ppt jembatan 2
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jembatanTRANSCRIPT
7/21/2019 PPT Jembatan 2
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Data Perencanaan
•
Bentang total : 32 m• Lebar jembatan : 1 + 8 + 1,5 + 8 + 1 = 19,5 m
• Lebar trotoar : 2 x 1 m
• Lebar median : 1,5 m
•
Mt baja : B! 3"• Mt beton : #$c 25 MPa
• Mt tlangan : #% 2&' MPa
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Data Ba(an
1) *trtr ata :a) -iang andaran, lantai trotoir,
.lat lantai jembatan• Mt beton /#0c : 25 MPa
• Mt baja /#% : 2&' MPa
b) Dia#ragma• Mt beton /#0c : 35 MPa
• Mt baja /#% : 2&' MPa
c) Beton .rategang• Mt beton /#0c : ' MPa
• Mt baja /#% : &'' MPa
2) Bangnan baa(
a) 4btment
• Mt beton /#0c : 3
•
Mt baja /#% : &'
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• -iang *andaran
-iang dari .rol baja /*-63"B = 5' mm 7 = 18" cm&
= 1'' mm = 3",5 cm3
t1 = 5 mm = 9,3 g;m
t2 = " mmBeban (oriontal = 1'' x 2 = 2'' g
Momen maimm %ang terjadi Mmax = 2'' x 5'1'''' g)cm
ce tegangan = = 2," g ; cm2 <1''
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• Dinding andaran
Matan (oriontal = 1'' g ; mleta = 9' cm dari trotoir
P = x L = 1'' x 2,'' = 2'' g
a%a momen am.ai jng trotoir / (
= 9' + 25 = 115 cm = 1,15 mM = P x ( = 2'' x 1,15 = 23' gm
Mn = M ; ? = 23' ; ',8 = 28",5 gm = 28"5'''
Di .aai tlangan ? 12 6 "5 / 4 ter.aang 15'8
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• -rotoarDirencanaan :
• Lebar /b = 1,' m
• -ebal /t = ',25 m
Pembebanan :
a) Beban Mati
1 = 1,2 x ',"5 x ',25 x 25'' = 52,5 g;m
2 = 1,2 x ',2 ' x 2,'' x 25'' = 12'' g;m
3 = 1,2 x ',25 x 1,'' x 25'' = "5' g;m
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• -rotoar
b) Beban id.A1 = 1, x 55' = 88' g;m
A2 = 1, x 125 = 2'' g;m
A3 = 1, x 1'' = 1' g;m
c) Beban ter.at
P1 = 1,2 x 2 x 2 x 5,'8 = 2&,38& gP2 = 1,2 x ',5' x 9,3 = 5,58 g
P3 = 1,2 x ','2 x ',15x ',15 x "85'= &,239 g
d) Momen /ter(ada. tit i 4
MA1 = 88' x 1,'' x 1,25 = 11'' gm
MA2 = 2'' x 1,'' x ',&5 = 9' gm
MA3 = 1' x 1,'' x 1,35 = 21 gm
Mg1 = 52,5 x 1,'' x 1,8"5 = 1'5&,9 gm
Mg2 = 12'' x 1,'' x 1,'' = 12'' gm
Mg3 = "5' x 1,'' x 1,25 = 93",5 gm
M.1 = 2&,38& x 1,"8815 = &3,'2 gm M.21',&25 gm
M.3 = &,239 x 1,8"5 = ",95 gm
M = &',2'' gm
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• -rotoar
e) Penlangan trotoarMn = M ; ? = &',2'' ; ',8= 5825,25' gm = 582525''@mm
Di .aai tlangan ? 12 6 "5/ 4 = 15'8 mm2
-langan .embagi = ',2 x 4tlangan tama = ',2 x15'8 = 3'1, mm2
!adi tlangan %ang dignaan
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• Plat Lantai !embatan
•
C Beban Mati /AD= Berat endiri .elat + Berat a.al + Berat air (ja
= ( x b x B! beton + t x b x B! a.al + t( x b x B!
= ',2 x 1 x 25'' + ',1 x 1 x 22'' + ','5 x 1 x1'
= ""' g;m = ","' @;m
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Beban 4ibat Matan E-E .ada Lan>endaraan
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Beban roda : - = 1'' @
Bidang roda : bx = 5' + 2 /1' + 1' = = ',9 m
b% = 3' + 2 /1' + 1' = "' cm = ',"Bidang onta : bx% = '," x ',9 = ',3'
Matan - diebaran : - = 1'' ; ',3' =158,"@;m2
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• Momen .ada aat 1 / at roda berada .ada tenga(.lat
Mxm = ',1538 x 158,"3' x '," x ',9 = 15,38 @m
M%m = ','885 x 158,"3' x '," x ',9 = 8,85 @m
• Momen .ada aat 2 / da roda berdeatan dengan antara a e a
Mxm = 7 77 = 21,111 2,"19 = 18,392 @m
M%m = 7 77 = 13,222 1,1"' = 12,'52 @m
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4ibat beban ementara
• Beban ementara adala( beban angin %ang be
.ada endaraan ebear A = 15' g;m2 .ada ar(oriFontal etinggi 2 /da meter dari lantai jem
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• Geai .ada roda = / 2 x 5 x 1x 15' ; 1,"5 = g = 8,5"1 @
• Beban angin diebaran = 8,5"1 : / '," x ',9 =13,'5 @;m2
Di tinja aibat beban 1 / at roda :
Mxm = ',1538 x 13,'5 x '," x ',9 = 1,318 @
M%m = ','885 x 13,'5 x '," x ',9 = ',"59 @
Momen .ada aat 2 / da roda :
Mxm = 7 77 = 1,8'9 ',233 = 1,5" @m
M%m = 7 77 = 1,133 ',1'' = 1,'33 @m
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>ombinai .embebanan :
•
4ibat beban mati + beban H-I .ada aat 1 roda beratenga( .lat
Mxm = 1,9&' + 15,38 + 1,318 = 18,38 @m
M%m = ',&'' + 8,85 + ',"59 = 1',''9 @m
• 4ibat beban mati + beban H-I .ada aat 2 roda berd
Mxm = 1,9&' + 18,392 + 1,5"= 21,9'8 @mM%m = ',&'' + 12,'52+ 1,'33 = 13,&85 @m
• Momen deain di .aai momen %ang terbear
Mxm = 21,9'8 @m
M%m = 13,&85 @m
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Penlangan Plat Lantai
•
Penlangan la.anganara( x
Mn = M ; ? = 219'8)1'3 ;',8 = 2"385)1'3 @mm
Di .aai tlangan ? 12 6
125 / 4 = 9'5 mm2 • Penlangan la.angan
ara( %
Mn = M ; ? = 13&85)1'3 ;
',8 = 185,25)1'3 @mm
• Penlangan tm.an ara( x
M = 1;1')AD ) L2 = 1;1' ) ","' ) 2,''2 =
3,'8)1' @mmMn = M ; ? = 3,'8)1' ; ',8 = 3,85)1' @m
Di .aai tlangan ? 12 6 125 / 4 = 9'5 mm
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Balo Prategang
Direncanaan :• Mt beton .rategang / #0c = ' MPa
• Berat jeni beton / B! = 25'' g;m3
• Mt baja / #% = J K 13 mm #% = 2&' MPa
J 13 mm #% = &'' MPa
• -%.e abel .rategang = ncoated *eNen*tre6relieNed
ig( rade Lo Gelaxation 4*-M 46&1
• Pengangran = *item Ore%inet
• Daar .erencanaan = Partial Pretreing
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Penairan -inggi Balo
• tinggi balo
= 1;1" x 32
= 1,89 m 1,9' m
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Per(itngan momen ineria BaPrategang
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•
Penentan cgc balo .rategang Qb/. = C 4) Q ; C 4 = "3218,125 ; "235 =cm
Qt/. = 19' 93,'5 = 9,95 cm
• Penentan bata inti balo .rategang
>t/. = 7x ; / 4 ) Qb/.
= 329'318',8'; / "235 x 93,'5 = &8,8"5 cm
>b/. = 7x ; / 4 x Qt/.
= 329'318',8'; / "235 x 9,95 = &,9'8 cm
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elagar >om.oitDirencanaan :
• Mt beton gelagar .rategang : #0c = ' MPa
• Mt beton .elat lantai : #0c = 25 MPa
• Modl elatiita beton / R = &"3' S#0c
R .lat = &"3' S25
R balo= &"3' S'
• 4nga eiNalen / n = R balo ; R .lat = 1,55
•
Mencari lebar be menrt 44*-Be = L;& = 32''' ; & = 8''' mm
Be = 12 x t.lat = 12 x 2'' = 2&'' mm
Be = jara antar gelagar = 2''' mm
Diambil nilai terecil 2''' mm
• be ; n = 2''' ; 1,55 = 129',32 mm = 129,'32 cm
• 4 = 129,'32 x 2' = 258',& cm2
Per(itngan momen ineria Ba
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Per(itngan momen ineria Ba>om.oit
• Penentan cgc balo om.oit
Qb/c = C 4) Q ; C 4 = 11893&5,"5; 9815,&5 = 12
Qt/c = 21' 121,18 = 88,832 cm
• Penentan bata inti balo om.oit
>t/c = 7x ; / 4 ) Qb/c
= 5&"&"29,39; / 9815,&5 x 121,18 = &,'31 cm
>b/c = 7x ; / 4 x Qt/c
= 5&"&"29,39; /9815,&5 x 88,832 = 2,"8" cm
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Pembebanan Balo Prategang
1) Berat endiri balo / AD1
Momen .ada jara x dari 4 : a%a Lintang .adadari 4 :
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2) Beban mati tamba(an / AD2
Momen .ada jara x dari 4 : a%a Lintang .a
dari 4 :
3 beban ter.at
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3) beban ter.atdia#ragma
• Berat dia#ragma P = b x ( x t x B!beton=2,''x1,'8'x',2x25''=1'8'
1',8' @• Ban%an%a dia#ragma %ang di.aang ada " ba(
PD1 = >M* x P = 1,2 x 1',8' = 12,9 >@
Mencari reai tm.an :
C MB = '
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Beban (id. / Beban lajr D
• Beban terbagi rata ebear A ton .er m0 .er jalr
nt .ia elebar / * 2,'' m
A0 = / A ; 2,"5 x * x T
= / 2,133 ; 2,"5 x 2,'' x 1,'' = 1,5"3 ton;m = 15,
AL1 = 1, U 15,"3 @;m0 = 25,18 @;m0
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Momen .ada jara x dari 4 : a%a Lintang .addari 4 :
Beban gari ebear P
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Beban gari ebear P.er jalrP = 12 ton, nt .ia elebar / * 2,' m
P0 = /P;2,"5 x > x * x T = /12;2,"5 x 1,2&39 x 2,= 1',8558 ton = 1'8,558 @ V PL1 = 1, U @ = 1"3,928 @
Momen .ada jara x dari 4 : a%a Lintang .ada
Beban *ender .ada Balo
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Beban *ender .ada BaloPrategang• 4ibat rem dan trai
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• -egangan bata betonberdaaran PB7 W"1
# Wc = ' MPa
# Wci = tegangan beton .adamr 1& (ari =
',88)' MPa = 52,8 53 MPa
• -egangan tegangan ijin
>ondii aal :
# ci = ', ) # Wci = ', ) 53 = 32 MPa
# ti = 6',5 )S # Wci = 6',5 )S 53 = 63,&' MPa
>ondii a(ir
# c = ',&5 # Wc = ',&5) ' = 2" MPa
# t = 6',5 S # Wc = 6',5 S ' = 6&,33" MPa
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*t= XML + /1 G MDY ; /#c + G ) #ti
= X &11,'& x 1' + /1 ',85 2""8,112 x 1'/ 2" + ',85 ) 3,&'
= 1"'811," mm3 K *t .enam.ang
*b= X ML + /1 G MD Y ; / #t + G ) #ci
= X &11,'& x 1' + /1 ',85 2""8,112 x 1'/ &,33" + ',85 ) 32
= 159&2&25",2 mm3 K *b .enam.ang
-egangan6tegangan aibat beban (id. dan beban mt t d b ( di(it d b
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erat ata dan baa( di(itng dengan rm ebagaZMDto. = MDg; *t/. = 2""8,112 x 1' ; 339382,98,18 MPa ZMDbot = MDg; *b/. = 2""8,112 x 1'353'",531 x 1'3 = 6",85 MPa
ZMD-to. = MDtot ; *t/c = 5558,2"2 x 1' ; 1295= 9,'19 MPa ZMD-bot = MDtot ; *b/c = 5558,2"2 x&5182&,982 x 1'3 = 612,3'2 MPa ZMLto. = ML ; *t/c&11,'& x 1' ; 1295,135 x 1'3 = ",&82 MPa ZML ; *b/c = &11,'& x 1' ; &5182&,982 x 1'3 MPa
-egangan .enam.ang di(itng berdaaran & a
• >a 7 : #c = 2" MPa #ci = 32 MPa
• >a 77 : #ti = 63,&' MPa #t = 6&,33" MPa
• >a 777 : #c = 2" MPa #t = 6&,33" MPa
• >a 7[ : #ti = 63,&' MPa #ci = 32 MPa
a e er ngan eganganP
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a e er ngan eganganPenam.ang
Di t
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Diagram tegangan .enam.angdalam & a
-egangan %ang terjadi .ada gari netral .enam.abeton /cgc da.at di(itng dengan memaai.erbandingan egitiga)
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Berdaaran tabel terebt, ita da.at menggambaran dengan mem.lot e ebagai #ngi -i etia. titi di daera(
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dengan mem.lot e ebagai #ngi -i, etia. titi di daera( memberian deain %ang bai erta memen(i .er%aratategangan ijin)
ambar daera( aman -i dan e
e max = %b \ ]tendon 6 ]tl)begel 6 ]tl)tama 6 .ent.
= 93,'5 8,;2 1,' 2,2 & = 81,55 cm
Di.ili( -i = 9&'' @ dengan e = "'' mm K e max = 815,5 m
on ro-
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on ro -egangan
1) a%a .ratean dan beratendiri eaat etela(tran#er tegangan .ratean
= 1,"9' MPa K 63,&' MPao ^
= 23,"&& MPa K 32 MPa o ^
2) *etela( e(ilangan tegangan dan .elat dicor
= 1',9&' MPa _ 6&,33" MPa o ^
= 11,1&2 MPa K 2" MPa o ^
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3) *etela( beban (id. beerja.ada balo om.oit
= 1,"3" MPa K 2" MPa o ^
= ',93" MPa _ 6&,33" MPao ^
&) -egangan .ada erat terata dan terb
= ",&82 K ',&5)25 = 11,25 MPa
= 5,"9" K 11,25 MPa
ambar diagram tegangan %an
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ambar diagram tegangan %anterjadi
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*etela( beban (id. beerja dan terjadi e(ilangan • Pada erat terbaa( aan terjadi tegangan :
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• Pada erat terbaa( aan terjadi tegangan :
• Pada erat terata aan terjadi tegangan :
-race dari e1, e2, e3 dan e& da.at dili dan aan memberaman bagi tendon) Bata bata ini da.at dieder(anaan me& aja
La% t -endon
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La% t -endonPrategang
• Bent la% ot tendon memanjang adala( .arab
nt menentan .oii tendon dignaan .er.arabola :
Q = 4`2 + B` +
d%;dx = '
Maa 24` + B = '
nt x = 1''', maa B = 632''' 4
nt % = "'' mm, maa :
Q = 4`2 32''' 4` V dida.at nilai 4 = 6','''''2= ','8"&88
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-endon• Di.aai ncoated *eNen6ire *tre relieNed #or
oncrete ig(grade6Lo Gelaxation 4*-M &.engangran item Ore%inet
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Menentan leta maing maingtendon
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tendon
a) Poii tendon .ada end bloc ;tm.an /x = '
leta cgc .ada tm.an = 93',5mm dari erat baa(
jara tendon e erat baa( /a =525 mm
jara antar tendon /b = 325 mm
Maa :
-endon 1 = 525 mm
-endon 2 = 85' mm
-endon 3 = 11"5 mm -
b) Poii tendon .ada tenga( bentan
e .ada tenga( bentang = "'' mm
leta cg .ada tenga( bentang = 93',5 baa(
Mial diambil a = 23',5 mm maa :
-endon 1 = 23',5 mm
-endon 2 = 23',5 mm
-endon 3 = 23',5 mm
-endon & = 23',5 mm
c) Poii tendon .ada jara x meterdari tm.an
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dari tm.an
-endon 1
Q = 4`2 + B` +
nt x = ' V % = 525 mm , maa c= 525 mm
d%;dx = ' Maa 24` + B = '
nt % = 23',5 mm, maa : Q =4`2 32''' 4` + 525 , dida.atnilai 4 = 1,15'39'25)1'6 V B =6','38125 Peramaan .arabolatendon 1 adala( :
Q = 1,15'39'25)1'6`2
' '38125 ` + 525
-endon 2
Q = 4`2 + B` +
nt x = ' V % = 85' mm , maa c = 85
d%;dx = ' Maa 24` + B = '
nt % = 23',5 mm, maa : Q = 4`2 34 = 2,&199218"5)1'6 V B = 6','""&3"5
Peramaan .arabola tendon 2 adala( : Q','""&3"5 ` + 85'
-endon 3Q = 4`2 + B` +
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Q = 4`2 + B` +
nt x = ' V % = 11"5 mm ,maa c = 11"5 mm
d%;dx = ' Maa 24` + B = 'nt % = 23',5 mm, maa : Q= 4`2 32''' 4` + 11"5 ,dida.at nilai 4 =3,89&53125)1'6 V B =
6',118'25
Peramaan .arabola tendon 3adala( : Q = 3,89&53125)1'6`2 ',118'25 ` + 11"5
-endon & Q = 4`2 + B` +
nt x = ' V % = 15'' mm , maa c = 15''
d%;dx = ' Maa 24` + B = '
nt % = 23',5 mm, maa : Q = 4`2 32''= &,95898&3"5)1'6 V B = 6',1588"5
Peramaan .arabola tendon & adala( : Q = &,',1588"5 ` + 15''