Post on 22-Mar-2018
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Foundations with Df equal to 3 to 4 times the width may be defined as shallow foundations
TWO MAIN CHARACTERISTICS β¦ Safe against overall shear failure β¦ Cannot undergo excessive displacement, or
settlement
ULTIMATE BEARING CAPACITY
the load per unit area of the foundation at which shear failure in soil occurs
GENERAL SHEAR FAILURE
Long rectangular model footing of width B at the surface of a
dense sand.
The triangular wedge-shaped zone of soil marked I is pushed down and in turn presses the zones marked II and III sideways. The soil on the both sides of the foundation will bulge out and the slip surface will extend to the ground surface.
LOCAL SHEAR FAILURE
Medium dense sand or clayey soil of medium compaction
Movement of the foundation will be accompanied by sudden jerks
The triangular wedge-shaped zone of soil marked I is pushed down but unlike in general shear failure, the slip surface end somewhere inside the soil.
PUNCHING SHEAR FAILURE
Fairly loose soil
Soil will not extend to the ground surface
GENERAL SHEAR FAILURE Applies to dense granular soil and to firmer saturated
cohesive soils subject to undrained loading (the U-U and C-U shearing conditions apply)
PUNCHING SHEAR FAILURE Applies to compressible soils, such as sands having low-to-
medium relative density, and for cohesive soils subject to slow loading (the C-D shearing conditions apply)
Vesic1963
Laboratory load-bearing tests on circular and rectangular plates supported by a sand at various relative densities of compaction, Dr.
Drβ₯ about 70%
(general shear failure)
Vesic 1973
Relationship for the mode of bearing capacity failure of foundations resting on sands.
Vesic1973
Dr = relative density of sand Df = depth of foundation measured from the ground surface
B* = 2π΅πΏ
π΅+πΏ
where B= width of foundation L= length of foundation
General range of S/B with the relative density of compaction of sand.
General shear failure β ultimate load may occur at settlements of 4 to 10% of B.
Local or punching shear failure β ultimate load may occur at settlements of 15 to 25% of B.
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 Eq.(3.3)
where πβ²= cohesion
πΎ = unit weight of soil
q = πΎDf
ππΆ, ππ, ππΎ= bearing capacity factors that are nondimensional and are functions only of the soil friction angle ΙΈβ
Eq.(3.4)
Eq.(3.5)
Eq.(3.6)
where KpπΎ= passive pressure coefficient
Square foundation
ππ’ = 1.3πβ²ππΆ + πππ + 0.4πΎπ΅ππΎ Eq.(3.7)
Circular foundation
ππ’ = 1.3πβ²ππΆ + πππ + 0.3πΎπ΅ππΎ Eq.(3.8)
LOCAL SHEAR FAILURE Strip foundation
ππ’ =2
3πβ²πβ²πΆ + ππβ²π +
1
2πΎπ΅πβ²πΎ Eq.(3.9)
Square foundation ππ’ = 0.867πβ²πβ²πΆ + ππβ²π + 0.4πΎπ΅πβ²πΎ Eq.(3.10) Circular foundation ππ’ = 0.867πβ²πβ²πΆ + ππβ²π + 0.3πΎπ΅πβ²πΎ Eq.(3.11)
ΙΈβ²
= tanβ1(2
3tan ΙΈ
β²)
ππππ =ππ’
πΉπ Eq.(3.12)
πππ‘ π π‘πππ π πππππππ π ππ π πππ =πππ‘ π’ππ‘ππππ‘π πππππππ πππππππ‘π¦
πΉπ Eq.(3.13)
ππππ‘(π’) = ππ’ β π Eq.(3.14)
where
ππππ‘(π’) = net ultimate bearing capacity
π = πΎDf
So,
ππππ(πππ‘) =ππ’βπ
πΉπ Eq.(3.15)
The factor of safety as defined by Eq.(3.15) should be at least 3 in all cases.
A square foundation is 2mx2m in plan. The soil supporting the foundation has a friction angle of ΙΈβ = 25Λ and πβ²= 20kN/m2. The unit weight of soil, πΎ, is 16.5kN/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 1.5m and that general shear failure occurs in the soil.
Solution
From Eq.(3.7) ππ’ = 1.3πβ²ππΆ + πππ + 0.4πΎπ΅ππΎ
From Table 3.1, for ΙΈβ = 25Λ, ππΆ = 25.13
ππ = 12.72
ππΎ = 8.34
Thus,
ππ’ = 1.3 20 (25.13) + (1.5π₯16.5)(12.72) + 0.4(16.5)(2)(8.34) = 1078.29kN/m2
So, the allowable load per unit area of the foundation is
ππππ =ππ’
πΉπ=
1078.29
3= 359.5KN/m2
Thus, the total allowable gross load is
Q= ππππ(B2) = 359.5(2x2) = 1438KN
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 0β€D1β€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
qcu
2
1'
If 0β€dβ€B,
q=Ξ³Df
Ξ³ in the last term becomes
* The preceding modifications are based on the assumption that there is no seepage force in the soil.
back
BNqNNcq
qcu
2
1'
)'(' B
d
If dβ₯B,
*The water will have
no effect on the
ultimate bearing
capacity.
back
BNqNNcq
qcu
2
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.
Wherein:
cβ= cohesion
q= effective stress at the level of the bottom of the foundation
Ξ³= unit weight of soil
B= width of foundation (or diameter for circular foundation),
Fcs, Fqs, FΞ³s = shape factors
Fcd, Fqd, FΞ³d = depth factors
Fci, Fqi, FΞ³i = load inclination factors
Nc, Nq, NΞ³ = bearing capacity factors
idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu
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)
idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu
2
1'
Ex3.3
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
qd
qdcd
d
qd
f
cd
f
F
B
DF
Nc
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
qd
qdcd
d
qd
f
cd
f
F
radiansB
DF
Nc
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
mD
FS
mkN
mkNc
f5.1
3
/5.16
/20'
25'
3
2
Square foundation 2mx2m
General shear failure Reqd: Allowable gross load
idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu
2
1'
87.1025tan)166.10(2'tan)1(2
72.2025cot)166.10('cot)1(
66.10)2/2545(tan
)2/'45(tan
25tan2
'tan2
q
qc
q
NN
NN
e
eN
Solution
Solution: Bearing Capacity Factors: *table 3.3 can also be used Load inclination Factors Since load is vertical, Fci, Fqi, Fyi=1
Shape Factors
Fcs=1+ (2/2)(10.66/20.72) = 1.514
Fqs=1 + (2/2)tan25 = 1.466
FΞ³s =1-0.4(2/2)= 0.6
Depth Factors (Df/B = 1.5/2 = 0.75)
L
BF
L
BF
N
N
L
BF
s
qs
c
q
cs
4.01
'tan1
1
1
233.12
5.1)25sin1)(25tan2(1)'sin1('tan21
257.125tan72.20
233.11233.1
'tan
1
'
1
22
d
f
qd
qd
qdcd
f
F
B
DF
Nc
FFF
For
B
D
kNxQ
mkNFS
mkNq
q
FFFBNFFFqNFFFNcq
u
all
u
u
idsqiqdqsqcicdcscu
8.1830)22(7.457
/7.4573
2.1373
/2.13737.1079.4766.788
)116.088.1025.165.0(
)1233.1466.1666.105.165.1(
)1257.1514.172.2020(
2
1'
2
2
sizefootingqd
FS
kNQall
mD
mD
mkN
mkN
BxBsquare
f
sat
:Re
3
2.667
61.0
22.1
34'
/55.18
/5.16
;
1
3
3
idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu
2
1'
dsqdqsqFFBNFFqNqu
'
2
1
3
1
/2.667 2
22
FS
mkNBB
u
all
all
all
dsqdqsqFFBNFFqN
'
2
1
Since there is no cohesion, Becomes Eq1 Eq2 Bearing Capacity Factors Table 3.3 For Οβ = 34
Nq = 29.44 Ny= 41.06
Case I q=D1Ξ³+D2(Ξ³sat-Ξ³w) q= (0.61)(16.5) + (0.61)(18.55-9.81)= 15.4KN/m2
1
05.11
4)34sin1(34tan21)'sin1('tan21
6.04.014.01
67.134tan1'tan1
22
d
f
qd
s
qs
F
BBB
DF
L
BF
L
BF
mBerrorandtrialBy
BBB
FS
B
EqnsCombine
B
B
Bxx
FS
mkNxq
FFBNFFqNFS
qqEqn
u
all
all
all
u
all
dsqdqsq
u
all
123.1,
89.35265
38.2522.667
:2&1
89.35B
265252.38
)1)(6.0)(06.41()81.955.18(2
1
05.1167.144.294.15
3
1
/4.15)81.955.18(61.05.1661.0
)'2
1(
3
1.2
2
2
2
Soil Compressibility Factors
Eqn 3.19
Is modified to
Eqn 3.27
Wherein Fcc, Fqc and FΞ³c are compressibility factors
idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu
2
1'
cdsqcqdqsqcccdcscFFFBNFFFqNFFFNcqu
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 = shear modulus of the soil q = effective overburden pressure at a depth of Df + B/2
'tan'' qc
GsIr
2
'45cot45.03.3exp
2
1)(
L
BcrIr
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 then However if Ir< Ir(cr), then Figure 3.12 shows the variation of
)(crIrIr
'sin1
)2)(log'sin07.3('tan6.04.4exp
Ir
L
BFqccF
'tan
1
log60.012.032.0
,0
.'/))30.3((
Nq
FqcFqcFcc
IrL
BFcc
For
andIrwseeEqFqccF
1 cFFqcFcc
Given: β¦ Shallow foundation β¦ B= 0.6m β¦ L= 1.2m β¦ Df=0.6m β¦ Soil characteristics:
Ο'=25 cβ=48Kn/m2
Ξ³=18KN/m3
Modulus of elasticity, Es= 620 KN/m2
Poissonβs ratio, ΞΌs=0.3
Required: β¦ Calculate the ultimate
bearing capacity.
Solution: Rigidity Index
)'tan'')(1(2
)1(2
'tan''
qc
EsIr
EsGs
qc
GsIr
s
s
29.4)25tan2.168.4)(3.01(2
620
/2.162
6.06.018
2'
2
Ir
mkNB
Dfq
Critical Rigidity Index
41.622
2545cot
2.1
6.045.03.3exp
2
1)(
2
'45cot45.03.3exp
2
1)(
crIr
L
BcrIr
Since Ir(cr)> Ir, use Eqs/ 3.30 and 3.32
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
Fcc
thereforeseeTableNcFor
Nc
FqcFqcFcc
and
xFqccF
Ir
L
BFqccF
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
cdsqcqdqsqcccdcscFFFBNFFFqNFFFNcqu
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
qd
qdcd
F
B
DF
Nc
FFF
cdsqcqdqsqcccdcscFFFBNFFFqNFFFNcqu
2
1'
232.549347.018.088.106.0182
1
]347.0311.1233.166.10)186.0[(
)279.0343.1257.172.2048(
m
kN
qu
When foundations are subjected to moments in addition to the vertical load, the distribution of pressure on the soil is not uniform.
Where Q is the total vertical load and M is the moment on the foundation.
LB
M
BL
LB
M
BL
2min
2max
6
6
B
e
BL
B
e
BL
Q
Me
LB
M
BL
LB
M
BL
61
61
6
6
min
max
2min
2max
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 = ultimate load-carrying capacity
Q
QFS
ult