Post on 30-Apr-2018
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
ππ’ = πβ²ππΆ + πππ +1
2πΎπ΅ππΎ
where,
πβ² is the cohesion
is the unit weight of soil
q is the equivalent surcharge load equal to πΎDf
ππΆ, ππ, ππΎ are bearing capacity factors that are nondimensional and are
functions only of the soil friction angle ΙΈβ.
where,
KpπΎ is the passive pressure coefficient
Modified for:
Square Foundation ππ’ = 1.3πβ²ππΆ + πππ + 0.4πΎπ΅ππΎ
Circular Foundation ππ’ = 1.3πβ²ππΆ + πππ + 0.3πΎπ΅ππΎ
LOCAL SHEAR FAILURE
Strip Foundation
ππ’ =2
3πβ²πβ²πΆ + ππβ²π +
1
2πΎπ΅πβ²πΎ
Square Foundation ππ’ = 0.867πβ²πβ²πΆ + ππβ²π + 0.4πΎπ΅πβ²πΎ
Circular Foundation ππ’ = 0.867πβ²πβ²πΆ + ππβ²π + 0.3πΎπ΅πβ²πΎ
ΙΈβ²
= tanβ1(2
3tanΙΈ
β²)
ππππ =ππ’
πΉπ
πππ‘ π π‘πππ π πππππππ π ππ π πππ =πππ‘ π’ππ‘ππππ‘π πππππππ πππππππ‘π¦
πΉπ
ππππ‘(π’) = ππ’ β π
where,
ππππ‘(π’) is the net ultimate bearing capacity
π = πΎπ·π
So,
ππππ(πππ‘) =ππ’ β π
πΉπ
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 β¦ Case I
β¦ Case II
β¦ 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 β€ D1 β€ Df,
q = D1Ξ³ + D2(Ξ³sat - Ξ³w)
where,
Ξ³sat = sat. unit wt. of soil
Ξ³w = unit wt. of water
Ξ³ in Β½Ξ³BNΞ³ becomes Ξ³β
where Ξ³β= Ξ³sat - Ξ³w
back
q = Ξ³Df
BNqNNcq qcu2
1'
If 0 β€ d β€ B,
q = Ξ³Df
Ξ³ 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 β₯ 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,
cβ is the cohesion
q is the effective stress at the level of the bottom of the foundation
Ξ³ is the unit weight of soil
B is the width of foundation (or diameter for circular foundation)
Fcs, Fqs, FΞ³s are shape factors
Fcd, Fqd, FΞ³d are depth factors
Fci, Fqi, FΞ³i are load inclination factors
Nc, Nq, NΞ³ are bearing capacity factors
idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2
1'
Ξ± = 45 + Οβ/2
Nq = tan2 (45 + Οβ/2) eΟtan Οβ
β¦ Reissner (1924)
Nc = (Nq β 1) cot Οβ
β¦ Prandtl (1921)
NΞ³ = 2(Nq + 1) tan Οβ
β¦ 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 FΞ³c 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 FΞ³c, 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 πΌπ β₯ πΌπ(ππ) then πΉππ = πΉππ = πΉπΎπ = 1.
However if πΌπ < πΌπ(ππ), then
Figure 3.12 shows the variation of πΉπΎπ = πΉππ with β β² and πΌπ .
'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 β > 0,
For β = 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: Οβ = 25Β°, cβ = 48 kN/mΒ², Ξ³ = 18 kN/mΒ³, modulus of elasticity (Es) = 620 kN/mΒ², and Poissonβs ratio (ΞΌs) =
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 πΌπ(ππ) > πΌπ,
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 Οβ = 25Β°, ππ = 20.72, ππ = 10.66, πππ ππΎ = 10.88.
ππ’ = 48 20.72 1.257 1.343 0.279 + 0.6π₯18 10.66 1.233 1.311 (0.347) + 0.5 18 0.6 10.88 0.8 1 0.347
ππ = πππ. ππ ππ΅/ππ
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.
Bβ = effective width = B β 2e
Lβ = effective length = L
Step 2: Use the general bearing capacity equation.
To evaluate the shape factors, use the effective dimensions (Bβ,
Lβ) 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
ππ’ππ‘ = ππ’(π΅β²π₯πΏβ²) = ππ’π΄β² (Aβ is the effective area)
Step 4: The factor of safety against bearing capacity failure is
πΉπ =ππ’ππ‘
π
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: Οβ = 40Β°, cβ = 0, and Ξ³ = 16.5 kN/mΒ³. If the load
eccentricity is 0.2 m, determine the ultimate load per unit length of the
foundation. (ππ’ππ‘ = 5,260 ππ)
Note that,
ππ΅ =ππ¦
ππ’ππ‘ πππ ππΏ =
ππ₯
ππ’ππ‘
ππ’ππ‘ = ππ’π΄β² = ππ’(π΅β²π₯πΏβ²)
In determining the effective are Aβ, effective width Bβ, and effective
length Lβ, five possible cases may arise (Highter and Anders, 1985).
idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2
1'
CASE 1: ππΏ
πΏβ₯
1
6 πππ
ππ΅
π΅β₯
1
6
π΄β² =1
2(π΅1)(πΏ1)
π΅1 = π΅ 1.5 β3ππ΅
π΅
πΏ1 = πΏ 1.5 β3ππΏ
πΏ
The effective length Lβ is the larger of
the two dimensions π΅1 and πΏ1. So the effective width is Bβ=Aβ/Lβ.
CASE 2: ππΏ
πΏ<
1
2 πππ 0 <
ππ΅
π΅<
1
6
π΄β² =1
2πΏ1 + πΏ2 π΅
π΅β² =π΄β²
πΏ1 ππ πΏ2 (π€βππβππ£ππ ππ ππππππ)
πΏβ² = πΏ1ππ πΏ2 (π€βππβππ£ππ ππ ππππππ)
The magnitudes of πΏ1 πππ πΏ2 can be
determined from Figure 3.21b.
CASE 2: ππΏ
πΏ<
1
2 πππ 0 <
ππ΅
π΅<
1
6
CASE 3: ππΏ
πΏ<
1
6 πππ 0 <
ππ΅
π΅<
1
2
π΄β² =1
2π΅1 + π΅2 πΏ
π΅β² =π΄β²
πΏ
πΏβ² = πΏ
The magnitudes of π΅1 πππ π΅2 can be
determined from Figure 3.22b.
CASE 3: ππΏ
πΏ<
1
6 πππ 0 <
ππ΅
π΅<
1
2
CASE 4: ππΏ
πΏ<
1
6 πππ
ππ΅
π΅<
1
6
π΄β² = πΏ2π΅ +1
2π΅ + π΅2 (πΏ β πΏ2)
π΅β² =π΄β²
πΏ
πΏβ² = πΏ
The ratio π΅2/π΅ πππ π‘βπ’π π΅2 can be determined by using the ππΏ/πΏ curves that slope upward. Similarly, the ratio πΏ2/πΏ πππ π‘βπ’π πΏ2 can be determined by using the ππΏ/πΏ curves that slope downward.
CASE 4: ππΏ
πΏ<
1
6 πππ
ππ΅
π΅<
1
6
CASE 5: Circular Foundation
πΏβ² =π΄β²
π΅β²
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 ππΏ = 0.3 m and ππ΅= 0.15 m. The known soil characteristics are as follows: Οβ = 30Β°, cβ = 0, and Ξ³ = 18
kN/mΒ³. Assume two-way eccentricity, and determine the ultimate load.
(ππ’ππ‘ = 606 ππ)
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 ππΏ = 0.18 m and ππ΅= 0.12 m. The known soil characteristics are as follows: Οβ = 25Β°, cβ = 25 kN/mΒ², and
Ξ³ = 16.5 kN/mΒ³. Assume two-way eccentricity, and determine the
ultimate load. (ππ’ππ‘ = 1,670 ππ)