problem dinamica 2

7/27/2019 Problem dinamica 2

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HOMEWORK # 2

EXAMPLE 2

Forced vibrations response of a 3DOF model

Given a structure of RC frame consisting of 3 floors, represented by a 3DOF lumped

model of Fig. 1, and applying a P100/2006 type spectrum let us found the free vibrationsresponse using the stiffness method.

A. Initial Dataa. Geometrical characteristics:

cmbcol

25 4

3

2233 083.3255212

25*25cm I I colcol

cmhcol25 262525*25 cm Acol

cm Lcol

400

b. Material characteristics:

2270000

cm

daN E concrete 2.0 )(0025.0

3densityweight

cm

daN W concrete

Fig. 1. 3DOF model of a 3 stories RC frame

c. Evaluation of Mass matrix:

cm

sdaN m

2

3 3185.02 / 981

400*25*25*0025.0

cm

sdaN m

2

2 6371.0981

400*25*25*0025.0



cm

sdaN m

2

1 6371.0981

400*25*25*0025.0

The final Mass matrix is:

6371.000

06371.00003185.0

00

0000

1

2

3

m

mm

M

d. Computation of Stiffness Matrix:

The elastic force vector: uF el

K

1

2

3

T

T

T

F el

The stories shear force vector is: K T

The stories drift:1

ii

uu

The lateral stiffness of the story:

cmdaN L

EI

L

EI k

col

col

column col

col / 949.1647400

083.32552*270000*121212333

and the lateral elastic forces are:

)21(1)21()01( 21211

11 uuk uk uuk uk T T T II R

a

I R R

)32()12(321

12

uuk uuk T T T III R

II R

b

R

)23(313

uuk T T III R R

II R

b

II R

a

T T 11 (equal, but opposite in sign)

1

2

3

211

1322

23

0

0

u

u

u

k k k

k k k k

k k

T R

1

2

3

20

2

0

u

u

u

k k

k k k

k k

T R

=>



898.3295949.16470

949.1647898.3295949.1647

0949.1647949.1647

20

2

0

k k

k k k

k k

K

e. Identification of Damping Matrix:

The damping force vector is:

.

uF D

C

1

2

3

D

D

D

D

F

F

F

F

The stories shear force due to damping effects:

.

cF D

The stories velocity: 1

...

ii uu

The damping of the story:

cr cc 0c for free undamped vibrations.

0

0

0

1

.

2

.

3

.

213

3322

23

u

u

u

ccc

cccc

cc

F D

f. Description of Seismic Spectrum recommended in P100/2006, presented in Fig.1:

mT S F d I b )( (1)

where:

q

T aT S gd

)()(

;

For Iaºi the Corner Period is sT C 7.0 ; 1 I (for regular buildings);

22.196981*2.02.0 s

cm

gag ; 25.625.1*55 1 uq .



Fig. 1. Normalized Elastic Spectrum after P100/2006

B. Solution:

The equation of motion of a MDOF model is:

)(...

t uuu PKCM (2)

For the case of forced undamped vibrations the form of Eq. (1) becomes:

)()()(........

t umt umt umuu gz zgy ygx x KM(3)

where:. ..

( ), ( ), ( )u t u t u t are the relative displacement, velocities and accelerations with respect to the

ground;.. .. ..

, ,gx gy gz u u u are the components of uniform ground acceleration;

, ,x y z m m m are the unit acceleration loads.

The dynamic force equilibrium equation can be written as a set of N second order differential

equations:

J

j

j j t g f uu1

..

)(KM (4)

where :

All the time-dependent loads are represented by a sum of “ j” space vectors j f , which do not

depend on time and “J” time functions )(t g j , where N J with “N” being the number of

displacements.



The spectrum function (P100/2006) defined in Eq. (1) becomes:

)(2000T cF

SAPb (5)

mq

a

cg

I SAP 2000 (6)

The problem of forced vibrations supposes the following steps:

Step 1Transformation of modal equations:

)()( t t ÖY

u

(7)

)()(..

t t YÖu (8)

)()(....

t t YÖu (9)

Step 2 The condition of mass and stiffness orthogonality must be satisfied:

IMÖÖ T (10)

2ÙKÖÖ T (11)

Step 3Substitute Eqs. (7), (8) and (9) in Eq. (4) it will be obtained the following dynamic force

equilibrium equation:

J

j

j j t g f t t 1

..

)()()( YKÖYMÖ(12)



Pre-multiplying Eq. (12) with T Ö we’ll obtain:

J

j

j j

T T T t g f t t

1

..

)()()( ÖYKÖÖYMÖÖ(13)

Substitute Eqs. (10) and (11) in Eq. (13) and obtain:

J

j

j j

T t g f t t 1

2..

)()()( ÖYÙYI(14)

J

j

j jT t g f t t

1

2

..

)()()( ÖYÙY(15)

Step 4 Identification of the uncoupled modal equation, of the Eq. (15) in matrix form, using modal

damping:

J

j j j

T t g f t t t

1

2...

)()()(2)( YYY(16)

Step 5 In our example we apply the spectrum function only in the X-direction and Eq. (16) becomes:

)()()(2)(..

2...

t u f t t t gx j

T YYY(17)

where:

the damping proportional coefficient if 2c ;cr

moden

c

thcc ; and the critical damping ratio

is 05.0 ;

Step 6 We have the Modal Normalized Matrix obtained from the modal analysis:



511.0023.1511.0

886.000001.0886.0

022.1022.1023.1

(18)

We obtain the modal participating factors:

ãMÖáT (19)

0866.0

3262.0

2158.1

1

1

1

6371.000

06371.00

003185.0

511.0023.1511.0

886.000001.0886.0

022.1022.1023.1T

á

Step 7

Obtain the total effective mass, computed as2 :

Mode á á

SAP2000

2á tot%M tot%M

SAP2000

tot%M

accumulated SAP2000

%M tot

1 1.2158 -1.2163 1.4783 92.84% 93% 92.84% 93%

2 0.3262 0.3259 0.1064 6.68% 6.667% 99.52% 100%

3 0.0866 -0.0873 0.0075 0.48% 0.4786% 100.00% 100%

TOTAL - 1.5922 - -

Step 8 Obtain the uncoupled vibration equations:

)(2158.1)()(2)(..

1

2

1

.

111

..

1 t ut yt yt y gx

)(3262.0)()(2)(..

2

2

2

.

222

..

2 t ut yt yt y gx

)(0866.0)()(2)(..

3

2

3

.

333

..

3 t ut yt yt y gx

where:

05.0i is the critical damping ratio.



Step 9Obtain values of the displacement spectrum according to the periods of vibration (P100/2006):

2

2)()(

T

T S T S e De (20)

From P100/2006 we get:

q

T aT S T S gd e

)()()( 1

(elastic/design spectrum)

C BC B T T and T T T for T 1.0)( ,0

It results:

2

01

2)(

T

qaT S g De (21)

Step 10Compute the maximum displacement values for the uncoupled degrees of freedom:

Mode á iiDe î,TS

(cm)

iiDei î,TS*áy

(cm)

1 1.2158 0.123864 0.010727

2 0.3262 0.016551 0.005399

3 0.0866 0.007811 0.009497

ModeiT

(s) 0â

0 1

3 0.063 2.475

0.07 2.75

2 0.087 2.75

1 0.238 2.750.7 2.75

ModeiT

(s)

iiDe î,TS

(cm)

1 0.238 0.123864

2 0.087 0.016551

3 0.063 0.007811



Step 11Compute the maximum modal displacements in mode “i” of vibrations:

iaxi yim öu

, (22)

Eq. (22) can be written in the matrix form:

max modU H (23)

where:

1

2

3

0.010727 0 00 0

0 0 0 0.005399 0

0 0 0 0 0.009497

y

y

y

modH

The values of maxU are:

max

1.023 1.022 1.022 0.010727 0 0

0.886 0.00001 0.886 0 0.005399 0

0.511 1.023 0.511 0 0 0.009497

U

max

0.010973 0.005518 0.009706

0.009504 0.00000005 0.008414

0.005481 0.005523 0.004853

cm

U

Step 12Compute the maximum modal lateral forces multiplying the stiffness matrix of the structure by

the modal lateral displacements:

max maxF K u(24)

max

1647.949 1647.949 0 0.010973 0.005518 0.009706

1647.949 3295.898 1647.949 0.009504 0.00000005 0.008414

0 1647.949 3 295.898 0.005481 0.005523 0.004853

F

max

2.4218 9.0931 29.8613

4.2073 0.0087 51.7251

2.4041 18.2037 29.8613

daN

F



Step 13Compute the maximum modal story shear from:

n

jk

ik

i j F T

(25)

Story 1

modT 2

modT 3

modT

3 2.4218 -9.0931 29.8613

2 2.4218+4.2073=6.6291 -9.1018 -21.8638

1 6.6291+2.4041=9.0332 9.1019 7.9975

Step 14Compute the base shear for each mode:

modmod F1TT

(26)

1 2.4218 9.0931 29.8613

1 4.2073 0.0087 51.7251 9.0332 9.1019 7.9975

1 2.4041 18.2037 29.8613

T

daN

modT

Step 15Compute the overturning moment for each story:

n

jk

i

j jk

i

j F hh M 1

)((27)

Story1

modM 2

modM 3

modM 3 2.4218*0=0 0 0

2 2.4218*400=968.72 -3637.24 11944.52

1 2.4218*800+4.2073*400=3620.36 -7277.96 3199.04

0 2.4218*1200+4.2073*800+2.4041*400=7233.64 -3637.2 6398.08



Step 16Compute the maximum overturning moment at the base:

modmod FhMT

(28)

12 2.4218 9.0931 29.8613

8 4.2073 0.0087 51.7251 7233.64 3637.2 6398.08

4 2.4041 18.2037 29.8613

T

daNcm

modM

Step 17Obtain the modal spectral combination values using the square root of the squares (SRSS):

n

i

ir r 1

2

(29)

Maximum credible lateral displacements of the structure:

n

i

i

j

SRSS U U

1

2

(30)

max

0.010973 0.005518 0.009706

0.009504 0.00000005 0.008414

0.005481 0.005523 0.004853

cm

U

max 2 2 23 0.010973 0.005518 0.009706 0.015654U cm

max 2 2 22 ( 0.008414) 0.00000005 0.009504 0.012693U cm

max 2 2 21 0 .00 485 3 0. 005 52 3 0 .00 548 1 0 .0 09 17 0U cm

Modal forces should be separated by node and never combined using SRSS method so that we

want use combined forces in the computation of story shear and overturning moments.

Maximum credible base shear of the structure:

n

i

iSRSS T T

1

2

modmod (31)



max 2 2 29.0332 9.1019 7.9975 15.113T daN

Maximum credible overturning moment of the structure:

n

i

iSRSS M M

1

2

modmod (32)

max 2 2 27233.64 3637.2 6398.08 10319.409M daNcm

SAP2000 results from Spectral Analysis

Displacements Overturning moment Shear force

problem dinamica 2

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