Chapter 2-24.02

BITS PilaniDubai Campus

Chapter-‐2. Vibration analysis, spring mass and damping elements in a vibrating system, Types of damping

Dr. Millerjothi, BITS Pilani, Dubai Campus

Modelling of vibrating systems natural and undamped

Effect of damping on systems in vibration Other types of damping and energy loss

Chapter 2


➢ The basic vibration model of a simple oscillatory system consists of a mass, a mass less spring, and a damper.

➢ The mass is measured in kilograms in the SI system & ➢ In the English system the mass is, m= w/g lb.s2/in ➢ The spring supporting the mass is of negligible mass. ➢ Its force-deflection relationship is considered to be linear,

following Hooke's law, F = kx, where k, the stiffness measured in N/m or lb/in.

➢ The viscous damping, generally represented by a dashpot, is described by a force proportional to the velocity, or f = c x.

➢ The damping coefficient c is measured in N/m/s or lb/in/s.

Vibration Model


Consider a simple undamped spring-mass system shown below with natural frequency fn.

Equation of Motion - Natural Frequency


The spring force is equal to gravitational force, w acting on mass m. !By Newton’s second Law !But hence ……………1 !Circular frequency is defined as !Therefore equation of motion is written as !Which has the general solution as x = A Sin ωnt + B Cos ωnt !where A & B are 2 constants found from initial conditions of

displacement and velocity.

)( xkwfxm +Δ−=Σ=!!

kxxm −=!!

mk

n =2ω

02 =+ xx nω!!


!This results in

! Or !And the natural frequency is

tCosxtSinx

x nnn

ωωω

)0()0(+=

!

πτω 2=n km

πτ 2=

Δ===

gmkfn ππτ 2

1211


A 0.25kg mass is suspended by a spring having a stiffness of 0.1533 N/mm. Determine its natural frequency in cycles per second. Determine its statical deflection.

Solution The stiffness is k = 153.3 N/m The natural frequency is ! = 3.941 Hz !The statical deflection of the spring suspending the 0.25 kg mass

is obtained from the relationship mg = kΔ

Example 2.2.1

25.03.153

21

21

ππ==

mkf

mmkmgmmN

0.161533.0

81.925.0

/

=×

==Δ


Determine the natural frequency of the mass M on the end of a cantilever beam of negligible mass shown in Fig. 2.2.

Example 2.2.2


The deflection of the cantilever be ever am under a concentrated end force P is

!!Where EI is the flexural rigidity, thus the stiffness of the beam is and the natural frequency of the system becomes

Solution

kP

EIPlx ==3

3

33 lEIk =

33

21

MlEIfn π

=


Example 2.2.3


�12

�13

�14

�15

�16

�17

�18

�19

�20

�21

�22

�46

�47

�48

�49

�50

�51

�52

�53

�54

�55

�59

�60


The following data are given for a vibrating system with viscous damping: w = 10 lb, k =30 lb/in., and c = 0.12 lb/in./s. Determine the logarithmic decrement and the ratio of any two successive amplitudes.

Example


The undamped natural frequency of the system in radians per second is

!!!The critical damping coefficient cc and damping factor ξ are

Solution

sradmk

n /0.3410/38630 =×==ω

sinlbmc nc /76.134

1038622 =××== ω

0681.076.112.0

===cccζ


The logarithmic decrement, !!!!The amplitude of any two consecutive cycles is

429.00681.010681.02

12

22 =−

×=

−=

π

ζ

πζδ

54.1429.0

2

1 === eexx δ

�64

�65


Figure 2.3 shows a uniform bar pivoted about point o with springs of equal stiffness k at each end. The bar is horizontal in the equilibrium position with spring forces P1 and P2 Determine the equation of motion and its natural frequency.

Example 2.2.4


Under rotation, the spring force on the left is decreased while that on the right is increased. With J0 as the moment of inertia of the bar about O, the moment equation about O is

!!However !in the equilibrium position, and hence we need to consider only the

moment of the forces due to displacement θ, which is

θθθ !!OJbkbPmgcakaPMO =+−+−=Σ )()( 21

021 =−+ bPmgcaP

θθ !!OJkbkaMO =−−=Σ )( 22


Thus the equation of motion can be written as !!!!And by inspection the natural frequency of oscillation is

0)( 22

=+

+ θθOJbak!!

On J

bak )( 22 +=ω

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Chapter 2-24.02