Download Fundamentals of Vibration - Department of Mechanical Engineering

Dr. Mostafa Ranjbar
Fundamentals of Vibration
1
-
Outline
• Why vibration is important?
• Definition; mass, spring (or stiffness)
dashpot
• Newton’s laws of motion, 2nd order ODE
• Three types of vibration for SDOF sys.
• Alternative way to find eqn of motion:
energy methods
• Examples
2
-
Why to study vibration
•
•
•
•
•
3
Vibrations can lead to excessive deflections
and failure on the machines and structures
To reduce vibration through proper design of
machines and their mountings
To utilize profitably in several consumer and
industrial applications
To improve the efficiency of certain
machining, casting, forging & welding
processes
To stimulate earthquakes for geological
research and conduct studies in design of
nuclear reactors
-
Why to study vibration
• Imbalance in the gas or diesel engines
• Blade and disk vibrations in turbines
• Noise and vibration of the hard-disks in
your computers
• Cooling fan in the power supply
• Vibration testing for electronic
packaging to conform Internatioal
standard for quality control (QC)
• Safety eng.: machine vibration causes
parts loose from the body
4
-
Stiffness
• From strength of materials (Solid Mech) recall:
Force
fk
103 N
x0
g
0
20 mm
x
Displacement
5
x1
x2
x3
-
Free-body diagram and equations of motion
• Newton’s Law:
m&x&(t ) = −kx(t )
y
kk
x(t)
x
fk
m
c
c=0
Friction-free
Friction-free
surface
mg
fc
surface
m&x&(t ) + kx(t ) = 0
N
x(0) = x0 , x& (0) = v0
6
-
2nd Order Ordinary Differential Equation
with Constant Coefficients
Divide by m : &x&(t ) + ω n2 x(t ) = 0
k
ωn =
: natural frequency in rad/s
m
x(t ) = A sin(ω nt + φ )
7
-
Periodic Motion
x(t)
Max velocity
Initial displacement
Amplitude, A
Time, t
Phase = φ
2π
ωn
period
8
-
Periodic Motion
x(t)
x(t) [mm]
A 1.5
1
0.5
0
Time,
t
Time [s]
2
4
-0.5
-1
-A -1.5
2π
2___
ωπn
ωn
9
6
8
10
12
-
Frequency
ωn is in rad/s is the natural frequency
ωn rad/s
ω cycles ωn
fn =
= n
=
Hz
2 π rad/cycle
2π s
2π
2π
s is the period
T=
ωn
We often speak of frequency in Hertz, but we
need rad/s in the arguments of the trigonometric
functions (sin and cos function).
10
-
Amplitude & Phase from the initial
conditions
x0 = Asin(ωn 0 + φ) = Asin φ
v0 = ωn A cos(ωn 0 + φ) = ωn A cos φ
Solving yields
⎛ ω n x0 ⎞
ω x + v , φ = tan ⎜
A=
⎟
ωn
v0 ⎠
⎝
14442444
3 1442443
1
2
n
Amplitude
11
2
0
2
0
−1
Phase
-
Phase Relationship between x, v, a
Displacement
A
x = A cos( ωDisplacement
nt + φ )
O
x (t) = A sin(vnt + f)
t
–A
vnA
Velocity
Velocity
x& = −ω n A sin(x ω
(t) = v tA cos(v
n + φt +)f)
O
•
n
n
t
–vnA
vn2A
Acceleration Acceleration
x (t) = –vn2A sin(vnt + f)
&x& = −ω A cos( ω n t + φ )
••
2
n
12
O
–v2nA
t
-
Example
For m= 300 kg and ωn =10 rad/s
compute the stiffness:
ωn =
k
2
⇒ k = mω n
m
= (300)10 kg/s
2
= 3 × 10 N/m
4
14
2
-
Other forms of the solution:
x(t) = Asin(ωn t + φ)
x(t) = A1 sin ω nt + A2 cos ωn t
x(t) = a1e
jω nt
+ a2e
− jω nt
Phasor: representation of a complex number in terms of a
complex exponential
Ref: 1) Sec 1.10.2, 1.10.3
2) http://mathworld.wolfram.com/Phasor.html
15
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Some useful quantities
A = peak value
1T
x = lim
x(t)dt
=
average
value
∫
T →∞
T0
1T 2
2
x = lim
x
(t)dt
=
mean
square
value
∫
T →∞
T0
2
x rms = x = root mean square value
16
-
Peak Values
max or peak value of :
displacement : xmax = A
velocity : x&max = ωA
acceleration : &x&max = ω A
2
17
-
Example Hardware store spring, bolt: m= 49.2x10
-3
kg,k=857.8 N/m and x0 =10 mm. Compute ωn and max
amplitude of vibration.
ωn =
k
=
m
857.8 N/m
= 132 rad/s
-3
49.2 × 10 kg
ωn
fn =
= 21 Hz
2π
1
1
2π
=
=
T=
ω n fn 21 cyles
x(t)max = A =
18
1
ωn
0.0476 s
sec
ω 2n x02 + v 20 = x 0 = 10 mm
-
Compute the solution and max velocity and
acceleration
v(t)max = ω n A = 1320 mm/s = 1.32 m/s
a(t)max = ω 2n A = 174.24 × 103 mm/s2
= 174.24 m/s 2 ≈ 17.8g!
⎛ ω n x0 ⎞ π
φ = tan
= rad
⎝ 0 ⎠ 2
x(t) = 10sin(132t + π / 2) = 10 cos(132t) mm
−1
19
-
Derivation of the solution
Substitute x(t ) = ae λt into m&x& + kx = 0 ⇒
mλ2 ae λt + kae λt = 0 ⇒
mλ2 + k = 0 ⇒
k
k
λ = ± − =±
j = ±ω n j ⇒
m
m
x(t ) = a1eω n jt and x(t ) = a2 e −ω n jt ⇒
x(t ) = a1eω n jt + a2 e −ω n jt
21
-
Damping Elements
‰Viscous Damping:
Damping force is proportional to the velocity
of the vibrating body in a fluid medium such
as air, water, gas, and oil.
‰Coulomb or Dry Friction Damping:
Damping force is constant in magnitude but
opposite in direction to that of the motion of
the vibrating body between dry surfaces
‰Material or Solid or Hysteretic Damping:
Energy is absorbed or dissipated by material
during deformation due to friction between
internal planes
22
-
Viscous Damping
‰Shear Stress (τ ) developed in the fluid layer at a
distance y from the fixed plate is:
du
(1.26 )
τ =μ
dy
where du/dy = v/h is the velocity gradient.
•Shear or Resisting Force (F) developed at the bottom
surface of the moving plate is:
Av
(1.27 )
F = τA = μ
= cv
h
where A is the surface area of the moving plate.
c=
24
μA
h
is the damping constant
and
c=
μA
h
Viscous Damping
(1.28)
is called the damping constant.
‰If a damper is nonlinear, a linearization process
is used about the operating velocity (v*) and the
equivalent damping constant is:
dF
c=
dv v*
25
(1.29)
-
-
Linear Viscous Damping
• A mathematical
form
• Called a dashpot or
viscous damper
• Somewhat like a
shock absorber
• The constant c has
units: Ns/m or kg/s
f c = cx& (t )
26
-
Spring-mass-damper systems
• From Newton’s law:
y
kk
x(t)
m
c
c
x
fkf
Friction-free
Friction-free f
surface
fc
k
mg
c
surface
N
m&x&(t ) = − f c − f k
= −cx& (t ) − kx(t )
m&x&(t ) + cx& (t ) + kx(t ) = 0
x(0) = x0 , x& (0) = v0
27
-
Derivation of the solution
Substitute x(t ) = ae λt into m&x& + cx& + kx = 0 ⇒
mλ2 ae λt + cλae λt + kae λt = 0 ⇒
mλ2 + cλ + k = 0 ⇒
2
k
k
λ1λ,2==±−ζω
− n =±
±ωn jζ= ±−
ω n1j ⇒
m
m
x(t ) = a1e λ1t and x(t ) = a2 e λ2t ⇒
x(t ) = a1e λ1t + a2 e λ2t
28
-
Solution (dates to 1743 by Euler)
Divide equation of motion by m
&x&(t ) + 2ζω n x& (t ) + ω n2 x(t ) = 0
where ω n = k
m
and
c
ζ =
= damping ratio (dimensionless)
2 km
29
-
Let x(t) = ae λt & subsitute into eq. of motion
λ2 ae λt + 2ζω n λ aeλt + ω 2n ae λt = 0
which is now an algebraic equation in λ :
λ1,2 = −ζω n ± ω n ζ − 1
2
from the roots of a quadratic equation
Here the discriminant ζ 2 − 1, determines
the nature of the roots λ
30
-
Three possibilities:
1) ζ = 1 ⇒ roots are equal & repeated
called critically damped
ζ = 1 ⇒ c = ccr = 2 km = 2mω n
x(t ) = a1e −ω nt + a2te −ω nt
Using the initial conditions :
a1 = x0 , a2 = v0 + ω n x0
31
Sec. 2.6
-
Critical damping continued
• No oscillation occurs
• Useful in door mechanisms, analog gauges
Displacement
Displacement (mm)
x(t) = [x0 + (v0 + ωn x0 )t]e
0.8
0.6
0.4
0.2
0.0
–0.2
0.5 1.0 1.5 2.0 2.5 3.0 3.5
Time (sec)
Time, t
32
− ω nt
-
2) ζ > 1, called overdamping - two distinct real roots :
λ1, 2 = −ζω n ± ω n ζ 2 − 1
x(t ) = e
−ζω n t
where a1 =
a2 =
33
(a1e
−ω n t ζ 2 −1
+ a2 e
ω n t ζ 2 −1
)
− v0 + (−ζ + ζ 2 − 1)ω n x0
2ω n ζ 2 − 1
v0 + (ζ + ζ 2 − 1)ω n x0
2ω n ζ 2 − 1
-
Displacement
34
Displacement (mm)
The overdamped response
0.4
1
0.2
3: x0 =-0.3, v0 = 0
2
0.0
3
–0.2
–0.4
1.1:x0x ==0.3,
0.3, vv0==00
0
2. x0 = 0,
v0 0 = 1
–0.3, vv0 0==10
3.2:x0x0==0.0,
0
1
2
3
4
Time,Time
t
(sec)
5
6
-
3) ζ < 1, called underdamped motion - most common
Two complex roots as conjugate pairs
write roots in complex form as :
λ1,2 = −ζωn ± ω n j 1 − ζ 2
where
35
j = −1
-
Underdamping
x(t) = e
− ζωnt
(a1e
jω nt 1−ζ 2
+ a2e
− j ωnt 1−ζ 2
)
= Ae − ζωnt sin(ωd t + φ)
ωd = ω n 1 − ζ 2 , damped natural frequency
A=
1
(v0 + ζω n x0 )2 + (x0ωd ) 2
ωd
⎛ x0 ω d ⎞
φ = tan ⎜
⎟
⎝ v0 + ζω n x0 ⎠
−1
36
-
Underdamped-oscillation
Displacement (mm)
Displacement
37
1.0
0.0
–1.0
10
15
Time
Time,
(sec)
t
• Gives an oscillating
response with
exponential decay
• Most natural systems
vibrate with and
underdamped response
• See textbook for details
and other
representations
-
Example consider the spring in Ex., if c = 0.11 kg/s,
determine the damping ratio of the spring-bolt system.
m = 49.2 × 10−3 kg, k = 857.8 N/m
ccr = 2 km = 2 49.2 × 10 −3 × 857.8
= 12.993 kg/s
c
0.11 kg/s
ζ=
=
= 0.0085 ⇒
ccr 12.993 kg/s
the motion is underdamped
and the bolt will oscillate
38
-
Example
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 a
damping ratio of ζ = 0.224.
Calculate the response of the tip if the leg bone to v0
(t=0)= 0.6 m/s and x0(t=0)=0
This correspond to the vibration induced while landing on your
feet, with your knees locked from a height of 18 mm) and plot
the response. What is the maximum acceleration
experienced by the leg assuming no damping?
39
-
Solution:
V0=0.6, X0=0, ζ
= 0.224
ωn =
20 cycles 2π rad
= 125.66 rad/s
s cycles
1
ω d = 125.66 1− (.224) = 122.467 rad/s
2
(0.6 + (0.224 )(125.66)(0)) + (0)(122.467)2
2
A=
A=
1
(v0 + ζω n x0 ) 2 + ( x0ω d ) 2
ωd
⎛ x 0ω d ⎞
⎟⎟
φ = tan ⎜⎜
⎝ v0 + ζω n x0 ⎠
−1
40
122.467
= 0.005 m
⎛ (0)(ω d ) ⎞
⎟=0
φ = tan ⎜
⎝ v0 + ζω n (0 )⎠
-1
⇒ x( t ) = 0.005e −28.148t sin(122.467t )
-
Use undamped formula to get max acceleration:
2
⎛v ⎞
A = x02 + ⎜⎜ 0 ⎟⎟ , ω n = 125.66, v0 = 0.6, x0 = 0
⎝ ωn ⎠
0.6
v
m
A= 0 m=
ωn
ωn
⎛ 0.6 ⎞
⎟⎟ = (0.6 ) 125.66 m/s 2 = 75.396 m/s 2
max(&x&) = − ω n2 A = − ω n2 ⎜⎜
⎝ ωn ⎠
(
maximum acceleration =
41
)
75.396 m/s2
g = 7.68g' s
2
9.81 m/s
-
Plot of the response:
Displacement
Displacement (mm)
x(t ) = 0.005e −28.148t sin(122.467t )
5
4
3
2
1
0
-1
-2
-3
-4
0
42
Time, t
Time (s)
-5
0.02
0.04
0.06
0.08
0.1
0.12
0.14
-
Example
Compute the form of the response of an
underdamped system using the Cartesian form
sin( x + y ) = sin x cos y − cos x sin y ⇒
x(t ) = Ae −ζω nt sin(ω d t + φ ) = e −ζω nt ( A1 sin ω d t + A2 cos ω d t )
x(0) = x0 = e 0 ( A1 sin(0) + A2 cos(0)) ⇒ A2 = x0
x& = −ζω n e −ζω nt ( A1 sin ω d t + A2 cos ω d t )
+ ω d e −ζω nt ( A1 cos ω d t − A2 sin ω d t )
v0 = −ζω n ( A1 sin 0 + x0 cos 0) + ω d ( A1 cos 0 − x0 sin 0)
⇒ A1 =
v0 + ζω n x0
ωd
⇒
⎛ v + ζω n x0
⎞
x(t ) = e −ζω nt ⎜⎜ 0
sin ω d t + x0 cos ω d t ⎟⎟
ωd
⎝
⎠
43
Eq. 2.72
-
MODELING AND ENERGY
METHODS
An alternative way to determine the
equation of motion and an alternative way
to calculate the natural frequency
44
-
Modelling
• Newton’s Laws
∑F
∑M
xi
45
= m&x&
0i
= I 0θ&&
-
Energy Methods
∫ Fdx = ∫ m&x&dx ⇒
Potential Energy
6
474
8
1
work done = U1 − U 2 = m x& 2
2
2
1
= T2 − T1
123
Kinetic Energy
⇒ T + U = constant
d
or
(T + U ) = 0
dt
Alternate method of getting the eq. of motion
46
-
Rayleigh’s Method
• T1+ U1= T2+ U2
• Let t1 be the time at which m moves through
its static equilibrium position, then
• U1=0, reference point
• Let t2 be the time at which m undergoes its
max displacement (v=0 so T2=0), U2 is max
(T1 must be max ),
• Thus
Umax=Tmax
47
Ref: Section 2.5
-
Example
The effect of including the mass of the
spring on the value of the frequency.
y
y +dy
m s, k
l
m
x(t)
53
Ex. 2.8
-
⎫
⎪⎪
⎬ assumptions
y
velocity of element dy : vdy = x& (t ),⎪
⎪⎭
l
mass of element dy :
Tspring
ms
dy
l
2
l
1 m ⎡y ⎤
= ∫ s ⎢ x& ⎥ dy (adds up the KE of each element)
2 0 l ⎣l ⎦
=
1 ⎛ ms ⎞ 2
⎜ ⎟ x&
2⎝ 3 ⎠
⎡1 ⎛ m ⎞ 1 ⎤
m ⎞
1 2
1⎛
mx& ⇒ Ttot = ⎢ ⎜ s ⎟ + m⎥ x& 2 ⇒ Tmax = ⎜ m + s ⎟ω n2 An2
2
2⎝
3 ⎠
⎣2 ⎝ 3 ⎠ 2 ⎦
1
= kA2
2
Provides some simple
k
⇒ ωn =
design and modeling
ms
m+
guides
3
Tmass =
U max
54
Ex. 2.8
Effect of the spring mass = add 1/3 of its mass to the
main mass
-
What about gravity?
kΔ
mg − kΔ = 0, from FBD,
and static equilibrium
m
k
+x(t)
0
m
Δ
+x(t)
mg
1
2
U spring = k (Δ + x)
2
U grav = − mgx
1 2
T = mx&
2
55
-
d
Now use (T + U ) = 0
dt
d ⎡1 2
1
2⎤
⇒ ⎢ mx& − mgx + k (Δ + x) ⎥ = 0
dt ⎣ 2
2
⎦
⇒ mx&&x& − mgx& + k (Δ + x) x&
⇒ x& (m&x& + kx) + x& ( kΔ − mg ) = 0
1
424
3
0 from static equ.
⇒ m&x& + kx = 0
56
-
More on springs and stiffness
l
k=
m
x(t)
60
EA
l
• Longitudinal motion
• A is the cross sectional area
(m2)
• E is the elastic modulus
(Pa=N/m2)
• l is the length (m)
• k is the stiffness (N/m)
-
Torsional Stiffness
Jp
GJ p
k=
l
0
J
61
• Jp is the polar moment of
inertia of the rod
• J is the mass moment of
inertia of the disk
• G is the shear modulus, l
is the length
θ(t)
Sec. 2.3
-
Example
compute the frequency of a shaft/mass
system {J = 0.5 kg m2}
∑ M = Jθ&& ⇒ Jθ&&(t ) + kθ (t ) = 0
k
⇒ θ&&(t ) + θ (t ) = 0
J
GJ p
πd 4
k
, Jp =
⇒ ωn =
=
lJ
J
32
For a 2 m steel shaft, diameter of 0.5 cm ⇒
ωn =
(8 ×1010 N/m 2 )[π (0.5 ×10 − 2 m) 4 / 32]
=
lJ
(2 m)(0.5kg ⋅ m 2 )
GJ p
= 2.2 rad/s
62
-
Transverse beam stiffness
• Strength of materials and
experiments yield:
f
m
x
64
3EI
k= 3
l
3EI
ωn =
3
ml
-
Samples of Vibrating Systems
• Deflection of continuum (beams, plates,
bars, etc) such as airplane wings, truck
chassis, disc drives, circuit boards…
• Shaft rotation
• Rolling ships
• See text for more examples.
65
-
Example : effect of fuel on
frequency of an airplane wing
E, I
• Model wing as transverse
beam
• Model fuel as tip mass
• Ignore the mass of the wing
and see how the frequency
of the system changes as
the fuel is used up
m
l
x(t)
66
-
Mass of pod 10 kg empty 1000 kg full
I = 5.2x10-5 m4, E =6.9x109 N/m, l = 2 m
• Hence the
natural
frequency
changes by an
order of
magnitude
while it empties
out fuel.
ω full =
ω empty =
3EI
3(6.9 × 109 )(5.2 × 10−5 )
=
3
ml
1000 ⋅ 23
= 11.6 rad/s = 1.8 Hz
3EI
3(6.9 × 10 9 )(5.2 × 10 −5 )
=
3
ml
10 ⋅2 3
= 115 rad/s = 18.5 Hz
Pod= a streamlined external housing that enclose engines or fuel
67
-
Does gravity effect frequency?
• Static equilibrium:
kx
0
δ
+x
mg
68
∑ F = 0 = −kδ + mg
• Dynamic equation :
∑ F = m&x& = −k ( x + δ ) + mg
m&x& + kx + kδ − mg = 0
m&x& + kx = 0 !
-
Static Deflection
δ , Δ = distance spring is stretched or
compressed under the force of
gravity by attaching a mass m to it.
mg
Δ = δ = δs =
k
Many symbols in use including xs and x0
69
-
Combining Springs
• Equivalent Spring
1
series : kAC = 1
1
+
k1
k2
parallel : kab = k1 + k2
70
-
Use these to design from
available parts
• Discrete springs available in standard
values
• Dynamic requirements require specific
frequencies
• Mass is often fixed or + small amount
• Use spring combinations to adjust wn
• Check static deflection
71
-
Example
Design of a spring mass system
using available springs: series vs parallel
k2
k1
m
k3
k4
72
• Let m = 10 kg
• Compare a series and parallel
combination
• a) k1 =1000 N/m, k2 = 3000 N/m,
k3 = k4 =0
• b) k3 =1000 N/m, k4 = 3000 N/m,
k1 = k2 =0
-
Case a) parallel connection :
k3 = k4 = 0,k eq = k1 + k2 = 1000 + 3000 = 4000 N/m
⇒ ω parallel =
keg
m
=
4000
= 20 rad/s
10
Case b) series connection :
1
3000
=
= 750 N/m
k1 = k2 = 0, keq =
(1 k3 ) + (1 k4 ) 3 + 1
⇒ ω series =
keg
m
=
750
= 8.66 rad/s
10
Same physical components, very different frequency
Allows some design flexibility in using off-the-shelf components
73
-
Free Vibration with Coulomb Damping
‰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. Thus, the
friction force F is given by:
F = μN = μW = μmg
(2.106)
where N is normal force,
μ is the coefficient of sliding or kinetic friction
μ is usu 0.1 for lubricated metal, 0.3 for nonlubricated
metal on metal, 1.0 for rubber on metal
‰ Coulomb damping is sometimes called constant
damping
74
-
Free Vibration with Coulomb Damping
• Equation of Motion:
Consider a single degree of freedom system with
dry friction as shown in Fig.(a) below.
Since friction force varies with the direction of
velocity, we need to consider two cases as
indicated in Fig.(b) and (c).
75
-
Free Vibration with Coulomb Damping
Case 1. When x is positive and dx/dt is positive or
when x is negative and dx/dt is positive (i.e., for
the half cycle during which the mass moves from
left to right) the equation of motion can be
obtained using Newton’s second law (Fig.b):
m&x& = −kx − μN
or
m&x& + kx = − μN
Hence,
μN
x(t ) = A1 cos ωnt + A2 sin ωnt −
k
(2.107)
(2.108)
where ωn = √k/m is the frequency of vibration
A1 & A2 are constants
76
-
Free Vibration with Coulomb Damping
Case 2. When x is positive and dx/dt is negative or
when x is negative and dx/dt is negative (i.e., for
the half cycle during which the mass moves from
right to left) the equation of motion can be derived
from Fig. (c):
− kx + μN = m&x& or m&x& + kx = μN
(2.109)
The solution of the equation is given by:
μN
x(t ) = A3 cos ωnt + A4 sin ωnt +
k
(2.110)
where A3 & A4 are constants
77
-
Free Vibration with Hysteretic Damping
Consider the spring-viscous damper arrangement
shown in the figure below. The force needed to
cause a displacement:
cωX
F = kx + cx&
(2.122)
cω X
For a harmonic motion
of frequency ω and
x(t )
amplitude X,
x(t ) = X sin ωt (2.123)
∴
F (t )
− cωX
F (t ) = kX sin ωt + cXω cos ωt
= kx ± cω X 2 − ( X sin ωt ) 2
85
= kx ± cω X 2 − x 2
(2.124)
2
− x2
-
Free Vibration with Hysteretic Damping
When F versus x is plotted, Eq.(2.124) represents
a closed loop, as shown in Fig(b). The area of the
loop denotes the energy dissipated by the
damper in a cycle of motion and is given by:
2π / ω
(kX sin ωt + cXω cos ωt )(ωX cos ωt )dt
ΔW = ∫ Fdx = ∫
0
= πωcX 2
(2.125)
Hence, the damping
coefficient:
c=
86
Fig.2.36 Hysteresis loop
h
ω
(2.126)
where h is called the hysteresis
damping constant.
-
Free Vibration with Hysteretic Damping
Eqs.(2.125) and (2.126) gives
ΔW = πhX 2
(2.127)
Complex Stiffness.
For general harmonic motion, x = Xeiωt , the force
is given by
F = kXeiωt + cωiXe iωt = (k + iωc) x
(2.128)
Thus, the force-displacement relation:
F = (k + ih) x
(2.129)
where
87
h⎞
⎛
k + ih = k ⎜1 + i ⎟ = k (1 + iβ )
k⎠
⎝
(2.130)
-
2.6.4 Energy dissipated in Viscous Damping:
In a viscously damped system, the rate of change
of energy with time is given by:
dW
⎛ dx ⎞
2
= force × velocity = Fv = −cv = −c⎜ ⎟
dt
⎝ dt ⎠
2
(2.93)
The energy dissipated in a complete cycle is:
2
2π
⎛ dx ⎞
ΔW = ∫
c⎜ ⎟ dt = ∫ cX 2ωd cos 2 ωd t ⋅ d (ωd t )
0
t =0
⎝ dt ⎠
= πcωd X 2
(2.94)
( 2π / ω d )
94
-
Energy dissipation
Consider the system shown in the figure below.
The total force resisting the motion is:
F = − kx − cv = − kx − cx&
(2.95)
If we assume simple harmonic motion:
x(t ) = X sin ωd t
(2.96)
Thus, Eq.(2.95) becomes
F = −kX sin ωd t − cωd X cos ωd t
(2.97)
The energy dissipated in a complete cycle will be
ΔW = ∫
2π / ω d
t =0
=∫
2π / ω d
t =0
+∫
2π / ω d
t =0
95
Fvdt
kX 2ωd sin ωd t ⋅ cos ωd t ⋅ d (ωd t )
cωd X 2 cos 2 ωd t ⋅ d (ωd t ) = πcωd X 2
(2.98)
-
Energy dissipation and Loss Coefficient
Computing the fraction of the total energy of the
vibrating system that is dissipated in each cycle of
motion, Specific Damping Capacity
⎛ 2π ⎞⎛ c ⎞
πcωd X 2
ΔW
⎟⎟⎜
(2.99)
=
= 2⎜⎜
⎟ = 2δ ≈ 4πζ = constant
1
W
⎝ ωd ⎠⎝ 2m ⎠
mω d2 X 2
2
where W is either the max potential energy or the max
kinetic energy.
The loss coefficient, defined as the ratio of the
energy dissipated per radian and the total strain
energy:
(ΔW / 2π ) ΔW
loss coefficient =
96
W
=
2πW
(2.100)