Formulas Din Mecanica Si Vibratii

8/8/2019 Formulas Din Mecanica Si Vibratii

1/29

Formulas in Mechanical Vibration page 1

Mechanical Vibrations

This is a compilation of useful definitions and formulas for mechanicalvibrations. It is under continuous development.

Lule 2002-09-04

Lars-Erik Lindgren ([email protected]) &Jan-Olov Aidanp ([email protected])


2/29


3/29


Content

1. INTRODUCTION ...............................................................................................................................................5

2. NOTATIONS AND DEFINITIONS .........................................................................................................................53. BASIC PRINCIPLES FOR SOLUTION OF EQUATION OF MOTION............................................................................ 74. SINGLE DEGREE OF FREEDOM SYSTEM ...........................................................................................................9

4. 1 .1 Free Vibration .................................................................................................................................................9

4. 1 .2 Free vibrations of underdamped system, 1. ..............................................................................................9

4. 2 Forced Vibration...................................................................................................................................104. 2 .1 Harmonic force applied on underdamped SDOF system...............................................................................104. 2 .2 Rotating unbalance in underdamped system .................................................................................................114. 2 .3 Harmonic base motion of underdamped system............................................................................................114. 2 .4 Transmissibility for base motion and force excitation...................................................................................124. 2 .5 Shock loading and arbitrary loading applied to a damped SDOF system, Impulse.......................................124. 2 .6 Arbitrary periodic loading applied to a damped SDOF system, Fourier series..............................................13

4. 2 .7 Arbitrary loading applied to an damped SDOF system, Laplace Transform .................................................144. 2 .8 Random loading applied to an damped SDOF system, Fourier Transform ...................................................15

5. MULTIPLE DEGREE OF FREEDOM SYSTEM ....................................................................................................16

5. 1 Free Vibration.......................................................................................................................................165. 2 Forced Vibration...................................................................................................................................18

5. 3 Modal Analysis.....................................................................................................................................19

6. LAGRANGES EQUATIONS ..............................................................................................................................197. CONTINUOUS SYSTEM ...................................................................................................................................20

7. 1 Wave Equation .....................................................................................................................................20

7. 2 Bending vibration of beam ...................................................................................................................218. DAMPING ......................................................................................................................................................22

9. APPENDIX A. LAPLACE TRANSFORMS ...........................................................................................................1910. APPENDIX B. MOMENTS OF INERTIA ...........................................................................................................21

11. APPENDIX C. MATHEMATICAL FORMULAS ..................................................................................................2312. APPENDIX D. BENDING VIBRATION OF FOR BEAM......................................................................... ..............24


4/29


5/29


version 2002-09-06

1. Introduction

Basic assumptions if not stated otherwise are those of linear systems. Thus we assume

small displacements and rotations, linear spring and viscous damping.

It is important to use consistent units. Note that units must be consistent with Newtonssecond law, F=ma. See the Table below for two common choices. Note that radian isalways used for angles. This is a nondimensional quantity.

Table 2.1 Consistent units for mechanical vibrationforce N MPa=N/mm2

mass kg tonne=1000kglength m mm

time s sdensity kg/m3 tonne/mm3

Whether inertia forces are important or not for a design depends on the relation betweenthe frequencies of the loading and the natural frequencies of the structure. If the time ofload application is greater than about three times the natural period of a structure, then the

loading can be specified as being static. Then inertia can be ignored. This is called aquasistatic problem. If the time of load application is less than about half the natural periodof vibration, then it is an impact or shock, i.e. the loading is dynamic.

2. Notations and definitions

The following notations are used if not otherwise stated.

Boldface is used to denote a vector or a matrix. Thus k is a stiffness matrix and F is a force

vector. Subscripts denote components of matrices or vectors. A prime () denotes

derivative w.r.t. to coordinate and a dot (.) means derivative w.r.t .time.

m massJ moment of inertia

k stiffnessc dampingn natural angular frequency

F forceM moment or masst time

angular frequency angle

f frequencyT periodA amplitude

x displacementx0 initial displacement


6/29


version 2002-09-06

v velocity

v0 initial velocitya acceleration

ui eigenvector

vi orthonormal eigenvector

The following abbreviations are usedSDOF - Single Degree Of Freedom

MDOF - Multiple Degree Of FreedomsFEM - Finite Element Method

Definitions

Resonance - when loading frequency equals the natural frequencyx =

lim

T

1

Tx()d

0

T

average or mean value

x2=

lim

T

1

Tx

2()d

0

T

mean square value

x rms = x2

root mean square value (rms-value)

dB = 10 log10x

x ref

2

= 20log10x

xref

decibel


7/29

Formulas in Mechanical Vibration page 7version 2002-09-06

3. Basic principles for solution of equation of motion

Analytical solution of Newtons second law, F=ma, can be performed in several ways.There is no method that is the best for all cases. They are illustrated below for the simple

case of one particle given a constant net force in x-direction. However, quite often energymethods are simple as they reduce vector field problems to scalar problems. This is not

obvious for this simple case. They can also form the basis for numerical procedures.

F

m

Figure 2.1 Mass accelerated by a constant force F in x-direction.

1. Newtons second law solved as an Ordinary Differential Equation (ODE)

200

00

212

2

=)0(and0conditionsinitial

2

tm

Ftvxx

vxxx

ctctm

Fx

Ftxm

++=

=

++=

=

&

&&

)(

)(

A Free Body Diagram is drawn in order to find all forces acting on the body. Acorresponding kinetic diagram can be drawn that corresponds to the left hand side ofNewtons second law. Note that this is a vector equation in the general case.

2. Change of momentum

0

0

0

0

00

=Ft

momentuminchangetheequalsimpulsthe

vtm

Fv

mvtmv

mvtmvFdt

dxmFdt

t

tt

+=

=

=

)(

)(

)( &&


8/29


version 2002-09-06

The velocity is obtained. It must be integrated in order to get displacement. Note that this is

a vector equation in the general case.

3. Energy method

[ ]

( )

200

20

20

02

000

00

2

2

energykineticthechangesworkthe

2

vm

xxFtv

vtvm

xxF

vm

vddt

dvm

dvd

substitutedvmFd

dxmFd

ttxx

xx

+=

=

==

===

=

)()(

)()(

&

&&

The velocity is obtained as function of coordinate. Note that this is a scalar equation also in

the general case. Lagranges equations in chapter is an energy formulation that can generateequations from a scalar equation, the Lagrangian, see chapter 6.

Energy methods are usually used for creating approximate solutions or formulatingapproximate computational methods like FEM.

Numerical procedures are often required for more complex problems, for eg severalunknowns or nonlinearity. Simple analytic models can serve as a first rough estimategiving the basic properties of the design. It may be advantageous to use numerical packages

like Matlab for models that can be limited to some, 3-100, unknowns. Special packages,often Finite Element or Rigid Body Dynamics codes, are used for larger problems.

Sometimes a more complex analytical model can be useful and a symbolic manipulationpackage like Maple can be applied.


9/29


version 2002-09-06

4. Single Degree Of Freedom system

4. 1 .1 Free Vibration

The typical model for SDOF is shown tothe right. There is a large variety ofphysical problems that also can be

modelled as SDOF, for eg torsion ofshaft.The gravity can be accounted forseperately and vibration can be solved as

the displacement x(t) from the staticequilibrium position.

k

c

m

x(t)

m

kn = is the natural frequency of the system. It is the frequency of free

vibration for an undamped system.

ncr mkmc 22 == is called the critical damping.

crc

c= is the damping factor.

4. 1 .2 Free vibrations of underdamped system, 1.

The homogenous solutions is of the formttt

hnnn eeaeax

+=

12

11

22

and

initial conditions give


10/29


tt

n

nt

n

n

fnnn ee

xv

e

xv

x

++

+

++

=1

2

02

01

2

02

0 22

12

1

12

1

.

4. 2 Forced Vibration

The loading cases below ranges fromsingle harmonics, arbitrary periodic,

arbitrary to random loading.Loading via base motion is also included.

m

tFtf

)()( = is the loading per unit

mass.

k

c

m

F(t)

x(t)

4. 2 .1 Harmonic force applied on underdamped SDOF system

The loading is assumed to be F0 cos(t)

n

r

= is ratio of loading frequency and natural frequency.

The steady state solution is xs = Xcos(t )which gives

( ) ( )( )

( ) ( )( )

+=

+= t

rrt

rrk

F

x statics coscos222222

0

2121

and

=

2

1

1

2

r

r tan , which is the phase shift between displacement and load.

It is often convenient to plot the nondimensional amplitude

( ) ( )2220

2

0 21

1

rrf

X

F

Xk n

+

==

The undamped case, =0, can not be obtained from above.


11/29


4. 2 .2 Rotating unbalance in underdamped system

The motion is restrained to occur only in

x direction. This is not shown in thefigure.The mass M includes m as it is the total

mass.This is the same as harmonic loading with

F0=me2.

For eg the nondimensional amplitude is

( ) ( )

( ) ( )222

2

2222

21

21

1

rr

r

me

MX

or

rrme

Xk

+

=

+

=

k

c

M

y(t)

x(t)

m

e

t

The undamped case, =0, can not be obtained from above.The solution can be extended to synchronous whirl with M=m and the force above appliedboth in the x- and y-directions independently.

4. 2 .3 Harmonic base motion of underdamped system

The base motion is assumed to bey = Y sin(t)

r=

nis ratio of loading frequency and

natural frequency.

k

c

m

y(t)

x(t)

The steady state solution is xp = Xsin(t )which gives

( )

( ) ( )

( )

+

+= t

rr

rYxs sin

222

2

21

21

and

( ) ( )

+=

22

31

21

2

rr

r

tan , which is the phase shift.


( )

( ) ( )2222

21

21

rr

r

Y

X

+

+=

The relative motion between base and mass if sometimes important, like for eg in the caseof accelerometer or seismometer. Then we introduce zp= x

py = Zsin(t ) that gives


12/29


( ) ( )( )

+

= t

rr

rYzs sin

222

2

21

and

( )

= 21 1

2

r

r tan , which is the phase shift.


( ) ( )2222

21 rr

r

Y

Z

+

=

The undamped case, =0, can not be obtained from above.

4. 2 .4 Transmissibility for base motion and force excitation

The transmissibility can be defined for the two cases above. One has to be careful andnotice the difference between them even if there are some similarities. Note the differencebetween reduce forces or vibrations.

Transmissibility of force for harmonic load on mass is a measure of how much of theloading on the mass that affects the base. The force on the base is FT. It is

( )

( ) ( )2222

0 21

21

rr

r

F

FTR T

+

+== .

Transmissibility of vibrations of base is a measure on how much the base vibrations is

affecting the mass.

( )

( ) ( )2222

21

21

rr

r

Y

XTR

+

+== .

So in this respect the isolator will do the same job for the two cases. However, if the forceaffecting the mass due to base motion is of interest, then the following should be used.

( )

( ) ( )2222

2

21

21

rr

rr

kY

FTR T

+

+

==

4. 2 .5 Shock loading and arbitrary loading applied to a damped SDOF system, Impulse

The loading is assumed to be F0(t). The application of loading is assumed to be so

short that it gives the system a momentum due to an impulse. Thus the velocity is changedinstantaneously without any change in displacement. Assuming zero initial displacementand zero initial velocity gives

=

t)-(


13/29


( )tem

th dt

d

n

sin)

=1

( is the unit impulse response function of an underdamped

SDOF. It gives x(t) for the system when loaded with an initial unit impulse.

F= mv 0 is the impulse. It is expressed in the velocity increase that can be observed.

The general defintion of impulse if

= dttFF )(The solution above can, with superposition, be used to find the displacement to an arbitraryloading. The superposition or convolution integral is

=t

dthFtx

0

)()()(

The unit impulse response function, h(t), for the system is required. In general, it is more

convenient to use the Laplace Transform to find the motion of the system as the integralmay be quite elaborate.

The solution of the convolution integral for a damped SDOF loaded by an initial step loadis

=

=

2

1

2

0

1

1

11

tan

)cos()( tek

Ftx d

tn

4. 2 .6 Arbitrary periodic loading applied to a damped SDOF system, Fourier series

Any periodic function, F(t), can be represented by an infinite series of the form

( )

T

tnbtnaa

tF

n

TnTn

2

where

2

T

1

0

=

++=

=

)sin()cos()(

and the coefficents are computed by

1,2...=n2

1,2...=n2

2

0

0

0

0

=

=

=

T

Tn

T

Tn

T

dttnwtFT

b

dttnwtFT

a

dttFT

a

)sin()(

)cos()(

)(

Note that 2

0a

is the average force, F .


14/29


The solution of the equation of motion is then written asx(t) = x h + xpwhere the homogenous solution is given in the sections about free vibrations and theparticular solution is written as

xp (t) = x1 + xcn (t) +xsn (t)( )n =1

where the different parts are solutions to the equations below.

)sin(

)cos(

tnbkxxcxm

tnakxxcxm

k

F

k

ax

akxxcxm

Tnsnsnsn

Tncncncn

=++

=++

===++

&&&

&&&

&&&

220

10

111

The solutions to the two latter equations can be constructed from the solution for aharmonic loaded system.

Note that it is the sum of particular and homogenous solutions, i.e. the total solution, that

should fulfil initial conditions. Thus, first put together the general solution and finallyapply initial conditions to find unknown coefficients.

4. 2 .7 Arbitrary loading applied to an damped SDOF system, Laplace Transform

The definition of the Laplace Transform and its properties together with a table ofcomputed transformations are given in Appendix A.Applying the transform to the equation of motion changes the problem of solving an

ordinary differential equation into an algebraic problem as shown below

( ) ( ) [ ]

( )

kcsms

cxxsvmsFX

sFtFLkXxcxsvX

tFkxxcxm

++

+++=

==+

=++

2000

0002 -sXsm

)(

)()(

)(&&&

Finding the Laplace Transform of the load and applying the formula above gives theLaplace Transform of the motion. The inverse Laplace Transform (also from table) gives

[ ])()( sXLtx 1= .

The special case of zero initial displacement and velocity gives)()()( sFsHsF

kcsmsX =

++=

2

1

The function H(s) is called the receptance transfer function (also called compliance oradmittance)

kcsmssF

sXsH

++==

2

1

)(

)()( .


15/29


version 2002-09-06

Other transfer function that are used in vibration measuring are given in the table below.

ResponseMeasurement

Transfer Function Formula Inverse TransferFunction

Acceleration Inertance s2H(s) Apparent massVelocity Mobility sH(s) Impedance

Displacement Receptance H(s) Dynamic stiffness

4. 2 .8 Random loading applied to an damped SDOF system, Fourier Transform

The Fourier Transform is defined in a way similar to the Laplace Transform. It is veryuseful in measurements due to the Fast Fourier Transform (FFT) that makes the calculationvery fast and possible to perform digitally with the Digital Fourier Transform (DFT). It is

not a tool for analysis like the Laplace Transform.

The purpose of the Fourier Transform is to study the amplitude, energy etc as a function of

frequency instead of time.

The mean value, x , is assumed to be zero for random vibration. It can be accounted for

separately as a static mean value added to the motion.

Autocorrelation contains the information of how fast a signal is changing. It is defined as

+=T

xx dttxtxTT

R

0

1)()(

lim)(

It can be seen that R xx (0) = x2

. It can be shown that the Fourier Transform gives the

distribution of energy, as it is related to the amplitude in square. This transform is called

the Power Spectral Density (PSD). It is written as

= dteRS tixxxx

)()(

2

1

The same quantities can be defined for the loading. The relation between the PSD of theload and the motion is

)()()( ffxx SHS2

= .

The mean square value can be computed as

dSx xx

= )(

2

Measuring the input, force, and the output, motion, of the system makes it possible tocompute the transfer function. The system can be described as a black box where the

determination of H() is a way to find out what is inside.

The response computed from measurements will in the ideal case of a damped SDOF be as

in the figure below. The parameter identification can be performed on the responsespectrum.


16/29


10-1

100

10110

-4

10-3

10-2

10-1

100

101

Figure. Magnitude of compliance transfer function versus frequency.

The static displacement gives

kH

1

0

)(

lim

The peak gives the following information

2

2

12

11

21

=

=

kHpeak

npeak

These three equations determines the parameters of the system.

5. Multiple Degree Of Freedom system

5. 1 Free Vibration

The typical model for MDOF is shown below. Most equations are given for 2-DOF

models. The general relations are also valid for any number of degrees of freedom (N).

k1

c1

m1

x1(t)

k2

c2

m2

x2(t)

Figure. 2-DOF model without external load.


17/29


version 2002-09-06

M is mass matrix, C is damping matrix, K is stiffness matrix, X is displacement vector and

F is load vector.

The equation of motion for the 2-DOF model above with no load is

=

++

++

0

0

0

0

2

1

22

221

2

1

22

221

2

1

2

1

x

x

kk

kkk

x

x

cc

ccc

x

x

m

m

&

&

&&

&&

A more general notation valid for any 2-DOF with lumped mass-matrix is

=

+

+

0

0

2

1

2221

2211

2

1

2221

1211

2

1

2221

1211

x

x

kk

kk

x

x

cc

cc

x

x

mm

mm

&

&

&&

&&

This is a coupled system of second order ordinary differential equations.

Determination of eigenmodes and natural frequencies

Harmonic motion is assumed which gives XX 2=&& . This is applied to the equations ofmotion for the undamped system

0KXXM =+&&

The requirement for a non-trivial solution, X0, requires det(K 2M) = 0 , which is

called the characteristic equation.

Solving by setting in a natural frequency gives the shape of the corresponding eigenmodes.The scaling of this eigenvector is arbitrary.

For the 2-DOF model above this gives the characteristic equation

( )( ) 0212

212214

21 =+++ kkkkmkmmm

The solution gives the two natural frequencies

+

++

+=

21

21

2

2

2

1

21

2

2

1

2122

21 4

2

1

mm

kk

m

k

m

kk

m

k

m

kk

This gives the ratio between the amplitudes of the corresponding eigenvectors by solving

the original system of equations with an eigenvalue inserted.

1,2=i

1

222

2

2

1

=

=

ii

i

mk

k

X

X

u

Determination of free vibrations for given initial conditions, undamped systemThe initial conditions are

0

0

VX

XX

=

=

&

Writing the motion as a linear combination of modal vibrations gives a motion of the form

+= iiii tA u)sin(X

Matching this with the initial conditions gives the solution for the unknown coefficients.


18/29


version 2002-09-06

5. 2 Forced Vibration

Undamped 2-DOF problem with load on one mass, vibration absorber, is shown below.

x1(t)

F1(t)k2k1

c1

m1

c2

m2

x2(t)

Figure. Vibration absorber, m2, on a loaded primary mass m1.

The equation of motion for the 2-DOF model above, but without damping, with is

=

++

00

0 1

2

1

22

221

2

1

2

1 F

x

x

kk

kkk

x

x

m

m

&&

&&

The loading is assumed to be harmonic F=F0sin(t)

Assuming )sin( tX

X

x

x

=

2

1

2

1and inserting into equation of motion give

( )( )( )

+=

20

2220

22

222

21212

1 1

kF

mkF

kmkmkkX

X

We introduce the following variables1

1211

m

k= and

2

2222

m

k= .

Then we can define the design variables1

2

m

m= and

11

22

= .

The natural frequencies of the system can be written in these variables.

( ) ( ) ( )

++++=

2422

22

22

2

2

22

1

1121112

1


19/29


version 2002-09-06

5. 3 Modal Analysis

It is possible to uncouple the equation of motions by analysing the vibration in terms of the

participating natural modes. Representing the motion as a linear combination of theeigenvectors, mode shapes, can decouple the undamped equations of motion. Proportionaldamping must be assumed in the case of damping. See chapter ??? about damping.

The modal analysis is presented in nondimensional form but can be performed without thisnormalization of the equations.

Nondimensional equations of motion is created by the transformation

XMQQMX // 2121 inversethewith == . Inserting into the equations of motion and

premultiplying with 21/M gives

F~

QK~

QI

FMQKMMQMMM /////

=

=

+

+ 2121212121

&&

&&

I is the unity matrix and K is the spectral matrix.The normalized eigenvectors, v, to the nondimensional form are set as columns in a matrix

[ ]Nv.vvP 21=

We apply one more transformation QPRPRQ T== inversethewith . Inserting this into

the nondimensional equations of motion and premultiplying with PT gives

F~

RRI

F~

PKPRPRPP

=

=

+

+

&&

&& TTT

The equations looks like below

=

+

NNNN f

f

f

r

r

r

r

r

r

~.

~

~

.

.

....

.

.

.

.

....

.

.

21

2

1

2

22

2

12

1

00

00

00

100

010

001

&&

&&

&&

The uncoupled equations, modal equations, can be solved as SDOF problems when given

transformed intial conditions. The modal coordinates, r, must be transformed back to X

using the inverse transformations given above,

Damping can be included to give damped SDOF equations. Then damping is assumed to be

proportional. Modal dampfactors may be obtained from measurements also.

6. Lagranges equations

Is an alternative to set up the equations of motion that can be easier than those methodsdescribed in the introduction. It can generate the N equations based on scalar functions for

kinetic, T, and potential energy, U.

N independent generalized coordinates, qi, are required to define the motion uniquely for a

N-DOF problem. More coordinates can be used if it is convenient for the problem but theyshould then be followed by constraint equations. The same number of constraints are then

required as the number of superfluous coordinates.


20/29


version 2002-09-06

Corresponding generalized forces, Qi, are defined. They are related to the coordinates via

the change in virtual work, W, that is produced for a virtual displacement.

ii

q

WQ

=

They are zero for a conservative system. System is defined as the structure including

applied forces.

The Lagrange formulation states that the equation of motion can be derived from

1,2....N=iiiii

Qq

U

q

T

q

T

dt

d=+

&

7. Continuous system

7. 1 Wave Equation

The one-dimensional wave equation is

2

2

2

22

t

w

x

wc

=

It governs several physical systems as shown in the table below.

Table. Physical problems for one-dimensional wave equation.

Problem type Variable w Other variables Wave speed c

Free lateral

vibrations of string

lateral displacement is tension in string

is density (kg/m)

=c

Free longitudinalvibrations in bar

axial displacement E is Youngs modulus

is density (kg/m3)

Ec =

Free torsional

vibrations in massiveshaft

angular rotation G is shear modulus

is density (kg/m3)

Gc =

It is assumed that w(x,t)=X(x)T(t).

The solution of the spatial equation is outlined below.

022

2

=+ )()(

xXx

xX

The general solution is X=asin(x)+bcos(x). Applying boundary conditions at the endsx=0 and x=L and looking for non-trivial solutions for the coefficients gives the

characteristic equation for the eigenvalue problem.

An infinite number of eigenmodes are then found.Xn (x) = an cos(nx )+ b nsin(nx) n = 1, 2...

It can be seen from the temporal equation that the corresponding natural frequencies will be

n = nc


21/29


version 2002-09-06

The solution of the temporal equation is outlined below.

0222

2

=+ )()(

tTct

tT

The general solution is Tn=Ansin(nct)+Bncos(nct). This gives infinte number of

independent solutions and we will have

( )( )

=

++=1n

nnnnnnnn xbxactBctAtxw )cos()sin()cos()sin(),(

Applying initial conditions gives the remaining unknown coefficients.

7. 2 Bending vibration of beam

The equation of motion for a free vibrating Bernoulli beam is

A

EIc

x

wc

t

w

=

=+ 04

42

2

2

Assuming separation of variables gives a temporal equation that together with the fourboundary conditions defines the eigenvalue problem. The general solution to this is

EI

A

c

xaxaxaxaxX

2

2

24

4321

==

+++= )cosh()sinh()cos()sin()(

Solution for some modes and different boundary conditions are given in Appendix C.


22/29


version 2002-09-06

8. Damping

Damping are nonconservative forces that dissipates energy. Linear, viscous damping isdefined as usually assumed. Equivalent damping for other cases are defined as the damping

that should be used in the linear, viscous damping model in order to get the same energyloss per cycle. Different sources of damping are given in the table below.

Table. Source of damping.

Name ),( xxFd & ceq Source

Linear, viscous damping xc & c Slow fluid

Air damping 2xxa &&)sgn(

3

8 Xa Fast fluid

Coulumb damping )sgn(x&

X

4 Sliding friction

Displacement squared damping 2xxd )sgn( &3

4dX Materialdamping

Solid damping xxb &&)sgn(

b2 Materialdamping

Proportional damping is defined as being proportional to stiffness and mass. In matrix form

for a MDOF problem C = M+K. The coefficients are not the same as in the table

above.

Damping can be measured several ways. One option is to compute the logaritmicdecrement which is the natural logaritm of the amplitude of any two successive amplitudes.

21

2

=

+=

)(

)(ln

Ttx

tx

Thus the dampfactor can be computed

224

+=

9. 9. Appendix A. Laplace transforms

Definition of Laplace transform of a function f(t)

[ ]

==

0

dtetfsFtfLst)()()(

This gives

[ ][ ]

.

)()()()(

)()()(

etc

fsfsFstfL

fssFtfL

00

0

2 &&&

&

=

=

The inverse transform is written as


23/29


version 2002-09-06

[ ] )()( tfsFL =1

The Table A.1 below can be used for computing transforms and inverse transforms.

Table A.1 Laplace transforms for functions with initial zero conditions and t>0.

Eq. # F(s) f(t)

(1) (t)

(2)s

1 H(t)

(3) ,...),( 211

=ns

n

)!( 1

1

n

tn

(4)))(( bsas ++

1 ( )btat eeab

1

(5))( 22

+ssin t

(6))( 22 +s

s cos t

(7))( 22

1

+ss( )t

cos1

12

(8))( 22 2

1

++ ss1


24/29


version 2002-09-06

(16)1

(s2 +2 )2( )ttt

cossin

32

1

(17)222 )( +s

st

t

sin

2

(18)222

22

)(

)(

+

s

s t cos t

(19)

))((

)(22

221

2

22

21

++

ss

tt 11

22

11

sinsin

(20)

))((

)(22

221

2

22

21

++

ss

s tt 12 coscos

(21)22

++ )( as

teat sin

(22)22 ++

+

)( as

as teat cos


25/29


version 2002-09-06

10. 10. Appendix B. Moments of inertia

Definition of mass moments of inertia

= dVrI 2 ,

where r is the orthogonal distance from the axis of rotation. It is a measure of resistance torotational acceleration of a body.

Definition of radius of gyration

m

Ik=

Transfer of axesThe moment of inertia of a body about a centroidal axis can be found from that of a

parallell axis through the mass center

I= I+ md2

,where I is the moment of inertia about a axis through the mass center and d is the distancebetween the axes.

Appended excerpt at the end of this collection of formulas Table B.1 gives the massmoment of inertia for some bodies. Note that these can be combined to form those of other

bodies.

Table B.1 Properties of homogeneous solids

Body Mass

center

Mass moments of inertia

1. Circular cylinder

x

z

L

r

-

2

22

2

1

3

1

4

1

mrI

mLmrI

zz

xx

=

+=

Semicylinder

3

4rx =

2

22

2

1

3

1

4

1

mrI

mLmrI

zz

xx

=

+=


26/29


version 2002-09-06

x

z

L

r

y

Cylindrical shell

z

x

L

r

m=2rh

h=thicknessr= mean radius

Ixx=

1

2mr2 +

1

3mL 2

Izz= mr2

Half cylindrical shell

y

z

L

rm=rhh=thicknessr=mean radius

x

rx

2=

2

22

41

3

1

2

1

mrI

mLmrI

zz

yy

=

+=

Sphere - 2

5

2mrIzz =

Rectangular parallellepiped - ( )

( )

( )22

22

22

3

1

3

1

3

1

HBmI

LBmI

LHmI

zz

yy

xx

+=

+=

+=


27/29


version 2002-09-06

x

yz

L

H

B

11. Appendix C. Mathematical formulas

Some formulas often used, but easily forgotten, in this context are given here.

Quadratic equations

ax2+ bx + c = 0 x =

b b2 4ac

2a

Inverse of 2 by 2 matrix

a b

c d

1

=1

ad bc

d b

c a

TrigonometricsDsin(x)=cos(x)

Dcos(x)=-sin(x)

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

cos(x)=-sin(x-90)

sin(x)=cos(x-90)acos(x )+ bsin(x) = rcos(x )

asin(x) + bcos(x) = rsin(x + )

r= a 2 + b2 , = arctan ba

-2Im(a)=B,2 )Re(

)cos()sin(

aA

tBtAeaaetiti

=

+=+


28/29


version 2002-09-06

The Heaviside function or step function is defined as

=

t1


29/29

12. 12. Appendix D. Bending vibration of beam

The solutions for the equation for beam bending vibration in Chapter 7.

Frequencies and mode shapes for some beam configurations. Beam length is L.Configuration Weighted frequencies (L)

and characteristic equation.

Mode shape

x

free-free

0 (rigid body mode)

4.73004074

7.85320462

10.9956078

14.1371655

17.2787597( )

2

12 +n n>5

( )xx

xx

sinsinh

coscosh

+

+ 0.9825

1.0008

0.9999

1.0000

0.9999

1 for n>5

x

clamped-free

1.875104074.69409113

7.85475744

10.99554073

14.13716839

( )2

12 n n> 5

( )xx

xx

sinsinh

coscosh

0.73411.0185

0.9992

1.0000

1.0000

1 for n>5

x

clamped-pinned

LLL = coshcos

3.92660231

7.06858275

10.21017612

13.35176878

16.49336143( )

4

14 +n n> 5

( )xx

xx

sinsinh

coscosh

1.0008

1 for n>1

x

clamped-sliding

LL tanhtan =

2.36502037

5.49780392

8.63937983

11.78097245

14.92256510( )14 n n> 5

( )xx

xx

sinsinh

coscosh

0.9825

1 for n>1

x

clamped-clamped

0=+ LL tanhtan

4.73004074

7.8532046210.9956079

14.1371655

17.2787597

( )

2

12 +n n> 5

( )xxxx

sinsinh

coscosh

0.9825

1.0008

1 for n>2

x

pinned-pinned

1=LL coshcos

n

0=L

xn sin -

Documents

Formulas Din Mecanica Si Vibratii