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8/11/2019 ChewMA1506-14 Ch2
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MA1506
Mathematics IIChapter 2
More Applications of ODEs
1Chew T S MA1506-14 Chapter 2
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In Chapter two,
we are interested in propertiesof solutions
We have learnt how to solve those ODEsin Chapter one, so we are not interested in
how to solve those ODE here
You dont have to memorizethose complicated solutions
But you need to know how to
derive the properties from the solutions
2Chew T S MA1506-14 Chapter 2
My presentation is different from the L. N.. However
the content remains unchanged.
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In this chapter , first we study animportance system called Harmonic
Oscillatorwhich is an application of 2nd
order linear ODE
The ODE for harmonic oscillator is given
by
'' ' ( )mx bx kx F t + + =
3Chew T S MA1506-14 Chapter 2
Introduction
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Notation
In this chapter
may be denoted by or
may be denoted by or
dx
dt'x
2
2
d x
dt x
''x
4
x
Chew T S MA1506-14 Chapter 2
Introduction
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We shall consider four types of HO(1)Simpleharmonic oscillator (pp 1-11)
where m (mass) and k (spring constant)
are positivenumbers
Its motion is periodic, called simple
harmonic motion (SHM)
5Chew T S MA1506-14 Chapter 2
0mx kx+ =
Introduction
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(2)Dampedharmonic oscillator (pp 11-16)
In real oscillator, friction (damping) slows
down the motion of the system. The
frictional force is given by
m >0 b >0 k > 0
springconstant
Dampingconstant
6
bx Chew T S MA1506-14 Chapter 2
0mx bx kx+ + =
Introduction
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(3) Forced harmonic oscillator without
damping (pp17-23)
The system is a simple harmonic oscillator
driven by EXTERNALLY applied force F(t)
( )mx kx F t + =
7Chew T S MA1506-14 Chapter 2
Introduction
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m, k >0, k called spring constant
9
The equation of simple harmonic motion
is
Chew T S MA1506-14 Chapter 2
0mx kx+ =
2.1 The simple harmonic oscillator (motion)
I have changed the subtitle The harmonic oscillator
to The simple harmonic oscillator (motion),
since it is more precise.
I shall introduce the general theory before talking aboutPendulum.
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Simple harmonic motion SHM can serve
as a mathematical model of a variety of
motions, such as a pendulum with small
amplitudes and a mass on a spring
See
http://en.wikipedia.org/wiki/Simple_harmonic_motion
10Chew T S MA1506-14 Chapter 2
2.1 Simple harmonic oscillator(cont)
http://en.wikipedia.org/wiki/Simple_harmonic_motionhttp://en.wikipedia.org/wiki/Simple_harmonic_motionhttp://en.wikipedia.org/wiki/Simple_harmonic_motionhttp://en.wikipedia.org/wiki/Simple_harmonic_motion8/11/2019 ChewMA1506-14 Ch2
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For SHM equationIt is typical to define the quantity
and write this equation
as
11Chew T S MA1506-14 Chapter 2
0mx kx+ =
/k m=0mx kx+ =
2 0x x+ =
(cont) 2.1 Simple harmonic oscillator
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The general soln of the equation is
which can be written as
the phase-amplitude form(see Appendix 1)
( ) cos( )x t A t =
amplitude phase orphase angle
12
( ) cos( ) sin( )x t C t D t = +
Chew T S MA1506-14 Chapter 2
See Chapter 1
(cont) 2.1 Simple harmonic oscillator
cos( )
cos( ) cos( ) sin( ) sin( )
A t
A t A t
= +
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Note that
Hence SHM is periodic with period
and amplitude A
cos( )x A t = cos( 2 )A t = + 2
cos( ( ) )A t
= +
2 2 mk
=
13Chew T S MA1506-14 Chapter 2
(cont) 2.1 Simple harmonic oscillator
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Some Technical Terms
Period T=
The time for a single oscillation (cycle)
Chew T S MA1506-14 Chapter 2 15
2 2 mk
=
Frequency f = the reciprocal of the
period T =
The number of cycles per unit time
1
2
k
m
2.1 Simple harmonic oscillator
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(cont) Some technical terms
Angular frequency=
The number of cycles per unit time
Amplitudeis the maximal
displacement from the equilibriumposition
Chew T S MA1506-14 Chapter 2 16
2 kfm
=
2
=
2.1 Simple harmonic oscillator
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Why ?
17
cos( )A t
sin( )A t cos( )A t +
In fact it can also be written
as one of the following forms
Chew T S MA1506-14 Chapter 2
2.1 Simple harmonic oscillator
sin( )A t +
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Two constants A andare determined by initial conditions
Suppose
Thenand
18
0(0)x x= 0'(0)x v=
0 cos( )x A = 0 ( )sin( )v A =
So we can find A and , for example,
22 00
vA x
= +
Chew T S MA1506-14 Chapter 2
(cont) 2.1 Simple harmonic oscillator
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An equilibrium soln (point) is said to be
stable if any soln with an initial point close
tothe equilibrium soln stays close to the
equilibrium soln (point)
20Chew T S MA1506-14 Chapter 2
(cont) 2.1 Simple harmonic oscillator
A solution x(t) of ODE is said to be an
equilibrium solution (equilibrium point)if x(t) is a constant function(never move)
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Chew T S MA1506-14 Chapter 2 21
(cont) 2.1 Simple harmonic oscillator
2
0x x+ =Hence zerosolution of SMH
is an equilibrium solution (equilibrium point)
This zero solution is stable
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Chew T S MA1506-14 Chapter 2 22
2x x=
We remark that in the above discussion of SHM
Minus is crucial
What happens if2
x x= (NOT SHM)
(cont) 2.1 Simple harmonic oscillator
Zero function is a solution, but it is NOT stable
Why
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Chew T S MA1506-14 Chapter 2 23
t tx Ce De
= +
(cont)
The general soln of
is
Instead of 0 = 0, 0 = 0
(which give zero solution), we assume
initial condition
(0) , (0) 0x x= =
2.1 Simple harmonic oscillator
2x x
=
where is small. Hence this new initial
condition is close to zero solution
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Chew T S MA1506-14 Chapter 2 24
(cont)
1 ( )2
t tx e e = +
2.1 Simple harmonic oscillator
The solution of 2x x=
is
(0) , (0) 0x x= =
Exponential function grows very quickly,so this solution does not stay close to zero soln
Hence zero solution is NOT stable
canbe written as=cosh()
2
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2.1.2 Pendulum with small amplitude
(small angle) (pp1-11)
rigid
25Chew T S MA1506-14 Chapter 2
0> 00, k > 0, b >0
40Chew T S MA1506-14 Chapter 2
2.2 Damped, Unforced Oscillators pp 11-16
0mx bx kx+ + =
Damping constant spring constant
kx restoring forcedamping force , e.g., friction,
air resistancebx
Again I have changed the subtitle, forced
replaced by unforced
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2.2.1 Damped, Unforced Oscillators
(2 real roots)always negative real roots
Overdamping
No oscillation42Chew T S MA1506-14 Chapter 2
General
soln
3 2 0x x x+ + = 2
1 2( )
t t
x t c e c e
= +
Goes to zero rapidlyOverdamping
2 4
2
b b mk
m
=
2
, , all positive
so bigger than
4
b m k
b
b mk
2 4 0b b mk + 0, k > 0, b >0
http://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.html
Damped
Mass Spring
Oscillator
Textbook
p196
53Chew T S MA1506-14 Chapter 2
0mx bx kx+ + =
http://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.html8/11/2019 ChewMA1506-14 Ch2
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2.2.4 Damped pendulum Example 2
with air
resistance
54Chew T S MA1506-14 Chapter 2
0g
m mL + = 0gm S mL
+ + =
S
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2.3 Forced Oscillators ( pp17-27)
General solution of
is (See appendix 3)
55Chew T S MA1506-14 Chapter 2
where
m
0 cosmx kx F t + =
0
2 2
/( ) cos( ) cos( )
F mx t A t t
= +
km
=
0cosF t
( ) ( ) ( )h px t x t x t= +
springexternal motor
Assume
Forced without damping Oscillators ( pp17-23)
2 3 Forced Oscillators
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2.3 Forced Oscillators
get
56Chew T S MA1506-14 Chapter 2
Assume initial condition
(See appendix 4)
02 2/( ) cos( ) cos( )F mx t A t t = +
(0) 0, (0) 0x x= =
[ ]0
2 2
/
( ) cos( ) cos( )
F m
x t t t =
Forced without damping Oscillators ( pp17-23)
2 3 Forced Oscillators
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Small angular frequency
when close to
57Chew T S MA1506-14 Chapter 2
[ ]02 2/( ) cos( ) cos( )F mx t t t = 0
2 2
2 /
( ) sin sin2 2
F m
x t t t
+
=
( )A t
We shall use the above form to discuss
properties of forced oscillator
2.3 Forced Oscillators
Forced without damping Oscillators ( pp17-23)
2 3 Forced OscillatorsF d ith t d i O ill t ( 17 23)
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where
58Chew T S MA1506-14 Chapter 2
Red curve
Blue curve
Red curve
( ) ( ) sin
2
x t A t t +
=
0
2 2
2 /( ) sin
2
F mA t t
=
2.3 Forced OscillatorsForced without damping Oscillators ( pp17-23)
Beating 2 3 F d O ill t
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Beating
if our ear is exposed to two sounds
Chew T S MA1506-14 Chapter 2 59
we only hear the above term A(t)
1st sound 2nd sound
2.3 Forced Oscillators
( )sin[ ]2
A t t + =
where
[ ]02 2
/( ) cos( ) cos( )
F mx t t t
=
0
2 2
2 /( ) sin
2
F mA t t
=
( t) B ti 2 3 F d O ill t
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(cont) Beating
Chew T S MA1506-14 Chapter 2 60
A fast signal
is modulated by a slower one
This behavior is called beatinginphysics
sin[( ) ]2
t +
http://www.school-for-champions.com/science/sound_beat.htm
2.3 Forced Oscillators
0
2 2
2 /( ) sin
2
F mA t t
=
2 3 Forced Oscillators
http://www.school-for-champions.com/science/sound_beat.htmhttp://www.school-for-champions.com/science/sound_beat.htm8/11/2019 ChewMA1506-14 Ch2
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61Chew T S MA1506-14 Chapter 2
L Hospital
Rule
0
sin2 / 2
lim ( ) lim
tF m
A t
=
+
0
2F t
m=
or
sin2 2
t t
2.3 Forced Oscillators
Forced without damping Oscillators ( pp17-23)
2 3 Forced Oscillators
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62Chew T S MA1506-14 Chapter 2
When external frequency is close
to natural frequency , A(t) tends
to a st. line
0
2
F t
m
2.3 Forced Oscillators
Forced without damping Oscillators ( pp17-23)
2 3 Forced OscillatorsForced without damping Oscillators ( pp17 23)
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63Chew T S MA1506-14 Chapter 2
lim 0 sin( )2F t
x tm
=
( ) ( ) sin2
x t A t t +
=
2.3 Forced OscillatorsForced without damping Oscillators ( pp17-23)
We can understand the above result more
by looking at the following case:when
0cosmx kx F t + =
=
The particular solution of
is of the from (sin+ cos)The solution of
0cosmx kx F t + =
(0) 0, (0) 0x x= =with is 0 sin( )2
F tx t
m
=
2.3 Forced Oscillators
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Oscillations go out of control whenclose to . It is called resonance
64Chew T S MA1506-14 Chapter 2
blue curve in slide 57
tends to green st. line
when tends to
Red curveGreen curve
2.3 Forced Oscillators
Forced without damping Oscillators ( pp17-23)
0
2 sin()
0
2
2 3 Forced Oscillators
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Resonance
If the external force has a frequencyclose to the natural frequency of the system,
the resulting amplitudes can be very large even
for small external amplitudes.
It may cause violent motions and even disasters
in bridges and buildings
Chew T S MA1506-14 Chapter 2 65
2.3 Forced Oscillators
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Chew T S MA1506-14 Chapter 2 66
Collapse of the Tacoma Narrow Bridge.
http://www.youtube.com/watch?v=3mclp
9QmCGs
2 3 Forced Oscillators
http://www.youtube.com/watch?v=3mclp9QmCGshttp://www.youtube.com/watch?v=3mclp9QmCGshttp://www.youtube.com/watch?v=3mclp9QmCGshttp://www.youtube.com/watch?v=3mclp9QmCGshttp://www.youtube.com/watch?v=3mclp9QmCGs8/11/2019 ChewMA1506-14 Ch2
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Resonance
Avoiding resonance disasters is a major concern
in every building and bridge construction project.
As a countermeasure, a tuned mass damper can
be installed to avoid disaster.
The Taipei 101 building relies
on a 730-ton pendulum
a tuned mass damper to avoid resonance.
Chew T S MA1506-14 Chapter 2 67
2.3 Forced Oscillators
(cont)
Taipei 101 Tuned Mass Damper
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Chew T S MA1506-14 Chapter 2 68
Taipei 101 Tuned Mass Damper
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A tuned mass damper is a device mounted in
structures to prevent damage caused by
vibration
Chew T S MA1506-14 Chapter 2 69
http://en.wikipedia.org/wiki/File:Tuned_mass_damper.gif
http://en.wikipedia.org/wiki/File:Tuned_mass_damper.gifhttp://en.wikipedia.org/wiki/File:Tuned_mass_damper.gif8/11/2019 ChewMA1506-14 Ch2
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Liquid tuned mass damper
Chew T S MA1506-14 Chapter 2 70
2.3 Forced OscillatorsForced without damping Oscillators ( pp17 23)
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has two important phenomena
beating, resonance
71Chew T S MA1506-14 Chapter 2
2.3 Forced Oscillators
Forced , NO damped, Oscillator
0cosmx kx F t + =
We have learnt
Forced without damping Oscillators ( pp17-23)
2.3 Forced Oscillators
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Forced Damped Oscillators (pp 24-27)
72Chew T S MA1506-14 Chapter 2
Hence a particular solution is of the form
0 cos( )mx bx kx F t + + =
(see Appendix 5)
sin cospx B t C t = +
2
0 0
2 2 2 2
( )cos( ) sin( )
( ) ( )p
F k m t F b t
x t k m b
+
= +
2.3 Forced Oscillators
sint or cost never appear in see Section 2.2
We can find A and B, so
2.3 Forced OscillatorsForced Damped Oscillators (pp 24-27)
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+ Gen Sol of
73Chew T S MA1506-14 Chapter 2
General soln is
The 2ndpart (damped oscillation)
tends to zero rapidly, see Section 2.2
Hence 2ndpart called transient soln
2
0 0
2 2 2 2
( )cos( ) sin( )( )
( )
F k m t F b t x t
k m b
+= +
0mx bx kx+ + =
o ced a ped Osc a o s (pp )
2.3 Forced OscillatorsForced Damped Oscillators (pp 24-27)
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http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html
74Chew T S MA1506-14 Chapter 2
So when t big enough, the general soln
becomes
called steady-state soln (response)
2
0 0
2 2 2 2
( )cos( ) sin( )( )
( )
F k m t F b t x t
k m b
+=
+
p (pp )
2.3 Forced OscillatorsForced Damped Oscillators (pp 24-27)
http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html8/11/2019 ChewMA1506-14 Ch2
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Oscillation at
angular
frequency
http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html
75Chew T S MA1506-14 Chapter 2
X(t) can be written as
0
22 2 2 2
2
( / )cos( )
( )( )
F m t
x t b
m
=
+
/k m=where
when t big enough
p (pp )
2.3 Forced OscillatorsForced Damped Oscillators (pp 24-27)
http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html8/11/2019 ChewMA1506-14 Ch2
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Although the steady-state oscillation has
the same frequency as the external force
but it is NOT in phase with the external
Force. (compare with )
The amplitudes of the steady-state
soln and the external force are also
different
Chew T S MA1506-14 Chapter 2 76
cos t cos( )t
p (pp )
2.3 Forced OscillatorsForced Damped Oscillators (pp 24-27)
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77Chew T S MA1506-14 Chapter 2
0
2
2 2 2 2
2
( / )cos( )( )
( )
F m tx tb
m
= +
( )cos( )A t =
0
2
2 2 2 2
2
( / )( )
( )
F mAb
m
=
+
where =amplitude
p (pp )
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2.4 Conservation pp28-29
79Chew T S MA1506-14 Chapter 2
We shall prove it in next slide
21( )2
dx xdx
=
We need the following formula in this section
dxx
dt=Recall
2.4 Conservation
( t)
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Chain rule
80Chew T S MA1506-14 Chapter 2
2
12
d dxdt dt
=
2
2
1 22
dx d x dt dt dt dx
=
2
2d x xdt
= =
2
21 1( )2 2
d d dxxdx dx dt
=
(cont)
dtdx
2.4 Conservation
In this section we shall look at
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81Chew T S MA1506-14 Chapter 2
21
2 mx
21
2
kx
Consider SHM
We shall show that E is constant
Kinetic energy
+ potential energy
mx kx= Let E=
In this section, we shall look at
conservative systems
(cont)2.4 Conservation
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Chew T S MA1506-14 Chapter 2 82
simple harmonic motion
0mx kx+ =
2
2
1( )
2
1( )
2
dmx m x
dx
dm x
dx
=
=
First
21 1 (2 )2 2
d kx k x kxdx
= =
(cont)
from slide 80
Next
( t)2.4 Conservation
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83Chew T S MA1506-14 Chapter 2
2 21 1 02 2
d mx kxdx + =
0mx kx+ =From
get
(cont)
and previous slide
(cont)2.4 Conservation
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Chew T S MA1506-14 Chapter 2 84
2 21 1
( )2 2m x kx E + =
Hence
Hence the kinetic energy + potential energy
of the system remains constant
This system is called conservative system
(cont)
kinetic energy potential energy
where E
is aconstant
2.5 EULERs equation (Cantilevered Beams) pp30-37
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2.5 EULER s equation (Cantilevered Beams) pp30 37
Chew T S MA1506-14 Chapter 2 85
Beamlong, thin object
Cantilevered Beam (supported at only one end)
beam must bend
x
y
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3.7 Cantilevered Beams
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In this section, we will discuss bending of
a cantilevered beam
Why bending?
It is due to the weight of the beam and
other forces acting on the beam, called
load W(x)=force per unit length at point x
OUR Convention is that W(x) is positive in
UPWARD direction
Chew T S MA1506-14 Chapter 2 87
y
xW(x)
3.7 Cantilevered Beams
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Chew T S MA1506-14 Chapter 2 88
Load=weight of
springboard Load=weight of bamboo
stick+clothes
Load=weight of
springboard +swimmers
3.7 Cantilevered Beams
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Bending depends on W(x), stiffness, and
shape of the cross-section.
The stiffnessis measured by a constant E
called Youngs modulus.
The cross-sectionis measured by a
constant called the second moment of
area.Chew T S MA1506-14 Chapter 2 89
I
3.7 Cantilevered Beams
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Chew T S MA1506-14 Chapter 2 90
2 2
2 2 ( )
d d yEI W x
dx dx
=
Recall the directionof load W(x)
The equation of the beam is given by 4th
order ODE, Eulers equation, (proof omitted)
y
xW(x)
4
4
( )d y W x
EIdx
=
3.7 Cantilevered Beams
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Chew T S MA1506-14 Chapter 2 91
y
x
4
4
( )d y W x
EIdx =
( )W x
Find max deflection 3.7 Cantilevered Beams
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Find max deflection
Chew T S MA1506-14 Chapter 2 92
L
Assume: uniform mass
( )W x =
4
4
( )d y W x
EIdx=
Then
( )y L =
y
x
Recall W(x)=force per unit length at point x
for all x
3.7 Cantilevered Beams
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Chew T S MA1506-14 Chapter 2 93
To find i.e. to find ( )y L
we need to solve4
4( )d y W x
EIdx=
This is 4th order ODE,
we need FOUR conditions
(0) 0y =
00
x
dy
dx = =
3
3 0x L
d y
dx=
=
2
2 0x L
d y
dx=
=
Here are the conditions
EI=
3.7 Cantilevered Beams
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L
Chew T S MA1506-14 Chapter 2 94
y
x(0) 0y =
00
x
dy
dx = = No moment at L
No shear force at L
2
20
x L
d y
dx ==
3
3 0
x L
d y
dx=
=
Proof omitted
Proof omitted
3.7 Cantilevered Beams
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Chew T S MA1506-14 Chapter 2 95
integrate
both sides get
4
4
d y
EIdx
=3
3
d y xA
EIdx
= +
LA
EI
=3
3 0
x L
d y
dx=
=By get
Now solve this 4thorder ODE with four conditions
3.7 Cantilevered Beams
3d y x L
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Chew T S MA1506-14 Chapter 2 96
3
d y x L
EI EIdx
= +
2 2
2 2
d y x LxB
EI EIdx
= + +
2 2 2
2 2
L L LB
EI EI EI
= =
2
2 0
x L
d y
dx=
=By
get
integrate
both sides get
3.7 Cantilevered Beams
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Chew T S MA1506-14 Chapter 2 97
0
0x
dy
dx ==
Byget
2 2 2
2
2 2
d y x Lx L
EI EI EIdx
= +
3 2 2
6 2 2
dy x Lx L xC
dx EI EI EI
= + +
0C=
integrate
both sides get
3.7 Cantilevered Beams
3 2 2
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Chew T S MA1506-14 Chapter 2 98
3 2 2
6 2 2
dy x Lx L x
dx EI EI EI
= +
By (0) 0y =get
4 3 2 2
24 6 4x Lx L xy DEI EI EI
= + +
0D=
Integrate both sides get
3.7 Cantilevered Beams
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Chew T S MA1506-14 Chapter 2 99
Beam equation
y
x
4 3 24 1 1 1
2 12 3 2
L x x xy EI L L L
= +
3.7 Cantilevered Beams
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Cantilever deflection
formula
Chew T S MA1506-14 Chapter 2 100
4 3 24 1 1 1
2 12 3 2
L x x xy
EI L L L
= +
4 1 1 1
( )
2 12 3 2
Ly L
EI
= = +
4
8
L
EI
=
R k
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Chew T S MA1506-14 Chapter 2 101
Remark:
In the above example, we assume that
( )W x = for all x
In tutorial and past year exam questions,we will consider the following cases:
( ) 2 1 x
W x
L
=
( ) 2 x
W x
L
=
( ) cos2
xW x
L
=
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2.6 Plug flow reactor (PFR) model pp38-48
The plug flow reactor(PFR) model is used todescribe chemical reactions in continuous,
flowing systems.
PFR is like a long tube into which you pushsome mixture of chemicals which move through
the tube while they react with each other.
Chew T S MA1506-14 Chapter 2 102
3.6 PFR model
A
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Assume
Velocity uof flow is constant
Cross-sectional areaA is constant
Temp constant
No mixing upstream or downstream----everything that happens in a small
region of the PER is controlled by
chemical reactions IN that region,and by mixture of chemicals following
in and out ( see next two slides)Chew T S MA1506-14 Chapter 2 103
Now we shall use the following example
to illustrate the idea of PFR model
3.6 PFR model
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to illustrate the idea of PFR model
Chew T S MA1506-14 Chapter 2 104
x
H2O
O2
H2
Assumption: A lot of Oxygen pumped in at velocity u
Hydrogen pumped in at velocity u
Keep pumping
H2
O2
H2O
H2
O2
H2O
H2
O2
3.6 PFR modelQ:At point x, how many molecules of 2H
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Let C(x) be concentration of at point x.
Chew T S MA1506-14 Chapter 2 105
2
H
i.e., C(x)= # of molecules /cubic meter at x2
H
are passing by second?
Then in a time t Let N be the number of 2H
that pass by
Since ( )
volume
volume
number ofnumber of
moleculesmolecules
per
=
we have ( ) ( ( ))N C x A u t = So ( )dN
C x A udt
=
3.6 PFR model
Now consider a small piece of tube called plug
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Chew T S MA1506-14 Chapter 2 106
2H
2H
Now consider a small piece of tube, called plug,
Let the length of plug is x
x
flowing in at x,
at a rate
( )
dN
C x A udt =
flowing out at x+x,
at a rate
( )
dN
C x x Audt = +
2H molecules are being destroyed inside the plug
at a rate ( 2)( )( )r A x
( 2)( )( )r A xWhy What is r ?
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Chew T S MA1506-14 Chapter 2 107
( )( )( )
r is rate per volume at which the chemical reaction
2H2+ O2= 2H2O
happens in each unit of volume
Why minus 2 ? Because we are losing 2H
and the 2 because each reaction costs 2 molecules
Hence ( )( )r A x is the # of reactions happensin the plug per second
being destroyed inside the plug is2
HSo in each second, the # of
( 2)( )( )r A xmolecules
x
A Volume ofplug isA x
fl i i flowing out
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Chew T S MA1506-14 Chapter 2 108
( )C x A u ( )C x x Au+( 2)( )( )r A x
flowing in
at a rate
flowing out
at a rate2H 2H
2H destroyed
at rate
Hence ( ) 2 ( )C x Au rA x C x x Au = +So ( ) ( ) 2
C x x Au C x AurA
x
+ =
Hence ( ) ( ) 2C x x u C x u rx
+ =
( )2
dC xu r
dx =
3.6 PFR model
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It is known that the rate of chemical reaction
r depends on concentration C(x) of andtemperature
Chew T S MA1506-14 Chapter 2 109
2H
We assume temperature is constant
Hence r=k C(x) where k is a constant
( )implies 2 ( )
dC xu kC x
dx
= ( )
2dC x
u r
dx
= So
3.6 PFR model
( ) 2dC x k
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Chew T S MA1506-14 Chapter 2 110
Solve the above ODE, get
How to find B? Let x=0, we get
2
( )
kx
u
C x Be
=
2 0
(0)k
uC Be B
= =
( ) 2( )
dC x k C x
dx u =
Therefore2
( ) (0)
kx
C C
3.6 PFR model
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Therefore
Chew T S MA1506-14 Chapter 2 111
( ) (0) uC x C e=
END
( ) 2 ( )C x Au rA x C x x Au = +
Remark: In the above, we assumearea A of cross section is constant
and we have
If A is not constant, then we will have
( ) ( ) 2 ( ) ( ) ( )C x A x u rA x x C x x A x x u = + +
In this case, following the same idea as above, we will have
( ) ( ) 2 ( )d C x A xu rA xdx
= ( ) ( )implies 2 ( ) ( )d C x A xu kC x A xdx
=
Appendix1
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pp
Chew T S MA1506-14 Chapter 2 112
Hence
cos( )
cos( ) cos( ) sin( ) sin( )
A t
A t A t
= +
By the above formula, get
cos( )
sin( )
C A
D A
=
= where
cos( )A t
cos( ) cos cos sin sin + = +
cos( ) sin( )C t D t = +
Appendix 2 (Defintions of cosh and sinh)
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Appendix 2 (Defintions of cosh and sinh)
Chew T S MA1506-14 Chapter 2 113
Formulae
cosh2
x x
e ex
+= sinh2
x x
e ex
=
2 2(cosh ) (sinh ) 1x x =
sinhcosh
d xxdx =
cosh
sinh
d x
xdx =
Appendix 2 (cont) Graphs of sinhx, coshx, tanhx
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Chew T S MA1506-14 Chapter 2 114
sinhxcoshxtanhx
http://www.graphmatica.com
sinhx
tanhx
In slide 36, we compute t using cosh function as follows:
Appendix 2 (cont)
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Chew T S MA1506-14 Chapter 2 115
Suppose now
1/ cosh (2)t L g =
p g
( ) 2t =
( )( ) 2 cosh /t g L t = = Hence
Find time t such that
2/ 100 / secg L=
9.8L= centimeters
Hence
11 cosh (2) 0.132sec100
t =
Appendix 2 (cont)
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Chew T S MA1506-14 Chapter 2 116
Suppose we use the following exp. function instead of cosh
( / ) ( / )
2g L t g L t e e = +
Then
Let We get 4 =+ 1
Solve the above eq , get the value x,
and then get the value t (choose positive t).
Hence using cosh is easier.
2 3x=
( ) ( )/ /2
2
g L t g L t e e
= +
( )/g L tx e=
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Wsin cosx B t C t = +
(cont)
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Chew T S MA1506-14 Chapter 2 118
We guess
Then subst. into
0cosmx kx F t + =
get
sin cosp
x B t C t = +
0
20,
FB C
k m= =
0
2 2
/F m
=
0
2 2/cos( ) cosF mx A t t
= +
,p px x
Assume
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Appendix 4 (cont)0cosmx kx F t + =
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By the initial conditions in previous slide,
get
120Chew T S MA1506-14 Chapter 2
0
2 2
/F m
A = 0=
0
So0
2 2
/
[cos cos ]
F m
x t t =
Appendix 5
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121Chew T S MA1506-14 Chapter 2
0 cos( )mx bx kx F t + + = p
xWe shall find a particular soln
sin cospx B t C t = +We guess
Then subst.
into 0 cos( )mx bx kx F t + + =
, ,p p p
x x x
Appendix 5 (cont.)
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pp ( )
122Chew T S MA1506-14 Chapter 2
0
2 2 2 2( )
F bB
k m b
=
+
2
0
2 2 2 2
( )
( )
F k mC
k m b
=
+
get
Appendix 5 (cont)
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pp ( )
so2
0 0
2 2 2 2 2 2 2 2
( )sin cos
( ) ( )p
F b F k mx t t
k m b k m b
= +
+ +
END
Chapter 2