A
load take down has revealed that the characteristic load at the column positions
is 3500 kN. A small site investigation has been undertaken and the
following reported:
* medium dense sand from ground level to 30m depth

• a peak angle of shearing resistance of ?�p= 34o (c� = 0)
• groundwater is at 2.0m depth
• average N below base of 29 blows

Consider a 1.2m deep square pad footing:

V / A = q’ Nq Sq + 0.5 ?’ B’ N? S?- sq = 1.56, S?= 0.7 for a square or
circle (EC7: Appendix D)

• q = 20 x 1.2 = 24 kPa,
• ?’ = 10 kPa (as water is so close to base)
• Nq = 30 (EC7: Appendix D or charts)
• N? = 38 (EC7: Appendix D or charts)Try a 2.5m x 2.5m pad
• (V / A) ult = (24 x 30 x 1.56) + (0.5 x 10 x 2.5 x 38 x 0.7 ) = 1456 kPa
• (V / A)max = 1456 / 2.5 = 580 kPa (Apply overall FoS of 2.5)
• (V / A)chara = 3500 / (2.5 x 2.5) = 560 kPa therefore OK.

Use a 2.5m square pad

Check settlement:

? = q B (1 – ?2) Is / E- q = 500 kPa

• B = 2.5 m
• ? = 0.25
• Is= 1.2 (centre of a square foundation)
• Ev = 2000 N = 2000 x 29 = 58000 kPa
• ? = (560 x 2.5 x (1 – 0.252) x 1.2) / 58000 = 27.2 mmConsider a 0.9m diameter
pile 25m long (contract pile tests will be done):
- ?’v base = 25 x 10 = 250 kPa
(assume water to ground level)
• ?’v ave = (0 + 250) / 2 = 125 kPa
• Ks = 0.7 (assume Ko = 1.0)
• ?s = ?’ = 34o
• Nq = 50 (Berezantsev curves)
• qs = ?’v ave Ks tan ?s = 125 x 0.7 x tan 34 = 59.0 kPa
• Qs = qs x As = 59.0 x 0.9 x ? x 25 = 4170 kN
• qb = Nq ?’v base = 50 x 250 = 12500 kPa
• Qb = 12500 x 0.92 x ? x 0.25 = 7952 kN
• (Qs+ Qb ) / 2.5 = (4170 + 7952) / 3.0 = 4040 kN
• Qs / 1.2 = 4170 / 1.2 = 3475 kN*Now: max pile load = 0.25 fcu A or fcu =(3500
/ 0.64) x 4 =21875 kPa ( e.g use C30 concrete)*Check settlement using
equivalent raft:
- Load applied at 2/3 pile depth = 25 x 2/3 = 16.6m
• Load spread a 1:4, so equivalent diameter = (2 x (16.6 / 4)) + 0.9) = 9.2
• equivalent stress = 3500 / (9.22 x ? x 0.25) = 52.7 kPa
• ? = 0.25
• Is= 1.0 (centre of a circular foundation)
• Ev = 2000 N = 2000 x 29 = 58000 kPa
• ? = (52.7 x 9.2 x (1 – 0.252) x 1.0) / 58000 = 7.8 mm