SOLVED PROBLEMS IN FOUNDATION ENGINEERING
Problem # (22)
Consider a precast concrete pile 12 – m long in a homogenous sandy soil layer. The pile
cross section is 305 mm × 305 mm, the dry unit weight of sand is, γd = 16.8 kN / m3, and the soil
friction angle Ф = 35o.
Calculate the ultimate point load that the pile can carry.
305 m m
X
X
12 m
305 m m
S e c tio n X - X
Solution:
For Ф = 35o , Nq* = 120. Thus;
q’ = γd L = 16.8 × 12 = 201.6 kPa
Qp = Ap q’ Nq* = (0.305)2 × 201.6 × 120 = 2250.46 kN
OR
q1 = 50 Nq* tan Φ = 50 × 120 × tan 35 = 4201.25
Qp = Ap q1 = (0.305)2 × 4201.25 = 390.82 kN
Thus the point ultimate load of the pile that is Qp is:
Qp = 390.82 kN
57
PROF. MOHAMMED AWAD
SOLVED PROBLEMS IN FOUNDATION ENGINEERING
Problem # (23):
Consider a pre-cast concrete pile in Problem # (21):
1. Estimate the ultimate skin frictional resistance Qs
2. Calculate the total ultimate load of the pile, Q
Assume that for the given dry unit weight the relative density of the sand Dr = 100 %
7 6 .8 6
O
Q s (1 )
L ' = 1 5 D = 4 .6 m
Q s (2 )
12 m
Z
Solution:
The unit skin friction is estimated as: f = K σ v’ tan δ
Where: σ v’ = overburden pressure = γ z (See Figure Above)
For depth z from 0 to 15 D : σ v’ = 16.8 z (kPa)
For depth z ≥ 15 D : σ v’ = 16.8 ( 15 * 0.305 ) = 76.86 kPa (Constant)
K = lateral earth pressure = 0.5 + 0.008 Dr = 0.5 + ( 0.008 * 100 ) = 1.3
Q s  1   PL ' f av   4  0 . 305  4 . 6 
 1 . 3  76 . 86  tan  0 . 6  35  
 107 . 64 kN


2


Qs (2) = P (L – L’) f = [ 4 × 0.305 ] [ 12 – 4.6 ] [1.3×76.86× tan(0.6× 35)] = 346.27 kN
Qs = Qs (1) + Qs (2) = 107.64 + 346.27 = 453.91 kN
Total load = Q pile = (problem # 21) + Q s = 390.82 + 453.91 = 844.73 kN
58
PROF. MOHAMMED AWAD
SOLVED PROBLEMS IN FOUNDATION ENGINEERING
Problem # (24):
Estimate the total load that the pile group shown below can carry. Note that; every single
pile is identical to that in problems # (22) and # (23).
P ile C a p
12 m
S o il s u r f a c e
4 p ile s
5 p ile s
1 .5 m
d = 1 .5 m
Solution:
Total load that the single pile can carry Q = 844.73 kN (Problem # (21) and # (22))
m=5
n=4
so, m × n = 20 piles
QT = 844.73 × 20 = 16894.6 kN
The efficiency of the group:
 
2 m  n  2   d  4 D
pmn
 
2  5  4  2   1 . 5    4  0 . 305
 4  0 . 305   5  4

= 0.91 (or 91 %)
Therefore; the total load that can be carried by the pile group is;
QTG = 16894.6 × 0.91 = 15374.09 kN.
59
PROF. MOHAMMED AWAD