microsoft powerpoint - 3-introduction to dynamic analysis

8/10/2019 Microsoft Powerpoint - 3-Introduction to Dynamic Analysis

1/37

INTRODUCTION TO

DYNAMIC ANALYSIS

Civil Engineering Department

Petra Christian University


2/37

2

Simple Structures Structures can be idealized as a Single Degree of Freedom (SDOF)System, consist of a concentrated mass m supported by a masslessstructure with stiffness k in lateral direction.

one-story building

a heavy, stiff, concrete roof

Light columns

tank,full of water

Reinforcedconcrete

column

water tankIdealization/ modelling

m

k/2

viscous

dampingrigidbase

massless,elasticcolumns

Rigid mass Viscous damping element is

added to model the energydissipation caused by variousmechanism such as thermaleffect, friction at connections,

cracks in concrete, etc.

k c

m


3/37

3

Equation of MotionWhere:P(t) = external force

fs(t) = elastic restoring force= k u(t)

k = lateral stiffness of the structure

fd(t) = damping force

= c (t)c = viscous damping coefficient

m

k/2c

P(t)

u(t)

Using Newtons second law of motion, we obtain:

P(t) fs(t) fd(t) = m (t)

mP(t)

fs/2 fs/2fd

u(t)(t)(t)

or:m + c + k u = P(t)

=d2udt2

Note:


4/37

4

Effective Earthquake Force

c

u(t)

ug(t)

ground displacement

Initial

conditionfs/2 fs/2fd

m

Using Newtons second law of motion, we obtain:

0 fs(t) fd(t) = m (t)

Where: ut(t) = total (absolute) displacement of the mass mut(t) = ug(t) + u(t)and fs(t) = k u(t) ; fd(t) = c (t)Then,

m + c + k u = - m g(t)

Effective earthquake force


5/37

5

Effective Earthquake Force cont.

moving base

u(t)

Identicaldeformation

u(t)c

m

u(t)

stationary base

P(t) = - mg(t)

g(t)


6/37

6

Undamped Free Vibrationm + k u = 0m (t) + c (t) + k u(t) = 0

Partial differential equation, because:

(t) = and (t) =du(t)dt d2

u(t)dt2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 0.2 0.4 0.6 0.8 1

t (sec)

u(t)

a

b

c

d

e

Tn = 2/n

b da

u0

c

u0

e

u0


7/37


8/37

8

The various natural frequency and period of structures:


9/37

9

Damped Free Vibrationm (t) + c (t) + k u(t) = 0

or(t) + 2n (t) + n2 u(t) = 0

Where: = =

= The critical damping ratio

cr= critical damping coefficient= 2(km) = 2mn

c2mn

ccr

1. The solution of P(t) = 0, exactly the same with the case of Undamped

Free Vibration (see p.7)


10/37

10

Free vibration of system with four levels of damping: = 0.02, 0.05, 0.1 and 0.2


11/37

11

2. For Damped Non-Free Vibration ( P(t) 0)

The solution is the summation of homogenous and particularsolution of partial differential equation:

u(t) = uh(t) + up(t)

uh(t) see the solution of Undamped Free Vibration (see p.7)up(t) usually following the function of P(t)

For example: a. for harmonic function,

P(t) = Po sin (t)then: up(t) = sin (t -)where :

=

= arc tg (2 / (1 -2)) = /

Pok

1

{(1-2

)2

+ (2)2

}


12/37

12

For example: b. for non-harmonic function (but still in periodic),the loading should be transformed into Fourier-function. The solution is superposition ofhomogenous and particular solution of eachfunction, i.e.

u(t) = uh(t) + up1(t) + up2(t) + + upn(t)

c. for un-periodic loading (Impulse Loading) use Duhamel Integration

d. for un-steady function (Earthquake) use Step-by-Step Direct Integration


13/37

13

Example: Undamped Free Vibration

10 cm

A 50 m3 water tank is supported by 30x30 cm2 concrete column at10 m height. The structure is a simple model of SDOF system,resists 10 cm ground displacement at its base. Calculate the top

maximum displacement! Assumed there is no structural damping.Answer:Mass, m = 50 tonStiffness, k = 12EI/L3 = 12*2E6/103 = 24000 ton/m

Natural circular frequency, = (k/m) = 21.9 rad/secNatural frequency, f = /(2) = 3.49 cycle/secNatural period = 1/f = 0.286 sec

Initial condition: initial displacement, u(0) = 0.1 minitial velocity, (0) = 0 m/sec


14/37

14

Displacement:u(t) = e-t cos (Dt -)

= = = 0

e-t = 1

= uo2 + = 0.12 + = 0.1 m

D = n (1-2

) = n = 21.9 rad/sec

= arc tg = arc tg = 0

u(t) = 0.1 cos (21.9 t)

c2mn

02mn

o + uonD

2 0+ 0D

2

o + uonuoD

0+ 00.1 D


15/37

15

How is the response during the first one second if we consider

the critical damping ratio, = 0.05; 0.1; 0.15; and 0.2


16/37

16

Displacement History

-0.15

-0.1

-0.05

00.05

0.1

0.15

0

0.

05

0.

1

0.

15

0.

2

0.

25

0.

3

0.

35

0.

4

0.

45

0.

5

0.

55

0.

6

0.

65

0.

7

0.

75

0.

8

0.

85

0.

9

0.

95 1

t (sec)

displacemen

,u(t)

= 0

= 0.05 = 0.1

= 0.15

= 0.2


17/37

17


18/37

18


19/37

19


20/37

20

Generalized SDOF SystemsConsidering 3 examples of complex systems:

ma

mb

kc

kd

ce

u(x,t)

ug(t)

(a)An assemblage ofrigid bodies

L

x

u(x,t)

ug(t)

(b)A system withdistributed mass

and flexibility

(c)A multi-storyshear building

ug(t)

k4

k3

k2k1

u4

u3

u2

u1

m4

m3

m2

m1

If the deformation of complex systems can be (approximately) expressed as:

u(x,t) = (x) . z(t)

where: (x) : a shape function (dimensionless) or a mode shapez(t) : a single generalized displacement

or for the case (c): ui(t) = i . z(t) where i : a shape factor at the ith story


21/37

21

Then, it can be shown that the governing equation of motion of each of

these complex system is in the form of:

m z(t) + c z(t) + k z(t) = - L g(t) . (1)

Where m, c, k, and L are defined as the generalized mass, generalized

damping, generalized stiffness, and generalized force factor of thesystem, respectively.

For the system (a):

ug(t)

ma

mb

kc

kd

ce

xb

xa

xdxe

xc

)()(

)(2

)(2

)(2

)(2

)(2

~

~

~

~

xxbxca

xddxcc

xee

xbbxaa

mmL

mkk

cc

mmm

+=

+=

=

+=


22/37

22

L

x

u(x,t)

ug(t) ug(t)

k4k3

k2k1

u4

u3

u2

u1

m4

m3

m2

m1

=

=

=

L

L

L

dxxxmL

dxxxEIk

dxxxmm

0

0

2

0

2

)().(~

)(").((~

)().(~

For the system (b): For the system (c):

=

=

=

=

==

==

N

jjj

column

j

N

j

jjj

N

j

jj

mL

h

EIkkk

mm

1

31

2

1

1

2

~

;12

where)(~

case)in this4(N~

heightstory:h

Therefore, given the vibration shape and the distribution of mass andflexibility of a complex system, it is possible to evaluate all of these

generalized properties of the system.


23/37

23

Dividing the equation (1) by gives:m~

factor)less(dimensionm~/L~~

ratiodamping(modal)the:~~

2/~

sytemSDOFdgeneralizetheoffrequencynaturalthe:~/~

:Where

(2)....)(~

2

2

2

=

=

=

=++

mkc

mk

tuzzz

n

gnn

&&&&&

The equation (2) says that thegeneralized displacement z(t) of thegeneralized SDOF system due toground motion ug(t) is identical tothe displacement response u(t) of a

simple SDOF system (having thesame n and ) to ground motion ).(

~tug

c

m

u(t) = z(t) of the

generalizedSDOF system.

).(~

tug

n,


24/37

24

Suppose that the response spectrum of the ground motion ug(t) isavailable. It is then possible to estimate peak earthquake response of thegeneralized SDOF system:

Peak value of z(t) = ADzn

o2

~~

==

where D and A are the deformation and pseudo-acceleration ordinates,respectively, of the spectrum at the modal period Tn = 2/n for the

modal damping ratio ,Peak displacements:

j

o

D

xDxu

.~

u

or)(.~

)(

jo =

=


25/37

25

Equivalent static forces (external static forces that would causedisplacement uo(x)) for the system (b) can be derived from elementarybeam theory as:

[ ]

)().(.~

(x).um(x).(x)f

elyalternativor

")(").()(

o

2

so xxmA

xuxEIxf

n

oso

==

= fso(x) uo(x)

Thus, the shear and bending moment at height above the base are:

=

====

==

==

L

oboobo

L

x

L

xsoo

L

x

L

x

soo

dxxxmxLwhere

.A.L)(MMALVV

dmxAdfxxM

dmAdfxV

0).().(.

~

:

~~0and.

~.

~)0(

aretowertheofbaseat themomentbendingandshearThe

)().().(.~

)().()(

)().(.~

).()(


26/37

26

For the case of shear building in system (c):

ug(t)

u4o

u3o

u2o

u1o

m4

m3

m2

m1

f4o

f3o

f2o

f1o 4Ncasein this

,...,2,1for

.~

=

=

=

Nj

mAf jjjo

The shear force Vio in the ith

story is:==

==N

j

jj

N

j

joio mAfV11

..~

The overturning moment at the ith story is:( ) ( )

==

==N

j

jjij

N

j

joijio mhhAfhhM11

..~

Where hi is the height of the ith floor above the base.


27/37

27

The shear and overturning moment at the base are:

where:

.~.

~.

.~.

~

1

1

ALfhM

ALfV

N

j

jojbo

N

j

jobo

==

==

=

=

=

=N

j

jjj mhL1

..~

Noted here that the generalized factors depend only on thevibration shape and the mass distribution of the complex system.

~

and,~

,~ LL


28/37

28

Example 1:

A uniform cantilever tower of length Lhas mass per unit length = m andflexural rigidity EI. Assuming the shapefunction (x) = 1 cos (x / 2L),

formulate the equation or motion forthe system excited by ground motion,and determine its natural frequency.

m,EIL

x

(x) = 1 cos x2L

1. Determine the generalized properties

mLdxL

xmL

LEIdxLx

LEIk

mLdxL

xmm

L

L

L

363.02

cos1

/04.32

cos4

~

227.02

cos1~

0

32

0

2

2

0

2

=

=

==

=

=

The computed is close tothe stiffness of the tower

under a concentrated lateralforce at the top (3EI/L)

k~


29/37

29

2. Determine the natural vibration frequency.

m

EI

Lm

kn 2

66.3~

~==

This approximate result is close to the exact natural frequency, (i.e.

(3.516/L2

)(EI/m). The error is only 4%.

3. Formulate the equation of motion.

)(6.0

:ismotionofequationtheThen,

6.1)227.0/(363.0~/~~

2 tuzz

mLmLmL

gn &&&& =+

===


30/37


31/37

31

1. Determine the chimney properties.

210

444

22

222

ft-kip10469.5:rigidityFlexural

ft507,105)5.2225(4

:areaofmomentSecond

/ftsec-kip738.12.32

1.373150:lengthMass/foot

1.373)5.2225(:areasectional-Cross

600:Length

xEI

I

xm

ftA

ftL

=

==

==

==

=

2. Determine the natural period. From Example 1,

sec49.32

rad/sec80.166.3

~

~

2

==

===

n

n

n

T

m

EI

Lm

k

3. Determine the peak value of z(t). For Tn = 3.49 sec and = 0.05, thedesign spectrum gives A/g = 0.25(1.80/3.49) = 0.129. Thecorresponding deformation is D = A/n2 = 15.3 in. The peak value of

z(t) is: zo = 1.6D = 1.6 x 15.3 = 27.6 in


32/37

32

4. Determine the peak displacements uo(x) of the tower (fig.b)

in2

cos16.27)()(

==L

xzxxu oo

5. Determine the equivalent static forces.

fig.c)in(shownkips/ft2

cos158.11

129.02

cos1)738.1)(6.1()()()(

=

==

L

x

gLxAxxmxfo

6. Compute the shears and bending moments. Static analysis of thechimney subjected to external forces fo(x) gives the shear forces andbending moments. The results are presented in fig. d and e. If we wereinterested only in the forces at the base of the chimney, they could be

computed directly. In particular, the base shear is

kips25180749.0

129.0)6.1)(363.0(~~

==

==

mLg

gmLALVbo

This is 7.49% of the total weight of the chimney.


33/37

33

Elastic pseudo-acceleration design spectrum for ground motions withgo = 1g, go = 48 in/sec, and ugo = 36 in; = 5%


34/37

34

Example 3:

The uniform five-story frame with rigid beamsshown in figure is subjected to groundacceleration g(t). All the floor masses are m,and all stories have the same height h and

stiffness k. Assuming the displacements toincrease linearly with height above the base(fig. b), formulate the equation of motion for thesystem and determine its natural frequency.

1. Determine the generalized properties.

mmmL

k

kkk

mmmm

j

jj

j

jjj

j

jj

35

54321~

55

11111

)(

~

5

11

5

54321~

5

1

2

222225

1

2

1

2

222225

1

2

=++++

==

=++++

==

=++++

==

=

=

=


35/37

35

2. Formulate the equation of motion. Substituting for and givesm~ L~

11

15~= and the equation of motion becomes

3. Determine the natural vibration frequency

)(11

152 tuzz gn &&&& =+

where z is the lateral displacement at the location where j = 1, in thiscase the top of the frame.

mk

mkn 302.0

5/115/ ==

This is about 6% higher than n = 0.285 (k/m) , the exact frequencyof the system.


36/37


37/37

microsoft powerpoint - 3-introduction to dynamic analysis

Documents