tb pelabuhan mila.xls

7/25/2019 TB Pelabuhan MILA.xls

1/39

I. KRITERIA PERENCANAAN

1. Lokasi Perencanaan Dermaga

Lokasi Perencanaan Dermaga terletak 45 m dari garis pantai dengan elevasi dasar laut = -21.2

2. Kondisi Alam

Angin : Bean angin 4! ertiup se"a"ar dengan dermagaPasang surut : #levasi muka air terenda$ %L&'( = 2.4! m

#levasi muka air tertinggi %)&'( = 4.5! m

3. Data-data Kapal

* +ross ,onage = 2!!!! ton

* verall Lengt$ = 2! m

* /olded Breat$ = 20.5! m

* /olded Drat = .5 m

* 3ull Load Drat = 1! m

4. Beban yang bekerja

* Bean mati %&D( = Berat sendiri konstruksi

* Bean $idup %&L( = * Bean merata = .!!* Bean crane = 45.!! ton

* Bean truck = 5.!! ton

. Konstr!ksi

Digunakan dermaga $ar catalk

Lear dermaga %L( = a 4 c 6 kedalaman 7 085 m

dimana :

a = oversteak depan = 1.5! m

= oversteak tenga$ = 4.25 m

c = oversteak elakang = 1.5! m

se$ingga

L = 1.5! 4. 4825 1.5!

= 2!.!! m

Dari tael diperole$ :

* 9ntuk kapal dengan oot 2!!!! ton maka pan"ang dermaga = 2! m

* Pan"ang $ar = 0!

* Pan"ang catalk = %2! - 0! ( 2 = ! m

". #ketsa Dermaga

kgm2

tonm2

APP;A


2/39

m


3/39

II.PERENCANAAN BANGUNAN ATAS

A. P$%$&'A&AA& P$LA( LA&(A)

* ,eal Pelat Lantai = 2 cm

* ,eal Lapisan Aus = 1 cm

* = 24!!

* = 05!

A. Pembebanan

A.1.1

Berat sendiri = !.2 @ 24!! = 02

Berat lapisan aus = !.!1 @ 05! = 0.5

Bean mati = 02 0.5 = 05!.5

/omen permeter lear menurut ''?> C1 %tael 14(

L=

4= 1.!L = 4 m L@ 4

L@ = 4 m

/>@ = !.!!1 @ @ @ 1 = 02.24 kgm

/> = !.!!1 @ @ @ C = 4.12 kgm

/t = !.!!1 @ @ L@2 @ C1 = 1!C2.02 kgm

/ti@ = 12 . /i@ = 1.12 kgm

/ti = 12 . /i = 24.1 kgm

A.1.2

= 25!!

/>@ = !.!!1 @ @ @ 1 = 124! kgm

/> = !.!!1 @ @ @ C = 15! kgm

/t = !.!!1 @ @ @ C1 = 4! kgm

/ti@ = 12 . /i@ = 2! kgm

/ti = 12 . /i = 0! kgm

A.1.2Beban bergerak terp!sat

a. Bean


4/39

. Bean ,ruck

* ,ekanan gandar ma@ = ! ton

* ,ekanan roda ma@ = !.5 @ ! = 15 ton

* Luas idang kontak = !.2 @ !.5

* Luas idang penearan = !. @ 1.5

/omen ang ter"adi pada plat akiat ean crane dan ean truck

Dari tael konstruksi eton >ndonesia %ole$ >r. 'utami(

/>@ /> /t@ /t

-!.!2 -!.!10 !.!2 !.1

-!.!10 -!.!2 !.1 !.!2

!.1! !.1! -!.55 -!.55

!.C! !.C! 1.!5 1.!5

ditentukan dengan persamaan :

/ = @

= = p

arena ean terpusat dan ergerak maka momen-momen ma@ diperole$ didemisioner s :

'>@ = F!84 - %2 !84 %@L@( !82 %L( - !8 %@L@( %L(G L@

'> = F!84 - %1 !82 %@L@( !84 %L( - !8 %@L@( %L(G L

't@ = F!8 - %2 !81 %@L@( - !81 %L( !81 %@L@( %L(G L@

't = F!8 - %1 - !81 %@L@( !81 %L( !81 %@L@( %L(G L

dimana :

= !.1 6 = !.1 % untuk u"ung ter"epit(

a. Akibat beban crane

ondisi > : satu ua$ out ringer crane ditenga$ plat

L@ = L = 4 m

@ = = !.05 m

P = ! ton =

= !!!! kg

= = !

Diperole$ momen-momen seagai erikut :

/>@ = 4510.15C kgm '>@ = -.5C2105

/> = 4510.15C kgm '> = -2.2C2105

/t@ = -22.C5 kgm 't@ = -5.55C05/t = -22.C5 kgm 't = -1.45C05

m2

m2

an

a1

a2

a

a4

9ntuk meng$itung momen-momen akiat ean terpusat8 koeisien-koeisien a18 a

28 a

8 dan a

4dapat

a1 %@L@( a2 %L( a

%@L@( %L( a4

/uatan E %kgm2(

c1

c2

c1

c2

2! cm

5! cm

5! cm

! cm

1 cm

2 cm

Plat dermaga

>@ = 4 m

> = 4 m

!845 !805


5/39

/omen seenarna :

/>@'>@ = -5.222!10

/>'i = -1C0!.0511C440

/t@'t@ = 115.54005540

/t't = 411.CCC41

L@ = L = 4 m

@ = = !.05 m

= %245 2@45 2 - 2@14( =

= !.05 m

= %245 - 2@45 2 - 2@14( =

= !.05 m

Diperole$ momen-momen s :

/>@ = 1.405 kgm '>@ = -0.022

/> = 1.505C kgm '> = -2.0522

/t@ = -2.21! kgm 't@ = -5.!C2

/t = -24.C02 kgm 't = -!.C02

/omen seenarna :

/>@'>@ = -10.1CC111C041 kgm

/>'i = -!4.C41425 kgm

/t@'t@ = 500.4!1!!5 kgm

/t't = 2C4.241CCC0 kgm

a. Akibat beban tr!ck

ondisi > : ;oda truck terletak di tenga$ plat lantai dermaga

@ = !.0

= !.5

L@ = L = 4 m

P = 25 on = 25!!!

= = !.1

Diperole$ momen-momen seagai erikut :

/>@ = C5.5C5C kgm '>@ = -.2CC

/> = 4!!.5!0 kgm '> = -2.50C/t@ = -505!.0110 kgm 't@ = -5.!C

/t = -54.054 kgm 't = -1.452C

/omen seenarna :

/>@'>@ = -5C.25440!5

/>'i = -10!C.4!2111214

/t@'t@ = 1!25.2101002

/t't = C1.4205404

@= %1!! 2@5! 2 2@14( = 1.0= %1!! - 2@5! 2 - 2@14( = !.2

ondisi >> : 'aat kedua roda erdekatan dengan "arak antara a5ke a

5min 2845 m

c1=c

2

@1

@2

@2

2

c1

c2

ondisi >> : 'aat kedua roda erdekatan dengan "arak antara a5ke a

5min 2845 m

!805

!845

!805

!845

!825 !85

!85

18!


6/39

/t't = 44.55C50C2


7/39

(abel omen

). Akibat Beban 'rane

/

ondisi Bean

3ormula

/omen 'eenarna

> %gm( >> %gm( > %gm( >> %gm/>@ 4510.15C 1.405 />@'>@ -5.222!10 -10.1CC1

/> 4510.15C 1.505C />'> -1C0!.0511C440 -!4.C4

/t@ -22.C5 -2.21! /t@'t@ 115.54005540 500.4!1!

/t -22.C5 -24.C02 /t't 411.CCC41 2C4.241C

'>@ -.5C2105 -0.022

/ = /> - />>

/>@ = -5!C.!2C0!1- . - . - .

't@ -5.55C05 -5.!C2 /t@ = !.1411C

't -1.45C05 -!.C02 /t = 140.54CC2

)). Akibat Beban (r!ck

/ondisi Bean

3ormula/omen 'eenarna

> %gm( >> %gm( > %gm( >> %gm

/>@ C5.5C5C 25!.421 />@'>@ -5C.25440!5 -4!1.15

/> 4!!.5!0 4.CC />'> -10!C.4!2111214 -12C.5!

/t@ -505!.0110 -511!.0!2C /t@'t@ 1!25.2101002 C.4!412

/t -54.054 -44C.!5 /t't C1.4205404 44.55

'>@ -.2CC -0.!C00

/ = /> - />>

/>@ = -1C4.0!015

'> -2.50C -2.C00 /> = -4C.CC2C

't@ -5.!C -5.0421 /t@ = 15.C10!0!

't -1.452C -1.C!21 /t = 5.055221

%$KAP)(/LA#) 0$&-0$& K%)()#

?Bean ang dipikul /omen ang ter"adi %kgm(

ole$ plat lantai /i@ /i /t@ /t

> Bean mati %ED( 02.24 4.12 1.124 24.15

>> Bean $idup %EL( 124! 15! 2! 0!

>>> Bean crane -5!C.!2C0!120 -15.024CC1C !.1411C44 140.54C>H Bean truck -1C4.0!015!0 -4C.CC2CC2 15.C10!0!0! 5.0552

ominasi Pemeanan

1 > >> 112.24 2!2.12 !.124 1!14.1

2 > >>> -1.010!120 -C0.414CC1C 0C4.2 11.

> >H 100.5C124C24 -1.5C2CC2 22.!42 !2.C

9ntuk momen rencana diamil momen teresar dalam perencanaan plat lantai :

/>@ = 112.24 kgm

/> = 2!2.12 kgm

/t@ = !.124 kgm

/t = 11.11 kgm


8/39

A. Perencanaan (!langan Plat Lantai

* Diamil data-data a$an :

/utu eton c = 25 /pa = 25!

/utu a"a = 4!! /pa = 4!!!

* Dimensi tin"auan plat :

= Lear tin"auan = 1!!! mm

$ = ,eal plat = 2! mm

P = Penutup eton = 4! mm

= Perkiraan tul. utama ara$ @ = 2! mm

= Perkiraan tul. utama ara$ = 2! mm

* ,inggi eekti d dalam ara$ @ adala$ :

d@ = 2! - 4! -

= 2! mm

* ,inggi eekti d dalam ara$ @ adala$ :

d = 2! - P - -

= 21! mm

* /omen lapangan dalam ara$ @

/>@ = 112.24 kgm

= = !400.200

/enurut uku graik dan tael perencanaan eton ertulang tael -2!

= !.!!02

= !.!!25 %tael 0(

= !.!4!4 %tael (

I I

As . L@ =

= !.!!02 @ 1 @ !.2 @ = 15

A = 14n = As . L@ A

= 15 14 = 5.2055 = atang

Jarak tulangan untuk lear tin"auan 1!!! m adala$ :

1!!!= 1.0 = 15! mm

* /omen lapangan dalam ara$

/i = 2!2.12 kgm

= = 45CC.4C


= !.!!0

As . L@ =

= !.!!0 @ 1 @ !.21 @ = 14!0

A = 14

n = As . L@ A

= 14!0 14 = 4.41 = 5 atang


1!!!= 2!! = 2!! mm

5

* /omen "epit tak terduga dalam ara$ @

/t@ = !.124 kgm

= = 152.C41


= !.!!1

As . t@ =

= !.!!1 @ 1 @ !.2 @ = A = 0.5

n = As . L@ A

= 0.5 = 4.0C!C2 = 5 atang


kgcm2

kgcm

D

D

12D@

12D D@

/ud2 112824 %18! . !82(2 kgm2

min

ma@

min

ma@

. . d . 1!

1! mm2

Digunakan tulangan 2! 6 mm2

2!-15!

/ud2 2!2812 %18! . !821(2 kgm2

. . d . 1!

1! mm2


2!-2!!

/ud2 !8124 %18! . !82(2 kgm2

. . d . 1!

1! mm2



9/39

* /omen "epit tak terduga dalam ara$

/t@ = 11.11 kgm

= = 4201.451!0


= !.!!15

As . t@ =

= !.!!15 @ 1 @ !.21 @ = 15

A = 0.5

n = As . L@ A

= 15 0.5 = 4.!12055 = 5 atang


1!!!= 2!! = 2!! mm

5

B. P$%$&'A&AA& PLA( LA&(A)

B. Perencanaan Balok emanjang Ara

B.1.1Akibat beban mati *D,a. Bean Plat Lantai %Ep(

E = 05!.5

E = !.5 @ E @ L@ = 15!1 kgm

Bean segitiga diratakan %K( :

K = !.5 @ E @ L@ = !!2 kE

= = !.5 @ K = 15!1 kE

/ma@ =

= 25!1.0 gm

/ma@ =

Ep =/ma@

=25!1.0

= 1251 kgm2

. Berat sendiri alok %E(

E = !84 . % !85 - !82 ( . c

= 0!! kgm

se$ingga D = Ep E

= 125!. 0!! = 4C51 kgm

= 4C.5! ?m

B.1.2Akibat beban id!p merata *L,

EL = 25!!

E = !.5 @ EL @ L@

= !.5 @ 25!! @ 4

= 5!!! kgm

Bean segitiga diratakan %K( :

K = !.5 @ E @ L@= !.5 @ 5!!! @ 4 = 1!!!! kg

= = !.5 @ K = 5!!! kg

/ma@ =

= . kgm

/ud2 11811 %18! . !821(2 kgm2

. . d . 1!

1! mm2


1!-2!!

kgm2

;A

;B

;A

%!85 . >@( -!85 . 15!1 . 1 %!85 . >@(

1 . Ep . L@2

1 . L@2

Ditaksir ukuran alok 4! @ 5 cm2

kgm2

;A

;B

;A

%!85 . >@( -!85 . E 1 %!85 . >@(

L@ = 4 m

E

L@ = 4 m

;A ;

B

4! cm

5 cm

E

;A

;B


10/39

/omen ang menentukan :

=

= 11 @ &u @ = C1.42! ?m

=

= 114 @ &u @ = 1!4.4!C52 ?m=

= 11! @ &u @ = 14.20 ?m

=

= 11 @ &u @ = C1.42! ?m

Perit!ngan (!langan

Diamil :

$ = ,inggi alok = 5! mm

P = Penutup eton = 4! mm

= Diameter tul. utama = 25 mm

= Diameter tul. sengkang = mm

d = ,inggi eekti

= $ - P - -

= 5! - 4! - - !85 . 25

= 5C.5 mm = 5C! mm

%encana Pen!langan Balok

x ,ulangan tumpuan A dan #

/u = C1.42! ?m

= = 5.50

= !.!!25

= !.!!!5

= !.!2!

I I

Luas tul Ast. == !.!!25 @ ! @ !.5C

= 554.

0!.5

x ,ulangan Lapangan AB dan D#

/u = 1!4.4!C52 ?m

= = 05!.5C2!0

= !.!!2C

Luas tul Ast. =

= !.!!2C @ !.4 @ !.5C @

= 4.4

.125

x ,ulangan Lapangan B dan D

/u = 14.20 ?m

= = 1!5!.512!4C

= !.!!42

Luas tul Ast. =

= !.!!42 @ !.4 @ !.5C @

= CC1.2

1!5C.05

x ,ulangan Lapangan B< dan AB

/>D#

L2

/B

/D

L2

/>B


11/39

Kontrol Lebar Balok

= 4!! mm

2 penutup alok = 2 @ 4! = !

2 tul. 'engkang = 2 @ = 1

= 4 @ 1! = 4!

= 4 @ 22 = 4 "arak antara = 4 @ 4! = 1!

= 4 mm

sarat

4!! 7 4

Kontrol %etak Balok

dimana :

= Perandingan lear retak pada penampang tak ertulang ter$adap lear retak

penampang ertulang

s = !8 . = !8 . 4!! = 24!

A = % 2 . dc . ( n 6 dc = $ - d = 5! - 5C! = ! mm 6 n = 5 ua$

= % 2 . ! . 4!! ( 5= C!!

se$ingga :

= 11 . 182 . 24! . ! . C!!

= 225.2221

=

= !!!!!

sarat :

7 !!!!! 7 225.2221

41!

422

total

total

= 11 . . s . dc . A

!8 . 1!


12/39


13/39


14/39


15/39


16/39


17/39


18/39


19/39


20/39


21/39

Perencanaan (!langan eser Ara

/utu eton c = 25 /pa % 25! (

/utu a"a = 4!! /pa % 4!! (

9kuran alok : = 4!! @ 5! 6 d = 5C! mm

Dari per$itungan seelumna diperole$ &u = C1.42! ?m

x/eng$itung gaa lintang [email protected]

/enurut ''?> C1 pasal .1. erlaku untuk alok ang ditentukan dengan koe. momen %tael 12(8 esar gaa lintang.

Hu = 1.15 &u@ ln2 %entang terluar tumpuan ter"epit(

Hu = &u @ ln2 %untuk seluru$ tumpuan lain(

Bentuk diagram gaa lintang ditentukan secara:

- ,umpuan A dan #

Hma@ = &u @ ln2 = C1.42! @4

= 12.410 ?2

-

Hma@ = 1.15 @ &u @ln2= 1.15 @ C1.42! @4

= 21!.20C10 ?2

-

Hma@ = 1.15 @ &u @ln2= 1.15 @ C1.42! @4

= 21!.20C10 ?2

-

Hma@ = 1.15 @ &u @ln2= 1.15 @ C1.42! @4

= 21!.20C10 ?2

/enurut ''?> ,15 C1 . ! pasal .4.1.2

+aa lintang ang ter"adi ole$ diproduksi sampai $arga d = !.5C m dari muka tumpuan

/aka gaa lintang ang ter"adi :

- ,umpuan A dan #

Hu = 12.410 - !.5C @ C1.42! = 12.C! ?- ,umpuan B8 < dan D

Hu = 21!.20C10 - !.5C @ C1.42! = 15.2C ?

21!.20 21!. 21!.

12.

15.! 15.! 15.!

12.C!

12.C

15.! 15.! 15.!

21!.20 21!. 21!.

x/encari nilai Hu dimana Hu M Hc

/enurut tael 15 untuk mutu eton c = 25 /pa 6 Hc = !.5 /pa

- ,umpuan A dan #

Hu =Hu

=12C!.05

= !.542!!042 /pad 4!! @ 5C!

karena Hu 7 Hc !.542!1 7 !.5 /pa maka digunakan tulangan geser

-

Hu = Hu = 152C.25 = !.2415 /pad 4!! @ 5C!

karena Hu 7 Hc !.2414 7 !.5 /pa maka digunakan tulangan geser

-

kgcm2

kgcm2

mm2

,umpuan Bkidan


22/39

x/enentukan pan"ang se$ingg Hu 7 Hc

- ,umpuan A dan #

Lokasi dengan Hu = Hc = !.5 /pa didapat dengan persamaan :

=Hma@ - Hc

6 Hc = Hc @ @ d

&u = !.5 @ 4!! @ 5C!

= 12.42 - 11= 11 ?

C1.42!= !.0!C2 m

Dengan nilai Hu sepan"ang !.5C m dari tumpuan konstan dan sepan"ang :

!.0!C - !.5C = !.11C m8 nilai Hu menurun

=% Hu - Hc ( d

=% !.542 - !.5 ( 4!! @ 5C!

!8 !. @ 4!!

= 45.4!0

sengkang ang memadai adala 2 penampang ganda = 1!! atau - 105

-

=21!.2 - 11

= 1.!!C20 mC1.42!

Dengan nilai Hu sepan"ang !.5C m dari tumpuan konstan dan sepan"ang :

1.!!C - !.5C = !.41C m8 nilai Hu menurun

=. - . @

= 15C.0!001!. @ 4!!

sengkang ang memadai adala 5 1! penampang ganda = C atau 1! - 15!

xPerencanaan ,ulangan +eser /iring

Hs =Av @ @ % ( d

s

dimana :

s = !.5 @ d = !.5 @ 5C! = 2C5 mm

Av =% s (

=4!! @ 2C5

= C. mm @ 4!!

= 45

se$ingga

Hs = C. @ 4!! @ % sin 45 cos 45 ( 5C! = 11125!.42C5

kontrol

Hs =% 2 c (

@ @ d

=% 2 25 (

@ 4!! @ 5C!

= 0.0 ?

sarat :

Hs Hs 0.0 11125!.4

"adi :v = . p u geser =

ontrol ukuran Agregat maksimum ter$adap "umla$ tulangan

BPerencanaan Balok elintang Ara y

,aksiran ukuran alok :

4! @ 5

Assengk.

mm2

mm2

,umpuan Bkidan C1dimana :

&u = 1.2 @ % 1.2 !.CC4 1!.052 ( 1. @ % 4! 55 ( = 1.!5 tm

se$ingga :

Hu = 1.15 @ 1.!5C @ !.5 = C. ton

= C. - % !.5C @ .120155 ( = C4.0455 ton

Hu = =C4.0455

= !.!21!4 @ d 1!! @ 45

karena :

7 2.1! /pa 7 !.5 /pa 8 maka tidak diperlukan tulangan geser

,ulangan tekan A = 2! Q @ A

= 2! Q

Digunakan tulangan ougel praktis 1! NNN

tm2

tm2

/ult.

$2. 2 . o . k

E1

Emin

E2

Emin

Emin

E . $ . 2 . o . k

a"a

mm2

mm2

Huseenarna

Huseenarna

tonm2

u

-

! 4! !

!

4!

!

! 4! !

5!

5!


29/39

D. P$%$&'A&AA& 'A(8ALK

x Per$itungan Lantai Dasar

Dipakai kau kelas >> mutu B %tidak terlindung(

=N

@N

@ ! @ !.05N N

= 05

x Pemeanan ang ter"adi %E(

Berat sendiri kau :

!.!5 @ !.2 @ C!! = C kgm

Bean $idup = 1!! kgm

Bean total %E( = 1!C kgm

x& =

N@ @ $ N =

N@ 2! @ N N = .

N N

/ =N

@ E @ % N @ N a N ( =N

@ 1!C @ % ! N @ N @ 1! N (N N

= 105! kgcm

=/

=105!

= C.1

.E = 1!C kgm sarat :

M C.1 M 05 S

x Perencanaan Proil Ba"a

Digunakan proil D>? 2! dengan data-data seagai erikut :

$ = 2! cm 3 = 2.0

= 2! cm + = 4.C

c = 1 cm &@ = 5C5

d = 1. cm >k = 5C5

$t = 1. cm L = 15E = 1!C kgm

Ktot = K +

= 1!C 4.C

= 10.C kgm

/ =N

@ E @ =/

=4C!.C

= .22!!N 5C5

=N

@ 10.C @ 15 NN sarat :

= 4C!.C kgm M .22!! M 14!! S

sketsa proil a"a D>? 2!

it

kgcm2

cm2

tot kgcm

2

tot

it kgcm

2

L2 tot kgcm2

tot kgcm

2

1! ! 1!

a 1! 1!!

a

$t

t

t

d d$


30/39


31/39

. P$%$&'A&AA& D0LP9)&

x +aa-gaa vertikal ang eker"a :

1. Berat sendiri dolp$in

= % 4 @ 4 ( @ 1.25 @ 2.4 = 4 ton

2. Berat sendiri tiang pancang

= % !.5 @ !.5 ( @ ! @ 2.4 = 1 ton

. +aa tarik kapal seesar 5 ton

mementuk sudut 2 ter$adap $oriontal.

mementuk sudut ! ter$adap depan dolp$in

x Per$itungan reaksi tiang iin %P(

P = Daa dukung tiang - erat tiang

= 0!4.2 - 1 = .2 ton

P = +aa geser ma@. tiang - erat tiang

= 1414.24 - 1 = 1C.24 ton

= P sin 25 = 5 sin 25 = 2.0 ton

= P cos 25 = 5 cos 25 = 5!.05 ton

= P cos ! = 5 cos ! = 4.C5 ton

4 @ = ! @ = 0.5

; 2 = ! 2 0.5 2 ; = !.C2

sin =0.5

= !.24254 = 14.!!.C2

3 tekan dolp$in = .2 @ .2 = 1C.5 ton

3 tarik dolp$in = 1C.24 @ 1C.24 = 25.41 ton

o

o

tekan

tarik

PH

o o

P)

o o

P)

o o

o

%1cos 148!o(

%1cos 148!o(

4

!84 !85 !84 !85 !84 !85 !84 !85 !84

4


32/39

.1. Perit!ngan (iang Pancang

3.1.1. ,iang pancang ang memikul ean vertikal$oriontal T momen

Pvt = 5! 1.05 2.0 = C2.42 ton

=

= C2.42 @ 1.5 C2.42 @ !.5

C2.42 @ 1.5 C2.42 @ !.5

= 241.1C1 tm

;eaksi momen dua ara$ ter$adap tiang P

ang mena$an

P = U Un

= 5! 1.05 2.0 = 22.0 ton

=

= C2.42 1 C2.42 1 = 14.4 tm

= 2 % ( = .C m

= 4 % ( = 4 m

=22.0

241.1C1

-14.4

= .!044 ton .C 4

=22.0

-241.1C1

-14.4

= -45.11 ton .C 4

3.1.2.

P = % (

= % .!04 sin 14.! 4.C5 cos 14.! (

sin % 14.! 14.! (

= C5.!1 ton

sarat :

P P

.2 C5.!1 RR.. %S(

3.1..

P = % (

= % 45.1!CC sin 14.! 4.C5 cos 14.! (sin % 14.! 14.! (

= 114.!!5 ton

sarat :

P P

1C.24 114.!!5 RR.. S

.1. Pen!langan Dolpin

/ = 241.1C1 tm = 2.0C ?m

$ = 4 m

= 1.25 m

d = 125! - 1!! - !.5 25 = 110.5 mm = 1.105 m

P = 1!! mm

= 25 mm

/u = 1.5 @ 2.0C = 55!.!15 ?m

/u 55! !15

/1

P1d

1P

1d

1P

1d

1P

1d

1

cos

cos

H /1.d1 /2.

d2 2

H

/2

P1

1P

2

2

d2 d12 d

22 d

2 d

42

2 12

22

Ptekan

Ptarik

Ptekan

maksimum ter"adi saat P tekan eker"a sama dengan P)

tk. ma@P

tk'in

1 P)


33/39

Dipili$ tulangan geser = !.2 @ 05.05

= 05!.05 1! 1!!mm2


34/39

PERENCANAAN PELABUHAN

III PERENCANAAN BANGUNAN ATAS

A. P$%$&'A&AA& P0&DA#) ()A& PA&'A&

A.1. Per$itungan gaa vertikal pada tiang pancang

Bean vertikal untuk 1 tiang pancang %>( :

Plat = N @ 4 @ !.2 @ 2.4 = 1!.052 ton

Lapisan aus = N @ 4 @ !.!1 @ 0.5 = 1.25 tonBalok = N @ % N @ !.4 @ !.5 ( @ 2.4 = 4.CC2 ton

Poer = % N @ N @ !.5 !. @ !. @ !.5 @ N ( @ 2.4 = 1.2 ton

,iang = !.5 @ !.5 @ 25 @ 2.4 = 15 ton

= .2 ton

Bean di atas dermaga

Bean $idup merata = .!!

= N @ N @ .!! = 4 ton

Bean terpusat dari truck = 5.!! ton

diperkiraan di atas dermaga %4 @ 4( terdapat dua truck8 maka :

= N @ 5.!! = 0! ton

Bean terpusat dari crane = 45.!! ton

satu gandar crane diletakkan di atas tiang % "arak gandar = 2.45 m (

= 45 % 1.55 N ( 45 = 2.405 ton

ominasi Pemeanan Pada ,iang Pancang

ominasi > = = .2 4 0! = 151.2 ton

ominasi >> = = .2 4 2.4 = 144.!C5 ton

A.2. Per$itungan gaa dukung tana$

a.Lapisan >

D3 = %kedalaman lapisan tana$( = 10 m

= !.

? =

c = !.

= ! : diperole$ : ?c = 0.2 6 ?E = 22.5 6 = 2!maka :

= 1. ?c t D3 ?E !.4 B

= 1. @ 0.2 @ !. !. @ 10 @ 22.5 !.4 @ !. @ !.5 @ 2!

= 24.4!

.Lapisan >>


= !.5

? = 1!

c = 1.5

= 5 : diperole$ : ?c = 50. 6 ?E = 41.4 6 = 44

maka :

= 1. ?c c D3 ?E !.4 B

= 1. @ 50. @ 1.5 !.5 @ 2 @ 41.4 !.4 @ !.5 @ !.5 @ 44

= 00.

c. Lapisan >>>


= !.05

? = 4!

c = N

= 4! : diperole$ : ?c = C5. 6 ?E = 1.2 6 = 114

maka :

= 1. ?c c D3 ?E !.4 B

Bean vertikal %v1(

tonm2

v2

v

v4

v1

v2

v

v1

v2

v4

Dari data soil investigationdiperole$ :

o

?

Eult ?

tm2

o ?

Eult ?

tm2

o ?

Eult ?


35/39


= 1. @ C5. @ 2 !.05 @ 1 @ 1.2 !.4 @ !.05 @ !.5 @ 114

= 11.

A.. Per$itungan gaa geser maksimum pada tiang pancang

3s = 9 @ @ i

dimana :

3s = gaa geser ma@. tiang %ton(

9 = keliling tiang pancang %m(

= kedalaman %teal( riap lapisan tana$ %m(

i =

Lapisan,eal tiap lapisan

Jenis tana$ ?i Li.i

%m( %tm(

> % 1! - 20 ( 10 Pasir $alus 1!2

>> % 20 - 5! ( 2 Lempung keras 1! 1! 2!

>>> % 5! - ( 1 Lempung keras ! ! 54!

02

9 = N @ !.5 N @ !.5 = 2

se$ingga :

3s = 2 @ 02 = 1044

Kp = A @

= % !.5 @ !.5 ( @ 11.

= 4!.40 ton

A.4. Per$itungan daa dukung iin tiang pancang

x Berdasarkan kekuatan a$an tiang pancang

= c @ A

dimana :

c = tegangan iin tekan = !. @ c = !. @ 25! = 2.5

= 2.5 @ % 5! @ 5! ( = 2!25! kg = 2!.25 ton

x ,er$adap daa dukung tana$

% Kp 3s ( '3

dimana :

'3 = angka kemanan = %untuk keadaan gempa(

se$ingga :

% 4!.40 1044 ( 151.2 ton

C4.2 ton 151.2 ton RRRR %S(

A.5. Daa dukung tana$ pondasi

a.Lapisan >

D3 = 10

= !.

? =

c = !.

= ! : diperole$ : ?c = 0.2 6 ?E = 22.5 6 = 2!

maka :

= 1. ?c t D3 ?E !.4 B

= 1. @ 0.2 @ !. !. @ 10 @ 22.5 !.4 @ !. @ !.5 @ 2!

= 24.4!

.Lapisan >>

D3 = 2 = !.5

? = 1!

tm2

Li

Li

"umla$ rata-rata nilai ? tiap lapisan %tm2(

%tm2(

Li.i

m2

,ekanan u"ung tiang %End Bearing File(

Eult.

Ktiang

kgcm2

Ktiang

H

o ?

Eult ?

tm2


36/39


c = 1.5

= 5 : diperole$ : ?c = 50. 6 ?E = 41.4 6 = 44

maka :

= 1. ?c t D3 ?E !.4 B

= 1. @ 50. @ 1.5 !.5 @ 2 @ 41.4 !.4 @ !.5 @ !.5 @ 44

= 00.

c. Lapisan >>>D3 = 1 = !.05

? = 4!

c = N

= 4! : diperole$ : ?c = C5. 6 ?E = 1.2 6 = 114

maka :

= 1. ?c t D3 ?E !.4 B

= 1. @ C5. @ 2 !.05 @ 1 @ 1.2 !.4 @ !.05 @ !.5 @ 114

= 11.

A.. Daa dukung tiang pancang

Lapisan,eal tiap lapisan

Jenis tana$ ?i Li.i

%m( %tm(> % 1! - 20 ( 10 Pasir $alus 1!2

>> % 20 - 5! ( 2 Lempung keras 1! 1! 2!

>>> % 5! - ( 1 Lempung keras ! ! 54!

02

3s = 2 @ 02 = 1044

Kp = A @

= % !.5 @ !.5 ( @ 11.

= 4!.40 ton

ontrol ter$adap gaa vertikal ang eker"a

% 4!.40 1044 ( N 151.2 ton

C4.2 ton 151.2 ton RRRR %S(

A.0. Per$itungan gaa $oriontal pada tiang pancang

x +aa $oriontal akiat gaa tarik ender

;@. 3ender ) = 2.1220 ton dan

diteruskan pada konstruksi dan menear

+aa $oriontal dipikul ole$ 4 tiang miring8

"adi gaa $oriontal per 1 unit tiang adala$:

)3 = 2.122 4 = 0.!45 ton

x +aa tarikan kapal pada ollard

+aa tarik pada ollard = 0! ton

gaa tarik ini dipikul ole$ 4 tiang miring8 maka gaa $oriontal per 1 unit tiang adala$:

) = 0! 4 = 10.5 ton

o ?

Eult ?

tm2

o ?

Eult ?

tm2

%tm2(

Li.i

Eult.

dengan sudut 45o

)

45o 45o

!

)


37/39


x +aa $oriontal akiat rotasi %momen trosi ter$adap pusat dermaga(

Ditin"au dermaga seagai satu kesatuan struktur dimana gaa akiat tumukan kapal

dianggap menimulkan rotasi ter$adap pusat konstruksi.

Analisa persamaan :

)i = % ) n ( % @i ( ) e

dimana :

)i = gaa $oriontal pada tiang

) = gaa $oriontal akiat reaksi ender

n = "umla$ tiang miring

@i = "arak antara pusat tiang ang ditin"au ter$adap pusat dermaga

= "umla$ "arak antara tiang kuadrat ter$adap pusat erat dermaga

x +aa $oriontal akiat gempa

3 = @

dimana :

3 = gaa $oriontal akiat gempa

= erat sendiri konstruksi

= koe. +empa = !.!0 @ E @ 1.2 = !.!4 E

se$ingga :

3 = !.!4 E @ E = !.!4 @ !.212 = !.250! ton

3) = !.250 4 = 0.54452 ton

x +aa $oriontal ma@. ang ter"adi saat gempa8 dimana gaa ini dilimpa$kan pada tiang miring"adi ean ma@. pada tiang miring adala$ :

= = 14.!

v = C2.42

) = 0.545

) = )

= 0.545 14.! = 0.C! ton

=v )

sin % (

=C2.42 sin 14.! 0.C!1 cos 14.!

sin % 14.! 14.! (

= 0.5C ton %tekan(

=v )

sin % (

=C2.42 sin 14.! - 0.54452 cos 14.!

sin % 14.! 14.! (

= .!!C ton %tekan(

A.. Penulangan tiang pancang

Penulangan tiang pancang didasarkan pada keutu$an pada saat pengangkatan pada tiang

pancang

pan"ang tiang pancang = 20 cos 14.!= 20.2C15 m = 2 m

@i

@i

1

2

o

cos1

cos o

P1

sin1

cos2

1

2

o o

o o

P2

sin1

cos2

1

2

o o

o o

o

H

)

1

P1

2

P2

m

)


38/39


= N % N N N 2 1 2 2! N 24 N 2 N 2 2 N ( = C12! m

se$ingga :

)i = % 2.12 1! ( % C12! ( 2.1220 = .2!00 ton

x +aa $oriontal akiat gempa

Bean-ean ang diper$itungkan

-

Lapisan aus = @ 2! @ !.!!1 @ 0.5 = 5.52

Plat lantai = @ 2! @ !.2 @ 2.4 = 4.4

Balok ara$ melintang = !.4 @ !.5 @ 2! @ 1! @ 2.4 = 124.

Balok ara$ meman"ang = !.4 @ !.5 @ 11 @ 2.4 = 240.1

Balok ender = !.5 @ 1.2 @ 4 @ 1.5 @ 2.4 = C.5!4

Poer = % 1 N !.5 !. !. !.5 4 ( 2.4 = 1.2

Berat sendiri konstruksi %E1( = 02.1 ton

-

= !.5 N 2! 2.5 = 1!! ton diper$itungkan seesar 5!Q

Berat total %Et( = = 02.1 1!! = C02.1 ton

E = % !.5 !.5 ( 2.4 = !. tm

=N

E l 2N

=N

!. .5 2 -.05 tmN

=N

-1

E % ( 2N 2 2

=N

.4C0 4.05 -1

!. .25 2N 2

= 12.2125 tm

= = -.05 tm

3aktor ke"ut / = 25 Q

/ = 21!.5 kgm

= 1.5 @ 21!.5 = C15.05 kgm = .414 kgm

= !.5 m

d = 5!! - 4! - N - % !.5 25 ( = 4C.5 mm

=

.414

= C0.0!051 k?m!.5 !.4C5 = !.!!1

As = d 1! = !.!!1 @ !.5 @ !.4C5 @ 1! N = 51.

Digunakan 5 12 = 55.2

Digunakan tulangan ougel praktis 15!

A.C. ,ulangan geser tiang pancang %tiang miring(

Av =% s (

=5!! @ 21C.05

= C1. mm N @ 4!!

= 45

se$ingga

@i

Berat sendiri konstruksi %E1(

Bean $idup %E2(

E2

E1

E2

/1

/1

;A

l2

l2

/

/1

/9

/9

d2

mm2

mm2

-

o

@1

@2

@2

@2

@

@

@4

@4

@5

@5

@

@

@0

@0

@

@C

@

@C

/1

/2

/


39/39


Hs =C1.5 @ 4!! @ % sin 45 cos 45 ( 4C.5

= 1!5C!.15 ?21C.05

kontrol

Hs =% N c (

@ @ dN

=% N 25 (

@ 5!! @ 4C.5

= 025!! ?

sarat :

Hs Hs 025!! 1!5C!.15

"adi :

Av = C1.525 dipili$ tul geser N 1! = 150

A.1!. ontrol daa dukung a$an

Pd = Ac c As

= 5! N 25 5.1 15!

= 12054 kg

= 120.54 ton

sarat :

Pd 120.54 0.5C!12 ton RRR. %S(

A.11. ontrol tegangan selama pemancangan

Digunakan alat pemancang tpe $ammer-vuloon ?o. !81!

- Berat $ammer %( = 5!!! l = 220! kg

- ,ingi "atu$ %$( = >n = C1.44 cm

- #isiensi alat pemancang = !.

- oe. retriusi = !.4

Pv = . $

6 ' = !.5N % ' !.25 (

=220! . C1.44

= 412.4 kg = 4.124 tonN % !.5 !.25 (

sarat :

Pv 4.124 M 120.54 ton RRR. S

A.12. ,egangan ang ter"adi selama pemancangan

= 122.1 = 12.2C

n = 22

As = Ac % n - N ( = 5! % 22 - N ( 12.2C = 205.!C

P = erat tiang = 105! kg

e = % P ( % P ( = % 220! !.4 N 105! ( % 220! 105! (= !.25!01

= 5!.!04!00

sarat :

5!.!04!00!C M 25! RRRR. S

mm2 mm2

a

P1

Pd

Asengk mm2 cm2

Asengk

e2

.

ma@ kgcm2

ma@ kgcm2

tb pelabuhan mila.xls

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