Transcript ME440 - Dan Negrut

ME 440
Intermediate Vibrations
Th, Feb. 5, 2009
Section 2.2 and 2.6
© Dan Negrut, 2009
ME440, UW-Madison
Before we get started…
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Last Time:
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Started Chapter 2 – free response of a 1DOF system (Sec. 2.2)
Discussed how to derive the EOM
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N2L, Energy Methods, Lagrange’s Equations and Hamilton Principle
Mentioned the equivalent mass approach
Today:
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HW Assigned: 2.106, 2.108, 2.109 (due on Feb. 12)
Topics covered:
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How to solve EOMs once you obtain them
Some analytical considerations regarding the nature of the response
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Undamped, underdamped, critically damped, and overdamped
2
Short Excursion: A Word on the Solution of
Ordinary Differential Equations
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Classical analytic techniques
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Laplace transforms
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Numerical solution
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Usually found using MATLAB, or some other software package
(Maple, EES, Sundials, etc.)
MATLAB demonstrated later in this lecture (or next…)
3
ODE vs. IVP
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How is the concept of ODE related to that of IVP?
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Ordinary Differential Equation (ODE)
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Typically, has an infinite number of solutions
Initial Value Problem (IVP)
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Is an ODE plus a set of initial conditions (ICs):
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The solution is indicated to assume at time T=0 a certain given value
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The solution for the IVP’s that we’ll deal with is UNIQUE
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It makes sense: only having the rate of change of a variable cannot tell
you the value of the variable as a function of time
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ODE: Infinite Number of Solutions
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5
0.1t
ODE Problem: y  0.1y 100e
Initial Condition: y0 =[-1000:50:1000]
ODE vs. IVP (Contd.)
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Example: consider a simple first order ODE
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It has an infinite number of solutions:
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However, if you specify an Initial Condition (IC) at time t=0, for instance,
x(0) =2.5, then there is a UNIQUE solution that satisfies both the ODE
and the imposed IC:
Remember:
1. IVP = ODE + IC
2. IVP has a UNIQUE solution (unlike on ODE)
6
x(t)
End short excursion. Back to regular business:
Introducing wn & 
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k
m
c
EOM:
General form of the EOM for a one degree of freedom system:
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Compare the boxed forms of the EOM to understand how the
natural frequency wn and damping ratio  are defined:
7
Some Quick Remarks
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EOM looks like (in standard form):
In Chapter 2, no applied external force present, that is, the right hand
side is zero
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In fact, mass moves due to presence of nonzero initial conditions (ICs):
You always have a set of two initial conditions
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One initial displacement and one initial velocity
This is because we’re dealing with an IVP for a second order ODE
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Motion Taxonomy
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In standard form, EOM looks like:
Motion taxonomy exclusively based on value of :
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Undamped motion for  = 0
Case 1
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Overdamped motion if >1
Case 2A
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Critically damped motion if =1
Case 2B
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Underdamped motion if 0<<1
Case 2C
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Case 1: Undamped Motion (=0)
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Try a solution of this type:
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Assume ICs at time t=0 are:
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That is,
and
are given to you (you know them)
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Based on chosen expression of x(t) and IC values, you get that
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That is,
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Case 1: Undamped Motion (=0)
~ Concluding Remarks ~
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You can add the two harmonics to obtain only one harmonic
and a phase angle that capture the response:
Solution could have been obtained by starting with an exponential
form and substituting back into ODE
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This requires the solution of the Characteristic Equation (CE, see ME340):
Then look for solution of the form (“exponential form”):
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Short Excursion:
HW Problem 2.35
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Requires you to compute the natural frequency of a flywheel system
(undamped system)
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In longitudinal direction
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With respect to torsional motion (vibration)
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In transversal direction
In the end, it all boils down to computing the equivalent mass meq and
equivalent spring constant keq, since then
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Short Excursion Problem 2.35 (Cntd):
Computing Equivalent Spring
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The tricky keq is for transversal motion
All information needed is in provided below
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Short Excursion Problem 2.35 (Cntd):
Computing Equivalent Spring
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You need the same info for the torsional motion and axial motion
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This info available in the collection of tables provided to you
Now Back To Original Business
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Case 2: Damped Motion
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EOM
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Trial Solution
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Characteristic Equation
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Roots
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Case 2, Possible Subcases…
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Case A: Overdamped (“Supercritical”)
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Case B: Critical Damping
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Nature of Motion: Aperiodic
Case C: Underdamped (“Subcritical”)
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Nature of Motion: Aperiodic
Nature of Motion: Periodic
Subcase 2A:  > 1
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Solution can be found to assume form
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Constants A1 and A2 found based on initial conditions at time t=0:
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Then, one gets
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Subcase 2A:  > 1 (Contd)
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Bringing the expression of the solution x(t) to a “nicer” form
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This is only cosmetics…
Use trick of the trade. Express A1 and A2 using two yet unknown variables B1 and B2 as
follows:
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Next, recall the definition of the hyperbolic sine and cosine:
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Then x(t) can be equivalently expressed as
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Subcase 2A:  > 1 (Contd)
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An overdamped system does not oscillate, see picture next slide
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Such a system dissipates energy due to presence of damper
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As such, it should come to rest
Theoretically, takes infinite amount of time to reach rest
For mass-spring-damper system, plot on next slides displays time
evolution of
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The generalized coordinate x(t) (upper plot)
Time derivative, that is, the velocity of the mass (lower plot)
Results obtained for (units used: SI)
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m=2
c = 12
k=8
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Overdamped Response
Response to Initial Conditions
2.5
2
To: Out(1)
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1.5
clear
m = 2;
c = 12;
k = 8;
posIC = 2.;
velIC = 5.;
1
0.5
Amplitude
Matlab Code:
0
6
Amat = [ 0 1 ; -k/m -c/m]
Cmat = eye(2)
4
To: Out(2)
sys = ss(Amat, [], Cmat, [])
2
icConds = [posIC, velIC]
initial(sys, icConds)
0
-2
0
1
2
3
4
Time (sec)
5
6
7
8
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Subcase 2B:  = 1
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Characteristic Equation has a double root. Solution assumes form
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Use initial conditions to get constants A1 and A2. Final form of x(t):
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Note that for critical damping, m, c, and k should be such that the
following condition holds (it leads to =1):
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Subcase 2B:  = 1 (Cntd)
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An overdamped system does not oscillate, see picture on next slide
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Parameters used (units used: SI)
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Since damper is present in system, energy dissipation occurs
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m=2
c=8
k=8
It takes infinite amount of time to come to rest
However, it gets within any neighborhood of the equilibrium configuration
faster then any overdamped system you compare against
For mass-spring-damper system, plot on next slide displays
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The generalized coordinate x(t) (upper plot)
Time derivative, that is, the velocity of the mass (lower plot)
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Critical Damping Response
Response to Initial Conditions
3
System: sys
Output: Out(1)
Time (sec): 0.774
Amplitude: 1.91
To: Out(1)
2
0
6
Amat = [ 0 1 ; -k/m -c/m]
Cmat = eye(2)
4
To: Out(2)
Matlab Code:
clear
m = 2;
c = 8;
k = 8;
posIC = 2.;
velIC = 5.;
System: sys
Output: Out(1)
Time (sec): 1.63
Amplitude: 0.638
1
Amplitude
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sys = ss(Amat, [], Cmat, [])
icConds = [posIC, velIC]
initial(sys, icConds)
2
0
-2
0
0.5
1
1.5
2
2.5
Time (sec)
3
3.5
4
4.5
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Subcase 2C:  < 1
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Solution can be found to assume form
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Nomenclature:
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wd – Damped Natural Frequency
Don’t rush like last time to get A1 and A2 based on initial conditions…
Rather, used the same trick with the B1 and B2 variables to massage
the solution x(t) a bit
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Subcase 2C:  < 1 (Contd)
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Solution then is equivalently expressed as
This is a better time to use ICs to get the constants B1 and B2 associated
with your solution…
Solution can also be equivalently expressed as
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Quick Remarks
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Note that we get the undamped case by setting in expression of x(t) on
previous slide the value of the damping ratio to be =0
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For the underdamped solution, amplitude of successive oscillations is
decreasing exponentially. Specifically, like
In typical structural systems,  ¿ 0.2.
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Kind of intuitive, no damping corresponds to case when   0
For all purposes, wd ¼ wn
Periods of successive cycles are the same; i.e., we have periodic motion
with the following attributes:
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Quick Remarks (Contd)
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Graphical representation of system response, underdamped case
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Underdamped Example
Response to Initial Conditions
3
2
To: Out(1)
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1
clear
m = 2;
c = 2;
k = 8;
posIC = 2.;
velIC = 5.;
0
-1
Amplitude
Matlab Code:
5
To: Out(2)
Amat = [ 0 1 ; -k/m -c/m]
Cmat = eye(2)
sys = ss(Amat, [], Cmat, [])
0
-5
0
icConds = [posIC, velIC]
initial(sys, icConds)
2
4
6
Time (sec)
8
10
12
28
Root Locations
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