SDEE: Lecture 6
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Transcript of SDEE: Lecture 6
StructuralDynamics
& EarthquakeEngineering
Dr AlessandroPalmeri
Structural Dynamics& Earthquake Engineering
Lectures #6 and 7: State-space equation of motionand Transition matrix for SDoF oscillators
Dr Alessandro Palmeri
Civil and Building Engineering @ Loughborough University
Tuesday, 25th February 2014
StructuralDynamics
& EarthquakeEngineering
Dr AlessandroPalmeri
State-Space Formulation
The equation of motion for a SDoF oscillator reads:
u(t) + 2 ζ0 ω0 u(t) + ω20 u(t) =
1m
f (t) (1)
and can be posed in the alternative matrix form:
y(t) = A · y(t) + b f (t) (2)
where y(t) is the array of the state variables (displacementand velocity) for the oscillator:
y(t) =
{u(t)u(t)
}(3)
while:
A =
[0 1−ω2
0 −2 ζ0 ω0
], b =
{0
1/m
}(4)
StructuralDynamics
& EarthquakeEngineering
Dr AlessandroPalmeri
Duhamel’s Solution
Let us consider a scalar first-order inhomogeneous ODE:
y(t) = A y(t) + b f (t) (5)
The integral solution of Eq. (5) can be formally written as:
y(t) = Θ(t) y(0) +
∫ t
0Θ(t − τ) b f (τ) dτ (6)
where y(0) is the initial condition at time t = 0, while thetransition function Θ(t) is so defined:
Θ(t) = eA t (7)
StructuralDynamics
& EarthquakeEngineering
Dr AlessandroPalmeri
Duhamel’s Solution
This integral solution can be extended to systems with manystate variables as:
y(t) = Θ(t) · y(0) +
∫ t
0Θ(t − τ) · b f (τ) dτ (8)
where the array y0 = y(0) collects the initial conditions attime t = 0, and the transition matrix Θ(t) is evaluated as theexponential matrix of [A t ]:
Θ(t) = eA t (9)
StructuralDynamics
& EarthquakeEngineering
Dr AlessandroPalmeri
Step-by-Step Numerical Solution
For t = ∆t , and assuming a linear variation of the forcingterm f (t) in the time interval [0,∆t ]:
f (t) = f0 +f1 − f0
∆tt (10)
one can mathematically prove that the Duhamel’s integralgives:
y1 = Θ(∆t) · y0 + Γ0(∆t) · {b f0}+ Γ1(∆t) · {b f1} (11)
where y1 = y(∆t) collects the state variables at t = ∆t ,while the integration matrices Γ0(∆t) and Γ1(∆t) can becomputed from the transition matrix Θ(∆t) and the matrix ofcoefficients A.
StructuralDynamics
& EarthquakeEngineering
Dr AlessandroPalmeri
Step-by-Step Numerical Solution
That is:Γ0(∆t) = [Θ(∆t)− L(∆t)] · A−1 (12)
Γ1(∆t) = [L(∆t)− I2] · A−1 (13)
in which I2 is the 2-dimensional identity matrix, while theloading matrix L(∆t) is given by:
L(∆t) =1
∆t[Θ(∆t)− I2] · A−1 (14)
StructuralDynamics
& EarthquakeEngineering
Dr AlessandroPalmeri
Step-by-Step Numerical Solution
Moreover:
Θ(∆t) = e−ζ0ω0∆t
(C + ζ0ω0ω0
S)
1ω0
S
−ω20ω0
S(
C − ζ0ω0ω0
S) (15)
in which C = cos(ω0 ∆t), S = sin(ω0 ∆t) and
ω0 =√
1− ζ20 ω0, while:
A−1 =
[−2 ζ0
ω0− 1ω2
0
1 0
](16)
StructuralDynamics
& EarthquakeEngineering
Dr AlessandroPalmeri
Step-by-Step Numerical Solution
The incremental solution offered by Eq. (11) for the timeinterval [0,∆t ] can be extended to a generic time instanttn = n ∆t as:
yn+1 = y(tn+1) =Θ(∆t) · yn
+ Γ0(∆t) · {b f (tn)}+ Γ1(∆t) · {b f (tn+1)}
(17)
for n = 1,2,3, · · ·