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8/6/2019 Chapter 3 Mod
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Foundation EngineeringChapter (3) Bearing capacity of shallow foundation
Foundation EngineeringECIV 4352
Lecture (3) Bearing capacity of shallow foundation Ultimate bearing capacity (qu)
Shear failure Allowable bearing capacity (qall)
Shear failure qallshould be adequate to prevent excessive settlement and shear failure
Types of shear failure:1- General shear failure: For Dense sand.
2- Local shear failure: For Medium compaction soil.
3- Punching shear failure: For loose soil
1
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Foundation EngineeringChapter (3) Bearing capacity of shallow foundation
Calculation of ultimate bearing capacity of shallow foundations without eccentricity:
1- Terzaghis theory:Assumption for Terzaghis theory: The foundation is considered to be shallow if ( )BDf , in recent studies the
foundation is considered to be shallow if ( 4/ BDf . Other wise it is consideredto be deep foundation.
Foundation is considered to be strip if ( )00.0/ LB . The soil from ground surface ( ) to the bottom of the foundation (
) is replaced by stress fDq = .
For General shear failure:Type of foundation Ultimate bearing capacity qu
Strip Footing BNqNcNq qcu2
1++=
Square footing BNqNcNq qcu 4.03.1 ++=
Circular footing BNqNcNq qcu 3.03.1 ++=
c: Cohesive.fDq =
B: Foundation width (Diameter if circular).NNN qc ,, : Bearing capacity factors given from table 3.1 P.158 as function of angle of
friction .
For Local shear failure:Type of foundation Ultimate bearing capacity qu
Strip Footing '''2
1
3
2
BNqNcNq qcu ++=
Square footing ''' 4.0867.0 BNqNcNq qcu ++=
Circular footing ''' 3.0867.0 BNqNcNq qcu ++=
''' ,,
NNN qc : Factors for bearing capacity given from table 3.2 P.160Or from table 3.1 P158 but replace ' instead of :
= tan
3
2tan 1'
2- Meyerhofs equations (General bearing capacity equation):Terzagi equations neglect:
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Foundation EngineeringChapter (3) Bearing capacity of shallow foundation
Rectangular footings. Inclination of loads. Shear strength of soil above the foundation.
Meyerhofs equation takes in consideration theses variables:idsqiqdqsqcicdcscu
FFFBNFFFqNFFFcNq
5.0++=
NNN qc ,, : Table 3.4 P.168
b eS h o
f a c t o r s .nI n c l i n a t i o,,
f a c t o r s .D e p t h,,
f a c t o r s .S h a p e,,
iq ic i
dq dc d
sq sc s
FFF
FFF
FFF
Shape Factors:
L
BF
L
BF
N
N
L
BF
s
qs
c
q
cs
4.01
tan1
1
=
+=
+=
Depth Factors: Case I: 1/ BDf
( )( )
1
sin1tan21
4.01
2
=
+=
+=
d
f
qd
f
cd
F
B
DF
B
DF
Depth Factors: Case II: 1/ >BDf
( )( )
1
tansin1tan21
tan4.01
12
1
=
+=
+=
d
f
qd
f
cd
F
B
D
F
B
DF
Inclination Factors
3
The Term is in radian
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Foundation EngineeringChapter (3) Bearing capacity of shallow foundation
2
2
1
901
=
==
i
qici
F
FF
Effect of water table in bearing capacity equations:
4
Case I: 0Dw1Df Case II: 0Dw2B
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Foundation EngineeringChapter (3) Bearing capacity of shallow foundation
Case I) Water table is located at depth Dw1 so that 0 Dw1 Df:
( )1
'
1 wfwDDDq += ( )
wsat =='
( )
Case II) Water table is located at depth Dw2 below the foundation so that 0 Dw2 B:fDq = .( )
( )'2' +==B
Dw ( (
Case III) Water table is located at depth Dw2 below the foundation so that Dw2 > B:No changes in equations.
Factor of safety:Ultimate bearing capacity
.
( )( )
( )
load.Ultimate
capacitybearingallowableNet
capacitybearingallowableGross
capacitybearingultimateNet
capacitybearingultimateGross
GrossQ
q
q
qqq
q
u
netall
all
unetu
u
=
( )( )
FS
FS
FS
FS
allunetu
netall
uall
=
==
=
FS = (3 4) for bearing capacity
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Foundation EngineeringChapter (3) Bearing capacity of shallow foundation
Example 1)Determine the size of square footings to carry gross allowable load (295 KN) given that:
00.0
.35
./15.18
00.1
3
3
=
=
=
=
=
C
mKN
D
FS
f
Use Terzagi equations assuming general shear failure.295KN
Df=1.0
0
B
C=0.00
=35
=18.15KN/m3
Solution
92cmB:errorandBy trial684.22814.2
41.4515.184.044.41115.180885
41.45,44.41,75.5735At
4.03.1
:footingsquareFor
.885
3295
.295
23
2
22
2
==+
++=
====
++=
===
===
BB
BB
NNN
BNqNcNq
BBFSqq
BA
AreaqQ
qc
qcu
allu
allall
allall
Example 2)
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Foundation EngineeringChapter (3) Bearing capacity of shallow foundation
Determine the net allowable load that foundation can carry (no inclination), UseMeyerhof equation given that:
./50
.25
./10
00.2
4
2
3
mKNC
mKN
D
FS
w
f
== =
==
Solutionidsqiqdqsqcicdcscu FFFBNFFFqNFFFcNq 5.0++=
88.10,66.10,72.2025At ====
NNN qc
The water table is at depth = 1m
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Foundation EngineeringChapter (3) Bearing capacity of shallow foundation
( )( )
1
313.1
2
225sin125tan21
4.12
24.01
12/2/
2
=
=+=
=
+=
==
d
qd
cd
f
F
F
F
BD
Inclination factors:Due to absence of inclined load, the inclination factor is 1 every where.
( )
( )( )
( ) ( ) KNAqQ
mKNFS
mKNqqq
mKNq
q
netallnetall
netu
netall
unetu
u
u
1.3716635.619
/35.6194
4.2477
/4.24772.266.2503
./6.2503
733.088.1024.95.01313.1311.166.102.2614.1343.172.2050
2
2
2
===
===
===
=
++=
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P
e
BL
Pe
B
Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
Foundation EngineeringECIV 4352
Lecture (4) Bearing capacity of shallow foundation (cont.) Eccentrically loaded foundation:
** :
P
q = P /( B x L )
B x L
** :
One way eccentricity
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P
B
M = P e
q m a x
q m i n
Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
1- For e
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
3- For e > B/6: There will be tension stresses on the foundation which is prohibitedin design, so we will neglect the tension stress and calculate qmax as follow:
qmax qmax
3(B/2- e)
)2(3
4max
eBL
Pq
=
Derivation:
2........................2
32323
00.0
00.0
:
1................................2
1max
=
=
=
=
==
=
eB
XeBX
eB
PX
N
M
PNF
Stabilityfor
LXqN
edgeright
y
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
)2(3
4
2
23*
2
1
2
1
:2&1
maxmax
max
eBL
Pq
eBLqP
LXqPN
From
=
=
==
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B
L
e
B / 2 - eB / 2 - e
B - 2 e2 e
Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
:
( ) ( )
BB
eLL
LBB
LBA
LL
eBB
used
=
=
=
=
=
=
'
'
'''
'''
'
'
2
:(L)ofplanein theismomenttheIf
,.min
2
The equation used to calculate the bearing capacity is Meyerhof's or Terzagi's equation: idsqiqdqsqcicdcscu FFFNBFFFqNFFFcNq
''5.0++=
To find shape factors: use '' , LBTo find depth factors: use B, Lin last term is related to soil below the foundation.To find the gross ultimate load Qu:
''AqQ uu =
max
'
max :qforadequateissafetyoffactor
capacity)bearingAgainst(
q
qFSthatCheck
Q
QFS
u
all
u
=
=
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
Example 3)Determine the size of square footing (B x B) that subjected to vertical load (100,000Ib)and moment (25,000Ib.ft), the soil profile is given below:
Use FS=6, w=62.4 Ib/ft3Qal l=100,000Ib
B
M=25,000Ib.f td=100Ib/f t3
C=0.00
= 3 0
sat=120Ib/f t3
C = 0 . 0 0
= 30
4ft
Solution1- Find the eccentricity (e):
ftP
Me 25.0
000,100
000,25===
2- Find the effective areaA':
( )
( )( ) BBBBABL
BLBB
BL
BB
LBA
used
5.05.0
.
5.0),.(min
5.0
2'
'
'''
'
'
'''
==
=
==
=
=
=
3- Find 'uq :
idsqiqdqsqcicdcscu FFFNBFFFqNFFFcNq ''
5.0++=
4.22,4.1803At ===
NNq
./6.574.62120
./4001004
3'
2
ftIb
ftIbq
===
==
Shape factors:
BB
BF
BB
BF
s
qs
2.06.0
5.04.01
2886.0577.130tan
5.01
+==
=+=
Depth factors:
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
1
154.11
4)30sin1(30tan21
1/DAssume
2
f
=
+=+=
==
>==
( )
( )
mL
AB
mLBA
mLBL
mL
mB
675.0908.1
2879.1
2879.1908.135.15.02
1
908.1),max(
908.1)182.0(35.12
35.1)2.0(35.15.1
'
''
2
11
'
11
'
1
1
===
===
==
====
idsqiqdqsqcicdcscu FFFNBFFFqNFFFcNq ''
5.0++=
Foe sand: c=0.0003.48,3.3353At ===
NNq
Shape factors:
858.0908.1
675.04.01
248.135tan908.1
675.01
==
=+=
s
qs
F
F
Depth factors:
1
167.15.1
1)35sin1(35tan21
166.05.1/1/D
2
f
=
=+=
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
Foundation EngineeringECIV 4352
Lecture (5) Bearing capacity of shallow foundation (Cont.)Bearing capacity of layered soilIn practice the foundation may be based on layered soil profile, or in some times we haveto replace adequate thickness of weak soil by stronger one.Calculating the ultimate bearing capacity for such case have some different as that will beshown below.
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
For continuous foundation:
11
12
1
tan1
2qH
B
K
H
DH
B
Hcqq s
fa
bu
+++=
( ) ( ) ( ) fqc DqBNNqNcq 11111111 5.0 =++=
( )1cN , ( )1qN and ( )1N are given from the same tables but use 1=
uq : Ultimate bearing capacity.
bq : bearing capacity of bottom layer.
In this case ( ) ( ) ( ) 1222222 5.0 HDqBNNqNcq fqcb +=++=
( )2cN , ( )2qN and ( )2N are given from the same tables but use 2=
ac : AdhesionH: Thickness of top soil layer below bottom of foundation.
B: Width of foundation.sK : Punching shear coefficient.
To find sK To find ac
1- find 12 / qq
( ) ( )
( ) ( )22222
11111
5.0
5.0
BNNcq
BNNcq
c
c
+=
+=
2- Go to figure 3.21 P.190 to find sK
1- find 12 / qq2- Go to figure 3.21 P.190 to find
1/ cca
When the top layer height (H) is relatively large, the failure surface will occur in the toplayer and so uq = 1qOtherwise there will be punching shear failure in the top layer and then general shearfailure in the bottom layer.
For rectangular foundation:
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )1111111111
222221222
11
12
1
5.0
5.0
tan211
21
sqsqfcsc
sqsqfcscb
sfa
bu
FBNFNDFNCq
FBNFNHDFNCq
qHB
k
H
D
L
BH
B
Hc
L
Bqq
++=
+++=
+
++
++=
Read the special cases from text book P.191 192
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
Example 1)Rectangular footing 1.5m x 1m based on the shown soil profile; find the grossallowable load that foundation can carry.
Special case III
112
2114.52.01 qD
B
Hc
L
BC
L
Bq
f
a
u+
++
+=
In this case 9.0/4.0120/48// 11212 === CcCCqq a from figure 3.22./1081209.0 2mKNca ==
2/4.65618.161
110825.1
114814.55.1
12.01 mKNqu =+
++
+=
2
111 /8.71518.1612014.55.1
12.0114.52.01 mKNDC
L
Bq f =+
+=+
+=
uq < OKq 1 , otherwise use uq = 1q
( ) KNQ
mKNq
all
all
15.2465.111.164
/1.1644
4.656 2
==
==
Read Example 3.9 P.194
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
Bearing Capacity of foundation on top of slopeFor continuous shallow foundation:
qcqu BNCNq 5.0+=
For pure sand C = 0.00 qu BNq 5.0=
For pure clay = 00.0 cqu CNq =
To find cqN :
1- Find stability numberC
HNs
=
2- Find b/B and BDf /
3- Go to figure A shown below: If B < H: use curves with 00.0=sN
If HB : use curves with calculated sN .To find qN :
1- Find b/B and BDf /2- Go to figure B shown below to find qN .
b: Distance from edge of foundation tothe top of slope.H: Height of top of slope.
: Angle of slope with horizontal
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
Fig.A Finding Ncq
Fig.B Finding N q
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Foundation EngineeringChapter 3: Bearing capacity of shallow foundation
Example 2)For the shown soil profile determine the ultimate bearing capacity of the continuousfoundation.
Solution= 00.0C qu BNq 5.0=
B=1.5m