CEGCEG--5065C Soil Dynamics5065C Soil Dynamics

Lecture #07Lecture #07

Free Vibrational Systemso (1) Undamped Free Vibrations

o (2) Damped Free Vibrationso (2) Damped Free Vibrations

Luis A. Prieto-Portar 2009

Case (1). A free-vibration system without dampening.

The soil subgrade reaction q is the foundation load W over an area A. In the lumped parameter system, the displacement zs of the soil is proportional to the load W, or

s

W lbk

z inch=

q W lb

expressed as an equality by using the spring constant k,

The coefficient of sub-grade reaction ks is,

3ss s

q W lbk

z Az in= =

When the foundation is disturbed from its static equilibrium, the foundation-soil system will vibrate. When damping is not present c=0, and the equation of motion is simply

0 0� �+ = + =� �� �

�� ��k

mu ku or u um

1 2

The solution to this first-order differential equation is of the form,

The parameter is the undamped natural circular frequency of

the system, and thus the natural period ando

k ku C sin t C cos tm m

km

km

ωωωω

= +

= frequency are,

2 1m kωωωωππππ

1 2

2 12 and

2 2

The solution in terms of the circular frequency is,

oo o

o

o o

m kT fk m

u C sin t C cos t

ωωωωππππ ππππω π πω π πω π πω π π

ω ωω ωω ωω ω

= = = =

= +

( ) ( )

( ) ( )

1 2 2 2

1 2 1 1

The boundary conditions reveals the value of the constants. At initialconditions, the time 0 with initial displacement and velocity

0 0

0 0

o o

o o

oo o o o

t u u ,

u C sin C cos C C u

uu C cos C sin C Cω ω ωω ω ωω ω ωω ω ω

ωωωω

=

= + = ∴ =

= − = ∴ =

�

��

Therefore, the equaton of motion becomes,o

oo o o

o

uu sin t u cos tω ωω ωω ωω ω

ωωωω= +�

The displacement of an undamped free vibrating system is,

oωωωω

The maximum displacement Z is called the single amplitude. The peak-to-peak amplitudeis 2Z and is also referred to as the double amplitude. The time required for the sinusoidal displacement to repeat itself is called the period T, which is given by,

2 12

1 1 12 2 2 2

= = =

� � � � � �= = = =� � � � � �� � � � � �

n

n

nn

s s

T and the frequency of oscillation f is, fT

k Wg gThe undamped natural frequency fm z W z

ωωωωππππω πω πω πω π

ωωωωπ π π ππ π π ππ π π ππ π π π

The plots for the velocity and the acceleration of the foundation can be found from the The plots for the velocity and the acceleration of the foundation can be found from the first and second derivatives of the displacement z,

2 2

2

= −= − − = − +

= − − = − +

�

��

n

n n n n

n n n n

u Acos( t )u ( A )sin( t ) A cos( t / ) and

u A cos( t ) A cos( t )

ω αω αω αω αω ω α ω ω α πω ω α ω ω α πω ω α ω ω α πω ω α ω ω α π

ω ω α ω ω α πω ω α ω ω α πω ω α ω ω α πω ω α ω ω α π

The displacement u of the foundation,

At different timeintervals,0

0

22 0

= −

= = − =

= = =

+= = =

+

n

n

n

u Acos( t )

t u Acos( ) Acos

t u Acos( ) A

/t u Acos( / )

ω αω αω αω α

α αα αα αα αααααωωωωπ απ απ απ α ππππ

ωωωωπ απ απ απ α

3 23 2 0

22

Thes

+= = = −

+= = =

+= = =

n

n

n

t u Acos( ) A

/t u Acos( / )

t u Acos( ) A etc...

π απ απ απ α ππππωωωωπ απ απ απ α ππππ

ωωωωπ απ απ απ α ππππωωωω

e results are shown on the next slide...

use the analogy of a pendulum.

( )2

2

1

An alternative form of this solution is,

sin where,

the amplitude is

and the phase angle is tan−

= +

� �= +� �

� �

� �= � �

� �

�

�

o

oo

o

o o

o

u A t

uA u

uu

ω φ

ω

ωϕ

Example 1.

Find the natural frequency of vibration of a foundation with a mass m supported by a soil that has experienced a static deflection of 0.5 inches under that mass.

21 1 32 2 ft/sec 12 inches/feet4 4 cycles per second

2 2 0 5 inchesns

g ( . )( )f .u ( . )π ππ ππ ππ π

� � � �= = =� � � �� � � �

Example 2.

Find the period of oscillation and the natural frequency of vibration of a foundation with a weight of 60 kN and a soil (spring) constant of 103 kN/m.

3

2

1 1 102 04 cycles persecond

2 2 60 9 81

1 10 5 seconds

2 04

n

n

k kN / mf .m kN / . m / s

T .f .

π ππ ππ ππ π� � � �= = =� � � �� � � �

= = =

Example 3.

A small water tower can be simplified as a point weight of 10 kips supported on a slender steel column (the figure idealized below). Assume that it is an un-damped system and that a horizontal wind load of 5 kips applied horizontally to the tower produces a lateral displacement of 0.04 inches. What is the natural circular frequency and period of the system, and plot its time history when the wind stops.

The stiffness of the column is,

5125 /

0.04= = =

∆windF kips

k kip inchinches

( )( )( )2

o

o

The natural circular frequency is,

125 / 12 / 32.2 /

1 10

70 / sec

The natural period is,

2 2

= = =

=

= =

o

oo

kip in in ft ft sk k gm W kips

radians

T

radianT

ω

ω

π πω

0.1seconds70

=skips

( ) ( ) ( )

The conditions of the problem give the initial displacement andvelocity, therefore

0.04 and 0

Therefore, the equation of motion is,

sin cos 0.04 cos 70

= =

= + =

�

�

o o

oo o o

o

u inches u

uu t t u t in tω ω

ω

Case (2). A free-vibration system with viscous damping.

We saw before that a free vibration imposed upon an un-damped foundation-soil system will continue to vibrate without end. In reality, all vibrations will gradually decrease in amplitude with time due to damping. The figure below shows the modeling of the damping through the use of a dashpot with a coefficient c.

In reality, all systems have some damping, which results from friction, heat, air resistance, etc. The equation of motion is therefore,

2

2 2

2 2

0

Dividing by and setting / yields,

0 0

This is commonly written as,

20

o

o o

mu cu ku

m k m

c k cu u u or u u u

m m k

cu u u

ω

ω ω

ω ω

+ + ==

+ + = + + =

+ + =

�� �

�� � �� �

�� �2 2 0

2

The denominator of the velocity constant is known as the critic

o ou u ukm

ω ω+ + =�� �

2

al dampingcoefficient ; the is defined as,

2 22Thus, the equation of motion with dampi

ng is,

2 0

c

o

c o

o o

c

cc c c

damping

c m kkm

u u

ratio

u

n ξωξ

ω

ξω ω

= = = =

+ + =�� �

The solution to the damped free vibration differential equation depends on the value of the damping ratio �.

1) When � < 100% the system is under-damped. In earthquake engineering, all cases of seismic events are under-damped.

2) When � = 100% the system is critically damped.

3) When � > 100% the system is over-damped.

The solution to the differential equation for the under-damped condition is of the following form,

( ) ( )2 21 2sin 1 cos 1o t

o ou e C t C tξω ω ξ ω ξ− � �= − + − � �

Notice that the bracket portion is multiplied by an exponential that gradually gets smaller with time and eventually becomes zero.

The damped natural circular frequency �d permits a different mode of solution,

[ ]1 2sin coso td du e C t C tξω ω ω−= +[ ]

( ) ( ) ( )

( )

1 2

1 2

01 2 2

01

where the coefficients and can be found from initial conditions.

The initial displacement is,

sin 0 cos 0

The initial velocity is,

cos

o

o

d d

o

o

o

o d

C C

u

u e C C C

u

u e C

ξω

ξω ω

−

−

� �= + =� �

=

�

� ( ) ( ) ( ) ( ) ( )02 1 2

1 2

0 sin 0 sin 0 cos 0od d d o d d

o d o

C e C C

u C C

ξωω ω ω ξω ω ωω ξω

−� � � �− − +� � � �

= −�

( )

( )

1 2

Therefore,

and

The solution for a damped free vibration system is,

sin coso

o o oo

d

o o otd o d

d

u uC C u

u uu e t u tξω

ξωω

ξωω ω

ω−

+= =

� �+= +� �

� �

�

�

Example 4.

A small billboard tower has mass of 1,000 kg, a stiffness of 20,000 N/m and has a natural damping of 1,000 N-m/sec.

An earthquake displaces the billboard’s foundation laterally 10 mm with an initial velocity of 50 mm/sec.

(a) Find the system’s damped natural frequency, and(b) Plot the time history of response of the billboard’s mass.

( )( )

The undamped natural frequency is,

1 1 20,000 /0.71

2 2 2 1,000

The damping ratio is,1,000 / sec

0.122 2 20,000 / 1,000

The damped natural frequency is,

= = = =

−= = =

o

oo

d

f

k N mf hertz

m kg

c N m

km N m kg

f

ωπ π π

ξ

ξ

( )21 0.71 1 0.1= − = −d of f hzξ ( )

( )

( )

2

o

2 0.70

The undamped natural circular frequency is,

2 2 0.71 4.5 radians/sec

The damped natural circular frequency is,

2 2 0.70 4.4 radians/sec

=

= = =

= = =

o

o

d

d d

hertz

f hz

f hz

ωω π π

ωω π π

( )

( )( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )

0.12 4.4

0.53

Thus, the displacement response is,

sin cos

0.05 0.12 4.5 0.01sin 4.4 1 cos 4.4

4.4

cos 4.4 0.01sin 4.4

o o o otd o d

d

t

t

u uu e t u t

u e t t

u e t t

ξω ξωω ω

ω−

−

−

� �+= +� �

� �

� �− += +� �

� �

� �= −� �

�

References.

1. Das, B., “Principles of Soil Dynamics”, PWS-Kent Publishing Co., Boston, 1993;

2. Richart F.E., Hall J.R., Woods R.D., “Vibrations of Soils and Foundations”, Prentice-Hall Inc., New Jersey, 1970;

3. Humar J.L., “Dynamics of Structures”, Prentice-Hall, New Jersey, 1990;

4. Prakash S., “Soil Dynamics”, McGraw-Hill, New York, 1981;