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Session 5 6
BEARING CAPACITY OF SHALLOWFOUNDATION
Course : S0484/Foundation Engineering
Year : 2007
Version : 1/0
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SHALLOW FOUNDATION
Topic: General
Terzaghi Model
Meyerhoff Model Brinch Hansen Model
Influence of multi layer soil
Influence of ground water elevation Shallow Foundation Bearing by N-SPT
value
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TYPES OF SHALLOW FOUNDATION
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TYPES OF SHALLOW FOUNDATION
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TERZAGHI MODEL
Assumptions: Subsoil below foundation structure is
homogenous
Shallow foundationDf
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TERZAGHI MODEL
FAILURE ZONES:1. ACD : TRIANGULAR ZONES2. ADF & CDE : RADIAL SHEAR ZONES3. AFH & CEG : RANKINE PASSIVE ZONES
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STRIP FOUNDATION
qult = c.Nc + q.Nq + 0.5..B.N SQUARE FOUNDATION
qult = 1.3.c.Nc + q.Nq + 0.4..B.N
CIRCULAR FOUNDATIONqult = 1.3.c.Nc + q.Nq + 0.3..B.N
( )
( )
tan1cos2
1
24cos.2
1
24cos.2
cot
2
2
tan2/4/32
2
tan2/4/32
=
+
=
+
=
pyK
N
eNq
eNc
Where:
c = cohesion of soil
q = . Df ; Df = the thickness of foundationembedded on subsoil
= unit weight of soil
B = foundation width
Nc, Nq, N = bearing capacity factors
TERZAGHI MODEL(GENERAL FAILURE)
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BEARING CAPACITY FACTORS
GENERAL
FAILURE
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BEARING CAPACITY FACTORS
GENERALFAILURE
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TERZAGHI MODEL(LOCAL FAILURE)
STRIP FOUNDATION
qult = 2/3.c.Nc + q.Nq + 0.5..B.N SQUARE FOUNDATION
qult = 0.867.c.Nc + q.Nq + 0.4..B.N
CIRCULAR FOUNDATION
qult = 0.867.c.Nc + q.Nq + 0.3..B.N
( )
( )
'tan1'cos2
1
2
'
4cos.2
1
2
'
4cos.2
'cot
2
2
'tan2/'4/32
2
'tan2/'4/32
=
+
=
+
=
pyKN
eNq
eNc
= tan-1 (2/3. tan)
Where:
c = cohesion of soil
q = . Df ; Df = the thickness of foundationembedded on subsoil
= unit weight of soil
B = foundation width
Nc, Nq, N = bearing capacity factors
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BEARING CAPACITY FACTORS
LOCAL FAILURE
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BEARING CAPACITY FACTORS
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GROUND WATER INFLUENCE
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GROUND WATER INFLUENCE
CASE 10 D1 < Df q = D1.dry + D2 .
CASE 20 d B q =dry.Df
the value of in third part of equation is
replaced with
= + (d/B).(dry - )
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FACTOR OF SAFETY
FS
FS
unet
netall
uall
)(
)( =
=
Where:
qu = gross ultimate bearing capacity of shallow foundation
qall = gross allowable bearing capacity of shallow foundation
qnet(u) = net ultimate bearing capacity of shallow foundation
qall = net allowable bearing capacity of shallow foundation
FS = Factor of Safety (FS 3)
f
uunet
Dq
qqq
.
)(
=
=
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NET ALLOWABLE BEARING CAPACITY
PROCEDURE:
1. Find the developed cohesion and the angle of friction
2. Calculate the gross allowable bearing capacity (qall)according to terzaghi equation with cd and d as theshear strength parameters of the soil
3. Find the net allowable bearing capacity (qall(net))
shear
dFS
cc =
=
shear
dFS
tantan 1 FSshear = 1.4 1.6
Ex.: qall = cd.Nc + q.Nq + .B.N
Where Nc, Nq, N = bearing capacity factor for the friction angle, d
qall(net) = qall - q
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EXAMPLE PROBLEM
A square foundation is 5 ft x 5 ft in plan. The soilsupporting the foundation has a friction angle of = 20o and c = 320 lb/ft2. The unit weight of soil,, is 115 lb/ft3. Assume that the depth of thefoundation (Df) is 3 ft and the general shearfailure occurs in the soil.Determine:
- the allowable gross load on the foundation with
a factor of safety (FS) of 4.- the net allowable load for the foundation withFSshear = 1.5
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EXAMPLE SOLUTION
Foundation Type: Square Foundation
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EXAMPLE SOLUTION
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GENERAL BEARING CAPACITY EQUATION
idsqiqdqscicdcsu FFFNBFFFNqqFFFNccq .....).5.0(........ ++=
Df
Meyerhof s Theory
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BEARING CAPACITY FACTOR
( )
tan)1(2
cot1
245tan tan.2
+=
=
+=
NqN
NqNc
eNq
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SHAPE, DEPTH AND INCLINATION FACTOR
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EXAMPLE 2
2 m
GWL
dry = 13 kN/m3
sat = 18 kN/m3
c = 1 kg/cm2
= 20o
P = 73 ton
Tank
Foundation
Determine the size (diameter) circle foundation of tank structure as
shown in the following picture
With P is the load of tank, neglected the weight of foundation and usefactor of safety, FS = 3.5.
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EXAMPLE 3
DETERMINE THE FACTOR OF SAFETY FOR:
-CASE 1 : GWL LOCATED AT 0.3m (MEASURED FROM THE SURFACEOF SOIL)
-CASE 2 : GWL LOCATED AT 1.5m (MEASURED FROM THE SURFACEOF SOIL)
dry= 13 kN/m3B = 4m
SQUARE FOUNDATION
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ECCENTRICALLY LOADED FOUNDATIONS
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ECCENTRICALLY LOADED FOUNDATIONS
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ONE WAY ECCENTRICITY
Meyerhofs step by step procedure: Determine the effective dimensions of the foundation as :
B = effective width = B 2eL = effective length = L
Note: if the eccentricity were in the direction of the length of the foundation, the value
of L would be equal to L-2e and the value of B would be B. The smaller of the two dimensions (L and B) is the effective width of the
foundation Determine the ultimate bearing capacity
to determine Fcs, Fqs, Fs use effective length and effective width
to determine Fcd, Fqd, Fd use B
The total ultimate load that the foundation can sustain isQult = qu.B.L ; where BxL = A (effective area)
The factor of safety against bearing capacity failure isFS = Qult/Q
Check the factor of safety against qmax, or,
FS = qu/qmax
idsqiqdqscicdcsu FFFNBFFFNqqFFFNccq ......5,0........ ++=
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EXAMPLE PROBLEM
A Square foundation is shown in the following figure.
Assume that the one- way load eccentricity e = 0.15m.Determine the ultimate load, Qult
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EXAMPLE SOLUTION
With c = 0, the bearing capacity equation becomes
TWO WAY ECCENTRICITY
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TWO-WAY ECCENTRICITY
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TWO-WAY ECCENTRICITY CASE 1
O CC C C S 2
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TWO-WAY ECCENTRICITY CASE 2
TWO WAY ECCENTRICITY CASE 3
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TWO-WAY ECCENTRICITY CASE 3
TWO WAY ECCENTRICITY CASE 4
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TWO-WAY ECCENTRICITY CASE 4
BEARING CAPACITY OF LAYERED SOILS
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BEARING CAPACITY OF LAYERED SOILS
STRONGER SOIL
UNDERLAIN BYWEAKER SOIL
BEARING CAPACITY OF LAYERED SOILS
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BEARING CAPACITY OF LAYERED SOILS
H
B
K
H
DH
B
Hcqq s
fabu 1
12
1
tan21
2
++
+=
)2(2)2(22
)1(1)1(11
21
2
1
BNNcq
BNNcq
c
c
+=
+=
( ) )2()2(2)2()2(1)2()2(22
1sqsqfcscb FBNFNHDFNcq +++=
BEARING CAPACITY OF LAYERED SOILS
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BEARING CAPACITY OF LAYERED SOILS
tsfa
bu qHB
K
H
DH
B
Hcqq
++
+= 1
12
1
tan21
2
tsfa
bu qHB
K
H
D
L
BH
B
Hc
L
Bqq
+
++
++= 1
12
1
tan211
21
Rectangular Foundation
( )
)1()1(1)1()1(1)1()1(1
)2()2(2)2()2(1)2()2(2
2
1
2
1
sqsqfcsct
sqsqfcscb
FBNFNDFNcq
FBNFNHDFNcq
++=
+++=
)2(2)2(22
)1(1)1(11
21
2
1
BNNcq
BNNcq
c
c
+=
+=
BEARING CAPACITY OF LAYERED SOILS
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BEARING CAPACITY OF LAYERED SOILS
SPECIAL CASES TOP LAYER IS STRONG SAND AND BOTTOMLAYER IS SATURATED SOFT CLAY (2 = 0)
TOP LAYER IS STRONGER SAND AND BOTTOM
LAYER IS WEAKER SAND (c1 = 0 , c2 = 0) TOP LAYER IS STRONGER SATURATED CLAY (1
= 0) AND BOTTOM LAYER IS WEAKERSATURATED CLAY (2 = 0)
Find the formula for the abovespecial cases
BEARING CAPACITY FROM N SPT VALUE
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BEARING CAPACITY FROM N-SPT VALUE
A square foundation BxB has to
be constructed as shown in thefollowing figure. Assume that =105 lb/ft3, sat = 118 lb/ft
3, Df= 4ft and D1 = 2 ft. The gross
allowable load, Qall, with FS = 3is 150,000 lb. The field standardpenetration resistance, NF valuesare as follow:
Determine the sizeof the foundation
SOLUTION
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SOLUTION
Correction of standard penetration number(Liao and Whitman relationship)
SOLUTION
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SOLUTION