1. LOS AGENTES NO ESTATALES: UNA CATEGORÍA DE DIFÍCIL
2.2. La aplicación del estándar de diligencia debida
2.2.2. La imputabilidad por infracción del estándar de diligencia debida en
2.2.2.2. La relación entre la obligación general de garantía y la diligencia debida en la
Bearing Piles and Groups – October 2010
Ultimate Capacity
The results above apply only to Whitaker’s experiments. For a generally-applicable rule:
Pile Group Capacity is the lesser of:
1. the sum of the failure loads of the individual piles Pult = n(Qs + Qb)
2. the bearing capacity (including side friction) of a block of soil defined by the perimeter of the pile group.
Pult = cuNcscdcBgLg + 2(Bg + Lg)Lc
If the bottom of the pile cap is in contact with the supporting soil, then the first of these can be increased by the failure capacity of this contact surface.
1. The sum of the failure loads of the individual piles plus the failure bearing capacity of the reminder of the pile cap.
Pult = n(Qs + Qb) + cuNcscdc(
BcLc – nAp)
End
Figure 10.15 Figure 10.16
Individual Pile Failure
Pile Cap
Block Failure
Pile Cap
Laterally-Loaded Piles Ultimate lateral resistance
Near the surface, passive pressures may be developed.
At depth, local flow of soil around the pile limits the lateral resistance.
Broms simplification for piles in clay
For a pile with a diameter D, ignore the passive pressure down to a depth of 1.5D.
Then take a limiting lateral pressure of 9cu.
L H
9(c u / m ) D 1.5D
D
L e
H a H a
L H A
B
Providing the pile is strong enough, the whole pile will rotate about the point A.
Normally the pile is not strong enough and the horizontal force is limited by the moment capacity of the pile at B.
Example
Calculate the ultimate moment capacity required for a long pile of 600 mm diameter if a lateral force of 100 kN (short-term) is to be applied 1 m above ground level.
Assume the clay has cu = 100 kN/m2. Partial safety factor on loads = 1.6 Partial safety factor on cu = 1.6 Design load = 1.6*100 = 160 kN Ignore top 1.5*0.6 = 0.9 m.
Bearing Piles and Groups – October 2010
Depth to resist lateral force = 160/337
= 0.475 m
160 kN
337 kN/m 0.9 m 0.6 m
1.0 m
160 kN 0.475 m
Depth to point of maximum moment
= 1.0 + 0.9 + 0.475 = 2.375 m Maximum moment
= 160*2.375 - 160*0.475/2
= 342 kNm
Note: Serviceability should also be checked because the passive pressures develop only with substantial movements.
Bearing Piles and Groups – October 2010
Example to show the implications of design to EC7.
In the following case, design the length of pile required i) to BS 8004, with an overall FoS of 2.5
Assume the top metre of the clay does not support load Adhesion factor = 0.45, bearing capacity factor Nc = 9
Revision Sheet Pile Design
1. a) A closed end steel tubular pile, 0.6m dia, is driven into stiff clay with a penetration of 35m. The undrained shear strength of the clay is 130 kN/m2 and the submerged unit weight is 13 kN/m3. Assuming a bearing capacity factor, Nc, of 9, determine the allowable pile working load assuming an overall factor of safety of 2.5.
(2414 kN) Relevant charts and expressions are given on Fig. Q6(a) and (b)
(9 marks) With reference to Figs Q6(a) and (b), explain why
(i) the peak adhesion factor p reduces as the soil strength ratio increases (ii) the length factor F reduces as the length/diameter ratio increases
(5 marks) 2. A 750 mm dia bored pile is to support a dead load of 900 kN and a variable load of
300 kN. Ground conditions involve two layers of clay. The upper layer is 8m thick and has an undrained shear strength of 50 kN/m2. The lower layer is of considerable thickness and has an undrained shear strength of 120 kN/m2.
The top metre of the shaft does not support any load.
The adhesion factor is 1.0 for the upper clay layer and 0.5 for the lower clay.
The bearing capacity factor, Nc, is 9 Determine the required pile length :
(i) in accordance with BS 8004 assuming factors of safety of 1.5 and 3.0 are applied to the shaft load and base load respectively (8 marks)
(ii) in accordance with Eurocode 7, to the ultimate limit state, assuming case C to be critical. Relevant clauses and tables from EC 7 are provided
(8 marks) (i) 13.2 m
(ii) 17.2m
Bearing Piles and Groups – October 2010
3. A precast concrete pile, 0.4m dia., is to be driven through a deposit of stiff clay 6m thick and into a thick deposit of dense sand. The water table lies at 2m below ground level. Properties of the soils and relevant parameters are as follows :
CLAY
Undrained shear strength increases from 90 kN/m2 at ground level to 126 kN/m2 at the base of the deposit.
Bulk unit weight b = 21 kN/m3 Adhesion factor = 0.35
SAND
Angle of friction = 37 Bulk unit weight b = 19 kN/m3 kstan = 1.5
For a working load of 1600 kN, determine the length of pile required assuming:
(i) no adhesion occurs over the top metre of the pile (ii) the factor of safety on adhesion in the clay is 2.0
(iii) the factor of safety on skin friction in the sand is 2.5 and on the base it is 3.0 (iv) there is a critical depth within the sand only, and measured below the top of the
sand (Use the Meyerhof curve for determination) Design charts, tables and expressions are given in Fig Q5
(15 marks) (13.5m)
Elastic Pile Settlement Example:
12m long concrete pile, diameter 450mm Working load = 600 kN
Load is carried 65% by skin friction and 35% by end bearing E value of soil beneath base is estimated to 250 MN/m2
E value for concrete is estimated to be 10,000 MN/m2 Using equation:
Bearing Piles and Groups – October 2010
Example of Pile Capacity from Static Test: to EC7
Three CFA piles tested. Estimate that 80% of capacity is in shaft friction and 20% in end bearing. Use the EC7 method to determine the Design Ultimate Bearing Capacity a pile similar to those tested.
Static Load Test Results:
Pile Number 1 2 3
Test Load at Failure 145 kN 120 kN 115 kN
Calculation:
Average test strength = 127 kN
Minimum strength = 115 kN
From the table of values
Characteristic Bearing Resistance is the lesser of:
127/1.3 = 98 kN (adopt this)
115/1.1 = 105 kN
Estimate of shaft friction resistance is 0.8 98 = 78 kN Estimate of end-bearing resistance is 0.2 98 = 20 kN Ultimate Design Bearing Resistance = 78/1.45 + 20/1.3 = 69 kN