Module 4 (CE-409: Introduction to Structural Dynamics and Earthquake Engineering)

University of Engineering and Technology Peshawar, Pakistan

CE-409: Introduction to Structural Dynamics and Earthquake Engineering

MODULE 4: UNDAMPED & DAMPED FREE VIBRATIONS IN

S.D.O.F SYSTEMS

Prof. Dr. Akhtar Naeem Khan & Prof. Dr. Mohammad Javed [email protected] [email protected]

1

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Undamped Free Vibration

We will first consider the case where there is no load acting on theStructure i.e. p(t)=0. This case is known as free vibration.The trivial solution for this case is where there is no displacement of the structure at t=0 i.e. u(t) = 0

0ku um =+There are several ways of obtaining a solution to this second order differential equation. The simplest is to assume that the solution of the equation is of following form

stGe u(t) =


nωm

k

m

ks ii ±=±=−±=

0)(Ge k) m(s st2 =+

0k ms2 =+

Substituting this solution into the equation of motion (given on previous slide) result in:

By rearranging:

Since , only possibility is that

By solving one get

The variable ωn is known as the natural circular frequency and the units are radians/second.

0Gest ≠

0kGe )Gem(s stst2 =+




0k ms2 =+Inserting in

and solving we get:

The above after further simplification results in:

This is the solution of equation of motion of undamped free vibration in the form of Simple Harmonic Motion with an angular

velocity ωn

nωs i±=

tω2

tω1

nn eGeG u(t) -ii +=

t)BSin(ω t)ACos(ω u(t) nn +=


Initial Conditions

The constants A and B can be found by evaluating the solution at two different times or more commonly from the velocity anddisplacement at time t = 0

t)BSin(ω t)ACos(ω u(t) nn +=

BSin0ACos0u(0) u(t) 0,At +===t

u(0) A =⇒


Initial Conditions

t)ω(Sinω

(0)ut)u(0)Cos(ω u(t) n

nn

+=

Using the value of A & B in equation given in slide 4 results in:

t)Cos(ωBωt)Sin(ωAω- (t)uSimilarly, nnnn +=

Cos0BωSin0Aω-(0)u (t)u,0At nn +=== t

nω(0)uB =⇒


Amplitude of displacement during undamped free vibration

[ ]2

n

2o ω

(0)uu(0) u

+=

It can be determined that the amplitude (i.e. peak value) of displacement during undamped free vibration is:

uo(0)u

(0)u


Element forces fs(t) from displacements u(t)

One of the most appealing approach for calculating element forces in a structural system is to use Equivalent Static force approach.

According to this approach, at any instant of time t the equivalent static force fs is the external force that will produce the

deformation u at the same t in the stiffness component of structure (i.e., the system without mass and damping). Thus

k.u(t)(t)fs =Where k is the lateral stiffness of the structure. Element forces or stresses can be determined at each instant of time by the static analysis of the structure subjected to the force fs.


Element forces fs(t) from displacements u(t)

Variation of displacement with time in an undamped system Variation of corresponding

equivalent static forces with time, fs(t).

fs(t) = k.u(t)

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Natural frequencies and Periods of free vibration

The Natural Circular Frequency ωn is not a convenient measure for

most engineers. The preferred usage is the Natural Frequency fn

which is usually measured in cycles/second (cps) or Hertz

However, in earthquake engineering the preferred measure of thedynamic characteristic of structures is the Natural Period of free

vibration Tn measured in units of seconds.

2π

ωf n

n =

nnn ω

2π

f

1T ==

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Natural Period of free vibration

a b c d e

Amplitude,uo

a

b

c

d

e

nn ω

2πT =

(0)u

Time, t (sec.)

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Alcoa Building, San Francisco

Steel structure, 26 stories.

Periods of vibration:

Transverse (east-west): 2.21 sec

Longitudinal (north-south): 1.67 sec

Torsional: 1.12 sec

Natural period of free vibration

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Golden Gate Bridge, San Francisco

Steel structure, center span of 4200 feet.

Periods of vibration:

Transverse: 18.2 sec

Vertical : 10.9 sec

Longitudinal: 3.81 sec

Torsional: 4.43 sec


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Rule of thumb:

Natural Time Period (sec) ≈ Number of Stories / 10

e.g. a 15 story building has a fundamental period of approximately 1.5 seconds. This applies primarily to moment frame buildings. This rule is very rough, but is good for understanding the general ballpark.


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Problem M4.1A beam shown in Figure is pulled for ¼ inch in the downward directionand then suddenly released to vibrate freely. Determine natural time Periodof the system and develop and solve the equation of motion for vibrationsresulting at free end. Also develop the equation showing variation in the Equivalent static forceswith time. What will be the amplitude of equivalent static force? Ignore theself weight of beam as well as damping effect. Take E = 29,000 ksi and

I = 150 in4. δst = Deflection due to 1000 lb static load

.

1000 lb

¼ "10'

δst

Static Equilibrium position

CE-409: MODULE 4 (Fall-2013)

Solution (M4.1)

16m

k

k

p(t)

ku+

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Variation of displacement with time (Problem M4.1)

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Variation of Equivalent static Forces with time (Problem M4.1)


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Problem M4.2:Considering free vibration, solve the equation of motion developed for the frame given in Problem M3.3. Also develop an equation showing variation with the Equivalent static forces. What will be the amplitude of equivalent static force if the Displacement and velocity (which occur in the same direction ?) at the start of free vibration are 0.003 ft and 0.2 ft/sec., respectively

.

I, 15ftI, 10ft

p(t)

20 ft

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Solution (M4.2)

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Variation of Equivalent static forces with time (Problem M4.2)


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Response of damped systems to free vibrations

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The equation of free vibration for damped free vibration has the form

0kuucum =++

The solution to this equation will be taken in the same form as for the undamped form i.e

stGe u(t) =

Viscously Damped Free Vibration

Substituting this value in equation of motion result in:

0kGe )c(sGe )Gem(s ststst2 =++

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By rearranging we get: 0)k)(Gescm(s st2 =++

Since 0Gest ≠ 0kscms2 =++⇒

22

22

m4

km4

m4

c

2m

c-

2m

4kmcc-sor −±=−±=


By rearranging:

2

nnn

n 2mω

c-1ωω

2mω

cs

±−=

Using the relation k = mωn2

2n2

2

ωm4

c

2m

c- s −±=

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crn c 2mω = is known as Critical damping coefficient

ζ=n 2mω

c where (Greek alphabet for Zeta) is known as

2nn -1ωωs ζζ ±−=

Where is known as damped natural frequency and represented by ωD

2n -1ω ζ

Dn ωωsor ±−= ζ

damping ratio or fraction of critical damping, which when substituted in the equation mentioned on previous slide gives:

CE-409: MODULE 4 (Fall-2013) 30

n2mωc >

n2mωc =


There are 3 forms of solution available depending on the magnitude of the damping coefficient c

1

2

3 n2mωc <

Critically damped system

Over damped system

Under damped system

CE-409: MODULE 4 (Fall-2013) 31

n2mωc =Viscously Damped Free Vibration

Critically damped system

The structure is said to be Critically damped.

There is no vibration in the response. The structure returns to its initial position without vibrating about the zero position but in the shortest time

An example of critically damped system

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n2mωc >


Over damped system

The structure is said to be Over damped.

Like critically damped systems, there is no vibration. However, the structure returns to its initial position slowly as compared to critically damped system.

Critically damped and over damped systems are of no interest to the civil or structural engineer.

An automatic door close is an example of an over damped system

CE-409: MODULE 4 (Fall-2013) 33

Free vibration of under damped, critically damped , and over damped systems

CE-409: MODULE 4 (Fall-2013) 34

n2mωc <Viscously Damped Free Vibration

Underdamped system

The structure is said to be under-damped

The structure again returns to its origin but now vibrates.

This is the only case that is of interest to civil or structural engineers as in all our structures the level of damping is very small, usually less than 5% of critical damping.

CE-409: MODULE 4 (Fall-2013) 35

Approximate Damping Ratios

Working Stress Level (1/2 yield point) ζ

Welded steel, Prestressed concrete, Well reinforced concrete (slight cracking) 2-3%

Reinforced concrete with considerable cracking 3-5%

Bolted or riveted steel, Timber 5-7%

At or just below yield point

Welded steel, Prestressed concrete (without loss of prestress) 5-7%

Reinforced concrete 7-10%

Bolted or riveted steel, Bolted timber 10-15%

Nailed timber 15-20%

CE-409: MODULE 4 (Fall-2013) 36

Response of underdamped systems to free vibrations

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Underdamped systems

The solution for the under-damped system have the form

[ ]t)(Sin B. t)os(CA.e u(t) DDt- n ωωζω +=

This is similar to the case for undamped free vibration except that the frequency is slightly smaller and there is a decay of the response with time. Again the constants A and B may be found from the solution at two different times or from the initial conditions at time t = 0

++= t)(Sinω

u(0)(0)ut)(Cos)0(ue u(t) D

D

nD

t- n ωζωωζω

2nD -1ω ω where ζ=

CE-409: MODULE 4 (Fall-2013) 38

Damped Free Vibration

Problem M4.3For the beam’s data given in problem M 4.1, develop and solve theequation of motion for vibrations resulting at free end. Also develop an equation showing variation in the Equivalentstatic forces with time. Take ζ = 2.5%

.

1000 lb

¼ "10'

δst

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Solution (M4.3)

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Variation of Equivalent static Forces with time (Problem M4.3)

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Decay of Response Due To DampingThe decay observed in the response of a structure to some initial disturbance can be used to obtain a measure of the amount of viscous damping ,c, present in the structure.

Consider two successive positive peaks ui and ui+1in the response

shown in the figure and which occur at times ti and ti+1.

ui ui+1

ti

ti+1

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Decay of Response Due To Damping

+

= D

n

ω

2π

1i

i e u

uζω

It can be derived that:

Taking the natural logarithm of both sides we get the so-called logarithmic decrement of damping, δ, defined by the following equation.

ω

2π

u

ulnδ

D

n

1i

i ζω=

=

+

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2nD -1ω ω Since ζ=


and; -1ω

2πδ

2n

n

ζζω=⇒

For Civil engineering systems,ζ is usually less than 0.1 andsystems dampedlightly for 2π δ and 1ζ-1 2 ζ=≈

2-1

2π δ

ζζ=⇒

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These values may be improved by taking the differences in

the peak response values not at two successive peaks but over a

range of j peaks.


u

ulnjδ

1j

1

=

+

u1

uj+1

j cycles

u

uln

2π

1j

1j

1

=⇒

+ζ

ζ2π u

uln

j

1 δ

1j

1 =

=⇒

+

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Problem M4.4 :A free vibration test was conducted on an empty water tank shown in figure. A force of 60 kips, applied through a cable attached to the tank, displace the tank by 2 " in Horizontal direction. The cable is suddenly cut and the resulting vibration is recorded. At the end of 5 cycles, which complete in 2.55 sec., the amplitude of displacement is 0.9 ". Ignore the vertical vibration of tank and compute the following:a) Damping ratiosb) Natural period of undamped vibrationc) Stiffness of structuresd) Weight of tanke) Damping coefficientf) Number of cycles to reduce the displacement amplitude to 0.5 ″ Solution: Refer to class notes.

60 k

2 "

60o

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Solution (M4.4)

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1. A spring- mass system has a natural frequency of 10 Hz. When the stiffness of spring is reduced by 800 N/m, the frequency is reduced by 45%. Find the mass and stiffness of actual system. 2. The natural frequency of a beam supporting a lumped mass, m, is 2 cycles per second (cps). When an additional mass of 50 kg is added, the natural frequency is reduced to 1.75 cps. Determine the stiffness and lumped mass, m.

3. The lateral stiffness of a cantilever beam, supporting a lumped mass of 25kg at the free end, is 5 kN/m. The beam, when hit with a hammer, start vibrating with an initial velocity of 0.1 m/s. Determine the displacement at free end from equilibrium position at the end of first second. Take viscous damping coefficient as 150 N.s/m.

Problems

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4. A body of mass 1.25 kg is suspended from a spring with stiffness of 2kN/m. A dashpot attached to the spring-mass system require a force of 0.5 N to move with a velocity of 50 mm/s. Determine :(a) Time required by the spring-mass system to complete one cycle during free vibration. (b) Amplitude of displacement of the body after 10 cycles when the body was released after an initial displacement of 20 mm. and (c) Displacement after 1.25s from the start of free vibration.

5. An underdamped shock absorber is to be designed for a motor cycle of mass 200 kg. The vibration starts occurring in motor cycle in vertical direction when it reaches a road bump. Find necessary stiffness and damping ratio of the absorber if the damped period of vibration is to be 2 s as well that the ratio of amplitudes of displacement in successive peaks is to be 16.

Home Assignment M4H1

Solve problems 1,4 and 5

Module 4 (CE-409: Introduction to Structural Dynamics and Earthquake Engineering)

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