WEEK-2:

Free Vibration of SDOF systems

Free vibration of SDOF systems

•Natural frequency

•Damping Ratio

•Equation of motion

•Standart form of equation of motion

•Assume following initial conditions

•Assume a harmonic solution 0eq eq eqm x c x k x

Free vibration of SDOF systems

•Equation of motion •Harmonic solution

• Types of Free Vibrations

Free Vibration of SDOF Systems (Undamped Case)

•General harmonic solution

•Applying initial conditions

Example:

Example:

Damped Free Vibrations with Viscous Damping

1.Underdamped vibration

•By applying initial conditions and expanding with cos and sine terms gives

Underdamped vibration

Underdamped vibration

Example: Governing equation of motion of an underdamped single degree of

freedom system is given as

a) Find the natural frequency

b) Find the damping ratio

c) Find the damped natural frequency of the system

Example: Governing equation of motion of an underdamped single degree of

freedom system shown below is given as

Find all vibration parameters

Solution:

2.Critically damped vibrations

•By applying initial conditions:

1 2( ) ntx t e B B t

If the initial conditions are

opposite and

0

0 0

0n

x

x x

then the response overshoots the

equilibrium position before

eventually decaying and

Approaching equilibrium from the

direction opposite that of the initial

position.

Example:

M=0

3.Overdamped vibrations

•By applying initial conditions:

The response of a system that is

overdamped is similar to a critically

damped system.

An overdamped system has more

resistance to the motion than critically

damped systems.

Therefore, it takes longer to reach a

maximum than a critically damped system,

but the maximum is smaller. An

overdamped system also takes longer than

a critically damped system to return to

equilibrium.

Example:

b)

(3.29)

or 3.29

(3.48)

(3.53)

Coulomb damping

Equation of motion:

Initial conditions:

Response:

b) 0t /n

a) /nt 2/n

Coulomb damping

/mg kWhen motion ceases a constant displacement from equilibrium of is maintained.

Example:

Hysteretic (Structural) Damping

A system undergoing periodic vibration has the following load-displacement diagram.

Energy dissipated per cycle is independent of frequency and proportional to the

amplitude

Hysteris loop

2E khX

h: Hysteretic damping coefficient

2

h

1

2

ln ln(1 )X

hX

Logarithmic decrement for hysteritic damping

For small h: h

Equivalent viscous damping coefficient: eq

n

kc h

Problem

The area under the hysteresis curve is approximated by

counting the squares inside the hysteresis loop. Each square

represents (1 104 N)(0.002 m)=20 N m of dissipated energy.

There are approximately 38.5 squares inside the hysteresis

loop resulting in 770 N m dissipated over one cycle of motion

with an amplitude of 20 mm.

keq= the slope of the force deflection curve=5 106 N/m.

=0.123

=0.0613

h =0.385

2

h

100 /n

kr s

m

2 6 3 2

770

(5 10 )(20 10 )

Eh

kX

The response of this structure with hysteretic damping is approximately the same

as the response of a simple mass-spring-dashpot system with a damping ratio of

0.0613 and a natural frequency of 100 rad/s. Then using underdamped free

vibration response equation, with