Ejemplos de Dinámica de maquinaria

7/24/2019 Ejemplos de Dinmica de maquinaria

1/19

1 Lpez Cruz Elizabeth

2 Lozada Crdenas Omar

3 Mayorga Basalda onathan

!

"eterminamos dimensiones# de la $gura anterior%a=64mm

b=38mm

c=19

mm


2/19

d=20mm

e=19mm

radios=5mm

"e nuestras $guras bus&amos sus Centros de 'ra(edad%

XCG1=C

2=19

2=9.5mm

YCG1=

abb2

2r2c

ab2r d=

(64 ) (38 )( 382)2(52 )(19 )(64 ) (38 )2(5 )

20 [ mm ]=1.069mm

ZCG 1=0

"eterminamos el (olumen y la masa en este segmento%

V1=abc2r2c=(64 ) (38 ) (19 )2(52 )(19 )=4.322x104 mm3

m=V1=(7.797x106

kg /mm3 ) (4.322x104 mm3 )=0.337Kg

"eterminamos los momentos de masa sobre los e)es lo&ales a tra(*s de los C'%

ma=a b c =(64 ) (38 ) (19 )(7.797x106

kg /mm3 ) ma=0.360Kg

Iax=ma

12( a2+b2 )=0.360 kg

12(642+382 )Iax=1.66410

4Kg m

2

Iay

=ma

12

( a2+c2 )=0.360 kg12

(642+192 )Iay

=1.339104Kg m2

Iaz=ma

12(b2+c2 )=0.360 kg

12(382+192 )Iaz=5.42110

5Kg m

2

+ara el &ilindro,

mb= r2

c =(52)(19)(7.797x106 kg /mm3)mb=0.012Kg

Ibx=mb r2

2=(0.012 kg)(5

2)2

Ibx=1.455107Kgm

2

Iby=mb

12(3 r2+c2 )=0.012 kg

12(352 )+ (192 )Iby=4.22910

7Kg m

2

Ibz=mb

12(3 r2+c2 )=0.012 kg

12(352 )+(192 )Iby=4.22910

7Kg m

2

"eterminamos los momentos de masa sobre los e)es globales%


3/19

Ixx1=Iax+ma[(YCG1 )2+ (ZCG 1 )2 ]+2 (Ibx+mb e 2 )

Ixx1=1.664104Kg m2+0.360Kg [(1.069mm )2+ (0 )2 ]+2 (1.455107Kg m2+(0.012kg) 19mm2

Ixx1=1.581104

Kg m2

Iyy1=Iay+ma[(XCG1 )2+( ZCG1 )2 ]+2(Iby+mb (c2 )

2

)

Iyy1=1.339104Kg m2+0.360kg [(9.5mm )2+ (0 )2 ]+2(4.229107Kg m2+(0.012 kg) (192)

2

)Iyy1=1.63410

4Kg m

2

Izz1=Iaz+ma [(XCG 1 )2+ (YCG1 )2 ]+2 (Ibz )

Izz1=5.421105Kg m2+0.360kg [(9.5mm)2+(1.069mm)2 ]+2 (4.229107Kg m2)

Izz1=8.631105

Kg m2

-hora para el segmento de la pieza donde se en&uentra el ngulo de ./0%eniendo &omo dimensiones,a=radio

1=13mm

b=32mm

c=64 mm

d=18mm

e=Radio2=32mm

"eterminamos la lo&aliza&in del &entro de gra(edad en &oordenadas globales%

Xcg2=e

4

3 b

3a3

b2a2

=32 4

332

3133

322132

[mm ]Xcg2=16.825mm

Zcg2=0mm

"eterminamos el (olumen y la masade este segmento%


4/19

"etermine los momentos de masa sobre los e)es lo&ales sealados en el dibu)o delsegmento%

0.013

0.032

3( 2)+0.0642

( 2)+3

Ix=M

2

12(3a2+3b2+c2 )=0.335kg

12

Ix=2.144104

Kg m2

0.013

0.032

3( 2)+0.0642

( 2)+3

Iy=M

2

12(3a2+3b2+c2 )=0.335kg

12

Ix=2.144104

Kg m2

Iz=M

2

12(a2+b2 )=0.335 kg

12(0.0132+0.0322 )

Ix=2104

Kg m2

"etermine los momentos de masa sobre los e)es globales%

I xx2=Ix+M2 d2=2.144104Kg m2+0.335kg0.018m

Ixx2=3.23104

Kg m2

I yy2=Iax+M2 e2Iyy2=4.77110

4Kg m

2

I zz2=Iaz+M2(d2+e2)Izz2=5.06110

4Kg m

2

$nalmente &on el prisma re&tangular


5/19

a=64mm

b=19mm

c=38mm

d=40.5mm

e=51mm

"eterminamos la lo&aliza&in del &entro de gra(edad en &oordenadas globales%Xcg

3=e Xcg

3=51mm

Ycg3=dYcg

3=40.5mm

Zcg3=0

"etermine el (olumen y la masa de este segmento%

V3=a b c=(64 ) (19) (38)[ mm ] V3=4.621104

mm3

M3= V3=(7.797x 106

kg /mm3) (4.621104 mm3 )M3=0.360Kg

"eterminamos los momentos de masa sobre lo e)es lo&ales a tra(*s de los &entros degra(edad%

Ix=M

3

12(a2+b2)=0.360kg

12(642+192 )Ix=1.33910

4Kg m

2

Iy=M

3

12

( a2+c2 )=0.360 kg

12

(642+382)Ix=1.664104

Kg m2

Iz=M

3

12(b2+c2)=0.360 kg

12(192+382 )Iz=5.42110

5Kg m

2


Ix x3=Ix+M3[(YCG3 )2+ (ZCG3 )2 ]=1.339104Kg m2+0.360Kg [ (40.5mm)2+(0 )2 ]

Ixx3=7.25104

Kg m2

Iyy3=Iy+M3 [(XCG 3)2

+( ZCG3)2

]=1.339104

Kg m2

+0.360Kg [(51mm )2

+(0 )2

]


6/19

Ixx3=1.104103

Kg m2

Izz3=Iz+M3[ (XCG3 )2+(YCG3 )2 ]=1.339104Kg m2+0.360Kg [(40.5mm )2+(40.5mm )2 ]

Izz3=5.061104

Kg m2

En la parte ! del &ilindro"imensiones

a=19mm

b=40.5mm

c=70mm

d=32mm

"etermine la lo&aliza&in del &entro de gra(edad en &oordenadas globales%

XCG4=c+4 r

3 =70+

4323

XCG4=83.581mm

YCG4=b YCG 4=40.5mm

ZCG4=0

"etermine el (olumen y la masa de este segmento%

V4=a

r2

2 =19

322

2 V4=3.056

10

4

mm

3

M4=V 4=(7.797x106

kg /mm3 ) (3.056104mm3 )M4=0.238Kg

"etermine los momentos de masa sobre los e)es lo&ales del segmento%

Ix=M

4

12(3 r2+a2)=0.238 kg

12(30.0322+0.0192 )Ix=6.8210

5Kg m

2

Iy=M

4

2 r

2=0.238 kg

20.0322Ix=1.22110

4Kg m

2


7/19

Iz=M

4

12(3 r2+a2 )=0.238 kg

12(30.0322+0.0192 )Iz=6.8210

5Kg m

2


Ixx4=Ix+M4 [ (Ycg4 )2+( Zcg4)2 ]=6.82105Kg m2+0.0238 kg [(40.5mm)2+(02)]

Ixx4=4.592104Kg m2

83.581mm

Iyy4=Iy+M4 [ (Xcg4 )2+(Zcg4)2 ]=1.221104Kg m2+0.238

Iyy4=1.787103

Kg m2

40.5mm2

83.581mm2+

Izz4=Iz+M4 [(Xcg4 )2+(Ycg4)2 ]=6.82105Kg m2+0.238kg

Izz4=2.124103

Kg m2

La siguiente parte es un &ilindro negati(o 4ranura5 &uyas dimensiones se muestrandeba)o%

a=70mmb=40.5mm

c=19mm

d=radio=13mm

"etermine la lo&aliza&in del C' 4&entro de gra(edad5%Xcg

5=a Xcg

5=70mm

Ycg5=bY cg

5=40.5mm

Zcg5=0

"etermine el (olumen y la masa de este segmento%m

13m 19mm( 2 )V

5=1.009104mm3

V

5=c r2=


8/19

M5= V5=(7.797x 106

kg/mm3 ) (1.009104 mm3 )M5=0.079Kg

"etermine los momentos de masa sobre los e)es lo&ales del segmento%

Ix=M

5

12(3 r2+c2 )=0.079 kg

12(30.0132+0.0192)

Ix=5.691106

Kg m2

Iy=M

5

2r2=

0.079kg2

(0.0132 )Ix=6.649106

Kg m2

Iz=M

5

12(3 r2+c2 )=0.079 kg

12(30.0132+0.0192)Ix=5.69110

6Kg m

2


0.0405mm

Ixx5=Ix+M5 [ (Ycg5 )2+ (Zcg5 )2 ]=5.691106Kg m2+(0.079kg)

Ixx5=1.348104

Kg m2

0.070mm2+(02)

Iyy5=Iy+M5 [ (Xcg5 )2+ (Zcg5 )2 ]=6.649106Kgm2+(0.079 kg)

Iyy5=3.922104Kg m

2

40.5mm2+(0.0405mm2)

Izz5=Iz+M5 [(Xcg5 )2+(Ycg5 )2 ]=5.691106Kg m2+(0.079kg)

Iz z5=5.203104

Kg m2

a En&ontrar la lo&aliza&in de su &entroide re6erido al punto B%

Xcg=

i

Xcgi Mi

i

Mi

0.337+0.335+0.360+0.238+(0.079)[ kg ]

(9.5mm0.337 kg)+(16.825mm0.335Kg )+(51mm0.360kg )+(83.581mm0.238 kg)+(70mm0

Xcg=34.919mm


9/19

Ycg=

i

Ycgi Mi

i

Mi

0.337+0.335+0.360+0.238+(0.079 )[kg ]

((1.069

mm)0.337

kg )+ (33.175

mm0.335

Kg )+(40.5

mm0.360

kg )+ (40.5

mm0.238

kg )+(40.5

m

Ycg=26.688mm

Zcg=

i

Zcgi Mi

i

Mi

0mm

0.337+0.335+0.360+0.238+(0.079)[kg ]

( 0.337 kg )+(0mm0.335Kg )+(0mm0.360 kg)+(0mm0.238kg )+(0mm0.079Kg)

Zcg=0

b En&ontrar su momento de masa de iner&ia Ixx sobre el e)e 7 a tra(*s del

punto B%

Ixx=i

I xxi

Ixx=1.581104

Kg m2

+3.23104

Kg m2

+7.25104

Kg m2

+4.592104

Kg m2

+(1.348104

K

Ixx=1.531103Kg m2

& En&ontrar su momento de masa de iner&ia Iyy sobre el e)e a tra(*s del punto

B%

Iyy=i

I yyi=

Iyy=1.634104Kg m2+4.771104Kg m2+1.104103Kg m2+1.787103Kg m2+(3.92210

Iyy=2.976103Kg m2


10/19


11/19

En :ourbar% "atos del programa,

La matriz


12/19

Me&anismo simulado


13/19


14/19

teniendo:

Datos iniciales:

(A - B) a:= 0.80in

(C - D) c:=1.55in

Link 2 angle: 82 :=

49deg

(B - C) b := l.23in

(A - D) d := 2.40in

Distancia desde el punto de fuerza hasta el perno D:

RP4 := 1.00in 84 := Odeg

!uerza e"ercida: FP4 := 2000/bf (perpendicular al enlace#)

Distancia del $ango al pernoA: Rn := 4.26in

Solucion:

1% Lo ;ue (amos a ha&er es introdu&ir los (alores ;ue anteriormente pudimos identi$&ar enuna tabla ;ue se program en E9&el 2/13 para determinar los enla&es 3 y ! )unto &onsus ngulos as< &omo &al&ular el ngulo de la 6uerza de prensado &on respe&to al mar&ode &oordenadas

3 :=

34.039deg BFP4 :=

84 + 90deg

3 :=

123.518degBFP4

% 2l3.518deg


15/19

2% "eterminaremos la distan&ia hasta el &entro de gra(edad en el L8C= en &ada enla&e m(il

&nlace 2 ' #: Rcc2 := 0. a Rcc2 = 0.#00 in Rc4 := 0. e Rcc4 = o. in

&nlace *: Rc3 := 0.5b Rc3 = 0.+1 in

2. Cal&ular las &omponentes 7 e de los (e&tores de posi&ion.

R12! :=Rcc2c"#$e2 + 180deg)

R12% :=Rcc2#in$82 + 180deg)

R32! := Rcc2c"#$ 02)

R12% := Rcc2#in$

82)

R23! := Rc&c"#$e3 +

180deg'R23% :=Rcin$03 +

180deg) R43! := $b -

Rc3'c"#$e3'

R43% :% $b -Rc&'#in$e3'

R14% :=

Rcc4#in$e4'

R(! :=Rcc4c"#$e4 + 180deg)

R14% := Rcc4#in$04 +180deg)

Rn% := $Rn + Rc2'#in$e2 +

180deg)


16/19

R12! % --0.2+2 inR12% %

--0.*02 in

R32! = 0.2+2 inR12%

= 0.*02 in R23! =

--0.1, inR13% =

--0.*## in

R43! % 0.(10 in

R43% = 0.*## in

R34! = --0.#28 in

R34% = 0.+#+ in

R)4! = 0.#28 in

R14% = --0.+#+ in

Rn% = -*.1 in


17/19

Rn! := $Rn +Rc2'c"#$e2 + 180-

deg)

RP4! := $RP4*

Rca4'c"#$e4' RP4% := $RP4

-

Rca4'#in$e4'

Rn! = -*.0

inRP4! =

-0.12# in

RP4% = 0.188 in

3% &ne#+e ,a#" -a"# a calc(la la# c","nen+e# en !

FP4! :=FP4c"#$

BFP4'FP4% :=

FP4#in$ BFP4'

FP4! = *166.4/bf

F,4% = *1104.4/bf

. 8ustituir estos (alores dados y &al&ulados en la e&ua&in matri&ial modi$&ado t*rminosde masa e iner&ia

O o O o o o 1 o

O o , o o o o 1

*R12% R)2! *R32%

R32! Rn! *Rn%

in in in in in in

o o -1 o o o

o o o -1 o o

R13% *R23! *R43% R43!

C- o o o o o oin in in in

o o o o -1 o o o o

o o o o o -1 o o o

o o o o R34% *R34! *R14% R(! o oin in in in

o o o o o o o o lan(02 + ) -1

F .-

o

o

o

o

o

o

- 1*FP4!lbf

- 1R:=C F

- 1

*FP4%*lbf

-1 -1(*RP4!FP4% +RP4%*FP4! )lbf in

o

F12! :=

R+lbfF32! :=

R&

lb

fF12! = 1029 lbfF32! =

*1069/bf

o o o o

o

o o

o

o


18/19


19/19

F43! :=

R5

lbfF 14!

:=Ri lbf

F43! % -10+

lbfF(! = 8

lbf

F43% :=

R6

lbfF14% :=

R8

lbf

F43% =

*22/bfF(% %

*82 lbf

Fand% :=R10

lbf Fand% = *34.9/bf

+. Calculando las fuerzas.

in at/:

in at:

in at

C: in

atD:

. Calculando la fuerza del

$ango.

F11 :=&F12/ + F11/

F32 := & F32/ +

F31/ F43 := &F43/

+F43/

F14:=&F(/F(/

F& 2 =

128lbf F32 =

120lbf

F43 % 1290/bf

F14 = 10/bf

Fand = *.l lbf

.

JFI = 0.+#

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