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




NIM : 12.21.103




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




1221103

r !%8 al

40 I




1221103

k7m atau




1221103

enjadi !




1221103

1%0 m




1221103




1221103




1221103

3 bag !




1221103




1221103




1221103




1221103

08%0040 kg




1221103

)*

,!00

50%/,; kg




1221103

K




1221103

kg




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




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

cm




12.21.103

,!4%/4 kg

,40,/,%/5 kg




12.21.103

cm

cm




12.21.103

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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