Pile Design and Construction Practice, Fifth edition

Total base resistance =
Total pile resistance 19297  7952 27 249 kN
Factor of safety = which is satisfactory
If the rock socket skin friction were to be only half the calculated value, no load would be
transferred to the pile base. Therefore, the pile head settlement will be caused by compres-
sion in the rock socket only.
From Section 5.5 the modulus ratio of a cemented mudstone is 150, and for a mass factor
of 0.2 the deformation modulus of the rock mass is 0.2 150  4.5 135 MN/m^2. In
Figure 4.36 the modulus ratio Ec/Edis 20 103 /135 148 and for L/B 7/1.5 4.7 the
influence factor Iis 0.25. The ratio D/B for a recessed socket is 4/1.5 2.7. There the reduction
factor from Figure 4.37 is about 0.8. Hence from equation 4.49:

Pile head settlement =

Checking the calculated shaft friction from equation 4.45 and take bas 0.25,

fs 0.25 0.53 MN/m^2 which agrees closely with the value from equation 4.44.

If the socket is grooved to an average depth of 25 mm over shortened socket length of
5.0 m with the grooves at vertical intervals of 0.75 m then:
In equation 4.47, and total length of travel 4.7  6.67
31.35 m
From equation 4.47, RF 0.025  31.33/0.75  5.0 0.21
From equation 4.46, fs 0.8(0.21)0.45 4.5 1.78 MN/m^2
Total shaft friction on 5 m socket length 1.78   1.5  5 42 MN
Factor of safety in shaft friction 42/9 4.7, therefore grooving the socket would
theoretically provide a much shorter socket length than the 7 m required for an ungrooved shaft.

Example 4.8

A tubular steel pile with an outside diameter of 1067 mm is driven with a closed end to near
refusal in a moderately strong sandstone (average quc = 20 MN/m^2 ) overlain by 15 m of soft
clay. Core drilling in the rock showed a fracture frequency of 5 joints per metre. Calculate
the maximum working load which can be applied to the pile and the settlement at this load.
Only a small penetration below rockhead will be possible with sandstone of this quality,
and the rock will be shattered by the impact. Hence, frictional support both in the soft clay and
the rock will be negligible compared with the base resistance.
Pile driving impact is likely to open joints in the rock hence the base resistance should not
exceed the unconfined compression strength of the intact rock.

Total ultimate pile resistance = base resistance=

For a safety factor of 2.5 allowed load = 17.9

2.5

7.2 MN

4

1.067^2  20 17.9 MN

r 0.775 – 0.75 0.025 m

4.5

0.8 9  103 0.25

1.5 135

9 mm

27 249

9000

3.0

4

1.5^2  45007952 kN

236 Resistance of piles to compressive loads