Best numerical problem group pile capacity (usefulsearch.org) (useful search)

Q8- A square pile group of piles penetrates through a filled up
soil of 3m depth. The pile diameter is 250mm and pile spacing is
0.75 m. The cohesion of the material is 18 kN/m2 and the unit weight
of soil is 15 kN/m2 .Draw plan and sectional elevation of the pile
group
and compute the negative skin friction on the group.
Solution-:
B= 3×0.75 +0.25 =2.5m
m=0.4
•Qun= n(m c p Lf)
= 16(0.4× 18× 3.41× 0.25×3)
= 271.4kN
•Qug= c (4B) Lf + Lf B2
= 18×4×2.5 × 3 +15× 3(2.5)2
=540 + 281.3
= 821.3kN
Therefore, greater of the above two= 821.3kN
Hence negative skin friction = 821.3 kN

Q9- 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
Solution-:
B=2 ×0.9 +0.3 = 2.1m
c=qu/2= 1.5/2 = 0.75kg/cm2= 7.5t/m2
•Piles acting individually
Qun= n.mc .As
= 9×0.9×7.59( 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

Q10- A square group of piles was driven into soft clay extending
to a
large depth . The dia and length of piles are 30cm and 9 m . if the
ucc=9t/m2 and the pile spacing is 100cm c/c. what is the capacity
of the
group? Assume adhesion factor =0.75.
Solution-:
B=2×100 +30 =23ocm=2.3m
Pile acting individually
Qun=( Ap .rp + As. rf) n
Where Ap=π/4(.3)2= 0.07069m2
rp= 9 cu=9×4.5
=40.5t/m2
As= 3.14×(0.3) ×9=8.4823m2
rf= m cu=0.75×4.5
=3.375 t/m2
Qun=9(0.07069×40.5 + 8.4823× 3.375)
= 283.4 t

Q11-A n-pile group has to be proportioned in a uniform
pattern in soft clay with equal spacing in all directions.
Assuming any value of c , determine the optimum value
of spacing of pile in the group.
Take n=25 and m=0.7 .
Neglect the end bearing effect and assume that each pile is
circular in section.
Sol - The optimum spacing of individually piles is based on premises
that the efficiency of the pile group is unity . That is , the total load
carried by group action.
Let the c/c spacing of each pile =s , Diameter of each pile =d
N=25
Width of each pile group= 4s + d ; Length of pile = L
Load carried by group action is
Qug= C{ 4( 4s+ d) L }
=4cL( 4s +d)
……(i)

Load carried by pile acting individually is
Qun =n (mc) (πdL)
=25 *0.7 c (3.14) dL
=55dL .c
(ii)
Equating (i) n(ii) for the optimum spacing
4cL (4s +d) = 55dL .c
16s + 4d = 55d
S=( 51/16) * d
Ans= S=3.19d

Q12-In a 16 pile group , the pile dia is 45cm and c/c spacing
of the square group is
1.5m . if c=50kN/m2 , determine whether the failure would
occur with the pile acting
individually ,or as a group ? Neglect bearing at the tip of the
pile . All piles are 10m
long. Take m=0.7 for shear moblisation around each pile.
Sol- n=16 , d=45cm ; L=10m
Width of group =B= (150*3) +45
=495cm= 4.95m
•For the group
Qug=c* perimeter *length
=c * 4BL
= 50 * 4* 4.95* 10
=9900kN
•For the pile acting individually
Qug =n Qup= n ( mcAp)
Ap=3.14 * 0.45 *10= πdL
Qug = 16* 0.7* 50* 3.14* 0.45* 10
=7917kN
Which is lesser than the load carried by the group action .
hence the foundation will
fail by pile individually.

Q13-200mm dia , 8m long piles are used as foundation for a
column in a uniform deposites , medium clay ( UCC= 100kN/m2)
and adhesion factor =0.9. there are nine piles arranged in a
pattern of 3m×3m . for a group efficiency =1. Find the spacing bt
the pile.
Solution -:
B= 2x + 0.2
C= qu/2 =100/2
=50kN/m2
•Pile acting alone
Qun= n mc .As
=9×0.9×50(3.14×0.2×8)
=2035.8kN
•Pile acting in group
Qug=c As= c (4BL)
=50× 4(2x + .2) 8
3200(x +0.1)
Efficiency = Qug / Qun=1
3200(x +0.1) = 2035.8
Ans= spacing, x= 0.536m

Q-14 A circular well of 4.5m external diameter and
0.75 m steining thickness is embedded upto a depth
of 12m in a uniform sand deposit. The angle of
shearing resistance of sand and the submerged unit
weight are 30⁰ and 1t/mᶟ. The well subjected to a
resultant horizontal force of 50t and a total moment
of 400tm at the scour level. Assuming the well to be
a light well ,compute the allowable total equivalent
resting force due to earth pressure. Safety factor=2.
Sol- Height of the point of application of horizontal
load above scour level
H=M/Q = 400/50=8m
Total height H₁=H+D=8+12=20m
2D₁=3H₁ + 9Hi²-2D(3H₁-D)
eₓ=0.15m; effective width B’=B-2eₓ = 2-0.3 = 1.7m

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