This document discusses shallow foundations and their bearing capacity. It defines shallow foundations as those that transfer loads to the soil at the base of the structure. The document then outlines Terzaghi's equations for calculating the ultimate bearing capacity of soils, including factors for cohesion, internal friction angle, soil unit weight, and foundation geometry. It also discusses factors of safety used to determine allowable bearing capacities and considerations for groundwater effects. Examples are provided to demonstrate calculating ultimate bearing capacities.

1. School of Environmental Engineering
UNIVERSITI MALAYSIA PERLIS
EAT 314/4 Geotechnical Engineering
Soil Bearing Capacity of Shallow
Foundations
AIN NIHLA KAMARUDZAMAN
ainnihla@unimap.edu.my
Ext: 8968

2. By: ANK
EAT314/4 Geotechnical Engineering
Course Outcome
NO. COURSE OUTCOME (CO) EAT 314/4
1 Ability to analyze soil bearing capacity and design for shallows
foundations based on types of soil.
2 Ability to describe type of deep foundation and its installation.
3 Ability to describe and design various concrete retaining walls
based on lateral earth pressure.
4 Ability to conduct slope stability analysis and landslide
investigations.
5 Ability to discuss common sampling methods for subsoil
exploration and report.

3. By: ANK
EAT314/4 Geotechnical Engineering
Outline
Introduction
Ultimate Bearing Capacity for Shallow
Foundation
Terzaghi’s Ultimate Bearing Capacity
Equations
Factor of Safety
Effect of Ground Water Table (GWT)
General Bearing Capacity Equation.
Allowable Bearing Pressure in Sand
Settlement Consideration.

4. By: ANK
EAT314/4 Geotechnical Engineering
Introduction
Foundations are the building component
which transfers building loads to the soil.
There are two basic types of foundations:
SHALLOW - Shallow foundations transfer
the load to soil at the base of the
substructure.
DEEP - Deep foundations transfer loads far
below the substructure.

5. By: ANK
EAT314/4 Geotechnical Engineering
SHALLOW vs. DEEP
Building
Building
Rock
Types of Foundation

6. By: ANK
EAT314/4 Geotechnical Engineering
Shallow Foundation
Shallow Foundation System
i) Spread Foundation (footing)
ii) Mat or Raft Foundation
Characteristics of shallow foundations are;
Cost (affordable)
Construction Procedure (simple)
Material (mostly concrete)
Labour (doesn’t need expertise)

7. By: ANK
EAT314/4 Geotechnical Engineering
Spread Foundation (footing)
Also known as a footer or footing.
It’s an enlargement at the bottom of a
column or bearing wall that spreads the
applied structural loads over a sufficiently
large soil area.
Each column & each bearing wall has its
own spread footing, so each structure may
include dozens of individual footings.

8. By: ANK
EAT314/4 Geotechnical Engineering
Spread Foundation (footing) …Cont.
Spread Footing

9. By: ANK
EAT314/4 Geotechnical Engineering
Spread Foundation (footing) …Cont.

10. By: ANK
EAT314/4 Geotechnical Engineering
Mat or Raft Foundation
ü A foundation system in which essentially the
entire building is placed on a large
continuous footing.
ü It is a flat concrete slab, heavily reinforced
with steel, which carries the downward
loads of the individual columns or walls.
ü Raft foundations are used to spread the
load from a structure over a large area,
normally the entire area of the structure.

11. By: ANK
EAT314/4 Geotechnical Engineering
Mat or Raft Foundation (Cont.)

12. By: ANK
EAT314/4 Geotechnical Engineering
Mat or Raft Foundation (Cont.)

13. By: ANK
EAT314/4 Geotechnical Engineering
Design Criteria:
To perform satisfactory, shallow foundation
must have two main criteria:
They have to be safe against overall
shear failure in the soil that supports them.
(Safety factor usually between 2.5 to 3.0)
They cannot undergo excessive
displacement or settlement.
(Settlement of individual footing on sand
50 mm or 75 mm for footing on clay)

14. By: ANK
EAT314/4 Geotechnical Engineering
Ultimate Bearing Capacity for Shallow
Foundation
Definition:
Bearing capacity is ability of a soil to bear the
loads transmitted by a footing.
Ultimate bearing capacity is reach when the
impose foundation pressure is in equilibrium
with resisting soil pressure.
When the pressure exceed the ultimate soil
bearing capacity value, the foundation
pronounced fail in shear.

15. By: ANK
EAT314/4 Geotechnical Engineering
Ultimate Bearing Capacity for Shallow
Foundation
Ultimate Bearing Capacity (qult) is the
maximum pressure which can be carried
by the soil immediately below foundation.
The theory is developed based on three
modes of failure;
a) General shear failures – for soils (dense or hard
state)
b) Local shear failures – for soils (medium density
or firm state)
c) Punching shear failure – for soils (loose or soft
state)

16. By: ANK
EAT314/4 Geotechnical Engineering
Failure modes of Shallow Foundation
General Shear Failure
Local Shear Failure
Punching Shear Failure
bulge
bulge

17. By: ANK
EAT314/4 Geotechnical Engineering
Terzaghi’s Ultimate Bearing Capacity
Equations
Terzaghi (1943) – formulated for strip
foundation – modified from Prandlt (early 1920)
According to Terzaghi:
Shallow foundation – ratio between the depth of
embedment (Df) and the width of foundation (B) is
less than 1.
The weight of the soil above the base of foundation
is;
f
D
q g
=
1
<
B
Df

18. By: ANK
EAT314/4 Geotechnical Engineering
Terzaghi’s Ultimate Bearing Capacity
Equations....Cont.
Figure: Derivation of Terzaghi’s Bearing Capacity Equation
B
I
II II
III III

19. By: ANK
EAT314/4 Geotechnical Engineering
Terzaghi’s Ultimate Bearing Capacity
Equations....Cont.
The failure mechanisms of the soil due to
foundation load is defined in three failure
zone:
Zone 1: The triangular active zone ADC
immediately under the foundation.
Zone 2: The Radial shear zones of ADF and CDE.
Zone 3: Two Triangular Rankine Passive zones AFH
and CEG
Note: The weight of the foundation and the soil in
zone 1 will pushed zone 2 to the sides AND zone 3 to
the surface of the soil resulting in the bulge of the
soil surface.

20. By: ANK
EAT314/4 Geotechnical Engineering
Terzaghi’s Ultimate Bearing Capacity
Equations....Cont.
Terzaghi’s assumption:
The angles CAD and ACD is equal to the soil
friction angle, Φ. (that is, α = Φ)
By replacing the weight of the soil above the
foundation base by an equivalent surcharge, q,
the shear resistance of the soil along the failure
surface GI and HJ was neglected.
Remember:
f
D
q g
=

21. By: ANK
EAT314/4 Geotechnical Engineering
Terzaghi’s Ultimate Bearing Capacity
Equations....Cont.
The ultimate bearing capacity equation, qult (kPa)
Q (vertical load causing a general
shear failure of the soil)
qult
Df
Ground surface
B
L
B
Q
Area
Q
q ult
ult
ult
´
=
=
G.W.T

22. By: ANK
EAT314/4 Geotechnical Engineering
Terzaghi’s Ultimate Bearing Capacity
Equations....Cont.
For a uniform vertical loading of a strip footing, Terzaghi
(1943) assumed a general shear failure in order to
develop the ultimate bearing capacity equation:
Where;
c = cohesion of the soil underlying the footing (kPa or kN/m2)
, = unit weight of the soil (kN/m3)
= Distance from the ground surface to the bottom of the
footing (m)
B = width of the footing (m)
L = Length of the footing (m)
γ
q
C
ult γBN
2
1
qN
cN
q +
+
= …..Eq. 1
f
D
g
=
q
f
D
g

23. By: ANK
EAT314/4 Geotechnical Engineering
Terzaghi’s Bearing
Capacity Factors

24. By: ANK
EAT314/4 Geotechnical Engineering
Terzaghi’s Ultimate Bearing Capacity
Equations....Cont.
If the shape factors were considered, the
equation was modified to;
For square foundation (B x B in size):
….Eq. 2
For circular foundation (Diameter = B):
….Eq. 3
γ
q
C
ult γBN
4
.
0
qN
cN
3
.
1
q +
+
=
γ
q
C
ult γBN
3
.
0
qN
cN
3
.
1
q +
+
=

25. By: ANK
EAT314/4 Geotechnical Engineering
Terzaghi’s Ultimate Bearing Capacity
Equations....Cont.
For foundation that exhibit local shear failure
mode in soils, Terzaghi suggested modification
to Eq. 1 by replacing;
The cohesion c à c’
Where;
The angle of internal friction Φ à Φ’
Where;
c
3
2
c'=
f
f tan
3
2
'
tan =

26. By: ANK
EAT314/4 Geotechnical Engineering
Example #1:
A square footing (2.25 m x 2.25 m) is
placed at depth of 1.5 m in a sand with
the shear strength parameters c’ = 0
and Φ’ = 38˚. Determine the ultimate
bearing capacity of the foundation. The
unit weight of the sand is 18 kN/m3.
Given: Df = 1.5 m
B = 2.25 m
3
kN/m
18
=
g

27. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #1
For a square footing on sand; using Eq. 2.
But, the cohesion of sand, c = 0, then
For Φ = 38˚, by using Table 1 (Terzaghi Bearing
Capacity Factors), we get
Nq = 61.5 and Nγ = 82.3
γ
q
C
ult γBN
4
.
0
qN
cN
3
.
1
q +
+
=
γ
q
γ
q
ult γBN
4
.
0
N
γBN
4
.
0
qN
q +
=
+
= f
D
g

28. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #1
Then, the ultimate bearing capacity is
2
3
3
ult
γ
q
f
ult
γ
q
ult
kN/m
2993.76
m)(82.3)
)(2.25
kN/m
0.4(18
m)(61.5)
1.5
kN/m
(18
q
γBN
4
.
0
N
D
q
γBN
4
.
0
qN
q
=
+
´
=
+
=
+
=
g

29. By: ANK
EAT314/4 Geotechnical Engineering
Factor of Safety, Fs
Factor of safety, Fs of about 3 or more is
applied to the ultimate soil bearing capacity to
arrive at the value of the allowable bearing
capacity.
There are two basic definition of the allowable
bearing capacity of shallow foundation:
Gross allowable bearing capacity, qall
Net allowable bearing capacity, qult(net)

30. By: ANK
EAT314/4 Geotechnical Engineering
Gross allowable bearing capacity, qall
The gross allowable bearing capacity can be
calculated as
qall is the allowable load (Qall) per unit area to
which the soil under the foundation should be
subjected to avoid any chance of bearing
capacity failure.
Fs
q
q ult
all =
Area
Q
q all
all =

31. By: ANK
EAT314/4 Geotechnical Engineering
Net allowable bearing capacity, qu(net)
The net allowable bearing capacity, qult(net) is
the allowable load per unit area of the
foundation in excess of the existing vertical
effective stress at the level of the foundation.
The vertical effective stress at the foundation
level is equal to .
So, the net ultimate load is
Hence,
f
D
q g
=
q
q
q ult
ult(net) -
=
Fs
γD
q
Fs
q
q
Fs
q
q f
ult
ult
ult(net)
all(net)
-
=
-
=
=

32. By: ANK
EAT314/4 Geotechnical Engineering
Net allowable bearing capacity, qu(net)
In the case of shallow footing – there is no
significant difference in the factor of safety
obtained in terms of net or gross pressure.
The consideration of net pressure is very
important for the case of design of mat or
raft foundation.

33. By: ANK
EAT314/4 Geotechnical Engineering
Example #2:
A strip of wall footing 1 m wide is supported by
a stiff clay layer with undrained shear strength
of 140 kPa. Unit weight of soil is 20 kN/m3.
Depth of footing is 0.6 m. Ground water was
not encountered during subsurface
exploration. Determine the allowable wall
load for a factor of safety 3.
Given: B = 1 m, Df = 0.6 m
= 20 kN/m3
g

34. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #2
For strip footing, using Eq. 1
Shear strength parameters: for undrained
condition (fully saturated), Φ = 0˚ and c = 140
kN/m2
By using Table 1 (Terzaghi Bearing Capacity
Factors), for Φ = 0˚, we get
Nc = 5.7, Nq = 1.0 and Nγ = 0.0
γ
q
C
ult γBN
2
1
qN
cN
q +
+
=

35. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #2
Thus, the ultimate bearing capacity is
Note: 1 kPa = 1 kN/m2
kN/m
810
1.0)
m
0.6
kN/m
(20
5.7)
kN/m
140
(
0
N
cN
γBN
2
1
qN
cN
q
2
3
2
q
C
γ
q
C
ult
=
´
´
+
´
=
+
+
=
+
+
=
f
D
g

36. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #2
With safety factor, FS = 3
The gross allowable bearing capacity is
The allowable wall load,
wall
of
length
m
per
kN
270
m
1
kN/m
270
B
q
Q
2
all
all
=
´
=
´
=
2
2
ult
all kN/m
270
3
kN/m
810
Fs
q
q =
=
=

37. By: ANK
EAT314/4 Geotechnical Engineering
Example #3:
A square footing 1.5 x 1.5 m in plan is placed
at depth of 1 m in a soil with friction angle,
Φ = 20˚ and c = 15.2 kPa. The unit weight of
the soil is 17.8 kN/m3. Determine the allowable
gross load for a factor of safety 3. Assume
general shear failure occurs in the soil.
Given: size: B x L = 1.5 m x 1.5 m,
Df = 1 m
= 17.8 kN/m3
g

38. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #3
For square footing, using Eq. 2
Shear strength parameters: Φ = 20˚ and
c = 15.2 kN/m2
By using Table 1 (Terzaghi Bearing Capacity
Factors), for Φ = 20˚, we get
Nc = 17.7, Nq = 7.4 and Nγ = 4.4
γ
q
C
ult γBN
4
.
0
qN
cN
3
.
1
q +
+
=

39. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #3
Thus, the ultimate bearing capacity
2
3
3
2
γ
q
C
γ
q
C
ult
kN/m
46
.
528
4.4)
m
1.5
kN/m
17.8
(0.4
7.4)
m
1
kN/m
8
.
17
(
)
7
.
17
kN/m
15.2
1.3
(
γBN
4
.
0
N
cN
3
.
1
γBN
4
.
0
qN
cN
3
.
1
q
=
´
´
´
+
´
´
+
´
´
=
+
+
=
+
+
=
f
D
g

40. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #3
With safety factor, FS = 3
The gross allowable load is
( )
kN
396.35
m
1.5
m
1.5
3
kN/m
528.46
footing
the
of
Area
footing
the
of
Area
Q
2
all
=
´
÷
÷
ø
ö
ç
ç
è
æ
=
´
=
´
=
Fs
q
q
ult
all

41. By: ANK
EAT314/4 Geotechnical Engineering
Example #4:
From Example #3, calculate total gross load if
local shear failure occurs in the soil.
Solution:
Find shear strength parameters;
kPa
10.13
kPa)
(15.2
3
2
c
3
2
c' =
=
=
!
!
!
13.6
20
tan
3
2
tan
20
tan
3
2
tan
3
2
'
tan 1
'
=
÷
ø
ö
ç
è
æ
=
®
=
= -
f
f
f

42. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #4
For square footing, using Eq. 2
By using Table 1 (Terzaghi Bearing Capacity
Factors), for Φ’ = 14˚, we get
Nc = 12.1, Nq = 4.0 and Nγ = 1.9
γ
q
C
ult γBN
4
.
0
qN
N
c'
3
.
1
q +
+
=

43. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #4
Thus, the ultimate bearing capacity
Note: 1 kPa = 1 kN/m2
2
3
3
2
γ
q
C
γ
q
C
ult
kN/m
84
.
250
)
9
.
1
m
1.5
kN/m
17.8
(0.4
)
0
.
4
m
1
kN/m
8
.
17
(
)
1
.
12
kN/m
13
.
0
1
1.3
(
γBN
4
.
0
N
N
c'
3
.
1
γBN
4
.
0
qN
N
c'
3
.
1
q
=
´
´
´
+
´
´
+
´
´
=
+
+
=
+
+
=
f
D
g

44. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #4
With safety factor, FS = 3
The gross allowable load is
( )
kN
188.13
m
1.5
m
1.5
3
kN/m
250.84
footing
the
of
Area
footing
the
of
Area
Q
2
all
=
´
÷
÷
ø
ö
ç
ç
è
æ
=
´
=
´
=
Fs
q
q
ult
all

45. By: ANK
EAT314/4 Geotechnical Engineering
Effect of Ground Water Table (GWT)
The presence of ground water table will
influence the bearing capacity of footing.
Groundwater reduces the density of soil due
to buoyancy.
When groundwater is present, the density of
the soil needs to be modified.
Depending on the position of water table,
the second and third terms in the bearing
capacity equation (Eq. 1 à Eq. 3) may
require modification.

46. By: ANK
EAT314/4 Geotechnical Engineering
Effect of Ground Water Table (GWT)
The effect of groundwater table can be categorized
into four conditions;
Case 1:
GWT at the ground surface (Fully submerged)
Case 2:
GWT above foundation base
Case 3:
GWT at the foundation base
Case 4:
GWT at a depth D below the foundation base.

47. By: ANK
EAT314/4 Geotechnical Engineering
Case 1: GWT at the ground surface
If the groundwater table is located
at the soil surface, then,
The magnitude of q in the second
term of the bearing capacity
equation should be calculated as;
Where,
The unit weight of soil, in the
second and third term of the
bearing capacity equations should
be replaced by .
w
sat g
g
g -
=
' = effective unit weight of soil
D
q '
g
=
g
'
g
Note:
w
g
gsat = saturated unit weight of soil
= unit weight of water
3
kN/m
9.81
=
Df = D
GL GWT
sat
g

48. By: ANK
EAT314/4 Geotechnical Engineering
Case 2: GWT above foundation base
If groundwater table is located at
a distance D above the bottom
of the foundation,
The magnitude of q in the second
term of the bearing capacity
equation should be calculated
as;
Where,
w
sat g
g
g -
=
' = effective unit weight of soil
( ) D
D
D
q f '
g
g +
-
=
g
'
g
§ The unit weight of soil, in the third term of the bearing capacity
equations should be replaced by .
GWT
g
sat
g
GL

49. By: ANK
EAT314/4 Geotechnical Engineering
Case 3: GWT at the foundation base
If the groundwater table is at
the bottom of the foundation,
The magnitude of q in the
second term of the bearing
capacity equation is equal
to;
However, the unit weight of
soil, in the third term of the
bearing capacity equations
should be replaced by .
f
D
q g
=
g
'
g
GL
GWT
g
sat
g

50. By: ANK
EAT314/4 Geotechnical Engineering
Case 4: GWT at a depth D below the
foundation base.
When the groundwater table is at a
depth D below the bottom of the
foundation,
The magnitude of q in the second
term of the bearing capacity
equation is equal to;
The magnitude of in the third term
of the bearing capacity equations
should be replaced by .
f
D
q g
=
( )
[ ] ( )
( )
B
D
For
B
D
For
'
1
>
=
£
-
+
=
g
g
g
g
g
av
av D
B
D
B
g
av
g
GL
GWT
g
sat
g

51. By: ANK
EAT314/4 Geotechnical Engineering
Example #5:
A square footing (2.25 m x 2.25 m) is placed at
depth of 1.5 m in a sand with the shear
strength parameters c’ = 0 and Φ’ = 38˚.
Determine the ultimate bearing capacity of
the foundation if water table exists at the
ground surface. The unit weight of the sand is
18 kN/m3 and the saturated unit weight of the
sand is 20 kN/m3.
Given: Df = 1.5 m
B = 2.25 m

52. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #5
GL GWT
Df = D = 1.5 m
B = 2.25 m
3
kN/m
20
=
sat
g

53. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #5
For a square footing on sand; using Eq. 2.
But, the cohesion of sand, c = 0, and the groundwater
table exist at the ground surface, then Eq. 2 was
modified to,
Where, assumed , then
γ
q
C
ult γBN
4
.
0
qN
cN
3
.
1
q +
+
=
γ
q
ult BN
γ'
4
.
0
N
'
q +
= D
g
3
3
kN/m
19
.
10
kN/m
)
81
.
9
20
(
'
=
-
=
-
= w
sat g
g
g
3
kN/m
81
.
9
=
w
g

54. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #5
For Φ = 38˚, by using Table 1 (Terzaghi Bearing
Capacity Factors), we get
Nq = 61.5 and Nγ = 82.3
Then, the ultimate bearing capacity is
2
3
3
ult
γ
q
ult
kN/m
1694.8
m)(82.3)
)(2.25
kN/m
0.4(10.19
m)(61.5)
1.5
kN/m
(10.19
q
BN
γ'
4
.
0
DN
'
q
=
+
´
=
+
= g

55. By: ANK
EAT314/4 Geotechnical Engineering
General Bearing Capacity Equation
After the development of Terzaghi’s bearing
capacity equation, several investigators
worked in this area and refined the solution.
For examples; Meyerhoff (1951, 1963), Hansen
(1961) and etc.
Meyerhoff argued that bearing capacity of
foundation is not only affected by the shape
of foundation but by others factors as well.

56. By: ANK
EAT314/4 Geotechnical Engineering
General Bearing Capacity Equation
The soil-bearing capacity equation for a strip
footing given in (Eq. 1) can be modified for general
use by incorporating the following factors:
Shape factor: to determine the bearing
capacity of rectangular and circular footings.
Depth factor: to account for the shearing
resistance developed along the failure surface in
soil above the bottom of the footing.
Inclination factor: To determine the bearing
capacity of a footing on which the direction of
load application is inclined at a certain angle to
the vertical.

57. By: ANK
EAT314/4 Geotechnical Engineering
General Bearing Capacity Equation
Meyerhoff (1951,1963) was modified the general
bearing capacity formula to account all the factors as
follows;
Where;
s = the shape factor,
d = the depth factor,
i = the load inclination factor, and
B and L = the dimension of footing
g
g
g
g
g i
d
s
BN
i
d
s
qN
i
d
s
cN
BL
Q
q
q
q
q
c
c
c
C
2
1
qult +
+
=
=
Nc, Nq and Nγ = Bearing capacity factors Table 2 (Meyerhoff and
Brinch & Hansen)

58. By: ANK
EAT314/4 Geotechnical Engineering
Shape factors (De Beer, 1970)
Shape factors for rectangular footing:
(B = width of footing, L = length of footing)
÷
ø
ö
ç
è
æ
-
=
÷
ø
ö
ç
è
æ
+
=
÷
÷
ø
ö
ç
ç
è
æ
÷
ø
ö
ç
è
æ
+
=
L
B
4
.
0
1
tan
L
B
1
s
N
N
L
B
1
s
q
c
q
c
g
f
s

59. By: ANK
EAT314/4 Geotechnical Engineering
Shape factors (De Beer, 1970)
Shape factors for square and circular
footing:
÷
÷
ø
ö
ç
ç
è
æ
+
=
c
q
c
N
N
1
s
6
.
0
=
g
s
f
tan
1
sq +
=

60. By: ANK
EAT314/4 Geotechnical Engineering
Example #6
A foundation is designed on a soil with
and .
The shear strength parameters of the soil are,
c = 80 kPa and Φ = 15˚. The depth of
embedment is 1.2 m and the size of
foundation is 1.5 x 2 m. Determine the ultimate
bearing capacity of the foundation and the
allowable load if factor of safety is 3.
3
kN/m
16.8
=
b
g 3
kN/m
17.6
=
sat
g

61. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #6
Given:
Df = 1.2 m, B = 1.5 m, L = 2 m
Use general bearing capacity equation;
From Table 2 (Meyerhoff Bearing Capacity
Factors); for Φ = 15˚, we get
Nc = 10.98, Nq = 3.94 and Nγ = 1.13
g
g
g
g s
BN
s
N
D
s
cN q
q
f
c
C
2
1
qult +
+
=

62. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #6
Calculate the shape factors;
7
.
0
m
2
m
1.5
4
.
0
1
L
B
4
.
0
1
2
.
1
15
tan
m
2
m
1.5
1
tan
L
B
1
s
27
.
1
10.98
3.94
m
2
m
1.5
1
N
N
L
B
1
s
q
c
q
c
=
÷
ø
ö
ç
è
æ
-
=
÷
ø
ö
ç
è
æ
-
=
=
÷
ø
ö
ç
è
æ
+
=
÷
ø
ö
ç
è
æ
+
=
=
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
+
=
÷
÷
ø
ö
ç
ç
è
æ
÷
ø
ö
ç
è
æ
+
=
g
f
s
!

63. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #6
Use, (There are no groundwater effect)
Then, the ultimate bearing capacity is
2
3
3
2
ult
kN/m
85
.
1220
7
.
0
13
.
1
m
5
.
1
kN/m
8
.
16
2
1
)
2
.
1
3.94
m
1.2
kN/m
8
.
16
(
)
27
.
1
98
.
10
kN/m
(80
2
1
q
=
÷
ø
ö
ç
è
æ
´
´
´
´
+
´
´
´
+
´
´
=
+
+
= g
g
g
g s
BN
s
N
D
s
cN q
q
f
c
C
3
kN/m
16.8
=
b
g

64. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #6
With safety factor, Fs = 3
The allowable load is
( )
kN
1220.85
m
2
m
1.5
3
kN/m
1220.85
footing
the
of
Area
footing
the
of
Area
Q
2
all
=
´
÷
÷
ø
ö
ç
ç
è
æ
=
´
=
´
=
Fs
q
q
ult
all

65. By: ANK
EAT314/4 Geotechnical Engineering
Depth factors (Hansen, 1970)
The depth of embedment influences the shear
strength at failure plane.
This factor can be neglected if the soil above
foundation base is not stable or not compacted.
( )
1
d
sin
1
tan
B
D
2
1
d
B
D
0.4
1
d
γ
2
f
q
f
c
=
-
÷
ø
ö
ç
è
æ
+
=
÷
ø
ö
ç
è
æ
+
=
f
f
:
1
For £
B
Df

66. By: ANK
EAT314/4 Geotechnical Engineering
Depth factors (Hansen, 1970)
( )
1
d
B
D
tan
sin
1
tan
2
1
d
B
D
tan
0.4
1
d
γ
f
1
2
q
f
1
-
c
=
÷
ø
ö
ç
è
æ
-
+
=
÷
ø
ö
ç
è
æ
+
=
-
f
f
:
1
For >
B
Df

67. By: ANK
EAT314/4 Geotechnical Engineering
Example #7:
Do Example #6 by taking into account the depth
factor.
Solution:
Use general bearing capacity equation;
From Table 2 (Meyerhoff Bearing Capacity Factors);
for Φ = 15˚, we get
Nc = 10.98, Nq = 3.94 and Nγ = 1.13
g
g
g
g
g d
s
BN
d
s
N
D
d
s
cN q
q
q
f
c
c
C
2
1
qult +
+
=

68. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #7
Shape factor;
Depth factor;
7
.
0
s
,
2
.
1
s
,
27
.
1
s γ
q
c =
=
=
1
8
.
0
m
5
.
1
B
m;
1.2
Df <
=
®
=
=
B
Df
( )
( )
( ) ( )
1
d
24
.
1
15
sin
1
15
tan
8
.
0
2
1
sin
1
tan
B
D
2
1
d
32
.
1
8
.
0
0.4
1
B
D
0.4
1
d
γ
2
2
f
q
f
c
=
=
-
+
=
-
÷
ø
ö
ç
è
æ
+
=
=
+
=
÷
ø
ö
ç
è
æ
+
=
!
!
f
f

69. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #7
Then, the ultimate bearing capacity is
2
3
3
2
ult
kN/m
1600.71
1
7
.
0
13
.
1
m
5
.
1
kN/m
8
.
16
2
1
)
24
.
1
2
.
1
3.94
m
1.2
kN/m
8
.
16
(
)
32
.
1
27
.
1
98
.
10
kN/m
(80
2
1
q
=
÷
ø
ö
ç
è
æ
´
´
´
´
´
+
´
´
´
´
+
´
´
´
=
+
+
= g
g
g
g
g d
s
BN
d
s
N
D
d
s
cN q
q
q
f
c
c
C

70. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #7
With safety factor, Fs = 3
The allowable load is
( )
kN
1600.71
m
2
m
1.5
3
kN/m
1600.71
footing
the
of
Area
footing
the
of
Area
Q
2
all
=
´
÷
÷
ø
ö
ç
ç
è
æ
=
´
=
´
=
Fs
q
q
ult
all

71. By: ANK
EAT314/4 Geotechnical Engineering
Inclination Factor
Footing may be subjected to inclined load.
The effect of load inclination is proposed by
Meyerhoff (1963) and Hanna and Meyerhoff (1981).
Where, is the angle of loading with vertical axis
2
γ
2
q
c
1
i
90
1
i
i
÷
÷
ø
ö
ç
ç
è
æ
-
=
÷
ø
ö
ç
è
æ
-
=
=
f
a
a
Q (Load)
a
f
D
B
a

72. By: ANK
EAT314/4 Geotechnical Engineering
Example #8:
A foundation of size 2 x 2 m carrying a column
load that form an angle of 10˚ to the vertical.
The depth of the foundation is 2 m. The
internal friction angle is 34˚ and the unit weight
of the soil is 20.8 kN/m3. Find the allowable
column load for a factor of safety 4.
Given: Df = 2 m
B = L = 2 m
3
kN/m
20.8
=
g

73. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #8
For c = 0 then,
The general bearing capacity equation;
From Table 2 (Meyerhoff Bearing Capacity Factors);
for Φ = 34˚, we get
Nq = 29.44 and Nγ = 31.15
g
g
g
g
g
g i
d
s
BN
i
d
s
N
D
i
d
s
cN q
q
q
q
f
c
c
c
C
2
1
qult +
+
=
0

74. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #8
For square footing; shape factor
Inclination factors:
6
.
0
=
g
s
67
.
1
34
tan
1
tan
1
sq =
+
=
+
= !
f
Q (Load)
!
10
=
a
m
2
=
f
D
2 x 2 m
50
.
0
34
10
1
1
i
79
.
0
90
10
1
90
1
i
2
2
γ
2
2
q
=
÷
ø
ö
ç
è
æ
-
=
÷
÷
ø
ö
ç
ç
è
æ
-
=
=
÷
ø
ö
ç
è
æ
-
=
÷
ø
ö
ç
è
æ
-
=
f
a
a

75. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #8
For depth factors:
1
1
m
2
B
m;
2
Df £
=
®
=
=
B
Df
( )
( ) ( )
1
d
26
.
1
34
sin
1
34
tan
1
2
1
sin
1
tan
B
D
2
1
d
γ
2
2
f
q
=
=
-
+
=
-
÷
ø
ö
ç
è
æ
+
=
!
!
f
f

76. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #8
Then, the ultimate bearing capacity is
2
3
3
ult
kN/m
2230.22
50
.
0
1
6
.
0
15
.
31
m
2
kN/m
8
.
20
2
1
)
79
.
0
26
.
1
67
.
1
29.44
m
2
kN/m
8
.
20
(
2
1
q
=
÷
ø
ö
ç
è
æ
´
´
´
´
´
´
+
´
´
´
´
´
=
+
= g
g
g
g
g
g i
d
s
BN
i
d
s
N
D q
q
q
q
f

77. By: ANK
EAT314/4 Geotechnical Engineering
Solution: Example #8
With safety factor, Fs = 4
The allowable load is
( )
kN
2230.22
m
2
m
2
4
kN/m
2230.22
footing
the
of
Area
footing
the
of
Area
Q
2
all
=
´
÷
÷
ø
ö
ç
ç
è
æ
=
´
=
´
=
Fs
q
q
ult
all

78. THANK YOU