Chp.8 pile foundations lecture 3

15
8.8 Pile groups 8.8.1Group Efficiency: in most cases, piles are used in groups (min. two or three piles) to transmit the structural load to the soil. A pile cap is constructed over group piles to spread the load to the several piles, it can be in contact with the ground as in most cases, or well above the ground, as in case of offshore platforms.

Transcript of Chp.8 pile foundations lecture 3

Page 1: Chp.8 pile foundations  lecture 3

8.8 Pile groups

8.8.1Group Efficiency: in most cases, piles are used in groups (min. two or three piles) to transmit the structural load to the soil. A pile cap is constructed over group piles to spread the load to the several piles, it can be in contact with the ground as in most cases, or well above the ground, as in case of offshore platforms.

Page 2: Chp.8 pile foundations  lecture 3
Page 3: Chp.8 pile foundations  lecture 3

8.8.2 Determining the load-bearing When the piles are placed close to each other, a reasonable assumption is that the stresses transmitted by the piles to the soil will overlap reducing the load-bearing capacity of the piles. Ideally the piles in a group should be spaced so that:

• the load-bearing capacity of the group is not less than the sum of

the bearing capacity of individual piles. • The minimum center- to center pile spacing ”d” or “S”, is 2.5D to

3.5D (D:diameter of pile) or 2H to 3H(H:diagonal of rectangular shape or HP pile) for vertical loads.

• For lateral loads larger spacings are usually more efficient. The efficiency of the load-bearing capacity of a group Of piles is defined as :

Page 4: Chp.8 pile foundations  lecture 3

We may write it as:

Where: Pg: perimeter of group cross section= 2(n1+n2-2)d+4D D: diameter of a pile; n1,n2 = number of piles Favg: average unit frictional resistance(skin friction)

Pg

Hence,

Case when d>>

Converse –labarre Equation:

Page 5: Chp.8 pile foundations  lecture 3

A- Piles in sand 1.Driven piles: with d≥ 3D Qg(u) = ∑ Qu

2. Bored piles: with d=3D Qg(u) =(2/3 to ¾) ∑ Qu

B- Piles in clay The ultimate load bearing capacity of pile groups in clay may be estimated as follows: Step1: Determine

Where:

Step2: Determine the ultimate capacity of the pile group block with dimensions LgxBgxL

The skin resistance,

Point-bearing capacity,

Page 6: Chp.8 pile foundations  lecture 3

Step3 compare the two obtained values of ∑Qu, the lower value is Qg(u)

Page 7: Chp.8 pile foundations  lecture 3

8.9 Elastic Settlement of Pile Groups

For pile groups in sand and gravel, Meyerhof (1976) suggested the following empirical relation for elastic settlement:

where q (kN/m2)= Qg/(Lg .Bg) N: average corrected standard penetration number within (Bg) below

the pile tip deep below the tip of the piles) I : influence factor = 1 - L/8Bg ≥ 0.5 L : length of embedment of piles Similarly, the pile group settlement is related to the cone penetration resistance as:

Where qc is the average cone penetration resistance within Bg below pile tip

Page 8: Chp.8 pile foundations  lecture 3

8.10 Consolidation Settlement of Group Piles

The consolidation settlement of a pile group can be estimated by assuming an approximate distribution method that is commonly referred to as the 2:1 method. The calculation procedure involves the following steps: 1. let Qg be The group net total load of. If the pile cap is below the

original ground surface, Qg equals the total load of the superstructure on the piles minus the effective weight of soil above the pile group removed by excavation: Qgnet = Qg(gross)-γ.Df

2. Assume that the load Qg is transmitted to the soil beginning at a depth of 2L/3 from the top of the pile (as shown in Figure), The load Qg spreads out along 2:1 method 3. Calculate the effective stress increase caused at the middle of each soil layer by the load Qg:

Page 9: Chp.8 pile foundations  lecture 3

4.Calculate the settlement of each layer caused by the increased stress:

5. Calculate the total consolidation settlement of the pile group by:

Page 10: Chp.8 pile foundations  lecture 3

8.11 Design of PILE CAPS

• Unless a single pile is used, a cap is necessary to spread the vertical and horizontal loads and any overturning moments to all the piles in the group.

• The cap is usually of reinforced concrete, poured on the ground unless the soil is expansive.

• Caps for offshore structures are often fabricated from steel shapes. • The pile cap has a reaction that is a series of concentrated loads (the piles); and the design considers the column loads and

moments, any soil overlying the cap (if it is below the ground surface), and the weight of the cap.

1-Load distribution on the piles below cap • For a concentric axial load on the cap, each pile carries an equal

amount of the load. For n piles carrying a total load Q, the load Pp per pile is:

• If pile cap is noncentrally loaded or loaded with a load Q and a moment:

Page 11: Chp.8 pile foundations  lecture 3

where Mx, My = moments about x and y axes, respectively x, y = distances from y and x axes to any pile ∑x2, ∑ y2 = moment of inertia of the group, computed as: I = Io + Ad2

2- Cap Design by Circulage Method In this method, it is assumed that the column load is resisted by a truss or number of trusses developed inside the concrete cap. The forces in the compression members are taken by concrete and those of tension are resisted by steel reinforcement that tie or circulage the pile heads in order to prevent their lateral movements.

Page 12: Chp.8 pile foundations  lecture 3

2.1 Conditions No interior piles should exist , the distance from all piles to the center of pile cap should be equal.

2.2 Procedure 1- Determination of the effective depth This effective depth of the pile cap is the largest of: a- Minimum effective depth dmin = 40Φ pile steel bars

b- Punching shear d = Q/qp.b0 Q= column service load qp= allowable concrete punching stress = 8-10 kg/cm2 b0 = punching perimeter taken at d/3 distance from each column

face

Page 13: Chp.8 pile foundations  lecture 3

C- Angle α such that 45 ≤α ≤60

Two corrections must be carried out: a- the column load is not concentrated but distributed as shown in the

figure b- yct equals 8/9 d For the new triangle force: d = 9/8 d’ S’ = x –Dequivalent/4 ; where b is the column width; Dequiv: equivalent diameter of column tan α = d’/S’

Then dmin corresponds to α = 45

Plan

Page 14: Chp.8 pile foundations  lecture 3

In case of 2-pile cap

X = S/2

In case of 3-pile cap

In case of 3-pile cap

Page 15: Chp.8 pile foundations  lecture 3

Use 1/3 of circulage steel as an additional reinforcement to be placed around the cap in depth Shrinkage steel and As’ Bottom mesh of 0.15% Ac Vertical side reinforcement 4 Φ 16/m Top mesh 0.1% Ac

4Φ16/m’

2- Circulage Steel

T = Rp/tan α where Rp is the reaction transmitted under the pile Ts = T/2cos (β/2) As = Ts/fs ; As is the circulage steel