two way eccentricity
TRANSCRIPT
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ECGD 4122ECGD 4122 Foundation EngineeringFoundation Engineering
Lecture 6
Lecture 6
Faculty of Applied Engineering and Urban PlanningFaculty of Applied Engineering and Urban Planning
Civil Engineering DepartmentCivil Engineering Department
2
2ndnd
Semester 2008/2009Semester 2008/2009
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ContentContent
Eccentrically Loaded FoundationsEccentrically Loaded Foundations
Foundations with twoFoundations with two--way eccentricityway eccentricity
Field Load TestField Load Test
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Stress ConceptStress Concept
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Stress DistributionStress Distribution Rigid FootingsRigid Footings
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Effect of EccentricityEffect of Eccentricity
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+
=
B
e
BL
Qq
6
1max
Q = total vertical load on the foundationQ = total vertical load on the foundationM = moment on the foundationM = moment on the foundation
ee = M/Q= M/Q
Effect of EccentricityEffect of Eccentricity
-
=
B
e
BL
Qq 6
1min
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MeyerhofMeyerhofs Principles Principle
OfOf
Effective Area MethodEffective Area Method
Effect of EccentricityEffect of Eccentricity
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Step 1:
Step 1:
Ultimate Load with EccentricityUltimate Load with Eccentricity
Find Eccentricity,Find Eccentricity,Q
Me=
If e B/6
+=
Be
BLQq
6
1max
-=B
e
BL
Qq
6
1min
If e > B/6
( )eBLQ
q
2
3
4
max+
=
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Step 2:
Step 2:
B
= B
2e
L
= L
If eccentricity is in the direction of B
If eccentricity is in the direction of L
B
= BL
= L
2e
Ultimate Load with EccentricityUltimate Load with Eccentricity
Determine the Effective DimensionsDetermine the Effective Dimensions
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Ultimate Load with EccentricityUltimate Load with Eccentricity
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Step 3:
Step 3: Use general equation to calculate the soilUse general equation to calculate the soil
ultimate bearing capacityultimate bearing capacity
q'u= c
Nc Fcs Fcd Fci + q Nq Fqs Fqd Fqi
+ 0.5 B
Fs Fd Fi
DONT use B
to calculate:
Fcd , Fqd , Fd
USE B
to calculate:Fcs , Fqs , Fs
Ultimate Load with EccentricityUltimate Load with Eccentricity
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Step 4:
Step 4: Determine the total ultimate load:Determine the total ultimate load:
Qult = qult A
= qult B
L
Step 5:
Step 5: Factor of safety against bearing failure:Factor of safety against bearing failure:
Q
QFS ult=
Ultimate Load with EccentricityUltimate Load with Eccentricity
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Example 1
Example 1
For the footing shown below, determine theFor the footing shown below, determine the
ultimate vertical loading with an 0.15 multimate vertical loading with an 0.15 m
eccentricity using the general formula. Take Neccentricity using the general formula. Take Nqq
==
18.4, N
18.4, Ngg
= 22.4
= 22.4..
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Example 1Example 1 -- SolutionSolution
462.
1
30tan
5.
1
2.
1
1
tan
1
/
6.
12)
18)(
7.
0(
5.
0'
5.
1
2.
1)
15.
0)(
2(
5.
1
2
'
'
'
2
1
''
'
'
=
+=
+=
==
=++=
==
=-=
-=
f
g
g gggg
L
B
F
mkN
Dq
FFFNBFFFqNFFFNcq
mLL
m
eBB
qs
idsqiqdqsqcicdcscult
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Example 1Example 1 -- SolutionSolution
68.
0
5.
1
2.
1
4.
0
1
4.
0
1
1
90
1
135.
1
30tan)
30sin
1(
5.
1
7.
0
2
1
tan)sin
1(
2
1
'
'
2
2
'
2'
=
-=
-=
=
-=
=-
+=
-
+=
L
B
F
F
B
DF
s
qi
f
qd
g
b
ff
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Example 1Example 1 -- SolutionSolution
2
''''
2
'
2
'
/
59.
988)
5.
1)(
2.
1)(
21.
549(
/
21.
549
)
1)(
1)(
68.
0)(
4.
22)(
2.
1)(
18)(
5.
0(
)
1)(
135.
1)(
462.
1)(
4.
18)(
6.
12(
0
1
1
1
mkN
LBqQ
mkN
q
F
F
ultult
ult
i
d
==
=
=
+
+=
=
-=
=
f
bg
g
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Two Way EccentricityTwo Way Eccentricity
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ult
xL
ult
y
B
Q
Me
Q
Me
=
=
'' AqQ ultult =
Two Way EccentricityTwo Way Eccentricity
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DONT use B
and L
to calculate:Fcd , Fqd , Fd
USE B
and L
to calculate:
Fcs , Fqs , Fs
Two Way EccentricityTwo Way Eccentricity
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'
''
3
5.
1
3
5.
1
2
1
'
1
1
1
1
L
AB
L
eLL
B
eBB
LBA
L
B
=
-=
-=
=
Case I: eCase I: eLL/L/L 1/6 and e
1/6 and eBB/B/B 1/61/6
Two Way EccentricityTwo Way Eccentricity
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Case II: eCase II: eLL
/L < 0.5 and 0 < e
/L < 0.5 and 0 < eBB
/B < 1/6/B < 1/6
( )
larger)(the'
larger)(the
''
2
1
'
2
1
2
1
2
1
LorLL
LorL
AB
LLBA
=
=
+=
Two Way EccentricityTwo Way Eccentricity
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Two Way EccentricityTwo Way Eccentricity
0
10
20
30
40
50
60
70
80
90
1st
Qtr
2nd
Qtr
3rd
Qtr
4th
Qtr
East
WestNorth
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Case III:Case III: eeLL
/L < 1/6 and 0
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Two Way EccentricityTwo Way Eccentricity
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Case IV: eCase IV: eLL
/L < 1/6 and e
/L < 1/6 and eBB
/B < 1/6/B < 1/6
( )( )
LL
L
AB
BBLLBLA
=
=
+-+=
'
''
2
1
'
2
2
2
Two Way EccentricityTwo Way Eccentricity
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Two Way EccentricityTwo Way Eccentricity
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Example 2
Example 2
Refer to Example 1 and the figure shown belowRefer to Example 1 and the figure shown below
to determine the ultimate vertical loading usingto determine the ultimate vertical loading using
the general formula.the general formula.
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Example 2Example 2 -- SolutionSolution
mL
m
LLBA
mL
LL
mLLL
chartFrom
IICaseBe
Le
B
L
28.
1
20.
1)
32.
0
28.
1)(
5.
1)(
5.
0(
)(
5.
0
32.
0)
5.
1)(
21.
0(
21.
0/
28.
1)
5.
1)(
85.
0(
85.
0/
:
1.
0
5.
1/
15.
0/
2.
0
5.
1/
3.
0/
'
2
2
1'
2
2
1
1
=
=+=
+=
==
@
==@
==
==
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Example 2Example 2 -- SolutionSolution
424.
1
30tan
28.
1
94.
0
1
tan
1
/
6.
12)
18)(
7.
0(
5.
0'
94.
0)
28.
1/()
20.
1(
/
'
'
'
2
1
''
1
''
=
+=
+=
==
=
++=
==
=
f
g
g gggg
L
BF
mkN
Dq
FFFNBFFFqNFFFNcq
m
LAB
qs
idsqiqdqsqcicdcscult
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Example 2Example 2 -- SolutionSolution
706.
0
28.
1
94.
0
4.
0
1
4.
0
1
1
90
1
135.
1
30tan)
30sin
1(
5.
1
7.
0
2
1
tan)sin
1(
2
1
'
'
2
2
'
2'
=
-=
-=
=
-=
=-
+=
-
+=
L
B
F
F
B
DF
s
qi
f
qd
g
b
ff
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Example 2Example 2 -- SolutionSolution
2
'''
2
'
2
'
/
20.
610)
20.
1)(
50.
508(
/
50.
508
)
1)(
1)(
706.
0)(
4.
22)(
94.
0)(
18)(
5.
0(
)
1)(
135.
1)(
424.
1)(
4.
18)(
6.
12(
0
1
1
1
mkN
AqQ
mkN
q
F
F
ultult
ult
i
d
===
=
+
+=
=
-=
=
f
bg
g
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Field Load TestField Load Test
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Field Load TestField Load Test
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Clayey SoilClayey Soil
)(u)(u PF qq =
Sandy SoilSandy Soil
P
F
PFB
Bqq
)(u)(u=
qqu(F)u(F) Ultimate capacity for proposed foundationUltimate capacity for proposed foundation
qqu(P)u(P) Ultimate bearing capacity for test plateUltimate bearing capacity for test plateBB FF Width of foundationWidth of foundation
BB PP Width of test plateWidth of test plate
P
FPFB
BSS )()( =
2
)()(
2
+
=PF
FPF
BB
BSS
Field Load TestField Load Test
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Example 3
Example 3
Refer to the accompanied figure for the resultsRefer to the accompanied figure for the results
of a plate load test on a sandy soil. Define theof a plate load test on a sandy soil. Define the
dimensions of a square footing to carry a load ofdimensions of a square footing to carry a load of
2500 kN with a permissible settlement of 25 mm.2500 kN with a permissible settlement of 25 mm.
The test plate is 1
The test plate is 1 in diameter.in diameter.
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Example 3
Example 3
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Example 3Example 3 -- SolutionSolution