Civil Engineering Reference
In-Depth Information
where the ground acceleration vector
g
ðÞin Eq. (3.48) has been expressed in terms of the three
earthquake ground motion components, i.e.
<
=
g
x
ðÞ
g
y
ðÞ
g
z
ðÞ
g
ðÞ=
h
=
ha
ðÞ
ð3
:
49Þ
:
;
similar to what was presented in Eq. (3.13).
To simplify Eq. (3.48), let
,
,
0 I
−
M
−1
K
−
M
−1
C
0
−
h
0
M
−1
K
A
=
H
=
G
=
ð3
:
50Þ
where
A
is again the 2
n
×2
n
state transition matrix in the continuous form,
H
is again the 2
n
×3
ground motion transition matrix in the continuous form, and
G
is the 2
n
×
n
inelastic displace-
ment transition matrix in the continuous form. Then, Eq. (3.48) becomes
z
ðÞ=
Az
ðÞ+
Ha
ðÞ+
Gx
00
ðÞ
ð3
:
51Þ
Solving for the first-order linear differential equation in Eq. (3.51) gives
Þ
z
t
ðÞ+
e
A
t
ð
t
t
o
z
ðÞ=
e
A
t
−
t
o
ð
e
−
A
s
Ha
ðÞ+
Gx
00
ðÞ
½
ds
ð3
:
52Þ
where
t
o
is the time of reference when the integration begins, which is typically the time when
the states
z
(
t
o
) are known, such as the initial conditions.
To integrate Eq. (3.52) numerically, let
t
k
+1
=
t
,
t
k
=
t
o
, and
Δ
t
=
t
k
+1
−
t
k
, and the subscript
k
denotes the
k
th time step, then it follows from Eq. (3.52) that
z
k
+1
=
e
A
Δ
t
z
k
+
e
A
t
k
+1
ð
t
k
+1
t
k
e
−
A
s
Ha
ðÞ+
Gx
00
ðÞ
½
ds
ð3
:
53Þ
Using the delta function approximation for the variables in the integral, where the ground
acceleration vector and the inelastic displacement vector take the form:
a
ðÞ=
a
k
δ
s
−
t
k
ð
Þ
Δ
t
,
t
k
≤
s
<
t
k
+1
ð3
:
54aÞ
x
00
ðÞ=
x
0
k
δ
s
−
t
k
ð
Þ
Δ
t
,
t
k
≤
s
<
t
k
+1
ð3
:
54bÞ
Substituting Eq. (3.54) into Eq. (3.53) and performing the integration gives
z
k
+1
=
e
A
Δ
t
z
k
+
Δ
t
e
A
Δ
t
Ha
k
+
Δ
t
e
A
Δ
t
Gx
0
k
ð3
:
55Þ
where
z
k
,
a
k
, and
x
0
k
are the discretized forms of
z
(
t
),
a
(
t
), and
x
00
(
t
), respectively. Let
F
d
=
e
A
Δ
t
,
H
d
=
e
A
Δ
t
H
Δ
t
,
G
d
=
e
A
Δ
t
G
Δ
t
ð3
:
56Þ
