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Chapter 3: Dynamic Analysis

Part 1: General

Part 2: Earthquake/1D

Part 3: Earthquake/nD

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Part 1: General

Contents:

1D mathematical spring model

3D example #1

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Undamped free response for SDOF systems

If you displace mass m a distance u from equilibrium position and release it, vibration occurs

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1D mathematical spring model

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1D mathematical spring model:

Free body versus kinetic diagrams

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1D mathematical spring model:

Equilibrium equation

2

2

2

2

( )

( )

u F t Ku m

t 

um Ku F t  

t 

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1D mathematical spring model:free response

 A and B are found frominitial conditions

02

2

kudt 

ud m

02

2

  um

k 

dt 

ud 

mk 

udt 

ud 

n

n

/

0

2

2

2

2

 

 

t  Bt  Aunn

       cossin  

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T n

= natural period of vibration

ωn =natural circular frequency of vibration

 f n =1/T n= natural cyclic frequency of vibration

in hertz (cycles per second)

Undamped free response for SDOF systems:

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Undamped Forced Vibrations

..

2

2   2

( ) sin

sin

1cos sin sin

1 ( )

1 11 ( )   1 ( )

n

n

n

n

 F t P t 

m u ku P t  

 P u C t D t t  

k 

 DMF 

 

 

 

 

 

 

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The ratio

If ratio →∞ response is zero, high frequency loading is

not felt, or flexible structure!

If ratio →1 response →∞ resonance loading

If ratio →0 static response, low frequency loading or rigid

structure

n

n

 

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3D example #1

RC flat plate structure shown next page

No superimposed loads

E=20GPa, μ=0.2, ρ=2.5t/m3

Force 16KN applied in y-direction (weak direction)

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Modal Analysis

Mode

number

periodModal mass

particip. ratio

participation

direction

11.0461Uy

20.531Ux

30.411Rz

40.050.95Uz

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1-D Analogy for 3-D model 

33 7 3

2

(0.2)(0.1)4*12 / 4 *12 *(2*10 ) /(3)

12

592 /

4* 4* 0.4* 2.5 0.2*0.1*3*2.5*4 / 2 16.3

592 /16.3 36.3,

2 / sec

Y 

Y 

 K EI L

 K KN m

 M ton

T      

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Dynamic Analysis

1. Do static solution (u=0.0272m) + Modal analysis

(fundamental period 1.046sec)

2. Use sine function (20 cycles) with periods 0.1, 0.5,

1, 5 and 50 seconds

50510.50.1Period

(sec)

27.834.158824.62.85Displ.

mm

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Part 2: Earthquakes

Understand 1D first

Go from 1D to nD

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1D analysis

Ground acceleration

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1D analysis

Derivation of equations

..

..   .. ..

* *

..   ..* *

..   ..* *

( )

,

( )

( )

 g 

 g  g 

 g 

 g 

m u k u u

u u u u u u

m u u k u

m u k u m u F t  

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1D analysis

ConclusionsGround acceleration is the same as forces applied to the structure(ignoring the sign).

To verify the result resolve previous example with groundacceleration 1m/sec2 in the y-direction (equivalent to 16KN force):

Use sine function (20 cycles) with periods 0.1, 0.5, 1, 5 and 50seconds

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3D example # 2

RC flat plate structure shown next page

Circular columns 50cm diameter.

No superimposed loads

E=25GPa, μ=0.2, ρ=2.5t/m3

Find column reaction in x-direction if:

1. structure is subjected to earthquake acceleration0.3g in x-direction:

Sinusoidal with period: 0.026, 0.26, 2.6sec

2. structure is subjected to Elcentro earthquake 3. structure is subjected to response spectrum twice

0.3g

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3D example # 2 :

Modal analysis

Mode

number

periodModal mass

particip. ratio

participation

direction

10.2661Ux

20.2661Uy

30.1721Rz

40.1530.80Uz

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3D example # 2 :

Solution

1. structure is subjected to earthquake acceleration 0.3g inx-direction:

Uniform for 1-sec: Rx=79.2KN

Sinusoidal with period: 0.026, Rx=6.23KN (almost zero

for short periods) 0.26, Rx=1597KN (resonance)

2.6sec, Rx=45KN (almost static for long periods)

2. structure is subjected to Elcentro earthquake, Rx=113KN

3. structure is subjected to response spectrum twice 0.3g:Rx=71KN

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Homework # 5

RC flat plate structure: slab 50cm thickness,

columns 20cmX60cm

No superimposed loads

E=25GPa, μ=0.2, ρ=2.5t/m3

Find displacement of first floor if structure is

subjected to uniform earthquake acceleration 0.3g

in y-direction

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Homework # 5

Modal Analysis

Mode

number

periodModal mass

particip. ratio

participation

direction

10.93791Uy

20.3561Rz

30.3331Ux

40.0420.83Uz

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Homework # 5

Modeling as SDOF

3

(12*12*.5 16*3*0.6*.2)*2.5 194.4

12*16* 8888 /

2 / 0.93sec

Y 

Y 

 M t 

 EI 

 K KN m L

T M K  

Homework: do analogical solutions for modes 2-4

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Homework # 5

1D analogy

Uy=0.131m SAP

194.4*0.3*9.8*2

0.1298888

194.4 2(1 cos ) (1 cos )

8888 0.940.95 0

0.47 0.133

 F 

u m K 

 F t u t 

 K 

t u

t u

  

 

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