Numerical problem and solution on pile capacity (usefulsearch.org) ( usefulsearch.org) (useful search)

1. Numerical problem and solution
on pile capacity
(usefulsearch.org)

2. PILE FOUNDATION

3. Que 1:- 200 mm diameter, 8 m long piles are used as foundation for a column in a
uniform deposit of medium clay (qu=100 KN/m2). The spacing between the piles is
500 mm. There are 9 piles in the ground arranged in a square pattern. Calculate the
ultimate pile load capacity of the group. Assume adhesion factor=0.9.
Sol:- Unconfind compressive strength,qu= 100 KN/m2
Cu = qu/2 = 50KN/m2
Ultimate capacity of a single pile is given by
Qu = cuNcAp + αcuAs
= 50 × 9 × π/4 × (0.2)2 + 0.9 × 50 × π × 0.2 × 8
= 240.333 KN
Ultimate load capacity of the pile group of 9 pile on the basis of individual action
= 9 × 240.333 = 2163 KN
Width of the pile group = 2 × 0.5 + 0.2 = 1.2 m
Ultimate load capacity of the group by block failure is given by
Qug = cubNcAb + PbLcu
Qug = 50 × 9 × 1.2 × 1.2 + 4 × 1.2 × 8 × 50
= 2568 KN

4. Que 2:-A group of nine piles, 12 m long and 250 mm in diameter, is to be arranged
in a square form in a clay with an average unconfined compressive strength of
60 KN/m2. Work out the centre to centre spacing of the piles for a group efficiency
factor of 1. Neglect bearing at the tip of the piles.
Sol:- Unconfined compressive strength, qu = 60 KN/m2
Undrained cohesion, cu = qu/2 = 30 KN/m2
Therefore, α = 0.9
Ultimate load capacity of the group of piles on the basis of individual action
= n × Qu
= 9 α cu As
= 9 × 0.9 × 30 × π × 0.25 × 12
= 2290 KN
Ultimate load capacity of the piles in group action
Qu = PbLcu
= 4 (2 s + 0.25) × 12 × 30
= (2880 s + 360) KN
For a group efficiency factor (η) of 1, Qg = nQs
i.e., 2880 s + 360 = 2290
s = 0.67 m or 670 mm

6. Que3:- A group of 9 pile , 10m long is used as a foundation for a bridges pier. The
piles used are 30cm diameter with centre to centre spacing of 0.9 m. The subsoil
consist of clay with unconfined compressive strength of 1.5 kg/cm2 . Determine the
efficiency neglecting the bearing action . Adhesion factor α=0.9
Sol:-
B = 2 × 0.9 + 0.3 = 2.1m
C = qu/2 = 1.5/2 = 0.75kg/cm2= 7.5t/m2
(a) Piles acting individually
Qun= n.mc .As
= 9 × 0.9 × 7.5 ×( 3.41 × 0.3 × 10)
= 572.6 t
(b) Piles acting in a group
Qug = c × (4BL)
= 7.5 × 4 × 2.1 × 10
= 630 t
Therefore, Efficiency for pile = 630/ 572.6 = 1.1

7. Que 4:- A 12 m long, 300 mm diameter pile is driven in a uniform deposit of sand (φ’ = 400). The water table is at
great depth and not likely to rise. The average dry unit weight of sand is 18 KN/m3. Using Nq values of Berezantzev,
Calculate the safe load capacity of the pile with a factor of safety of 2.5.
Sol:- For concrete piles, adopt, ɗ = 3/4 φ’ = 300 and
K = 2.0 for dense sand
For Lcr / D = 15, the critical length of the pile = 15 × 0.3 = 4.5 m
Limiting vertical effective stress, Ϭ’ at 4.5 m = 18 × 4.5 = 81 KN/m2
The ultimate pile load capacity Qu is given by
Qu = qpuAb + fsAs
Qpu = Ϭ’Nq and fs(av) = Ϭ’av K tanɗ
Skin friction resistance over length 4.5 m:
Average Ϭ’ = 81 / 2 = 40.5 KN/m2
Fs(av) = Ϭ’av × 2 × tan 300
= 46.8 KN/m2
Skin friction resistance = 46.8 × π × 0.3 × 4.5
= 198 KN
Skin friction resistance over the remaining length 7.5 m
Ϭ’av = 81 KN/m2
Fs(av) = 81 × 2 × tan 300
= 93.5 KN/m2
= 93.5 × π × 0.3 × 7.5
= 661 KN
Qf = 198 + 661 = 859 KN
Φ = 400 and L / D = 12 / 0.3 = 40
Nq = 137
qpu = 81 × 137 = 11097 KN/m2
Qpu = 11097 × (π × 0.32) / 4 = 784 KN
Qu = Qpu + Qf
= 784 + 859 = 1643 KN
Qa = Qu / 2.5 = 1643 / 2.5 = 651 KN

8. QUE 13:- A jetty is sited on 4.5m of very soft clayey soil overlying stiff to very stiff clay. The
average shear strength from 5 to 13m is 140kN/m2.Fissured shear strength at 13 m is 145
kN/m2. For structural reasons , it is desire to use 500mm steel tube piles . Check the average
shear strength from 5 to 13m = 140 kN/m2.
Sol:-
Qup= Rp + Rf
= Ap.rp + As. rf
Ap= area of cross section of pile
= π/4(0.5)2
= 0.1963m2
rp= c Nc = 145 × 9 = 1305
As = surface area of pile = πdL= 3.41 × 0.5 × 8
= 12.55m2
rf = mc =0.5 × 140
=70kN/m2
Qup = 0.1963 × 1305 + 12.5664 × 70
=1135.8kN
Actual load Qa = 400kN
Available FOS =1135.8/400 = 2.84
Desire factor of safety=3.
SAFE