Guia 2 Ejercicios Resueltos de Metodo de Las Fuerzas

(1)

¾

Ejercicio Nº1:

Ehormigon= 300.000 kg/cm

2

1t/m

B

A

C

D

20/30 20/50 20/30

Sistema Isostático Fundamental

1t/m

B

A

C

D

20/30 20/50 20/30

X =1tm

1

X =1t

2

(2)

Diagrama de Cuerpo Libre

P

/Cargas P

0

1t/m

B

C

D

20/30 20/50 20/30

R

A V

R

A H

R

B

A

Reacciones y Momento Flector

P

/Cargas P

0

El cálculo para este estado dio los siguientes resultados:

REACCIONES FLECTOR

M

A

= 0

M

C

=0

M

D

=1,5tm

M

B

=0

Diagrama de Momentos Flectores

P

/Cargas P

0

B

C

_D

A

(3)

Diagrama de Cuerpo Libre

P

/Carga

X

₁

=

1 tm

B

A

C

D

20/30 20/50 20/30

X =1tm

1

R

A H

R

A V

R

B

Reacciones y Momento Flector

P

_{/Cargas P}

₁

El cálculo para este estado dio los siguientes resultados:

REACCIONES FLECTOR

M

A

= -1tm

,

M

C

= -1tm

,

M

D

= -0,33tm

M

B

= 0

Diagrama de Momentos Flectores

P

_{/Cargas P}

₁

B

C

D

A

(4)

Diagrama de Cuerpo Libre

P

/Carga

X

₂

=

1 t

B

C

D

20/30 20/50 20/30

X =1t

2

R

A H

R

A V

R

B

Reacciones y Momento Flector

P

_{/Cargas P}

₂

El cálculo para este estado dio los siguientes resultados:

REACCIONES FLECTOR

M

A

= 0

,

M

C

= -2tm

,

M

D

= -1,67tm

M

B

= 0

Diagrama de Momentos Flectores

P

_{/Cargas P}

₂

B

C

D

A

-2tm

(5)

Cálculo de las Deformaciones

δ

ij

Se trabajara con la siguiente expresión:

_⎥

⎦

⎤

⎢

⎣

⎡

⋅

=

∫

dx

I

E

M

I

E

δ

b j i 0 b ij

Lo que es lo mismo:

=

∫

⋅

dx

ij j i ij

α

M

δ

a) Cálculo de las Inercias:

(

)

3 4 AC

45000

12

30 ,

0

20 ,

0 I

=

I

_DB

=

m

⋅

m

=

cm

(

)

4 3 CD

208333

,

33

12

50 ,

0

20 ,

0 I

=

m

⋅

m

=

cm

Adoptando como

I

₀

=

I

_AB

=

45000

cm

4

b) Cálculo de los Coeficientes “

α

ij

”:

0 ij ij

I

α

=

1,00

α

45000

I

α

₄ _AB 4 0 AB AC

=

=>

=

cm

4,63

α

45000

33 ,

208333

I

α

₄ _BE 4 0 BE CD

=

=>

=

cm

Sistema de Ecuaciones a utilizar

0 δ

X

δ

X

δ

0 δ

X

δ

X

δ

22 2 21 1 20 12 2 11 1 10

=

⋅

+

⋅

+

=

⋅

+

⋅

+

(6)

a) Cálculo del

δ

10

:

Barra

l

(m) α

ij

Diagramas Integración

Resoluci

_{ón (t}

2

_m)

CD 3,00

4,63

-0,33 tm -1 tm 1,5tm +

(

₂

₀

_.

₃₃

₁

)

3 ₄

_,

₆₃

50 ,

1

6

1 _⋅

_⋅

₊

_⋅

−

_{- 0,27}

-0,33 tm

-1 tm

1,125tm

(

₀

_.

₃₃

₁

)

3 ₄

_,

₆₃

125 ,

1

3

1 _⋅

_⋅

₊

_⋅

−

_-0,32

DB 2,12

1,00

-0,33tm -+ 1,50tm

12 ,

2

5 ,

1

33 ,

0

3

1 _⋅

_⋅

−

- 0,35

10 0 b

I

δ

E

⋅

- 0,94

b) Cálculo del

δ :

₁₁

Ba

rra

l

(m) α

ij

Diagramas Integración

Resol

ución

(t

2

m)

AC 2,00 1,00

-1 tm

1 ⋅

2 2,00

(7)

CD 3,00 4,63

-0,33 tm

-1 tm

-0,33 tm

-1 tm

63 ,

4

3 )

1

2

33 .

0

1

33 ,

0

1

33 ,

0

33 ,

0

2 (

6

1 ⋅

⋅

+

⋅

+

⋅

+

⋅

0,31

DB 2,12 1,00

--0,33tm

-12

,

2

33 ,

0

33 ,

0

3

1 _⋅

_⋅

_0,077

11 0 b

I

δ

E

⋅

2,387

Cálculo del

δ

₁₂

=

δ

₂₁

:

Barr

a

l (m)

α

ij

Diagramas Integración

Resolució

_{n (t}

2

_m)

AC 2,00 1,00

-1 tm -1 tm -2tm

-2

2

1

2

1 _⋅

_⋅

_2,00

CD 3,00 4,63

-0,33 tm -1 tm -2tm -1,67tm

(

)

63 ,

4

3

2

1

2

67 ,

1

2

33 ,

0

67 ,

1

33 ,

0

2

6

1 ⋅

⋅

+

⋅

+

⋅

+

⋅

0,80

DB 2,12 1,00

-0,33tm --1,67tm

-12

,

2

67 ,

1

33 ,

0

3

1 _⋅

_⋅

_0,39

12 0 b

I

δ

E

⋅

3,19

(8)

Cálculo del

δ

₂₀

:

Barra

_{(m) α}

l

ij

Diagramas Integración

Resolución

_(t

2

_m)

CD 3,00

4,63

-1,67tm 1,5tm +

(

4 ,

63

3

5 ,

1 )

2

67 ,

1

2

6

1 _⋅

₊

_⋅

−

_-0,86

-2tm -1,67tm 1,125tm

(

)

63 ,

4

3

125 ,

1

67 .

1

2

3

1 _⋅

+

⋅

−

- 0,89

DB 2,12

1,00

-1,67tm -1,5tm +

3

1 ,

67

1 ,

5

2 ,

12

1 _⋅

_⋅

−

- 1,77

20 0 b

I

δ

E

⋅

- 3,52

Cálculo del

δ :

₂₂

Barra

_{(m) α}

l

ij

Diagramas Integración

Resolución

_(t

2

_m)

AC 2,00

1,00

--2tm

-2

2

3

1 _⋅

_⋅

_2,66

CD 3,00

4,63

-2tm -1,67tm -2tm -1,67tm

(

)

3 ₄

_,

₆₃

2

67 ,

1

2

67 ,

1

67 ,

1

67 ,

1

2

6

1 ⋅

⋅

+

⋅

+

⋅

+

⋅

2,19

DB 2,12

1,00

-1,67tm --1,67tm

-12

,

2

67 ,

1

67 ,

1

3

1 _⋅

_⋅

_1,97

* 22 0 b

I

δ

E

⋅

6,82

(9)

Reemplazando los valores obtenidos en el sistema de ecuaciones queda:

0 ,820

6 X

,180

3 X

3,520

0 ,180

3 X

,387

2 X

,940

0

2 1 2 1

=

⋅

+

⋅

+

−

=

⋅

+

⋅

+

−

=>

88 ,

0 X

77 ,

0 X

2 1

=

−

=

Cálculo de las Reacciones Finales

2 2 1 1 0 Final

X

R

=

R

+

⋅

R

+

⋅

R

t

88 ,

0

1

88 ,

0 R

H A

=

⋅

=

t

R

V_A

=

2 +

(

−

0 .

77 )

⋅

0 ,

22 +

0,88

⋅

0 ,

11 =

1 ,

93 t

07 ,

1 ,11)

0 (

,88

0 )

22 ,

0 (

)

77 ,

0 (

1 R

V B

=

+

−

⋅

−

+

⋅

−

=

-0,88t

-1)

(

0,88

R

H B

=

⋅

=

Diagrama de Cuerpo Libre Final

B

C

D

20/30 20/50 20/30

R

A H

R

A V

R

B

1t/m

R

HB

= 1,93t

= 0,88t

= 1,07t

V

= 0,88t

M

A

= 0,77t

Cálculo del Momento Flector Final

2 2 1 1 0 Final

M

X

M

X

M

=

+

⋅

+

⋅

tm

77 ,

0

88 ,

0 )

1 (

77 ,

0

0 M

_A

=

+

−

⋅

−

+

⋅

=

tm

M

_C

=

0 +

(

−

0 .

77 )

⋅

(

−

1 )

+

0,88

⋅

(

−

2 )

=

−

1 tm

,28

0 ,67)

1 (

,88

0 )

33 ,

0 (

)

77 ,

0 (

5 ,

1 M

_D

=

+

−

⋅

−

+

⋅

−

=

tm

0 M

_B

=

(10)

Diagrama del Momento Flector Final

C

A

B

D

0,77

-1tm

0,28tm

0tm

¾

Ejercicio Nº2:

Estructura con Tensor:

Datos:

E

b

= 210 t/cm

2

(11)

Sistema Isostático Fundamental

Diagrama de Cuerpo Libre

P

/Cargas

P

0

Cálculo de Reacciones

P

/Cargas

P

0

=>

⋅

+

⋅

=

=>

⋅

+

⋅

+

⋅

−

=>

=

∑

t

_m

m

t

m

10

12

2

10

8

6

1 R

12 P

6

8 q

10 R

0 M

_A _D _D

t

7,20

R

_D

=

( )

₋

⋅

_=>

⋅

=

=>

⋅

+

⋅

−

⋅

=>

=

∑

t

_m

m

t

m

10

2

10

2

8

1 R

2 P

2

8 q

10 R

0 M

2 A 2 A D

t

2,80

R

_A

=

Cálculo de los Momentos Flectores

P

/Cargas

P

0

0 M

0A

=

tm

m

5

60

2 R

M

0_B

=

_A

⋅

=

,

( )

tm

m

8

80

2

4 q

6 R

M

2 A 0 C

=

,

⋅

−

⋅

=

tm

m

4

2 P

M

0D

=

−

⋅

=

−

0 M

0_E

=

(12)

( )

CD 2 2 BC

2

8

4

1

8 l

q

f

tm

m

t

f

=

⋅

=

⋅

=

( )

tm

m

t

f

8

1

8 l

q

2 2 BD

=

⋅

=

⋅

=

Diagrama de Momentos Flectores

P

/Cargas

P

0

Diagrama de Cuerpo Libre

P

/Carga

X

₁

=

1 tm

Cálculo de Reacciones

P

/Carga

X

₁

=

1 tm

=>

=

=>

−

⋅

=>

=

∑

tm

_m

m

10

1

10 X

R

X

10 R

0 M

1 D 1 D A

t

0,10

R

_D

=

=>

=

=>

−

⋅

=>

=

∑

tm

_m

m

10

1

10 X

R

X

10 R

0 M

1 A 1 A D

t

0,10

R

_A

=

Cálculo de los Momentos Flectores

P

/Carga

X

₁

=

1 tm

tm

1 M

1_A

=

tm

m

0

80

8 R

M

1 _D B

=

⋅

=

,

(13)

tm

m

0

40

4 R

M

1C

=

D

⋅

=

,

0 M

1D

=

Diagrama de Momentos Flectores

P

/Carga

X

₁

=

1 tm

Diagrama de Cuerpo Libre

P

_/Carga

t

1 X

2

=

Cálculo de Reacciones

P

_/Carga

t

1 X

2

=

=>

⋅

=

⋅

=

=>

⋅

−

⋅

=>

=

∑

t

_m

m

10

6

1

10

6 X

R

6 X

10 R

0 M

2 D 2 D A

t

0,60

R

_D

=

=>

⋅

=

⋅

=

=>

⋅

−

⋅

=>

=

∑

t

_m

m

10

4

1

10

4 X

R

4 X

10 R

0 M

2 A 2 A D

t

0,40

R

_A

=

Cálculo de los Momentos Flectores

P

/Carga

X

₂

=

1 t

0 M

2_A

=

tm

m

0

80

2 R

M

2 _A B

=

⋅

=

,

tm

m

2

40

4 R

M

2_C

=

_D

⋅

=

,

0 M

2

=

(14)

Diagrama de Momentos Flectores

P

/Carga

X

₂

=

1 t

Diagrama de Esfuerzos Normales

P

/Carga

X

₂

=

1 t

Cálculo de las Deformaciones

δ

ij

Se trabajara con la siguiente expresión:

_⎥

⎦

⎤

⎢

⎣

⎡

⋅

+

⋅

=

∫

dx

∫

dx

t e j i b j i 0 b ij

A

E

N

I

E

M

I

E

δ

Lo que es lo mismo:

∫

⋅

+

⋅

=

dx

t e j i 0 b ij j i ij

A

E

N

I

E

α

M

δ

Nota: El primer sumando corresponde a la estructura de hormigón y el segundo al tensor

de acero.

Cálculo de las Inercias:

(

)

3 4 AB

45000

12

30

0

20

0 I

=

,

m

⋅

,

m

=

cm

(

)

3 4 BE

208333

33

12

50

0

20

0 I

=

,

m

⋅

,

m

=

,

cm

Adoptando como

I

₀

=

I

_AB

=

45000

cm

4

(15)

0 ij ij

I

α

=

1,00

α

_AB

=

=>

=

₄4 0 AB AB

45000

I

α

cm

4,63

α

,

BE

=

=>

=

₄ 4 0 BE BE

45000

33 208333

I

α

cm

Sistema de Ecuaciones a utilizar

0 δ

X

δ

X

δ

0 δ

X

δ

X

δ

22 2 21 1 20 12 2 11 1 10

=

⋅

+

⋅

+

=

⋅

+

⋅

+

Cálculo del

δ

₁₀

:

Barra l

(m)

α

ij

Diagramas Integración

Resolución

_(t

2

_m)

AB 3,60

1,00

−

1 ₆

⋅

(

2 ⋅

0 ,

8 +

1 )

⋅

5 ,

6 ⋅

3 ,

6 - 8,736

BD 8,00

4,63

(

)

63

4

8

0

4

6

5

2

6

1 ,

,

+

⋅

−

⋅

- 1,659

63

4

8

0

8

3

1 ,

,

⋅

- 3,686

10 0 b

I

δ

E

⋅

- 14,081

(16)

Cálculo del

δ :

₁₁

Barra l

(m)

α

ij

Diagramas Integración

Resolución

_(t

2

_m)

AB 3,60

1,00

1 ₆

⋅

(

2 +

2 ⋅

0 ,

8 +

2 ⋅

0 ,

8 ⋅

0 ,

8 )

⋅

3 ,

6 2,928

BD 8,00

4,63

1 ₃

⋅

0 ,

8 ⋅

0 ,

8 ⋅

8 ₄

_,

₆₃

0,369

11 0 b

I

δ

E

⋅

3,297

Cálculo del

δ

₁₂

=

δ

₂₁

:

Barra l (m)

α

ij

Diagramas Integración

Resolución

_(t

2

_m)

AB 3,60

1,00

1 ₆

⋅

(

2 ⋅

0 ,

8 +

1 )

⋅

0 ,

8 ⋅

3 ,

6 1,248

BC 4,00

4,63

(

)

4 ₄

₆₃

4

2

4

0

2

8

0

4

0

4

2

8

0

8

0

8

0

2

6

1 ,

,

⋅

+

⋅

+

⋅

+

⋅

0,783

CD 4,00

4,63

1 ₃

⋅

0 ,

4 ⋅

2 ,

4 ⋅

4 ₄

_,

₆₃

0,276

12 0 b

I

δ

E

⋅

2,307

(17)

Cálculo del

δ

₂₀

:

Barra l (m)

α

ij

Diagramas Integración

Resolución

_(t

2

_m)

AB 3,60

1,00

−

1 ₃

⋅

0 ,

8 ⋅

5 ,

6 ⋅

3 ,

6 - 5,376

BC 4,00

4,63

(

)

4 ₄

₆₃

4

2

8

2

8

0

8

4

2

6

5

8

0

6

5

2

6

1 ,

,

⋅

+

⋅

+

⋅

+

⋅

−

- 1,843

(

)

63

4

2

4

2

8

0

3

1 ,

,

+

⋅

−

- 10,321

CD 4,00

4,63

(

)

63

4

2

4

8

2

6

1 ,

,

+

⋅

−

⋅

- 4,700

63

4

2

3

1 ,

,

⋅

−

- 1,382

20 0 b

I

δ

E

⋅

- 23,622

(18)

Cálculo del

δ :

₂₂

Barra l

(m)

α

ij

Diagramas Integración

Resolución

_(t

2

_m)

AB 3,60 1,00

1 ₃

⋅

0 ,

8 ⋅

0 ,

8 ⋅

3 ,

6 0,768

BC 4,00 4,63

(

)

4 ₄

_,

₆₃

4 ,

2

4 ,

2

8 ,

0

4 ,

2

4 ,

2

8 ,

0

8 ,

0

8 ,

0

2

6

1 ⋅

⋅

+

⋅

+

⋅

+

⋅

2,400

CD 4,00 4,63

1 ₃

⋅

2 ,

4 ⋅

2 ,

4 ⋅

4 ₄

_,

₆₃

1,659

* 22 0 b

I

δ

E

⋅

4,827

Tensor 2,00

--- ** 22

2

00 ,

2

1

1 ⋅

⋅

=

δ

2

10

6

10

5

4

10

1 δ

I

E

** 22 0 b

⋅

⎟

⎠

⎞

⎜

⎝

⎛

⋅

=

⋅

,

0,1500

22 0 b

I

δ

E

⋅

4,977

Reemplazando los valores obtenidos en el sistema de ecuaciones queda:

0 4,977

X

2,307

X

23,622

0 2,307

X

3,297

X

14,081

2 1 2 1

=

⋅

+

⋅

+

−

=

⋅

+

⋅

+

−

=>

t

tm

4,095

X

1,406

X

2 1

=

(19)

Cálculo del Momento Flector Final

2 2 1 1 0 Final

M

X

M

X

M

=

+

⋅

+

⋅

tm

1,406

M

_A

=

tm

1,20

4,095

0,80

1,406

0,80

5,60

M

_B

=

−

⋅

−

⋅

=

tm

1,59

4,095

,40

2 1,406

0,40

0

8

8 M

_C

= ,

−

⋅

−

⋅

=

−

tm

4,00

M

_D

=

−

Diagrama del Momento Flector Final