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```HOMEWORK 1
1. Derive equation of motion of SDOF using energy method
2. Find amplitude A and tanΦ for given x0, v0
3. Find natural frequency of cantilever, l=400mm, Φ=5mm, E=2e11Pa, m=2.7kg.
Confirm with SW Simulation
4. Work with exercises in chapter 19 – blue book
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RING
Energy method
Ring.SLDASM
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ωn, x0, v0 fully define free undamped vibration
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MODULE 02
DAMPED VIBRATION
Inman (3rd edition) section 1.3
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SINGLE DEGREE OF FREEDOM SYSTEM WITHOUT DAMPING
stiffness
mass
Notice that mass and stiffness
are completely separated.
& kx  0
mx&
This is the equation of motion of a single degree of freedom system with no damping. It states
that inertial forces are equal and opposite to stiffness forces.
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SINGLE DEGREE OF FREEDOM SYSTEM WITH DAMPING
Notice that mass, damping and
stiffness are completely separated.
Viscous damper, damping force is
proportional to velocity
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mx
Equation of motion
We assume solution in the form of:
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x  ae t
x  a e  t
x  a 2 e  t
Characteristic equation
Critical damping
Determinant of
characteristic equation
Damping ratio (modal damping)
n 
k
m
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underdamped
 1
 1
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Equation is linear, therefore the sum of solutions is also solution
underdamped
 1
or, using Euler’s relations:
Damped natural frequency
Two forms of solution of free under- damped vibration
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underdamped
 1
ωd- natural frequency of damped oscillations
A and Ф are calculated from initial conditions
Inman p. 24
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overdamped
 1
This is a non-oscillatory motion
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critically
 1
Damping is critical
Damping
iswhen:
critical when:
c
ζ=
=1
2 km
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FREE VIBRATIONS WITH VISCOUS DAMPING
Depending on the sign of ζ we have three cases
ζ<1
system is under damped, motion is oscillatory with an
exponential decay in amplitude
ζ=1
system is critically damped, at most one overshot of system
resting position is possible
ζ>1
system is over damped, motion is exponentially decaying
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nt
atop  Ae
abottom   Aent
Damped vibration amplitude decay.xls
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m =10kg
k = 1000N/m
c = 20Ns/m
SDOF damped.SLDASM
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Linear damper 20Ns/m
Modal damping 10%
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0.62, 56.48
t1 = 0.62s
a1 = 56.48
t2 =1.23s
a2 = 30.30
1000
n 
 10
10
1.23, 30.30
Ae  10 x 0.62 56.48

 1.86
 10 x1.23
Ae
30.30
e6.1  1.86
6.1  ln1.86  0.62
  0.10
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swing arm.SLDASM
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L1=0.2m
m =0.56kg
Assume that beam
mass is negligible
kL=2000N/m
cL=10Ns/m
L2=0.1m
Is this system under damped, critically damped or over damped?
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System is underdamped, modal damping is 7.4%
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Linear damper 10Ns/m
Modal damping 7.4%
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ζ = 5%
ζ = 100%

ζ = 10%
ζ = 200%
ζ = 50%
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TYPICAL VALUES OF DAMPING RATIO
SYSTEM
Damping ratio  (% of critical damping)
METALS
0. 1%
CONTINUOUS METAL STRUCTURES
2% - 4%
METAL STRUCTURES WITH JOINTS
3% - 7%
ALUMINUM/STEEL TRANSMISSION LINES
~0.4%
SMALL DIAMETER PIPING SYSTEMS
1% - 2%
LARGE DIAMETER PIPING SYSTEMS
2% - 3%
SHOCK ABSORBERS
30%
RUBBER
~5%
LARGE BUILDINGS DURING EARTHQUAKE
1%-5%
PRE-STRESSED CONCRETE STRUCTURES
2% - 5%
REINFORCED CONCRETE STRUCTURES
4%-7%
TIMBER
5% - 12%
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MODAL SUPERPOSITION METHOD
Structure vibrating in a given mode can be considered as the Single Degree of Freedom (SDOF)
system. Structure can be considered a series of SDOF. For linear systems the response can be
found in terms of the behavior in each mode and these summed for the total response. This is the
Modal Superposition Method used in linear dynamics analyses.
A linear multi-DOF system can be viewed as a combination of many single DOF systems, as can be
seen from the equations of motion written in modal, rather than physical, coordinates. The dynamic
response at any given time is thus a linear combination of all the modes. There are two factors
which determine how much each mode contributes to the response: the frequency content of the
forcing function and the spatial shape of the forcing function. Frequency content close to the
frequency of a mode will increase the contribution of that mode. However, a spatial shape which is
nearly orthogonal to the mode shape will reduce the contribution of that mode.
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MODAL SUPERPOSITION METHOD
The response of a system to excitation can be found by summing up the response of multiple SODFs.
Each SDOF represents the system vibrating in a mode of vibration deemed important for the vibration response.
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TIME AND FREQUENCY
RESPONSE ANALYSIS
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TIME RESPONSE ANALYSIS
In Time Response analysis the applied load is an explicit function of time, mass and damping
properties are both taken into consideration and the vibration equation appears in its full form:
Where:
[ M ] mass matrix
[ C ] damping matrix
[ K ] stiffness matrix
[ F ] vector of nodal loads
[ d ] unknown vector of nodal displacements
Dynamic time response analysis is used to model events of short duration. A typical example
would be analysis of vibrations of a structure due to an impact load or acceleration applied to the
base (called base excitation). Results of Time Response analysis will capture both the response
during the time when load is applied as well as free vibration after load has been removed.
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TIME RESPONSE ANALYSIS
F(t)
Force excitation or base excitation is a function
of time. Solution is performed in time domain i.e.
data of interest (displacement, stresses) are
computed as functions of time.
F(t)
F(t)
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12
10
4
8
3
6
2
4
1
2
0
0
0
2
4
6
8
10
0
2
time
Examples of load time history in Time Response analysis.
4
6
8
10
time
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FREQUENCY RESPONSE ANALYSIS
Force excitation or base excitation is not directly a function of
time, rather it is a function of the excitation frequency. Solution
is performed in frequency domain i.e. data of interest
(displacement, stresses) are computed as functions of
frequency.
Force amplitude does
not have to be constant
time
Example of load time history that can be used in Frequency Response analysis.
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TIME RESPONSE DUE TO IMPULSE LOAD
force
force
time
time
Δt
Different ways to illustrate a unit impulse
An impulse applied to a SDOF is the same as applying the initial conditions of zero displacement
and initial velocity v0 = F Δt /m (Δt here is duration of “square” impulse, Δt is very short)
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TIME RESPONSE DUE TO IMPULSE LOAD
F t
Fˆ
A


d md md
v0
0
To model impulse load we set:
x 0= 0
v 0 = F Δt /m
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TIME RESPONSE DUE TO IMPULSE LOAD
Analytical solution of unit impulse excitation problem.
Response of a system due to an impulse at t = 0
Where
Fˆ
is force impulse
Inman p. 195
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TIME RESPONSE DUE TO IMPULSE LOAD
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TIME RESPONSE DUE TO IMPULSE LOAD
Impulse 4.77Ns
A force is considered to be an impulse if its duration Δt is very short compared with the period T=1/f
Here Δt = 0.0075, T=0.03
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TIME RESPONSE DUE TO IMPULSE LOAD
The critical damping for SDOF is:
ccr = 2 km = 4000
Ns
m
To define damping as 5% of critical damping we can either enter it as 200 in the
Spring-Damper Connector window or as 0.05 in the Global Damping window.
Damping definition. Damping can be defined explicitly (left) or as a fraction of critical damping (right).
The entries in both windows define the same damping.
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TIME RESPONSE DUE TO IMPULSE LOAD
Location for Time
History graph
Results of Dynamic Time analysis; displacement time history
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In class exercise to demonstrate:
Equivalence of explicit damping and modal damping (study 01, study 02)
Equivalence of short impulse load to initial velocity (study 03, study 04)
SDOF damped
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01
02
Explicit damping 80Ns/m
Equivalent modal damping 0.02
Results are identical
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03
04
Impulse 4.77Ns
Initial velocity 0.477m/s
Results are identical except for the very
beginning (can’t be seen in these graphs)
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HOMEWORK 2
1.
SDOF m=10kg, k=1000N/m is critically damped. Find a combination n of x0 and
v0 that will make the SDOF cross the resting position. Prepare plot in Excel
2.
SDOF m=10kg, k=1000N/m performs damped vibration. After 10s the
displacement amplitude is 5% of the original amplitude. Find linear damping
3.
Swing arm problem
4.
SDOF m=10kg, k=1000N/m is at rest. At t=0 it is subjected an impulse 10Ns.
What is the amplitude of displacement, velocity and acceleration?
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