perencanaan jembatan rangka baja

7/16/2019 Perencanaan Jembatan Rangka Baja

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Perencanaan JembatanSertin N. M. Mooy

NIM : 12.21.103

BAB III

DATA PERENCANAAN & PEMBEBANAN

3.1 DATA PERENCANAAN

a. Gambar Perencanaan

Gambar 3.1 Potongan Memanjang Jembatan

Gambar 3.2 Potongan Melintang Jembatan

b. Data Struktur Adapun data struktur perencanaan jembatan sebagai berikut :

- Kelas Jembatan = Kelas 1

- Panjang Jembatan = 50 m

- Lebar Lantai Kendaraan = 8 m

- Lebar Trtar = ! " 1 m

- Tipe Jembatan = #angka $aja




NIM : 12.21.103

- Jarak antar gel% &elintang = 5 m

- Jarak antar gel% &emanjang = ! m

- &utu Pr'il $aja()*+ = ,!0 &Pa

- pr'il baja = !10000 &pa

- &utu $etn ('.c+ = ,5 &Pa

- )* Tulangan Pls = !/0 &Pa

- )* Tulangan ulir = ,0 &pa

3.2 PEMBEBANAN

a. Beban Plat Lantai Kendaraan

2raianTebal $erat Jenis Panjang )ak

( m + ( m + $eb

$erat 3endiri Aspal 0%050 !!00%000 1%000 1%

$erat 3endiri $etn 0%!50 !400%000 1%000 1%

$erat Air ujan 0%050 1000%000 1%000 1%

$eban &ati (6d+

= 8,%000 Kg7m

$eban idup 9T9 adala beban gandar truk maksimum sebesar 100 k

sebesar !%0 ($&3 bagian ! alaman ! - !;+

= 100 " !%0

= !00 k

= !0000 kg

b. Beban Trotoar

2raianTebal $erat Jenis Panjang )aktr

( m + ( m + $eban

$erat 3endiri $etn 0%550 !400%000 1%000 1%,

$erat Air ujan 0%050 1000%000 1%000 1%!

$eban &ati (6d+

= 1;;/ Kg7m

$eban idup (6l+

Knstruksi trtar arus memperitungkan beban idup (6+ sebesar 5

sebesar 500 dengan 'aktr beban : !%0 ($&3 bagian ! ala

= 500 " 1 " !%0

$eban &ati (6d+

( Kg7m, +

6d *ang selanjutn*a disebut 6u

lt

$eban idup (6l+

Tu

$eban &ati (6d+

( Kg7m, +

6u1

kg7m²

6u!




NIM : 12.21.103

= 1000 kg7m

= <

= 1;;/%00 < 1000%00

= !;;/%00 Kg7m

$eban Kerb

3epanjang bagian atas lantai trtir arus diperitungkan teradap beban *ang bekerja

secara rintal (6+ sebesar = 15 k7m atau sebesar 1500%00

bagian ! alaman ! - /; +

= 1%0 " 1500%00

= 1500%00 kg

3.3 PERHITUNGAN STATIKA

a. Kondisi Pembebanan I

Gambar 3.3 Kondisi Pembebanan I

>engan pembebanan sebagai berikut :

Pu = 1500%00 Kg

= !;;/%00 Kg7m

= 8,%000 Kg7mTu = !0000 Kg

Momen Primer dan Reaksi Tumpuan

Batang A-A'

&aka 6utr

6u1

6u!

Pu

6utr

6ult




NIM : 12.21.103

A1%0

A'

&men Primer :

=

= ( 1500%00 " 0%55 + < ( !;;/%00 "

= !!1, Kg%m

#eaksi tumpuan :

# AA. = " 1%0

= !;;/%00 " 1%0

= !;;/ Kg

Batang A-B

A B

1%8 0%!0

&men Primer :

=-

1" " !%0 ² -

Tu " 0%!

1! L?

=

-

1

" 8, " ! ² -

!0000 " 0%!

1! !%00

= - /8;%////; kgm

= 1" " !%0 ² <

Tu " 1%80 ?

1! L?

= 1" 8, " ! ² <

!0000 " 1%8 ?

1! !%00

= ,5/;%///; kgm

#eaksi tumpuan :

= " !%0 " 1%0 < Tu " 0%!!%00

=8, " !%0 " 1%0 < !0000 " 0%!

!%00

= !8, Kg

=" !%0 " 1%0 < Tu " 1%8

!%00

tr

&@AA.

( Pu

" 055 + < ( 6utr " 10 " 05 +

6utr

Tu

qut

&@A$ 6u

lt

&@$A 6u

lt

# A$ 6ult

# $A

6ult




NIM : 12.21.103

=8, " !%0 " 1%0 < !0000 " 1%8

!%00

= 188, Kg

Batang B - C

B C

1%55 0%45

&men Primer :

=-

1" " !%0 ² -

Tu " 0%45

1! L?

=-

1" 8, " !%0 ² -

!0000 " 0%45

1! !%00

= - 18;%041; kgm

= 1" " !%0 ² <

Tu " 1%55 ?

1! L?

= 1" 8, " ! ² <

!0000 " 1%55 ?

1! !%00

= 5;,,%!1; kgm

#eaksi tumpuan :

=" !%0 " 1%0 < Tu " 0%45

!%00

=8, " !%0 " 1%0 < !0000 " 0%45

!%00

= 548, Kg

=" !%0 " 1%0 < Tu " 1%55

!%00

=8, " !%0 " 1%0 < !0000 " 1%55

!%00

= 1/48, Kg

Tu

qut

&@$B 6u

lt

&@B$ 6u

lt

# $B

6ult

# B$

6ult




NIM : 12.21.103

Batang C - D

C D

0%45 1%55

&men Primer :

=-

1" " ! ?

- Tu " 1%55

1! L?

=-

1" 8, " ! ? -

!0000 " 1%55

1! !

= - 5;,,%!1; kgm

= 1 " " ! ? < Tu " 0%45 ?1! L?

= 1" 8, " ! ? <

!0000 " 0%45 ?

1! ! ?

= 18;%041; kgm

#eaksi Tumpuan :

=" !%0 " 1 < Tu " 1%55

!%00

=

8, " !%0 " 1 < !0000 " 1%55

!%00

= 1/48, kgm

=" !%0 " 1 < Tu " 0%45

!%00

=8, " !%0 " 1 < !0000 " 0%45

!%00

= 548, kgm

Batang D - E

D E

0%! 1%8

&men Primer :

Tu

qut

&@B> 6u

lt

&@>B 6ult

# B>

6ult

# >B

6ult

Tu

qut




NIM : 12.21.103

=-

1" " ! ?

- Tu " 1%8

1! L?

=-

1" 8, " ! ? -

!0000 " 1%8

1! !

= - ,5/;%///; kgm

= 1" " ! ? <

Tu " 0%! ?

1! L?

= 1" 8, " ! ? <

!0000 " 0%! ?

1! ! ?

= /8;%////; kgm

#eaksi Tumpuan :

=" !%0 " 1%0 < Tu " 1%80

!%00

=8, " !%0 " 1%0 < !0000 " 1%80

!%00

= 188, kgm

=" !%0 " 1%0 < Tu " 0%!0

!%00

=8, " !%0 " 1%0 < !0000 " 0%!0

!%00

= !8, kgm

Batang E - E'

E E'

1%0

= - -= -

1500%00 " 0%55 -1

" !;;/!

= -!!1, kgm

#eaksi tumpuan :

= " 1%0

= !;;/%00 " 1%0

&@> 6u

lt

&@> 6u

lt

# >

6ult

# >

6ult

qutr

&@. (Pu " 055+ ((17!+ " 6utr " 1?+

# .

6utr




NIM : 12.21.103

= !;;/ Kg

Tabel mmen primer dan reaksi tumpuan pada kndisi pembebanan 1

Titik Jin &men Primer 3atuan #eaksi Tumpuan 3atuan

1 A. - A - kg%m - k

! A - A. !!1, kg%m !;;/ k

, A - $ -/8;%/////////; kg%m !8, k

4 $ - A ,5/;%/////////; kg%m 188, k

5 $ - B -18;%041//////; kg%m 548, k

/ B - $ 5;,,%!1//////; kg%m 1/48, k

; B - > -5;,,%!1//////; kg%m 1/48, k

8 > - B 18;%041//////; kg%m 548, k

> - -,5/;%/////////; kg%m 188, k

10 - > /8;%/////////; kg%m !8, k

11 - . -!!1, kg%m !;;/ k

1! . - - kg%m - k




NIM : 12.21.103

&men Lapangan

=48 " 0%5 -

1" !;;/%00 " 1 ²

!110/%5 Kg%m

=!881%5/!5 " 1%8 -

1" 8,%00 " 1%8 ²

!

= 1,81%,5!5 Kg%m

=,54%0 " 1%55 -

1" 8,%00 " 1%55 ²

!

&AA.

&A$

&$B




NIM : 12.21.103

= !5,1%/, Kg%m

=1801! " 0%45 -

1" 8,%00 " 0%45 ²

!

= !5,1%/, Kg%m

=

1084%4 " 0%!0 -

1

" 8,%00 " 0%!0 ²!

= 1,81%,5!5 Kg%m

=48%0 " 0%50 -

1" !;;/%00 " 1%00 ²

!

= 110/%5 Kg%m

b. Kondisi Pembebanan II

Gambar 3.4 Kondisi Pembebanan II

>engan data pembebanan sebagai berikut :Pu = 1500 Kg

= !;;/ Kg7m

= 8, Kg7m

Tu = !0000 Kg

Batang A - A'

&B>

&>

&.

6utr

6ult




NIM : 12.21.103

A1%0

A'

&men Primer :

=

= ( 1500%00 " 0%55 + < ( !;;/%00 " 1%0

= !!1, Kg%m

#eaksi tumpuan :

# AA. = " 1%0

= !;;/%00 " 1%0

= !;;/ Kg

Batang A - B

A B

0%5 1%5

&men Primer :

=

-

1

" " ! ?

- Tu " 1%5 ?

1! L?

=-

1" 8, " ! ? -

!0000 " 1%5 ?

1! ! ?

= - 55!%///; kgm

= 1" " ! ? <

Tu " 0%5 ? "

1! L?

= 1" 8, " ! ? <

!0000 " 0%5 ? "

1! ! ?= !!0!%///; kgm

#eaksi Tumpuan :

=" !%0 " 1%0 < Tu " 1%5

!%00

=8, " !%0 " 1%0 < !0000 " 1%5

!%00

qutr

&@AA.

( Pu

" 055 + < ( 6utr " 10 " 05 +

6utr

Tu

qut

&@A$ 6u

lt

&@$A 6u

lt

# A$

6ult




NIM : 12.21.103

= 158, kgm

=" !%0 " 1%0 < Tu " 0%5

!%00

=8, " !%0 " 1%0 < !0000 " 0%5

!%00

= 58, kgm

Batang B - C

B C

0%!5 1%;5

&men Primer :=

-1

" " ! ?- Tu " 1%;5 ?

1! L?

=-

1" 8, " ! ? -

!0000 " 1%;5 ?

1! ! ?

= - 4155%;1; kgm

= 1" " ! ? <

Tu " 0%!5 ? "

1! L?

= 1" 8, " ! ? <

!0000 " 0%!5 ? "

1! ! ?

= 8;4%541/; kgm

#eaksi Tumpuan :

=" !%0 " 1%0 < Tu " 1%;5

!%00

=8, " !%0 " 1%0 < !0000 " 1%;5

!%00

= 1848, kgm

=" !%0 " 1%0 < Tu " 0%!5

!%00

=8, " !%0 " 1%0 < !0000 " 0%!5

!%00

= ,48, kgm

# $A

6ult

Tu

qut

&@$B 6u

lt

&@B$ 6u

lt

# $B

6ult

# B$

6ult




NIM : 12.21.103

Batang C - D

C D

1%;5 0%!5

&men Primer :

=-

1" " !%0 ² -

Tu " 0%!5 ²

1! L?

=-

1" 8, " !%0 ² -

!0000 " 0%!5 ²

1! !%00 ²

= - 8;4%541/; kgm= 1

" " !%0 ² <Tu " 1%;5 ? "

1! L?

= 1" 8, " ! ² <

!0000 " 1%;5 ? "

1! !%00 ?

= 4155%;1; kgm

#eaksi tumpuan :

=" !%0 " 1%0 < Tu " 0%!5

!%00

=8, " !%0 " 1%0 < !0000 " 0%!5

!%00

= ,48, Kg

=" !%0 " 1%0 < Tu " 1%;5

!%00

=8, " !%0 " 1%0 < !0000 " 1%;5

!%00

= 1848, Kg

Batang D - E

D E

1%5 0%50

Tu

qut

&@B> 6u

lt

&@>B 6u

lt

# B>

6ult

# >B

6ult

Tu

qut




NIM : 12.21.103

&men Primer :

=-

1" " !%0 ² -

Tu " 0%50 ²

1! L?

=-

1" 8, " !%0 ² -

!0000 " 0%50 ²

1! !%00 ²

= - !!0!%///; kgm

= 1" " !%0 ² <

Tu " 1%50 ? "

1! L?

= 1" 8, " ! ² <

!0000 " 1%50 ? "

1! !%00 ?

= 55!%///; kgm

#eaksi tumpuan :

=" !%0 " 1%0 < Tu " 0%50

!%00

=8, " !%0 " 1%0 < !0000 " 0%50

!%00

= 58, Kg

=" !%0 " 1%0 < Tu " 1%50

!%00

=8, " !%0 " 1%0 < !0000 " 1%50

!%00

= 158, Kg

Batang E - E'

E E'

1%0

= - -

= -1500%00 " 0%55 -

1" !;;/%00

!

= -!!1, kgm

&@> 6u

lt

&@> 6u

lt

# >

6ult

# >

6ult

qutr

&@.

(Pu " 055+ ((17!+ " 6u

tr" 1?+




NIM : 12.21.103

#eaksi tumpuan :

= " 1%0

= !;;/%00 " 1%0

= !;;/ Kg

Tabel mmen primer dan reaksi tumpuan pada kndisi pembebanan !

Titik Jin &men Primer 3atuan #eaksi Tumpuan 3atuan

1 A. - A - kg%m - k

! A - A. !!1, kg%m !;;/ k

, A - $ -55!%/////////; kg%m 158, k

4 $ - A !!0!%/////////; kg%m 58, k

5 $ - B -4155%;1//////; kg%m 1848, k

/ B - $ 8;4%541//////; kg%m ,48, k

; B - > -8;4%541//////; kg%m ,48, k 8 > - B 4155%;1//////; kg%m 1848, k

> - -!!0!%/////////; kg%m 58, k

10 - > 55!%/////////; kg%m 158, k

11 - . -!!1, kg%m !;;/ k

1! . - - kg%m - k

&men Lapangan

# .

6utr




NIM : 12.21.103

=48 " 0%5 -

1" !;;/%00 " 1 ²

!

110/%5 Kg%m

=

150,5%40! " 0%5 -

1

" 8,%00 " 0%5 ²!

= 5181%8!5 Kg%m

=!008;% " 0%!5 -

1" 8,%00 " 0%!5 ²

!

= 88,%0/// Kg%m

=18;8 " 1%;5 -

1" 8,%00 " 1%;5 ²

!

= 88,%0/// Kg%m

= /,0%/ " 1%50 - 1 " 8,%00 " 1%50 ²!

= 5181%8!5 Kg%m

=48%0 " 0%50 -

1" !;;/%00 " 1%00 ²

!

= 110/%5 Kg%m

c. Kondisi Pembebanan III

&AA.

&A$

&$B

&B>

&>

&.




NIM : 12.21.103

Gambar 3. Kondisi Pembebanan III

>engan pembebanan sebagai berikut :

Pu = 1500 Kg

= !;;/ Kg7m

= 8, Kg7m

Tu = !0000 Kg

Batang A - A'

A1%0

A'

&men Primer :

=

= ( 1500%00 " 0%55 + < ( !;;/%00 " 1%0

= !!1, Kg%m

#eaksi tumpuan :

# AA. = " 1%0

= !;;/%00 " 1%0

= !;;/ Kg

Batang A - B

A B

1%1 0%0

&men Primer :

=-

1" " !%0 ² -

Tu " 0%0 ²

1! L?

=-

1" 8, " !%0 ² -

!0000 " 0%0 ²

1! !%00 ²

6utr

6ult

qutr

&@AA.

( Pu

" 055 + < ( 6utr " 10 " 05 +

6utr

Tu

qut

&@A$ 6u

lt




NIM : 12.21.103

= - 4;8!%///; kgm

= 1" " !%0 ² <

Tu " 1%10 ? "

1! L?

= 1" 8, " ! ² <

!0000 " 1%10 ? "

1! !%00 ?

= 5;;!%///; kgm

#eaksi tumpuan :

=" !%0 " 1%0 < Tu " 0%0

!%00

=8, " !%0 " 1%0 < !0000 " 0%0

!%00

= 8, Kg

=" !%0 " 1%0 < Tu " 1%10

!%00

=8, " !%0 " 1%0 < !0000 " 1%10

!%00

= 118, Kg

Batang B - C

B C

0%85 1%15

&men Primer :

=-

1" " ! ?

- Tu " 1%15 ?

1! L?

=-

1" 8, " ! ? -

!0000 " 1%15 ?

1! ! ?= - 548%!1; kgm

= 1" " ! ? <

Tu " 0%85 ? "

1! L?

= 1" 8, " ! ? <

!0000 " 0%85 ? "

1! ! ?

= 448!%041; kgm

&@$A 6u

lt

# A$

6ult

# $A

6ult

Tu

qut

&@$B 6u

lt

&@B$ 6u

lt




NIM : 12.21.103

#eaksi Tumpuan :

=" !%0 " 1%0 < Tu " 1%15

!%00

=8, " !%0 " 1%0 < !0000 " 1%15

!%00

= 1!48, kgm

=" !%0 " 1%0 < Tu " 0%85

!%00

=8, " !%0 " 1%0 < !0000 " 0%85

!%00

= 48, kgm

Batang C - D

C D

1%15 0%85

&men Primer :

=-

1" " !%0 ² -

Tu " 0%85 ²

1! L?

=-

1" 8, " !%0 ² -

!0000 " 0%85 ²

1! !%00 ²

= - 448!%041; kgm

= 1" " !%0 ² <

Tu " 1%15 ? "

1! L?

= 1" 8, " ! ² <

!0000 " 1%15 ? "

1! !%00 ?

= 548%!1; kgm#eaksi tumpuan :

=" !%0 " 1%0 < Tu " 0%85

!%00

=8, " !%0 " 1%0 < !0000 " 0%85

!%00

= 48, Kg

# $B

6ult

# B$

6ult

Tu

qut

&@B> 6u

lt

&@>B 6u

lt

# B>

6ult




NIM : 12.21.103

=" !%0 " 1%0 < Tu " 1%15

!%00

=8, " !%0 " 1%0 < !0000 " 1%15

!%00

= 1!48, Kg

Batang D - E

D E

0% 1%1

&men Primer :=

-1

" " ! ?- Tu " 1%10 ?

1! L?

=-

1" 8, " ! ? -

!0000 " 1%10 ?

1! ! ?

= - 5;;!%///; kgm

= 1" " ! ? <

Tu " 0%0 ? "

1! L?

= 1

" 8, " ! ? <

!0000 " 0%0 ? "

1! ! ?

= 4;8!%///; kgm

#eaksi Tumpuan :

=" !%0 " 1%0 < Tu " 1%10

!%00

=8, " !%0 " 1%0 < !0000 " 1%10

!%00

= 118, kgm

= " !%0 " 1%0 < Tu " 0%0

!%00

=8, " !%0 " 1%0 < !0000 " 0%0

!%00

= 8, kgm

# >B

6ult

Tu

qut

&@> 6u

lt

&@> 6u

lt

# >

6ult

# > 6ult




NIM : 12.21.103

Batang E - E'

E E'

1%0

= - -

= -1500%00 " 0%55 -

1" !;;/%00

!

= -!!1, kgm

#eaksi tumpuan :

= " 1%0= !;;/%00 " 1%0

= !;;/ Kg

Tabel mmen primer dan reaksi tumpuan pada kndisi pembebanan ,

Titik Jin &men Primer 3atuan #eaksi Tumpuan 3atu

1 A. - A - kg%m - k

! A - A. !!1, kg%m !;;/ k

, A - $ -4;8!%/////////; kg%m 8, k

4 $ - A 5;;!%/////////; kg%m 118, k

5 $ - B -548%!1//////; kg%m 1!48, k

/ B - $ 448!%041//////; kg%m 48, k

; B - > -448!%041//////; kg%m 48, k

8 > - B 548%!1//////; kg%m 1!48, k

> - -5;;!%/////////; kg%m 118, k

10 - > 4;8!%/////////; kg%m 8, k

11 - . -!!1, kg%m !;;/ k

1! . - - kg%m - k

qutr

&@.

(Pu " 055+ ((17!+ " 6u

tr" 1?+

# . 6utr




NIM : 12.21.103

&men Lapangan

= 48 " 0%5 - 1 " !;;/%00 " 1 ²!

110/%5 Kg%m

=;;8%4,;5 " 1%1 -

1" 8,%00 " 1%1 ²

!

= 5;;0%5//, Kg%m

=1,/1%5 " 0%85 -

1" 8,%00 " 0%85 ²

!

= 4;00%541, Kg%m

=8!;5 " 1%15 -

1" 8,%00 " 1%15 ²

!

= 4;00%541! Kg%m

=141/;%/ " 0%0 -

1" 8,%00 " 0%0 ²

!

= 5;;0%5//, Kg%m

=48%0 " 0%50 -

1" !;;/%00 " 1%00 ²

!

= 110/%5 Kg%m

&AA.

&A$

&$B

&B>

&>

&.




NIM : 12.21.103




NIM : 12.21.103

r Jumla

an

14,%000

;80%000

/0%000

8,%000

dengan 'aktr beban

Jumla

1;1/%000

/0%000

1;;/%000

kpa atau

an !-,!+

( Kg7m +

( Kg7m +




NIM : 12.21.103

kg7m ($&3




NIM : 12.21.103

1%0 " 0%5 +

² " 1%80

² " 1%8

²

" 0%!0

" 0%!

?




NIM : 12.21.103

² " 1%55

² " 1%55

²

" 0%45

" 0%45

?




NIM : 12.21.103

? " 0%45

? " 0%45

?

" 1%55

" 1%55




NIM : 12.21.103

? " 0%!0

? " 0%!0

?

" 1%8

" 1%8

%00 " 1?




NIM : 12.21.103




NIM : 12.21.103

- !!1,

- !415%8;5




NIM : 12.21.103

- 54;,%8;5

- !415%8;5




NIM : 12.21.103

" 0%5 +

" 0%5

" 0%5

1%5

1%5




NIM : 12.21.103

" 0%!5

" 0%!5

1%;5

1%;5




NIM : 12.21.103

" 1%;5

" 1%;5

0%!5

0%!5




NIM : 12.21.103

" 1%50

" 1%5

0%50

0%5

" 1?




NIM : 12.21.103




NIM : 12.21.103

- !!1,

- 4108%!0

- 88%,,!

- 4108%!0




NIM : 12.21.103

" 0%5 +

" 1%10

" 1%1




NIM : 12.21.103

0%0

0%

" 0%85

" 0%85

1%15

1%15




NIM : 12.21.103

" 1%15

" 1%15

0%85

0%85




NIM : 12.21.103

" 0%0

" 0%0

1%1

1%1




NIM : 12.21.103

" 1?

an




NIM : 12.21.103

- !!1,

- /58!%1!

- 41/5%1!5

- /58!%1!




NIM : 12.21.103

Tabel B#C33 Peritungan &men Kndisi Pembebanan D

Titik A $ B >

$atang AA. A$ $A $B B$ B> >B >

A%> 1 0%5 0%5 0%5 0%5 0%5 0%5

&P !!1,%000 -/8;%//; ,5/;%//; -18;%04! 5;,,%!! -5;,,%!! 18;%04! -,5/;%//;

-15!5%,,, -;/!%//;

-!!/%0 -45,%; -45,%; -!!/%0

5/%;4; 11,%45 11,%45 5/%;4;

40,%4/ 80/%, 80/%,

5/0%,!

!!/%0 11,%45

-4!%5/1 -85%1!1 -85%1!1 -4!%5/1

-0%!!; -180%454 -180%454 -0%!!;

-11;%/;/ -!,5%,5! -!,5%,5!

58%8,8

4!%5/1 !1%!80

1;%!,; ,4%4;, ,4%4;, 1;%!,;

!5%110 50%!!0 50%!!0 !5%110

-!0%8; -41%;4 -41%;410%4,

-1;%!,; -8%/18

-4%1!, -8%!4/ -8%!4/ -4%1!,

/%!;; 1!%555 1!%555 /%!;;

-4%1, -8%,85 -8%,85

!%0/

4%1!, !%0/1




NIM : 12.21.103

-!%085 -4%1/ -4%1/ -!%085




NIM : 12.21.103

1%5/ ,%1, ,%1, 1%5/

-0%1/ -1%8,, -1%8,,

0%458

!%085 1%04!

-0%/5, -1%,0/ -1%,0/ -0%/5,

0%,! 0%;85 0%;85 0%,!

-0%!1, -0%4!5 -0%4!5

0%10/

0%/5, 0%,!/

-0%180 -0%,5 -0%,5 -0%180

0%08 0%1/ 0%1/ 0%08

-0%051 -0%10! -0%10!

0%0!/

0%180 0%00

-0%04; -0%04 -0%04 -0%04;

0%0!5 0%04 0%04 0%0!5

-0%01, -0%0!5 -0%0!5

0%00/

0%04; 0%0!,

-0%01! -0%0!4 -0%0!4 -0%01!

0%00/ 0%01! 0%01! 0%00/

-0%00, -0%00/ -0%00/

0%00!

0%01! 0%00/

-0%00, -0%00/ -0%00/ -0%00,




NIM : 12.21.103

0%00! 0%00, 0%00, 0%00!

-0%001 -0%00! -0%00!

0%000

0%00, 0%00!

-0%001 -0%00! -0%00! -0%001

0%000 0%001 0%001 0%000

0%000 0%000 0%000

0%000

0%001 0%000

0%000 0%000 0%000 0%000

0%000 0%000 0%000 0%000

0%000 0%000 0%000

0%000

!!1,%000 -!!1,%000 !415%8;5 -!415%8;5 54;,%8;5 -54;,%8;5 !415%8;5 -!415%8;5




NIM : 12.21.103

> .

1

/8;%//; -!!1,%000

40,%4/

11!1%8/4

-11;%/;/

11;%/;/

-!0%8;!0%8;

-4%1,

4%1,




NIM : 12.21.103




NIM : 12.21.103

-0%1/

0%1/

-0%!1,

0%!1,

-0%051

0%051

-0%01,

0%01,

-0%00,

0%00,




NIM : 12.21.103

-0%001

0%001

0%000

0%000

0%000

0%000

!!1,%000 -!!1,%000



Tabel B#C33 Peritungan &men Kndisi Pembebanan DD

Titik A $ B >


A%> 1 0%5 0%5 0%5 0%5 0%5 0%5

&P !!1,%000 -55!%//; !!0!%//; -4155%;! 8;4%54! -8;4%54! 4155%;! -!!0!%//;

,;,%//; 18/%8,,

!0%8!, 41%/4/ 41%/4/ !0%8!,

-5%!0/ -10%411 -10%411 -5%!0/

-48/%80 -;,%/0 -;,%/0-1/!/%,4,

-!0%8!, -10%411

,%04 ;%80 ;%80 ,%04

1!0%;/ !41%5,8 !41%5,8 1!0%;/

,;/%,4 ;5!%;8; ;5!%;8;

-188%1;

-,%04 -1%5!

-!%;04 -5%408 -5%408 -!%;04

-8/%/;! -1;,%,45 -1;,%,45 -8/%/;!

/8%;1; 1,;%4,5 1,;%4,5-,4%,5

!%;04 14%85!

1;%55 ,5%10 ,5%10 1;%55

-!1%//8 -4,%,,/ -4,%,,/ -!1%//8

14%00; !8%01, !8%01,

-;%00,

-1;%55 -8%;8



;%//1 15%,!, 15%,!, ;%//1



-5%41; -10%8,4 -10%8,4 -5%41;

,%105 /%!10 /%!10

-1%55,

-;%//1 -,%8,1

!%,1! 4%/!4 4%/!4 !%,1!

-1%,54 -!%;0 -!%;0 -1%,54

0%;!; 1%45, 1%45,

-0%,/,

-!%,1! -1%15/

0%/!8 1%!55 1%!55 0%/!8

-0%,, -0%/;; -0%/;; -0%,,

0%1;5 0%,51 0%,51

-0%088

-0%/!8 -0%,14

0%1/, 0%,!/ 0%,!/ 0%1/,

-0%085 -0%1/ -0%1/ -0%085

0%04, 0%08/ 0%08/

-0%0!!

-0%1/, -0%08!

0%04! 0%08, 0%08, 0%04!

-0%0!1 -0%04! -0%04! -0%0!1

0%011 0%0!1 0%0!1

-0%005

-0%04! -0%0!1

0%010 0%0!1 0%0!1 0%010



-0%005 -0%011 -0%011 -0%005

0%00, 0%005 0%005

-0%001

-0%010 -0%005

0%00, 0%005 0%005 0%00,

-0%001 -0%00, -0%00, -0%001

0%001 0%001 0%001

0%000

-0%00, -0%001

0%001 0%001 0%001 0%001

0%000 -0%001 -0%001 0%000

0%000 0%000 0%000

0%000

!!1,%000 -!!1,%00 4108%!0 -4108%!0 88%,, -88%,, 4108%!0 -4108%!0



> .

1

55!%//; -!!1,%000

-48/%80-,!5!%/8;

,;/%,4

-,;/%,4

/8%;1;-/8%;1;

14%00;

-14%00;



,%105

-,%105

0%;!;

-0%;!;

0%1;5

-0%1;5

0%04,

-0%04,

0%011

-0%011



0%00,

-0%00,

0%001

-0%001

0%000

0%000

!!1,%000 -!!1,%000




NIM : 12.21.103

Tabel B#C33 Peritungan &men Kndisi Pembebanan DDD

Titik A $ B >


A%> 1 0%5 0%5 0%5 0%5 0%5 0%5

&P !!1,%000 -4;8!%//; 5;;!%//; -548%!! 448!%04! -448!%04! 548%!! -5;;!%//;

!5/%//; 1!84%8,,

-!;;%,0! -554%/04 -554%/04 -!;;%,0!

/%,!/ 1,8%/51 1,8%/51 /%,!/

-/1%!,8 -1!!%4;5 -1!!%4;5

-1!54%!15

!;;%,0! 1,8%/51

-51%4 -10,%88 -10,%88 -51%4

!8%,08 5/%/1/ 5/%/1/ !8%,08

,0/%4;; /1!%5, /1!%5,

-15,%!,8

51%4 !5%;

-1,%5;/ -!;%15, -!;%15, -1,%5;/

-;,%!!5 -14/%450 -14/%450 -;,%!!5

5/%/1/ 11,%!,! 11,%!,!-!8%,08

1,%5;/ /%;88

1/%/0 ,,%!18 ,,%!18 1/%/0

-18%,0/ -,/%/1, -,/%/1, -18%,0/

11%/54 !,%,0; !,%,0;

-5%8!;

-1/%/0 -8%,05




NIM : 12.21.103

/%/5, 1,%,05 1,%,05 /%/5,




NIM : 12.21.103

-4%5;; -%15, -%15, -4%5;;

!%/01 5%!0! 5%!0!

-1%,00

-/%/5, -,%,!/

1%;/ ,%51 ,%51 1%;/

-1%144 -!%!88 -!%!88 -1%144

0%/11 1%!!! 1%!!!

-0%,0/

-1%;/ -0%88

0%5,, 1%0// 1%0// 0%5,,

-0%!8/ -0%5;! -0%5;! -0%!8/

0%148 0%!/ 0%!/

-0%0;4

-0%5,, -0%!/;

0%1,8 0%!;/ 0%!;/ 0%1,8

-0%0;! -0%14, -0%14, -0%0;!

0%0,/ 0%0;, 0%0;,

-0%018

-0%1,8 -0%0/

0%0,5 0%0;0 0%0;0 0%0,5

-0%018 -0%0,/ -0%0,/ -0%018

0%00 0%018 0%018

-0%005

-0%0,5 -0%018

0%00 0%018 0%018 0%00




NIM : 12.21.103

-0%004 -0%00 -0%00 -0%004

0%00! 0%004 0%004

-0%001

-0%00 -0%004

0%00! 0%004 0%004 0%00!

-0%001 -0%00! -0%00! -0%001

0%001 0%001 0%001

0%000

-0%00! -0%001

0%001 0%001 0%001 0%001

0%000 -0%001 -0%001 0%000

0%000 0%000 0%000

0%000

!!1,%000 -!!1,%00 /58!%1! -/58!%1, 41/5%1!5 -41/5%1!5 /58!%1! -/58!%1,




NIM : 12.21.103

> .

1

4;8!%//; -!!1,%000

-/1%!,8

-!508%4!

,0/%4;;

-,0/%4;;

5/%/1/-5/%/1/

11%/54

-11%/54




NIM : 12.21.103




NIM : 12.21.103

!%/01

-!%/01

0%/11

-0%/11

0%148

-0%148

0%0,/

-0%0,/

0%00

-0%00




NIM : 12.21.103

0%00!

-0%00!

0%001

-0%001

0%000

0%000

!!1,%000 -!!1,%000



Perencanaan Jembatan

Kondisi Pembebanan I

!!1, !415%8;500014,/ 54;,%8;4!8! !415%8;5000018

A' 1 A ! B ! C ! D

!;;/ !8, 188,%0 548,%00 1/48,%0 1/48, 548, 188,

110/%5 110/%5 1!0;%,;5 1!0;%,;5 !;,/%,;5 !;,/%,;5 1!0;%,;5

!!1, 1!0;%,;5 1!0;%,;5 !;,/%,;5 !;,/%,;5 1!0;%,;5 1!0;%,;5 110/%5

48 !881%5/!5 1084%4 ,54%00 1801!%0 1801!%00 ,54%0 1084%44

;8;0%5/!4!8! !,0,8%4,;50018 ,/0!4%00 !,0,8%4,;500054




Kondisi Pembebanan II

!!1, 4108%!0 88%,,!85,04 4108%!0

A' 1 A ! B ! C ! D

!;;/ 158, 58,%0 1848,%00 ,48,%0 ,48, 1848, 58,

110/%5 110/%5 !054%08!1 !054%08!1 44%1//4, 44%1//4, !054%08!1

!!1, !054%08!1 !054%08!1 44%1//4, 44%1//4, !054%08!1 !054%08!1 110/%5

48 150,5%4018 /,0%/ !008;%, 18;8%1 18;8%0; !008;% /,0%/0

!00!4%401;85, !;018%5!/;851;4 ,;5/%14 !;018%5!/;8555!




Kondisi Pembebanan III

!!1, /58!%1! 41/5%1!50001;0, /58!%1!

A' 1 A ! B ! C ! D

!;;/ 8, 118,%0 1!48,%00 48,%0 48, 1!48, 118,

110/%5 110/%5 ,!1%0/!5 ,!1%0/!5 !08!%5/!5 !08!%5/!5 ,!1%0/!5

!!1, ,!1%0/!5 ,!1%0/!5 !08!%5/!5 !08!%5/!5 ,!1%0/!5 ,!1%0/!5 110/%5

48 ;;8%4,;5 141/;%/ 1,/1%50 8!;4%5 8!;4%50 1,/1%5 141/;%5/

1!;8;%4,;5001; !;85%0/!45;4 1/54%00 !;85%0/!48;!



Perencanaan JembatanSertin N. M. Mo

NIM : 12.21.1

asil Peritungan &men

Tumpuan LapanganKndisi 1 Kndisi ! Kndisi , &aksimum

< - < - < - < -

1 A !!1, !!1, !!1, !!1,

! $ !415%88 4108%!0 /58!%1! /58!%1!

, B 54;,%8;5 88%,,! 41/5%1!5 54;,%8;

4 > !415%8;5 4108%!0 /58!%1! /58!%1!

5 !!1, !!1, !!1, !!1,

; AA. 110/%5 110/%5 110/%5 110/%5

8 A$ 1,81%,5!5 5181%8!5 5;;0%5//, 5;;0%5//,

$B !5,1%/, 88,%0/// 4;00%541, 4;00%541,

10 B> !5,! 88,%0/// 4;00%541! 4;00%541!

11 > 1,81%,5 5181%8!5 5;;0%5//, 5;;0%5//,

1! . 110/%5 110/%5 110/%5 110/%5

maksimum /58!%1!5 Tumpuan

Pembebanan ,

&aksimum 5;;0%5//, Lapangan

Pembebanan ,




NIM : 12.21.103

BAB I!

PERHITUNGAN PENU"ANGAN P"AT

#.1 P"AT TUMPUAN

>ari asil peritungan berdasarkan pembebanan D dan DD diperle mmen tumpuan terbesar :

&ma" Tumpuan = /58!%1!5 Kg%m

>igunakan tulangan dengan diameter > = 1/ mm tebal selimut betn = 40

= !50 mm

d = - tebal selimut - E >

= !50 - 40 - 1" 1/

!

= !0! mm

&u = /58!%1!5 Kg%m&men nminal (&n+ :

&n=

&u=

/58!%1!5=

/5%8!1!5 " 10 /

Ø 0%8 0%8

= 8!!;/5/!%44/8 mm

Ke'isien Taan (#n+ :

#n = &n=

8!!;/5/!%44/;;

1000%00 " !0! !

= !%01/,84;!4

Perbandingan tegangan (m+ :

m = '*=

!/0

0%85 '.c 0%85 " ,5

= 8%;,458

F = 0%85 ( 0%008 ( '.c - ,0 ++

= 0%85 ( 0%008 ,5 - ,0 ++

= 0%81

#asi penulangan keseimbangan7rasi tulangan *ang memberikan kndisi regangan

:=

0%85 " F "'.c /00

'* /00 < '*

= 0%85 " 0%81 ",5 /00

!/0 /00 < !/0

= 0%0/4//!,4

b%d²

*ang seimbang (G b+G

b

G b




NIM : 12.21.103

= 0%;5 "

= 0%;5 " 0%0/4//!,4

= 0%0484/;5;/

=

1%4

=

1%4

'* !/0

= 0%005,84/154

#asi penulangan perlu 7 rasi tulangan tarik *ang memberikan kndisi regangan pada suatu

=1

1 - 1 -! Rn m

m !"

= 11 - 1 -

! " !%01/,84; " 8%;,458

8%;,458 !/0

= 0%0080,;/!//0%0080,;/!// H = 0%005,84/ maka dipakai G = 0%0080,;/

Luas penampang tulangan tarik *ang dibutukan (As perlu + :

As perlu = " b " d

= 0%0080,;/, " 1000%00 " !0!

= 1/!,%/0058 mm ²

=1

" π " 1/ !4

=

1

"

22

" 1/ !4 ;

= !01%14!85; mm?

Jumla tulangan (n+ =As perlu

=1/!,%/00/

= 8%0;18;; ≈ $!01%14!8/

Jarak tulangan (s+ =1000

=1000

= 111%11111 mm ≈ 1%%n

As ada = " 1 " π " 1/ !4

= 1808%/4

Tulangan Bagi

As bagi = !0I " As perlu

= !0I " 1/!,%/00581

= ,!4%;!011/

Gma"

G b

$atasan rasi penulangan minimum (Gmin

+ :

Gmin

penampang plat (G perlu

+ :

G perlu

G perlu

= Gmin

G

As ɸ 1/

As ɸ 1/

&aka dipakai tulangan ɸ 1/ - 100

mm²




NIM : 12.21.103

=1

" π " 10 !4

=1

"22

" 10 !4 ;

= ;8%5;14!8/ mm?


=,!4%;!01!

= 4%1,!8015 ≈ ;8%5;14!


=1000

= !00 mmn 5

As ada = 5 "1

" π " 10!

4

= ,!%5 mm?

#.2 PENU"ANGAN P"AT "APANGAN

>ari asil peritungan berdasarkan pembebanan D dan DD diperle mmen tumpuan terbesar :

&ma" Lapangan = 5;;0%5//!5 Kg%m

>igunakan tulangan dengan diameter > = 1/ mm tebal selimut betn = 40

= !50 mm

d = - tebal selimut - E >

= !50 - 40 - 1

" 1/!

= !0! mm

&u = 5;;0%5//!5 Kg%m

&men nminal (&n+ :

&n=

&u=

5;;0%5//!5=

5;%;05//!5 " 10 /

Ø 0%8 0%8

= ;!1,!0;8%1!;,4! mm

Ke'isien Taan (#n+ :

#n = &n = ;!1,!0;8%1!;,4!1000%00 " !0! !

= 1%;/;;/;80/

Perbandingan tegangan (m+ :

m = '*=

!/0

0%85 '.c 0%85 " ,5

= 8%;,458

As ɸ 10

As ɸ 1/

&aka dipakai tulangan ɸ 10 - !00

b%d²




NIM : 12.21.103

F = 0%85 ( 0%008 ( '.c - ,0 ++

= 0%85 ( 0%008 ( ,5 - ,0 ++

= 0%81

#asi penulangan keseimbangan7rasi tulangan *ang memberikan kndisi regangan

:

=0%85 " F "

'.c /00

'* /00 < '*

= 0%85 " 0%81 ",5 /00

!/0 /00 < !/0

= 0%0/4//!,4

= 0%;5 "

= 0%;5 " 0%0/4//!,4= 0%0484/;5;/

=1%4

=1%4

'* !/0

= 0%005,84/154

#asi penulangan perlu 7 rasi tulangan tarik *ang memberikan kndisi regangan pada suatu

=

1

1 - 1 -

! Rn m

m !"

= 11 - 1 -

! " 1%;/;;/8 " 8%;,458

8%;,458 !/0

= 0%00;014054

0%00;014054 H = 0%005,84/ maka dipakai G = 0%00;0141

Luas penampang tulangan tarik *ang dibutukan (As perlu + :

As perlu = " b " d

= 0%00;0141 " 1000%00 " !0!

= 141/%84;!/ mm ²

=1

" π " 1/ !4

=1

"22

" 1/ !4 ;

= !01%14!85; mm?

As perlu 141/%84;,

*ang seimbang (G b+

G b

G b

Gma"

G b

$atasan rasi penulangan minimum (Gmin

+ :

Gmin

penampang plat (G perlu

+ :

G perlu

G perlu

= Gmin

G

As ɸ 1/




NIM : 12.21.103

!01%14!8/

%


=1000

= 1!5 mm ≈ 1%%n 8

As ada = 8 "1

" π " 1/!

4

= 1/0;%/8

Tulangan Bagi

As bagi = !0I " As perlu

= !0I " 141/%84;!/,

= !8,%,/45,

=1

" π " 10 !

4

=1

"22

" 10 !4 ;

= ;8%5;14!8/ mm?


=!8,%,/45

= ,%/0/5!0, ≈ #;8%5;14!


=1000

= !50 mm ≈ 2%%n 4

As ada = 4 "1

" π " 10!

4

= ,14 mm?

As ɸ 1/

&aka dipakai tulangan ɸ 1/ - 100

mm²

As ɸ 10

As ɸ 1/

&aka dipakai tulangan ɸ 10 - !00




NIM : 12.21.103




NIM : 12.21.103

mm




NIM : 12.21.103




NIM : 12.21.103

mm




NIM : 12.21.103



BAB !

PERENCANAAN GE"AGAR

.1 PERATAAN BEBAN

a. Beban Tie A

1 ! 0%5

0%5 4%0 0%5

1 =1

" 0%5 " 0%5 = 0%1!5!

! =1

" 4%0 " 0%5 = 1%0!



#a = #b = 1 < !

= 0%1!5 < 1%0

= 1%1!5



= ! #a - (1 " (1

" 0%5 <1

4%0 ++ <, !

= ! " 1%1!5 - ( 0%1!5 (0%5

<4%0

++ <, !

= 0%;1///;

=1

" "8

=1

" " 5%0!

8

= ,%1!5

&1 = &!

0%;1//; = ,%1!5

= 0%,1,,,,,

b. Perataan beban Tie B

1 ! 1%0

1%0 ,%0 1%0

1 =1

" 1%0 " 1%0 = 0%5!

! =1

" ,%0 " 1%0 = 1%5!

#a = #b = 1 < !

= 0%5 < 1%5

= !%000

= ! #a - (1 " (1

" 1%0 <1

,%0 ++ <, !

= ! " !%000 - ( 0%5 (1%0

<,%0

++ <, !

= 1%58,,,,,

=1

" "8

1 !

&1

&! l²

&1

&! l²



8%

= ,%1!5

&1 = &!

1%58,,,, = ,%1!5

= 0%/!////;

c. Perataan Beban Tie C

=1

" 0%5 " 0%5

0%5!

= 0%1!5

#a = #b = = 0%1!5

1%0

&1 = ( #a "1

+ - ( "1

"1

! , !

= ( 0%1!5 "1

+ - ( 0%1!5 "1

"1

! , !

= 0%041////;

=1

" "8

=1

" " 1%0!

8

= 0%1!5

&1 = &!

0%041///; = 0%1!5

= 0%,,,,,,,

d. Perataan Beban Tie D

=1

" 1%0 " 1%0

1%0!

= 0%5

#a = #b = = 0%5

!%0

&1 = ( #a "!

+ - ( "1

"1

! , !

&! l²



= ( 0%5 "!

+ - ( 0%5 "1

"1

! , !

= 0%,,,,,,,,



=1

" "8

=1

" " !%0!

8

= 0%5

&1 = &!

0%,,,,,,, = 0%5

= 0%//////;

.2 PERENCANAAN GE"AGAR MEMAN(ANG

>iketaui data sebagai berikut :

Jarak antar gelagar memanjang = ! m

Jarak antar gelagar melintang = 5 m

a. Per!itungan Pembebanan "Beban #ati$

Akibat berat lantai trtar (untuk gelagar tepi+

= Perataan beban A " 6 plat trtar (beban mati trtar+

= 0%,1,,,,, " 1;;/

= 55/%48 kg7m

Akibat berat lantai kendaraan (untuk gelagar tenga+

= Perataan beban $ " 6 plat lantai kendaraan

0%/!////; " 8, " !

1!,!%0!/; kg7m

b. Per!itungan Pembebanan "Beban %idu &D&$

$eban > menjadi penentuan dalam peritungan gelagar memanjang bentang sedang sampai benta

panjang dan lebar melintang lajur kendaraan *akni !%;5 m ($uku $&3

L = 50 m H L = ,0 m maka

6 = 8%00 " 0%5 <15

L

= 8%00 " 0%5 <15

50%00

= /%400 Kpa

= /40%00

&uatan terbagi rata 'aktr beban = !

6 = /40%00 " !

= 1!80 kg7m

&! l²

6u

6u

kg7m²



$eban terpusat P *ang ditempatkan tegak lurus ara lalu lintas pada jembatan adala =

k7m = 4400 kg7m dengan 'aktr beban = ! &aka beban garis

sepanjang gelagar adala :



= 4400 " !

= 8800 kg

>iketaui L = 50 m maka nilai >LA 0%4 ($&3 bag ! : ! - !+

k = 1 < >LA

= 1 < 0%4

= 1%4

Perbandingan beban idup gelagar :

Melagar Tepi

= beban idup trtar " Perataan beban tipe A " 'aktr

= 500 " 0%,1,,,,, " !

= ,1,%,,,,, kg7m

Melagar Tenga

=

6

" ! " Perataan $eban Tipe $!%;5

=1!80

" ! " 0%/!////;!%;5

= 58,%,/; kg7m

=P

"!%0

<!%0

!%;5 ! !

=8800

"!%0

<!%0

!%;5 ! !

= /400 kg

c. Per!itungan Statika

&erupakan peritungan mmen *ang terjadi ditenga gelagar memanjang *akni :

Megalar tepi

- Akibat beban mati

= beban mati akibat berat lantai kendaraan untuk gelagar tepi

= 55/%48 kg7m

#a = #b =A $

=

5%0=

1

Pu

)aktr beban dinamis 7 ke'isien kejut untuk bentang L ≤ 50 m maka nilai >LA

6u

6u

Pu

6u

6u

²



8

=1

" 55/%48 " 5%0!

8

= 1;,%0 kg%m

u u



- Akibat beban idup

= beban idup trtar untuk gelagar tepi

= ,1,%,,,,, kg7m

#a = #b =

A $

=

5%0=

=1

" "8

=1

" ,1,%,,,,, " 5%0!

8

= ;%! kg%m

Megalar Tenga

- Akibat $eban mati

= beban mati akibat berat lantai kendaraan untuk gelagar tenga

= 1!,!%0!/; kg7m

#a = #b =A $

=

5%0=

=1

" "8

=1

" 1!,!%0!/; " 5%0!

8

= ,850%1 kg%m

- Akibat beban idup

Pu = $eban garis idup gelagar tenga = /400 kg

6u = $eban merata idup gelagar tenga = 58,%,/; kg7m

Pu

#a =1

Pu <A $ !

1

6u

6u

&u

6u l ²

6u

6u

&u

6u l ²

6u



5%0!

= 4/58%4!4! kg

=1

" " <1

" " l8 4

=1

" 58,%,/; " 5%0 ² <1

" /400

8 4

= 8!,%0,0, kg%m

>engan demikian diperle mmen ttal :

= 1;,%0 < ;%!

= !;18%! kg%m

= ,850%1 < 8!,%0,0,

= 1,/;,%1 kg%m

d. Perencanaan dimensi gelagar memanang

>igunakan pr'il N) ,0 " ,00 " 10 " 1/

M = 10/%;/ kg7m

A = 1,/

D" = ,8;00

t' D* = ;!10

i" = 1/%8; cm

tO i* = ;%!8 cm

r = 8%1; cm

3" = 185

3* = 481

b = ,00 mm

b = ,0 mm

tO = 10 mm

t' = 1/ mm

= ,!00 (tegangan ijin baja+

&u =1

" M " " 'aktr beban baja8

=1

" 10/%;/ " 5%0 ² x 1%18

= ,//%8;5 kg%m

&u ttal = ,//%8;5 < 1,/;,%1

= 14040%1 kg%m

&u

6u l ² P

u

2ntuk gelagar tepi &u1

2ntuk gelagar tenga &u!

cm²

cm⁴

cm⁴

cm³

cm³

kg7cm²

l ²



= 1404010%11,/,/4

S)arat *+,an *r/0

≥ &u

>imana :

ɸ = )aktr resistensi = 0% (untuk balk lentur+

&n = Kekuatan mmen nminal

&u = &men beban la*an ter'aktr

ɸ&n



=

= ɸ " 1%1! " " '*

>imana :

&p = Kekuatan mmen plastis

1%1! = ke'isien penampang plastis untuk pr'il N)

'* = ,!0 &pa = ,!00

maka

= 0.9 " 1%1! " 185 " ,!00

= /40!81/ kgcm

3eingga diperle

≥ &u

/40!81/ kgcm > 1404010%11,/,/4 kgcm K

Kntrl Plat $adan

/%,/

= /%,/!10000

'* ,!0

= 1/!%!/5!

=

,0= ,

tO 10

>engan demikian

→

Q /%,/

tO '*

→ , Q 1/!%!/5! seingga tidak perlu pengaku

Kntrl Meser

Ru gelagar tenga = ,080%1 < 4/58%4!4! < " 10/%;/

= 80,!%1 kg

kn = 5 < 5 dikarenakan tidak ada pengaku maka kn =

1%10kn "

= 1%105 " !10000

'* ,!0

= /,%01041/

→

Q 1%10kn "

tO '*

→ , Q /,%01041/ K

Rn = 0%/ " '* " AO

= 0%/ " ,!00 " (( , - ! " 1%/ + " 1%0 +

= /8;,/ kg

ɸ &n ɸ &p

3 #

kg7cm²

ɸ &p

ɸ &n

(½

( a 7 +²



Ru Q ɸ Vn

80,!%1 kg Q 0% " /8;,/

80,!%1 kg Q /18/!%4 K

Kntrl Lendutan

=1

" L (>engan L = 5 m atau L = 500 m +!40

=1

" 500!40

= !%08,,,,, cm

= <,84 % % D" 48 % % D"

=

5 " ( 1!%,!0!//; < 5%8,,/; < 1%0/;/ + " 500 ⁴

,84 " !%1 " " ,8;00

/400 " 500 ³

48 " !%1 " 10⁶ " ,8;00

= 0%1!4;/, < 0%!050;;;

= 0%,;554 cm

Q

0%,;554 cm Q !%08,,,,,, cm K

e. Per!itungan S!ear Connector

→ L = 500 m

Q1

" L4

Q1

" 5004

Q 1!5 cm

Q 3 (dengan 3 =jarak antar gelagar memanjang+

Q !00 cm

Q1

" 3 kiri <1

" 3 kanan! !

Q 1" !00 <

1" !00

! !

Q !00 cm

1!5 cm

s (mdulus elastisitas baja+ = !10000 &pa = !%1 "

' ijin

' ada

5 % u % L⁴ P % L³

10⁶

' ada

' ijin

Peritungan b e''

be''

be''

b e''

3eingga diambil be''

sebesar =



c (mdulus elastisitas betn+ = 4;00 '.c

= 4;00 ,5

= !;805%5;5 &Pa

n =s

=!10000

= ;%55!44!,c !;805%5;5

=1!5

= 1/%550,; cmn ;%55!44!,

b e''



Peritungan Ma*a Meser rintal (R+

- B ma" = ga*a geser *ang disumbangkan le betn

= 0%85 " '.c " Ac

= 0%85 " ,5 " ( 1/5%50,; " !50 +

= 1!,0;/

- T ma" = ga*a geser *ang disumbangkan le pr'il baja

= As " '*

= 1,/00 " ,!0

= 4,5!000

Karena B ma" Q T ma" maka dipakai R = B ma" = 1!,

Peritungan Jumla 3tud

1/ mm maka :

Asc = !01 mm?

= 100 cm

'* = !/0 &pa

c = 4;00 '.c

= 4;00 ,5

= !;805%5;5 &pa

n = Kekuatan geser 1 stud

= 0%5 " Asc " '.c " c " rs Q Asc "

= 0%5 " !01 " ,5 " !;805%5;5 " 1 Q !01 "

= 14,%0/ H 5!!/0

Karena n H Asc " 'u maka *ang digunakan adala nilai Asc " 'u

Jumla stud (n+ =R

=1!,0;/

Asc " 'u 5!!/0

= !,%554840; ≈ !4 bua

Jarak antar stud *ang dipasang memanjang *akni :

>ipakai stud dengan Ø



3 =L

=500

= !0%8, cmn !4

8 t' = 8 " 1%/

= 1!%8 cm

Karena 3 %= !0%8 cm H 8 % t' = 1!%8 cm maka dipakai jarak antar

n =L

=500

= 50 bua3 10



( ! "1

" 4%0 + S4

( 1%0 "4%0

+ S4

( ! "1

" ,%0 + S4

( 1%5 ",%0

+ S4



" 1%0 +

" 1%0 +

" !%0 +



" !%0 +



g

bag ! 1! : !-!1+



44

*ang bekerja pada



= 0%4

beban

1" " 5%0

!

1" 55/%48 " 5

!

1,1%! kg

6u



1" " 5%0

!

1" ,1,%,,,,, " 5

!

;8,%,,,,, kg

1" " 5%0

!

1" 1!,!%0!/; " 5

!

,080%0//; kg

6u " l

6u

6u



% %

" 5%0



" 5 " 1%1 +

5



<

10? kg7cm?



;/

'u

!/0



stud = 10 cm



.2 PERENCANAAN GE"AGAR ME"INTANG

>iketaui data sebagai berikut :

Jarak antar gelagar memanjang = ! m

Jarak antar gelagar melintang = 5 m

a. Per!itungan Pembebanan

Akibat beban mati (lantai Kendaraan+

= Perataan beban > " 6u lantai kendaraan " !

= 0%//////; " 8, " !

= 1,10%///; kg7m

Akibat beban mati (trtar+

= Perataan beban B " 6u trtar " !

= 0%,,,,,,, " 1;;/ " !

= 1184 kg7m

$erat gelagar memanjang < plat < aspal (N) ,0 " ,00+ 'aktr beban baja =

M = 10/%;/ kg7m

Pu = M " l " 'aktr beban baja

= 10/%;/ " 5 " 1%1

= 58;%18 kg

Pu tenga (ttal+ =58;%18

< ,080%0//;!

= ,,;,%/5//; kg

Pu tepi (ttal+ =58;%18

< 1,1%!!

= 1/84%; kg

)aktr beban dinamik 7 ke'isien kejut

adala 40 I% &aka dengan demikian untuk jembatan bentang (L+ =

>LA = 40 I = 0%4

k = 1 < >LA

= 1 < 0%4

= 1%4

$eban idup terbagi rata

L = 50 m H L = ,0 m maka :

6 = 8%00 " 0%5 <15

L

15

6u

6u

3esuai gra'ik 'aktr beban dinamis ( $&3 bag ! : ! - ! + diperle nilai >LA untuk bentang



% %50%00

= /%400 Kpa

= /40%00 kg7m²



6" (Perataan beban tipe > " ! + " ! "

!%;5

=/40%00

" ( 0%///////; " ! + " ! " 100I!%;5

= /!0%/0/0/ kg7m

6" (Perataan beban tipe > " ! + " ! "

!%;5

=/40%00

" ( 0%///////; " ! + " ! " 50I!%;5

= ,10%,0,0, kg7m

$eban idup (garis+

P = 44 k7m

= 4400 kg7m

P

" k " ! " 100I!%;5

=4400

" 1%4 " ! " 100I!%;5

= 4480 kg7m

P" k " ! " 50I

!%;5

=4400

" 1%4 " ! " 50I!%;5

= !!40 kg7m

$eban idup 9>9 dengan 'aktr beban = ! lalu lintas rencana arus mempu

<

= /!0%/0/0/ < 4480

= 5100%/0/1 kg7m

<

= ,10%,0,0, < !!40

= !550%,0, kg7m

$eban idup terpusat *ang diterima gelagar memanjang

P tepi =6u " l

!

= ( 500 " perataan beban tipe A " jarak e'ekti' +

!

= ( 500 " 0%,1,,,,, " ! + " 5

!%00

= ;8,%,, kg

6100I

=

650I

=

P 100I =

P50I

=

>100I

= 6100I

P100I

>50I

= 650I

P50I



p" ( Perataan beban tipe $ " ! + " k "

!%;5P

tenga =



= /40%00" ( 0%/!/////; " ! + " 1%4 " ! <

1

!%;5 !

= 401/%;1;/ kg

$eban truck 9T9 (beban gandar+

$eban diambil sebesar (T+ = 10 tn dengan 'aktr beban = ! dan le

Tu = 10 " !

= !0 tn

= !0000 kg

$eban idup trtar dengan 'aktr beban = !

6 = 5 kPa

= 500 " ! " ( perataan beban tipe B " ! +

= 500 " ! " ( 0%,,,,,,, " ! +

= ///%////; kg7m

b. Per!itungan Statika

&erupakan peritungan mmen *ang terjadi ditenga-tenga bentang gelagar melintang

&men akibat berat lantai kendaraan dan berat lantai trtar

1184%0 kg7m

1,10%////; kg7m

1%0 8%0

#a = - ( 1184%0 " 1%0 + < ( 1,10%////; " 4 +

= 4058%///; kg

= #a " 5 - " 1%0 " 4%5 -

= 4058%///; " 5 - 1184%0 " 1%0 " 4%5 - 1,10%

= 4480 kg7m

&men akibat pembebanan gelagar memanjang (beban mati+

1/84%; kg ,,;,%/5/; kg ,,;,%/5/; kg ,,;,%/5//; kg ,,;,%/5///; kg

1%0 8%0

&u1

6u 1

6u !



#a = 1/84%; " ! < ,,;,%/5/; " /

!

= 11805%;/ kg

= - 1/84%; " 5 < 11805%;/ " 4 - ,,;,%/5//;

!

= 1855;%15 kgm

&u!



&men akibat beban idup terpusat *ang diterima gelagar memanjang

;8,%,, kg 401/%;1;/ kg 401/%;1;/ kg 401/%;1;58 kg 401/%;1;5;/ kg

1%0 8%0

#a = ;8,%,,,,, " ! < 401/%;1;/ " /

!

= 1!8,,%48/ kg

= - ;8,%,,,,, " 5 < 1!8,,%48/ " 4 - 401/%;1;58

!

= !,,1; kgm

&men akibat beban idup 9>9

!550%,0,0, kg7m 5100%/0/1 kg7m !550%,0,0,

1%0 0%;5 /%5 0%;5

#a = !550%, " 0%;5 < 5100%/0/1 " ,%!5

= 1848%; kg

= 1848%; " 4 - !550%,0, " 0%;5 " ,%/, - 510

= 4008;%/ kg%m

&men akibat beban idup trtar

///%////; kg7m ///%/////;

1%0 8%0

#a = ///%; " 1%0 < ///%////; " 1%0

!%0

= ///%; kg

&u,

&u4



= ///%; " 1%0 " 5 - ///%; " 4

= ///%; kg%m

&men akibat beban truck 9T9

!0000 kg !0000 kg !0000 kg !0000

1%0 1%1 1%;5 !%, 1%;5 1%

#a =Tu " 4

=!0000 " 4%0

= 40000 kg! !

= 40000%0 " 4%0 - !0000 " !%0 - !0000

= ;000%0 kgm

Karena mmen akibat beban truck 9T9 H mmen akibat beban idup 9>9 maka diambil mme

akibat mmen beban truck = ;000%0 kgm

3eingga mmen ttal *ang bekerja pada gelagar melintang *akni :

= &u1 < &u! < &u, < &u5 <

= 4480 < 1855;%15 < !,,1;%0 < ///%; <

= 1!/0!0%; kg%m

c. Perencanaan dimensi gelagar memanang

>igunakan pr'il N) ;00 " ,00 " 1, " !4

M = 185 kg7m

A = !,5%5

D" = !01000

t' D* = 10800

i" = !%, cm

tO i* = /%;8 cm

r = !%8 cm

3" = 5;/0

3* = ;!!

b = ,00 mm

b = ;00 mm

tO = 1, mm

t' = !4 mm

= ,!00 (tegangan ijin baja+

1

&u5

&u/

&u ttal

cm²

cm⁴

cm⁴

cm³

cm³

kg7cm²

²



8

=1

" 185 " 5%0 ² x 1%18

= /,5%,;5 kg%m

= /,5%,;5 < 1!/0!1

= 1!//5/%;, kg%m

= 1!//5/;!%/!8;88

&uttal



S)arat *+,an *r/0

≥ &u

>imana :

ɸ = )aktr resistensi = 0% (untuk balk lentur+

&n = Kekuatan mmen nminal

&u = &men beban la*an ter'aktr

=

= ɸ " 1%1! " " '*

>imana :

&p = Kekuatan mmen plastis

1%1! = ke'isien penampang plastis untuk pr'il N)

'* = ,!0 &pa = ,!00

maka

= 0.9 " 1%1! " 5;/0 " ,!00

= 185;45/ kgcm

3eingga diperle

≥ &u

185;45/ kgcm > 1!//5/;!%/!8; kgcm K

Kntrl Plat $adan

/%,/

= /%,/!10000

'* ,!0

= 1/!%!/51;

=

;00= 5,%85

tO 1,

>engan demikian

→

Q /%,/

tO '*

→ 5,%85 Q 1/!%!/5! seingga tidak perlu pengaku

Kntrl Meser

Ru gelagar tenga = 4058%; < 11805%;/ < 1!8,,%48/0/ < //

185 " 1%1 " 1, 7 ! < 40000

= ;0/8;%, kg

kn = 5 < 5 dikarenakan tidak ada pengaku maka kn =

1%10kn "

= 1%105 " !10000

'* ,!0

ɸ&n

ɸ &n ɸ &p

3 #

kg7cm²

ɸ &p

ɸ &n

( a 7 +²



= /,%0104158

→

Q 1%10kn "

tO '*

→ 5,%85 Q /,%01041/ K

Rn = 0%/ " '* " AO

= 0%/ " ,!00 " (( ;0 - ! " !%4 + " 1%, +

= 1/!;,%! kg

Ru Q ɸ Vn

;0/8;%, kg Q 0% " 1/!;,%!

;0/8;%, kg Q 14/4/5%!8 K

Kntrl Lendutan

=1

" 6u " l?8

1!//5/;!%/!8;88 = 1" 6u " 1000

?

8

6u = 101%,!5,81 kg7m

=1

" L (>engan L = 1, m atau L = 1,00!40

=1

" 1,00!40

= 5%41////; cm

=,84 % % D"

=5 " 101%,!5,81 " 1,00 ⁴

,84 " !%1 " " !01000

= 5%0,1// cm

Q

5%0,1// cm Q 5%41/////; cm K

d. Per!itungan S!ear Connector

→ L = 1,00 m

Q1

" L4

Q1

" 1,004

Q ,!5 cm

Q 3 (dengan 3 =jarak antar gelagar melintang+

&uttal

' ijin

' ada

5 % u % L⁴

10⁶

' ada

' ijin

Peritungan be''

be''

be''



Q 500 cm

Q1

" 3 kiri <1

" kanan

! !

Q 1" 500 <

1"

500

! !

Q 500 cm

,!5 cm

s (mdulus elastisitas baja+ = !10000 &pa = !%1 "

c (mdulus elastisitas betn+ = 4;00 '.c

= 4;00 ,5

= !;805%5;5 &Pa

n =s

=!10000

= ;%55!44!,c !;805%5;5

be''

3eingga diambil be''

sebesar =



=,!5

= 4,%0,!4,;5 cmn ;%55!44!,

Peritungan Ma*a Meser rintal (R+

- B ma" = ga*a geser *ang disumbangkan le betn

= 0%85 " '.c " Ac

= 0%85 " ,5 " ( 4,0%,!4,; " !50 +

= ,!005,;%54

- T ma" = ga*a geser *ang disumbangkan le pr'il baja

= As " '*

= !,550 " ,!0

= ;5,/000

Karena B ma" Q T ma" maka dipakai R = B ma" = ,!005,;%5,;

Peritungan Jumla 3tud

1/ mm maka :

Asc = !01 mm?

= 100 cm

'* = !/0 &pa

c = 4;00 '.c

= 4;00 ,5

= !;805%5;5 &pa

n = Kekuatan geser 1 stud

be''

>ipakai stud dengan Ø



= 0%5 " Asc " '.c " c " rs Q Asc "

= 0%5 " !01 " ,5 " !;805%5;5 " 1 Q !01 "

= 14,%0/ H 5!!/0

Karena n H Asc " 'u maka *ang digunakan adala nilai Asc " 'u

Jumla stud (n+ =R

=,!005,;%5,/8!58

Asc " 'u 5!!/0

= /1%!4!585 ≈ /! bua

Jarak antar stud *ang dipasang memanjang *akni :

3 =L

=1,00

= !1 cmn /!

8 t' = 8 " !%4

= 1%! cm

Karena 3 %= !1 cm H 8 % t' = 1%! cm maka dipakai jarak antar s

n =

L

=

1,00

= 8/%/; bua3 15



1%1

50 m diperle nilai

≤ 50m



100I

50I

*ai lebar !%;5

" 5



! <1

" Pu!



" /400

ar gandar = 1%;5 m

1%0

" 4 " !

////; " 4 " !

,,;,%/5/; kg 1/84%; kg

1%0



" 4 - ,,;,%/5//; "



401/%;1;/ kg ;8,%,,,,,, kg

1%0

" 4 - 401/%;1;58 "

kg7m

1%0

%/1 " ,%!5 " 1%/,

kg7m

1%0



kg

1 1%0

" 1%15

n

&u/

;000%0



%; <

5



m +



10? kg7cm?



'u

!/0

tud = 15 cm



Perancangan JembatanSertin n. M. Mooy

1221103

.3 GE"AGAR INDUK

a. Per!itungan Pembebanan

Melagar induk direncanakan menggunakan pr'il baja N)

$eban &ati

- $erat sendiri gelagar induk

>idalam mengitung berat sendiri gelagar induk pen*usun tidak menggunakan rumus

pendekatan tetapi menggunakan bantuan s'tOare analisa struktur *akni :

3TAA>P#C !004 (3el' Oeigt+

-

n " " L

= 5 " 10/%;/ " 50

= !//0 kg

-

n " " L

= 11 " 185 " 10

= !0,50 kg

-

6 " a " L

= 8, " 8 " 50

= ,,!00 kg

-

! " ( 6 " a " L +

= ! " ( 1;;/ " 1%0 " 50 +

= 1;;/00 kg

-

! " ( 6 " n " L +

= ! " ( 5%08 " !%0 " 50 +

= 101/ kg

-

>idalam mengitung berat sendiri ikatan angin pen*usun tidak menggunakan rumus

pendekatan tetapi menggunakan bantuan s'tOare analisa struktur *akni :

$erat sendiri gelagar memanjang (M!+

M!

u = M!

$erat sendiri gelagar melintang (M,+

M,

u = M,

$erat lantai kendaraan (M4+

M4

u =

$erat lantai trtar (M5+

M5

u =

$erat sendiri pipa sandaran > = ;/%, mm dan t = !%8 mm (M/+

M/u

=

$erat sendiri ikatan angin (M;+ dengan 'aktr beban 11




1221103

3TAA>P#C !004 (3el' Oeigt+




1221103

3eingga ttal beban *ang bekerja *akni :

< < < <

= !//0 < !0,50 < ,,!00 < 1;;/00 < 101/

= /1885/ kg

- $eban mati *ang dipikul le tiap gelagar induk

M =!

=/1885/

= ,04!8 kg!

- $eban mati *ang diterima tiap titik buul tenga

M

10

=,04!8

= ,04!%8 kg10

- $eban mati *ang diterima tiap titik buul tepi

P

!

=,04!%8

= 154;1%4 kg!

$eban idup Lantai kerndaraan

- Ke'isien Kejut

>iketaui panjang bentang jembatan adala = 50 m% >ari gamba

! - ! buku $&3 bagian ! didapat nilai ke'i'ien kejut (>LA sebesar

= 0%4

k = 1 < >LA

= 1 < 0%4

= 1%4

- $eban terbagi rata

$erdasarkan buku $&3 bag ! alaman ! - !! untuk jembatan dengan panjang L

= 50 m H L = ,0 m maka :

6 = 8%00 " 0%5 <15

L

Mu = M!

u M,

u M4

u M5

u M/

u

Mu ttal

Ptenga

=

Ptepi

=




1221103

= 8%00 " 0%5 <15

50%00

= /%400 Kpa

= /40%00

= 6 " 5%5 " 100I!%;5

=/40%00

" 5%5 " 100I!%;5

= 1!80 kg7m

=6

" ! " 0%;5 " 50I!%;5

=/40%00

" ! " 0%;5 " 50I!%;5

= 1;4%54545 kg7m

$eban *ang diterima tiap gelagar induk

M = " L!

=1!80 < 1;4%54545

" 50!

= ,/,/,%/,/,/,/4 kg

$eban *ang diterima tiap titik buul tenga

M

10

=,/,/,%/,/

= ,/,/%,/,/ kg10

$eban *ang diterima tiap titik buul tepi

P

!

=,/,/%,/,/

= 1818%1818 kg!

$eban garis (idup lantai kendaraan+

$erdasarkan buku $&3 bag ! al ! - !! beban garis diambil P = 44

kg7m²

61

61

6ttal

Ptenga

=

Ptepi

=




1221103

= 4400 kg7m dengan lebar lantai kendaraan 8%0 m dan dibagi

lajur%

=P

" 5%5 " 100I " k !%;5

= 4400 " 5%5 " 100I " 1%4!%;5

= 1!,!0 kg

P1




1221103

=P

" ! " 0%;5 " 50I " k !%;5

=4400

" ! " 0%;5 " 50I " 1%4!%;5

= 1/80 kg$eban *ang diterima gelagar induk

P =<

!

=1!,!0 < 1/80

!

= ;000 kg

$eban *ang diterima tiap titik buul

P = ;000 kg


,/,/%,/,/ < ;000

= 10/,/%,/,/,/,/,/ kg


1818%1818 < ;000

= 8818%1818181818 kg

$eban idup Trtar

$erdasarkan buku $&3 bag ! al ! - ,1 beban idup trtar diambil sebesar

P = 5 kPa = 500 kg7m? dengan lebar lantai trtar =

P = 500 " 1%0 " 50 " !

= 50000 kg

$eban *ang diterima tiap gelagar induk

P =P

!

=50000

!

= !5000 kg


P

P!

P1

P!

Ptenga

=

Ptepi

=

=




1221103

n

=!5000

10

= !500 kg

tenga




1221103

$eban *ang diterima tiap titik buul tepi

P

!

=!500

= 1!50 kg

!

Ma*a #em

>iketaui :

Panjang Jembatan = 50 m

$erdasarkan gambar !% buku $&3 bagian ! al ! - ,1 didapatkan ga*a rem sebesar

M = !50 k = !5000 kg

Ma*a rem *ang diterima tiap gelagar induk

P =M

!

=!5000

!

= 1!500 k

Ma*a rem *ang diterima tiap titik buul tenga

P

n

=1!500

10

= 1!50 kg

Ma*a rem *ang diterima tiap titik buul tepi

P

!

=1!50

= /!5 kg

!

$eban Angin

Ptepi

=

Ptenga

=

Ptepi

=




1221103

= 0%001! " BO " ( RO + ?

= 0%001! " 1%! " ( ,0 + ?

= 1%!/ k

= 1!%/ kg

= 0%000/ " BO " ( RO + ? " Ab

>imana :

RO = Kecepatan angin rencana ( ,0 m7det +

BO = Ke'isien seret (untuk bangunan atas rangka BO = 1%! + $

1! al ! - 44

Ab = Luas ke'isien bagian samping jembatan (m?+

- Peritungan bagian samping jembatan

/

5%0 5%0

=1

" 5%0 " /%0!

= 15%0 m?

= 5%0 " /%0

= ,0%0 m?

- Peritungan Ma*a Angin Pada Dkatan Angin Atas dan $aOa

= 1!%/ kg

= 0%000/ " BO " ( RO + ? " Ab

= 0%000/ " 1%! " ( ,0 + ? " 15 " ,0I

= !%1/ k

TN1

TN!

AbA

Ab$

TN1

TN!




1221103

= !1%/ kg

= 0

" / = " ( 0%05 < 0%!5 < 1 + <

0%5 " /

= 1!%/ " ( 0%05 < 0%!5 < 1 + <!1%/ " 0%5 " /

/ = 104,%!8 kg

= 1;,%88 kg

=1;,%88

kg!

= 8/%4 kg

&A

$

TN1

TN!

"

$

$ tenga

$ tepi




1221103

< - - = 0

= < -

= 1!%/ < !1%/ - 1;,%88

= !4;%,! kg

= !4;%,! kg!

= 1!,%// kg

b. #odel Statika

3kema pembebanan akibat beban mati

= 154;1%4 kg

= ,04!%8 kg

3kema pembebanan akibat beban idup

= 8818%1818 kg

= 10/,/%4 kg

3kema pembebanan akibat beban trtar

$

A

TN1

TN!

A tenga

TN1

TN!

A

A tepi

Ptepi

Ptenga

Ptepi

Ptenga




1221103

= 1!50 kg

= !500 kg

3kema pembebanan akibat beban rem

= /!5 kg

= 1!50 kg

3kema pembebanan akibat beban angin atas

= 8/%4 kg

= 1;,%88 kg

Analisa pembebanan jembatan selanjutn*a menggunakan prgram bantu 3TAA>P#C

R8i *ang dilampirkan pada bagian akir lapran ini

Ptepi

Ptenga

Ptepi

Ptenga

Ptepi

Ptenga




1221103

c. Perencanaan Dimensi Gelagar Induk

1 $atang tekan ($atang 15+

>imensi batang pr'il N) 400"400"45";0

>igunakan baja dengan )* = ,!00 kg7cmM = /05 kg7m d = 400 mm

Ag = ;;0%1 cm? b' = 400 mm

L = 40 cm tO = 45 mm

D" = !8000 t' = ;0 mm

D* = 4400

(BM% 3almn J Jnsn 3truktur $aja dan >esain Perilaku Jilid 1 1! : ,4!+

>ari asil analisa 3TAA>P#C U8i ga*a aksial ter'aktr Pu *akni : ;;

&engitung radius girasi ( r +

r" = D"

Ag

= !8000

;;0%1

= 1%/;1,/8 cm

r* = D*

Ag

= 4400

;;0%1

= 11%0;1/5! cm

= K % L )*

r V? %

>imana

K % L= rasi kerampingan e'ekti'

r

K = )aktr panjang e'ekti' (sendi - sendi = 1+

L = Panjang batang *ang ditinjau = 500 cm

r* = radius girasi ara sumbu *

cm⁴

cm⁴

3*arat kekakuat nminal batang tekan berdasarkan L#)> ɸc % Pn ≥ Pu

&engitung Parameter Kerampingan $ λ%&

λc




1221103

r" = radius girasi ara sumbu "

)* = tegangan lele baja ,!00 kg7cm?

D = &men inersia

= &dulus elastisitas baja !10000 &pa

!100000 kg7cm?= 1 " 500 ,!00

11%0;1/5!!!! ,%14 ? " !100000

= 0%5/14!;

&engitung tegangan kritis penampang ()cr+

≤ 1%5 → )cr = 0%/58 W ( Xc +W ? "

= 0%/58 W ( 0%5/14!; +W ? "

= !804%45440,0! kg7cm?

&aka c % Pn Y Puɸ

c % )cr % Agɸ ≥ Pu

0%85 " !804%454 " ;;0%1 ≥ ;;08%0040 kg

18,5;80%/4;; kg ≥ ;;08%0040 kg K

! Perencanaan $atang Tarik ($atang 5+

>imensi batang pr'il N) 400"400"45";0

>igunakan baja dengan )* = ,!00 kg7cm

M = /05 kg7m

Ag = ;;0%1 cm?

L = 500 cm

D" = !8000

D* = 4400

>ari asil analisa 3TAA>P#C U8i ga*a aksial ter'aktr Pu *akni: //

Lebar lubang baut = 1%1 < 0%1 = !%01 cm

Bek rasi kerampingan

L≤ ,00

r

>imana :

L = panjang $atang = 500 cm

r = radius girasi terkecil

λc

λc

cm⁴

cm⁴




1221103

L=

500= 45%1/0,;8 ≤ ,00

r 11%0;1/5!

&engitung luas nminal

An = Ag - 4 % (lebar lubang baut % Tebal 'lens+

= ;;0%1 - 4 " ( !%01 " 0%; += ;/4%4;! cm?

Peren%anaan desain kekuatan ba'an terdiri atas 2 kriteria( "akni )

>idasarkan pada pelelean penampang brut

Tn = )* % Ag

>imana :

= )aktr resistensi = 0%

)* = tegangan lele baja = ,!00

Ag = Luas Penampang brut

Tn = 0% " ,!00 " ;;0%1

= !!1;888%0 kg

>idasarkan pada retakan penampang bersi

Tn = )u Ae

>imana :

= )aktr resistensi = 0%;5

)* = tegangan lele baja = 4800

Ag = Luas e'ekti' penampang

= 0%85 " An

Tn = 0%;5 " 4800 " ( 0%85 " ;/4%4;! +

= !,,!84%,! kg

>ari asil ! kriteria diatas diambil kekuatan desain *ang lebi kecil *aitu :

Tn = !!1;888%00

&aka Tn = !!1;888%00 kg H Tu = //50%/,; kg

.# GE"AGAR ME"INTANG ATAS

Perencanaan >imensi batang Tarik ($atang Tarik 5+

>imensi batang pr'il N) 150 " 150 " ; " 10

>igunakan baja dengan )* = ,!00 kg7cm

M = ,1%51 kg7m

ɸt % ɸt %

ɸt

kg7cm²

ɸt %

ɸt % ɸt %

ɸt

kg7cm²

ɸt %

ɸt %

ɸt %




1221103

Ag = 40%1 cm?

L = 1000 cm

D" = 1/40

D* = 5/,

>ari asil analisa 3TAA>P#C U8i ga*a aksial ter'aktr Pu *akni: ,81%!85Lebar lubang baut = 1%1 < 0%1 = !%01 cm


r" = D"

Ag

= 1/40

40%1

= /%,51,5, cm

r* = D*

Ag

= 5/,

40%1

= ,%;4/855 cm

Bek rasi kerampingan

L≤ ,00

r

>imana :

L = panjang $atang = 1000 cm

r = radius girasi terkecil

L=

1000= !//%881! ≤ ,00

r ,%;4/855

&engitung luas nminal

An = Ag - ! % (lebar lubang baut % Tebal 'lens+

= 40%1 - ! " ( !%01 " 1 +

= ,/%08 cm?

Peren%anaan desain kekuatan ba'an terdiri atas 2 kriteria( "akni )

>idasarkan pada pelelean penampang brut

Tn = )* % Ag

>imana :

cm⁴

cm⁴

ɸt % ɸt %




1221103

= )aktr resistensi = 0%

)* = tegangan lele baja = ,!00

Ag = Luas Penampang brut

Tn = 0% " ,!00 " 40%1

= 115488%0 kg >idasarkan pada retakan penampang bersi

Tn = )u Ae

>imana :

= )aktr resistensi = 0%;5

)* = tegangan lele baja = 4800

Ag = Luas e'ekti' penampang

= 0%85 " An

Tn = 0%;5 " 4800 " ( 0%85 " ,/%08 +

= 110404%80 kg

>ari asil ! kriteria diatas diambil kekuatan desain *ang lebi kecil *aitu :

Tn = 110404%80

&aka Tn = 110404%80 kg H Tu = ,81%!85 kg K

. IKATAN ANGIN PR4I" "12%12%11

Kekuatan Tekan $atang ($atang 118+

>imensi $atang Pr'il L1!01!011

M = 1%4 kg7m

Ag = !5%4 cm?

L = 55 cm

D" = ,41

D* = ,41

Tebal plat siku (d+ = 11 mm = 1%1 cm

>ari asil analisa 3TAA>P#C U8i ga*a aksial ter'aktr Pu *akni : ,!81%;


r = D

Ag

= ,41

ɸt

kg7cm²

ɸt %

ɸt % ɸt %

ɸt

kg7cm²

ɸt %

ɸt %

ɸt %

cm⁴

cm⁴

3*arat kekakuat nminal batang tekan berdasarkan L#)> ɸc % Pn ≥ Pu




1221103

!5%4

= ,%//40411 cm

Lebar lubang baut = 1%1 < 0%1 = !%01 cm

= K % L )*r V? %

= 1 " 55 ,!00

,%//4041054; ,%14 ? " !100000

= 1%8//4

&engitung tegangan kritis penampang ()cr+

≤ 1%5 → )cr = 0%/58 W ( Xc +W ? " )*

= 0%/58 W ( 1%8//4 +W ? " ,!00

= ;0%85//,;1 kg7cm?

&aka c % Pn Y Puɸ

c % )cr % Agɸ ≥ Pu

0%85 " ;0%85; " !5%4 ≥ ,!81%;;00 kg

15,!8%8/1184/ kg ≥ ,!81%;;00 kg K

&engitung Parameter Kerampingan $ λ%&

λc

λc

λc




1221103




1221103

r !%8 al

40 I




1221103

k7m atau




1221103

enjadi !




1221103

1%0 m




1221103




1221103

3 bag !




1221103




1221103

08%0040 kg




1221103

)*

,!00

50%/,; kg




1221103

K




1221103

kg




1221103



Perancangan JembatanSertin N. M. Mooy

12.21.103

BAB !I

PERENCANAAN SAMBUNGAN

5.1 SAMBUNGAN GE"AGAR MEMAN(ANG & ME"INTANG

>irencanakan menggunakan baut A 40 dengan diameter (>+ = ; 7 8

150 ksi = 10,4!%5 7mm? $ 1 ksi * +,.-

Jarak tepi baut (L+ = 1%5d - ,d dan jarak antar baut (L+ = ,d - ;d ( Dr% 3

Teri dan Pen*elesaian 3al-3al Knstruksi $aja D al 14+ atau lebi besar dari pada *ang diitung

dari pers*aratan dan jarak minimum *ang ditentukan le tabel ,%; (B%M% 3almn J%% Jnsn 3truktur

>esain $aja dan Perilaku Jilid D 1! : 1,/+

a. Kuat Geser Gelagar #emanang

3ambungan berdasarkan kekuatan batas7kapasitas penampang singga memungkinkan sambungan

lebi kuat dari pada batang% Kuat geser gelagar memanjang adala :

Ru = /18/!%4 kg

b. Luas Baut

Ab =1

" V " >?4

=1

" ,%14 " !%!!!5?

4

= ,%8;;51!41 cm?

c. (arak Baut

lubang baut =ɸ !%!!!5 < 0%1

= !%,!!5 cm

Jarak tepi baut = 1%5 d s7d , d= 1%5 " !%,!!5 s7d , " !%,!!5

= ,%48,;5 s7d /%/;5 → diambil L = 4 cm

Jarak antr baut = , d s7d ; d

= , " !%,!!5 s7d ; " !%,!!5

= /%/;5 s7d 1/%!5;5 → diambil L = ; cm

kekuatan tarik baut ()u b+ =




12.21.103

d. Sambungan irisan tunggal "ada gelagar melintang$

- Kekuatan tarik desain :

ɸ ( 0%;5 " + " Ab

= 0%;5 " ( 0%;5 " 10,4!%5 + " ,%8;;51!41

= !!558%0 kg

- Kekuatan geser desain :

$an*akn*a bidang geser *ang terliat adala 1 karena merupakan sambungan irisan tunggal

seingga m = 1

ɸ ( 0%/0 " + " Ab % m

= 0%/5 " ( 0%/0 " 10,4!%5 + " ,%8;;51!41 " 1

= 15/40%!,;1 kg

- Kekuatan tumpu desain :

Peritungan kekuatan tumpu desain pada perumusann*a mempertimbangkan ketebalan plat *g

akan disambung% >alam al ini ketebalan plat *ang diperitungkan adala ketebalan gelagar

melintang *aitu = 1%, cm

ɸ ( !%4 " d " t " )u +

= 0%;5 " ( !%4 " !%,!!5 " 1%, " 4800 +

= !/08/%,! kg

- Kekuatan nminal :

Tn = 0%/ " )* " Aug

= 0%/ " ,!00 " ( 1%, " ( ;0 - ! % !%4 + +

= 1/!;,%!0 kg

Tn H Tu = /18/!%4 kg

- &men ultimate

&u = Pu " O (dimana O adala jarak titik *ang dilemakan+

= /18/!%4 " 5

= ,0,1! kgcm- Jumla baut :

n =/ " &u

# " P

>imana : &u = &men ultimate

# =

P = Jarak minimum sumbu baut ; cm

ɸ #n = )u b

ɸ #n = )u b

ɸ #n =

ɸ #n (kekuatan desain *ang menentukan+




12.21.103

n =/ " ,0,1!

≈ 4%1! ≈ 4 bua15/40%!,;1 " ;

- Ketebalan plat *ang digunakan adala :

t =P

ɸ )u " L

t =/18/!%4 7 4

0.75 " 4800 " 4

= 1%0;4 cm

&aka *ang digunakan plat pen*ambung siku L 0%0%1/ dengan tebal

1 Kntrl kekuatan tarik desain H beban tarik ter'aktr baut

>imana :

%#n =ɸ Kekuatan tarik desain *ang menentukan

= !!558%0 kg

= $eban tarik ter'aktr baut

= &u " Z

=,0,1! " !5

4 ? < 11 ? < 18 ? < !5 ?

= ;1!0%44! kg

%#n =ɸ !!558%0 kg H = ;1!0%4418 kg K

! Kekuatan geser desain H beban geser ter'aktr baut

#nɸ H

%#n =ɸ Kekuatan geser desain *ang menentukan

= 15/40%! kg

= Pu

n

= /18/!%44

= 154/5%/ kg

%#n =ɸ 15/40%! H = 154/5%/ kg K

e. Sambungan irisan ganda "ada gelagar memanang$

- Kekuatan tarik desain (berdasarkan L#)> al 100+ :

ɸ % #n H #ut

#ut

∑ Z ?

#ut

#ut

#ut

#ut




12.21.103

ɸ ( 0%;5 " + " Ab

= 0%;5 " ( 0%;5 " 10,4!%5 + " ,%8;;51!41

= !!558%0 kg


$an*akn*a bidang geser *ang terliat adala ! karena merupakan sambungan irisan tunggal

seingga m = !

ɸ ( 0%/0 " + " Ab % m

= 0%/5 " ( 0%/0 " 10,4!%5 + " ,%8;;51!41 " !

= ,1!80%4;4! kg




melintang *aitu = 0 cm

ɸ ( !%4 " d " t " )u +

= 0%;5 " ( !%4 " !%,!!5 " 1%0 " 4800 +

= !00//%4 kg

- Kekuatan nminal :

Tn = 0%/ " )* " Aug

= 0%/ " ,!00 " ( 1%0 " ( , - ! % 1%/ + +

= /8;,/%00 kg

Tn H Tu = /18/!%4 kg

- &men ultimate


= /18/!%4 " 5

= ,0,1! kgcm

- Jumla baut :

n =/ " &u

# " P>imana : &u = &men ultimate

# =


n =/ " ,0,1!

= ,% ≈ 4 bua,1!80%4;4! " ;


ɸ #n = )u b

ɸ #n = )u b

ɸ #n =





12.21.103

t =P

ɸ )u " L

t =/18/!%4 7 4

0.75 " 4800 " 4

= 1%0;4 cm



>imana :


= !!558%0 kg


= &u " Z

=,0,1! " !5

4 ? < 11 ? < 18 ? < !5 ?

= ;1!0%44! kg

%#n =ɸ !!558%0 kg H = ;1!0%4418 kg K


#nɸ H


= ,1!80%5 kg

= Pu

n

=/18/!%4

4

= 154/5%/ kg

%#n =ɸ ,1!80%5 H = 154/5%/ kg K

ɸ % #n H #ut

#ut

∑ Z ?

#ut

#ut

#ut

#ut




12.21.103




12.21.103

5.2 SAMBUNGAN GE"AGAR ME"INTANG DAN GE"AGAR INDUK

>irencanakan menggunakan baut A 40 dengan diameter (>+ = ; 7 8

150 ksi = 10,4!%5 7mm? $ 1 ksi * +,.-

Jarak tepi baut (L+ = 1%5d - ,d dan jarak antar baut (L+ = ,d - ;d ( Dr% 3

Teri dan Pen*elesaian 3al-3al Knstruksi $aja D al 14+ atau lebi besar dari pada *ang diitung

dari pers*aratan dan jarak minimum *ang ditentukan le tabel ,%; (B%M% 3almn J%% Jnsn 3truktur

>esain $aja dan Perilaku Jilid D 1! : 1,/+

a. 3ambungan berdasarkan kekuatan batas7kapasitas penampang singga memungkinkan sambungan

lebi kuat dari pada batang% Kuat geser gelagar memanjang adala :

Ru = 1/4;;,%44 kg

b. Luas Baut

Ab =1

" V " >?4

=1

" ,%14 " !%!!!5?

4

= ,%8;;51!41 cm?

c. Sambungan irisan tunggal "ada gelagar induk$- Kekuatan tarik desain :

ɸ ( 0%;5 " + " Ab

= 0%;5 " ( 0%;5 " 10,4!%5 + " ,%8;;51!41

= !!558%0 kg



kekuatan tarik baut ()u b+ =

ɸ #n = )u b




12.21.103

seingga m = 1

ɸ ( 0%/0 " + " Ab % m

= 0%/5 " ( 0%/0 " 10,4!%5 + " ,%8;;51!41 " 1

= 15/40%!,;1 kg




melintang *aitu = ; cm

ɸ ( !%4 " d " t " )u +

= 0%;5 " ( !%4 " !%!!!5 " ; " 4800 +

= 1,441/%8 kg

- Kekuatan nminal :

Tn = 0%/ " )* " Aug

= 0%/ " ,!00 " ( ; " 40 +

= 5,;/00%00 kg

Tn H Tu = 1/4;;,%44 kg

Aug adala luas sa*ap gelagar induk

- &men ultimate


= 1/4;;,%44 " 5

= 8!,8/;%! kgcm

- Jumla baut :

n =/ " &u

# " P


# =


n = / " 8!,8/;%! = /%;144;! ≈ ; bua15/40%!,;1 " ;


t =P

ɸ )u " L

t =1/4;;,%44 7 ;

0.75 " 4800 " 4

ɸ #n = )u b

ɸ #n =





12.21.103

= 1%/,4/5;14 cm



>imana :


= !!558%0 kg


= &u " Z

=8!,8/;%! " 58

4 ? < 11 ? < !8 ? < ,5 ? <

= 4;5/%0;/! kg

%#n =ɸ !!558%0 kg H = 4;5/%0;/!0! kg K


#nɸ H


= 15/40%! kg

= Pu

n

=1/4;;,%44

;

= 14/;%/, kg

%#n =ɸ 15/40%! H = 14/;%/, kg K

e. Sambungan irisan ganda "ada gelagar memanang$

- Kekuatan tarik desain (berdasarkan L#)> al 100+ :

ɸ ( 0%;5 " + " Ab= 0%;5 " ( 0%;5 " 10,4!%5 + " ,%8;;51!41

= !!558%0 kg


$an*akn*a bidang geser *ang terliat adala ! karena merupakan sambungan irisan tunggal

seingga m = !

ɸ ( 0%/0 " + " Ab % m

ɸ % #n H #ut

#ut

∑ Z ?

#ut

#ut

#ut

#ut

ɸ #n = )u b

ɸ #n = )u b




12.21.103

= 0%/5 " ( 0%/0 " 10,4!%5 + " ,%8;;51!41 " !

= ,1!80%4;4! kg




melintang *aitu = 0 cm

ɸ ( !%4 " d " t " )u +

= 0%;5 " ( !%4 " !%!!!5 " 1%, " 4800 +

= !4/,%1! kg

- Kekuatan nminal :

Tn = 0%/ " )* " Aug

= 0%/ " ,!00 " ( 1%, " ( ;0 - ! % !%4 + +

= 18,081%/0 kg

Tn H Tu = 1/4;;,%44 kg

- &men ultimate


= 1/4;;,%44 " 5

= 8!,8/;%! kgcm

- Jumla baut :

n =

/ " &u

# " P


# =


n =/ " 8!,8/;%!

= ,% ≈ ; bua,1!80%4;4! " ;


t = Pɸ )u " L

t =1/4;;,%44 7 ;

0.75 " 4800 " 4

= 1%/,4/5;14 cm



ɸ #n =





12.21.103

>imana :


= !!558%0 kg


= &u " Z

=8!,8/;%! " 58

4 ? < 11 ? < !8 ? < ,5 ? <

= 4;5/%0;/! kg

%#n =ɸ !!558%0 kg H = 4;5/%0;/!0! kg K


#nɸ H


= ,1!80%5 kg

= Pu

n

=1/4;;,%44

;

= !,5,%0/, kg

%#n =ɸ ,1!80%5 H = !,5,%0/!8/ kg K

ɸ % #n H #ut

#ut

∑ Z ?

#ut

#ut

#ut

#ut




12.21.103




12.21.103

5.3 SAMBUNGAN GE"AGAR INDUK

a. Per!itungan Kekuatan Baut

>igunakan baut A 40 dengan diameter d = , 7 4 inc = 1%

150 ksi = 10,4!%5 Jarak tepi b

s7d , d dan jarak antar baut = , s7d ; d

b. Luas Baut

Ab =1

" V " >?4

=1

" ,%14 " 1%05?

4

= !%848;84/, cm?

c. Kekuatan geser desain$an*akn*a bidang geser *ang terliat adala 1 karena merupakan sambungan irisan tunggal

seingga m = 1

ɸ ( 0%/0 " + " Ab % m

= 0%/5 " ( 0%/0 " 10,4!%5 + " !%848;84/, " 1

= 1140%;8/4 kg

Kekuatan tarik baut )u b = 7mm²

ɸ #n = )u b




12.21.103

d. Kekuatan tumu desain


akan disambung% >alam al ini ketebalan plat *ang diperitungkan adala ketebalan plat

simpul *aitu = !%!5 cm

ɸ ( !%4 " d " t " )u +

= 0%;5 " ( !%4 " 1%05 " !%!5 " 4800 +

= ,;0,,%! kg

e. Per!itungan Kebutu!an Baut

- Jint 1

= - ,/5!1/%14 kg

2

1 1 = < 1!0,!8%84 kg

Jumla baut *ang diperlukan

1!0,!8%84= 10%4;1;/; ≈ 1/ bua

1140%;8/44,;84

Tebal plat pen*ambung *ang digunakan :

Jarak ujung minimum untuk baut diameter ,74 inc adala !5%4 mm (BM% 3almn J Jns

3truktur $aja >esain dan Perilaku Jilid 1 1! : 1,/+

3*arat Jarak ujung = 1%5 d s7d , d

= 1%5 " 1%05 s7d , " 1%05

= !%85;5 s7d 5%;15

digunakan L = 4 cm

Ketebalan plat *ang digunakan adala :

t =P

ɸ )u " L

t = 1!0,!8%84 7 1/0.75 " 4800 " 4

= 0%5!!!/05 cm

dalam perencanaan digunakan plat dengan ketebalan = !%!5

Rn<

/b

0u # t 2

≥ 1!0,!8%84 7 1/ 1%05

ɸ #n =

3!0

31

31

=

Jarak antar baut ≥




12.21.103

0.75 " 4800 " !%!5 !

≥ 1%880/,, cm.

3*arat jarak antar baut (L+ = , d s7d ; d

= , " 1%05 s7d ; " 1%05

= 5%;15 s7d 1,%,,5 mm

>alam perencanaan digunakan jarak antar baut L = 1! cm


,/5!1/%14= ,1%;8,,8 ≈ ,/ bua

1140%;8/44,;84





= 1%5 " 1%05 s7d , " 1%05

= !%85;5 s7d 5%;15

digunakan L = 4 cm


t =P

ɸ )u " L

t =

,/5!1/%14 7 ,/

0.75 " 4800 " 4

= 0%;0450/44 cm


Rn<

/b

0u # t 2

≥ ,/5!1/%14 7 ,/<

1%05

0.75 " 4800 " !%!5 !

≥ !%!0455 cm.3*arat jarak antar baut (L+ = , d s7d ; d

= , " 1%05 s7d ; " 1%05

= 5%;15 s7d 1,%,,5 mm


3!0

=





12.21.103

Gambar +.4 ambungan Gelagar Induk Joint 1

- Jint !

,5584!%05 kg !8

1!0,!8%84 kg

2


1!0,!8%84= 10%4;1;/; ≈ !0 bua

1140%;8/44,;84



3truktur $aja >esain dan Perilaku Jilid 1 1! : 1,/+3*arat Jarak ujung = 1%5 d s7d , d

= 1%5 " 1%05 s7d , " 1%05

= !%85;5 s7d 5%;15

digunakan L = 4 cm


P

3!1

= < 3!!

= -

31= < 3

!= <

31

=




12.21.103

ɸ )u " L

t =1!0,!8%84 7 !0

0.75 " 4800 " 4

= 0%41;8084; cm


Rn<

/b

0u # t 2

≥ 1!0,!8%84 7 !0<

1%05

0.75 " 4800 " !%!5 !

≥ 1%/5!;0/ cm.


= , " 1%05 s7d ; " 1%05

= 5%;15 s7d 1,%,,5 mm



,40,/,%/5= !%/!05;1 ≈ 40 bua

1140%;8/44,;84





= 1%5 " 1%05 s7d , " 1%05

= !%85;5 s7d 5%;15

digunakan L = 4 cm


t =P

ɸ )u " L

t = ,40,/,%/5 7 400.75 " 4800 " 4

= 0%50011 cm


Rn<

/b

0u # t 2

≥ ,40,/,%/5 7 40 1%05


3!

=





12.21.103

0.75 " 4800 " !%!5 !

≥ !%00,0051 cm.


= , " 1%05 s7d ; " 1%05

= 5%;15 s7d 1,%,,5 mm



,5584!%05= ,0%/;58 ≈ ,/ bua

1140%;8/44,;84





= 1%5 " 1%05 s7d , " 1%05

= !%85;5 s7d 5%;15

digunakan L = 4 cm


t =P

ɸ )u " L

t =

,5584!%05 7 ,/

0.75 " 4800 " 4

= 0%/8/4!,;1 cm


Rn<

/b

0u # t 2

≥ ,5584!%05 7 ,/<

1%05

0.75 " 4800 " !%!5 !

≥ !%1;!8088 cm.3*arat jarak antar baut (L+ = , d s7d ; d

= , " 1%05 s7d ; " 1%05

= 5%;15 s7d 1,%,,5 mm



!8,!4%/4

3!1

=


=




12.21.103

1140%;8/44,;84%





= 1%5 " 1%05 s7d , " 1%05

= !%85;5 s7d 5%;15

digunakan L = 4 cm


t =P

ɸ )u " L

t =!8,!4%/4 7 ,/

0.75 " 4800 " 4

= 0%54;/414 cm


Rn<

/b

0u # t 2

≥ !8,!4%/4 7 ,/<

1%05

0.75 " 4800 " !%!5 !

≥ 1%!/1;85 cm.


= , " 1%05 s7d ; " 1%05

= 5%;15 s7d 1,%,,5 mm


3!1





12.21.103

Gambar +. ambungan Gelagar Induk Joint 2

- Jint 1!

12

,/5!1/%14 kg ,5584!%


,/5!1/%14= ,1%;8,,8 ≈ ,/ bua

1140%;8/44,;84Tebal plat pen*ambung *ang digunakan :




= 1%5 " 1%05 s7d , " 1%05

= !%85;5 s7d 5%;15

311

= <

3!0

= - 3!1

= <

3!0

=




12.21.103

digunakan L = 4 cm


t =P

ɸ )u " L

t =,/5!1/%14 7 ,/

0.75 " 4800 " 4

= 0%;0450/44 cm


Rn<

/b

0u # t 2

≥ ,/5!1/%14 7 ,/<

1%05

0.75 " 4800 " !%!5 !

≥ !%!0455 cm.


= , " 1%05 s7d ; " 1%05

= 5%;15 s7d 1,%,,5 mm



,5584!%05= ,0%/;58 ≈ ,/ bua

1140%;8/44,;84





= 1%5 " 1%05 s7d , " 1%05

= !%85;5 s7d 5%;15

digunakan L = 4 cm


t =P

ɸ )u " L

t =,5584!%05 7 ,/

0.75 " 4800 " 4

= 0%/8/4!,;1 cm



3!1

=




12.21.103

Rn<

/b

0u # t 2

≥ ,5584!%05 7 ,/<

1%05

0.75 " 4800 " !%!5 !

≥ !%1;!8088 cm.


= , " 1%05 s7d ; " 1%05

= 5%;15 s7d 1,%,,5 mm



!;;4;/%,8= !4%14;;!8 ≈ ,! bua

1140%;8/44,;84





= 1%5 " 1%05 s7d , " 1%05

= !%85;5 s7d 5%;15

digunakan L = 4 cm


t =P

ɸ )u " L

t =!;;4;/%,8 7 ,!

0.75 " 4800 " 4

= 0%/0!1/!!8 cm


Rn<

/b

0u # t 2≥ !;;4;/%,8 7 ,!

<1%05

0.75 " 4800 " !%!5 !

≥ !%0!,010; cm.


= , " 1%05 s7d ; " 1%05

= 5%;15 s7d 1,%,,5 mm


311

=





12.21.103


Gambar +.+ ambungan Gelagar Induk Joint 12

5.# SAMBUNGAN BATANG IKATAN ANGIN

a. Per!itungan Kekuatan Baut

>igunakan baut A 40 dengan diameter d = 1 7 ! inc = 1%

150 ksi = 10,4!%5 Jarak tepi bs7d , d dan jarak antar baut = , s7d ; d

b. Luas Baut

Ab =1

" V " >?4

1 ?

Kekuatan tarik baut )u b

= 7mm²




12.21.103

4% %

= 1%!//1!/5 cm?

c. Kekuatan geser desain


seingga m = 1

ɸ ( 0%/0 " + " Ab % m

= 0%/5 " ( 0%/0 " 10,4!%5 + " 1%!//1!/5 " 1

= 510;%01/! kg

d. Kekuatan tumu desain


akan disambung% >alam al ini ketebalan plat *ang diperitungkan adala ketebalan plat

simpul *aitu = 1 cm

>iameter lubang = 1%!; < 0%1 = 1%,; cm

ɸ ( !%4 " d " t " )u +

= 0%;5 " ( !%4 " 1%,; " 1%00 " 4800 +

= 118,/%8 kg

e% Per!itungan Sambungan

- Jint 4,

1!50/%;! kg 1405%/!

#3

1!50/%;! kg 1405%/!


ɸ #n = )u b

ɸ #n =

3100

= - 310!

= -

310,

= - 3101

= -




12.21.103

1!50/%;!= !%448! ≈ 4 bua

510;%01/1;!,;5





= 1%5 " 1%,; s7d , " 1%,;

= !%055 s7d 4%11

digunakan L = , cm


t =P

ɸ )u " L

t =1!50/%;! 7 4

0.75 " 4800 " ,

= 0%!850;41 cm

dalam perencanaan digunakan plat dengan ketebalan = 1%00

Rn<

/b

0u # t 2

≥ 1!50/%;! 7 4<

1%,;

0.75 " 4800 " 1 !

≥ 1%55,5!!! cm.


= , " 1%,; s7d ; " 1%,;

= 4%11 s7d %5 mm

>alam perencanaan digunakan jarak antar baut L = 5 cm


1!50/%;!= !%448! ≈ 4 bua

510;%01/1;!,;5Tebal plat pen*ambung *ang digunakan :




= 1%5 " 1%,; s7d , " 1%,;

= !%055 s7d 4%11

310,

=


3100

=




12.21.103

digunakan L = , cm


t =P

ɸ )u " L

t =1!50/%;! 7 4

0.75 " 4800 " ,

= 0%!850;41 cm


Rn<

/b

0u # t 2

≥ 1!50/%;! 7 4<

1%,;

0.75 " 4800 " 1 !

≥ 1%55,5!!! cm.


= , " 1%,; s7d ; " 1%,;

= 4%11 s7d %5 mm



1405%/!= !%;5,000 ≈ 4 bua

510;%01/1;!,;5





= 1%5 " 1%,; s7d , " 1%,;

= !%055 s7d 4%11

digunakan L = , cm


t =P

ɸ )u " L

t =1405%/! 7 4

0.75 " 4800 " ,

= 0%,!54541; cm



3101

=




12.21.103

Rn<

/b

0u # t 2

≥ 1405%/! 7 4<

1%,;

0.75 " 4800 " 1 !

≥ 1%//1,/!5 cm.


= , " 1%,; s7d ; " 1%,;

= 4%11 s7d %5 mm



1405%/!= !%;5,000 ≈ 4 bua

510;%01/1;!,;5





= 1%5 " 1%,; s7d , " 1%,;

= !%055 s7d 4%11

digunakan L = , cm


t =P

ɸ )u " L

t =1405%/! 7 4

0.75 " 4800 " ,

= 0%,!54541; cm


Rn<

/b

0u # t 2≥ 1405%/! 7 4

<1%,;

0.75 " 4800 " 1 !

≥ 1%//1,/!5 cm.


= , " 1%,; s7d ; " 1%,;

= 4%11 s7d %5 mm


310!

=





12.21.103


Gambar +. ambungan Ikatan ngin Joint 43




12.21.103

inci = !%!!!5 cm

kg%m5&

udirman Dndra &sc




12.21.103




12.21.103

1%/ cm




12.21.103




12.21.103

1%/ cm




12.21.103

Gambar+.1ambun

ganGel. MemanjangdanGe

l. Melint an g




12.21.103

Gambar+.2 Jarak ntar 6aut




12.21.103

inci = !%!!!5 cm

kg%m5&

udirman Dndra &sc




12.21.103




12.21.103

1%/ cm

44 < 51 < 58




12.21.103

1%/ cm




12.21.103

44 < 51 < 58




12.21.103

Gambar+.3ambun gan

Gel.

ind ukdan

Gel.

Melint an




12.21.103

05 mm

aut L = 1%5




12.21.103

cm




12.21.103

,!4%/4 kg

,40,/,%/5 kg




12.21.103

cm

cm




12.21.103

cm




12.21.103

!;;4;/%,8 kg

05 kg




12.21.103

cm

cm




12.21.103

cm




12.21.103

; mm

aut L = 1%5




12.21.103

kg

kg




12.21.103

cm




12.21.103

cm

cm




12.21.103

cm




12.21.103



BAB !II

PERENCANAAN PER"ETAKAN

6.1 PER"ETAKAN SENDI

a.

>irencanakan :

l = L < 40

= 50 < 40

= 0 cm

b = 45 cm

Pu = ,;/!0%0 kg

)* = ,!00

= 1"

, " Pu " l

! b " ɸ " '*

= 1"

, " ,;/!0%0 " 0

! 45 " 0.9 " ,!00

= 1,%;,; ≈ 14cm

b.

&u = 1" Pu " l

8

= 1" ,;/!0%0 " 0

8

= 4!,!,5!%!/!5 kg7cm

N = Mu

ɸ " '*

= 4!,!,5!%,0% " ,!00

= 14/%5//8

Tebal Bantalan S )

kg7cm²

31

Tebal Bantalan S *

cm³

2ntuk arga 3! 3

, dan 3

4dipakai tabel muller $reslaO :



umber 7 8.J tru"k( K.8.9.: ;an /er ;een( oemargono( Jembatan 7 24-

>iambil = 4

b= 4%!

3! a " 3,

>ipakai jumla rusuk (a+ = 4 bua

= 4

3!

b= 4%!

a " 3,

3, =

b

=

45

= !%/;85;14 cm ≈ , cm4%! " a 4%! " 4

&encari nilai dipakai rumus :

N = 0%!!51 " a " " 3,

= 0%!!51 " 4 " " ,

= !%;01! " ?

14/%5//8 = !%;01! " ?

? = 14/%5//8

!%;01! = !,%,!4;1! cm ≈ !5cm

&aka :

= 4 → 3! =

!5≈ /%!5 cm

3! 4

34 =

=!5

= 4%1; cm ≈ 5 cm/ /

²

²

cm³



34 =

=!5

= !%;8 cm ≈ , cm

c. Garis Tenga! Sumbu Sendi

1" =

0%8 " P

! ɸ " )* " L

=0%8 " ,;/!0%0

0.9 " ,!00 " 0

= 1%1/11,! cm

= 1%1/11,! " !

= !%,!!!;8, cm

; cm

=1

"

4

=1

" 4

= 1%;5 cm ≈ ! cm

= < ( ! " +

= ; < ( ! " ! +

= 11 cm

Gambar .1 Penumpu <ngsel Gambar .2 Kurs

d 1

d 1

untuk d1 minimum diambil

d 3

d 1

d 2

d 1

d 3



6.2 PER"ETAKAN R"

a. Panang emiris

>irencanakan :

l = L < 40

= 50 < 40

= 0 cm

b = 45 cm

Pu = ,;/!0%0 kg

b. Tebal Bantalan

=1

", " Pu " l

! b " ɸ " !"

=1

", " ,;/!0%0 " 0

! 45 " 0.9 " ,!00

= 1,%;,; cm ≈ 14 cm

c. Diameter +ol

= 0%;5 " 10 ⁶P

l " ( ɸ " 'u

)u = tgangan putus untuk A5! = 8500

= 0%;5 " 10 ⁶ x,;/!0%0

0 " ( 0.9 " 8500

= 5,%5;04,4 ≈ 50cm

d. Tebal bibir rol

= diambil sebesar !%5 cm

e. Tinggi total rol

= < ! "

= 50 < ! " !%5

31

d4

+ ²

kg7cm²

d4

+ ²

d/

d5

d4

d/



= 55 cm

Gambar .3 Perletakan Rol



iPenumpu <ngsel

perencanaan jembatan rangka baja

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Pengenalan Jembatan Baja - · PDF fileBentuk anatomi dari jembatan rangka baja dan elemen-elemennya terlihat dalam gambar berikut, Gambar 18 : Bentuk anatomi jembatan rangka

PERANCANGAN STRUKTUR JEMBATAN …e-journal.uajy.ac.id/1516/1/0TS12436.pdf · jembatan baja (warren truss), ... Jembatan rangka (truss bridge) ... Langkah – Langkah Pelaksanaan Tugas

Gambar Standar Rangka Baja Bangunan Atas Jembatan Kelas a Dan B

Metode Pelaksanaan Rangka Baja Jembatan

Jembatan Rangka Baja-1

STUDI FORENSIK JEMBATAN RANGKA BAJA SOEKARNO – HATTA MALANG DITINJAU DARI ASPEK KEAMANAN DAN TINGKAT LAYAN JEMBATAN

6˜˜ ˜e˜im˜u˜˜n˜ - UAJY Repositorye-journal.uajy.ac.id/7700/7/TS613645.pdf · Perancangan jembatan berupa konstruksi rangka baja dengan sistem rangka baja pelengkung dengan

STUDI PERENCANAAN JEMBATAN MENGGUNAKAN …repository.untag-sby.ac.id/5628/8/Jurnal.pdfJembatan rangka baja Desa Ipi Kabupaten Morowali ini direncanakan dengan jembatan rangka tipe

SEMINAR TUGAS AKHIR MODIFIKASI PERENCANAAN … · Maka dipilih jembatan dengan busur rangka baja. PERMASALAHAN 1. Bagaimana prosedur perencanaan busur rangka batang baja jembatan?

Perkuatan Jembatan Rangka Baja Australia Dengan Metode Prategang Eksternal

struktur jembatan rangka baja

Kak Ded Jembatan Rangka Baja Long Pipa Ok

Perhitungan Jembatan Baja