MAE 340 – Vibrations

Multi-Degree-of-Freedom Systems

Section 4.1

Examples of Multi-Degree-of-Freedom (MDOF) Systems

Swaying Building

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k3

m3

k4

m4

x4

x3

Fcos ωt

k2

m2

k1

m1

x2

x1

k ck c

Moving Vehicle

x1

x3

x2

Flying Aircraft

Two DOF Spring-Mass System

• Given:

� m1 = 9 kg

� m2 = 1 kg

� k1 = 24 N/m

� k2 = 3 N/m

• Find:

� ωn1, ωn2

� x1 (t ), x2 (t )

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k1m1

x1k2

x2

m2

� x1,0 = 1 mm

� x2,0 = 0 mm

� v1,0 = 0 m/s

� v2,0 = 0 m/s

Two DOF Spring-Mass System

• Sol’n:

Step 1: Derive system differential equations

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Mass 1 Mass 2

Two DOF Spring-Mass SystemWrite out coupled equations as matrix equation

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With variables:

With numbers:

Solving Differential EquationStep 2: Solve matrix differential equation

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Assume a solution:

)sin()sin(

)sin(

)(

)()(

2

1

2

1

2

1φω

φω

φω+

=

+

+=

= t

u

u

tu

tu

tx

txtx

n

n

n

)sin()( φω += tutxn

=

=

)(

)()(

2

1

tx

txtx

&&

&&&&

Solving Differential EquationSubstitute back into system differential equation:

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Simplify:

=

0

0u is a trivial solution; we are not interested

in it.

Solving Differential EquationIf we could solve for then we

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could get

The only way to get a non-trivial solution would be if it were impossible to solve for

( ) 12 −+− KM

nω

( )

==+−=

−

0

000

12KMu

nω

but this is still the trivial solution!

( ) 12 −+− KM

nω

When is a matrix not invertible??

Solving Differential Equation

When the determinant is zero!!

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Need to find ωn that satisfy:

Recall that the determinant of a matrix is computed as follows:

( ) 0det2 =+− KMn

ω

Solving Differential EquationPutting in the numbers:

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Mode 1: ωn1 =

Substituting ωn1 into equation from slide 7:

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Mode 1These equations are not linearly independent. There is no single solution. Instead there is a set of solutions. But each solution must satisfy:

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We could use a scaling factor (s):

This is the “Mode Shape”.

Mode 1Usually s is chosen so that the largest u value is 1:

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Therefore, the equation of motion, considering only the first mode is:

This is the “Normalized Mode Shape”.

Mode 2: ωn2 =

Substituting ωn2 into equation from slide 7:

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Mode 2

The normalized mode shape is:

The equation of motion, considering only the second mode is:

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Both Modes Together

The overall equation of motion, considering both modes is:

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The unknowns (A1, Φ1, A2, Φ2) must be determined from the initial conditions.

=)0(x

=)(tx&

Solving for A1, Φ1, A2, Φ2

• Applying the initial conditions:

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=)(tx

=)0(x&

Solving for A1, Φ1, A2, Φ2

• Need to solve the nonlinear equations:

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Solving for A1, Φ1, A2, Φ2

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Solving for A1, Φ1, A2, Φ2

• Therefore, the overall equation of motion is:

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