1. Chapter Three- Examples 1
Example (1): For the footing shown in the
figure, find the ultimate bearing capacity,
using Terzaghi's Eq. Assume general shear
failure.
Solution
Terzaghi's Equation; from Table (3.1) ; qult = cNcsc + ̅Nq + 0.5γBNγsγ
To choose shape factors; = >> 10;
Hence, the footing can be considered as continuous (strip); sc = sγ = 1
Bearing capacity factors, using Table (3.2); for ϕ = 25ᵒ
Nc = 25.1 , Nq = 12.7 , Nγ = 9.7
̅ = 1 × 18.2 = 18.2 kN/m2
Hence; qult = 16 × 25.1 × 1 + 18.2 × 12.7 + 0.5 × 18.2 × 1 × 9.7 × 1 = 721.01 kN/m2
Example (2): A (1.2m) square footing is
shown in the figure. Determine the gross
and net allowable load that the footing can
carry for a safety factor of (3), assuming;
a- General shear failure of soil.
b- Local shear failure of soil.
(Note: use Terzaghi's Equation)
γ = 18.2 kN/m3
c = 16 kPa
ϕ = 25 ᵒ
1
m
B =1 m , L = 20 m
γ = 17.3 kN/m3
c = 9.6 kPa
ϕ = 20 ᵒ
0.9
m
1.2 m
P
6. Chapter Three- Examples 6
Example (6) (H.W.): A (0.75 m × 1.5 m )
rectangular footing is shown in the figure.
Determine the magnitude of the ultimate
load (Pult) applied eccentrically for bearing
failure in the supporting soil. Use the
general bearing capacity equation.
Solution
Hansen's Equation; from Table (3.1);
qult = cNcscdcicgcbc + ̅Nqsqdqiqgqbq + 0.5γ ́ Nγsγdγiγgγbγ
since c = 0, hence; qult = ̅Nqsqdqiqgqbq + 0.5γ ́ Nγsγdγiγgγbγ
iq = gq = bq = iγ = gγ = bγ = 1
Bearing capacity factors, using Table (3.4); for ϕ = 30ᵒ ; Nq = 18.4 , Nγ = 15.1
́ = L – 2ey= 1.5 – 2 × 0.12 = 1.26 m
́ = B – 2ex = 0.75 – 2 × 0.06 = 0.63 m
Shape and depth factors; from Table (3.5);
dq = 1 + 2 tan ϕ (1 – sin ϕ)2
= 1 + 2 tan30 (1 – sin 30)2
= 1.231
dγ = 1
sq = 1 +
́
́
tan ϕ = 1 + tan 30 = 1.289
sγ = 1 – 0.4
́
́
= 1 – 0.4 × = 0.8
̅ = 0.6 × 18.08 = 10.85 kPa
qult = 10.85 × 18.4 × 1.289 × 1.231 + 0.5 × 18.08 × 0.63 × 15.1 × 0.8 × 1 = 385.58 kPa
Pult = qult × Af = 385.58 × 0.63 × 1.26 = 306.07 kN
1.5
m
0.06 m
0.12
m
γ = 18.08 kN/m3
c = 0 kPa
ϕ = 30 ᵒ
0.75 m
×
0.6
m
7. Chapter Three- Examples 7
Example (7): for the rectangular
footing shown find the allowable
bearing capacity and the factor of
safety using Hansen's equation.
Use the following data;
P = 3000 kN , Mx = 600 kN.m ,
My = 1500 kN.m , c = 0 ,
tri = 30o
, γ = 16 kN/m3
,
γsat = 18.8 kN/m3
Solution
ex = = = 0
ey = = = 0.2 m
e = = 3.6 m
e = = 6 m
γ γ γ = = 8.99 m
̅ = = 12.5 kN/m3
̅ Df = 16 × 1.5 = 24 kN/m2
N-factors; for = 30o
[from Table 3.4] ; Nq = 18.4 , Nγ = 15.1
Shape factors; From table (3.5)
tan = tan30 = 1.35
1.5
m
2.0
m
W.T.
P
2.0
m
2.0
m
2.5 m 3.5 m
x
y
Mx
My
Mx , My
9. Chapter Three- Examples 9
Example (8) (H.W.): A foundation (1.5 m ×
2.0m) is located at depth (Df = 1.0 m) in a clay
layer (cu1 = 120 kPa , γ = 16.8 kN/m3
). A softer
clay layer (cu2 = 48 kPa , γ = 16.2 kN/m3
) is
located at a depth (d = 0.75 m) measured from
the bottom of the foundation. Determine the
allowable load (Pa) the foundation can carry with
a factor of safety (F = 3).
Solution
̅
= 120 kN/m2
̅ = 16.8 × 1 = 16.8 kN/m2
0.4 , b = 0.75 , 1.0
From (Fig. 3.11), Nc = 3.6
́ = = 0.15
́ = = 0.267
= 628.9 kN/m2
209.6 kN/m2
Pa = qa × B × L = 209.6 × 1.5 × 2 = 628.9 kN
d
10. Chapter Three- Examples 11
Example (9): Check the adequacy of the
footing shown in the figure.
Use F = 3 and γw = 10 kN/m3
).
Solution
To find γt for soil layer I, and γsat for al soil layers I, II, and III;
γt)I = γ = = 17.25 kN/m3
γsat)I = γ = × 10 = 19.4 kN/m3
In the same manner, γsat)II = 18.7 kN/m3
,
and γsat)III = 19.5 kN/m3
Assume that the effective depth equals (B = 1.5 m);
Hence; HIII = 1.5 - 0.5 = 1 m
A - Using Hansen equation with average shear parameters;
From Table (3.1) ; qult = 5.14 su (1 + ́ + ́ - ́ - ́ - ́ ) + ̅
(qult)net = 5.14 su (1 + ́ + ́ - ́ - ́ - ́ )
́ = ́ = ́ = 1 ; su = cav
cav =
∑
∑
= = 73.33 kPa
III
II
I
Soil
Property
2..2
2..2
2..
GS
0.82
0.0
0.8
e
0..1
0..0
0.12
w
80
.0
10
c ( kPa)
0
0
.2
ϕᵒ
1.2 m × 2 m
0.8
m
0.4
m
W.T
.
P = 300 kN working load
0.5
m
Soil I
Soil II
Soil III
12. Chapter Three- Examples 12
Example (10): For the soil-footing geometry shown, find the allowable bearing capacity if
(F = 2) for sand and (F = 3) for clay. Use Hansen's equation
Solution
For the sand layer,
qult = ̅Nqsqdq + 0.5γBNγsγdγ (all other factors = 1.0)
For = 34o
, from Table (3.4) ;
Nq = 29.4
Nγ = 28.7
Shape factors; From table (3.5)
tan = tan34 = 1.67
= = 0.60
depth factors; From table (3.5)
tan (1 sin)2
= 1 + 2 tan34 (1 – sin34)2
× = 1.20
dγ = 1
qult = (1.5 × 17.25) × 29.4 × 1.67 × 1.2 + 0.5 × 17.25 × 2.0 ×28.7 × 0.60 × 1.0
= 1821.54 kN/m2
For the clay layer,
= cu1. Nc (1 + ́ + ́ ) + ̅ (all other factors = 0)
́ = = 0.2
́ = = 0.3 D ˃ B [note: in rad]
= 622.19 kN/m2
Punching contribution (per meter length)
P = ̅
B × L = 2 m × 2 m
1.5
m
W.T
.
P
0.6
m
Sand
C = 0 , = 34o
γ = 17.25 kN/m3
clay
su = qu/2 = 75 kPa
su = undrained shear strength
qu = unconfined shear strength
13. Chapter Three- Examples 13
P = (1.5 × 17.25) × 0.6 + × 17.25 × (0.6)2
= 18.63 kN/m
= 1 sin34 = 0 .44
p = 2 (L + B) = 2 (2 + 2) = 8m
Af = 2 × 2 = 4 m2
= 633.25 kN/m2
< qult = 1821.54 kN/m2
The maximum footing pressure is controlled by the clay layer giving qult = 633.25 kN/m2
qa = = = 211.10 kN/m2
Example (11): The (S.P.T.) results from a soil boring located adjacent to a planned
foundation for a proposed workhouse are shown below. If spread footings for the project
are to be found (1.2 m) below surface ground, what foundation size should be provided to
support (1800 kN) column load. Assume that (25.4 mm) settlement is tolerable. Water
table is encountered at (7.5 m). Use Peck, Hansen, and Thornburn method.
S.P.T. depth
(m)
Blow count.
Nfield
0.3 9
1.2 10
2.4 15
3.6 22
4.8 19
6 29
7.5 33
10 27
γ = 17 kN/m3
1.2
m
B
1800 kN
W.T.
7.5
m
γ́ = 10 kN/m3
14. Chapter Three- Examples 14
Solution
Find pressure ( ́ ᵒ ) at each depth and correct Nfield values;
depth
(m)
Blow count.
Nfield
́ ᵒ
(kPa)
́ ᵒ
(MPa)
CN
(from Fig. 2.12)
́
(Nfield × CN)
0.3 9 ---- ---- ---- -----
1.2 10 1.2 × 17 = 20.4 0.0204 1.55 15
2.4 15 2.4 × 17 = 40.8 0.0408 1.28 19
3.6 22 3.6 × 17 = 61.2 0.0612 1.15 25
4.8 19 4.8 × 17 = 81.6 0.0816 1.05 20
6 29 6 × 17 = 102 0.1020 0.95 27
7.5 33 7.5 × 17 = 127.5 0.1275 0.90 30
10 27 7.5 × 17 + 2.5 × 10 = 152.5 0.1525 0.85 23
Assume that the effective depth equals to ( B )
Try B = 2.4 m ; and take values of ́ in the influence zone to find Nav ;
Nav = 20
= = 0.5
From Figure (3.15) ; (qa)net = 0.211 MPa = 211 kPa
(qa)net = ;
L = = = 3.55 m
So, we can use (2.5 m × 3.6 m)
15. Chapter Three- Examples 15
Example (12): For the rectangular footing
(B × 5m), the average value of SPT
number within the footing influence zone
is Nav. = 14. Find;
(a) The corrected value of N considering
the effect of overburden pressure only.
(b) The footing width (B) and the
allowable bearing capacity using
Meyerhof approach.
Solution
́ = 18.6 × 1.5 = 27.9 kN/m2
CN = 0.77 log
́
= 0.77 log = 1.43
Nc = CN . = 1.43 × 14 = 20
Pw = DL + LL = 3000 + 1500 = 4500 kN
B
( m )
qa
from Fig. (3.16)
(kPa)
=
(kPa)
2 330 450
3 300 300
So, take B = 3m
Sand
γ = 18.6 kN/m3
1.5
m
B
DL = 3000 kN
LL = 1500 kN
G.L.
.Nav = 14
16. Chapter Three- Examples 16
Example (13) (H.W.): For the
square footing shown, if the
average value of the cone
resistance within the rupture zone
is qc = 8.25 MPa, find the footing
dimensions and the allowable B.C.
using Meyerhof approach for the
following two cases;
( a ) when the W.T. (5m) depth
below G.L.
( b ) when the W.T. at G.L.
Solution
Pw = DL + LL = 1000 + 500 = 1500 kN
{
r
( ) r
}
case
B
( m )
qa = ( )
(kPa)
=
(kPa)
a
2.0 218.21 375
2.5 206.98 240
3.0 199.65 166.67
2.75 202.96 198.35
b 4 0.5 (190.7) = 95.34 93.75
So, for case ( a ); B = 2.75 m
and for case ( b ); B = 4.0 m
Medium sand
( qc )av. = 8.25 MPa
5.0
m
B
DL = 1000 kN
LL = 500 kN
G.L.
W.T. (a)
W.T. (b)
17. Chapter Three- Examples 17
Example (14): A (1.2m × 1.2m × 0.6m) footing is placed at (1.8m) depth in a soil having
γ = 7.3 kN/m3
, ϕ = 20ᵒ, and c = 19.2 kPa). Estimate the allowable uplift force
for (F = 2.5).
Solution
To find the type of foundation; shallow or deep;
= = 1.5
From Table (3.7), to find limiting embedment ration for ϕ = 20ᵒ;
= 2.5 > 1.5
So, the footing behaves as shallow footing.
To find the ultimate uplift force for shallow rectangular footing, the equation to be used is;
Tu = c.D B + L + γ.D2
(2sf .B + L – B) Ku.tanϕ + W
From Table (3.7), for ϕ = 20ᵒ; m = 0.05, and maximum Sf = 1.12
Sf = 1 + m. = 1 + 0.05 × 1.5 = 1.075 < 1.12
Ku = Kᵒ = (1 – sinϕ) √
For normally consolidated clay; OCR = 1;
Hence; Ku = Kᵒ = (1 – sin20) √ = 0.66
W = 1.2 × 1.2 × 0.6 × 24 + (1.8 – 0.6) × 1.2 × 1.2 × 17.3 = 50.63 kN
Tu = 2 × 19.2 × 1.8 (1.2 + 1.2) + 17.3 × (1.8)2
(2 × 1.075 × 1.2 + 1.2 – 1.2) 0.66 tan20
+ 50.63 = 251.3 kN
Tall = = = 100.5 kN
