Static analysis of pile foundation

1. Static Analysis of Pile Foundation

2. Driven Pile Design and Construction
Process (1)

3. Driven Pile Design and Construction
Process (2)

4. Driven Pile Design and Construction
Process (3)

5. Static analysis of piles (1)
• Use soil strength and compressibility properties
to determine pile capacity and performance.
• Determine the most cost effective pile type and
estimate the number of piles and the required
pile lengths for the design of sub-structure
elements.
• Construction procedures can have a significant
influence on the behavior of pile foundations.
Adequate construction techniques are necessary
if static methods could lead to successful designs
of deep foundation.

6. Static analysis of piles (2)
• Static load tests, wave equation analysis or
dynamic monitoring for construction control
should be used to confirm the results of a static
design method.
• Two static analysis are required for a design (in
some cases)
– First static analysis to determine the number and
length of piles.
– Second static analysis to determine the total driving
resistance the pile will encounter during installation.

8. Scour hole around bridge pier
(Briaud 2013)

9. Local and channel degradation scour

11. Factor of safety (1)
• The factors of safety of static analyses range from
2 to 4.
• Most of static analyses methods recommended a
factor of safety of 3.0
• Use of high factor of safety leads to pile
installation problems.
• Construction control methods have significant
influence on pile capacity.
• The factor of safety used in a static analysis
calculation should be based upon the
construction control method specified.

12. Factor of safety (2)

13. Load transfer
• The ultimate pile capacity, Qu, of a pile in homogeneous soil:
– Qu = Rs + Rt
– Qu = fs As + qt At
– fs is the unit shaft resistance of the shaft surface area, As, and qt is the
unit toe resistance over the pile toe area, At.
– Displacement is needed to mobilize the shaft resistance, and above
equations for pile bearing capacity assume that both pile toe and the
pile shaft have moved sufficiently with respect to the adjacent soil to
simultaneously develop the ultimate shaft and toe resistances.
– The displacement needed to mobilize the shaft resistance is smaller
than that required to mobilize the toe resistance.
– Maximum frictional resistance along the pile shaft will be fully
mobilized when the relative displacement between the soil and the
pile is about 5 to 10 mm (0.2 to 0.3 in.), irrespective of the pile size
and length. But the maximum point resistance will not be mobilized
until pile tip has moved about 10 to 25% of the pile width (or
diameter) (Das, 2011).

14. Typical load transfer mechanism for
piles (1)
Ultimate
Condition

15. Typical load transfer mechanism for
piles (2)

16. Bearing capacity of a single pile
• Unit toe bearing, qp (or total bearing
resistance force Qp)
– For shallow foundation,
– For pile foundation,
Relatively, D is
little, and can be
dropped without
introducing a
serious error.

17. The total point bearing Qp
Qp
= Ap
qp
= Ap
(c'
Nc
*
+q'
Nq
*
)

18. The total skin friction Qs
Qs
= åpDLf
= perimeter of the pile section
= incremental pile length over which p and f
are constant
= unit friction resistance at any depth z
p
DL
f

19. ASD and LRFD Design Method
• ASD (Allowable Stress Design)
𝑇ℎ𝑒 𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝐿𝑜𝑎𝑑, 𝑄𝑎𝑙𝑙 =
𝑄𝑢(𝑜𝑟 𝑅𝑛) = 𝑄𝑝(𝑜𝑟 𝑅𝑝) + 𝑄𝑠 (𝑜𝑟 𝑅𝑠)
𝐹. 𝑆.
• LRFD (Load and Resistance Factor Design)
(More details can be seen page 14 “
Design and Construction of Driven Pile
Foundations – Vol. I “)

20. Resistance Factors for Static Analysis Methods
(modified from AASHTO 2014)

21. Ultimate bearing capacity of piles
driven in cohesionless soils

22. Methods of Static Analysis for Piles in Cohesionless Soils

23. Meyerhof’s method for tip resistance
Qp
• In sand, c’ = 0, therefore,
• The tip resistance qp in sand
increases with the depth of
embedment in the bearing
stratum and reach a maximum
value when the embedment
ratio L/D reaches (L/D)cr (16~18).
• The limiting point resistance

25. Tip resistance versus L/D

26. Tip resistance versus N60

27. Frictional resistance (Qs) in
cohesionless soil
• The total frictional resistance
• The unit skin friction increases linearly with
depth until a depth of L’, and remains constant
thereafter.
• L’ ranges at (15~20) pile diameters,
conservatively taken as L’= 15D
• From z = 0 to L’, 𝑓 = 𝐾𝜎0
′
𝑡𝑎𝑛𝛿′
• From z = L’ to L, 𝑓 = 𝑓𝑧=𝐿′

28. Unit skin friction distribution
=15D
𝑓 = 𝐾𝜎0
′
𝑡𝑎𝑛𝛿′
𝑓 = 𝐾𝜎0
′
𝑡𝑎𝑛𝛿′
(𝜎0
′=15D × 𝛾)

30. The SPT method
(cohesionless soils)

31. Meyerhof (1976) method based on SPT
data –tip resistance (total bearing) (kPa) (1)
• For piles driven into sands and gravels
𝑁𝐵
′
= 𝑁60, qt = qp

32. Meyerhof (1976) method based on
SPT data –tip resistance (kPa) (2)
• The limiting value of 400N’B is reached when the
embedment depth into the bearing stratum
reaches 10 pile diameters, and the effect of
overlying stratum becomes irrelevant.
• The above equation applies when the pile toe is
located near the interface of two strata with a
weaker stratum overlying the bearing stratum.
• For pile driven in a uniform cohesionless stratum,
the unit toe resistance is:

33. Meyerhof (1976) method based on
SPT data –tip resistance (kPa) (3)
• The average corrected SPT N' value, N’B, is
recommended to be used by averaging N'
values within the zone extending 3 diameters
below the pile toe.
• For pile driven in a non-plastic silts stratum,
the unit toe resistance is:

34. Meyerhof (1976) method based on SPT
data –tip resistance (kPa) (4)
(Textbook)
• For pile driven in a homogeneous granular
soil, the unit toe resistance is
𝑞𝑝 (𝑜𝑟𝑞𝑡 ) = 0.4𝑝𝑎𝑁60
𝐿
𝐷
≤ 4𝑝𝑎𝑁60
 𝑁60=the average value of the standard
penetration number near the pile point (about
10D above and 4D below the pile point)
 pa = atmospheric pressure (≅
100
𝑘𝑁
𝑚2 𝑜𝑟 2000 𝑙𝑏𝑠/𝑓𝑡2)

35. Meyerhof (1976) method based on
SPT data –skin friction(kPa)
• The unit shaft resistance fs, of driven displacement
piles (e.g., closed-end pipe piles and precast concrete
piles:
fs = 2N’ ≤ 100 kPa
• The unit shaft resistance fs, of driven non-displacement
piles (e.g., H-pile)s:
fs = N’ ≤ 100 kPa
• N’ (N60) is the average corrected SPT resistance value,
in blows per 300 mm (1 ft), along the embedded length
of pile. Typically, the soil profile is delineated into 3 to 6
m (10 to 20 ft) thick layers, and the average unit shaft
resistance is calculated for each soil layer.

36. Example: Meyerhof’s method for Qp
and Qs
Consider a 20 m long concrete pile with a cross
section of 0.407m × 0.407m fully embedded in
sand. For the sand, given: unit weight 𝛾 =
18
𝑘𝑁
𝑚3 ; and soil fricitonal angle ∅′
= 350
.
– Estimate the end bearing 𝑄𝑝and shaft resistance
𝑄𝑠

37. • 𝑄𝑝 = 𝐴𝑝𝑞′
𝑁𝑞
∗
≤ 𝐴𝑝 0.5𝑝𝑎𝑁𝑞
∗
𝑡𝑎𝑛∅′
• ∅′
= 350
→ 𝑁𝑞
∗
=143 (from the table or chart)
• 𝑞′
=𝛾𝐿=(18)(20)=360 kN/m2
• 𝐴𝑝𝑞′
𝑁𝑞
∗
=(0.407×0.407)(360)(143) = 8528kN
• The capping value
𝐴𝑝 0.5𝑝𝑎𝑁𝑞
∗
𝑡𝑎𝑛∅′
=(0.407×0.407)(0.5)(100)(1
43)(tan35o)=829 kN
• So, the total end bearing (tip resistance)
𝑄𝑝=829 kN

38. Example: Qs and 𝑄𝑢(cont’d)
• 𝑄𝑠 = 𝑄𝑠1 + 𝑄𝑠2
• 𝑄𝑠1 = 𝐿′ × 𝑝 × 𝑓15𝐷 = 6.105 ×
4 × 0.407 ×
70.65
2
= 351.1 𝑘𝑁
• 𝑄𝑠2 = 𝐿 − 𝐿′ × 𝑝 × 𝑓15𝐷 =
20 − 6.105 × 4 × 0.407 ×
70.65 = 1,598.18 𝑘𝑁
• 𝑄𝑠 = 𝑄𝑠1 + 𝑄𝑠2 = 351.1 +
1598.18 = 1949.28 𝐾𝑁
• The total bearing capacity 𝑄𝑢 =
𝑄𝑝 + 𝑄𝑠 =829+1949.28 =
2,778.28 kN
𝑓 = 𝐾𝜎0
′
𝑡𝑎𝑛𝛿′
=15D=15× 0.407 =
6.105 𝑚
𝑓15𝐷 = 𝐾𝜎0
′
𝑡𝑎𝑛𝛿′
𝜎0
′
= 15𝐷𝛾 = 15 × 0.407 × 18
= 109.89 kN/m2
𝐾 =1.5
𝛿′
=
2
3
× 35𝑜
= 23.3𝑜
𝑓15𝐷 =1.5× 109.89 ×
tan 23.3𝑜
= 70.65kN/m2
L = 20 m

39. Example: Pile driven cohesionless soils
For the soil profile
as shown on the
right, perform the
Meyerhof method
pile capacity
calculation for an
embedded length
of 10 meters.
Assume that
scour has not
occurred.

40. Compute the
average
corrected SPT
value for each
layer 𝑁′
Compute unit
skin friction fs
(kPa) using the
equation for
driven
displacement
piles

41. Compute ultimate shaft resistance
(kN)

42. Compute average SPT 𝑁′
value, 𝑁𝑜
′
and 𝑁𝐵
′

43. Compute the ultimate toe resistance

44. Compute the ultimate pile capacity, Qu (kN)
and the allowable design load, Qa (kN)
Note: Factor of safety
should selected based
on the construction
control method to be
specified, as
recommended in the
previous slide.

45. Ultimate Capacity of Piles in Cohesive
Soils

46. Methods of Static Analysis for Piles in
Cohesive Soils

47. Total Stress - α-Method
• Total Stress - α-Method
– The unit shaft resistance, fs, is equal to the adhesion, ca, which is
the shear stress between pile and soil at failure.
– fs = ca = α cu
– α is an empirical adhesion factor for reduction of the average
undrained shear strength cu
– α depends on the nature and strength of the clay, pile
dimension, method of pile installation, and time effects
– The values of α vary within wide limits and decrease rapidly
with increasing in shear strength
– The unit toe resistance, qp = cu Nc
– Nc is a dimensionless bearing capacity factor, and it depends on
the pile diameter and the depth of embedment, and usually
taken as 9 for deep foundations.

48. Variation
of Alpha
Table
12.11
(page 482)

49. Adhesion Values for Piles in
Cohesive Soils (after
Tomlinson, 1979)

52. Effective Stress Method - β -Method
• Effective Stress Method - β -Method

54. Chart for estimating Nt
Coefficients versus Soil
Type φ' Angle (after
Fellenius, 1991)

55. Example:
Perform a
static pile
capacity
calculation
using the SPT
and 𝛼 methods
for an
embedded
length of 13
meters

56. Properties of the soil layers
• Layer 1 (Extremely dense sand and Gravel):
𝑁1
′
= 135
– ∅1 = 36𝑜(Usually, if the corrected blow count N60
is greater than 50, the frictional angle is 43o. In
soil layers with greater than 50% gravel, the φ
angle for shaft resistance calculations should be
limited to: 36° for hard angular gravel, and 32° for
soft rounded gravel.

57. Properties of the soil layers (cont’d)
• Layer 2 (depth 3 to 6 m, stiff silty clay), the
average undrained shear strength 𝐶𝑢2 =
91+120
2
= 106 𝑘𝑃𝑎
• Layer 3 (depth 6 to 15 m, very stiff silty clay),
the average undrained shear strength 𝐶𝑢3 =
139+154+158+156+158+163
6
= 155 𝑘𝑃𝑎

58. Shaft resistances
• Layer 1
– 𝑓𝑠1 = 2N’ ≤ 100 kPa, 2 × 135 =
270 𝑘𝑃𝑎, 𝑠𝑜, 𝑓𝑠1 = 100 𝑘𝑃𝑎
– 𝑅𝑠1 = 𝐴𝑠1 × 𝑓𝑠1 = 4 × 0.356m × 1𝑚 × 100 =
142.4 𝑘𝑁

59. Shaft resistances (cont’d)
• For layers 2 and 3, use the 𝛼 − 𝑚𝑒𝑡ℎ𝑜𝑑
• Layer 2
– 𝑓𝑠2 = 𝛼𝐶𝑢2; (D/b) =
3.0 𝑚
0.356 𝑚
= 8.43 → 𝛼 = 1.0
– 𝑅𝑠2 = 𝐴𝑠2𝑓𝑠2 = 4 × 0.356 × 3 × 1.0 × 106 = 453 𝑘𝑁
• Layer 3
– 𝑓𝑠3 = 𝛼𝐶𝑢1; (D/b) =
9.0 𝑚
0.356 𝑚
= 25.28 → 𝛼 = 0.35
– 𝑅𝑠2 = 𝐴𝑠2𝑓𝑠2 = 4 × 0.356 × 9 × 1.0 × 155 = 696 𝑘𝑁
• The total shaft resistance 𝑅𝑠 = 𝑅𝑠1 + 𝑅𝑠2+ 𝑅𝑠3
– 𝑅𝑠=142 + 453 + 696 = 1,291𝑘𝑁

60. The toe resistance Qp – Meyerhof
method
• Estimate the friction angle for the zone from pile toe to 3 diameter
below pile toe (1.065m)
– 𝑁𝑡𝑜𝑒
′
= 33= ∅𝑡𝑜𝑒
′
= 35𝑜
• 𝑁𝑞
∗
= 143
• The overburden pressure 𝑞′
= 5.1 × 2.0 + 10.6 × 1.0 + 9.8 ×
3.0 + 10.4 × 9.0 = 140.2
𝑘𝑁
𝑚2
• 𝑞𝑝 = 𝑞′
𝑁𝑞
∗
= 140.2 × 143 = 20,048.6
𝑘𝑁
𝑚2
• Capping value 𝑞𝑙 = 0.5𝑝𝑎𝑁𝑞
∗
tan ∅𝑡𝑜𝑒
′
= 0.5 × 100 × 143 ×
tan 35𝑜 = 5,006.48
𝑘𝑁
𝑚2
• Since 𝑞𝑝> 𝑞𝑙, so 𝑞𝑝= 𝑞𝑙=5,006.48 kN/m2
• Qp = qpAp =5006.48× 0.356 × 0.356 = 634.50 𝑘𝑁

61. The ultimate pile capacity Qu and the
allowable design load Qa
• Qu=Rs + Qp=1291+634.5=1925.5 kN
• Qa=
Qu
𝐹.𝑆.
=
1925.5𝑘𝑁
𝐹.𝑆.
• The factor of safety should be selected based
on the construction control method to be
specified, as recommended in the table as
shown in the table discussed previously.

62. The CPT Method

63. Methods Based on Cone Penetration Test (CPT)
Data (cohesive and cohesionless soils)
• two main approaches for using CPT data to pile design,
indirect methods and direct methods.
• Indirect methods use CPT derived soil parameters such as
soil friction angle and undrained shear strength along with
bearing capacity and / or cavity expansion theories.
• Direct methods use cone resistance values to determine
unit shaft and toe resistances.
• The CPT methods use total stress rather than effective
stress values.
• The methods were developed based on pile types and soil
conditions in a local area and may therefore not perform as
well outside of that locality.

64. Shaft Resistance from the Nottingham and
Schmertmann Method—Cohesionless soils (1)
• For the cohesionless soils, the ultimate shaft
resistance, Rs, may be derived from unit sleeve
friction of the CPT using the following expression.

66. Shaft Resistance from the Nottingham and
Schmertmann Method—Cohesionless soils (2)
• If the sleeve friction is not available, the ultimate
shaft resistance, Rs, may be obtained from tip
resistance of CPT test.

67. Shaft Resistance from the Nottingham and
Schmertmann Method—Cohesive soils
• For the cohesive soils, the ultimate shaft
resistance, Rs, may be derived from unit sleeve
friction of the CPT using the following expression.

69. Tip Resistance from the Nottingham
and Schmertmann
• An elaborate averaging scheme is used to weight the
cone tip resistance, from 8 pile diameters above the
pile toe to as much as 3.75 diameters below the pile
toe, favoring the lower cone tip resistance, qc, values
local within the depth range.

72. CPT method homework
• Complete the homework problem using the
DOTD-adopted software (PileConeAnalysis) at
https://www.ltrc.lsu.edu/research_products.html
– (pile_CPT22.zip. You might need to try to install any of
the three versions, and see which one works with your
computer: Pile_CPT22.zip - Windows 7; Pile_CPT22.zip
- Windows 2000; Pile_CPT22.zip - Windows XP)
• Load up CPT data file CPT 192.txt (uploaded to
moodle).