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 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 End Bearing Piles 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
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
6 May 2015 Deep Foundations Pile Applications Low Weight Large Distributed Weight Very Large Concentrated Weight Soft to Firm Clay Dense Sand
TYPES OF PILE 1. Concrete Piles - Cast In Place Concrete Piles - Precast Concrete Piles 2. Steel Piles - H Piles - Cylindrical 3. Timber Piles 4. Composite Piles
Common Driven Pile Types
CAST IN PILES
Loads applied to Piles V 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. M H
Modes of failure The soil is always failure by punching shear. The failure mode of pile is always in buckling failure mode.
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
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
(unreinforced
dr. isam jardaneh / foundation engineering 61303 / 2010
dr. isam jardaneh / foundation engineering 61303 / 2010
Two methods are mainly exist: 1- Driven Method 2- Bored Cast in-situ Piles Method
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
dr. isam jardaneh / foundation engineering 61303 / 2010
Qall = [qu design] Ap/FS
example Find elastic settlement Se Solution Se = Se1 + Se2 + Se3 SAND γ=16kn/m³ Ф=35 Es=30000KN/m² µ=0.3 Qwp=98KN Qws=240KN 338KN 12m 305mm
Se1= -------------------------- = 0.00148 m = 1.48mm [98+(0.6x240)] 12 [(0.305)² (21x10⁶)]
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