Vibrations 10

Vibrations and Aeroelasticity

Vibrations

© 2006 Confidential and proprietary to Infosys Technologies Ltd

Session Objective

•

To illustrate the principles of Structural Vibrations

•

To illustrate application of Vibration Principles in the design of aircraft Structures


1.1 Introduction

(1) Some Basic TermsVibration

: Any to and fro motion is called VibrationEx: Sea waves, Simple pendulum

Simple Harmonic Motion: Motion of the projection Q of a point P moving on a circle at a uniform angularvelocity ω

-

on the diameter.

y = A sin өө

= ω

t

y = -

ω

A cos өy = -

ω2

A sin ө= -

ω2

yy + ω2

y = 0

* PQ y

x

Aөω

...

..


1.1 Introduction

Amplitude of Vibration : APeriod : τ

Time required for repeating a cycle

Frequency

: f No of cycles per second = 1/ τω

= 2 π

f

Phase angle: angle between 2 vectors rotating at same speed

x0 = Amplitude

= period

t

xy A

τ


1.1 Introduction

(2) Some useful Vibrations

: Musical notes; Sieving machineSome harmful Vibrations: Earth quake; Rotor unbalance

(3) Types of Vibrations(i) Free vibration: Vibration that takes place after an externally

applied disturbance seizes.(ii) Forced vibration: Vibration that takes place during the

application of an external dynamic load(a) Sinusoidal vibrations : Ex; Motor unbalance(b) Transient vibrations –

vary in magnitude with time and act

for a specific time: Ex; Rocket motor firing(c) Random vibrations –

characterized by statistical quantities:

Ex; earth quake


1.1 Introduction

(4) Important Components in Vibration:(a) Stiffness K : Elasticity of the elements(b) Mass M : Density of the material and objects(c) Damping C : Due to intermolecular action in material,

friction in joints, specifically dampers(d) Dynamic load : time dependent

(5) Classification of Structures for the Present Study(a) Single degree of freedom systems(b) Two degree of freedom systems(c) Continuous systems: Ex beams, plates, shells(d) Multi degree of freedom systems


1.1 Introduction

(6) Importance of Vibrations in Aircraft Design

The airplane structure being light weight and flexible is very

susceptible to vibrations and noise transmission.The noise and vibration have to be reduced in commercial planes if

they have to be competitive with other transport.The source of noise are generally the propeller, engines and

aerodynamic forces. It is better to isolate such disturbances from getting transmitted to airframe.

Equipment mounted in airplanes must be prevented from excessive vibrations which can lead to malfunction or strength or fatigue failures.

Aerodynamic forces can induce high amplitude vibrations which can damage the structural components like nose wheel doors. They

can also result in increasing loads on deflected surfaces like wings causing instability and failures


1.2 Free Vibration of Structures1.2.1 Single Degree of Freedom Systems(1)

Equation of Equilibrium

c

mk x

k (x-0)..

m x

Elastic force

.c x

Inertial forceViscous force

Total internal force = kx + cx + mx = External Nodal Force = 0...


1.2 Free Vibration of Structures

(2) Natural FrequencyIf we neglect damping

x + ωn2

x = 0 where ωn

=√(k/m) radians/secis the circular natural frequency of the systemfn

= 1/(2 π). √(k/m) cycles/secAlternatively:If W is the weight of the mass m and gravity g acts in the

direction x,the static deflection of the weight is given by

δ

= W/k where W = mgfn

= 1/(2 π). √(g/ δ)This expression is seen to be a good approximation for natural frequency of even quite complex structures also, if δ

is interpreted as the maximum deflection under 1g load.

..



Torsional Natural Frequencyө

+ ωt2

ө

= 0 where ө

is the angular motion of an inertia of I attached to a torsional spring of stiffness kt

and ωt

=√(kt

/I) radians/secis the circular torsional natural frequency of the system ft

= 1/(2 π). √(kt

/I) cycles/secAlternatively:If T is the torque given to a mass moment of inertia I at an angular acceleration α,the static rotation of the inertial body is given by

ψ

= T/kt

where T = Iαft

= 1/(2 π). √(α/ψ)

..



(3) Free Vibration ResponseConsidering damping, the general solution of the second

order differential equation is obtained as:(i) Solve auxiliary equation

m λ2

+ λ

c + k = 0 for 2 roots λ1, λ2 = (a + ib) where

a = -c/2mb = √

[(c/2m)2

–

(k/m)] (ii) Then x = C1 eλ1t

+ C2 eλ2

t or in generalx = eat [A cos (bt) + B sin (bt)]

This equation can be solved for x given two initial

conditions,for ex; x = 0 at t =0 and x = v0 at t =0, then A = 0 and B = v0/b

x = (v0/b) eat

sin (bt)

.



(4) Free Vibration Response –

Effect of dampingc is called critical damping cc

when (c/2m)2

= (k/m) or Cc

= 2m ωnThe damping ratio ξ

is defined as (c/cc

)In terms of ξ

the equilibrium equation reduces to

x + 2 ξ ωn

x + ωn2

x = 0Consider different damping scenarios:

(i) Under damped system: ξ

< 1The system oscillates at damped natural frequency ωd

with logarithmically decreasing amplitudes in successive cycles

ωd = ωn

√

(1 -

ξ

2)

This is usually the form of damping in many structures.

.. .


1.2 Free Vibration of Structures(ii) Critically damped system: ξ

= 1It is threshold of a situation where the system comes to rest slowly without oscillating. Critical damping is useful in door closures, measuring devices, pick up heads of turn tables etc.

(iii) Overdamped system: ξ

> 1In this case also the system comes to rest without oscillating but takes more time than when critically damped.Such high damping is useful in recoil mechanism of field guns and military hardware to absorb the shock during the operation of equipment.

(iv) Undamped system: ξ

= 0This is a hypothetical situation. In this case the system oscillates at a constant amplitude for an indefinite period.



Free vibration response of mass-without damping (Top) and under damping (Bottom)



Normalised response of mass with Overdamping (top)and critical damping (bottom) with time



1.2.2 Two Degree of Freedom Systems(1)

Equations of Equilibrium

m1k1

k2 (x2 –

x1)

..m2 x2

Elastic force

Inertial force

m2k2

x1 x2

k1 (x1 –

0)

m1 x1..

At m1 : m1 x1 + k1 (x1 –

0) + k2 (x1 –

x2) = 0

At m2: m2 x2 + k2 (x2 –

x1) = 0

k2 (x1 –

x2)

..

..



(2) Natural FrequenciesIn a matrix form the equations of equilibrium are written as

K X + M

X = 0where stiffness matrix K and mass matrix M are given by

k1 + k2 -k2 m1 0K = M =

-k2 k2 m2

The solution is given by [K -

ωn2

M] X = 0Where ωn

2

is a diagonal matrix of eigen values or natural frequencies and X is matrix whose columns are corresponding eigen vectors.

..



For non trivial solution K -

ωn2

M = 0From which the eigen values are determined.

( k1 + k2 ) -

ωn2 -k2

= 0-k2 k2 -

ωn2

ω12 , ω2

2 = (1/2) [ a + √(a2

– b2] wherea = [ α2

2

+ α12

(1+ k2/k1)] and b = 2 α1

α2

α12

= k1/m1 and α22

= k2/m2 are individual frequencies of separate spring-mass sytemsThus a two degree of freedom system has two natural frequencies



Example problem:Let k1 = 10 kgf/mm ; k2 = 20 kgf/mm ; m1 = 5 kg and m2 = 20 kgand g = 10**4 mm/sec2

α12

= k1/m1 = 10*10**4/5 = 2*10**4α2

2

= k2/m2 = 20*10**4/20 = 10**4 a = [ α2

2

+ α12

(1+ k2/k1)] = 7*10**4b = 2 α1

α2

= 2√

2*10**4ω1

2 , ω22 = (1/2) [ a + √(a2

– b2]= (1/2) [7*10**4 + √(49*10**8 –

8*10**8)]= 0.3*10**4; 6.7*10**4

f1

= 8.7 Hzf2

= 41.2 Hz



Alternative method of frequency calculationIf W1 and W2 are the weights of the masses m1 and m2; then

under 1 g load the displacements are calculated as

k1 + k2 -k2 x1 W1=

-k2 k2 x2 W2

The maximum displacement δ

= x2 = (W1/k1) + (1/k1 + 1/k2)W2The natural frequency f1

= 1/(2 π). √(g/ δ)For the example problem δ

= 3.5 mm and f1

= 8.5 Hz



Mode 1:

Mode 2:x

xm2

m1

m2

m1



1.2.3 Continuous Systems-

Beams(1) Equation of Equilibrium

If

we consider a unit length of a beam of uniform cross section and

subjected to and external load of w/unit length, it is in equilibrium between internal and external load as given by the equation;

EI d4y / dx4

= w where y is the lateral deflection. If m is the mass per unit length of the beam and vibrating at

frequency of ωn

,

instead of external load w, we can consider the inertial force( m y ) = - mωn

2

yto be acting in addition to the elastic force. Therefore we get

EI d4y / dx4 - mωn2

y = 0 or d4y / dx4

-

β4

y = 0 where β4

= (mωn2) / (EI) is called

frequency parameter

..



(2) Natural Frequencies(i) Solve auxiliary equation

λ4

-

β4

= 0 for 4 roots λ1 = β; λ2 = -β;

λ1 = iβ; λ2 = -iβ

(ii) Then y = A cos (βx) + B sin (βx) + C cosh (βx) + D sinh (βx)We need 4 boundary conditions to solve for the constants. Example Problem:Consider a simply supported beam of length l.At x = 0, y= 0 it gives A + C = 0

x = 0, d2y/dx2

= 0 it gives A -

C = 0From the above 2 eqs A = C = 0

x = l, y = 0 it gives B sin (βl) + D sinh (βl) = 0x = l, d2y/dx2

= 0 it gives B sin (βl) -

D sinh (βl) = 0



From the above 2 equations we get B sin (βl)

= 0 , the general solution of which is βl = nπ

where n = 1,2,3,4 -----. The natural frequencies are given by

ωn

= ( n2π2/l2) √

(EI/m)It can be seen that the natural frequencies of simply supported beam are in the ratio 12, 22, 32, 42

--f1

= 1.5708√

(EI/Ml3) where M is mass of the beamFirst natural frequency from static deflection:The maximum deflection of simply supported beam with a udl of w is δ

= (5/384) (Wl3/EI) from which f1

= 1.395√

(EI/Ml3) The ratio of exact to approximate frequency is 1.13A scale factor of 1.25 on f1

is found to give best approximation for any general structure with distributed mass.



First natural frequency from Raleigh’s MethodThe potential energy of the whole beam

PE = (EI/2) ∫

(d2y/dx2)2dx integrated between x=0 and lThe kinetic energy of the whole beam

KE = (1/2) mv2 =

(1/2) m ωn2

∫y2dx integrated between x=0 and lThe maximum PE is equal to maximum KE as no energy is lost.

ωn2 = (EI/m) ∫

(d2y/dx2)2dx / ∫y2dx

The natural frequency is not very sensitive to deflection curve. A function which satisfies boundary conditions is usually chosen to get reasonably accurate first frequency estimate.



Example of simply Supported BeamLet y = x (l –x)

The potential energy of the whole beam PE = (EI/2) ∫

(d2y/dx2)2dx integrated between x=0 and l

= 2EI. lThe kinetic energy of the whole beam

KE = (1/2) m ωn2

∫y2dx integrated between x=0 and l= (1/60) m ωn

2

l5

ωn2 = 120 (EI/ml4) ; f1

= 1.743√

(EI/ml4)



Typical mode shapes of different beams



Fundamental Bending frequency of Aircraft Wing with Single EngineBurgess proposed an approximate method where the wing curvature is assumed to be constant.

d2y/dx2

= CRaleigh’s method and step by step integration is performed for solution. This is found to give reasonably accurate estimate. Usually C is taken as unity.Consistent units are: length-

in, weight -

lbs


1.2 Free Vibration of StructuresExample problem of wing bending frequency

Station Distance I(d2y/dx2)2 dy/dx/20 y/400 y2/160000 w wy2/160000

1 220 2 11 66 4336 0.53 2298

2 200 6 10 55 3025 0.72 2178

3 180 22 9 45 2025 0.91 1843

4 160 52 8 36 1296 1.1 1426

5 140 96 7 28 784 1.29 1011

6 120 156 6 21 441 1.48 653

7 100 234 5 15 225 1.67 376

8 80 329 4 10 100 1.86 186

9 60 422 3 6 36 2.05 74

10 40 480 2 3 9 2.24 20

11 20 546 1 1 1 2.43 2

12 0 620 0 0 2.62 0

Total 2965 10067

f = 1/2/π √ (386*10*7*2965)/(10067*16000) = 13.42 hz

More exact method gives 13.36 Hz



1.2.4 Multi Degree of Freedom Systems(1)

Equation of EquilibriumM X + K X = 0

(2) Natural Frequencies and ModesFor a non trivial solution

K -

ωn2

M = 0Methods:* For small problems and many/ all frequencies and vectors

-

Determinant expansion -

Given’s method* For large problems and few frequencies and vectors

-

Subspace iteration-

Lanczo’s method

..


•

A free structure has 6 rigid body modes ψ

and many elastic modes φ. A constrained structure has only elastic modes.

•

The elastic modes are orthogonal with respect to stiffness and mass matricesφTKφ

= k

generalised stiffnesses and φTMφ

= m

generalised

masses•

The elastic modes are normalised such that

(a) The maximum component is unity or(b) The rms value of the vector is unity or(c) The generalised stiffness/ mass is unity

•

The rigid body modes can be normalised with unity in each direction at a base point.

•

The rigid body mass is given by ψTMψ

= M0

referred to base



1.3 Forced Vibration of Structures

1.3.1 Single Degree of Freedom Systems(1)

Equation of Equilibrium

c

mk x

k (x-0)..

m x

Elastic force

.c x


Total internal force = kx + cx + mx = External Nodal Force = F ...

F (ω, t)


1.3 Forced Vibration of Structures(2) Sinusoidal Load –

(a) Steady State Force VibrationLet input Force = F = F0

eiωt

ω

is excitation frequencyF0 is magnitude of ForceStructure vibrates at same frequency as excitationLet x = A eiωt be structural response

x = Aiω

eiωt

x = -ω2

A eiωt

Substituting in equation of equilibriumA [ -ω2m + icω

+ ωn

2m ] = F0A = F0

/ [ (ωn2-ω2) m + icω]

= δ

/ [ (1 –

r2) + i r(2ξ) ] where δ

is static deflectionF0

/k

; r is frequency ratio (ω/ωn

); and ξ

damping ratio

..

.



•

The displacement increases to a maximum as the excitation frequency reaches close to natural frequency, r=1. This situation is called resonance. The peak depends only on damping. The amplification of displacement over static displacement Q (magnification) is given by 1/2ξ.

•

Below the natural frequency, r<1, the displacement is in the same direction as the applied force. After the natural frequency, r>1, the displacement is exactly in opposite direction to applied force.



(b) Steady State Base VibrationLet input Force = F = F0

eiωt

c

mk x

k (x-xb

)..

m x

Elastic force

.c x


Total internal force = k(x-xb

) + cx + mx = 0...

xb

(ω, t)


1.3 Forced Vibration of StructuresLet xr

= x-xbkxr

+ cxr

+ mxr = -mxbLet input base displacement = xb

= X0

eiωt

ω

is excitation frequencyX0 is magnitude of base displacement

Then Xr

response of mass relative to base is given byXr

= r2X0

/ [ (1 –

r2) + i r(2ξ) ]This curve shows that when base excitation frequency is higher than √2 times natural frequency, r> √2, the (relative) response of the mass will be less than base imposed displacement. This situation is called “isolation”.

On the other hand if the excitation frequency is lower than natural frequency, r<1, the mass displacement will be amplified over base displacement. At resonance r=1 it reaches a peak value of 1/2ξ

. ....


Magnification of mass displacement with frequency ratio



1.3 Forced Vibration of Structures(3) Transient Load

The response of the mass subjected to a transient load F(t) is given by Duhamel’s integral over 0 to t

x = (1/mωn

)

∫e -ξ ωn(t-t1)

F(t1) sin ωn

(t-t1)dt1If we consider specific cases of shock loads like triangular or square pulses, the following can be observed.-

if the shock frequency is low compared to natural frequency

the peak response occurs during the shock and responses are high.-

if the shock frequency is high compared to natural frequency

the peak response occurs after the shock is over and the responses (residual shock) are generally benign.

-

in any case the maximum amplification is about 2


1.3 Forced Vibration of Structures(4) Random Load

Random vibration encountered in practice is mostly in the form of a base input acceleration power spectral density P specified as a function of ω

within a band ω1

and ω2

.The solution consists of the following steps.

1. Obtain acceleration response as magnification factor Q as a function of ω

for a constant unit magnitude sine input.

2. Square the Q vs ω

plot3. Multiply the Q vs ω

plot by input plot P vs ω. This plot R vs ω

gives the psd of acceleration response as a function of ω.4. The area under the curve R vs ω

between ω1

and ω2

gives the mean square value of the acceleration response, which ispredominantly dictated by the response close to resonance. The square root of the area gives rms value of acceleration.



•

If the input is a white noise P g2/hz (psd constant with ω), then the root mean square value of acceleration response Rrms

is given byRrms

= √

(π/2 * P * Qmax

* fn )

Example ProblemAn aircraft package is subjected to a white noise of 0.1 g2/hz . If the natural

frequency of the package is 4 hz and damping is 5% find the rms acceleration response.Here P = 0.1 g2/hz ; Qmax = 1/2ξ

= 10; fn = 4.0HzPeak acceleration psd response = 100*0.1 = 10 g2/hzRrms

= √

(π/2 * 0.1* 10.0 * 4.0

) = 2.5g



1.3.2 Two Degree of Freedom Systems(1)

Equations of Equilibrium

m1k1

k2 (x2 –

x1)

..m2 x2

Elastic force

Inertial force

m2k2

x1 x2

k1 (x1 –

0)

m1 x1..

At m1 : m1 x1 + k1 (x1 –

0) + k2 (x1 –

x2) = F1 (ω,t)

At m2: m2 x2 + k2 (x2 –

x1) = 0

k2 (x1 –

x2)

..

..

F1 (ω, t)


1.3 Forced Vibration of Structures(2) Sinusoidal Load –

Steady State Force VibrationLet input Force acting on mass 1 = F1 = F0

eiωt

ω

is excitation frequencyF0 is magnitude of ForceStructure vibrates at same frequency as excitationLet x1

= A eiωt be structural responsex2

= B eiωt

x1

= -ω2

A eiωt

x2

= -ω2

B eiωt

α12

= k1/m1 and α22

= k2/m2 are individual frequencies of separate spring-mass systems

..

..



Substituting in equations of equilibrium, we getA = F0

(α22

-

ω2

) / [(α22

-

ω2

) {(α12

m1

+ α22

m2 ) -

ω2

m1

} -

α24

m2

] andB = F0

α22

/ [(α22

-

ω2

) {(α12

m1

+ α22

m2 ) -

ω2

m1

} -

α24

m2

] * It is interesting to observe that when ω

= α2

A = 0 and B = -

F0

/k2This situation is called “tuned dynamic absorber”.The main mass m1

can be protected from excessive vibration-

in fact it can be made to stand still-

if an auxiliary spring mass system k2

,m2

is attached, tuned to the excitation frequency ω

of the sinusoidal force acting on the main mass. The auxiliary mass can vibrate with large amplitude but it is less vulnerable.

* If the two individual component frequencies are equal, α1

= α2when the excitation frequency equals first component frequency ω

= α1then always A = 0, that is the main mass is at rest.



ω

= α1Ratio of excitation freq/ 2nd

component frequency

Displacement response of m1

for 2 dampings



1.3. 3 Multi Degree of Freedom Systems(1)

Equation of EquilibriumM X + CX + K X = F (ω,t)

(2) Sinusoidal Load –

Steady State Force VibrationMode superposition methodThe displacement vector at any excitation frequency is given by

X = φηwhere η

is a vector of scale factors on each mode. Usually a few modes are selected with natural frequency above 1.5 times frequency of excitation.Substituting in equation of equilibrium and pre multiplying by φT,and using the orthogonality properties of eigen vectors,

mη

+ cη

+ kη

= φTF = fwhere m, c, and k are diagonal matrices and c = 2 m ξ ωn

.. .

.. .



The equations get decoupled and become single degree of freedom systems in modal coordinates η

* Useful when a few eigen vectors can represent the response* Equilibrium equations are converted into modal coordinates* Eigen value analysis is first step before response analysis

(2) Transient Load -

Direct Integration method* Conditionally stable algorithm

-

Central difference method* Unconditionally stable algorithms

-

Wilson -ө

method- Newmark -β

method* No transformation of equilibrium equations

Solved step by step in time domainEigenvalue analysis is not carried outUseful when the load is changing too frequently with time.


1.4 Aircraft Vibration

* The aircraft natural modes fall in 2 categories-

Symmetric and antisymmetric about a vertical plane through the center line of fuselage

* Heavy masses like engines form nodes in the modes* The fundamental symmetric mode is generally wing bending * The torsional frequency of wing is more important from flutter

and is kept higher than bending frequency. Typical Frequencies of Aircraft

Fighter BomberWing bending 10 Hz 3 HzWing torsion+bending 25 Hz 5 HzFuselage bending 13 Hz 9 HzTail plane bending 16 Hz 11 Hz


Typical vibration modes of(a) Single engine plane (b) Twin

engine plane

1.4 Aircraft Vibration


1.5 Aircraft Vibration Testing

Objective of testing: Finding natural frequencies, modes and damping.

Method: Mounting some kind of mechanical or electrical vibrator (even an unbalanced motor) (weight is small not to alter member character) at reasonably rigid location, to shake the member with a periodic force whose frequency can be varied. Responses are picked up through accelerometers at salient points. Planes may allowed to stand on partly deflated tires; seaplanes are suspended on slings.

Wing bending mode: Vibrator is placed near outer end of the wing near flexural axis, at about 33% of chord.

Wing torsion mode: Vibrator is placed off the flexural axis, usually near trailing edge.

Aileron, fin, stabilizer etc., bending mode: Vibrator is placed near trailing (far away from hinge axis) edge.


References

(1)

Freeberg, C. R., Aircraft Vibration and Flutter, John Wiley & Sons, London, 1944.

(2)

Broadbent, E. G., The Elementary Theory of Aeroelasticity, Bunhill

Publications Ltd., HOLborn, 1954.

(2) Srinivasan, P., Mechanical Vibration Analysis, Tata McGraw hill Publishing Company Ltd., New Delhi, 1982.


Summary

•

Discuss the points covered for the day

Vibrations 10

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