Chapter 2-24.02
-
Upload
vishnu-pradeep -
Category
Documents
-
view
240 -
download
0
Transcript of 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
Dr. Millerjothi, BITS Pilani, Dubai Campus
➢ 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
Dr. Millerjothi, BITS Pilani, Dubai Campus
Consider a simple undamped spring-mass system shown below with natural frequency fn.
Equation of Motion - Natural Frequency
Dr. Millerjothi, BITS Pilani, Dubai Campus
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!!
Dr. Millerjothi, BITS Pilani, Dubai Campus
!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
Dr. Millerjothi, BITS Pilani, Dubai Campus
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
Dr. Millerjothi, BITS Pilani, Dubai Campus
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
Dr. Millerjothi, BITS Pilani, Dubai Campus
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
=
Dr. Millerjothi, BITS Pilani, Dubai Campus
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
Dr. Millerjothi, BITS Pilani, Dubai Campus
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
Dr. Millerjothi, BITS Pilani, Dubai Campus
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
Dr. Millerjothi, BITS Pilani, Dubai Campus
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
Dr. Millerjothi, BITS Pilani, Dubai Campus
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