lec vibration 3 2014 ready 1

NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE

Friday, April 21, 2023

SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING

Free Vibration of Undamped SDOF SystemsLecture Part 2 – L3 1/26

Free Vibration of SDOF Systems

SDOF Systems Revisited

(a) Building Frame

(b) The Space Needle





Free Vibration of A SystemA system is said to undergo free vibration when it oscillates under only an initial disturbance, with NO external forces acting on it after the initial disturbance.

Vibration of an SDOF System:

1. Free Vibration

2. Forced Vibration

3. Resonance





x

m

k

x

t

Typical Response of Free Vibration (Undamped)

T

X

tXx nsinT

where n 2

Natural Frequency of Vibration : nFor an SDOF system of mass m and stiffness k, its Natural Frequency is defined as,

sradmk

n / Hzf nn

2





Derivation of Free Vibration ResponseFor a known SDOF system (values of mass m and stiffness k are given), one can derive vibration response x(t), under any known initial disturbance.

x

mk

00 00 xxxx &kmGiven , tx

x

kmm mkx

FBDO +

maF xmxk 02 xx n





Solve x from

02 xx n (1) 00 00 xxxx &with

From Math Course, general solution of (1) becomes,

tBtAtx nn sincos A, B constants to be determined from initial conditions.

Quick Check! tBtAtx nn sincos If

tBtAtx nnnn cossin Then

tBtAtx nnnn sincos 22 Then

tBtAx nnnnn sincos 222

02 xx n (1)





00 00 xxxx &Determine A, B from given initial conditions:

tBtAtx nn sincos

000 nn BAx sincos A

tBtAtx nnnn cossin

000 nnnn BAx cossin nB

0xA

n

xB

0The final solution becomes,

tx

txtx nn

n

sincos 00

(2)mk

n





Alternative form of Solution:

x

m

k

x

tT

X

tXx nsin

tx

txtx nn

n

sincos 00

(2)

From trigonometry,

tXtx nsin (3)2

020

n

xxX

where

0

01

x

x n

tan





Extra:tbta cossin

t

ba

bt

ba

aba

2222

22 cossin

cos 22 ba

aLet

,cossin22

2

ba

b1Then

a

btan

tbta cossin ttba 22 cossinsincos

,sin tba 22

a

b1tan





Ex 1 – The column of water tank shown is 100m high and is made of reinforced concrete (E=2.6x1010N/m2) of tubular cross section of d1=2.5m and d2=3m. The tank weighs 2.5x105kg with water.

(i) Find the natural frequency of transverse vibration (neglecting mass of the column);

(ii) If the tank is initially displaced to the right by 0.5m, find its subsequent vibration.





As far as transverse vibration is concerned, the tank can be modeled as an SDOF system,

x

mk

3

3

l

EIk

sradmk

n /..

.80

1052

10615

5

3

4410

100

52364

10623 ..

mN /. 51061

(i) The natural frequency becomes,

(ii) Initial conditions are:

0500 00 xmxtAt .:





tXx nsin2

020

n

xxX

0

01

x

x n

tan

28050

tx .sin.

50050 222

020 ..

n

xxX

080501

0

01 ..tantan

x

x n

2

#.cos. tx 8050





M

k1

l

l/2

m v0

Ex 2 – A pendulum is shown with the link assumed to be massless.

(i) Find the natural frequency of vibration of the system;

(ii) If at t=0, a bullet of mass m with speed v0 hits the pendulum, determine the subsequent motion (assume the bullet stays with the pendulum after the impact).





O

M

k1

O

Mg

21l

k

OO IM

21 lM

2

l

2

lklgM cossin

assumed to be small!

small1 cos,sin

21 22

lMll

klgM

04

1

Mk

lg

Mk

lg

n 41

(i)

(ii) After the impact, the total mass of the system will be M+m,

mMk

lg

n

41The new natural frequency is:





Recall the general solution in this case,

ttt nn

n sincos 0

0

Conservation of momentum:

M

k1

l

l/2

m v00

v1

10 vmMvm 01 vmM

mv

lv1

0 lmM

vm

0

t = 0

00

tt nn

sin0

#sin tmM

klg

mMk

lg

mM

vmt

4

4

1

1

0

and





Ex 3 -The crate, of mass m=250kg, hangs from a helicopter using steel cables as shown. Determine the equivalent stiffness of the crate-cable system and compute its natural frequency . nThe cross-sectional area A of the cables selected is A=1.9mm2. If the rotor blades of the helicopter rotate at 300rpm, comment on the choice of the cables and how to improve the situation.





mAD 7071.05.05.0 22

mOD 1213.27071.02 22

mN

l

EAk

POPO /393300

1

109.110207 69

mN

l

EAk

ODOD /185405

1213.2

109.110207 69

Stiffnesses of cable segments:





coss

cos/FFs

Vertical Stiffness:Imagine a force F is applied, the spring force becomes,

Vertical stiffness ky by definition,

N

Fs

F

s

N

Imagine the tip of the spring moves upward by , the actual elongation of the spring becomes,

2

s2

s

ss

sy

kF

FFk

coscos

cos/

cos





,220 bal

Why ?

s

coss

a

b

x

y

2222s baba

22222 bab2ba

1ba

b21ba

22

222

22 bal

1ba2

b21ba

22

222

22 ba

b

cos





o468.192

707.0

OA

ADtg

mN16481146819185405

46819kkko2

o2OD

2sy

/.cos

.coscos

The total stiffness of the 4 springs in parallel:

m/N6592441648114k4k yyt

kPO and kyt in series, the total equivalent stiffness becomes

m/N246337393300659244

393300659244

kk

kkk

POyt

POyteq





s/rad39.31250

246337

m

keqn

s/rad415.3110260/300

The natural frequency becomes,

The rotor passing frequency:

The rotor speed is too close to the natural frequency of the crate-cable system, there will be a lot of vibration generated for the crate-cable system, severely interfering the operation of the system. To improve the situation, the natural frequency of the crate-cable system should be raised and one way of achieving this is to increase the cable cross-sectional area A.





Ex 3** – A block of mass M moves on an ideal frictionless surface. A pendulum is attached to the block through a massless link as shown. Determine the natural frequency of vibration of the system.

This is a difficult problem. We first of all need to establish the EOM. Consider first the conservation of linear momentum in x direction for the system.

x

x

ll

The x direction velocity component of the pendulum is,

coslxvpx





Assume very small as usual (sin=, cos=1), we have,

lxlxvpx cos

Apply conservation of linear momentum in x direction:

0lmxmxM0mvxM px

lmxmM

lmxmM

)(1mM

mlx





x

l l2

lxapt cos

Tangential acceleration component,

Tangential component of force,

x

x

T

mg

te

te

Fix a translational reference on M, then the acceleration of the pendulum has 3 components – relative acceleration , tangential acceleration , and normal acceleration as shown.

xl

l2

On the other hand, the pendulum is subjected 2 forces, and .mg T

sinmgFpt





Apply Newton’s 2nd Law in tangential direction, we have

lxmmg

maF ptpt

cossin

lxg cossin

Again since is assumed small, we have,

)2(lxg

l l2

x

x

T

mg

te

te

)3(lmM

mlg

)4(0

gmM

Ml

)5(l

g

M

mMn





SUMMARY of SDOF Free Vibration

Natural Frequency of Vibration

sradmk

n / Hzmk

fn 21

Natural Period of Vibrationnn

2

f

1T

s

Equation of Motion

02 xx n0kxxm

T





General solution for free vibrations given 00 xx ,

tBtAtx nn sincos 0xA n

xB

0

tx

txtx nn

n

sincos 00

(2)

202

0

n

xxX

0

01

x

x n

tan

Alternatively,

tXtx nsin (3)

lec vibration 3 2014 ready 1

Documents