sistemas amortiguados de dos grados de libertad

8/19/2019 Sistemas Amortiguados de Dos Grados de Libertad

1/16

16 de Febrero del 2016

DEPARTAMENTO DE CIENCIAS DE LA ENERGIA Y MECANICA

CARRERA DE INGENIERIA MECATRÓNICA

VIBRACIONES

NRC DE LA ASIGNATURA 162!

CONSULTA

VIBRACIÓN LIBRE AMORTIGUADA PARA SISTEMAS VIBRATORIOS DE 2 GRADOS DELIBERTAD

DOCENTE ING" #AIME EC$EVERRIA YANE%

AUTOR

o &AVIER FREIRE


2/16

VIBRACIÓN LIBRE AMORTIGUADA PARA SISTEMAS VIBRATORIOS DE 2GRADOS DE LIBERTAD

U' ()(*e+, -.e *e'/, do( .ero( o ,r*.l,(3 -.e re(e'*e' .' +o4)+)e'*o de do( /r,do(

de l)ber*,d3 5 -.e ,de+( e7)(*,' ele+e'*o( de ,+or*)/.,)8'3 lo( .,le( ro4o-.e' -.e,d, .ero o ,r*.l, de(r)b, .' +o4)+)e'*o de ,+or*)/.,+)e'*o3 5, (e, (obre,+or*)/.,do3 r*),+e'*e ,+or*)/.,do o (.b ,+or*)/.,do9 (e le o'oer o+o ()(*e+,(4)br,*or)o( de 2 /r,do( de l)ber*,d o' ,+or*)/.,+)e'*o"

U'o de lo( ()(*e+,( +( o+.'e( de e(*e *)o 5 o' el -.e (e e7l),r, (. d)'+), de:.')o',+)e'*o o o+or*,+)e'*o3 e( el ()(*e+, M,(,;Re(or*e o' ,+or*)/.,+)e'*o"

D)


3/16

P,r, l, +,(, m2

Ao+od,'do l,( e.,)o'e( 1 5 2 ob*e'e+o(

m1´ x

1+ (c1+c2 ) ´ x1−c2 ´ x2+( k 1+k 2 ) x1−k 2 x2=0m

2´ x

2−c

2´ x

1+c

2´ x

2+k

2 x

2−k

2 x

1=0

L,( do( e.,)o'e( ,'*e( ob*e')d,( (e l,( e7re(, +ed),'*e .', e.,)8' +,*r)),l3 de(r)*,de l, ()/.)e'*e :or+,"

[m1 00 m2] [´ x1´ x

2]+[c1+c2 −c2−c

2 c

2 ][ ´ x1´ x

2]+[k 1+ k 2 −k 2−k

2 k

2 ][ x1 x

2]=[00](3)

L, e.,)8' ?3 .ede e7re(,r e' .' ,r*er +,*r)),l +( /e'er,l o+o

M ´ X +C ´ X + KX =0

Do'de

M =[m1 00 m2] C =[c1+c2 −c2−c

2 c

2 ] K =[

k 1+k

2 −k

2

−k 2

k 2 ]

´ X = ´ x

1

´ x2

´ X = ´ x

1

´ x2

X = x

1

x2

M=M,*r)= de +,(, 5@o I'er),C=M,*r)= de ,+or*)/.,+)e'*oK=M,*r)= de r)/)de=

C,be re,l,r -.e l,( +,*r)e( M, C 5 K .ede' e(*,r *,'*o ,ol,d,( o+o de(,ol,d,("

F externas=∑ F inerciales

k 2 ( x2− x1 )−c2 ( ´ x2−´ x1 )=m2 ´ x2(2)


4/16

P,r, e'o'*r,r l, (ol.)8' de l, e.,)8' ?3 el roed)+)e'*o , (e/.)r e( +.5 ()+)l,r ,le+le,do ,r, ()(*e+,( de .' (olo /r,do de l)ber*,d3 do'de (e ,(.+e l, (ol.)8' ,r, el()(*e+, d)'+)o3 ,r, e(*e ,(o (e ,(.+e' (ol.)o'e( ,r, l,( do( e.,)o'e(d):ere'),le( 1 5 2 o ,r, l, e.,)8' +,*r)),l ?3 o' lo .,l ob*e'e+o( l,( )/.,ld,de( de o())8'3 4elo)d,d 5 ,eler,)8' de ,d, .', de l,( +,(,(3 d)


5/16

{s2[m1 00 m2]+s [c

1+c

2 −c

2

−c2 c2 ]+[k

1+k

2 −k

2

−k 2 k 2 ]}es t [ X

1

X 2]=[0

0]

{ s2 M +s C + K } es t X =0

D,do -.e es t

X rere(e'*, l, (ol.)8' e' :or+, +,*r)),l 5 (,b)e'do -.e

x1 (t )= X 1 es t

5 x2 (t )= X 2 es t

'o .ede' (er ero3 5, -.e de (erlo el ()(*e+, 'o e(*,r,

e' 4)br,)8'3 -.ed, o+o '), o()b)l)d,d de .+l)r l, )/.,ld,d"

s2

[m

1 0

0 m2]+

s

[c

1+c

2 −c

2

−c2

c2 ]+[

k 1+k

2 −k

2

−k 2

k 2 ]=[

0

0]s

2 M +s C + K =0

[s2

m1+s ( c1+ c2)+ k 1+ k 2 −s c2−k 2

−s c2−k

2 s

2m

2+s c

2+ k

2]=[00]

S,,+o( el de*er+)','*e de l, +,*r)= de l, e7re()8' ,'*er)or3 ob*e')e'do

(s2 m1+s ( c1+c2 )+k 1+k 2 ) (s2m2+s c2+k 2 )−(−s c2−k 2 )(−s c2−k 2)=0

m1

m2

s4+[m1 c2+m2 (c1+c2) ] s3+[ m1 k 2 +c2 (c1+ c2 )+m2 (k 1+k 2 )−c22 ] s2 +[k 2 ( c1+c2 )+c2 (k 1 +k 2 ) ] s+[ (k 1 +k 2

L, e7re()8' ,'*e( +o(*r,d, e( o'o)d, o+o l, e.,)8' ,r,*er(*), del ()(*e+,3 e'd)


6/16

?" Do( r,e( .ede' (er re,le( 'e/,*)4,(3 5 l,( o*r,( do( r,e( (o' o+le,(o'./,do"

Raíces Reales negativa.- C,d, .', de l,( r,e( *e'dr ,(o),d, .', (ol.)8')'dee'd)e'*e 5 l, (ol.)8' *o*,l (er l, (.+, de l,( (ol.)o'e( )'dee'd)e'*e( de ,d, r,="

x1

(t )= X 11

es1

t + X 12

es2

t + X 13

es3

t + X 14

es4

t

x2

(t )= X 21

es1

t + X 22

es2t + X

23e

s3

t + X 24

es4

t

D,do -.e l, e.,)8' ,r,*er(*), del ()(*e+, *)e'e r,e(3 e7)(*)r' :,*ore( de :or+,o +odo( :.'d,+e'*,le( de 4)br,)8'"

β i=( X

1

X 2)s=si=

X 1 i

X 2 i

L, e7re()8' re(.l*,'*e de l, (ol.)8' *o+,'do e' .e'*, l, o'()der,)8' del :,*or de:or+, e(

x1

(t )= β1 X

21e

s1

t + β2 X

22e

s2

t + β3 X

23e

s3

t + β4 X

24e

s4

t

x2

(t )= X 21

es

1t + X

22e

s2t + X

23e

s3

t + X 24

es

4t

Co'(*,'*e( X 21 3 X 22 3 X 23 5 X 24

E(*e ,(o (e ,(e+e, , .' ()(*e+, de .' /r,do de l)ber*,d (obre ,+or*)/.,do3 5, -.e l,(ol.)8' ob*e')d, e( .', :.')8' e7o'e'),l 5 'o re(e'*,r, o()l,)o'e("

Raíces son complejos conjugados.- Co+o e' el ,(o ,'*er)or l, (ol.)8' *o*,l3 (er l,(.+, de l,( (ol.)o'e( ,r),le( o )'dee'd)e'*e("

x1

(t )= X 11

es

1t + X

12e

s2

t + X 13

es

3t + X

14e

s4

t

x2

(t )= X 21

es

1t + X

22e

s2t + X

23e

s3

t + X 24

es

4t


7/16

P,r, e'o'*r,r l, e7re()8' :)',l de l, re(.e(*,3 ,r, e(*e *)o de r,e(3 (e *o+,r,' e'.e'*, l,( ()/.)e'*e( o'()der,)o'e("

R,e( For+.l, de E.ler F,*or de :or+,

s1=− p

1+ i q

1

s2=− p

1−i q

1

s3=− p

2+iq

2

s4=− p

2−i q

2

ei q

1t =cos q

1t +i sin q

1t

e−i q

1t =cosq

1t −isinq

1t

ei q

2t =cosq

2t +i sinq

2t

e−i q

2t =cosq

2t −isinq

2t

β i=( X 1 X 2)

s=si

= X

1 i

X 2 i

B,('do(e e' l,( o'()der,)8' de l, *,bl, ,'*er)or (e ob*)e'e' l,( e7re()o'e( d l,Sol.)8'

x1

(t )=e− p1 t [ ( β1 X 21+ β2 X 22) cos q1 t +i ( β1 X 21− β2 X 22 )sin q1t ]+e− p2 t [ ( β3 X 23+ β 4 X 24 ) cosq2 t +i ( β3 X 23−

x2

(t )=e− p1 t [ ( X 21+ X 22 )cosq1t +i ( X 21− X 22 ) sinq1 t ]+e− p2 t [ ( X 23+ X 24 )cosq2 t +i ( X 23− X 24 ) sinq2 t ]

Co'(*,'*e( X 21 3 X 22 3 X 23 5 X 24

qq

(¿¿2t +∅12)

(¿¿1 t +∅11)+C 12 e− p

2t sin¿

x1 (t )=C

11e− p

1t sin¿

q

q

(¿¿2 t +∅22)

(¿¿2 t +∅21)+C 22e− p

2t sin¿

x2

(t )=C 21

e− p

1t sin¿

Co'(*,'*e( C 11 3 C 21 3 C 12 3 C 22 3 ∅11 3 ∅12, ∅21 5 ∅22


8/16

Dos raíces reales negativas y dos raíces complejas conjugadas.- E(*e ,(o )'4ol.r, lo(do( ,'*e( 4)(*o(

L, (ol.)8' e( l, (.+, de l,( (ol.)o'e( ,r),le( o )'dee'd)e'*e("

x1

(t )= X 11

es1 t + X 12

es2 t + X 13

es3 t + X 14

es 4 t

x2

(t )= X 21

es

1t + X

22e

s2t + X

23e

s3

t + X 24

es

4t

P,r, e'o'*r,r l, e7re()8' :)',l de l, re(.e(*,3 ,r, e(*e *)o de r,e(3 (e *o+,r,' e'.e'*, l,( ()/.)e'*e( o'()der,)o'e("

R,e( For+.l, de E.ler F,*or de :or+,s1= Realnegativo

s2= Real negativo

s3=− p+i q

s4=− p−i q

ei q t =cosq t + isinq t

e−i qt =cos q t −i sin q t

β i=( X 1 X 2)

s=si

= X

1 i

X 2 i

B,('do(e e' l,( o'()der,)8' de l, *,bl, ,'*er)or (e ob*)e'e' l,( e7re()o'e( d l,Sol.)8'

x1

(t )= β1 X

21e

s1

t + β2 X

22e

s2

t +e− pt [ ( β3 X 23+ β4 X 24 ) cos qt + i ( β3 X 23− β4 X 24 ) sin qt ]

x2

( t )= X 21

es1

t + X 22

es2t +e− pt [ ( X 23+ X 24 )cosqt +i ( X 23− X 24 ) sinqt ]

Co'(*,'*e( X 21 3 X 22 3 X 23 5 X 24

q(¿¿ 2 t +∅11)

x1

(t )= β1 X

21e

s1

t + β2 X

22e

s2

t +C 11

e− p

2t

sin ¿


9/16

q

(¿¿ 2t +∅22) x

2(t )= X

21e

s1

t + X 22

es

2t +C

22e

− p2

t sin ¿

Co'(*,'*e( X 21 3 X 22 3 C 11 3 C 22 3 ∅11 3 5 ∅22

Ejercicios

1. Hallar la ecuación diferencial matricial de movimiento del sistema amortiguado

de dos grados de libertad que se muestra en la figura e identificar cada una de las

matrices

Resolución

Se 4, , ,(.+)r -.e θ2 >θ1 5 θ́2>θ́1 3


10/16

M,(, m1

M,(, m1

{ ka (θ2−θ1 )+ ca (θ́2−θ́1 )−m1 gl θ1=m1 l

2

θ́1

−ka (θ2−θ1 )−ca (θ́2−θ́1 )−m2 gl θ2=m2 l2

θ́2

{m1 l2

θ́1+c a2 θ́

1−c a2 θ́

2+(k a2+m1 gl )θ1−k a

2θ

2=0

m2

l2

θ́2+c a2 θ́

2−c a2 θ́

1+(k a2+m2 gl )θ2−k a

2θ

1=0

[m

1l2

0

0 m2l2

][θ́1

θ́2

]+

[ c a

2 −c a2

−c a2

c a

2

][θ́1

θ́2

]+

[k a

2+m1

gl −k a2

−k a2

k a2

+m2 gl

][θ1

θ2

]=

[0

0

] M =[m1 l

20

0 m2

l2] C =[ c a2 −c a2−c a2 c a2 ] K =[

k a2+m

1gl −k a2

−k a2 k a2+m2

gl]

M externas=∑ M inerciales

a (θ2 −θ1 )+ca (θ́2−θ́1 )−m

1 glθ 1=m1l2

θ́ 1

M externas=∑ M inerciales

ka (θ2−θ1 )−ca (θ́2−θ́1 )−m2 gl θ2=m2l2 θ́2


11/16

Θ́=[θ́1θ́2] Θ́=[θ́

1

θ́2] Θ= θ

1

θ2

2. Para el ejercicio anterior si k=1000Nm!c=10 Ns/m a=0!"m! b=0.#m!

m1=0.$kg % m#=0."kg! determine en que caso se encuentra el sistema.

[m1 l2

0

0 m2

l2][ θ́1θ́

2]+[ c a

2 −c a2

−c a2 c a2 ][θ́1θ́2]+[k a

2+m1

gl −k a2

−k a2 k a2+m2

gl ][θ1θ2]=[00]Sol.)8'

θ1 (t )=Θ1 es t

θ2 (t )=Θ

2e

s t Θ (t )=[θ1 (t )θ2

(t )]=

[Θ1e

s t

Θ2

es t ]

=es t [Θ1Θ

2]=es t Θ

[m1 l2

0

0 m2

l2][θ1 s

2e

s t

θ2

s2

es t ]+[ c a

2 −c a2

−c a2 c a2 ][θ1 s es t

θ2

s es t ]+[k a

2+m1 gl −k a2

−k a2 k a2+m2

gl ] [θ1es t

θ2

es t ]=[00 ]

s2

est

[m

1l2

0

0 m2l

2

][Θ

1

Θ2

]+ sest

[

c a2 −c a2

−c a2

c a

2

] [

Θ1

Θ2

]+ est

[k a

2+m1

gl −k a2

−k a2

k a

2

+m2 gl

] [Θ

1

Θ2

]=

[

0

0

]s

2e

st M Θ +s es t C Θ+es t K Θ=0

{s2[m1 l2

0

0 m2

l2]+s [ c a

2 −c a2

−c a2 c a2 ]+[k a2+m

1gl −k a2

−k a2 k a2+m2

gl]}es t [Θ1Θ2]=[00]

{ s2 M +s C + K } es t Θ=0

s2[m1l

20

0 m2

l2]+ s [ c a

2 −c a2

−c a2 c a2 ]+[k a2+m

1gl −k a2

−k a2 k a2+ m2

gl ]=[00 ]


12/16

s2 M +s C + K =0

[s2

m1

l2+c a2 s +k a2+m

1gl −c a2 s−k a2

−c a2 s−k a2 s2m2l2+c a2 s+k a2+m2 gl

]=[

0

0

]

[0.125 s2+0.9 s+154.5523 −0.9 s−90−0.9 s−90 0.075 s2+ 0.9 s+153.5715]=[00]

Re(ol4)e'do l, e.,)8' ,r,*er)(*), (e *)e'e -.e l,( r,)e( del ()(*e+, (o'

−7.7920+55.8060 i

−7.7920−55.8060 i

−1.8080+7.0216 i

−1.8080−7.0216 i

". Hallar la ecuación diferencial matricial de movimiento del sistema amortiguado

de dos grados de libertad que se muestra en la figura e identificar cada una de lasmatrices

Re(ol.)8'


13/16

−c1 ( ´ y−aθ́ )−k

1( y −aθ )−c

2( ´ y +b θ́ )−k

2( y +bθ )=m ´ y

c1( ´ y −a θ́ )+ k

1( y −aθ )−c

2( ´ y +b θ́ )−k

2( y +bθ )= I θ́

Re,o+od,'do l,( e7re()o'e( ,'*er)ore(

m ´ y +( c1+c2 ) ´ y−( c1a−c2 b) θ́+(k 1+k 2) y−(k 1a−k 2 b)θ=0

I θ́+(c1a2+ c

2b2 ) θ́−(c1 a−c2 b ) ´ y +(k 1 a

2+ k 2

b2 )θ−(k 1a−k 2 b) y=¿

[m 00 I ][ ´

yθ́ ]+[ (

c1+c2 ) −(c1 a−c2b )−( c1a−c2b ) (c1a

2+ c2

b2 ) ][

´ yθ́ ]+[

(k 1+k 2) −(k 1 a−k 2 b)−(k

1a−k

2b) (k 1a

2+ k 2

b2 ) ] [

yθ ]=[

0

0]

M =[m 00 I ] C =[ ( c1+c2 ) −( c1a−c2b )−(c1 a−c2 b) (c1 a2+c2b2 ) ] K =[ (k 1+ k 2) −(k 1a−k 2 b)−(k

1a−k

2b) (k 1 a

2+k 2

b2 ) ]

v́ =[ ´ yθ́ ] v́ =[ ´ yθ́ ] v=[ yθ ]

&. Para el ejercicio anterior! si m=1000kg! l=&m! a=b=#m! I =1300 kg m2

!

k 1=50000 N /m

!k

2=70000 N /m

%c

1=c

2=10 Ns/m

. 'etermine la


14/16

res(uesta del sistema como función del tiem(o! con las condiciones iniciales

siguientes.

y (0 )=0.03m , ´ y (0)=0 , θ (0)=0 , θ́ (0 )=0

[m 00 I ][ ´ yθ́ ]+[ (c1+c2 ) −(c1 a−c2b)−( c1a−c2b ) (c1a2+ c2b2 ) ][ ´ yθ́ ]+[ (k

1+k

2) −(k

1a−k

2b)

−(k 1

a−k 2

b) (k 1a2+ k

2b2 ) ] [ yθ ]=[00]

M =[m 00 I ] C =[ ( c1+c2 ) −( c1a−c2b )−(c1 a−c2 b) (c1 a2+c2b2 ) ] K =[ (k

1+k

2) −(k

1a−k

2b)

−(k 1

a−k 2

b) (k 1 a2+k

2b2 ) ]

M =

[1000 0

0 1300

C =

[20 0

0 80

] K =

[120000 40000

40000 480000

]s

2 M +s C + K =0

[ s2 m+s (c1 +c2 )+(k 1+k 2) −(c1a−c2 b ) s−(k 1 a−k 2 b)

−s (c1 a−c2 b )−(k 1 a−k 2b) s2 I +s (c1 a2+c2b2)+(k 1 a2 +k 2 b2 )]=[00]

mI s4 [ I (c1+c2 )+ m (c1a2+ c2 b2 ) ] s3+[ m (k 1 a2+k 2 b2 )+ I (k 1+k 2)+ (c1+ c2) (c1a2+ c2 b2 ) ] s2+[ (c1+ c2 )(k 1 a

13∗105 s4+ 1.06∗105 s3+63.6∗107 s2+ 192∗105 s +560∗108=0

R,e(s

1=−0.0104+10.7 i

s2=−0.0104−10.7 i

s3=−0.0304+19.3 i

s4=−0.0304−19.3 i


15/16

Al),'do el :,*or de :or+,

β i=−m12 si

2 +c12 si+k 12m

11si

2+ c11

si+ k 11

β1=−8.279−0.01417 i

β2=−8.279+0.01417 i

β3=0.106−4.645∗10−4

i

β4=0.106+4.645∗10−4

i

Sol.)o' e' el *)e+o

y (t )=e−0.01041 t [ ( β1 X 21+ β2 X 22 ) cos10.7 t +i ( β1 X 21− β2 X 22 ) sin10.7 t ]+e−0.0104 t [ ( β3 X 23+ β4 X 24 ) cos19.3

θ (t )=e−0.01041 t [ ( X 21+ X 22) cos10.7 t +i ( X 21− X 22 ) sin 10.7 t ]+e−0.0104 t [ ( X 23 + X 24 )cos 19.3t +i ( X 23− X 24 ) s

C,+b)o de oe:))e'*e( ,r, :,)l)*,r l, re(ol.)8'"

1= X

21+ X

22

2=i ( X 21− X 22 )

3= X

23+ X

24

4=i ( X 23− X 24 )

M,'eo del :,*or de :or+, ,r, e'o'*r,r lo( oe:))e'*e( β i=! i " # i

β1 X 21 + β2 X 22=! 1 1+ # 1 2

i ( β1 X 21− β2 X 22)=! 1 1−# 1 2

β3 X

23+ β

4 X

24=!

3

3+#

3

4

β3 X

23− β

4 X

24=!

3

3−#

3

4


16/16

Do'de!

1=!

2=−8.279 3 # 1=# 2=−0.01417 i 3 ! 3=! 4=0.106 5

# 3=# 4=−4.645∗10−4 i

y (t )=e−0.01041 t [ (! 1 1+# 1 2 )cos10.7 t + i (! 1 1−# 1 2)sin 10.7 t ]+e−0.0104t [ ( ! 3 3+ # 3 4 )cos19.3 t + i (! 3

θ (t )=e−0.01041 t [ 1cos10.7 t + 2sin 10.7 t ]+e−0.0104 t [ 3cos19.3 t +4 sin 19.3 t ]

Co' l,( do( :.')o'e( e(*,ble)d,( o' lo( '.e4o( oe:))e'*e(3 '),+e'*e *e'e+o(

)'8/')*,(3 or lo -.e () e4,l.,+o( l,( o'd))o'e( )')),le( lo/r,re+o( e'o'*r,r lo(4,lore( de lo( oe:))e'*e("

1=−0.358∗10−2

2=−0.105∗10−4

3=0.358∗10−2

4=0.915∗10−5

BibliografíaShabana, A. (1995). Theory of Vibration . Springer.

sistemas amortiguados de dos grados de libertad

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