chapter1 part 1

College of Engineering © Eng. Vib, 3rd Ed.

EG-260 - Dynamics 1

Professor S Adhikari [email protected]

@ProfAdhikari, Hash tag: #EG260

Swansea University

Following: Engineering Vibration D J Inman, 3rd Edition

Pearson Education


2/43

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

OPENED AND CLOSED WITHIN A FEW HOURS BECAUSE OF UNDESIRABLE VIBRATION

(Show movies)


3/43

Millennium Bridge Vibration

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


4/43 4/43

Modeling and Degrees of Freedom

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

The goal of this course is to understand such phenomenon and how to prevent it

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


5/43

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

α= =∑ l 20 0 0, J J mM


6/43

The problem is one dimensional, hence a scalar equation results J0α(t) = −mgl sinθ(t)⇒ m2 θ (t)+mgsinθ(t)

restoringforce

= 0

This is a second order, nonlinear ordinary differential equation We can linearise the equation by using the approximation

Here the over dots denote differentiation with respect to time t

sinθ ≈θ

⇒ m2 θ (t)+mgθ(t) = 0⇒ θ (t)+ gθ(t) = 0

Requires knowledge of θ(0) and θ(0)

the initial position and velocity.


7/43

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

•  From strength of materials recall:

A plot of force versus displacement:

FBD:

linear

nonlinear

⇒ =experiment kf kx


8/43

Free-body diagram and equation of motion

mx(t) = −kx(t)⇒ mx(t)+ kx(t) = 0x(0) = x0, x(0) = v0

• Newton’s Law:

Again a 2nd order ordinary differential equation (1.2)


9/43

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

m k

x Displacement

Mass Spring

Stiffness (k)

Mass (m)

= − ( )kf kx t

fm =ma(t) =mx(t)

Proportional to displacement

Proportional to acceleration

statics

dynamics


10/43

Examples of Single-Degree-of-Freedom Systems

m

θ

l =length

Pendulum

θ (t)+ glθ(t) = 0

Gravity g

Shaft and Disk

θ

J θ (t)+ kθ(t) = 0

Torsional Stiffness

k Moment of inertia

J


11/43

Solution to 2nd order DEs

ω φ= +( ) sin( ) (1.3)nx t A t

Lets assume a solution:

Differentiating twice gives:

x(t) =ωnAcos(ωnt +φ) (1.4)x(t) = −ωn

2Asin(ωnt +φ) = -ωn2x(t) (1.5)

Substituting back into the equations of motion gives:

ω ω φ ω φ

ω ω

− + + + =

− + = =

2

2

sin( ) sin( ) 0

0 or

n n n

n n

m A t kA t

km km

Natural frequency

t

rad/s

x(t)


12/43

Summary of simple harmonic motion

t

x(t)

x0 Slope here is v0

nωφ

Period

T =2πω n

Amplitude A

Maximum Velocity

Anω

ω ω ωπ π π

= = = rad/s cycles Hz

2 rad/cycle 2 s 2n n n

nf


13/43

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(ωn0+φ) = Asin(φ) v0 = x(0) =ωnAcos(ωn0+φ) =ωnAcos(φ)


14/43

Initial Conditions Determine the Constants of Integration

Solving these two simultaneous equations for A and ϕ gives:

A = 1ωn

ωn2x0

2 + v02

Amplitude

, φ = tan−1 ωnx0

v0

"

#$

%

&'

Phase

v0ωn

x0

1ω n

ω n2x0

2 + v02

t

x(t)

x0

Slope here is v0

nωφ

φ


15/43

Thus the total solution for the spring mass system becomes:

x(t) =ωn2x0

2 + v02

ωn

sin ωnt + tan−1ωnx0

v0

⎛

⎝⎜⎞

⎠⎟(1.10)

Called the solution to a simple harmonic oscillator and describes oscillatory motion, or simple harmonic motion.

Note (Example 1.1.2)

x(0) =ωn2x0

2 + v02

ωn

ωnx0ωn2x0

2 + v02= x0

as it should


16/43

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 the correct phase.

+

+

-

_ +

+

-

-


17/43

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

k =mωn2 = 30 kg( ) 10 cylce

seci2π rad

cylce!

"#

$

%&= 1.184×105 N/m

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


18/43

Section 1.2 Harmonic Motion

T =2π radωn rad/s

=2π ωn

s (1.11)

The natural frequency in the commonly used units of hertz:

The period is the time elapsed to complete one complete cylce

fn =ωn

2π=

ωn rad/s2π rad/cycle

=ωn cycles

2π s=ωn

2π Hz (1.12)

For the pendulum:

ω n =

gl

rad/s, T = 2π lg

s

ωn =kJ

rad/s, T = 2π Jk

s

For the disk and shaft:


19/43

Relationship between Displacement, Velocity and Acceleration

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

0

1

x

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

0

20 v

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

0

200

Time (sec)

a

A=1, ωn=12

ω φ= +( ) sin( ) nx t A t

x(t) =ωnAcos(ωnt +φ)

x(t) = −ωn2Asin(ωnt +φ)

Note how the relative magnitude of each increases for ωn>1

Displacement

Velocity

Acceleration


20/43

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.

€

ωn =km

=857.8 N/m

49.2×10-3 kg=132 rad/s

fn =ωn

2π= 21 Hz

T =2πωn

=1fn

=1

21cyles sec0.0476 s

x(t)max = A =1ωn

ωn2x0

2 + v02 = x0 =10 mm

0

Note: common Units are Hertz

To avoid Costly errors use fn when working in Hertz and ωn when in rad/s

Units depend on system


21/43

Compute the solution and max velocity and acceleration:

€

v(t)max =ωnA =1320 mm/s = 1.32 m/sa(t)max =ω

n

2A =174.24×103 mm/s2

=174.24 m/s2 ≈ 17.8g!

φ = tan−1 ωnx0

0⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

π2

rad

x(t) =10sin(132t +π / 2) =10cos(132t) mm

g = 9.8 m/s2

2.92 mph

90°

~0.4 in max


22/43

Does gravity matter in spring problems?

Let Δ be the deflection caused by hanging a mass on a spring (Δ = x1-x0 in the figure) Then from static equilibrium: mg = kΔNext sum the forces in the vertical for some point x > x1 measured from Δ

mx = −k x +Δ( )+mg = −kx +mg− kΔ=0

⇒ mx(t)+ kx(t) = 0

So no, gravity does not have an effect on the vibration (note that this is not the case if the spring is nonlinear)

Δ


23/43

Example 1.2.2 Pendulums and measuring g

•  A 2 m pendulum swings with a period of 2.893 s

•  What is the acceleration due to gravity at that location?

π π= =

⇒

l2 2

2 2 2

2

4 4 2 m 2.893 s

= 9.796 m/s

gT

g

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

ππ

ω= =

l2 2n

Tg


24/43

Review of Complex Numbers and Complex Exponential (See Appendix A)

ℜ

ℑ

b

a

A

θ= + = jc a jb Ae

A complex number can be written with a real and imaginary part or as a complex exponential

Where

θ θ= =cos , sina A b AMultiplying two complex numbers:

θ θ+= 1 2( )1 2 1 2

jc c A A eDividing two complex numbers:

θ θ−= 1 2( )1 1

2 2

jc A ec A


25/43

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

ω ω

ω φ

ω ω−

= +

= +

= +

1 2

1 2

( ) sin( )( ) sin cos

( ) n n

n

n nj t j t

x t A tx t A t A t

x t a e a e

All of the following solutions are equivalent:

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.

Sometimes called the Cartesian form

Called the magnitude and phase form

Called the polar form

• Each is useful in different situations • Each represents the same information • Each solves the equation of motion


26/43

Derivation of the solution

Substitute x(t) = aeλt into mx + kx = 0 ⇒mλ 2aeλt + kaeλt = 0 ⇒

mλ 2 + k = 0 ⇒

λ = ± −km=±

kmj = ±ωn j⇒

x(t) = a1eωn jt and x(t) = a2e

−ωn jt ⇒

x(t) = a1eωn jt + a2e

−ωn jt

This approach will be used again for more complicated problems

(1.18)


27/43

Is frequency always positive? From the preceding analysis, λ = + ωn then

ω ω−= +1 2( ) n njt jtx t a e a eUsing the Euler relations* for trigonometric functions, the above solution can be written as (recall Window 1.4)

( )ω φ= +( ) sin (1.19)nx t A t

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

e jx = cos x + j sin x


28/43

Calculating root mean square (RMS) values

→∞

→∞

=

=

=

=

∫

∫

0

2 2

0

2

peak value1lim ( ) = average value

1lim ( ) = mean-square value

= root mean square value

T

T

T

T

rms

A

x x t dtT

x x t dtT

x xProportional

to energy

Not very useful since for a sine function the

average value is zero

May need to be limited due to physical constraints

Also useful when the vibration is random

(1.21)


29/43

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

10 100 0

10 log 20 log (1.22)x xdBx x

For example: if an acceleration value was 19.6m/s2 then relative to 1g (or 9.8m/s2) the level would be 6dB,

( )⎛ ⎞= =⎜ ⎟

⎝ ⎠

2

10 1019.610 log 20 log 2 69.8

dB

Or for Example 1.2.1: The Acceleration Magnitude is 20log10(17.8)=25dB relative to 1g.


30/43

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.

Damper (c)

fc = −cv(t) = −cx(t) x

fc

Mostly a mathematically motivated form, allowing a solution to the resulting equations of motion that predicts reasonable (observed) amounts of energy dissipation.


The Role of Damping

31/43


The Role of Damping

32/43


33/43

Differential Equation Including Damping

M

k x Displacement

c

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.

mx(t) = −kx(t)− cx(t) or mx(t)+ cx(t)+ kx(t) = 0 (1.25)

To solve this for of the equation it is useful to assume a solution of the form (again):

€

x(t) = aeλt


34/43

Solution to DE with damping included (dates to 1743 by Euler)

x(t)= λaeλt x(t)= λ 2aeλt

The velocity and acceleration can then be calculated as:

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

aeλt (mλ2 + cλ + k) = 0 (1.26)Divide equation by m, substitute for natural frequency and assume a non-trivial solution

aeλt ≠ 0 ⇒ (λ2 + cmλ +ωn

2 ) = 0


35/43

Solution to our differential equation with damping included:

ζ = c2 km

(1.30)

For convenience we will define a term known as the damping ratio as:

The equation of motion then becomes:

€

(λ2 + 2ζωnλ +ωn2) = 0

Solving for λ then gives,

λ1,2 = −ζωn ±ωn ζ 2 −1 (1.31)

Lower case Greek zeta NOT ξ as some like to use


36/43

Possibility 1. Critically damped motion

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

€

λ1,2 = −1ωn ±ωn 12 −1 = −ωn

ζ=1⇒ c = ccr = 2 kmdefinition of criticaldamping coefficient

= 2mωn

The solution then takes the form

x(t) = a1e−ωnt + a2te

−ωnt

Solving for λ then gives,

A repeated, real root

Needs two independent solutions, hence the t in the second term


37/43

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

x = (a1 + a2t)e−ωnt

⇒ a1 = x0

v = (−ωna1 −ωna2t + a2 )e−ωnt v0 = −ωna1 + a2

⇒ a2 = v0 +ωnx0

0 1 2 3 4 -0.1 0

0.1 0.2 0.3 0.4 0.5 0.6

Time (sec)

Dis

plac

emen

t (m

m)

k=225N/m m=100kg and ζ = 1 x 0 =0.4mm v 0 =1mm/s x 0 =0.4mm v 0 =0mm/s x 0 =0.4mm v 0 =-1mm/s

•  No oscillation occurs •  Useful in door

mechanisms, analog gauges


38/43

Possibility 2: Overdamped motion An overdamped case occurs when ζ >1. Both of the roots of the equation are again real.

€

λ1,2 = −ζωn ±ωn ζ 2 −1

x(t) = e−ζωn t(a1e−ωn t ζ 2−1 + a2e

ωn t ζ 2−1)

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

€

a1 =−v0 + ( −ζ + ζ2 −1)ωnx0

2ωn ζ 2 −1

a2 =v0 + (ζ + ζ 2 −1)ωnx0

2ωn ζ2 −1

0 1 2 3 4 -0.1 0

0.1 0.2 0.3 0.4 0.5 0.6

Time (sec) D

ispl

acem

ent (

mm

)

k=225N/m m=100kg and ζ =2 x 0 =0.4mm v 0 =1mm/s x 0 =0.4mm v 0 =0mm/s x 0 =0.4mm v 0 =-1mm/s

Slower to respond than critically damped case


39/43

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.

€

λ1,2 =−ζωn ±ωn j 1−ζ2

x(t) = e−ζωn t(a1ejωn t 1−ζ 2 + a2e

− jωnt 1−ζ 2 ) = Ae−ζωn t sin(ωdt +φ)

ωd =ωn 1−ζ 2 (1.37)

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


40/43

Constants of integration for the underdamped motion case

As before A and ϕ can be calculated from initial conditions (t=0),

€

A =1ωd

(v0 +ζωnx0)2 + (x0ωd)2

φ = tan−1 x0ωd

v0 +ζωnx0

⎛

⎝ ⎜

⎞

⎠ ⎟

0 1 2 3 4 5 -1

-0.5

0

0.5

1

Time (sec)

Dis

plac

emen

t

•  Gives an oscillating response with exponential decay

•  Most natural systems vibrate with and underdamped response

•  See Window 1.5 for details and other representations


41/43

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. m = 49.2 × 10−3 kg, k = 857.8 N/m

ccr = 2 km = 2 49.2 ×10−3 ×857.8 = 12.993 kg/s

ζ = cccr

=0.11 kg/s

12.993 kg/s= 0.0085 ⇒

the motion is underdamped and the bolt will oscillate


42/43

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?


43/43

Solution:

ωn =201

cycless

2π radcycles

= 125.66 rad/s

ωd =125.66 1− .224( )2= 122.467 rad/s

A =0.6 + 0.224( ) 125.66( ) 0( )( )2

+ 0( ) 122.467( )2

122.467= 0.005 m

φ = tan-1 0( ) ωd( )v0 +ζω n 0( )⎛

⎝ ⎜ ⎞

⎠ ⎟ = 0

⇒ x t( )= 0.005e−28.148t sin 122.467t( )


44/46

Use the undamped formula to get maximum acceleration:

A= x02 +

v0

ωn

!

"##

$

%&&

2

, ωn =125.66, v0 = 0.6, x0 = 0

A=v0

ωn

m =0.6ωn

m

max x( ) = −ωn2A = −ωn

2 0.6ωn

!

"##

$

%&& = 0.6( ) 125.66 m/s2( ) = 75.396 m/s2

maximum acceleration = 75.396 m/s2

9.81 m/s2 g = 7.68g's


45/46

Here is a plot of the displacement response versus time


46/46

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 xsin y+ cos xcos y⇒x(t) = Ae−ζωnt sin(ωdt +φ) = e−ζωnt (A1 sinωdt + A2 cosωdt)x(0) = x0 = e

0 (A1 sin(0)+ A2 cos(0))⇒ A2 = x0

x = −ζωne−ζωnt (A1 sinωdt + A2 cosωdt)

+ωde−ζωnt (A1 cosωdt − A2 sinωdt)

v0 = −ζωn (A1 sin0+ x0 cos0)+ωd (A1 cos0− x0 sin0)

⇒ A1 =v0 +ζωnx0

ωd

⇒

x(t) = e−ζωnt v0 +ζωnx0

ωd

sinωdt + x0 cosωdt#

$%

&

'(

chapter1 part 1

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