Basic dynamics of structures in Civil Engineering

Dynamics

1. Basic structural dynamics I
Wind loading and structural response -Lecture 10
Dr. J.D. Holmes

2. Basic structural dynamics I
•Topics :
•Revision of single degree-of freedom vibration theory
•Response to sinusoidal excitation
Refs. : R.W. Clough and J. Penzien ‘Dynamics of Structures’ 1975
R.R. Craig ‘Structural Dynamics’ 1981
J.D. Holmes ‘Wind Loading of Structures’ 2001
•Multi-degree of freedom structures –Lect. 11
•Response to random excitation

3. Basic structural dynamics Ixc
Equation of free vibration :
Example : mass-spring-damper system :
k
c
m
x0kxxcxm 
mass ×acceleration = spring force + damper force
-kx
-c(dx/dt)xm kx-
equation of motion
•Single degree of freedom system :

4. Basic structural dynamics I
•Single degree of freedom system :
Equation of free vibration :
Example : mass-spring-damper system :
Ratio of damping to critical c/c
c :
k
c
m
x0kxxcxm  mk2
c

often expressed as a percentage

5. Basic structural dynamics I
•Single degree of freedom system :
Damper removed :
k
m
x(t)
Equation of motion :0kxxm  xmkx 
is an equivalent static force (‘inertial’ force) xm
Undamped natural frequency :

2m
k
2π
1
n
1
1 
Period of vibration, T :11
12
T
n




6. Basic structural dynamics I
•Single degree of freedom system :
Initial displacement = X
o
Free vibration following an initial displacement :
k
c
m
xm
k2
1 )1cos(.)(
2
1
1




teCtx
t 2
2
0
-1
X

C 2
2
-1
tan




7. Basic structural dynamics I
•Single degree of freedom system :
Free vibration following an initial displacement :-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5
time/T
amplitude

8. Basic structural dynamics I
•Single degree of freedom system :
Free vibration following an initial displacement :-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5
time/T
amplitude t
eC

1
.


9. Basic structural dynamics I
•Single degree of freedom system :
Response to sinusoidal excitation :
Equation of motion :tsinFF(t)kxxcxm
o
Steady state solution :   tx sin H(n) /kF(t)
o
k
c
mF(t)
= 2n
H(n) = 2
1
2
2
2
1
n
n
4ζ
n
n
1
1


























Frequency of
excitation

10. Basic structural dynamics I
•Single degree of freedom system :
Critical damping ratio –
damping controls amplitude at
resonance
0.1
1.0
10.0
100.0
0 1 2 3 4
n/n1
H(n)
=0.01
=0.05
=0.1
=0.2
=0.5
At n/n
1=1.0, H(n
1) = 1/2 Then,2kζ
F
x
0
max

Dynamic amplification factor, H(n)

11. Basic structural dynamics II
•Response to random excitation :
Consider an applied force with spectral density S
F(n) :dn (n).SH(n).(1/k)dn (n)Sσ
F
2
0
2
x
0
2
x 


k
c
mF(t)
Spectral density of displacement :
|H(n)|
2
is the square of the dynamic amplification factor (mechanical admittance) (n).SH(n).(1/k) (n)S
F
22
x

Variance of displacement :
see Lecture 5

12. Basic structural dynamics II
•Response to random excitation :
Special case -constant force spectral density S
F(n) = S
ofor all n (‘white
noise’):dn H(n).S
k
1
σ
2
o
2
2
x 








0
The above ‘white noise’ approximation is used widely in wind
engineering to calculate resonantresponse -with S
o taken as S
F(n
1)dn.S
k
1
2
1
n
n2
4ζ
2
2
1
n
n
1
1
0
2






























0 














4
Sπn
.
k
1
σ
o1
2
2
x

13. End of Lecture
John Holmes
[email protected]