DEEP FOUNDATIONS PILES
Pile foundation used to support structure when poor quality soil bearing capacity failure excessive settlement
piles END BEARING PILES SKIN FRICTION PILES
End-bearing pile pile driven until it comes to rest on a hard impenetrable layer of soil or rock
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
Friction pile load of the structure must come from the skin friction or adhesion between surface of the pile and the soil
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
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
TYPES OF PILE Concrete Piles - Cast In Place Concrete Piles - Precast Concrete Piles Steel Piles - H Piles - Cylindrical Timber Piles Composite Piles
Common Driven Pile Types
CAST IN PILES
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.
Modes of failure The soil is always failure by punching shear. The failure mode of pile is always in buckling failure mode.
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
Procedure for drilling pile
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 61303 / 2010
When to use Pile Foundations The following list identifies some of the conditions that require pile foundations (Vesic 1977): dr. isam jardaneh / foundation engineering 61303 / 2010
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Types of Piles and Their Structural Characteristics dr. isam jardaneh / foundation engineering 61303 / 2010
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dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
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(unreinforced
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dr. isam jardaneh / foundation engineering 61303 / 2010
Installation of Piles Two methods are mainly exist: 1- Driven Method 2- Bored Cast in-situ Piles Method
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 61303 / 2010
Equations for Estimating Single Pile Capacity dr. isam jardaneh / foundation engineering 61303 / 2010
Qall = [qu design] Ap / FS
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
Se1= -------------------------- = 0.00148 m = 1.48mm [98+(0.6x240)] 12 [(0.305)² (21x10⁶)]
Se2 = ----------------------(1-0.3²)(0.85) = 0.0083m = 8.3mm qwp = Qwp/Ap = 98/(0.305)² = 1053.5KN/m] Se2 = ----------------------(1-0.3²)(0.85) = 0.0083m = 8.3mm (1053.5)(0.305) (30000)
Se3 = (-------------------)(0.305/30000)(1-0.3²)(4.2)=0.64mm 240 (4x0.305)(12)
Se = Se1 + Se2 + Se3 Se = 1.48 + 8.3 + 0.64 = 10.42mm