Chapter 2 Free Vibration of Single Degree of Freedom
-
Upload
senthilcae -
Category
Documents
-
view
801 -
download
20
Transcript of Chapter 2 Free Vibration of Single Degree of Freedom
Free vibration of single degree
of freedom (SDOF)Chapter 2
Introduction
• A system is said to undergo free vibration when it oscillates only under an initial disturbance with no external forces acting after the initial disturbance
Introduction - SDOF
• One coordinate (x) is sufficient to specify the position of the mass at any time
• There is no external force applied to the mass
• Since there is no element that causes dissipation of energy during the motion of the mass, the amplitude of motion remains constant with time, undamped system
Introduction - SDOF
• If the amplitude of the free vibration diminished gradually over time due to the resistance the resistance offered by the surrounding medium, the system are said to be damped
• Examples: oscillations of the pendulum of a grandfather clock, the vertical oscillatory motion felt by a bicyclist after hitting a road bump, and the swing of a child on a swing under an initial push
Free Vibration of an Undamped
Translation System• Equation of Motion using Newton’s Second Law
▫ Select a suitable coordinate to describe the position of the mass or rigid body
▫ Determine the static equilibrium configuration of the system and measure the displacement of the mass or rigid body
▫ Draw the free body diagram of the mass or rigid body when a positive displacement and velocity are given
▫ Apply Newton’s second law of motion
FV of an undamped …
• Newton’s second law
• Applied to a undamped SDOF system
xmtF
)(
JtM )( For rigid body undergoing rotational motion
FV of an undamped …
xmkxtF
)(
0 kxxm
FV of an undamped …
• Equation of Motion using other methods
▫ D’Alembert’s Principle
▫ Principle of Virtual Displacements
▫ Principle of Conservation of Energy
0 kxxm
• Spring-Mass System in Vertical Position
0 kxxm
stkmgW
Wxkxm st
FV of an undamped …Solution!
• The solution can be found assuming,
substituting
stCetx )(
0)()(2
2
stst CekCedt
dm 02 kmsC
nim
ks
21
21
m
kn
characteristic equation eigenvalues
02 kms
FV of an undamped …Solution!
• The general solution,
titi nn eCeCtx
21)(
tite ti sincos using,
tAtAtx nn sincos)( 21
A and C can be determine from the initial conditions
FV of an undamped …Solution!
• The initial conditions at t =0,
• The solution is
01)( xAtx
tx
txtx n
n
n
sincos)( 00
02)( xAtx n
FV of an undamped …Harmonic
• Previous equations are harmonic function of time
• The motion is symmetric about the equilibrium position of the mass
• The velocity is maximum and the acceleration is zero at the equilibrium position
• At the extreme displacement the velocity is zero and the acceleration is a maximum
• The quantity is the natural frequency n
FV of an undamped …Harmonic
sin2 AA cos1 AA
amplitudex
xAAAn
2/12
02
0
2
2
2
1
21
anglephasex
x
A
A
n
0
01
1
21 tantan
tAtAtx nn sincos)( 21
FV of an undamped …Harmonic
2/12
02
00
n
xxA
0
01
0 tanx
x n
tAtx ncos)(
Substituting, the solution can be written
using the relation,001 cosAA 002 sinAA
00 sin)( tAtx n
• Obtain the free response of
a) in the form
b) in the form
Initial condition are x(0) = 0.05 m and x(0) = -0.3 m/s
Example - Harmonic
)(1282 tfxx
tAtAx nn cossin 21
tAx nsin
Example - solution
ttttxa 8cos05.08sin0375.08cos05.08sin8
3.0)
062.08
3.005.0)
2
22
Ab
806.0062.0
05.0sin 605.0
)8(062.0
3.0cos
rad214.20927.927.)333.1(tan605.0
806.0tan 11
mtx 214.28sin062.0
FV of an undamped …Harmonic
2/1
m
kn
1)Note the following aspect,
stst
mgWk
2/1
st
n
g
2/1
2
1
st
n
gf
2/1
21
gf
st
n
n
FV of an undamped …Harmonic
tAtx ncos)(
2cossin)(
tAtAtx nnnn
tAtAtx nnnn coscos)( 22
2)
FV of an undamped …Harmonic
3) If the initial displacement is zero,
tx
tAx
tx n
n
n
n
sin
2cos)( 00
If the initial velocity is zero,
txtx ncos)( 0
4)The response of a SDOF system can be represent in the displacement ( ) – velocity ( ), known as the state space or phase plane
FV of an undamped …Harmonic
xx
tAtx ncos)(
tAtx nn sin)(
A
x
A
y
12
2
2
2
A
y
A
x
• The column of the water tank shown is 300 ft. high and is made of reinforced concrete with a tubular cross section of inner diameter 8 ft. and outer diameter 10 ft. The tank weighs 6 x 105 lb with water. By neglecting the mass of the column assuming the Young’s modulus of reinforce concrete as 4 x 106 psi, determine the following:
Example - Harmonic
• the natural frequency and the natural time period of transverse vibration of the water tank
• the vibration response of the water tank due to an initial transverse displacement of 10 in.
• the maximum values of the velocity and acceleration experienced by the water tank
Example
Example – solution
3
3
l
EIPk
a) The stiffness of the beam,
l= 3600 in. , E = 4 x 106 psi, 444444
0 10600961206464
inddI i
inlbl
EIk /1545
3600
)10600)(104(333
46
3
sec/9977.0106
4.38615455
radm
kn
sec2977.62
n
n
Example – solution
b) Using the initial displacement of x0 =0 and the initial velocity =0,
00 sin)( tAtx n
inxx
xAn
100
2/12
02
00
20tan 01
0
nx
tttx 9977.0cos102
9977.0sin10)(
Example – solution
c) The velocity and acceleration can be determine by differentiating,
29977.0cos9977.010)(
ttx
sec/977.99977.0100max inAx n
29977.0sin9977.010)(
2 ttx
222
0max sec/9540.99977.010 inAx n
• A simply supported beam of square cross section 5 mm x 5 mm and length 1 m, carrying a mass of 2.3 kg at the middle, is found have a natural frequency of transverse vibration of 30 rad/s. Determine the Young’s modulus of elasticity of the beam.
Example - Harmonic
Example – solution
3
192
l
EIk
41033 105208.010510512
1mI
m
kn
I
lmE n
192
32
2
3
192nm
l
EIk
29
10
32
/100132.207)105208.0(192
0.10.303.2mNE
• An industrial press is mounted on a rubber pad to isolated it from its foundation . If the rubber pad is compressed 5 mm by the self-weight of the press, find the natural frequency of the system
Example - Harmonic
mst
3105
Hzradm
g
st
n 0497.7sec/2945.44105
81.92/1
3
2/1
• An air-conditioning chiller unit weighing 2,000 lb is to be supported by four air springs. Design the air springs such that the natural frequency of vibration of the unit lies between 5 rad/s and 10 rad/s
Example - Harmonic
Example – solution
sradn /5.7
4.386
2000m
m
keq
n
inlbmk neq /1491.2915.74.386
2000 22
inlbkinlbk /78.72/1491.2914
Example
• An electrical switch gear is supported by a crane through a steel cable of length 4 m and diameter 0.01 m. If the natural time period of the axial vibration of the switch gear is found to be 0.01 s, find the mass of the switch gear
Example – solution
mN /)10(064.4)10(07.201.044
1 6112
l
AEk stiffness cable
nn
nf
211.0
m
kn
20
1.0
2
kgk
mn
53.1029)20(
)10(0644.42
6
2
Example
• A bungee jumper weighing 160 lb ties one end of a elastic rope of length 200 ft and stiffness 10 lb/in to bridge and the other to himself and jumps from the bridge. Assuming the bridge to be rigid, determine the vibratory motion of the jumper about his static equilibrium position
Example – solution
sec/88.361,1
)12(200)4.386(22mgh
jumper theofvelocity
2
21
in
ghvormv
inlbkinch
lbm /10,
sec
4.386
160 2
88.362,1)0(,0)0(
:position mequilibriu about the
00 txxtxx
)sin()(:jumper theof response 00 tAtx n
Example – solution
ink
mxxxx
nn
12.2774.386
160
10
1361A
where,
00
2
2
00
21
0tan
where,
01
0
n
n
x
x
Free Vibration of an Undamped
Torsional System• If a rigid body oscillate about a specific reference
axis, the resulting motion is called torsionalvibration
• The displacement of the body is measured in terms of a angular coordinate
• The restoring moment may be due to the torsion of an elastic member or to the unbalanced moment of a force or couple
l
GIM o
t
32
4dIo
l
Gd
l
GIMk ot
t32
4
• Equation of Motion▫ The equation of the angular
motion of the disc about its axis can be derived by using Newton’s second laws
00 tkJ
0J
ktn
t
nk
J02 02
1
J
kf t
n
• Important aspects of this system
▫ If the cross section of the shaft supporting the disc is not circular, an appropriate torsionalspring constant is to be used
▫ The polar mass moment of inertia of a disc is given by
▫ The general solution
g
WDDhJ
832
24
0
tAtAt nn sincos)( 21
01 A
nA /02
Example
• The figure shows a spacecraft with four solar panel. Each panel has the dimension of 5 ft. x 3 ft. x 1 ft. with a density of 0.1 lb/in3 , and is connected to the body by aluminum rods of length 12 in. and diameter 1 in. Determine the natural frequency of vibration each panel about the axis of the connecting rod
Example – solution
5820.1386.4
0.283)112)(12)(3(5panel a of massm
98.17036112
5820.1
12
m axis- xabout the panel theof inertia momentof mass
22
22
0
baJ
444
0 098175.0132
32
rod of inertia momentofpolar ind I
radinlbl
GIkt /101089.3
12
098175.0108.3 48
0
sec/4841.13
21
0
radJ
ktn
Example
• Find the equation of motion of the uniform rigid bar OA of length l and mass m shown in the figure. Also find its natural frequency
Example – solution
llkaakkJ t 210
:motion ofEquation
2
2
2
03
1
212
1 where ml
lmmlJ
03
1 2
2
2
1
2 lkakkml t
21
2
2
2
2
13
ml
lkakktn
Rayleigh’s Energy Method
• Uses the energy method to find the natural frequencies of a single degree of freedom systems
• The principle of conservation of energy, in the context of an undamped vibrating system, can be restated as
• Subscript 1 denote the time when the mass is passing through its static equilibrium position (U1=0)
2211 UTUT
Rayleigh’s Energy Method
• Subscript 2 indicate the time corresponding to the maximum displacement of the mass (T2=0)
• If the system is undergoing harmonic motion, then T1 and U2 denote the maximum values
21 00 UT
maxmax UT
Example
• Find the natural frequency of the transverse vibration of the water tank considered in the first example by including the mass of the column
Example
32
3
max
2
32
36
xlxl
y
xlEI
Pxxy
The maximum kinetic energy of the beam,
dxxyl
mT
l2
0
max2
1
Example
32
3
max 32
xlxl
yxy
dxxlxl
y
l
mT
l
0
232
2
3
maxmax 3
22
2
max
7
6
2
maxmax
140
33
2
1
35
33
42yml
l
y
l
mT
Example
2
maxmax2
1ymT eq
mmeq140
33
eqeff mMM
meq denotes the equivalent mass of the cantilever at the free end, its maximum kinetic energy
The total effective mass acting at the end, M is the mass of the water tank
mM
k
M
k
eff
n
140
33
Free Vibration with Viscous Damping
• The viscous damping force F is proportional to the velocity,
• c is the damping constant or coefficient of viscous damping
• The negative sign indicates that the damping force is opposite to the direction of velocity
xcF
Free Vibration with Viscous Damping
kxxcxm
0 kxxcxm
FV with Viscous Damping - Solution
0 kxxcxm
02 kcsms
stCetx
m
k
m
c
m
c
m
mkccs
22
2,1222
4
tstseCeCtx 21
21 The general solution
FV with Viscous Damping• Critical Damping Constant and Damping Ratio
▫ The critical damping is defined as the damping constant for which the radical becomes zero
▫ The damping ratio is defined as the ratio of the damping ratio to the critical damping constant
02
2
m
k
m
cc
nc mkmm
kmc 222
ccc /
FV with Viscous Damping• Critical Damping Constant and Damping Ratio
▫ the solution,
nc
c m
c
c
c
m
c
22
ns 12
2,1
tt nn
eCeCtx
1
2
1
1
22
FV with Viscous Damping▫ The nature of the roots and hence the behavior of
the solution depends upon the magnitude of damping;
▫ Case 1: Underdamped
s vibrationundamped toleads,0
nis 2
1 1
overdamped critical, d,underdampe :cases three,0
mkmcorccor c /2/1
nis 2
2 1
expressed becan roots theand negative is1 condition, for this 2
FV with Viscous Damping
titi nn
eCeCtx
22 1
2
1
1
tCtCetx nn
tn 2'
2
2'
1 1sin1cos
Case 1: Underdamped
n
n xxCandxC
2
00'
20
'
1
1
For the initial condition, 00 )0()0( xtxandxtx
t
xxtxetx n
n
nn
tn
2
2
002
0 1sin1
1cos
the solution,
nd 21
The frequency of damped vibration is
FV with Viscous Damping
2'
2
2'
10 CCXX
'
2
'
1
1 /tan CC
The constants are,
'1'
2
1
0 /tan CC
tXetx n
tn 21sin
0
2
0 1cos
teXtx n
tn
the solution can be expressed as ,
FV with Viscous Damping▫ Case 2: Critical damped
nc
m
css
221
mkmcorccor c /2/1
equal, are and roots two thecase In this 21 ss
the solution, tnetCCtx
21)(
For the initial condition,00 )0()0( xtxandxtx
01 xC 002 xxC n
t
nnetxxxtx
000)(
zero odiminish t eventually lmotion wil the,0 as 0
Since . isequation by the drepresentemotion the
te
aperiodic
tn
FV with Viscous Damping▫ Case 3: Overdamped
mkmcorccor c /2/1
distint, and real are and roots two thecase In this 21 ss
the solution,
tt nn
eCeCtx
1
2
1
1
22
)(
012
1 ns 012
2 ns
FV with Viscous Damping▫ Case 3: Overdamped
For the initial condition,00 )0()0( xtxandxtx
12
1
2
0
2
0
1
n
n xxC
12
1
2
0
2
0
2
n
n xxC
mely with tiexponental diminishesmotion the, negative are roots the
Since . isequation by the drepresentemotion the aperiodic
FV with Viscous Damping
FV with Viscous Damping
• The nature of the roots with varying values of damping can be shown in a complex plane. The semicircle represent the locus of the roots for different values of damping ratio in the range of 0 to 1
FV with Viscous Damping
• A critically damped system will have the smallest damping required for aperiodic motion: hence the mass returns to the position of rest in the shortest possible time without overshooting
• The figure represent the phase plane or state space of a damped system
FV with Viscous Damping
• Logarithmic Decrement▫ Represent the rate at which the amplitude of a free
damped vibration decreases. It is defined as the natural logarithm of the ratio of any successive amplitude
n
n
n
eteX
teX
x
x
n
t
n
t
020
010
2
1
cos
cos2
1
m
c
x
x
dn
ndn2
2
1
2
1
2ln
222
1
damping smallfor
22 22
or
FV with Viscous Damping
• Energy Dissipated in Viscous Damping▫ The rate of change of energy with time is given
▫ The negative sign denotes that energy dissipate with time. Assuming a simple harmonic motion
2
2 velocity force
dt
dxccvFv
dt
dW
tdcXdtdt
dxcW ddd
t
n
22
0
2
2/2
0cos
2XcW d
FV with Viscous Damping• Energy Dissipated in Viscous Damping
▫ The fraction of the total energy of the vibrating system that is dissipated in each cycle of motion is called the specific damping capacity
▫ Another quantity used to compare damping capacity of engineering materials is called loss coefficient and is define as the ratio of energy dissipated per radian and the total energy
constant422
22
22
21
2
m
c
Xm
Xc
W
W
dd
d
W
W
W
W
2
2/tcoefficien loss
FV with Viscous Damping
• Torsional Systems with Viscous Damping
tcT
d,underdampe for the case;linear in the as
00 tt kcJ
is,motion ofequation the
nd 21
0J
ktn
00 22 Jk
c
J
c
c
c
t
t
n
t
tc
t
Example- FV viscous damping• The human leg has a measured natural
frequency of around 20 Hz when in its rigid (knee-locked) position in the longitudinal direction ( i.e., along the length of the bone) with damping ratio of ξ = 0.224. Calculate the response of the tip of the leg bone to an initial velocity of v0 = 0.6m/s and zero initial displacement ( this would correspond to the vibration induced while landing on your feet, with your knee locked from a height of 18 mm. What is the maximum acceleration experience by the leg assuming no damping?
Example- FV viscous damping• Highway crash barrier are design to absorb a
vehicle’s kinetic energy without bringing the vehicle to such an abrupt stop that the occupants are injured. Knowledge of the barrier’s materials provide the spring constant k and the damping coefficient c; the mass m is the vehicle mass. For this application , t=0 denotes the time at which the moving vehicle contacts the barrier at x=0; thus v(0) is the speed of the vehicle at the time of contact and x(0) =0. The applied force is zero. Most of the barrier’s resistance is due to the term cv, and it stops resisting after the vehicle comes to rest; so the barrier does reverse the vehicle motion. Cont……
Example- FV viscous damping• A particular barrier’s construction gives
k=18000 N/m and c = 20000 N s/m. A vehicle 1800 kg vehicle strikes the barrier at 22 m/s. Determine how long it takes for the vehicle to come to rest, how far the vehicle compress the barrier, and the maximum deceleration of the vehicle
Example- FV viscous damping• An underdamped shock absorber is to be design
for a motorcycle of mass 200 kg. When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as shown. Find the necessary stiffness and damping constants of the shock absorber if the damped period is to be 2 s and the amplitude x1 is to be reduced to one-fourth in one half cycle (i.e., x1.5 = x1/4 ). Also find the minimum initial velocity that leads to a maximum displacement of 250 mm
Example- Solution• Approach: We use the equation for the logarithm
decrement in term of the damping ratio, equation for the damped period of vibration, time corresponding to maximum displacement for an underdamped system, and envelop passing through the maximum points of an underdamped system
22
1
1
27726.2)16ln(ln
x
x
16/4/,4/ 15.1215.1 xxxxx
4037.0
Example - Solution
21
222
nd
d
sradn /4338.34037.012
2
2
msNmc nc /54.13734338.320022
msNcc c /4981.55454.13734037.0
mNmk n /2652.23584338.320022
Example - Solution
2
1 1sin td
sec3678.0
9149.0sin 1
1
t
tnXex
21
The displacement of the mass attain its maximum value at time t,
9149.04037.01sinsin2
11 ttd
The envelop passing through the maximum points,
Example - Solution
tnXex
21
The envelop passing through the maximum points,
3678.0433.44037.024037.0125.0 Xe
mX 4550.0
The velocity of the mass can be obtained by differentiating,
tXetx d
tn sin
Example - Solution
The velocity of the mass,
ttXetx nddn
tn cossin
2
0 10 nd XXxtx
sm /4294.1
4037.01)4338.3)(4550.0(2
Example
• The maximum permissible recoil distance of a gun is specified as 0.5 m. If the initial recoil velocity is to be 8 m/s and 10 m/s, find the mass of the gun and the spring stiffness of the recoil mechanism. Assume that a critically damped dashpot is used in the recoil mechanism and the mass of the gun has to be at least 500 kg
Example - Solution
1000 Eetxxxtxt
nn
20
2
00 Etxtxxetx nn
tn
Let tm = time at which x=xmax and v=0 occur. Here x0 = 0 and v0 =0 initial recoil velocity. By setting v(t)=0,
3
1
0
0
00
0 Ex
x
xx
xt
nnnn
m
Example - Solution
n
t
m
exetxx mn
1
00max
7178.25.0max0 nn exx
smx /10 when, 0
sec/3575.77178.2*5.0/10 radn
kg, 500 isgun of masswhen
mNmk n /403.066,275003575.722
FV with Coulomb Damping
• In vibrating structure, whenever the components slide relative to each other, dry friction damping appears internally
• Coulomb’s law of dry friction states that, when two bodies are in contact, the force required to produce sliding is proportional to the normal force acting in the plane of contact
mgWNF
FV with Coulomb Damping
• Equation of Motion
▫ Case 1
When x is positive and dx/dt is positive or when x is negative and dx/dt is positive (half cycle)
Nkxxm
FV with Coulomb Damping
• Equation of Motion
▫ Case 1
This is a second-order nonhomogeneous DFQ. The solution is
k
NtAtAtx nn
sincos 21
k
NxA
301 02 A
k
Nxxt
4/2/ position, extreme At the 02nn
0)0(/2)0( 0 txandkNxtx
FV with Coulomb Damping
FV with Coulomb Damping
• Equation of Motion
▫ Case 2
When x is positive and dx/dt is negative or when x is negative and dx/dt is negative (half cycle)
Nkxxm
FV with Coulomb Damping▫ Case 2
The solution is
k
NtAtAtx nn
sincos 43
k
NxA
03 04 A
k
Nxx
2/ tposition, extreme At the 01n
0)0()0( 0 txandxtx
FV with Coulomb Damping
k
Nk
Nx
r
2
cycles ofnumber
0
kNxn /at stopsmotion
FV with Coulomb Damping▫ The previous solutions can be expressed as a
single equation
▫ Where sgn(y) is called the signum function and it is define as
0)sgn( kxxmgxm
10
11
11
yfor
yfor
yfor
FV with Coulomb Damping
• The equation of motion is nonlinear with Coulomb damping, while it is linear with viscous damping
• The natural frequency of the system is unaltered with the addition of Coulomb damping, while it is reduced with addition of viscous damping
• The motion is periodic with Coulomb damping, while it can be nonperiodic in a viscouly damped (overdamped) system
FV with Coulomb Damping
• The system comes to rest some time with Coulomb damping, whereas the motion theoretically continues forever with viscous and hysteresis damping
• The amplitude reduces linearly with Coulomb damping, whereas it reduces exponentially with viscous damping
FV with Coulomb Damping
• In each successive cycle, the amplitude of motion is reduced by the amount 4μN/k, so the amplitude at the end of any two consecutive cycles are related:
The slope of the enveloping straight lines is
kNXX mm /41
k
N
k
N n
n
224
FV with Coulomb Damping
Example
• A metal block, placed on a rough surface, is attached to a spring and is given an initial displacement of 10 cm from its equilibrium position. It is found that the natural time period of motion is 1 s and the amplitude reduces by 5 cm in each cycle. Find (a) the kinetic coefficient between the metal block and the surface and (b) the number of cycles of motion executed by the block before it stops.
FV with Coulomb Damping
• Torsional Systems with Coulomb Damping▫ If a constant frictional torque acts on a torsional
system, the equation governing the angular oscillations can be derived similar to that of the linear,
torquedampingconstant thedenoted T
TkJ t 0
0J
ktn
t
rk
Tr
2
amplitude, the
0
TkJ t 0
k
Tk
T
r2
ceased,motion
0
FV with Hysteretic Damping
• The damping caused by the friction between the internal planes that slip or slide as the material deforms is called hysteretic damping
• This causes a loop to be formed in the stress-strain of force-displacement curve
• The energy loss in one loading and unloading cycle is equal to the area enclosed by the loop
• The energy loss per cycle is independent of the frequency but proportional to the square of the amplitude
FV with Hysteretic Damping
FV with Hysteretic Damping
constant damping hysteretic theish and ct coefficien Damping
hc
,dissipatedenergy The 2hXW
relation,nt displaceme-force The
xihkF
stiffness,complex
ikihk 1
FV with Hysteretic Damping
decrement, logarithm hysteresis The
1lnln1j
j
X
X
12
2
1j
j
X
X
FV with Hysteretic Damping
ratio damping viscousequivalent The
k
heq
2
is,constant damping equivalent The
hkmkmkcc eqceq
22
isfrequency The
m
k
k
heq
22