CE 620: STRUCTURAL DYNAMICS SEMESTER I: 2011

CE:620/Dr Durgesh Rai/S-I-2011/Page 1

Classical Methods of Finding Fundamental Frequency DUNKERLEY’S FORMULA

Frequency Equation:

2 0k mω− =% %

Premultiplying by leads to 1k−%

12

1 0I k mω

−− + =% % %

(A)

D%

1k f− = ≡% %

Flexibility matrix whose coefficients are , ija , 1.....i j n=

For lumped mass with diagonal mass matrix, equation (A) becomes

11 12 1 1

21 22 2 22

1 2

0 11 01 0

0 0 1

n

n

n n mn n

a a a ma a a m

a a a mω

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥ =⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦ ⎣ ⎦

K

K

M L O O

K

⎥ (B)

That is

11 1 12 2 12

21 1 22 2 22

1 2 2 21

1

10

1

ω

ω

ω

− +

− +=

−

L

L

M M L M

L

n n

n n

n nn

a m a m a m

a m a m a m

a m a m a mn n

(C)

Expand to get nth degree polynomial in 21ω⎛ ⎞⎜ ⎟ ⎝ ⎠

( )1

11 1 22 221 0ω ω

−⎛ ⎞ ⎛ ⎞− + + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Ln n

nn na m a m a m 21

=L (D)

If roots of equation (D) are 21

1ω

, 22

1ω

,… 21

nω, then it can be written in factored form as follows.

2 2 2 2 2 21 2

1 1 1 1 1 1 0ω ω ω ω ω ω

⎛ ⎞⎛ ⎞⎛ ⎞− − −⎜⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜

⎝ ⎠⎝ ⎠ ⎝ ⎠L

n=⎟⎟ (E)

CE 620: STRUCTURAL DYNAMICS SEMESTER I: 2011

CE:620/Dr Durgesh Rai/S-I-2011/Page 2

or 1

2 2 2 2 21 2

1 1 1 1 1 ... 0ω ω ω ω ω

−⎛ ⎞⎛ ⎞ ⎛ ⎞− + + + + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠L

n n

n (F)

Comparing equations (F) & (D)

11 1 22 22 2 21 2

1 1 1..... .... nn nn

a m a m a mω ω ω

+ + + = + + + (G)

If higher frequencies iω are larger the fundamental frequency 1ω and

21

1

i2

1ω ω

<< for i= 1, 2…n

Equation (G) can be approximated as

11 1 22 221

1 ... nn na m a m a mω

+ +

Each term on RHS is reciprocal of concerned frequency and is representing contribution to 21

1ω

by each mass when other masses are absent, hence,

2 2 21 11 22

1 1 1 1....ω ω ω ω

= + + +nn

2

where,

1 121ω

⎛ ⎞ ⎛= =⎜ ⎟ ⎜⎝ ⎠ ⎝

inin

in i i

ka m m

2⎞⎟⎠

; i = 1…n

CE 620: STRUCTURAL DYNAMICS SEMESTER I: 2011

CE:620/Dr Durgesh Rai/S-I-2011/Page 3

EXAMPLE I m2

Shaker

a22

Frequency of beam when vibrated by shaker of mass 10 kg is 10 Hz. Adding another mass of 10 kg frequency reduces to 8 Hz. Determine true natural frequency of beam.

Let natural frequency of structure plus exciter = 1ω

Let natural frequency of structure by itself = 11ω

Natural frequency of exciter mounted on the structure in absence of other masses = 22ω .

2 2 21 11 22

1 1 1ω ω ω

∴ = + = 22 2211

1 a mω

+

For two loads cases:

( ) ( )

( ) ( )

222 211

222 211

1 1 10 ( )2 10 2

1 1 20 ( )2 8 2

π π

π π

= + ××

= + ××

kg a af

kg a bf

Eliminate a22 to get f11: f11 = 15.1 Hz

and a22 = 0.014 mm/N or a22 = 70 kN/m