BC14
•
Eccentric Loading:
i.e.
load not at footing center – overturning moment
o
Overturning moment equivalent to
eccentric load at distance e from
footing centerline
o
Consider footing of width B and length L
o
Examine pressure distribution under footing
o
Assume footing is rigid and that soil has
linear behavior. Use beam analogy.
o
Separate flexural and uniform load components
BL
Q
A
Q
q
uniform
=
=
2
3
flex
LB
Qe
6
)
12
/
LB
(
)
2
/
B
)(
Qe
(
I
Mc
q
=
=
=
o Sum the components:

=
+
=
±
=
±
=
B
e
6
1
BL
Q
q
and
B
e
6
1
BL
Q
q
LB
Qe
6
BL
Q
q
q
q
min
max
2
flex
uniform
this equation is valid for
e
≤
B/6
at which point q
min
= 0
(To prevent tipping, most designers maintain e
≤
B/6.)
o If
(e > B/6)
then part of overall pressure diagram is in tension.
Since soil doesn’t withstand sustained tension, the pressure
distribution becomes triangular and the footing begins to
lift at the heel.
Since the load Q must act through the
centroid of the pressure diagram:
(
)
0
q
and
L
2e

B
3
4Q
L
e

2
B
3
2Q
2
xL
Q
q
e
2
B
3
x
and
q
2
xL
Q
min
max
max
=
=
=
=

=
=
o
Can design footing for q
max
using this approach
but
the soil does not behave linearly, failure
at the toe will redistribute the load, the footing is not rigid, the column connection carrying
moment is not rigid, and therefore the actual pressure distribution varies from the above.
Q
Q
M
e
=
B
e
B/2
x
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BC15
o
Meyerhof (1953) proposed an approximation to the above analysis using an “effective area”
for the footing based on an “effective width”, B’, or an “effective length”, L’, (depending on
the direction of the eccentricity).
The analysis shown here is for 1D eccentricity (e
L
or e
B
).
Analysis Procedure:
1. Calculate
X = B  2e
B
and
Y = L – 2e
L
2. Assign
X and Y to B’ and L’.
B’ is the shortest dimension and L’ is the longest.
3. Note that e
max
= 0.5 because this will give B’ (or L’) = 0.
Usual design limit is e
≤
B/6 (or e
≤
L/6) to keep resultant in
center 1/3 of footing (best approximation by this method).
4. Use BC equation as follows:
q
u
=
cN
c
λ
cs
λ
cd
λ
ci
+ qN
q
λ
qs
λ
qd
λ
qi
+ ½
γ
B’N
γ
λ
γ
s
λ
γ
d
λ
γ
i
use B = B’ with N
γ
term
use B’ and L’ for calculation of shape factors
λ
cs
,
λ
qs
,
λ
γ
s
use actual B for calculation of depth factors
λ
cd
,
λ
qd
,
λ
γ
d
5. Use the effective area to calculate the bearing capacity Q
u
= q
u
A’ = q
u
B’L’
The concept of effective area can be applied to circular footings and 2D eccentricity.
These problems are solved for effective dimensions so that the resultant load acts at the
centroid of the effective area.
(This is true of the above 1D problems which are much
easier to calculate.)
For clays (
φ
≈
0) the bearing capacity decreases linearly with e.
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 Fall '08
 Staff
 QU, Trigraph, effective area, footing, square footing

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