= 7.5 𝑃𝑖𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 ⇒ 𝐵 = 600𝑚𝑚

40. Ex2 (cont.)
Deep Foundations- Single Pile Capacity
6 m Sand γ=17 kN/m3 ϕ'=28˚
Clay γ=18 kN/m3 cu=30 kPa
Sand γ=19 kN/m3 ϕ' =38˚
Clay γ=18 kN/m3 cu=150 kPa
6 m
6 m
• Embedment length>3 m or 5B= 3m
L=12+3=15 m
•
𝐿𝑝
𝐵
=
15
0.6
= 25 ≥ 12
<50
• 𝐹𝑜𝑟 𝑡ℎ𝑒 𝑠𝑎𝑛𝑑 𝑙𝑎𝑦𝑒𝑟 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑜𝑝
𝑄𝑠 = 𝐾𝑠σ′𝑣0 tan δ′ 𝐴𝑠
ϕ = 0.75ϕ′ = 21.8˚
𝐾0 = 1 − 𝑠𝑖𝑛ϕ′
= 0.53 𝑎𝑛𝑑 𝐾𝑠 = 𝐾0 0.85 = 0.45
𝑄𝑠 = 𝐾𝑠σ′𝑣0 tan δ′ 𝐴𝑠 = 0.45 51 tan 21.8 π(0.6)6 = 103.8 𝑘𝑁
• 𝑄𝑢𝑙𝑡 = 𝑄𝑠,𝑠𝑎𝑛𝑑1 + 𝑄𝑠,𝑐𝑙𝑎𝑦 + 𝑄𝑠,𝑠𝑎𝑛𝑑2 + 𝑄𝑏 − 𝑊
𝑝

41. Ex2 (cont.)
Deep Foundations- Single Pile Capacity
6 m Sand γ=17 kN/m3 ϕ'=28˚
Clay γ=18 kN/m3 cu=30 kPa
Sand γ=19 kN/m3 ϕ' =38˚
Clay γ=18 kN/m3 cu=150 kPa
6 m
6 m
L=15 m
• 𝑄𝑢𝑙𝑡 = 𝑄𝑠,𝑠𝑎𝑛𝑑1 + 𝑄𝑠,𝑐𝑙𝑎𝑦 + 𝑄𝑠,𝑠𝑎𝑛𝑑2 + 𝑄𝑏 − 𝑊
𝑝
𝑄𝑠 = α(𝑐𝑢)𝐴𝑠 α = 1.0 𝑓𝑜𝑟 𝑐𝑢 ≤ 30
𝑄𝑠 = α(𝑐𝑢)𝐴𝑠 = 1 30 π 0.6 6 = 339.3 𝑘𝑁
• For clay layer
• 𝐹𝑜𝑟 𝑠𝑎𝑛𝑑 𝑙𝑎𝑦𝑒𝑟 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚
𝑄𝑠 = 𝐾𝑠σ′𝑣0 tan δ′ 𝐴𝑠
𝑄𝑠 = 𝐾𝑠σ′𝑣0 tan δ′ 𝐴𝑠 = 0.33 163.5 tan 28.5 π 0.6 3 = 165.7 𝑘𝑁
𝑄𝑏 = 𝐴𝑏 𝑁𝑞σ′𝑞 = π0.32
120 177 = 6005.5 kN
• 𝑊
𝑝 = π 0.3 2 6 25 + 9 15 = 80.6 𝑘𝑁
𝑄𝑢𝑙𝑡 = 103.8 + 339.3 + 165.7 + 6005.5 − 80.6 = 6533.7 𝑘𝑁
𝑄𝑎𝑙𝑙 =
∑𝑄𝑠 + 𝑄𝑏
3
− 𝑊
𝑝 = 2124.2 ≥ 2000 𝑘𝑁
120
γ𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒′ = 25 − 10

42. -Load vs settlement: Load controlled test, to obtain ultimate load
 Load increment= 25% of working load
 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
δ𝑝𝑖𝑙𝑒 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 =
(𝑄𝑤𝑝+0.6𝑄𝑤𝑠)𝐿𝑝
𝐴𝑏𝐸𝑝𝑖𝑙𝑒
𝑄𝑤𝑝 = 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑙𝑜𝑎𝑑
𝑄𝑤𝑠 = 𝑠ℎ𝑎𝑓𝑡 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑙𝑜𝑎𝑑 @𝑤𝑜𝑟𝑘𝑖𝑛𝑔

43. -Load vs settlement
 Use net settlement values
Deep Foundations- Pile Axial Load Test
δ𝑢𝑙𝑡 = 0.012 𝐵𝑟𝑒𝑓 + 0.1
𝐵
𝐵𝑟𝑒𝑓
+
𝑄𝑢𝐿𝑝
𝐴𝑏𝐸𝑝
K M
K
M
Net settlement (mm)
𝑄𝑢𝑙𝑡 = 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑙𝑜𝑎𝑑 (𝑘𝑁)
𝐵 = 𝑃𝑖𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 (𝑚𝑚)
𝐵𝑟𝑒𝑓 = 𝑅𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑝𝑖𝑙𝑒 𝑤𝑖𝑑𝑡ℎ 𝑚𝑚
= 300 mm
𝐿𝑝 = 𝑃𝑖𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ (𝑚𝑚)
𝐴𝑏 = 𝐸𝑛𝑑 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑎𝑟𝑒𝑎 𝑚𝑚2
𝐸𝑝 = 𝑃𝑖𝑙𝑒 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 (𝐺𝑃𝑎)
δ𝑢𝑙𝑡
𝑄𝑢𝑙𝑡

44. -Ex3: Lp= 20 m, BxB=406x406 mm, Ep=30 GPa (C30 concrete), Qult ?
Deep Foundations- Pile Axial Load Test
δ𝑢𝑙𝑡 = 0.012 𝐵𝑟𝑒𝑓 + 0.1
𝐵
𝐵𝑟𝑒𝑓
+
𝑄𝑢𝐿𝑝
𝐴𝑏𝐸𝑝
K=3.735
Net settlement (mm)
δ𝑢𝑙𝑡 = 0.012 300 + 0.1
406
300
+
𝑄𝑢20000
(406)(406)(30)
𝑄𝑢𝑙𝑡 = 1460 𝑘𝑁

45. -Downdrag force acting on pile: force exerted by settling soil on pile (δsoil> δpile)
Deep Foundations- Negative Skin Friction
 Placement of fill (granular) settlement of the fill by selfweight
additional consolidation of soft clayey layers beneath fill
 Lowering of GWT σ' and settlement increases
additional
settlement
for the pile

46. -Downdrag force acting on pile
Deep Foundations- Negative Skin Friction
• 𝑄𝑝 = 𝑃0 + 𝑄𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 + 𝑄𝑑𝑜𝑤𝑛𝑑𝑟𝑎𝑔
𝑄𝑑𝑜𝑤𝑛𝑑𝑟𝑎𝑔 = 0.25 σ′𝑣(π𝐵𝑙𝑐𝑙𝑎𝑦 𝑙𝑎𝑦𝑒𝑟)
• For clay layer upon consolidation
• For sand layer upon settling
𝑄𝑑𝑜𝑤𝑛𝑑𝑟𝑎𝑔 = 𝐾0σ′𝑣 tan δ′ (π𝐵𝑙𝑠𝑎𝑛𝑑)
δ′ = (0.6)ϕ′
Resistance
Load
Force
P0
Qb
𝐹𝑆 =
𝑄𝑠 + 𝑄𝑏
𝑃0 + 𝑄𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 + 𝑄𝑑𝑜𝑤𝑛𝑑𝑟𝑎𝑔
≥ 2.0
𝑄𝑠 + 𝑄𝑏

47. Ex2 (cont.): 2 m height of a very large fill (γ=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 γ=17 kN/m3 ϕ'=28˚
6 m
6 m
Clay γ=18 kN/m3 cu=30 kPa
Sand γ=19 kN/m3 ϕ' =38˚
Clay γ=18 kN/m3 cu=150 kPa
• 𝐹𝑆 =
𝑄𝑠+𝑄𝑝
𝑃0+𝑄𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡+𝑄𝑑𝑜𝑤𝑛𝑑𝑟𝑎𝑔
≥ 2.0
𝐹𝑆 =
6614.3
2000 + 80.6 + 𝑄𝑑𝑜𝑤𝑛𝑑𝑟𝑎𝑔
𝑄𝑑𝑜𝑤𝑛𝑑𝑟𝑎𝑔 = 0.25 σ′𝑣(π𝐷𝑙𝑐𝑙𝑎𝑦 𝑙𝑎𝑦𝑒𝑟)
σ′𝑣0 =
102 + 150
2
+ (2)(20) = 166 𝑘𝑃𝑎
Fill
𝑄𝑑𝑜𝑤𝑛𝑑𝑟𝑎𝑔 = 0.25 166 π 0.6 6 = 469 𝑘𝑁
𝐹𝑆 =
6614.3
2000 + 80.6 + 469
= 2.59

48. Piles working under tension:
Deep Foundations- Uplift Capacity
• 𝐹𝑆 =
𝑄𝑠+𝑄𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡
𝑄𝑢𝑝𝑤𝑎𝑟𝑑
≥ 3.0~6.0
 Ship building docks  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
 (# of piles) x Qult (single pile) > Qult (group) under same displacement
 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 = (η) x (# of piles) x Qult (single pile)
η (efficiency)=1−ψ
[ 𝑚 − 1 𝑛 + 𝑛 − 1 𝑚]
90(𝑚)(𝑛)
ψ in degrees = tan−1
𝐵
𝑠
B
B
B
B B
𝑚 = 𝑛𝑜 𝑜𝑓 𝑟𝑜𝑤𝑠
𝑛 = 𝑛𝑜 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛𝑠
s = spacing b/w piles measured from center to center

51. -Group capacity:
Deep Foundations- Piles in Group
Terzaghi-Peck Block Approach
 shallow spread foundation
 Crucial for soil profiles consisting of soft clay layers (low cu values)
 𝑄𝑢𝑙𝑡 = ∑𝑄𝑠 + 𝑄𝑏
𝑄𝑠 = 2 𝐿𝑏𝑙𝑜𝑐𝑘 + 𝐵𝑏𝑙𝑜𝑐𝑘 ∑𝐷𝑖α(𝑐𝑢) or 𝑄𝑠 = 2 𝐿𝑏𝑙𝑜𝑐𝑘 + 𝐵𝑏𝑙𝑜𝑐𝑘 ∑𝐷𝑖𝐾𝑠σ′𝑣0 tan δ′
clayey soil granular soil
𝑄𝑏 = (𝐿𝑏𝑙𝑜𝑐𝑘𝐵𝑏𝑙𝑜𝑐𝑘) 𝑠𝑐𝑁𝑐𝑐𝑢 𝑄𝑏 = (𝐿𝑏𝑙𝑜𝑐𝑘𝐵𝑏𝑙𝑜𝑐𝑘) 𝑁𝑞σ′𝑞𝑠𝑞
clayey soil granular soil
𝐷 = 𝐿𝑝

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
𝐺𝑟𝑜𝑢𝑝 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ
s=1.8 m
• 𝑄𝑢𝑙𝑡,𝑠𝑖𝑛𝑔𝑙𝑒 𝑝𝑖𝑙𝑒 = 𝑄𝑠 + 𝑄𝑏
σ′𝑣0 (𝑘𝑃𝑎)
114
228
𝑄𝑏 = 𝐴𝑏 𝑠𝑐𝑁𝑐𝑐𝑢 = π 0.3 2 9 60 = 152.7 𝑘𝑁
𝑄𝑠 = α(𝑐𝑢)𝐴𝑠 = 1.16 −
60
185
60 π 0.6 12 = 1134.2 𝑘𝑁 α = 1. 16 −
𝑐𝑢
185
𝑓𝑜𝑟 30 < 𝑐𝑢 ≤ 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
• Q𝑢𝑙𝑡,𝑠𝑖𝑛𝑔𝑙𝑒 𝑝𝑖𝑙𝑒 = 𝑄𝑠 + 𝑄𝑏 = 152.7 + 1134.2 = 1286.9 𝑘𝑁
• Qult, group = (η) x (# of piles) x Qult (single pile)
η (efficiency)=1−ψ
[ 𝑚 − 1 𝑛 + 𝑛 − 1 𝑚]
90(𝑚)(𝑛)
ψ in degrees = tan−1
𝐵
𝑠
= tan−1
0.6
1.8
= 18.4˚
η =1−18.4
[ 3 − 1 4 + 4 − 1 3]
90(4)(3)
= 0.71
𝑚 = 3
𝑛 = 4
• 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
Terzaghi−Peck Block 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ
𝑄𝑠 = 2 𝐿𝑏𝑙𝑜𝑐𝑘 + 𝐵𝑏𝑙𝑜𝑐𝑘 α(𝑐𝑢)
 𝑄𝑢𝑙𝑡,𝑏𝑙𝑜𝑐𝑘 = 𝑄𝑠 + 𝑄𝑏
𝑄𝑠 = 2[ 1.8 3 + 1.8 2 12 0.84 60 = 10886.4 kN
α = 1. 16 −
𝑐𝑢
185
𝑓𝑜𝑟 30 < 𝑐𝑢 ≤ 150
𝑄𝑏 = 𝐿𝑏𝑙𝑜𝑐𝑘𝐵𝑏𝑙𝑜𝑐𝑘 𝑠𝑐𝑁𝑐𝑐𝑢 = 𝐿𝑏𝑙𝑜𝑐𝑘𝐵𝑏𝑙𝑜𝑐𝑘 9𝑐𝑢
𝑄𝑏 = 5.4 3.6 9 60 = 10497.6𝑘𝑁
 𝑄𝑢𝑙𝑡,𝑏𝑙𝑜𝑐𝑘 = 𝑄𝑠 + 𝑄𝑏 = 10886.4 + 10497.6 = 21384 𝑘𝑁
 𝑄𝑢𝑙𝑡,𝑏𝑙𝑜𝑐𝑘 = min 𝐺𝑟𝑜𝑢𝑝 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ, 𝑇𝑒𝑟𝑧𝑎𝑔ℎ𝑖 − 𝑃𝑒𝑐𝑘 𝑏𝑙𝑜𝑐𝑘 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ = 10694.4 𝑘𝑁