1. science
2. engineering

3
3.1
Shallow Foundations: Ultimate
Bearing Capacity
Introduction
To perform satisfactorily, shallow foundations must have two main characteristics:
1. They have to be safe against overall shear failure in the soil that supports them.
2. They cannot undergo excessive displacement, or settlement. (The term excessive is
relative, because the degree of settlement allowed for a structure depends on several
considerations.)
The load per unit area of the foundation at which shear failure in soil occurs is called the
ultimate bearing capacity, which is the subject of this chapter.
3.2
General Concept
Consider a strip foundation with a width of B resting on the surface of a dense sand or
stiff cohesive soil, as shown in Figure 3.1a. Now, if a load is gradually applied to the
foundation, settlement will increase. The variation of the load per unit area on the foundation (q) with the foundation settlement is also shown in Figure 3.1a. At a certain
point—when the load per unit area equals qu—a sudden failure in the soil supporting the
foundation will take place, and the failure surface in the soil will extend to the ground
surface. This load per unit area, qu, is usually referred to as the ultimate bearing capacity of the foundation. When such sudden failure in soil takes place, it is called general
shear failure.
If the foundation under consideration rests on sand or clayey soil of medium compaction (Figure 3.1b), an increase in the load on the foundation will also be accompanied by an increase in settlement. However, in this case the failure surface in the soil will
gradually extend outward from the foundation, as shown by the solid lines in Figure
3.1b. When the load per unit area on the foundation equals qu(1), movement of the foundation will be accompanied by sudden jerks. A considerable movement of the foundation is then required for the failure surface in soil to extend to the ground surface (as
shown by the broken lines in the figure). The load per unit area at which this happens
is the ultimate bearing capacity, qu. Beyond that point, an increase in load will be
133
134 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Load/unit area, q
B
qu
Failure
surface
in soil
(a)
Settlement
Load/unit area, q
B
qu(1)
qu
Failure
surface
(b)
Settlement
Load/unit area, q
B
qu(1)
qu
Failure
surface
(c)
qu
Surface
footing
Settlement
Figure 3.1 Nature of bearing capacity failure in soil: (a) general shear failure: (b) local shear failure;
(c) punching shear failure (Redrawn after Vesic, 1973) (Vesic, A. S. (1973). “Analysis of Ultimate
Loads of Shallow Foundations,” Journal of Soil Mechanics and Foundations Division, American
Society of Civil Engineers, Vol. 99, No. SM1, pp. 45–73. With permission from ASCE.)
accompanied by a large increase in foundation settlement. The load per unit area of
the foundation, qu(1), is referred to as the first failure load (Vesic, 1963). Note that a
peak value of q is not realized in this type of failure, which is called the local shear
failure in soil.
If the foundation is supported by a fairly loose soil, the load–settlement plot will
be like the one in Figure 3.1c. In this case, the failure surface in soil will not extend
to the ground surface. Beyond the ultimate failure load, qu, the load–settlement plot
will be steep and practically linear. This type of failure in soil is called the punching
shear failure.
Vesic (1963) conducted several laboratory load-bearing tests on circular and rectangular plates supported by a sand at various relative densities of compaction, Dr. The variations of qu(1)> 12gB and qu> 12gB obtained from those tests, where B is the diameter of a
circular plate or width of a rectangular plate and g is a dry unit weight of sand, are shown
in Figure 3.2. It is important to note from this figure that, for Dr > about 70%, the general
shear type of failure in soil occurs.
On the basis of experimental results, Vesic (1973) proposed a relationship for
the mode of bearing capacity failure of foundations resting on sands. Figure 3.3
3.2 General Concept
0.2
0.3
0.4
Punching
shear
700
600
500
Relative density, Dr
0.5
0.6
0.7
0.8
135
0.9
General
shear
Local shear
400
300
qu
qu(1)
100
90
80
70
60
50
qu
1 B
2
and
1 B
2
200
1 B
2
40
Legend
Circular plate 203 mm (8 in.)
Circular plate 152 mm (6 in.)
Circular plate 102 mm (4 in.)
Circular plate 51 mm (2 in.)
Rectangular plate 51 305 mm
(2 12 in.)
30
qu(1)
1 B
2
20
Reduced by 0.6
Small signs indicate first failure load
10
1.32
1.35
1.40
1.45
1.50
Dry unit weight, d
Unit weight of water, w
1.55
1.60
Figure 3.2 Variation of qu(1)>0.5gB and qu>0.5gB for circular and rectangular plates on the
surface of a sand (Adapted from Vesic, 1963) (From Vesic, A. B. Bearing Capacity of Deep
Foundations in Sand. In Highway Research Record 39, Highway Research Board, National
Research Council, Washington, D.C., 1963, Figure 28, p. 137. Reproduced with permission of the
Transportation Research Board.)
shows this relationship, which involves the notation
Dr 5 relative density of sand
Df 5 depth of foundation measured from the ground surface
2BL
B* 5
B1L
where
B 5 width of foundation
L 5 length of foundation
(Note: L is always greater than B.)
(3.1)
136 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
0.2
0
Relative density, Dr
0.4
0.6
0.8
1.0
0
1
Punching shear
failure
Local shear
failure
General
shear
failure
Df /B*
2
3
4
Df
B
5
Figure 3.3 Modes of foundation failure in sand (After Vesic, 1973) (Vesic, A. S. (1973).
“Analysis of Ultimate Loads of Shallow Foundations,” Journal of Soil Mechanics and
Foundations Division, American Society of Civil Engineers, Vol. 99, No. SM1, pp. 45–73.
With permission from ASCE.)
For square foundations, B 5 L; for circular foundations, B 5 L 5 diameter, so
B* 5 B
(3.2)
Figure 3.4 shows the settlement S of the circular and rectangular plates on the surface of a
sand at ultimate load, as described in Figure 3.2. The figure indicates a general range of
S>B with the relative density of compaction of sand. So, in general, we can say that, for
foundations at a shallow depth (i.e., small Df> B*), the ultimate load may occur at a
foundation settlement of 4 to 10% of B. This condition arises together with general shear
failure in soil; however, in the case of local or punching shear failure, the ultimate load
may occur at settlements of 15 to 25% of the width of the foundation (B).
3.3
Terzaghi’s Bearing Capacity Theory
Terzaghi (1943) was the first to present a comprehensive theory for the evaluation of the
ultimate bearing capacity of rough shallow foundations. According to this theory, a foundation is shallow if its depth, Df (Figure 3.5), is less than or equal to its width. Later investigators, however, have suggested that foundations with Df equal to 3 to 4 times their
width may be defined as shallow foundations.
Terzaghi suggested that for a continuous, or strip, foundation (i.e., one whose widthto-length ratio approaches zero), the failure surface in soil at ultimate load may be assumed
to be similar to that shown in Figure 3.5. (Note that this is the case of general shear failure,
as defined in Figure 3.1a.) The effect of soil above the bottom of the foundation may also be
assumed to be replaced by an equivalent surcharge, q 5 gDf (where g is a unit weight of
soil). The failure zone under the foundation can be separated into three parts (see Figure 3.5):
3.3 Terzaghi’s Bearing Capacity Theory
0.3
0.2
25%
0.4
Relative density, Dr
0.5
0.6
Punching
shear
0.7
Local shear
0.8
General
shear
20%
S
B
Rectangular
plates
Circular plates
15%
10%
5%
Circular plate diameter
203 mm (8 in.)
152 mm (6 in.)
102 mm (4 in.)
51 mm (2 in.)
51 305 mm (2 12 in.)
Rectangular plate (width B)
0%
1.35
1.40
1.45
1.50
Dry unit weight, d
Unit weight of water, w
1.55
Figure 3.4 Range of settlement of circular and rectangular plates at ultimate
load (Df>B 5 0) in sand (Modified from Vesic, 1963) (From Vesic, A. B.
Bearing Capacity of Deep Foundations in Sand. In Highway Research Record
39, Highway Research Board, National Research Council, Washington, D.C.,
1963, Figure 29, p. 138. Reproduced with permission of the Transportation
Research Board.)
B
J
I
qu
Df
H
45 /2
A
45 /2
F
C
D
q Df
G
45 /2 45 /2
E Soil
Unit weight Cohesion
c
Friction angle Figure 3.5 Bearing capacity failure in soil under a rough rigid
continuous (strip) foundation
137
138 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
1. The triangular zone ACD immediately under the foundation
2. The radial shear zones ADF and CDE, with the curves DE and DF being arcs of
a logarithmic spiral
3. Two triangular Rankine passive zones AFH and CEG
The angles CAD and ACD are assumed to be equal to the soil friction angle fr.
Note that, with the replacement of the soil above the bottom of the foundation by an
equivalent surcharge q, the shear resistance of the soil along the failure surfaces GI and
HJ was neglected.
Using equilibrium analysis, Terzaghi expressed the ultimate bearing capacity in
the form
qu 5 crNc 1 qNq 1 12 gBNg
(continuous or strip foundation)
(3.3)
where
cr 5 cohesion of soil
g 5 unit weight of soil
q 5 gDf
Nc, Nq, Ng 5 bearing capacity factors that are nondimensional and are functions only of
the soil friction angle fr
The bearing capacity factors Nc, Nq, and Ng are defined by
Nc 5 cot fr
e2 (3p>42fr>2)tan fr
C
fr
p
2 cos ¢ 1 ≤
4
2
2
Nq 5
2 1 5 cot fr(Nq 2 1)
S
e2 (3p>42fr>2)tan fr
fr
2 cos ¢ 45 1 ≤
2
(3.4)
(3.5)
2
and
1 Kpg
Ng 5 ¢
2 1≤ tan fr
2 cos2 fr
(3.6)
where Kpg 5 passive pressure coefficient.
The variations of the bearing capacity factors defined by Eqs. (3.4), (3.5), and (3.6)
are given in Table 3.1.
To estimate the ultimate bearing capacity of square and circular foundations,
Eq. (3.1) may be respectively modified to
qu 5 1.3crNc 1 qNq 1 0.4gBNg
(square foundation)
(3.7)
3.3 Terzaghi’s Bearing Capacity Theory
139
Table 3.1 Terzaghi’s Bearing Capacity Factors—Eqs. (3.4), (3.5), and (3.6) a From
Kumbhojkar (1993)
a
f9
Nc
Nq
N ga
f9
Nc
Nq
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
5.70
6.00
6.30
6.62
6.97
7.34
7.73
8.15
8.60
9.09
9.61
10.16
10.76
11.41
12.11
12.86
13.68
14.60
15.12
16.56
17.69
18.92
20.27
21.75
23.36
25.13
1.00
1.10
1.22
1.35
1.49
1.64
1.81
2.00
2.21
2.44
2.69
2.98
3.29
3.63
4.02
4.45
4.92
5.45
6.04
6.70
7.44
8.26
9.19
10.23
11.40
12.72
0.00
0.01
0.04
0.06
0.10
0.14
0.20
0.27
0.35
0.44
0.56
0.69
0.85
1.04
1.26
1.52
1.82
2.18
2.59
3.07
3.64
4.31
5.09
6.00
7.08
8.34
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
27.09
29.24
31.61
34.24
37.16
40.41
44.04
48.09
52.64
57.75
63.53
70.01
77.50
85.97
95.66
106.81
119.67
134.58
151.95
172.28
196.22
224.55
258.28
298.71
347.50
14.21
15.90
17.81
19.98
22.46
25.28
28.52
32.23
36.50
41.44
47.16
53.80
61.55
70.61
81.27
93.85
108.75
126.50
147.74
173.28
204.19
241.80
287.85
344.63
415.14
N ga
9.84
11.60
13.70
16.18
19.13
22.65
26.87
31.94
38.04
45.41
54.36
65.27
78.61
95.03
115.31
140.51
171.99
211.56
261.60
325.34
407.11
512.84
650.67
831.99
1072.80
From Kumbhojkar (1993)
and
qu 5 1.3crNc 1 qNq 1 0.3gBNg
(circular foundation)
(3.8)
In Eq. (3.7), B equals the dimension of each side of the foundation; in Eq. (3.8), B equals
the diameter of the foundation.
For foundations that exhibit the local shear failure mode in soils, Terzaghi suggested
the following modifications to Eqs. (3.3), (3.7), and (3.8):
qu 5 23crN cr 1 qN qr 1 12gBN gr
(strip foundation)
qu 5 0.867crN cr 1 qN qr 1 0.4gBN gr
(square foundation)
(3.10)
qu 5 0.867crN cr 1 qN qr 1 0.3gBN gr
(circular foundation)
(3.11)
(3.9)
140 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Table 3.2 Terzaghi’s Modified Bearing Capacity Factors Ncr, Nqr, and Ngr
f9
N 9c
N 9q
N 9g
f9
N 9c
N 9q
N 9g
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
5.70
5.90
6.10
6.30
6.51
6.74
6.97
7.22
7.47
7.74
8.02
8.32
8.63
8.96
9.31
9.67
10.06
10.47
10.90
11.36
11.85
12.37
12.92
13.51
14.14
14.80
1.00
1.07
1.14
1.22
1.30
1.39
1.49
1.59
1.70
1.82
1.94
2.08
2.22
2.38
2.55
2.73
2.92
3.13
3.36
3.61
3.88
4.17
4.48
4.82
5.20
5.60
0.00
0.005
0.02
0.04
0.055
0.074
0.10
0.128
0.16
0.20
0.24
0.30
0.35
0.42
0.48
0.57
0.67
0.76
0.88
1.03
1.12
1.35
1.55
1.74
1.97
2.25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
15.53
16.30
17.13
18.03
18.99
20.03
21.16
22.39
23.72
25.18
26.77
28.51
30.43
32.53
34.87
37.45
40.33
43.54
47.13
51.17
55.73
60.91
66.80
73.55
81.31
6.05
6.54
7.07
7.66
8.31
9.03
9.82
10.69
11.67
12.75
13.97
15.32
16.85
18.56
20.50
22.70
25.21
28.06
31.34
35.11
39.48
44.45
50.46
57.41
65.60
2.59
2.88
3.29
3.76
4.39
4.83
5.51
6.32
7.22
8.35
9.41
10.90
12.75
14.71
17.22
19.75
22.50
26.25
30.40
36.00
41.70
49.30
59.25
71.45
85.75
Ncr, Nqr, and Ngr, the modified bearing capacity factors, can be calculated by using
the bearing capacity factor equations (for Nc, Nq, and Ng, respectively) by replacing fr
by fr 5 tan21 ( 23 tan fr). The variation of Ncr, Nqr, and Ngr with the soil friction angle fr
is given in Table 3.2.
Terzaghi’s bearing capacity equations have now been modified to take into account
the effects of the foundation shape (B>L), depth of embedment (Df ), and the load inclination. This is given in Section 3.6. Many design engineers, however, still use Terzaghi’s
equation, which provides fairly good results considering the uncertainty of the soil conditions at various sites.
3.4
Factor of Safety
Calculating the gross allowable load-bearing capacity of shallow foundations requires the
application of a factor of safety (FS) to the gross ultimate bearing capacity, or
qall 5
qu
FS
(3.12)
3.4 Factor of Safety
141
However, some practicing engineers prefer to use a factor of safety such that
Net stress increase on soil 5
net ultimate bearing capacity
FS
(3.13)
The net ultimate bearing capacity is defined as the ultimate pressure per unit area of the
foundation that can be supported by the soil in excess of the pressure caused by the
surrounding soil at the foundation level. If the difference between the unit weight of concrete used in the foundation and the unit weight of soil surrounding is assumed to be
negligible, then
qnet(u) 5 qu 2 q
(3.14)
where
qnet(u) 5 net ultimate bearing capacity
q 5 gDf
So
qall(net) 5
qu 2 q
FS
(3.15)
The factor of safety as defined by Eq. (3.15) should be at least 3 in all cases.
Example 3.1
A square foundation is 2 m 3 2 m in plan. The soil supporting the foundation has a
friction angle of fr 5 25° and cr 5 20 kN>m2. The unit weight of soil, g, is
16.5 kN>m3. Determine the allowable gross load on the foundation with a factor of
safety (FS) of 3. Assume that the depth of the foundation (Df ) is 1.5 m and that general
shear failure occurs in the soil.
Solution
From Eq. (3.7)
qu 5 1.3crNc 1 qNq 1 0.4gBNg
From Table 3.1, for fr 5 25°,
Nc 5 25.13
Nq 5 12.72
Ng 5 8.34
Thus,
qu 5 (1.3) (20) (25.13) 1 (1.5 3 16.5) (12.72) 1 (0.4) (16.5) (2) (8.34)
5 653.38 1 314.82 1 110.09 5 1078.29 kN>m2
142 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
So, the allowable load per unit area of the foundation is
qu
1078.29
5
< 359.5 kN>m2
qall 5
FS
3
Thus, the total allowable gross load is
Q 5 (359.5) B2 5 (359.5) (2 3 2) 5 1438 kN
3.5
■
Modification of Bearing Capacity Equations
for Water Table
Equations (3.3) and (3.7) through (3.11) give the ultimate bearing capacity, based on the
assumption that the water table is located well below the foundation. However, if the water
table is close to the foundation, some modifications of the bearing capacity equations will
be necessary. (See Figure 3.6.)
Case I. If the water table is located so that 0 # D1 # Df, the factor q in the bearing
capacity equations takes the form
(3.16)
q 5 effective surcharge 5 D1g 1 D2 (gsat 2 gw )
where
gsat 5 saturated unit weight of soil
gw 5 unit weight of water
Also, the value of g in the last term of the equations has to be replaced by gr 5 gsat 2 gw.
Case II. For a water table located so that 0 # d # B,
(3.17)
q 5 gDf
In this case, the factor g in the last term of the bearing capacity equations must be replaced
by the factor
g 5 gr 1
Groundwater
table
d
(g 2 gr)
B
(3.18)
D1
Case I
Df
D2
B
d
Groundwater table
Case II
sat saturated
unit weight
Figure 3.6 Modification of bearing
capacity equations for water table
3.6 The General Bearing Capacity Equation
143
The preceding modifications are based on the assumption that there is no seepage force in
the soil.
Case III. When the water table is located so that d $ B, the water will have no effect on
the ultimate bearing capacity.
3.6
The General Bearing Capacity Equation
The ultimate bearing capacity equations (3.3), (3.7), and (3.8) are for continuous, square,
and circular foundations only; they do not address the case of rectangular foundations
(0 , B>L , 1). Also, the equations do not take into account the shearing resistance
along the failure surface in soil above the bottom of the foundation (the portion of the failure surface marked as GI and HJ in Figure 3.5). In addition, the load on the foundation
may be inclined. To account for all these shortcomings, Meyerhof (1963) suggested the
following form of the general bearing capacity equation:
qu 5 crNcFcsFcdFci 1 qNqFqsFqdFqi 1 12 gBNgFgsFgdFgi
(3.19)
In this equation:
cr 5
q5
g5
B5
Fcs, Fqs, Fgs 5
Fcd, Fqd, Fgd 5
Fci, Fqi, Fgi 5
Nc, Nq, Ng 5
cohesion
effective stress at the level of the bottom of the foundation
unit weight of soil
width of foundation (5 diameter for a circular foundation)
shape factors
depth factors
load inclination factors
bearing capacity factors
The equations for determining the various factors given in Eq. (3.19) are described briefly
in the sections that follow. Note that the original equation for ultimate bearing capacity is
derived only for the plane-strain case (i.e., for continuous foundations). The shape, depth,
and load inclination factors are empirical factors based on experimental data.
Bearing Capacity Factors
The basic nature of the failure surface in soil suggested by Terzaghi now appears to have been
borne out by laboratory and field studies of bearing capacity (Vesic, 1973). However, the
angle a shown in Figure 3.5 is closer to 45 1 fr>2 than to fr. If this change is accepted, the
values of Nc, Nq, and Ng for a given soil friction angle will also change from those given in
Table 3.1. With a 5 45 1 fr>2, it can be shown that
Nq 5 tan2 ¢ 45 1
fr p tan fr
≤e
2
(3.20)
144 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
and
Nc 5 (Nq 2 1) cot fr
(3.21)
Equation (3.21) for Nc was originally derived by Prandtl (1921), and Eq. (3.20) for Nq was
presented by Reissner (1924). Caquot and Kerisel (1953) and Vesic (1973) gave the relation for Ng as
Ng 5 2 (Nq 1 1) tan fr
(3.22)
Table 3.3 shows the variation of the preceding bearing capacity factors with soil friction
angles.
Table 3.3 Bearing Capacity Factors
f9
Nc
Nq
Ng
f9
Nc
Nq
Ng
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
5.14
5.38
5.63
5.90
6.19
6.49
6.81
7.16
7.53
7.92
8.35
8.80
9.28
9.81
10.37
10.98
11.63
12.34
13.10
13.93
14.83
15.82
16.88
18.05
19.32
20.72
1.00
1.09
1.20
1.31
1.43
1.57
1.72
1.88
2.06
2.25
2.47
2.71
2.97
3.26
3.59
3.94
4.34
4.77
5.26
5.80
6.40
7.07
7.82
8.66
9.60
10.66
0.00
0.07
0.15
0.24
0.34
0.45
0.57
0.71
0.86
1.03
1.22
1.44
1.69
1.97
2.29
2.65
3.06
3.53
4.07
4.68
5.39
6.20
7.13
8.20
9.44
10.88
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
22.25
23.94
25.80
27.86
30.14
32.67
35.49
38.64
42.16
46.12
50.59
55.63
61.35
67.87
75.31
83.86
93.71
105.11
118.37
133.88
152.10
173.64
199.26
229.93
266.89
11.85
13.20
14.72
16.44
18.40
20.63
23.18
26.09
29.44
33.30
37.75
42.92
48.93
55.96
64.20
73.90
85.38
99.02
115.31
134.88
158.51
187.21
222.31
265.51
319.07
12.54
14.47
16.72
19.34
22.40
25.99
30.22
35.19
41.06
48.03
56.31
66.19
78.03
92.25
109.41
130.22
155.55
186.54
224.64
271.76
330.35
403.67
496.01
613.16
762.89
Shape, Depth, Inclination Factors
Commonly used shape, depth, and inclination factors are given in Table 3.4.
3.6 The General Bearing Capacity Equation
145
Table 3.4 Shape, Depth and Inclination Factors (DeBeer (1970); Hansen (1970); Meyerhof (1963);
Meyerhof and Hanna (1981))
Factor
Shape
Depth
Relationship
Reference
B Nq
Fcs 5 1 1 a b a b
L Nc
B
Fqs 5 1 1 a b tan fr
L
B
Fgs 5 120.4 a b
L
Df
1
B
For ␾ ⫽ 0:
Df
Fcd 5 1 1 0.4 a b
B
DeBeer (1970)
Hansen (1970)
Fqd ⫽ 1
F␥d ⫽ 1
For ␾⬘ ⬎ 0:
Fcd 5 Fqd 2
1 2 Fqd
Nc tan fr
Fqd 5 1 1 2 tan fr (1 2 sin fr) 2 a
Df
B
b
F␥d ⫽ 1
Df
B
1
For ␾ ⫽ 0:
Df
Fcd 5 1 1 0.4 tan 21 a b
B
('')''*
radians
Fqd ⫽ 1
F␥d ⫽ 1
For ␾⬘ ⬎ 0:
Fcd 5 Fqd 2
1 2 Fqd
Nc tan f9
Df
Fqd 5 1 1 2 tan fr (1 2 sin fr) 2 tan 21 a b
B
('')''*
radians
F␥d ⫽ 1
Inclination
Fci 5 Fqi 5 a1 2
Fgi 5 a1 2
b
f9
b° 2
b
90°
b
␤ ⫽ inclination of the load on the
foundation with respect to the vertical
Meyerhof (1963); Hanna and
Meyerhof (1981)
146 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Example 3.2
Solve Example Problem 3.1 using Eq. (3.19).
Solution
From Eq. (3.19),
qu 5 c9NcFcsFcdFci 1 qNqFqsFqdFqt 1
1
gBNgFgsFgdFgt
2
Since the load is vertical, Fci ⫽ Fqi ⫽ F␥i ⫽ 1. From Table 3.3 for ␾⬘ ⫽ 25°, Nc ⫽ 20.72,
Nq ⫽ 10.66, and N␥ ⫽ 10.88.
Using Table 3.4,
B Nq
2 10.66
Fcs 5 1 1 a b a b 5 1 1 a b a
b 5 1.514
L Nc
2 20.72
B
2
Fqs 5 1 1 a b tanf9 5 1 1 a b tan 25 5 1.466
L
2
B
2
Fgs 5 1 2 0.4a b 5 1 2 0.4a b 5 0.6
L
2
Fqd 5 1 1 2 tanfr (1 2 sinfr) 2 a
Df
B
b
5 1 1 (2) (tan 25) (1 2 sin 25) 2 a
Fcd 5 Fqd 2
1 2 Fqd
Nc tanf9
5 1.233 2 c
1.5
b 5 1.233
2
1 2 1.233
d 5 1.257
(20.72) (tan 25)
Fgd 5 1
Hence,
qu ⫽ (20)(20.72)(1.514)(1.257)(1)
⫹(1.5 ⫻ 16.5)(10.66)(1.466)(1.233)(1)
1
1 (16.5) (2) (10.88) (0.6) (1) (1)
2
⫽ 788.6 ⫹ 476.9 ⫹ 107.7 ⫽ 1373.2 kN/m2
qall 5
qu
1373.2
5
5 457.7 kN>m2
FS
3
Q ⫽ (457.7)(2 ⫻ 2) ⫽ 1830.8 kN
■
3.6 The General Bearing Capacity Equation
147
Example 3.3
A square foundation (B 3 B) has to be constructed as shown in Figure 3.7. Assume
that g 5 16.5 kN>m3, gsat 5 18.55 kN>m3, fr 5 34°, Df 5 1.22 m, and D1 5 0.61 m.
The gross allowable load, Qall, with FS 5 3 is 667.2 kN. Determine the size of the
footing. Use Eq. (3.19).
D1
Water
table
; ; c 0
sat
c 0
Df
BB
Figure 3.7 A square foundation
Solution
We have
qall 5
Qall
B2
5
667.2
kN>m2
B2
(a)
From Eq. (3.19) (with cr 5 0), for vertical loading, we obtain
qall 5
qu
1
1
5 ¢ qNqFqsFqd 1 grBNgFgsFgd ≤
FS
3
2
For fr 5 34°, from Table 3.3, Nq 5 29.44 and Ng 5 41.06. Hence,
Fqs 5 1 1
B
tan fr 5 1 1 tan 34 5 1.67
L
Fgs 5 1 2 0.4 ¢
B
≤ 5 1 2 0.4 5 0.6
L
Fqd 5 1 1 2 tan fr (1 2 sin fr) 2
Df
B
5 1 1 2 tan 34 (1 2 sin 34) 2
4
1.05
511
B
B
Fgd 5 1
and
q 5 (0.61) (16.5) 1 0.61 (18.55 2 9.81) 5 15.4 kN>m2
148 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
So
qall 5
1
1.05
B(15.4) (29.44) (1.67) ¢1 1
≤
3
B
1
1 ¢ ≤ (18.55 2 9.81) (B) (41.06) (0.6) (1) R
2
5 252.38 1
(b)
265
1 35.89B
B
Combining Eqs. (a) and (b) results in
667.2
265
5 252.38 1
1 35.89B
2
B
B
By trial and error, we find that B < 1.3 m.
3.7
■
Case Studies on Ultimate Bearing Capacity
In this section, we will consider two field observations related to the ultimate bearing
capacity of foundations on soft clay. The failure loads on the foundations in the field will
be compared with those estimated from the theory presented in Section 3.6.
Foundation Failure of a Concrete Silo
An excellent case of bearing capacity failure of a 6-m diameter concrete silo was provided by Bozozuk (1972). The concrete tower silo was 21 m high and was constructed
over soft clay on a ring foundation. Figure 3.8 shows the variation of the undrained shear
strength (cu) obtained from field vane shear tests at the site. The groundwater table was
located at about 0.6 m below the ground surface.
On September 30, 1970, just after it was filled to capacity for the first time with
corn silage, the concrete tower silo suddenly overturned due to bearing capacity
failure. Figure 3.9 shows the approximate profile of the failure surface in soil. The
failure surface extended to about 7 m below the ground surface. Bozozuk (1972)
provided the following average parameters for the soil in the failure zone and the
foundation:
•
•
•
•
Load per unit area on the foundation when failure occurred < 160 kN>m2
Average plasticity index of clay (PI) < 36
Average undrained shear strength (cu) from 0.6 to 7 m depth obtained from field vane
shear tests < 27.1 kN>m2
From Figure 3.9, B < 7.2 m and Df < 1.52 m.
cu (kN/m2)
40
60
20
0
80
100
1
Depth (m)
2
3
4
5
Figure 3.8 Variation of cu with depth
obtained from field vane shear test
6
50
Original position
of foundation
1.46 m
0
22 2 1m
m
30
0.9
4
6
Paved apron
Original
ground surface
7.2
Depth below paved apron (m)
1
2
Collapsed silo
50
Upheaval
22
45
m
1.
60
8
10
12
Figure 3.9 Approximate profile of silo failure (Adapted from Bozozuk, 1972)
149
150 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
We can now calculate the factor of safety against bearing capacity failure. From Eq. (3.19)
qu 5 crNcFcsFcdFci 1 qNcFqsFqdFqi 1 12 gB NgFgsFgdFgi
For f 5 0 condition and vertical loading, cr 5 cu, Nc 5 5.14, Nq 5 1, Ng 5 0, and
Fci 5 Fqi 5 Fgi 5 0. Also, from Table 3.4,
Fcs 5 1 1 a
7.2
1
ba
b 5 1.195
7.2 5.14
Fqs 5 1
Fcd 5 1 1 (0.4) a
1.52
b 5 1.08
7.2
Fqd 5 1
Thus,
qu 5 (cu ) (5.14) (1.195) (1.08) (1) 1 (g) (1.52)
Assuming g < 18 kN>m3,
qu 5 6.63cu 1 27.36
(3.23)
According to Eqs. (2.34) and (2.35a),
cu(corrected) 5 l cu(VST)
l 5 1.7 2 0.54 log 3PI(%)4
For this case, PI < 36 and cu(VST) 5 27.1 kN>m2. So
cu(corrected) 5 51.7 2 0.54 log 3PI(%)46cu(VST)
5 (1.7 2 0.54 log 36) (27.1) < 23.3 kN>m2
Substituting this value of cu in Eq. (3.23)
qu 5 (6.63) (23.3) 1 27.36 5 181.8 kN>m2
The factor of safety against bearing capacity failure
FS 5
qu
181.8
5
5 1.14
applied load per unit area
160
This factor of safety is too low and approximately equals one, for which the failure occurred.
Load Tests on Small Foundations in Soft Bangkok Clay
Brand et al. (1972) reported load test results for five small square foundations in soft
Bangkok clay in Rangsit, Thailand. The foundations were 0.6 m ⫻ 0.6 m, 0.675 m ⫻
0.675 m, 0.75 m ⫻ 0.75 m, 0.9 m ⫻ 0.9 m, and 1.05 m ⫻ 1.05 m. The depth of of the
foundations (Df) was 1.5 m in all cases.
Figure 3.10 shows the vane shear test results for clay. Based on the variation of
cu(VST) with depth, it can be approximated that cu(VST) is about 35 kN/m2 for depths between
zero to 1.5 m measured from the ground surface, and cu(VST) is approximately equal to
24 kN/m2 for depths varying from 1.5 to 8 m. Other properties of the clay are
•
•
•
Liquid limit ⫽ 80
Plastic limit ⫽ 40
Sensitivity ⬇ 5
3.7 Case Studies on Ultimate Bearing Capacity
151
cu (VST) (kN/m2)
0
10
20
30
40
1
2
Depth (m)
3
4
5
6
7
8
Figure 3.10 Variation of cu(VST) with depth for
soft Bangkok clay
Figure 3.11 shows the load-settlement plots obtained from the bearing-capacity tests on
all five foundations. The ultimate loads, Qu, obtained from each test are shown in Figure 3.11
and given in Table 3.5. The ultimate load is defined as the point where the load-settlement plot
becomes practically linear.
From Eq. (3.19),
qu 5 c9NcFcsFcdFci 1 qNqFqsFqdFqi 1
1
gBNgFgsFgdFgi
2
For undrained condition and vertical loading (that is, ␾ ⫽ 0) from Tables 3.3 and 3.4,
•
•
•
•
•
•
Fci ⫽ Fqi ⫽ F␥i ⫽ 1
c⬘ ⫽ cu, Nc ⫽ 5.14, Nq ⫽ 1, and N␥ ⫽ 0
B Nq
1
Fcs 5 1 1 a b a b 5 1 1 (1) a
b 5 1.195
L Nc
5.14
Fqs ⫽ 1
Fqd ⫽ 1
Df
1.5
Fcd 5 1 1 0.4 tan 21 a b 5 1 1 0.4 tan 21 a b
B
B
(3.24)
152 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Load (kN)
0
40
0
200
160
120
80
Qu (ultimate load)
Settlement (mm)
10
20
B = 0.675 m
30
B = 1.05 m
B = 0.6 m
B = 0.75 m
B = 0.9 m
40
Figure 3.11 Load-settlement plots obtained from bearing capacity tests
(Note: Df /B ⬎ 1 in all cases)
Thus,
qu ⫽ (5.14)(cu)(1.195)Fcd ⫹ q
(3.25)
The values of cu(VST) need to be corrected for use in Eq. (3.25). From Eq. (2.34),
cu ⫽ ␭cu(VST)
From Eq. (2.35b),
␭ ⫽ 1.18e⫺0.08(PI) ⫹ 0.57 ⫽ 1.18e⫺0.08(80 ⫺ 40) ⫹ 0.57 ⫽ 0.62
From Eq. (2.35c),
␭ ⫽ 7.01e⫺0.08(LL) ⫹ 0.57 ⫽ 7.01e⫺0.08(80) ⫹ 0.57 ⫽ 0.58
Table 3.5 Comparison of Ultimate Bearing Capacity—Theory versus Field Test Results
B
(m)
(1)
Df
(m)
(2)
0.600
0.675
0.750
0.900
1.050
1.5
1.5
1.5
1.5
1.5
Fcd
(3)
qu(theory)‡‡
(kN/m2)
(4)
1.476
1.459
1.443
1.412
1.384
158.3
156.8
155.4
152.6
150.16
‡
Qu(field) (kN)
(5)
qu(field)‡‡‡
(kN/m2)
(6)
qu(field) 2 qu(theory)
(%)
qu(field)
60
71
90
124
140
166.6
155.8
160.6
153.0
127.0
4.98
⫺0.64
2.87
0.27
⫺18.24
Eq. (3.24); ‡‡Eq. (3.26); ‡‡‡Qu(field)/B2 ⫽ qu(field)
‡
(7)
3.8 Effect of Soil Compressibility
153
So the average value of ␭ ⬇ 0.6. Hence,
cu ⫽ ␭cu(VST) ⫽ (0.6)(24) ⫽ 14.4 kN/m2
Let us assume ␥ ⫽ 18.5 kN/m2. So
q ⫽ ␥Df ⫽ (18.5)(1.5) ⫽ 27.75 kN/m2
Substituting cu ⫽ 14.4 kN/m2 and q ⫽ 27.75 kN/m2 into Eq. (3.25), we obtain
qu(kN/m2) ⫽ 88.4Fcd ⫹ 27.75
(3.26)
The values of qu calculated using Eq. (3.26) are given in column 4 of Table 3.5. Also,
the qu determined from the field tests are given in column 6. The theoretical and field values of qu compare very well. The important lessons learned from this study are
1. The ultimate bearing capacity is a function of cu. If Eq. (2.35a) would have been
used to correct the undrained shear strength, the theoretical values of qu would have
varied between 200 kN/m2 and 210 kN/m2. These values are about 25% to 55%
more than those obtained from the field and are on the unsafe side.
2. It is important to recognize that empirical correlations like those given in Eqs.
(2.35a), (2.35b) and (2.35c) are sometimes site specific. Thus, proper engineering
judgment and any record of past studies would be helpful in the evaluation of bearing capacity.
3.8
Effect of Soil Compressibility
In Section 3.3, Eqs. (3.3), (3.7), and (3.8), which apply to the case of general shear failure,
were modified to Eqs. (3.9), (3.10), and (3.11) to take into account the change of failure
mode in soil (i.e., local shear failure). The change of failure mode is due to soil compressibility, to account for which Vesic (1973) proposed the following modification of Eq. (3.19):
qu 5 crNcFcsFcdFcc 1 qNqFqsFqdFqc 1 12 gBNgFgsFgdFgc
(3.27)
In this equation, Fcc, Fqc, and Fgc are soil compressibility factors.
The soil compressibility factors were derived by Vesic (1973) by analogy to the
expansion of cavities. According to that theory, in order to calculate Fcc, Fqc, and Fgc, the
following steps should be taken:
Step 1. Calculate the rigidity index, Ir, of the soil at a depth approximately B>2
below the bottom of the foundation, or
Ir 5
Gs
cr 1 qr tan fr
where
Gs 5 shear modulus of the soil
q 5 effective overburden pressure at a depth of Df 1 B>2
(3.28)
154 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Step 2.
The critical rigidity index, Ir(cr), can be expressed as
fr
1
B
Ir(cr) 5 bexp B ¢ 3.30 2 0.45 ≤ cot ¢ 45 2 ≤ R r
2
L
2
Step 3.
(3.29)
The variations of Ir(cr) with B>L are given in Table 3.6.
If Ir $ Ir(cr), then
Fcc 5 Fqc 5 Fgc 5 1
However, if Ir , Ir(cr), then
Fgc 5 Fqc 5 exp b ¢ 24.4 1 0.6
(3.07 sin fr) (log 2Ir )
B
≤ tan fr 1 B
Rr
L
1 1 sin fr
(3.30)
Figure 3.12 shows the variation of Fgc 5 Fqc [see Eq. (3.30)] with fr and Ir. For f 5 0,
Fcc 5 0.32 1 0.12
B
1 0.60 log Ir
L
(3.31)
For fr . 0,
Fcc 5 Fqc 2
1 2 Fqc
(3.32)
Nq tan fr
Table 3.6 Variation of Ir(cr) with ␾⬘ and B/L
Ir(cr)
␾⬘
(deg)
B/L ⴝ 0
B/L ⴝ 0.2
B/L ⴝ 0.4
B/L ⴝ 0.6
B/L ⴝ 0.8
B/L ⴝ 1.0
0
5
10
15
20
25
30
35
40
45
13.56
18.30
25.53
36.85
55.66
88.93
151.78
283.20
593.09
1440.94
12.39
16.59
22.93
32.77
48.95
77.21
129.88
238.24
488.97
1159.56
11.32
15.04
20.60
29.14
43.04
67.04
111.13
200.41
403.13
933.19
10.35
13.63
18.50
25.92
37.85
58.20
95.09
168.59
332.35
750.90
9.46
12.36
16.62
23.05
33.29
50.53
81.36
141.82
274.01
604.26
8.64
11.20
14.93
20.49
29.27
43.88
69.62
119.31
225.90
486.26
3.8 Effect of Soil Compressibility
1.0
1.0
500
100 250
0.8
250
0.8
50
50
0.6
10
0.4
100
F c Fqc
F c Fqc
155
25
5
2.5
0.2
0.6
25
10
0.4
5
2.5
0.2
Ir 1
0
500
Ir 1
0
0
10
20
30
50
40
Soil friction angle, (deg)
L
(a)
1
B
0
10
20
30
50
40
Soil friction angle, (deg)
L
(b)
5
B
Figure 3.12 Variation of Fgc 5 Fqc with Ir and fr
Example 3.4
For a shallow foundation, B 5 0.6 m, L 5 1.2 m, and Df 5 0.6 m. The known soil
characteristics are as follows:
Soil:
fr 5 25°
cr 5 48 kN>m2
g 5 18 kN>m3
Modulus of elasticity, Es 5 620 kN>m2
Poisson’s ratio, ms 5 0.3
Calculate the ultimate bearing capacity.
Solution
From Eq. (3.28),
Ir 5
Gs
cr 1 qr tan fr
However,
Gs 5
Es
2 (1 1 ms )
So
Ir 5
Now,
qr 5 g ¢Df 1
Es
2 (1 1 ms ) 3cr 1 qr tan fr4
B
0.6
≤ 5 18 ¢0.6 1
≤ 5 16.2 kN>m2
2
2
156 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Thus,
Ir 5
From Eq. (3.29),
620
5 4.29
2 (1 1 0.3) 348 1 16.2 tan 254
fr
1
B
Ir(cr) 5 bexp B ¢3.3 2 0.45 ≤ cot ¢45 2 ≤ R r
2
L
2
1
25
0.6
5 bexp B ¢ 3.3 2 0.45
≤ cot ¢45 2 ≤ R r 5 62.41
2
1.2
2
Since Ir(cr) . Ir, we use Eqs. (3.30) and (3.32) to obtain
Fgc 5 Fqc 5 exp b ¢ 24.4 1 0.6
5 exp b ¢ 24.4 1 0.6
1 c
(3.07 sin fr)log(2Ir )
B
≤ tan fr 1 B
Rr
L
1 1 sin fr
0.6
≤ tan 25
1.2
(3.07 sin 25)log(2 3 4.29)
R r 5 0.347
1 1 sin 25
and
Fcc 5 Fqc 2
1 2 Fqc
Nc tan fr
For fr 5 25°, Nc 5 20.72 (see Table 3.3); therefore,
1 2 0.347
Fcc 5 0.347 2
5 0.279
20.72 tan 25
Now, from Eq. (3.27),
qu 5 crNcFcsFcdFcc 1 qNqFqsFqdFqc 1 12gBNgFgsFgdFgc
From Table 3.3, for fr 5 25°, Nc 5 20.72, Nq 5 10.66, and Ng 5 10.88. Consequently,
Nq B
10.66 0.6
Fcs 5 1 1 ¢ ≤ ¢ ≤ 5 1 1 ¢
≤¢
≤ 5 1.257
Nc
L
20.72 1.2
Fqs 5 1 1
B
0.6
tan fr 5 1 1
tan 25 5 1.233
L
1.2
Fgs 5 1 2 0.4 ¢
B
0.6
≤ 5 1 2 0.4
5 0.8
L
1.2
Fqd 5 1 1 2 tan fr (1 2 sin fr) 2 ¢
Df
5 1 1 2 tan 25 (1 2 sin 25) 2 ¢
Fcd 5 Fqd 2
1 2 Fqd
Nc tan fr
5 1.311 2
B
≤
0.6
≤ 5 1.311
0.6
1 2 1.311
5 1.343
20.72 tan 25
3.9 Eccentrically Loaded Foundations
157
and
Fgd 5 1
Thus,
qu 5 (48) (20.72) (1.257) (1.343) (0.279) 1 (0.6 3 18) (10.66) (1.233) (1.311)
(0.347)1( 12 ) (18) (0.6) (10.88) (0.8) (1) (0.347) 5 549.32 kN , m2
3.9
■
Eccentrically Loaded Foundations
In several instances, as with the base of a retaining wall, foundations are subjected to
moments in addition to the vertical load, as shown in Figure 3.13a. In such cases, the distribution of pressure by the foundation on the soil is not uniform. The nominal distribution
of pressure is
qmax 5
Q
6M
1 2
BL
BL
(3.33)
qmin 5
Q
6M
2 2
BL
BL
(3.34)
and
Q
Q
e
M
B
B
BL
For e < B/6
qmin
qmax
L
For e > B/6
qmax
(a)
Figure 3.13 Eccentrically loaded foundations
2e
B
(b)
158 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
where
Q 5 total vertical load
M 5 moment on the foundation
Figure 3.13b shows a force system equivalent to that shown in Figure 3.13a. The
distance
e5
M
Q
(3.35)
is the eccentricity. Substituting Eq. (3.35) into Eqs. (3.33) and (3.34) gives
qmax 5
Q
6e
¢1 1 ≤
BL
B
(3.36)
qmin 5
Q
6e
¢1 2 ≤
BL
B
(3.37)
and
Note that, in these equations, when the eccentricity e becomes B>6, qmin is zero. For
e . B>6, qmin will be negative, which means that tension will develop. Because soil cannot
take any tension, there will then be a separation between the foundation and the soil underlying it. The nature of the pressure distribution on the soil will be as shown in Figure 3.13a.
The value of qmax is then
qmax 5
4Q
3L(B 2 2e)
(3.38)
The exact distribution of pressure is difficult to estimate.
Figure 3.14 shows the nature of failure surface in soil for a surface strip foundation
subjected to an eccentric load. The factor of safety for such type of loading against bearing capacity failure can be evaluated as
FS 5
Qult
Q
(3.39)
where Qult 5 ultimate load-carrying capacity.
The following sections describe several theories for determining Qult.
Qult
e
B
Figure 3.14 Nature of
failure surface in soil
supporting a strip foundation subjected to
eccentric loading
(Note: Df 5 0; Qult is ultimate load per unit length
of foundation)
3.10 Ultimate Bearing Capacity under Eccentric Loading—One-Way Eccentricity
3.10
159
Ultimate Bearing Capacity under Eccentric
Loading—One-Way Eccentricity
Effective Area Method (Meyerhoff, 1953)
In 1953, Meyerhof proposed a theory that is generally referred to as the effective area method.
The following is a step-by-step procedure for determining the ultimate load that the
soil can support and the factor of safety against bearing capacity failure:
Step 1. Determine the effective dimensions of the foundation (Figure 3.13b):
Br 5 effective width 5 B 2 2e
Lr 5 effective length 5 L
Note that if the eccentricity were in the direction of the length of the foundation, the value of Lr would be equal to L 2 2e. The value of Br would
equal B. The smaller of the two dimensions (i.e., Lr and Br) is the effective
width of the foundation.
Step 2. Use Eq. (3.19) for the ultimate bearing capacity:
qur 5 crNcFcsFcdFci 1 qNqFqsFqdFqi 1 12 gBrNgFgsFgdFgi
(3.40)
To evaluate Fcs, Fqs, and Fgs, use the relationships given in Table 3.4 with
effective length and effective width dimensions instead of L and B, respectively. To determine Fcd, Fqd, and Fgd, use the relationships given in Table
3.4. However, do not replace B with Br.
Step 3. The total ultimate load that the foundation can sustain is
Ar &
$'%'
Qult 5
(3.41)
q ru (Br) (Lr)
where Ar 5 effective area.
Step 4. The factor of safety against bearing capacity failure is
Qult
FS 5
Q
Prakash and Saran Theory
Prakash and Saran (1971) analyzed the problem of ultimate bearing capacity of eccentrically and vertically loaded continuous (strip) foundations by using the one-sided failure
surface in soil, as shown in Figure 3.14. According to this theory, the ultimate load per unit
length of a continuous foundation can be estimated as
1
Q ult 5 B ccrNc(e) 1 qNq(e) 1 gBNg(e) d
2
(3.42)
where Nc(e), Nq(e), N␥(e) ⫽ bearing capacity factors under eccentric loading.
The variations of Nc(e), Nq(e), and N␥(e) with soil friction angle ␾⬘ are given in
Figures 3.15, 3.16, and 3.17. For rectangular foundations, the ultimate load can be given as
1
Q ult 5 BL ccrNc(e)Fcs(e) 1 qNq(e)Fqs(e) 1 gBNg(e)Fgs(e) d
2
where Fcs(e), Fqs(e), and F␥s(e) ⫽ shape factors.
(3.43)
160 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
60
40
e/B = 0
Nc (e)
0.1
0.2
20
0.3
0.4
f9 5 408
0
0
10
20
Friction angle, (deg)
30
40
e/B
Nc(e)
0
0.1
0.2
0.3
0.4
94.83
66.60
54.45
36.3
18.15
Figure 3.15 Variation of Nc(e) with fr
Prakash and Saran (1971) also recommended the following for the shape factors:
Fcs(e) 5 1.2 2 0.025
L
(with a minimum of 1.0)
B
Fqs(e) 5 1
(3.44)
(3.45)
and
Fgs(e) 5 1.0 1 a
2e
B
3 e
B 2
2 0.68b 1 c0.43 2 a b a b d a b
B
L
2 B
L
(3.46)
Reduction Factor Method (For Granular Soil)
Purkayastha and Char (1977) carried out stability analysis of eccentrically loaded continuous foundations supported by a layer of sand using the method of slices. Based on that
analysis, they proposed
3.10 Ultimate Bearing Capacity under Eccentric Loading—One-Way Eccentricity
161
60
40
e/B = 0
Nq (e)
0.1
20
0.2
0.3
f9 5 408
0.4
0
0
10
20
Friction angle, (deg)
30
40
e/B
Nq(e)
0
0.1
0.2
0.3
0.4
81.27
56.09
45.18
30.18
15.06
Figure 3.16 Variation of Nq(e) with fr
Rk 5 1 2
qu(eccentric)
qu(centric)
(3.47)
where
Rk ⫽ reduction factor
qu(eccentric) ⫽ ultimate bearing capacity of eccentrically loaded continuous
foundations
qu(centric) ⫽ ultimate bearing capacity of centrally loaded continuous foundations
The magnitude of Rk can be expressed as
e k
Rk 5 aa b
B
where a and k are functions of the embedment ratio Df /B (Table 3.7).
(3.48)
162 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
60
40
e/B = 0
N(e)
0.1
20
0.2
f9 5 408
0.3
0.4
0
0
10
20
Friction angle, (deg)
30
40
e,B
Ng(e)
0
0.1
0.2
0.3
0.4
115.80
71.80
41.60
18.50
4.62
Figure 3.17 Variation of Ng(e) with fr
Table 3.7 Variations of a and k [Eq. (3.48)]
Df/B
a
k
0.00
0.25
0.50
1.00
1.862
1.811
1.754
1.820
0.73
0.785
0.80
0.888
Hence, combining Eqs. (3.47) and (3.48)
e k
qu(eccentric) 5 qu(centric) (1 2 Rk ) 5 qu(centric) c1 2 aa b d
B
(3.49)
1
qu(centric) 5 qNqFqd 1 gBNgFgd
2
(3.50)
where
3.10 Ultimate Bearing Capacity under Eccentric Loading—One-Way Eccentricity
163
The relationships for Fqd, and F␥d are given in Table 3.4.
The ultimate load per unit length of the foundation can then be given as
Qu ⫽ Bqu(eccentric)
(3.51)
Example 3.5
A continuous foundation is shown in Figure 3.18. If the load eccentricity is 0.2 m,
determine the ultimate load, Qult, per unit length of the foundation. Use Meyerhof’s
effective area method.
Solution
For c⬘ ⫽ 0, Eq. (3.40) gives
qu9 5 qNqFqsFqdFqi 1
1
grBrNgFgsFgdFgi
2
where q ⫽ (16.5) (1.5) ⫽ 24.75 kN/m2.
Sand
40
c 0
16.5 kN/m3
1.5 m
2m
Figure 3.18 A continuous foundation with load
eccentricity
For ␾⬘ ⫽ 40°, from Table 3.3, Nq ⫽ 64.2 and N␥ ⫽ 109.41. Also,
B⬘ ⫽ 2 ⫺ (2)(0.2) ⫽ 1.6 m
Because the foundation in question is a continuous foundation, B⬘/L⬘ is zero. Hence,
Fqs ⫽ 1, F␥s ⫽ 1. From Table 3.4,
Fqi ⫽ F␥ i ⫽ 1
Fqd 5 1 1 2 tan fr(1 2 sin fr) 2
Df
B
5 1 1 0.214a
1.5
b 5 1.16
2
F␥d ⫽ 1
and
qur 5 (24.75) (64.2) (1) (1.16) (1)
1
1 a b (16.5) (1.6) (109.41) (1) (1) (1) 5 3287.39 kN>m2
2
Consequently,
Qult ⫽ (B⬘)(1)(qu⬘) ⫽ (1.6)(1)(3287.39) ⬇ 5260 kN
■
164 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Example 3.6
Solve Example 3.5 using Eq. (3.42).
Solution
Since c⬘ ⫽ 0
1
Qult 5 B cqNq(e) 1 gBNg(e) d
2
e
0.2
5
5 0.1
B
2
For ␾⬘ ⫽ 40° and e/B ⫽ 0.1, Figures 3.16 and 3.17 give Nq(e) ⫽ 56.09 and N␥ (e) ⬇ 71.8.
Hence,
Qult ⫽ 2[(24.75)(56.09) ⫹ (12)(16.5)(2)(71.8)] ⫽ 5146 kN
■
Example 3.7
Solve Example 3.5 using Eq. (3.49).
Solution
With c⬘ ⫽ 0,
1
qu(centric) 5 qNqFqd 1 gBNgFgd
2
For ␾⬘ ⫽ 40°, Nq ⫽ 64.2 and N␥ ⫽ 109.41 (see Table 3.3). Hence,
Fqd ⫽ 1.16 and F␥d ⫽ 1 (see Example 3.5)
1
qu(centric) 5 (24.75) (64.2) (1.16) 1 (16.5) (2) (109.41) (1)
2
5 1843.18 1 1805.27 5 3648.45 kN>m2
From Eq. (3.48),
e k
Rk 5 aa b
B
For Df /B ⫽ 1.5/2 ⫽ 0.75, Table 3.7 gives a ⬇ 1.75 and k ⬇ 0.85. Hence,
Rk 5 1.79a
0.2 0.85
b
5 0.253
2
Qu ⫽ Bqu(eccentric) ⫽ Bqu(centric)(1 ⫺ Rk) ⫽ (2)(3648.45)(1 ⫺ 0.253) ⬇ 5451 kN
■
3.11 Bearing Capacity—Two-way Eccentricity
3.11
165
Bearing Capacity—Two-way Eccentricity
Consider a situation in which a foundation is subjected to a vertical ultimate load Qult and
a moment M, as shown in Figures 3.19a and b. For this case, the components of the
moment M about the x- and y-axes can be determined as Mx and My, respectively. (See
Figure 3.19.) This condition is equivalent to a load Qult placed eccentrically on the foundation with x 5 eB and y 5 eL (Figure 3.19d). Note that
My
eB 5
(3.52)
Qult
and
eL 5
Mx
Qult
(3.53)
If Qult is needed, it can be obtained from Eq. (3.41); that is,
Qult 5 qur Ar
where, from Eq. (3.40),
qur 5 crNcFcsFcdFci 1 qNqFqsFqdFqi 1 12 gBrNgFgsFgdFgi
and
Ar 5 effective area 5 BrLr
As before, to evaluate Fcs, Fqs, and Fgs (Table 3.4), we use the effective length Lr and
effective width Br instead of L and B, respectively. To calculate Fcd, Fqd, and Fgd, we do
Qult
M
(a)
BL
B
y
eB
Mx
L
Qult
M
Qult
x
Qult
eL
My
B
(b)
(c)
(d)
Figure 3.19 Analysis of foundation with two-way eccentricity
166 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
not replace B with Br. In determining the effective area Ar, effective width Br, and effective length Lr, five possible cases may arise (Highter and Anders, 1985).
Case I. eL>L $ 16 and eB>B $ 16. The effective area for this condition is shown in
Figure 3.20, or
Ar 5 12B1L1
(3.54)
where
B1 5 B ¢1.5 2
3eB
≤
B
(3.55)
L1 5 L ¢1.5 2
3eL
≤
L
(3.56)
and
The effective length Lr is the larger of the two dimensions B1 and L1. So the effective width is
Br 5
Ar
Lr
(3.57)
Case II. eL>L , 0.5 and 0 , eB>B , 16. The effective area for this case, shown in
Figure 3.21a, is
Ar 5 12 (L1 1 L2 )B
(3.58)
The magnitudes of L1 and L2 can be determined from Figure 3.21b. The effective width is
L1 or L2
Ar
(whichever is larger)
(3.59)
Lr 5 L1 or L2
(whichever is larger)
(3.60)
Br 5
The effective length is
Case III. eL>L , 16 and 0 , eB>B , 0.5. The effective area, shown in Figure 3.22a, is
Ar 5 12 (B1 1 B2 )L
Effective
area
B1
eB
eL
Qult
L1
L
B
Figure 3.20 Effective area for the case of eL>L > 16
and eB>B > 16
(3.61)
3.11 Bearing Capacity–Two-way Eccentricity
167
Effective
area
B
eB
L2
eL
Qult
L1
L
(a)
0.5
eB /B 0.4
0.167
0.1
0.08
0.06
eL /L
0.3
0.2
0.2
0.4
0.6
L1 /L, L2 /L
(b)
2
2
0.0
0
For
obtaining
L2 /L
01
0
0.
4
0.0
6
8
0.0
eB /B 0.0
0.10
0.12
0.14
0.16
0.1
0.0
4
0.0
0.
0
0.8
1
1.0
For
obtaining
L1 /L
Figure 3.21 Effective area
for the case of eL>L , 0.5 and
0 , eB>B , 16 (After Highter
and Anders, 1985) (Highter,
W. H. and Anders, J. C. (1985).
“Dimensioning Footings
Subjected to Eccentric Loads,”
Journal of Geotechnical
Engineering, American Society
of Civil Engineers, Vol. 111,
No. GT5, pp. 659–665. With
permission from ASCE.)
The effective width is
Ar
L
(3.62)
Lr 5 L
(3.63)
Br 5
The effective length is
The magnitudes of B1 and B2 can be determined from Figure 3.22b.
Case IV. eL>L , 16 and eB>B , 16. Figure 3.23a shows the effective area for this case. The
ratio B2>B, and thus B2, can be determined by using the eL>L curves that slope upward.
Similarly, the ratio L2>L, and thus L2, can be determined by using the eL>L curves that
slope downward. The effective area is then
Ar 5 L2B 1 12 (B 1 B2 ) (L 2 L2 )
(3.64)
168 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
B1
eB
eL
Qult
L
Effective
area
B2
B
(a)
0.5
eL /L 0.4
0.167
0.1
0.08
0.06
eB /B
0.3
0.2
0.2
0.4
0.6
B1 /B, B2 /B
(b)
2
2
0.0
0
For
obtaining
B2 /B
01
0
0.
4
0.0
6
8
0.0
eL /L 0.0
0.10
0.12
0.14
0.16
0.1
0.0
4
0.0
0.
0
0.8
1
1.0
For
obtaining
B1 /B
Figure 3.22 Effective area for the case of eL>L , 16 and 0 , eB>B , 0.5 (After Highter
and Anders, 1985) Highter, W. H. and Anders, J. C. (1985). “Dimensioning Footings
Subjected to Eccentric Loads,” Journal of Geotechnical Engineering, American Society
of Civil Engineers, Vol. 111, No. GT5, pp. 659–665. With permission from ASCE.)
The effective width is
Ar
L
(3.65)
Lr 5 L
(3.66)
Br 5
The effective length is
Case V. (Circular Foundation) In the case of circular foundations under eccentric
loading (Figure 3.24a), the eccentricity is always one way. The effective area Ar and the
effective width Br for a circular foundation are given in a nondimensional form in Table 3.8.
Once A9 and B9 are determined, the effective length can be obtained as
Lr 5
Ar
Br
3.11 Bearing Capacity–Two-way Eccentricity
169
B
L2
eB
eL
L
Qult
Effective
area
B2
(a)
For obtaining B2 /B
0.16
0.14
0.12
0.20
0.10
0.15
0.08
eB /B
1
0.
0.06
0.0
0.10
0.1
4
8
0.04
0.0
6
0.05
0.04
0.02 eL /L
eL/L 0.02
For obtaining L2/L
0
0
0.2
0.4
0.6
B2 /B, L2 /L
(b)
0.8
eR
Qult
R
Figure 3.24 Effective area for circular foundation
1.0
Figure 3.23 Effective area for the
case of eL>L , 16 and eB>B , 16
(After Highter and Anders, 1985)
(Highter, W. H. and Anders, J. C.
(1985). “Dimensioning Footings
Subjected to Eccentric Loads,” Journal
of Geotechnical Engineering,
American Society of Civil Engineers,
Vol. 111, No. GT5, pp. 659–665. With
permission from ASCE.)
170 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Table 3.8 Variation of Ar>R2 and Br>R with
eR>R for Circular Foundations
eR9 , R
A9 , R 2
B9 , R
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
2.8
2.4
2.0
1.61
1.23
0.93
0.62
0.35
0.12
0
1.85
1.32
1.2
0.80
0.67
0.50
0.37
0.23
0.12
0
Example 3.8
A square foundation is shown in Figure 3.25, with eL 5 0.3 m and eB 5 0.15 m. Assume two-way eccentricity, and determine the ultimate load, Qult.
Solution
We have
eL
0.3
5
5 0.2
L
1.5
and
eB
0.15
5
5 0.1
B
1.5
This case is similar to that shown in Figure 3.21a. From Figure 3.21b, for eL>L 5 0.2
and eB>B 5 0.1,
L1
< 0.85;
L
L1 5 (0.85) (1.5) 5 1.275 m
L2
< 0.21;
L
L2 5 (0.21) (1.5) 5 0.315 m
and
From Eq. (3.58),
Ar 5 12 (L1 1 L2 )B 5 12 (1.275 1 0.315) (1.5) 5 1.193 m2
3.11 Ultimate Bearing Capacity–Two-way Eccentricity
Sand
18 kN/m3
30
c 0
0.7 m
1.5 m 1.5 m
eB 0.15 m
1.5 m
eL 0.3 m
Figure 3.25 An eccentrically loaded
foundation
1.5 m
From Eq. (3.60),
Lr 5 L1 5 1.275 m
From Eq. (3.59),
Br 5
Ar
1.193
5
5 0.936 m
Lr
1.275
Note from Eq. (3.40) with cr 5 0,
qur 5 qNqFqsFqdFqi 1 12gBrNgFgsFgdFgi
where q 5 (0.7) (18) 5 12.6 kN>m2.
For fr 5 30°, from Table 3.3, Nq 5 18.4 and Ng 5 22.4. Thus from Table 3.4,
Fqs 5 1 1 ¢
Br
0.936
≤ tan fr 5 1 1 ¢
≤ tan 30° 5 1.424
Lr
1.275
Fgs 5 1 2 0.4 ¢
Br
0.936
≤ 5 0.706
≤ 5 1 2 0.4 ¢
Lr
1.275
Fqd 5 1 1 2 tan fr(1 2 sin fr) 2
Df
B
511
and
Fgd 5 1
(0.289) (0.7)
5 1.135
1.5
171
172 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
So
Qult 5 Arqur 5 Ar(qNqFqsFqd 1 12gBrNgFgsFgd )
5 (1.193) 3(12.6) (18.4) (1.424) (1.135)
1(0.5) (18) (0.936) (22.4) (0.706) (1)4 < 606 kN
■
Example 3.9
Consider the foundation shown in Figure 3.25 with the following changes:
eL ⫽ 0.18 m
eB ⫽ 0.12 m
For the soil, ␥ ⫽ 16.5 kN/m3
␾⬘ ⫽ 25°
c⬘ ⫽ 25 kN/m2
Determine the ultimate load, Qult.
Solution
eL
0.18
5
5 0.12;
L
1.5
eB
0.12
5
5 0.08
B
1.5
This is the case shown in Figure 3.23a. From Figure 3.23b,
B2
< 0.1;
B
L2
< 0.32
L
So
B2 ⫽ (0.1)(1.5) ⫽ 0.15 m
L2 ⫽ (0.32)(1.5) ⫽ 0.48 m
From Eq. (3.64),
1
1
Ar 5 L2B 1 (B 1 B2 ) (L 2 L2 ) 5 (0.48) (1.5) 1 (1.5 1 0.15) (1.5 2 0.48)
2
2
5 0.72 1 0.8415 5 1.5615 m2
Br 5
Ar
1.5615
5
5 1.041m
L
1.5
L9 5 1.5 m
From Eq. (3.40),
1
qur 5 crNcFcs Fed 1 qNqFqsFqd 1 gBrNgFgsFgd
2
3.12 Bearing Capacity of a Continuous Foundation Subjected to Eccentric Inclined Load
173
For ␾⬘ ⫽ 25°, Table 3.3 gives Nc ⫽ 20.72, Nq ⫽ 10.66 and N␥ ⫽ 10.88. From Table 3.4,
Fcs 5 1 1 a
Br Nq
1.041 10.66
ba b 5 1 1 a
ba
b 5 1.357
Lr Nc
1.5
20.72
Fqs 5 1 1 a
Br
1.041
b tanfr 5 1 1 a
b tan 25 5 1.324
Lr
1.5
Fgs 5 1 2 0.4 a
Br
1.041
b 5 120.4 a
b 5 0.722
Lr
1.5
Df
0.7
Fqd 5 1 1 2 tan fr(1 2 sinfr) 2 a b 5 1 1 2 tan 25(12sin 25) 2 a b 5 1.145
B
1.5
Fcd 5 Fqd 2
1 2 Fqd
9
Nc tan f
5 1.145 2
1 2 1.145
5 1.16
20.72 tan 25
Fgd 5 1
Hence,
qru 5 (25) (20.72) (1.357) (1.16) 1 (16.5 3 0.7) (10.66) (1.324) (1.145)
1
1 (16.5) (1.041) (10.88) (0.722) (1)
2
5 815.39 1 186.65 1 67.46 5 1069.5 kN>m2
Qult ⫽ A⬘qu⬘ ⫽ (1069.5)(1.5615) ⫽ 1670 kN
3.12
■
Bearing Capacity of a Continuous Foundation
Subjected to Eccentric Inclined Loading
The problem of ultimate bearing capacity of a continuous foundation subjected to an
eccentric inclined load was studied by Saran and Agarwal (1991). If a continuous foundation is located at a depth Df below the ground surface and is subjected to an eccentric load
(load eccentricity 5 e) inclined at an angle b to the vertical, the ultimate capacity can be
expressed as
1
(3.67)
Qult 5 B ccrNc(ei) 1 qNq(ei) 1 gBNg(ei) d
2
where Nc(ei), Nq(ei), and Ng(ei) 5 bearing capacity factors
q 5 gDf
The variations of the bearing capacity factors with e>B, fr, and b derived by Saran and
Agarwal are given in Figures 3.26, 3.27, and 3.28.
174 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
100
60
=0
80
50
= 10
40
Nc (ei)
Nc (ei)
60
e/B = 0
40
30
0.1
0.2
20
20
0.3
e/B = 0
0.1
0.2
10
0
0
10
20
30
Soil friction angle, (deg)
0.3
40
(a)
0
0
10
20
30
Soil friction angle, (deg)
40
(b)
40
= 20
30
30
20
e/B = 0
Nc(ei)
Nc (ei)
= 30
0.1
20
e/B = 0
0.2
10
10
0.3
0.1
0.2
0.3
0
0
0
10
20
30
Soil friction angle, (deg)
40
(c)
Figure 3.26 Variation of Nc(ei) with fr, e>B, and b
0
10
20
30
Soil friction angle, (deg)
(d)
40
3.12 Bearing Capacity of a Continuous Foundation Subjected to Eccentric Inclined Load
175
50
80
= 10
40
60
=0
Nq (ei)
Nq (ei)
30
40
e/B = 0
20
e/B = 0
0.2
0.1
0.2
20
0.1
10
0.3
0.3
0
0
0
10
20
30
Soil friction angle, (deg)
0
40
10
20
30
Soil friction angle, (deg)
40
(b)
(a)
30
30
= 20
= 30
e/B = 0
10
Nq (ei)
20
Nq(ei)
20
0.1
10
0.2
e/B = 0
0.3
0.1
0.2
0.3
0
0
0
10
20
30
Soil friction angle, (deg)
0
40
10
20
30
Soil friction angle, (deg)
40
(d)
(c)
Figure 3.27 Variation of Nq(ei) with fr, e>B, and b
Example 3.10
A continuous foundation is shown in Figure 3.29. Estimate the ultimate load, Qult per
unit length of the foundation.
Solution
With cr 5 0, from Eq. (3.67),
1
Qult 5 cqNq(ei) 1 gBNg(ei) d
2
176 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
160
=0
120
= 10
80
e/B = 0
N (ei)
N (ei)
80
40
e/B = 0
40
0.1
0.1
0.2
0.2
0.3
0.3
0
0
0
10
20
30
Soil friction angle, (deg)
10
40
20
30
Soil friction angle, (deg)
40
(b)
(a)
30
60
= 20
20
40
N(ei)
N (ei)
= 30
e/B = 0
0.1
20
e/B = 0
0.1
10
0.2
0.3
0.2
0.3
0
0
20
30
Soil friction angle, (deg)
40
30
35
Soil friction angle, (deg)
40
(d)
(c)
Figure 3.28 Variation of Ng(ei) with fr, e>B, and b
B 5 1.5 m, q 5 Dfg 5 (1) (16) 5 16 kN>m2, e>B 5 0.15>1.5 5 0.1, and b 520°.
From Figures 3.27(c) and 3.28(c), Nq(ei) 5 14.2 and Ng(ei) 5 20. Hence,
Qult 5 (1.5) 3(16) (14.2) 1 ( 12 ) (16) (1.5) (20) 4 5 700.8 kN , m
Problems
177
Qult
20
16 kN/m3
35
c 0
1m
0.15 m
1.5 m
■
Figure 3.29
Problems
3.1
For the following cases, determine the allowable gross vertical load-bearing capacity
of the foundation. Use Terzaghi’s equation and assume general shear failure in soil.
Use FS 5 4.
Part
a.
b.
c.
3.2
3.3
3.4
3.5
3.6
B
1.22 m
2m
3m
Df
0.91 m
1m
2m
f9
25°
30°
30°
g
c9
2
28.75 kN>m
0
0
Foundation type
3
17.29 kN>m
17 kN>m3
16.5 kN>m3
Continuous
Continuous
Square
A square column foundation has to carry a gross allowable load of 1805 kN
(FS 5 3). Given: Df 5 1.5 m, g 5 15.9 kN>m3, fr 5 34°, and cr 5 0. Use
Terzaghi’s equation to determine the size of the foundation (B). Assume general
shear failure.
Use the general bearing capacity equation [Eq. (3.19)] to solve the following:
a. Problem 3.1a
b. Problem 3.1b
c. Problem 3.1c
The applied load on a shallow square foundation makes an angle of 15° with the vertical. Given: B 5 1.83 m, Df 5 0.9 m, g 5 18.08 kN>m3, fr 5 25°, and cr 5
23.96 kN>m2. Use FS 5 4 and determine the gross allowable load. Use Eq. (3.19).
A column foundation (Figure P3.5) is 3 m 3 2 m in plan. Given: Df 5 1.5 m,
fr 5 25°, cr 5 70 kN>m2. Using Eq. (3.19) and FS 5 3, determine the net
allowable load [see Eq. (3.15)] the foundation could carry.
For a square foundation that is B 3 B in plan, Df 5 2 m; vertical gross allowable
load, Qall 5 3330 kN, g 5 16.5 kN>m3; fr 5 30°; cr 5 0; and FS 5 4. Determine
the size of the foundation. Use Eq. (3.19).
178 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
17 kN/m3
1m
1.5 m
Groundwater level
sat 19.5 kN/m3
3m2m
Figure P3.5
3.7
For the design of a shallow foundation, given the following:
Soil:
fr 5 25°
cr 5 50 kN>m2
Unit weight, g 5 17 kN>m3
Modulus of elasticity, Es 5 1020 kN>m2
Poisson’s ratio, ms 5 0.35
Foundation:
L 5 1.5 m
B51m
Df 5 1 m
Calculate the ultimate bearing capacity. Use Eq. (3.27).
3.8 An eccentrically loaded foundation is shown in Figure P3.8. Use FS of 4 and determine the maximum allowable load that the foundation can carry. Use Meyerhof’s effective area method.
3.9 Repeat Problem 3.8 using Prakash and Saran’s method.
3.10 For an eccentrically loaded continuous foundation on sand, given B ⫽ 1.8 m, Df ⫽
0.9 m, e/B ⫽ 0.12 (one-way eccentricity), ␥ ⫽ 16 kN/m3, and ␾r ⫽ 35°. Using the
reduction factor method, estimate the ultimate load per unit length of the foundation.
(Eccentricity
in one direction
only) 0.1 m
0.8 m
Qall
17 kN/m3
c 0
32
1.5 m 1.5 m
Centerline
Figure P3.8
3.11 An eccentrically loaded continuous foundation is shown in Figure P3.11. Determine
the ultimate load Qu per unit length that the foundation can carry. Use the reduction
factor method.
3.12 A square footing is shown in Figure P3.12. Use FS 5 6, and determine the size of
the footing. Use Prakash and Saran theory [Eq. (3.43)].
References
179
Qu
16.5 kN/m3
Groundwater table
0.61 m
1.22 m
sat 18.55 kN/m3
c 0
35
0.61 m
1.52 m
Figure P3.11
450 kN
70 kN•m
16 kN/m3
c 0
30
1.2 m
BB
Water table
sat 19 kN/m3
c 0
30
Figure P3.12
3.13 The shallow foundation shown in Figure 3.19 measures 1.2 m 3 1.8 m and is subjected to a centric load and a moment. If eB 5 0.12 m, eL 5 0.36 m, and the depth of
the foundation is 1 m, determine the allowable load the foundation can carry. Use a
factor of safety of 3. For the soil, we are told that unit weight g 5 17 kN>m3, friction
angle fr 5 35°, and cohesion cr 5 0.
References
BOZOZUK, M. (1972). “Foundation Failure of the Vankleek Hill Tower Site,” Proceedings, Specialty Conference on Performance of Earth and Earth-Supported Structures, Vol. 1, Part 2,
pp. 885–902.
BRAND, E. W., MUKTABHANT, C., and TAECHANTHUMMARAK, A. (1972). “Load Test on Small Foundations in Soft Clay,” Proceedings, Specialty Conference on Performance of Earth and EarthSupported Structures, American Society of Civil Engineers, Vol. 1, Part 2, pp. 903–928.
CAQUOT, A., and KERISEL, J. (1953). “Sur le terme de surface dans le calcul des fondations en milieu
pulverulent,” Proceedings, Third International Conference on Soil Mechanics and Foundation
Engineering, Zürich, Vol. I, pp. 336–337.
DE BEER, E. E. (1970). “Experimental Determination of the Shape Factors and Bearing Capacity
Factors of Sand,” Geotechnique, Vol. 20, No. 4, pp. 387–411.
HANNA, A. M., and MEYERHOF, G. G. (1981). “Experimental Evaluation of Bearing Capacity of Footings
Subjected to Inclined Loads,” Canadian Geotechnical Journal, Vol. 18, No. 4, pp. 599–603.
180 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
HANSEN, J. B. (1970). A Revised and Extended Formula for Bearing Capacity, Bulletin 28, Danish
Geotechnical Institute, Copenhagen.
HIGHTER, W. H., and ANDERS, J. C. (1985). “Dimensioning Footings Subjected to Eccentric Loads,”
Journal of Geotechnical Engineering, American Society of Civil Engineers, Vol. 111, No.
GT5, pp. 659–665.
KUMBHOJKAR, A. S. (1993). “Numerical Evaluation of Terzaghi’s Ng,” Journal of Geotechnical Engineering, American Society of Civil Engineers, Vol. 119, No. 3, pp. 598–607.
MEYERHOF, G. G. (1953). “The Bearing Capacity of Foundations Under Eccentric and Inclined
Loads,” Proceedings, Third International Conference on Soil Mechanics and Foundation Engineering, Zürich, Vol. 1, pp. 440–445.
MEYERHOF, G. G. (1963). “Some Recent Research on the Bearing Capacity of Foundations,” Canadian Geotechnical Journal, Vol. 1, No. 1, pp. 16–26.
PRAKASH, S., and SARAN, S. (1971). “Bearing Capacity of Eccentrically Loaded Footings,” Journal
of the Soil Mechanics and Foundations Division, ASCE, Vol. 97, No. SM1, pp. 95–117.
PRANDTL, L. (1921). “Über die Eindringungsfestigkeit (Härte) plastischer Baustoffe und die Festigkeit von Schneiden,” Zeitschrift für angewandte Mathematik und Mechanik, Vol. 1, No. 1,
pp. 15–20.
PURKAYASTHA, R.D., and CHAR, R. A. N. (1977). “Stability Analysis of Eccentrically Loaded Footings,” Journal of Geotechnical Engineering Div., ASCE, Vol. 103, No. 6, pp. 647–651.
REISSNER, H. (1924). “Zum Erddruckproblem,” Proceedings, First International Congress of Applied
Mechanics, Delft, pp. 295–311.
SARAN, S., and AGARWAL, R. B. (1991). “Bearing Capacity of Eccentrically Obliquely Loaded Footing,” Journal of Geotechnical Engineering, ASCE, Vol. 117, No. 11, pp. 1669–1690.
TERZAGHI, K. (1943). Theoretical Soil Mechanics, Wiley, New York.
VESIC, A. S. (1963). “Bearing Capacity of Deep Foundations in Sand,” Highway Research Record
No. 39, National Academy of Sciences, pp. 112–153.
VESIC, A. S. (1973). “Analysis of Ultimate Loads of Shallow Foundations,” Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 99, No. SM1,
pp. 45–73.