Module4 s dynamics- rajesh sir

Structural Analysis - III
Structural Dynamics
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKN
1

Module IVModule IV
Structural dynamics
• Introduction – degree of freedom – single degree of freedom
li t ti f ti D’Al b t’ i i llinear 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 lDiscrete (Lumped) parameter model
– ordinary differential equations
Distributed (Continuous) parameter model
5
( ) p
– partial differential equations

Types of prescribed loadingsyp p g
Periodic 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 SDOFg 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
k
c
m
my
cy Damping force
Inertial force
y
k
cy
my F
Massm
( )
ky
F t
Spring force
p g
Force of excitation
ky
y
( ) displacement at any timey t t→( ) p y
velocity at any time
dy
y t
dt
= →
( )k F t+ +
Eqn. of dynamic equilibrium
2
2
acceleration at any time
d y
y t
dt
= →
( )my cy ky F t+ + =

Springs
Springs in series Springs in parallel
12

Free vibration
Undamped 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 my
D Alembert s principle: 0my ky+ =
Differential equation
f
13
of motion

• Effect of gravity
ky
k
0ky
0y
0y y+
( )0k y y+
W
W
W
W
my
Static deflection
( )W k
W
0W ky= W
Static deflection
Vibration
( )0W k y y my− + =
S ti b
0 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 k
m
ω =
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⇒ = ( )
0
tan
y
v
α
ω
=
0
0
,y A
v Bω
⇒ =
=
0
i
v
t t
( )0v ω
0
0 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,
2
2 0v
C
⎛ ⎞
⎜ ⎟Wh called the amplitude of vibration
16
2 0
0C y
ω
⎛ ⎞
= + ⎜ ⎟
⎝ ⎠
Where , called the amplitude of vibration

y
2
2 0
0
v
C y
⎛ ⎞
= + ⎜ ⎟
⎝ ⎠
0v
y
0C y
ω
+ ⎜ ⎟
⎝ ⎠0y
α
t
α
ω 2
T
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 0
0
C t t
t t
ω α ω α
ω α α ω
+ ⇒ +
⇒ + = ⇒ = −

Note:
2
⎛ ⎞ β
( ).siny C tω α= +
Note:
0y
2
2 0
0
v
C y
ω
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
α
β
( )
0 0
cos .sin sin .cos
cos sin
C t t
y v
C t t
ω α ω α
ω
ω ω
= +
⎛ ⎞
= +⎜ ⎟
0v ω
y
0
0
cos . sin .
cos sin
C t t
C C
v
y t t
ω ω
ω ω
= +⎜ ⎟
⎝ ⎠
= +
( )
0
0
tan
y
v
α
ω
=
0y
ω
18

2
f
ω
π
=Cyclic frequency Units: cycles/second
1 2
T
f
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
3
Pl
L
3
Pl
EI
Δ =
50 N
3
4 200 300⎛ ⎞×
50 N
1 3
3P EI
k
l
= =
Δ
4
3
200 300
3 2.5 10
12
4218.75N mm 4218750N m
2000
⎛ ⎞×
× × ×⎜ ⎟
⎝ ⎠= = =
1 1 1
4218750 20k
= + 2
19.9kg s
= 1.98 rad sek
ω = =
4218750 20
19.9N m
e
e
k
k∴ =
( )50 / 9.
1.98 ra
8 g
d s
1 km
ω

Problem 2: Calculate the natural frequency in side sway and natural
period of vibration for the frame in figure If the initial displacement isperiod 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
6
30 10 N×
SDOF, Undamped free vibration
12
30 10 MPaAB CDEI EI= = ×
6 2
5
2
30 10 kg.m s
30.58 10 kg
9.81 m s
m
×
= = ×
12 12EI EI 12 1 1
12 30 10
⎛ ⎞
3 3
12 12AB CD
AB CD
EI EI
k
l l
= +
12
3 3
1 1
12 30 10
1000 800
⎛ ⎞
= × × × +⎜ ⎟
⎝ ⎠
1063125 N/mm=
2
3 kg m/s
3 kg.m/s
1063125 10
m
= ×

3 2
5
1063125 10 kg s
=
30.58 10 k
18.645 r d
g
a s
k
m
ω
×
= =
×30.58 10 kgm ×
2 2
0.337
18.645
Time period T s
π π
ω
= = =
( ).siny C tω α= +Displacement of SDOF, undamped free vibration ( )
2
2 v⎛ ⎞ ( )
0
0
tan
y
v
α
ω
=
2 0
0
v
C y
ω
⎛ ⎞
= + ⎜ ⎟
⎝ ⎠
Where amplitude
( )0v ω
( )
25
18.645
25 18 645
= =
( )25 18.645
0
86.93 1.517radα∴ = =

2 22 2
2 20
0
25
25
18.645
25.036 mm
v
C y
ω
⎛ ⎞ ⎛ ⎞
= + = + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
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 ialong 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
4
3
4
cos cos
5
t tω α ω α+ = +→ ( ) ( )1
2 1
4
cos cos
5
t tω α ω α− ⎡ ⎤
⇒ + = +⎢ ⎥⎣ ⎦

6 4⎛ ⎞⎡ ⎤
( )
( )
( )1
1
1
6 4
10 sin cos cos
s
2
in 5
t
t
ω α
ω α
−
→
⎛ ⎞⎡ ⎤
= +⎜ ⎟⎢ ⎥+ ⎣ ⎦⎝ ⎠
0
1 0.4228 24.24t radω α+ = =With trial and error,
14 6229C cm∴ =Amplitude
1,
14.6229C cm∴ =Amplitude,
0.3749rad 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
( )2
max 30.06
2 2
my m C Nmω= = =Maximum Kinetic Energy,

Damped free vibrationp
One DOF, with damping, no external forces (only initial displacement
condition)
0my cy ky+ + =
y
2
0pt pt pt
C C kC
pt
y Ce=Solution is of the form
2
0pt pt pt
mCp e cCpe kCe∴ + + =
2
0mp cp k⇒ + + =
In general, the roots of the above equation are 2
1 2,
c c k
p p
⎛ ⎞
= − ± −⎜ ⎟
⎝ ⎠
General solution is:
1 2,
2 2
p p
m m m
⎜ ⎟
⎝ ⎠
1 2
1 2
p t p t
y C e C e= +
1 2,C C To be determined from initial conditions

2
k⎛ ⎞
2
c k
m m
⎛ ⎞
−⎜ ⎟
⎝ ⎠
Final form of the solution depends on the sign of
Case 1:
2
c k⎛ ⎞
0
2
c k
m m
⎛ ⎞
− =⎜ ⎟
⎝ ⎠
2c km⇒ =
2 k
m
ω =
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
crc
p p
m
⇒ = −
2
0
2
c k
m m
⎛ ⎞
− =⎜ ⎟
⎝ ⎠
Now, Equal roots

1 2
1 2
p t p t
C e C teTwo independent solutions and
2
crc
t− 2
crc
t
m
y C te
−
d
c
2
1 1
m
y C e∴ =
2
2 2
m
y C te=and
( ) 2
1 2
crc
t
m
y C C t e
−
= +Hence,
F ib ti ith iti l d i
Free vibration response with critical damping

Case 2:
2
0
c k⎛ ⎞
>⎜ ⎟
2 2
0crc c⎛ ⎞ ⎛ ⎞
⇒ > ⇒ >⎜ ⎟ ⎜ ⎟0
2m m
− >⎜ ⎟
⎝ ⎠
0
2 2
cr
crc 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 c
c m
ζ
ω
= = 1crc c ζ> ⇒ >
2crc mω

( )y t
t
Free vibration response of critically damped andFree vibration response of critically damped and
overdamped systems
30

Case 3: 2
0
c k⎛ ⎞
<⎜ ⎟
2 2
0crc c
c c
⎛ ⎞ ⎛ ⎞
⇒ < ⇒ <⎜ ⎟ ⎜ ⎟ 1ζ⇒ <0
2m m
− <⎜ ⎟
⎝ ⎠
0
2 2
cr
crc c
m m
⇒ − < ⇒ <⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
1ζ⇒ <
• i.e., Damping in the system is smaller than critical damping
(Underdamped system)
• Oscillatory motiony
• Complex roots for the equation
2
c k c⎛ ⎞
1 2,
2 2
c k c
p p i
m m m
⎛ ⎞
= − ± − ⎜ ⎟
⎝ ⎠
iζ ±
nω ω=
2
c c
c m
ζ
ω
= =
n Diζω ω= − ±
2
k c⎛ ⎞
⎜ ⎟
2cr nc mω
Damped natural
2
D
m m
ω
⎛ ⎞
= − ⎜ ⎟
⎝ ⎠
2
i 1 ζ
Damped natural
frequency
2
i.e., 1D nω ω ζ= −

( ) ( )n D Dt i t i t
t A Aζω ω ω− −
+( ) ( )1 2
n D Di t i t
y t e Ae A eζ ω ω
= +
cos sinDi t
D De t i tω
ω ω= ±But
( ) ( ) ( )1 2 1 2cos sinnt
D Dy t e A A t i A A tζω
ω ω−
∴ = + + −⎡ ⎤⎣ ⎦
( ) [ ]i e cos sinnt
y t e A t B tζω
ω ω−
= +( ) [ ]i.e., cos sinn
D Dy t e A t B tζ
ω ω= +

Applying initial conditionsApplying initial conditions,
( ) 0 0
cos sinnt nv y
y t e y t tζω ζω
ω ω− ⎡ ⎤+
+⎢ ⎥( ) 0 0
0 cos sinn n
D D
D
y
y t e y t tζ ζ
ω ω
ω
= +⎢ ⎥
⎣ ⎦
0 0
0
tan n
D
v y
y
ζω
α
ω
+
=
( ) ( )cosnt
y t Ce tζω
ω α−
=
0D yω
( ) ( )cos Dy t Ce tω α= −Or,
2
2 0 0
0
nv y
C y
ζω
ω
⎛ ⎞+
= + ⎜ ⎟
⎝ ⎠
Where
33
Dω⎝ ⎠

( ) ( )cosnt
y t Ce tζω
ω α−
= −( ) ( )cos Dy t Ce tω α= −
Oscillatory motion, with an exponentially decaying amplitude ofy , p y y g p
nt
Ce ζω−
2π
2
T
π
ω
=
y
2
2
Damped natural period
1 D d i l t l f
d
T
p
π
ζ
= =
t
ny 1ny +
2
1 Damped circular natural frequencydp p ζ= − =
nt
y 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 ( )( ) 0
cos( ) 1
y t
tω α
=
− =Point of tangency ( )
2
T
π
ω
=
y
2
1 Damped circular natural frequency
d
d
T
p
p p ζ
= =
= − =
t
ny 1ny +
p q ydp p ζ
nt
y Ce ζω−
=
36

( )y t
0v nt
y Ce ζω−
=
( )0y( )0y
nt
y 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 =
2
i.e., 1 0.98D n n nω ω ζ ω ω= − =
Hence, for structures, damped natural frequency is practically
same as undamped natural frequencysame 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 totaldampers. 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
m
ω =
k/2 k/2
8000N
20 kg
m
= 20 r= ad s
k/2 k/2
Damping ratio,
2 n
c
m
ζ
ω
=
130 N130
20 202
Ns m
kg rad s
=
× × 0.1625=
Damped natural frequency,
2
19.734 rad/s1D nω ω ζ= − =

( ) ( )cosnt
y t Ce tζω
ω α−
=We have ( ) ( )cos Dy t Ce tω α= −We have,
for damped free vibration.
2
2 0 0 nv y
C
ζω⎛ ⎞+
+ ⎜ ⎟h
0 0
tan nv y ζω
α
+
=and2 0 0
0
n
D
y
C y
ζ
ω
= + ⎜ ⎟
⎝ ⎠
where
0
tan
D y
α
ω
and
Initial conditions: 00 100 0.10,vy = == mm/s m/s
2
0.1 0
0
D
C
ω
⎛ ⎞+
∴ = + ⎜ ⎟
⎝ ⎠
0.005=
0.1 0
t
+ 0
90 1 571 dα⇒
D⎝ ⎠
40
tan
0
α = = ∞
0
90 1.571radα⇒ = =

( ) ( )0.1625 20
0.005 cos 19.734 1.571t
y t e t− ×
= −Hence,
( )3.25
0.005 cos 19.734 1.571t
e t−
= −
( ) ( ) ( )3.25
0 005 19 734sin 19 734 1 571 3 25cos 19 734 1 571t
y t e t t−
= ⎡ ⎤⎣ ⎦
and,
( ) ( ) ( )0.005 19.734sin 19.734 1.571 3.25cos 19.734 1.571y t e t t= − − − −⎡ ⎤⎣ ⎦
( )3.25
0 005 19 734 1 571−
Displacement at t=1s
( ) ( )3.25
1
0.005 cos 19.734 1.571t
y e=
= −
4
1 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, 0.05Initial conditions m/sec: y v= =
2
1Damped natural fre uencyq D nω ω ζ∴ = −
2
10 1 0.1 9.9499 rad s= − =

0E i f i k 0Equation of motio :n my cy ky+ + =
0i.e.,
c k
y y y+ + = 2
2 0y y yξω ω⇒ + + = 2 100 0y y y⇒ + + =
Solution of the equation of motion
, y y y
m m
2 0n ny y yξω ω⇒ + + y y y
( ) ( )cos
Solution of the equation of motion
i.e., Displacement, nt
Dy t Ce tζω
ω α−
= −
2
2 0 0
0where, n
D
v y
C y
ζω
ω
⎛ ⎞+
= + ⎜ ⎟
⎝ ⎠
2
0.05 0
0 0.005
9.9499
+⎛ ⎞
= + =⎜ ⎟
⎝ ⎠Dω⎝ ⎠ ⎝ ⎠
1 0 0
tan nv y ζω
α − ⎛ ⎞+
= ⎜ ⎟
1 10.05 0
tan tan
0 2
π
α − −+⎛ ⎞
= = ∞ =⎜ ⎟
⎝ ⎠
( ) 0.005 cos 9.9499t
y t e t
π− ⎛ ⎞
∴ = −⎜ ⎟
0D yω⎜ ⎟
⎝ ⎠
0 2⎝ ⎠
( ) 0.005 cos 9.9499
2
y t e t∴ ⎜ ⎟
⎝ ⎠

Logarithmic decrement
2
n
T
π
ω
=
y
Logarithmic decrement
( )1
11
cosnt
DCe ty
ζω
ω α−
−
2
2
1 Damped circular natural frequency
d
d
T
p
p p
π
ζ
= =
= − =
t
ny 1ny +
nt
y Ce ζω−
=( )
( )2
11
2 2cosn
D
t
D
y
y Ce tζω
ω α−
=
−
( )1
11
cosnt
DCe ty
ζω
ω α−
−( )
( )
1
11
2
2
1cos 2
n
D
D
t
D
y
y
Ce t
π
ζω
ω
ω π α
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
=
+ −( )1D
22 ππ ζ
2
2
1
nn
nD
e e
π ζωζω
ω ζω −
= =

⎛ ⎞ 2 ζ1
2
2
2
ln
1
n
n
y
y
π
ζω
ω ζ
⎛ ⎞
=⎜ ⎟
−⎝ ⎠
2
2
1
πζ
ζ
=
−
(say)δ=
2π
ζ
2
21
2
n
D
y
e e e
y
π
ζω
ω πζ δ
= = =
2πζ
2y
δ δ
2
2
1
πζ
ζ
δ=
− ( )
2 2 22
δ δ
ζ
ππ δ
⇒ =
+
2i e δ πζ (Logarithmic decrement)
( )
45
2. .,i e δ πζ= (Logarithmic decrement)

A practical way to determine dampingA practical way to determine damping
- Logarithmic decrement/ exponential decay method
1
2
y
e
y
δ
=
2y
1 1 2 3
1 2 3 4 1
ny y y y y
y y y y y
=After n cycles,
1 2 3 4 1n ny y y y y+ +
. . . n
e e e e eδ δ δ δ δ
= =
11
ln
y
δ
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
∴
1nn y +
⎜ ⎟
⎝ ⎠

Problem 5:
SDOF damped free vibrationSDOF, damped free vibration
8000 kgm =
12 12EI EI
3 3
12 12AB CD
AB CD
EI EI
k
l l
= +
6
24 6 10× ×
6
3
24 6 10
5.332 10 N m
3
× ×
= = ×

Undamped natural frequency n
k
m
ω =
6
5.332 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= =
2
1D d t l f ζ 2
1Damped natural fre uencyq D nω ω ζ∴ = −
2
25 82 1 0 04 25 8 rad s= =25.82 1 0.04 25.8 rad s= − =
48

2Logarithmic decrement, δ πζ=
0
2 0.04 0.251 ln
y
y
δ π
⎛ ⎞
∴ = × = = ⎜ ⎟
⎝ ⎠1y⎝ ⎠
0 30y0
1 0.251
30
23.33 mm
y
y
e eδ
∴ = = =
0
3 3 3 0 251
30
14.12 mm
y
y δ
= = =
1
2 0.251
23.33
18.15 mm
y
y
e eδ
= = =
3 3 3 0.251
14.12 mmy
e eδ ×
0 30
10 98 mm
y
y = = =
0
2 2 0.251
30
18.15 mm
y
e eδ ×
= = =
4 4 4 0.251
10.98 mmy
e eδ ×
= = =
0 30
8 54 mm
y
y = = =5 5 5 0.251
8.54 mmy
e eδ ×
= = =
0 30
6 6 mm
y
y = = =
6 6 6 0.251
6.6 mmy
e 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 icolumns 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 frequency
2. Logarithmic decrementg
3. Number of cycles and time required for the amplitude to
reduce from an initial value of 3mm to 0.3 mm.
F
k =
Δ
8000
2666.67
3
N
N mm
mm
= =Stiffness,
n
k
m
ω =1. Undamped natural frequency
m
3
2666.67 10
20000 9 81
N m
k
×
= 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 reduce3. 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
ln
n
y
n y
δ
+
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
1 3
0.314 ln
0.3n
⎛ ⎞
⇒ = ⎜ ⎟
⎝ ⎠
7.333cy en cl s⇒ =
51

4 Ti i d f th lit d t d f i iti l4. 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×=
2
7 333
π 2
7 333
π
7.333
nω
×= 7.333
36.15
= ×
1 2 4 d1.274 seconds=
52

Forced vibration
• External forces cause vibration
Forced vibration
Response of undamped system to harmonic excitation
y
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
2
m Y kY Fω− + =
Let
A 0m Y kY Fω + =
F0
2
F
Y
k mω
=
−
0 0
2 2
F k F k
= =2 2
2
1 1
nk m
ω ω
ω
− −
η
ω
= Frequency ratio
n
η
ω Frequency ratio
( )0
2
1
sty
=
( )F k St ti d fl ti
2
1 η− ( )00 stF k y= Static deflection

( ) 0
cos sin sin
F k
y 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 sin
1
n 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= = = =
0
2
0,
1
F k
A B
η
η
−
∴ = =
−
( ) 0 0
sin sin
F k F k
y t t t
η
ω ω
−
= +( ) 2 2
sin sin
1 1
ny t t tω ω
η η
= +
− −
( ) ( )0
2
sin sin
1
i.e., n
F k
y t t tω η ω
η
= −
1 η−

( )
( )
( )
( )2
0
O sir, n sin
1
st
ny
y
t t tω η ω
η
= −
−
• The above represents a superposition of two harmonic
responses of different frequencies
• The result is NOT harmonic
57

( )
( )
y t
y( )0sty
( )0 0y = ( )0 00 nv y F kω= =

( )
( )
( )0
2
sin sin
s
n
t
y
y
t t tω η ω= −( ) ( )2
1
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+ + =
B
0 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 eqn
0my cy ky+ + =
( ) [ ]cos sinnt
c 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 thSubstituting this back in the eqn of motion and equating the
coefficients of sin cos ,t tω ωand
2
C 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− + + =

( )
( ) ( )
2
0
1 2 22
F k m
C
k
ω−
=
+ ( ) ( )
0
2 2 22
F c
C
k m c
ω
ω ω
−
=
+( ) ( )2
k m cω ω− + ( ) ( )k m cω ω− +
( )
( ) ( )
( )20
2 22
sin cosp
F
y t k m t c tω ω ω ω⎡ ⎤= − −⎣ ⎦( )
( ) ( )
( )2 22
p
k m cω ω
⎣ ⎦
− +
62

( )
( ) ( )
( ) ( )
[ ]
2 22
0
2 22
cos sin sin cosp
F k m c
y t t t
k
ω ω
φ ω φ ω
ω ω
− +
= −
+( ) ( )k m cω ω− +
22
k mω−
φ− 2
tan
c
k m
ω
φ
ω
=
−
( ) ( )
2 22
k m cω ω− +
cω−
( )
( ) ( )
( )0
2 22
sinp
F
y t t
k m c
ω φ
ω ω
= −
− +
( ) ( )

( ) ( )0
sinp
F k
y t tω φ= −( )
( ) ( )
( )2 22
s
1 2
py t tω φ
η ζη− +
( )
( )2
0
sin
sty
tω φ= −
( ) ( )
( )2 22
1 2
φ
η ζη− +
( )( )
( ) ( ) ( )
( )2
0
22
1
sin
1 2t
p
s
y
y
t
tω φ
η ζη
= −
− +( ) ( )

( ) ( ) ( )( ) ( ) ( ), c py t y t y t= +Total response
[ ]
( )
( )0
cos sin sinn stt
y
e A t B t tζω
ω ω ω φ−
+ +[ ]
( )
( ) ( )
( )0
2 22
cos sin sin
1 2
n
D De A t B t tζ
ω ω ω φ
η 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 2
Where, amplitude of steady-state vibration
sty
Y
η ζη
=
− +
( ) ( ) ( )
2 22
0
,
1
1 2
Dynamic amplification factor,
t
D
Y
y η ζη
=
+( ) ( ) ( )2
0 1 2sty η ζη− +
,nω ω=When resonance happens.
1Y Y Y= =
( )0F k
=
Hence, resonant amplitude,
1n
Y Y Yηω ω ==resonant 2ζ

tortionfactmplificatnamicamDyn
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
2
tan
cω ζη
φ = =
2 2
tan
1k m
φ
ω η
= =
− −

Problem 4:
SDOF, damped vibration with harmonic excitation
2
800 kg.m s
81 55 kgm = =
SDOF, damped vibration with harmonic excitation
2
81.55 kg
9.81 m s
m
5
48 48 2 10 6000EI × × ×
d/
3 3
48 48 2 10 6000
0.9 N mm
4000
EI
k
L
× × ×
= = =
15 rad/secω =0 sin 20sin15F t tω = 0 20F Newtons=

900 N/k 15900 N/m
3.32 rad/sec
81.55 kg
n
k
m
ω = = =
15
4.52
3.32n
ω
η
ω
= = =
5% 0.05ζ = =
Amplitude of steady state vibration,
( )
( ) ( )
0
max 2 22
1 2
F k
y Y
η ζη
= =
− +( ) ( )1 2η ζη− +
( )20 900( )
( ) ( )
2 22
20 900
1 4.52 2 0.05 4.52
=
− + × ×
3
1.143 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== ( )20 900
0.222 m
2 0.05
= =
×
0
2
F k
ζ
=
72

Vibration isolationVibration isolation
A method for protecting equipment from vibrating foundation
O f f b hOR for protecting structure from vibrating machinery
2 Vib ti hi1. 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 22
0 0 0, 1 2Where F y k c y kω ξη= + = +
tan 2
c
k
ω
β ζη= =
( )0 sinF tω β+
( )y t
( )
( )y t
Equivalent to
( )
q
( )sy t
75

F k
( )
( ) ( )
( ) ( )0
2 22
sin sin
1 2
F k
y t t Y tω β φ ω β φ
η ζη
∴ = + − = + −
− +
T
Amplitude of respo
Transmissibility
nse
=RT
Amplitude
Transmissibi
of support
l
d
ity
isplace
=
ment
( ) ( )
0
2 22
0
0 1 2
R
F kY
T
yy η ζη−
∴ =
+
=
( ) ( )0y η ζη
( )
2
( )
( ) ( )
2
2 220
1 2
1 2
i.e., R
Y
T
y
ζη
η ζη
+
= =
− +
( )
20
0 1 2
F
y
k
ζη= +∵
( ) ( )1 2η ζη+
1 RT= −Degree of isolation

TR
n
ω
η
ω
=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
( ) ( )
0
2 22
=
F k
Y
( )F t cy ky= +
Force transmitted to foundation, ( ) ( )
2 22
1 2η ζη− +
( ) ( ) ( )cos sinF t c Y t kY tω ω φ ω φ= − + −
( ) ( )
22
sinY k c tω ω φ β= + − + tan 2
c
k
ω
β ζη= =
( )
22
Y k cω= +Max. force transmitted

Max force transmittedMax. force transmitted
Force Transmissibility =
Max. force of excitation
Here excitation is due to vibrating machine
( )
22
0F
Y k cω
=
+
∴Force Transmissibility
( ) ( )
( )
22
0
2 22 01 2
=
k cF k
F
ω
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
=
( )
2
1 2ζη+
Force Transmissibility
Max. force of excitation
RT
3500 N
( )
( ) ( )
2 22
1 2
1 2
Also, RT
ζη
η ζη
+
=
− +
3 2
3
2
20 10 kg.m s
2.0387 10 kg
9.81 m s
m
×
= = ×
10 2 62.83 rad/sω π= × =
k
2
2
2 62.83 8047990.264
=
ω
η
⎛ ⎞
= =⎜ ⎟
n
k
m
ω =
3
=
2.0387 10n k k
η
ω
= =⎜ ⎟
×⎝ ⎠
10% 0.1ζ = =

( )
2 8047990.264
1 2 0 1+ × ×( )
( )
2
2
1 2 0.1
300
3500 8047990.264 8047990.264
1 2 0.1
R
kT
+ × ×
= =
⎛ ⎞
− + ×⎜ ⎟
⎝ ⎠
( )1 2 0.1
k k
+⎜ ⎟
⎝ ⎠
2
2 28047990.264 321919.61 321919.61
300 1 3500 1
⎡ ⎤⎛ ⎞ ⎛ ⎞
⎢ ⎥⎜ ⎟ ⎜ ⎟
2 2
300 1 3500 1
k k k
⎛ ⎞ ⎛ ⎞
× − + = × +⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
( ) ( )
22 2
300 8047990.264 321919.61 3500 321919.61k 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 18
90000 1 419 1 5 829 1 00 0k k− × + × =
7887098.515k∴ = N m
90000 1.419 1 5.829 1 00 0k k× + ×

SummarySummary
Structural dynamics
• Introduction – degree of freedom – single degree of freedom
li t ti f ti D’Al b t’ i i llinear 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
83

Module4 s dynamics- rajesh sir

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