PERENCANAAN JEMBATAN RANGKA

7/24/2019 PERENCANAAN JEMBATAN RANGKA

1/20

PERENCANAAN JEMBATAN RANGKA (RBB) BENTANG 32 M


2/20

BEBAN BANGUNAN ATAS

Data Jembatan :

1. BENTANG JEMBATAN ( L ) = 32 mete2. LEBAR JEMBATAN ( B ) = ! mete3. T"NGG" JEMBATAN ( # ) = $ mete

%. BENTANG LAPANGAN JEMBATAN ( a ) = % mete&. TEGANGAN LENTUR BET'N ( ) = & *+,-m2$. MUTU BET'N ( -/ ) = 2& M0a. JEN"S BAJA BJ 3 ( ) = 2% M0a. TEKANAN AS R'DA GANDAR = 2 T4n!. TEKANAN R'DA = 1 T4n1. BERAT ASPAL = 2 *+,m311. BERAT BET'N BERTULANG = 2& *+,m312. BEBAN #"DUP = & *+,m2


3/20

A.PERHITUNGANBEBAN MATI

2

3

2

3

2

Beban mati setiap 1mtebal aspal = 0.05 m

berat aspal = 2000 kg/m

berat q = 2000 x 0.05 = 100 kg/m

Tebal beton = 0.2 m

Berat beton = 2500 kg/m

Berat q = 2500 x 0.2 = 500 kg/m

Jumlah berat q 2

2

2

aspal dan beton = 100500 = !00 kg/m

10beban lain lain 10" dari berat q aspal beton = x!00 = !0 kg/m100

#adi berat total q = !00!0 = !!0 kg/m

$ebar plat %ang di tin#au = 1m

Beban pada plat lantai ke 2

2

2

ndaraan = !!0 kg/m x 1 m = !!0 kg/m

&omen q total '

$ = 1m

&omen = 1/( x q x $

= 1/( x !!0kg/m x 1m = (2.5 kg.m

B.PERHITUNGAN BEBAN HIDUP

1. KEKUATAN LANTAI BETON JEMBATAN


4/20

Beban 5T6 (beban47a)

BIDANG KONTAK :


5/20

ha = 0.05 m

hb = 0.2 m

hn = ha )hb/2*

= 0.05 )0.2/2*

= 0.15 m

x = )2xhn*0.2

= )2x0.15*0.2

= 0.5

+ = )2xhn*0.5

= )2x0.15*0.5

= 0.(

$,- B = x +

= 0.2

5 x 0.(

= 0.m

BEBAN TERSEBAR :

2

Tek. roda = 10000 kg

Tek. roda 10000Tekanan )q* = = = 25000 kg/m

$uas B 0.

aktor ke#ut)* = 1)20/)50$*

= 1 )20/)501*

= 1.342

$ebar plat %ang di ti

2

2

2

n#au = 1m

Beban pada $antai kendaraan = 25000 kg/m x 1m = 25000 kg/m

$=1m )#arak antara balok ba#a*

&omen beban tersebar '

& = x1/(xqxl

& = 1.342x1/(x25000kg/m x 1m

& = 350 kg.m


6/20

Momen Total (Beban T anbe!at"en#!#$

&omen T = 350 kg.m

&omen Berat sendiri = (2.5 kg.m

&omen total = )350(2.5*x100 = 3250 kg.6m

$ebar pelat lantai /1m )b* = 100 6m

Tinggi beton lantai )hb* = 20 6moeisien = 1/! = 0.1!7

8x = 1/!xbxhb2

2

3

2tot

i#in

2 2

8x = 1/!x100x20

8x = !!!!.!7 6m

& 3250%g ter#adi = = = !!.4 kg/6m

8x !!!!.!7

%arat #ika %g ter#adi 9 dari :;

#adi !!.4 kg/6m 9 75 kg/6m :;

%. KEKUATAN TROTOAR


7/20

2

2

2

max

Beban hidup )q* = 500 kg/m

$ )$ebar Trotoar*= 1m

$ebar plat %ang di tin#au = 1m

Beban


8/20

a. GELAGAR MEMANJANG


9/20

Beban an+ 78 *ena*an 0a7a +e9a+a memanan+ a8t; :

BebanMat8

Te788 7a8 beat en788 a0a9< bet4n 7an beban 9a8n9a8n

Beban #87;0

Beban >87;0 0a7a +e9a+a embatan 78nata*an 7en+an beban 5D6 ata;beban a9;< an+ te788 7a8 beban teba+8 ata 5?6 t4n 0e mete 0anan+

0e a9;< 7an beban +a8 5P6 t4n 0e a9; 9a9;98nta teeb;t.

Saat Beban D? :

A0ab89a embatan L @ 3m ma*a : ? = 22 *+,m

A0ab89a embatan 3 @ L@ $m ma*a : ? = (2.2 (1.1,$) (L3))

1

A0ab89a embatan L $m ma*a : ? = 1.1 (13,L) 1

( ) ( )( )

$=32m

q = 2.2 1.1/!0 x 3230 x1000 = 21!3. kg/m

Jadi beban hidup terbagi rata adalah '

21!3.q> = 7(!.7 kg/m/#alur

2.75=

BebanGa8

P = 12 *+,a9;

Bebanteba+8 ata (P) = ( P*+ ) , 2.& mete

= 12,2.&

= %3$3.$% *+,m,a9;

Beaten788

A0a9 = .& m 2 *+,mF = 1 *+,m2

bet4n = .2 m 2& *+,mF = & *+,m2


10/20

= $ *+,m2

9a8n9a8n 1 = 1 $ *+,mH = $ *+,m2

T4ta9 = $$ *+,m2

Jaa*Beban = 1 m

U79 = $$ *+,m2 1 m = $$ *+,m

Ja78BebanPa7aBa94* = $$ *+,m BS Ba94* Baa (%!.! *+,m) = !.!

*+,m

Ge9a+amemanan+78-4bamema*a8 :

P4I9 = " = 2 2

Beat = %!.! *+,m = %2 -mF

K4e*e;t ( * ) = 1 (2,( & L )) = 1 (2,(&

%)) =1.3

M4men ma bebanmeata (D?) = * 1, ? LH

= 1.3 1, $.$ %H =

21&$.&*+.m; = ?; L = ? 9

= $.$ % =

1&3.3% *+

M4men ma beban+a8 (D0) = * O %3$3.$% %= 1.3 O %3$3.$% % =

&!.2 *+.m; = P = P

= %3$3.$% =

211.2 *+

M4men ma beaten788 (BS) = 1, ? LH

= 1, !.! %H =

1%1!. *+.m; = ?; L = ? 9= !.! % =

1%1!.*+

T4ta9M4men = 21&$.& &!.2 1%1!. =

!&&%.& *+.m

= !&.&%1

KN.m


11/20

T4ta9 ; = 1&3.3% 211.2 1%1!. =

&1%.!$ *+

= &1.& KN

Men;;t LRD ( 0e>8t;n+anba94*9ent; )

M4mentea*t4en-ana = M;

= !&.%&1 KN.m

Gaa +eea*t4en-ana = ;

= &1.& *N

U*;an 0enam0an+ be7aa*an *;at 9ent; tan0a 0en+a;> te*;*

Seba+a8ba94*0en7e*e;a8a;m8;Mn = .! Q = 2% N,mm2

Q M; , .! = !&.%&1 KNm , .! 2% N,mm2

= !&.%&1 KNm 1 = !&%&1 Nmm= !&%&1 Nmm , .! 2% N,mm2= %%1.!3 Cm3

Q %2 %%1.!3 -mF.'KS 3!%.$ -mF ( Q, 1.12 )

Ce*0enam0an+ "

" 2 2 12

" = %2 -m% = .$2 -m = 13 mm

" = 1$ -m% = &.2 -m

t = t = 12

Ce**e9an+8n+an

1. P9at aa0( LB )

200 170 170(.33 10.47.......


12/20

2) * 200 2)12 13* 1!(0 1!(0=1(.75 10(........ a7a0 te*;* +ee

n = .$ A

= .$ (( 2 2( 12 13 )) ( $.& ) 2%

= 1%.% *N

;n = .! 1%.% *N = 12$.3$ *N

;n?; = 12$.3$ *N? &1.& KN ..'K

Met47aD8t8b;8M = A .7 .

= ( 2 12 ) ( 2 12) 2% N,mm2= 1.2! KN m M;.'*

Met47e"ntea*8M;,Mn .$2& ;,Mn@ 1.3&!&.&%1 . 1$, (.!)(1.12)(3!%.$. 13) (2%) .$2& . 1&.3, 12$.3$= 1. .= 1. @ 1.3&..'K

Cekdefeksiakibatbebantetap

5. . 5 7.(!7 000

3( 3( 200000 720 10

2.77(

0001!.!!7......... %arat deleksi terpenuhi.

20 20

q l x x

EI x x x

mm ijin

Lijin OK

= =

= +e9a+ame98ntan+ = b;a> Panan+embatan (9) = 32 m Bentan+9a0an+an (L) = !m ( 9eba )

Pe>8t;n+an789a*;*an;nt;*t8a09ebaaa*ebea = %m


14/20

a*t4*e;t

K4e*e;t( * ) = 1 (2,( & L )) = 1 (2,(& !)) =1.33!

Beban7a9ama9;

BebanMeata (D?) = L @ 3m ma*a : ? = 22 *+,m , 2.& =

*+,m

BebanGa8 (D0) = %3$3.$% 1.33! , &.& = 1$2.3% *+,m

T4ta9 D? D0 = 1$2.3% = 1$2.3% *+,m

Beban9;aa9;

BebanMeata (D?) = L @ 3m ma*a : ? = 22 *+,m , 2.& =

*+,m


15/20

BebanGa8 (D0) = * ((2 .&),2.&) (P,2)),(2 .&)

= 1.33! ((2 .&),2.&) (12,2)),(2 .&)

= 2!21.%& *+,m

*.BEBAN AKHIR AKIBAT BEBAN HIDUP

BebanDa9ama9; T4ta9 D? D0 = 1$2.3% = 1$2.3%*+,m

BebanL;aa9;J;m9a> T4ta9 D? D0= 2!21.%& = 321.%& *+,m

Beban#87;00a7at4t4a = * & *+,m2 L

= 1.33! & % = 2$ *+,m

D.BEBAN AKHIR AKIBAT BERAT SENDIRI

A0a9 = .& m 2 *+,mF %m = % *+,m Bet4n = .2 m 2& *+,mF %m = 2 *+,m T4t4a = .2 m 2& *+,mF %m = 2 *+,m

= %% *+,m

9a8n9a8n 1 (t4t4a) = 1 %% *+,m = %%

*+,m

9a8n9a8n 1 (Lt *en7aaan) = 1 (%2) = 2%

*+,m

T4ta9 A*>8Bebant4t4a = %% % = % *+,m

T4ta9 A*>8Beban Lt *en7aaan = % 2 2% = 2$%

*+,m

T4ta9 A*>8BebanGe9a+aMemanan+ = %!.! *+,m %m = 1!!.$ *+


16/20

SKEMA PEMBEBANAN ARAH MELINTANG


17/20

Ce*0enam0an+ "

I+, - / & / - / 1& / %1 0)m

" = 2!2 -m% = 33 -m = 2 mm

" = 11 -m% = $.$2 -m Q = 2! -m3

t = 1%mm t = 2$ mm S = 2! 1.12 = 1$%.-m3

1. BebanPaabalo0te2#t!otoa!

L (0anan+Ba94*te08) = % m

Beban#87;0 = & *+,m2

Bebanbet4nLanta8*en7aaan 2 -m = 2& *+,m3 .2m = &*+,m2

Bebanbet4nT4t4a 2 -m = 2& *+,m3 .2m = & *+,m2

LebaT4t4a = 1 m

T4ta9 Beban0a7aba94*te08t4t4a= (&&&) 1 = 1& *+,m

; = ? L = 1& %= 3 *+ = 3 *N

%. BebanPaabalo0ten)a3

T4ta9 ; = 1&3.3% 211.2 1%1!. = &1%.!$ *+.m= &1.&KN

Unt40PembebananBalo0Mel#ntan)"eba)a#Be!#04t :

Gaa +eeBebanPa7aBa94*Te08 2 = 2 3 = $*N

Gaa +eeBebanPa7aBa94* Ten+a> 2 = 2 &1.& = 13.& *N

ReaksiPerletakanAkibatBebanTerpusat


18/20

) 1 2 3 5 ! 7 (*

2

)!0 103.5 103.5 103.5 103.5 103.5 103.5 !0*

2

71 370.52

P P P P P P P PRA RB

RA RB

RA RB kN

+ + + + + + += =

+ + + + + + += =

= = =

ReaksiPerletakanAkibatBebanMerata

) . *

2

)1.(5 4*(.325

2

q lRA RB

xRA RB kN

= =

= = =

MomenMaksimumAkibatBebanTerpusat

M01 M0 = A.1 = 3.& 1 = 3.& *Nm

M02 M0 = A.2 P1.1 = (3.& 2) ($ 1) = $1 *Nm

M03 M0$ = A.3 P1.2 P2.1 = (3.& 3) ($ 2) (13.& 1)= *Nm

M0% M0& = A.% P1.3 P2.2 P3.1 = (3.& %)($ 3)(13.&

2)(13.& 1)

= !!1.& *Nm

MomenMaksimumAkibatBebanMerata

M = 1, L2= 1, 1.& !2= 1.31 *Nm

T4ta9 M4men = M BebanTe0;at M BebanMeata


19/20

= !!1.& 1.31 = 11.231 *Nm

T4ta9 Gaa Gee = BebanTe0;at BebanMeata

= 3.& .32& = 3.2& *N

Pe!3#t4n)anDe"a#nBalo0 Ba5a Men4!4t LR,D

M4mentea*t4en-ana = M; =11.231*Nm

Gaa +eea*t4en-ana = ; = . a*8batbebante0;at .

a*8batbeaten788

= 3.&*N .32&*N = 3.2& *N

U*;an0enam0an+be7aa*an*;at9ent;tan0a0en+a;>te*;*

Seba+a8ba94*0en7e*e;a8a;m8;Mn = .! Q = 2% N,mm2

Q M; , .! = 11.231KNm , .! 2% N,mm2

= 11.231 KNm 1 = 11231Nmm= 11231Nmm , .! 2% N,mm2 = %$ Cm3

Q&$ %$-mF.'KS $%&1.2 -mF ( Q 1.12 )

Ce**e9an+8n+an

1. P9at aa0( LB )

300 170 170!.25 10.47.......


20/20

n = .$ A

= .$ (&!$ 13) 2%

= 111&.12*N

;n = .! 111&.12*N = 1%.1%1 *N

;n?; = 1%.1%1*N? 3.2&KN ..'K

Met47aD8t8b;8M = A .7 .

=( 3 2% ) ( 2%) 2% N,mm2

= 11$.12 KN m M; = 11.231.'*

Met47e"ntea*8M;,Mn .$2& ;, Mn@ 1.3&11.231 1$, (.!)(1.12)(&$ 13) (2%) .$2& 3.2& ,

$%&1.2 13

= .2%!$ .3$= .2&13 @ 1.3&..'K

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PERENCANAAN JEMBATAN RANGKA