emparrillado plano 3

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 MA TRIZ DE RIGIDEZ PARA EMP ARRILLADOS A= C= E= G= B= D= F= H= A B A B A C D - A G F K= B D E -B F H - A -B - A -B A G F - A C D B F H - B D E BARRA PARALELA A "X" 0 0 0 K 0 0 G F K= 0 0 0 0 0 0 0 0 0 0 BARRA PARALELA A "Y" 0 0 0 0 K= 0 0 0 0 0 0 0 0 0 0 0 0  =/    _1=12 ^3  /  _2 _2      _1^2 +      _2^2   _3 〖−    _1 _2     _1   _2 (   _1    _1    _ 〖−〗  − 〖−〗 _ 〖−〗

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7/18/2019 Emparrillado Plano 3

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MATRIZ DE RIGIDEZ PARA EMPARRILLADOS

A= C= E= G=

B= D= F= H=

A B A B

A C D -A G F

K = B D E - B F H

- A - B - A - B

A G F - A C DB F H - B D E

BARRA PARALELA A "X"

0 0

0 K 0 0 G F

K = 0 0

0 00 0 0 0

0 0

BARRA PARALELA A "Y"

0 0

0 0

K = 0 0 0 0

0 0

0 0

0 0 0 0

 =/    _1=12 ^3 /

 _2 _2    _1^2 +     〖 _2^2  〗 _3

〖−  〗  _1 _2   〖 _1 〗 _2 (−

   _1

   _1

  ____________

〖−〗 ____________

 − 

〖−〗 ____________

〖−〗 ____________

7/18/2019 Emparrillado Plano 3

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   _2=6 /

  _3 )

〖−〗

7/18/2019 Emparrillado Plano 3

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METODO DE RIGIDEZ

Dado ! #a$$%!!ado d !a &'($a) * #%d+

Determinar los desplazamienos y giros del extremo del voladizo

Determinar los diagramas de es!"erzos

, Mod!o d C.!/(!o

#e en"meran nodos$ tal %omo se indi%a en la &g y se determina los grados de li'er

,, Ma$%1 d $%'%d1

  ((((((21*(((

= =1,22,((( 1(

16((.( 32

1(6(((( ((((( -16((.((( -1(6((((

((((( 1,22,(( ((((( (((((

-16((.((( ((((( 32(1(((( 16((.(((

-1(6(((( ((((( 16((.((( 1(6((((

((((( -1,22,(( ((((( (((((

-16((.((( ((((( 16((.((( 16((.(((

 =/    _1=12 ^3 /    _2=6 /

  =    _1=

   _2=    _3=

  =

  _=___________

7/18/2019 Emparrillado Plano 3

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16*2 m 61,3((2

11...1(2(*1 m 2*31*2

61,3(( 11.6 ((((( -61,3((

11.6 2*31*3 ((((( -11.6

((((( ((((( 16*2(( (((((

-61,3(( -11.6 ((((( 61,3((

11.6 131.1 ((((( -11.6

((((( ((((( -16*2(( (((((

, MATRIZ GLOBAL

13,3(( -11.6 -16((.(((

K= -11.6 2,36((*3 (((((-16((.((( ((((( 336.2((

,2 E*ado I#e introd"%en !"erzas de empotramiento a &nes de 'lo/"ear los desplazamientos n

 _ 

   _2=    _3=

  _=___________

7/18/2019 Emparrillado Plano 3

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-1

(

- = (,

-1

(

-(,

n este estado no se prod"%en desplazamientos pero s 4ay re!"erzos

,3 E*ado IIl ve%tor de %argas nodales es ig"al y op"esto a las !"erzas de empotramiento per! 

3,

5 = -122.

-(,

-21

-122.

- = (

-21 m

122.

(

2 CALC4LO DE DESPLAZAMIENTOS

os desplazamientos en ese estadoo se %al%"lan mediante la matriz de rigidez y el

.,-(3 777

8 = 1,-(32.-(3 777

7/18/2019 Emparrillado Plano 3

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777

3 CALC4LO DE ESF4ERZOS EN BARRAS

9 &nes de determinar los diagramas$ se determinan las !"erzas de 'orde$ s"perponi

(*22,.,

(321(,2

-(*163-*(22,6

-(321(,

-626.(1**

7/18/2019 Emparrillado Plano 3

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-32(2*

-.*1*.

-(*163

-(*22,6

-(321(,

(*163*

DIAGRAMAS FINALES

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7/18/2019 Emparrillado Plano 3

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 EMPARRILLADOS PLANOS

3 m

3. m

12 t:m

= 3(((((( t:m2

(1

'= (1. m

4= (* m

ad del pro'lema

2(.(((((( 5a ((((((2.

((

1((

((((( -16((.(((

-1,22,(( (((((

((((( 16((.(((

((((( 16((.((( BARRA 2-,

1,22,(( (((((

((((( 32(1((((

  _1=

  _2=

/=

=

^2    _3=* /

= =

7/18/2019 Emparrillado Plano 3

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,1.*.1, m

.1*2, m

11.6 (((((

131.1 (((((

((((( -16*2((

-11.6 ((((( BARRA 3-2

2*31*3 (((((

((((( 16*2((

  dales0

7/18/2019 Emparrillado Plano 3

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  %to

  ve%tor de %argas

7/18/2019 Emparrillado Plano 3

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endo stados y

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