PPT Jembatan 2

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



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 /#% : &'



Per(itngan *trtr 4ta



• -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''



• 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



• -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



• -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



• -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



• 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



Beban 4ibat Matan E-E .ada Lan>endaraan



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



• 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



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



• 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



>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



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



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



Penairan -inggi Balo

• tinggi balo

= 1;1" x 32

= 1,89 m 1,9' m



Per(itngan momen ineria BaPrategang



•

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



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



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



Pembebanan Balo Prategang

1) Berat endiri balo / AD1

Momen .ada jara x dari 4 : a%a Lintang .adadari 4 :



2) Beban mati tamba(an / AD2

Momen .ada jara x dari 4 : a%a Lintang .a

dari 4 :

3 beban ter.at



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 = '



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



Momen .ada jara x dari 4 : a%a Lintang .addari 4 :

Beban gari ebear P



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



Beban *ender .ada BaloPrategang• 4ibat rem dan trai



4nalia a%a Pratean



• -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



*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



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



a e er ngan eganganPenam.ang

Di t



Diagram tegangan .enam.angdalam & a

-egangan %ang terjadi .ada gari netral .enam.abeton /cgc da.at di(itng dengan memaai.erbandingan egitiga)



Berdaaran tabel terebt, ita da.at menggambaran dengan mem.lot e ebagai #ngi -i etia. titi di daera(



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-



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 ^



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



ambar diagram tegangan %anterjadi



*etela( beban (id. beerja dan terjadi e(ilangan • Pada erat terbaa( aan terjadi tegangan :



• 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



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



Penentan !mla(-endon



-endon• Di.aai ncoated *eNen6ire *tre relieNed #or

oncrete ig(grade6Lo Gelaxation 4*-M &.engangran item Ore%inet



Menentan leta maing maingtendon



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



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` +



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''

PPT Jembatan 2

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