Environmental Engineering Reference
In-Depth Information
Groundwater level
Sand
6 m
Basement
0.5 m
1 m
Concrete wall
10 × 8 m
Concrete base
Figure 10.6
Basement subjected to uplift pressure.
Solution
The first stage in solving this example is to calculate the characteristic angle of shear-
SOF values in the vertical and horizontal directions are δ
v
= 2.0 m and δ
h
= 50 m. The
extents of the failure surfaces are L
v
= 6 m vertically and L
h
= 2(8+10) = 36 m horizon-
tally, hence using
Equations 10.16
and
10.17
together with
Equation 10.13
,
the variance
reduction factors are Γ
2
= .
0 2779
and Γ
2
= .
0 6529
so that for the 2D failure plane
v h
. . Assuming a normal distribution and substituting for COV
inher
and
and the ξ value is obtained using
Equation 10.19
or the graph in
Figure 10.5,
it is 1.098,
which is a more conservative value. Adopting ξ = 1.075, the value of φ
′
is obtained as
follows:
ΓΓΓ
2
=
2
2
=
0 1815
φ
′
=
tan
−
1
(tan
′
/
ξ
)
=
tan
−
1
(tan.
3/175 82
00
)
=
.
°
k
mean
As this is an ultimate limit state due to uplift failure, the recommended UPL γ
M
value in
Table 10.8
for tan ϕ′ = 1.25, so that the design angle of shearing resistance is
φ
′
=
tan
−
1
(tan
φ γ
′
/
)
=
tan
−
1
(tan.
282/1 25
.)
=
23 2
.
°
d
k
Μ
Assuming the angle of shearing resistance between the wall and the ground δ
=
2/3
φ
′
d
d
and using the graph in Eurocode 7 for the horizontal component of the coefficient of active
earth pressure coefficient of wall, K
a;d
= 0.377 for φ
d
′
= .. The area of the side walls
A = 6 × (2 × 10 + 2 × 8) = 216 m
2
. Hence, the design frictional resistance on the side walls
against uplift is
23 2
R5 K
=
0
.
;
σδ
′
tan
A5
=
0
.
×
0
.
377
×
(.
18
0
−
9816
.)
×
×
tan.
23 2
×
216
=
8
58 kN
d
a dv
d
To check the margin of safety in the wall frictional resistance provided by the above
calculation, the characteristic, that is, unfactored, wall frictional resistance is calculated
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