Excel Gudang

PT TATA BUMI RAYA

PERENCANAAN STRUKTUR BAJA Proyek : PT INDONESIA TRI SEMBILAN SEMBIL AN Pekerjaan : KANTOR PABRIK Lokasi : NGORO - MOJOKERTO

PERENCANAAN KOLOM W Profil kolom Jarak antar kuda-kuda Bentang kuda-kuda Sudut kemiringan ording Jarak antar gording %ta&

= = = = = = =

WF-250.125.5.8 6 m 23 m 15 o ! 125.50.20.3"2 10# $m $m 'in$oated 0.35mm

PEMBEBANAN Be(an )ang diterima gording * - Berat ata& = +"5 , 1.0# , 6 - Berat gording = #"51 , 6

= =

a ain-lain 10 

= +2.06 / 10

=

- Berat air uan - Be(an &ekera

= 0"005 , 1000 , 6 , 1.0# = 25 , 100 otal P

= = =

#5.00 28.1# +3.1# +.31 80.#6 31.20 #0.00 151.66

kg kg kg kg kg kg kg kg

Berat 4endiri WF = 25.+ kgm akan diitung otomati4 ole &rogram S%P-0 Selanutn)a untuk men$ari ga)a-ga)a dalam dari rangka ( atang" digunakan &rogram S%P0

6.00

23.00

23.00

23.00

PT TATA BUMI RAYA

PERENCANAAN STRUKTUR BAJA

PT TATA BUMI RAYA


PEMBEBANAN PADA LANTAI KANTOR LANTAI 2 : Be(an &ada (alok WF (era4al dari lantai multi&lek 7 (alok ka)u 612 *

200

12 P

12 P

60

- (e ( erat ka)u 612 - (erat multi&lek 18mm - (e(an idu&

= 0"06/0"12/1000/2 = 0"018/0"6/2/800 = 250/0"6/2

= = =

P

= =

1#.#0 1+.28 300.00 331.68 60.00 31.68

12 P

=

15.8# kg

- (e(an lain-lain

kg kg kg kg kg kg

Selanutn)a untuk men$ari ga)a-ga)a dalam dari rangka ( atang" digunakan &rogram S%P0 Berat 4endiri (alok WF akan diitung otomati4 ole &rogram S%P-0

PT TATA BUMI RAYA


PT TATA BUMI RAYA


KONTROL KEK!ATAN PROIL KOLOM W ari a4il 9ut&ut S%P-0 dida&at * :omen ma, teradi &ada element no.  * - :omen ma, ;:< = 1"+1.6# kg kg.m - a)a ekan ;P< = - #"36+.2+ kg - &anang &anang tekuk ;lk< = 320.00 $m - WF W F-200.100.5"5.8 % = 2+.16  = 18# i, = 8.2# i) = 2.22 !ek tekuk * l , = lk  i, terada& 4um(u , * 320.00 = = 8.2# l , = lk  i, terada& 4um(u , * 320.00 = = 2.22

$m2 $m3 $m $m

38.8

1##.1

> ;menekuk teada& 4um(u )<

!ek terada& 4)arat PPBB? &4 #..1 utk &o rtal dimana uungn)a (ergo)ang P : n, ;1< s = w, 0.85 q @ s 7 %  n, - 1 P : n, ;2< s = w) 0 . 8 5 q @ s 7 %  n, - 1 P : ;3< s = q @ s 7 %  karena 4um(u lentur ;4( ,< tegak luru4 td 4um(u tekuk ;4( )< maka faktor am& am&lifika4i n, =1 n, - 1 q = am(il 1 l,= 38.8 A w, = 1.13 l)= 1##.1 A w) = #.01 4)arat PPBB? * ;1<

s =

= = ;2<

;3<

s =

#"36+.2+ 1+"16# 7 0.85 1 2+.16 18# 181.+0 7 82+.66 1"00 1"00. .36 36 kg$ kg$m2 m2 @ 1600 1600 kg$ kg$m2 m2 ;ok ;ok<<

1.13

= =

#"36+.2+ 1+"16# 7 0.85 1 2+.16 18# 6##.80 7 82+.66 1"#+ 1"#+2. 2.#6 #6 kg$ kg$m2 m2 @ 1600 1600 kg$ kg$m2 m2 ;ok ;ok<<

= =

#"36+.2+ 1+"16# 7 1 2+.16 18# 160.80 7 +3.+2 1"13 1"13#. #.52 52 kg$ kg$m2 m2 @ 1600 1600 kg$ kg$m2 m2 ;ok ;ok<<

#.01

s =

PT TATA BUMI RAYA


PT TATA BUMI RAYA


KONTROL KEK!ATAN PROIL BALOK W-"#$ ari a4il 9ut&ut S%P-0 dida&at * :omen ma, teradi &ada element no. 21 * - :omen ma, ;:< = 2"236.0# kg kg.m - a)a ekan ;P< = 100.00 kg kg - WF-250.125.5.8 = lua4 ;%< = 32.68 $m2  = 285 $m3

s =

P %

s < siin =

: 

100.00 32.68

7

223"60# 285

3.06

7

+8#.58

= =

7

+8+.6# kg$m2 1600 kg$m2 --------A 9 C

KONTROL KEK!ATAN PROIL BALOK W-"$$ ari a4il 9ut&ut S%P-0 dida&at * :omen ma, teradi &ada element no. #1 * - :omen ma, ;:< = 2"#6+.23 kg kg.m - a)a ekan ;P< = 850.52 kg kg - WF-250.125.5.8 = lua4 ;%< = 2+.16 $m2  = 18# $m3

s =

= = s < siin =

P %

7

: 

850.52 2+.16

7

2#6"+23 18#

31.32

7

1"3#0.8

1"3+2.20 kg$m2 1600 kg$m2 --------A 9 C

PT TATA BUMI RAYA

PT TATA BUMI RAYA

PERHITUNGAN STRUKTUR Proyek : R!KO KEBRAON Pekerjaan : A%a& Baja Lokasi : Ke'raon( S)ra'aya

PERENCANAAN GORDING Data teknik : 6.00 m 1.25 m

- Jarak antar kuda-kuda ; l < = - Jarak antar gording =

= 11 o = 25.00 kgm2 = !EP 125.50.20.3"2 ?, = 181 $m#  W, = ?) = 26.6 $m $m#  W) =

- Sudut kemiringan ata& - Berat ata& a4(e4 - Dkuran gording

Beban gording : - Berat 4endiri gording - Berat ata& a4(e4 * 11.3 , 1"25 - Berat air uan = 0"02,1000,1"25

- Be(an &ekera P

=

2 $m3 8.02 $m $m3

= = =

6.31 1#.13 25 #5.# #5.## #

kgm kgm kgm kgm kgm

100 kg ditenga (entang

Momen pada gording : G, = G $o4 a G) = G 4in a

akibat beban mati : &anan &anang g (entan (entang g gordin gording g l, = G, = #5.## $o411 = G) = #5.## 4in 11 = :, = 18 ;G,< . l,2

=

6.00 6.00 m ##.60 kgm 8.6+ kgm 200.+0 kg.m

untuk ara )" di&a4ang 2 ( &enggantung gording" 4eingga (entang gording l) = 2.00 m :) = 18 ;G)< . l)2 = #.33 kg.m akibat beban hidup : :, = 1# ;100 $o4 a < l, = :) = 1# ;100 4in a < l) =

1#+.2# kg.m .5# kg.m

:, total = :) total =

1#+.2# = .5# =

200.+0 #.33

7 7

3# 3#+.5 kg.m 13.88 kg.m

hal - 5

PT TATA BUMI RAYA

PERHITUNGAN STRUKTUR Proyek : R!KO KEBRAON Pekerjaan : A%a& Baja Lokasi : Ke'raon( S)ra'aya

ek Tegangan

s

:, W,

= =

:) W)

7

3#+ 3#+#. #.51 511+ 1+# # 138+ 138+.5 .515 153 3 7 2 8.02 1"1.81 7

= =

1"3+2.82

1+3.01

@ 1"600 -----A 9

ek Lendutan 5 G, l# 38# H?,

d, = = =

7

1 P, l3 #8 H?,

28"00"#+"38.0# 1#5"58"#00"000.00

= =

1.801 3.1#22 $m 5 G) l# 38# H?)

d) = =

7

7

7

1.1621

1 P) l3 #8 H?)

=

63"552"536.3 21"#50"2#0"000.00

7

= =

0.0323 7 0.083 $m

0.056

d,

d = =

152"6#+"16.30 2"681"280"000.00

d) ;

=

d =

7

21"203"1#+"162.#+ 18"2##"800"000.00

3. 3 .1#22

7

0.083 <

1.+8 $m 1250 ;600< =

dtotal @ d

2.# $m

------A 9

hal - 6

PERENCANAAN PONDASI ( Untuk Gudang bentang 10 m') Pem(e(anan &ada 1 titik kolom * - a)a ak4ial &ada kolom ;dari out&ut S%P-0< - Berat &il$a& = 0.6 / 0.6 / 0.2 / 2#00 - Berat 4loof 20,#0 = 0.2 / 0.# / 2#00 / 6m - Berat kolom &ede4tal = 0.2 / 0.3 / 2#00 / 1m - Berat dinding (ata )ang menum&ang &ada 4loof = 250kgm2 / 6m / 6m

= = = = =

1"+31.13 1+2.80 1"152.00 1##.00

kg kg kg kg kg

"000.00 kg 12"1.3 kg

Berda4arkan rekomenda4i Ponda4i dari P %4ia Iarda * i&akai tiang &an$ang 20,20 $m" dengan kedalaman 4e(agai (erikut * Eo

oka4i

e&t ;m<

all ;ton<

D4ulan ata Bumi Ka)a

1

S.1

11

15.6

all A 12"1.3 >..9

2

S.2

10

1+.6

all A 12"1.3 >..9

3

S.3

6.5



all @ 12"1.3 >.!not >. !not OK 4e(aikn)a di&akai tiang &an$ang dengan kedalaman 10 m

PERENCANAAN PONDASI ( Untuk Gudang bentang 23 m') Pem(e(anan &ada 1 titik kolom * - a)a ak4ial &ada kolom ;dari out&ut S%P-0< - %ki(at momen kolom = 10 P - Berat &il$a& = 0.6 / 0.6 / 0.2 / 2#00 - Berat 4loof 20,#0 = 0.2 / 0.# / 2#00 / 6m - Berat kolom &ede4tal = 0.2 / 0.3 / 2#00 / 1m - Berat dinding (ata )ang menum&ang &ada 4loof = 250kgm2 / 6m / 6m

= = = = = = = =

2"1+. 218.00 1+2.80 1"152.00 1##.00

kg kg kg kg kg kg

"000.00 kg 12"866.+ kg 12.8+ ton

Berda4arkan rekomenda4i Ponda4i dari P %4ia Iarda * i&akai tiang &an$ang 20,20 $m" dengan kedalaman 4e(agai (erikut * Eo

oka4i

e&t ;m<

all ;ton<

1

S.1

11

15.6

all

A

12.8+ >.. 9

2

S.2

10

1+.6

all

A

12.8+ >.. 9

3

S.3

6.5



all

@

12.8+ "! not OK

D4ulan ata Bumi Ka)a

4e(aikn)a di&akai tiang &an$ang dengan kedalaman 10 m

B# $% & 'e $() )

*

2

$

+

,

(

%

-

)

1.000

1.000

1.000

1.000

1.000

1.000

1.000

1.000

1.000

*)

1.000

1.000

1.000

1.000

1.000

1.000

1.000

1.000

1.000

2)

1.000

1.00#

1.011

1.01+

1.02#

1.031

$)

1.066

1.0+3

1.081

1.088

1.06

1.103

+)

1.1##

1.152

1.161

1.16

1.1+8

1.18+

,)

1.23#

1.2##

1.25#

1.26#

1.2+#

1.28#

()

1.33

1.351

1.363

1.3+5

1.38+

1.3

%)

1.#65

1.#+8

1.#2

1.50+

1.521

1.536

-)

1.616

1.632

1.650

1.66+

1.685

1.+03

.)

1.801

1.822

1.8##

1.866

1.888

1.11

*))

2.036

2.062

2.00

2.118

2.1#+

2.1++

**)

2.3#0

2.3+5

2.#21

2.#6#

2.508

2.552

*2)

2.++

2.826

2.8+3

2.20

2.68

3.016

*$)

3.262

3.312

3.363

3.#1#

3.#65

3.51+

*+)

3.+83

3.83+

3.82

3.#+

#.002

#.058

*,)

#.3#2

#.#01

#.#5

#.518

#.5++

#.63+

*()

#.#1

5.003

5.065

5.128

5.11

5.25#

*%)

5.5+8

5.6#3

5.+10

5.++6

5.8#3

5.11

*-)

6.253

6.323

6.33

6.#63

6.53#

6.605

*.)

6.6+

+.0#1

+.115

+.18

+.26#

+.33

2))

+.+20

. 1.000 1.000

PER/IT0N1AN TR0KT0R #EMBATAN IT


1ambar 3ket3 4embatan :

balok , t=90cm t=90cm

2.30

30.10

2.30

HE%I JH:B%%E JH:B%%E

balok , t=90cm 7 5.00 7 #.10

kolom 40x40

kolom 40x40

jalan 7- 0.00 2.30

30.10

2.30

PHE%:P%E JH:B%%E

railling pagar 7 6.20

plat lantai t=12cm 7 5.00

7 #.10

0.#0 0.15

1.50 2.00

0.#0 0.15

P99E%E JH:B%%E JH:B%%E

PERENANAAN PELAT LANTAI

11


Pembebanan per m5 pada ba6ok ge6agar utama : Be(an :ati * - Berat &elat lantai = 0.12 , 2.3 , 2#00 kgm3 - Berat 4$reed = 2.3 , #0 kgm2 - Berat air uan = 0.01 , 2.3 , 1000 kgm3 - Berat &agar  railing = 0. , 2 , 250 kgm2

= = = =

662.#0 2.00 23.00 #50.00

kg kg kg kg kg kg

1"22+.#0 kg Be(an Iidu& *

= 2.3 , #00

om(ina4i &em(e(anan * u = 1"2 B.:ati 7 1"6 B.Iidu& B.Iidu& = 1"#+ 1"#+2. 2.88 88 7 = 2"# 2"##. #.8 88 kg

=

20.00 kg

1"#+ 1"#+2. 2.00 00

Be(an dinami4 aki(at getaran kendaraan alan ra)a ;20< * t = 1"2 .u = 3"53 3"533. 3.8 86 kg

Men7ari Momen :

10.50

Peritungan :omen * :ma, = 18 G l2 =

18

Pem(e4ian &elat * - te(al &elat = - (eton de$king = - te(al efektif d = - :utu (eton f$ = -225 = - :utu (aa f) = D-2# = m

10.5 2 =

3"533.86

120 20 100 1+.8 2#0

#8"+00.5 kg.m

mm mm mm :&a :Pa

= f$  ;0"85 .f)< = 2#0  ;0.85 . 1+"8< = 15.+8

Kn = :  ;f ( d2< = ; 3.H#<  ;0"85 1000 . 1002< = 5+.252 r

= ;1m< / ;1 = ;115"+8< ;115"+8< / ;1 =

; 1 - ;;2 Kn m<f)<< m<f)<< < 1 - ;;2 1"16 . 15"+8<2#0<< 15"+8<2#0<<

Hrr*502

%&erlu = r .( .d = 0"00508 . 1000 . 100 = 10 - 100 ;%a$t = i&a4ang tulangan f 10

Hrr*502 mm2

+85 mm2<

PERENANAAN PONDAI

12


jembatan

kolom 40x40

P*

P2

2.30

30.10

2.30

Pembebanan pada kaki ko6om 8P* dan P29 : Be(an P1 ;# kolom< * - Berat &elat lantai - Berat (alok 30,0 - Berat kolom #0,#0 - Berat (alok 30,50 - Berat tangga

= = = = =

2##.88 , ;30.1  2< 0.3 , 0. , 2 , 2#00 , ;30.1  2< 0.# , 0.# , # , 5 , 2#00 0.3 , 0.5 , ;2.3 , 8< , 2#00 2.3 , 0.15 , # , 2#00

= = = = =

##"320.## 1"50#.80 +"680.00 6"62#.00 3"312.00

kg kg kg kg kg kg

=

81"##1.2# kg

Be(an dinami4 aki(at getaran kendaraan alan ra)a ;20< * t = 1"2 .u = +"+ +"+2 2.# .#  kg

Diren7anakan memakai gabungan ponda3i u66 p6ate ; ponda3i 3trau33 :

kolom 40x40

1.85

balok sloof 30x50 pelat t=20cm 6.00

2.30

strauss dia.30  l=!m 1.85

0.35

2.30

0.35

2.30

0.35

3.00

0.35

3.00

HE%I P9E%S?

P99E%E P9E%S?

Perhitungan Da
=

3

tm2

13


- ua4 ta&ak &onda4i ;%< = 6.00 , 3.00 = - Berat Jem(atan ; P < = - a)a dukung &onda4i full &late =

18 +.+3

m2 ton

P%

= +.+3 18 =

5.#3 tm2

A 3 tm2

ekurangan da)a dukung 4e(e4ar = = #.65 - 3 =

2.#3 tm2

atau = 1.65 tm2 , 3m , 6m =

+$!%$ ton

ekurangan ter4e(ut akan di(e(ankan &ada &onda4i 4trau44 4e(agai (erikut * i&akai &onda4i 4trau4 dia.30 - 6m -%

= 0.25/3.1#/30/30

=

+06.5 $m2

-

= 3.1#/30

=

#.2 $m

- iangga& nilai !onu4 dan JIP 4e(e4ar * - !n = +.5 kg$m2 - JIP = 250 kg$m - a)a dukung &onda4i 4trau44 P=

%. !n . JIP 7 3 5

P=

528.+5 7 3

23550 5

P =

1+66.25 7

#+10

P =

6"#+6.25 kg

=

(!+- ton

- Jumla 4trau44 )ang di&akai =

#3.+3 6.#8

=

6.+5

==> dipakai - titik ponda3i 3trau33

PERENANAAN KOLOM +)?+) ari a4il out-&ut S%P0 di&erole P =

30"2##.# kg

1#


: = 5+#.50 kg.m &an &anan ang g teku tekuk k = 500 500 $m

eo1 =

: = P

5+#.50 = 30"2##.#

eo2 =

1 LL t = = 1.333 $m LL LL

eo = eo1 7 eo2 = eo 3.23 = t #0

0.02

=

1. 1.0 $m

3.23 $m

= 0.081 $m

dari ta(el PB?+1 10.6.2 " untuk (aa kera4 kera4 -----A !2 = !1 =

6.+5

1

e1 = !1 !2 ; lk  100.t <2 . t = =

1

6.+5 ; 500  #.21 $m

e2 = 0.15 t

=

<2

#0

6

=

13.#5 $m

eu 7 ; 12 t < 13.#5 7 LL 33.#5 $m

P . eau =

30"2##.# ,

Pembe3ian ko6om : - le(ar kolom = - (eton de$king = - te(al efektif d = - :utu (eton f$ = -225 = - :utu (aa f) = D-2# = m

#0

6 $m

eu = eo 7 e1 7 e2 = 3.23 7 #.21 7 eau = = =

100

#00 20 380 1+.8 2#0

33.#5 = =

1"011"+#0.8+ kg kg.$m 10"11+.#1 kg.m

mm mm mm :&a :Pa

= f$  ;0"85 .f)< = 2#0  ;0.85 . 1+"8< = 15.+8

Kn = :  ;f ( d2<

15


= ; 1011+.#H#<  ;0"85 #00 . 3802< = 2.061 r

= ;1m< / ;1 -

; 1 - ;;2 Kn m<f)<< m<f)<< <

= ;115"+8< ;115"+8< / ;1 =

1 - ;;2 2"061 . 15"+8<2#0<< 15"+8<2#0<<

0.00 0.002 26 6

%&erlu = r .( .d = i&a4ang tulangan

0.0026 - d @@

#00

; %a$t =

380 =

1"#08.0+ mm2

226+ mm2<

8d 1 d10-150

#00

#00

16

PERENANAAN #EMBATAN PK9MH 9%S?

* ?! :%E% D% * J.BDEDK%E - SDK%B%M%

Potongan :

10"5

Denah

15"00

10"5 Pembebanan : Be(an :ati * - Berat 4endiri &elat 20$m - Berat a4&al - Berat air uan

Be(an Iidu& * - Be(an Iidu&

= 0.2/2#00/2 = 0.0+/2300/2 = 0"01 / 1000 /2

= = =

60 kgm2 322 kgm2 20 kgm2 1302 1302 kgm kgm2 2

= #00 /2

=

800 kg kgm2

om(ina4i &em(e(anan = 1.2 B.:ati 7 1.6 B.Iidu&  = 1.2 1302 7 1.6 800 = Be(an :erata &ada (alok elagar *  = 28#2.# / 2 = - (erat 4endiri (alok = 0.# / ;0. - 0.2
28#2.# kgm2 568#.8 kgm 6+2 kgm ($,(!- kgm5

1+  #0


* ?! :%E% D% * J.BDEDK%E - SDK%B%M%

Be(an endaraan *

P = 10000 kg

- Berat kendaraan ru$k = 10000  2 - Be(an keut roda (elakang 20 20 = 0" 0"2 / 5000

= =

5"000 kg kg 1"000 kg ())) kg

Momen pada Ba6ok 1e6agar : P=

= l =

10.5

6"000 kg  = 6356.8

kgm

m

: ma, = 18  l2 7 1# P l = = = =

18

2 7 1# 6356.8 10.5 6"000 8+ 8 +"60#.65 7 15"+50 103"35#.65 kg.m 103"35#.65 H# E.mm

Data ba6ok ge6agar - le(ar = - tinggi = - (eton de$king = - te(al efektif d = - :utu (eton f$ = -300 = - :utu (aa f) = D-32 =

#00 00 25 8+5 2#.61 320

10.5

mm mm mm mm :&a :Pa

Pembe3ian Ba6ok : m = f)  ;0"85 .f$< 320 = 0.85 2#.61 = 15.30 Kn

= :  ;f ( d2< 1"033"5#6"500.00 = 0.85 #00 8+5 2 = 3.+0#

18  #0


* ?! :%E% D% * J.BDEDK%E - SDK%B%M%

=

r

;1m< / ;1 -

1 ; 1 - 1 15.30 0.0138813+

= =

r

(

=

0"85/f$/b1 f)

=

r min

2

15.30

3.+0# 320

600 600 7 f)

0.85

=

r ma,

; 1 - ;;2 Kn m<f)<< <

2#.61 320

0.85

600 600 7

320

0.03623+8

= = =

0"+5 r ( 0"+5 0.03623+8 0.02+1+8#2 A r =

0.0138813+

= =

1"#  f) 0.00 .00#38

0.01 .013881 8813+

@ r =

%&erlu = r .( .d =

0. 0.01388

Dipa3ang tu6angan : % = %

=

#00

8+5

=

#"858.#8 mm2

*)

D

2,

; %a$t =

#"08.+# mm2<

+

D

2,

; %a$t =

1"63.50 mm2<

PER/IT0N1AN PONDAI Berat Jem(atan * - Berat &elat (eton - Berat %4&al - Berat Balok #0,0 - Berat Balok 30,60 - Berat air uan - Berat railing - Berat a(utment ;2(< - Be(an Iidu& - Berat endaraan

= 0"2/15/10"5/2#00 = 0"2/15/10"5/2300 = 0"#/0"+/8/10"5/2#00 = 0"3/0"#/+/15/2#00 = 0"01/15/10"5/1000 = 2"2/16/0"+5/2/2#00 = #00/15/10"5 = 10000/2

= = = = = = = = = =

+5"600.00 +2"#50.00 56"##8.00 30"2#0.00 1"5+5.00 1"500.00 126"+20.00 63"000.00 20"000.00 ##+"533.00

kg kg kg kg kg kg kg kg kg kg

Ken$ana Ponda4i *

1  #0


* ?! :%E% D% * J.BDEDK%E - SDK%B%M%

i&akai &onda4i ?%E P%E!%E dia.#0 - 12m -% = 0.25/3.1#/#0/#0 = - = 3.1#/#0 = - Eilai !onu4 dan JIP 4e(e4ar * - !n = 15 - JIP = 300

1256 $m $m2 125.6 $m

kg$m2 kg$m

- a)a dukung dukung 1 tiang &onda4i * %. !n 3

P=

7

. JIP 5

188#0 3

= =

7

6"280.00 7

=

3+680 5 +"536

13"816.00 kg

- 1 tiang menerima (e(an * =

##+"533.00 36

=

12"# 12"#31 31

kg @ P= 13.8 13.816 16 kg >.. >.. 9

20  #0


* ?! :%E% D% * J.BDEDK%E - SDK%B%M%

21  #0

PERENANAAN PELAT LANTAI RAMP 0P PK9MH 9%S?

* ?! :%E% D% * J. - SDK%B%M%

Denah :

DP

l, = # 9WE

l) = 15 Pembebanan : Be(an :ati * - Berat 4endiri &elat 12.5$m - Berat fini4ing ;Pela&i4 ranit< - Berat air uan

= 0.125/2#00

= = =

300 50 10 360

kgm2 kgm2 kgm2 kgm2

Be(an Iidu& - Be(an Iidu& gedung &arkir untuk lantai (aa

=

800

kgm2

om(ina4i &em(e(anan = 1.2 B.:ati 7 1.6 B.Iidu&  = 1.2 36 3 60 7 1.6 800

=

*%*2

kgm2

=

10"000 kg

= 0"01 / 1000

Be(an eut - Berat kendaraan ru$k

P = 5000 kg

- Be(an keut roda (elakang 20 20 = 0" 0"2 / 2 / 5000 - Be(an keut  m2 = 60  ;# / 8<

= =

60 $)

kg kgm2

Be(an otal =

7

1+12

30

=

*%+2

kgm2

Momen
=

:l) = -:t) -:t) = 0"035 .  .l,2 =

3.+5 0.062

1+#2.0

#

2

0.035

1+#2.0

#

2

= = = =

1"+28.06 kg.m 1"+28.06 H# E.m +5.52 kg.m +5.52 H#

E.m

Pembe3ian pe6at 6antai : - le(ar &elat = - te(al &elat = - (eton de$king = - te(al efektif d = - :utu (eton f$ = - :utu (aa f) = Pembe3ian pe6at arah C : m = f)  ;0"85 .f$< 2#0 = 0.85 Kn

25 2#0

:&a :Pa

25

1.+3H70+ 0.85 1000

= =

110 2

1.6802 ;1m< / ;1 -

; 1 - ;;2 Kn m<f)<< <

1 ; 1 11.2 = 0.00+30 = 1"# 1"#  f)

= r min

mm mm mm mm

= 11.2 = :  ;f ( d2< =

r

-300 = D-2# =

1000 125 15 110

1-

2

11.2

1.6802 2#0

= 0.00 0.0058 583 3 %&erlu = r .( .d =

0.00+30

Dipa3ang tu6angan :

dia.

*)

-

*))

; %a$t =

dia.

*)

-

-)

; %a$t =

d

1000

Pembe3ian pe6at arah  : = :  ;f ( d2< Kn .+6H706

110

=

803.20 mm2 +85 $m2< @ % &erlu ---A Eot 9 C 82 $m2< A % &erlu ---A OK 

r

=

0.85 1000 0.#85

=

;1m< / ;1 -

110 2

; 1 - ;;2 Kn m<f)<< <

1 ; 1 11.2

=

1-

2

0.#85 2#0

11.2

= 0.00# 00#0## r min

= 1"# 1"#  f) = 0.00 0.0058 583 3

%&erlu = r .( .d =

0.00583

Dipa3ang tu6angan :

dia.

*)

-

*%,

; %a$t =

##8.80 $m2< A % &erlu ---A Eot 9 C

dia.

*)

-

*2)

; %a$t =

65#.50 $m2< A % &erlu ---A 9 C

d

1000

110

=

6#1.6+ mm2

2#00 1600

PT TATA BUMI RAYA PERENANAAN PELAT LANTAI

PT TATA BUMI RAYA PERENANAAN PELAT LANTAI PK9MH

* %P%E%E HE?S

9% 9%S? S?

* !D !DB B I9D I9DSH SH K%I K%I% % F%:? %:?M M - SDK SDK%B %B% %M% B%K% B%K% 

Pelat dia4um4ikan 4e(agai un4ur 4ekunder )ang tere&it ela4ti4 &ada ke em&at 4i4in)a" dan akan mem(e(ani (alok-(alok di4ekelilingn)a. Denah : ;diam(il &elat ukuran ter(e4ar<

l, = #

l) = 8 Pembebanan : Be(an :ati * - Berat 4endiri &elat 12.5$m - Berat 4$reed 7 granit

= 0.125/2#00 = 2# 7 21

= =

300 #5

kgm2 kgm2

- Berat air uan

= 0"01 / 1000

=

10 355

kgm2 kgm2

= =

300 60

kgm2 kgm2

360

kgm2

06

kgm2

Be(an Iidu& - Be(an dinami4 20 = 0"2 / 300 om(ina4i &em(e(anan = 1.2 B.:ati 7 1.6 B.Iidu& 

=

1.2 355

7

1.6 360

=

Momen
:l) = -:t) = 0"035 .  .l, =

Pembe3ian pe6at 6antai : - le(ar &elat

m

0.035

06.0

#

2

#

2

=

1000

mm

- te(al &elat

=

125

mm

- (eton de$king

=

15

mm

110 20.11

mm :&a

2#0

:Pa

- te(al efektif d = - :utu (eton f$ =

-300 =

- :utu (aa f)

D-2#

=

=

= f)  ;0"85 .f$< = =

2#0 0.85 1#.0#

20.11

28

=

88.+5 kg.m

= =

88.+5 H# E.mm 50+.36 kg.m

=

50+.36 H#

E.mm


* %P%E%E HE?S

9% 9%S? S?


Kn

= :  ;f ( d2< 8.H706

=

0.85

= =

r

1000

110 2

0.8+38 ;1m< / ;1 -

1

=

1#.0#

; 1 - ;;2 Kn m<f)<< <

; 1 -

1-

2

0.8+38 2#0

1#.0#

= 0.003+3 r

min

= 1"# 1"#  f) =

0.00 0.0058 583 3

%&erlu = r .( .d =

0.00583

1000

Dipa3ang tu6angan : : dia. d

8

*) -

110

-

*2,

=

6#1.6+ mm2

; %a$t =

10

2

628.32 $m2< A % &erlu ---A OK 


* %P%E%E HE?S

9% 9%S? S?


Pembe3ian pe6at arah  : Kn = :  ;f ( d2< = = =

r

5.0+H706 0.85 1000 0.#33 ;1m< / ;1 -

; 1 - ;;2 Kn m<f)<< <

1 ; 1 1#.0#

=

110 2

1-

2

0.#33 2#0

1#.0#

= 0.002086 r

min

= 1"# 1"#  f) = 0.00 0.0058 583 3

%&erlu = r .( .d =

0.00583

1000

Dipa3ang tu6angan : : dia. d

8

*) -

110

-

2,)

=

6#1.6+ mm2

; %a$t =

10

30

31#.16 $m2< A % &erlu ---A OK 

1.00

1.10

1.20

1.30

1.#0

1.50

1.60

1.+0

1.80

1.0

:t,

36

#2

#6

50

53

56

58

5

60

61

:l,

36

3+

38

38

38

3+

36

36

35

35

2.00

2.10

2.20

2.30

2.#0

2.50

62

62

62

63

63

63

35

3#

3#

3#

3#

3#

PERENANAAN PELAT LANTAI

PERENANAAN PELAT LANTAI PK9MH

* %P%E%E HE?S

9%S?

* !DB I9DSH K%I% F%:?M - SDK%B%M% B%K%

KONTROL LEND0TAN PELAT LANTAI Data tekni3 : - (entang &anang - (entang &endek - te(al &elat - mutu (eton - mutu (aa

= = = = =

+20 360 13 250 2#

 D

$m $m $m ----A s/(k = ----A s/au =

250 kg kg$m2 2080 kg kg$m2

- &em(e4ian &elat ter&a4ang = d10 - 100 ;%a$t =

+.85 +.85 $m2< $m2<

Lendutan i4in : f =

1 250

lo

1 250

=

360

=

1.## $m

Lendutan ter4adi : f =

fe fe =

g. k1 . k2 7 & g7& sa .lo2

a ; 1- x< . 10-6 dimana *

g

=

g & & k1 k2

= = = = =

sa = sa =

(e(an angka &anang ;(e(an idu& )ang (er4ifat teta& 7 (e(an mati< #00 kg kgm2 7 ; 0.13 , 2#00 < = +12 kg kgm2 (e(an mati = 0.13 , 2#00 = 312 kgm2 (e(an angka &endek = 0 kgm2 (e(an idu& = #00 kg k gm2 am(il = 0.6 1850 1850 kg$ kg$m2 m2

lo

=

360 $m ;)ang terke$il antara l, dan l)<

a

=

20.8 ;(aa &olo4<

x

=

( .  s(k 2.5 +.85# 2080 100 13 225

= fe

=

=

= = =

=

0.1 0.1#

sa .lo2

a ; 1- x< . 10-6 10-6 0.1#0 < 10

1850 360 2 20.8 ; 1 -

=

1.3#0 $m

=

f

s/au

2.5 %

fe

g. k1 . k2 7 & g7& 1.3#0

0.831

+12 288

0.6 7

$m @ f =

H HED%E M%E HKJ%? =

7 #00

0

1.## $m $m

0.831

!:

23"+60"000 1+8.6

PT TATA BUMI RAYA PERENANAAN PONDAI PK9M PK9MH H 9%S? 9%S?

* KD:% KD:%I I ?E ?E % % * S%DK S%DK% % KHHE KHHE!M !M - H?E H?E%E %E  - SDK% SDK%B% B%M M%

Diren7anakan ponda3i ponda3i pe6at 6a4ur :

P*

10 20 80

Pembebanan pada kaki ko6om dengan beban terbe3ar :

3"00

3"00

3"00

Be(an :ati * - Berat &elat laur - Berat (alok 20,#0 - Berat kolom 20,#0 - Berat &elat lantai 12$m - Berat keramik - Berat dinding (ata - Be B erat &lafond - Be Berat kuda-kuda - Berat &enutu& ata&

- Berat lain-lain ;10< Be(an Iidu& * BHB%E 9%

3"00

= = = = = = = = =

0"8/0"25/12/2#00 0"2/0"#/12/2#00 0"2/0"#/#/2/2#00 6/6/0"12/2#00 6/6/;2172#< ;673
= 0"1/ = 6/ 6/6/250

35"511.00

= = = = = = = = = =

5"+60.00 2"30#.00 1"536.00 10"368.00 1"620.00 "6+5.00 6#8.00 1"800.00 1"800.00 35"5 5"511.00 .00

kg kg kg kg kg kg kg kg kg kg

=

3"551.10 kg

=

"000.00 kg

=

#8 #8"062.10 kg

35



Perhitungan Da
= = =

0.# 6"000.00 #8"062.10

kg$m2 m2 kg

P% #8"062.10 6"000

=

0.5 0.50 kg$ kg$m2 m2

A

0"# 0"# kg$m g$m2 2

;not ;not 9< 9<

Diren7anakan memakai memakai gabungan ponda3i pe6at 6a4ur ; ponda3i 3trau33 :

10 20

800

ekurangan da)a dukung 4e(e4ar = = 0.5 - 0"# =

0.10 kg$m2

atau = 0"1 kg$m $m2 , 80$m , 1200$m =

"60 "600.00 .00 kg

ekurangan ter4e(ut akan di(e(ankan &ada &onda4i 4trau44 4e(agai (erikut * i&akai &onda4i 4trau4 dia.30 - 8m -%

= 0.25/3.1#/30/30

=

+06.5 $m2

-

= 3.1#/30

=

#.2 $m

- Eilai !onu4 dan JIP &ada daera etintang 4e(e4ar * - !n = 5 kg$m2 - JIP = 30 kg$m

36



- a)a dukung &onda4i 4trau44 %. !n . JIP 7 3 5

P=

P=

3532.5 3

7

2826 5

P =

11++.5

7

565.2

P =

1"+#2.+0 kg

- Jumla 4trau44 )ang di&akai =

==>

"600.00 1"+#2.+0

=

5.51

dipakai ( titik ponda3i 3trau33 untuk untuk pe6at 6a4ur 3epan4ang 3epan4ang *2 m5 atau * titik 3trau3 3etiap 2m5

3+



PERENANAAN KOLOM +)?+) ari a4il out-&ut S%P0 di&erole P = 30"2##.# kg : = 5+#.50 kg.m &an &anan ang g teku tekuk k = 500 500 $m

eo1 =

: = P

eo2 =

1 LL t = = 1.333 $m LL LL

eo = eo1 7 eo2 = eo 3.23 = t #0

5+#.50 = 30"2##.#

0.02

=

1. 1.0 $m

3.23 $m

= 0.081 $m

dari ta(el PB?+1 10.6.2 " untuk (aa kera4 -----A !2 = !1 =

6.+5

1

e1 = !1 !2 ; lk  100.t <2 . t = =

1

6.+5 ; 500  #.21 $m

e2 = 0.15 t

=

#0

<2

#0

6 $m

eu = eo 7 e1 7 e2 = 3.23 7 #.21 7 eau = = =

100

6

=

13.#5 $m

eu 7 ; 12 t < 13.#5 7 LL 33.#5 $m

P . eau =

30"2##.# ,

33.#5 = =

1"011"+#0.8+ kg kg.$m 10"11+.#1 kg.m

38



Pembe3ian ko6om : - le(ar kolom = - (eton de$king = - te(al efektif d = - :utu (eton f$ = -225 = - :utu (aa f) = D-2# = m

#00 20 380 1+.8 2#0

mm mm mm :&a :Pa

= f$  ;0"85 .f)< = 2#0  ;0.85 . 1+"8< = 15.+ 15.+8 8

Kn = :  ;f ( d2< = ; 1011+.#H#<  ;0"85 #00 . 3802< = 2.061 r

=

;1m< / ;1 -

=

;115"+8< ;115"+8< / ;1 -

=

; 1 - ;;2 Kn m<f)<< < 1 - ;;2 2"061 . 15"+8<2#0<< 15"+8<2#0<<

0.00 0.002 26 6

%&erlu = r .( .d = i&a4ang tulangan

0.0026 - d @@

#00

; %a$t =

380 =

1"#08.0+ mm2

226+ mm2<

8d 1 d10-150

#00

#00

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