Civil Engineering Reference
In-Depth Information
x
a(t)
0.4
x
401
Step
1
PHL
(
Ω
(t),
Ω
y
= 236.9)
Figure 8.2
A normalized SDOF system and loads history.
ω
0
=39
:
74;
ω
00
= 125
:
67;
ω
=12
:
56;
Ω
y
= 236
:
9
ð8
:
21Þ
Assume that the system shown in Figure 8.2 is subjected to a set of horizontal displacements, in
which the maximum displacement is 0.4 m and the incremental displacement is 0.001 m so that
the whole displacement set is divided into 400 displacement steps. In order to be consistent with
the mass-normalized load
¨
g
in the right item of Eq. (8.12), the latter resistant load at step
k
is
written as the form of
a
(
k
)
, in which
k
= 1 to 401.
Solution
Step
1
:(
k
=1,
x
=0)
Since the initial state of the system is static, there are
a
ðÞ
=0;
ðÞ
=0
Ω
ð8
:
22Þ
Step
2
:(
k
=2,
x
= 0.001 m)
Assuming the mass-normalized SDOF system at step
k=
2 is linear yields
00
ðÞ
=0
Δθ
ð8
:
23Þ
Δ
a
ðÞ
=
ω
2
Δ
x
1
= 157
:
92 × 0
:
001 = 0
:
158 N
=
kg
ð8
:
24Þ
ðÞ
=
ω
0
2
ΔΩ
Δ
x
ðÞ
= 1579
:
2×0
:
001 = 1
:
579 N-m
=
kg
ð8
:
25Þ
The trial responses at step
k
= 2 is expressed by
a
ðÞ
=
a
ðÞ
+
Δ
a
ðÞ
=0+0
:
158 = 0
:
158 N
=
kg
ð8
:
26Þ
ðÞ
=
Ω
2
ðÞ
=0+1
:
579 = 1
:
579 N-m
=
kg
1
Ω
ðÞ
+
ΔΩ
ð8
:
27Þ
