Practica de Puentes

FACULTAD DE INGENIERA ESCUELA PROFESIONAL DE INGENIERA CIVIL X CICLO

1

DISEO DE UN PUENTE DE CONCRETO ARMADO

DISEO DE LA FRANJA DE BORDE

DATOS

Acero Corrugado Grado 60, fy = 4200 kg/cm2

Resistencia cilndrica del concreto, fc= 280 kg/cm2

PERFIL LONGITUDINAL

PERFIL TRANSVERSAL

3 m 10 m

0,45 m 7,2 m 0,45 m

1,8 m 1,8 m

asfalto 2''

franja de

borde


2

= = {0.45 + 0.30 + (3.18

2) = 2.34

1.80 }

= 1.80 .

1. PREDIMENSIONAMIENTO

= = 1.2( + 3

30)

= 1.2(10 + 3

30) = 0.52 0.55

2. DIBUJO DE LAS LINEAS DE INFLUENCIA DE LAS SECCIONES MAS ESFORZADAS

a. MOMENTO POSITIVO (Al centro del tramo BC)

0,5

5 m

1 m

1.50

2.50

B C

A

3 m 10 m


3

b. MOMENTO NEGATIVO

3. DISEO DE FRANJA DE BORDE

3.1. METRADO DE CARGA PERMANENTE:

DC:

=1.800.552.40

1.80= 1.32

=0.85

1.80 = 0.47 /

= . + . = . /

DW:

=(1.800.45)0.05082.25

1.80= 0.0857 /

= . /

B C

A

3.00

3 m 10 m


4

3.2 CLCULO DE LOS MOMENTOS PRODUCIDOS POR LA CARGA

PERMANENTE:

a) MOMENTO POSITIVO:

= 1

2 (3 1.50) +

1

2 (10 2.50) = 10.25 /

= 10.25 1.79 = 18.35 /

= 10.25 0.09 = 0.88 /

b) MOMENTO NEGATIVO:

1.50

2.50

B C

A

3 m 5 m 5 m

WDC = 1.79 Tn/mWDW = 0.09 Tn/m

B C

A

3.00

3 m 10 m

DC

WDW = 0.09 Tn/m


5

= 1

2 (3 3) = 4.5/

= . . = . /

= . . = . /

3.3 CLCULO DE LOS MOMENTOS PRODUCIDOS POR LA CARGA VIVA

VEHICULAR

a) MOMENTO POSITIVO:

CAMION:

1 =2.5

5 (5 4.30) = 0.35

=14.52 (2.5 + 0.35) + 3.63 (0.35)

2

= . /

1.50

2.50

B C

A

3 m 5 m 5 m

4,3 m 4,3 m

14.52 Tn14.52 Tn3.63 Tn

Y1Y1


6

TANDEM

2 =2.5

5 (5 1.20) = 1.90

=. (. + . )

= . /

CARRIL:

1.50

2.50

B C

A

3 m 5 m 5 m

11.30 Tn11.30 Tn11.30 Tn

1,2 m 1,2 m

Y2 Y2

1.50

2.50

B C

A

3 m 5 m 5 m

Carril= 0.952 Tn/m


7

=. (

. )

= . /

b) MOMENTO NEGATIVO:

CAMION:

= .

= . /

TANDEM:

B C

A

3.00

3 m 10 m

14.52 Tn

B C

A

3.00

3 m 10 m

11.35 Tn

1,2 m


8

= . ( + . )

= . /

CARRIL:

=

( ) .

= . /

c) DISTRIBUCIN TRANSVERSAL DEL MOMENTO

Sabiendo que Eb = 1.80 m.:

B C

A

3.00

3 m 10 m

Carril= 0.952 Tn/m

B C

A

MTa = -27.24 Tn.m/carril

MLn = -2.12 Tn.m/carril

MTa = 24.97 Tn.m/carril

MLn = 5.95 Tn.m/carril


9

B C

A

MDC = 8.06 Tn.m/m

MDW = 0.39 Tn.m/mMLL = 16.31 Tn.m/mMIM = 4.99 Tn.m/m

MDC = 18.35 Tn.m/m

MDW = 0.88 Tn.m/mMLL = 17.18 Tn.m/mMIM = 4.58 Tn.m/m

MOMENTO POSITIVO

=24.97+5.95

1.80= 17.18 . /

= 0.33 (24.97

1.80) = 4.58 . /

MOMENTO NEGATIVO

=27.24+2.12

1.80= 16.31 . /

= 0.33 (27.24

1.80) = 4.99 . /

3.4 CLCULO DE LOS MOMENTOS DE DISEO ESTADO LMITE DE

RESISTENCIA

+ = . [. . + . . + . (. + . )]

+ = .

= . [. . + . . + . (. + . )]

= .

4. CLCULO DEL REFUERZO POSITIVO Y NEGATIVO


10

4.1 REFUERZO PRINCIPAL POSITIVO

:

= 55 (1.5 + 2.5 + 1.27)

= 49.73

= 62.34

= 2 (2

0.85 100)

= 49.73 49.732 (262.34105

0.850.9280100)

= 6.34

:

=

. . ( )

=. ^

. ( . .

)= .

: Seccin crtica de Momento Positivo

a) Armadura Mxima:

0.42 , = 0.75

S: = 6.24

=

1=

6.24

0.85= 7.34

=

7.34

49.73= 0.15 0.42 . !

b) Armadura Mnima:

En funcin de una resistencia mnima de la flexin:

1.2 1.33

, = 2

=(. )


11

=100553

12= 1386458.3 4

= 2280 = 33.47

2

=55

2= 27.5

=33.471386458.3

27.5105= 16.80

1.2 = 1.216.8 = 20.2 1.33 = 1.3362.34 = 82.91

:

Mmin = 20.20 Tn. m Asmin = 10.96 2

As = 35.38 2 > Asmin = 10.96 2 ok!

:

=0.75

=

0.751000550

420100= 9.82 2

= 35.88 2 > = 9.82 2 !

c) Separacin:

= .100

:

= 5.10 100

35.38= 14.42

1.5 55 = 82.5

45

1.52.54 = 3.81

4

@ .

=5.10100

14= 36.43 2

d) Verificando la separacin del refuerzo longitudinal ( Control de

fisuracin)

=

=

(. )13

0.6

= 30000

()

= 1.0 [118.35 + 10.88 + 1.0(17.18 + 4.58)]

= 40.99 = 16.80

, :


12

=( )

; = .

= . . (

3)

=

( 3 )

=

=

2106

15000280= 7.97 8

() (

2) . . ( ) = 0

(

2) (2) + (. ) . . = 0

100

22 + (836.43) 836.4349.73 = 0

= 14.35

Entonces:

=2

( 3 ) . .

= 240.99105

(49.73 14.35

3 ) 10014.35= 127.10

2

= 127.10

2

2=

280

2= 140

2

=

( 3 )

=40.99105

36.43 (49.73 14.35

3 )= 2503.35

2 = 4200

2

=

(. )13

0.60

: = 1423.77 = 105.562

=30000

(3.77105.56)13

= 4078.57

2 . =

= .

=

, !


13

4.2 REFUERZO PRINCIPAL NEGATIVO

:

= 55 (1.5 + 2.5 + 1.27)

= 49.73

= 47.94

= 2 (2

0.85 100)

= 49.73 49.732 (247.94105

0.850.9280100)

= 4.73

:

=

. . ( )

=. ^

. ( . .

)= .

: Seccin crtica de Momento Negativo

a) Armadura Mxima:

0.42 , = 0.75

S: = 4.73

=

1=

4.73

0.85= 5.56

=

5.56

49.73= 0.11 0.42 . !

b) Armadura Mnima:

En funcin de una resistencia mnima de la flexin:

1.2 1.33

, = 2

=(. )


14

=100553

12= 1386458.3 4

= 2280 = 33.47

2

=55

2= 27.5

=33.471386458.3

27.5105= 16.80

1.2 = 1.216.8 = 20.2 1.33 = 1.3347.94 = 63.76

:

Mmin = 20.20 Tn. m Asmin = 10.96 2

As = 26.78 2 > Asmin = 10.96 2 ok!

:

=0.75

=

0.751000550

420100= 9.82 2

= 26.78 2 > = 9.82 2 !

c) Separacin:

= .100

:

= 5.10 100

26.78= 19

1.5 55 = 82.5

45

1.52.54 = 3.81

4

@ .

=5.10100

19= 26.84 2

d) Verificando la separacin del refuerzo longitudinal ( Control de

fisuracin)

=

=

(. )13

0.6

= 30000

()

= 1.0 [18.06 + 10.39 + 1.0(16.31 + 4.99)]

= 29.75 = 16.80

, :


15

=( )

; = .

= . . (

3)

=

( 3 )

=

=

2106

15000280= 7.97 8

() (

2) . . ( ) = 0

(

2) (2) + (. ) . . = 0

100

22 + (826.84) 826.8449.73 = 0

= 12.62

Entonces:

=2

( 3 ) . .

= 229.75105

(49.73 12.62

3 ) 10012.62= 103.57

2

= 103.57

2

2=

280

2= 140

2

=

( 3 )

=29.75105

26.84 (49.73 12.62

3 )= 2434.84

2 = 4200

2

=

(. )13

0.60

: = 1923.77 = 143.262

=30000

(3.77143.26)13

= 3683.83

2 . =

= .

=

, !


16

4.3 ARMADURA DE DISTRIBUCIN

=

=

= {1750.

1000.50

}

ASDAS

AT

Practica de Puentes

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