Lectures Part#1 - Department of Mechanical Engineering
Lectures Part#1 - Department of Mechanical Engineering
Lectures Part#1 - Department of Mechanical Engineering
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Dr. Mostafa Ranjbar<br />
Fundamentals <strong>of</strong> Vibration<br />
1
-<br />
Outline<br />
• Why vibration is important?<br />
• Definition; mass, spring (or stiffness)<br />
dashpot<br />
• Newton’s laws <strong>of</strong> motion, 2 nd order ODE<br />
• Three types <strong>of</strong> vibration for SDOF sys.<br />
• Alternative way to find eqn <strong>of</strong> motion:<br />
energy methods<br />
• Examples<br />
2
-<br />
Why to study vibration<br />
• Vibrations can lead to excessive deflections<br />
and failure on the machines and structures<br />
• To reduce vibration through proper design <strong>of</strong><br />
machines and their mountings<br />
• To utilize pr<strong>of</strong>itably in several consumer and<br />
industrial applications<br />
• To improve the efficiency <strong>of</strong> certain<br />
machining, casting, forging & welding<br />
processes<br />
• To stimulate earthquakes for geological<br />
research and conduct studies in design <strong>of</strong><br />
nuclear reactors<br />
3
-<br />
Why to study vibration<br />
• Imbalance in the gas or diesel engines<br />
• Blade and disk vibrations in turbines<br />
• Noise and vibration <strong>of</strong> the hard-disks in<br />
your computers<br />
• Cooling fan in the power supply<br />
4<br />
• Vibration testing for electronic<br />
packaging to conform Internatioal<br />
standard for quality control (QC)<br />
• Safety eng.: machine vibration causes<br />
parts loose from the body
-<br />
Stiffness<br />
• From strength <strong>of</strong> materials (Solid Mech) recall:<br />
Force<br />
f k<br />
10 3 N<br />
x 0<br />
x 1<br />
x 2<br />
x 3<br />
g<br />
0<br />
20 mm<br />
Displacement<br />
x<br />
5
Free-body diagram and equations <strong>of</strong> motion<br />
-<br />
• Newton’s Law:<br />
k<br />
k<br />
x(t)<br />
y<br />
x<br />
mx &&(<br />
t)<br />
= −kx(<br />
t)<br />
c=0<br />
c<br />
m<br />
Friction-free<br />
surface<br />
surface<br />
f k<br />
f c<br />
mg<br />
N<br />
mx &&(<br />
t)<br />
+<br />
kx(<br />
t)<br />
=<br />
0<br />
x(0)<br />
=<br />
x<br />
x&<br />
0<br />
, (0)<br />
=<br />
v<br />
0<br />
6
-<br />
2nd Order Ordinary Differential Equation<br />
with Constant Coefficients<br />
Divide by<br />
ω<br />
n<br />
=<br />
x(<br />
t)<br />
=<br />
2<br />
m : && x(<br />
t)<br />
+ ω x(<br />
t)<br />
=<br />
k<br />
: natural frequency in rad/s<br />
m<br />
Asin(<br />
ω t + φ)<br />
n<br />
n<br />
0<br />
7
-<br />
Periodic Motion<br />
x(t)<br />
Max velocity<br />
Initial displacement<br />
Amplitude, A<br />
Phase = φ<br />
2π<br />
ω n<br />
period<br />
Time, t<br />
8
-<br />
Periodic Motion<br />
x(t)<br />
x(t) [mm]<br />
A 1.5<br />
1<br />
0.5<br />
0<br />
-0.5<br />
-1<br />
2 4 6 8 10 12<br />
Time, t<br />
Time [s]<br />
-A<br />
-1.5<br />
___ 2π<br />
ω 2π n<br />
ω n<br />
9
-<br />
Frequency<br />
ω n<br />
is in rad/s is the natural frequency<br />
ω<br />
f n<br />
= n<br />
rad/s<br />
2π rad/cycle = ω n<br />
cycles<br />
2π s<br />
T = 2π s is the period<br />
ω n<br />
= ω n<br />
2π Hz<br />
We <strong>of</strong>ten speak <strong>of</strong> frequency in Hertz, but we<br />
need rad/s in the arguments <strong>of</strong> the trigonometric<br />
functions (sin and cos function).<br />
10
-<br />
Amplitude & Phase from the initial<br />
conditions<br />
x 0<br />
= Asin(ω n<br />
0 + φ) = Asinφ<br />
v 0<br />
= ω n<br />
Acos(ω n<br />
0 + φ) = ω n<br />
Acosφ<br />
Solving yields<br />
A = 1 ⎛<br />
ω 2<br />
ω<br />
n<br />
x 2 2<br />
ω<br />
0<br />
+ v 0<br />
, φ = tan −1 n<br />
x 0<br />
⎞<br />
⎜ ⎟<br />
144 n<br />
v 24443<br />
⎝ 0 ⎠<br />
1442<br />
443<br />
Amplitude<br />
Phase<br />
11
-<br />
Phase Relationship between x, v, a<br />
x<br />
Displacement<br />
= A cos( ω t + φ )<br />
n<br />
Displacement<br />
x (t) = A sin(v n t + f)<br />
A<br />
O<br />
t<br />
–A<br />
x&<br />
= −ω Asin( ω t + φ )<br />
n<br />
Velocity<br />
Velocity<br />
•<br />
x (t) = v n A cos(v n t + f)<br />
n<br />
v n A<br />
O<br />
t<br />
&& x<br />
Acceleration<br />
Acceleration<br />
••<br />
x (t) = –v 2 n A sin(v n t + f)<br />
2<br />
= −ω A cos( ω t + φ )<br />
n<br />
n<br />
–v n A<br />
v n 2 A<br />
O<br />
–v 2 nA<br />
t<br />
12
-<br />
Example For m= 300 kg and ω n =10 rad/s<br />
compute the stiffness:<br />
ω n<br />
=<br />
k<br />
m ⇒ k = mω 2<br />
n<br />
= (300)10 2 kg/s 2<br />
= 3 ×10 4 N/m<br />
14
-<br />
Other forms <strong>of</strong> the solution:<br />
x(t) = Asin(ω n<br />
t + φ)<br />
x(t) = A 1<br />
sinω n<br />
t + A 2<br />
cosω n<br />
t<br />
x(t) = a 1<br />
e jω n t + a 2<br />
e − jω n t<br />
15<br />
Phasor: representation <strong>of</strong> a complex number in terms <strong>of</strong> a<br />
complex exponential<br />
Ref: 1) Sec 1.10.2, 1.10.3<br />
2) http://mathworld.wolfram.com/Phasor.html
-<br />
Some useful quantities<br />
A = peak value<br />
1<br />
x = lim<br />
T →∞<br />
T<br />
x 2<br />
= lim<br />
T→∞<br />
1<br />
T<br />
T<br />
∫<br />
0<br />
x(t)dt = average value<br />
T<br />
∫ x 2 (t)dt = mean - square value<br />
0<br />
x rms<br />
= x 2 = root mean square value<br />
16
-<br />
Peak Values<br />
max or peak value <strong>of</strong><br />
:<br />
displacement<br />
:<br />
x<br />
max<br />
=<br />
A<br />
velocity:<br />
x&<br />
max<br />
=<br />
ω<br />
A<br />
acceleration :<br />
&& x<br />
max<br />
=<br />
ω<br />
2<br />
A<br />
17
-<br />
Example Hardware store spring, bolt: m= 49.2x10 -3<br />
kg,k=857.8 N/m and x 0 =10 mm. Compute ω n and max<br />
amplitude <strong>of</strong> vibration.<br />
ω n<br />
=<br />
k<br />
m =<br />
857.8 N/m<br />
49.2 × 10 -3 kg<br />
= 132 rad/s<br />
f n<br />
= ω n<br />
= 21 Hz<br />
2π<br />
T = 2π = 1 1<br />
=<br />
ω n<br />
f n 21 cyles sec<br />
0.0476 s<br />
x(t) max<br />
= A = 1 ω n<br />
ω n2<br />
x 02<br />
+ v 02<br />
= x 0<br />
=10 mm<br />
18
Compute the solution and max velocity and<br />
acceleration<br />
-<br />
v(t) max<br />
= ω n<br />
A = 1320 mm/s = 1.32 m/s<br />
a(t) max<br />
= ω n2<br />
A = 174.24 × 10 3 mm/s 2<br />
=174.24 m/s 2 ≈17.8g!<br />
⎛ ω<br />
φ = tan −1 n<br />
x 0 ⎞<br />
⎝ 0 ⎠ = π 2 rad<br />
x(t) =10sin(132t + π /2)=10 cos(132t) mm<br />
19
21<br />
-<br />
Derivation <strong>of</strong> the solution<br />
jt<br />
jt<br />
jt<br />
jt<br />
n<br />
t<br />
t<br />
t<br />
n<br />
n<br />
n<br />
n<br />
e<br />
a<br />
a e<br />
t<br />
x<br />
e<br />
a<br />
t<br />
x<br />
a e<br />
t<br />
x<br />
j<br />
j<br />
m<br />
k<br />
m<br />
k<br />
k<br />
m<br />
kae<br />
ae<br />
m<br />
kx<br />
mx<br />
ae<br />
t<br />
x<br />
ω<br />
ω<br />
ω<br />
ω<br />
λ<br />
λ<br />
λ<br />
ω<br />
λ<br />
λ<br />
λ<br />
−<br />
−<br />
+<br />
=<br />
⇒<br />
=<br />
=<br />
⇒<br />
= ±<br />
=±<br />
−<br />
= ±<br />
⇒<br />
=<br />
+<br />
⇒<br />
=<br />
+<br />
⇒<br />
=<br />
+<br />
=<br />
2<br />
1<br />
2<br />
1<br />
2<br />
2<br />
)<br />
(<br />
)<br />
(<br />
and<br />
)<br />
(<br />
0<br />
0<br />
0<br />
into<br />
)<br />
(<br />
Substitute<br />
&<br />
&
Damping Elements<br />
-<br />
Viscous Damping:<br />
Damping force is proportional to the velocity<br />
<strong>of</strong> the vibrating body in a fluid medium such<br />
as air, water, gas, and oil.<br />
Coulomb or Dry Friction Damping:<br />
Damping force is constant in magnitude but<br />
opposite in direction to that <strong>of</strong> the motion <strong>of</strong><br />
the vibrating body between dry surfaces<br />
Material or Solid or Hysteretic Damping:<br />
Energy is absorbed or dissipated by material<br />
during deformation due to friction between<br />
internal planes<br />
22
Viscous Damping<br />
-<br />
24<br />
Shear Stress (τ ) developed in the fluid layer at a<br />
distance y from the fixed plate is:<br />
du<br />
τ = μ ( 1.26)<br />
dy<br />
where du/dy = v/h is the velocity gradient.<br />
•Shear or Resisting Force (F) developed at the bottom<br />
surface <strong>of</strong> the moving plate is:<br />
Av<br />
F = τA<br />
= μ =<br />
h<br />
( 1.27)<br />
where A is the surface area <strong>of</strong> the moving plate.<br />
c =<br />
μA<br />
h<br />
cv<br />
is the damping constant
and<br />
c<br />
=<br />
μA<br />
h<br />
Viscous Damping<br />
( 1.28)<br />
is called the damping constant.<br />
If a damper is nonlinear, a linearization process<br />
is used about the operating velocity (v*) and the<br />
equivalent damping constant is:<br />
-<br />
c =<br />
dF<br />
dv<br />
v*<br />
( 1.29)<br />
25
-<br />
Linear Viscous Damping<br />
f c<br />
=<br />
cx(t) &<br />
• A mathematical<br />
form<br />
• Called a dashpot or<br />
viscous damper<br />
• Somewhat like a<br />
shock absorber<br />
• The constant c has<br />
units: Ns/m or kg/s<br />
26
-<br />
Spring-mass-damper systems<br />
• From Newton’s law:<br />
k<br />
c<br />
k<br />
c<br />
m<br />
x(t)<br />
y<br />
Friction-free<br />
surface<br />
x<br />
Friction-free<br />
surface<br />
f k<br />
f c<br />
f k<br />
f c<br />
mg<br />
N<br />
m&&<br />
x(<br />
t)<br />
= − f<br />
= −cx&<br />
( t)<br />
− kx(<br />
t)<br />
mx &&(<br />
t)<br />
+ cx&<br />
( t)<br />
+ kx(<br />
t)<br />
= 0<br />
x(0)<br />
=<br />
c<br />
x<br />
−<br />
f<br />
&<br />
k<br />
0<br />
, x(0)<br />
=<br />
v<br />
0<br />
27
28<br />
-<br />
Derivation <strong>of</strong> the solution<br />
t<br />
t<br />
t<br />
t<br />
n<br />
t<br />
t<br />
t<br />
t<br />
e<br />
a<br />
a e<br />
t<br />
x<br />
e<br />
a<br />
t<br />
x<br />
a e<br />
t<br />
x<br />
j<br />
j<br />
m<br />
k<br />
m<br />
k<br />
k<br />
c<br />
m<br />
kae<br />
ae<br />
c<br />
ae<br />
m<br />
kx<br />
cx<br />
mx<br />
ae<br />
t<br />
x<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
2<br />
)<br />
(<br />
)<br />
(<br />
and<br />
)<br />
(<br />
0<br />
0<br />
0<br />
into<br />
)<br />
(<br />
Substitute<br />
λ<br />
λ<br />
λ<br />
λ<br />
λ<br />
λ<br />
λ<br />
λ<br />
ω<br />
λ<br />
λ<br />
λ<br />
λ<br />
λ<br />
+<br />
=<br />
⇒<br />
=<br />
=<br />
⇒<br />
= ±<br />
=±<br />
−<br />
= ±<br />
⇒<br />
=<br />
+<br />
+<br />
⇒<br />
=<br />
+<br />
+<br />
⇒<br />
=<br />
+<br />
+<br />
= &<br />
&&<br />
1<br />
2<br />
,2<br />
1 −<br />
±<br />
−<br />
= ζ<br />
ω<br />
ζω<br />
λ<br />
n<br />
n
-<br />
Solution (dates to 1743 by Euler)<br />
Divide equation <strong>of</strong> motion by m<br />
2<br />
&& x(<br />
t)<br />
+ 2ζω x&<br />
n<br />
( t)<br />
+ ωn<br />
x(<br />
t)<br />
= 0<br />
where ω = k<br />
n<br />
and<br />
m<br />
c<br />
ζ = = damping ratio (dimensionless)<br />
2 km<br />
29
-<br />
Let x(t) = ae λt<br />
& subsitute into eq. <strong>of</strong> motion<br />
λ 2 ae λt + 2ζω n<br />
λae λt + ω n 2 ae λt = 0<br />
which is now an algebraic equation in λ :<br />
λ 1,2<br />
=−ζω n<br />
± ω n<br />
ζ 2 −1<br />
from the roots <strong>of</strong> a quadratic equation<br />
Here the discriminant ζ 2 −1, determines<br />
the nature <strong>of</strong> the roots λ<br />
30
-<br />
Three possibilities:<br />
1) ζ = 1⇒<br />
roots are equal & repeated<br />
called critically damped<br />
ζ = 1⇒<br />
c =<br />
Using the initial<br />
a<br />
1<br />
x(<br />
t)<br />
=<br />
=<br />
x<br />
0<br />
,<br />
a<br />
a e<br />
2<br />
c<br />
1<br />
cr<br />
=<br />
v<br />
= 2 km =<br />
−ω<br />
t<br />
n<br />
0<br />
+<br />
a<br />
2<br />
te<br />
conditions :<br />
+ ω x<br />
n<br />
0<br />
−ω<br />
t<br />
n<br />
2mω<br />
n<br />
31<br />
Sec. 2.6
-<br />
Critical damping continued<br />
• No oscillation occurs<br />
• Useful in door mechanisms, analog gauges<br />
x(t) = [x 0<br />
+ (v 0<br />
+ ω n<br />
x 0<br />
)t]e −ω nt<br />
Displacement<br />
Displacement (mm)<br />
0.8<br />
0.6<br />
0.4<br />
0.2<br />
0.0<br />
–0.2<br />
0.5 1.0 1.5 2.0 2.5 3.0 3.5<br />
Time (sec)<br />
Time, t<br />
32
33<br />
-<br />
1<br />
2<br />
1)<br />
(<br />
1<br />
2<br />
1)<br />
(<br />
where<br />
)<br />
(<br />
)<br />
(<br />
1<br />
roots :<br />
two distinct real<br />
called overdamping -<br />
1,<br />
2)<br />
2<br />
0<br />
2<br />
0<br />
2<br />
2<br />
0<br />
2<br />
0<br />
1<br />
1<br />
2<br />
1<br />
1<br />
2<br />
1,2<br />
2<br />
2<br />
−<br />
−<br />
+<br />
+<br />
=<br />
−<br />
−<br />
+<br />
−<br />
+<br />
−<br />
=<br />
+<br />
=<br />
−<br />
±<br />
= −<br />
><br />
−<br />
−<br />
−<br />
−<br />
ζ<br />
ω<br />
ω<br />
ζ<br />
ζ<br />
ζ<br />
ω<br />
ω<br />
ζ<br />
ζ<br />
ζ<br />
ω<br />
ζω<br />
λ<br />
ζ<br />
ζ<br />
ω<br />
ζ<br />
ω<br />
ζω<br />
n<br />
n<br />
n<br />
n<br />
t<br />
t<br />
t<br />
n<br />
n<br />
x<br />
v<br />
a<br />
x<br />
v<br />
a<br />
e<br />
a<br />
a e<br />
e<br />
t<br />
x<br />
n<br />
n<br />
n
-<br />
The overdamped response<br />
Displacement (mm)<br />
Displacement<br />
0.4<br />
1<br />
0.2<br />
2<br />
0.0<br />
–0.2<br />
–0.4<br />
0 1<br />
1. x 0 = 0.3,<br />
2. x 0 = 0,<br />
3. x 0 = –0.3,<br />
3<br />
v 0 = 0<br />
v 0 = 1<br />
v 0 = 0<br />
1: x 0 =0.3, v 0 = 0<br />
2: x 0 =0.0, v 0 = 1<br />
3: x 0 =-0.3, v 0 = 0<br />
2 3 4 5 6<br />
Time (sec)<br />
Time, t<br />
34
-<br />
3) ζ < 1, called underdamped motion - most common<br />
Two complex roots as conjugate pairs<br />
write roots in complex form as :<br />
λ 1,2<br />
=−ζω n<br />
± ω n<br />
j 1− ζ 2<br />
where j =<br />
−1<br />
35
-<br />
Underdamping<br />
x(t) = e − ζω n t (a 1<br />
e jω n t 1−ζ2 + a 2<br />
e − jω n t 1−ζ2 )<br />
= Ae −ζω nt<br />
sin(ω d<br />
t + φ)<br />
ω d<br />
= ω n<br />
1− ζ 2 , damped natural frequency<br />
A = 1 (v 0<br />
+ ζω n<br />
x 0<br />
) 2 + (x 0<br />
ω d<br />
) 2<br />
ω d<br />
⎛ x<br />
φ = tan −1 0<br />
ω d<br />
⎞<br />
⎜ ⎟<br />
⎝ v 0<br />
+ ζω n<br />
x 0<br />
⎠<br />
36
-<br />
Underdamped-oscillation<br />
Displacement<br />
Displacement (mm)<br />
1.0<br />
0.0<br />
–1.0<br />
10 15<br />
Time<br />
(sec)<br />
Time, t<br />
• Gives an oscillating<br />
response with<br />
exponential decay<br />
• Most natural systems<br />
vibrate with and<br />
underdamped response<br />
• See textbook for details<br />
and other<br />
representations<br />
37
-<br />
Example consider the spring in Ex., if c = 0.11 kg/s,<br />
determine the damping ratio <strong>of</strong> the spring-bolt system.<br />
m = 49.2 × 10 −3 kg, k = 857.8 N/m<br />
c cr<br />
= 2 km = 2 49.2×10 −3 ×857.8<br />
ζ = c<br />
c cr<br />
=<br />
= 12.993 kg/s<br />
0.11 kg/s<br />
12.993 kg/s = 0.0085 ⇒<br />
the motion is underdamped<br />
and the bolt will oscillate<br />
38
Example<br />
The human leg has a measured natural frequency <strong>of</strong> around<br />
20 Hz when in its rigid (knee locked) position, in the<br />
longitudinal direction (i.e., along the length <strong>of</strong> the bone) with a<br />
damping ratio <strong>of</strong> ζ = 0.224.<br />
-<br />
Calculate the response <strong>of</strong> the tip if the leg bone to v 0<br />
(t=0)= 0.6 m/s and x 0 (t=0)=0<br />
This correspond to the vibration induced while landing on your<br />
feet, with your knees locked from a height <strong>of</strong> 18 mm) and plot<br />
the response. What is the maximum acceleration<br />
experienced by the leg assuming no damping?<br />
39
-<br />
Solution:<br />
V 0 =0.6, X 0 =0, ζ = 0.224<br />
ω n<br />
= 20<br />
1<br />
cycles<br />
s<br />
2π rad<br />
cycles<br />
= 125.66 rad/s<br />
A =<br />
1<br />
ω<br />
d<br />
φ = tan<br />
−1<br />
( v<br />
⎛<br />
⎜<br />
⎝ v<br />
ω d<br />
=125.66 1− (.224) 2 = 122.467 rad/s<br />
A =<br />
2<br />
2<br />
0<br />
+ ζω<br />
n<br />
x0<br />
) + ( x0ω<br />
d<br />
)<br />
0<br />
x<br />
0<br />
ω<br />
d<br />
+ ζω x<br />
n<br />
0<br />
⎞<br />
⎟<br />
⎠<br />
( 0.6 + ( 0.224)125.66<br />
( )()<br />
0 ) 2 + ()122.467 0<br />
122.467<br />
( )<br />
⎛ 0 ()ω<br />
φ = tan -1 ⎜<br />
d<br />
⎝ v 0<br />
+ζω n<br />
0<br />
()<br />
( ) 2<br />
⎞<br />
⎟ = 0<br />
⎠<br />
⇒ xt ()= 0.005e −28.148t sin( 122.467t)<br />
= 0.005 m<br />
40
Use undamped formula to get max acceleration:<br />
-<br />
A =<br />
max<br />
x<br />
2<br />
0<br />
2<br />
⎛ v ⎞<br />
0<br />
+<br />
⎜<br />
⎟ , ωn<br />
= 125.66, v0<br />
⎝ωn<br />
⎠<br />
v0<br />
0.6<br />
A = m = m<br />
ω ω<br />
⎛ 0.6 ⎞<br />
⎜ ⎟<br />
⎝ ωn<br />
⎠<br />
= 0.6, x<br />
2<br />
2<br />
2<br />
2<br />
(&&<br />
x) = −ω<br />
A = −ω<br />
⎜ ⎟ = ( 0.6)( 125.66 m/s ) = 75.396 m/s<br />
n<br />
n<br />
n<br />
n<br />
0<br />
= 0<br />
maximum acceleration =<br />
75.396 m/s 2<br />
9.81 m/s 2 g = 7.68g' s<br />
41
-<br />
Plot <strong>of</strong> the response:<br />
Displacement<br />
Displacement (mm)<br />
5<br />
x(<br />
t)<br />
=<br />
0.005e<br />
−28.148t<br />
sin(122.467t)<br />
4<br />
3<br />
2<br />
1<br />
0<br />
-1<br />
-2<br />
-3<br />
-4<br />
-5<br />
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14<br />
Time, t<br />
Time (s)<br />
42
43<br />
-<br />
Example Compute the form <strong>of</strong> the response <strong>of</strong> an<br />
underdamped system using the Cartesian form<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
+<br />
=<br />
⇒<br />
+<br />
=<br />
⇒<br />
−<br />
+<br />
+<br />
= −<br />
−<br />
+<br />
+<br />
= −<br />
=<br />
⇒<br />
+<br />
=<br />
=<br />
+<br />
=<br />
+<br />
=<br />
⇒<br />
−<br />
=<br />
+<br />
−<br />
−<br />
−<br />
−<br />
−<br />
t<br />
x<br />
t<br />
x<br />
v<br />
e<br />
t<br />
x<br />
x<br />
v<br />
A<br />
x<br />
A<br />
x<br />
A<br />
v<br />
t<br />
A<br />
t<br />
A<br />
e<br />
t<br />
A<br />
t<br />
A<br />
e<br />
x<br />
x<br />
A<br />
A<br />
A<br />
e<br />
x<br />
x<br />
t<br />
A<br />
t<br />
A<br />
e<br />
t<br />
Ae<br />
t<br />
x<br />
y<br />
x<br />
y<br />
x<br />
y<br />
x<br />
d<br />
d<br />
d<br />
n<br />
t<br />
d<br />
n<br />
d<br />
n<br />
d<br />
d<br />
t<br />
d<br />
d<br />
d<br />
t<br />
n<br />
d<br />
d<br />
t<br />
d<br />
t<br />
n<br />
n<br />
n<br />
n<br />
n<br />
ω<br />
ω<br />
ω<br />
ζω<br />
ω<br />
ζω<br />
ω<br />
ζω<br />
ω<br />
ω<br />
ω<br />
ω<br />
ω<br />
ζω<br />
ω<br />
ω<br />
φ<br />
ω<br />
ζω<br />
ζω<br />
ζω<br />
ζω<br />
ζω<br />
cos<br />
sin<br />
)<br />
(<br />
sin 0)<br />
cos0<br />
(<br />
cos0)<br />
sin 0<br />
(<br />
)<br />
sin<br />
cos<br />
(<br />
)<br />
cos<br />
sin<br />
(<br />
cos(0))<br />
sin(0)<br />
(<br />
(0)<br />
)<br />
cos<br />
sin<br />
(<br />
)<br />
sin(<br />
)<br />
(<br />
sin<br />
cos<br />
cos<br />
sin<br />
)<br />
sin(<br />
0<br />
0<br />
0<br />
0<br />
0<br />
1<br />
0<br />
1<br />
0<br />
1<br />
0<br />
2<br />
1<br />
2<br />
1<br />
0<br />
2<br />
2<br />
1<br />
0<br />
0<br />
2<br />
1<br />
&<br />
Eq. 2.72
-<br />
MODELING AND ENERGY<br />
METHODS<br />
An alternative way to determine the<br />
equation <strong>of</strong> motion and an alternative way<br />
to calculate the natural frequency<br />
44
-<br />
Modelling<br />
• Newton’s Laws<br />
∑<br />
F<br />
xi<br />
=<br />
mx &&<br />
∑<br />
M = I & θ 0 i 0<br />
45
-<br />
Energy Methods<br />
∫<br />
Fdx<br />
work done =<br />
⇒ T<br />
=<br />
+ U<br />
∫<br />
mxdx &&<br />
=<br />
⇒<br />
64748<br />
U −U<br />
Potential Energy<br />
1<br />
constant<br />
2<br />
=<br />
1<br />
2<br />
mx&<br />
2<br />
2<br />
1<br />
=<br />
T<br />
−T<br />
2 1<br />
123<br />
Kinetic Energy<br />
or<br />
d<br />
dt<br />
( T<br />
+ U )<br />
=<br />
0<br />
Alternate method <strong>of</strong> getting the eq. <strong>of</strong> motion<br />
46
-<br />
Rayleigh’s Method<br />
• T 1 + U 1 = T 2 + U 2<br />
• Let t 1 be the time at which m moves through<br />
its static equilibrium position, then<br />
• U 1 =0, reference point<br />
• Let t 2 be the time at which m undergoes its<br />
max displacement (v=0 so T 2 =0), U 2 is max<br />
(T 1 must be max ),<br />
•Thus U max =T max<br />
Ref: Section 2.5<br />
47
-<br />
Example The effect <strong>of</strong> including the mass <strong>of</strong> the<br />
spring on the value <strong>of</strong> the frequency.<br />
y<br />
y +dy<br />
m s , k<br />
l<br />
m<br />
53<br />
Ex. 2.8<br />
x(t)
-<br />
ms<br />
mass <strong>of</strong> element dy : dy<br />
l<br />
velocity <strong>of</strong> element dy : v<br />
T<br />
T<br />
U<br />
spring<br />
mass<br />
max<br />
=<br />
=<br />
=<br />
=<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
l<br />
∫<br />
0<br />
⎛<br />
⎜<br />
⎝<br />
kA<br />
m<br />
3<br />
mx&<br />
m<br />
l<br />
2<br />
2<br />
s<br />
s<br />
⎞<br />
⎟x&<br />
⎠<br />
2<br />
⇒ T<br />
tot<br />
2<br />
⎡ y ⎤<br />
⎢<br />
x&<br />
⎣ l ⎥<br />
⎦<br />
dy<br />
⎡1<br />
= ⎢<br />
⎣2<br />
dy<br />
(adds up the KE <strong>of</strong><br />
⎛<br />
⎜<br />
⎝<br />
=<br />
m<br />
3<br />
s<br />
⎫<br />
⎪<br />
⎬ assumptions<br />
y<br />
x&<br />
( t),<br />
⎪<br />
l<br />
⎭<br />
⎞<br />
⎟ +<br />
⎠<br />
⇒ ω<br />
1<br />
2<br />
n<br />
⎤<br />
m⎥x&<br />
⎦<br />
=<br />
2<br />
⇒ T<br />
k<br />
m<br />
m +<br />
3<br />
each element)<br />
s<br />
max<br />
=<br />
1<br />
2<br />
⎛<br />
⎜m<br />
+<br />
⎝<br />
m<br />
3<br />
s<br />
⎞ 2<br />
⎟ω<br />
n<br />
A<br />
⎠<br />
2<br />
n<br />
Provides some simple<br />
design and modeling<br />
guides<br />
54<br />
Ex. 2.8<br />
Effect <strong>of</strong> the spring mass = add 1/3 <strong>of</strong> its mass to the<br />
main mass
-<br />
What about gravity?<br />
kΔ<br />
mg − kΔ = 0, from FBD,<br />
and static equilibrium<br />
m<br />
k<br />
m<br />
+x(t)<br />
0<br />
Δ<br />
mg<br />
+x(t)<br />
U<br />
U<br />
spring<br />
grav<br />
1<br />
= k(<br />
Δ +<br />
2<br />
= −mgx<br />
x)<br />
2<br />
T<br />
=<br />
1<br />
2<br />
mx&<br />
2<br />
55
56<br />
-<br />
0<br />
0<br />
)<br />
(<br />
)<br />
(<br />
)<br />
(<br />
0<br />
)<br />
(<br />
2<br />
1<br />
2<br />
1<br />
0<br />
)<br />
(<br />
Now use<br />
0 from static equ.<br />
2<br />
2<br />
=<br />
+<br />
⇒<br />
=<br />
Δ −<br />
+<br />
+<br />
⇒<br />
Δ +<br />
+<br />
−<br />
⇒<br />
=<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎣<br />
⎡<br />
Δ +<br />
+<br />
−<br />
⇒<br />
=<br />
+<br />
kx<br />
mx<br />
mg<br />
k<br />
x<br />
kx<br />
mx<br />
x<br />
x<br />
x<br />
k<br />
mgx<br />
mxx<br />
x<br />
k<br />
mgx<br />
mx<br />
dt<br />
d<br />
U<br />
T<br />
dt<br />
d<br />
&&<br />
4243<br />
1<br />
&<br />
&&<br />
&<br />
&<br />
&<br />
&&&<br />
&
-<br />
More on springs and stiffness<br />
l<br />
x(t)<br />
m<br />
k = EA l<br />
• Longitudinal motion<br />
• A is the cross sectional area<br />
(m 2 )<br />
• E is the elastic modulus<br />
(Pa=N/m 2 )<br />
• l is the length (m)<br />
• k is the stiffness (N/m)<br />
60
-<br />
Torsional Stiffness<br />
k = GJ p<br />
l<br />
0<br />
• J p is the polar moment <strong>of</strong><br />
inertia <strong>of</strong> the rod<br />
• J is the mass moment <strong>of</strong><br />
inertia <strong>of</strong> the disk<br />
• G is the shear modulus, l<br />
is the length<br />
J p<br />
J<br />
θ(t)<br />
61<br />
Sec. 2.3
-<br />
Example compute the frequency <strong>of</strong> a shaft/mass<br />
system {J = 0.5 kg m 2 }<br />
∑<br />
⇒ && k<br />
θ ( t)<br />
+ θ ( t)<br />
= 0<br />
J<br />
4<br />
k GJ<br />
p πd<br />
⇒ ωn<br />
= = , J<br />
p<br />
=<br />
J lJ<br />
32<br />
For a 2 m steel shaft, diameter <strong>of</strong> 0.5 cm ⇒<br />
ω<br />
n<br />
M<br />
=<br />
= J && θ ⇒ J && θ ( t)<br />
+ kθ<br />
( t)<br />
= 0<br />
GJ<br />
lJ<br />
p<br />
=<br />
=<br />
(8×<br />
10<br />
2.2 rad/s<br />
10<br />
2<br />
−2<br />
N/m )[ π (0.5×<br />
10 m)<br />
2<br />
(2 m)(0.5kg⋅m<br />
)<br />
4<br />
/ 32]<br />
62
-<br />
Transverse beam stiffness<br />
f<br />
x<br />
m<br />
• Strength <strong>of</strong> materials and<br />
experiments yield:<br />
k = 3EI<br />
l 3<br />
ω n<br />
=<br />
3EI<br />
ml 3<br />
64
-<br />
Samples <strong>of</strong> Vibrating Systems<br />
• Deflection <strong>of</strong> continuum (beams, plates,<br />
bars, etc) such as airplane wings, truck<br />
chassis, disc drives, circuit boards…<br />
• Shaft rotation<br />
• Rolling ships<br />
• See text for more examples.<br />
65
-<br />
Example : effect <strong>of</strong> fuel on<br />
frequency <strong>of</strong> an airplane wing<br />
E, I m<br />
l<br />
• Model wing as transverse<br />
beam<br />
• Model fuel as tip mass<br />
• Ignore the mass <strong>of</strong> the wing<br />
and see how the frequency<br />
<strong>of</strong> the system changes as<br />
the fuel is used up<br />
x(t)<br />
66
-<br />
Mass <strong>of</strong> pod 10 kg empty 1000 kg full<br />
I = 5.2x10 -5 m 4 , E =6.9x10 9 N/m, l = 2 m<br />
• Hence the<br />
natural<br />
frequency<br />
changes by an<br />
order <strong>of</strong><br />
magnitude<br />
while it empties<br />
out fuel.<br />
ω full<br />
=<br />
ω empty<br />
=<br />
3EI<br />
ml = 3(6.9 ×109 )(5.2 ×10 −5 )<br />
3 1000 ⋅ 2 3<br />
= 11.6 rad/s = 1.8 Hz<br />
3EI<br />
ml = 3(6.9 ×10 9 )(5.2 ×10 −5 )<br />
3 10 ⋅2 3<br />
= 115 rad/s =18.5 Hz<br />
Pod= a streamlined external housing that enclose engines or fuel<br />
67
-<br />
Does gravity effect frequency?<br />
• Static equilibrium:<br />
kx<br />
mg<br />
0<br />
δ<br />
+x<br />
∑<br />
∑<br />
F = 0 = −kδ<br />
+ mg<br />
• Dynamic equation :<br />
F = mx &&<br />
mx && + kx = 0 !<br />
= −k(<br />
x<br />
+ δ ) +<br />
mx && + kx + kδ<br />
− mg = 0<br />
mg<br />
68
-<br />
Static Deflection<br />
δ , Δ = distance spring is stretched or<br />
compressed under the force <strong>of</strong><br />
gravity by attaching a mass m to it.<br />
Δ = = = mg<br />
δ δ<br />
s<br />
k<br />
Many symbols in use including x s and x 0<br />
69
-<br />
Combining Springs<br />
• Equivalent Spring<br />
series : k AC<br />
=<br />
1<br />
1<br />
k1<br />
+ 1 k 2<br />
parallel : k ab<br />
= k 1<br />
+ k 2<br />
70
-<br />
Use these to design from<br />
available parts<br />
• Discrete springs available in standard<br />
values<br />
• Dynamic requirements require specific<br />
frequencies<br />
• Mass is <strong>of</strong>ten fixed or + small amount<br />
• Use spring combinations to adjust w n<br />
• Check static deflection<br />
71
-<br />
Example Design <strong>of</strong> a spring mass system<br />
using available springs: series vs parallel<br />
k 1<br />
k 2<br />
k 4<br />
m<br />
k 3<br />
• Let m = 10 kg<br />
• Compare a series and parallel<br />
combination<br />
• a) k 1 =1000 N/m, k 2 = 3000 N/m,<br />
k 3 = k 4 =0<br />
• b) k 3 =1000 N/m, k 4 = 3000 N/m,<br />
k 1 = k 2 =0<br />
72
-<br />
Case a) parallel connection :<br />
k 3<br />
= k 4<br />
= 0,k eq<br />
= k 1<br />
+ k 2<br />
=1000 + 3000 = 4000 N/m<br />
⇒ ω parallel<br />
=<br />
k eg<br />
m = 4000 = 20 rad/s<br />
10<br />
Case b) series connection :<br />
1<br />
k 1<br />
= k 2<br />
= 0, k eq<br />
=<br />
(1 k 3<br />
) + (1 k 4<br />
) = 3000 = 750 N/m<br />
3 +1<br />
⇒ ω series<br />
=<br />
k eg<br />
m = 750<br />
10<br />
= 8.66 rad/s<br />
Same physical components, very different frequency<br />
Allows some design flexibility in using <strong>of</strong>f-the-shelf components<br />
73
-<br />
Free Vibration with Coulomb Damping<br />
Coulomb’s law <strong>of</strong> dry friction states that, when<br />
two bodies are in contact, the force required to<br />
produce sliding is proportional to the normal<br />
force acting in the plane <strong>of</strong> contact. Thus, the<br />
friction force F is given by:<br />
F<br />
= μ N = μW<br />
= μmg<br />
(2.106)<br />
where N is normal force,<br />
μ is the coefficient <strong>of</strong> sliding or kinetic friction<br />
μ is usu 0.1 for lubricated metal, 0.3 for nonlubricated<br />
metal on metal, 1.0 for rubber on metal<br />
Coulomb damping is sometimes called constant<br />
damping<br />
74
-<br />
Free Vibration with Coulomb Damping<br />
• Equation <strong>of</strong> Motion:<br />
Consider a single degree <strong>of</strong> freedom system with<br />
dry friction as shown in Fig.(a) below.<br />
Since friction force varies with the direction <strong>of</strong><br />
velocity, we need to consider two cases as<br />
indicated in Fig.(b) and (c).<br />
75
-<br />
Free Vibration with Coulomb Damping<br />
Case 1. When x is positive and dx/dt is positive or<br />
when x is negative and dx/dt is positive (i.e., for<br />
the half cycle during which the mass moves from<br />
left to right) the equation <strong>of</strong> motion can be<br />
obtained using Newton’s second law (Fig.b):<br />
m&&<br />
x = −kx<br />
− μN<br />
or mx &<br />
+ kx<br />
Hence,<br />
x( t)<br />
= A1<br />
cosωnt<br />
+ A2<br />
=<br />
sinω<br />
t<br />
−μN<br />
n<br />
−<br />
μN<br />
k<br />
where ω n = √k/m is the frequency <strong>of</strong> vibration<br />
A 1 & A 2 are constants<br />
(2.107)<br />
(2.108)<br />
76
-<br />
Free Vibration with Coulomb Damping<br />
Case 2. When x is positive and dx/dt is negative or<br />
when x is negative and dx/dt is negative (i.e., for<br />
the half cycle during which the mass moves from<br />
right to left) the equation <strong>of</strong> motion can be derived<br />
from Fig. (c):<br />
− kx + μ N = mx && or mx &<br />
+ kx = μN<br />
(2.109)<br />
The solution <strong>of</strong> the equation is given by:<br />
μN<br />
x( t)<br />
= A3<br />
cosω nt<br />
+ A4<br />
sinωnt<br />
+<br />
k<br />
(2.110)<br />
where A 3 & A 4 are constants<br />
77
-<br />
Free Vibration with Hysteretic Damping<br />
Consider the spring-viscous damper arrangement<br />
shown in the figure below. The force needed to<br />
cause a displacement:<br />
F = kx + cx&<br />
(2.122)<br />
cωX<br />
cω<br />
For a harmonic motion<br />
<strong>of</strong> frequency ω and<br />
amplitude X,<br />
x( t)<br />
= X sinωt<br />
(2.123)<br />
x(t)<br />
F(t)<br />
− cωX<br />
X<br />
2<br />
− x<br />
2<br />
∴<br />
F(<br />
t)<br />
=<br />
kX<br />
sinωt<br />
+<br />
cXω<br />
cosωt<br />
=<br />
kx<br />
±<br />
cω<br />
X<br />
2<br />
− ( X<br />
sinωt)<br />
2<br />
85<br />
=<br />
kx<br />
±<br />
cω<br />
X<br />
2<br />
−<br />
x<br />
2<br />
(2.124)
-<br />
86<br />
Free Vibration with Hysteretic Damping<br />
When F versus x is plotted, Eq.(2.124) represents<br />
a closed loop, as shown in Fig(b). The area <strong>of</strong> the<br />
loop denotes the energy dissipated by the<br />
damper in a cycle <strong>of</strong> motion and is given by:<br />
ΔW<br />
=<br />
∫<br />
Fdx<br />
=<br />
∫<br />
2π<br />
/ ω<br />
0<br />
= πωcX<br />
Fig.2.36 Hysteresis loop<br />
( kX sinωt<br />
+ cXω<br />
cosωt)( ωX<br />
cosωt)<br />
2<br />
Hence, the damping<br />
coefficient:<br />
c =<br />
h<br />
ω<br />
dt<br />
(2.125)<br />
(2.126)<br />
where h is called the hysteresis<br />
damping constant.
-<br />
Free Vibration with Hysteretic Damping<br />
Eqs.(2.125) and (2.126) gives<br />
2<br />
ΔW = πhX<br />
(2.127)<br />
Complex Stiffness.<br />
iωt<br />
For general harmonic motion, x = Xe , the force<br />
is given by<br />
F<br />
=<br />
kXe<br />
iωt<br />
+ cωiXe<br />
iωt<br />
= ( k + iωc)<br />
x<br />
(2.128)<br />
87<br />
Thus, the force-displacement relation:<br />
where<br />
k<br />
F = ( k + ih)<br />
x<br />
⎛ h ⎞<br />
+ ih = k⎜1<br />
+ i ⎟ = k(1<br />
+ iβ )<br />
⎝ k ⎠<br />
(2.129)<br />
(2.130)
-<br />
2.6.4 Energy dissipated in Viscous Damping:<br />
In a viscously damped system, the rate <strong>of</strong> change<br />
<strong>of</strong> energy with time is given by:<br />
dW<br />
dt<br />
= force×<br />
velocity<br />
=<br />
Fv = −cv<br />
The energy dissipated in a complete cycle is:<br />
ΔW<br />
(2π<br />
/ ωd<br />
) ⎛<br />
= ∫ c<br />
t =<br />
⎜<br />
0<br />
⎝<br />
2<br />
= πcω<br />
X<br />
d<br />
dx<br />
dt<br />
⎞<br />
⎟<br />
⎠<br />
2<br />
dt<br />
=<br />
∫<br />
2π<br />
0<br />
cX<br />
2<br />
2<br />
ω<br />
⎛<br />
= −c⎜<br />
⎝<br />
d<br />
cos<br />
2<br />
dx<br />
dt<br />
⎞<br />
⎟<br />
⎠<br />
2<br />
ω t ⋅d(<br />
ω t)<br />
d<br />
d<br />
(2.93)<br />
(2.94)<br />
94
-<br />
95<br />
Energy dissipation<br />
Consider the system shown in the figure below.<br />
The total force resisting the motion is:<br />
F = −kx<br />
− cv = −kx<br />
− cx&<br />
(2.95)<br />
If we assume simple harmonic motion:<br />
x( t)<br />
= X sinω<br />
t (2.96)<br />
Thus, Eq.(2.95) becomes<br />
F<br />
= −kX<br />
sinω<br />
t<br />
d<br />
d<br />
− cω<br />
X<br />
cosω<br />
t<br />
(2.97)<br />
The energy dissipated in a complete cycle will be<br />
ΔW<br />
=<br />
=<br />
+<br />
∫<br />
∫<br />
2π<br />
/ ω<br />
t=<br />
0<br />
2π<br />
/ ω<br />
t=<br />
0<br />
∫<br />
2π<br />
/ ω<br />
t=<br />
0<br />
d<br />
d<br />
d<br />
Fvdt<br />
kX<br />
cos<br />
d<br />
ω sinω<br />
t ⋅cosω<br />
t ⋅d(<br />
ω t)<br />
2<br />
cω<br />
X<br />
d<br />
d<br />
2<br />
2<br />
d<br />
ω t ⋅d(<br />
ω t)<br />
d<br />
d<br />
d<br />
d<br />
d<br />
= πcω<br />
X<br />
d<br />
2<br />
(2.98)
-<br />
Energy dissipation and Loss Coefficient<br />
Computing the fraction <strong>of</strong> the total energy <strong>of</strong> the<br />
vibrating system that is dissipated in each cycle <strong>of</strong><br />
motion, Specific Damping Capacity<br />
2<br />
ΔW<br />
πcω<br />
⎛ 2 ⎞<br />
d<br />
X π ⎛ c ⎞<br />
=<br />
1 = 2<br />
⎜<br />
⎟⎜<br />
⎟ = 2δ<br />
≈ 4πζ<br />
= constant<br />
W<br />
2<br />
mω<br />
X<br />
2 ⎝ ωd<br />
⎠⎝<br />
2m<br />
⎠<br />
d<br />
2<br />
where W is either the max potential energy or the max<br />
kinetic energy.<br />
The loss coefficient, defined as the ratio <strong>of</strong> the<br />
energy dissipated per radian and the total strain<br />
energy:<br />
( ΔW<br />
/ 2π<br />
) ΔW<br />
loss coefficient = =<br />
π<br />
W<br />
2<br />
W<br />
(2.99)<br />
(2.100)<br />
96