diseño de puente a escala

8/18/2019 Diseño De PUENTE A ESCALA

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DISEÑO DE UN PUENTE ATIRANTADO DE UNCLARO DE 1.3m

1.DESCRIPCION GENERALel puente a diseñar será del tipo atirantado de un luz libre de 1.3m, consta de12 tirantesa cada lado teniendo así un total de 24 tirantes, la escala paranuestro diseño será:

escala geométrica : 0.0125

escala de carga: 0.0125

1.1. camión de diseño:

para nuestro caso utilizaremos el camin de diseño L!"3 las dimensiones semuestran

en la !gura

"obrecarga #$%&3, 'orma (("#)* $+-, eui/alente a un camin de 32.toneladasdel camin de diseño se obtiene ue la carga dediseño es:


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diseño es:


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).DISEÑODEL

PUENTE).1.condicion:ca&*a+,n-,a$ a $a mi-ad de$c$a&o

' e?

146.012

& 40.37 0.10.0020

6720.776&7

47&&0.432&415

0.06211436

270.471

22 34.46 0.10.0020

6720.4&301

25360.60022205

0.0015154

37.675

&6 30.61 0.10.0020

6720.2222

&570.&364121&

0.07276414

4

103.13

402 30.61 0.1

0.0020

672

%0.2222

&57

0.&36

4121&

0.0727

6414

511&.53

66 34.46 0.50.0104

337

%0.4&301

25360.60022205

0.014030306

7131.&6

21 40.37 20.041

344

%0.776&7

47&&0.432&415

0.0041053

46.012

& 40.37 2

0.041

344

0.776&7

47&&

0.432

&415

0.0041

053

670.471

22 34.46 0.50.0104

3370.4&301

25360.60022205

0.014030306

7.675 0.0020 0.2222 0.&36 0.0727


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(+(C3 1

0.&364122

)+(@* $ 8;

0ABA 27 130

01333.3

27 ABA 130

/ector dezpla

Ci>cm

>cm d e> e?

1 7 %0.007 0 0.0072 0.004&&

2 17 %0.013 0 0.0127 0.01502

3 27 %0.0270 0 0.02703 0.02535

4 &0 %0.0336 0 0.03364 0.032&7

5 60 %0.0374 0 0.03736 0.03175

7 0 %0.037 0 0.0377& {d3}= 0.022

70 %0.0351 0 0.03507 0.02707

6 50 %0.0316 0 0.0317 0.0273

& 40 %0.021 0 0.0206 0.0273

10 27 %0.016 0 0.0161 0.0162211 17 %0.0116 0 0.01163 0.0102&

12 7 %0.0045 0 0.00450 0.00334


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(+(C4 1

0.&364122

)+(@* $ 8;

0ABA 40 130

01333.3

40ABA 130

/ector dezpla

Ci

>c

m

>c

m d e> e?1 7 %0.0064 0 0.00642 0.00727

2 17 %0.021& 0 0.021&0 0.01&07

3 27 %0.0336 0 0.03364 0.032&7

4 40 %0.0471 0 0.04714 0.044&4

5 60 %0.050 0 0.0500 0.04411

7 0 %0.0516 0 0.05164 {d4}= 0.03653 70 %0.0500 0 0.04&&& 0.0317

6 50 %0.0457 0 0.0455 0.03&75

& 40 %0.03&0 0 0.03&02 0.03600

10 27 %0.021 0 0.0206 0.0273

11 17 %0.011 0 0.0114 0.014&1

12 7 %0.0075 0 0.00752 0.00465


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(+(C5 1

0.6002221

)+(@* $ 8;

0ABA 50 130 01333.3

50ABA

130

/ector dezpla

Ci>cm

>cm d e> e?

1 7%

0.00&&0.0000

00.00&

&0.005&

4

2 17%

0.020650.0000

00.0206

50.0161

4

3 27

%

0.03250

0.0000

0

0.0325

0

0.0317

5

4 40%

0.045300.0000

00.0453

00.0441

1

5 50%

0.0506&0.0000

00.0506

&0.0442

6

7 0%

0.052660.0000

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1

70%

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30.0363

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6 50%

0.04310.0000

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0 0000 0 0 0 0 03&7


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8/15

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30.00

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0.00

50.01

50.06

60.03

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0.02

70.02

60.02

70.01

60.01

00.00

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0.00

70.01

&0.03

30.10

6 0.0440.03

&0.03

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{flex}=

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0.03& 0.03&

0.040

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70.03

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3

0.4553

1.763


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10/15

/ectordezpla

Ci >cm d e> e?

1 71.470

6 0 %1.471

%1.06

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%3.2

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& 0 %5.7&3

%5.544

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%7.&7

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%7.01

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11/15


12/15

(+(

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13/15

0.517342

&4.&&7256

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1.660002

5616.12

07

3.&7104

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5.765373&5.&&4

41

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DC-iseñoE

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6& 1

CALCULO DE DEZPLASAMIENTOS

ara determinar los desplazamientos en los puntos de controlPa&a $a +&ime&a condición de ca&*a c,ando e$ cen-&oide de$ ca&&o coincide cone$ cen-&oide de$ +,en-e :ara esto se utiliz el método de energía


15/15

6 16 26

0 6 16

1( ( 2 .0493*X)*( 0.01*X)* ( 2 .0493*X 0.4553*(X 6))*( 0.01*X 0.0022*(X 6))* ( 2 .0493*X 0.4553*(X 6) 1.63*(X 16))*( 0.01*X 0.0022*(X 6) 0.002*(X 16))*

!01333.333

( 2.0493*X 0.4553*(X

dx dx dx= − − + − + − − + − + − + − + − − + − + −

+ − +

∫ ∫ ∫ 40

26

50

40

6) 1.63*(X 16) 3.496!*(X 26))*( 0.01*X 0.0022*(X 6) 0.002*(X 16) 0.01!1*(X 26))*

( 2.0493*X 0.4553*(X 6) 1.63*(X 16) 3.496!*(X 26) 5.5245*(X 40))*( 0.01*X 0.0022*(X 6) 0.002*(X 16)

dx− + − + − − + − + − + −

+ − + − + − + − + − − + − + − +

∫

∫ 60

50

0.01!1*(X 26) 0.02!*(X 40))*

( 2.0493*X 0.4553*(X 6) 1.63*(X 16) 3.496!*(X 26) 5.5245*(X 40) 2!.6665*(X 50))*( 0.01*X 0.0022*(X 6) 0.002*(X 16) 0.01!1*(X 26) 0.02!*(X 40) 0.1354*(X 50))*

(

dx

dx

− + −

+ − + − + − + − + − + − − + − + − + − + − + −

+

∫ !0

60

2.0493*X 0.4553*(X 6) 1.63*(X 16) 3.496!*(X 26) 5.5245*(X 40) 2!.6665*(X 50) 93.60!4*(X 60))*( 0.01*X 0.0022*(X 6) 0.002*(X 16) 0.01!1*(X 26) 0.02!*(X 40) 0.1354*(

( 2.0

− + − + − + − + − + − + − − + − + − + − + − +

+ −

∫ 6 16 26

0 6 16

493*X)*( 0.01*X)* ( 2 .0493*X 0.4553*(X 6))*( 0.01*X 0.0022*(X 6))* ( 2 .0493*X 0.4553*(X 6) 1.63*(X 16))*( 0.01*X 0.0022*(X 6) 0.002*(X 16))*

( 2.0493*X 0.4553*(X 6) 1.63*(X 16)

dx dx dx− + − + − − + − + − + − + − − + − + − +

+ − + − + −

∫ ∫ ∫ 40

26

50

40

3.496!*(X 26))*( 0.01*X 0.0022*(X 6) 0.002*(X 16) 0.01!1*(X 26))*

( 2.0493*X 0.4553*(X 6) 1.63*(X 16) 3.496!*(X 26) 5.5245*(X 40))*( 0.01*X 0.0022*(X 6) 0.002*(X 16) 0.01!1*(X 26) 0.0

dx+ − − + − + − + −

+ − + − + − + − + − − + − + − + − +

∫

∫ 60 25

1450

2!*(X 40))*

( 2.0493*X 0.4553*(X 6) 1.63*(X 16) 3.496!*(X 26) 5.5245*(X 40) 2!.6665*(X 50))*( 0.01*X 0.0022*(X 6) 0.002*(X 16) 0.01!1*(X 26) 0.02!*(X 40) 0.1354*(X 50))* )

0.11

dx

dx E

δ

−

+ − + − + − + − + − + − − + − + − + − + − + − +

=

∑∫

440 0.4549 0.5999cm+ =

diseño de puente a escala

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