Dynamics eg260 l1

EG-260 - Dynamics 1
Professor S Adhikari
S.Adhikari@swansea.ac.uk
@ProfAdhikari, Hash tag: #EG260

Swansea University
Following: Engineering Vibration
D J Inman, 3rd Edition
Pearson Education
College of Engineering 1/43
© Eng. Vib, 3rd Ed.

Motivation The Millennium Bridge
a Recent Vibration Problem (2000-1)

OPENED AND CLOSED WITHIN A FEW HOURS
College of Engineering BECAUSE OF UNDESIRABLE VIBRATION 2/43
© Eng. Vib, 3rd Ed. (Show movies)

Millennium Bridge Vibration

Lack of consideration of dynamic loads and vibration
caused this to relatively new bridge to vibrate wildly


Modeling and Degrees of Freedom
The bridge has many degrees of freedom, we will start with one
and work towards many.

Lack of consideration of dynamic loads and vibration
caused this to relatively new bridge to vibrate wildly

College of Engineering
The goal of this course is to understand such 4/43
© Eng. Vib, 3rd Ed. phenomenon and how to prevent it

Example 1.1.1 The Pendulum
•  Sketch the structure or
part of interest
•  Write down all the
forces and make a
“free body diagram”
•  Use Newton’s Law
and/or Euler’s Law to
find the equations of
motion
∑M 0 = J 0α , J 0 = ml 2


The problem is one dimensional,
hence a scalar equation results

J 0α (t) = −mgl sin θ (t) ⇒ m 2θ (t) + mgsin(t) = 0
θ
 
restoring
force
Here the over dots denote differentiation with respect to time t

This is a second order, nonlinear ordinary differential equation

We can linearise the equation by using the approximation sin θ ≈ θ

(t) + mgθ (t) = 0 ⇒ θ (t) + g θ (t) = 0
⇒ m θ 2 


Requires knowledge of θ (0) and θ (0)

the initial position and velocity.

Next consider a spring mass system
and perform a static experiment:

•  From strength of
materials recall:
FBD:

nonlinear

linear
A plot of force versus displacement:

experiment ⇒ f k = kx 7/43

Free-body diagram and equation
of motion

• Newton’s Law:
m = −kx(t) ⇒ m + kx(t) = 0
x(t) x(t)

x(0) = x0 , x (0) = v0
(1.2)
Again a 2nd order ordinary differential equation

Stiffness and Mass
Vibration is caused by the interaction of two different forces
one related to position (stiffness) and one related to
acceleration (mass).
Proportional to displacement
Stiffness (k)

Displacement
f k = −kx (t ) x
statics k
Mass (m) m

fm = ma(t) = m
x(t)
Mass Spring
dynamics

Proportional to acceleration

Examples of Single-Degree-of-
Freedom Systems
Pendulum Shaft and Disk

Torsional
l =length Stiffness
k
θ

Moment
Gravity g of inertia
m J
θ

(t) + g θ (t) = 0
θ
l 
Jθ (t) + kθ (t) = 0

Solution to 2nd order DEs
Lets assume a solution: x(t)
x (t ) = A sin(ωnt + φ ) (1.3)
t
Differentiating twice gives:


x (t) = ω n A cos(ω n t + φ ) (1.4)
 = −ω n Asin(ω n t + φ ) = -ω n x(t)
x(t) 2 2
(1.5)
Substituting back into the equations of motion gives:
2
−mωn A sin(ωnt + φ ) + kA sin(ωnt + φ ) = 0
2 k Natural
−mω + k = 0
n or ωn =
m frequency
College of Engineering rad/s 11/43

Summary of simple harmonic
motion
x(t) Period
Amplitude
2π A
T=
ωn
x0 Slope
here is v0
t

Maximum
φ Velocity
ωn ωn A
ωn rad/s ωn cycles ωn
fn = = = Hz
2π rad/cycle 2π s 2π

Initial Conditions
If a system is vibrating then we must assume that something
must have (in the past) transferred energy into to the system
and caused it to move. For example the mass could have
been:
• moved a distance x0 and then released at t = 0 (i.e. given
Potential energy) or
• given an initial velocity v0 (i.e. given some kinetic energy) or
• Some combination of the two above cases
From our earlier solution we know that:

x0 = x(0) = Asin(ω n 0 + φ ) = Asin(φ )

v0 = x (0) = ω n A cos(ω n 0 + φ ) = ω n A cos(φ )

Initial Conditions Determine the
Constants of Integration
Solving these two simultaneous equations for A and ϕ gives:

1 2 2 2
"
−1 ω n x0
%
A= ω n x0 + v0 , φ = tan $ '
ωn v &
   # 0 
  
Amplitude Phase

Slope
x(t) here is v0 1 2 2 2
ω n x0 + v0
x0 ωn
x0
t
φ
φ
ωn v0
College of Engineering ωn 14/43

Thus the total solution for the
spring mass system becomes:

2 2 2
ω n x0 + v0 ⎛ −1 ω n x0 ⎞
x(t) = sin ⎜ ω nt + tan (1.10)
ωn ⎝ v0 ⎟ ⎠
Called the solution to a simple harmonic oscillator
and describes oscillatory motion, or simple harmonic motion.

Note (Example 1.1.2)
2 2 2
ω n x0 + v0 ω n x0
x(0) = = x0
ωn 2 2 2
ω n x0 + v0

College of Engineering as it should 15/43

A note on arctangents
•  Note that calculating arctangent from a
calculator requires some attention. First, all
machines work in radians.

•  The argument atan(-/+) is in a different quadrant
then atan(+/-), and usual machine calculations
will return an arctangent in between -π/2 and
+π/2, reading only the atan(-) for these two
different cases.
+ +

φ

-

_ +
+
φ

- -
In MATLAB, use the atan2(x,y) function to get
College of Engineering the correct phase. 16/43

Example 1.1.3 wheel, tire suspension
m = 30 kg, ωn = 10 hz, what is k?

⎛ cylce 2π rad ⎞
2
k = mω = ( 30 kg ) ⎜ 10
n g ⎟ = 1.184 × 10 5 N/m
⎝ sec cylce ⎠

There are of course more complex models of suspension systems
and these appear latter in the course as our tools develope


Section 1.2 Harmonic Motion
The period is the time elapsed to complete one complete cylce
2π rad 2π
T= = s (1.11)
ω n rad/s ω n
The natural frequency in the commonly used units of hertz:

ωn ω n rad/s ω n cycles ω n
fn = = = = Hz (1.12)
2π 2π rad/cycle 2π s 2π
For the pendulum:
g l
ωn = rad/s, T = 2π s
l g
For the disk and shaft:

k J
ωn = rad/s, T = 2π s
J k

Relationship between Displacement,
Velocity and Acceleration
A=1, ωn=12
1
Displacement
x (t ) = A sin(ωnt + φ ) 0

x
-1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Velocity 20


x (t) = ω n A cos(ω n t + φ ) 0
v

-20
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Acceleration 200

 = −ω n Asin(ω n t + φ )
x(t) 2
0
a

-200
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (sec)
College of Engineering Note how the relative magnitude of each increases for ωn>1 19/43

Example 1.2.1Hardware store spring, bolt: m= 49.2x10 -3

kg, k=857.8 N/m and x0 =10 mm. Compute ωn and the max amplitude
of vibration.
Note: common
Units are Hertz
k 857.8 N/m
ωn = = -3
= 132 rad/s
m 49.2 ×10 kg
ωn To avoid Costly errors use fn
fn = = 21 Hz when working in Hertz and ωn
when in rad/s
2π
2π 1 1
T= = = 0.0476 s
ωn fn 21 cyles
sec
1 0
2 2 2
x(t) max = A = ω x + v = x0 = 10 mm
n 0 0
ωn
College of Engineering Units depend on system 20/43

Compute the solution and max velocity and
acceleration:

v(t)max = ωn A = 1320 mm/s = 1.32 m/s 2.92 mph

2 3 2
a(t) max = ω n A = 174.24 ×10 mm/s
2
=174.24 m/s ≈ 17.8g!
g = 9.8 m/s2
−1 ⎛ω n x0 ⎞ π
φ = tan ⎜ ⎟ = rad 90°
⎝ 0 ⎠ 2
x(t) = 10 sin(132t + π / 2) = 10 cos(132t) mm
~0.4 in max


Does gravity matter in spring
problems?
Δ

Let Δ be the deflection caused by
hanging a mass on a spring
(Δ = x1-x0 in the figure)

Then from static equilibrium: mg = kΔ
Next sum the forces in the vertical for some point x > x1 measured
from Δ

m = −k ( x + Δ ) + mg = −kx + mg − 
x kΔ
 
=0

⇒ m + kx(t) = 0
x(t)
So no, gravity does not have an effect on the vibration
© Eng. Vib, 3rd Ed. (note that this is not the case if the spring is nonlinear)

Example 1.2.2 Pendulums and
measuring g
•  A 2 m pendulum 2π l
T= = 2π
swings with a period ωn g
of 2.893 s
2 2
•  What is the 4π 4π
g= 2 l = 2m
acceleration due to T 2 2
2.893 s
gravity at that
⇒ g = 9.796 m/s2
location?

This is g in Denver, CO USA, at 1638m
and a latitude of 40°

Review of Complex Numbers and Complex
Exponential (See Appendix A)
A complex number can be written with a real and imaginary
part or as a complex exponential

c = a + jb = Ae jθ
Where
ℑ
a = A cos θ , b = A sin θ a

Multiplying two complex numbers:
j (θ1 +θ2 ) A
c1c2 = A1 A2 e b

Dividing two complex numbers: ℜ
c1 A1 j (θ1 −θ2 )
= e
c2 A2

Equivalent Solutions to 2nd order
Differential Equations (see Window 1.4)

All of the following solutions are equivalent:

x (t ) = A sin(ωnt + φ ) Called the magnitude and phase form

x (t ) = A1 sin ωnt + A2 cos ωnt Sometimes called the Cartesian form

jωn t − jωn t
x (t ) = a1e + a2 e Called the polar form

The relationships between A and ϕ, A1 and A2, and a1 and a2
can be found in Window 1.4 of the course text, page 17.

• Each is useful in different situations
• Each represents the same information
College of Engineering • Each solves the equation of motion 25/43

Derivation of the solution
Substitute x(t) = ae λt into m + kx = 0 ⇒
x
mλ 2 ae λt + kae λt = 0 ⇒
2
mλ + k = 0 ⇒
k k
λ = ± − =± j = ±ω n j ⇒
m m
x(t) = a1eωn jt and x(t) = a2 e−ωn jt ⇒
x(t) = a1eωn jt + a2 e−ωn jt (1.18)

This approach will be used again for more complicated problems

Is frequency always positive?
From the preceding analysis, λ = + ωn then
ωn jt −ωn jt
x (t ) = a1e + a2 e
Using the Euler relations* for trigonometric functions, the
above solution can be written as (recall Window 1.4)

x (t ) = A sin (ωnt + φ ) (1.19)
It is in this form that we identify as the natural frequency
ωn and this is positive, because the + sign being used up
in the transformation from exponentials to the sine
function.
* http://en.wikipedia.org/wiki/Euler's_formula
ix
e = cos x + i sin x

Calculating root mean square
(RMS) values
May need to be limited due
to physical constraints
Not very useful since for
A = peak value a sine function the
1
T average value is zero
x = lim ∫ x (t )dt = average value
T →∞ T
0
T
2 1 2
x = lim ∫ x (t )dt = mean-square value (1.21)
T →∞ T
0
Proportional
xrms = x 2 = root mean square value to energy

Also useful when the vibration is random

The Decibel or dB scale
It is often useful to use a logarithmic scale to plot vibration
levels (or noise levels). One such scale is called the decibel or
dB scale. The dB scale is always relative to some reference
value x0. It is define as:
2
⎛ x ⎞ ⎛ x ⎞
dB = 10 log10 ⎜ ⎟ = 20 log10 ⎜ ⎟ (1.22)
⎝ x0 ⎠ ⎝ x0 ⎠
For example: if an acceleration value was 19.6m/s2 then relative
to 1g (or 9.8m/s2) the level would be 6dB,
2
⎛ 19.6 ⎞
10 log10 ⎜ ⎟ = 20 log10 ( 2 ) = 6 dB
⎝ 9.8 ⎠
Or for Example 1.2.1: The Acceleration Magnitude
College of Engineering is 20log10(17.8)=25dB relative to 1g. 29/43

1.3 Viscous Damping
All real systems dissipate energy when they vibrate. To
account for this we must consider damping. The most simple
form of damping (from a mathematical point of view) is called
viscous damping. A viscous damper (or dashpot) produces a
force that is proportional to velocity.
Mostly a mathematically motivated form, allowing
a solution to the resulting equations of motion that predicts
reasonable (observed) amounts of energy dissipation.
Damper (c)


fc = −cv(t) = −cx (t) x

fc


The Role of Damping


Differential Equation Including
Damping
For this damped single degree of freedom system the force acting
on the mass is due to the spring and the dashpot i.e. fm= - fk - fc.
Displacement
x
k
m = −kx(t) − cx (t)
x(t) 
M
or
c
m + cx (t) + kx(t) = 0 (1.25)
x(t) 

To solve this for of the equation it is useful to assume a
solution of the form (again):
λt
x(t) = ae

Solution to DE with damping
included (dates to 1743 by Euler)
The velocity and acceleration can then be calculated as:
x(t) = λ ae λt

 = λ 2 ae λt
x(t)

If this is substituted into the equation of motion we get:

aeλt (mλ 2 + cλ + k) = 0 (1.26)
Divide equation by m, substitute for natural frequency and
assume a non-trivial solution

λt 2c 2
ae ≠ 0 ⇒ (λ + λ + ω n ) = 0
m

Solution to our differential
equation with damping included:
For convenience we will define a term known as the
damping ratio as:
c
ζ= (1.30) Lower case Greek zeta
2 km NOT ξ as some like to use

The equation of motion then becomes:
2 2
(λ + 2ζωn λ + ω ) = 0
n
Solving for λ then gives,

2
λ1,2 = −ζω n ± ω n ζ − 1 (1.31)

Possibility 1. Critically damped motion

Critical damping occurs when ζ =1. The damping coefficient
c in this case is given by:

ζ =1 ⇒ c = ccr = 2 km = 2mω n
   
definition of critical
damping coefficient
Solving for λ then gives,
2
λ1,2 = −1ωn ± ωn 1 −1 = −ωn
A repeated, real root
The solution then takes the form

x(t) = a1e−ω n t + a2te−ω n t
Needs two independent solutions, hence the t
in the second term 36/43

Critically damped motion
a1 and a2 can be calculated from initial conditions (t=0),

x = (a1 + a2t)e−ω n t
k=225N/m m=100kg and ζ =1
⇒ a1 = x0 0.6
x0=0.4mm v0=1mm/s
0.5
v = (−ω n a1 − ω n a2t + a2 )e−ω n t x0=0.4mm v0=0mm/s
x0=0.4mm v0=-1mm/s

Displacement (mm)
0.4
v0 = −ω n a1 + a2
0.3
⇒ a2 = v0 + ω n x0 0.2

•  No oscillation occurs 0.1

•  Useful in door 0
mechanisms, analog
-0.1
gauges 0 1 2 3 4
Time (sec)


Possibility 2: Overdamped motion
An overdamped case occurs when ζ >1. Both of the roots of the
equation are again real.
k=225N/m m=100kg and ζ =2
2 0.6
λ1,2 = −ζωn ± ωn ζ −1 x0=0.4mm v0=1mm/s
0.5 x0=0.4mm v0=0mm/s
2 2
x(t) = e−ζωn t (a1e−ωn t ζ −1
+ a2 eω n t ζ −1
) x0=0.4mm v0=-1mm/s

Displacement (mm)
0.4

a1 and a2 can again be calculated 0.3
from initial conditions (t=0), 0.2

−v0 + ( − ζ + ζ 2 −1)ωn x0 0.1

a1 = 0
2ωn ζ 2 −1
-0.1
0 1 2 3 4
2
v0 + (ζ + ζ −1)ωn x0 Time (sec)
a2 =
2ωn ζ 2 −1 Slower to respond than
College of Engineering critically damped case 38/43

Possibility 3: Underdamped motion
An underdamped case occurs when ζ <1. The roots of the
equation are complex conjugate pairs. This is the most
common case and the only one that yields oscillation.
2
λ1,2 = −ζωn ± ωn j 1− ζ
−ζωn t jωn t 1−ζ 2 − jω n t 1−ζ 2
x(t) = e (a1e + a2 e )
−ζω n t
= Ae sin(ω d t + φ)

The frequency of oscillation ωd is called the damped natural
frequency is given by.

ω d = ω n 1− ζ 2 (1.37)


Constants of integration for the
underdamped motion case
As before A and ϕ can be calculated from initial conditions (t=0),
1
A= (v0 + ζωn x0 )2 + (x0ω d ) 2
ωd
⎛ x0ω d ⎞
−1
φ = tan ⎜ ⎟
⎝ v0 + ζωn x0 ⎠ 1

•  Gives an oscillating
0.5
response with exponential

Displacement
decay
0
•  Most natural systems vibrate
with and underdamped
-0.5
response
•  See Window 1.5 for details
-1
and other representations 0 1 2 3 4 5
College of Engineering Time (sec) 40/43

Example 1.3.1: consider the spring of 1.2.1, if c = 0.11
kg/s, determine the damping ratio of the spring-bolt system.
−3
m = 49.2 × 10 kg, k = 857.8 N/m
−3
ccr = 2 km = 2 49.2 × 10 × 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

Example 1.3.2
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 an
initial velocity of v0 = 0.6 m/s and zero initial
displacement (this would correspond to the
vibration induced while landing on your feet,
with your knees locked form a height of 18
mm) and plot the response. What is the
maximum acceleration experienced by the
leg assuming no damping?

Solution:
20 cycles 2π rad
ωn = = 125.66 rad/s
1 s cycles
2
ω d = 125.66 1− (.224) = 122.467 rad/s
2
(0.6 + (0.224 )(125.66)(0)) + (0)(122.467)2
A= = 0.005 m
122.467

⎛ (0)(ω d ) ⎞
φ = tan ⎜ -1
⎟ = 0
⎝ v0 + ζω n (0 )⎠
⇒ x(t ) = 0.005e −28.148t sin(122.467t )


Use the undamped formula to get maximum
acceleration:
2
2
!v $
A = x0 + # 0 & , ω n = 125.66, v0 = 0.6, x0 = 0
#ω &
" n%
v0 0.6
A= m= m
ωn ωn
! 0.6 $
x () n #ω &
" n%
n ( )( )
max  = −ω A = −ω # & = 0.6 125.66 m/s 2 = 75.396 m/s 2
2 2

maximum acceleration = 75.396 m/s2
2
g = 7.68g's
9.81 m/s

Here is a plot of the displacement
response versus time


Example 1.3.3 Compute the form of the response of an
underdamped system using the Cartesian form of the solution given in
Window 1.5.

sin(x + y) = sin x sin y + cos x cos y ⇒
x(t) = Ae−ζωnt sin(ω d t + φ ) = e−ζωnt (A1 sin ω d t + A2 cosω d t)
x(0) = x0 = e 0 (A1 sin(0) + A2 cos(0)) ⇒ A2 = x0

x = −ζω n e−ζωnt (A1 sin ω d t + A2 cosω d t)
+ ω d e−ζωnt (A1 cosω d t − A2 sin ω d t)
v0 = −ζω n (A1 sin 0 + x0 cos0) + ω d (A1 cos0 − x0 sin 0)
v0 + ζω n x0
⇒ A1 = ⇒
ωd

−ζω nt
# v0 + ζω n x0 &
x(t) = e % sin ω d t + x0 cosω d t (
$ ωd '

Dynamics eg260 l1

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