SDEE: Lecture 6

8

Click here to load reader

Transcript of SDEE: Lecture 6

Page 1: 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

Page 2: SDEE: Lecture 6

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)

Page 3: SDEE: Lecture 6

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)

Page 4: SDEE: Lecture 6

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)

Page 5: SDEE: Lecture 6

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.

Page 6: SDEE: Lecture 6

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)

Page 7: SDEE: Lecture 6

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)

Page 8: SDEE: Lecture 6

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, · · ·