Taller #1 Dinamica Estructural
Transcript of Taller #1 Dinamica Estructural
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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
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". 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$4+(&0(,*+$ ,
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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$$ %&. So'e i(ue"!a.
• @%%0$/ ,% 0)&%$ , ,%/+('( ( E ) .
E=3900√ f ' c=3900√ 21 MPa E=17872 MPa
• @%%0$/ % (*,4'( ( I ) , % /,''()*.
I =b∗h3
12=
0.5m∗0.5m3
12= I =0.005208m2
L 4((,( K )
, &* 6( ,05$+4 ,* &* ,=+4,0$ ; /(05%,0,*+,5$; ,* ,% $+4$ ,=+4,0$ ,/+ 5$4
5¿¿
¿37∗¿
K =768∗ E∗ I
7 L3 = K =
768∗17872∗0.005208¿
• @%%0$/ % :4,'&,*'( *+&4% ,* (rad / s).
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K
M =¿ω=√
81700000 N .m
100 Kg. =ω=903.88 rad /s
ω=√ ¿
• E* '('%$/ 5$4 /,&*$.
f = ω
2∗π =f =
903.88 rad /s2∗π
=f =143.85 Hz
• P,4($$ , *+&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).
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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($$ *+&4% ,* /,&*$/.
T =2∗π
ω =T =
2∗π 165.02 rad /s
=T =0.038 s
1.+ Solución de la fecuencia na!ual !eniendo en cuen!a !oda" la"a"a".
L 0/ +$+% ,% /(/+,0 /,4( 100 Kg. de la carga+3000 Kg . viga=3100 Kg .
• @%%0$/ ,% 0)&%$ , ,%/+('( ( E ) .
E=17872 MPa
• @%%0$/ % (*,4'( ( I ) , % /,''()*.
I =0.005208m2
L 4((, ( K ) , &* 6( ,05$+4 ,* &* ,=+4,0$ ; /(05%,0,*+,
5$; ,* ,% $+4$ ,=+4,0$ ,/+ 5$4
5¿¿¿37∗¿
K =768∗ E∗ I
7 L3 = K =
768∗17872∗0.005208¿
• @%%0$/ % :4,'&,*'( *+&4% ,* (rad / s).
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K
M =¿ω=√
81700000 N . m
3100 Kg. =ω=162.34 rad /s
ω=√ ¿
• E* '('%$/ 5$4 /,&*$.
f = ω
2∗π =f =
162.34 rad /s2∗π
=f =25.83 Hz
• P,4($$ ,* /,&*$/.
T =2∗π
ω =T =
2∗π 162.34 rad /s
=T =0.0038 s
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). SOLUCIÓN DEL PROBLEMA #)
DATOSL 0
E2!!!!! MP
S,''()* '&4 %! '0
@%%0$/ % (*,4'( , % /,''()*.
0.5m¿¿¿3
0.5m∗¿ I =
b∗h3
12= I =¿
L 4((, ( K ) , &* 6( ,05$+4 ,* &* ,=+4,0$ ; ,* 6$%($ ,* ,% $+4$
,=+4,0$ ,/+ 5$4
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5¿
¿¿3¿¿
K =3 EI
L3 = K =
3∗200000∗0.00508
¿
@%%0$/ % 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)$$$.
T,*(,*$ %$/ 6%$4,/ '%'&%$/ /, 54,',, 0$,%4 ,% /(/+,0 ,* SAP2!!!
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S, '%'&% ,% 5,4($$ T ,* ,% 54$40 T =0.07196 s
S, '%'&% ,% ,/5%0(,*+$ u0=−0.00198m
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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$/ % (*,4'( ( I ) , % /,''()*.
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I =0.005208m2
• @%%0$/ % 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.
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$#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
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2label-t seg!,label-"eloci#a# m$s!,
legen#-analit.,calc.
+.1 ,a-ca" de la" e"(ue"!a" del "i"!ea.
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. 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(,*+$.
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@%%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]
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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 @
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F4,'&,*'(, 9 @
0.) C8lculo / &a-ca de la e"(ue"!a en fecuencia del "i"!ea.
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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
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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