1. DEEP FOUNDATIONS
PILES

2. Pile foundation used to support structure when
poor quality soil
bearing capacity failure
excessive settlement

3. piles
END BEARING PILES
SKIN FRICTION PILES

4. End-bearing pile pile driven until it comes to rest on a hard impenetrable layer of soil or rock

5. End Bearing Piles
End bearing pile rests on a relative firm soil . The load of the structure is transmitted through the pile into this firm soil or rock because the base of the pile bears the load of the structure, this type of pile is called end bearing pile
PILES
SOFT SOIL
ROCK

6. Friction pile load of the structure must come from the skin friction or adhesion between surface of the pile and the soil

7. Friction Piles SOFT SOIL
If the firm soil is at a considerable depth, it may be very expensive to use end bearing piles. In such situations, piles are driven through the soil for some distance. Piles transmit the load of structure to the soil by means of skin friction
between the
Soil and the
piles.
Friction Piles
SOFT SOIL

8. Very Large Concentrated
Pile Applications
Very Large Concentrated
Weight
Large Distributed
Weight
Low
Weight
Soft to
Firm Clay
Dense Sand
20 September 2018
Deep Foundations
Strong Rock

9. TYPES OF PILE Concrete Piles - Cast In Place Concrete Piles
- Precast Concrete Piles
Steel Piles
- H Piles
- Cylindrical
Timber Piles
Composite Piles

10. Common Driven Pile Types

11. CAST IN PILES

12. Loads applied to Piles V M H
Combinations of vertical, horizontal and moment loading may be applied at the soil surface from the overlying structure
For the majority of foundations the loads applied to the piles are primarily vertical
For piles in jetties, foundations for bridge piers, tall chimneys, and offshore piled foundations the lateral resistance is an important consideration
The analysis of piles subjected to lateral and moment loading is more complex than simple vertical loading because of the soil-structure interaction.

13. Modes of failure The soil is always failure by punching shear.
The failure mode of pile is always in buckling failure mode.

14. Drilling Pile
Usually used for those tall buildings or massive industrial complexes, which require foundations which can bear the load of thousands of tons, most probably in unstable or difficult soil conditions. The method of drilling bored pile is different from RC Square pile or spun pile which are using driving method, the 

15. Procedure for drilling pile

18. Introduction
Piles are structural members that are made of steel, concrete, or timber.
They are used to build pile foundations, which are deep and cost more than shallow foundations.
Despite the cost, the use of piles often is necessary to ensure structural safety.
dr. isam jardaneh / foundation engineering / 2010

19. When to use Pile Foundations The following list identifies some of the conditions that require pile foundations (Vesic 1977):
dr. isam jardaneh / foundation engineering / 2010

20. dr. isam jardaneh / foundation engineering 61303 / 2010

21. dr. isam jardaneh / foundation engineering 61303 / 2010

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23. dr. isam jardaneh / foundation engineering 61303 / 2010

24. dr. isam jardaneh / foundation engineering 61303 / 2010

25. Types of Piles and Their Structural Characteristics
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26. dr. isam jardaneh / foundation engineering 61303 / 2010

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33. dr. isam jardaneh / foundation engineering 61303 / 2010

34. dr. isam jardaneh / foundation engineering 61303 / 2010

35. (unreinforced

38. dr. isam jardaneh / foundation engineering 61303 / 2010

39. dr. isam jardaneh / foundation engineering 61303 / 2010

44. Installation of Piles Two methods are mainly exist: 1- Driven Method
2- Bored Cast in-situ Piles Method

48. Load Transfere Mechanism
The load on the pile is resisted by
Qs = side friction (skin friction), and
Qp = end (point, tip) bearing
dr. isam jardaneh / foundation engineering / 2010

50. Equations for Estimating Single Pile Capacity
dr. isam jardaneh / foundation engineering / 2010

68. Qall = [qu design] Ap / FS

74. Se = Se1 + Se2 + Se3 example 338KN Find elastic settlement Se SAND
γ=16KN/m³
Ф=35
Es=30000KN/m²
µ=0.3
Qwp=98KN
Qws=240KN
Solution
12m
Se = Se1 + Se2 + Se3
305mm

75. Se1= -------------------------- = 0.00148 m = 1.48mm
[98+(0.6x240)] 12
[(0.305)² (21x10⁶)]

76. Se2 = ----------------------(1-0.3²)(0.85) = 0.0083m = 8.3mm
qwp = Qwp/Ap = 98/(0.305)² = KN/m]
Se2 = (1-0.3²)(0.85) = m = 8.3mm
(1053.5)(0.305)
(30000)

77. Se3 = (-------------------)(0.305/30000)(1-0.3²)(4.2)=0.64mm
240
(4x0.305)(12)

78. Se = Se1 + Se2 + Se3
Se = = 10.42mm