Chapter 7. Free and Forced Response of Single-Degree-of-Freedom Linear Systems 7.1 Introduction...

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Transcript of Chapter 7. Free and Forced Response of Single-Degree-of-Freedom Linear Systems 7.1 Introduction...

Chapter 7. Free and Forced Response of Single-Degree-of-Freedom Linear Systems

7.1 Introduction •Vibration: System oscillates about a certain equilibrium position. •Mathematical models: (1) Discrete-parameter systems, or lumped systems. (2) Distributed-parameter systems, or continuous systems. Usually a discrete system is a simplification of a continuous system through a suitable “lumping” modelling. •Importance: performance, strength, resonance, risk analysis, wide engineering applications

Single-Degree-of-Freedom (Single DOF) linear system•Degree of freedom: the number of independent coordinates required to describe a system completely.

Single DOF linear system: Two DOF linear system:

)()()()( tFtkxtxctxm =++ &&&

⎩⎨⎧

=+−=−+

)()(2)()(

0)()(2)(

212

211

tFtkxtkxtxm

tkxtkxtxm

&&

&&

System response

•Defined as the behaviour of a system characterized by the motion caused by excitation.•Free Response: The response of the system to the initial displacements and velocities. •Forced Response: The response of the system to the externally applied forces.

7.2 Characteristics of Discrete System Components

The elements constituting a discrete mechanical system are of three types:The elements relating forces to displacements, velocities and accelerations.

• Spring: relates forces to displacements

X1 X2

FsFs

)( 12 xxkFs −=

Fs

X2-X1

Slope K is the spring stiffness, its unit is N/m.

Fs is an elastic force known as restoring force.

0

• Damper: relates force to velocity

1x& 2x&

c

Fd Fd

Fd

12 xx && −

Slope C is the viscous damping coefficient, its unit is N·s/m

)( 12 xxcFd &&−=

Fd is a damping force that resists an increase in the relative velocity 12 xx &&−

0

The damper is a viscous damper or a dashpot

• Discrete Mass: relates force to acceleration

m Fm

x&&

Fm

x&&0

Slope m, its unit is Kg

xmF &&=

Note: 1. Springs and dampers possess no mass unless otherwise stated 2. Masses are assumed to behave like rigid bodies

• Spring Connected in Parallel

k1

k2

x1 x2

Fs Fs

)( 1211 xxkFs −=

)( 1222 xxkFs −=

)( 12 xxkF eqs −=21 sss FFF +=

21 kkkeq +=

• Spring Connected in Seriesx1

x2x0

k1 k2

Fs Fs

)( 101 xxkFs −=

)( 022 xxkFs −=

)( 12 xxkF eqs −=

∑=

=n

iieq kk

1

1

21

)11

( −+=kk

keq 1

1

)1

( −

=∑=n

i ieq kk

7.3 Differential Equations of Motion for First Order and Second Order Linear Systems

• A First Order System: Spring-damper systemk

c

x(t)

F(t)

Free body diagram:

F(t)Fs(t)

Fd(t)

0)()()( =−− tFtFtF ds

)(txkFs =

)(txcFd &=

)()()( tFtxktxc =+&

• A Second Order System: Spring-damper-mass system

F(t)

k

c

m

x(t) Free body diagram:

F(t)Fs(t)

Fd(t)m

)()()()( txmtFtFtF ds &&=−− )()()()( tFtxktxctxm =++ &&&

7.4 Harmonic Oscillator

F(t)

k

m

x(t) Second order system:

)()()()( tFtxktxctxm =++ &&&

m

ktxtx nn ==+ 22 0)()( ωω&& (1)

Undamped case, c=0:

)cos()( φω −= tCtx n is called Phase angleφWith initial conditions 00 )0(,)0( vxxx == &

tv

txtx nn

n ωω

ω sincos)( 00 +=

2020 )(

w

vxC +=

nx

v

ωφ

0

01tan−=and

Solution: tBtAtx nn ωω sincos)( +=

rad2/90 πφ == o

radπφ == o180

0=φ

Period(second): Natural frequency:Hertz(Hz)

Example:

,/2,1,10 sradHzfmA nn πω ===

n

Tωπ2

=T

f nn

1

2==

πω

7.5 Free Vibration of Damped Second Order Systems

• A Second Order System: Spring-damper-mass system

F(t)

k

c

m

x(t) Free body diagram:

F(t)Fs(t)

Fd(t)m

• Express it in terms of nondimensional parameters:

Viscous damping factor:

(1.7.1)

• The solution of (1.7.1) can be assumed to have the form, stAetx =)(

We can obtain the characteristic equation

With solution:

)()()()( tFtxktxctxm =++ &&&

0)()(2)( 2 =++ txtxtx nn ωωζ &&&

nm

c

ωζ

2=

02 22 =++ nn ss ωωζ ns

sωζζ )1( 2

2

1 −±−=

)()()()( txmtFtFtF ds &&=−−

ns

sωζζ )1( 2

2

1 −±−=

The locus of roots plotted as a function of ζ

nis

sωζ ±==

2

1,0

s1 ,s2 are complex conjugates.

Undamped case, the motion is pure oscillation

Overdamped case, the motion is aperiodic and decay exponentially in terms of

Critical damping

Underdamped Case

(1)

(2)

(3)

(4)

,10 << ζ

ns ωζ −== 2,1,1

−∞→→> 21 ,0,1 ssζ

,1>ζ Overdamped case

,1=ζ Critical damping

tnn

tsts netAtAeAeAtx ζωωζωζ −−−+−=+= )]1(exp)1(exp[)( 22

2121

21

Figure 1.7.3

Underdamped Case

where, the frequency of the damped free vibration

as

tnn

netiAtiAtx ζωωζωζ −−−+−= )]1(exp)1(exp[)( 22

21

ttiti ndd eeAeAtx ζωωω −−+= )()( 21

)cos()( φωζω −= − teAtx dtn

nd ωζω 21−=

0)(, →∞→ txt

,10 << ζ

7.6 Logarithmic Decrement • Experimentally determine the damping of a system from the decay of the vibration amplitude during ONE complete cycle of vibration:

Let , We obtain

Introduce logarithmic decrement

for small damping,

• For any number of complete cycles:

)cos(

)cos(

2

1

2

1

2

1

φωφω

ζω

ζω

−−

= −

teAteA

xx

dt

dt

n

n

d

TTttωπ2

,12 =+=

tee

e

x

xnTt

t

n

n

ζωζω

ζω

== +−

)(2

1

1

1

22

1

1

2ln

ζπζζωδ−

=== Txx

n

22)2( δπ

δζ+

=

πδζ2

TjjT

j

j

j

nn eex

x

x

x

x

x

x

x ζωζω ===++

)(13

2

2

1

1

1 L1

1ln1

+

=jxx

7.7 Energy Method

Total energy of a spring-mass system on a horizontal plane:

22

2

1

2

1kxxmVT +=+ &

0)( =+=+ xkxxxmVTdtd

&&&&

0=+ kxxm&&

maxmax VT = 2max

2max )(

2

1)(

2

1xkxm n =ω

mk

n =ω