Civil Engineering Reference
In-Depth Information
In a time domain solution it is necessary to perform a stochastic simulation of the
stationary flow components
()
()
()
()
ˆ
n
v
(see Eqs. 9.45
and 9.50). Such a simulation procedure is presented in Appendix A.3. Thus, the solution
is given by
ut
,
vt
and
wt
contained in
−
⎡
1
()
() (
)
()
r
t
=+
r
r
t
= −
K
K
R
+
R
t
⎤
(9.90)
tot
ae
⎣
⎦
Since
()
()
()
ut
,
vt
and
wt
are stochastic, this solution will also be stochastic, i.e. it
()
t
may be necessary to perform several calculations of
R
such that statistics may be
()
t
r
solutions (in general at least ten). Also, the
simulations must be performed over a sufficiently large time window (e.g. T=10 min.).
Thus, a time domain solution may numerically be quite demanding.
In a stochastic solution based on the covariance properties of the stationary turbulence
components
performed on a representative set of
()
()
()
()
ˆ
n
t
ut
,
vt
and
wt
contained in
v
the solution is given by (see Eq.
6.2 and Fig. 6.2.a)
r
=+
r
r
=+
r σ
k
(9.91)
tot
max
p
r
max
where
r
is given in Eq. 9.86,
k
is a peak factor defined in Chapter 2.4 and
σ
is a
vector containing all the standard deviations of the chosen set of displacement degrees of
freedom in the system.
σ
may be extracted from the square root of the vector contained
on the diagonal of the covariance matrix
2
1
⎡
Cov
Cov
Cov
⎤
σ
"
"
"
1
i
1
j
1
N
r
⎢
⎥
⎢
# %#
#
#
⎥
⎢
⎥
2
Cov
Cov
"
σ
"
#
⎢
⎥
i
1
i
ij
⎢
⎥
T
⎡ ⎤
Cov
=⋅
E
r r
⎢
(9.92)
#
# % #
#
rr
⎣ ⎦
⎥
⎢
⎥
2
Cov
Cov
"
"
σ
#
j
1
ji
j
⎢
⎥
⎢
⎥
#
% #
"""""
⎢
⎥
2
Cov
⎢
σ
⎥
⎣
⎦
N
1
N
r
r
where
N
is the total number of global degrees of freedom in the system. Since
() (
−
1
)
()
r
t
=−
KKR
then
t
ae
