9 - multi dof - free vibrations

7/30/2019 9 - Multi DOF - Free Vibrations

1/28

Week 9

1

Part 1. Mainly based on the text book by D.J. Inman


2/28

Week 9

2

1. Undamped 2-DOF System

We review again the steps to solve the free undamped vibration ofa 2-DOF system

Focus on vectors and matrices

Easy to extend to more than 2-DOF systems


3/28

Week 9

3

Undamped 2-DOF System

A 2 Degree-of-Freedom system has

Two equations of motion!

Two natural frequencies (as we shall see)!


4/28

Week 9

4


The dynamics of a 2 DOF system consists of2 homogeneous and

coupled equations

Free vibrations, so homogeneous eqs.

Equations are coupled: Both have x1 and x2; if only one mass

moves, the other follows

In this case the coupling is due to k2.

Mathematically and Physically

Ifk2 = 0, no coupling occurs and can be solved as two

independent SDOF systems


5/28

Week 9

5


Two coupled, second -order, ordinary differential equations withconstant coefficients

Needs 4 constants of integration to solve

Thus 4 initial conditions on positions and velocities

1 10 1 10 2 20 2 20(0) , (0) , (0) , (0)x x x x x x x x


6/28

Week 9

6

Solution by Matrix Methods

The two equations can be written in the form of a single matrix equation:

(4.4), (4.5)

(4.6), (4.9)

The initial conditions can also be written in vector form:

10 10

20 20

(0) , and (0)x x

x x

x x


7/28

Week 9

7

Approach to a Solution

We assume a solution of the vector form:

Remember that the scalar ejt represents harmonic motion

(Euler Form):

Substitute into the vector EoM

The scalar ejt 0 for any time t

(4.15)

(4.16)

(4.17)


8/28

Week 9

8


This changes the differential equation of motion into algebraic vector

equation:

(4.17)

This is two algebraic equationswith three unknowns:

2

12

2

If the inv - exists : which is the

static equilibrium position. For motion to occur

- does not exist

or det - (4.19)

M K

M K

M K

u 0

u 0

0


9/28

Week 9

9


The determinant results in 1 equation in one unknown , calledthe characteristic equation.

For our specific system:

det -2M K

0

det2m1 k1 k2 k2

k2 2m2 k2

0

m1m

2

4

(m

1k

2 m

2k

1 m

2k

2 )

2

k

1k

2 0

(4.20)

(4.21)

Eq. (4.21) is quadratic in 2 so we have foursolutions:

2 2

1 2 1 2and and


10/28

Week 9

10


With known, use equation (4.17) again to calculate the correspondingvectors u1 and u2 .

This yields vector equation for each squared frequency:

2

1 1

2

2 2

( ) (4.22)

and

( ) (4.23)

M K

M K

u 0

u 0

Each of these matrix equations represents two equations in the twounknowns components of the vector u, but the coefficient matrix is

singular (what does this mean?) so each matrix equation results in

only one independent equation.


11/28

Week 9

11

Example: calculating u and

Related to Examples 4.1.5 & 4.1.6 in the text book of D.J. Inman.

m1 = 9 kg, m2 = 1 kg, k1 = 24 N/m and k2 = 3 N/m

The characteristic equation becomes:

Each value of2 yields an expression foru:


12/28

Week 9

12


112

1 1

12

2

1 1

11

12

11 12 11 12

For =2, denote then we have

(- )

27 9(2) 3 03 3 (2) 0

9 3 0 and 3 0

u

u

M K

u

u

u u u u

u

u 0

See that we have 2 equations with 2 unknowns but DEPENDENT!(the 2nd equation is -3 times the first).

When solving, only the direction of vectors u can be determined,

not the magnitude as it remains arbitrary why?


13/28

Week 9

13


1111 12

12

2

1

1 1results from both equations:

3 3

only the direction, not the magnitude can be determined!

This is because: det( ) 0.

The magnitude of the vector is arbitrary. To see this suppose

t

uu u

u

M K

1

2

1 1 1

2 2

1 1 1 1

hat satisfies( ) , so does , arbitrary. So

( ) ( )

M K a a

M K a M K

u

u 0 u

u 0 u 0

W k 9


14/28

Week 9

14


Likewise for the second value of2:

212

2 2

22

21

21

22

21 22 21 22

For = 4, let then we have

(- )

27 9(4) 3 0

3 3 (4) 0

19 3 0 or

3

u

u

M K

u

u

u u u u

u

u 0

Also here we get only the direction of vectors u.

What to do about the magnitude?

W k 9


15/28

Week 9

15


A numerical value for each element of the vectoru may be obtainedby arbitrarily fixing one of the elements.

u12 1 u1 1

3

1

u22 1 u2 1

3

1

Choose:

Choose:

Thus the solution to the algebraic matrix equation is:

1,3 2, has mode shape u1 1 3

1

2, 4 2, has mode shape u2 1

3

1

W k 9


16/28

Week 9

16

Time Response

We have computed four solutions:

Since linear, we can combine as:

determined by initial conditions.

(4.24)

(4.26)

W k 9


17/28

Week 9

17

Physical Interpretation of All That Math!

Each of the TWO masses is oscillating at TWO natural

frequencies1 and 2

The relative magnitude of each sine term, and hence of the

magnitude of oscillation ofm1

and m2

is determined by the value

ofA1u1 and A2u2

The vectors u1 and u2 are called mode shapesbecause they

describe the relative magnitude of oscillation between the two

masses

What is a mode shape?

Week 9


18/28

Week 9

18

Mode Shape

What is a mode shape?

Remember that A1, A2, 1 and 2 are determined by the initial

conditions

If we choose the initial conditions so that A2

= 1

= 2

=0

Then:

Thus each mass oscillates at (one) frequency 1 withmagnitudes proportional to u11

stmode shape

likewise for the 2ndmode shape

x(t) x1(t)

x2 (t)

A1

u11

u12

sin1t A1u1 sin1t

Week 9


19/28

Week 9

19

Mode Shape

A graphic look at mode shapes:If initial conditions correspond to mode 1 or 2, then the response is

purely in mode 1 or mode 2.

Week 9


20/28

Week 9

20

Example: compute the time response

Related to Example 4.1.7 in the text book of D.J. Inman.

Given the initial conditions, calculate the solution of the system of

Examples 4.1.5 and 4.1.6 (previous slides).

Form of solution:

We have found:

Take the positive values

Week 9


21/28

Week 9

21


Is written as:

Differentiating:

(4.26)

Week 9


22/28

Week 9

22


At t = 0 this yields

Displacement:

Velocity:

4 equations in 4 unknowns:

Week 9


23/28

Week 9

23


Solution of the 4 simultaneous equations:

So, the final solution is:

Remember that the time response was in the form:

x1(t) 0.5 cos 2t 0.5cos2t

x2 (t) 1.5 cos 2t 1.5cos2t

Week 9


24/28

Week 9

24


Final solution: x1(t) 0.5 cos 2t 0.5cos2tx2 (t) 1.5 cos 2t 1.5cos2t

(4.34)

The initial conditions gives responses that are combinations of the

two modes. Each of both are harmonic, but their summation is not.

Figure 4.3

Week 9


25/28

Week 9

25

Solution as a Sum of Modes

x(t) a1u1 cos1t a2u2 cos2t

Determines how the first

frequency contributes to the

response

Determines how the second

frequency contributes to the

response

Week 9


26/28

Week 9

26

Things to Note

Two degrees of freedom implies two natural frequencies

Each mass oscillates with these two frequencies present in the

response and beats could result

Frequencies are not those of two component systems

The above method is not the most efficient way to calculate natural

frequencies.

1 2 k1

m1

1.63, 2 2 k2

m2

1.732

Week 9


27/28

Some Matrix and Vector Reminders

Week 9

27

1

2 2

1 2

1 2 2

1 1 2 2

2

1

0

0

0 0 for every value of except 0

T

T

T

a b d bA A

c c c aad cb

x x

mM M m x m x

m

M M

x x

x x

x x x

Then Mis said to be positive definite

Week 10


28/28

ee 0

28

3 Approaches to Compute Mode Shapes

and Frequencies

(i) 2Mu Ku (ii) 2u M1Ku (iii) 2v M

12KM

12v

i. Is the Generalized Symmetric Eigenvalue Problem,easy for hand computations, inefficient for computers

ii. Is the Asymmetric Eigenvalue Problem very

expensive computationally see in Rao, Kelly

iii. Is the Symmetric Eigenvalue Problem the cheapestcomputationally

This was discussed today

9 - multi dof - free vibrations

Documents