Civil Engineering Reference
In-Depth Information
(
)
(
)
(
)
PTP
1,
x T
exp
=
=
−
κ
(2.39)
x
κ
max
max
ax
x
Introducing Eqs. 2.33 (with
=
) into 2.38 and solving for
, then the
max
max
following is obtained
1
2
{
}
()
x
2ln
f
0
T
2ln
=⋅
σ
⋅
⎡
⎤
−⋅
κ
⎣
⎦
max
x
x
(2.40)
⎧
⎫
ln
⎪
κ
⎪
()
2ln
f
0
T
1
≈⋅
σ
⋅
⎡
⋅
⎤
−
⎨
⎬
⎣
⎦
x
x
()
2ln
⋅
⎡
f
0
⋅
T
⎤
⎪
⎪
⎣
⎦
⎩
x
⎭
n
(
)
where the approximation
has been applied, assuming that
1
−≈−⋅
x
1
nx
()
⎦
is large as compared to
ln
κ
. Thus, observing that
x
=
0
corresponds
ln
⎡
f
0
⋅
T
⎤
⎣
max
to
, while
x
=∞
corresponds to
, the mean value of
may be
κ =∞
κ =
0
x
max
max
estimated from
∞
dP
∞
dP
⎛
⎞
⎛
⎞
⎛
d
⎞
κ
x
x
max
max
x
∫
x
dx
∫
x
dx
=
⋅
=
⋅
⋅
⎜
⎟
⎜
⎟
⎜
⎟
max
max
⎜
⎟
max
max
⎜
⎟
⎝
max
dx
d
dx
κ
⎠
⎝
⎠
⎝
⎠
max
max
0
0
0
∞
(
)
(
)
∫
x
exp
d
∫
x
exp
d
=
⋅
⎡
−
−
κκ
⎤
=
⋅
−
κκ
⎣
⎦
max
max
∞
0
∞
⎡
⎤
(
)
∫
ln
κ
⋅
exp
−
κ
d
κ
⎢
⎥
∞
⎢
⎥
()
(
)
∫
0
⇒
x
=⋅
σ
2ln
⋅
⎡
f
0
⋅
T
⎤
⋅
exp
−
κ
d
κ
−
(2.41)
⎣
⎦ ⎢
⎥
max
x
x
()
2ln
f
0
T
⋅
⎡
⋅
⎤
⎣
⎦
⎢
x
⎥
0
⎢
⎥
⎣
⎦
x
Thus, the mean value of
is given by
max
⎧
⎫
γ
⎪
⎪
()
x
2ln
⎡
f
0
T
⎤
(2.42)
=⋅
σ
⋅
⋅
+
⎨
⎬
⎣
⎦
max
x
x
()
2ln
f
0
T
⋅
⎡
⋅
⎤
⎪
⎪
⎣
⎦
⎩
x
⎭
∞
=−
(
)
∫
ln
exp
0.5772
where
γ
κ
⋅
−
κ
≈
is the Euler constant. Similarly, it may be shown
0
x
that the variance of
is given by
max
2
π
2
2
σ
=
⋅
σ
(2.43)
x
x
()
max
12 ln
f
0
T
⋅
⎡
⋅
⎤
⎣
⎦
