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Transcript
Chapter 10
Elasticity & Oscillations
Elasticity and Oscillations
• Elastic Deformations
• Hooke’s Law
• Stress and Strain
• Shear Deformations
• Volume Deformations
• Simple Harmonic Motion
• The Pendulum
• Damped Oscillations, Forced Oscillations, and
Resonance
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
2
Elastic Deformation of Solids
A deformation is the change in size or shape of an
object.
An elastic object is one that returns to its original size
and shape after contact forces have been removed. If
the forces acting on the object are too large, the object
can be permanently distorted.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
3
Hooke’s Law
F
F
Apply a force to both ends of a long wire. These forces
will stretch the wire from length L to L+L.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
4
Stress and Strain
Define:
L
strain 
L
The fractional
change in length
F
stress 
A
Force per unit crosssectional area
Stretching ==> Tensile Stress
Squeezing ==> Compressive Stress
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
5
Hooke’s Law
Hooke’s Law (Fx) can be written in terms of stress
and strain (stress  strain).
F
L
Y
A
L
The spring constant k is now
YA
k
L
Y is called Young’s modulus and is a measure of an
object’s stiffness. Hooke’s Law holds for an object
to a point called the proportional limit.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
6
A steel beam is placed vertically in the basement of a building to keep the
floor above from sagging. The load on the beam is 5.8104 N and the length
of the beam is 2.5 m, and the cross-sectional area of the beam is 7.5103 m2.
Find the vertical compression of the beam.
Force of
ceiling
on beam
Force of
floor on
beam
F
L
Y
A
L
 F  L 
L    
 A  Y 
For steel Y = 200109 Pa.
4
2.5 m
 F  L   5.8 10 N 

4


L      

1
.
0

10
m


3
2 
9
2
 A  Y   7.5 10 m  200 10 N/m 
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
7
Example (text problem 10.7): A 0.50 m long guitar string, of cross-sectional
area 1.0106 m2, has a Young’s modulus of 2.0109 Pa.
By how much must you stretch a guitar string to obtain a tension of 20 N?
F
L
Y
A
L
0.5 m
 F  L   20.0 N 

L      
6
2 
9
2 
 A  Y   1.0 10 m  2.0 10 N/m 
 5.0 10 3 m  5.0 mm
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
8
Beyond Hooke’s Law
Elastic Limit
If the stress on an object exceeds the elastic limit, then
the object will not return to its original length.
Breaking Point
An object will fracture if the stress exceeds the
breaking point. The ratio of maximum load to the
original cross-sectional area is called tensile strength.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
9
Ultimate Strength
The ultimate strength of a material is the
maximum stress that it can withstand before
breaking.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
10
An acrobat of mass 55 kg is going to hang by her teeth from a steel wire
and she does not want the wire to stretch beyond its elastic limit. The
elastic limit for the wire is 2.5108 Pa.
What is the minimum diameter the wire should have to support her?
F
stress   elastic limit
A
Want
F
mg
A

elastic limit
elastic limit
2
mg
D
 
elastic limit
2

D
MFMcGraw
4mg
 1.7 10 3 m  1.7 mm
  elastic limit
Chap 10c - Elas & Vibrations - Revised 4-3-10
11
Shear Deformations
A shear deformation
occurs when two forces
are applied on opposite
surfaces of an object.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
12
Stress and Strain
Define:
Shear Force F
Shear Stress 

Surface Area A
displaceme nt of surfaces x
Shear Strain 

separation of surfaces
L
Hooke’s law (stressstrain) for shear deformations is
F
x
S
A
L
MFMcGraw
where S is the shear
modulus
Chap 10c - Elas & Vibrations - Revised 4-3-10
13
Example (text problem 10.25): The upper surface of a cube of
gelatin, 5.0 cm on a side, is displaced by 0.64 cm by a tangential
force. The shear modulus of the gelatin is 940 Pa.
What is the magnitude of the tangential force?
F
F
x
S
A
L
F
From Hooke’s Law:
x
F  SA
L



 0.64 cm 
 940 N/m 0.0025 m 
  0.30 N
 5.0 cm 
MFMcGraw
2
2
Chap 10c - Elas & Vibrations - Revised 4-3-10
14
Volume Deformations
An object completely submerged in a fluid will be
squeezed on all sides.
F
volume stress  pressure 
A
The result is a volume
strain;
MFMcGraw
V
volume strain 
V
Chap 10c - Elas & Vibrations - Revised 4-3-10
15
For a volume deformation, Hooke’s Law is
(stressstrain):
V
P   B
V
where B is called the bulk modulus. The bulk modulus is
a measure of how easy a material is to compress.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
16
An anchor, made of cast iron of bulk modulus 60.0109 Pa and a volume of
0.230 m3, is lowered over the side of a ship to the bottom of the harbor
where the pressure is greater than sea level pressure by 1.75106 Pa.
Find the change in the volume of the anchor.
V
P   B
V
VP
0.230 m 3 1.75 106 Pa
V  

B
60.0 109 Pa
 6.7110 6 m 3

MFMcGraw

Chap 10c - Elas & Vibrations - Revised 4-3-10

17
Deformations Summary Table
Tensile or
compressive
Shear
Volume
Stress
Force per unit
cross-sectional
area
Shear force divided by
the area of the surface
on which it acts
Pressure
Strain
Fractional
change in length
Ratio of the relative
displacement to the
separation of the two
parallel surfaces
Fractional
change in
volume
Constant of
Young’s modulus Shear modulus (S)
proportionality (Y)
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
Bulk Modulus
(B)
18
Simple Harmonic Motion
Simple harmonic motion (SHM)
occurs when the restoring force
(the force directed toward a stable
equilibrium point) is proportional
to the displacement from
equilibrium.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
19
The Spring
The motion of a mass on a spring is an example of
SHM.
Equilibrium
position
y
x
x
The restoring force is F = kx.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
20
Equation of Motion & Energy
Assuming the table is frictionless:
F
Classic form for SHM
Also,
MFMcGraw
x
  kx  max
k
a x t    xt 
m
1 2
1 2
E  t   K  t   U  t   mv  t   kx  t 
2
2
Chap 10c - Elas & Vibrations - Revised 4-3-10
21
Spring Potential Energy
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
22
Simple Harmonic Motion
At the equilibrium point x = 0 so a = 0 too.
When the stretch is a maximum, a will be a maximum too.
The velocity at the end points will be zero, and
it is a maximum at the equilibrium point.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
23
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
24
What About Gravity?
When a mass-spring system is oriented vertically,
it will exhibit SHM with the same period and
frequency as a horizontally placed system.
The effect of gravity is cancelled out.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
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Spring Compensates for Gravity
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
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Representing Simple Harmonic Motion
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
27
A simple harmonic oscillator can be described
mathematically by:
xt   A cos t
x
vt  
  A sin t
t
v
at  
  A 2 cos t
t
Or by:
xt   A sin t
x
vt  
 A cos t
t
v
at  
  A 2 sin t
t
MFMcGraw
where A is the amplitude of
the motion, the maximum
displacement from
equilibrium, A = vmax, and
A2 = amax.
Chap 10c - Elas & Vibrations - Revised 4-3-10
28
Linear Motion - Circular Functions
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
29
Projection of Circular Motion
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
30
The Period and the Angular Frequency
The period of oscillation is
where  is the angular frequency of
the oscillations, k is the spring
constant and m is the mass of the
block.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
T
2

.

k
m
31
The period of oscillation of an object in an ideal mass-spring
system is 0.50 sec and the amplitude is 5.0 cm.
What is the speed at the equilibrium point?
At equilibrium x = 0:
1 2 1 2 1 2
E  K  U  mv  kx  mv
2
2
2
Since E = constant, at equilibrium (x = 0) the
KE must be a maximum. Here v = vmax = A.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
32
Example continued:
The amplitude A is given, but  is not.
2
2


 12.6 rads/sec
T
0.50 s
and v  Aω  5.0 cm 12.6 rads/sec   62.8 cm/sec
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
33
The diaphragm of a speaker has a mass of 50.0 g and responds to a
signal of 2.0 kHz by moving back and forth with an amplitude of
1.8104 m at that frequency.
(a) What is the maximum force acting on the
diaphragm?
2
2




F

F

ma

m
A


mA
2

f

4

mAf

max
max
2
2
The value is Fmax=1400 N.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
34
Example continued:
(b) What is the mechanical energy of the diaphragm?
Since mechanical energy is conserved, E = Kmax = Umax.
U max
K max
1 2
 kA
2
1 2
 mvmax
2
The value of k is unknown so use Kmax.
K max
1 2
1
1
2
2
 mvmax  m A   mA2 2f 
2
2
2
The value is Kmax= 0.13 J.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
35
Example (text problem 10.47): The displacement of an
object in SHM is given by:
yt   8.00 cmsin 1.57 rads/sec  t 
What is the frequency of the oscillations?
Comparing to y(t) = A sint gives A = 8.00 cm
and  = 1.57 rads/sec. The frequency is:
 1.57 rads/sec
f 

 0.250 Hz
2
2
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
36
Example continued:
Other quantities can also be determined:
The period of the motion is
2
2
T

 4.00 sec
 1.57 rads/sec
xmax  A  8.00 cm
vmax  A  8.00 cm 1.57 rads/sec   12.6 cm/sec
amax  A 2  8.00 cm 1.57 rads/sec   19.7 cm/sec 2
2
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
37
The Pendulum
A simple pendulum is constructed by attaching a
mass to a thin rod or a light string. We will also
assume that the amplitude of the oscillations is
small.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
38
The pendulum is best
described using polar
coordinates.
The origin is at the pivot
point. The coordinates are
(r, φ). The r-coordinate
points from the origin
along the rod. The φcoordinate is perpendicualr
to the rod and is positive in
the counterclock wise
direction.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
39
2nd
Apply Newton’s
Law to the pendulum
bob.
 F  mg sin   ma
2
v
 Fr  T  mg cos   m r  0
If we assume that φ <<1 rad, then sin φ  φ and cos φ 1, the angular
frequency of oscillations is then:
 F  mg sin   ma  mL
mg sin   mL
  ( g / L)sin 

  ( g / L)
The period of oscillations is
MFMcGraw
T
Chap 10c - Elas & Vibrations - Revised 4-3-10
g
L
2

 2
L
g
40
Example (text problem 10.60): A clock has a pendulum that
performs one full swing every 1.0 sec. The object at the end of the
string weighs 10.0 N.
What is the length of the pendulum?
L
T  2
g
Solving for L:
MFMcGraw


gT 2 9.8 m/s 2 1.0 s 
L

 0.25 m
2
2
4
4
Chap 10c - Elas & Vibrations - Revised 4-3-10
2
41
The gravitational potential energy of a pendulum is
U = mgy.
Taking y = 0 at the lowest point of the swing, show that y = L(1-cos).

Lcos
L
L
y  L(1  cos  )
y=0
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
42
The Physical Pendulum
A physical pendulum is any rigid object that is free to
oscillate about some fixed axis. The period of
oscillation of a physical pendulum is not necessarily the
same as that of a simple pendulum.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
43
The Physical Pendulum
http://hyperphysics.phy-astr.gsu.edu/HBASE/pendp.html
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
44
Damped Oscillations
When dissipative forces such as friction are not
negligible, the amplitude of oscillations will decrease
with time. The oscillations are damped.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
45
Graphical representations of damped oscillations:
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
46
Damped Oscillations
Overdamped: The system returns to equilibrium without
oscillating. Larger values of the damping the return to
equilibrium slower.
Critically damped : The system returns to equilibrium as
quickly as possible without oscillating. This is often
desired for the damping of systems such as doors.
Underdamped : The system oscillates (with a slightly
different frequency than the undamped case) with the
amplitude gradually decreasing to zero.
Source: Damping @ Wikipedia
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
47
Damped Oscillations
The larger the damping the more difficult it is to assign
a frequency to the oscillation.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
48
Forced Oscillations and Resonance
A force can be applied periodically to a damped oscillator
(a forced oscillation).
When the force is applied at the natural frequency of the
system, the amplitude of the oscillations will be a
maximum. This condition is called resonance.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
49
Tacoma Narrows Bridge
Nov. 7, 1940
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
50
Tacoma Narrows Bridge
Nov. 7, 1940
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
51
Tacoma Narrows Bridge
The first Tacoma Narrows Bridge opened to traffic on July 1, 1940. It
collapsed four months later on November 7, 1940, at 11:00 AM (Pacific
time) due to a physical phenomenon known as aeroelastic flutter caused
by a 67 kilometres per hour (42 mph) wind.
The bridge collapse had lasting effects on science and engineering. In
many undergraduate physics texts the event is presented as an example of
elementary forced resonance with the wind providing an external periodic
frequency that matched the natural structural frequency (even though the
real cause of the bridge's failure was aeroelastic flutter[1]).
Its failure also boosted research in the field of bridge aerodynamics/
aeroelastics which have themselves influenced the designs of all the
world's great long-span bridges built since 1940. - Wikipedia
http://www.youtube.com/watch?v=3mclp9QmCGs
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
52
Chapter 10: Elasticity & Oscillations
•
•
•
•
•
•
•
•
•
•
MFMcGraw
Elastic deformations of solids
Hooke's law for tensile and compressive forces
Beyond Hooke's law
Shear and volume deformations
Simple harmonic motion
The period and frequency for SHM
Graphical analysis of SHM
The pendulum
Damped oscillations
Forced oscillations and resonance
Chap 10c - Elas & Vibrations - Revised 4-3-10
53
Extra
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
54
Aeroelasticity
Aeroelasticity is the science which studies the
interactions among inertial, elastic, and aerodynamic
forces. It was defined by Arthur Collar in 1947 as "the
study of the mutual interaction that takes place within
the triangle of the inertial, elastic, and aerodynamic
forces acting on structural members exposed to an
airstream, and the influence of this study on design."
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
55
Aeroelastic Flutter
Flutter
Flutter is a self-feeding and potentially destructive vibration where aerodynamic forces on an object couple with a
structure's natural mode of vibration to produce rapid periodic motion. Flutter can occur in any object within a strong
fluid flow, under the conditions that a positive feedback occurs between the structure's natural vibration and the
aerodynamic forces. That is, that the vibrational movement of the object increases an aerodynamic load which in turn
drives the object to move further. If the energy during the period of aerodynamic excitation is larger than the natural
damping of the system, the level of vibration will increase, resulting in self-exciting oscillation. The vibration levels can
thus build up and are only limited when the aerodynamic or mechanical damping of the object match the energy input,
this often results in large amplitudes and can lead to rapid failure. Because of this, structures exposed to aerodynamic
forces - including wings, aerofoils, but also chimneys and bridges - are designed carefully within known parameters to
avoid flutter. It is however not always a destructive force; recent progress has been made in small scale (table top) wind
generators for underserved communities in developing countries, designed specifically to take advantage of this effect.
In complex structures where both the aerodynamics and the mechanical properties of the structure are not fully
understood flutter can only be discounted through detailed testing. Even changing the mass distribution of an aircraft or
the stiffness of one component can induce flutter in an apparently unrelated aerodynamic component. At its mildest this
can appear as a "buzz" in the aircraft structure, but at its most violent it can develop uncontrollably with great speed and
cause serious damage to or the destruction of the aircraft.
In some cases, automatic control systems have been demonstrated to help prevent or limit flutter related structural
vibration.
Flutter can also occur on structures other than aircraft. One famous example of flutter phenomena is the collapse of the
original Tacoma Narrows Bridge.
MFMcGraw
Chap 10c - Elas & Vibrations - Revised 4-3-10
56