foundations design examples

1. Foundation Engineering 1
Example (1): Estimate the
allowable group capacity for the
(3 × 5) pile group shown with
spacing s = 0.75 m in both
directions. Use F.S. = 3.
Solution
( a ) Single pile action
Qu = Qb + Qs
( i ) Point bearing capacity;
Qb = Ab ( ̅. ́ ) ≤ Ab (11000) kN
̅ = (15 ‒ 9.8) × 4 + (20 ‒ 9.8) × 9 = 112.6 kPa
From Fig. (8.5); = = 30
ϕ = 36ᵒ  ́ = 60
̅. ́ = 112.6 × 60 = 6756 < 11000 kPa
Qb = (0.3)2
× 6756 = 477.6 kN
( ii ) Skin friction capacity;
 For the first soil layer;
Qs1 = As1 . fs1
fs1 = k1 . ̅ . tanδ1 ≤ 100 kPa
From Table (8.3); for concrete pile; δ1 = 0.75 × 28 = 21ᵒ
̅ = (15 ‒ 9.8) × 2.5 = 13 kPa
fs1 = 0.6 × 13 × tan 21 = 2.99 kPa < 100 kPa o.k.
Qs1 = ( π × 0.3 × 3 ) × 2.99 = 8.5 kN
 For the second soil layer;
Qs2 = As2 . fs2
fs2 = k2 . ̅ . tanδ2 ≤ 100 kPa
From Table (8.3); δ2 = 0.75 × 36 = 27ᵒ
̅ = (15 ‒ 9.8) × 4 + (20 ‒ 9.8) × 4.5 = 66.7 kPa
0.75m 0.75m
W.T. G.L.
9
m
3
m
1
m
(Concrete piles with 0.3m dia.)
Loose sand
ϕ = 28 ᵒ
γsat = 15 kN/m3
Ks = 0.6
Sand
)med - dense)
ϕ = 36 ᵒ
γsat = 20 kN/m3
Ks = 0.7

2. Foundation Engineering 2
fs2 = 0.7 × 66.7 × tan 27 = 23.79 kPa < 100 kPa o.k.
Qs2 = ( π × 0.3 × 9 ) × 23.79 = 201.8 kN
⸫ Qu = Qb + Qs1 + Qs2 = 477.6 + 8.5 + 201.8 = 687.9 kN
Gu = 15 × 687.9 = 10318.5 kN
( b ) Block action
B = (3 ‒ 1) × 0.75 + 0.3 = 1.8 m
L = (5 ‒ 1) × 0.75 + 0.3 = 3.3 m
Ab = B × L = 1.8 × 3.3 = 5.94 m2
As = 2 (B + L) × ∆ L = 2 (1.8 + 3.3) × ∆ L = 10.2 ∆ L
( i ) Point bearing capacity;
Gb = Ab . ̅. ́
From Fig. (8.5); = = 5
ϕ = 36ᵒ  ́ = 70
Gb = 5.94 × 112.6 × 70 = 46819.1 kN
( ii ) Skin friction capacity;
 For the first soil layer;
Gs1 = As1 . fs1
For block action; δ1 = 1 = 28ᵒ
fs1 = 0.6 × 13 × tan28 = 4.15 kPa
⸫ Gs1 = (10.2 × 3) × 4.15 = 127.0 kN
 For the second soil layer;
Gs2 = As2 . fs2
For block action; δ2 = 2 = 36ᵒ
fs2 = 0.7 × 66.7 × tan36 = 33.92 kPa
⸫ Gs2 = (10.2 × 9) × 33.92 = 3113.9 kN
⸫ Gu = Gb + Gs1 + Gs2 = 46819.1 + 127 + 3113.9 = 50060.0 kN
Single pile action controls
⸫ Gu = 10318.5 kN
Ga = = = 3439.5 kN

3. Foundation Engineering 3
Example (2 - HW): Estimate the required number
of 350 mm diameter, 9m length bored piles to
withstand 2000 kN column load for the profile
shown in the figure. Suggest a reasonable layout
with 1m spacing and check the block action. Take
FS = 2.6.
Solution
Qu = Qb + Qs
Qb = Ab (c. ́ + ̅. ́ ) ; for  = 0; ̅. ́ = 0 and ́ = 9
Qb = × 125 × 9 = 108.238 kN
Qs = Qs1 + Qs2 + Qs3 = As1 . fs1 + As2 . fs2 + As3 . fs3
fs = α . c + k . ̅ . tanδ ≤ 100 kPa
for  = 0; k . ̅ . tanδ = 0
For bored piles in clay, Skempton suggested;  = 0.45
fs1 = 0.45 × 100 = 45 kPa < 100 kPA o.k.
fs2 = 0.45 × 110 = 49.5 kPa < 100 kPA o.k.
fs3 = 0.45 × 125 = 56.25 kPa < 100 kPA o.k.
As1 = As2 = As3 = π (0.35) × 3 = 3.299 m2
Qs = 3.299 (45 + 49.5 + 56.25) = 497.324 kN
Qu = 108.238 + 497.324 = 605.562 kN
Qa = = 232.908 kN
N = = 8.6 ; say 9 piles
Use square group (3×3) piles with a spacing of (1m);
3m
3m
3m
 = 0
 = 0
 = 0
c = 100 kPa
c = 110 kPa
c = 125 kPa

4. Foundation Engineering 4
Check group action;
a = b = 2 × 1 + 0.35 = 2.35 m
Qu = Qb + Qs
Qb = Ab (c . ́ )
From Fig. (3.7); for = = 3.82 ; Nc = 8.95
Qb = (2.35)2
× 125 × 8.95 = 6178.3 kN
Qs = c1 . As1 + c2 . As2 + c3 . As3
As1 = As2 = As3 = (4 × 2.35) × 3 = 28.2 m2
Qs = 28.2 (100 + 110 + 125) = 9447 kN
Qu = 6178.3 + 9447 = 15625.3 kN
Qa = = 6009.7 kN ˃ 2000 kN OK
b
a

5. Foundation Engineering 5
Example (3 - HW): Compute the
efficiency of the 3 × 6 pile group
shown. Use the following additional
data;
d = 0.4 m
Qu = 600 kN
Gu = 36000 kN
Solution
( i ) Converse-Labarre equation;
( ) [ ]
= 21.8ᵒ
m = 3
n = 6
( ) [ ] = 0.64 or 64%
( ii ) Poulos and Davis equation;
N = 18 ,
Qu = 600 kN
Gu = 36000 kN
= 1.09
Eg = 0.92 or 92%
1
m
1m

6. Foundation Engineering 6
Example ( 4 ): Estimate the consolidation settlement for the pile group shown
Solution
Lp = 9.0 m , ho = 0
h = = = 3 m
4 m
4 m × 3 m
W.T.
G.L.
Dense sand
EL. 106 m
EL. 104 m
EL. 95 m
2300 kN
EL. 102 m
EL. 94 m
EL. 89 m
EL. 92 m
Clay ( I )
γsat = 19.2 kN/m3
Cc = 0.23
eo = 0.8
γ = 17 kN/m3
Clay ( II ) γsat = 18.2 kN/m3
Cc = 0.34 , eo = 1.08
Clay ( III ) γsat = 20 kN/m3
Cc = 0.2 , eo = 0.7
4 m
4 m × 3 m
W.T.
G.L.
2
m
Dense sand
2300 kN
4
m
2
m
2
m
3
m
3
m
 𝜎𝑜 , ∆𝜎
 𝜎𝑜 , ∆𝜎
 𝜎𝑜 , ∆𝜎
 𝜎𝑜 , ∆𝜎
4
m
Clay ( I )
γsat = 19.2 kN/m3
Cc = 0.23
eo = 0.8
γ = 17 kN/m3
Clay ( II ) γsat = 18.2 kN/m3
Cc = 0.34 , eo = 1.08
Clay ( III ) γsat = 20 kN/m3
Cc = 0.2 , eo = 0.7

7. Foundation Engineering 7
Sc = ∑ , (Sc)i = log
Layer Z
Area (m2
)
A = (4 + z) (3 + Z)
∆ =
(kN/m2
) (kN/m2
)
1 1 20 115 4×17 + 5 (19.2  9.8) = 115
2 3 42 54.8 115 + 2 (19.2  9.8) = 133.8
3 5 72 31.9 133.8 + 1 (19.2  9.8) + 1 (18.2  9.8) = 151.6
4 7.5 120.75 19.0 151.6 + 1 (18.2  9.8) + 1.5 (20  9.8) = 175.3
Sc1 = log = 7.7 cm
Sc2 = log = 3.8 cm
Sc3 = log = 2.71 cm
Sc4 = log = 1.6 cm
Sc = ∑ = 15.81 cm

8. Foundation Engineering 8
Example ( 5 ): the pile group
shown is subjected to the following
loads;
V = 6000 kN , mx = 4800 kN.m ,
H = 800 kN , my = 3000 kN.m .
Compute the vertical and horizontal
forces on the corner piles
Solution
Center of gravity of piles;
̅
∑
∑
, ̅
∑
∑
̅ = 5.5 m
̅ = 4.33 m
Locate the coordinate system at
the c.g. of piles.
ex = 5.0 – 5.5 = -0.5 m
ey =5.0 – 4.33 = 0.67 m
∑ ∑
x2
= 3(6.5)2
+ 3(1.5)2
+ 3( 2.5)2
+ 3( 5.5)2
= 243 m2
y2
= 4(3.67)2
+ 4(0.67)2
+ 4( 4.33)2
= 130.67 m2
Mx = mx + V . ey = 4800 + 6000 × 0.67 = 8820 kN.m
My = my + V . ex = 3000 + 6000 × (-0.5) = 0
3
m
1
5
m
5 m
4 m 5 m
y
x
c.g.
2
3 4
my
mx
3 m
3
m
1
5
m
5 m
4 m 5 m
5.5 m
y
x
c.g.
2
3 4
my
mx
3 m
4.33
m

9. Foundation Engineering 9
= 500 + 67.5 y
500 + 67.5 (3.67) = 747.7 kN
500 + 67.5 ( 4.33) = 207.7 kN
= = 66.67 kN