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Foundation Analysis
Part 2
Bearing capacity failure in soil under a rough rigid continuous (strip)
foundation
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
Continuous or Strip Foundation
𝑞𝑢 = 𝑐′𝑁𝐶 + 𝑞𝑁𝑞 +1
2𝛾𝐵𝑁𝛾
where,
𝑐′ is the cohesion
is the unit weight of soil
q is the equivalent surcharge load equal to 𝛾Df
𝑁𝐶, 𝑁𝑞, 𝑁𝛾 are bearing capacity factors that are nondimensional and are
functions only of the soil friction angle ɸ’.
where,
Kp𝛾 is the passive pressure coefficient
Modified for:
Square Foundation 𝑞𝑢 = 1.3𝑐′𝑁𝐶 + 𝑞𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾
Circular Foundation 𝑞𝑢 = 1.3𝑐′𝑁𝐶 + 𝑞𝑁𝑞 + 0.3𝛾𝐵𝑁𝛾
LOCAL SHEAR FAILURE
Strip Foundation
𝑞𝑢 =2
3𝑐′𝑁′𝐶 + 𝑞𝑁′𝑞 +
1
2𝛾𝐵𝑁′𝛾
Square Foundation 𝑞𝑢 = 0.867𝑐′𝑁′𝐶 + 𝑞𝑁′𝑞 + 0.4𝛾𝐵𝑁′𝛾
Circular Foundation 𝑞𝑢 = 0.867𝑐′𝑁′𝐶 + 𝑞𝑁′𝑞 + 0.3𝛾𝐵𝑁′𝛾
ɸ′
= tan−1(2
3tanɸ
′)
𝑞𝑎𝑙𝑙 =𝑞𝑢
𝐹𝑆
𝑁𝑒𝑡 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑛 𝑠𝑜𝑖𝑙 =𝑛𝑒𝑡 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦
𝐹𝑆
𝑞𝑛𝑒𝑡(𝑢) = 𝑞𝑢 − 𝑞
where,
𝑞𝑛𝑒𝑡(𝑢) is the net ultimate bearing capacity
𝑞 = 𝛾𝐷𝑓
So,
𝑞𝑎𝑙𝑙(𝑛𝑒𝑡) =𝑞𝑢 − 𝑞
𝐹𝑆
The factor of safety should be at least 3 in all cases.
The bearing capacity equation is modified when the
water table is in the proximity of the foundation.
Bearing Capacity Equation
Modified Bearing Capacity Equation ◦ Case I
◦ Case II
◦ Case III
next
back
BNqNNcq
BNqNNcq
BNqNNcq
qcu
qcu
qcu
3.0'3.1
4.0'3.1
2
1'
GENERAL SHEAR FAILURE
(Continuous or Strip Foundation)
(Square Foundation)
(Circular Foundation)
If 0 ≤ D1 ≤ Df,
q = D1γ + D2(γsat - γw)
where,
γsat = sat. unit wt. of soil
γw = unit wt. of water
γ in ½γBNγ becomes γ’
where γ’= γsat - γw
back
q = γDf
BNqNNcq qcu2
1'
If 0 ≤ d ≤ B,
q = γDf
γ in the last term is
* The preceding modifications are based on the assumption that there is no seepage force in the soil.
back
BNqNNcq qcu2
1'
)'('_
B
d
If d ≥ B,
*The water will have no effect on the ultimate bearing capacity.
back
BNqNNcq qcu2
1'
SHAPE: The bearing capacity eqns do not address the case of rectangular foundations (0 < B/L < 1). Wherein L > B.
DEPTH: The eqns also do not take into account the shearing resistance along the failure surface in soil above the bottom of the foundation.
LOAD INCLINATION: The load on the foundation may be inclined.
where,
c’ is the cohesion
q is the effective stress at the level of the bottom of the foundation
γ is the unit weight of soil
B is the width of foundation (or diameter for circular foundation)
Fcs, Fqs, Fγs are shape factors
Fcd, Fqd, Fγd are depth factors
Fci, Fqi, Fγi are load inclination factors
Nc, Nq, Nγ are bearing capacity factors
idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2
1'
α = 45 + ϕ’/2
Nq = tan2 (45 + ϕ’/2) eπtan ϕ’
◦ Reissner (1924)
Nc = (Nq – 1) cot ϕ’
◦ Prandtl (1921)
Nγ = 2(Nq + 1) tan ϕ’
◦ Caquot and Kerisel (1953), Vesic (1973)
idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2
1'
Ex 3.3
Ex 3.4
Shape Factors
Reference: DeBeer (1970)
L
BF
L
BF
N
N
L
BF
s
qs
c
q
cs
4.01
'tan1
1
Depth Factors
Reference: Hansen (1970)
1
)'sin1('tan21
'tan
1
'
1
1
4.01
1
2
d
f
qd
c
qd
qdcd
d
qd
f
cd
f
F
B
DF
N
FFF
For
F
F
B
DF
For
B
D
1
tan)'sin1('tan21
'tan
1
'
1
1
tan4.01
1
12
1
d
f
qd
c
qd
qdcd
d
qd
f
cd
f
F
radiansB
DF
N
FFF
For
F
F
radiansB
DF
For
B
D
Inclination Factors
Reference: Meyerhof (1963);
Hanna and Meyerhof (1981)
'
1
901
2
i
qici
F
FF
inclination of the load
on the foundation with
respect to the vertical
General Bearing Capacity Equation
is modified to (Vesic, 1973)
where Fcc, Fqc and Fγc are 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 Fγc, the
following steps should be taken:
idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2
1'
cdsqcqdqsqcccdcscu FFFBNFFFqNFFFNcq 2
1'
Step 1. Calculate the rigidity index, Ir, of the soil at a depth
approximately B/2 below the bottom of the foundation, or
where,
Gs is the shear modulus of the soil
q' is the effective overburden pressure at a depth of Df + B/2
'tan'' qc
GI s
r
2
'45cot45.03.3exp
2
1)(
L
BI crr
Step 2. The critical rigidity index, Ir(cr), can be expressed as
The variations of Ir(cr) with B/L are given in Table 3.6.
Step 3. If 𝐼𝑟 ≥ 𝐼𝑟(𝑐𝑟) then 𝐹𝑐𝑐 = 𝐹𝑞𝑐 = 𝐹𝛾𝑐 = 1.
However if 𝐼𝑟 < 𝐼𝑟(𝑐𝑟), then
Figure 3.12 shows the variation of 𝐹𝛾𝑐 = 𝐹𝑞𝑐 with ∅′ and 𝐼𝑟 .
'sin1
)2)(log'sin07.3('tan6.04.4exp
rqcc
I
L
BFF
rcc IL
BF log60.012.032.0
'tan
1
c
qc
qcccN
FFF
For ∅ > 0,
For ∅ = 0,
1. For a shallow foundation, B = 0.6 m, L = 1.2 m, and Df = 0.6 m. The known soil characteristics are as follows: ϕ’ = 25°, c’ = 48 kN/m², γ = 18 kN/m³, modulus of elasticity (Es) = 620 kN/m², and Poisson’s ratio (μs) =
0.3. Calculate the ultimate bearing capacity.
Solution:
Rigidity Index
)'tan'')(1(2
)1(2
'tan''
qc
EI
EG
qc
GI
s
sr
s
ss
sr
29.4)25tan2.1648)(3.01(2
620
/2.162
6.06.018
2' 2
rI
mkNB
Dfq
Critical Rigidity Index
41.622
2545cot
2.1
6.045.03.3exp
2
1
2
'45cot45.03.3exp
2
1
)(
)(
crr
crr
I
L
BI
Since 𝐼𝑟(𝑐𝑟) > 𝐼𝑟,
279.025tan72.20
347.01347.0
,);3.3(72.20,25'
'tan
1
347.025sin1
))29.42)(log(25sin07.3(25tan
2.1
6.06.04.4exp
'sin1
)2)(log'sin07.3('tan6.04.4exp
cc
c
c
qc
qccc
qcc
rqcc
F
thereforeseeTableNFor
N
FFF
and
xFF
I
L
BFF
Shape Factors Depth Factors
Table 3.3
8.02.1
6.04.014.01
233.125tan2.1
6.01'tan1
257.172.20
66.10
2.1
6.011
L
BF
L
BF
N
N
L
BF
s
qs
c
q
cs
cdsqcqdqsqcccdcscu FFFBNFFFqNFFFNcq 2
1'
1
311.16.0
6.0)25sin1(25tan21
)'sin1('tan21
343.125tan72.20
311.11311.1
'tan
1
2
2
d
f
qd
c
qd
qdcd
F
B
DF
N
FFF
From Table 3.3, for ϕ’ = 25°, 𝑁𝑐 = 20.72, 𝑁𝑞 = 10.66, 𝑎𝑛𝑑 𝑁𝛾 = 10.88.
𝑞𝑢 = 48 20.72 1.257 1.343 0.279 + 0.6𝑥18 10.66 1.233 1.311 (0.347) + 0.5 18 0.6 10.88 0.8 1 0.347
𝒒𝒖 = 𝟓𝟒𝟗. 𝟑𝟐 𝒌𝑵/𝒎𝟐
When foundations are
subjected to moments in
addition to the vertical load, the distribution of
pressure on the soil is not
uniform. The nominal
distribution of pressure is,
where Q is the total vertical load and M
is the moment on the foundation.
LB
M
BL
LB
M
BL
2
2
6
6
min
max
B
e
BL
B
e
BL
Q
Me
LB
M
BL
LB
M
BL
61
61
6
6
min
max
min
max
2
2
When e= B/6, qmin=0
When e> B/6 , qmin <0
Which means tension will develop.
Soil cannot take any tension
There will be a separation of the foundation and the soil underlying it.
qmax = 4Q/ 3L(B-2e)
The exact distribution of failure is difficult to estimate.
The factor of safety for such type of loading against
bearing capacity failure can be evaluated as
where Qult is the ultimate load-carrying capacity
Q
QFS ult
Effective Area Method (Meyerhoff, 1953)
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.
B’ = effective width = B – 2e
L’ = effective length = L
Step 2: Use the general bearing capacity equation.
To evaluate the shape factors, use the effective dimensions (B’,
L’) instead of B and L. To determine the depth factors, use B and L.
Step 3: The total ultimate load that the foundation can sustain is
𝑄𝑢𝑙𝑡 = 𝑞𝑢(𝐵′𝑥𝐿′) = 𝑞𝑢𝐴′ (A’ is the effective area)
Step 4: The factor of safety against bearing capacity failure is
𝐹𝑆 =𝑄𝑢𝑙𝑡
𝑄
2. A continuous foundation, supported by sand, has a width of 2 m and the depth of foundation is 1.5 m. The known soil characteristics are as follows: ϕ’ = 40°, c’ = 0, and γ = 16.5 kN/m³. If the load
eccentricity is 0.2 m, determine the ultimate load per unit length of the
foundation. (𝑄𝑢𝑙𝑡 = 5,260 𝑘𝑁)
Note that,
𝑒𝐵 =𝑀𝑦
𝑄𝑢𝑙𝑡 𝑎𝑛𝑑 𝑒𝐿 =
𝑀𝑥
𝑄𝑢𝑙𝑡
𝑄𝑢𝑙𝑡 = 𝑞𝑢𝐴′ = 𝑞𝑢(𝐵′𝑥𝐿′)
In determining the effective are A’, effective width B’, and effective
length L’, five possible cases may arise (Highter and Anders, 1985).
idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2
1'
CASE 1: 𝑒𝐿
𝐿≥
1
6 𝑎𝑛𝑑
𝑒𝐵
𝐵≥
1
6
𝐴′ =1
2(𝐵1)(𝐿1)
𝐵1 = 𝐵 1.5 −3𝑒𝐵
𝐵
𝐿1 = 𝐿 1.5 −3𝑒𝐿
𝐿
The effective length L’ is the larger of
the two dimensions 𝐵1 and 𝐿1. So the effective width is B’=A’/L’.
CASE 2: 𝑒𝐿
𝐿<
1
2 𝑎𝑛𝑑 0 <
𝑒𝐵
𝐵<
1
6
𝐴′ =1
2𝐿1 + 𝐿2 𝐵
𝐵′ =𝐴′
𝐿1 𝑜𝑟 𝐿2 (𝑤ℎ𝑖𝑐ℎ𝑒𝑣𝑒𝑟 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟)
𝐿′ = 𝐿1𝑜𝑟 𝐿2 (𝑤ℎ𝑖𝑐ℎ𝑒𝑣𝑒𝑟 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟)
The magnitudes of 𝐿1 𝑎𝑛𝑑 𝐿2 can be
determined from Figure 3.21b.
CASE 2: 𝑒𝐿
𝐿<
1
2 𝑎𝑛𝑑 0 <
𝑒𝐵
𝐵<
1
6
CASE 3: 𝑒𝐿
𝐿<
1
6 𝑎𝑛𝑑 0 <
𝑒𝐵
𝐵<
1
2
𝐴′ =1
2𝐵1 + 𝐵2 𝐿
𝐵′ =𝐴′
𝐿
𝐿′ = 𝐿
The magnitudes of 𝐵1 𝑎𝑛𝑑 𝐵2 can be
determined from Figure 3.22b.
CASE 3: 𝑒𝐿
𝐿<
1
6 𝑎𝑛𝑑 0 <
𝑒𝐵
𝐵<
1
2
CASE 4: 𝑒𝐿
𝐿<
1
6 𝑎𝑛𝑑
𝑒𝐵
𝐵<
1
6
𝐴′ = 𝐿2𝐵 +1
2𝐵 + 𝐵2 (𝐿 − 𝐿2)
𝐵′ =𝐴′
𝐿
𝐿′ = 𝐿
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.
CASE 4: 𝑒𝐿
𝐿<
1
6 𝑎𝑛𝑑
𝑒𝐵
𝐵<
1
6
CASE 5: Circular Foundation
𝐿′ =𝐴′
𝐵′
3. A square foundation (1.5 m x 1.5 m), supported by sand, has its bottom 0.7 m below the ground level, with 𝑒𝐿 = 0.3 m and 𝑒𝐵= 0.15 m. The known soil characteristics are as follows: ϕ’ = 30°, c’ = 0, and γ = 18
kN/m³. Assume two-way eccentricity, and determine the ultimate load.
(𝑄𝑢𝑙𝑡 = 606 𝑘𝑁)
4. A square foundation (1.5 m x 1.5 m), supported by sand, has its bottom 0.7 m below the ground level, with 𝑒𝐿 = 0.18 m and 𝑒𝐵= 0.12 m. The known soil characteristics are as follows: ϕ’ = 25°, c’ = 25 kN/m², and
γ = 16.5 kN/m³. Assume two-way eccentricity, and determine the
ultimate load. (𝑄𝑢𝑙𝑡 = 1,670 𝑘𝑁)