Tugas Jembatan Gantung

8/16/2019 Tugas Jembatan Gantung

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SERTIN N. M. MOOY1221103

PERENCANAAN JEMBATANGANTUNG

PERENCANAAN JEMBATAN GANTUNG

NIM

Gambar 1. Penampang Memanjang Jembatan

Gambar 2. Penampang Melintang Jembatan

Diketahui data perencanaan :

● Tipe Jembatan : Deckstay tanpa pengaku

● Panjang : 43.0 m

● ebar : !."0 m

● Tinggi : "." m

● #. J. $ayu : %&0 kg'm(

● ) $ayu : *0.0 kg'cm+

● #eban hidup : ,00.0 kg'm+


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A. Plat deck kayu benua 4 / 20 cm

4

!0

Gambar 3. Penampang Melintang Plat Deck

#eban mati - 0.04 0.! %&0 - /.&4 kg'm

#eban hidup - 0.! ,00.0 - !0 kg'm

- !/.&4 kg'm

- !/.&4 kg'm4

1 # 2

,,0 ,,0

M M

Gambar 4. Model Statika perhitungan Plat Deck

-,

&

-,

!/.&4 ,.,!

&

- 4.!,0& kg.m

1 - 2 -,

!

-,

!/.&4 ,., 4.!,0&

! ,.,

- ,,.4&4 kg

# - ! ,

5 ! !

- ! ,

!/.&4 ,., 5 ! 4.!

! ,.

- 3&.!& kg

M - Mma l 4.!,0& ,,0) -

6,

b h² - ,

!0 4²

* *

Mmax ²

Mmax

M


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- &.*&4//" 7 *0.0 …

B Pe"encanaan Gela#a" Meman$an# !ayu Benua% & / '2 cm

,!

&

Gambar . Penampang Melintang Gelagar Memanjang

a) Gelagar Memanjang Tengah (Middle Beam)

#eban mati - 0.0& 0.,! %&0 - %.40&

#eban deck - 3&.!& 0.! - ,%,.4

- !00.&0&

- !00.&0& kg'm

,

1 # 2

!00 !00

Gambar . Model Statika perhitungan Gelagar Memanjang

-,

&

-,

!00.&0& !.00!

&

- ,00.404 kg.m

1 - 2 -,

5!

-,

!00.&0& !.00 5,00.404

! !.00

- ,"0.*0* kg

# - ! ,

5 ! !

kg'cm² kg'cm²

Mmax Mmax

Mmax ²

Mmax


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- ! ,

!00.&0& ! 5 ! !

- "0!.0! kg

M-

Mma l ,00.404 ,00

) -

6

,

b h

² - ,

& ,!* *

- "!.!%3/" 7 *0.0

b) Gelagar Memanjang Tepi (Side Beam)

#eban mati - 0.0& 0.,! %&0 - %.40&

#eban deck - ,,.4& 0.! - "/.4!

- **.&!&

- **.&!& kg'm

,

1 # 2

!00 !00

Gambar 1. Model Statika perhitungan Gelagar Memanjang

-,

&

-,

**.&!& !.00!

&

- 33.4,4 kg.m

1 - 2 -,

5!

-,

**.&!& !.00 533.4,4

! !.00

- "0.,!, kg

# - ! ,

5 ! !

- ! ,

**.&!& ! 5 ! !

- ,*/.0/ kg

M - Mma l 33.4,4 ,00) -

6,

b h² - ,

& ,!* *

kg'cm² kg'cm²

Mmax Mmax

Mmax ²

Mmax


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- ,/.403,3 7 *0.0

C PERENCANAAN GE(AGAR ME()NTANG *Ba$a + '00 , -0

Diketahui : - ,0.*

$8ntr89 9endutan :

0.00,

!"0

!."- 0.0, 7 0.00,

!"0

1 - ,*/ kg # - "0! kg 2 - ,*/ kg

1 # 2," ,,0 ,,0 ,"

PERENCANAAN ANGER / PENGGANTUNG

, ; 1 5 # 5 2 <

&

-,

; ,*/.0/ 5 "0!.0! 5 ,*/.0/ < !."²

&

- ergergeh

,% dengan kuat tarik : !3

a) 19 mm

● Pembebanan :

!eban mati "

=e9agar memanjang - 3 %.40& -

P9at deck - , 0.04 %&0 -

=e9agar me9intang - ;!3 ,0.*< ' 43 ->anger ;diasumsikan< - ! -

$abe9 utama - ,.4, -

kg'cm² kg'cm²

kg'm²

max - ²

Digunakan kabe9 baja dengan diameter Ø

Main cable (Kabel utama) Ø


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#erat untuk jembatan - !

brt kabe9 ;bs.deck 5 bs ge9 memanjang te

bs ge9 memanjang tepi 5 bs ge9 me9intang 5

kabe9 utama<

panjang bentang jem

- !

,.4, ; !/.&4 5 !00.&0&

,0.* 5 ! 5 ,.4,

43.0

- !0.!%/ kg'm

?ub t8ta9 - !&.!!4 5 3%.! 5 ".**%/*/ 5 ! 5 !0.

- %".3% kg'm

ain9ain - , kg'm

!eban hidup"

9 -/"0

-/"0

- ,/.44,&* kg'mP. bentang 43.0

#otal "

d 5 9 - %".3% 5 ,/.44,&*

- ,,!.&3 kg'm

tan @ - 4n -4

½ bentang

-4

½ 43.0

- 0.,&*04*",,*

α - arc tan 0.,&*04*"

- ,0."3%,& ⁰

> -P

-,,!.&3 43.0 ²

- ,303%.,/& A & !

Ʈ ->

-,303%.,/

- ,3!*!.%,0,*, kgc8s ,0."3%,&4

Maka untuk satu kabe9 :

,Ʈ - Ʈ

-,3!*!.%,0,*0*

- **3,.4""0&03, kg! !

● $apasitas beban kabe9 B ,% - !3., t8n ,Ʈ - *.*3

● Cakt8r keamanan :

n - -!3,00

- 3.4&33%& ,,Ʈ **3,.4""0&03,

²

c8s α

)ijin

kabe9


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b) Backstay (Kabel Belakang) 19 mm

30 ⁰ ,0.,/%3* ⁰

Ʈ

> - ".

,Ʈ t8ta9

$88rdinat kabe9 :

$88rdinat - ,." m

?hag - ! m

$88rdinat y -4 shag jarak ;bentang jarak<

-4 ! ,." ; 43 ,." <

43.0 ²

- 0.!*%33"

- ,." 0.!*%33"- 0.,/%""/

,."y - 0.!*%33" α - ,0.,/%3* ⁰

,Ʈ ->

-,303%.,/

- ,"0"*.33,44&& kgc8s 30

Maka untuk satu kabe9 :

,Ʈ - Ʈ

-,"0"*.33,44&&

! !

- /"!&.,*"/!44! kg 7 ) ijin - !3,00 kg

Cakt8r keamanan :

n - -!3,00

- 3.0*&4// ,,Ʈ /"!&.,*"/!44!

c) Pyln ! terhadap kntrl buckling

; **3,.4* sin ,0.,/%3* 5 ; /"!&.,**

- ,,/,.%/& 5 3/*4.0&3

- 4%3*.0*, kg

Direncanakan menggunakan ba98k bet8n dengan ukuran "0 "0

$ !"0 - !"4%.!%0"3!44

bentang ²

tan α -

c8s α

)ijin

kabe9

t8ta9

- ⁰ <

kg'cm²


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Cbr - "0 "0

- !"00

- I -

- 0.& b

- 0.& "0

- 40 cm

E -k

-""0

- ,3./"40

) tekuk -P λ

Cbr

-4%3*., ,3./"

!"00

- !/.,4&33*0*43 7 !"4%.!%0"3!44

E PERENCANAAN PNA1)

=aya Fertika9 - ! 4%3*.0*,

- %&/!.,!!!0"!, kg

Tegangan ijin tanah n8rma9 ;) ijin tanah < - 0.*0

Dic8ba untuk p8ndasi dengan ukuran ! ! , meter

#erat p8ndasi :

= - ! ! , !.4

- %.* t8n

) -P 5 6

40000

-%&/!.,!! 5 %*00

40000

- 0.4&*&03 7 0.*0

E PERENCANAAN B(! ANG!UR

Diketahui :

,." ,

!.3 t'm(

> - !.0 m

=

!.0

- /"!&.! sin 30

- 3/*4., kg

- /"!&.! c8s 30

- *",%.* kg

#erat b98k angkur :

cm²

Imin

Imin

kg'm² kg'm²

kg'cm²

kg'cm² kg'cm²

ϒ tanah

- t'm

ϒ bet8n

-

,

,


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= - ,." !.0 !.0 !.4

- ,4.4 t8n

$8eAisien stabi9itas gesek bet8n terhadap tanah - 0.3

G - ,4400 3/*4.0&

- ,0*3".%! kg

$p - tan+ ; 4" 5!

<!

- tan+ ; 4" 530

<!

- 3.0

Pp - ½ $p

- ½ ,." !.0 ² " 3.0

- %.0 kg

> - %000 5 ; ,0*3".%! 0.3 <

- ,!,%0.//",4,3 kg

?C -,!,%0.//",4,3

*",%.*

- ,.&*%&/ ,

ϒ tanah >²


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: ,! !, ,0


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5

0&

,

ax


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…………..!

kg'm

kg'm 5

kg'm

!

Mmax


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,00.404

!

²

………………..!

kg'm

kg'm 5

kg'm

!

33.4,4

!

²

Mmax


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………………..!

………………..!

., t8n

!&.!!4 kg'm

3%.! kg'm

".**%/*/ kg'm! kg'm

,.4, kg'm 5


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/*."03// kg'm

gah 5

bs hanger 5

atan

5 **.&!& 5

<

%/

kg

4"" t8n .....!

………....!


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" m

………....!

………....!

sin 30 H <

cm dengan mutu


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………….....!

………….....!

T,

>, !.0


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………….....!

Tugas Jembatan Gantung

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