k 0 x x m ••+ = z multiply by x • : k 0 x x x x m z integration of the Eq

(1)

Wave propagation (overview)

Rayleigh`s method [Linear conservative system (no damping)]

-

The equation of Energy for a vibrating system

z the original Eq. : k 0

x x

m

••+ =

z multiply by x

• : k 0

x x x x m

•• • •

+ =

z integration of the Eq. : 1 ² 1 ² 2( ) 2

x k x c

m

• + = ( c: integral constant )

→

1 ² 1 ²

( )

2 m x 2 k x c m

• + =

Instantaneous Instantaneous kinetic Energy potential Energy

z When the displacement x is a maximum Ⅹ

The velocity x 0 → all the dynamic energy in the system is potential

• =

→

¹ ²

2kX =cm=P

z When the displacement x is zero, the velocity x

• will be a maximum X

•

→

all the d.e. is kinetic

→

¹ ² 2k X cm

• =

- Thus, 1 ² 1 ² 1 ² 1 ²

( ) ( )

2m x 2kx 2kX 2m X P

• •

+ = = =

sin x=X wt

cos ( )

x wX wt X wX

• •

= =

(2)

- Rayleigh`s principles

z The fundamental natural frequency as calculated from the assumed shape of a dynamic deflection curve of a system will be equal to or higher than the system`s true natural frequency

(upper bound nature ↔ Southwell-Dunkerley Method)

z Small departures from the shape of the true dynamic – deflection curve will not be critical in the determination of the system`s w_n

(justifies the use of the static deflection curve)

Ex. 1. Natural frequency of the spring-mass system (spring w/ its weight)

- maximum KE(1 ² 2m X( )

• )

Spring: the displacement of spring at distance x

- x ₀cos _n

z Z w

= L t ( ← assumed the extension of the spring is linear ) - Velocity of element dx, x _n ₀sin _n

z w Z

= −L w t - Max. KE of the element ( mass = (w^s)

g dx⁾ d(KE)max =

2 ws

gdx⁽

2

(x _n 0) Lw Z ^{← [}

2 0

1 ( ) ]

2

s

n

w x

dx w Z

g L

⋅ ⋅

(3)

- For the whole spring

2 2

0 max

0

( ) ( )

2

L

s n

w w Z

KE x dx

g L

=

∫

2 2

0

1 2 3

s n

w Lw Z

= g - mass :

2 2

max 0

1 1

( ) ( ) ( )

2 2

m n

KE m Z W w Z

g

= • =

- The total KE = ² ₀²

1

1( 3 )

2

m s

n

w w L g w Z +

- maximum PE (1 ² 2kX )

2 2

max 0

1 1

( )

2 2

PE = kX = kZ

- (KE)_max =(PE)_max

2 2 2

0 0

1

1( 3 ) 1

2 2

m s

n

W W L

w Z kZ

g

+ =

1 3

n

m s

w kg

W w L

∴ =

+

(4)

Ex. 2 Cantilever beam

cos _n z=y w t

n sin n

z w y w

• = − t

Assume the dynamic deflection curve of the beam as a static defection curve of a weightless beam w/ the concentrated load P acting at its end.

- (PE)_max

→ ₀ ³

3 y PL

= EI →

( k

_{e q}

)

at the free end is (

3

3 L

EI ⁾

→ _max 1 ² 1 ₀² 3 ₃ ₀

( )

2 2 2

EI 2

PE kX ky

= = = L y

- (KE)_max

→ ² ²

0

1 1

( ) ( )

2 2

L n

m X W w y dx g

• =

∫

2 2 3 2

0 0

( ) [3( ) ( ) ]

2 2

L

w yn

w x x

g L L dx

=

∫

−

2 2

0

1 33( )

2 140 ⁿ

wL w y

= g

- _n 3.56 gEI₄

w = wL (cf. ( _{n exact}) 3.515 gEI₄

w = wL )

(5)

Logarithmic Decrement

-

¹

2

log_e x

δ = x , x1&x₂: successive peak amplitudes

And ₀

exp( 2)sin( )

1

nd

z Z ξw t w t φ ξ

= − +

−

If x₁ is the amplitude at w t_nd ₁, then x₂, at (w t_nd ₁+2π)

→

1 2

exp log 1

( 2

exp 1

nd

e

nd

w t w t )

ξ

δ ξ ξ π

ξ

−

= − − +

−

2

log exp 2

e 1

πξ

= ξ

−

2

2 1

πξ

= ξ

−

≒2πξ(if ξ <<1)

※ quite often ξ≤0.1 in practice - In this case (ξ <<1)

0 1 1

1 2

n n

x x x

x x x e

− δ

= = = = =

e

²^{π ξ}

0 0 1 1

1 2

& ⁿ ( ⁿ

n n

x x x x

x x x x e

− δ

= ⋅ ⋅ ⋅ = )

→ 1 ⁰

log_e

n

x

n x

δ =

(6)

Determination of viscous Damping

- In a free vibration test,

1 2

1 log

2 2 ^e

x x ξ δ

π π

= =

or 1 ⁰

2 2 log^e _n x

n x

ξ δ

π π

= =

- In a forced vibration test, [ varying frequencies obtain a resonance curve ]

0

2 2 2

1

(1 ) (2 )

st

Z N

r r

δ ⁼ − + ξ ⁼ (magnification factor)

If 1

n

r w

=w = (at resonance)

0

max

1 ( )

st 2

Z N

δ ⁼ ξ ^, ⁽∵ξ<<¹⁾ n practice

(Fig 2.20) - When the magnification factor of motion is 1 1

(2

2 )

ξ , the frequency ratio, r is,

2 2 2 2

1/ 2 1

2ξ ⁼ (1−r ) +4ξ r

→r₂² −r₁² =4ξ 1−ξ²≒ 4ξ (ifξ <<1)

And

2 2

2 2 2 1 2 1 2 1

2 1 2 ( )(

n n

)

n

f f f f f f

r r

f f f

− + −

− = =

≒ 2 ( ² ¹)

n

f f f

⋅ − - refer to Fig 2.20

→ ( ² ¹

4 2

n

) f f

ξ = ⋅ f⁻ → 1 ² ¹

( )

2 _n

f f ξ = ⋅ f⁻

※ called ‘bandwidth’ method

(7)

Ground Acceleration

-

∑

F =ma

( ^g) ( _g)

c x x k x x m x

• • •

− − − − = ^•

→m x c x x( g) k x( xg) 0

•• • •

+ − + − = ….①

- let x_d = −x x_g₁ Then, x^d x x^g

• • •

= − x^d x x^g

•• •• ••

= − →x^d x^d x^g

•• •• ••

= − ….②

- Sub. Eq. ② into ①

( ^d ^g) _d _d

m x x c x kx

•• •• •

+ + + =0

→ m x^d c x_d kx_d m xg ← Govern Eq. ….③

•• • ••

+ + = −

- If x_g =x t_g( )=X_gsinwt

→x tg( ) w X² gsin

•• = − wt

→m x^d c x_d kx_d mw X² _gsinwt

•• •

+ + =

- Solution :

sin( )

d d

x = X wt−φ

2

2 2 2

( ) ( )

g d

X mw X

k mw cw

= − + ^,

1

tan cw 2

k mw φ = ⁻

−

2

2 2 2

( / ) /

(1 ) (2 )

n

d g

X X w w

r ξr

= − + ^….④,

1 2

tan 2 1

r r φ = ⁻ ξ

−

(8)

The Governing Eq. of the system of vibration measuring instrument is same with Eq. ③

- Displacement pickup

→ For large value of w w/ _n(=r)

d / g

X X =1 for all values of damping (refer to Fig. 2.23)

→ i.e. the resulting relative motionXd = Xg(Xbase) Picked up by the instrument

- Acceleration pickup

→ Rearrange Eq. ④

2 2 2 2 2 2 2

1 1

(1 ( / ) ) [2 ( / ) ]

d

b n n n n

X

w X ⁼ w − w w + ξ w w ⁼w D For ξ =0.69 → D =1 when w w/ _n is small

w/wn 0.0 0.1 0.2 0.3 0.4

D ^1.000 ^0.9995 ^0.9989 ^1.0000 ^1.0053

→ i.e. X_d ∝w X² _b (max. acceleration of the vibrating base)

2 2

1

d n

X w Xb

= w ⋅ x= X_bsinwt

bcos x wX wt

• =

2 _bsin x w X wt

••= − → ( )x _max w X² _b

•• =

(9)

System w/ Two-degrees of Freedom

- mass 1

1 1 0sin 1 1 2( 1 2)

m z F wt k z k z z

•• = − − − ….①

- mass 2

2 2 2( 1 2

m z k z z

•• = − )

t

…. ①’

- assume( vibrating frequency = exciting frequency )

∵

1 1sin

z =Z w

2 2sin

z =Z wt - Sub. These into Eq. ①, ①’

2

1 1 2 1 2 2

(−m w + +k k Z) −k Z =F₀ 0

….②

2

2 2 2 2 1

(−m w +k Z) −k Z = ….②’

- Let the natural circular frequencies of the systems 1 & 2 be

1 11

1

w k

= m , 22 ² 2

w k

= m

& 0 ⁰ 1

Z F

= k (i.e. the static deflection of the mass m1 due to F₀ ) - Dividing Eq. ② & ②` by k₁&k₂ respectively,

1 2 2

1 2

2 11 1

[1 k ( w) ] k

Z Z Z0

k w k

+ − − = ….③

1

22

[1 ( w )] 0

Z w

− + − Z₂ = ….③’

(10)

- From Eq. ③’, we get

1

2 2

1 ( / 22) Z Z

= w w

−

- Sub. Into Eq. ③

2 2 2 1

1 0

1 11 1 2

22

[1 ( ) ]

k w k Z

Z Z

k w k w

w

+ − − =

−

- Therefore,

2

1 22

2 2 2 2

0

22 1 11 1

1 ( )

[1 ( ) ][1 ( ) ]

w

Z w

w k w

Z

w k w

−

=

− + − −k

k

….④

Similarly, we get

2

2 2 2 2

0

22 1 11 1

1

[1 ( ) ][1 ( ) ]

Z

w k w

Z

w k w

=

− + − k

− k

w

….④’

- From Eq. ④

If w₂₂ = → Z₁=0

then → ₂ ₀ ¹ ⁰

2 2

F Z Z k

k k

= − = −

※ the system 2 is used as a vibration absorber for the main system 1

(11)

- the natural frequencies of the system

(consider the free vibration → R.H.S of ③ = 0 )

From Eq. ③ & ③’

1 2 1 2

2

2 2 1 11

/ 1 ( )

[1 / ( / ) ] ₂₂

Z k k w

Z = k k w w = w

+ − −

or ² ² ²

1 11 22 2

[1 k ( w) ][1 ( w ) ] k 0

k w w k

+ − − − ¹ = [ Frequency equation ]

or, since ² ² ¹ ² ²² ²

1 2 1 1 11

( )

k k m m w

k =m k m =μ w where, ²

1

m μ = m

2 4 2 2

22 22

11 22 11 22

[w ] ( w ) [1 (1 )(w ) ]( w ) 1

w w − + +μ w w + =0

(12)

Spectral Response

- Response quantities : the quantities that are examined as a result of a forcing function (may or may not be a mathematical function)

Ex. Displacement, velocity, acceleration, stress, strain…..

- maximum response of a multidegree freedom system

2 2

11 1 22 2

( ) ( ) ( _nn _n

X = A X + A X + + A X )²

where, A A₁₁, _22, : the mode participation factor (A₁₁ > A₂₂ > ) X1 : the maximum response of the 1^st mode

Mode participation factor depends on

1) type of system 2) boundary condition 3) type of response quantity

- e.g., Longitudinal vibration of Prismatic Bar

2

1,3,5.. 2

4 1

cos cos _n

n

Pl n x

u w

AE

n

l ^t

π π

∞

=

= −

∑

1,3,

4 1

n cos _n

n

P n x

A l w t

π

5..

n

si σ π

∞

=

∑

(13)

)

- Response spectrum (Fig.2.25) : max. response of SDF system vs. natural period (of frequency) under different damping ← (N = f w c( _n, )