1. Modeling Dynamic Systems
• Basic Quantities From Earthquake Records
• Fourier Transform, Frequency Domain
• Single Degree of Freedom Systems (SDOF)
Elastic Response Spectra
• Multi-Degree of Freedom Systems, (MDOF)
Modal Analysis
• Dynamic Analysis by Modal Methods
• Method of Complex Response
4. Acceleration vs. Time
Acceleration vs. Time
4.0000E-01
3.0000E-01
2.0000E-01
Accel (g)
1.0000E-01
0.0000E+00
-1.0000E-01
-2.0000E-01
-3.0000E-01
-4.0000E-01
0.00
10.00
20.00
30.00
40.00
50.00
Time (sec)
60.00
70.00
80.00
90.00
5. Acceleration vs. Time, t=16.00 tot=16 to 20 sec
vs Time 20.00 seconds
Acceleration
4.0000E-01
3.0000E-01
2.0000E-01
Accel (g)
1.0000E-01
0.0000E+00
-1.0000E-01
-2.0000E-01
-3.0000E-01
-4.0000E-01
16.00
16.50
17.00
17.50
18.00
Time (sec)
18.50
19.00
19.50
20.00
6. Harmonic Motion
t = time
A = amplitude of wave
ω = frequency (radians / sec) SDOF Response
1.00E-02
8.00E-03
6.00E-03
X=A sin(ωt-φ)
Displ. (m)
4.00E-03
Amplitude
2.00E-03
0.00E+00
φ = phase lag (radians )
Mass = 10.132 kg
Damping = 0.00
Spring = 1.0 N/m
ωn=√k/m=0.314 r/s
Drive Freq = 0.0
Drive Force = 0.0 N
Initial Vel. = 0.0
m/s
Initial Disp. = 0.01 m
-2.00E-03
-4.00E-03
-6.00E-03
Period=1/Frequency
-8.00E-03
-1.00E-02
0.000
5.000
10.000
15.000
20.000
time (sec)
25.000
30.000
35.000
40.000
7. Fourier Transform
2π s
ωS =
N ∆t
N /2
(t ) = Re ∑ X s e iωS t
x
s =0
1 N −1 e −iωS k∆t
,
∑ xk
N k =0
=
X S N −1
−iω k∆t
2
k e S ,
x
N∑
k =0
e
− iωS k∆t
N
s = 0,1, 2,...,
2
N
2
N
for 1 ≤ s <
2
for s = 0, s =
= cos(ωS k∆t ) − i sin(ωS k∆t )
Mag X S = ℜX + ℑX
2
S
2
S
ℑX S
φ = tan
ℜX
S
−1
8. Fourier Transform; El Centro
Fourier Transform of El Centro Accleration Record
0.008
0.007
0.006
Magnitude
0.005
0.004
0.003
0.002
0.001
0
0
20
40
60
Circular Frequency, v
80
100
120
9. Earthquake Elastic Response Spectra
P0 sin(ωt )
x
xt
m
c
k/2
m
x
k/2
c
k/2
k/2
xg
(a)
m + cx + kx = P0 sin(ω t )
x
m + mg + cx + kx = 0 or
x
x
ωn =
k
m
(b)
D = c / ccrit
c crit = km
m + cx + kx = − mg = Pearthquake (t )
x
x
undamped systems; ωd =
k
(1 − D 2 ) damped systems
m
10. Duhamel's Integral
t
p(τ)
dx (t ) = e
−ξ (1) ( t −τ )
t
1
x(t ) =
mω D
p (τ )dτ
sin ω D (t − τ )
mω D
p(τ ) e −ξω (t −τ ) sin ω D (t − τ ) dτ
∫
0
x(t ) = A(t ) sin ω D t − B(t ) cos ω D t
t
t
1
eξωτ
1
eξωτ
A(t ) =
∫ p(τ ) eξωt cos ωD τ dτ B(t ) = mωD ∫ p(t ) eξωt sin ωD τ dτ
mωD 0
0
A
∆τ 1 A
A
A(t ) =
∑ (t ) ∑ (t ) = ∑ (t − ∆τ ) + p(t − ∆τ ) cos ωD (t − ∆τ )
mωD ζ ζ
2
2
exp(−ξω∆τ ) + p(t ) cos ωD t
12. Multi-Degree of Freedom
x3
m3
c3
k3 /2
x2
k1/2
k3/2
c2
k2/2
m1
c1
k1/2
y3
y2
m + cx + kx = p(t)
x
y4
y1
y5
θ1
(a)
k12 k1N x1
k 22 k 2 N x2
ki 2 kiN xi
kij = force corresponding to coordinate i
due to unit displacement of coordinate j
cij = force corresponding to coordinate i
due to unit velocity of coordinate j
mij = force corresponding to coordinate i
due to unit acceleration of coordinate j
m2
k2/2
x1
f S 1 k11
f k
S 2 21
=
f Si ki1
θ2
θ3
(b)
θ4
θ5
13. Modal Analysis
m + kx = p(t)
x
mΦX + kΦX = p(t )
T
T
T
φ n mΦX + φ n kΦX = φ n p(t)
T
T
T
φ n mφ n X n + φ n kφ n X n = φ n p(t)
M n X n + K n X n = Pn (t )
14. Modal Damping
M n X n + C n X n + K n X n = Pn (t )
+ 2ξ ω X + K X = Pn (t )
Xn
n n
n
n
n
Mn
T
M n ≡ φ n mφ n
T
C n ≡ φ n cφ n
T
K n ≡ φ n kφ n
c = a 0 m + a1k
C nb = φ T c b φ n = ab φ T m[m −1 k ]b φ n
n
n
T
Pn (t ) ≡ φ n p(t )
15. FEM Frequency Domain
[ M ]{ u} + [ K ]{ u} = { p} e
iωt
{ u} = { U} e then
{[ K ] − ω 2 [ M ]}{ U} = {p}
i ωt
16. Finite Elements
u1
u7
G1,ρ1,ν1
u2
u8
[ K1 ] = fn(G1 , ρ1 ,ν 1 )
[ m1 ] = fn( ρ1 )
ui = ai x + bi y + c
k1,1
k
2,1
k
7 ,1
k8,1
k1, 2
k 2, 2
k1, 7
k 2, 7
k7, 2
k8, 2
k7,7
k8, 7
k1,8 u1
k 2,8 u2
k7 ,8 u7
k8,8 u8
ε = constant
σ = constant
17.
[ M ]{ u} + [ K ]{ u} = { p} eiωt
m1
m2
m3
m4
u1 k1,1 k1, 2
u k
k
2 2,1 2, 2
u3 + k3,1 k3, 2
k 4, 2
u4
m5 u5
k1,3
k 2,3
k 2, 4
k 3, 3
k 3, 4
k 4,3
k 4, 4
k 5, 3
k 5, 4
u1 p1
u2 p2
k3,5 u3 = p3 eiωt
k 4,5 u4 p4
k5,5 u5 p5
if { u} = { U} e iωt then { u} = −ω 2 { U} e iωt and
{[ K ] − ω [ M ]}{ U} = {p} given ω, {p}, solve for { U}
2
[ K ], { U} are complex − valued
(
G* = G 1 − 2 D 2 + 2iD 1 − D 2
)
18. Method of Complex Response
• Given earthquake acceleration vs. time, ü(t)
• FFT => ω1, ω 2 , ω 3...ωn ; {p}1 ,{p}2 ,{p}3,{p}n
N /2
• Recall that
(t ) = Re ∑ X s e iωS t
x
s =0
{ [ K ] − ω [ M ] } { U} = {p}
2
• Solve
• FFT-1 => ü (t)
19. 212,428 nodes, 189,078 brick elements and 1500 shell elements
Circular boundary to reduce reflections