Pile Cap Design using Strut-and-tie Method – Worked Example

This article presents a worked example on the design of a pile cap to EN 1992-1-1-2004 (Eurocode 2). The pile cap to be designed is to transmit column loads to piles. A geotechnical test has been carried out on the site located at Greenwich, South-East London. Due to the occurrence of London clay, which is highly unstable, on the upper 6m layer on the site, it is recommended that a deep foundation in the form of piles to a depth of 9m is adopted. The recommended pile size is 600mm with safe working load of 400KN. The characteristics strength of concrete and reinforcement for the pile cap are 25MPa and 500MPa respectively. The dimension of the column, and the load acting on it are given below:

Column Size = 300 x 300mm

Dead Load = 1205KN

Live Load = 291KN

Estimate the number of piles required

Calculate the number of piles to support the column load.

Total service load on the column = 1205 + 291 = 1496KN

No of piles = Total service load/capacity of a pile = 1496/400 = 3.74

Adopt 4 number of piles

Determine the size and Dimension of the Pile cap

The distance between each pile shall be kØ

K =3, Ø = 600mm

kØ = 3 x 600 = 1800

Edge distance = 150mm

Since the piles are four then they will be arranged in a square array so that they can be symmetrically distributed under the column.

The length of each side of pile cap = Pile dist + 2 x Ø/2 + 2 x edge distance

= 1800 + 2 x 300 + 2 x 150 = 2700mm

Take the height of the pile cap as 2.3Ø = 2.3 x 600 = 1380

Adopt 1350mm as the height of the pile

The layout of the pile cap is shown below

 

Determine the Main Reinforcement using Strut-and-Tie Method

Since the length/depth ≤ 2, strut-and-tie method shall be adopted

The tensile force in the pile cap = NL/4d (Click overview of pile cap design for more explanation of concept)

N = Ultimate axial load = 1.35 x 1205 + 1.5 x 291 = 2063KN

L = 1800/2 = 900

Assume 12mm bar as main bar, cover = 50mm

d = 1350 – 50mm – 12/2 = 1294mm

Tensile force in the pile cap (T) = $
\frac{\text{2063}\times \,\,10^{\text{3}}\times \,\,900}{\text{4}\times \,\,1294}
$ = 358713N

Area of steel = T/fyd

Fyd = 500/1.15 = 434.8MPa

Area of steel = 358713/434.8 = 825mm

Since there are two roles of piles in the pile cap, the Area of reinforcement = 2 x 825 = 1650mm²

Check minimum area of reinforcement

Asmin = 0.26 x fctm/fyk x b x d  ≥ 0.0013bd

Fctm = 0.3 x fck^(2/3) = 0.3 x 25^(2/3) = 2.57MPa

Asmin = 0.26 x 2.57/500 x 2700 x 1294

Asmin = 4669mm²

Since the minimum area of reinforcement (4669mm²) is greater than area of steel required (1650mm²), then provide minimum area of steel.

Provide 15Y20 @ 180c/c (4713mm²) at the bottom along the two directions.

Top Face Reinforcement

Provide minimum reinforcement on the top face in both directions

Since minimum reinforcement is already calculated for the bottom face, hence:

Provide 15Y20 @ 180c/c (4713mm²) at the top face along the two directions.

 

Check the Pile Cap for Vertical Shear

Vertical shear shall be provided at Ф/5 within the pile diameter (Click overview of pile cap design for more explanation of concept)

Shear at critical section = Reaction of two piles within the critical section

Shear at critical section = 2063/2 = 1031.5

Calculate a_v to allow for shear enhancement

a_v = 1800/2 – 600/2 – 300/2 + 600/5 = 570

since av ≤ 0.5d (647), a_v = 647

Effective shear =  V_ED  x a_v/2d = $
\frac{\text{1031.5}\times \,\,647}{\text{2}\times \,\,1294}
$  =257.875KN

 

Calculate the shear resistance without shear reinforcement

v_{Rdc  x bwd    = {0.12K(100ρLfck)^1/3} bwd ≥ vmin bwd}

K =    (1 +√200/d)    ≤   2.0

K =    (1 +√200/1294) = 1.4

ρL = Asl/bwd ≤   0.02

ρ_l = (4713) / (2700 × 1294) = 0.0013

v_{Rdc  = {0.12 x 1.4 (100 x 0.0013 x 25)^1/3} ≥ vmin} 

v_Rdc = 0.5 N/mm²

Calculate the minimum resistance without reinforcement.

_{vmin = {0.035K³/2fck^1/2}}

_{vmin = 0.035  x 1.4^3/2  x 25^1/2}

_vmin = 0.2 N/mm²

Since v_Rdc (0.5N/mm²) > v_min (0.2 N/mm²); hence v_Rdc passes minimum reinforcement requirement

Convert the shear capacity to resistance and compare against design shear force.

V_Rdc = v_Rdc  x bd

V_Rdc = 0.5 x 2700 x 1294 x 10^ (-3) = 1969.8KN

Since V_ED < V_Rdc the section is adequate for one way shear

 

Check the Pile Cap for Punching Shear

The punching shall be checked at the column perimeter and at perimeter 2d from the column face

Punching shear at column perimeter

Shear force acting on column = 2063KN

Calculate the maximum resistance of the compressive struct

v_RDmax = 0.3 (1 -f_ck/250)f_cd

f_cd = fck/1.5 = 25/1.5 = 16.67 N/mm2

v_RDmax = 0.3 (1 -25_/250)16.67

v_RDmax =4.5 N/mm²

Shear resistance (V_RDmax) = v_RDmax ud

where u is the perimeter of the column

Shear resistance (v_RDmax) = 4.5 x 4(300) x 1294 x 10 ^ (-3)

V_{RDmax =} 6987KN

V_RDmax > V_Ed, the section passes punching check

Check Punching shear at basic control perimeter

Since the piles at spaced at not more than 3 x pile diameter, then punching shear check at basic control perimeter is not necessary.

Circumferential Reinforcement

The circumferential reinforcement should be provided in accordance with surface reinforcement for deep beams by providing 0.2% of reinforcement according to the UK national annex. This will however result to large area of reinforcement. We shall instead provide the circumferential reinforcement following the guidance of IStructE EC2 design manual.

Hence, we shall Y12@150mm c/c

 

Click here to study a Worked Example on the design of this same pile cap using bending theory

Detailing