Perhitungan Struktur Gudang Baja

Pre - Eliminary Design 1 Perencanaan Atap Perencanaan Atap Merencanakan Pola Beban Data Perencanaan Perencanaan Dimensi Gording Perencaan Penggantung Gording Perencanaan Gording Ujung

Perencanaan Ikatan Angin

1.1 Merencanakan Pola Beban Pola Beban Diambil dari peraturan Pembebanan Indonesia untuk gedung 1983 Merencanak an Pola Beban

Beban Mati

Beban Penutup Atap

Beban Hidup

Beban Profil

Beban Pengikat dll

Beban Terbagi Rata

Beban Terpusat

Beban Angin

Beban Tekanan Angin

Beban Angin Hisap

1.1.1 Merencanakan Beban Mati ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung ) a. Atap Berat asbes : 10.3 kg/m2 Berat Profil : Menyesuaikan Perencanaan Berat Pengikat dll : 10 % dari Berat Total 1.1.2 Merencanakan Beban Hidup ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung ) a. Beban Hidup Terbagi Rata ( Atap ) : α= 25 0 q = (40 - 0.8 α) = ambil q =

20

20 kg/m2

kg/m2

≤

20

kg/m2

b. Beban Hidup Terpusat ( Atap ) P=

100 kg

1.1.3 Merencanakan Beban Angin ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung ) a. Beban Tekanan Angin Bangunan Jauh dari Pantai -> asumsi Tekanan Angin :

Koefisien Angin (C) tekan = (0.02 α - 0.4)

=

0.1

Angin Tekan = C x W

=

3

kg/m2

Angin Hisap = 0.4 x W

=

12

kg/m2

30

1.2 Data - Data perencanaan Data Atap Jenis Tebal Berat Lebar Gelombang Kedalaman Gelombang Jarak Miring Gording Jarak Kuda-Kuda (L) Sudut Kemiringan Atap

1.3 Perencanaan Dimensi Gording

: : : : : : : :

Asbes Gelombang 5 mm 10.3 kg/m2 110 mm 57 mm 110 cm 400 cm 0.44 rad =

25

0

kg/m2

1.3.1 Perencanaan Profil WF untuk Gording Dengan ukuran : WF 100 x A = 11.85 cm2 W= 9.3 kg/m a= 100 mm bf = 50 mm iy = 1.12 cm

50

Mutu Baja = BJ 37 fu = 3700 kg/cm2 = fy = 2400 kg/cm2 =

x tf = Ix = Iy = tw = ix =

7 187 14.8 5 3.98

x mm cm4 cm4 mm cm

7 Zx = Zy = h=

41.8 cm3 8.94 cm3 70 mm {=D - 2 x (tf + r)}

370 Mpa 240 Mpa

1.3.2 Perencanaan Pembebanan 1.3.2.1 Perhitungan Beban Beban Mati Berat Gording Berat Asbes Gelombang = = alat Pengikat dll 10 % =

5

0.1

w 10.3

x x

x

20.63

Beban Hidup Beban Terbagi Rata = (40 - 0.8 α) = qL = jarak gording horisontal x q

=

=

9.3 kg/m

= = =

11.33 kg/m 20.63 kg/m 2.06 kg/m

qD

=

22.69 kg/m

l 1.1 Berat Total

40

-

20 q

= =

0.997

x

20.00

=

20 20

19.94 kg/m

Beban Hidup Terpusat, PL

=

Beban Angin Tekanan Angin Angin Tekan Angin Hisap q = jrk gording horisontal x angin hisap =

= = = =

30 3 12 11.96

+

19.94

3

kg/m

0.997

x

12.00

==>

22.69 qw =

Beban Mati + Beban Hidup > dari Beban Angin Hisap : Beban Angin Hisap

tidak perlu diperhitungkan

kg/m2 kg/m2

100 kg

kg/m2 kg/m2 kg/m2 (menentukan = q) kg/m >

11.96

1.3.2.2 Perhitungan Momen Akibat Beban thp Sbx dan Sby Beban Mati MXD = 1/8 (qD x cosα) L2 = 0.13 x ( 22.69 x 0.91

x

16

)=

41.13 kgm

MYD = 1/8(qDxsinα xL/3)2 =

0.13

x(

22.69

x

0.42

x

1.78

)=

2.13 kgm

Beban Hidup Terbagi Rata MXLD = 1/8 (qL x cosα) L2 =

0.13

x(

19.94

x

0.91

x

16

)=

36.25 kgm

MYL = 1/8(qLxsinαxL/3) =

0.13

x(

19.94

x

0.42

x

1.78

)=

1.87 kgm

Beban Hidup Terpusat MXL = 1/4 (qL x cosα) L =

0.25

x(

100

x

0.91

x

4

)=

90.63 kgm

MYL = 1/4(qL x sinα)(L/3) =

0.25

x(

100

x

0.42

x

1.33

)=

14.09 kgm

2

Beban Angin Terbagi Rata MXW = 1/8 x qw x L =

0.13

x

3

x

16

=

6

kgm

1.3.3.3 Besar Momen Berfaktor ( Mu = 1.2 M D + 1.6 ML + 0.8 MW ) * Mu Beban Mati ,Beban Angin dan Beban Hidup Terbagi Rata Sumbu X Sumbu Y MD = MD = 41.13 kgm 2.13 kgm ML = Mw =

36.25 kgm 6 kgm

ML =

MUX =

1.2

x

41.13

+

1.6

x

36.25

+

0.8

x

6

=

### kgm

MUY =

1.2

x

2.13

+

1.6

x

1.87

+

0.8

x

0

=

5.55 kgm

1.87 kgm

* Mu Beban Mati, Beban Angin dan Beban Hidup Terpusat Sumbu X Sumbu Y MD = MD = 41.13 kgm 2.13 kgm ML = Mw =

90.63 kgm 6 kgm

ML =

MUX =

1.2

x

41.13

+

1.6

x

90.63

+

0.8

x

6

=

### kgm

MUY =

1.2

x

2.13

+

1.6

x

14.09

+

0.8

x

0

=

25.1 kgm

50

14.09 kgm

1.3.3 Kontrol Kekuatan Profil 1.3.3.1 Penampang Profil Untuk Sayap bf 170 ≤ 2 tf fy 50 170 ≤ 2 7 240 3.57 ≤ 10.97 OK

Untuk Badan h 1680 ≤ tw fy 70 1680 ≤ 5 240 14.0 ≤ 108.4 OK

Penampang Profil Kompak, maka Mnx = Mpx 1.3.3.2 Kontrol Lateral Buckling Jarak Baut Pengikat / pengaku lateral = L B = E fy 200000 240

500

mm

=

=

56.90

cm

=

Mpx

LP =

1.76

x

iY

x

=

1.76

x

1.12

x

LB

<

LP

maka :

Mnx

41.8

x

2400

=

### Kgm

x

2400

Ternyata :

Mnx = Mpx = Zx . Fy = Mny = Zy ( satu sayap ) * fy = 1/4 x tf x bf 2 x fy = 0.25 x 0.7 = 105 kgm

x

52

=

cm

10500 kgcm

1.3.3.3 Persamaan Iterasi Mux + φb . Mnx

Muy φb . Mny

≤

1

Beban Mati , Beban Angin dan Beban Hidup Terbagi Rata 112.164 5.554 + ≤ 1 0.9 x ### 0.9 x 94.5 0.12 + 0.07 ≤ 1 0.19 ≤ 1 OK Beban Mati , Beban Angin dan Beban Hidup Terpusat 199.170 25.097 + ≤ 0.9 x ### 0.9 x 105 0.2 + 0.27 ≤ 1 0.46 ≤ 1 OK

1

1.3.3.4 Kontrol Lendutan Profil L 180

400 180

=

2.22 cm

L4 Ix

=

5

x 384

0.43 x

x 0.91 2000000

x x

400 4 187

=

5

x 384

0.43 x

x 0.42 2000000

x x

### 4 14.8

L3 Ix

=

1

x 48

100 x

x 0.91 2000000

x x

400 3 187

L3 Iy

=

1

x 48

30 x

x 0.42 2000000

x x

### 3 14.8

Lendutan Akibat Beban Angin merata (3) qW cos α L4 5 fx = x 384 E x Ix = 0.02 cm

=

5

x 384

0.03 x

x 0.91 2000000

x x

400 4 187

=

5

x 384

0.03 x

x 0.42 2000000

x x

### 4 14.8

0.02

+

)2

Lendutan Ijin f =

=

Lendutan Akibat Beban Merata (1) qD + L cos α 5 fx = x 384 E x = 0.34 cm fy = =

5 x 384 0.03 cm

qD + L E

sin α (L/3)4 x Iy

Lendutan Akibat Beban Terpusat (2) cos α 5 P fx = x 384 E x = 0.32 cm sin α 5 P fy = x 384 E x = 0.02 cm

fy =

5 384

=

x

qW E

sin α (L/3)4 x Iy

0 cm

Lendutan total yang terjadi ftot = = (

fx2 + fy2 0.34

(fx1 + fx2 + fx3)2 + (fy1 + fy2 + fy3)2

= +

0.32

≤

+

0.02 ) 2 + (

0.03

+

0

ftot =

≤ 0.69 cm

<

fijin =

2.22 cm

OK

1.4 Perencanaan Penggantung Gording 1.4.1 Data Penggantung Gording Jarak Kuda - Kuda (L) Jumlah Penggantung Gording Jumlah Gording Jarak Penggantung gording

= = = =

400 2 9 ###

cm buah buah cm

1.4.2 Perencanaan Pembebanan Beban Mati Berat Sendiri Gording Berat Asbes gelombang Alat Pengikat dll 10 %

RD = =

qD 22.69

=

0.1

sinα 0.42

x x

x x

x

L/3 1.33

Beban Hidup Beban Terbagi Rata = (40 - 0.8 α) = qL = jarak gording horisontal x q RL = =

qL 19.94

x x

Beban Terpusat = RL = =

PL 100.0

sinα 0.42

x x

PL sinα 0.42

= x x

qW 2.99

x x

sinα 0.42

kg/m kg/m kg/m kg/m

=

22.69 kg/m

12.787

40

-

0.8 q

= =

0.997

x

20.00

=

=

=

100 kg

=

42.262

kg

11.235

x x

=

0.997

L/3 1.33

=

=

1.4.3.1 Penggantung Gording Tipe B

x (

102.29

20 20

kg/m2 kg/m2

19.94 kg/m kg

kg

=

1.4.3 Perhitungan Gaya 1.4.3.1 Penggantung Gording Tipe A RA = 1.2 RD + 1.6 RL + 0.8 RW = 1.2 x 12.8 + 1.6 = 102.29 kg RA total = Ra x jumlah Gording

9.3 11.33 20.63 2.06

=

L/3 1.33

Beban Angin Angin Tekan = q qW = jarak gording horisontal x q RW = =

20.63 qD

= = = =

x

3.00 1.685

=

3

kg/m2

2.99 kg/m

kg

11.2

+

42.3

x

9

=

) +

0.8

920.60

x

kg

1.7

panjang miring gording L / 3

arctan β =

RB =

β= RA

=

110 133.33

=

0.83

=

0.670 cm2

39.52 o 920.60

=

sin β

sin

39.52

=

1446.607

kg

1.4.4 Perencanaan Batang Tarik Pu =

RB =

1446.607 kg fu = 3700 kg/cm2 fy = 2400 kg/cm2

BJ 37

1.4.4.1 Kontrol Leleh Pu = φ . fy . Ag ; dengan φ = 0.9 Ag perlu =

ϕ

Pu

= fy

0.9

1446.607 x 2400

Tidak Menentukan 1.4.4.2 Kontrol Putus Pu = φ . fu . 0,75 Ag ; dengan φ = Ag perlu =

Pu fu

ϕ

0.75 =

0.75

0.75

x

1446.607 3700

x

0.75

=

0.7 cm2 Menentukan

Ag perlu = 1/4 . π . d Ag d = ==>

2

Pakai d =

x

4

π 10

= = Cek :

0.7

4

=

### cm

(jarak penggantung gording)2 + (panjang miring gording)2 133.33 2 ### cm

d

>

1

>

1

>

+

Panjang Rb 500 172.85 500 0.35

110 2

OK

1.5 Perencanaan Ikatan Angin Atap 1.5.1 Data Perencanaan Ikatan Angin Atap Tekanan Angin W = 30 kg/m2 Koefisien Angin C tekan = 0.9 Koefisien Angin C hisap a1 = α=

x

π

mm

1.4.5 Kontrol Kelangsingan Jarak Penggantung Gording = Panjang Rb =

=

300 cm 0.44 rad

= =

0.4 a2 = 200 cm 25 0

0.94 cm

1.5.2 Perhitungan Tinggi Ikatan Angin ( h ) h1 = 9 m h2 =

9

+

2

x

tg

0.44

=

9.93

m

h3 =

9

+

4

x

tg

0.44

=

10.87

m

h4 =

9

+

6

x

tg

0.44

=

11.8

m

h5 =

9

+

9

x

tg

0.44

=

13.2

m

1

x

9

=

121.5

kg

1.5.3 Perhitungan Gaya - Gaya yang Bekerja R = 1/2 . W . C . a . h R1 = 0.50 x 30 x 0.9 x R2 =

0.50

x

30

x

0.9

x

2

x

9.93

=

###

kg

R3 =

0.50

x

30

x

0.9

x

2

x

10.87

=

###

kg

R4 =

0.50

x

30

x

0.9

x

2.5

x

11.8

=

###

kg

R5 =

0.50

x

30

x

0.9

x

3

x

13.2

=

###

kg

268.18 kg

+

293.36

+

###

+

###

)

Rtotal = ( R1+R2+R3+R4+(R5/2)) = 121.5 + = 1348.454

###

1.5.4 Perencanaan Dimensi Ikatan Angin 1.5.4.1 Menghitung gaya Normal 2 tg φ = = 0.5 4 φ = 26.57 0 R1 = 121.5 kg Rtotal = 1348.454 kg Gaya Normal Gording Akibat Angin Dimana untuk angin tekan C = dan untuk angin hisap C = Chisap Rtotal x N = Ctekan 0.4 x 1348.454 = = 599.31 kg 0.9 1.5.4.2 Menghitung gaya Pada Titik Simpul Pada Titik Simpul A ΣV = 0 Rtotal + S1 = 0 ===> S1 = - Rtotal ===> ΣH = 0 S2 =

S1 =

0.9 0.4

### kg

0

Pada Titik Simpul B EV = 0 R1 + S1 +S3 Cos ϕ = 0 S3 = S3 =

-(

R1 cos

-1643.458

S1

ϕ

)

fu =

-(

121.5 cos 26.57

kg

1.5.5 Perencanaan Batang Tarik Pu = S3 x 1.6 x 0.75 = -1643.46 BJ 37

=

3700 kg/cm

2

x

1.6

x

0.75

=

-1972.150 kg

fy =

2400 kg/cm2

1.5.5.1 Kontrol Leleh Pu = φ . fy . Ag ; dengan φ = 0.9 Ag perlu =

ϕ

Pu

= fy

0.9

1972.150 x 2400

=

0.913 cm2 Tidak Menentukan

1.5.5.2 Kontrol Putus Pu = φ . fu . 0,75 Ag ; dengan φ = Ag perlu =

ϕ

Pu fu

0.75 =

0.75

0.75

x

1972.150 3700

x

0.75

=

0.95 cm2 Menentukan

Ag perlu = 1/4 . π . d2 Ag d = ==>

Pakai d =

x

4

π 11

=

0.95

x

4

π

=

1.1 cm

mm

1.5.6 Kontrol Kelangsingan Jarak kuda-kuda = 400 cm Panjang S3 = = = Cek :

(jarak kuda-kuda)2 + (jarak miring gording)2 400 2 + 110 2 ### cm

d

>

1.1

>

1.1

>

Panjang S3 500 414.85 500 0.83

OK

1.6 Perencanaan Gording Ujung 1.6.1 Perencanaan Pembebanan Mntx , Mnty dan Gaya Normal Akibat Angin Gording Ini adalah Balok Kolom. Akibat beban mati dan beban hidup Menghasilkan Momen Lentur Besaran Diambil Dari Perhitungan Gording Mntx = MUX (1.2 D + 1.6 L + 0.8 W) x 0.75

=

199.170

x

0.75

=

149.377

kgm

= 25.097 x Nu = 1.6 x Rtotal (dari ikatan angin atap) x 0.75 = 1618.144

0.75

=

18.823

kgm

Mnty = MUY (1.2 D + 1.6 L + 0.8 W) x 0.75

kg

1.6.2 Perencanaan Profil Gording Ujung WF 100 x A = 11.85 cm2 W = 9.3 kg/m a = 100 mm bf = 50 mm iy = 1.12 cm

50 x 5 tf = 7 mm Ix = 187 cm4 Iy = 14.8 cm4 tw = 5 mm ix = 3.98 cm

Mutu Baja = BJ 37 fu = 3700 kg/cm2 =

370 Mpa

x 7 Zx = 41.8 Zy = 8.9 h = 70

cm3 cm3 mm

{=D - 2 x (tf + r)}

fy = 2400 kg/cm2

=

240 Mpa

1.6.3 Kontrol Tekuk Profil Lkx =

400

Ncrbx

=

cm

50

Ncrby

=

λx =

π2 . E . A λx 23157.64 cm

==>

λy 117366.49

= π2

λy =

φ

11.85

=

Pu

=

Pn

0.85

x

Lkx iy

π2

=

2

Ag x fy ω

=

100.5

2000000

x

11.85

100.5 2

=

50 1.12

x

=

2000000

44.64 x

11.85

44.64 2

kg

Tekuk Kritis adalah arah X, Karena λx > λy Pn =

400 3.98

kg

π2 . E . A

=

Lkx ix

=

2

= Lky =

==>

x 2.29

2400

ω= =

1618.144 x 12437.136

2.29 12437.136

=

kg

0.15

<

0.2

(Pu = Nu)

Pakai Rumus = Pu 2

x

φc . Pn

+

Mux φb

x

+

Mnx

1.6.4 Perhitungan Faktor Pembesaran Momen Gording dianggap tidak bergoyang, maka : Mux = Mntx . Sbx

Cmx Nu 1 - ( ) Ncrbx Untuk elemen Beban Tranversal, ujung sederhana Cmx = 1 1 Sbx = = 1618.144 1 - ( ) 23157.64 Sbx = Sbx =

1.08 1.08

Muy = Mnty * Sby

Sbx =

>

1.01

Sby =

>

1

1.08

1

Cmy Nu 1 - ( ) Ncrby Untuk elemen Beban Tranversal, ujung sederhana Cmy = 1 1 Sby = 1618.144 1 - ( ) 117366.49 Sby =

≥

1

≥

=

1.01

1

Muy φb

x

Mny

≤

1

Sby =

1.01

1.6.5 Perhitungan Momen Ultimate Sbx dan Sby Mux = Sbx . Mntx = 1.08 x 149.377 Muy = Sby . Mnty = 1.08 x 18.823

1.6.6 Perhitungan Persamaan Interaksi Mnx = 1003 kgm

2

x

2

x

Pu φc x 1618.144 0.85 x

Pn

+

12437.136

Mny = Mux

φb +

x

= =

105

160.599 20.237

kgm kgm

kgm +

Mnx 160.599 0.9 x 1003 0.47 ≤ 1 OK

Muy φb +

≤ Mny 20.237 0.9 x 105 x

1 ≤

1

Pre - Eliminary Design 2 Perencanaan Dinding 2.1 Data - Data perencanaan Data Dinding : Jenis Tebal Berat Kedalaman Gelombang Jarak Kolom Dinding (L) Jarak Gording Lt Dasar Jarak Gording Lt 1

: : : : : : :

Seng Gelombang 4 mm 4.15 kg/m2 25 mm 400 cm 125 cm 100 cm

2.2 Perencanaan Regel Balok ( Dinding Samping ) 2.2.1 Perencanaan Profil WF untuk Regel Balok Dinding Dengan ukuran : WF 100 x A = 11.85 cm2 W= 9.3 kg/m a = 100 mm bf = 50 mm iy = 1.12 cm

50 tf = Ix = Iy = tw = ix = r=

Mutu Baja = BJ 37 fu = 3700 kg/cm2 = fy = 2400 kg/cm2 =

x

5 7 mm 187 cm4 14.8 cm4 5 mm 3.98 cm mm

x Zx = Zy = h= Sx =

7 41.8 cm3 8.94 cm3 70 mm 37.5 mm

370 Mpa 240 Mpa

2.2.2 Perencanaan Pembebanan 2.2.2.1 Perhitungan Beban Beban Mati Lantai Dasar Berat Gording Berat Seng Gelombang = 4.15

x

1.25

alat Pengikat dll 10 %

x

14.49 Berat Total 15.94 x 1.78

= = = = = =

9.3 kg/m 5.19 kg/m 14.49 kg/m 1.45 kg/m 15.94 kg/m 3.54 kg/m

Berat Total x 1.78

= = = = = =

9.3 kg/m 4.15 kg/m 13.45 kg/m 1.35 kg/m 14.8 kg/m 3.29 kg/m

30 1.25 30 1.25

= = = = =

30 27 33.75 12 15

Berat Total = 0.1

Myd = 1/8 x q x (L/3)

= 0.13

x

Lantai 1 Berat Gording Berat Seng Gelombang

= 4.15

x

alat Pengikat dll 10 %

= 0.1

Myd = 1/8 x q x (L/3)2

= 0.13

2

{=D - 2 x (tf + r)}

1 Berat Total

Beban Angin Lantai Dasar Tekanan Angin Angin Tekan ( C = 0.9 ) = q = Angin Tekan x Jarak Gording = Angin Hisap ( C = 0.4 ) q = Angin hisap x Jarak Gording =

13.45 x

0.9 27 0.4 12

14.8

x x x x

kg/m2 kg/m2 kg/m kg/m2 kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) : Mxw = 1/8 x q x (L)2 = 0.13 x 33.75 x N = q x Jarak Gording = 15 x 1.25

16

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) : Mxw = 1/8 x q x (L)2 = 0.13 x 15 x N = q x Jarak Gording = 33.75 x 1.25

16

Lantai 1 Tekanan Angin Angin Tekan ( C = 0.9 ) = q = Angin Tekan x Jarak Gording = Angin Hisap ( C = 0.4 ) q = Angin hisap x Jarak Gording =

0.9 27 0.4 12

x x x x

30 1 30 1

16

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) Mxw = 1/8 x q x (L)2 = 0.13 x 12 x N = q x Jarak Gording = 27 x 1

16

=

67.5 kgm 18.75 kg

(Tarik)

= =

30 kgm 42.19 kg

(Tekan)

= = = = =

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) Mxw = 1/8 x q x (L)2 = 0.13 x 27 x N = q x Jarak Gording = 12 x 1

2.2.3 Kombinasi Pembebanan Lantai Dasar 1. U = 1.4 D Muy = 1.4 x 3.54

= =

30 27 27 12 12

kg/m2 kg/m2 kg/m kg/m2 kg/m

= =

54 kgm 12 kg

(Tarik)

= =

24 kgm 27 kg

(Tekan)

4.96 kgm

2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha ) Akibat Beban Angin yg Tegak Lurus Dinding (tarik) : Mux = 1.2 x 0 + 1.3 x 67.5 = 87.75 kgm Muy = 1.2 x 3.54 + 1.3 x 0 = 4.25 kgm Nu = 1.2 x 0 + 1.3 x 18.75 = 24.38 kg Akibat Beban Angin yg Tegak Lurus Gevel (tekan) : Mux = 1.2 x 0 + 1.3 x 30 = 39 kgm Muy = 1.2 x 3.54 + 1.3 x 0 = 4.25 kgm Nu = 1.2 x 0 + 1.3 x 42.19 = 54.84 kg Lantai 1 1. U = 1.4 D Muy = 1.4 x 3.29 = 4.6 kgm 2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha ) Akibat Beban Angin yg Tegak Lurus Dinding (tarik) : Mux = 1.2 x 0 + 1.3 x 54 = 70.2 kgm Muy = 1.2 x 3.29 + 1.3 x 0 = 3.95 kgm Nu = 1.2 x 0 + 1.3 x 12 = 15.6 kg

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) : Mux = 1.2 x 0 + 1.3 x 24 = 31.2 kgm Muy = 1.2 x 3.29 + 1.3 x 0 = 3.95 kgm Nu = 1.2 x 0 + 1.3 x 27 = 35.1 kg 2.2.4 Kontrol Kekuatan Profil 2.2.4.1 Penampang Profil Untuk Sayap Untuk Badan bf 170 h ≤ ≤ 2 tf fy tw 50 170 70 ≤ ≤ 2 7 240 5 3.57 ≤ 10.97 14.0 ≤ OK OK Penampang Profil Kompak, maka Mnx = Mpx 2.2.4.1 Kontrol Lateral Buckling Jarak Baut Pengikat / pengaku lateral = L B = LP =

1.76

x

iY

x

= 1.76

x

1.12

x

LB

<

Ternyata :

Mnx = Mpx = Zx . Fy = 41.8 1.5 Myx = 1.5 Sx fy = 1.5 x ===> Mnx < 1.5 Myx Mny = Zy ( satu sayap ) * fy = 1/4 x tf x bf 2 x fy = 0.25 x 0.7 = 105 kgm

x

2.2.5 Perhitungan Kuat Tarik 2.2.5.1 Kontrol Kelangsingan λp ≤ 300 Lk 400 = = λ= ix 3.98

500

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

=

50

1680 fy 1680 240 108.4

mm

E fy 200000 = 56.90 cm 240 LP maka : Mnx = Mpx x 37.5

2400 x 2400

52

= =

x

2400

OK

100.5

<

300

2.2.5.2 Berdasarkan Tegangan Leleh φ Nn = φ .Ag . fy = 0.85 x 11.85

x

2400

### Kgm 1350 Kgm

=

10500 kgcm

λp

2.2.5.3 Berdasarkan Tegangan Putus φ Nn = φ .Ae . fu = = 0.75 x 0.85 x Ag x fu = 0.75 x 0.85 = 0.75 x 0.85 = ### kg Tidak Menentukan 2.2.5.4 Kontrol Kuat Tarik

cm

x x

= 24174 kg Menentukan

Ag 11.85

x x

fu 3700

Lantai Dasar φ Nn > Nu 24174 > 54.84 OK Lantai 1 φ Nn > Nu 24174 > 1404 OK 2.2.6 Perhitungan Kuat Tekan 2.2.6.1 Kontrol Kelangsingan λp ≤ 200 Lkx 400 = = λpx = ix 3.98 Lky 50 = = λpy = iy 1.12

100.5

<

200

OK

44.64

<

200

OK

2.2.6.2 Berdasarkan Tekuk Arah X fy λx 100.5 = x λc = 3.14 E π 0.25 < < 1.2 λc 1.43 = ω= 1.6 - 0.67 λc 1.6 fy = 0.85 x φ Nn = φAg ω 2.2.6.3 Berdasarkan Tekuk Arah Y fy λy = 44.64 λc = x 3.14 E π λc 0.25 < < 1.2 1.43 = ω= 1.6 - 0.67 λc 1.6 fy = 0.85 x φ Nn = φAg ω

2.2.7 Perhitungan Pembesaran Momen Ab x fy Ncr = λc 2 11.85 x 2400 Ncrbx = = 1.108 2 11.85 x 2400 Ncrby = = 0.492 2

2400 2000000

=

1.43 0.67

x

11.85

x

1.11 2400 1.67

2400 2000000

=

1.43 0.67

x

11.85

x

23156.27

0.49 2400 1.13

1.11

= =

= =

kg

1

Lantai Dasar 1

Sbx = 1 Sby =

- (

24.38 23156.27 1

###

kg

0.49

117359.57 kg

2.2.7.1 Komponen Struktur Ujung Sederhana Cm = Cmx Sbx = ≥ 1 Nu 1 - ( ) Ncrbx

1.67

=

1.001

(Tarik)

=

1.000

(Tarik)

)

1.13 ###

kg

Sby = 1

- (

Sbx = 1

- (

1

- (

Sby =

24.38 ) 117359.57 1 54.844 ) 23156.27 1 54.844 ) 117359.57

=

1.000

(Tarik)

=

1.002

(Tekan)

=

1.000

(Tekan)

=

1.001

(Tarik)

=

1.000

(Tarik)

=

1.002

(Tekan)

=

1.000

(Tekan)

Lantai 1 1

Sbx = 1

- (

1

- (

Sby =

15.6 ) 23156.27 1 15.6 ) 117359.57 1

Sbx = 1

- (

1

- (

Sby =

35.1 ) 23156.27 1 35.1 ) 117359.57

2.2.8 Kontrol Gaya Kombinasi 2.2.8.1 Angin Dari Arah Tegak Lurus Dinding (tarik) Lantai Dasar Nu 24.38 = = 0 < φ . Nn 24174 Nu 2 2

φ . Nn

x 24.375 x 24174

Lantai 1 Nu = φ . Nn

15.6 24174 Nu

2 2

φ . Nn x 15.600 x 24174

+ +

Mux φ 87.8 0.9 0.14

=

+ +

x

Sbx

x x x < OK

Mnx 1.001 1003 1

0

<

Mux

x

Sbx

φ 70.2 0.9 0.12

x x x < OK

Mnx 1.001 1003 1

2.2.8.2 Angin Dari Arah Tegak Lurus Gevel (tekan) Lantai Dasar Nu 54.84 = = 0 < 0.2 φ . Nn 24174 Nu Mux x Sbx + φ . Nn φ 2 x x Mnx

0.2

+ +

0.2

+ +

OK Muy φb 4.250 0.9

x

Sby

x x x

Mny 1.000 105

Muy φb 3.945 0.9

x

Sby

x x x

Mny 1.000 105

Muy φb

x

Sby

x

Mny

<

1

<

1

<

1

<

1

<

1

OK

OK +

2

54.844 x 24174

Lantai 1 Nu = φ . Nn 2 2

39.0 0.9 0.09

+

35.1

= 24174 Nu φ . Nn x 35.100 x 24174

x x < OK

1.002 1003 1

<

0.2

Mux

x

Sbx

φ 31.2 0.9 0.08

x x x < OK

Mnx 1.002 1003 1

0 + +

+

4.250 0.9

x x

1.000 105

Muy φb 3.945 0.9

x

Sby

x x x

Mny 1.000 105

+ +

2.3.2 Perencanaan Profil WF untuk Regel Horizontal Gevel Dengan ukuran : WF 100 x 50 x 5 x 7 A = 11.85 cm2 tf = 7 mm Zx = 41.8 cm3 W= 9.3 kg/m Ix = 187 cm4 Zy = 9 cm3 a = 100 mm Iy = 14.8 cm4 h= 70 mm bf = 50 mm tw = 5 mm Sx = 37.5 mm iy = 1.12 cm ix = 3.98 cm 41.8 r= 8.94 Mutu Baja = BJ 37 fu = 3700 kg/cm2 = 370 Mpa fy = 2400 kg/cm2 = 240 Mpa

x

1.25

alat Pengikat dll 10 %

= 0.1

x

14.49

Myd = 1/8 x q x (L/3)2

= 0.13

x

15.94

Lantai 1 Berat Gording Berat Seng Gelombang

= 4.15

x

1

alat Pengikat dll 10 %

= 0.1

x

13.45

Berat Total Berat Total x 1

Berat Total

Myd = 1/8 x q x (L/3)

2

Beban Angin Lantai Dasar Tekanan Angin Angin Tekan ( C = 0.9 )

= 0.13

x

= 0.9

14.8

x

Berat Total x 1

30

1

<

1

<

1

OK

2.3 Perencanaan Regel Horizontal Gevel 2.3.1. Data - Data perencanaan tambahan Jarak Kolom Dinding (L) : 300 cm Jarak Gording Lt Dasar : 125 cm Jarak Gording Lt 1 : 100 cm

2.3.3 Perencanaan Pembebanan 2.3.3.1 Perhitungan Beban Beban Mati Lantai Dasar Berat Gording Berat Seng Gelombang = 4.15

<

= = = = = =

9.3 kg/m 5.19 kg/m 14.49 kg/m 1.45 kg/m 15.94 kg/m 1.99 kg/m

= = = = = =

9.3 kg/m 4.15 kg/m 13.45 kg/m 1.35 kg/m 14.8 kg/m 1.85 kg/m

= =

30 27

kg/m2 kg/m2

q = Angin Tekan x Jarak Gording = 27 Angin Hisap ( C = 0.4 ) 0.4 q = Angin hisap x Jarak Gording = 12

x x x

1.25 30 1.25

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) : Mxw = 1/8 x q x (L)2 = 0.13 x 33.75 x N = q x Jarak Gording = 15 x 1.25

9

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) : Mxw = 1/8 x q x (L)2 = 0.13 x 15 x N = q x Jarak Gording = 33.75 x 1.25

9

Lantai 1 Tekanan Angin Angin Tekan ( C = 0.9 ) = q = Angin Tekan x Jarak Gording = Angin Hisap ( C = 0.4 ) q = Angin hisap x Jarak Gording =

0.9 27 0.4 12

x x x x

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) : Mxw = 1/8 x q x (L)2 = 0.13 x 27 x N = q x Jarak Gording = 12 x 1

9

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) : Mxw = 1/8 x q x (L)2 = 0.13 x 12 x N = q x Jarak Gording = 27 x 1

9

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) : Mux = 1.2 x 0 + 1.3 x 16.88 = 21.94 kgm Muy = 1.2 x 1.99 + 1.3 x 0 = 2.39 kgm Nu = 1.2 x 0 + 1.3 x 42.19 = 54.84 kg Lantai 1 1. U = 1.4 D Muy = 1.4 x 1.85 = 2.59 kgm 2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha ) Akibat Beban Angin yg Tegak Lurus Dinding (tarik) : Mux = 1.2 x 0 + 1.3 x 30.38 = 39.49 kgm Muy = 1.2 x 1.85 + 1.3 x 0

33.75 kg/m 12 kg/m2 15 kg/m

= =

37.97 kgm 18.75 kg

(Tarik)

= =

16.88 kgm 42.19 kg

(Tekan)

= = = = =

30 1 30 1

2.3.3.2 Kombinasi Pembebanan Lantai Dasar 1. U = 1.4 D Muy = 1.4 x 1.99 = 2.79 kgm 2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha ) Akibat Beban Angin yg Tegak Lurus Dinding (tarik) : Mux = 1.2 x 0 + 1.3 x 37.97 = 49.36 kgm Muy = 1.2 x 1.99 + 1.3 x 0 = 2.39 kgm Nu = 1.2 x 0 + 1.3 x 18.75 = 24.38 kg

= = =

30 27 27 12 12

kg/m2 kg/m2 kg/m kg/m2 kg/m

= =

30.38 kgm 12 kg

(Tarik)

= =

13.5 kgm 27 kg

(Tekan)

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

= Nu = =

2.22 1.2 15.6

kgm x kg

0

+

1.3

x

12

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) : Mux = 1.2 x 0 + 1.3 x 13.5 = 17.55 kgm Muy = 1.2 x 1.85 + 1.3 x 0 = 2.22 kgm Nu = 1.2 x 0 + 1.3 x 27 = 35.1 kg 2.3.4 Kontrol Kekuatan Profil 2.3.4.1 Penampang Profil Untuk Sayap Untuk Badan bf 170 h ≤ ≤ 2 tf fy tw 50 170 70 ≤ ≤ 2 7 240 5 3.57 ≤ 10.97 14.0 ≤ OK OK Penampang Profil Kompak, maka Mnx = Mpx

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

+

0.5

x

0

1680 fy 1680 240 108.4

2.3.4.1 Kontrol Lateral Buckling Jarak Baut Pengikat / pengaku lateral = L B = 500 mm = E LP = 1.76 iY x x fy 200000 = 1.76 x 1.12 x = 56.90 cm 240 LB LP maka : Mnx Ternyata : < = Mpx Mnx = Mpx = Zx . Fy = 41.8 1.5 Myx = 1.5 Sx fy = 1.5 x ===> Mnx < 1.5 Myx Mny = Zy ( satu sayap ) * fy = 1/4 x tf x bf 2 x fy = 0.25 x 0.7 = 105 kgm

x 37.5

x

2.3.5 Perhitungan Kuat Tarik 2.3.5.1 Kontrol Kelangsingan λp ≤ 300 Lk 300 = = λ= ix 3.98

2400 x 2400

52

= =

x

2400

OK

75.38

<

300

2.3.5.2 Berdasarkan Tegangan Leleh φ Nn = φ .Ag . fy = 0.85 x 11.85

x

2400

50

### Kgm 1350 Kgm

=

10500 kgcm

= 24174 kg Menentukan

2.3.5.3 Berdasarkan Tegangan Putus φ Nn = φ .Ae . fu =

= 0.75 x 0.85 x Ag x fu = 0.75 x 0.85 = 0.75 x 0.85 = ### kg Tidak Menentukan

x x

Ag 11.85

cm

x x

fu 3700

2.3.5.4 Kontrol Kuat Tarik Lantai Dasar φ Nn > Nu 24174 > 54.84 OK Lantai 1 φ Nn > Nu 24174 > 1404 OK 2.3.6 Perhitungan Kuat Tekan 2.3.6.1 Kontrol Kelangsingan λp ≤ 200 Lkx 300 = = λpx = ix 3.98 Lky 50 = = λpy = iy 1.12

75.38

<

200

OK

44.64

<

200

OK

2.3.6.2 Berdasarkan Tekuk Arah X fy λx 75.38 = x λc = 3.14 E π λc 0.25 < < 1.2 1.43 = ω= 1.6 - 0.67 λc 1.6 fy = 0.85 x φ Nn = φAg ω 2.3.6.3 Berdasarkan Tekuk Arah Y fy λy = 44.64 x λc = E π π λc 0.25 < < 1.2 1.43 = ω= 1.6 - 0.67 λc 1.6 fy = 0.85 x φ Nn = φAg ω 2.3.7 Perhitungan Pembesaran Momen Ab x fy Ncr = λc 2 11.85 x 2400 Ncrbx = = 0.831 2 11.85 x 2400 Ncrby = = 0.492 2

2400 2000000

=

0.83

1.43 0.67

x

11.85

x

0.83 2400 1.37

2400 2000000

=

x

11.85

x

41166.71

=

0.49 2400 1.13

= =

kg

117366.49 kg

2.3.7.1 Komponen Struktur Ujung Sederhana Cm = Cmx Sbx = ≥ 1 Nu 1 - ( ) Ncrbx

1

Lantai Dasar 1

Sbx = 1

- (

24.38

= )

1.001

1.37 ###

kg

0.49

1.43 0.67

=

(Tarik)

1.13 ###

kg

1

- (

1

- (

Sby =

Sbx = 1

- (

1

- (

Sby =

) 41166.71 1 24.38 ) 117366.49 1 54.844 ) 41166.71 1 54.844 ) 117366.49

=

1.000

(Tarik)

=

1.001

(Tekan)

=

1.000

(Tekan)

=

1.000

(Tarik)

=

1.000

(Tarik)

=

1.001

(Tekan)

=

1.000

(Tekan)

Lantai 1 1

Sbx = 1

- (

1

- (

Sby =

15.6 ) 41166.71 1 15.6 ) 117366.49 1

Sbx = 1

- (

1

- (

Sby =

35.1 ) 41166.71 1 35.1 ) 117366.49

2.3.8 Kontrol Gaya Kombinasi 2.3.8.1 Angin Dari Arah Tegak Lurus Dinding (tarik) Lantai Dasar Nu 24.375 = = 0 < φ . Nn 24174 Nu Mux x Sbx + φ . Nn φ 2 x x Mnx 24.375 49.4 x 1.001 + 2 x 24174 0.9 x 1003 0.08 < 1 OK Lantai 1 Nu 15.6 = = 0 < φ . Nn 24174 Nu Mux x Sbx + φ . Nn φ 2 x x Mnx 15.600 39.5 x 1.000 + 2 x 24174 0.9 x 1003 0.07 < 1 OK 2.3.8.2 Angin Dari Arah Tegak Lurus Gevel (tekan) Lantai Dasar Nu 54.84 = = 0 < 0.2 φ . Nn 24174 Nu Mux x Sbx + φ . Nn φ 2 x x Mnx

0.2 + +

0.2 + +

OK Muy φb 2.390 0.9

x

Sby

x x x

Mny 1.000 105

Muy φb 2.219 0.9

x

Sby

x x x

Mny 1.000 105

Muy φb

x

Sby

x

Mny

<

1

<

1

<

1

<

1

<

1

OK

OK +

2

54.844 x 24174

Lantai 1 Nu = φ . Nn 2 2

21.9 0.9 0.05

+

35.1

= 24174 Nu φ . Nn x 35.100 x 24174

x x < OK

1.001 1003 1

<

0.2

Mux

x

Sbx

φ 17.6 0.9 0.04

x x x < OK

Mnx 1.001 1003 1

0 + +

2.4 Perencanaan kolom Gevel 2.4.1 Data Perencanaan Panjang Beban Atap Regel 5 Panjang Beban Atap Regel 2

= =

3 3

m m

7 6

x x

1.000 105

Muy φb 2.219 0.9

x

Sby

x x x

Mny 1.000 105

Panjang Cantilever = Jarak Kuda-kuda =

1 4

m m

<

1

<

1

<

1

OK + +

qw regel 5 = panjang x angin tekan = 3 x 27 = qw regel 2 = panjang x angin tekan = 3 x 27 =

Lebar Beban Atap Regel 5 = 2.5 m Lebar Beban Atap Regel 2 = 2 m Tinggi Regel 5 = Tinggi Regel 2 =

2.390 0.9

+

m m

81

kg/m

81

kg/m

Regel 5 Luas atap yg Dipikul oleh Regel 5 ( A1 ) = Lebar Beban Atap Regel 5 x Pjg Beban Atap Regel 5 = 3 x 2.5 = 7.5 m2 Luas Dinding Regel 5 ( A2 ) = Pjg Beban Atap Regel 5 x Tinggi Regel 5 = 2.5 x 7 = 17.5 m2 Regel 2 Luas atap yg Dipikul oleh Regel 2 ( A3 ) =Lebar Beban Atap Regel 2 x Pjg Beban Atap Regel 2 = 3 x 2 = 6 m2 Luas Dinding Regel 2 ( A4 ) = Pjg Beban Atap Regel 2 x Tinggi Regel 2 = 2 x 6 = 12 m2 2.4.2 Perencanaan Pembebanan 2.4.2.1 Beban Mati Regel 5 ND atap = A1 x qD atap

=

7.5

x

20.63

=

### kg

ND Dinding = A2 x qD Dinding

=

17.5

x

4.15

=

72.63 kg

ND Gording = Jml Gording . w Gording

=

7

x

9.3

=

65.1 kg

Regel 2 ND atap = A3 x qD atap

=

6

x

20.63

=

### kg

ND Dinding = A4 x qD Dinding

=

12

x

4.15

=

49.8 kg

ND Gording = Jml Gording . w Gording

=

6

x

9.3

=

55.8 kg

2.4.2.2 Beban hidup

Regel 5 NL atap = A1 x qL atap

=

7.5

x

20

=

150 kg

Regel 2 NL atap = A2 x qL atap

=

6

x

20

=

120 kg

2.4.2.3 Beban Angin Regel 5 Mw = 1/8 x qw x (h)2

=

0.13

x

81

x

7

2

=

### kgm

Regel 2 Mw = 1/8 x qw x (h)2

=

0.13

x

81

x

6

2

=

364.5 kgm

=

3.5

cm

=

2215.767

2.4.3 Syarat Kekakuan Regel 5 h 700 Y= = 200 200 5 q Ix = x 384 E 5 4.96 x 384 0.02 ===> Ix Profil yg Dipakai > Pakai Profil : WF 175 A = 51.21 W = 40.2 a = 175 bf = 175 iy = 4.38

x cm2 kg/m mm mm cm

175 tf = Ix = Iy = tw = ix = r=

Mutu Baja = BJ 37 fu = 3700 kg/cm2 = fy = 2400 kg/cm2 =

L4 x x Y x 7 4 x 3.5 2215.767 cm4

x 11 2880 984 7.5 7.5 12

7.5 mm cm4 cm4 mm cm cm

x Zx = Zy = h= = Sx =

11 ### ### 175 136 2050

cm3 cm3 mm mm

2

cm4

x(

11

+

12

)

370 Mpa 240 Mpa

Nd Profil Nd total

= 7 x 40.2 = 281.4 kg = Nd atap + Nd (Dinding+Gording ) + Nd Profil = ### + ### + 281.4 = ### kg NL Total = NL atap = 150 kg Mw = ### kgm U = ( 1.2D + 1.6L+ 1.6W ) x 0.75 Nu = ( 1.2 x ### + 1.6 x 150 ) x 0.75 Mntx = 1.6 x Mw x 0.75 = 1.6 x ### Regel 2 h 600 Y= = 200 200 5 q Ix = x 384 E 5 3.65 x 384 0.02 ===> Ix Profil yg Dipakai >

x L4 x Y x 6 x 3 1025.156

Pakai Profil : WF 150

x

x

100

=

3

6

= x

cm

4

=

cm4

x

9

1025.156

cm4

### kg 0.75 =

###

kg

A= W= D= Bf = iy =

26.84 21.1 148 100 2.37

cm2 kg/m mm mm cm

tf = 9 mm Ix = 1020 cm4 Iy = 151 cm4 tw = 6 mm ix = 6.17 cm r = 11 cm

Mutu Baja = BJ 37 fu = 3700 kg/cm2 = fy = 2400 kg/cm2 =

Zx = Zy = h= = Sx =

### 45.88 150 116 138

cm3 cm3 mm mm

2

x(

9

+

11

)

437.4

kg

≤

1

370 Mpa 240 Mpa

Nd Profil Nd total

= 6 x 21.1 = 126.6 kg = Nd atap + Nd (Dinding+Gording ) + Nd Profil = ### + 105.6 + 126.6 = ### kg NL Total = NL atap = 120 kg Mw = 364.5 kgm U = ( 1.2D + 1.6L+ 1.6W ) x 0.75 Nu = ( 1.2 x ### + 1.6 x 120 ) x 0.75 Mntx = 1.6 x Mw x 0.75 = 1.6 x 364.5 2.4.4 Kontrol Tekuk Regel 5 untuk arah x : Lkx = 700 cm Lkx 700 λx = = = 93.33 ix 7.5 λx fy 93.33 2400 = x = 1.03 λc = 2000000 E π π π2 . E . A π2 x 2000000 Ncrbx = = 2 λx 93.33 2 = 116040.87 kg untuk Arah y : Lky = 100 cm Lky 100 λy = = = 22.83 iy 4.38 fy λy 22.83 2400 = x = 0.25 λc = 2000000 E π π π2 . E . A π2 x 2000000 Ncrby = = 2 λy 22.83 2 = 1939245.26 kg Tekuk Kritis Adalah Arah ====> X karena λx > λy λc 0.25 < < 1.2 1.43 1.43 = = ω= 1.6 - 0.67 λc 1.6 0.67 x 1.03 Pn = Ag . fy = 51.21 x 2400 = 122904 kg Pu 696.47 = = 0.01 < 0.2 φ . Pn 0.85 x 122904 Pakai Rumus : Pu Mux + + φc . Pn φb 2 x x Mnx Batang Dianggap Tidak Bergoyang Maka : Cmx Sbx = ≥ 1 Nu 1 - ( )

;Cm =

1

= x

### kg 0.75 =

x

51.21

x

51.21

1.57

Muy φb

x

Mny

1

- (

) Ncrbx 1 Sbx = = 696.5 1 - ( ) 116040.87 Mux = Mntx . Sbx Mux = ### x 1.006 = ### kgm

1.006

≥

1

Regel 2 untuk arah x : Lkx = 600 cm Lkx 600 λx = = = 97.24 ix 6.17 fy λx 97.24 2400 λc = = x = 1.07 2000000 E π π π2 . E . A π2 x 2000000 Ncrbx = = 2 λx 97.24 2 = 56024.77 kg untuk Arah y : Lky = 100 cm Lky 100 λy = = = 42.19 iy 2.37 fy λy 42.19 2400 = x = 0.47 λc = 2000000 E π π π2 . E . A π2 x 2000000 Ncrby = = λy 2 42.19 2 = 297583.57 kg λy Tekuk Kritis Adalah Arah ====> X karena λx > λc 0.25 < < 1.2 1.43 1.43 = = ω= 1.6 - 0.67 λc 1.6 0.67 x 1.07 Pn = Ag . fy = 26.84 x 2400 = 64416 kg Pu 464.38 = = 0.01 < 0.2 φ . Pn 0.85 x 64416 Pakai Rumus : Pu Mux + + φc . Pn φb 2 x x Mnx Batang Dianggap Tidak Bergoyang Maka : Cmx Sbx = ≥ 1 Nu 1 - ( ) Ncrbx 1 Sbx = = 464.4 1 - ( ) 56024.77 Mux = Mntx . Sbx Mux = 437.4 x 1.008 = ### kgm 2.4.5 Menentukan Mnx Regel 5 * Penampang Profil Untuk Sayap : bf 170 ≤ 2 tf fy

1.008

;Cm =

1

≥

1

Untuk Badan : h 1680 ≤ tw fy

x

26.84

x

26.84

λ

0.25 < c < 1.2

1.62

Muy φb

x

Mny

≤

1

175 2

≤

11 7.95

≤ OK

170 240 10.97

136 7.5 18.1

≤ ≤ OK

1680 240 108.4

Penampang Profil Kompak, maka Mnx = Mpx * Kontrol Lateral Buckling Lateral Bracing = L B = 1000 LP =

mm

x

iY

x

= 1.76

x

100

x

LB

<

LP

0

x

Mnx = Mpx = Zx . Fy = Mny = Zy ( satu sayap ) * fy = 1/4 x tf x bf 2 x fy = 0.25 x 0 = 0 kgm

x

Regel 2 Penampang Profil untuk Sayap

=

maka : Mnx p

02

###

cm

=

Mpx

=

0

x

p

Kgm

=

0

kgcm

untuk Badan

b 170 ≤ 2tf fy #REF! #REF! #REF!

100 cm

E fy 200000 2400

1.76

Ternyata :

=

h 1680 ≤ t fy 170

#REF! #REF! #REF!

≤ #REF! ≤ #REF!

#REF!

1680

≤ #REF! ≤ #REF!

#REF!

Penampang Profil Kompak, maka Mnx = Mpx Lateral BracingLb =

Lp = 1.76 * iy

E fy

100

cm

Lp = #REF! cm

Ternyata Lp > Lb

maka Mnx = Mpx

Mnx = Mpx = Zx. Fy = #REF! * #REF! = #REF! Kgm Mny = Zy ( 1 flen ) * fy = (1 / 4 * tf * bf 2) * fy = 0.25 x#REF! x#REF! x#REF! = #REF! kgcm = #REF! kgm 2.4.6 Persamaan Interaksi Pu

+

Mux

+

Muy

<

1

2

φc . Pn

x

+

φb

x

Mnx

+

Regel 5 #REF! 0.17 x x #REF!

+

598.94 0.9 x#REF!

+

#REF! #REF!

+

0 0.9 x#REF!

+ <

1

OK

Regel 2 #REF! 0.17 x#REF!

+

#REF!

+

#REF! 0.9 x#REF! #REF! #REF!

+

0 0.9 x#REF!

+ <

1

OK

2.5 Perencanaan Penggantung Gording Dinding Samping dan Gevel 2.5.1 Data Penggantung Gording Jarak Kuda - Kuda = 400 cm Jumlah Penggantung Gording = 2 buah Jumlah Gording Gevel = 7 buah Jumlah Gording Dinding= 3 buah Jarak Penggantung gording = ### cm Jarak antara Gevel = 300 cm Jarak Antar Gordng Horizontal 125 Dinding cm = Jarak Antar Gordng Horizontal 100 Gevel cm= 2.5.2 Perencanaan Pembebanan Dinding Samping Beban Mati Berat Sendiri Gording = Berat Seng Gelombang = = Alat Pengikat dll 10 =% 0.1 x 4.15 = = Ra = q * JarakKuda − Kuda

Gevel Beban Mati Berat Sendiri Gording Berat Seng Gelombang Alat Pengikat dll 10 =% 0.1 x 4.15

0 kg/m 4.15 kg/m 4.15 kg/m 0 kg/m 4.15 kg/m

=

16.6 kg

= = = = =

0 kg/m 4.15 kg/m 4.15 kg/m 0.42 kg/m 4.57 kg/m

φb

x

Mny

<

1

Ra = q * JarakGevel

=

13.7 kg

2.5.3 Perhitungan Gaya 2.5.3.1 Penggantung Gording Tipe A Dinding Samping Ra =

` 23.24 kg

Ra Total = Ra * jumlah Gording Ra

=

69.72 kg

=

` 19.17 kg

Gevel Ra

Ra Total = Ra * jumlah Gording Ra = ### kg 2.5.3.2 Penggantung Gording Tipe B Dinding Samping arctgnβ = 0.94 β = 0.75 β = 43.14 o

RB =

RA Sinβ

Rb

=

69.72 0.68

=

###

kg

###

kg

Gevel

arctgnβ = β=

1.4 0.95 β = 54.44 o

RB =

RA Sinβ

Rb

=

### 0.81

=

2.5.4 Perencanaan Batang Tarik Dinding Samping Pu = ### kg BJ 37 fu = 0 kg/cm2 fy = 0 kg/cm3 Gevel Pu = ### kg BJ 37 fu = 0 kg/cm2 fy = 0 kg/cm3 2.5.4.1 Kontrol Leleh Dinding Samping Pu = φ fy Ag dengan φ = 0.9 Ag perlu = Pu/φ fy =

d=

Ag * 4 π

### 0

Gevel Pu = φ fy Ag dengan φ = 0.9

=

### cm2 ###

Ag perlu = Pu/φ fy =

d=

Ag * 4 π

### 0

=

### cm2 ###

2.5.4.2 Kontrol Putus Dinding Samping Pu = φ fu 0.75 Ag dengan φ = 0.75 Ag Perlu = Pu φ fu 0.75

=

Ag = 1 / 4π d=2

### cm2

=

### x 4 3.14

### 0

=

### cm2 ###

d=

### cm

Pakai d = 10 mm

Dinding Samping Pu = φ fu 0.75 Ag dengan φ = 0.75 Ag Perlu = Pu φ fu 0.75

=

Ag = 1 / 4π d=2

### cm2

=

### x 4 3.14

### 0

=

### cm2 ###

d=

### cm

Pakai d = 10 mm

2.5.5 Kontrol Kelangsingan Dinding Samping Jarak Penggantung Gording ### cm =

PanjangRb = JrkPenggantungGording 2 + JrkantarGordingHorizontal 2 Panjang Rb = ### Panjang Rb =

+

15625

### cm

PanjangRb d≥ 500

1

>

### 500

1

>

0.37

OK Gevel Jarak Penggantung Gording 100 cm =

PanjangRb = JrkPenggantungGording 2 + JrkantarGordingHorizontal 2 Panjang Rb = 10000 Panjang Rb =

+

10000

### cm

PanjangRb d≥ 500

1

>

### 500

1

> OK

0.28

2.6 Perencanaan Ikatan Angin Dinding 2.6.1 Data Perencanaan Ikatan Angin Dinding Tekanan Angin W =0 kg/m2 Koefisien Angin C0.9 = a1 = 300 cm a2 = 200 cm = 00 α= 0 2.6.2 Perhitungan Tinggi Ikatan Angin ( h ) h1 = 9m h2 = 9 + 2 x tg x 0.44 = h3 = 9 + 4 x tg x 0.44 = h4 = 9 + 6 x tg x 0.44 = h5 = 9 + 9 x tg x 0.44 =

9.93 10.87 11.8 13.2

m m m m

2.6.3 Perhitungan Gaya - Gaya yang Bekerja R = 1/2 W C a h R1 = 0.50 x 0 x#REF! x 1 x 9 = R2 = 0.50 x 0 x#REF! x 2 x 9.93 = R3 = 0.50 x 0 x#REF! x 2 x10.87 = R4 = 0.50 x 0 x#REF! x 2.5 x 11.8 = R5 = 0.50 x 0 x#REF! x 3 x 13.2 =

#REF! #REF! #REF! #REF! #REF!

kg kg kg kg kg

Rtotal = ( R1+R2+R3+R4+(R5/2)) = #REF!

kg

2.6.4 Perencanaan Dimensi Ikatan Angin

tg φ

= =

1 4 0.25

φ

=

0.24 rad

R1 Rtotal

= =

2.6.4.1 Menghitung gaya Pada Titik Simpul Pada Titik Simpul A ΣV = 0 Rtotal + S1 = 0 S1 = - Rtotal S1 = #REF! kg ΣH = 0 S2 = 0

Pada Titik Simpul B EV = 0 R1 + S1 +S3 Cos Φ = 0

S3 =

− ( R1 − S1 ) Cosφ

S3 = #REF! kg 2.6.5 Perencanaan Batang Tarik Pu = #REF! kg Pu = S 3 *1.6 * 0.75 BJ 37 fu = 0 kg/cm2 fy = 0 kg/cm3 2.6.5.1 Kontrol Leleh Pu = φ fy Ag dengan φ = 0.9 Ag perlu = Pu/φ fy =

#REF! 0

=

#REF! cm2 #REF!

2.6.5.2 Kontrol Putus Pu = φ fu 0.75 Ag dengan φ = 0.75 Ag Perlu = Pu φ fu 0.75

=

Ag = 1 / 4π d=2

#REF! cm2

Ag * 4 = π

#REF! x 4 3.14

d=

#REF! 0

=

#REF! #REF!

d = #REF! cm

#REF! kg #REF! kg

Pakai d = 12 mm 2.6.6 Kontrol Kelangsingan Jarak Kuda - Kuda 400 = cm

PanjangS 3 = JrkantarKuda − Kuda 2 + Jrkantar 2 Re gelHorizontal 2 Panjang S3 =

0

Panjang S3 =

+

0

0 cm

d≥

PanjangS 3 500

1.2

>

0 500

1.2

> OK

0

Start

Masukkan Data - Data Perencanaan Bondex dan Balok Anak : Panjang Bentang Beban Bondex Yang Dipikul Balok Anak = ? Panjang Balok Anak = ? Berat Sendiri Beton = ? Berat Sendiri Bondex = ? Berat Spesi per cm Tebal = ? Berat Tegel = ? Beban Berguna = ?

Hitung Pembebanan terhadap Balok Anak : Beban Mati Beban Hidup

Hitung Tebal Lantai Bondex Tebal Lantai Bondex Dicari dengan Menggunakan Tabel yang ada dengan memperhitungkan Beban Berguna yang akan Disalurkan Bondex ke Balok Anak sebagai Dasar Perencanaan. T=?

Hitung Luasan Tulangan Negatif Bondex Luasan Tulangan Negatif Bondex Dicari dengan Menggunakan Tabel yang ada dengan memperhitungkan Beban Berguna yang akan Disalurkan Bondex ke Balok Anak sebagai Dasar Perencanaan. A=?

Asumsikan Diamter Tulangan Negatif Bondex : φ = ? Mm

Hitung Banyaknya Tulangan Yang Diperlukan Tiap 1 m : A/As = ? Hasilnya Dibulatkan Keatas

Hitung Jarak Tulangan Tarik : Jarak Tulangan Tarik = Jarak Tulangan yang Diperlukan ( 1 m ) Dibagi dengan Banyaknya Tulangan yang diperlukan dengan Jarak yang Telah Ditetapkan Diatas

Perencanaan Pembebanan Beban Mati Beban Hidup

Hitung qU, Mu Max dan Du Max : qU = 1.2 qD + 1.6 qL

Mu max =

1 qu l 2 8

Du max =

1 qu l 2

PERHITUNGAN Ix PROFIL MINIMUM Dimana Y ijin = L/360

Ix >

5 ( qD + qL ) * l 4 384 EY

Pilih Profil Baja Dimana Ix-nya Harus > Ix Minimum : A = ? ; W = ? ; a = ? ; bf = ? ; iy = ? ;tf = ? ; Ix = ? Iy = ? ; tw = ? ; Zx = ? ; Zy = ? ; h = ? ; fu = ? ; Fy = ?

Perencanaan Pembebanan + Berat Profil Beban Mati Beban Hidup

Hitung qU, Mu Max dan Du Max ( Berat Profil Dimasukkan ) : qU = 1.2 qD + 1.6 qL

Mu max =

1 qu l 2 8

Du max =

1 qu l 2

KONTROL LENDUTAN BALOK Dimana Y ijin = L/360

Y max =

Perbesar Profil

KO

5 (qD + qL) * l 4 384 EIx

Y mak < Y ijin OK

KONTROL LOKAL BUCKLING Hitung λp, λr Penampang Sayap dan λp, λr Penampang Badan : Sayap Badan 170

λp =

fy

370

λr =

f y − fr

b 2tf

b ≤ λp 2tf

b 2tf

λr =

2550

fy

fy

1

2

Sayap

Badan

KO

λp ≤

≤ λr

Profil Tak Kompak

OK

Mn = Mp − ( Mp − Mr )

KO

KO

Profil Langsing

Profil Langsing

Mn = Mr (λr / λ ) 2

Profil Tak Kompak

λ − λp λr − λp

h ≤ λr t

OK

λ − λp λr − λp

Mn = Mr (λr / λ ) 2

Mn = Mp − ( Mp − Mr )

h ≤λ p t OK

Profil Kompak Mnx = Mnp

KO

λp ≤

1680

h t

OK

Profil Kompak Mnx = Mnp

λp =

Mnx Sayap > Mnx Badan

2 OK

KO

Ambil Mnx Badan Local Buckling

Ambil Mnx Sayap Local Buckling

KONTOL LATERAL BUCKLING Hitung λp dan λr daripada Lateral Buckling

E fy

Lp = 1.76 * iy

Lr = ry (

X1 2 ) 1+ 1+ X 2 fL fL G=

X1 =

E 2(1 + µ)

π Sx

X 2 = 4(

EGJA 2

S 2 Iw ) GJ Iy

Jarak Lateral Bracing λb : λb = ?

KO

KO

λ p ≤ λb ≤ λr

λb ≤ λ p

OK

OK Bentang Pendek Mnx = Mpx

Bentang Menengah

Mn = Cb( Mr + ( Mr − Mp )(

Lr − L ) ≤ Mp Lr − Lp

Bentang Panjang

Mn = Mcr = Cb

π πE E _ I y GJ + ( ) 2 I y I w ≤ Mp L L

Mnx Local Buckling > Mnx Lateral Buckling OK

KO

Ambil Mnx Lateral Buckling

Ambil Mnx Local Buckling

Hitung : 0.9 Mnx

Perbesar Profil

KO

0.9 Mnx > Mu max OK

KONTROL KUAT RENCANA GESER h Hitung

tw

Vn = 0.6 fy Aw

OK

h 1100 ≤ tw fy

KO

1100

≤

h 1370 ≤ tw fy

Vn =

900000 Aw ( h )2 tw

fy

KO

Hitung 0.9 Vn

Perbesar Profil

KO

0.9 Vn > Vu Max OK

Profil Dapat Dipakai

OK

Vn = 0.6 f y Aw

1100t w h fy

Pre - Eliminary Design 3 Perencanaan Bondex dan Balok Anak 3.1 Data - Data perencanaan Beban Hidup : Beban Finishing : Beban Berguna :

400 Kg/m2 90 Kg/m2 490 Kg/m3

Berat Beton Kering : 2400 kg/m3 Panjang Bentang Beban Bondex yang Dipikul Oleh Balok Anak Panjang Balok Anak : 4 m

:

3

m

288 Kg/m2 10.1 Kg/m2 42 Kg/m2 48 Kg/m2 388.1 Kg/m2 400 Kg/m2 90 Kg/m2 490 Kg/m2

3.2 Perencanaan Pelat Lantai Bondex 3.2.1 Data Perencanaan Berat Sendiri Beton Berat Sendiri Bondex Berat Spesi per cm Tebal Berat Tegel

= = = =

3.2.2 Perencanaan Pembebanan Beban Mati Berat Beton = 2400 Berat Bondex Berat Spesi 2 Cm = 21 Berat Tegel 2 Cm = 24

2400 kg/m3 10.1 kg/m2 21 kg/m2 24 kg/m2

*

0.12

* *

2 2 qD

= = = = =

qL

= = =

Beban Hidup Beban Hidup Lantai gudang Beban Finishing

3.2.3 Perencanaan Tebal Lantai Beton dan Tulangan Negatif 3.2.3.1 Perencanaan Tebal Lantai qL = 490 kg/m2 Beban Berguna yang Dipakai = Jarak Antar Balok = Jarak Kuda - Kuda =

500 kg/m2 300 cm 400 cm

Dari Tabel Brosur ( Bentang Menerus dengan Tulangan Negatif ),didapat : t = 12 mm A = 3.57 cm2/m 3.2.3.2 Perencanaan Tulangan Negatif Direncanakan Tulangan Dengan φ = As =

10 mm 0.79 mm2

Banyaknya Tulangan Yang diperlukan Tiap 1 m =

Jarak Tulangan Tarik =

A As

200

= = = cm

3.57 0.79 4.55 5

Buah Buah

Pasang Tulangan Tarik φ10 - 200 3.3 Perencanaan Dimensi Balok Anak 3.3.1 Perencanaan Pembebanan Beban Mati ( D ) Bondex = 3 Plat Beton = 3 Tegel + Spesi = 3

10.1 0.12 90

= = = =

2400

qD

Beban Hidup ( L ) qL =

3

490

=

3.3.3 Perhitungan qU , Mu Max dan Du Max qU = 1.2 qD + 1.6 qL qU = 1.2 1164.3

1470 kg/m

1.6

1470

=

3749.16 Kg/m 7498.32 Kgm

Mu max =

1 qu l 2 8

=

0.13

3749.16

16

=

Du max =

1 qu l 2

=

0.5

3749.16

4

=

3.3.4 Perhitungan Ix Profil Yang Diperlukan Y= L = 400 360 360

30.3 kg/m 864 kg/m 270 kg/m 1164.3 kg/m

=

1.11

11.64 2100000

14.7 1.11

7498.32 Kg

5 ( qD + qL ) * l 4 Ix > 384 EY Ix

>

5 384

Ix

>

3763.29

(

)

2.56E+10

cm4

3.3.5 Perencanaan Profil WF untuk Balok Anak 250 A= W= a= bf = iy = Mutu Baja = fu = fy =

x

37.66 cm2 29.6 kg/m 250 mm 125 mm 2.79 cm

125 tf = Ix = Iy = tw = ix =

x

6

x

9 mm 4050 cm4 294 cm4 6 mm 10.4 cm

9 Zx = Zy = h= r=

351.86 cm3 72.02 cm3 208 mm 12 mm 351.86 72.02

BJ 37 3700 kg/cm2 2400 kg/cm2

3.3.6 Perencanaan Pembebanan + Beban Profil Beban Mati ( D ) Bondex = 3 10.1 Plat Beton = 3 0.12 Tegel + Spesi = 3 90 Berat Profil =

2400

= = = =

30.3 kg/m 864 kg/m 270 kg/m 29.6 kg/m

qD

Beban Hidup ( L ) qL =

3

490

=

=

1470

3.3.7 Perhitungan qU , Mu Max dan Du Max ( Berat Profil Dimasukkan ) qU = 1.2 qD + 1.6 qL qU = 1.2 1193.9 1.6 1470 =

Mu max =

1 qu l 2 8

=

0.13

3784.68

16

=

Du max =

1 qu l 2

=

0.5

3784.68

4

=

400 360

=

1.11

11.94 2100000

14.7 4050

3.3.8 Kontrol Lendutan Balok Y= L = 360

Y max =

1193.9 kg/m

3784.68 Kg/m 7569.36 Kgm

7569.36 Kg

5 ( qD + qL) * l 4 384 EIx

=

5 384

=

1.04

(

<

)

2.56E+10

1.11

OK 3.3.9 Kontrol Kuat Rencana Momen Lentur 3.3.9.1 Kontrol Penampang untuk Sayap

untuk Badan

b 170 ≤ 2tf fy 125 18 6.94

h 1680 ≤ t fy 170 15.49 10.97

≤ ≤

208 6 34.67

≤ ≤

OK

OK

Penampang Profil Kompak, maka Mnx = Mpx Mp = fy * Zx = 2400 * = 844466.4 kgcm = 8444.66 kgm

351.86

3.3.9.2 Kontrol Lateral Buckling Jrk Pengikat Lateral :

Lp = 1.76 * iy

E fy

1000 mm

Lp =

=

141.75 cm

100

cm

1680 15.49 108.44

Lp = 1.76 * iy

E fy

Ternyata

Lp > Lb

maka

Mnx = Mpx = Zx. Fy = 351.86 Mny = Zy ( 1 flen ) * fy = (1 / 4 * tf * bf 2) * fy x 156.25 x = 0.25 x 0.9 = 843.75 kgm 0.9 Mp =

0.9

*

8444.66

0.9 Mp 7600.2

> > OK

Mu 7569.36

*

2400

2400

=

=

3.3.9.3 Kontrol Kuat Rencana Geser

h 1100 ≤ tw fy 208 6

<

1100 15.49

34.67

< Plastis

71

Vn = 0.6 fy Aw = 0.6 2400 = 21600 Kg Vu 7569.36 7569.36

< < < OK

ФVn 0.9 19440

0.6

21600

Mnx = Mpx

25

7600.2 kgm

=

8444.66 Kgm

84375 kgcm

4 Perencanaan Tangga Baja 4.1 Data Perencanaan Tinggi tangga Lebar injakan (i) Panjang Tangga Lebar Pegangan Tangga

= = = =

250 cm 28 cm 600 cm 10 cm

4.2 Perencanaan Jumlah Injakan Tangga 4.2.1 Persyaratan - Persyaratan Jumlah Injakan Tangga 60 cm 25 o

< <

( 2t + I ) a

< <

65 cm 40 o

Dimana : t = tinggi injakan (cm) i = lebar injakan (cm) a = kemiringan tangga 4.2.2 Perhitungan Jumlah Injakan Tangga Tinggi tanjakan (t) Jumlah Tanjakan

= = = =

65 18.5 cm 250 = 18.5 14 buah

Jumlah injakan (n)

=

14

Lebar Bordes Lebar Tangga

= =

600 200

a

= 392

28

/

2

13.51 buah

buah

− −

392 20

32.54 0 cm

180 cm

180 cm

4.3 Perencanaan Pelat Tangga 4.3.1 Perencanaan Tebal Pelat Tangga Tebal Pelat Tangga = 4 mm Berat Jenis Baja = 7850 kg/m3 Tegangan Leleh Baja = 2400 kg/m2 4.3.2 Perencanaan Pembebanan Pelat Tangga

= = =

208 180

cm cm

0.57 rad 208

cm

Beban Mati Berat Pelat = Alat Penyambung (10 %)

x

0

1.8

x 7850 qD

Beban Hidup qL = 500

x

1.8

=

= = =

56.52 kg/m' 5.65 kg/m' 62.17 kg/m'

900 kg/m'

4.3.3 Perhitungan M D dan ML

MD =

1 qDl 2 8

MD

ML =

=

0.13

x 62.17 x

0.08

=

0.61

kgm

=

0.13

x

0.08

=

8.82

kgm

1 qLl 2 8

MD

900

x

4.3.4 Perhitungan Kombinasi Pembebanan M U MU = 1.4 MD Mu = 1.4

x

0.61

=

MU = 1.2 MD + 1.6 ML Mu = 1.2 x 0.61

+

0.85 kgm Tidak Menentukan 1.6

x 8.82

=

14.84 kgm Menentukan

4.3.5 Kontrol Momen Lentur

Zx =

1 2 bh 4

=

0.25

x

180

x

0.16

=

7.2

cm3

=

0.9

x

7.2

x 2400

=

15552

kgcm

kgm

> > OK

Mu 14.84

kgm

=

28 360

=

0.08

1 bh 3 12

=

0.08

=

0.96

cm4

0.96

cm4

φMn = φ Zx * fy φMn = Syarat ->

155.52 kgm φMn 155.52

4.3.6 Kontrol Lendutan f =

Ix = Ix =

L 360

5 ( qD + qL) * l 4 Y max = 384 EIx

x

180

x

0.06

Y max =

5 ( qD + qL) * l 4 384 EIx

=

5 384

=

0.04

(

<

0.62 2100000 x

9 0.96

)

6.15E+05

0.08

OK Ambil Pelat Tangga dengan Tebal =

4

mm

4.4 Perencanaan Penyangga Pelat Injak 4.4.1 Perencanaan Pembebanan Beban Mati Berat Pelat = Berat Baja Siku =

0.14 45

x x

0 45

x 7850 7 x

Alat Penyambung ( 10 % )

x

0.14

4.4 kg/m' 4.6 kg/m' 9 kg/m' 0.9 kg/m' 9.9 kg/m'

= =

qD Beban Hidup qL = 500

= =

=

70

kg/m'

. 4.4.2 Perhitungan M D dan ML

MD =

1 qDl 2 8

MD

ML =

=

0.13

x

9.9

x

3.24

=

4.01

kgm

=

0.13

x

70

x

3.24

=

28.35

kgm

1 qLl 2 8

MD

4.4.3 Perhitungan Kombinasi Pembebanan M U MU = 1.4 MD Mu = 1.4

x

4.01

=

MU = 1.2 MD + 1.6 ML Mu = 1.2 x 4.01

+

5.61 kgm Tidak Menentukan 1.6

x 28.35

=

50.17 kgm Menentukan

4.4.4 Kontrol Momen Lentur Dari Perhitungan Sap 2000 Version 8.2.3 Didapat untuk Profil Siku 45x45x7 : Zx = 6.14 cm3 ( Modulus Plastis ) φMn = φ Zx * fy φMn = Syarat ->

=

0.9

kgm

> >

x

6.14

x 2400

Mu 50.17

kgm

132.62 kgm φMn 132.62

=

13262.4

kgcm

OK

4.4.5 Kontrol Lendutan f =

L 360

=

180 360

=

0.5

Dari Tabel Profil Baja Didapat : Ix =

10.42

Y max =

cm4

5 ( qD + qL) * l 4 384 EIx

=

5 384

=

0.5

(

<

0.1 0.7 2100000 x 10.42

)

1.05E+09

0.5

OK Ambil Profil Baja Siku Sama Kaki

45

x

45

x

7

4.5 Perencanaan Pelat Bordes 4.5.1 Perencanaan Tebal Pelat Bordes Tebal Pelat Tangga Berat Jenis Baja Tegangan Leleh Baja Lebar Pelat Bordes

= = = =

8 7850 2400 2

mm kg/m3 kg/m2 m

4.5.2 Perencanaan Pembebanan Pelat Bordes Beban Mati Berat Pelat = 0.01 x 2 Alat Penyambung (10 %)

x 7850 qD

Beban Hidup qL = 500

x

2

=

= = =

125.6 kg/m' 12.56 kg/m' 138.16 kg/m'

1000 kg/m'

4.5.3 Perhitungan M D dan ML

MD =

1 qDl 2 8

MD

ML = MD

=

0.13

x 138.16 x

0.48

=

8.3

kgm

=

0.13

x 1000 x

0.48

=

60.09

kgm

1 qLl 2 8

4.5.4 Perhitungan Kombinasi Pembebanan M U

MU = 1.4 MD Mu = 1.4

x

8.3

=

MU = 1.2 MD + 1.6 ML Mu = 1.2 x

8.3

+

11.62 kgm Tidak Menentukan 1.6

x 60.09

=

106.1 kgm Menentukan

4.5.5 Kontrol Momen Lentur

Zx =

1 2 bh 4

=

0.25

x

200

x

0.64

=

32

cm3

=

0.9

x

32

x 2400

=

69120

kgcm

kgm

> > OK

Mu 106.1

kgm

=

69.33 360

=

0.19

1 bh 3 12

=

0.08

=

8.53

cm4

8.53

cm4

φMn = φ Zx * fy φMn = Syarat ->

691.2 kgm φMn 691.2

4.5.6 Kontrol Lendutan f =

Ix = Ix =

L 360

Y max =

x

x

0.51

1.38 2100000 x

10 8.53

200

5 ( qD + qL) * l 4 384 EIx

=

5 384

=

0.19

(

<

0.19

OK ambil Tebal Pelat Bordes =

8

mm

4.6 Perencanaan Balok Bordes 4.6.1 Perencanaan Balok Bordes dengan Profil I

)

2.31E+07

100 A= W= a= bf = iy =

x

100

21.9 cm2 17.2 kg/m 100 mm 100 mm 2.47 cm

x

tf = Ix = Iy = tw = ix =

4.6.2 Perencanaan Pembebanan Beban Mati Berat Pelat = 0.01 x Berat Profil I =

0.69

6

x

8 mm 383 cm4 134 cm4 6 mm 4.18 cm

Zx = Zy = h=

x 7850

= = = = =

qD

1 q D L2 8 1 PD = q D L 2

Beban Hidup

1 q L L2 8 1 PL = q L L 2

ML =

84.18 cm3 40.61 cm3 84 mm 84.18 40.61

Alat Penyambung ( 10 % )

MD =

8

43.54 kg/m' 17.2 kg/m' 60.74 kg/m' 6.07 kg/m' 66.82 kg/m'

=

0.13

x 66.82 x

4.33

=

36.13

kgm

=

0.5

x 66.82 x

4.33

=

144.54

kg

qL =

500

x

0.69

=

346.67 kg/m'

. =

0.13

x 346.67 x

4.33

=

187.48

kgm

=

0.5

x 346.67 x

4.33

=

749.91

kg

4.6.3 Perhitungan Kombinasi Pembebanan MU = 1.4 MD Mu = 1.4 = 50.59 kgm x 36.13 Pu = 1.4 = 202.35 kgm x 144.54 Tidak menentukan MU = 1.2 MD + 1.6 ML Mu = 1.2 + 1.6 x 36.13 x 187.48 Pu = 1.2 + 1.6 x 144.54 x 749.91 4.6.4 Kontrol Kekuatan Profil 4.6.4.1 Penampang Profil

fy =

untuk Sayap

= =

343.32 kgm 1373.3 kgm Menentukan

2400 kg/m2 untuk Badan

b 170 ≤ 2tf fy 100 16 6.25

h 1680 ≤ t fy 170 15.49 10.97

≤ ≤

84 6 14

≤ ≤

OK

OK

Penampang Profil Kompak, maka Mnx = Mpx 4.6.4.2 Kontrol Lateral Buckling Jarak Baut Pengikat :

Lp = 1.76 * iy

E fy

250 mm

=

25 cm

1680 15.49 108.44

E fy

Lp = 1.76 * iy

Lp =

Ternyata

125.49 cm

Lp > Lb

maka

Mnx = Mpx = Zx. Fy = 84.18 Mny = Zy ( 1 flen ) * fy = (1 / 4 * tf * bf 2) * fy x 100 x = 0.25 x 3.2 = 1920 kgm

Mnx = Mpx

*

2400

2400

=

=

2020.42 Kgm

192000 kgcm

4.6.5 Kontrol Momen Lentur Zx =

84.18

cm3

φMn = φ Zx * fy φMn = Syarat ->

x 84.18 x 2400

=

0.9

kgm

> > OK

Mu 343.32

kgm

180 360

=

0.5

=

181837.44

1818.37 kgm φMn 1818.37

4.6.6 Kontrol Lendutan f =

Ix =

L 360

=

84.18

cm4

Y max =

5 ( qD + qL) * l 4 384 EIx

=

5 384

=

0.32

(

<

0.67 3.47 2100000 x 84.18

)

1.05E+09

0.5

OK

4.7 Perhitungan Balok Induk Tangga 4.7.1 Data - Data Perencanaan h min = I sin α = 28 x

sin

32.54

=

15.06

cm

kgcm

4.7.2 Perencanaan Balok Induk Dengan Menggunakan Profil WF 250 A= W= a= bf = iy =

Syarat -->

x

32.68 cm2 25.7 kg/m 250 mm 125 mm 2.79 cm

h 25

> > OK

125 tf = Ix = Iy = tw = ix =

x

5

x

8 mm 3540 cm4 255 cm4 5 mm 10.4 cm

8 Zx = Zy = h= r =

310.45 cm3 63.71 cm3 210 mm 12 310.45 63.71

hmin 15.06

4.7.3 Perencanaan Pembebanan

4.7.3.1 Perencanaan Pembebanan Anak Tangga Beban Mati Berat Pelat = 0 x 1.04 x 7850 Berat Profil siku = 4.6 2 x x 0.9 Berat Sandaran Besi Berat Profil WF = 32.68 / cos

0.28 32.54

= = = =

32.66 kg/m' 29.57 kg/m' 15 kg/m' 38.76 kg/m'

mm

Alat Penyambung (+ 10 %)

= =

qD1 Beban Hidup qL1 =

500

Beban q1 Total = = =

x

1.04

1.2 qD + 1.6 qL 1.2 x 127.59 985.1 kg/m'

=

520

+

1.6

kg/m'

x

4.7.3.2 Perencanaan Pembebanan Bordes Beban Mati Berat Profil WF = Berat Pelat Bordes = Berat Profil I =

0.01 17.2

x x

1 1

x

x 7850

= =

Alat Penyambung (+ 10 %)

= =

Pd

Beban Hidup qL2 =

500 kg/m2

520

= 0.69

25.7

= =

jadi q2 total = 1.2 qD + 1.6 qL = 1.2 x 25.7 = 830.84 kg/m'

+

1.6

x

jadi P total = 1.2 PD + 1.6 PL = 1.2 x 66.82 = 634.85 kg

+

1.6

x 346.67

500

kg/m'

43.54 kg 17.2 kg 60.74 kg 6.07 kg 66.82 kg

500 x 0.69 346.67 kg

PL2

4.7.4 Perhitungan Gaya - Gaya pada Tangga

115.99 kg/m' 11.6 kg/m' 127.59 kg/m'

x

1

Lab = Lbc =

Σ Ma = 0

1 1 2 ( q 1 l ab ) + ( p (3l ab + 1.5l bc )) + (q 2 l cb ( l cb + l ab )) − ( Rc(l ab + l bc )) = 0 2 2 492.552 15.366 + 634.845 ( 11.76 + 3.120 ) + 1728.147 ( 1.040 +

3.92 m 2.08 m

3.92

)

RC

6

Rc =

4264.48 kg

ΣV=0

Rva = q1l ab + q 2 l bc + 3P − Rc Rva = ( 985.10 3.92 ) +( 830.84 Rva = 3229.81 kg

2.08 ) + 1904.54 _ 4264.48

B

C +

+ A 5092.12 5294.72

kgm

kgm

3229.81

RAh Bidang M Pers :

Mx1 = Mx1 =

=

0

RVA x 3229.81 x

X1 X1

-

x 0.5 492.55 x

q1 X12

x

X12

dMx1 =

0

985.1

dX1 X1 Xmax X1

= = =

0 3.28 3.92

m m m

X1 X1 MA = Mmax= MB = B

4.65 A

a=

32.5444

= =

3229.81 3.28

m

0 5294.72 5092.12

Kgm Kgm Kgm

tangga tangga

C

Rav cos a

3.92 m

Rav sin a

2.08 m

Rav X1

X2

-532.68 -2944 2722.65

kg

kg -4264.48 kg

Bidang D Permisalan gayaDari kiri : searah jarum jam gaya dianggap positif X= 0 m DA = Rva cos a x 32.544 = 3229.81 cos = 2722.65 kg X= 3.92 m Dbkiri = Rva cos a x = -532.68 kg

q1

x

LAB cos a

Dbkanan = P = 634.85 = -2944 X= 6 m Dc = = -4264.48

x

LBC 2.08 kg

-

RC 4264.48

RC kg

726.51

kg

+

q1

+

-

-1737.49

kg

Bidang N NA = = = NBkiri =

-RVA -3229.81 -1737.49

sin a sin 32.544 kg

-RVA

sin

a

L1

=

726.51

kg sin a

NBkanan -C =

0

4.7.5 Kontrol Kekuatan Profil 4.7.5.1 Penampang Profil

fy =

untuk Sayap

2400 kg/m2 untuk Badan

b 170 ≤ 2tf fy 125 16 7.81

h 1680 ≤ t fy 170 15.49 10.97

≤ ≤

210 5 42

≤

1680 15.49 108.44

=

7450.68 Kgm

≤

OK

OK

Penampang Profil Kompak, maka Mnx = Mpx 4.7.5.1 Kontrol Lateral Buckling Jarak Baut Pengikat :

Lp = 1.76 * iy

250 mm

E fy

=

Lp =

Ternyata

25 cm

0 cm

Lp > Lb

maka

Mnx = Mpx = Zx. Fy = 310.45 Mny = Zy ( 1 flen ) * fy = (1 / 4 * tf * bf 2) * fy = 0.25 x 2.56 x 0.64 x = 9.83 kgm

Mnx = Mpx

*

2400

2400

=

983.04 kgcm

4.7.5 Kontrol Momen Lentur Zx =

310.45

cm3

φMn = φ Zx * fy φMn = Syarat ->

x 310.45 x 2400

=

0.9

kgm

> > OK

Mu 5294.72

kgm

600 360

=

1.67

6705.61 kgm φMn 6705.61

4.7.6 Kontrol Lendutan f =

Ix =

L 360

=

3540

cm4

=

670561.2

kgcm

Y max =

5 ( qD + qL) * l 4 384 EIx

=

5 384

=

1.53

(

<

OK Profil yang Dipakai untuk Balok induk Adalah Profil WF 250 x 125

1.53 2100000 x

5.2 3540

)

1.30E+11

1.67

x

5

x

8

5. Pembebanan 5.1 Perencanaan Beban Atap Beban Mati Berat Gording = 9.3 Berat Asbes Gelombang =

x

4 11.33 x

= = = = = =

4

Alat Pengikat dll (+10 %) Berat Profil Kuda - Kuda = Beban Hidup Ph =

0

19.94 x

P Ultimate = 1.2 P D+ 1.6 PL =

Pm 0.44 Pmtot

cos

+

kg kg kg kg kg kg

90.77

4 108.93

37.2 45.32 82.52 8.25 90.77 0

127.61

=

79.76

kg

=

236.53

kg

5.2. Perencanaan Beban Angin (Gudang Tertutup) - 0,4

(- 0,02α - 0,4)

0,9

W

- 0,4

=

30 kg/m2

Beban Tekan Atap

=

0.1

x

30

x

4

=

12

kg/m

Beban Sedot Atap

=

-0.4

x

30

x

4

=

-48

kg/m

BebanTiup Kolom

=

0.9

x

30

x

4

=

108

kg/m

Beban Sedot Atap

=

-0.4

x

30

x

4

=

-48

kg/m

5.3 Perencanaan Beban Akibat Plat Lantai, Kolom Memanjang dan Melintang P4 I I

P3

memanjang

I I

P3

600

P2

P5 I I

P6

P6

P1

600

melintang

P2 I I 600 I

P3 P4

I V

B 400

400

400

.

C A B

A

( Balok Induk Melintang ) 0 x 0 A= W= a= bf = iy =

B

0 cm2 0 kg/m 0 mm 0 mm 0 cm ( Balok Anak ) 250 x

A= W= a= bf = iy = C

x tf = Ix = Iy = tw = ix =

125

37.66 cm2 29.6 kg/m 250 mm 125 mm 2.79 cm

tf = Ix = Iy = tw = ix =

( Balok Induk Memanjang) 0 x 0 A= W= a= bf = iy =

Beban Mati - Bondex - Beton - Beban Finishing

0 cm2 0 kg/m 0 mm 0 mm 0 cm

=

0 0 mm 0 cm4 0 cm4 0 mm 0 cm

x

x

9 mm 4050 cm4 294 cm4 6 mm 10.4 cm

tf = Ix = Iy = tw = ix =

0

2400 qM

0 cm3 0 cm3 0 mm 0 mm 0 0 9

Zx = Zy = h= r=

x

0 mm 0 cm4 0 cm4 0 mm 0 cm

x

0 Zx = Zy = h= r=

6

x

0.12

x

351.86 cm3 72.02 cm3 208 mm 12 mm 351.86 72.02 0

Zx = Zy = h= r=

= = = =

0 cm3 0 cm3 0 mm 0 mm 0 0 10.1 288 90 388.1

kg/m2 kg/m2 kg/m2 kg/m2

Beban Hidup qL

=

400

kg/m2

5.3.1 Perencanaan Pembebanan Portal Melintang 5.3.1.1 Beban P1 Beban Mati Berat Pelat = Balok induk melintang = Balok anak =

Beban Hidup Ph1 = 5.3.1.2 Beban P2 Beban Mati Berat Pelat = Balok induk memanjang = Balok induk melintang =

Beban Hidup Ph2 = 5.3.1.3 Beban P3 Beban Mati Berat Pelat = Balok induk melintang = Balok anak =

Beban Hidup Ph3 = 5.3.1.4 Beban P4 Beban Mati Berat Pelat = Balok induk memanjang = Berat Dinding = Balok induk melintang =

Beban Hidup Ph4 =

388.1 x 29.6

x

3 0 4

x x

4 3 Pm1

= = = =

4657.2 0 118.4 4775.6

kg kg kg kg

x

3

x

4

=

4800

kg

388.1 x

3 0 0

x x x

4 4 6 Pm2

= = = =

4657.2 0 0 4657.2

kg kg kg kg

x

3

x

4

=

4800

kg

388.1 x

3 0 6

x x

4 6 Pm1

= = = =

4657.2 0 177.6 4834.8

kg kg kg kg

400

400

29.6

400

x

x

388.1 x 4.15

x

400

x

3

x

4

=

4800

kg

1.5 0 4.5 6

x x x x

4 4 4 0 Pm4

= = = = =

2328.6 0 74.7 0 2403.3

kg kg kg kg kg

1.5

x

4

=

2400

kg

x x x

2 6 2 Pm5

= = = =

4657.2 0 0 4657.2

5.3.2 Perencanaan Pembebanan Portal Memanjang 5.3.2.1 Beban P5 * Beban Beban Mati Mati Berat Pelat Balok induk melintang = Balok induk memanjang =

388.1 x

6 0 0

kg kg kg kg

* Beban Beban Hidup Hidup Ph5 =

x

6

x

2

=

4800

388.1 x

4 0 0

x x x

6 4 6 Pm5

= = = =

9314.4 0 0 9314.4

x

4

x

6

=

9600

400

5.3.2.2 Beban P6 * Beban Beban Mati Mati - Berat Pelat = - Balok induk memanjang = - Balok induk melintang = * Beban Beban Hidup Hidup Ph5 =

400

5.3.3 Beban Portal - Portal Melintang P1 Pm1 = Ph1 =

4775.6 kg 4800 kg

P2

Pm3 Ph3

= =

4657.2 kg 4800 kg

P3

Pm2 Ph2

= =

4834.8 kg 4800 kg

P4

Pm4 Ph4

= =

2403.3 kg 2400 kg

5.3.4 Beban - beban Portal Memanjang P5 Pm5 = Ph5 = P6

Pm6 Ph6

4657.2 kg 4800 kg

= =

9314.4 kg 9600 kg

5.4 Perencanaan Beban Gempa ( Arah X )

F2

2

4 4.5

F1 5

6 Data Gempa: - Zone Gempa = - tanah lunak C = -I =

6

6

6 0.95 1

9

kg

kg kg kg kg kg

4

F 1

4 4

6

6

6

18

Kolom 0 A= W= a= bf = iy =

x 0 cm2 0 kg/m 0 mm 0 mm 0 cm

0

x tf = Ix = Iy = tw = ix =

5.4.1 Perencanaan Beban Lantai (W1) 5.4.1.1 - BebanBeban Mati: Mati Berat Plat = 388.1 x Balok Induk Memanjang = Balok Induk Melintang = Berat Dinding = 8.3 x Kolom = 0 x 0

5.4.1.2 Beban Hidup Balok + Plat

Beban Lantai ( W1 tot)

= = = = =

0

x

0 mm 0 cm4 0 cm4 0 mm 0 cm

18 0 0 4 4.5 4.5

x x x x x x

4 4 18 4.5 2 1

0 Zx = Zy = h= r=

= = = = = = =

0 cm3 0 cm3 0 mm 0 mm 0 0

27943.2 0 0 149.4 0 0 28092.6

Ph1 + 2 x Ph2 + 2 x Ph3 + 2 x Ph4 4800 + 9600 + 9600 + 4800 28800 Kg 28092.6 56892.6

5.4.2 Perencanaan Beban Atap ( W2) 5.4.2.1 Beban Mati Berat Atap = 22.69 Cosα x Cosα x Balok Kuda-kuda = 0 x x Kolom = 0 2 x Berat Dinding = 8.3 x 2

+ Kg

4 18 2 4

28800

x

18

= = = = =

1802.8 Kg 0 Kg 0 Kg 14.3 Kg 1817.1 Kg

Kg Kg Kg Kg Kg Kg Kg

5.4.2.2 Beban Hidup Ph1 =

Beban Atap ( W2 tot)

19.94 x

4

x

18

= =

1817.1 3252.7

+ Kg

1435.59

= = =

W1 56892.6 60145.3

+ + Kg

W2 3252.7

x x

H3/4

=

1435.59

=

1

kg

5.4.3 Berat Total Wt Berat Total ( W tot )

5.4.4 Perencanaan Gaya Gempa T

Tanah Lunak C R

= = =

0.09 0.09 0.44

= =

0.95 4.5

V = (C x I x Wt )/R

9 3/4

I

0.95 x 1 12697.34 Kg

= =

Lantai

=

W1

x

h1

Atap

=

W2

x

H

Σ Wi Hi

F1

56892.6 284463 3252.7 29274.27

x Kgm x Kgm

=

313737.27

Kgm

W1 h1 S Wi hi

x

V

=

284463 313737.27

x

12697.34

11512.57

Kg

=

W2 h2 S Wi hi

x

V

=

29274.27 313737.27

x

12697.34

=

1184.77

Kg

4.5

= = = =

=

=

F2

x 60145.3

5 9

5.4.5 Gambar Perencanaan Beban Akibat Gempa ( Arah X )

F2 1184.77 kg 4 F1 11512.57 kg 5

6

6

6

5.5 Perencanaan Beban Gempa ( Arah Y ) Data Gempa: - Zone Gempa = - tanah lunak C = -I = Balok Kuda-Kuda 0 x A= W= a= bf = iy =

A= W= a= bf = iy =

x 0 cm2 0 kg/m 0 mm 0 mm 0 cm

0.95 1

0

0 cm2 0 kg/m 0 mm 0 mm 0 cm Kolom 0

6

x tf = Ix = Iy = tw = ix =

0

0 0 mm 0 cm4 0 cm4 0 mm 0 cm

x tf = Ix = Iy = tw = ix =

5.5.1 Perencanaan Beban Lantai (W1) 5.5.1.1 - BebanBeban Mati: Mati Berat Plat = 388.1 x Balok Induk Memanjang =

x Zx = Zy = h= r=

0

x

0 mm 0 cm4 0 cm4 0 mm 0 cm

6 0

x x

0

40 40

0 Zx = Zy = h= r=

= =

0 cm3 0 cm3 0 mm 0 mm 0 0

0 cm3 0 cm3 0 mm 0 mm 0 0

93144 0

Kg Kg

Balok Induk Melintang Berat Dinding Kolom

5.5.1.2 Beban Hidup Beban Hidup Merata

Beban Lantai ( W1 tot)

= = =

x x

0 6 4.5 4.5

x x x x

40 4.5 2 4.5

= = = = =

250 x 60000

40 Kg

x

6

Kg

8.3 0 0

= = = =

93368.1 153368.1

5..2 Perencanaan Beban Atap ( W2) 5.5.2.1 Beban Mati Berat Atap = 22.69 x 6 x Cosα Balok Kuda-kuda = 0 x Kolom = 0 2 x Berat Dinding = 8.3 6

5.5.2.2 Beban Hidup Ph1 =

Beban Atap ( W2 tot)

+ Kg

x x x x

40 40 2 2

19.94 x

6

0 224.1 0 0 93368.1

Kg Kg Kg Kg Kg

60000

= = = = =

x

40

5446.32 Kg 0 Kg 0 Kg 16.3 Kg 5462.62 Kg

=

4785.31

I

=

1

x163616.03

4.5

= = = =

153368.1 766840.5 10247.93 92231.33

= =

5462.62 10247.93

+ Kg

4785.31

= = =

W1 153368.1 163616.03

+ + Kg

W2 10247.93

kg

5.4.3 Berat Total Wt Berat Total ( W tot )

5.4.4 Perencanaan Gaya Gempa T

Tanah Lunak C R

= = =

0.09 0.09 0.44

= =

0.95 4.5

V = (C x I x Wt )/R

= =

x x

H3/4 9 3/4

0.95 x 1 34541.16 Kg

Lantai

=

W1

x

h1

Atap

=

W2

x

H

x Kgm x Kgm

5 9

Σ Wi Hi

F1

=

W1 h1 S Wi hi

x

V

=

766840.5 859071.83

x

34541.16

=

F2

=

30832.77

W2 h2 S Wi hi

x

V

=

92231.33 859071.83

x

34541.16

3708.39

Kgm

Kg

=

=

859071.83

Kg

5.4.5 Gambar Perencanaan Beban Akibat Gempa ( Arah Y )

F2 3708.39 kg

P4

F1 30832.77 kg

P3

P2

P2

P1

P3

Kombinasi Pembebanan * Beban Mati + Beban Hidup Pm Ph

= =

90.77 kg 79.76 kg

200

P4

P3

P2

P1

P2

P3

P4

P4

P3

P2

P2

P1

6

P3

6

P4

6

Pm1 = Ph1 =

4775.6 kg 4800 kg

Pm2 = Ph2 =

4657.2 kg 4800 kg

Pm3 = Ph3 =

4834.8 kg 4800 kg

Pm4 = Ph4 =

2403.3 kg 2400 kg

* Beban Mati + Beban Hidup + Beban Angin

q =

12

P4

q =

kg/m

P3

q =

P2

P2

P1

108 kg/m 6

-48

kg/m

P3

q = 6

P4

-48

kg/m

6

* Beban Mati + Beban Hidup + Beban Gempa

1184.77 kg P4

P3

P2

P1

P2

P3

P4

11512.57 kg

6

6

6

melintang

P4

6 Perencanaan Dimensi Struktur Utama 6.1 Kontrol Dimensi Kuda -Kuda Dari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 9: (U-G)

Beban Ultimate - Beban Gempa Mutx = -3679.74 Kgm Muty = 0 Kgm Nu = -2677.93 Kg Vu = -1184.35 kg Ma = Mb = Ms =

-3679.4 Kgm -466.19 Kgm 301.67 Kgm

6.1.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 : 200 A= W= a= bf = iy =

x

200

63.53 cm2 49.9 kg/m 200 mm 200 mm 5.02 cm

Mutu Baja = fu = fy =

x

tf = Ix = Iy = tw = ix =

8

12 mm 4720 cm4 1600 cm4 8 mm 8.62 cm

x

12 Zx = Zy = h= Sx =

513.15 cm3 216.32 cm3 150 mm 37.5 mm 513.15 216.32

BJ 37 3700 kg/cm2 2400 kg/cm2

6.1.2 Kontrol Lendutan f ijin =

L 360

=

993.2 360

=

2.76

cm

0.1

-3679.4

-466.19

5 L2 f = ( Ms − 0.1( Ma − Mb)) 48 EI f=

5 48

f=

98.64 200

301.67 4720

0.68 cm f

< OK

f ijin

6.1.3 Kontrol Tekuk untuk arah x : kcx = 1.2 L= 993.2

(jepit-rol tanpa putaran sudut) cm

Lkx =

λx =

1191.84

Lkx ix

=

138.26 cm

π 2 EA λx 2

Ncrbx =

cm

Ncrbx = Ncrbx =

untuk arah y : kcy = 0.8 L= 110 Lky = 88

λy =

Lky iy

Ncrby =

(MENENTUKAN) 9.87

x 2000000 x 63.53

19117.07 65593.62 kg

(jepit-Sendi) cm cm

=

17.53 cm

π 2 EA λy 2

Ncrby =

9.87

x 2000000 x 63.53 307.3

Ncrby =

### kg

Maka dipakai λx karena λx > λy

λc =

λx π

fy E

λc =

138.26 3.14

λc = λc

>

ω = 1.25 λc2

Pn =

1.52

1.2 ω=

Ag fy w

Pu = ϕcPn

=

2677.93 0.85 x 52474.9

2.91

63.53

=

X Batang Dianggap Tidak Bergoyang Maka :

Cmx = Sbx =

* 2.91

2400

0.06

Pu Mux Muy + + ≤1 2ϕcPn ϕbMnx ϕbMny

Pakai Rumus =

Sbx =

2400 2000000

Cmx ≥1 Nu 1− ( ) Ncrbx 1 1

1 -

2677.93

<

=

52474.9 kg

0.2

65593.62 Sbx =

1.04

Sbx =

1.04

Mux = Mutx * Sbx Mux = -3679.74 x

<

1

1.04

=

-3836.36 kgm

Y Batang Dianggap Tidak Bergoyang Maka :

Sby =

Cmy ≥1 Nu 1− ( ) Ncrby

Cmy = Sby =

1 1

1 -

Sby =

1.04

<

Sby =

1.04

Muy = Muty * Sby Muy = 0

2677.93 65593.62

x 1.04

1

=

0 kgm

6.1.4 Menentukan Mnx 6.1.4.1 Kontrol Penampang profil untuk Sayap

b 170 ≤ 2tf fy 200 24 8.33

untuk Badan

h 1680 ≤ t fy ≤ ≤

OK

170 15.49 10.97

150 8 18.75

Kgm

≤ ≤

OK

1680 15.49 108.44

Penampang Profil Kompak, maka Mnx = Mpx Lateral Bracing

Lb =

E fy

Lp = 1.76 * iy

110

Lp =

Ternyata

cm

255.05 cm

Lp > Lb

maka

Mnx = Mpx = Zx. Fy = 513.15 Mny = Zy ( 1 flen ) * fy = (1 / 4 * tf * bf 2) * fy x 400 x = 0.25 x 1.2 = 2880 kgm

Mnx = Mpx

*

2400

2400

=

=

12315.65 kgm

288000 kgcm

6.1.5 Persamaan Interaksi

Pu Mux Muy + + ≤1 2ϕcPn ϕbMnx ϕbMny

2677.93 1.7 x 52474.9 0.03

+

3836.36 x 12315.65 0.9

+

0.35

+

+ 0.38

< OK

6.1.6 Kontrol Kuat Rencana Geser

h 1100 ≤ tw fy 150 2400

<

1100 15.49

0.06

< Plastis

71

Vn = 0.6 fy Aw = 0.6 2400 = 23040 Kg Vu 1184.35

< <

ФVn 0.9

0.8

23040

0

0.9

20

x 2880 0

1

1184.35

< OK

20736

6.2 Kontrol Dimensi Kolom Dari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 13: (U-G) Sbx -->

Beban Ultimate - Beban Gempa Mutx = 10123.25 Kgm Nu = -28026.45 Kg Vu = 3834.45 kg Max = 9049.01 Kgm Mbx = 10123.25 Kgm Msx = 537.12 Kgm

Sby -->

Muty = Nu = Vu =

-8137 Kgm -27466 Kg 3149 kg

Max = Mbx = Msx =

8137 Kgm 7611.48 Kgm 262.96 Kgm

6.2.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 : 300 A= W= a= bf = iy =

x

300

119.8 cm2 94 kg/m 300 mm 300 mm 7.51 cm

Mutu Baja = fu = fy =

tf = Ix = Iy = tw = ix =

BJ 37 3700 kg/cm2 2400 kg/cm2

f=

25.00 200

< OK

f =

700 kg/cm2

500 360

=

1.39

cm

537.12

0.1

9049.01

10123.25

500 360

=

1.39

cm

0.1

8137

7611.48

f ijin

=

25.00 200

262.96 6750

0.04 cm f

< OK

6.2.3 Kontrol Tekuk untuk arah x : kcx = 0.8 L= 500 Lkx = 400

Ncrbx =

1464.75 cm3 652.5 cm3 234 mm 1360 cm3 1464.75 652.5

5 L2 ( Ms − 0.1( Ma − Mb)) 48 EI

5 48

f=

Lkx ix

Zx = Zy = h= Sx =

20400

6.2.1.2 Kontrol Lendutan Arah Y f ijin = L 360

λx =

15

0.04 cm f

f=

x

5 L2 ( Ms − 0.1( Ma − Mb)) 48 EI

5 48

f=

=

10

15 mm 20400 cm4 6750 cm4 10 mm 13.1 cm

fr =

6.2.1 Kontrol Lendutan 6.2.1.1 Kontrol Lendutan Arah X f ijin = L 360

f =

x

=

π 2 EA λx 2

(jepit-Sendi) cm cm 30.53 cm

f ijin

π 2 EA Ncrbx = λx 2

Ncrbx =

9.87

932.35 ### kg

Ncrbx = untuk arah y : kcy = 0.8 L= 500 Lky = 400

λy =

Lky iy

Ncrby =

x 2000000 x 119.8

(jepit-Sendi) cm cm

=

53.26 cm

π 2 EA λy 2

Ncrby = Ncrby =

(MENENTUKAN)

9.87

x 2000000 x 119.8

2836.87 833529.23 kg

Maka dipakai λy karena λy > λx

λc =

λy π

fy E

λc =

53.26 3.14

λc = 0.25

ω=

<

Ag fy w

Pu = ϕcPn

0.59 <

ω=

1.43 1,67-0,67λc Pn =

λc

=

28026.45 0.85 x256656.17

1.2

1.12

119.8

=

X Batang Dianggap Tidak Bergoyang Maka :

Cmx = Sbx =

Sbx =

* 1.12

2400

0.13

Pu Mux Muy + + ≤1 2ϕcPn ϕbMnx ϕbMny

Pakai Rumus =

Sbx =

2400 2000000

Cmx ≥1 Nu 1− ( ) Ncrbx 1 1

1 -

1.01

<

28026.45 ### 1

<

=

256656.17 kg

0.2

Sbx =

1.01

Mux = Mutx * Sbx Mux = 10123.25 x

1.01

=

10236.37 kgm

Y Batang Dianggap Tidak Bergoyang Maka :

Sby =

Cmy ≥1 Nu 1− ( ) Ncrby

Cmy = Sby =

1 1

1 -

Sby =

1.01

<

Sby =

1.01

Muy = Muty * Sby Muy = -8137

28026.45 ### 1

x 1.01

=

-8227.92 kgm

6.2.4 Menentukan Mnx 6.2.4.1 Kontrol Penampang profil untuk Sayap

untuk Badan

b 170 ≤ 2tf fy 300 30 10

h 1680 ≤ t fy 170 15.49 10.97

≤ ≤

234 10 23.4

OK

≤ ≤

OK

Penampang Profil Kompak, maka Mnx = Mpx Lateral Buckling

Lp = 1.76 * iy

Kgm

Lb =

E fy

Lp =

500

cm

381.56 cm

Lr = iy . [x1/(fy-fr)] . Sqrt{1+[sqrt(1+(x2 . fl2))]} J = Σ(1/3).b.(t^3) =

76.5 cm4

Iw = Iy.((h^2)/4)

### cm6

=

x1 = [π/s]*[sqrt((EGJA)/2)] =

197694.62 kg/cm2

1680 15.49 108.44

x2 = 4.[(S/GJ)^] = Lr =

4.01E-07 cm2/kg

1372.41 cm

Lp

<

Lb

<

Lr

(INELASTIC BUCKLING)

Mnx = Cb.[Mr+[(Mp-Mr).((Lr-Lb)/(Lr-Lp))]]

Cb =

12,5 Mmax 2,5Mmax + 3Ma + 4Mb + 3Mc

Cb =

Mr = Mp = My = 1,5 . My

Sx(fy-fr) = Zx . fy = Sx . fy = =

Mnx =

2312000 kgcm 3515400 kgcm 3264000 kgcm 4896000 kgcm

### kgcm

Mnx=Mp=

2.25

< 1,5My > Mp

> Mp

3515400 kgcm

Mny = Zy * fy = 652.5 x 2400 = 1566000 kgcm = 15660 kgm

6.2.5 Persamaan Interaksi

Pu Mux Muy + + ≤1 2ϕcPn ϕbMnx ϕbMny

28026.45 1.7 x256656.17 0.06

+

+

10236.37 x 35154 0.9 0.32

+

+ 0.97

< OK

6.2.6 Kontrol Kuat Rencana Geser

h 1100 ≤ tw fy

8227.92 0.9 x 15660 0.58 1

234 2400

<

1100 15.49

0.1

< Plastis

71

Vn = 0.6 fy Aw = 0.6 2400 = 43200 Kg Vu 3834.45 3834.45

< < < OK

ФVn 0.9 38880

1

30

43200

6.3 Kontrol Dimensi Balok Melintang Dari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 5 : (U-G)

Beban Ultimate - Beban Gempa Mutx = -15837.38 Kgm Muty = 0 Kgm Nu = -8457.09 Kg Vu = 9218.69 kg Ma = 970.4 Kgm Mb = 15837.88 Kgm Ms = 11818.69 Kgm

6.3.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 : 340 A= W= a= bf = iy = Mutu Baja =

x

101.5 cm2 79.7 kg/m 340 mm 250 mm 6 cm BJ 37

250 tf = Ix = Iy = tw = ix =

x 14 mm 21700 cm4 3650 cm4 9 mm 14.6 cm

9

x

14 Zx = Zy = h= Sx =

1360.02 cm3 439.2 cm3 276 mm 1280 cm3 1360.02 439.2

fu = fy =

3700 kg/cm2 2400 kg/cm2

fr =

700 kg/cm2

6.3.1 Kontrol Lendutan f ijin =

f =

f=

L 360

500 360

=

1.39

cm

0.1

970.4

15837.88

5 L2 ( Ms − 0.1( Ma − Mb)) 48 EI

5 48

25.00 200

f=

=

11818.69 21700

0.8 cm f

< OK

f ijin

6.3.3 Kontrol Tekuk untuk arah x : kcx = 1 L= 600 Lkx = 600

λx =

Lkx ix

Ncrbx =

=

(Sendi-Sendi) cm cm 41.1

π 2 EA λx 2

Ncrbx =

9.87

1688.87 ### kg

Ncrbx = untuk arah y : kcy = 1 L= 600 Lky = 600

λy =

Lky iy

x 2000000 x 101.5

(Sendi-Sendi) cm cm

=

100

π 2 EA Ncrby = λy 2

Ncrby = Ncrby =

(MENENTUKAN)

9.87

x 2000000 x 101.5

10000 200341.15 kg

Maka dipakai λy karena λy > λx

λc =

λy π

fy E

λc =

100 3.14

λc = 0.25

<

λc

2400 2000000 1.1

<

1.2

ω=

1.43 1,67-0,67λc Pn =

ω=

Ag fy w

Pu = ϕcPn

=

8457.09 0.85 x158629.17

1.54

101.5

=

* 1.54

2400

0.06

Pu Mux Muy + + ≤1 2ϕcPn ϕbMnx ϕbMny

Pakai Rumus =

X Batang Dianggap Tidak Bergoyang Maka :

Sbx =

Cmx ≥1 Nu 1− ( ) Ncrbx

Cmx = Sbx =

1 1

1 -

Sbx =

1.01

<

1

Sbx =

1.01

1.01

=

Mux = Mutx * Sbx Mux = 15837.38 x

8457.09 ###

15951.1 kgm

Y Batang Dianggap Tidak Bergoyang Maka :

Sby =

Cmy = Sby =

Cmy ≥1 Nu 1− ( ) Ncrby 1 1

1 -

Sby =

1.01

<

Sby =

1.01

Muy = Muty * Sby Muy = 0

x 1.01

8457.09 ### 1

=

0 kgm

<

=

158629.17 kg

0.2

6.3.4 Menentukan Mnx 6.3.4.1 Kontrol Penampang profil untuk Sayap

untuk Badan

b 170 ≤ 2tf fy 250 28 8.93

h 1680 ≤ t fy 170 15.49 10.97

≤ ≤

276 9 30.67

OK

Kgm

≤ ≤

1680 15.49 108.44

OK

Penampang Profil Kompak, maka Mnx = Mpx Lateral Buckling

Lb =

E fy

Lp = 1.76 * iy

Lp =

600

cm

304.84 cm

Lr = iy . [x1/(fy-fr)] . Sqrt{1+[sqrt(1+(x2 . fl2))]} J = Σ(1/3).b.(t^3) = Iw = Iy.((h^2)/4)

=

53.32 cm4 969768.5 cm6

x1 = [π/s]*[sqrt((EGJA)/2)] = x2 = 4.[(S/GJ)^] = Lr = Lp

161407.01 kg/cm2 3.60E-09 cm2/kg

806.68 cm <

Lb

<

Lr

(INELASTIC BUCKLING)

Mnx = Cb.[Mr+[(Mp-Mr).((Lr-Lb)/(Lr-Lp))]] Cb = Cb =

Mr = Mp = My = 1,5 . My

Sx(fy-fr) = 2176000 kgcm Zx . fy = 3264057.6 kgcm Sx . fy = 3072000 kgcm 4608000 kgcm =

< 1,5My > Mp

12,5 Mmax 2,5Mmax + 3Ma + 4Mb + 3Mc 1.81

Mnx =

### kgcm

> Mp

Mnx=Mp= 3264057.6 kgcm Mny = Zy ( 1 flen ) * fy = (1 / 4 * tf * bf 2) * fy x = 0.25 x 1.4 = 5250 kgm

625

x 2400

=

525000 kgcm

6.3.5 Persamaan Interaksi

Pu Mux Muy + + ≤1 2ϕcPn ϕbMnx ϕbMny

8457.09 1.7 x158629.17 0.03

+

15951.1 x 32640.58 0.9

+

0.54

+

+ 0.57

< OK

6.3.6 Kontrol Kuat Rencana Geser

h 1100 ≤ tw fy 276 2400

<

1100 15.49

0.12

< Plastis

71

Vn = 0.6 fy Aw = 0.6 2400 = 44064 Kg Vu 9218.69 9218.69

< < < OK

ФVn 0.9 39657.6

0.9

44064

0

0.9

34

x 5250 0

1

6.3 Kontrol Dimensi Balok Memanjang Dari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 5 : (U-G)

Beban Ultimate - Beban Gempa Mutx = 6219 Kgm Muty = 0 Kgm Nu = -28073 Kg Vu = -2663 kg Ma = Mb = Ms =

6219 Kgm 4435 Kgm 891 Kgm

6.3.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 : 300 A= W= a= bf = iy =

x

200

72.38 cm2 56.8 kg/m 300 mm 200 mm 4.71 cm

Mutu Baja = fu = fy =

x

tf = Ix = Iy = tw = ix =

BJ 37 3700 kg/cm2 2400 kg/cm2

8

12 mm 11300 cm4 1600 cm4 8 mm 12.5 cm

fr =

x

12 Zx = Zy = h= Sx =

843.55 cm3 248.32 cm3 240 mm 771 cm3 843.55 248.32

700 kg/cm2

6.3.1 Kontrol Lendutan f ijin =

f =

f=

=

500 360

=

1.39

cm

0.1

6219

4435

5 L2 ( Ms − 0.1( Ma − Mb)) 48 EI

5 48

f=

L 360

25.00 200

891 11300

0.08 cm f

6.3.3 Kontrol Tekuk

< OK

f ijin

untuk arah x : kcx = 1 L= 600 Lkx = 600

λx =

Lkx ix

(Sendi-Sendi) cm cm

=

48

π 2 EA λx 2

Ncrbx =

Ncrbx =

2304 Ncrbx =

untuk arah y : kcy = 1 L= 600 Lky = 600

λy =

Lky iy

x 2000000 x 72.38

9.87

620069.3 kg

(Sendi-Sendi) cm cm

=

127.39

π 2 EA Ncrby = λy 2

Ncrby = Ncrby =

(MENENTUKAN)

x 2000000 x 72.38

9.87

16227.84 88036.35 kg

Maka dipakai λy karena λy > λx

λc =

λy π

fy E

λc =

127.39 3.14

λc = 0.25 ω=

λc

<

1.43 1,67-0,67λc Pn =

Pu = ϕcPn

1.4 <

ω=

Ag fy w

0.85

Pakai Rumus =

=

28073 x88538.49

1.2 1.96

72.38

=

* 1.96

2400

0.37

Pu Mux Muy + + ≤1 2ϕcPn ϕbMnx ϕbMny

X Batang Dianggap Tidak Bergoyang Maka :

Sbx =

2400 2000000

Cmx ≥1 Nu 1− ( ) Ncrbx

<

=

88538.49 kg

0.2

Sbx =

Cmx ≥1 Nu 1− ( ) Ncrbx

Cmx = Sbx =

1 1

1 -

Sbx =

1.05

<

Sbx =

1.05

Mux = Mutx * Sbx Mux = -6219

28073 620069.3 1

x 1.05

=

-6513.91 kgm

Y Batang Dianggap Tidak Bergoyang Maka :

Sby =

Cmy ≥1 Nu 1− ( ) Ncrby

Cmy = Sby =

1 1

1 -

Sby =

1.05

<

Sby =

1.05

Muy = Muty * Sby Muy = 0

28073 620069.3

x 1.05

1

=

0 kgm

6.3.4 Menentukan Mnx 6.3.4.1 Kontrol Penampang profil untuk Sayap

untuk Badan

b 170 ≤ 2tf fy 200 24 8.33

h 1680 ≤ t fy ≤ ≤

170 15.49 10.97

240 8 30

OK

≤ ≤

OK

Penampang Profil Kompak, maka Mnx = Mpx Lateral Buckling

Kgm

Lb =

600

cm

1680 15.49 108.44

E fy

Lp = 1.76 * iy

Lp =

239.3 cm

Lr = iy . [x1/(fy-fr)] . Sqrt{1+[sqrt(1+(x2 . fl2))]} J = Σ(1/3).b.(t^3) = Iw = Iy.((h^2)/4)

=

53.32 cm4 969768.5 cm6

x1 = [π/s]*[sqrt((EGJA)/2)] =

226284.21 kg/cm2

x2 = 4.[(S/GJ)^] = Lr =

3.60E-09 cm2/kg

806.68 cm

Lp

<

Lb

<

Lr

(INELASTIC BUCKLING)

Mnx = Cb.[Mr+[(Mp-Mr).((Lr-Lb)/(Lr-Lp))]] Cb = Cb =

Mr = Mp = My = 1,5 . My

Sx(fy-fr) = 1310700 kgcm Zx . fy = 2024524.8 kgcm Sx . fy = 1850400 kgcm 2775600 kgcm =

Mnx =

### kgcm

12,5 Mmax 2,5Mmax + 3Ma + 4Mb + 3Mc 1.81

< 1,5My > Mp

> Mp

Mnx=Mp= 2024524.8 kgcm Mny = Zy ( 1 flen ) * fy = (1 / 4 * tf * bf 2) * fy x = 0.25 x 1.2 = 2880 kgm

400

x 2400

=

288000 kgcm

6.3.5 Persamaan Interaksi

Pu Mux Muy + + ≤1 2ϕcPn ϕbMnx ϕbMny

28073

+

6513.91

+

0

1.7

x 88538.49 0.19

0.9 +

x 20245.25 0.36

0.9 +

0.54

< OK

6.3.6 Kontrol Kuat Rencana Geser

h 1100 ≤ tw fy 240 2400

<

1100 15.49

0.1

< Plastis

71

Vn = 0.6 fy Aw = 0.6 2400 = 34560 Kg Vu 2663 2663

< < < OK

ФVn 0.9 31104

0.8

34560

30

x 2880 0

1

7 Sambungan A B

D

C

7.1 Sambungan Kuda - Kuda ( Detail A )

60 420

80 180 100 200

7.1.1 Data Perencanaan Sambungan Dari hasil perhitungan SAP diperoleh: Mu Pu

= =

1279.27 1152

kgm kg

=

Baut Tanpa Ulir ( Bor ) Diameter φ = Tebal Plat = 10 mm

8

7.1.2 Kontrol Kekuatan Baut 7.1.2.1 Perhitungan Ruv yang Diterima Setiap Baut Ruv = Pu = 1152 n 6

127927 kgcm

mm

=

192 kg

7.1.2.2 Perhitungan Kuat Geser Baut f Rnv = 0.75 0.5 = 0.75 0.5 = 772.44 kg

fu 4100

Ab 0.5

n 1

7.1.2.3 Perhitungan Kuat Tumpu Baut f Rn = 2.4 d = 2.4 0.8 = 5904 kg

tp 1

fu 4100

0.75 0.75

7.1.2.4 Perhitungan Kuat tarik Baut f Rnt = 0.75 0.75 = 0.75 0.75 = 1158.66 kg

fu 4100

Ab 0.5

7.1.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

(

Ruv 2 R ) + ( ut ) 2 ≤ 1 φRnv Rnt

192 772.44 Rut = T =

+

Rut 1158.66

<

1

870.66 kg

7.1.2.6 Kontrol Momen Sambungan Letak Garis Netral a: 60

2T

80

2T d3

180

d2

2T d1

100

a

200

a =

ΣT fy B

= =

d1 = d2 = d3 = Sdi =

870.66 x 2500 x 0.1 cm

9.9 cm 27.9 cm 35.9 cm 73.69 cm

6 20

f Mn =

0.9

fy

a2

B

+

S T di

20

+

128311.88

2 =

0.9

2500

0.01 2

=

128557.49 kgcm

> OK

127927 kgcm

7.2 Sambungan Kuda - Kuda dan Kolom ( Detail B )

Pu

60

Mu 420

80 180 100 200

7.2.1 Data Perencanaan Sambungan Dari hasil perhitungan SAP diperoleh: Mu Pu

= =

3240 1344.96

kgm kg

=

Baut Tanpa Ulir ( Bor ) Diameter φ = Tebal Plat = 10 mm

12

7.2.2 Kontrol Kekuatan Baut 7.1.2.1 Perhitungan Ruv yang Diterima Setiap Baut Ruv = Pu = 1344.96 n 6

324000 kgcm

mm

=

224.16 kg

7.2.2.2 Perhitungan Kuat Geser Baut f Rnv = 0.75 0.5 = 0.75 0.5 = 1737.99 kg

fu 4100

Ab 1.13

n 1

7.2.2.3 Perhitungan Kuat Tumpu Baut f Rn = 2.4 d = 2.4 1.2 = 8856 kg

tp 1

fu 4100

0.75 0.75

7.2.2.4 Perhitungan Kuat tarik Baut f Rnt = 0.75 0.75 = 0.75 0.75 = 2606.99 kg

fu 4100

Ab 1.13

7.2.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

(

Ruv 2 R ) + ( ut ) 2 ≤ 1 φRnv Rnt

224.16 1737.99 Rut = T =

+

Rut 2606.99

<

1

2270.75 kg

7.2.2.6 Kontrol Momen Sambungan Letak Garis Netral a: 60

2T

80

2T

180

2T

d3 d2 d1

100

a

200

a =

ΣT fy B

= =

d1 = d2 = d3 = Sdi =

f Mn =

0.9

2270.75 x 2500 x 0.27 cm

6 20

9.73 cm 27.73 cm 35.73 cm 73.18 cm

fy

a2

B

+

S T di

20

+

332357.74

2 =

0.9

2500

0.07 2

=

334028.37 kgcm

> OK

7.3 Sambungan Balok dan Kolom ( Detail C )

324000 kgcm

Pu 57 60

Mu 576

60 60 198 141

200 7.3.1 Data Perencanaan Sambungan Dari hasil perhitungan SAP diperoleh: Mu Pu

= =

12614 12203

kgm kg

=

Baut Tanpa Ulir ( Bor ) Diameter φ = Tebal Plat = 10 mm

18

7.3.2 Kontrol Kekuatan Baut 7.3.2.1 Perhitungan Ruv yang Diterima Setiap Baut Ruv = Pu = 12203 n 10

1261400 kgcm

mm

=

1220.3 kg

7.3.2.2 Perhitungan Kuat Geser Baut f Rnv = 0.75 0.5 = 0.75 0.5 = 3910.48 kg

fu 4100

Ab 2.54

n 1

7.3.2.3 Perhitungan Kuat Tumpu Baut f Rn = 2.4 d = 2.4 1.8 = 13284 kg

tp 1

fu 4100

0.75 0.75

7.3.2.4 Perhitungan Kuat tarik Baut f Rnt = 0.75 0.75 = 0.75 0.75 = 5865.72 kg

fu 4100

7.3.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

(

Ruv 2 R ) + ( ut ) 2 ≤ 1 φRnv Rnt

1220.3 3910.48

+

Rut 5865.72

<

1

Ab 2.54

Rut = T =

4035.27 kg

7.3.2.6 Kontrol Momen Sambungan

57 60 576

2T 2T

60 60

2T d4

2T 2T

198 141

d5

d3 d1

d2

a 200 ΣT fy B

a =

= =

d1 d2 d3 d4 d5

= = = = = Sdi =

f Mn =

0.9

4035.27 x 2500 x 0.81 cm

10 20

13.29 cm 33.09 cm 39.09 cm 45.09 cm 51.09 cm 181.66

fy

a2

B

+

S T di

20

+

###

2 =

0.9

2500

0.65 2

=

### kgcm

> OK

1261400 kgcm

7.4 Sambungan Balok dan Kolom ( Detail D ) Pu

Pu

Pu

Pu 57

Mu

Mu

576

60 60 60 198 141

7.4.1 Data Perencanaan Sambungan Dari hasil perhitungan SAP diperoleh: Mu Pu

= =

12976 7343

kgm kg

=

Baut Tanpa Ulir ( Bor ) Diameter φ = Tebal Plat = 10 mm

18

7.4.2 Kontrol Kekuatan Baut 7.4.2.1 Perhitungan Ruv yang Diterima Setiap Baut Ruv = Pu = 7343 n 10

1297600 kgcm

mm

=

734.3 kg

7.4.2.2 Perhitungan Kuat Geser Baut f Rnv = 0.75 0.5 = 0.75 0.5 = 3910.48 kg

fu 4100

Ab 2.54

n 1

7.4.2.3 Perhitungan Kuat Tumpu Baut f Rn = 2.4 d = 2.4 1.8 = 13284 kg

tp 1

fu 4100

0.75 0.75

7.4.2.4 Perhitungan Kuat tarik Baut f Rnt = 0.75 0.75 = 0.75 0.75 = 5865.72 kg

fu 4100

7.4.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

(

Ruv 2 R ) + ( ut ) 2 ≤ 1 φRnv Rnt

734.3 3910.48 Rut = T =

+

Rut 5865.72

4764.27 kg

<

1

Ab 2.54

7.4.2.6 Kontrol Momen Sambungan

57 60

2T 2T

60 60

2T

198 141

d4

2T 2T

d5

d3 d1

d2

a

200

ΣT fy B

a =

= =

d1 d2 d3 d4 d5

= = = = = Sdi =

f Mn =

0.9

4764.27 x 2500 x 0.95 cm

10 20

13.15 cm 32.95 cm 38.95 cm 44.95 cm 50.95 cm 180.94

fy

a2

B

+

S T di

20

+

###

2 =

0.9

2500

0.91 2

=

### kgcm

> OK

1297600 kgcm

7.5 Kontrol Kekuatan Sambungan Las 7.5.1 Perencanaan Tebal Las Efektif pada Sambungan 7.5.2 Perhitungan Tebal Las Efektif Pada Web

aeff max Di Web

aeff max Di end plate

=

0.71

x

=

0.71

x

=

5.32

mm

=

1.41

x

=

1.41

=

11.75

fu 70 3700 70

70.3 10 70.3

fu 70 4100 70

x

70.3 10 70.3

mm

7.5.3 Perhitungan Gaya yang Bekerja Pada Sambungan Las Misal te =

1

A

=

72.2

+

34.8

=

d2 3

x

1

x

2

=

72.2

+ 3

34.8

x

1

=

2544.22 cm3

Sx =

cm 107

cm2

x

2

fv =

Pu A

=

7343 107

=

68.63 kg/cm2

fh =

Mu Sx

=

1297600 2544.22

=

510.02 kg/cm2

f total

= = =

fv2 + 2 68.63 + 514.61 kg/cm2

fh2 510.02 2

7.5.4 Kontrol Kekuatan Las φ fn = φ fn

0.75

0.6

> OK

f total

70

70.3

=

2214.45 kg/cm2

=

514.61 2214.45

=

0.23 cm

7.5.5 Perhitungan Tebal efektif te perlu

=

f total f fn

7.5.6 Perhitungan Lebar Perlu a perlu

=

0.87 0.71

=

1.23 mm

< OK

5.89 mm

7.5.7 Perhitungan Tebal Efektif Dengan Lebar Minimum a minimum = te perlu =

4 4

mm x

0.71

=

2.83

mm

7.5 Sambungan Kolom Pondasi 7.5.1 Data Perencanaan

Rencana Panjang Plat Dasar kolom L Rencana Lebar Plat Dasar kolom B fc' beton Momen yang bekerja pada dasar kolom Mu Lintang yang bekerja pada dasar kolom Du M Normal yang bekerja pada dasar kolom Pu a a1 b c d s pelat

P

a1

a

a1

c b d1

B

b c L

7.5.2 Kontrol Pelat Landasan Beton ( Pondasi ) fc' beton = 20 Mpa Pu yang bekerja = Kekuatan nominal tumpu beton Pn = 0.85 fc' A

= 22617

200 kg/cm2 kg

40 cm 40 cm 20 Mpa 1093100 kgcm 3854 kg 22617 kg 300 mm 50 mm 150.0 mm 50 mm 300 mm 1600 kg/cm2

A = Pn = Pu

40 0.85 <= <= <=

22617 22617

x 200

40 1600

= =

f Pn 0.6 x 163200

1600 cm2 272000

kg

272000 OK

7.5.2 Perencanaan Tebal Pelat Baja Pondasi 7.5.2.1 Perhitungan Tegangan Yang Bekerja Akibat Adanya Eksentrisitas e =

M P A = 40 W = 1/6 B L 2

σ =

= x =

P A

=

+ 22617 1600

1093100 22617 40 = 1/6 40

M W +

σ maks =

116.61 kg/cm2

σ min = Jadi, q =

88.34 kg/cm2 116.61 kg/cm2

=

48.33 cm

1600 cm2 40 2 =

10666.67 cm3

1093100 10666.67

2 3 1 7.5.2.2 Perhitungan Momen Yang Bekerja ~ Daerah 1 M = 1/2 q L2 = 1/2 116.61 52 = 1457.67 kgcm ~ Daerah 2 a/b = a1 = Ma

Mb

~ Daerah 3

30 0.1

/

15 a2

= =

2 0.05

= =

a1 q b2 0.1 116.61

15 2 =

2623.81 kgcm

= =

a2 q b2 0.05 116.61

15 2 =

1206.95 kgcm

a/b

=

5

/

M3

= =

1/2 q a12 1/2

116.61

7.5.2.3 Perhitungan Tebal Pelat Baja s = 6 M t2

30

=

0.17

52 =

1457.67 kgcm

t=

6 M s plat

=

6 x

=

2.34

1457.67 1600

~

3

7.5.3 Perencanaan Diameter Angker 7.5.3.1 Perhitungan Tegangan Yang Bekerja Pada Angker

M P

40 y

x

M σ min

20

P σ max 1/3x

σ min x x =

=

C

1/3y

σ max B-x σ min B

=

88.34

<

x

40

cm

0.5

σ min + σ max

88.34 + 17.24 cm

= y =B - x S min

=

= = =

40

1.5 d 1.5

-

116.61

17.24

= 2

1.5 ( 2 x tf ) x 1.2

=

22.76 cm

3.6 cm

1/3 x =

5.75 cm

>

S min

1/3 y =

7.59 cm

>

S min

=

20

-

7.59

=

12.41 cm

-

7.59

-

5.75

=

26.67 cm

r = 20 - 1/3 y C = T =

40 M - P r C

=

1093100

-

22617

12.41

26.7 =

30424.65

kg

=

Pu

7.5.3.2 Perencanaan Diameter Angker φ fy Ag - Leleh: Pu = Ag perlu =

- Putus:

30424.65 0.9 4100

8.25 cm2 #DIV/0!

φ 0.75 fu Ag

Pu =

Ag perlu = 0.75 - A baut perlu =

=

30424.65 0.75

30424.65 1120

= 0 =

#DIV/0! cm2 #DIV/0!

27.16 cm2

#DIV/0!

Untuk tiap sisi A baut perlu = 27.16 / 2 sisi = 13.58 Direncanakan menggunakan angker D30 ( A = 7.07 cm2 ) Jumlah angker dalam 1 sisi = #DIV/0! / 7.07 = #DIV/0! ~ #DIV/0! buah Dipasang 3 D 30 A = 21.2 cm2 Abaut > Aperlu OK

cm2

7.5.4 Perencanaan Panjang Angker Kekuatan Baut untuk menerima beban tarik pada tiap sisi adalah: 30424.65 / #DIV/0! baut

=

#DIV/0! kg

2 sisi l =

#DIV/0! 2p

4.47

=

#DIV/0! cm

1.5

7.5.5 Perencanaan Sambungan Las

t plat

=

3

Profil Baja Bj 52 fu las E70 xx

fu = = =

Syarat tebal plat: a min = a max =

af max

3 t - 0.1

=

=

Akibat Pu :

70 70

a te b d Sx

fvp

0 kg/cm2

ksi x

70.3

kg/cm3

mm = =

3 2.9

1.41 x fu elemen t fu las 1.41 x 70 x 0 cm

=

Pakai

cm

0 70.3 =

= = = = = =

3 mm 0.707 a 30 cm 30 cm b * d + ( d2 / 3) 30 30

=

1200 cm3

=

Pu

-

0.1

x

3

cm

0 mm

=

+

2.12 mm

900 3

2 ( 2 b + d ) te = = Akibat Mu:

fhm

=

22617 2 2 30 592.41 kg/cm2 Mu Sx

=

= =

f total

= = = =

0.75 2214.45

x

fv2 + fh2 2 + 592.41 ### kg/cm2 1086.61 kg/cm2

0.6

910.92 <= <=

30

2.12

1093100 1200 910.92 kg/cm2

= f las

+

x

70

x

70.3

2

0.75 0.6 fu las 2214.45 kg/cm2

OK !!