01 sdof - spc408 - fall2016

SPC408 – Dynamics of Aerospace Structures#WikiCourses

http://WikiCourses.WikiSpaces.com

Single Degree of Freedom SystemsMaged Mostafa

Dynamics of Aerospace Structures

SPC408




Single degree of freedom systems




Objectives• Recognize a SDOF system• Be able to solve the free vibration equation

of a SDOF system with and without damping

• Understand the effect of damping on the system vibration

• Apply numerical tools to obtain the time response of a SDOF system




Single degree of freedom systems

• When one variable can describe the motion of a structure or a system of bodies, then we may call the system a 1-D system or a single degree of freedom (SDOF) system. e.g. x(t), q(t) Z(t), y(x).




Stiffness

• From strength of materials recall:




Newton’s Law

• Newton’s Law:

00 )0(,)0(0)()(

)()(

vxxxtkxtxm

tkxtxm




Solving the ODE

• The ODE is• The proposed solution:• Into the ODE you get the characteristic

equation:

• Giving:

0)()( tkxtxm taetx )(

02 tt aemkae

mk

2mkj




Solving the ODE (cont’d)

• The proposed solution becomes:

• For simplicity, let’s define:

• Giving:

tmkjt

mkj

eaeatx

21)(

mk

tjtj eaeatx 21)(




Let’s manipulate the solution!




Recall

ajSinaCose ja

bSinaCosbCosaSinbaSin




Manipulating the solution

• The solution we have:• Rewriting:

tjtj eaeatx 21)(

tjSintCosatjSintCosatx

2

1)(

tSinaajtCosaatx 2121)(

tSinAtCosAtx 21)(




Further manipulation tSinAtCosAtx 21)(

22

21 AAA

AASin

AACos 12 &

tSinCostCosSinAtx )(

tASintx )(




Different forms of the solution

tjtj eaeatx

tCosAtSinAtx

tASintx

21

21

)(

)(

)()(




NOTE!




Natural Frequency of Oscillation• In the previously obtained solution:

• The frequency of oscillation is • It depends only on the characteristics of the

oscillating system. That is why it is called the natural frequency of oscillation

tASintx )(

mk

n




Frequency

period theis s 2

Hz2s 2

cycles rad/cycle 2

rad/s frequency natural thecalled is rad/sin is

n

nnnn

n

T

f

We often speak of frequency in Hertz or RPM, but we need rad/s in the arguments of the trigonometric functions.




Recall: Initial Conditions




Amplitude & Phase from the ICs

Phase

0

01

Amplitude

2

202

0

0

0

tan ,

yields Solvingcos)0cos(

sin)0sin(

vxvxA

AAvAAx

n

n

nnn

n




Some useful quantities

peak value A

T

Tdttx

Tx

0

valueaverage = )(1lim

valuesquaremean root = 2xxrms

valuesquare-mean = )(1lim0

22

T

Tdttx

Tx




Peak Values

Ax

AxAx

2max

max

max

:onaccelerati

:velocity :ntdisplaceme

Maximum or peak (amplitude) values:




Samples of Vibrating Systems

• Deflection of continuum (beams, plates, bars, etc) such as airplane wings, truck chassis, disc drives, circuit boards…

• Shaft rotation• Rolling ships




Wing Vibration




Ship Vibration




Effective Stiffness of Structures




Bars• Longitudinal motion• A is the cross sectional

area (m2)• E is the elastic modulus

(Pa=N/m2)• l is the length (m)• k is the stiffness (N/m) x(t)

m

EAk

l




Rods

• Jp is the polar moment of inertia of the rod

• J is the mass moment of inertia of the disk

• G is the shear modulus, l is the length

Jp

J qt)

0

pGJk




Helical Spring

2R

x(t)

d = diameter of wire2R= diameter of turns n = number of turns x(t)= end deflectionG= shear modulus of spring material

3

4

64nRGdk




Beams

f

m

x

• Strength of materials and experiments yield:

3

3

3

3

mEI

EIk

n




Equivalent Stiffness




Example 1.5.2 Effect of fuel on frequency of an airplane wing

• Model wing as transverse beam

• Model fuel as tip mass• Ignore the mass of the

wing and see how the frequency of the system changes as the fuel is used up

x(t) l

E, I m




Mass of pod 10 kg empty 1000 kg fullI = 5.2x10-5 m4, E =6.9x109 N/m, l = 2 m

• Hence the natural frequency changes by an order of magnitude while it empties out fuel.

Hz 18.5=rad/s 115 210

)102.5)(109.6(33

Hz 1.8=rad/s 6.11 21000

)102.5)(109.6(33

3

59

3empty

3

59

3full

mEI

mEI




Example 1.5.5 Design of a spring mass system using available springs: series vs parallel

• Let m = 10 kg• Compare a series and

parallel combination• a) k1 =1000 N/m, k2 = 3000

N/m, k3 = k4 =0• b) k3 =1000 N/m, k4 = 3000

N/m, k1 = k2 =0

k1 k2

k3 k4

m




rad/s 66.810750

N/m 75013

3000)1()1(

1,0

:connection series b) Case

rad/s 2010

4000

N/m 400030001000,0:connection parallel a) Case

4321

2143

mk

kkkkk

mk

kkkkk

egseries

eq

egparallel

eq

Same physical components, very different frequencyAllows some design flexibility in using off the shelf components




Harmonic Excitation of Undamped Systems

• Consider the usual spring mass damper system with applied force F(t)=F0cost

• is the driving frequency• F0 is the magnitude of the applied force• We take c = 0 to start with




Equation of motion

mkmFf

tftxtx

tFtkxtxm

n

n

,/ where

)cos()()(

)cos()()(

00

02

0




Linear non-homogenous ODE:

• Solution is sum of homogenous and particular solution

• The particular solution assumes a form of forcing function (physically the input wins)

)cos()( tXtxp




Substitute into the equation of motion:

220

022

:yields solving

coscoscos

2

n

x

n

x

fX

tftXtXpnp




Thus the particular solution has the form:

)cos()( 220 tftx

np




General Solution

particular

nnn tftAtAtx

coscossin )( 22

0

shomogeneou

21




01

0220

2

)0(

)0(

vAx

xfAx

n

n

Apply the initial conditions to evaluate the constants

Solving for the constants and substituting into x yields

tftfxtvtxn

nn

nn

coscossin)( 220

220

00




Comparison of free and forced response• Sum of two harmonic terms of different

frequency• Free response has amplitude and phase

affected by forcing function• Our solution is not defined for n =




Example• Compute and plot the response for m=10 kg, k=1000 N/m,

x0=0,v0=0.2 m/s, F=23 N, =2n

rad/s 202 rad/s, 10kg 10N/m 1000

nn mk

Solution

m 02.0rad/s 10m/s 2.0 N/kg, 3.2

kg 10N 23 0

0 n

vmFf




Example (cont’d)

m 109667.7s/rad )20(10

N/kg 3.2 3222222

0

n

f

)20cos10(cos10667.710sin02.0)( 3 ttttx

Substituting into the general solution:




Response to Harmonic Excitation




Beat !

• What happens when the driving frequency is near the natural frequency?




Example• Given zero initial conditions a harmonic input of 10 Hz

with 20 N magnitude and k= 2000 N/m, and measured response amplitude of 0.1m, compute the mass of the system.

ttftx nn

coscos)( 220

Solution:




Example (cont’d)

• Using Trigonometric Identities:

• Using given data

termFast

n

termSlow

n

n

ttftx

2sin

2sin2)( 22

0

kg 45.0

1.0)20(2000

)/20(21.02222

0

mm

mf

n




Plotting the result




Twice Frequency and Beat




Resonance

• When excitation frequency is equal to the natural frequency, the previous solution fails.

• The particular solution of the ODE become:

)sin()( ttXtxp




Substituting!

20fX

Use the particular solution into the equation:

bound without grows

021 )sin(

2cossin)( ttftAtAtx

The total solution becomes:




Resonance




Summary• Write down the equation of motion using Newton’s law• Solve the equation of motion for a SDOF• Use initial conditions to determine the amplitude and

phase of vibration for a SDOF• Evaluate the effective stiffness of structural members• Analyzing the response of a SDOF to harmonic

excitations• Different types of response depend on the excitation

frequency• What happens during the beat and resonance

phenomena




1. The amplitude of vibration of an undamped system is measured to be 1 mm. the phase shift is measured to be 2 rad and the frequency 5 rad/sec. Calculate the initial conditions.

2. Using the equation:

evaluate the constant A1 and A2 in terms of the initial conditions

HW #1

tSinAtCosAtx 21)(




HW #1 (cont’d)

3. A vehicle is modeled as 1000 kg mass supported by a stiffness k=400 kN/m. When it oscillates, the maximum deflection is 10 cm. when loaded with the passengers, the mass becomes 1300 kg. calculate the change in the frequency, velocity amplitude, and acceleration if the maximum deflection remain 10 cm.

01 sdof - spc408 - fall2016

Engineering