Engineering Vibrations Review

University of Maryland B. Balachandran & E. Magrab

Review Problems For Exam#2


Chapter 3 Lagrange’s Equations of Motion

Time Response of SDOF Systems

Frequency Response of SDOF Systems

Stability of SDOF Systems


Problem 1

Gωt


Problem 1


Problem 1


Problem 2


Problem 2


Problem 2


Problem 3


Problem 3


Problem 3


Problem 4

The radius of the pulley is R =100 mmand its moment of inertia is I = 0.1 kg-m2. The mass m = 5 kg, and the springconstant is k = 135 N/m. The cable doesnot slip relative to the pulley. Thecoordinate x measures the displacementof the mass relative to the position inwhich the spring is unstretched.Determine x as a function of time if c =60 N-s/m and the system is released fromrest with x = 0.


Problem 4

Equation of Motion


Problem 4


Problem 4

The system Response is given by


Assume the mass of a car is m supported on springs with stiffness k. To analyzethe suspension’s behavior, consider the height of the road surface relative to itsmean height is h. If the car’s wheels remain on the road and its horizontalcomponent of velocity is v and the damping due to the suspension’s shockabsorbers is neglected, determine:

(a) magnitude of the car vertical steady-state vibration relative to the road.(b) velocity v at which resonance occurs

Problem 5

λ

h

vmk

x

y


Problem 5


Problem 5

Equation of Motion of Car( ) 0mx k x y+ − =

( ) ( )m x y k x y mymz kz my− + − = −

→ + = −

where ( )z x y= −

or2

2 2 2sinnv vz z h tπ πω

λ λ + = −


Problem 5


Problem 5


Problem 6

Consider the effect of dropping onto the floor a package thatresides inside a container that has a coefficient of restitution ε.If the container falls from a height h, determine the timeresponse of the motion of the package relative to the container

0 2V gh=


( ) ( ) 0mx c x y k x y+ − + − =

Equation of Motion

or mz cz kz my mg+ + = − =

Initial Conditions

0 0

(0) (0) (0) 0(0) (0) (0)

z x yz x y V Vε

= − == − = − −

Problem 6


m z cz k z mg+ + =

Equation of Motion

has the solutions

Homogeneous Solution

/pz mg k=Particular Solution

General Solution sin( ) /nth d hz Z e t mg kζω ω φ−= + +

( )sinnth h d hz Z e tζω ω φ−= +

Problem 6


Problem 6

At t=0, z(0)=0

0 sin /h hZ mg kφ= +

( ) 20 1 [ sin 1 cos ]h n h hV Zε ω ζ φ ζ φ− + = − + −

0 00, (0)At t z V Vε= = − −

(1)

(2)

Solve eqns. (1) & (2) simultaneously for Zh and ϕh


Problem 6

From eqns. (1) & (2):

( )/ sinh hZ mg k φ= −

( )2

0

1tan1h

nVg

ζφε ω

ζ

−=

+ +

(3)

(4)

and


Problem 6

( )2/ 1 /hZ mg k cζ= − −

( )2

0

1tan1h

nVg

ζφε ω

ζ

−=

+ +

(5)and

ϕh

21b ζ= −

( ) 01 nVa

gε ω

ζ+

= +

2 2c a b= +


Problem 7

An underdamped single degree-of-freedom systemis travelling with a velocity Vo when its base impactsa stationary rigid wall, as shown in the figure below.Determine:

(a) - the expression of the coefficient ofrestitution ε,

(b) – the amount of energy dissipated duringthe interval 0 < t < tvm.


Problem 7Barrier

Time

v(t)V0

-V(tvm)

0


Problem 7

The bumper remains in contact with the barrier if:

0kx cx+ >

The coeff. of Restitution is given by:

0

( )vmseparation relative velocity x tapproach relativevelocity V

ε −= =


Problem 7

22 0n nx x xζω ω+ + =

Equation of Motion

Homogeneous Solution ( )sinnth h d hx X e tζω ω φ−= +

Xh and ϕh are determined from the initial conditions x(t=0) = (0) and 0( 0)x t V= =


Problem 7

Constants Xh and ϕh

Then, at t=0:

( )sin 0 1 sin( ) 0nth h d h h h hx X e t Xζω ω φ φ φ−= + → = × × → =

& as:( ) ( )cos sinn nt t

h d h d h n dx X e t X e tζω ζωω ω ζω ω− −= −

at t=0:

00 d h h

d

VV X Xωω

= → = ( )0 sinnth d

d

Vx e tζω ωω

−=


Problem 7

The bumper leaves contact with the barrier if:

0 0 02

nkkx cx x x x xc

ωζ

+ = → + = → + =

or: 2

22

2 1tan( ) sin( ) 2 11 2d dt tζ ζ

ω ω ζ ζζ−

= → = −−

( ) 22

2n

vm ox t x V e ζωζ

−= − = −


Problem 7

0

( )vmseparation relative velocity x tapproach relativevelocity V

ε −= =

22

22

0

oV e eV

ζζε

−−−

= − =

Dimensionless Energy Dissipation2

2

2 2 4

4

2

1 12 2 11

2

o oDissipated

initialo

mV mV eEe

E mV

ζ

ζ

−

−−

= = −


Problem 7

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Damping Ratio

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Dim

ensi

onle

ss D

issi

pate

d En

ergy


Problem 8


Problem 8

Equation of Motion

Solution where

Where also


Problem 8


Problem 9

Derive the equation ofmotion of the spring-masssystem shown in the figure.

Determine the freeresponse of the mass subjected toan initial velocity.


Maxwell modelStandard linear solid model

If k1 ∞ leads to Kelvin-Voigt model

Problem 9


Equation of Motion

&

&

Problem 9


Applying the Laplace Transform, gives

The resulting dimensionless equations, are

Problem 9


Solution of the transformed equations, are

Force Transmitted to the Fixed Support

Problem 9


If γ → ∞

Problem 9


Problem 9%********Review of Exam 2***************%***********Problem 6******************syms s t z gF=(1+2*z*s)/(s^2+2*z*s+1);z=0.15;FS=subs(F);ft=ilaplace(FS,s,t)t=linspace(0,15,100);fs=subs(ft);plot(t,fs,'b',T,fgt,'r','linewidth',3)hold on%*************************************

z=0.15; g=1;n=[2*z*(1+g) g];d=[2*z g 2*z*(1+g) g];[R,P,K]=residue(n,d);for ij=1:1000

t=0.015*(ij-1);fgt(ij)=R(1,1)*exp(P(1,1)*t)...

+R(2,1)*exp(P(2,1)*t)...+R(3,1)*exp(P(3,1)*t);

T(ij)=t;end%************************************

plot(T,fgt,'r','linewidth',3)xlabel('t');ylabel('FB/(kVo/\omegan)');grid


0 5 10 15-1

-0.5

0

0.5

1

time

FB/(k

Vo/ ω

n)

γ=inf γ=1

Problem 9


O

θ

m

L

kDerive the equation ofmotion of the pendulumsystem shown in the figure.

Study the stability of thesystem as function of thedamping coefficient c of thedamper.

Problem 10

c

k

x


Problem 10

Kinetic Energy 2 212

T mL θ=

Potential Energy

( ) ( )

( )

2 2

2 2 2

1 1212

U k L x kx mgL cos

k L x mgL kx

θ θ

θ θ

= − + − −

= − − +

Rayleigh Dissipative Function ( )212

D c L xθ = −


Problem 10

Equations of Motion

( ) ( )2mL kL L x mgL cL L xθ θ θ θ+ − − = − −

( ) ( )k L x kx c L xθ θ− − + = −

d L L Ddt θθ θ

∂ ∂ ∂ − = − ∂∂ ∂

d L L Ddt x x x

∂ ∂ ∂ − = − ∂ ∂ ∂

&


Problem 10

Using the Laplace Transform

( ) ( )2 2mL s X kL L X mgL cLs L Xθ θ θ+ − − = − −

( ) ( )k L X kx cs L Xθ θ− − + = −

(1)

(2)

From eqn. (2)

2X cs kL cs k

θ+=

+(3)


Problem 10

From eqns. (1) & (3)( )2 kL mgc k c Xs s s

m mL m m Lθ

− + + = +

( ) ( )3 2 2 2 2 22 0n nn p n ps s sω ωω ω ω ω θ

ζ ζ

+ + − + − =

Let 2 22 n n pc , k / m, g / Lm

ζω ω ω= = =

( )22

kL mgc k c cs ks s sm mL m m cs k

θ θ− + + + = + +


Problem 10

( ) ( )3 2 2 21 11 1 2 0 p

n n

ss s s where ,sω

ζ ζ ω ω+ + −Ω + − Ω = Ω = =

Divide by , the characteristic equation becomes𝝎𝝎𝒏𝒏𝟑𝟑

Putting the characteristic equation into Root Locus form gives

( )( )( )

2 2

2 2

11 21 0 1 0

1

s N sgain

D ss sζ

+ −Ω + = → + =

+ −Ω


Problem 10

-1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0.8

0.140.280.420.560.680.8

0.91

0.975

0.140.280.420.560.68

0.91

0.975

0.20.40.60.811.21.4

Root Locus

Real Axis (seconds-1

)

Imag

inar

y A

xis

(sec

onds

-1)

System: sys

Gain: 1.41

Pole: -0.707

Damping: 1

Overshoot (%): 0

Frequency (rad/s): 0.707

MATLAB

w=.5;%w=Ω2

n=[1 0 0.5-w];d=[1 0 1-w 0];rlocus(n,d)

Best Design of damperζ = 1/gain

= 1/1.414 = 0.707


END

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