AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Direct Time-Integration Methods
These slides are based on the recommended textbook: M. Géradin and D. Rixen, “Mechanical
Vibrations: Theory and Applications to Structural Dynamics,” Second Edition, Wiley, John &
Sons, Incorporated, ISBN-13:9780471975465
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Outline
1 Stability and Accuracy of Time-Integration Operators
2 Newmark’s Family of Methods
3 Explicit Time Integration Using the Central Difference Algorithm
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Multistep Time-Integration Methods
Lagrange’s equations of dynamic equilibrium (p(t) = 0)
Mq̈ + Cq̇ + Kq = 0
q(0) = q0
q̇(0) = q̇0
First-order form
0 M
q̈
−M 0
q̇
0
+
=
M C
0
K
q
0
q̇
|
{z
} | {z } |
{z
} | {z } | {z }
AB
u̇
=⇒ u̇ = Au
−AA
where
u
A = A−1
B AA
Direct time-integration
AA242B: MECHANICAL VIBRATIONS
0
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Multistep Time-Integration Methods
General multistep time-integration method for first-order systems of
the form u̇ = Au
m
m
X
X
un+1 =
αj un+1−j − h
βj u̇n+1−j
j=1
j=0
where h = tn+1 − tn is the computational time-step, un = u(t n ), and
qn+1
un+1 =
q̇n+1
is the state-vector calculated at tn+1 from the m preceding state
vectors and their derivatives as well as the derivative of the
state-vector at tn+1
β0 6= 0 leads to an implicit scheme — that is, a scheme where the
evaluation of un+1 requires the solution of a system of equations
β0 = 0 corresponds to an explicit scheme — that is, a scheme where
the evaluation of un+1 does not require the solution of any system of
equations and instead can be deduced directly from the results at the
previous time-steps
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Multistep Time-Integration Methods
General multistep integration method for first-order systems
(continue)
m
m
X
X
αj un+1−j − h
βj u̇n+1−j
un+1 =
j=1
j=0
trapezoidal rule (implicit)
un+1 = un +
h
h
h
(u̇n + u̇n+1 ) ⇒ ( A − I)un+1 = −un − u̇n
2
2
2
backward Euler formula (implicit)
un+1 = un + hu̇n+1 ⇒ (hA − I)un+1 = −un
forward Euler formula (explicit)
un+1 = un + hu̇n ⇒ un+1 = (I + hA)un
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Numerical Example: the One-Degree-of-Freedom Oscillator
Consider an undamped one-degree-of-freedom oscillator
q̈ + ω02 q = 0
with ω0 = π rad/s and the initial displacement
q(0) = 1, q̇(0) = 0
exact solution
q(t) = cos ω0 t
associated first-order system
u̇ = Au
where
A=
−ω02
0
0
1
u = [q̇, q]T , and initial condition
u(0) =
0
1
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Numerical Example: the One-Degree-of-Freedom Oscillator
Numerical solution
T
T = 3s, h = 32
3
Exact solution
Trapezoidal rule
Euler backward
Euler forward
2
1
q
0
−1
−2
−3
−4
0
0.5
1
1.5
t
2
2.5
AA242B: MECHANICAL VIBRATIONS
3
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Stability Behavior of Numerical Solutions
Analysis of the characteristic equation of a time-integration method
consider the first-order system u̇ = Au
for this problem, the general multistep method can be written as
un+1 =
m
X
αj un+1−j − h
j=1
m
X
βj u̇n+1−j ⇒
j=0
m
X
[αj I − hβj A] un+1−j = 0,
α0 = −1
j=0
let µr be the eigenvalues of A and X be the matrix of associated
eigenvectors
m
P
the characteristic equation associated with [αj I − hβj A] un+1−j = 0 is
j=0
obtained by searching for a solution of the form
un+1−m
=
Xa
(decomposition on an eigen basis)
u(n+1−m)+1
=
λun+1−m = λXa
=
λun = · · · = λk un+1−k = · · · = λm Xa
(solution form)
..
.
un+1
where λ ∈ C is called the solution amplification factor
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Stability Behavior of Numerical Solutions
Analysis of the characteristic equation of a time-integration method
(continue)
Hence
m
X
[αj I − hβj A] λm−j Xa = 0
j=0
−1
Since X
leads to
AX = diag(µr ), premultiplying the above result by X−1
" m
#
X
m−j
[αj I − hβj diag(µr )] λ
a=0
j=0
=⇒
m
X
[αj − hβj µr ] λm−j = 0, r = 1, 2
j=0
hence, the numerical response un+1 = λm Xa remains bounded if each
solution of the above characteristic equation of degree m satisfies
|λk | < 1, k = 1, · · · , m
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Stability Behavior of Numerical Solutions
Analysis of the characteristic equation of a time-integration method
(continue)
the stability limit is a circle of unit radius
in the complex plane of µr h, the stability limit is therefore given by
writing λ = e iθ , 0 ≤ θ ≤ 2π
m
X
=⇒ µr h =
αj e i(m−j)θ
j=0
m
X
βj e i(m−j)θ
j=0
one-step schemes (m = 1)
µr h =
α0 e iθ + α1
−e iθ + α1
=
β0 e iθ + β1
β0 e iθ + β1
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Stability Behavior of Numerical Solutions
Analysis of the characteristic equation of a time-integration method
(continue)
one-step schemes (m = 1) (continue)
µr h =
α0 e iθ + α1
−e iθ + α1
=
iθ
β0 e + β1
β0 e iθ + β1
forward Euler: α1 = 1, β0 = 0, β1 = −1 ⇒ µr h = e iθ − 1
the solution is unstable in the entire plane except inside the circle of
unit radius and center −1
backward Euler: α1 = 1, β0 = −1, β1 = 0 ⇒ µr h = 1 − e −iθ
the solution is stable in the entire plane except inside the circle of
unit radius and center 1
1
1
2i sin θ
trapezoidal rule: α1 = 1, β0 = − , β1 = − ⇒ µr h = 1+cos
θ
2
2
the solution is stable in the entire left-hand plane
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Stability and Accuracy of Time-Integration Operators
Stability Behavior of Numerical Solutions
Analysis of the characteristic equation of a time-integration method
(continue)
application to the single degree-of-freedom oscillator
0 −ω02
q̈ + ω02 q = 0,
A=
1
0
the eigenvalues are µr = ±iω0
the roots µr h are located in the unstable region of the forward Euler
scheme ⇒ amplification of the numerical solution
the roots µr h are located in the stable region of the backward Euler
scheme ⇒ decay of the numerical solution
the roots µr h are located on the stable boundary of the trapezoidal
rule scheme ⇒ the amplitude of the oscillations is preserved
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
The Newmark Method
Taylor’s expansion of a function f
0
f (tn + h) = f (tn ) + hf (tn ) +
hs (s)
1
h2 00
f (tn ) + · · · +
f (tn ) +
2
s!
s!
Z
tn +h
f
(s+1)
Application to the velocities and displacements
Z tn+1
f = q̇, s = 0 ⇒ q̇n+1 = q̇n +
q̈(τ )dτ
tn
Z tn+1
f = q, s = 1 ⇒ qn+1 = qn + h q̇n +
q̈(τ )(tn+1 − τ )dτ
tn
AA242B: MECHANICAL VIBRATIONS
s
(τ )(tn + h − τ ) dτ
tn
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
The Newmark Method
Taylor expansions of q̈n and q̈n+1 around τ ∈ [tn , tn+1 ]
q̈n
=
q̈n+1
=
(tn − τ )2
+ ···
(1)
2
(tn+1 − τ )2
+ · · · (2)
q̈(τ ) + q(3) (τ )(tn+1 − τ ) + q(4) (τ )
2
q̈(τ ) + q(3) (τ )(tn − τ ) + q(4) (τ )
Combine (1 − γ) (1) + γ (2) and extract q̈(τ )
=⇒ q̈(τ ) = (1 − γ)q̈n + γq̈n+1 + q(3) (τ )(τ − hγ − tn ) + O(h2 q(4) )
Combine (1 − 2β) (1) + 2β (2) and extract q̈(τ )
=⇒ q̈(τ ) = (1 − 2β)q̈n + 2βq̈n+1 + q(3) (τ )(τ − 2hβ − tn ) + O(h2 q(4) )
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
The Newmark Method
tn+1
Z
Substitute the 1st expression of q̈(τ ) in
q̈(τ )dτ
tn
Z
=⇒
tn+1
tn+1
Z
q̈(τ )dτ
=
tn
(1 − γ)q̈n + γq̈n+1 + q
(τ )(τ − hγ − tn ) + O(h q
) dτ
(3)
3 (4)
(3)
2 (4)
tn
Z
=
(1 − γ)h q̈n + γh q̈n+1 +
tn+1
q
(τ )(τ − hγ − tn )dτ + O(h q
)
tn
Apply the mean value theorem
Z
=⇒
tn+1
"
q̈(τ )dτ
=
(1 − γ)h q̈n + γh q̈n+1 + q
(3)
(τ̃ )
tn
=
(1 − γ)h q̈n + γh q̈n+1 + (
Substitute the 2nd expression of q̈(τ ) in
(τ − hγ − tn )2
2
#t
n+1
3 (4)
+ O(h q
)
tn
1
2 (3)
3 (4)
− γ)h q (τ̃ ) + O(h q )
2
Z
tn+1
q̈(τ )(tn+1 − τ )dτ
tn
Z
=⇒
tn+1
tn
q̈(τ )(tn+1 − τ )dτ
=
(
1
1
2
2
3 (3)
4 (4)
− β)h q̈n + βh q̈n+1 + ( − β)h q (τ̃ ) + O(h q )
2
6
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
The Newmark Method
In summary
Z
tn+1
q̈(τ )dτ = (1 − γ)h q̈n + γh q̈n+1 + rn
tn
Z
tn+1
q̈(τ )(tn+1 − τ )dτ =
tn
1
− β h2 q̈n + βh2 q̈n+1 + rn0
2
where
rn
rn0
1
=
− γ h2 q(3) (τ̃ ) + O(h3 q(4) )
2
1
− β h3 q(3) (τ̃ ) + O(h4 q(4) )
=
6
and tn < τ̃ < tn+1
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
The Newmark Method
Hence, the approximation of each of the two previous integral terms
by a quadrature scheme leads to
q̇n+1
=
qn+1
=
q̇n + (1 − γ)h q̈n + γh q̈n+1
1
2
qn + h q̇n + h
− β q̈n + h2 βq̈n+1
2
where γ and β are parameters associated with the quadrature
scheme
AA242B: MECHANICAL VIBRATIONS
(3)
(4)
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
The Newmark Method
Particular values of the parameters γ and β
γ=
γ=
1
1
and β = leads to linearly interpolating q̈(τ ) in [tn , tn+1 ]
2
6
q̈n+1 − q̈n
q̈ln (τ ) = q̈n + (τ − tn )
h
1
1
and β = leads to averaging q̈(τ ) in [tn , tn+1 ]
2
4
q̈n+1 + q̈n
q̈av (τ ) =
2
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
The Newmark Method
Application to the direct time-integration of Mq̈ + Cq̇ + Kq = p(t)
write the equilibrium equation at t n+1 and substitute the expressions
(3) and (4) into it
=⇒ [M + γhC + βh2 K]q̈n+1 = pn+1 − C[q̇n + (1 − γ)hq̈n ]
1
− β h2 q̈n
− K qn + hq̇n +
2
if the time-step h is uniform, M + γhC + βh2 K can be factored once
solve the above system of equations for q̈n+1
substitute the result into the expressions (3) and (4) to obtain q̇n+1
and qn+1
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Consistency of a Time-Integration Method
A time-integration scheme is said to be consistent if
un+1 − un
= u̇(tn )
h→0
h
lim
The Newmark time-integration method is consistent


(1 − γ)q̈n + γq̈n+1
un+1 − un
q̈n


1
=
lim
= lim
q̇n
− β h q̈n + βh q̈n+1
q̇n +
h→0
h→0
h
2
Consistency is a necessary condition for convergence
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Stability of a Time-Integration Method
A time-integration scheme is said to be stable if there exists an
integration time-step h0 > 0 so that for any h ∈ [0, h0 ], a finite
variation of the state vector at time tn induces only a non-increasing
variation of the state-vector un+j calculated at a subsequent time
tn+j
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Stability of a Time-Integration Method
The application of the Newmark scheme to Mq̈ + Cq̇ + Kq = p(t)
can be put under the form
un+1 = A(h)un + gn+1 (h)
where A is the amplification matrix associated with the integration
operator
−1
A(h) = H−1
1 (h)H0 (h), gn+1 = H1 (h)bn+1 (h)


(1 − γ)hp
n + γhpn+1
M + γhC
γhK

bn+1 =  1
,
H
=
1
βh2 C
M + βh2 K
− β h2 pn + βh2 pn+1
2


(1− γ)hK
−M +(1 − γ)hC

1
H0 = −  1
− β h2 C − hM −M +
− β h2 K
2
2
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Stability of a Time-Integration Method
Effect of an initial disturbance
δu0 = u00 − u0
=⇒ δun+1 = A(h)δun = A2 (h)δun−1 = · · · = A(h)n+1 δu0
consider the eigenpairs of A(h)
(λr , xr )
then
δun+1 = An+1 (h)
2N
X
s=1
as xs =
2N
X
as λn+1
xs
s
s=1
where N is the dimension of the semi-discrete second-order
dynamical system
=⇒ δun+1 will be amplified by the time-integration operator only if
the moduli of an eigenvalue of A(h) is greater than unity
=⇒ δun+1 will not be amplified by the time-integration operator if all
moduli of all eigenvalues of A(h) are less than unity
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Stability of a Time-Integration Method
Undamped case
decouple the equations of equilibrium by writing them (for the
purpose of analysis) in the modal basis
q = Qy =
N
X
yi qai =⇒ ÿi + ωi2 yi = pi (t)
i=1
apply the Newmark scheme to the i-th modal equation recalled
above to obtain the amplification matrix


ωi2 h2
ωi2 h2
1 − γ 1+βω
−ωi2 h2 1 − γ2 1+βω
2 h2
2 h2
i
i

A(h) = 
2 2
h
1 ωi h
1
−
2
2
2
2
2 1+βω h
1+βω h
i
2
i
characteristic equation is λ − λ 2 − (γ +
ωi2 h2
1+βωi2 h2
1
)ξ 2
2
+ 1 − (γ − 12 )ξ 2 = 0
where ξ 2 =
characteristic equation has a pair of conjugate roots λ1 and λ2 if
2
1
4
γ+
− 4β ≤ 2 2 , i = 1, · · · , N
2
ωi h
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Stability of a Time-Integration Method
Undamped case (continue)
the eigenvalues λ1 and λ2 can be written as
λ1,2
=
ρe ±iφ
where
s
ρ
=
φ
=
1
ξ2
1− γ−
2
 q

ξ 1 − 14 (γ + 12 )2 ξ 2

arctan 
1 − 12 (γ + 21 )ξ 2
then, the Newmark scheme is stable if
ρ≤1⇒γ≥
but recall that this is assuming
2
1
4
γ+
− 4β ≤ 2 2 ,
2
ωi h
1
2
i = 1, · · · , N
=⇒ limitation on the maximum time-step
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Stability of a Time-Integration Method
Undamped case (continue)
the algorithm is conditionally stable if
γ≥
1
2
it is unconditionally stable if furthermore
1
β≥
4
2
1
γ+
2
1
1
the choice γ = and β = leads to an unconditionally stable
2
4
time-integration operator of maximum accuracy
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Stability of a Time-Integration Method
Undamped case (continue)
Stability of the Newmark scheme
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Stability of a Time-Integration Method
Damped case (C 6= 0)
consider the case of modal damping
then, the uncoupled equations of motion are
ÿi + 2εi ωi ẏi + ωi2 yi = pi (t)
where εi is the modal damping coefficient
1
1
consider the case γ = , β =
2
4
an analysis similar to that performed in the undamped case reveals
that in this case, the Newmark scheme remains stable as long as
ε<1
in general, damping has a stabilizing effect for moderate values of ε
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Amplitude and Periodicity Errors
Free-vibration of an undamped linear oscillator
2
ÿ + ω y = 0
and
y (0) = y0 , ẏ (0) = 0
A=
0
1
−ω02
0
the above problem has an exact solution y (t) = y0 cos ωt which can
be written in complex discrete form as yn+1 = e iωh yn ⇒ the exact
amplification factor is ρex = 1 and the exact phase is φex = ωh
the numerical solution satisfies
ẏn+1
un+1 =
= A(h)un
yn+1
let λ1,2 (β, γ) be the eigenvalues of A(h, β, γ)
2
when γ + 21 − 4β ≤ ω24h2 , λ1 and λ2 are complex-conjugate
i
λ1,2 (β, γ) = ρ(β, γ)e ±iφ(β,γ)
where
s
ρ=
1−
1
γ−
2
ξ2 ,
 q

1
1 2 2
 ξ 1 − 4 (γ + 2 ) ξ 
φ = arctan 
,
1 − 12 (γ + 12 )ξ 2
AA242B: MECHANICAL VIBRATIONS
2
ξ =
ω 2 h2
1 + βω 2 h2
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Amplitude and Periodicity Errors
Free-vibration of an undamped linear oscillator (continue)
amplitude error
ρ − ρex = ρ − 1 = −
1
2
1
γ−
ω 2 h2 + O(h4 )
2
relative periodicity error
∆ φ1
∆T
= 1 =
T
φ
1
φ
−
1
φex
1
φex
=
ωh
1
−1=
φ
2
β−
1
12
ω 2 h2 + O(h3 )
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Amplitude and Periodicity Errors
Algorithm
γ
β
Stability
limit
ωh
Purely explicit
Central difference
0
1
2
0
0
0
2
Fox & Goodwin
1
2
1
12
Linear acceleration
1
2
Average constant
acceleration
1
2
Table:
Amplitude
error
ρ−1
ω 2 h2
4
Periodicity
error
∆T
T
0
—
2 2
− ω24h
2.45
0
O(h3 )
1
6
3.46
0
ω 2 h2
24
1
4
∞
0
ω 2 h2
12
Time-integration schemes of the Newmark family
The purely explicit scheme (γ = 0, β = 0) is useless
The Fox & Godwin scheme has asymptotically the smallest phase
error but is only conditionally stable
1
1
The average constant acceleration scheme (γ = , β = ) is the
2
4
unconditionally stable scheme with asymptotically the highest
accuracy
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Total Energy Conservation
Conservation of total energy
dynamic system with scleronomic constraints
ns
X
d
(T + V) = −mD +
Qs q̇s
dt
s=1
1
1
T = q̇T Mq̇ and V = qT Kq
2
2
the dissipation function D is a quadratic function of the velocities
(m = 2)
1
D = q̇T Cq̇
2
external force component of the power balance
ns
X
Qs q̇s = q̇T p
s=1
integration over a time-step [tn , tn+1 ]
Z tn+1
t
[T + V]tn+1
=
(−q̇T Cq̇ + q̇T p)dt
n
tn
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Newmark’s Family of Methods
Total Energy Conservation
Conservation of total energy (continue)
note that because M and K are symmetric (MT = M and KT = K)
1
T
(q̇n+1 − q̇n ) M(q̇n+1 + q̇n )
2
1
T
+
(qn+1 − qn ) K(qn+1 + qn )
2
when time-integration is performed using the Newmark algorithm with
1
1
γ = , β = , the above variation becomes see (3) and (4)
2
4
t
[T + V]tn+1
= [Tn+1 − Tn ] + [Vn+1 − Vn ]
n
t
=
[T + V]tn+1
n
=
1
h
T
T
(qn+1 − qn ) (pn + pn+1 ) − (q̇n+1 + q̇n ) C(q̇n+1 + q̇n )
2
4
when applied to a conservative system (C = 0 and p = 0), preserves the total energy
Rt
t
= tnn+1 (−q̇T Cq̇ + q̇T p)dt and therefore
for non-conservative systems, [T + V]tn+1
n
both terms in the right-hand side of the above formula result from numerical
quadrature relationships that are consistent with the time-integration operator
Z t
Z t
n+1 T
n+1 T
pn + pn+1
1
T
q̇ dt
q̇ pdt ≈
= (qn+1 − qn ) (pn + pn+1 )
2
2
tn
tn
Z
Z t
tn+1
n+1 T
q̇n + q̇n+1
1
q̇n + q̇n+1
T
T
q̇ Cq̇dt ≈
q̇ dt C
= (qn+1 − qn ) C
2
2
2
tn
tn
=
h
T
(q̇n+1 + q̇n ) C(q̇n+1 + q̇n )
4
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Explicit Time Integration Using the Central Difference Algorithm
Algorithm in Terms of Velocities
Central Difference (CD) scheme = Newmark’s scheme with γ = 12 ,
β=0
q̇n+1
=
qn+1
=
q̈n + q̈n+1
)
2
h2
qn + hn+1 q̇n + n+1 q̈n
2
q̇n + hn+1 (
where hn+1 = tn+1 − tn
Equivalent three-step form
h2
start with qn = qn−1 + hn q̇n−1 + n q̈n−1
2
divide by hn
subtract the result from qn+1 divided by hn+1
account for the relationship (5)
=⇒ q̈n =
hn (qn+1 − qn ) − hn+1 (qn − qn−1 )
hn+ 1 hn hn+1
2
where hn+ 1
2
hn + hn+1
=
2
AA242B: MECHANICAL VIBRATIONS
(5)
AA242B: MECHANICAL VIBRATIONS
Explicit Time Integration Using the Central Difference Algorithm
Algorithm in Terms of Velocities
Case of a constant time-step h
q̈n =
qn+1 − 2qn + qn−1
h2
Efficient implementation
compute the velocity at half time-step
q̇n+ 1 = q̇(tn+ 1 ) = q̇n +
2
2
h
1
q̈n = (qn+1 − qn )
2
h
compute
q̈n =
1
(q̇ 1 − q̇n− 1 )
h n+ 2
2
Stability condition
ωcr h ≤ 2
where ωcr is the highest frequency contained in the model: this condition is also known as
the Courant condition, and
2
hcr =
ωcr
is referred to here as the maximum Courant stability time-step
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Explicit Time Integration Using the Central Difference Algorithm
Application Example: the Clamped-Free Bar Excited by an End Load
Clamped bar subjected to a step load at its free end
Model made of N = 20 finite elements with equal length l =
1
2
1
3
2
19
3
17
18
L
N
20
19
20
lumped mass matrix
Eigenfrequencies of the continuous system
r
r
π
EA
2r − 1 π EA
2r − 1 π
ωcontr = (2r − 1)
=
=
2 mL2
N
2 ml 2
N
2
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Explicit Time Integration Using the Central Difference Algorithm
Application Example: the Clamped-Free Bar Excited by an End Load
Finite element stiffness and mass matrices

ml
M=
2








2
2
 −1


EA 

K=

l 





2
0








2
..
.
0
2
1
−1
2
−1

−1
2
..
0
..
..
.
.
.
−1
0
−1
2
−1









−1 
1
(6)
Analytical frequencies of the discrete system
r
ωr = 2
EA
sin
ml 2
2r − 1
2N
π
2
2r − 1
2N
π
2
=
2 sin
,
⇒
ωcr < ωcr (N → ∞) = 2
r = 1, 2, · · · N
Critical time-step for the CD algorithm
ωcr hcr = 2 ⇒ hcr = 1
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Explicit Time Integration Using the Central Difference Algorithm
Application Example: the Clamped-Free Bar Excited by an End Load
h = 1, h = 0.707
Node 1
3
Node 10
25
h=1
h=0.707
2.5
h=1
h=0.707
20
Displacement
Displacement
2
1.5
1
0.5
15
10
5
0
0
−0.5
−1
0
50
100
−5
0
150
Time
Node 20
40
1
100
150
100
150
Time
Node 10
1.5
h=1
h=0.707
35
50
h=1
h=0.707
0.5
25
Velocity
Displacement
30
20
15
0
−0.5
10
−1
5
0
0
50
100
Time
150
−1.5
0
50
Time
AA242B: MECHANICAL VIBRATIONS
AA242B: MECHANICAL VIBRATIONS
Explicit Time Integration Using the Central Difference Algorithm
Application Example: the Clamped-Free Bar Excited by an End Load
h = 1.0012
Node 10
60
Node 20
100
h = 1.0012
80
h = 1.0012
40
Displacement
Displacement
60
20
0
−20
40
20
0
−20
−40
−40
−60
−60
0
50
100
Time
150
−80
0
50
100
Time
AA242B: MECHANICAL VIBRATIONS
150
AA242B: MECHANICAL VIBRATIONS
Explicit Time Integration Using the Central Difference Algorithm
Restitution of the Exact Solution by the Central Difference Method
For the clamped-free bar example, the CD method computes the
exact solution when h = hcr
Comparison of the exact solution of the continuous free-vibration bar
problem and the analytical expression of the numerical solution
denote by qj,n the value of the j-th d.o.f. at time tn
if qj,n is not located at the boundary, it satisfies see (6)
EA
ml
(qj,n+1 − 2qj,n + qj,n−1 ) +
(−qj−1,n + 2qj,n − qj+1,n ) = 0
h2
l
the general solution of the above problem is
qj,n = sin(jµ + φ) [a cos nθ + b sin nθ]
|
{z
} |
{z
}
spatial component
temporal component
comparing the above expression to the exact harmonic solution of
the continuous form of this free-vibration problem (which can be
derived analytically)
=⇒ nθ = ωt = ωnh ⇒
θ
= ωnum
h
AA242B: MECHANICAL VIBRATIONS
(7)
AA242B: MECHANICAL VIBRATIONS
Explicit Time Integration Using the Central Difference Algorithm
Restitution of the Exact Solution by the Central Difference Method
Comparison of the exact solution of the free-vibration bar problem
and the analytical expression of the numerical solution (continue)
introduce the exact expression for qj,n in the CD scheme
2[(1 − cos µ) − λ2 (1 − cos θ)]qj,n = 0
2
1
1
ml
1
where λ2 =
= 2 ⇒ 1 − cos θ = 2 (1 − cos µ)
EA h2
h
λ
make use of the boundary conditions in space q0,n = 0,
and plug (7)
in the last equation in (6) =⇒ φ = 0 and µr = 2rN−1 π2 , r ∈ N∗
=⇒1 − cos θr =
1
(1 − cos µr )
λ2
special case λ2 = 1 (h = hcr = 1) ⇒ θr = µr and
r
θr
2r − 1 π EA
2r − 1 π
ωnumr =
= µr =
=
h
N
2 ml 2
N
2
=⇒ the numerical frequency coincides with the r -th eigenfrequency
of the continuous system
AA242B: MECHANICAL VIBRATIONS