2.4 Time Domain Analysis
31
To summarize, the steady-state response to the input exp(i ν t) in Eq. 2.50 has
the same frequency as the input but different size and phase. The input-output
relationship is
f (t) = e
i ν t
⇒ x p (t) =
r d (ν)/ω
2
e
i (ν t−ϕ) which implies
f (t) = cos(ν t) ⇒ x p (t) =
r d (ν)/ω
2
cos(ν t − ϕ) and
f (t) = sin(ν t) ⇒ x p (t) =
r d (ν)/ω
2
sin(ν t − ϕ)
(2.51)
by the linearity of the equation of motion. These particular solutions are the steadystate solutions to the inputs f (t) = exp(i ν t), f (t) = cos(ν t), and f (t) = sin(ν t)
since the general solution of the homogeneous equation vanishes for large times (see
Eq. 2.13).
2.4.6.2 Dynamic Amplification Factor (DAF)
The form of the particular solution x p (t) = x st r d (ν) sin(ν t − ϕ) in Eq. 2.45 shows
that the dynamic amplification factor r d (ν) gives the scale of the dynamic response
relative to the static response x st , see Fig. 2.10 which shows r d (ν) for two values of
the damping ratio ζ over a broad range of ν/ω ratios.
The ratio ν/ω controls not only the scale of the particular solution but also its
phase. We consider three cases, ν/ω 1, ν/ω 1 and ν/ω 1. If ν/ω 1,
the DAF r d (ν) 1 and tan(ϕ) 0 which implies ϕ 0 so that x p (t) describes
a quasi-static response in phase with the forcing function. If ν/ω 1, the DAF
r d (ν) ∼ O(ω/ν) 2 and tan(ϕ) → 0 through negative values as ν/ω → ∞. This
means that, for large values of ν/ω, the DAF is small, and so is the system response
x p (t), and ϕ π , i.e., the solution x p (t) and the forcing functions are out of phase.
Fig. 2.10 DAF r d (ν) for
ζ = 0.05 and 0.1
0
0.5
1
1.5
2
2.5
3
0
1
2
3
4
5
6
7
8
9
10
ν/ω
r
d (ν)
ζ = 0.05
ζ = 0.1
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