(10.668 mm). Compute the bearing capacity of the pile by applying Redtenbacker's for-
mula and using a factor of safety of 3.
Calculation Procedure:
- Find the weight of the pile and the area of
the transformed section
The work performed in driving a pile into the soil is a function of the reaction of the soil
on the pile and the properties of the pile. Therefore, the soil reaction may be evaluated if
the work performed by the hammer is known. Let A = cross-sectional area of pile; E =
modulus of elasticity; h = height of fall of ram; L = length of pile; P = allowable load on
pile; R — reaction of soil on pile; s = penetration per blow; W= weight of falling ram; w =
weight of pile.
Redtenbacker developed the following equation by taking these quantities into consid-
eration: the work performed by the soil in bringing the pile to rest; the work performed in
compressing the pile; and the energy delivered to the pile:
R
2
L W
2
H
RS+
2AE=W^
(35)Finding the weight of the pile and the area of the transformed section, we get w =
16(16) (0.150)(45)/144 = 12 kips (53.4 kN). The area of a no. 7 bar = 0.60 m^2 (3.871
cm^2 ); n = 9; A = 16(16) + 8(9 - 1)0.60 = 294 m^2 (1896.9 cm^2 ).
- Apply Eq. 191 to find R; evaluate P
Thus, s = 0.42 in (10.668 mm); L = 540 in (13,716 mm); Ec = 3160 kips/m^2 (21,788.2
MPa.); W= 4.6 kips (20.46 kN); Wh = 17 ft-kips = 204 in-kips (23,052 J). Substituting
gives QA2R + 540#V[2(294)(3160)] = 4.6(204)7(4.6 + 12); R = 84.8 kips (377.19 kN);
P = RfI = 28.3 kips (125.88 kN).
CAPACITY OF A GROUP OF FRICTION PILES
A structure is to be supported by 12 friction piles of 10-in (254-mm) diameter. These will
be arranged in four rows of three piles each at a spacing of 3 ft (91.44 cm) in both direc-
tions. A test pile is found to have an allowable load of 32 kips (142.3 kN). Determine the
load that may be carried by this pile group.
Calculation Procedure:
- State a suitable equation for the load
When friction piles are compactly spaced, the area of soil that is needed to support an in-
dividual pile overlaps that needed to support the adjacent ones. Consequently, the capaci-
ty of the group is less than the capacity obtained by aggregating the capacities of the indi-
vidual piles. Let P = capacity of group; Pi = capacity of single pile; m = number of rows;
n = number of piles per row; d = pile diameter; s = center-to-center spacing of piles; B =
tan"
1
dls deg. A suitable equation using these variables is the Converse-Labarre equation
T.
=mn
~ (w)[m(n
~1} + n(m
~1)] (36)