ME 440Intermediate Vibrations

Th, Feb. 5, 2009Section 2.2 and 2.6

© Dan Negrut, 2009ME440, UW-Madison

Before we get started…

Last Time: Started Chapter 2 – free response of a 1DOF system (Sec. 2.2) Discussed how to derive the EOM

N2L, Energy Methods, Lagrange’s Equations and Hamilton Principle Mentioned the equivalent mass approach

Today: HW Assigned: 2.106, 2.108, 2.109 (due on Feb. 12) Topics covered:

How to solve EOMs once you obtain them Some analytical considerations regarding the nature of the

response Undamped, underdamped, critically damped, and overdamped

2

Short Excursion: A Word on the Solution of

Ordinary Differential Equations

Classical analytic techniques

Laplace transforms

Numerical solution 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

How is the concept of ODE related to that of IVP?

Ordinary Differential Equation (ODE) Typically, has an infinite number of solutions

Initial Value Problem (IVP) Is an ODE plus a set of initial conditions (ICs):

The solution is indicated to assume at time T=0 a certain given value

The solution for the IVP’s that we’ll deal with is UNIQUE

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

4

ODE: Infinite Number of Solutions

ODE Problem: Initial Condition: y0 =[-1000:50:1000]

0.10.1 100 ty y e

5

ODE vs. IVP (Contd.)

Example: consider a simple first order ODE

It has an infinite number of solutions:

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 + IC2. IVP has a UNIQUE solution (unlike on

ODE) 6

End short excursion. Back to regular business:

Introducing n &

EOM:

7

General form of the EOM for a one degree of freedom system:

Compare the boxed forms of the EOM to understand how the natural frequency n and damping ratio are defined:

Some Quick Remarks EOM looks like (in standard form):

8

In Chapter 2, no applied external force present, that is, the right hand side is zero

In fact, mass moves due to presence of nonzero initial conditions (ICs):

You always have a set of two initial conditions One initial displacement and one initial velocity This is because we’re dealing with an IVP for a second order ODE

Motion Taxonomy In standard form, EOM looks like:

9

Motion taxonomy exclusively based on value of :

Undamped motion for = 0 Case 1

Overdamped motion if >1 Case 2A

Critically damped motion if =1 Case 2B

Underdamped motion if 0<<1 Case 2C

Case 1: Undamped Motion (=0)

Try a solution of this type:

10

Assume ICs at time t=0 are: That is, and are given to you (you know them)

Based on chosen expression of x(t) and IC values, you get that

That is,

Case 1: Undamped Motion (=0)~ Concluding Remarks ~

You can add the two harmonics to obtain only one harmonic and a phase angle that capture the response:

11

Solution could have been obtained by starting with an exponential form and substituting back into ODE

This requires the solution of the Characteristic Equation (CE, see ME340):

Then look for solution of the form (“exponential form”):

Short Excursion: HW Problem 2.35

Requires you to compute the natural frequency of a flywheel system (undamped system)

In longitudinal direction

With respect to torsional motion (vibration)

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

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

You need the same info for the torsional motion and axial motion

This info available in the collection of tables provided to you

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Now Back To Original Business

Case 2: Damped Motion EOM

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Trial Solution

Characteristic Equation

Roots

Case 2, Possible Subcases… Case A: Overdamped (“Supercritical”)

Nature of Motion: Aperiodic

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Case B: Critical Damping Nature of Motion: Aperiodic

Case C: Underdamped (“Subcritical”) Nature of Motion: Periodic

Subcase 2A: > 1 Solution can be found to assume form

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Constants A1 and A2 found based on initial conditions at time t=0:

Then, one gets

Subcase 2A: > 1 (Contd)

Bringing the expression of the solution x(t) to a “nicer” form 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:

Then x(t) can be equivalently expressed as

Subcase 2A: > 1 (Contd)

An overdamped system does not oscillate, see picture next slide

Such a system dissipates energy due to presence of damper 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

The generalized coordinate x(t) (upper plot) Time derivative, that is, the velocity of the mass (lower plot) Results obtained for (units used: SI)

m = 2 c = 12 k = 8

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Overdamped Response

clearm = 2;c = 12;k = 8;posIC = 2.;velIC = 5.;

Amat = [ 0 1 ; -k/m -c/m]Cmat = eye(2)

sys = ss(Amat, [], Cmat, [])

icConds = [posIC, velIC]initial(sys, icConds)

Matlab Code:

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0

0.5

1

1.5

2

2.5

To:

Out

(1)

0 1 2 3 4 5 6 7 8-2

0

2

4

6

To:

Out

(2)

Response to Initial Conditions

Time (sec)

Am

plitu

de

Subcase 2B: = 1 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):

Note that for critical damping, m, c, and k should be such that the following condition holds (it leads to =1):

Subcase 2B: = 1 (Cntd)

An overdamped system does not oscillate, see picture on next slide

Parameters used (units used: SI) m = 2 c = 8 k = 8

Since damper is present in system, energy dissipation occurs 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 The generalized coordinate x(t) (upper plot) Time derivative, that is, the velocity of the mass (lower plot) 22

Critical Damping Response

0

1

2

3System: sysOutput: Out(1)Time (sec): 0.774Amplitude: 1.91

System: sysOutput: Out(1)Time (sec): 1.63Amplitude: 0.638

To:

Out

(1)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-2

0

2

4

6

To:

Out

(2)

Response to Initial Conditions

Time (sec)

Am

plitu

de

clearm = 2;c = 8;k = 8;posIC = 2.;velIC = 5.;

Amat = [ 0 1 ; -k/m -c/m]Cmat = eye(2)

sys = ss(Amat, [], Cmat, [])

icConds = [posIC, velIC]initial(sys, icConds)

Matlab Code:

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Subcase 2C: < 1 Solution can be found to assume form

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Nomenclature: d – 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

Subcase 2C: < 1 (Contd)

Solution then is equivalently expressed as

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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

Quick Remarks 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

Kind of intuitive, no damping corresponds to case when 0

For the underdamped solution, amplitude of successive oscillations is decreasing exponentially. Specifically, like

In typical structural systems, ¿ 0.2. For all purposes, d ¼ n

Periods of successive cycles are the same; i.e., we have periodic motion with the following attributes:

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Quick Remarks (Contd)

Graphical representation of system response, underdamped case

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Underdamped Example

clearm = 2;c = 2;k = 8;posIC = 2.;velIC = 5.;

Amat = [ 0 1 ; -k/m -c/m]Cmat = eye(2)

sys = ss(Amat, [], Cmat, [])

icConds = [posIC, velIC]initial(sys, icConds)

Matlab Code:

280 2 4 6 8 10 12

-5

0

5

To:

Out

(2)

Response to Initial Conditions

Time (sec)

Am

plitu

de

-1

0

1

2

3

To:

Out

(1)

Root Locations

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