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8/18/2019 Calculos Pueste Estructual Sii

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CALCULOS PUENTE TIPO BALTIMORE ANALISIS ESTRUCTURAL I

INTEGRANTES: Karen Aguilera Tania Pineda Harold Linares

PROYECTO FINAL ( PUENTE)

ACUÑA LEMUS NELSON ALEXANDER AGUILERA RODRIGUEZ KAREN LIZETH

BERMEJO BERMEJO LUIS DAVIDLINARES LUGO HAROLD ANDRES

PINEDA BERMUDEZ TANIA HANIRYS

UNIVERSIDAD COOPERATIVA DE COLOMBIAFACULTAD DE INGENIERÍA

INGENIERÍA CIVÍL ANÁLISIS ESTRUCTURAL

2!"

PROYECTO FINAL ( PUENTE)


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ACUÑA LEMUS NELSON ALEXANDER AGUILERA RODRIGUEZ KAREN LIZETHBERMEJO BERMEJO LUIS DAVIDLINARES LUGO HAROLD ANDRES

PINEDA BERMUDEZ TANIA HANIRYS

ING# SAULO ANDRES OLARTE BURITICÁ

UNIVERSIDAD COOPERATIVA DE COLOMBIAFACULTAD DE INGENIERÍA

INGENIERÍA CIVÍL ANÁLISIS ESTRUCTURAL

2!"


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O6;*.4' ?,-%

• C'&.%-, %'0 5''5.&.*'0 *:,.5'0 '6*.+'0 5%-08 -+&70 05''5,7 + &-,- ,75*.5- 5'&' +6 9-%%-, 3- 0*,35*3,- =5'&,+,&'0 %'0 9-5*',0 - %'0 53-%0 % ,'*'*.' 9-%%-#

O6;*.4'0 [email protected]'0

• D.01-, 3 3* >3 53&%- 5' 05.9.5-5.'0 0*-6%5.+-0 = 5' 363 935.'-&.*' 0*,35*3,-%

• C-%53%-, 537*' 0' % 5',,0'+,7 - 5-+- 055.: +% 3*• D*,&.-, % 0' >3 ,0.0* % 3*

PROCESO CONSTRUCTIVO

1. S%55.: +% +.01' 6-0-+' % 3* *.' B-%*.&',


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2. S 0*-6%5.,' &+.+-0 0? %'0 ,>3,.&.*'0 +% 5'53,0'#

3. D.01' SAP -,- 5-%53%-, %- +9%.: + 0*#


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4. M-*,.-%0#


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5. T'&- + &+.+-0 = 5',* + %'0 -%'0 + 6-%0'


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6. U.: + %-0 055.'0 5' ?-+.*8 0'%+-+3,- :.5-#


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7. E0*


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9. R03%*-+' 9.-%#


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REACCIONES

P/2

P/2 P/2


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∑ Fy=0

Ay+ My= P

My= P

2

∑ MM =0

− P (35 )+ Ay (70 )=0

Ay (70)= P35

Ay= P35

70

Ay= P

2


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Secciones:

∑ MB=0

P

2(5,9cm)+ ANy (5,9m)=¿

ANy (5,9 )=− P2(5,89m)

ANy=− P2 (5,9cm)

5,9

ANy=−0.5 P

ANy= AN sin 58°

AN = 0,5 P

sin58 °

∑ MN =0

P

2(5,9m )− AB (9,4cm )=0

P(5,9)2(9,4)

= AB

AB=0.313 P

AN =−0.589 P

AN

Ay

AB


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∑ MC =0

P

2(11,7)+ NUy (5,8)+ NUx(9,4)=0

P

2(11,7)+ NU 5,8 Ssin (58° )+ NU 9,4cos (5,8)

P

2(11,7)+ NU (9,89 )=0

NU =−

P

2(11,7)9,89

NU =−0,591 P

∑ MN =0

P

2(5,9)−BC (9,4 )=0

BC = P(5,9)2(9,4 )

BC =0,314 P


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NUy= NU sin (58 °)

NUx= NU cos (58)

NUx=0,591 Pcos (58)

NUx=0 ,313 p

NCx= NC cos (58)

∑ MB=0

P

2(5,9 )+ NCx (9,4 )+ NUx (9,4)=0

NC =−0,0008 Pcos(58° )

NC =−0,001 P

P

2(5,9 )+0,313 P+ NCx (9,4 )=0

NCx=−3,263 P

9,4

NCy=0,001 P∗sin (58)

NCy=−0,0008 P

NCx=−0,0008 P

N


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∑ Fy=0

P

2

− NUy+CU +COy=0

P

2−0,501+CU +COsen(58°)=0

CU =0,001 P−COsin (58°)

COy=CO sin (58)

COx=COcos (58)

∑ Fx=0

− NUx+CD+COx=0


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−(0,313 P )+CD+COcos (58 ° )=0

CD=0,313 P−CO (58°)

∑ MO=0

P

2(17,5 )−CD (9,4 )+CU (5,8 )− NUy (11,6)=0

0,001 P−COsin (58° )∗(58 )−0,501 p (11,6 )=00,313 P−COcos (58° ) (9,4 )+¿

8,75 P−¿

8,75 P−2,9422 P+4,98CO+5,8 x 10−3−4,918CO−5,812=0

0,0058 P+0,062CO=0

CO=−0,096 P

∑ Fx=0

−0,313 P+CD−0,1cos (58 ° )=0

CD=0,313 P+01cos (58° )

CD=0,36 P

∑ Fy=0

0,5 P−0,51 P+CU −0,1sin (58° )=0

CU =−0,5 P+0,51 P+01sin (58 ° )

CU =0,094 P


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∑ Mo=0

p

2(17,5 )−0,4 P (9,4 )+UVy (5,8)+UVx (9,2)=0


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8,75 P−3,76 P+UVsin28 (5,8 )+UVcos28 (9,

4,99 P+10,85UV =0

UV =−0,46 P

NODO D

∑ Fy=0

∑ Fx=0

DO=O

DE=O ,368

∑ Mc=0

P

2(11,7)+UVx (18,6 )+VOx (18,6 )=0

5,85 P−7,55 P+VOx (18,6 )=0

VOx (18,6 )=1,7 P

VOx=0,091 P

VOx=VOcos (58)

VO=0,17 P


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∑ Mv=0

P

223,3−0,36 P (24,8 )−OEy (5,8)−OEx (15,4 )=0

11,65 P−8,928 P−OEcos32 (5,8 )−OEsin32 (15,4 )=0

2,722−OE (13,08 )=0

OE=0,21 P

∑ ME=0

P

223,3−UVsin28 (11,6)−UVcos28 (18,6 )+OVcos 69(9,4)+OVsin69 (5,8 )=0

11,65 P−2,72 P−8,21 P+OV 3,37+OV 5,41=0

0,72 P=0V (8,78)


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INTEGRANTES: Karen Aguilera Tania Pineda Harold Linares0V =0,08 P

∑ MP=0

P

2(30)− EF (10,6 )+ EV (6,7)−OVx (1,2)−0Vy (12,5 )−UVy (14,2 )−UVx (8)=0

15 P− EF (10,6 )+ EV (6,7 )−0,083 P−1,983 P−3,067 P−3,249=0

6,618 P+ EV (6,7 )−10,6( EF )=0

∑ Fy=0

P

2(0,46 ) sin (28) P−0,17Ssin (69) p+ EPSIN (58)+0,12 P+ EV + EPsin (58 )=0

∑ Fx=0

EF + EPcos (58 )−0,46cos (28 ) P−0,17cos (69 ) P+ EF + EPcos (58 )−0,345=0

EV =−0,12 P− EPsin(58) EF =−0,406− EPcos(58)


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6,618 P−(−0,406− EPcos (58 ) ) (10,6 )−(0,342 P− EPsin (58 ) ) (6,7 )=0

6,618 P−3,657 P+5,617 EP+0,804+5,68 EP=0

EP=0,3

NODO E

∑ Fy=0

EV −0,3sin (58 )+0,21 sin (58 )=0

EV =0,08

NODO V

∑ Fy=0

0,46 sin (28 ) P+0,17 sin (69 ) P+0,07 P+VWsin (15 )−VPsin (65 )=0

∑ Fx=0

0,46cos (28 ) P+0,17cos (69 )+VWcos (15 )+VPcos (65 )=0

VW =−0,46cos (28 ) P−0,17 cos (69 )−VPcos (65 )cos15

VW =−0,46cos (28) P−0,17cos(69 )−0,07 P+VPsin (65)−0,376 P−VPsin(65)

sin 15


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VW =−1,453 P−3,502VP

VW =−0,484 P+0,438VP

−1,453 P−3,502VP=−0,484 P+0,438VP

0,969 P=3,064VP

VP=0,969

3,064

VP=−0,24 P

VW =−0,484 P+0,438(−0,24 )

VW =−0,58 P

NODO F

∑ Fy=0

∑ Fx=0

FG= EF

GE


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FG=0,56

∑ Mq=0

P

2(35)−0,58 Psin15 (11,7 )−0,58 Pcos15 (14,2 )+ PWsin74 (5)− PWcos74 (3,4 )− PGsin65 (5 )− PGcos

∑ Fx=0


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−0,58 Pcos (15 )+ PWcos (74 )+ PGcos (65 )+0,56 P+ Pcos (34 )=0

∑ Fy=0

−0,58 Psin (15 )+ PWsin (74)− PGsin (65 )+ Psin (34 )+ p

2=0

PG!"#$%"P&!"#"'P

P(!)"#*+

CÁLCULO DE ESFUERZO A COMPRESI$N#

! m"x c=101,5 #$cm

2

! m"x c= P

A

101,5 #$cm

2= P1,44 cm

2

P=146,16 %$


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CÁLCULO DE ESFUERZO A FLEXI$N#

! m"x c=101,5 #$

cm2

P=146,16 %$

3


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P=97,44 %$

PESO PUENTE: 300gr

CARGAaprox.: 100Kg


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CONCLUSI$N

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