FUNDAMENTALS OF
STRUCTURAL DYNAMICS
Original draft by
Prof. G.D. Manolis, Department of Civil Engineering
Aristotle University, Thessaloniki, Greece
Final draft - Presentation
Prof. P.K. Koliopoulos, Department of Structural Engineering,
Technological Educational Institute of Serres, Greece
• Topics :
• Revision of single degree-of freedom vibration theory
• Response to sinusoidal excitation
• Response to impulse loading
• Response spectrum
• Multi-degree of freedom structures
References :
R.W. Clough and J. Penzien ‘Dynamics of Structures’ 1975
A.K.. Chopra ‘Dynamics of Structures: Theory and Applications to Earthquake
Engineering’ 20011
G.D. Manolis, Analysis for Dynamic Loading, Chapter 2 in Dynamic Loading
and Design of Structures, Edited by A.J. Kappos, Spon Press, London, pp. 31-65, 2001.
Why dynamic analysis? Æ Loads change with time
Unit impulse
f(t)
Harmonic load
f(t)
1/ε
t
τ
Ground acceleration
ε
Single degree of freedom (sdof) system
u(t)
c
m
k
mass-spring-damper
system
f(t)
Μass m (kgr, tn), spring parameter k (kN/m), viscous
damper parameter c (kN*sec/m), displacement u(t) (m),
excitation f(t) (kN).
Definitions of restoring force parameter k
Dynamic equilibrium – D’Alembert’s principle
f(t) = fI(t) + fD(t) + fS(t)
Inertia force fI(t),
Damping force fD(t)
Restoring (elastic) force fS(t)
fI(t)
fD(t)
f(t)
fS(t)
Setting response parameters as: displacement u(t) (in m),
velocity u’(t) (in m/s) and acceleration u’’(t) (in m/s2),
then:
fI(t) = m u’’(t) , fD(t) = c u’(t) ,
fS(t) = k u(t) .
Shear plane frame - dynamic parameters
q
u
C
B
h
k
A
D
l
Rigid beam, mass less
columns. Total weight
(mass) accumulated in the
middle of the beam.
AB – Fixed end
CD – Hinged end
m = w/g = (ql)/g
k = fst(u=1) = VBA + VΓ∆ = 12EI/h3 + 3EI/h3 = 15EI/h3
Free vibration with no damping
u
q
B
C
h
k
A
l
D
FI
B
VΒΑ
C
VCD
No external force f(t). Oscillations due to initial conditions
at t = 0. Initial displacement u0 or/and initial velocity u’0
m u’’(t) + k u(t) = 0
u(t) = R1 sin ωt + R2 cos ωt = R sin(ωt+θ)
where R2 = R12 + R22 and tan θ = R2/R1
Natural frequency ω = [k/m]1/2 (rad/s),
Nat. period T = 2π/ω (sec)
u’0
2
R
u0
1
3
t(s)
5
R
4
To = 2π/ωο
1
2
3
Unrealistic – no decay
4
5
m
Free vibration
with damping
B
c
u(t)
Ι
∞
Γ
∆
A
Equation of motion Æ Homogeneous 2nd order-ODE:
m u’’(t) + c u’(t) + k u(t) = 0
Characteristic equation
(mr2 + cr + k) = 0
and roots: r1,2 = ±
c2
k
2 m
( 2m)
2
k
c
2 m
( 2m)
{
>0
=0
<0
[c/2m]2 – k/m = 0 Î
oscillat
ion
ccr = 2 k * m = 2mω0
ccr = critical damping
c
c
Critical damping ratio ξ =
=
c cr 2mω 0
For ξ < 1.0, and setting
ω d = ω0 1 - ξ 2
u(t) = e-ξω0t (R1 sin ωdt + R2 cos ωdt) = R e-ξω0t sin(ωdt+θ)
R
u 0 + u 0 ξω 0
2
2
2
R1 =
, R2 = u0, R = R 1 + R 2 , tan θ =
R1
ωd
u’0
Exponential decay R*exp(-ξωοt)
Undamped
u0
t(s)
Damped
T0 = 2π/ω0
Td = 2π/ωd
1
0.9
0.8
0.7
0.6
Range of damping for most
structures
ξ 0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
ωd/ω 0
Logarithmic decrement δ = 2πξ, relates the magnitude of
successive peaks
2 πξ
ln(Rj/Rj+n) = n
2 ≈ n*2πξ = nδ
1- ξ
Oscillation due to ground motion
fI = m ut’’(t)
fD = c u’(t)
u
m
Total displacement (ut),
ground displacement (ug),
relative displacement (u).
ut(t) = ug(t) + u(t)
Dynamic equilibrium:
fI + fD + fS = 0
ut
c
ug
fS = k u(t)
Equation of motion:
m ut’’(t) + c u’(t) + k u(t) = 0
k
Setting ut’’(t) = ag(t) + u’’(t), where ag(t) = ground
acceleration, the equation of motion becomes:
m u’’(t) + c u’(t) + k u(t) = - m ag(t) = fg(t)
The above is the equation of motion of a fixed-base frame
under an external dynamic force fg(t).
m
c
k
ug
fg(t) = - m ag(t)
m
=
k
c
Harmonic excitation
f0 sinϖt
m
f(t)
c
t (s)
k
Force with amplitude f0 and excitation frequency ω
Equation of motion Æ Non-homogeneous 2nd order-ODE:
m u( t ) + c u (t) + k u(t) = f0 sin ω t.
Two part solution Æ u(t) = uc(t) + up(t)
Complementary component (transient)
uc(t) = e-ξω0t (C1 sin ωdt + C2 cos ωdt)
Particular component (steady-state)
f0
up(t) =
k
1
(1 - β 2 ) 2 + (2 * βξ) 2
* sin( ω t-θ) = ρ sin( ω t-θ)
ω
where β =
= frequency ratio
ω0
2ξβ
Phase θ is determined via the relation: tan θ =
1 − β2
The steady-state peak ρ is related to the peak of the static
response ust (corresponding to static force fst = f0).
f0
ρ=
D(β,ξ) = ust D(β,ξ)
k
Dynamic amplification factor D(β,ξ), expresses the degree
of error, if an ‘equivalent’ static (instead of fully dynamic)
analysis is performed
D(β,ξ) =
6
1
(1 - β 2 ) 2 + (2 * βξ) 2
ξ=0
ξ=0,1
5
4
D(ξ,β) 3
ξ=0,2
2
ξ=0,5
ξ=1
1
0
0
0.5
1
1.5
β
2
2.5
3
Unit impulse excitation
m
u(t))
Ι∞
f(t)
f
c
1/ε
τ
ε
t
Due to infinitesimal duration ε, during impulse damping
and restoring forces are not activated. After impulse, the
system performs a damped free vibration with initial
conditions u(τ) = 0, u’(τ) = 1/m, (change of momentum
equal to applied force).
Unit impulse response function h(t-τ):
1
u(t) = h(t-τ) =
e--ξω(t-τ) sin[ωd(t-τ)]
mω d
h(t-τ)
1/m
h(t-τ )
τ
t
t
An impulse occurring at time τ, determines the response at
a later time (t ≥ τ). Due to damping, the influence of an
impulse weakens as the time interval increases (memory of
vibration).
Response to arbitrary excitation
f
R e s p o n s e to 1 s t im p u ls e
R e s p o n s e to 2 n d im p u ls e
R e s p o n s e to ν th im p u ls e
T o ta l re s p o n s e
In the limit, for infinitesimal time steps, the summation of
impulse responses becomes an integral - known as
Duhamel’s integral:
t
1
u(t) = ∫h(t - τ) f(τ ) dτ =
m ωd
0
t
∫
f(τ) e-ξωο(t-τ) sin[ωd(t-τ)]dτ
0
The above relation provides a means for determination of
the response of a single degree elastic system subjected to
arbitrary excitation (in analytical or digital form).
Earthquake response spectra
400
300
Athens 1999 (Splb1-L)
200
100
cm/s2 0
-100
s
-200
-300
-400
Equation of motion
m u’’(t) + c u’(t) + k u(t) = - m ag(t) = fg(t)
Duhamel
t
1 t
-ξωο(t-τ) sin[ω (t-τ)]dτ
y(t) = ∫h(t - τ) f g (τ ) dτ =
a
(τ)
e
g
d
∫
ω
d 0
0
For a system with
ξ = 5% και Το = 0.5 s
(ωο = 12.57 rad/s)
the response was
computed as Æ
Quasi-harmonic
response
3.0
2.0
1.0
ξ = 5%, Tο = 0.5 s
cm0.0
s
-1.0
-2.0
-3.0
For design purposes, only peak response parameters
(displacement, velocity, acceleration, moments, shear
forces ) are of interest. These peak values, express the
seismic demand.
The seismic demand for systems with different periods is
expressed via the response spectra.
3 .0
1 .0
0 .8
T = 0 .2
0 .6
s
0 .4
cm
cm
0 .0
s
- 0 .2
0 .0
s
- 1 .0
- 0 .4
- 0 .8
T = 0 .5 s
1 .0
0 .2
- 0 .6
umax= 2,32
2 .0
- 2 .0
umax=-0,75
- 1 .0
- 3 .0
4 .0
3 .0
umax=3,61
T = 1 .0 s
2 .0
1 .0
cm
0 .0
s
- 1 .0
- 2 .0
- 3 .0
- 4 .0
6 .0
5 .0
4 .0
Sd
(c m ) 3 .0
2 .0
1 .0
0 .0
0 .2
0 .4
0 .6
0 .8
1
1 .2
T (s )
1 .4
1 .6
1 .8
2
7
6
Sd (cm)
5
4
3
2
1
0
0
0.5
1
1.5
2
2.5
3
Τ (sec)
Displacement response
component SPLB1-L).
spectrum
Sd
(Athens
99,
The peak displacement values tend to increase with period
(more flexible or taller structures, exhibit larger
deflections).
Sv (cm/sec)
50
45
40
35
30
25
20
15
10
5
0
0
0.5
1
1.5
2
2.5
3
Τ (sec)
Velocity response spectrum Sv
The previously noticed trend is not observed in Sv. After
an initial rise, follows a relatively constant value range and
then a decrease for large periods.
1.2
1
Sa (g)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
Τ (sec)
Acceleration response spectrum Sa
Here, an initial increase of Sa is followed by a rapid
decrease for periods above 0.4 sec. (Flexible structures do
not oscillate rapidly Æ small values of acceleration).
Actual shape depends on rapture characteristics and local
soil conditions
If it is assumed that the response is quasi-harmonic with
frequency equal to the natural frequency, then:
u(t) = umaxsinωt, u’(t) = umaxωcosωt, u’’(t) = - umaxω2sinωt
Therefore, the following (approximate) relations between
response spectra are often implemented:
S v ≈ ω 0 * S d = P Sv ,
Sa ≈ ω02* Sd = PSa
where, P Sv = pseudo-spectral velocity and PSa = pseudospectral acceleration
These approximate relations enable us to present all 3
response spectra with one tri-partite logarithmic plot.
Design parameters of response spectra
m
Static equivalence
approach
Vb = fs = k*Sd
Mb = h* Vb
h
k
fs= k*Sd = m* PSa
Vb = Base shear
Mb = Base moment
νEI PSa
νEI
Column moment: Μc = 2 * Sd = 2 * ω 2
h
h
ο
where, ν = 3 for hinged-end, ν = 6 for fixed-end columns.
Spectral ‘static equivalence’ approach (exact - !) is not a
fully static analysis approach (false - X).
fg(t) = -m ag(t)
fs = -m PSa
(!)
=
Î
V b = fs
ug(t)
fg,max = -m Pga
fg(t) = -m ag(t)
=
ug(t)
(X)
Î
Vb ≠ fg,max
Two degree of freedom (2-dof) system
m2
u2(t) f2(t)
Ι∞
Rigid beams
Massless columns
Zero damping
k2
m1
u1(t)
Ι∞
k1
Two storey shear-frame
f1(t)
Dynamic equilibrium
fI2
fS2a
f2(t)
fS2b
fS2a
fI1
fS1a
fS2b
f1(t)
fS1b
fIj = inertia force j = mj * u j
fSj = fSja + fSjb = kj*(uj – ui) = restoring force due to
columns connecting levels j-1 and j.
fI2 + fS21 = f2(t) → m2 u 2 + k2 (u2-u1) = f2(t)
fI1 + fS12 + fS10 = f1(t) → m1 u1 + k2 (u1-u2) +k1 u1= f1(t)
System of coupled differential equations
Matrix notation
+ K U = F(t)
M U
⎡ u1 (t) ⎤
U = U(t) = displacement vector = ⎢
⎥
u
(t)
⎣ 2 ⎦
⎡ m1 0 ⎤
Μ = mass matrix = ⎢
⎥
0
m
2⎦
⎣
⎡ k1 + k 2
Κ = stiffness matrix = ⎢
⎣ −k 2
⎡ f1 (t) ⎤
F(t) = force vector = ⎢
⎥
⎣ f 2 (t) ⎦
−k 2 ⎤
⎥
k2 ⎦
Free vibration of undamped 2-dof system
Μ*U’’ + K*U = 0
⎡ u1 (t) ⎤ ⎡ φ1 cos(ωt − θ) ⎤ ⎡ φ1 ⎤
U(t) = ⎢
= ⎢
= ⎢ ⎥ cos(ωt-θ) =
⎥
⎥
⎣ u 2 (t) ⎦ ⎣φ 2 cos(ωt − θ) ⎦ ⎣ φ 2 ⎦
Φ cos(ωt-θ)
u1 (t) ⎤ ⎡ −ω2 φ1 cos(ωt − θ) ⎤
⎡ (t) = ⎢
=⎢ 2
= -ω2 Φ cos(ωt-θ)
U
⎥
⎥
u 2 (t) ⎦ ⎣ −ω φ 2 cos(ωt − θ) ⎦
⎣ M [-ω2 Φ cos(ωt-θ)] + K [Φ cos(ωt-θ)] = [0] →
{K - ω2 Μ} Φ cos(ωt-θ) = [0]
Unknowns are the amplitude vector Φ and the frequency
of free oscillation ω.
{K - ω2 Μ} Φ cos(ωt-θ) = [0]
Should be valid for any time instant Æ zero determinant
Κ − ω Μ = [0] →
2
k1 + k 2 − ω m1
−k 2
−k 2
k 2 − ω m2
2
2
= [0] →
ω4 (m1m2) – ω2 {(k1+k2)m2 + k2m1} + k1k2 = 0
This is the frequency equation. Setting ω2 = λ, we get two
solutions for λ and hence, two frequency values for free
vibration λ1 = ω12 and λ2 = ω22.
Therefore, a 2-dof system exhibits 2 natural frequencies,
ω1 and ω2.
Substituting ω1 and ω2 back into the matrix equation, the
two corresponding amplitude vectors (eigenvectors) can be
evaluated.
{K – ωj2 Μ} Φj cos(ωjt-θ) = [0] → {K – ωj2 Μ} Φj = [0]
The eigenvalue problem does not fix the absolute
amplitude of the vectors Φj , but only the shape of the
vector (relative values of displacement)
u2(t)
m
Example
Ι∞
k
2m
Ι∞
2k
u1(t)
Natural frequencies determination
Κ−ω Μ = 0 Æ
2
3k − 2ω2 m
−k
−k
k −ω m
2
=0Æ
2ω4 m2 – 5ω2 km + 2k2 = 0
Roots of quadratic equation ω12 = k/2m και ω22 = 2k/m,
with corresponding natural periods
Τ1 = 2π/ω1 = π
8m
,
k
Modal shapes calculation Æ
Τ2 = 2π/ω2 = π
2m
k
⎡ φ11 ⎤
Eigenvectors Φ1 = ⎢ ⎥ and Φ2 =
⎣ φ 21 ⎦
as:
2k
−
k
⎡
⎤
2
ω1 = k/2m Æ ⎢
⎥
⎣ −k k / 2⎦
⎡ φ12 ⎤
⎢ φ ⎥ , are computed
⎣ 22 ⎦
⎡ φ11 ⎤ ⎡ 0⎤
⎢ φ ⎥ = ⎢ 0⎥ Æ 2φ11 = φ21
⎣ 21 ⎦ ⎣ ⎦
− k − k ⎤ ⎡ φ12 ⎤ ⎡ 0⎤
⎡
ω22 = 2k/m Æ ⎢
= ⎢ ⎥ Æ φ12 = -φ22
⎥
⎢
⎥
⎣ − k − k ⎦ ⎣ φ 22 ⎦ ⎣ 0⎦
Setting (arbitrarily) φ21 = φ22 = 1.0, we get:
⎡ 0.5⎤
Φ1 = ⎢ ⎥ ,
⎣1.0 ⎦
⎡ −1⎤
Φ2 = ⎢ ⎥ , και Φ =
⎣1⎦
⎡ 0.5 −1⎤
⎢ 1
⎥
1
⎣
⎦
Orthogonality of modes
Eigenvectors are orthogonal with respect to mass and
stiffness matrices.
ΦjT M Φk = 0 and ΦjT Κ Φk = 0, για j ≠ k
Modal analysis
2
Set
U(t) =
∑ Φ q (t)
j=1
j
j
= Φ Q(t)
Substitute to the matrix equation of motion:
+ Κ ΦQ(t) = [0]
+ K U = [0] Æ Μ Φ Q(t)
MU
Pre-multiply all terms with ΦΤ :
+ Κ* Q(t) = [0]
+ ΦΤ ΚΦ Q(t) = [0] ÆΜ* Q(t)
ΦΤ ΜΦ Q(t)
The transformed matrix equation of free vibration, reads:
+ Κ* Q(t) = [0]
Μ* Q(t)
Due to orthogonality property the new matrices M* and
K* are diagonal.
⎡ m1*
Μ* = generalized mass matrix = ⎢
⎣0
0 ⎤
*⎥
m2 ⎦
⎡ k1*
Κ* = generalized stiffness matrix = ⎢
⎣0
0⎤
*⎥
k2 ⎦
Therefore, the original matrix equation is transformed
into a set of uncoupled sdof free vibration equations of the
form (for j = 1, 2):
m*j q j (t) + k*j qj(t) = 0 Æ q j (t) + ω2j qj(t) = 0
Modal decoupling
m*1
k*1
u2
m2
q1
k2
m1
u2(t) = u21(t) + u22(t) =
φ21 q1(t) + φ22 q2(t)
u1
k1
m*2
q2
k*2
Forced vibration of a damped multi degree of
freedom (mdof) system
Original (coupled) equation of motion:
+ K U = F(t)
+ C U
ΜU
Modal (decoupled) equation of motion:
+ ΦΤCΦ Q + ΦΤKΦ Q = ΦΤ F(t) Æ
ΦΤMΦ Q
+ C* Q + K* Q = F*(t)
Μ* Q
where, C* = generalized damping matrix and F* =
generalized force vector.
To ensure diagonalization of C*, here the assumption is
made that the damping matrix of the original system C can
be expressed as
C = α·Μ + β·Κ
Typical generalized (sdof) equation of motion:
m*j q j + c*j q j + k *j qj = f j* (t) Æ q j +2ξjωj q j +ωj2qj =
f j* (t)
m*j
= f j (t)
To be solved within the framework of sdof theory (1st part
of presentation).
Following the determination of generalized vector Q, the
original response vector U is computed as
ν
U(t) = Φ Q(t) = ∑ Φ jq j (t)
j=1
The contribution of first modes are much more important
than the contribution of higher modes.
Earthquake excitation of mdof systems
(Response spectrum analysis)
ν-storey
shear
plane frame under
ground
motion
ug(t)
The total displacement vector Ut(t), is composed by the
relative displacement vector U(t) and the ground motion.
Ut(t) = U(t) +[1]ug(t)
The matrix equation of motion of the original system is:
+ K U = - Μ [1] a (t) = F (t)
+ C U
ΜU
g
g
Firstly we compute ωj και Φj, and then we proceed to the
modal transformation
+ ΦΤCΦ Q + ΦΤKΦ Q = ΦΤ F (t) Æ
ΦΤMΦ Q
g
+ C* Q + K* Q = F*(t)
Μ* Q
Here, the generalized force vector is
F*(t) = ΦΤ Fg(t) = - ΦΤ Μ [1] ag(t)
The sdof generalized equations are
q j + 2ξ ω q j + ω 2q =
j j
j j
f j* (t)
m
*
j
ν
∑m φ
k
= - ag(t)
k =1
ν
kj
2
m
φ
∑ k kj
= - Γj ag(t)
k =1
The generalized force parameter Γj is known as modal
participation factor.
If the seismic action is expressed via the standard response
or design spectra, the corresponding spectral values of the
generalized response qj, are
Sd,j = Γj Sd(Tj,ξj),
Sv,j = Γj Sv(Tj,ξj),
Sa,j = Γj Sa(Tj,ξj)
Example of utilization of Greek Design spectrum (EAK)
for the estimation of modal spectral accelerations of a
mdof frame
The ordinates
of the design
spectrum
should
be
multiplied
by
the
corresponding
modal
participation
factors Γj.
The problem of combination of modal peak values
The following decomposition of physical response uj in
terms of generalized (modal) components qk is valid for
any instant of time.
ν
uj(t) =
∑φ
k =1
ν
jk
q k (t) = ∑ u jk (t)
k =1
where ujk(t) is the ‘contribution’ of k modal component
qk(t) to the response of the j degree of freedom uj(t) of the
original system.
However, if only spectral (peak) modal response quantities
are available
u jk= φj,k Γk Sd(Tk,ξk)
these do not occur at the same time and hence, cannot be
added to obtain the peak value of uj(t)
Modal contributions
u3k(t) (for k = 1,2,3)
to the response of
the top floor of a 3storey frame
( u 31 + u 32 + u 33 ) =
(4.91 + 1.56 + 0.10) =
6.57 > 5.16
Modal combination
rule SRSS
2
2
2
= 4.912 + 1.562 + 0.102 = 5.15 ≈ 5.16
u 31
+ u 32
+ u 33
THE END