Planilla Puentes

PROYECTO ACADEMICO: CIV - 3312

Realizado por: Giovanny air A!"#a C$

PRO%ECTO CIV - &&1" DISEÑO DE UN PUENTE SOBRE VIGAS METALICAS DATOS: Ec =

√

15250 f'c 15250

Es =

Luz de cálculo (m)

=

10

carga * 1.20

=

H2 0

f'c [g/cm2!

=

"00

f# [acero de refuerzo!

=

2$00

%# [acero es& es&ruc&u c&ural!

=

25"0

=

+%

+c [m!

=

12

,- de lumo

=

1

o de ga

210000 2100000 0 Kg/cm2 Kg/cm2

1.- VIGA DE WF. Dimensiones en mm! Es&mac de la al&ura de la secc. H=L/20

H=

5 0 0 mm

"# WF 1$$

e &alas se elge secc aro3mada

H= 24 24 e6u=

700 mm mm

24

"00

12 18.4

0.995 280."1 H= 700

571." 11.9

0.47$

"00

18.4

0.995 :E :E E;= E;=

 = 2*1. 2*1.84 84*" *"0 0 < 57.1 57.1"* "*1. 1.19 19 =

E;> E;=

 = 0.01$2 * 9$00

1$1. 1$1.82 82 cm2 cm2 =

141.$8 Kg/m

?E,:> E @:AE E;B Cc =

H/2 =

"0 c m

D>DE,> E ,E:?B 1

o=l=

1

∗1 . 1 7∗56.133 + 2 ∗ ∗30∗1 . 9 43 + 2 ∗30∗1 . 9 4∗( 29.031 )2 = 115250.808 cm4 12 12

PROYECTO 1 ACADEMICO: CIV - 3312 1 3 3 1 . 1 7 56.13 2 30 1 . 9 4 12 12

∗

∗

+ ∗ ∗ ∗


+ 2 ∗30∗1 . 9 4∗( 29.031 )2 =

o= 115250.81 cm4

ES'UEMA ESTRUCTURAL. Fdades e ?e&Gme&ros +c =

1200

$1." sf.=

5 (l&ura) 1$

"0 70

Dise(o )e B*+*n)*)o Ce++*)o. 15

%=

4540 Kg.

M =

15 5

=

25

0.$1" m&s.

P

X

E

E = 0.$* < 1.14" = D=

15

(D)

21."

d=

1

+

√

fc = 0.45*f'c =

1"5 g/cm2

fs = 0.50*f# =

1400 g/cm2

1

=

fs n∗ fc

=

205$.11" Kg*m

mesoame&o.

"0 odeB

1.98"

M b∗k

0.4"5 K= 1/2*fc**I

I= 1 J /" =

0.$55 K= 1/2*fc**I =

25.1" Kg/cm2

LuegoB d=

0.081 :ecur.

m&s

=

8.05 cm

=

2.54 cm

=

11.58 cm

F;:B  =

12 cm



15 cm



:DF:B s = D/fs*I*d =

USAR

18.00 cm2

D &,# , 1/ m =

18 cm2

PESO BARANDADO POR METRO LINEAL.

A = 0.15 ∗ 0.813

+ 0.15 ∗ 0.15 + ( 0.05 ∗ 0.15 ) /

6 = *2400 =

+ 0.213 ∗ 0.15 + ( 0.15 ∗ 0.15 ) / 2 =

2

0.1814

458."7 Kg/m

Dise(o )e 0* Los* )e T*0e+o: ?álculo de de carga or losa # asfal&o. Losa = 0.1$ * 2400=

4"2.00 Kg/m2

sfal&.=0.05*2000 =

100.00 Kg/m2 6=

5"2.00 Kg/m2

Momen2os 3o+ *+4* m5e+2*. a)Dome&o de &ramo or ?.D.

l se cálcula más adela&e cm

M

T

=

q∗l

or 1 m&.

2

16

=

209.$1 Kg*m

de aco de losa

)Dome&o e ao#o &ermedo or ?D. cm AP

M

=−

q∗l 11

2

or 1 m&.

=

J"02.29 Kg*m

de aco de losa

c)Dome&o e olado or ?D. 1 =

458."7 Kg/m

" =0.1$*0.85*2400 < 0.05*0.85*2000 =

505.4 Kg/m

D =1*1 < "*" =

910.81 Kg*m or me&ro de aco de losa

1

0.1$ 0."

0.85

a=

1.25

0.4$

Momen2os 3o+ *+4* 6i6*. a)Dome&o or carga a e &ramo re de carga H20

?am H20

Dodfcado= H24 J 44



$91$

$91$

$91$

4."

4."

O+i4in*0 ?arga or eMe =

145"0 Kg

?arga or rueda=

9275 Kg

Mo)i7i*)o 8 "$9 ?arga or eMe =

194"7 Kg

?arga or rueda= cv

M

T

=+

S+0.61 9.75

$91$ Kg odeB ;=l=

∗ P∗( 1 + 20 )∗( 1 + I )∗ FR

=

 = 15 / (; < "$ ) =

cv

M

T

=

2.500 m 9275

=

0."90 N 0."0

sumr =

0."

%: =

0.$

2$82.05 Kg*m or me&ro de aco de losa

) Dome&o or carga a e ao#o &ermedo. cv

M

T

=−

S + 0 . 6 1 9.75

cv AP

M

=

∗ P∗( 1 + 20 )∗( 1+ I )∗ FR 2$82.05 Kg*m

?) Dome&o or carga a e olado :. 1.2.11O :. 1.".2 () # ()

1.M6.B*+ 2.Dc :=

M M =

P E

"$/;.11& <4=m

X ∗( 1 +20 )∗(1+ I )

B :

=

=

0.4$

9275 Kg

E = 0.$* < 1.14" =

1.52"

0.1$ 0."

0.4$

".M6 PR

-&/&#.>1 <4=m

0.85

M6M8M+ -//?".;"# <4=m a=

1.25

0.4$

RESUMEN DE MOMENTOS DE DISEÑO. a) Dome&o e &ramo D = Dcm < Dc =

"088.$7 Kg*m

) Dome&o e ao#o &ermedo D = Dcm < Dc = c) Dome&o e olado

J"184."2 Kg*m



D = Dcm < Dc =

J7"0".9" Kg*m

VERIFICACION DE LA ALTURA UTIL ) d=

a) Dome&o e &ramo

d=

√

M b∗k

11.11 cm

:ecu.=

2.54 cm 1".75 cm

 1$ cm

>.

 1$ cm

>.

 1$ cm

>.

) Dome&o e ao#o &ermedo d=

11.29 cm

:ecu.=

2.54 cm 1".$1 cm

c) Dome&o e olado d=

15.44 cm

:ecu.=

2.54 cm 19.8$ cm

ACERO DE REFUER@O As 8 T+*mo: s = D/fs*I*d =

2"."2 cm2

USAR

D &,# , 1" m

As - A3oo In2e+m: s = D/fs*I*d =

= 2".79 cm2

USAR

2".9 cm2

D &,# , 1" m

As - Vo0*)o:

=

s = D/fs*I*d =

"4.12 cm2

ARMADURA DE DISTRIBUCION:

USAR

2".9 cm2

D &,# , ; m =

"5.7 cm2

;=

2.500 m&s.

r&. 1.".2 (e)

Es&a armadura será aralela al &rafco. (P) s = 122 / (;)Q0.5

odeB

(P) s = 122 / (;)Q0.5 =

99.170 P

,o&ese 6ue la armadura de ds&ruc má3mo erm&do será del 79P ?omoB ;e usaráB

99.170 P N 79 P 79P s(ds&r.)=

0.79*(s ma3. el &alero)

s má3mo del &alero =

2".79 cm2


s(ds&r.)=


15.$7

cm2

USAR : D 1," , ; m s = 15.8

cm2

USAR : D &,# , 1; m s = 15.$

cm2

CALCULOS. n =

Es 2100000 = = Ec 15250 √ f ' c

re de carga H20

9.85

=

$

?am H20 $91$

$91$

Dodfcado= H24 J 44 $91$

4."

4."

O+i4in*0 ?arga or eMe =

145"0 Kg

?arga or rueda=

9275 Kg

Mo)i7i*)o 8 "$9 ?arga or eMe =

194"7 Kg

?arga or rueda=

$91$ Kg

i. C050o )e se3*+*in )e 6i4*s 3o+ e5i0i+io )e momen2o en 0os*. C*so i a) l mome&o ress&e&e &ero de la losa D: = K**dQ2

K= 1/2*fc**I fc = 0.45*f'c =

1"5 g/cm2

fs = 0.50*f# =

1400 g/cm2 1

d

= 1$

2.5

1

+

fs n∗fc

I= 1 J /" = d=

15.5 cm

=

0.4"5

0.$55


=


1$ cm K= 1/2*fc**I =

25.1" Kg/cm2

D: = K**dQ2 =

70"90".1" Kg*cm

Momen2os A25*n2es: a) eso roo de la losa = 0.1$*2400*1.00*1.00 = eso del asfal&o

4"2 Kg/m2

= 0.05*2000*1.00*1.00 =

100 Kg/m2 ;um. 6 =

Dome&o = 1/$*;um.6 *LQ2 =

5"2 Kg/m2

77.5 LQ2

) Dome&o or carga a.

r&. 1"2 ( c ) caso  . . cv

M

cv

M

L + 0.61

=

9.75

L + 0.61

=

9.75

al &ráfco

P20∗1. 20∗ Fc

7265∗1.20∗0 . 8 =

915."2" (L<0.71)

c) Dome&o or mac&o. M I =

( 1 + I )∗ M cv

= 15/(10<"$)=

;

I =

15

L + 38

"1.25 P N "0P

sumr =

"0

d)Dome&o &o&al ac&ua&e cm

M T = M

+ M cv + M I

M T =6 6 . 5∗l 2 +( 1 . 3 0 )∗715. 323∗(l + 0 . 6 1 ) e)E6ulro de Dome&os D: = D ac&ua&e

66.50 ∗l

[o*m!

2

+( 1.30 )∗715.323∗( l +0.61 ) = 603703.13 2 0.0665∗l +0.930∗l + 0.56725 =6.0370313 2 l + 13.985 −82.25 =0 :esoledo la ecuac se &ee 6ueB l1 = l2 =

4.458 m&s

=

se desreca

,?H> E ?LRB ara +c =

+c =

12 m

12

,=

" rocas

4.47 m&s


a

l

l


l

l

+c < 2*0."0 = 2a < *l

a a = l/2

l=2*a

12.7 = l < *l l = 12.7/(1<)

Fsamos

l=

2.520 =

l=

2.500 m&s

,-de Agas=

4 2.520 m&s

 4.47 m

a= 1.25

m

5

2.59

70 l

l

l

l=

cm

2.5

2.59

Es6uema de dsosc ;e.=L/4=

2.5 m&s.

4 ramos log&udales 1

2.5 2 ".5"7 "

4 5 2.500

l

l

l

DISEÑO DE VIGAS Losa

= 0.1$*2400*2.5

=

=

10$0 Kg/m

sfal&o = 0.05*2000*2.5

=

=

250 Kg/m


Sarada = 1/7*0.1814*2400 Aga

= 0.01$2*9$00


=

=

=

97.57 Kg/m

=

141.$8 Kg/m

rr&.Aer&.= (1/5*5*2*"*2.59*(LJ2*L/10))/10 =

12."41 Kg/m

rr&.Horz.= (1/5*4*2*"*".5"7*(LJ2*L/10))/10 =

1".597 Kg/m

1/>#.&> <4,m CALCULO DE MOMENTOS. 1. D>DE,> >: E;> :>> A@B Dcm = 1/$*ga * LQ2 =

199".7$"44 Kg*m

Dcm = 1/$*ga * LQ2 =

199"7$."44 Kg*cm

Esfuerzos. f1 = Dcm/o*C& =

47.19 Kg/cm2

f2 = Dcm/o*C =

47.19 Kg/cm2 47.19

70 C&=

"0 47.19

2. D>DE,> >: L>; < %:@D < ;%L> < S:,B T = 10$0<250<97.57<12."41<1".597

=

14"2.499 Kg/m

D = 1/$*T*LQ2 =

19805.89 Kg*m

D = 1/$*T*LQ2 =

1980587.5 Kg*cm

Esfuerzos. f1 = D/+& =D/*#& =

477.10 Kg/cm2

f2 = D/+& =D/*# =

477.10 Kg/cm2

". D>DE,> >: ?:@ AA < D?>B %ac&or de carga ar&. 1.2.12 ( c ) # 1.".1.(1) ?osderado 6ue &odas las gas &raaMa or gualB %c = 0.587*; =

1.48

;=l

 = 15/ 2*10 =

0.95 N 0."0

sumr =

cv

M

( cv+ I )

= [ ( 1 + I )∗ Fc∗ M R ]∗1.20

0."



e &ala ,-2 araB

L=

10 m

;e &eeB

re de carga H20

D= D:=1/2*D=

cv

M cv

M

"9870 Kg*m

1$8$0 Kg*m

( cv+ I )

=

44119.112 Kg*m

( cv+ I )

=

4411911.2 Kg*cm

EE:D,?>, EL ,?H>  E, EL ,. a) 1/4 L de cálculo =

2.5 m Elegmos el meor

) s&aca ?L # ?L gas =

2.500 m

c) 12* & = 12 * 0.1$ =

2.17 m

2.17 m

C

=

217.000 cm '=/

' = 29

= $

1$

24 12 10.7

0.995

"8 78.00

2$."$

L.g.d.g

700

59.0

5$."$

1.2

0.47$

"0 

18.4

0.995 "0

DE,;>,>.

ITEM

ELEMENTO

?ocre&o

29*1$

Aga

lma Llea

S5m*2o+i*

A m"

% m

A=% m&

Io m#

4$7

78.00

""5"4

1"122 54$"5.$"41

1$1.82

"0

5459.4$95

115250.81 147489.17$

HH>.?1H

&;??1.#;; 1";&>".?1

Cc = C = ;um. *C /  =

5$."9$ cm

C& = H& J Cc =

18.722 cm

∑ ∑

I cdg = Io + Ado

2

cdg = Los Esfuerzos soB

A=)" m#

"28905.81 cm4

"$1&&&


∑ M = M

CM


CM LDAD

+ M

CV

+ M

;um. Dom =

(1+ I )

7"98797.1 Kg*cm

f1 = f& = ;um.Dom / cdg*C& =

"98.7$ Kg/cm2

f2 = f = ;um.Dom / cdg*C =

1128.58 Kg/cm2

ESFUER@OS LIMITES. ara el acero es&ruc&ural (J"7)B %# =

25"0 Kg/cm2

Esfuerzo dmsle (0.50 J 0.55)%# 0.50*%# =

1275 Kg/cm2

0.55*%# =

1"81.5 Kg/cm2

;e deduce 6ueB

7  11"?./?



1"81.5

Kg/cm2

>.

7  11"?./?



1275

Kg/cm2

>.

ANALISIS DE ESFUER@OS CORTANTES. AE:%??>, L ?>:E. ?L?FL> E %FE:R ?>:,E. a) or carga muer&aB g =141.$8<14"2.499= Q

cm

!"cc#$n

=

1 2

1594."9 Kg/m

∗g∗ L = 9$91.$584 Kg Tcm = 9$91.$584 Kg

) or carga a más mac&oB %c =

1.48

=

Tcm de la ala ,-2

0."

Tc = araB L =

Q

cv

!"cc#$n

1

= ∗(1 + I )∗ Fc∗Q cv∗1.20 2

Tc =

18281."5$ Kg.

17588 Kg 10 m



?>:E >L T = Tcm < Tc =

2917".219 Kg.

E;%FE:R> ?>:,E.

=

%

1.19

Q A ( Alm" )

=

41".757 Kg/cm2

57.1   adm = 0.40*%# =

1012 Kg/cm2

LF: ,E?E;: E, >C>.  = T/(0.50*1012) =

5".7$ cm2

H alma = /2 =

27.$4 cm2

27.$

28.1$

?>,E?>:E; E ?>:E.

1$

L/10

L/10

L = 10 50 ?>,E?>:E; E ?>:E 10 cm 120*50 c/"0



Planilla Puentes

Recommend Documents