Civil Engineering Reference
In-Depth Information
1
T
N
′
= 0.8 s,
a
t
n
= ln (1.25)
0.8
T
N
′
= 0.9 s,
a
t
n
= ln (1.50)
0.6
T
N
′
= 1.0 s,
a
t
n
= ln (1.75)
0.4
0.2
CDF of
D
d
(
x
, S
a
,
)
Q
CDF of (
Y
t
+
a
t
)
n
n
0
-5
-4
-3
-2
-1
0
1
2
3
y
16.6
Comparison of CDFs of
D
d
(
x
,
S
a
,
Θ
) and (
Y
t
n
+
α
t
n
).
variable to satisfy the additive form in Eq. (16.6). From the relation between
α
t
n
and
T
s
t
n
′
N
/
T
N
, and recalling that
α
0
=
0, we can write
Δ
α
=
ln[1
+
1.75(
T
′
N
/
T
N
−
1)]. From the data in Fig. 16.5, we compute
Ŷ
t
n
=
ln(100
Δ
D
/
H
)/
C
0
and
s
t
n
/
C
0
. Now, we develop a regression model
Z
t
n
(
Ŷ
t
n
).
Linear regression analysis with the following model form is performed to
develop the model:
=
Δ
α
ζ
(
y
) such that
Z
t
n
=
ζ
()
++
θ
ln
y
θ
σ
ε
y
y
>
≤
0
⎩
Z
δ
1
Z
δ
2
Z
δ
δ
Z
[
()
]
=
ln
ζ
y
0
[16.27]
0
where
θ
z
d1
,
θ
z
d2
and
θ
z
d
are the model parameters and
ε
z
d
is a standard normal
random variable. The values of
1.477
and 0.345 respectively. Figure 16.7 shows the plot of predicted versus
observed values of ln(
Z
t
n
) for the developed regression model. Next, we
need to develop models for
F
z
(
z
) and
F
ˆ
|
Z
(
y
|
Z
t
n
). The CDF
F
z
(
z
)
θ
z
d1
,
θ
z
d2
and
θ
z
d
are found to be 1.332,
−
=
P
[
ζ
(
Ŷ
t
n
)
≤
z
] and the conditional CDF
F
ˆ
|
Z
(
y
|
Z
t
n
) is computed as follows:
(
)
ˆ
⎧
PY
<
yZ
=
0
z
=≤
=>
0
,
,
y
0
t
t
n
n
⎪
⎪
1
z
0
y
0
(
)
=
FyZ
[16.28]
⎨
ˆ
t
((
)
YZ
n
ˆ
PY
<
yZ
=
zZ
,
>
0
z
,
otherwise
>>
0
y
0
⎪
⎪
t
t
t
n
n
n
0
⎩
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