01 SDOF - SPC408 - Fall2016

SPC408 – Dynamics of Aerospace Structures
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Single Degree of Freedom Systems
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Dynamics of Aerospace
Structures
SPC408

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Single degree of freedom
systems

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Objectives
• Recognize a SDOF system
• Be able to solve the free vibration equation
of a SDOF system with and without
damping
• Understand the effect of damping on the
system vibration
• Apply numerical tools to obtain the time
response of a SDOF system

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Single degree of freedom systems
• When one variable can describe the
motion of a structure or a system of
bodies, then we may call the system a 1-D
system or a single degree of freedom
(SDOF) system. e.g. x(t), q(t) Z(t), y(x).

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Stiffness
• From strength of materials recall:

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Newton’s Law
• Newton’s Law:
00 )0(,)0(
0)()(
)()(
vxxx
tkxtxm
tkxtxm







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Solving the ODE
• The ODE is
• The proposed
solution:
• Into the ODE you get
the characteristic
equation:
• Giving:
0)()(  tkxtxm 
t
aetx 
)(
02
 tt
ae
m
k
ae 

m
k
2

m
k
j

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Solving the ODE (cont’d)
• The proposed
solution becomes:
• For simplicity, let’s
define:
• Giving:
t
m
k
jt
m
k
j
eaeatx

 21)(
m
k

tjtj
eaeatx  
 21)(

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Let’s manipulate the solution!

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Recall
   ajSinaCose ja

         bSinaCosbCosaSinbaSin 

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Manipulating the solution
• The solution we have:
• Rewriting:
tjtj
eaeatx  
 21)(
    
    tjSintCosa
tjSintCosatx




2
1)(
       tSinaajtCosaatx  2121)( 
   tSinAtCosAtx  21)( 

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Further manipulation
2
2
2
1 AAA 
   
A
A
Sin
A
A
Cos 12
&  
        tSinCostCosSinAtx  )(
   tASintx )(

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Different forms of the solution
tjtj
eaeatx
tCosAtSinAtx
tASintx







21
21
)(
)(
)()(

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NOTE!

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Natural Frequency of Oscillation
• In the previously obtained solution:
• The frequency of oscillation is 
• It depends only on the characteristics of the
oscillating system. That is why it is called the
natural frequency of oscillation
   tASintx )(
m
k
n 

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Frequency
periodtheiss
2
Hz
2s2
cycles
rad/cycle2
rad/s
frequencynaturalthecalledisrad/sinis
n
nnn
n
n
T
f











We often speak of frequency in Hertz or
RPM, but we need rad/s in the arguments
of the trigonometric functions.

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Recall: Initial Conditions

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Amplitude & Phase from the ICs
    
Phase
0
01
Amplitude
2
2
02
0
0
0
tan,
yieldsSolving
cos)0cos(
sin)0sin(










v
xv
xA
AAv
AAx
n
n
nnn
n






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Some useful quantities
peak valueA


T
T
dttx
T
x
0
valueaverage=)(
1
lim
valuesquaremeanroot=2
xxrms 
valuesquare-mean=)(
1
lim
0
22


T
T
dttx
T
x

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Peak Values
Ax
Ax
Ax
2
max
max
max
:onaccelerati
:velocity
:ntdisplaceme







Maximum or peak (amplitude) values:

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Samples of Vibrating Systems
• Deflection of continuum (beams, plates,
bars, etc) such as airplane wings, truck
chassis, disc drives, circuit boards…
• Shaft rotation
• Rolling ships

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Wing Vibration

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Ship Vibration

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Effective Stiffness of
Structures

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Bars
• Longitudinal motion
• A is the cross sectional
area (m2)
• E is the elastic modulus
(Pa=N/m2)
• l is the length (m)
• k is the stiffness (N/m)x(t)
m

EA
k 
l

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Rods
• Jp is the polar
moment of inertia of
the rod
• J is the mass
moment of inertia of
the disk
• G is the shear
modulus, l is the
length
Jp
J qt)
0

pGJ
k 

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Helical Spring
2R
x(t)
d = diameter of wire
2R= diameter of turns
n = number of turns
x(t)= end deflection
G= shear modulus of
spring material
3
4
64nR
Gd
k 

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Beams
f
m
x
• Strength of materials
and experiments
yield:
3
3
3
3


m
EI
EI
k
n 



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Equivalent Stiffness

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Example 1.5.2 Effect of fuel on
frequency of an airplane wing
• Model wing as transverse
beam
• Model fuel as tip mass
• Ignore the mass of the
wing and see how the
frequency of the system
changes as the fuel is
used up
x(t)
l
E, I m

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Mass of pod 10 kg empty 1000 kg full
I = 5.2x10-5 m4, E =6.9x109 N/m, l = 2 m
• Hence the
natural
frequency
changes by an
order of
magnitude
while it
empties out
fuel.
Hz18.5=rad/s115
210
)102.5)(109.6(33
Hz1.8=rad/s6.11
21000
)102.5)(109.6(33
3
59
3empty
3
59
3full












m
EI
m
EI



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Example 1.5.5 Design of a spring mass
system using available springs: series vs parallel
• Let m = 10 kg
• Compare a series and
parallel combination
• a) k1 =1000 N/m, k2 = 3000
N/m, k3 = k4 =0
• b) k3 =1000 N/m, k4 = 3000
N/m, k1 = k2 =0
k1
k2
k3
k4
m

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rad/s66.8
10
750
N/m750
13
3000
)1()1(
1
,0
:connectionseriesb)Case
rad/s20
10
4000
N/m400030001000,0
:connectionparallela)Case
43
21
2143








m
k
kk
kkk
m
k
kkkkk
eg
series
eq
eg
parallel
eq


Same physical components, very different frequency
Allows some design flexibility in using off the shelf components

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Harmonic Excitation of
Undamped Systems
• Consider the usual spring mass damper system
with applied force F(t)=F0cost
•  is the driving frequency
• F0 is the magnitude of the applied force
• We take c = 0 to start with

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Equation of motion
m
kmFf
tftxtx
tFtkxtxm
n
n






,/where
)cos()()(
)cos()()(
00
0
2
0



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Linear non-homogenous ODE:
• Solution is sum of homogenous and
particular solution
• The particular solution assumes a form
of forcing function (physically the input
wins)
)cos()( tXtxp 

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Substitute into the equation of
motion:
22
0
0
22
:yieldssolving
coscoscos
2






n
x
n
x
f
X
tftXtX
pnp
    


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Thus the particular solution has
the form:
)cos()( 22
0
t
f
tx
n
p 
 


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General Solution
  
  
particular
n
nn t
f
tAtAtx 

 coscossin)( 22
0
shomogeneou
21



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01
022
0
2
)0(
)0(
vAx
x
f
Ax
n
n







Apply the initial conditions to evaluate the constants
Solving for the constants and substituting into x yields
t
f
t
f
xt
v
tx
n
n
n
n
n






coscossin)( 22
0
22
0
0
0










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Comparison of free and forced
response
• Sum of two harmonic terms of different
frequency
• Free response has amplitude and phase
affected by forcing function
• Our solution is not defined for n = 

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Example
• Compute and plot the response for m=10 kg, k=1000 N/m,
x0=0,v0=0.2 m/s, F=23 N, =2n
rad/s202rad/s,10
kg10
N/m1000
 nn
m
k

Solution
m02.0
rad/s10
m/s2.0
N/kg,3.2
kg10
N23 0
0 
n
v
m
F
f


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Example (cont’d)
m109667.7
s/rad)20(10
N/kg3.2 3
222222
0 



n
f
)20cos10(cos10667.710sin02.0)( 3
ttttx  
Substituting into the general solution:

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Response to Harmonic Excitation

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Beat !
• What happens when the driving frequency
is near the natural frequency?

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Example
• Given zero initial conditions a harmonic input of 10 Hz
with 20 N magnitude and k= 2000 N/m, and measured
response amplitude of 0.1m, compute the mass of the
system.
 tt
f
tx n
n


coscos)( 22
0



Solution:

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Example (cont’d)
• Using Trigonometric Identities:
• Using given data
    
termFast
n
termSlow
n
n
tt
f
tx 




 





 


2
sin
2
sin
2
)( 22
0 

 
kg45.0
1.0
)20(2000
)/20(2
1.0
2
222
0





m
m
mf
n 

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Plotting the result

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Twice Frequency and Beat

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Resonance
• When excitation frequency is equal to the
natural frequency, the previous solution
fails.
• The particular solution of the ODE
become:
)sin()( ttXtxp 

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Substituting!
2
0f
X 
Use the particular solution into the equation:
   boundwithoutgrows
0
21 )sin(
2
cossin)( tt
f
tAtAtx 

 
The total solution becomes:

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Resonance

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Summary
• Write down the equation of motion using Newton’s law
• Solve the equation of motion for a SDOF
• Use initial conditions to determine the amplitude and
phase of vibration for a SDOF
• Evaluate the effective stiffness of structural members
• Analyzing the response of a SDOF to harmonic
excitations
• Different types of response depend on the excitation
frequency
• What happens during the beat and resonance
phenomena

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1. The amplitude of vibration of an undamped
system is measured to be 1 mm. the phase
shift is measured to be 2 rad and the
frequency 5 rad/sec. Calculate the initial
conditions.
2. Using the equation:
evaluate the constant A1 and A2 in terms of
the initial conditions
HW #1

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HW #1 (cont’d)
3. A vehicle is modeled as 1000 kg mass
supported by a stiffness k=400 kN/m.
When it oscillates, the maximum
deflection is 10 cm. when loaded with the
passengers, the mass becomes 1300 kg.
calculate the change in the frequency,
velocity amplitude, and acceleration if the
maximum deflection remain 10 cm.

01 SDOF - SPC408 - Fall2016

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