Taller #1 Dinamica Estructural

8/18/2019 Taller #1 Dinamica Estructural

1/23

TALLER # 1

JONATAN MIGUEL VILORIA ROJAS

DINÁMICA ESTRUCTURAL Y DISEÑO SÍSMICO

FABIAN CONSUEGRA Ph. D.

DIRECCIÓN DE ESPECIALIZACIONES

ESPECIALIZACIÓN EN ANÁLISIS Y DISEÑO DE ESTRUCTURAS

FUNDACIÓN UNIVERSIDAD DEL NORTE

BARRANQUILLA-COLOMBIA

2 SEPTIEMBRE DE 2!"

TABLA DE CONTENIDO


2/23

". SOLUCIÓN DEL PROBLEMA # "

"."S$%&'()* +,*(,*$ ,* '&,*+ /$%$ % 0/ , "!! 1. /$34,(05&,/+.

".2S$%&'()* +,*(,*$ ,* '&,*+ /$%$ % 0/ , % 6(.".7S$%&'()* +,*(,*$ ,* '&,*+ +$/ %/ 0//.

2. SOLUCION DEL PROBLEMA # 22." M$,%'()* ,* SAP2!!!.

7. SOLUCION AL PROBLEMA # 77." G48'/ , % 4,/5&,/+ ,% /(/+,0 '$* MATLAB.

9. SOLUCION AL PROBLEMA # 9

9."E6%&'()* , %$/ ,:,'+$/ , %/ '$*('($*,/ (*('(%,/ ; ,%0$4+(&0(,*+$ ,* % 4,/5&,/+.9.2G48'/ '$03(*'($*,/ , &! ; 6! '$* 0$4+(&'()* , "<9.7G48'/ , '$03(*'($*,/ , &! ; 6! '$* 0$4+(&0(,*+$ ,


3/23

1. SOLUCIÓN DEL PROBLEMA # 1

DATOS

:H' 2"MP L2. 0 M"!! 1

1.1 Solución de la fecuencia na!ual !eniendo en cuen!a "olo la a"ade 1$$ %&. So'e i(ue"!a.

• @%%0$/ ,% 0)&%$ , ,%/+('( ( E ) .

E=3900√ f ' c=3900√ 21 MPa E=17872 MPa

• @%%0$/ % (*,4'( ( I ) , % /,''()*.

I =b∗h3

12=

0.5m∗0.5m3

12= I =0.005208m2

L 4((,( K )

, &* 6( ,05$+4 ,* &* ,=+4,0$ ; /(05%,0,*+,5$; ,* ,% $+4$ ,=+4,0$ ,/+ 5$4

5¿¿

¿37∗¿

K =768∗ E∗ I

7 L3 = K =

768∗17872∗0.005208¿

• @%%0$/ % :4,'&,*'( *+&4% ,* (rad / s).


4/23

K

M =¿ω=√

81700000 N .m

100 Kg. =ω=903.88 rad /s

ω=√ ¿

• E* '('%$/ 5$4 /,&*$.

f = ω

2∗π =f =

903.88 rad /s2∗π

=f =143.85 Hz

• P,4($$ , *+&4% ,* /,&*$/.

T =2∗π

ω =T =

2∗π 903.88rad /s

=T =0.00695 s

1.) Solución de la fecuencia na!ual !eniendo en cuen!a "olo la a"ade la *i&a.

T,*,0$/ K&, % ,*/( ,% '$*'4,+$ 5$4 ρ=2400 Kg

m3

m ( viga )=2400 Kg

m 3∗0.5m∗0.5m∗5m=m ( viga )=3000 Kg

• @%%0$/ ,% 0)&%$ , ,%/+('( ( E ) .

E=17872 MPa

• @%%0$/ % (*,4'( ( I ) , % /,''()*.

I =0.005208m2

L 4((, ( K ) , &* 6( ,05$+4 ,* &* ,=+4,0$ ; /(05%,0,*+,

5$; ,* ,% $+4$ ,=+4,0$ ,/+ 5$4

5¿¿

¿37∗¿

K =768∗ E∗ I

7 L3 = K =

768∗17872∗0.005208¿

• @%%0$/ % :4,'&,*'( *+&4% ,* (rad / s).


5/23

K

M =¿ω=√

81700000 N . m

3000 Kg. =ω=165.02rad / s

ω=√ ¿

• E* '('%$/ 5$4 /,&*$.

f = ω

2∗π =f =

165.02rad / s2∗π

=f =26.26 Hz

• P,4($$ *+&4% ,* /,&*$/.

T =2∗π

ω =T =

2∗π 165.02 rad /s

=T =0.038 s

1.+ Solución de la fecuencia na!ual !eniendo en cuen!a !oda" la"a"a".

L 0/ +$+% ,% /(/+,0 /,4( 100 Kg. de la carga+3000 Kg . viga=3100 Kg .

• @%%0$/ ,% 0)&%$ , ,%/+('( ( E ) .

E=17872 MPa

• @%%0$/ % (*,4'( ( I ) , % /,''()*.

I =0.005208m2

L 4((, ( K ) , &* 6( ,05$+4 ,* &* ,=+4,0$ ; /(05%,0,*+,

5$; ,* ,% $+4$ ,=+4,0$ ,/+ 5$4

5¿¿¿37∗¿

K =768∗ E∗ I

7 L3 = K =

768∗17872∗0.005208¿

• @%%0$/ % :4,'&,*'( *+&4% ,* (rad / s).


6/23

K

M =¿ω=√

81700000 N . m

3100 Kg. =ω=162.34 rad /s

ω=√ ¿

• E* '('%$/ 5$4 /,&*$.

f = ω

2∗π =f =

162.34 rad /s2∗π

=f =25.83 Hz

• P,4($$ ,* /,&*$/.

T =2∗π

ω =T =

2∗π 162.34 rad /s

=T =0.0038 s


7/23

). SOLUCIÓN DEL PROBLEMA #)

DATOSL 0

E2!!!!! MP

S,''()* '&4 %! '0

@%%0$/ % (*,4'( , % /,''()*.

0.5m¿¿¿3

0.5m∗¿ I =

b∗h3

12= I =¿

L 4((, ( K ) , &* 6( ,05$+4 ,* &* ,=+4,0$ ; ,* 6$%($ ,* ,% $+4$

,=+4,0$ ,/+ 5$4


8/23

5¿

¿¿3¿¿

K =3 EI

L3 = K =

3∗200000∗0.00508

¿

@%%0$/ % 0/ , % 344 /3(,*$ K&, ,*/( ,% ',4$ ,/

ρ=7850 Kg /m3

m=(0.5m )2∗5m∗7850 Kg

m3 =mbarra=9812.5 Kg.

m¿=( masa3 )=m¿=9812.5 Kg.

3=m¿ .=3270.83 Kg .

@%%0$/ u0

0=¿0.00128m

u0=

m¿.∗g K

=u0=

32.08kN

25000 kN .m=u¿

@%%0$/ % :4,'&,*'( *+&4% ,* rad /s

ω=

√ K

m¿. =

√25000000 N . m

3270.83 Kg . =ω=87.42 rad

s

f = ω

2∗π =

87.42rad

s

2∗π =f =13.913 Hz .

T =2∗π

ω =T =

2∗π 87.42rad / s

=T =0.07187 s

).1 Modelación en SAP)$$$.

T,*(,*$ %$/ 6%$4,/ '%'&%$/ /, 54,',, 0$,%4 ,% /(/+,0 ,* SAP2!!!


9/23

S, '%'&% ,% 5,4($$ T ,* ,% 54$40 T =0.07196 s

S, '%'&% ,% ,/5%0(,*+$ u0=−0.00198m


10/23

S, 5&,, $3/,464 K&, %$/ 6%$4,/ /$* 0&; 54,'($/ %$/ '%'&%$/0*&%0,*+,.

+. SOLUCIÓN AL PROBLEMA # +

S, 4,+$0* %$/ 6%$4,/ '%'&%$/ ,* ,% 54$3%,0 # " ,/5,'8'0,*+, $*,/, *%(* +$/ %/ 0// +*+$ , % 6( '$0$ , % '4 , "!! 1.

L 0/ +$+% ,% /(/+,0 /,4( 100 Kg. de la carga+3000 Kg . viga=3100 Kg .

• @%%0$/ ,% 0)&%$ , ,%/+('( ( E ) .

E=17872 MPa

• @%%0$/ % (*,4'( ( I ) , % /,''()*.


11/23

I =0.005208m2

• @%%0$/ % 4((, ( K ) .

5

¿¿¿37∗¿

K =768∗ E∗ I

7 L3 = K =

768∗17872∗0.005208¿

• @%%0$/ % :4,'&,*'( *+&4% ,* (rad / s).

K

M =¿ω=√

81700000 N . m

3100 Kg. =ω=162.34 rad /s

ω=√ ¿

• E* '('%$/ 5$4 /,&*$.

f = ω

2∗π =f =

162.34 rad /s2∗π

=f =25.83 Hz

• P,4($$ ,* /,&*$/.

T =2∗π

ω =T =

2∗π 162.34 rad /s

=T =0.0038 s

@%%0$/ ,% ,/5%0(,*+$ (*('(%

u

(¿¿ 0)¿

+,*(,*$ ,* '&,*+ % '4 ,

"!!! N &/+$ ,*'(0 , % 0/ M.

L '4 +$+% /,4( P=( Masaiga∗g )+(masa100 Kg.∗g )+(1000 N )e!"#!ces

P=29.4 kN +0.981 kN +1kN = P=31.38 kN

u0= P K = 31.38kN

81700k N . m=u0=0.0003841m

C%'&%0$/ % 4,/5&,/+ ,% /(/+,0 54+(,*$ ,% 4,5$/$ K&(+*$/3(+0,*+, % '4 M.


12/23

$#m# %ar"e delres%#s# ú0=0e!"#!ces lares%ues"a del sis"ema&uedar'a:

u=u0cos (ω" ) ( sereem%laza! l#s val#resde ω=162.34rad

s ) u0=0.0003841m

u=(0.0003841m)∗cos (162.34rad

s ∗" )

S, 48' % 4,/5&,/+ ,% /(/+,0 ,* MATLAB ,* &* (*+,46%$ , +(,05$ , 2/.

% Fórmula vibraciones libres sin amortiguamiento %

clear all, clc, close all

u0 = 0.0003841; % Desplazamiento inicial m!v0 = 0; % "eloci#a# inicial m$s!

= &'.83; % recuencia natural (z!

) = &*pi*; % recuencia angular ra#$s!

#t = 0.01;

t = 0+#t+&; % iempo s

u = u0*cos-)*t / v0$)*sin-)*t;

v = )*u0*sin-)*t / v0*cos-)*t;

igure,subplot-&11,plot-t,u,title-"ibraciones libres sin

amortiguamiento

2label-t seg!,label-Despl. m! subplot-&1&,plot-t,v,title-"ibraciones libres sin

amortiguamiento

2label-t seg!,label-"eloci#a# m$s!

vcal = #i-u$#t;

igure,plot-t,v,t-1+en#1,vcal,

title-omparación veloci#a#es


13/23

2label-t seg!,label-"eloci#a# m$s!,

legen#-analit.,calc.

+.1 ,a-ca" de la" e"(ue"!a" del "i"!ea.


14/23


15/23

. SOLUCIÓN AL PROBLEMA #

A '$*+(*&'($* /, 0&,/+4* +$/ %/ 48'/ , %/ 4,/5&,/+/ '$* +$/ %/

'$03(*'($*,/ 5$/(3%,/ , %$/ 6%$4,/ 54 u0 ; v0, '$* 0$4+(&'($* ,

"< ; < 4,/5,'+(60,*+, /, +$0) &* +(,05$ , 7! / 54 ,*,44 %$/48'$/.

.1 E*aluación de lo" efec!o" de la" condicione" iniciale" / elao!i&uaien!o en la e"(ue"!a.

D,/5&/ , *%(4 %/ 48'/ /, 5&$ $3/,464 54(0,40,*+, K&,%$/ 5$4',*+,/ , 0$4+(&0(,*+$ 04'* &* 4* (:,4,*'( ,*%/ 4,/5&,/+/ , %$/ /(/+,0/ /, 5&,, *$+4 :'(%0,*+, K&, ,%/(/+,0 '$* 0,*$/ 0$4+(&'()* ,* ,% +(,05$ K&, /, ,/+&() >7! /?*$ %%,$ % 4,5$/$ 0(,*+4/ K&, % 4,/5&,/+/ '$* &* < ,0$4+(&'()* '$* &* +(,05$ , >2! /? ; /, ,*'$*+43*54$=(00,*+, ,* ,/+$ , 4,5$/$.

L/ '$*('($*,/ (*('(%,/ , u0 ; v0 ,* %/ 4,/5&,/+/ 0&,/+4*

K&, '&*$ ,% ,/5%0(,*+$ (*('(% ,/ 0,*$4 K&, % 6,%$'( %'&46 ,% ,'4,0,*+$ %$4(+0('$ , % 0$4+(&'($* /, /,54 0/, %$/ 5('$/ ,% ,/5%0(,*+$ ,/+ /,54'()* 6 (/0(*&;,*$'$4, '$0$ %$/ 6%$4,/ ,% ,/5%0(,*+$ /, (&%* %$/ , %6,%$'(.E* %/ 4,/5&,/+/ $*, ,% ,/5%0(,*+$ *$ ,/ +* (:,4,*+, , %6,%$'( /, 5&,, $3/,464 K&, % ,*6$%6,*+, , % 0$4+(&'($*'$(*'(, +$+% 0,*+, '$* ,% 4,'4,0,*+$ ,% ,/5%0(,*+$.


16/23


17/23


18/23

@%%0$/ %$/ 6%$4,/ '$44,/5$*(,*+,/ %/ '$*/+*+,/ A ; B.

" =0 ) u0

esiguala :

u0=e−+ω ( 0)( ,si!ωd (0 )+-c#s ωd (0 ))+

P0

k

1

(1− 2)2

+(2 + )2((1− 2 )sin / (0 )−(2+ ) cos/ (0 ))

u0=1(-)+

P0

k

1

(1− 2)2

+ (2+ )2(−(2+ ))

-=u0+

P0

k

2+

(1− 2 )2

+(2+ )2

S, ,4(6 % :&*'()* 54 h%%4 ,% 6%$4 , % '$*/+*+, A

u=e−+ω" ( ,si! ωd " +-c#s ωd " )+ P

0

k

1

(1− 2 )2

+(2+ )2((1− 2 )sin/ " −(2+ ) cos/ " )

ú=e−+ω" [ ( ωd ,c#sωd " −ωd -sinωd " ) ]+(−+ωe−+ ω" ) [ ,si! ωd " −-c#sωd" ]+ P

0

k

1

(1− 2 )2

+ (2+ )2[/ (1−

ú0=ωd ,−+ω-+

P0

k

1

(1− 2 )2

+(2+ )2

[ / (1− 2) ]

,= 1

ωd[ ú

0++ω−

P0

k

[/ (1− 2 ) ](1− 2 )

2

+(2 + )2]


19/23

C$* %$/ 6%$4,/ , %/ '$*/+*+,/ $3+,*($/ /, 0$,%* ,* M+%3 +,*(,*$ ,*'&,*+ +$$/ %$/ 6%$4,/ , ,*+4 K&, /, 5(,* ,* ,% ,*&*'($.

0.1 ,a-ca" de e"(ue"!a del "i"!ea a cada fecuencia deeci!ación 234 cada $.) 56 7a"!a 56.

F4,'&,*'( , !.2 @


20/23


21/23

F4,'&,*'(, 9 @

0.) C8lculo / &a-ca de la e"(ue"!a en fecuencia del "i"!ea.


22/23

S, +$04$* %$/ ,/5%0(,*+$/ 0=(0$/ ,* ' 48' ,5,*(,*+, ,' &* , %/ :4,'&,*'(/ , ,='(+'()*.

0.+ Calculo del ao!i&uaien!o del "i"!ea con el 9!odo de:7alf;(o


23/23

S,* ,% 0+$$ ; % 48' /, +(,*, %$ /(&(,*+,

0ma1=¿ !.9"

0ma1

√ 2=0.2938

S,* % 48' ω1=11.23,ωd=12.86,ω2=14.5

ω2−ω

1

ω!=2+=

1

2

14.5−11.2312.86

=0.254

2+=

0.254des%e3am#s ) "e!em#s &ue+=

0.127

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Taller #1 Dinamica Estructural