TR335: Tutorial on deep foundations
1. A bored pile with an enlarged base is to be installed in
a stiff clay whose characteristic undrained shear strength
(Su) increases linearly by 10kPa/m depth to Su of 220kPa
at the pile base. The saturated unit weight of the clay is
21kN/m3 and the unit weight of concrete is 23.5kN/m3.
Diameters of the pile shaft and base are 1.05m and
3.00m respectively. The pile extends from a depth of 4m
to 22m, the top of the under-ream being at a depth of
20m. Past experience indicates that skin friction (shaft
adhesion) coefficient α of 0.70 is appropriate for the
clay. Determine the design/ allowable load of the pile
ensuring an overall load factor of 2.0 and a load factor of
3.0 under the base when shaft resistance is fully
mobilized.
TR335: Tutorial on deep foundations
G.L
A bored pile with an
enlarged base is to be 4.0m 𝛾𝑠𝑎𝑡 = 21𝑘𝑁/𝑚3
installed in a stiff clay
whose characteristic 40 Su (kPa)
undrained shear
strength (Su) increases
linearly by 10kPa/m
depth to Su of 220kPa
at the pile base.
The saturated unit
weight of the clay is
21kN/m3. 16.0m 1.0m
I
Diameters of the pile
shaft and base are 10.0kPa
1.05m and 3.00m
respectively.
Su(z) = 10z
The pile extends from 1.05m
a depth of 4m to 22m,
the top of the under-
ream being at a depth 2.0m II
of 20m. 220
3.0m z (m)
TR335: Tutorial on deep foundations
G.L
Q B = AB 9.0SuB + γsat d
4.0m 𝛾𝑠𝑎𝑡 = 21𝑘𝑁/𝑚3 π32
= 9x220 + 21x22
4
40 Su (kPa) = 17.261MN
20
QIs = αPs 10zdz
z=4
20
10z 2
= 0.7xπx1.05x
2 4
= 4.43MN
16.0m 1.0m
I
10.0kPa
22
QII
s =α Ps (z). Su (z)dz
z=20
Su(z) = 10z 22
1.05m = 2πα (0.4875z
20
2.0m II
220
3.0m z (m)
TR335: Tutorial on deep foundations
G.L
I II
Wpile = Wpile + Wpile
4.0m 𝛾𝑠𝑎𝑡 = 21𝑘𝑁/𝑚3
πx1.052
= x16x23.5 +
40 Su (kPa) 4
1 π32
x x3.08
3 4
16.0m 1.0m
I
The lesser of
10.0kPa QB + QS
Q allowable = −W
F
or
QB
Su(z) = 10z Q allowable = + QS − W
Fb
1.05m
=11.3MN
2.0m II
220
3.0m z (m)
TR335: Tutorial on deep foundations
2. Thirty six piles, 0.60m in diameter, are spaced at
2.40m centres in a 6x6 group. The piles extend between
depths of 3 and 18m in a deposit of stiff clay 28m thick
overlying a bedrock. The characteristic undrained
strength of the clay at a depth of 18m is 145kPa and the
average characteristic value over the pile length is
105kPa. The following have also been determined for
the clay, α = 0.40, Eu = 65MN/m2, mv = 0.08m2/MN and
γsat = 21kN/m3. If the pile group supports a load of
21MN determine;
(1) The overall load factor of safety if the pile group’s
efficiency is 80% and
(2) The total settlement of the group.
TR335: Tutorial on deep foundations
G.L
Thirty six piles, 0.60m in
diameter, are spaced at 2.40m 3.0m
centres in a 6x6 group.
The piles extend between
depths of 3 and 18m in a
deposit of stiff clay 28m thick
overlying a bedrock.
The characteristic undrained
strength of the clay at a depth
of 18m is 145kPa and the 15.0m
average characteristic value
over the pile length is 105kPa.
The following have also been
determined for the clay, α =
0.40, Eu = 65MN/m2, mv =
0.08m2/MN.
10.0m
BEDROCK
TR335: Tutorial on deep foundations
G.L Consider single pile analysis;
3.0m Q B = AB 9.0SuB + γsat d
π(0.6)2
= 9x145 + 18x21
4
= 476kN
𝑠ℎ𝑎𝑓𝑡
𝑄𝑠 = As α𝑆𝑢
= 0.4xπx0.6x15x105
= 1,187.5kN
πx0.62
15.0m Wpile = x15x23.5
4
=100kN
Consider the piles in a group;
QB + QS
Q allowable + Wpile group = xnxE
F
476 + 1,187.5
21,000 + 100x36 = x36x0.8
F
F = 2.0
10.0m
BEDROCK
TR335: Tutorial on deep foundations
G.L
3.0m Assume a minimum raft plan for the
6x6, 0.6m dia 2.4m spacing pile group;
Raft width (=length) will be;
5x2.4 + 0.6 = 12.6m
Qnet on founding ground;
= Qapplied – Qexcavated overburden
15.0m = 21,000 - 3x21x(12.6)2 = 11MN
(Assume concrete replaces equal soil weight)
Assume an equivalent raft 2/3pile length
from pile top and load shedding at 1:4;
Equivalent raft dimensions = 17.6m square
10.0m
BEDROCK
TR335: Tutorial on deep foundations
G.L
3.0m 12.6m
10m
15.0m
17.6m
10.0m
BEDROCK
TR335: Tutorial on deep foundations
G.L
3.0m 12.6m
10m
17.6m
11
𝑞𝑎𝑝𝑝𝑙𝑖𝑒𝑑 = 2
= 0.036𝑀𝑁/𝑚2
17.6
𝑠𝑡𝑜𝑡𝑎𝑙 = 𝑠𝑢𝑛𝑑𝑟𝑎𝑖𝑛𝑒𝑑 + 𝑠𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑖𝑜𝑛
2 3
𝑞𝑎𝑝𝑝 𝐵 1 − 𝜗 𝐼𝑝
𝑠𝑡𝑜𝑡𝑎𝑙 = + 𝑞𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑚𝑣 𝑛4 ∆𝑧𝑖
𝐸𝑢
𝑖=1
0.036𝑥17.6𝑥(1 − 0.52)𝑥0.82
=
65
+ 0.036𝑥0.08𝑥5𝑥 0.796 + 0.47 + 0.325
=6 + 22.9 = 28.9mm
