module4 s dynamics- rajesh sir

Structural Analysis - III

Structural Dynamics

Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur

Dept. of CE, GCE Kannur Dr.RajeshKN

1

Module IVModule IVStructural dynamics

• Introduction – degree of freedom – single degree of freedom li t ti f ti D’Al b t’ i i l linear systems – equation of motion – D’Alembert’s principle –damping – free response of damped and undamped systems –logarithmic decrement – response to harmonic and periodic excitation – vibration isolation.


2

IntroductionIntroduction

Dynamic LOAD

• Load whose magnitude, direction or position changes with time

P ib d l d Ti i ti f th l d i f ll kPrescribed load: Time-variation of the load is fully known

Analysis of response: Deterministic

Random load: Time-variation of the load is NOT fully known

Analysis of response: Non-deterministic


3

Dynamic RESPONSEDynamic RESPONSE

• Deflections and stresses are time-variant

Normally, dynamic load/displacement causes dynamic response.

P bl i d i l h h i d iProblem is dynamic only when the response is dynamic.


4

Classification of vibratory systemsClassification of vibratory systems

I Type of load/excitation Prescribed systemsI. Type of load/excitation Prescribed systems

Random systems

II. Linearity Linear – Linear differential equations of motion

Non-Linear – Non-linear differential equations

III. Type of mathematical model

Di t (L d) t d l Discrete (Lumped) parameter model – ordinary differential equations

Distributed (Continuous) parameter model


5

( ) p– partial differential equations

Types of prescribed loadingsyp p gPeriodic Non-periodic

• E g Sinusoidal simple harmonicE.g. Sinusoidal, simple harmonic • E.g. Impact• Any periodic load can be represented by a sum of simple harmonic componentsby a sum of simple harmonic components


6

Characteristics of a dynamic problemC a acte st cs o a dy a c p ob e

Ti i • Time-varying response – Non-unique solutions

• Presence of inertia forces– If motion is slow, inertia is neglected – problem is static


7

Degrees of freedomDegrees of freedom

• Number of independent coordinates required to specify the configuration of the system at any given time

• Single Degree of Freedom SDOF g g• Multiple Degrees of Freedom MDOF

Types of Vibrations

F F dFree Forced

Damped Undamped


8

Damped Undamped

Structures that can be modelled as SDOF systems


9

Structures that can be modelled as SDOF systems

Components of the basic dynamic systemp y y(Mass-Spring-Dashpot model of an SDOF system)

Dashpot Mass

Spring Smooth surface

• Damping

• Energy dissipating mechanism• Friction, viscosity etc.


10

Dynamic equilibrium - SDOF systemy q y

F

S i tiffk

y

Spring stiffness

Damping constant

M

kcm

mycy Damping force

Inertial forcey

kcy

my F

Massm

( )kyF t

Spring forcep g

Force of excitationky y

( ) displacement at any time y t t→( ) p y

velocity at any time dyy tdt

= →

( )k F t+ +

Eqn. of dynamic equilibrium


2

2 acceleration at any time d yy tdt

= →( )my cy ky F t+ + =

Springs

Springs in series Springs in parallel


12

Free vibrationUndamped free vibration

Free vibration

One DOF, no damping, no external forces (only initial displacement condition) y

• Formulation using Newton’s law

F ma=

i k

Newton’s law:

D’Alembert’s principle: 0my ky+ =

i.e., ky my− =

ky myD Alembert s principle: 0my ky+ =

Differential equation f


13

of motion

• Effect of gravity

kyk0ky

0y0y y+

( )0k y y+

WW

W

Wmy

Static deflection

( )W k

W

0W ky= W

Static deflectionVibration

( )0W k y y my− + =

S ti b0 0ky ky ky my− + =


14

0my ky⇒ + =• Same equation as above• Hence, gravity has no effect in vibration

Solution of differential equation of motion

0my ky+ =S l ti

cosLe st iny A t B tω ω= +

Solution:and.cos .siny A t y B tω ω= = will satisfy the eqn.

.cosLe st . iny A t B tω ω= +

Substituting in the above eqn.,

( ) ( )2 .cos .sin 0m k A t B tω ω ω− + + = 2 0m kω⇒ − + =

g q ,

2 kmω =

Natural frequency( Angular /circular natural frequency)

ω→ (denoted as a o) lsnω


15

( g / q y)Units: radians/second

To find y: y.cos .siny A t B tω ω= +

.sin .cosy A t B tω ω ω ω= − +

Initial conditions: 0 at 0y y t= =

at 0t0 at 0y v t= =

y A⇒ = ( )0tan y

vα

ω=

0

0

,y Av Bω⇒ ==

0 ivt t

( )0v ω

00 cos siny y t tω ω

ω∴ = +

( )siny C tω α+The above solution can also be written as ( ).siny C tω α= +The above solution can also be written as,

22 0vC ⎛ ⎞

⎜ ⎟Wh called the amplitude of vibration


16

2 00C y

ω⎛ ⎞= + ⎜ ⎟⎝ ⎠

Where , called the amplitude of vibration

y

22 00

vC y ⎛ ⎞= + ⎜ ⎟⎝ ⎠

0v

y

0C yω

+ ⎜ ⎟⎝ ⎠

0y

αt

αω 2T π

ω=

Undamped free vibration response

To find time when y=0 : ( ) ( )0 .sin sin 0C t tω α ω α= + ⇒ + =To find time when y 0 : ( ) ( )0 .sin sin 00

C t tt t

ω α ω αω α α ω

+ ⇒ +

⇒ + = ⇒ = −


Note:

2⎛ ⎞ β

( ).siny C tω α= +

Note:

0y

22 00

vC yω

⎛ ⎞= + ⎜ ⎟⎝ ⎠

α

β( )0 0

cos .sin sin .cos

cos sin

C t t

y vC t t

ω α ω α

ωω ω

= +

⎛ ⎞= +⎜ ⎟0v ω

y0

0

cos . sin .

cos sin

C t tC Cvy t t

ω ω

ω ω

= +⎜ ⎟⎝ ⎠

= +( )

0

0

tan yv

αω

=0yω


18

2f ω

π=Cyclic frequency Units: cycles/second

1 2Tf

π= =Time period Units: seconds

f ω


19

Problem 1: Determine the natural frequency of the system shown in figure, consisting of a weight of 50 N attached to a cantilever through the coil spring k2=20 N/m. The cantilever cross-section is 200x300 mm, Young’s modulus of elasticity E=2.5x104 MPa, L=2m.g y

3PlL

3PlEI

Δ =

50 N

34 200 300⎛ ⎞×

50 N

1 3

3P EIkl

= =Δ

4

3

200 3003 2.5 1012

4218.75N mm 4218750 N m2000

⎛ ⎞×× × ×⎜ ⎟

⎝ ⎠= = =

1 1 14218750 20k

= + 219.9kg s = 1.98 rad sekω = =


4218750 2019.9 N m

e

e

kk∴ =

( )

50 / 9.1.98 ra

8 gd s

1 kmω

Problem 2: Calculate the natural frequency in side sway and natural period of vibration for the frame in figure If the initial displacement is period of vibration for the frame in figure. If the initial displacement is 25 mm and the initial velocity is 25 mm/sec, what is the amplitude and displacement at t = 1 sec? Weight of the beam =

12

630 10 N×

SDOF, Undamped free vibration

1230 10 MPaAB CDEI EI= = ×

6 25

2

30 10 kg.m s 30.58 10 kg9.81 m s

m ×= = ×

12 12EI EI 12 1 112 30 10 ⎛ ⎞3 3

12 12AB CD

AB CD

EI EIkl l

= +12

3 3

1 112 30 101000 800⎛ ⎞= × × × +⎜ ⎟⎝ ⎠

1063125 N/mm=

23 kg m/s


3 kg.m/s1063125 10 m

= ×

3 2

5

1063125 10 kg s = 30.58 10 k

18.645 r dg

a skm

ω ×= =

×30.58 10 kgm ×

2 2 0.33718.645

Time period T sπ πω

= = =

( ).siny C tω α= +Displacement of SDOF, undamped free vibration ( )

22 v⎛ ⎞ ( )

0

0

tan yv

αω

=2 00

vC yω

⎛ ⎞= + ⎜ ⎟⎝ ⎠

Where amplitude( )0v ω

( )25 18.645

25 18 645= =


( )25 18.645086.93 1.517radα∴ = =

2 22 22 200

252518.645

25.036 mmvC yω

⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Hence, amplitude

( ).siny C tω α= +Displacement at t = 1 sec

( )25.036sin 18.645 1 24.207 m7 m1.51= × =+( )25.036sin 18.645 1 24.207 m7 m1.51+


Problem 3: A particle of mass 2g is making simple harmonic motion l i A di 6 d 10 f h ilib i along x-axis. At distances 6cm and 10cm from the equilibrium

position, the velocities of the particle are 5 cm/s and 4cm/s respectively. Find the time period of vibration, the amplitude and maximum kinetic energy.

( ).siny C tω α= +( )

( ).cosy C tω ω α= + maxy Cω=

( ) ( )1 16 sinC tω α+ →=

( ) ( )10 2iC

( ) ( )15 c 3osC tω ω α+ →=

( ) ( )4 4C( ) ( )210 s 2inC tω α+ →= ( ) ( )24 c 4osC tω ω α+ →=

( ) ( )( ) ( )16 sin1 C tω α=→ +

( )4 4 4⎡ ⎤


24

( )( ) ( ) ( )2 1

43

4cos cos5

t tω α ω α+ = +→ ( ) ( )12 1

4cos cos5

t tω α ω α− ⎡ ⎤⇒ + = +⎢ ⎥⎣ ⎦

6 4⎛ ⎞⎡ ⎤( ) ( ) ( )11

1

6 410 sin cos coss

2in 5

tt

ω αω α

−→⎛ ⎞⎡ ⎤= +⎜ ⎟⎢ ⎥+ ⎣ ⎦⎝ ⎠

01 0.4228 24.24t radω α+ = =With trial and error,

14 6229C cm∴ =Amplitude

1,

14.6229C cm∴ =Amplitude,

0.3749 rad sω =Natural frequency,q y

2 16.749T sπ ω= =Time period of vibration,

( )221 1 30 06C NM i Ki ti E

p


25

( )2max 30.06

2 2my m C Nmω= = =Maximum Kinetic Energy,

Damped free vibrationpOne DOF, with damping, no external forces (only initial displacement condition)

0my cy ky+ + =y

2 0pt pt ptC C kC

pty Ce=Solution is of the form

2 0pt pt ptmCp e cCpe kCe∴ + + =2 0mp cp k⇒ + + =

In general, the roots of the above equation are 2

1 2, c c kp p ⎛ ⎞= − ± −⎜ ⎟⎝ ⎠

General solution is:

1 2,2 2

p pm m m⎜ ⎟

⎝ ⎠1 2

1 2p t p ty C e C e= +


1 2,C C To be determined from initial conditions

2 k⎛ ⎞2c km m

⎛ ⎞ −⎜ ⎟⎝ ⎠

Final form of the solution depends on the sign of

Case 1:2c k⎛ ⎞ 0

2c km m

⎛ ⎞ − =⎜ ⎟⎝ ⎠

2c km⇒ =2 k

mω =

i e 2 2 2c km m kω ω= = =

This is defined as critical damping, ccr

i.e., 2 2 2crc km m kω ω

1 2,2

crcp pm

⇒ = −2

02c km m

⎛ ⎞ − =⎜ ⎟⎝ ⎠

Now, Equal roots


1 21 2 p t p tC e C teTwo independent solutions and

2crc t− 2

crc tmy C te

−d

c

21 1

my C e∴ = 22 2

my C te=and

( ) 21 2

crc tmy C C t e

−= +Hence,

F ib ti ith iti l d i


Free vibration response with critical damping

Case 2: 2

0c k⎛ ⎞ >⎜ ⎟2 2

0crc c⎛ ⎞ ⎛ ⎞⇒ > ⇒ >⎜ ⎟ ⎜ ⎟02m m

− >⎜ ⎟⎝ ⎠

02 2

crcrc c

m m⇒ − > ⇒ >⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

• i.e., Damping in the system is larger than critical damping(Overdamped system)

• Non-oscillatory motion, exponentially decaying to zero

• Two real, distinct roots for the equation

Damping ratio,2

c cc m

ζω

= = 1crc c ζ> ⇒ >2crc mω


( )y t

t

Free vibration response of critically damped and Free vibration response of critically damped and overdamped systems


30

Case 3: 2

0c k⎛ ⎞ <⎜ ⎟2 2

0crc c c c⎛ ⎞ ⎛ ⎞⇒ < ⇒ <⎜ ⎟ ⎜ ⎟ 1ζ⇒ <02m m

− <⎜ ⎟⎝ ⎠

02 2

crcrc c

m m⇒ − < ⇒ <⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠1ζ⇒ <

• i.e., Damping in the system is smaller than critical damping(Underdamped system)• Oscillatory motiony• Complex roots for the equation

2c k c⎛ ⎞1 2,

2 2c k cp p im m m

⎛ ⎞= − ± − ⎜ ⎟⎝ ⎠

iζ ±

nω ω=

2c c

c mζ

ω= =

n Diζω ω= − ±

2k c⎛ ⎞⎜ ⎟

2cr nc mω

Damped natural 2D m m

ω ⎛ ⎞= − ⎜ ⎟⎝ ⎠

2i 1 ζ

Damped natural frequency


2i.e., 1D nω ω ζ= −

( ) ( )n D Dt i t i tt A Aζω ω ω− −+( ) ( )1 2n D Di t i ty t e A e A eζ ω ω= +

cos sinDi tD De t i tω ω ω= ±But

( ) ( ) ( )1 2 1 2cos sinntD Dy t e A A t i A A tζω ω ω−∴ = + + −⎡ ⎤⎣ ⎦

( ) [ ]i e cos sinnty t e A t B tζω ω ω−= +( ) [ ]i.e., cos sinnD Dy t e A t B tζ ω ω= +


Applying initial conditionsApplying initial conditions,

( ) 0 0cos sinnt nv yy t e y t tζω ζωω ω− ⎡ ⎤++⎢ ⎥( ) 0 0

0 cos sinn nD D

D

yy t e y t tζ ζω ωω

= +⎢ ⎥⎣ ⎦

0 0

0

tan n

D

v yyζωα

ω+

=

( ) ( )cosnty t Ce tζω ω α−=

0D yω

( ) ( )cos Dy t Ce tω α= −Or,

22 0 00

nv yC y ζωω

⎛ ⎞+= + ⎜ ⎟

⎝ ⎠Where


33

Dω⎝ ⎠

( ) ( )cosnty t Ce tζω ω α−= −( ) ( )cos Dy t Ce tω α= −

Oscillatory motion, with an exponentially decaying amplitude of y , p y y g pntCe ζω−

2π

2T πω

=y

2

2 Damped natural period

1 D d i l t l fd

Tpπ

ζ

= =

tny 1ny +

21 Damped circular natural frequencydp p ζ= − =

nty Ce ζω−=


( )y t

t

Free vibration response of critically damped, overdamped and underdamped systemsp y


35

2 D d t l i dT π

Extremum point ( )( ) 0cos( ) 1y t

tω α=− =Point of tangency ( )

2T πω

=y

2

Damped natural period

1 Damped circular natural frequencyd

d

Tp

p p ζ

= =

= − =t

ny 1ny +

p q ydp p ζ

nty Ce ζω−=


36

( )y t0v

nty Ce ζω−=

( )0y ( )0y

nty Ce ζω−=y Ce

Effect of damping on free vibrationEffect of damping on free vibration


37

For structures, damping c ranges between 2 to 20% of ccr.

When c is 20% of ccr, % cr,

0.2ζ = 2i.e., 1 0.98D n n nω ω ζ ω ω= − =

Hence, for structures, damped natural frequency is practically same as undamped natural frequency same as undamped natural frequency


Problem 3: A machine of mass 20kg is mounted on springs and dampers. The total stiffness of the springs is 8kN/m and the total dampers. The total stiffness of the springs is 8kN/m and the total damping is 130 Ns/m. If the system is initially at rest and a velocity of 100mm/s is imparted to the mass, determine: 1) displacement and velocity of the mass as a function of time 2) displacement at t=1svelocity of the mass as a function of time, 2) displacement at t=1s.

N t l f k

Damped free vibration

8000N m

Natural frequency, n mω =

k/2 k/2

8000N 20 kg

m= 20 r= ad s

k/2 k/2Damping ratio, 2 n

cm

ζω

=

130 N13020 202

Ns mkg rad s

=× × 0.1625=


Damped natural frequency, 2 19.734 rad/s1D nω ω ζ= − =

( ) ( )cosnty t Ce tζω ω α−=We have ( ) ( )cos Dy t Ce tω α= −We have,

for damped free vibration.

22 0 0 nv yC ζω⎛ ⎞++ ⎜ ⎟h

0 0tan nv y ζωα +=and2 0 0

0n

D

yC y ζω

= + ⎜ ⎟⎝ ⎠

where0

tanD y

αω

and

Initial conditions: 00 100 0.10,vy = == mm/s m/s

20.1 00

D

Cω

⎛ ⎞+∴ = + ⎜ ⎟

⎝ ⎠0.005=

0.1 0t + 090 1 571 dα⇒

D⎝ ⎠


40

tan0

α = = ∞ 090 1.571radα⇒ = =

( ) ( )0.1625 200.005 cos 19.734 1.571ty t e t− ×= −Hence,

( )3.250.005 cos 19.734 1.571te t−= −

( ) ( ) ( )3.250 005 19 734sin 19 734 1 571 3 25cos 19 734 1 571ty t e t t−= ⎡ ⎤⎣ ⎦

and,

( ) ( ) ( )0.005 19.734sin 19.734 1.571 3.25cos 19.734 1.571y t e t t= − − − −⎡ ⎤⎣ ⎦

( )3.250 005 19 734 1 571−

Displacement at t=1s

( ) ( )3.251 0.005 cos 19.734 1.571ty e= = −

41 5 10 0 15m mm− == ×


41

1.5 10 0.15m mm== ×

Problem 4:Problem 4:

SDOF, damped free vibration

d d l f d10Undamped natural fr radequ sency nω =

10% 0.1Damping ratio ζ = =p g ζ

0 00, 0.05 Initial conditions m/sec: y v= =

21Damped natural fre uency q D nω ω ζ∴ = −


210 1 0.1 9.9499 rad s= − =

0E i f i k 0Equation of motio : n my cy ky+ + =

0i.e., c ky y y+ + = 22 0y y yξω ω⇒ + + = 2 100 0y y y⇒ + + =

Solution of the equation of motion

, y y ym m

2 0n ny y yξω ω⇒ + + y y y

( ) ( )cos

Solution of the equation of motioni.e., Displacement, nt

Dy t Ce tζω ω α−= −

22 0 00where, n

D

v yC y ζωω

⎛ ⎞+= + ⎜ ⎟

⎝ ⎠

20.05 00 0.0059.9499

+⎛ ⎞= + =⎜ ⎟⎝ ⎠Dω⎝ ⎠ ⎝ ⎠

1 0 0tan nv y ζωα − ⎛ ⎞+= ⎜ ⎟

1 10.05 0tan tan0 2

πα − −+⎛ ⎞= = ∞ =⎜ ⎟⎝ ⎠

( ) 0.005 cos 9.9499ty t e t π− ⎛ ⎞∴ = −⎜ ⎟

0D yω⎜ ⎟⎝ ⎠

0 2⎝ ⎠


( ) 0.005 cos 9.94992

y t e t∴ ⎜ ⎟⎝ ⎠

Logarithmic decrement 2

n

T πω

=yLogarithmic decrement

( )111 cosnt

DCe ty ζω ω α− −2

2 Damped natural period

1 Damped circular natural frequencyd

d

Tp

p p

π

ζ

= =

= − =t

ny 1ny +

nty Ce ζω−=( )( )2

11

2 2cosn

Dt

D

yy Ce tζω ω α−=

−

( )111 cosnt

DCe ty ζω ω α− −( )

( )1

112

2

1cos 2n

D

D

t

D

yy

Ce tπζω

ω ω π α⎛ ⎞

− +⎜ ⎟⎝ ⎠

=

+ −( )1D

22 ππ ζ2

21

nnnDe e

π ζωζωω ζω −= =


⎛ ⎞ 2 ζ1

22

2ln1

n

n

yy

πζωω ζ

⎛ ⎞=⎜ ⎟

−⎝ ⎠2

21πζζ

=−

(say)δ=

2πζ 221

2

nD

y e e ey

πζωω πζ δ= = =

2πζ

2y

δ δ2

21πζζ

δ=− ( )2 2 22

δ δζππ δ

⇒ =+

2i e δ πζ (Logarithmic decrement)

( )


45

2. .,i e δ πζ= (Logarithmic decrement)

A practical way to determine damping A practical way to determine damping - Logarithmic decrement/ exponential decay method

1

2

y ey

δ=2y

1 1 2 3

1 2 3 4 1

ny y y y yy y y y y

=After n cycles,1 2 3 4 1n ny y y y y+ +

. . . ne e e e eδ δ δ δ δ= =

11 ln yδ⎛ ⎞

= ⎜ ⎟⎝ ⎠

∴


1nn y +⎜ ⎟⎝ ⎠

Problem 5:

SDOF damped free vibrationSDOF, damped free vibration

8000 kgm =12 12EI EI

3 3

12 12AB CD

AB CD

EI EIkl l

= +

624 6 10× ×


63

24 6 10 5.332 10 N m3

× ×= = ×

Undamped natural frequency nkm

ω =65.332 10 25.82

8000 rad s×

= =

4% 0 04Damping ratio ζ = = 30 0Initial conditions : y v= =4% 0.04Damping ratio ζ 0 030, 0Initial conditions : y v= =

21D d t l f ζ 21Damped natural fre uency q D nω ω ζ∴ = −

225 82 1 0 04 25 8 rad s= =25.82 1 0.04 25.8 rad s= − =


48

2Logarithmic decrement, δ πζ=

02 0.04 0.251 ln yy

δ π⎛ ⎞

∴ = × = = ⎜ ⎟⎝ ⎠1y⎝ ⎠

0 30y01 0.251

30 23.33 mmyye eδ∴ = = =

03 3 3 0 251

30 14.12 mmyy δ= = =1

2 0.251

23.33 18.15 mmyye eδ= = =

3 3 3 0.251 14.12 mmye eδ ×

0 30 10 98 mmyy = = =0

2 2 0.251

30 18.15 mmye eδ ×= = =

4 4 4 0.251 10.98 mmye eδ ×= = =

0 30 8 54 mmyy = = =5 5 5 0.251 8.54 mmye eδ ×= = =

0 30 6 6 mmyy = = =


6 6 6 0.251 6.6 mmye eδ ×= = =

Problem 6: A platform of weight 20kN is supported by four equal l l d h f d i ll h l f A i columns clamped to the foundation as well as to the platform. A static

force of 8kN applied horizontally to the platform produces a displacement of 3mm. Damping in the structure is 5% of critical damping. Find:

1. Undamped natural frequency2. Logarithmic decrementg3. Number of cycles and time required for the amplitude to

reduce from an initial value of 3mm to 0.3 mm.

Fk =Δ

8000 2666.673

N N mmmm

= =Stiffness,

nkm

ω =1. Undamped natural frequency m

32666.67 1020000 9 81

N mk

×= 36.15= rad s


50

20000 9.81 kg

2δ πζ=2. Logarithmic decrement,

2 0.05 0.314π= × =

3. Number of cycles required for the amplitude to reduce 3. Nu be o cyc es equ ed o t e a p tude to educe from an initial value of 3mm to 0.3 mm

3y = mm 0 3y = mm1 3y = mm 1 0.3ny + = mm

11 y⎛ ⎞ 1 3⎛ ⎞1

1

1 lnn

yn y

δ+

⎛ ⎞= ⎜ ⎟

⎝ ⎠

1 30.314 ln0.3n

⎛ ⎞⇒ = ⎜ ⎟⎝ ⎠

7.333cy en cl s⇒ =


51

4 Ti i d f th lit d t d f i iti l 4. Time required for the amplitude to reduce from an initial value of 3mm to 0.3 mm

333 d f b7.333 Time period of vibration×=

27 333 π 27 333 π7.333nω

×= 7.33336.15

= ×

1 2 4 d1.274 seconds=


52

Forced vibration• External forces cause vibration

Forced vibration

Response of undamped system to harmonic excitationy

0 sinF tω

ky my 0 sinF tω


( ) sinF t F tω=Excitation (force): 0F Amplitude of excitation( ) 0 sinF t F tω=Excitation (force):

sinmy ky F tω+ A

ω Frequency of excitation

( ) ( ) ( )y t y t y t= +Solution is

0 sinmy ky F tω+ = A

( ) ( ) ( )c py t y t y t= +Solution is

( )( ) .cos .sinc n ny t A t B tω ω= +

( )

( )cy t Complimentary solution – Soln of homogeneous eqn 0my ky+ =

( )py t ( )my ky F t+ =Particular solution – Soln of non-homogeneous eqn


( ) siny t Y tω=Let ( ) sinpy t Y tω

A 2m Y kY Fω− + =

Let

A 0m Y kY Fω + =

F02

FYk mω

=−

0 02 2

F k F k= =2 2

21 1nk m

ω ωω

− −η ω= Frequency ratio

n

ηω Frequency ratio

( )021

sty=

( )F k St ti d fl ti


21 η− ( )00 stF k y= Static deflection

( ) 0cos sin sinF ky t A t B t tω ω ω= + +Hence total solution

L t i iti l diti ( ) ( )and0 0 0 0y y y v= = = =

( ) 2.cos .sin sin1n ny t A t B t tω ω ω

η= + +

−Hence, total solution,

Let initial conditions are: ( ) ( )0 0and0 0 0 0y y y v= = = =

020,

1F kA B ηη

−∴ = =

−

( ) 0 0sin sinF k F ky t t tη ω ω−= +( ) 2 2sin sin

1 1ny t t tω ωη η

= +− −

( ) ( )02 sin sin

1i.e., n

F ky t t tω η ωη

= −


1 η−

( )( ) ( ) ( )20O sir, n sin

1 st

nyy

t t tω η ωη

= −−

• The above represents a superposition of two harmonic responses of different frequencies

• The result is NOT harmonic


57

( )( )y ty( )0sty

( )0 0y = ( )0 00 nv y F kω= =


( ) ( ) ( )02 sin sins

nty

yt t tω η ω= −( ) ( )21 ny η

η−

( ),n y tω ω= = ∞→When Resonance

Amplitude is , but structure will fail before displacement reaches ∞∞


59

Response of damped system to harmonic excitation

( )y t

0 sinmy cy ky F tω+ + =

B0 sinF tω

( ) ( ) ( )

B

( ) ( ) ( )c py t y t y t= +Solution is

( )t C li t l ti S l f h ( )cy t Complimentary solution – Soln of homogeneous eqn0my cy ky+ + =

( ) [ ]cos sinntc D Dy t e A t B tζω ω ω−= +


• But damping will cause this part to die out -> Transient response

( )y t P ti l l ti S l f h ( )py t

( )my cy ky F t+ + =Particular solution – Soln of non-homogeneous eqn

Thi t i > St d t t

( ) 1 2sin cospy t C t C tω ω= +

•This part remains -> Steady state response

S b tit ti thi b k i th f ti d ti th Substituting this back in the eqn of motion and equating the

coefficients of sin cos ,t tω ωand

2C m C c C k Fω ω− − + =1 2 1 0C m C c C k Fω ω + =

2 0C m C c C kω ω+ +


2 1 2 0C m C c C kω ω− + + =

( )( ) ( )

20

1 2 22

F k mC

k

ω−=

+ ( ) ( )0

2 2 22

F cCk m c

ω

ω ω

−=

+( ) ( )2k m cω ω− + ( ) ( )k m cω ω− +

( )( ) ( )

( )202 22

sin cospFy t k m t c tω ω ω ω⎡ ⎤= − −⎣ ⎦( )

( ) ( )( )2 22p

k m cω ω ⎣ ⎦− +


62

( )( ) ( )

( ) ( )[ ]

2 220

2 22cos sin sin cosp

F k m cy t t t

k

ω ωφ ω φ ω

ω ω

− += −

+( ) ( )k m cω ω− +

22k mω−φ− 2tan c

k mωφω

=−

( ) ( )2 22k m cω ω− +cω−

( )( ) ( )

( )02 22

sinpFy t t

k m cω φ

ω ω= −

− +


( ) ( )

( ) ( )0 sinpF ky t tω φ= −( )

( ) ( )( )

2 22s

1 2py t tω φ

η ζη− +

( ) ( )2

0 sinstytω φ= −

( ) ( )( )

2 221 2φ

η ζη− +

( )( )( ) ( ) ( )

( )2

022

1 sin1 2t

p

s

yy

ttω φ

η ζη= −

− +( ) ( )


( ) ( ) ( )( ) ( ) ( ), c py t y t y t= +Total response

[ ] ( ) ( )0cos sin sinn stt ye A t B t tζω ω ω ω φ− + +[ ] ( )

( ) ( )( )0

2 22cos sin sin

1 2n

D De A t B t tζ ω ω ω φη ζη

= + + −− +Transient response

Steady state response


65

( )y t( )0sty

( )0 00 nv y F kω= =( )0 0y =


( ) ( )( ) ( )sinpy t Y tω φ= −

( )y( )

( ) ( )0

2 22,

1 2Where, amplitude of steady-state vibration sty

Yη ζη

=− +

( ) ( ) ( )2 220

,1

1 2Dynamic amplification factor,

t

DYy η ζη

=+( ) ( ) ( )20 1 2sty η ζη− +

,nω ω=When resonance happens.

1Y Y Y= =( )0F k

=Hence, resonant amplitude,


1n

Y Y Yηω ω ==resonant 2ζ

tor

tion

fact

mpl

ifica

tna

mic

am

Dyn

nη ω ω=Frequency ratio

Dynamic amplification factor as a function of frequency ratio


Dynamic amplification factor as a function of frequency ratio for various amounts of damping

nη ω ω=

Phase angle as a function of frequency ratio for various amounts of damping

2tan cω ζηφ = =


2 2tan1k m

φω η

= =− −

Problem 4:

SDOF, damped vibration with harmonic excitation

2800 kg.m s 81 55 kgm = =

SDOF, damped vibration with harmonic excitation

2 81.55 kg9.81 m s

m

548 48 2 10 6000EI × × ×

d/

3 3

48 48 2 10 6000 0.9 N mm4000

EIkL

× × ×= = =


15 rad/secω =0 sin 20sin15F t tω = 0 20F Newtons=

900 N/k 15900 N/m 3.32 rad/sec81.55 kgn

km

ω = = =15 4.52

3.32 n

ωηω

= = =

5% 0.05ζ = =

Amplitude of steady state vibration,

( )( ) ( )

0max 2 221 2

F ky Y

η ζη= =

− +( ) ( )1 2η ζη− +

( )20 900( )

( ) ( )2 22

20 900

1 4.52 2 0.05 4.52=

− + × ×31.143 10 m−= ×


71

R t lit d f t d t t ib ti

( )20 900F k

Resonant amplitude of steady state vibration,

resonant 1y Yη== ( )20 9000.222 m

2 0.05= =

×0

2F kζ

=


72

Vibration isolationVibration isolationA method for protecting equipment from vibrating foundation O f f b hOR for protecting structure from vibrating machinery

2 Vib ti hi 1. Vibrating foundation

iF t

2. Vibrating machinery

( )y t( )y t

0 sinF tω

( )y t

( )y

( )sy t


1. Response to support motion (Vibrating foundation) p pp ( g )

( ) 0 sinsy t y tω= Support motion - harmonic

( )y t → Total displacement of the mass including support motion

( )y t

including support motion

( ) ( )sy t y t− → Net displacement of the mass ( )y

Hence, equation of motion is:

( ) ( ) 0s smy c y y k y y+ − + − = ( )sy t

0 0sin cosmy cy ky ky t c y tω ω ω+ + = +


( )0 sinmy cy ky F tω β+ + = +

( ) ( )2 22 ( ) ( )2 220 0 0, 1 2Where F y k c y kω ξη= + = +

tan 2ckωβ ζη= =

( )0 sinF tω β+

( )y t

( )

( )y t

Equivalent to

( )

q

( )sy t


75

F k( )( ) ( )

( ) ( )02 22

sin sin1 2

F ky t t Y tω β φ ω β φη ζη

∴ = + − = + −− +

T Amplitude of respoTransmissibility nse = RTAmplitude

Transmissibi of support

ld

ity ispl

ace

= ment

( ) ( )0

2 220

0 1 2R

F kYTyy η ζη−

∴ =+

= ( ) ( )0y η ζη

( )2( )

( ) ( )

2

2 220

1 2

1 2i.e., R

YTy

ζη

η ζη

+= =

− +( )20

0 1 2F yk

ζη= +∵


( ) ( )1 2η ζη+1 RT= −Degree of isolation

T R

n

ωηω

=Frequency ratio


Transmissibility versus frequency ratio for vibration isolation

2. Force transmitted to foundation (Vibrating machinery)

Excitation (due to vibrating machine)

( )

0 sinF tω

( ) ( )siny t Y tω φ= −

0 sinmy cy ky F tω+ + =( )y t

( ) ( )siny t Y tω φ

( ) ( )02 22

= F kY

( )F t cy ky= +

Force transmitted to foundation, ( ) ( )2 221 2η ζη− +

( ) ( ) ( )cos sinF t c Y t kY tω ω φ ω φ= − + −

( ) ( )22 sinY k c tω ω φ β= + − + tan 2ckωβ ζη= =


( )22Y k cω= +Max. force transmitted

Max force transmitted Max. force transmitted Force Transmissibility = Max. force of excitation

Here excitation is due to vibrating machine

( )22

0FY k cω

=+

∴Force Transmissib ility( ) ( )

( )2202 22 01 2

=k cF k

Fω

η ζη

+

− +( ) ( )1 2η ζη+

2

1 cω⎛ ⎞⎜ ⎟ 2

( ) ( )2 22

1

1 2= k

η ζη

⎛ ⎞+ ⎜ ⎟⎝ ⎠

− +

( )

( ) ( )

2

2 22

1 2

1 2= = RT

ζη

η ζη

+

− +( ) ( )1 2η ζη+ ( ) ( )η ζη

Same as Transmissibility


for vibrating foundation

Problem 5:

Max. force transmitted Force Transmissibility = RT 300 N=

( )21 2ζη+

Force Transmissibility Max. force of excitationRT

3500 N

( )

( ) ( )2 22

1 2

1 2Also, RT

ζη

η ζη

+=

− +

3 23

2

20 10 kg.m s 2.0387 10 kg9.81 m s

m ×= = ×

10 2 62.83 rad/sω π= × =

k2 2

2 62.83 8047990.264=ωη⎛ ⎞

= =⎜ ⎟

nkm

ω =


3 =2.0387 10n k k

ηω

= =⎜ ⎟ ×⎝ ⎠ 10% 0.1ζ = =

( )2 8047990.2641 2 0 1 + × ×( )

( )2

2

1 2 0.13003500 8047990.264 8047990.2641 2 0.1

R

kT+ × ×

= =⎛ ⎞− + ×⎜ ⎟⎝ ⎠

( )1 2 0.1 k k

+⎜ ⎟⎝ ⎠

22 28047990.264 321919.61 321919.61300 1 3500 1

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ 2 2300 1 3500 1

k k k⎛ ⎞ ⎛ ⎞× − + = × +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

( ) ( )22 2300 8047990.264 321919.61 3500 321919.61 k k k⎡ ⎤× − + = × +⎣ ⎦

( ) ( )( )

2 2 2 2 2

2 2 2

300 2 300 8047990.264 300 321919.61 3500

300 8047990 264 3500 321919 61 0

k k+ − × × + × −

+ × × =( )300 8047990.264 3500 321919.61 0+ × − × =

2 12 1890000 1 419 1 5 829 1 00 0k k− × + × =


7887098.515k∴ = N m90000 1.419 1 5.829 1 00 0k k× + ×

SummarySummaryStructural dynamics

• Introduction – degree of freedom – single degree of freedom li t ti f ti D’Al b t’ i i l linear systems – equation of motion – D’Alembert’s principle –damping – free response of damped and undamped systems –logarithmic decrement – response to harmonic and periodic excitation – vibration isolation.


82

Reference Books

1 St t l D i M i P1. Structural Dynamics – Mario Paz

2. Fundamentals of Vibrations - Leonard Meirovitch

3 Th f Vib ti ith A li ti Willi T Th3. Theory of Vibration with Application – William T Thomson

4. Mechanical Vibrations - Tse, Morse Hinkle

5 St t l D i M i k S l5. Structural Dynamics – Manicka Selvam

6. Dynamics of Structures – Anil Chopra

Dept. of CE, GCE Kannur Dr.RajeshKN83

module4 s dynamics- rajesh sir

Engineering

c t t t t

t y b t

cos siny y t t

time d y y t dt

timey t t p y velocity

siny c t

time dy y t dt

cy ky f t