237

(3.154)

(3.153)

Lecture 16. SOLVING FOR TRANSIENT MOTIONUSING MODAL COORDINATES

Free MotionWith normalized eigenvectors, the matrix version of the modalEq.(3.125) is

= the modal force vector. Uncoupled, modal differential

equations:

The homogeneous version of Eq.(3.154)

have solutions:

238

Adding particular solutions corresponding to

specific right-hand forcing functions, yield the completesolution

The constants must be determined from the modal-

coordinate initial conditions.

To obtain modal-coordinate initial conditions:

Hence, the modal-coordinate initial conditions are defined by

Similarly, the modal-velocity initial conditions are defined by

239

(3.144)

(3.145)

(3.151)

For the prior example

with

Modal coordinate initial conditions:

240

(3.155b)

Modal velocity initial conditions:

With these initial conditions, we can solve Eqs.(3.132) for anyspecified modal force terms , obtaining a complete

modal-coordinate solution defined by . From

Eq.(3.123a) the solution for the physical coordinates can then bestated

Hence, the physical coordinate response vector is a linear sumof the modal solutions times their respective mode

241

(3.157)

shapes (or eigenvectors).

Example. With , Eqs. (3.155)

gives , and:

The corresponding physical-coordinate solution is

Any disturbance to the system from initial conditions (orexternal forces) will result in combined motion at the twonatural frequencies. Note that the physical initial conditions arecorrectly represented,

242

-1

-0.5

0

0.5

1

0 10 20 30 40

t

q_1

-1

-0.5

0

0.5

1

0 10 20 30 40

t

q_2

-2

-1

0

1

2

0 10 20 30 40

t

x_1

-1

-0.5

0

0.5

1

0 10 20 30 40

t

x_2

Figure 3.53 Solution from Eq.(3.157) for and for

243

( i )

Modal Transient Example Problem 1. Free UndampedMotion

The 2-mass model illustrated in a is just about to collide with awall. The right-hand spring will cushion the shock of thecollision. Before collision, the system’s model is

Once contact is established, the system looks like the model offrame b and is governed by the matrix differential equations ofmotion

244

( ii )

( ii )

The physical parameters are:

yielding

Starting from the instant of contact, the physical initialconditions are:

Solve for and the reaction forces

.

245

(iv)

Solution. Following the procedures of the preceding examples,the eigenvalues and natural frequencies are:

The matrix of unnormalized eigenvectors is

The matrix of normalized eigenvectors follows from as

Hence, the model is

From , the modal-coordinate initial

246

( v )

conditions are zero. From , the modal-

velocity initial conditions are

In terms of initial conditions, the solution to is

. Hence, the modal solutions

are:

From , the solution for the physical coordinates is

247

q_2 versus time

-1

-0.8-0.6

-0.4-0.2

0

0.20.4

0.60.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

q_2

q_1 versus time

-2

0

2

4

6

8

10

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

q_1

248

x_1 (m) versus time

-0.2

-0.10

0.10.2

0.3

0.40.5

0.60.7

0.8

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

x_1

x_2 versus time

-0.1

-0.050

0.050.1

0.15

0.20.25

0.30.35

0.4

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

x_2

249

f_s1=k_1 (x_1 - x_2 ) versus time

-1000

0

1000

2000

3000

4000

5000

6000

7000

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

f_s1

= k

_1 (

x_1

- x_

2 )

(N)

f_s2 = k_2 x_2 versus time

-1000

0

1000

2000

3000

4000

5000

6000

0 0.1 0.2 0.3 0.4 0.5 0.6

time (seconds)

f_s2

= k

_2 x

_2 (

N)

250

Do the numbers seem right? We can do a quick calculation tosee if the peak force and deflection seem to be reasonable. Suppose both bodies are combined, so that one body has a massof . Then a conservation of

energy equation to find the peak deflection is

The predicted peak deflection for is 0.35 m , which is on

the order of magnitude for the estimate, but lower. We wouldexpect the correct number to be lower, because two masses witha spring between them will produce a lower collision force thana single rigid body with an equivalent mass.

Note that the spring-mass system loses contact with the wallwhen changes sign at about . After this

time, the right spring is disengaged, and the model becomes

251

(3.23)

(3.24)

(3.25)

Damping and Modal Damping FactorsThe SDOF harmonic oscillator equation of motion is

yields the homogeneous equation

Assumed solution: Yh=Aest yields:

Since A … 0, and ,

For ζ<1, the roots are

is called the damped natural frequency. The

homogeneous solution looks like

252

where A1 and A2 are complex coefficients. The solution can bestated

where A and B are real constants.

MDOF vibration problems with general damping matricesmodeled by,

also have complex roots (eigenvalues) and eigenvectors.

The matrix of real eigenvectors, based on a symmetric stiffness and mass matrices, will only diagonalize a damping

matrix that can be stated as a linear summation of and; i.e.,

For a damping matrix of this particular (and very unlikely)form, the modal damping matrix is defined by

253

where and are the identity matrix and the diagonalmatrix of eigenvectors, respectively. With this damping-matrixformat, an n-degree-of-freedom vibration problem will havemodal differential equations of the form:

This is not a very useful or generally applicable result. Forlightly damped systems, damping is more often introduceddirectly in the undamped modal equations via :

The damping factors are specified for each modal

differential equation, based on measurements or experience.

254

(3.129)

Transient Modal Example Problem 2, Free-Motion withModal Damping

For the preceding collision problem, with viscous dampers andduring contact the vibration model is now

We will rework the problem using assumed modal damping of

255

(3.27)

10% for each mode; i.e., . Hence

The modal differential equation model now looks like:

The solution to the differential equation is

For the initial conditions , the constants A,

B are solved from , and from

The solution is

256

Applying this result to the present example gives,

and the modal coordinate solutions are:

The transformation to obtain the physical coordinates remainsunchanged as

257

q_1 versus time

-1

0

1

2

3

4

5

6

7

8

9

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

q_1

q_2 versus time

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

q_2

258

x_1 (m) versus time

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

x_1

x_2 versus time

-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

x_2

259

f_s1 = k_1 (x_1 - x_2) versus time

-1000

0

1000

2000

3000

4000

5000

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

f_s1

= k

_1 (

x_1

- x

_2)

(N)

f_s2 = k_2 x_2 versus time

-1000

0

1000

2000

3000

4000

5000

6000

0 0.1 0.2 0.3 0.4 0.5 0.6

t (seconds)

f_s2

= k

_2 x

_2 (

N)

Adding damping reduces the peak force in spring 2 from 5363 N to

260

4820 N.