basic dyn

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BASIC DYNAMICS 1

NEi Nastran and Femap

DYNAMICS ANALYSIS

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BASIC DYNAMICS 2

REVIEW OF FUNDAMENTALS

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BASIC DYNAMICS 3

Page

SINGLE DOF SYSTEM 1-3

CREATING A SPRING ELEMENT 1-5

CREATING A MASS ELEMENT 1-8

SETTING UP A MODAL ANALYSIS 1-12

SPRING ELEMENT TYPES 1-18

UNDAMPED FREE VIBRATIONS SDOF SYSTEM 1-25

DAMPED FREE VIBRATION SDOF 1-30 DAMPING WITH FORCED VIBRATION 1-34

TEXT REFERENCES ON DYNAMIC ANALYSIS 1-44

TABLE OF CONTENTS

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BASIC DYNAMICS 4

In this section we introduce the basics of Dynamic Analysis by considering aSingle Degree of Freedom (SDOF) problem

Initially a free vibration model is used to describe the natural frequency

Damping is then introduced and the concept of critical damping and theundamped solution is shown

Finally a Forcing function is applied and the response of the SDOF is explored interms of time dependency and frequency dependency and compared to theterms found in the equations of motion

SINGLE DOF SYSTEM

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BASIC DYNAMICS 6

SINGLE DOF SYSTEM

Creating a spring element : Step 2

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BASIC DYNAMICS 7

SINGLE DOF SYSTEM

Creating a spring element : Step 3

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BASIC DYNAMICS 8

SINGLE DOF SYSTEM

Creating a mass element : Step 1

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BASIC DYNAMICS 9

SINGLE DOF SYSTEM


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BASIC DYNAMICS 10

SINGLE DOF SYSTEM


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BASIC DYNAMICS 11

SINGLE DOF SYSTEM

Creating constraints: Step 1

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BASIC DYNAMICS 12

SINGLE DOF SYSTEM

Setting up Analysis: Step 1

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BASIC DYNAMICS 13

SINGLE DOF SYSTEM


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BASIC DYNAMICS 14

SINGLE DOF SYSTEM


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BASIC DYNAMICS 16

SINGLE DOF SYSTEM

Data defined in .NAS file: SOL, Case Control, Parameters and EIGRL

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BASIC DYNAMICS 17

SINGLE DOF SYSTEM

Data defined in .NAS file: CELAS2, CONM2, Parameters and EIGRL

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BASIC DYNAMICS 19

SINGLE DOF SYSTEM

If Spring/Damper with BUSH formulation is used in Femap a CBUSH element results:

Spring types

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BASIC DYNAMICS 20

SINGLE DOF SYSTEM

Spring types

Set up in Femap:

CELAS2 DOF Spring 1 DOF spring stiffness with no Property in Femap

CROD Spring/Damper 1 DOF element axial stiffness, links to PROD and MAT1 to define AE/L

CBUSH Spring/Damper 6 DOF element linked to PBUSH

with BUSH

formulation

Manually input in Nastran:

CELAS1 1 DOF spring element with links to PELAS property definition

Format:

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BASIC DYNAMICS 21

SINGLE DOF SYSTEM

Spring types

In general CBUSH is recommended in most cases. It will be used in more advanced

analysis later in the course.

The main advantages are:

All 6 DOF in a single element

Extendable to Frequency dependent and nonlinear

Arbitrary elastomeric center

No errors due to large rotation and other grounding effects Special purpose 1D version for shock mounts

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BASIC DYNAMICS 22

SINGLE DOF SYSTEM (Cont.)

SDOF System theory

m = mass (inertia)

b = damping (energy dissipation)

k = stiffness (restoring force)

p = applied force

u = displacement of mass

= velocity of mass

= acceleration of mass

m

k b

p(t)u(t)

u&&

u&

analysislinearintimewithconstantareand,

timeinvarycaninputloadtimeinvaryandresponses

bkm

puu

&&&

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BASIC DYNAMICS 23

The equation of motion is:

)()()()( tptkutubtum =++ &&&


Inertia ForceDamping Force

Stiffness Force Applied Force

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BASIC DYNAMICS 24

In undamped, free vibration analysis, the SDOF equationof motion reduces to:

Assume a solution of the form:

This form defines the response as being HARMONIC,combinations of sine and cosine shape responses with aresonant frequency of:

0)()( =+ tkutum&&

tBtAtu nn cossin)( +=

n


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BASIC DYNAMICS 25

UNDAMPED FREE VIBRATION SDOF SYSTEM

For a SDOF system the resonant, or natural frequency, is givenby:

We can solve for the constants A and B:

m

kn=

n

nn

nnnn

n

tuA

tBt

tBtAtu

u(tBtt

)0(,0

sincos)(

0,0

==

==

=

====

&

&

thus0)sin(When

:solutionatingDifferenti

)thus0)sin(When

tutu

tunn

n

cos)0(sin)0(

)( += &

UNDAMPED FREE VIBRATION SDOF SYSTEM

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BASIC DYNAMICS 26

UNDAMPED FREE VIBRATION SDOF SYSTEM(Cont.)

The response of the Spring will be harmonic, but the actual form of theresponse through time will be affected by the initial conditions:

If there is no response; no initial disturbanceor velocity

If response is a sine function magnitude

If response is a cosine function (180

phase change), magnitude

If response is phase and magnitudedependent on the initial values

0)0(and0)0( == uu &

0)0(and0)0( = uu &

n

u

0&

0)0(and0)0( = uu &

0u

0)0(and0)0( uu &

Depends on input velocity

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BASIC DYNAMICS 27

SINGLE DOF SYSTEM UNDAMPED FREE VIBRATIONS

The graph is from a transient analysis of a spring mass system with Initial

velocity conditions only

Time (Seconds)

Disp.

k= 100

m= 1

T = 1/f = 0.63 secs

Hz59.12/f

rad/s10

==

==

n

nmk

10=u&T

Amp

1.0/Amp 0 == nu&

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BASIC DYNAMICS 28


The results we saw from NEi Nastran had the correct frequency, but how do weget the displacement value?

n

u0

&

0u

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BASIC DYNAMICS 29


The displacement is indeterminate in an Eigenvalue Analysis

No loading or initial conditions are applied

The mathematical method balances Inertia energy and Stiffnessenergy for a particular unique (or eigen) value solution

It is finding a characteristic of the system, not a response

The displacement is an arbitrary scaling, the NEi Nastran solution it is

(1.0)

We will see later that this is an eigenvector

DAMPED FREE VIBRATION SDOF

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BASIC DYNAMICS 30

DAMPED FREE VIBRATION SDOF

If viscous damping is assumed, the equation of motion becomes:

There are 3 types of solution to this, defined as:

Critically Damped

Overdamped

Underdamped

A swing door with a dashpot closing mechanism is a good analogy

If the door oscillates through the closed position it is underdamped

If it creeps slowly to the closed position it is overdamped.

If it closes in the minimum possible time, with no overswing, it is

critically damped.

0)()()( =++ tkutubtum &&&

DAMPED FREE VIBRATION SDOF (C t )

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BASIC DYNAMICS 31

DAMPED FREE VIBRATION SDOF (Cont.)

For the critically damped case, there is no oscillation, just adecay from the initial conditions:

The damping in this case is defined as:

A system is overdamped when b > bcr

We are generally only interested in the final case - underdamped

ncr

mkmbb 22 ===

mbteBtAtu 2/)()( +=


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BASIC DYNAMICS 32


For the underdamped case b < bcr and the solution is the form:

represents the Damped natural frequency of the system

is called the Critical damping ratio and is defined by:

In most analyses is less than .1 (10%) so

)cossin()(

2/

tBtAetu ddmbt

+=

d

2

1 = nd

crb

b=

nd


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BASIC DYNAMICS 33

The graph is from a transient analysis of the previous spring mass system withdamping applied

Frequency and

period as before

Amplitude is afunction of damping

2%Damping

5% Damping

Time

Disp.


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BASIC DYNAMICS 34

DAMPING WITH FORCED VIBRATION

We now apply a harmonic forcing function: note that is the DRIVING or INPUT frequency

The equation of motion becomes

The solution consists of two terms: The initial response, due to initial conditions which decays rapidly in the presence of

damping

The steady-state response as shown:

This equation is described on the next page

tp sin

tptkutubtum sin)()()( =++ &&&

22

2

2

)/2()1(

)sin(/)(

n

n

tkptu

+

+=

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BASIC DYNAMICS 36

is defined as a phase lead in Nastran :

2

2

1

1

/2tan

n

n

=

DAMPING WITH FORCED VIBRATION (Cont.)

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BASIC DYNAMICS 37

Summary:

For

Magnification factor 1 (static solution)

Phase angle 360 (response is in phase with the force)

For

Magnification factor 0 (no response)

Phase angle 180 (response has opposite sign of force)

For

Magnification factor 1/2

Phase angle 270

1

n

>

1

n


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BASIC DYNAMICS 38

To explore the response of our spring mass system to theforcing function we will use a Frequency ResponseAnalysis

This method allows us to compare the response of thespring with the input force applied to the spring over a wide

range of input frequencies It is more convenient in this case than running multiple

Transient Analyses, each with different input frequencies

We will apply the input load as 1 unit of force over afrequency range from .1 Hz to 5 Hz

Damping is 1% of Critical


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BASIC DYNAMICS 39

Magnification Factor = 1/2 = 1/G = 50

Static Response = p/k = .01

Peak Response = .5 at 1.59 Hz

Note:

Use of a Log scale helps identify low

order response

Displacement

Frequency (Hz)


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BASIC DYNAMICS 40

There are many important factors in setting up a Frequency Response Analysisand we will cover these in a later section

For now, note the response is as predicted by the equation of motion

At 0 Hz result is p/k

At 1.59 Hz result is p/k factored by Dynamic Magnification

At 5 Hz result is low and becoming insignificant

The Phase change is shown here:

In phase up to 1.59 Hz

Out of phase180 Degrees after 1 .59 Hz


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BASIC DYNAMICS 41

We now try a Transient analysis with a unit force applied to the spring at 1.59

Hz

Again damping of 1% Critical is applied

The result is shown on the next page:

The response takes around 32 seconds to reach a steady-state solution

After this time the displacement response magnitude stays constant at .45

units

The theoretical value of .5 is not reached due to numerical inaccuracy (seelater) and the difficulty of hitting the sharp peak


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BASIC DYNAMICS 42

Transient analysis with a unit force applied to the spring at 1.59 Hz

Displacement

Time


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BASIC DYNAMICS 43

If we plot input and output at the steady-state period, we can see that the input signal isnot very accurate hence our problem finding the exact magnification factor

We can also see the phasing between input and output is around 90 degrees as expected

at resonance

Lead = 0.18 sec (approx)=103 degrees (approx 90)

T input = 1/1.59 = 0.629 sec

input

output


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BASIC DYNAMICS 44

TEXT REFERENCES ON DYNAMIC ANALYSIS

1. W. C. Hurty and M. F. Rubinstein, Dynamics of Structures, Prentice-Hall, 1964.

2. R. W. Clough and J. Penzien, Dynamics of Structures, McGraw-Hill, 1975.

3. S. Timoshenko, D. H. Young, and W. Weaver, Jr., Vibration Problems in Engineering,4th Ed., John Wiley & Sons, 1974.

4. K. J. Bathe and E. L. Wilson, Numerical Methods in Finite Element Analysis, Prentice-Hall, 1976.

5. J. S. Przemieniecki, Theory of Matrix Structural Analysis, McGraw-Hill, 1968.

6. C. M. Harris and C. E. Crede, Shock and Vibration Handbook, 2nd Ed., McGraw-Hill,1976.

7. L. Meirovitch, Analytical Methods in Vibrations, MacMillan, 1967.8. L. Meirovitch, Elements of Vibration Analysis, McGraw-Hill, 1975.

9. M. Paz, Structural Dynamics Theory and Computation, Prentice-Hall, 1981.

10. W. T. Thomson, Theory of Vibrations with Applications, Prentice-Hall, 1981.

11. R. R. Craig, Structural Dynamics: An Introduction to Computer Methods, John Wiley

& Sons, 1981.12. S. H. Crandall and W. D. Mark, Random Vibration in Mechanical Systems, Academic

Press, 1963.

13. J. S. Bendat and A. G. Piersel, Random Data Analysis and Measurement Techniques,2nd Ed., John Wiley & Sons, 1986.

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TABLE OF CONTENTS

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BASIC DYNAMICS 46

Page

2 DOF EQUATIONS OF MOTION 1-47

MULTI DEGREES OF FREEDOM 1-73

LANCZOS METHOD 1-83 MASS REPRESENTATION 1-84

RIGID BODY MOTION 1-86

WHY CALCULATE NORMAL MODES? 1-87

ACCURACY IN NORMAL MODES 1-89

NORMAL MODES CHECK LIST 1-101

Normal Modes Analysis

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BASIC DYNAMICS 47

Define the following problem in Femap and solve in NEi Nastran

K K K

DOF: 1 2 3 4

2MM

K = 1000 lbf/in

M = 20 lb

OVERVIEW

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BASIC DYNAMICS 48

In the previous section we looked at a SDOF problemof a spring mass system.

In this section we will look at Normal Modes analysisof Multi Degree of Freedom problems

The steps we will follow are:

Building a 2 DOF equation of motion using engineeringapproach.

Summarizing some important ideas about Normal Modesthat emerge.

Setting the same problem using a Matrix approach.

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2 DOF EQUATION OF MOTION

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BASIC DYNAMICS 50

We assume the motion ofx1 andx2 is harmonic so

We want to find what the frequency is, and the amplitudes. Now

Then putting the harmonic terms into the free body equations.

For the 1st mass: So

For the 2nd mass:

so

tx sin11

=tAx sin

22=

This means they vibrate at the same frequency but havedifferent amplitudes A.

tx sin1

2

1 =&& tAx sin

2

2

2 =&&

( )ttAktkAtA sinsinsinsin1211

2 +=

( ) 02 212 = kAAMk

( ) tkAttkt sinsinsinsin22212

2 +=

022 22

1 =+ AMkkA


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BASIC DYNAMICS 51

If we assemble these two equations in matrix form,we have:

So we have 3 unknowns; 2 and the pair of

amplitudes

( ) ( )

=

00

22

2

1

2

2

AA

MkkkMk

2

1

A

A


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BASIC DYNAMICS 52

We can solve this by using the determinant of theabove equation, letting 2=.

Two roots of the equation are found as 1 and 2. These roots are called Eigenvalues.

m

k634.0

2

11 ==

mk366.2

2

22 ==

So the two frequencies where the inertia and elastic terms

balance are1 and 2.


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BASIC DYNAMICS 53

The amplitudes are investigated by substituting backinto the equations of motion.

In turns out we can only solve for the ratio of theamplitudes.

This is an important physical point in the analysis ofnormal modes. We do not know the absoluteamplitudes, only relative amplitudes.

731.0

1

2

1 =AA 73.2

2

2

1 =AA


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BASIC DYNAMICS 54

If we arbitrarily call A2 = 1.00, then we can expressthe relative amplitudes as Mode Shapes or

Eigenvectors.

=

000.1

731.01

2

1

A

A

=

00.1

73.22

2

1

A

A

0.731 1.000 -2.731 1.000

Mode 1 Mode 2

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BASIC DYNAMICS 58

The equation of motion in matrix form is:

If we substitute in

And

Then

So

[ ]{ } [ ]{ } 0=+ xKx&&

{ } { } tiex =This means we have a mode shape,

{} , which varies sinusoidally with afrequency .

{ } { } tiex 2=&&

[ ]{ } [ ]{ } 02

=+ K

[ ] [ ]( ){ } 02 = KThis means we can find a mode

shape, {}, and frequency wherethe inertia terms and elastic termsbalance

{ } 020

01

21

122 =

mk The Eigenvalue problem

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BASIC DYNAMICS 60

So at , the motion is defined by:

is in balance at this first resonant or natural frequency.

And at , the motion is defined by:

is also in balance at this second resonant or naturalfrequency

m

k634.0

1= { }

=000.1

731.01

{ }

=000.1

731.22

m

k366.2

2=


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BASIC DYNAMICS 61

Let us add some values in and check out the numbers:

Let k = 1000 units of force / length

Let m = 20 units of mass

Then

Notice the conversion of Frequency from Radians/s to Cycles/s(Hertz)

Hzs

rads

m

k896.0629.5634.0

1 ===

Hzs

radsm

k731.1875.10366.2

2 ===

2

f=


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BASIC DYNAMICS 62

If we load this model with a time dependant set offorces at DOF 2 and 3 we will get a displacementresponse which is a combination of the two mode

shapes we calculated.

So

in our case

The scaling factors i for each mode shape i arecalled the Modal Displacements. We will return to

this when we apply loading in later sections.

( ){ } { } (t)tx in

i

i=

=1

( ){ } )()(

2211 tttx

+=

0.731 1.000 -2.731 1.000

1(t) 2(t)+( ){ }=tx


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BASIC DYNAMICS 63

Now let us model the system using NEi Nastran

K K K

DOF: 1 2 3 4

2MM

K = 1000 lbf/in

M = 20 lb

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The NEi Nastran Solution Case Control


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BASIC DYNAMICS 66

The NEi Nastran Solution , Case Control,

Parameter and EIGRL definition:

The NEi Nastran results showing first and second


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BASIC DYNAMICS 67

The NEi Nastran results showing first and second

natural frequencies


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BASIC DYNAMICS 68

The values agree with those we calculated earlier.The meaning of the Generalized Mass and Stiffnesswill be discussed in the next few pages.

Hzsradm

k896.0/629.5796.01 ===

Hzsradm

k731.1/875.10538.1

2

===

The NEi Nastran mode shape results:2 DOF EQUATION OF MOTION

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BASIC DYNAMICS 69

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BASIC DYNAMICS 71

Mass normalization is a useful way of normalizing anEigenvector because it can be thought of as a

universal standard. We scale {} so that for each mode:

If we wish to compare modes between differentanalyses, or even to test data, then it becomesmeaningful to compare the mass normalizedeigenvectors as:

{ }[ ]{ } [ ]IMT =

{ }[ ]{ } 0.1=iT

i M

{ }[ ]{ }21 MODELT

MODEL M

{ }[ ]{ }TESTT

MODEL M

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EXTENDING TO MULTI DOF PROBLEMS

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BASIC DYNAMICS73

Normal Modes of Multi Degree of Freedom (MDOF) Set up the example in Femap

Use mesh size of 1 inch

EXTENDING TO MULTI DOF PROBLEMS

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BASIC DYNAMICS74

Beam Section properties:

EXTENDING TO MULTI DOF PROBLEMS (Cont.)

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BASIC DYNAMICS75

Analysis Results:

First 10 Natural Frequencies range from 166 Hz to 2764 Hz

Generalized Mass is 1.0 in all modes due to Mass Normalization


Analysis Results:

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BASIC DYNAMICS76

It is very important to identify each mode by shape as well as by frequency. Only bydoing both will an unambiguous definition of the response be made.

For example the frequencies could easily shift to switch Mode 9 and 10. Only by

description or plot can we confirm which is which

Mode Frequency Description

1 166.75 xz plane bend 1

2 409.17 xy plane bend 1







9 2761.51 axial


XY plane Modes

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BASIC DYNAMICS 77

XZ Plane Modes

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BASIC DYNAMICS 78

Axial Mode

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BASIC DYNAMICS 79


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BASIC DYNAMICS 80

Remember, we defined the contribution of each mode as the modaldisplacement:

So

In the 2 DOF case

For the beam example, it may be possible to represent the response toloading in the XZ plane using the first three modes. The assumption is thatthe higher modes do not contribute significantly to the solution.

This is a significant advantage of modal methods, the response of thebeam, {x(t)}, has 222 physical degrees of freedom.

But we can represent the response by 3 Modal DOF, i

EXTREME CARE must be taken when assuming which modes contribute andwe will discuss this more in later sections.

As a taster consider the following over page

( ){ } { } ii i

tx =

=nto1

( ){ } 2211 +=tx


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BASIC DYNAMICS 81

Possible Problems with 3 modes only: If we have loading in the XY plane, the response will be represented by first

order bending and will therefore be limited in accuracy

If the XZ loading excites at multiple inputs as shown, then the first twomodes may not represent the response, and the third order mode may be

needed


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BASIC DYNAMICS 82

Repeat the example with different element sizes:

mesh size = 0.5 inch mesh size = 4 inch mesh size = 12 inch

Mode Frequency Description Frequency Description Frequency Description

12

3

4

5

67

8

10

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RIGID BODY MODES

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BASIC DYNAMICS 86

For every DOF in which a structure is not totally constrained, itallows a Rigid Body Mode (stress-free mode) or a mechanism.

There should be six Rigid Body Modes.

The natural frequency of each Rigid Body Mode should be zero.

A later section is devoted to discussion of Rigid Body Modes

WHY CALCULATE NORMAL MODES

We now consider some reasons to compute natural frequencies

d l d f

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BASIC DYNAMICS 87

and normal modes of structures To assess the dynamic characteristics of a structure.

For example, if a structure is going to be subject to rotational or cyclicloading input, to avoid excessive vibrations, it might be necessary tosee if the frequency of the input is close to one of the naturalfrequencies of the structure.

Passing blade frequencies of a helicopter

Rotational speed of an automobile wheel

Rotational speed of a lathe

Vortex shedding or flutter of bridge and deck structures

Assess the possible dynamic amplification of the loads. If a structure is loaded near a natural frequency with an input that

matches that frequency then the dynamic amplification can besignificant for a lightly damped structure, perhaps being an order ofmagnitude higher than an equivalent static loading

Dynamic response of aircraft structure due to landing loadscan exceed static loading

Dynamic response of Tacoma Narrows bridge, runawayloading

WHY CALCULATE NORMAL MODES (Cont.)

Use the Modal Data (natural frequencies and mode shapes) in a

b t d i l i

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BASIC DYNAMICS 88

subsequent dynamic analysis We will see later that we have a class of transient and frequency

response analysis methods that use modal techniques, usingModal data.

Assess requirements of subsequent dynamic analysis For Transient response, calculate time steps based on the highest

frequency of interest

For Frequency response, calculate the range of frequencies ofinterest

Guide the experimental analysis of structures Identify optimum location of accelerometers, etc.

Avoid overstressing of components

Evaluate the effect of design changes A normal modes analysis will give a clear indication of of frequency

shifts, changes in mode shapes to allow an early judgment oneffect of design changes to be made

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MESH DENSITY

Mesh Density

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BASIC DYNAMICS 90

Mesh Density The mesh must be fine enough to permit a representation of the of the

highest mode considered

In the case of the beam, we assumed the second order mode wassufficient. The mesh is adequate for this.

However, if the higher order mode shown is required, then the mesh isinadequate.

NORMAL MODES ANALYSIS WITH VARIOUS MESHSIZE

Rectangular plate.

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BASIC DYNAMICS 91

The coarse mesh (2 x 1)is unable to represent any higher order bending or torsionalmodes, so it drops these and finds higher frequency in plane extensional andshearing modes

Mode f Mesh 1a (10 x 4)

Description

Mode f Mesh 1b (2 x 1)

Description

Mode f Mesh 1c (50 x 20)

Description1 133.1 1st order Bending 1 120.1 1st Order Bending 1 133.6 1st order Bending

2 348.7 1st Order Torsion 2 395.7 1st Order Torsion 2 689.6 1st Order Torsion

3 821.4 2nd Order Bending 3 624.5 2nd Order Bending 3 832.8 2nd Order Bending

4 2043 2nd Order Torsion 4 1003. 2nd Order Torsion 4 2133. 2nd Order Torsion

5 2278. 3rd Order Bending 5 2144. 1st Order Shear 5 2332. 3rd Order Bending

6 2358. 1st Order Shear 6 8722. 2nd Order Shear 6 2358. 1st Order Shear

7 3705. 3rd Order Torsion 7 9988. 1st Order Extension 7 4051. 3rd Order Torsion

8 4344. 4th Order Bending 8 16667 2nd Order Shear II 8 4552. 4th Order Bending

9 4763. 1st Order AxialBending

9 20793 2nd Order Extension 9 5633. 1st Order AxialBending

10 5569. 2nd Order AxialBending

10 22799 2nd Order Shear III 10 6433. 4th Order Torsion

ELEMENT TYPE

Element Type

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Element Type The type of element chosen is very important in dynamic

analysis, in that it can control the stiffness representationand to a lesser extent the mass distribution of the structure.

Examples of poor choices are: Using TET4 elements to model solid structures. If they are used

in relatively thin regions that have plate or shell the results canbe very poor. TET10 or preferably HEXA are a better choice.

If RBE2 is used instead of an RBE3 on a flexible structure such

as a satellite platform then it may over stiffen the structure andinfluence the frequencies badly.

In the next workshop a classic structure is analyzed usingdifferent elements.

Bracket Example

Bath Tub Fitting ( Tension to Shear Load transfer )

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The geometry is found in the Femap examples folder.

Constrain as shown

Apply Material as shown

Mesh with TET10s; fine, coarse TET4s; fine , coarse

Bracket Example

Bath Tub Fitting ( Tension to Shear Load transfer )

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Fine mesh TET10 model is assumed as baseline.

A very coarse TET10 mesh has maximum errors of 15.8%

A very coarse TET4 mesh has maximum errors of 135%

A fine mesh TET4 has maximum errors of 7.7%

!"!10 !"!4

#ine co$rse % & co$rse % #ine % co$rse % & co$rse %

Mode 61101 D'F 1845 D'F di##erence 2187 D'F di##erence 10525 D'F di##erence 321 D'F di##erence 1216 D'F di##erence

1 73 808 1.8 18 15.80 854 7.6 1274 60.77 1867 135.53

2 873 885 1.46 1007 15.42 35 7.15 134 54.60 117 11.703 2628 2653 0.5 2805 6.73 2770 5.40 343 32.3 4760 81.13

4 34 4065 2.2 4503 14.03 420 6.57 5783 46.45 8115 105.4

5 566 5613 (1.47 5772 1.32 5884 3.30 7171 25.88 10708 87.8

6 5860 6078 3.71 6770 15.53 6277 7.12 8065 37.63 1081 87.38

7 610 660 0.72 704 2.66 7063 2.22 302 34.61 1232 87.16

8 65 7207 3.03 781 11.78 7324 4.71 535 36.31 1474 111.4

41 740 2.63 10562 11.2 25 4.58 1327 3.2 18702 7.0610 10046 10250 2.03 1124 12.43 1067 6.30 14778 47.10 20044 .53

We are interested in the first elastic mode of the tuning fork. The theoretical value is440 Hz, and is known as the A above middle C musical note.

As an aside, it is interesting to note that the vertical translation of the stem

Tuning Fork Example

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As an aside, it is interesting to note that the vertical translation of the stemis what excites an instrument or another object that the tuning fork is placedagainst.

Import the Femap geometry

Test the first natural frequency

Adjust the length of the tuning fork to get a better match

MASS DISTRIBUTION

Mass Distribution

A poor stiffness representation can influence a structure

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A poor stiffness representation can influence a structurebadly and a poor mass representation can also have thesame effect.

The mass values may be wrong due to user error. The

values can be checked in the Nastran output file

There are two forms of mass representation in Nastran lumped and coupled. Differences may occur in the analysisdepending on which is selected.

DETAIL OF JOINTS

Detail of Joints:

Is the joint flexibility correct?

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Is the joint flexibility correct? For example a corner of a formed sheet structure will have an

internal radius which increases its torsional stiffness. It may beimportant in this case to include the torsional stiffness via RODelement.

If bolts are used to connect components together then thebolt stiffness may play an important role in dynamic analysis.

CQUAD4

CQUAD4

CROD

DETAIL OF CONSTRAINTS

Detail of Constraints When we idealize a structure we always make assumptions about

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When we idealize a structure we always make assumptions aboutthe connection to an adjacent structure or to ground.

Hence if a panel is surrounded on all sides by reinforcing structure, dowe represent that as fully built in, simply supported, or model anequivalent edge stiffness using CELAS or CBUSH elements?

A particularly difficult case is where the connectivity is ill-defined,such as the push-fit and snap connectors of a typical car dashboardassembly.

Remember there is no such thing in nature as an infinitely stiffconnection or structure.

The mode shapes of the tower shown on the next slide aresignificantly changed by the fact that the connection to ground isnot rigid. Errors will occur if it is assumed fully fixed.

ANALYSIS OF A TOWER WITH SOFT GROUNDCONNECTION

Objectives: Account for the soil-base interaction using CBUSH

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Account for the soil base interaction using CBUSHelements.

Soil Stiffness modeled with CBUSH

Elements Tower Leg

RBE2

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CHECK LIST FOR NORMAL MODES PRIOR TODOING FURTHER ANALYSIS

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RBMs - are they as expected

Is the frequency range adequate (we will discuss this more inthe section on modal effective mass)

Are the modes clearly identified

Is Mesh Density adequate

Is the Element Type appropriate Is the Mass distribution correct

Is coupled vs. lumped mass important

Are the internal joints modeled correctly

Are the constraints modeled correctly Do the results compare with hand calcs, previous experience or

test

basic dyn

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