Pile Design and Construction Practice, Fifth edition

Unit negative skin friction at top of layer
� 0.30�� vo � 0.30 � 9.81 � 2.1 � 4 � 4.7 kN/m 2
Unit negative skin friction at groundwater level (see Figure 5.45)
� 0.30 � 9.81[(2.1 � 4) � (1.6 � 3)] � 8.8 kN/m 2
Unit negative skin friction at bottom of layer
Therefore total negative skin friction in soft clay:
Pile groups under compressive loading 301
� 0.30 � 9.81[(2.1 � 4) � (1.6 � 3) � (0.6 � 2.5)] � 43.3 kN/m 2
� � � 1.2 � [ 1
2(4.7 � 8.8) � 3 � 1
2(8.8 � 43.3) � 2.5] � 746 kN
Because the pile will settle due to yielding of the stiff clay when the full working load is
applied, the pile will move downwards relative to the lower part of the soft clay. Thus negative
skin friction will be developed only over about 80% of the length within the soft clay. Thus
approximate total negative skin friction in soft clay � 0.8�746 � 597 kN
The negative skin friction in the sand can be calculated using the coefficients for K s in
Table 4.10. Although the compacted sand fill is dense it will be loosened by pile boring to
give a coefficient K s of 1 and a value of � of 30�. From equation 4.16:
negative skin friction on pile in sand fill
� 1
2 � 1 � 9.81 � 2.1 � 4 � tan 30��� � 1.2 � 4 � 359 kN
Total negative skin friction on pile � 359 � 597 � 956 kN
Total applied load on piles in centre rows � 956 � 2200 � 3156 kN
The required pile penetration depth is calculated on the basis of the building loading, with
a check being made to ensure that the safety factor on the combined building load and
negative skin friction is adequate.
Required ultimate pile resistance for overall safety factor of 2 (Section 4.6) �
2 �2200 � 4400 kN
Take a trial penetration depth of 10 m into the stiff clay stratum. At the pile base level
c ub � 190 kN/m 2 and the average value of c u on the shaft is 140 kN/m 2 . Thus
Ultimate base resistance
� 1
4 � � � 1.2 2 � 9 � 190 � 1935 kN
Load to be carried in skin friction � 4 400 � 1 935 � 2465 kN
The adhesion factor for a straight-sided pile can be taken as 0.45. Therefore from
equation 4.7:
total load � 2 465 � 0.45 � 140 � � � 1.2 � l