Environmental Engineering Reference
In-Depth Information
100,000
90,000
80,000
70,000
X
~
N
(μ′
T
X
,σ′
T
X
)
Optimal design
60,000
50,000
40,000
μ
T
X
= μ′
μ
=50 kPa
σ
X
2
+σ′
μ
2
x
*
Opt
= 24 kPa
E
(
C
)
Min
= $12,275
σ′
T
X
=
÷
ææææ
σ
X
= 10 kPa
30,000
20,000
10,000
0
σ′
μ
= 7.5 kPa
σ′
T
X
= 12.5 kPa
0
20 40
Design undrained strength,
x
* (kPa)
60
80
100
Figure 13.23
Expected cost versus design strength for compacted fill.
(
σ
2
/
/
n
n
)
×
+
()
σ
′2
X
µ
σ
″2
=
(13.24)
µ
(
σ
2
)
()
σ
′2
X
µ
where
σ
µ
is the prior standard
deviation for the mean, 150 psf, and
x
is the sample mean of the undrained shear strength
measured in n QA/QC tests. Therefore, the information from the test will affect the design
µ
µ
is the prior mean value f
or
the mean, 1000 psf, and
′
′
C:Cost ($1000):
Implementation failure
Design OK
20-0.4
x
*
0
P
(
X
≥
x
*|
x
)
x
*
Te st
n
samples
E
(
C
|
x
*,
x
)=20-0.4
x
*
+
P
(
X
<
x
*|
x
)×100
Design fails
x
*: Design
strength
20-0.4
x
*
Measure
x
100
P
(
X
<
x
*|
x
)
Updated total
distribution of X
X
~
N
(μ′′
T
X
, σ′′
T
X
)
0.04
μ′′
T
X
= μ′′
m
x = 40 kPa, n = 10
0.03
σ′′
T
X
=
÷
ææææ
s
X
2
+σ′′
m
2
μ′′
T
X
= 41.5
kPa
σ′′
T
X
= 10.4
kPa
0.02
Prior total
distribution of X
0.01
μ′
T
X
= 50
kPa
σ′
T
X
= 12.5
kPa
0.00
0
20
40
60
80
100
X
(kPa)
Figure 13.24
Posterior decision tree to select design strength based on test data for compacted fill.
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