Program Bangunan Baja

PERENCANAAN GORDING

Data Penutup AtapJenis : asbes gelombangTebal : 6 mmBerat : 12.5Panjang asbes : 305 cmLebar asbes : 110 cmSudut Kemiringan : 18 °Overlap : 20 cmJarak gording diagonal : 142.5 cmJarak gording horizontal : 135.5256 cm = 1 mDirencanakan dengan menggunakan PROFIL C 150x65x20x3,2

A : 150 mm Cy 2.11 cm BJ : 41B : 65 mm Ix 332 fu : 410C : 20 mm Iy 53.8 fy : 250t : 3.2 mm ix 5.89 cm

Ag : 9.567 iy 2.37 cmW : 7.51 kg/m Sx 44.3 cm³

Cx : 0 cm Sy 12.2 cm³

1. PERHITUNGAN PEMBEBANANA. BEBAN MATI (D)

Berat Gording == 7.51= 7.51 kg/m

Berat asbes = x 1.15 x Jarak Miring Gording= 12.5 x 1,15 x 1.425= 20.48437 kg/m

Berat total = Berat Gording + Berat asbes= 7.51 + 20.48= 27.99437 kg/m

Alat-alat pengikat = 10% x Berat total= 10% x 27.99437= 2.799438

Berat + alat pengikat = 27.99437 + 2.799438= 30.79381 kg/m

= 30.7938125 kg/m

= ⅛ X x L²= ⅛ X 30.79381 x cos 18 x 6 ²

kg/m²

cm⁴cm⁴

cm²

Wgording

Wasbes

qD

MXD q cos α

= 131.78995218 kgm

= ⅛ X x (L/3)²= ⅛ X 30.79381 x sin 18 x (6/3)²= 4.757905692 kgm

B. BEBAN HIDUP (L)B.1 Beban Merata

q == 25.6

Dipakai : 20 karena > 20

= 20 x 1.355256= 27.105110714 kg/m

= ⅛ X x L²= ⅛ X 27.10511 x cos 18 x 6 ²= 116.00321476 kgm

= ⅛ X x (L/3)²= ⅛ X 27.10511 x sin 18 x (6/3)²= 4.1879699226 kgm

B.2 Beban Terpusat (P = 100 kg)= ¼ X P x L= ¼ X 100 X cos 18 x 6= 142.65847744 kgm

= ¼ X P x L/3= ¼ X 100 X sin 18 x 6 /3= 15.450849719 kgm

C. BEBAN ANGINDalam PPI 1983 diatur bahwa beban angin dihitung sebagai berikut :Koefisien arah angin :

= - 0,4 Tekanan Tiup angin = 40= -0.04 hisap (di pihak angin)

= C1 x w= -0.04 x 40= -1.6 kgm

= -0.4 (di belakang angin)

= C2 x w= -0.4 x 40= -16

Kombinasi beban yang diperoleh :Mx total = Beban mati (x) + Beban hidup (x) + Beban angin

= 131.78995218 + 142.6585 += 195.24842962 kgm

My total = Beban mati (y) + Beban hidup(y)

MYD q sin α

40 - 0,8 αkg/m²kg/m²

qL

MXL q cos α

MYL q sin α

MXD cos α

MYD sin α

C1 0,02 α x kg/m²

q1

C2

q2

⅛ x (q1 + q2) x L²

= 4.757905692 + 15.45085= 20.208755411 kgm

D. MOMEN BERFAKTORa) Beban mati + beban hidup merata

Mux = + 0,8W= 1,2 x 131.79 + 1,6 x 116 + 0= 343.75308624 kgm

Muy = + 0,8W= 1,2 x 4.757906 + 1,6 x 4.188 + 0= 12.410238707 kgm

b) Beban mati + beban hidup terpusat

Mux = + 0,8W= 1,2 x 131.79 + 1,6 x 143 + 0= 386.40150653 kgm

Muy = + 0,8W= 1,2 x 4.757906 + 1,6 x 15.45 + 0= 30.43084638 kgm

2. PERHITUNGAN GORDING

A : 150 mm Cy 2.11 cm BJ : 41

B : 65 mm Ix 332 fu : 4100

C : 20 mm Iy 53.8 fy : 2500t : 3.2 mm ix 5.89 cm

Ag : 9.567 iy 2.37 cmW : 7.51 kg/m Sx 44.3 cm³

Cx : 0 cm Sy 12.2 cm³

Zx = ½ A B t -= 37810.688= 37.810688 cm³

Zy = [(½.t.Cy²) + ((A - 2t).t.(Cy - ½ t)) ] + 2 [(½t (B - Cy)²) + (t.(C - t). (B - Cy - ½t))]= 20388.144= 20.388144 cm³

Karena profil C merupakan PENAMPANG TIDAK KOMPAK, maka dipakaiMn = Mp - (Mp - Mr) x

1,2 MxD + 1,6MXL

1,2 MxD + 1,6MXL

1,2 MxD + 1,6MXL

1,2 MxD + 1,6MXL

Dipakai nilai beban mati mati + beban hidup terpusat karena nilai kombinasi terbesar

cm⁴ kg/cm2

cm⁴ kg/cm2

cm²

½ B t² + (½ A t - t² ) + (C t - t²) ( ½ A - C/2 - t/2)mm³

mm³

λ - λpλ - λp

Mpx = Zx x fy= 37.810688 x ###= 94526.72 kg.cm

Mpy == 20.388144 x ###= 50970.36 kg.cm

Mrx = Sx ( fy - fr)= 44.3 x ( 2500 - ### )= 79740 kg.cm

Mry = Sy ( fy - fr)= 12.2 x ( 2500 - ### )= 21960 kg.cm

λp = 1,76 x

= 1,76 x

= 49.780317396

J =

=

= 3355.4432

= 0.33554432

=

= 113645.95245

Iw =

= 2773.51912

=

= 5.616123E-06

λr =

= 146.46273158

Mnx = Mp - (Mp - Mr) x

= 86560.435147 kgcm

Mny = Mp - (Mp - Mr) x

= 40585.982185 kgcm

Zy x fy

Σ 1/3 b.t³(1/3. A. T³) + 2/3. (B-t). T³ + 2/3.(C-t).t³

mm⁴cm⁴

x1

kg/cm²

cm6

x2

λ - λpλr - λp

λ - λpλr - λp

LATERAL BUCKLINGLp= 1,76 x iy x

Lp= 117.979352227413 cm

50 cm

Mn = Z x fy (Tidak Menentukan)

Mny = Zy (1 flens) x fy

== 8450 kgcm

Kontrol InteraksiMux + Muy ≤ 1

38640.1506525 + 3043.084638 ≤ 177904.3916322 76050.89613713296 ≤ 1

KONTROL LENDUTANa) Lendutan Izin

F =

= 600180

= 3.3333333333 cm

b) Lendutan yang Terjadi

LB=

Karena LB < Lp, maka termasuk Bentang Pendek, sehingga

(1/4 x tf x bf2) x fy

ø Mnx ø Mny

PERENCANAAN GORDING

II - 7

BAB II

PERHITUNGAN PENGGANTUNG GORDING

PERHITUNGAN PEMBEBANAN

Jarak Kuda-Kuda = 600 cm

Jumlah Penggantung Gording 2 buah

Jumlah Gording = 11 buah

Jarak Penggantung Gording = 200 cm

Luas horisontal atap (A) = 13.00 m x 2 m

= 26.00

Beban Mati (D)

A x= 25.999983 x ###

= #REF! kg

Nd =

= #REF! sin #REF! arc tan

= #REF! kg/m' = 34.1 °

Nd

= #REF! /sin 34.10 °= #REF! kg

m²

qd = qDL

qd sin aθ = ((1.425 x cos 18°)/2)

TD = / sin θ

Rtot

II - 8

Beban Hidup (L)

u Beban air hujan (H)

A x

= 25.999983 ### = #REF! sin ###

= #REF! kg = #REF! kg/m'

= #REF! sin 18

= #REF! kg

= #REF! /sin 34.10 °= #REF! kg

u Beban terpusat (P)

= #REF! sin ###

= #REF! kg

Beban Berfaktor

Pu = 1.2 TD + 1.6 TL

= #REF! kg

PERENCANAAN BATANG TARIK

Kuat Leleh

Pu = f . fy . Ag

Ag =Pu

=#REF!

= #REF! (Menentukan)0.9 . #REF!

Kuat Putus

Pu = f . 0.75 fu . Ag

Ag =Pu

=#REF!

= #REF!0.75 . 0.75 . ###

d =4 . Ag

=4 . #REF!

= ### cm = ### mmp p

Dipakai batang bulat dengan diameter 10 mm

qL = qLL qy = qL sin a

NLL = ql sin a

TL = NL / sin θ

Py = PL sin a

cm2

f . Fy

cm2

f . 0.75. fu

II - 9

KONTROL BATANG TARIK

u Kontrol leleh

Pn = f . fy . Ag

= 0.9 . #REF! . 0.25 . p . 1 2

= #REF! kg > #REF! kg #REF!

u Kontrol kelangsingan

L<= 500 L =

#REF!= #REF! cm

d sin 34.10

d >=#REF!

500

= #REF! cm ###

u Kontrol putus

Pn = 0.75 . f . fu . Ag

= 0.75 . 0.75 #REF! 0.25 . p . 1 2

= #REF! kg > #REF! kg #REF!

Ag = 0.785Sehingga dapat dipakai f 10 mm cm2

II - 10

BAB II

PERHITUNGAN PENGGANTUNG GORDING

PERENCANAAN REGEL DINDING

DATA PERENCANAANDirencanakan Menggunakan Profil C 150 x 50 x 20 x 3.2

A = 150 mm Cy = 1.54 cm BJ 41

B = 50 mm = 280 fu 410 MPa

C = 20 mm = 28.3 fy 250 MPa

t = 3.2 mm = 5.71 cm

Ag = 8.607 = 1.81 cm

W = 6.76 kg/m = 37.4 cm³

Cx = 0 cm = 8.19 cm³1. PERHITUNGAN PEMBEBANANRegel Dinding Bentang Kuda-Kuda (L) = 6 meterA. Beban Mati (D)Dinding = Berat Seng 0.5 mm x Koefisien Impitan

= 5 x 1.15

= 5.75

Regel = Berat Regel / Jarak Regel= 6.76 / 2.25

= 3.004444

Berat Total = Berat Regel + Berat Dinding Seng= 3.0044444444444 + 5.75

= 8.754444

Berat Alat-Alat Pengikat = 15% x Berat Total= 0.15 x 8.754

= 1.313167

= Berat Total + Berat Alat-Alat Pengikat= 8.7544444444444 + 1.31316667

= 10.067611111111

= 0

= 0.9 x W = 36

= 0.4 x W = 16

Bentang Kuda-Kuda = 6 meter > 5 meterPakai 2 Penggantung Regel , sehingga Lx= 6 m

Ly= 2 m

B. Beban Pada Regel

= Jarak Regel x= 2.25 x 10.0676111= 22.65213 kg/m

=

= 0.125 x 22.65213 x= 11.32606 kgm

= 0

= 0

Ix cm⁴

Iy cm⁴

ix

cm² iy

Sx

Sy

kg/m2

kg/m2

kg/m2

kg/m2

qDDinding

kg/m2

qL

qW1 kg/m2

qW2 kg/m2

qD qDDinding

MYD 1/8 x qD x (Ly)2

(2)2

MXD

MYL

= 0

= Jarak Regel x= 2.25 x 36= 81 kg/m

=

= 0.125 x 81 x= 364.5 kgm

= 0

C. Momen Berfaktor

Mux = + += 0 + 0 + 1.6 x 364.5 ) x 0.75= 437.4 kgm

Muy = + += 1.2 x 11.32606 + 0 + 0 ) x 0.75= 10.19346 kgm

2. PERHITUNGAN REGEL

A = 150 mm Cy = 1.54 cm BJ 41

B = 50 mm = 280 fu 4100

C = 20 mm = 28.3 fy 2500

t = 3.2 mm = 5.71 cm

Ag = 8.607 = 1.81 cm

W = 6.76 kg/m = 37.4 cm³

Cx = 0 cm = 8.19 cm³

Zx = (½ A B t -= 39984.77= 39.98477 cm³

Zy = [(½.t.Cy²) + ((A - 2t).t.(Cy - ½ t)) ] + 2 [(½t (B - Cy)²) + (t.(C - t). (B - Cy - ½t))]= 14099.9= 14.0999 cm³

Karena profil C merupakan PENAMPANG TIDAK KOMPAK, maka dipakai

Mn = Mp - (Mp - Mr) x

Mpx = Zx x fy= 39.98477 x 2500= 99961.92 kg.cm

Mpy == 14.0999 x 2500= 35249.76 kg.cm

Mrx = Sx ( fy - fr)

MXL

qW qW1

MXW 1/8 x qW x (Lx)2

(6)2

MYW

(1,2 MxD 1.6 MXL 1.6 MXW) x 0.75

(1,2 MyD 1.6 MyL 1.6 MyW) x 0.75

Ix cm⁴ kg/cm2

Iy cm⁴ kg/cm2

ix

cm² iy

Sx

Sy

½ B t² + (½ A t - t² ) + (C t - t²) ( ½ A - C/2 - t/2)) x 2mm³

mm³

λ - λpλr - λp

Zy x fy

= 37.4 x ( ### - 700 )= 67320 kg.cm

Mry = Sy ( fy - fr)= 8.19 x ( ### - 700 )= 14742 kg.cm

λp = 1,76 x

= 1,76 x

= 49.78032

J =

=

= 3027.763

= 0.302776

=

= 121286

Iw =

= 1458.933

=

= 4.916185911E-06

λr =

= 152.3827


= 82369.26 kgcm


= 23113.91 kgcm

LATERAL BUCKLINGLp= 1,76 x iy x

Lp= 90.1023745 cm

50 cm

Mn = Z x fy (Tidak Menentukan)

Mny = Zy (1 flens) x fy

== 5000 kgcm

Σ 1/3 b.t³(1/3. A. T³) + 2/3. (B-t). T³ + 2/3.(C-t).t³

mm⁴cm⁴

x1

kg/cm²

cm6

x2

λ - λpλr - λp

λ - λpλr - λp

LB=

Karena LB < Lp, maka termasuk Bentang Pendek, sehingga

(1/4 x tf x bf2) x fy

Kontrol InteraksiMux

+Muy

≤ 1

43740 + 1019.346≤ 174132.338 4500

0.816547156254224 ≤ 1

ø Mnx ø Mny

Besar Momen Lentur akibat beban mati dan beban hidup diambil dari perhitungan gording:

= = 289.8011

= = 22.82313

Nu =

Mntx 0.75 x (1.2 MxD+1.6 MxL)

Mnty 0.75 x (1.2 MyD+1.6 MyL)

kg/cm2

kg/cm2


kgm

kgm

PERENCANAAN IKATAN ANGINBeban Angin (W) = 40Koefisien Angin = 0.9

1. PERHITUNGAN IKATAN ANGINA. Perhitungan Tinggi Kolom Gevel

= 5 + 0 = 5 + 0 = 5 m

= 5 + 1 x 1.425 x sin 18 = 5 + 0.4403 = 5.44 m

= 5 + 2 x 1.425 x sin 18 = 5 + 0.8807 = 5.881 m

= 5 + 3 x 1.425 x sin 18 = 5 + 1.321 = 6.321 m

= 5 + 4 x 1.425 x sin 18 = 5 + 1.7614 = 6.761 m

= 5 + 5 x 1.425 x sin 18 = 5 + 2.2017 = 7.202 m

= 5 + 6 x 1.425 x sin 18 = 5 + 2.6421 = 7.642 m

= 5 + 7 x 1.425 x sin 18 = 5 + 3.0824 = 8.082 m

= 5 + 11 x tan 18 = 5 + 3.5741 = 8.574 m

B. Perhitungan Luas Dinding Pikul

= = 7.07467054

= = 15.9396982

= = 18.3268411

= = 20.7139839

= = 11.2869484

C. Perhitungan Gaya Kerja Pada Ikatan Angin AtapGaya Yang Bekerja Pada Ikatan Angin Atap Akibat Beban Angin

R = 0.5 x C x W x A

= 0.5 x 0.9 x 40 x 7.075 = 127.34406964057 kg

= 0.5 x 0.9 x 40 x 15.94 = 286.91456783542 kg

= 0.5 x 0.9 x 40 x 18.33 = 329.88313924113 kg

= 0.5 x 0.9 x 40 x 20.71 = 372.85171064684 kg

= 0.5 x 0.9 x 40 x 11.29 = 203.1650706763 kg

= 1320.1585580402 kg

PERHITUNGAN IKATAN ANGINGaya Pada Ikatan Angin (Rangka Batang) --> Metode Keseimbangan Titik BuhulAsumsi Semua Gaya Batang Tarik

RTotal

f = arctan(2.85/6.0) = 25.408 °

Titik Buhul B

∑V =0 → + RTotal = 0

f → = -1320.1585580402 Tekan

Titik Buhul A

∑V =0 → + + = 0

2.85 m → 127.34407 + 0.903 -1320.159 = 0

→ 0.903 = 1320.158558 + -127.3441

→ = 1320.540457 Tarik

kg/m2

h1

h2

h3

h4

h5

h6

h7

h8

h9

A1 0.5 x (h1+h2) x 1.425 x cos 18° m2

A2 0.5 x (h2+h4) x 2 x 1.425 x cos 18° m2

A3 0.5 x (h4+h6) x 2 x 1.425 x cos 18° m2

A4 0.5 x (h6+h8) x 2 x 1.425 x cos 18° m2

A5 0.5 x (h8+h9) x 1.425 x cos 18° m2

R1

R2

R3

R4

R5

RTotal

V1 S1 V1

V1 kg →

R1 R2 R1 S1 cosf V1

S1 +

S1

S1 kg →

6.0

m

A

B

PERENCANAAN DIMENSI IKATAN ANGIN (BATANG TARIK)

Pu = = 1320.540457 TarikKuat Leleh → f = 0.9

Pu =Ag = = 1320.540457 / 0.9 x 2500

→ Ag = 0.586907 → Menentukan

Kuat Putus → f = 0.75 Ae = 0.75 x Ag (Batang Bulat)Pu = =Ag = = 1320.540457 / 0.75 x 4100 x 0.75

→ Ag = 0.572592 → Tidak Menentukan

Ag Perlu Max = 0.5869068698

Ag =

d =

d = 4 x 0.5869068698 / 3.142 0.5d = 0.86444956 cm → Pakai Diameter (d) = 16 mm

KONTROL DIMENSI IKATAN ANGIN (BATANG TARIK)

Kontrol Kelangsingan → Panjang Ikatan Angin (L) = 0.5L/d <= 500 L = 6.642 m

664.248 / 1.6 <= 500 L = 664.2 cm415.1548092278 <= 500 → OK

Jadi, Ikatan Angin dengan Diameter (d) = 16 mm dapat dipakai

S1 kg →

f x fy x AgPu / (f x fy)

cm2

f x fu x Ae f x fu x 0.75 x AgPu / (f x fu x 0.75)

cm2

cm2

0.25 x p x d2

(4 x Ag / p )0.5

62 + 2.852

PERENCANAAN KOLOM GEVEL7 m 8 m 4.5 m 5 m

1. MENENTUKAN KEBUTUHAN MOMEN INERSIA IxJarak Antar Kuda-Kuda (b) = 6 mBerat Gording = 7.51 kg/mBerat Dinding Seng Gelombang = 5

Modulus Elastisitas (E) = 2000000

Angin --> Bentang Gevel Max x C x W C = 0.9 W = 40

2.868467 m x 0.9 x 40

103.2648 kg/m = 1.032648 kg/cm

Momen Akibat Angin h = Tinggi Kolom Gevel Max (Tengah) = 8.574 m

=

= 0.125 x 103.26 kg/m x 8.574117 m x 8.574 m

= 948.9452 kgm

Syarat Kekakuan (Lendutan Ijin)

= h / 200 = 857.41 cm / 200

= 4.287058 cm

Lendutan yang Terjadi

y =

4.287058 = 5 / 384 x 1.032648043 x 857.41167 / 2000000 / Ix

Ix ≥ 847.5382

2. PERHITUNGAN GEVELDATA PERENCANAANDirencanakan Menggunakan Profil Light Channel 250x50x50x4

A = 250 mm Cy = 0.88 cm BJ 41

B = 50 mm = 1050 fu 4100

C = 0 mm = 23.3 fy 2500

t = 4 mm = 8.81 cm

Ag = 13.47 = 1.32 cm

W = 10.6 kg/m = 83.7 cm³

Cx = 0 cm = 5.66 cm³

A. Beban Vertikal Jumlah Regel = 3 Buah

Berat Sendiri Gevel = = 10.6 kg/m x 8.574 m= 90.8856 kg

Berat Dinding =

= 5 x 1.15 x 2.868 m x 8.574 m= 141.4188 kg

Berat Regel Horizontal == 6.7600 kg/m x 2.87 m x 3= 58.1725 kg

Berat Alat Pengikat = 10% x (Berat Sendiri Gevel + Berat Dinding + Berat Regel Horizontal)= 0.1 x 90.8856 + 141.4188 + 58.1725= 29.0477 kg

L1 = L2 = H1 = H2 =

kg/m2

kg/cm2

qw = kg/m2

qw = kg/m2

qw =

Mw 0.125 x qw x h2

Mw

Mw

ymax

ymax

5 / 384 x qw x h4 / (EIx)

cm4

Ix cm⁴ kg/cm2

Iy cm⁴ kg/cm2

ix

cm² iy

Sx

Sy

WProfil x Tinggi Gevel Max

WSeng Gelombang x Koefisien Impitan x Bentang Gevel Max x Tinggi Gevel Max

kg/m2

WRegel x Bentang Gevel Max x Jumlah Regel

4

= 90.8856 + 141.4188 + 58.1725 + 29.0477

= 319.5246 kg

B. Beban BerfaktorBeban Aksial

Nu = = 1.2 x 319.5246 kg = 383.4295 kgMomen

Mntx = = 0.8 x 948.9452 kgm = 759.1561 kgm

C. Kontrol Kolom → kx = ky = 1a). Kontrol Penampang

Leleh Di Sebagian Penampang

b). Kontrol Kelangsingan Struktur (Tekuk Struktural)λx = kx L / ix = 1 x 857 cm / 8.81 cmλx = 97.32255 → Tidak Menentukan

λy = ky L / iy = 1 x 225 cm / 1.32 cmλy = 170.4545 → Menentukan

=

= 3.141593 2000000 x 13.5 / 97.32255

= 28071.81 kg

=

= 3.141593 2000000 x 13.5 / 170.454545

= 9151.245 kg

c). Kekuatan Nominal Kolom (Nn) → f = 0.85

= 170.4545

=

= 170.4545 3.141592654 x 2500 / 2000000

= 1.918288 → → w =

w = = 1.25 x 1.91828760545 = 4.59978417

Nn = = 13.47 x 2500 / 4.599784172

Nn = 7320.996 kg

= 383.4295 0.85 x 7320.9957 = 0.062

= 0.061616 ≤ 0.2 → Rumus Interaksi 2

D. Kontrol Baloka). Momen Berfaktor → Elemen Tidak Bergoyang → Terhadap Sumbu x

Mux = → Elemen dengan Beban Transversal, Ujung-Ujung Sederhana

δbx = ≥ 1 → Cm = 1

δbx = 1 1 - 383.4295 kg / 28071.8075 kgδbx = 1.013848 ≥ 1 → δbx = 1.01384803

Berat Total (ND) = Berat Sendiri Gevel + Berat Dinding + Berat Regel Horizontal + Berat Alat Pengikat

ND

ND

1.2 ND

0.8 MW

Profil Light Channel merupakan PENAMPANG TIDAK KOMPAK →

Ncrbx p2 E Ag / λx2

Ncrbx kg/cm2 cm2

Ncrbx

Ncrby p2 E Ag / λy2

Ncrby kg/cm2 cm2

Ncrby

lmax

lc lmax/p x fy/E

lc

lc lc ≥ 1.2 1.25lc2

1.25lc2

Ag fy / w cm2 kg/cm2

Nu / fNn

Nu / fNn

δbx Mntx

Cm / (1 - (Nu/Ncrbx))

22 x

22 x

/

2

/

/

Mux =Mux = 1.013848 x 759.1561 kgm = 769.668947 kgm

b). Perhitungan Modulus Plastis ZxZx =Zx = 2 25000 - 400 + 484 x 60.5Zx = 107764 = 108

Zy =Zy = 2 154.88 + 3395 + 242 x 4 x 6.8Zy = 13681.92 = 13.7

Karena profil C merupakan PENAMPANG TIDAK KOMPAK, maka dipakaiMn =

Mp - (Mp - Mr) x

Mpx = Zx x fy= 107.764 x 2500= 269410 kg.cm

Mpy == 13.68192 x 2500= 34204.8 kg.cm

Mrx = Sx ( fy - fr)= 83.7 x ( 2500 - 700 )= 150660 kg.cm

Mry = Sy ( fy - fr)= 5.66 x ( 2500 - 700 )= 10188 kg.cm

λp = 1,76 x

= 1,76 x

= 49.78032

J =

= (1/3. A. T³) + 2/3. (B-t). T³

= 7296= 0.7296

=

= 105243.6

Iw =

= 3411.353

=

δbx Mntx

2 x (1/2ABt - 1/2Bt2 + (1/2At-t2) (1/4A-1/2t))

mm3 cm3

2 x (1/2tCy2 + 1/2t (B-Cy)2) + (A-2t) t (Cy-1/2t)

mm3 cm3

λ - λpλ - λp

Zy x fy

Σ 1/3 b.t³

mm⁴cm⁴

x1

kg/cm²

cm6

x2

x

x

= 0.000012

λr =

= 158.2554


= 217364.5 kgcm = 2173.6448 kg.m


= 7487.055 kgcm

= 225 cm = 208.8971 cm

Lp = 1,76 x iy x Efy

= 65.71002 cm = 0.66 mKarena Lp < LB < LR maka disebut BENTANG MENENGAH

Mn =

XA = 3.0625 mXB = 3.625 mXC = 4.1875 m

Mi =MA = 871.5201 kgm Cb =MB = 926.3136 kgmMC = 948.4334 kgm = 1.03

Mmax = 948.9452 kgm

Mp = fy . Zx = 2500 x 108 = 269410 kg cmMR = (fy - fR) Sx = ( 2500 - 700) 83.7 = 150660 kg cmMy = fy . Sx = 2500 x 83.7 = 209250 kg cm

Mn =

= 1544.096 kgm

>> Persamaan Interaksi

0.0308082111 + 0.553843634 + 0 ≤ 10.5847 ≤ 1

Jadi, karena 0.584651846 ≤ 1 maka OK>> Lendutan Izin

y = h / 200y = 4.287058 cm

>> Re-kontrol Lendutan yang Terjadiy = 5 / 384 x qw x h4 / (EIx)y = 5 / 384 x 1.032648043 x 857 2000000 / 1050

cm4/kg2

λ - λpλr - λp

λ - λpλr - λp

LB LR

2,5 + ¼ . 2,25 =2,5 + 2/4 . 2,25 =2,5 + ¾ . 2,25 =

R . Xi - ½ q L²

⁴ /

= 3.460424 cmKarena lendutan yang terjadi 3,46 cm ≤ lendutan izin 4.287 cmmaka profil tersebut DAPAT DIPAKAI

Perlukah Impitan?

LB > LR

PERENCANAAN KOLOM GEVEL7 m 8 m 4.5 m 5 m

1. MENENTUKAN KEBUTUHAN MOMEN INERSIA IxJarak Antar Kuda-Kuda (b) = 6 mBerat Penutup Atap (Asbes Gelombang) = 12.5Berat Gording = 7.51 kg/mBerat Dinding Seng Gelombang = 5

Modulus Elastisitas (E) = 2000000

Angin --> Bentang Gevel Max x C x W C = 0.9 W = 40

2.868467 x 0.9 x 40

103.2648 kg/m = 1.032648 kg/cm

Momen Akibat Angin h = Tinggi Kolom Gevel Max (Tengah) = 8.574 m

=

= 0.125 x 103.26 x 8.57 x 8.574

= 948.9452 kgm

Syarat Kekakuan (Lendutan Ijin)y = h / 200y = 4.287058 cm


y =

4.287058 = 5 / 384 x 1.0326480428 x 857 / 2000000 / Ix

Ix ≥ 847.53822. PERHITUNGAN GEVELDATA PERENCANAANDirencanakan Menggunakan Profil Light Channel 250x50x50x4

A = 250 mm Cy = 0.88 cm BJ 41

B = 50 mm = 1050 fu 410 MPa

C = 0 mm = 23.3 fy 250 MPa

t = 4 mm = 8.81 cm

Ag = 13.47 = 1.32 cm

W = 10.6 kg/m = 83.7 cm³

Cx = 0 cm = 5.66 cm³

a) PEMBEBANAN1. Beban Vertikal

Beban sendiri gevel == 90.8856 kg

Berat dinding == 307.4321 kg

Berat alat penggantung = 10% x (beban sendiri gevel + berat dinding)= 39.8318 kg

Berat regel horizontal = Wregel x bentang gevel max x tinggi gel max= 166.2593 kg

berat gevel + berat dinding + berat alat penggantung + berat regel horizontal

L1 = L2 = H1 = H2 =

kg/m2

kg/m2

kg/cm2

qw = kg/m2

qw =

qw =

Mw 0.125 x qw x h2

Mw

Mw

5 / 384 x qw x h4 / (EIx)

cm4

Ix cm⁴

Iy cm⁴

ix

cm² iy

Sx

Sy

Wprofil x tinggi gevel max.

Wasbes gelombang x bentang gevel max. x tinggi gevel max.

ND =

4

= 604.4088 kg

2. Beban Berfaktor

Beban Aksial (Nu) == 725.29056886 kg

Momen (Mntx) = 1,2 Mw= 1138.7341935 kgm

b) KONTROL PROFIL

Lkx = kx . Lλk =

Lkx=

857.4= 1 x 8.5741166586 ix 8.81= 8.574117 m = 857.41 cm = 97.32

= . Ag

= 2000000. 9,567 = 28071.8075101

97.32255 ²

Lky = ky . Lλk =

Lky=

225= 1 x 2.25 iy 1.32= 2.25 m = 225 cm = 170.5

= . Ag

2000000. 9,567 = 9151.24458678

170.4545 ²

fy=

170.45 2500

E 2000000= 1.9183 m

≥ 1,2 maka= 4.5998

Nn = Agfy Nu

= 0.116553ω= 7321 kg

c) KONTROL PENAMPANG>> Menentukan Nilai Mux

Mux =

δbx = Mntx2 = 0,8 x Mw= 0,8 x 949

= 1 = 759.1561 kgm

1 -725.29056886

28071.80751= 1.026522

Mux = 1.026522 x 759.156129= 779.2906 kgm

1,2 ND

>> Menentukan Nilai λc

NCRbx

p2 . E

lx2

p2 x

NCRby

p2 . E

ly2

p2 x

lc =l

p p

Karena λc ω = 1,25 λc²

ø Nn

δbx .Mntx2

A = 250 mm Cy = 0.88 cm BJ = 41

B = 50 mm = 1050 fu = 4100

C = 0 mm = 23.3 fy = 2500

t = 4 mm = 8.81 cm

Ag = 13.47 = 1.32 cm

W = 10.6 kg/m = 83.7 cm³

Cx = 0 cm = 5.66 cm³

Zx = 2(½ A B t -= 107764= 107.764 cm³

Zy = [(½.t.Cy²) + ((A - 2t).t.(Cy - ½ t)) ] + 2 [(½t (B - Cy)²) = 12272.64= 12.27264 cm³

Karena profil C merupakan PENAMPANG TIDAK KOMPAK, maka dipakaiMn =

Mp - (Mp - Mr) x

Mpx = Zx x fy= 107.764 x 2500= 269410 kg.cm

Mpy == 12.27264 x 2500= 30681.6 kg.cm

Mrx = Sx ( fy - fr)= 83.7 x ( 2500 - 700 )= 150660 kg.cm

Mry = Sy ( fy - fr)= 5.66 x ( 2500 - 700 )= 10188 kg.cm

λp = 1,76 x

= 1,76 x

= 49.78032

J =

=

= 7125.333

= 0.712533

=

Direncanakan Menggunakan Profil Channel 250x50x50x4

Ix cm⁴ kg/cm2

Iy cm⁴ kg/cm2

ix

cm² iy

Sx

Sy

½ B t²) + (½ A t - t² )(¼ A - ½t) mm³

mm³

λ - λpλ - λp

Zy x fy

Σ 1/3 b.t³(1/3. A. T³) + 2/3. (B-t). T³ + 2/3.(C-t).t³

mm⁴cm⁴

x1

= 104005

Iw =

= 3411.353

=

= 0.000013

λr =

= 157.963 cm


= 324053 kgcm = 3241 kg.m


= 40111.74 kgcm

>> Lateral Bracing

= 225 cm = 208.5111 cm = 2.25 m

Lp = 1,76 x iy x Efy

= 65.71002 cm = 0.66 mKarena Lp < LB < LR maka disebut BENTANG MENENGAH

Mn =

XA = 3.0625 mXB = 3.625 mXC = 4.1875 m

Mi =MA = 871.5201 kgm Cb =MB = 926.3136 kgmMC = 948.4334 kgm = 1.03

Mmax = 948.9452 kgm

Mp = fy . Zx = 2500 x 108 = 269410 kg cmMR = (fy - fR) Sx = ( 2500 - 700) 83.7 = 150660 kg cmMy = fy . Sx = 2500 x 83.7 = 209250 kg cm

Mn =

= 1548.953 kgm

>> Persamaan Interaksi

kg/cm²

cm6

x2

λ - λpλr - λp

λ - λpλr - λp

LB LR

2,5 + ¼ . 2,25 =2,5 + 2/4 . 2,25 =2,5 + ¾ . 2,25 =

R . Xi - ½ q L²

0.058276 + 0.2672028284 + 0 ≤ 10.325479261 ≤ 1

Jadi, karena 0.325479261 ≤ 1 maka OK

>> Lendutan Iziny = h / 200y = 4.287058 cm

>> Re-kontrol Lendutan yang Terjadiy = 5 / 384 x qw x h4 / (EIx)y = 5 / 384 x 1.0326480428 x 857 2000000 / 1050

= 3.460424 cmKarena lendutan yang terjadi 3,46 cm ≤ lendutan izi 4.287058 cm

maka profil tersebut DAPAT DIPAKAI

⁴ /

PERENCANAAN GORDING IKATAN ANGIN Modulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. DATA PENUTUP ATAPJenis = Asbes GelombangTebal = 6 mmBerat = 12.5Panjang = 305 cmLebar = 110 cmSudut Kemiringan Atap = 18 °Overlap = 20 cmJarak Gording Diagonal = 142.5 cm = 1.425 mJarak Gording Horizontal = 135.5 cm = 1.355256 mBentang Antar Kuda-Kuda = 6 m

2. PERHITUNGAN GORDING IKATAN ANGINDATA PERENCANAANDirencanakan Menggunakan Profil Double Light Lip Channel Back To Back Arrangement C 150 x 130 x 20 x 2.3

A = 150 mm Cy = 0 cm Zx = 79 cm³

B = 130 mm = 496 Zy = 31 cm³

C = 20 mm = 145 BJ 41

t = 2.3 mm = 5.95 cm fu 4100

Ag = 14.02 = 3.22 cm fy 2500

W = 11 kg/m = 66.1 cm³

Cx = 0 cm = 22.3 cm³

A. Beban Mati (D)

Berat Gording = = 11 kg/m

Berat Asbes =

= 12.5 x 1.15 x 1.425 m= 20.484375 kg/m

Berat Total = Berat Gording + Berat Asbes= 11 kg/m + 20.48437 kg/m= 31.484375 kg/m

Berat Alat-Alat Pengikat = 10% x Berat Total= 0.1 x 31.48437 kg/m = 3.148438 kg/m

34.6328125 kg/m

kg/cm2 kg/cm2

kg/m2

Ix cm⁴

Iy cm⁴

ix kg/cm2

cm² iy kg/cm2

Sx

Sy

WGording

WAsbes x Koefisien Impitan x Jarak Gording Diagonal

kg/m2

Beban Mati (qD) =

A

C

=

= 0.125 x 34.6328125 kg/m x 0.951057 x 6= 148.21993 kgm

=

= 0.125 x 34.6328125 kg/m x 0.309017 x 2= 5.3510638 kgm

B. Beban Hidup (L)a). Beban Terbagi Rata (Hujan)

q = ≤ 20 = 40 - 0.8 x 18

q = 25.6 ≤ 20 → q = 20

= q x Jarak Gording Horizontal

= 20 x 1.3552555 m = 27.10511 kg/m

=

= 0.125 x 27.105110714412 kg/m x 0.951057 x 6= 116.00321 kgm

=

= 0.125 x 27.105110714412 kg/m x 0.309017 x 2= 4.1879699 kgm

b). Beban Terpusat (Manusia) → P = 100 kg

=

= 0.25 x 100 kg x 0.951057 x 6 m= 142.65848 kgm

=

= 0.25 x 100 kg x 0.309017 x 2 m= 15.45085 kgm

C. Beban Angin (W) → Tekanan Tiup Angin (W) Dekat Pantai = 40Koefisien Angin C

Angin Tekan → = = 0.02 x 18 - 0.4

= -0.04 → Hisap

Angin Hisap → = -0.4 → Hisap

=

= -0.04 x 40 x 1.425 m= -2.28 kg/m → Di Pihak Angin → Tidak Diperhitungkan

=

= -0.4 x 40 x 1.425 m= -22.8 kg/m → Di Belakang Angin → Tidak Menentukan

=

= 0.125 x 0 kg/m x 6= 0 kgm

MxD 1/8 x qD cosα x L2

m2

MyD 1/8 x qD sinα x (L/3)2

m2

40 - 0.8α kg/m2

kg/m2 kg/m2 kg/m2

qL

qL kg/m2

MxL1 1/8 x qL cosα x L2

m2

MyL1 1/8 x qL sinα x (L/3)2

m2

MxL2 1/4 x P cosα x L

MyL2 1/4 x P sinα x (L/3)

kg/m2

C1 0.02α-0.4

C1

C2

qW1 C1 x W x Jarak Gording Diagonal

kg/m2


kg/m2

MxW 1/8 x qw1 x L2

m2

2

2

2

2

2

D. Beban Berfaktora). Beban Mati (D) + Beban Hidup (L) Terbagi Rata (Hujan) + Beban Angin (W) <Lentur>

=

= 1.2 148 kgm 1.6 116.0032 kgm 0.8 0 kgm= 363.46906 kgm

=

= 1.2 5.351 kgm 1.6 4.18797 kgm= 13.122028 kgm

b). Beban Mati (D) + Beban Hidup (L) Terpusat (Manusia) + Beban Angin (W) <Lentur>

=

= 1.2 148 kgm 1.6 142.6585 kgm 0.8 0 kgm= 406.11748 kgm

=

= 1.2 5.351 kgm 1.6 15.45085 kgm= 31.142636 kgm

c). Beban Aksial (N) Dari Ikatan Angin

Nu = Pu = = 1320.1586 kg

d). Beban Berfaktor Untuk Beam Column

= 363.46906 kgm

= 13.122028 kgm

= 406.11748 kgm

= 31.142636 kgm

Nu = 1056.1268 kgE. Perhitungan Modulus Plastis Zx

Zx =Zx = 4 149.5 73.85 167.21 36.35 40.71 63.85Zx = 78871.968 = 78.871968

Zy =Zy = 4 149.5 32.5 167.21 1.15 40.71 63.85Zy = 30601.5 = 30.6015

Mux1 1.2 MxD + 1.6 MxL1 + 0.8 MxW

Muy1 1.2 MyD + 1.6 MyL1

Mux2 1.2 MxD + 1.6 MxL2 + 0.8 MxW

Muy2 1.2 MyD + 1.6 MyL2

RTotal Ikatan Angin

Mntx1

Mnty1

Mntx2

Mnty2

4 x (Bt/2 (0.5A - 0.5t) + (0.5At - t2) (0.25A - 0.5t) + (Ct - t2) (0.5A - 0.5C - 0.5t))

mm3 cm3

4 x (0.5Bt (0.25B) + (0.5At - t2) (0.5t) + (Ct - t2) (0.5B - 0.5t))

mm3 cm3

+x +x x

+x x

+x +x x

+x x

+x +x x

x + x + x

x x x+

x

x +

a) Kontrol Kelangsingan StrukturLkx = 600 cm

λkx = Lkx = 600ix 5.95

= 101 ( menentukan )

= . Ag

= ### 19.14 = 27215.05 kg100.84034 ²

Lky = 225 cmλky = Lky = 225

NCRbx p2 . E

lx2

p2 x

iy 3.22= 69.88

= . Ag

### . 19.14 = 56679.3 kg69.875776 ²

katakan Tekuk Kritis Arah X

fy = 100.8 410

E 2000000= 0.46 meter

1.43

= 1.1067416

Pu == 1056.1268 kg

Pn = Ag fy Pu = 0.239228

ω5193.8052 kg

≤ 0.2 maka NOT OKE

Gording ini dianggap tidak bergoyang >> Terhadap Sumbu XUntuk elemen beban transversal dan ujung beban sederhana Cm =1Mntx = 343.75309 kgm

δbx = Mux = δbx . Mntx= 1.040373 x 343.7531

= 1 = 357.6316 kgm1 - 1056

###= 1.0403735 (menentukan)

Mnty = 12.410239 kgm Mux = δbx . Mntx= 1.018987 x 12.41024

δby = = 12.64587 kgm

= 11 - 1056

###= 1.0189872

≤ 1

≤ 1

0.125 ≤ 1

NCRby p2 . E

ly2

p2 x

lc = l

p p

Karena 0.25 < λc < 1.2, maka ω =1.6 - 0.67λc

0.8 Rtotal

ø Pn

Karena nilai Pu / ø Pn

Pu2xφ . Pn

+{ Muxφb .Mnx

+Muyφ .Mny }

78.87197

30.6015

PELAT LANTAI BONDEX1. PELAT LANTAI BONDEX AA. Beban Mati (D)

Spesi 2 cm = = 0.02 m x 2100 = 42

Tegel 2 cm = = 0.02 m x 2400 = 48

Plafond = 11

Penggantung Plafond = 7

= 108

B. Beban Hidup (L)

Beban Hidup untuk Lantai Gudang = 400

C. Beban Berguna

= + = 108 + 400 = 508 ≈ 600Bentang Antar Balok Anak = 2.33 m ≈ 2.50 m → Tanpa Penyangga

Dari Tabel Perencanaan Praktis Pelat Bondex, untuk: Tebal Pelat Bondex = 0.75 mmBentang = 2.50 m Tanpa Penyangga → Bentang Menerus Dengan Tulangan Negatif

Beban Berguna = 600 , didapat: Tebal Pelat = 9 cm

Luas Tulangan Negatif = 3.25

12 mm

= = 0.25 x p x 12 x 12 = 113

= 1.131

As Negatif Perlu = 3.25

Jumlah Tulangan Perlu n = As Negatif Perlu / As 12Jumlah Tulangan Perlu n = 3 / 1.131 = 2.87 Buah/mJarak Antar Tulangan S = 1 m / 2.874 = 0.34799 m = 348 mmJarak Antar Tulangan S ≈ 300 mm

→ 1000 / 300 x 1.131 ≥ 3.25

→ 3.77 ≥ 3.25

2. PELAT LANTAI BONDEX BA. Beban Mati (D)

Spesi 2 cm = = 0.02 m x 2100 = 42

Tegel 2 cm = = 0.02 m x 2400 = 48

Plafond = 11


= 108

B. Beban Hidup (L)


C. Beban Berguna

= + = 108 + 400 = 508 ≈ 600Bentang Antar Balok Anak = 2.0 m → Tanpa Penyangga

Dari Tabel Perencanaan Praktis Pelat Bondex, untuk: Tebal Pelat Bondex = 0.75 mmBentang = 2.00 m Tanpa Penyangga → Bentang Menerus Dengan Tulangan Negatif

Beban Berguna = 600 , didapat: Tebal Pelat = 9 cm

Tebal Spesi x gSpesi kg/m3 kg/m2

Tebal Tegel x gBeton kg/m3 kg/m2

kg/m2

kg/m2

qD kg/m2

qL kg/m2

qTotal qD qL kg/m2 kg/m2

kg/m2

cm2/m

Diputuskan Memakai Tulangan Negatif dengan Diameter f =

As 1 f 12 0.25 x p x f 2

As 1 f 12 cm2

cm2/m

1 f

Kontrol As Aktual ≥ As Negatif Perlu cm2

cm2 cm2

Jadi, dipasang Tulangan Negatif f 12 - 300



kg/m2

kg/m2

qD kg/m2

qL kg/m2


kg/m2

Luas Tulangan Negatif = 2.00

12 mm

= = 0.25 x p x 12 x 12 = 113

= 1.131

As Negatif Perlu = 2

Jumlah Tulangan Perlu n = As Negatif Perlu / As 12Jumlah Tulangan Perlu n = 2 / 1.131 = 1.77 Buah/mJarak Antar Tulangan S = 1 m / 1.768 = 0.57 m = 565 mmJarak Antar Tulangan S ≈ 500 mm

→ 1000 / 500 x 1.131 ≥ 2

→ 2.262 ≥ 2

cm2/m


As 1 f 12 0.25 x p x f 2

As 1 f 12 cm2

cm2/m

1 f


cm2 cm2


PELAT LANTAI BONDEX

+

+

mm2

PERENCANAAN BALOK ANAKJarak Antar Kuda-Kuda (b) = 6 mModulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. MENENTUKAN KEBUTUHAN MOMEN INERSIA Ix DAN MODULUS PLASTIS ZxA. Berat Pelat LantaiPelat Bondex = 10.1

Pelat Beton = 0.09 m x 2400 = 216

Spesi 2 cm = 0.02 m x 2100 = 42

Tegel 2 cm = 0.02 m x 2400 = 48

Plafond = 11

Penggantung Plafond = 7 +

= 334.1

B. Beban Mati (D) → Bentang Balok Anak Max = 2.33 mPerkiraan Berat Sendiri Balok Anak = 60.00 kg/mBerat Pelat Lantai = 334.1 x 2.3333 m = 779.57 kg/m +Berat Total = 839.57 kg/mBerat Alat-Alat Pengikat 10 % = 0.1 x 60.00 kg/m = 6 kg/m +

= 845.57 kg/m

C. Beban Hidup (L)

400 x 2.33 m = 933.3

D. Beban Kombinasi (Ultimate)

1.2 + 1.6

1.2 x 845.57 kg/m + 1.6 x 933.33 kg/m

1014.68 kg/m + 1493 kg/m 2508 kg/m

E. Kontrol Lendutan Ijin

= L/360 = 600 / 360 = 1.66667 cm


y =

1.66667 = 5/384 x 8.46 + 9.3333 x 600 / 2E+06 xIx ≥ 9005.6813

F. Beban Berfaktor dan Kontrol Kuat Rencana Momen Lentur → f =

L = 6 m

= 2508.0133 kg/m

= 0.5 x 2508 kg/m x 6 m

= 7524.04 kg

= 0.125 x 2508 kg/m x 6

= 11286.06 kgm

Mp = Zx x fy = 2500 Zx → Misal: Mn = 0.9 Mp = 2250≤

1E+06 kgcm ≤ 0.9 x 2250 Zx

kg/cm2 kg/cm2

kg/m2

kg/m3 kg/m2

kg/m3 kg/m2

kg/m3 kg/m2

kg/m2

kg/m2

qLantai kg/m2

kg/m2

qD

Beban Hidup Lantai Gudang (qL) = kg/m2

qu = qD qL

qu =

qu = =

Lendutan Ijin ymax

5 / 384 x (qD+qL) x L4 / (EIx)

cm4

qu

qu

Vu 1/2 x qu x L =

Vu

Mumax 1/8 x qu x L2 = m2

Mumax

Mumax f Mn

kg/cm2 x cm3

4

2

Zx ≥ 557.3363

2. PERHITUNGAN BALOK ANAKDATA PERENCANAANDirencanakan Menggunakan Profil WF 300 x 200 x 8 x 12

W = 56.8 kg/m h = 160

d = 294 mm = 234 mm 823

= 200 mm = 11300 244

= 8 mm = 1600 BJ 41

= 12 mm = 12.5 cm fu 4100

r = 18 mm = 4.71 cm fy 2500

Ag = 72.38 = 771 cm³

A. Berat Pelat LantaiPelat Bondex = 10.1

Pelat Beton = 0.09 m x 2400 = 216

Spesi 2 cm = 0.02 m x 2100 = 42

Tegel 2 cm = 0.02 m x 2400 = 48

Plafond = 11


= 334.1

B. Beban Mati (D)Berat Sendiri Balok Anak = 56.80 kg/mBerat Pelat Lantai = 334.1 x 2.3333 m = 779.57 kg/m +Berat Total = 836.37 kg/mBerat Alat-Alat Pengikat 10 % = 0.1 x 56.80 kg/m = 5.68 kg/m +

= 842.05 kg/m

C. Beban Hidup (L)

400 x 2.33 m = 933.3


1.2 + 1.6

1.2 x 842.05 kg/m + 1.6 x 933.33 kg/m

1010.456 kg/m + 1493 kg/m 2503.8 kg/m

L = 6 m

= 2503.7893 kg/m

= 0.5 x 2504 kg/m x 6 m

= 7511.368 kg

= 0.125 x 2504 kg/m x 6

= 11267.052 kgm = 1126705.2 kgcm

cm3

d - 2(tf+r) Sy =

Zx =

bf Ix cm⁴ Zy =

tw Iy cm⁴

tf ix

iy

cm² Sx

kg/m2

kg/m3 kg/m2

kg/m3 kg/m2

kg/m3 kg/m2

kg/m2

kg/m2

qLantai kg/m2

kg/m2

qD


qu = qD qL

qu =

qu = =

qu

qu

Vu 1/2 x qu x L =

Vu


Mumax

2

E. Kontrol Penampang (Tekuk Lokal)Badan (Web)

≤ → 234 / 8 ≤ 1680 / 250

→ 29.25 ≤ 106.25 → OK

Sayap (Flange) 1≤ → 200 / 24 ≤ 170 / 250

→ 8.3333 ≤ 10.752 → OK

Profil WF diatas tergolong PENAMPANG KOMPAKMn = Mp = Zx fyMn = 823 cm³ x 2500 = 2057500 kgcm

F. Kontrol Tekul Lateral → Tegangan Residu (fr)= 70 MPa = 700Bentang Antar Balok Anak = 2.3333 m

= 6 m = 600 cm

Lp = = 1.76 x 4.71 cm x 2E+06 /

Lp = 234.46529 cm ≈ 234 cm

= fy - fr = 2500 - 700 = 1800

Konstanta Puntir Torsi (J)

J = =

J = 0.3333333 x 27 x 0.8 + 0.67 x 20 xJ = 4.608 + 23.04 = 27.648

h' = d - tf = 294 - 12 = 282 mm

= 0.25 x 1600 x 28.2

= 318096

=

= p / 771 x 2000000 800000 27.648 72.38

= 163035.3

=

= 4 x 771 / 8E+05 x 27.6 x 318096 /

= 0.0000009663 = 9.662724E-07

=

= 4.71 x 163035 / 2500 700 x 1 1 1E-06

= 742.89009 cm ≈ 743 cm

= 600 cm

Lp = 234.46529 cm → Lp < LB < LR → BENTANG MENENGAH

= 742.89009 cm

G. Kekuatan Nominal Momen Lentur (Mn)

Mn = Cb(MR+(Mp-MR) x (LR-LB)/(LR-Lp) = 2503.789333 kg/m

= 7511.368 kg

= 1126705.2 kgcm = 11267

= = 11267

L = 6 m

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

kg/cm2

kg/cm2

Panjang Balok Anak (LB)

1.76 iy E/fy

fL kg/cm2

∑ 1/3 bt3 1/3 x (d-2tf) x tw3 + 2 x 1/3 x bf x tf

3

cm4 cm4 cm4

Konstanta Puntir Lengkung (Iw) →

Iw 1/4 x Iy x h'2 = cm4 cm2

Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2

LR iy (x1/(fy-fr)) x 1+ 1+x2fL2

LR

LR

LB

LR

qu

qu Vu

Mumax

Mu(x=3m) = Mumax MB

3

2

/xx x

2

- + + x

x x x 2

1.5 7511 - 0.5 2503.79 1.5

= = = 8450.289 kgm

= 140838.15 / 28168 25351 45068 25351 = 1.1364 ≤ 2.3

Mp = Zx fy = 823 x 2500Mp = 2057500 kgcm = 20575 kgm

= = 771 x 2500 - 700

= 1387800 kgcm = 13878 kgm

Mn =

Mn = 1.1363636 x 13878 20575 13878 742.9 600 742.89 234.465 ≤Mn = 17909.269 kgm ≤ 20575 kgm

Mcr =

Mcr = 0.00595 70778880000000000 + 109662271.123215 1600 318096

Mcr = 2116990 kgcm = 21169.9039752 kgm ≤ 20575 kgm

Kuat Nominal Momen Lentur (Mn)TEKUK LOKAL TEKUK LATERAL

Mn = 2057500 kgcm Mn = Cb[MR + (Mp-MR) x (LR-LB) / (LR-Lp)Mn = 20575 kgm Mn = 17909.26852 kgm

= 17909.2685168645 kgm

H. Kontrol Kuat Momen Lentur → f = 0.9

≤ → 11267.052 kgm ≤ 0.9 x 17909.2685169

→ 11267.052 kgm ≤ 16118.34167 kgm OK

I. Kontrol Lendutan

= L/360 = 600 / 360 = 1.66667 cm


y =

y = 5/384 x 8.42 + 9.3333 x 600 / 2E+06 x

y = 1.3256433 cm ≤ = 1.66666667 cm OK

J. Kontrol Kuat Geser → f = 0.9

Badan (Web)≤ → 234 / 8 ≤ 1100 / 250

→ 29.25 ≤ 69.57 → PLASTIS

≤ → ≤ → ≤→ 7511.368 kg ≤ 0.9 0.6 2500 29.4

→ 7511.368 kg ≤ 31752 kg OK

Mu(x=1.5m)=Mu(x=4.5m) = 1.5 Vu - 0.5 qu (1.5)2 =

Mu(x=1.5m)=Mu(x=4.5m) MA MC

Koefisien Momen Lentur (Cb) = 12.5Mmax / (2.5Mmax+3MA+4MB+3MC) ≤ 2.30

Cb

cm3 kg/cm2

MR Sx (fy-fr) cm3 kg/cm2

MR

Cb [MR + (Mp-MR) (LR-LB) / (LR-Lp)] ≤ Mp

Cbp/L E Iy GJ + (pE/L)2 Iy Iw ≤ Mp

MnAktual

Mumax f Mn

Lendutan Ijin ymax

5 / 384 x (qD+qL) x L4 / (EIx)

ymax

h/tw 1100 / fy1/2

Vu f Vn Vu f 0.6 fy Aw Vu f 0.6 fy d tw

x x x 2

+ ++

+ - x - -/

x x x x

x x

4

PERENCANAAN BALOK ANAK

kg/m

Ix

0.9

Zx

cm³

cm³ 822.6

cm³ 244.32

kg/m

kg/cm2

kg/cm2

2500

1.2

2

1600

1800

kgm

kgm

3

/

x 2

20575

318096

TEKUK LATERALCb[MR + (Mp-MR) x (LR-LB) / (LR-Lp)

kgm

11300

0.8

f 0.6 fy d tw

x

Section IndexThickness

mm kg/m mm mm mm mm mm1 100 x 50 x 5 x 7 9.3 100 50 5 7 8 11.852 100 x 100 x 6 x 8 17.2 100 100 6 8 10 21.93 125 x 60 x 6 x 8 13.2 125 60 6 8 9 16.844 125 x 125 x 6.5 x 9 23.8 125 125 6.5 9 10 30.315 150 x 75 x 5 x 7 14 150 75 5 7 8 17.856 150 x 100 x 6 x 9 21.1 148 100 6 9 11 26.847 150 x 150 x 7 x 10 31.5 150 150 7 10 11 40.148 175 x 90 x 5 x 8 18.1 175 90 5 8 9 23.049 175 x 125 x 5.5 x 8 23.3 169 125 5.5 8 12 29.65

10 175 x 175 x 7.5 x 11 40.2 175 175 7.5 11 12 51.2111 200 x 100 x 4.5 x 7 18.2 198 99 4.5 7 11 23.1812 200 x 100 x 5.5 x 8 21.3 200 100 5.5 8 11 27.1613 200 x 150 x 6 x 9 30.6 194 150 6 9 13 39.0114 200 x 200 x 8 x 12 49.9 200 200 8 12 13 63.5315 200 x 200 x 12 x 12 56.2 200 204 12 12 13 71.5316 200 x 200 x 10 x 16 65.7 208 202 10 16 13 83.6917 250 x 125 x 5 x 8 25.7 248 124 5 8 12 32.6818 250 x 125 x 6 x 9 29.6 250 125 6 9 12 37.6619 250 x 175 x 7 x 11 44.1 244 175 7 11 16 56.2420 250 x 250 x 11 x 11 64.4 244 252 11 11 16 82.0621 250 x 250 x 8 x 13 66.5 248 249 8 13 16 84.722 250 x 250 x 9 x 14 72.4 250 250 9 14 16 92.1823 250 x 250 x 14 x 14 82.2 250 255 14 14 16 104.724 300 x 150 x 5.5 x 8 32 298 149 5.5 8 13 40.825 300 x 150 x 6.5 x 9 36.7 300 150 6.5 9 13 46.7826 300 x 200 x 8 x 12 56.8 294 200 8 12 18 72.3827 300 x 200 x 9 x 14 65.4 298 201 9 14 18 83.3628 300 x 300 x 12 x 12 84.5 294 302 12 12 18 107.729 300 x 300 x 9 x 14 87 298 299 9 14 18 110.830 300 x 300 x 10 x 15 94 300 300 10 15 18 119.831 300 x 300 x 15 x 15 106 300 305 15 15 18 134.832 300 x 300 x 11 x 17 106 304 301 11 17 18 134.833 350 x 175 x 6 x 9 41.4 346 174 6 9 14 52.6834 350 x 175 x 7 x 11 49.6 350 175 7 11 14 63.1435 350 x 250 x 8 x 12 69.2 336 249 8 12 20 88.1536 350 x 250 x 9 x 14 79.7 340 250 9 14 20 101.537 350 x 350 x 13 x 13 106 338 351 13 13 20 135.338 350 x 350 x 10 x 16 115 344 348 10 16 20 14639 350 x 350 x 16 x 16 131 344 354 16 16 20 166.640 350 x 350 x 12 x 19 136 350 350 12 19 20 173.941 350 x 350 x 19 x 19 156 350 357 19 19 20 198.442 350 x 350 x 14 x 22 159 356 352 14 22 20 20243 400 x 200 x 7 x 11 56.6 396 199 7 11 16 72.1644 400 x 200 x 8 x 13 66 400 200 8 13 16 84.1245 400 x 300 x 9 x 14 94.3 386 299 9 14 22 120.146 400 x 300 x 10 x 16 107 390 300 10 16 22 13647 400 x 400 x 15 x 15 140 388 402 15 15 22 178.548 400 x 400 x 11 x 18 147 394 398 11 18 22 186.849 400 x 400 x 18 x 18 168 394 405 18 18 22 214.450 400 x 400 x 13 x 21 172 400 400 13 21 22 218.751 400 x 400 x 21 x 21 197 400 408 21 21 22 250.752 400 x 400 x 16 x 24 200 406 403 16 24 22 254.953 400 x 400 x 18 x 28 232 414 405 18 28 22 295.4

Profile Code

Weight (W)

Depth of Section

(d)

Flange Width

(bf)

Corner Radius

r

Sectional Area (Ag)

Web (tw)

Flange (tf)

cm2

54 400 x 400 x 20 x 35 283 428 407 20 35 22 360.755 400 x 400 x 30 x 50 415 458 417 30 50 22 528.656 400 x 400 x 45 x 70 605 498 432 45 70 22 770.157 450 x 200 x 8 x 12 66.2 446 199 8 12 18 84.358 450 x 200 x 9 x 14 76 450 200 9 14 18 96.7659 450 x 300 x 10 x 15 106 434 299 10 15 24 13560 450 x 300 x 11 x 18 124 440 300 11 18 24 157.461 500 x 200 x 9 x 14 79.5 496 199 9 14 20 101.362 500 x 200 x 10 x 16 89.7 500 200 10 16 20 114.263 500 x 200 x 11 x 19 103 506 201 11 19 20 131.364 500 x 300 x 11 x 15 114 482 300 11 15 26 145.565 500 x 300 x 11 x 18 128 488 300 11 18 26 163.566 600 x 200 x 10 x 15 94.6 596 199 10 15 22 120.567 600 x 200 x 11 x 17 106 600 200 11 17 22 134.468 600 x 200 x 12 x 20 120 606 201 12 20 22 152.569 600 x 200 x 13 x 23 134 612 202 13 23 22 107.770 600 x 300 x 12 x 17 137 582 300 12 17 28 174.571 600 x 300 x 12 x 20 151 588 300 12 20 28 192.572 600 x 300 x 14 x 23 175 594 302 14 23 28 222.473 700 x 300 x 13 x 20 166 692 300 13 20 28 211.574 700 x 300 x 13 x 24 185 700 300 13 24 28 235.575 700 x 300 x 15 x 28 215 708 302 15 28 28 273.676 800 x 300 x 14 x 22 191 792 300 14 22 28 243.477 800 x 300 x 14 x 26 210 800 300 14 26 28 267.478 800 x 300 x 16 x 30 241 808 302 16 30 28 307.679 900 x 300 x 15 x 23 213 890 299 15 23 28 270.980 900 x 300 x 16 x 28 243 900 300 16 28 28 309.881 900 x 300 x 18 x 34 286 912 302 18 34 28 364

Moment of Inertia Radius of Gyration

Ix Iy ix iy Sx Sy Zx Zycm cm

187 14.8 3.98 1.12 37.5 5.91 42 9.3383 134 4.18 2.47 76.5 26.7 84 40.8413 29.2 4.95 1.32 66.1 9.73 74 15.4847 293 5.29 3.11 136 47 149 71.4666 49.5 6.11 1.66 88.8 13.2 98 20.5

1020 151 6.17 2.37 138 30.1 150 46.21640 563 6.39 3.75 219 75.1 240 1141210 97.5 7.26 2.06 139 21.7 152 33.41530 261 7.18 2.97 181 41.8 193 63.72880 984 7.5 4.38 330 112 360 1711580 114 8.26 2.21 160 23 170 35.21840 134 8.24 2.22 184 26.8 200 41.42690 507 8.3 3.61 277 67.6 296 1034720 1600 8.62 5.02 472 160 513 2434980 1700 8.35 4.88 498 167 553 2566530 2200 8.83 5.13 628 218 698 3313540 255 10.4 2.79 285 41.1 305 634050 294 10.4 2.79 324 47 352 72.46120 984 10.4 4.18 502 113 535 1718790 2940 10.3 5.98 720 233 781 3569930 3350 10.8 6.29 801 269 859 407

10800 3650 10.8 6.29 867 292 937 44211500 3880 10.5 6.09 919 304 1015 4666320 442 12.4 3.29 424 59.3 455 90.97210 508 12.4 3.29 481 67.7 522 104

11300 1600 12.5 4.71 771 160 823 24413300 1900 12.6 4.77 893 189 963 28816900 5520 12.5 7.16 1150 365 1241 55718800 6240 13 7.51 1270 417 1353 63120400 6750 13.1 7.51 1360 450 1465 68221500 7100 12.6 7.26 1440 466 1577 71323400 7730 13.2 7.57 1540 514 1669 77811100 792 14.5 3.88 641 91 689 13913600 984 14.7 3.95 775 112 841 17218500 3090 14.5 5.92 1100 248 1163 37721700 3650 14.6 6 1280 292 1360 44428200 9380 14.4 8.33 1670 534 1799 81433300 11200 15.1 8.78 1940 646 2070 97735300 11800 14.6 8.43 2050 669 2247 102240300 13600 15.2 8.84 2300 776 2493 117542800 14400 14.7 8.53 2450 809 2708 123947600 16000 15.3 8.9 2670 909 2927 137820000 1450 16.7 4.48 1010 145 1088 22223700 1740 16.8 4.54 1190 174 1286 26633700 6240 16.7 7.21 1740 418 1846 63338700 7210 16.9 7.28 1980 481 2116 72949000 16300 16.6 9.54 2520 809 2730 123256100 18900 17.3 10.1 2850 951 3046 143659700 20000 16.7 9.65 3030 985 3318 150566600 22400 17.5 10.1 3330 1120 3600 169570900 23800 16.8 9.75 3540 1170 3920 178778000 26200 17.5 10.1 3840 1300 4207 197292800 31000 17.7 10.2 4480 1530 4954 2325

Elastic Modulus of Section

Plastic Modulus of Section

cm4 cm4 cm3 cm3 cm3 cm3

119000 39400 18.2 10.4 5570 1930 6239 2935187000 60500 18.8 10.7 8170 2900 9468 4428298000 94400 19.7 11.1 12000 4370 14385 671328700 1580 18.5 4.33 1290 159 1393 24433500 1870 18.6 4.4 1490 187 1621 28946800 6690 18.6 7.04 2160 448 2287 68156100 8110 18.9 7.18 2550 541 2728 82241900 1840 20.3 4.27 1690 185 1836 28747800 2140 20.5 4.33 1910 214 2096 33256500 2580 20.7 4.43 2230 257 2462 39860400 6760 20.4 6.82 2500 451 2663 68971000 8110 20.8 7.04 2910 541 3100 82468700 1980 23.9 4.05 2310 199 2535 31177600 2280 24 4.12 2590 228 2863 35790400 2720 24.3 4.22 2980 271 3317 424

103000 3180 24.6 4.31 3380 314 3778 493103000 7670 24.3 6.63 3530 511 3782 785118000 9020 24.8 6.85 4020 601 4309 920137000 10600 24.9 6.9 4620 701 5017 1076172000 9020 28.6 6.53 4980 602 5414 928201000 10800 29.3 6.78 5760 722 6249 1108237000 12900 29.4 6.86 6700 853 7344 1314254000 9930 32.3 6.39 6410 662 7040 1027292000 11700 33 6.62 7290 782 7995 1207339000 13800 33.2 6.7 8400 915 9287 1416345000 10300 35.7 6.16 7760 688 8634 1076411000 12600 36.4 6.39 9140 843 10174 1314498000 15700 37 6.56 10900 1040 12221 1619

PERENCANAAN GORDING Modulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. DATA PENUTUP ATAPJenis = Asbes GelombangTebal = 6 mmBerat = 12.5Panjang = 305 cmLebar = 110 cmSudut Kemiringan Atap = 18 °Overlap = 20 cmJarak Gording Diagonal = 142.5 cm = 1.425 mJarak Gording Horizontal = 135.526 cm = 1.3552555 mBentang Antar Kuda-Kuda = 6 mJumlah Penggantung Gording = 2 Buah2. PERHITUNGAN GORDINGDATA PERENCANAANDirencanakan Menggunakan Profil Light Lip Channel C 150 x 65 x 20 x 2.8

A = 150 mm Cy = 2.12 cm Zx = 48 cm³

B = 65 mm = 295 Zy = 19 cm³

C = 20 mm = 48 BJ 41

t = 2.8 mm = 5.91 cm fu 4100

Ag = 8.44 = 2.39 cm fy 2500

W = 6.63 kg/m = 39.4 cm³

Cx = 0 cm = 11 cm³

A. Beban Mati (D)

Berat Gording = = 6.63 kg/m

Berat Asbes =

= 12.5 x 1.15 x 1.425 m= 20.484375 kg/m

Berat Total = Berat Gording + Berat Asbes= 6.63 kg/m + 20.48437 kg/m= 27.114375 kg/m


29.8258125 kg/m

=

= 0.125 x 29.8258125 kg/m x 0.951057 x 6= 127.64715 kgm

=

= 0.125 x 29.8258125 kg/m x 0.309017 x 2

kg/cm2 kg/cm2

kg/m2

Ix cm⁴

Iy cm⁴

ix kg/cm2

cm² iy kg/cm2

Sx

Sy

WGording


kg/m2

Beban Mati (qD) =

MxD 1/8 x qD cosα x Lx2

m2

MyD 1/8 x qD sinα x Ly2

m2

2

2

= 4.6083415 kgm


q = ≤ 20 = 40 - 0.8 x 18

q = 25.6 ≤ 20 → q = 20


= 20 x 1.35526 m = 27.10511 kg/m

=

= 0.125 x 27.105110714412 kg/m x 0.951057 x 6= 116.00321 kgm

=

= 0.125 x 27.105110714412 kg/m x 0.309017 x 2= 4.1879699 kgm


=

= 0.25 x 100 kg x 0.951057 x 6 m= 142.65848 kgm

=

= 0.25 x 100 kg x 0.309017 x 2 m= 15.45085 kgm


Angin Tekan → = = 0.02 x 18 - 0.4

= -0.04 → Hisap


=


=


= → = 0 kgm

= 0.125 x 0 kg/m x 6= 0 kgm


=

= 1.2 128 kgm 1.6 116.0032 kgm 0.8 0 kgm= 338.78172 kgm

=

40 - 0.8α kg/m2

kg/m2 kg/m2 kg/m2

qL

qL kg/m2

MxL1 1/8 x qL cosα x Lx2

m2

MyL1 1/8 x qL sinα x Ly2

m2

MxL2 1/4 x P cosα x Lx

MyL2 1/4 x P sinα x Ly

kg/m2

C1 0.02α-0.4

C1

C2


kg/m2


kg/m2

MxW 1/8 x qw1 x L2 MyW

m2

Mux1 1.2 MxD + 1.6 MxL1 + 0.8 MxW

Muy1 1.2 MyD + 1.6 MyL1 + 0.8 MyW

2

+x +x x

+x x

2

2

+ x

= 1.2 4.608 kgm 1.6 4.18797 kgm 0.8 0 kgm= 12.230762 kgm


=

= 1.2 128 kgm 1.6 142.6585 kgm 0.8 0 kgm= 381.43014 kgm

=

= 1.2 4.608 kgm 1.6 15.45085 kgm 0.8 0 kgm= 30.251369 kgm

E. Perhitungan Modulus Plastis Zx

Zx =Zx = 2 182 73.6 202.16 36.1 48.16 63.6Zx = 2 13395.2 7297.976 3062.976Zx = 47512.304 = 47.5123

Zy =Zy = 404.32 19.8 2 59.36 10.6 122.64 21.9 48.16 42.4Zy = 8005.536 2 629.216 2685.816 2041.984Zy = 18719.568 = 18.7196

F. Kontrol Tekuk Lokal → Tegangan Residu (fr)= 70 MPa = 700Profil Light Lip Channel C diatas tergolong PENAMPANG TIDAK KOMPAK

Mn = Mp - (Mp - Mr) x (λ-λp) / (λr-λp)

Mpx = Zx fy

= 48 x 2500= 120000 kgcm

Mpy = Zy fy

= 19 x 2500= 47500 kgcm

Mrx = Sx ( fy - fr)

= 39.4 x 2500 - 700= 70920 kgcm

Mry = Sy ( fy - fr)

= 11 x 2500 - 700

= 19800 kgcm

λp = = 1.76 x 2000000 / 2500

λp = 49.780317

= fy - fr = 2500 - 700 = 1800


Mux2 1.2 MxD + 1.6 MxL2 + 0.8 MxW

Muy2 1.2 MyD + 1.6 MyL2 + 0.8 MyW

2 x (Bt (0.5A - 0.5t) + (0.5At - t2) (0.25A - 0.5t) + (Ct - t2) (0.5A - 0.5C - 0.5t))

mm3 cm3

(At - 2t2) (Cy - 0.5t) + 2 x [Cyt (0.5Cy) + (Bt - Cyt) (0.5B - 0.5Cy)+ (Ct - t2) (B - Cy - 0.5t)]

mm3 cm3

kg/cm2

cm3 kg/cm2

cm3 kg/cm2

cm3 kg/cm2 kg/cm2

cm3 kg/cm2 kg/cm2

1.76 E/fy

fL kg/cm2

+x x

+x +x x

+x x

+

x + x + x

x + x x

xxxx + +

x + x ++ x + +

+ x

+ x

J = =

J = 0.10976 + 0.09103 + 0.025172

J = 0.2259593

h' = A - t = 150 - 2.8 = 147.2 mm

= 0.25 x 48 x 14.72

= 2600.1408

=

= p / 39.4 x 2000000 800000 0.225959 8.44 2

= 98488.366

=

= 4 x 39.4 / 800000 x 0.225959 x 2600.141 48

= 0.0000102936 = 1.029361279377E-05

λr =

λr = 98488.366 / 2500 700 x 1 1 0.0000102936 1800

λr = 143.31977

λx = Lx / ix

λx = 600 cm / 5.91 cm

λx = 101.52284

λy = Ly / iy

λy = 200 cm / 2.39 cm

λy = 83.682008

Mnx = Mpx - (Mpx - Mrx) x (λx-λp) / (λr-λp)= 120000 120000 70920 x 101.52284 49.78032 / 143.3198 49.78032= 92850.78 kgcm

Mny = Mpy - (Mpy - Mry) x (λy-λp) / (λr-λp)= 47500 47500 19800 x 83.682008 49.78032 / 143.3198 49.78032= 37460.633 kgcm

G. Kontrol Tekuk Lateral

= 0.38 m = 38 cm

Lp = = 1.76 x 2.39 2000000 / 2500

Lp = 118.97496 cm ≈ 119 cm

=

= 2.39 x 98488 2500 700 1 1 0.0000102936 1800

= 342.53424 cm ≈ 343 cm

= 38 cm

Lp = 118.97496 cm → LB < Lp → BENTANG PENDEK

= 342.53424 cm

∑ 1/3 bt3 1/3 At3 + 2/3 (B-t) t3 + 2/3 (C-t) t3

cm4



Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2

(x1/(fy-fr)) x 1+ 1+x2fL2

Panjang Pengaku Lateral (LB)

1.76 iy E/fy


LR

LR

LB

LR

2

/xx x

2

- + + x 2

/

- - - -

- - - -

- + + x 2/ x

Mnx = Mp = Zx fyMnx = 48 cm³ x 2500 = 120000 kgcm

Mny = Zy (1 Flens) fyMny =

Mny = 0.25 x 0.28 cm x 6.5 x 2500Mny = 7393.75 kgcm

H. Kekuatan Nominal Momen Lentur (Mn)TEKUK LOKAL TEKUK LATERAL

Mnx = 92850.78 kgcm Mnx = 120000 kgcmMnx = 928.5078 kgm Mnx = 1200 kgm

Mny = 37460.633 kgcm Mny = 7393.75 kgcmMny = 374.60633 kgm Mny = 73.9375 kgm

= 928.5077976 kgm

= 73.9375 kgm

I. Kontrol Persamaan Interaksi Momen Lentur → f = 0.9

Mux+

Muy≤ 1

a). Beban Mati (D) + Beban Hidup (L) Terbagi Rata (Hujan) + Beban Angin (W) <Lentur>

+ ≤ 1

338.7817236 kgm+

12.230761636259 kgm ≤ 1835.6570178 kgm 66.54375 kgm

0.405407621 + 0.1838003064 ≤ 1

0.589207927 ≤ 1 → OK


+ ≤ 1

381.4301439 kgm+

30.25136931012 kgm ≤ 1835.6570178 kgm 66.54375 kgm

0.456443416 + 0.4546087245 ≤ 1

0.911052141 ≤ 1 → OK

J. Kontrol Lendutan

= L/180 = 600 / 180 = 3.333333 cm

Lendutan yang Terjadia). Beban Mati (D) + Beban Hidup (L) Terbagi Rata (Hujan) + Beban Angin (W) <Lentur>

=

kg/cm2

(1/4 x t x B2) fy

cm2 kg/cm2

MnxAktual

MnyAktual

f Mnx f Mny

Mux1 Muy1

f Mnx f Mny

Mux2 Muy2

f Mnx f Mny

Lendutan Ijin fmax

fx 5 / 384 x (qD+qL) x cos α x Lx4 / (EIx)

2

4+x x x / x

= 5/384 0.30 0.27105 0.95106 600 2000000 295

= 1.5486252 cm

=

= 5/384 0.30 0.27105 0.30902 200 2000000 48

= 0.0381784 cm

f =

f = 2.39824 + 0.00146 0.5

f = 1.5490957 cm ≤ = 3.3333333 cm OK


=

= 5/384 0.30 0.95106 600 100 0.9510565 600 48 2000000 295

= 1.5366987 cm

=

= 5/384 0.30 0.30902 200 100 0.309017 200 48 2000000 48

= 0.0736503 cm

f =

f = 2.361443 + 0.00542 0.5

f = 1.5384626 cm ≤ = 3.3333333 cm OK

K. Kontrol Kuat Geser → f = 0.9

Badan (Web) → h = A - 2t = 144.4 mm = 14.44 cm

h/t ≤ → 144.4 / 2.8 ≤ 1100 / 250

→ 51.5714 ≤ 69.57011 → PLASTIS

≤ → ≤ ≤

Vn = 0.6 fy A t

Vn = 0.6 x 2500 x 15 cm x 0.28 cmVn = 6300 kg

a). Beban Mati (D) + Beban Hidup (L) Terbagi Rata (Hujan) + Beban Angin (W) <Lentur>

=

= 0.5 x 29.8258 kg/m x 6 m

= 89.477437 kg

=

= 0.5 x 27.1051 kg/m x 6 m

= 81.315332 kg

=

= 1.2 x 89.4774 kg + 1.6 x 81.31533 kg

= 237.47746 kg

≤ → 237.47745642918 kg ≤ 0.9 x 6300 kg

fx

fx

fy 5 / 384 x (qD+qL) x sin α x Ly4 / (EIy)

fy

fy

(fx2 +fy2)0.5

fmax

fx 5 / 384 x qD x cos α x Lx4 / (EIx) + 1 / 48 x P x cos α x Lx3 / (EIx)

fx

fx

fy 5 / 384 x qD x sin α x Ly4 / (EIy) + 1 / 48 x P x sin α x Ly3 / (EIy)

fy

fy

(fx2 +fy2)0.5

fmax

1100 / fy1/2

Vu f Vn Vu f 0.6 fy Aw → Vu f 0.6 fy A t

kg/cm2

VD1 1/2 x qD x L

VD1

VD1

VL1 1/2 x qL x L

VL1

VL1

Vu1 1.2 VD1 + 1.6 VL1

Vu1

Vu1

Vu1 f Vn

4+x x x / x

4+x x x / x

4x x /xx + x 3 / x

4x x /xx + x 3 / x

→ 237.47745642918 kg ≤ 5670 kg OK


=

= 0.5 x 29.8258 kg/m x 6 m

= 89.477437 kg

= P

= 100 kg

=

= 1.2 x 89.4774 kg + 1.6 x 100 kg

= 267.37292 kg

≤ → 267.372925 kg ≤ 0.9 x 6300 kg

→ 267.372925 kg ≤ 5670 kg OK

Jadi, Gording dengan Profil Light Lip Channel C 150 x 65 x 20 x 2.8 dapat dipakai

VD2 1/2 x qD x L

VD2

VD2

VL2

VL2

Vu2 1.2 VD2 + 1.6 VL2

Vu2

Vu2

Vu2 f Vn

KURANG EKONOMIS PILIH PROFIL LAIN

47.5123

18.71957

PERENCANAAN PENGGANTUNG GORDING Modulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. DATA PENUTUP ATAP DAN PENGGANTUNG GORDINGJenis = Asbes GelombangTebal = 6 mmBerat = 12.5Panjang = 305 cmLebar = 110 cmSudut Kemiringan Atap = 18 °Overlap = 20 cmJarak Gording Diagonal = 142.5 cm = 1.425 m BJ 41Jarak Gording Horizontal = 135.526 cm = 1.3552555 m fu 4100

Bentang Antar Kuda-Kuda = 6 m fy 2500Jumlah Penggantung Gording = 2 BuahJumlah Gording = 11 BuahJarak Penggantung Gording = 2 mSosoran Atap Melintang Horizontal = 2 m

7 m 8 m 4.5 m 5 m

2. PERHITUNGAN PENGGANTUNG GORDINGLuas Horizontal Atap (A) = 13 m x 2 m = 26

A. Beban Mati (D)

=

= 26 x 22.0075= 572.1955 kg

=

= 572.1955 kg x 0.30902= 176.81813 kg

θ =θ = 1.3552555357 m / 2 mθ = 34.122657 °

=

= 176.81813 kg / 0.56097= 315.20272 kg

B. Beban Hidup (L)Beban Terbagi Rata (Hujan)

=

= 26 x 20

kg/cm2 kg/cm2

kg/m2

kg/cm2

kg/cm2

L1 = L2 = H1 = H2 =

m2

qD A x qDGording

m2 kg/m2

ND qD sin α

tan-1 (Jarak Gording Horizontal / Jarak Penggantung Gording)

tan-1

TD ND / sin θ

qL A x qLHujan

m2 kg/m2

= 520 kg

=

= 520 kg x 0.30902= 160.68884 kg

θ =θ = 1.3552555357 m / 2 mθ = 34.122657 °

=

= 160.68884 kg / 0.56097= 286.45002 kg

C. Beban Berfaktor

Pu =

= 1.2 x 315.203 kg + 1.6 x 286.45 kg= 836.56329 kg

D. Perencanaan Dimensi Batang Tarik Penggantung Gording

a). Kontrol Leleh → f = 0.9

Pu ≤836.563295 kg ≤ 0.9 x Ag x 2500

Ag ≥ 0.3718059 → Menentukan

b). Kontrol Putus → f = 0.75 Ae = 0.75 Ag

Pu ≤836.563295 kg ≤ 0.75 x 0.75 Ag x 4100

Ag ≥ 0.3627375 → Tidak Menentukan

Ag Perlu Max = 0.3718059088792

Ag =

d =

d = 4 x 0.371805908879193 / 3.141593 0.5d = 0.6880392330029 cmd = 6.8803923 mm → Pakai Diameter (d) = 10 mm

E. Kontrol Dimensi Batang Tarik Penggantung Gording

Kontrol Kelangsingan → L/d ≤ 500

L = 1.425 + 2 2.455733 m

L/d ≤ 500

2455.733088 mm / 10 mm ≤ 500

245.5733088 ≤ 500 → OK

Jadi, Penggantung Gording dengan Diameter (d) = 10 mm dapat dipakai

NL qL sin α

tan-1 (Jarak Gording Horizontal / Jarak Penggantung Gording)

tan-1

TL ND / sin θ

1.2 TD + 1.6 TL

f Ag fy

cm2 kg/cm2

cm2

f Ae fu

cm2 kg/cm2

cm2

cm2

0.25 x p x d2

(4 x Ag / p )0.5

Panjang Penggantung Gording → 0.5 =2 2

2

- + + x 2

/

- - - -

- - - -

- + + x 2/ x

2

4+x x x / x

4+x x x / x

4x x /xx + x 3 / x

4x x /xx + x 3 / x

KURANG EKONOMIS PILIH PROFIL LAIN

(Batang Bulat)

PERENCANAAN REGEL DINDINGModulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. DATA DINDINGJenis = Seng GelombangTebal = 0.5 mmBerat = 5Jarak Antar Regel = 2.5 m

Bentang Antar Kuda-Kuda = 6 m

Jumlah Penggantung Regel Memanjang = 2 Buah

2. PERHITUNGAN REGEL DINDINGDATA PERENCANAANDirencanakan Menggunakan Profil Light Lip Channel C 150 x 65 x 20 x 2.3

A = 150 mm Cy = 2.12 cm Zx = 39 cm³

B = 65 mm = 248 Zy = 16 cm³

C = 20 mm = 41 BJ 41

t = 2.3 mm = 5.94 cm fu 4100

Ag = 7.01 = 2.42 cm fy 2500

W = 5.5 kg/m = 33 cm³

Cx = 0 cm = 9.4 cm³

A. Beban Mati (D)

Berat Dinding =

= 5 x 1.15 x 2.5 m= 14.375 kg/m

Berat Regel = = 5.5 kg/m

Berat Total = Berat Dinding Seng + Berat Regel= 14.375 kg/m + 5.5 kg/m= 19.875 kg/m

Berat Alat-Alat Pengikat = 10% x Berat Total= 0.1 x 19.875 kg/m= 1.9875 kg/m

21.8625 kg/m

= 0 kgm

=

= 0.125 x 21.8625 kg/m x 2= 10.93125 kgm

B. Beban Hidup (L)

= 0 kg/m

= 0 kgm= 0 kgm


Angin Tekan → = 0.9 → Tekan


kg/cm2 kg/cm

kg/m2

Ix cm⁴

Iy cm⁴

ix kg/cm2

cm² iy kg/cm2

Sx

Sy

WSeng 0.5 mm x Koefisien Impitan x Jarak Antar Regel

kg/m2

WRegel

Beban Mati (qD) =

MxD

MyD 1/8 x qD x Ly2

m2

qL

MxL

MyL

kg/m2

C1

C2

2

=

= 0.9 x 40 x 2.5 m= 90 kg/m

=

= -0.4 x 40 x 2.5 m= -40 kg/m

=

= 0.125 x 90 kg/m x 6= 405 kgm

= 0 kgm

D. Beban Berfaktor

Mux =

= 1.2 0 kgm 1.6 0 kgm 0.8 405 kgm= 324 kgm

Muy =

= 1.2 10.931 kgm 1.6 0 kgm 0.8 0 kgm= 13.1175 kgm

E. Perhitungan Modulus Plastis Zx

Zx =Zx = 2 149.5 73.85 167.2 36.35 40.71 63.85Zx = 2 11040.575 6078.0835 2599.3335Zx = 39435.984 = 39.435984

Zy =Zy = 334.42 20.05 2 48.76 10.6 100.74 21.9 40.71 42.7Zy = 6705.121 2 516.856 2206.206 1736.2815Zy = 15623.808 = 15.623808

F. Kontrol Tekuk Lokal → Tegangan Residu (fr)= 70 MPa = 700Profil Light Lip Channel C diatas tergolong PENAMPANG TIDAK KOMPAK


Mpx = Zx fy

= 39 x 2500= 97500 kgcm

Mpy = Zy fy

= 16 x 2500= 40000 kgcm

Mrx = Sx ( fy - fr)

qW1 C1 x W x Jarak Antar Regel

kg/m2

qW2 C2 x W x Jarak Antar Regel

kg/m2

MxW 1/8 x qW x Lx2

m2

MyW

1.2 MxD + 1.6 MxL + 0.8 MxW

1.2 MyD + 1.6 MyL + 0.8 MyW

2 x (Bt (0.5A - 0.5t) + (0.5At - t2) (0.25A - 0.5t) + (Ct - t2) (0.5A - 0.5C - 0.5t))

mm3 cm3

(At - 2t2) (Cy - 0.5t) + 2 x [Cyt (0.5Cy) + (Bt - Cyt) (0.5B - 0.5Cy)+ (Ct - t2) (B - Cy - 0.5t)]

mm3 cm3

kg/cm2

cm3 kg/cm2

cm3 kg/cm2

2

+x +x x

+x x + x

x + x + x

x + x x

xxxx + +

x + x ++ x + +

= 33 x 2500 - 700= 59400 kgcm

Mry = Sy ( fy - fr)

= 9.4 x 2500 - 700= 16920 kgcm

λp = = 1.76 x 2000000 / 2500

λp = 49.780317

= fy - fr = 2500 - 700 = 1800

Konstanta Puntir Torsi (J)J = =J = 0.060835 + 0.050858 + 0.0144J = 0.1260501

h' = A - t = 150 - 2.3 = 147.7 mm

= 0.25 x 41 x 14.77

= 2236.0672

=

= p / 33 x 2000000 800000 0.13 7.01 2

= 80040.827

=

= 4 x 33 / 8E+05 x 0.126 x 2236 41

= 0.0000233627 = 2.336266435E-05

λr =λr = 80040.827 / 2500 700 x 1 1 2.3363E-05 1800λr = 138.90245

λx = Lx / ixλx = 600 cm / 5.94 cmλx = 101.0101

λy = Ly / iy

λy = 200 cm / 2.42 cm

λy = 82.644628

Mnx = Mpx - (Mpx - Mrx) x (λx-λp) / (λr-λp)= 97500 97500 59400 x 101.0101 49.78 / 139 49.8= 75599.101 kgcm

Mny = Mpy - (Mpy - Mry) x (λy-λp) / (λr-λp)= 40000 40000 16920 x 82.64463 49.78 / 139 49.8= 31489.114 kgcm

G. Kontrol Tekuk Lateral

= 0.5 m = 50 cm

Lp = = 1.76 x 2.42 2000000 / 2500

Lp = 120.46837 cm ≈ 120 cm

cm3 kg/cm2 kg/cm2

cm3 kg/cm2 kg/cm2

1.76 E/fy

fL kg/cm2

∑ 1/3 bt3 1/3 At3 + 2/3 (B-t) t3 + 2/3 (C-t) t3

cm4



Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2

(x1/(fy-fr)) x 1+ 1+x2fL2


1.76 iy E/fy

2

xx x

2

- + + x 2

/

- - - -

- - - -

/

=

= 2.42 x 80040.83 2500 700 1 1 2.3363E-05 1800

= 336.14392 cm ≈ 336 cm

= 50 cm


= 336.14392 cm



Mny = 0.25 x 0.23 cm x 6.5 x 2500Mny = 6073.4375 kgcm

H. Kekuatan Nominal Momen Lentur (Mn)TEKUK LOKAL TEKUK LATERAL



= 755.99101297289 kgm

= 60.734375 kgm

I. Kontrol Persamaan Interaksi Momen Lentur → f = 0.9Mux

+Muy

≤ 1

324 kgm+

13.1175 kgm ≤ 1680.3919117 kgm 54.6609375 kgm

0.476196137 + 0.2399794185747 ≤ 1

0.716175556 ≤ 1 → OK

J. Kontrol Lendutan

Lendutan Ijin = L/180 = 600 / 180 = 3.3333333 cm


y =

y = 5/384 0.90 600 2000000 248

y = 3.061996 cm ≤ = 3.333333333333 cm OK

Jadi, Regel dengan Profil Light Lip Channel C 150 x 65 x 20 x 2.3 dapat dipakai


LR

LR

LB

LR

kg/cm2

(1/4 x t x B2) fy

cm2 kg/cm2

MnxAktual

MnyAktual

f Mnx f Mny

ymax

5 / 384 x qW x L4 / (EIx)

ymax

- + + x 2/ x

2

4x x / x

PERENCANAAN REGEL DINDING


= = 289.8011

= = 22.82313Nu =

39.43598

15.62381

kg/cm2

Mntx 0.75 x (1.2 MxD+1.6 MxL)

Mnty 0.75 x (1.2 MyD+1.6 MyL)


kgm

kgm

PERENCANAAN IKATAN ANGINBeban Angin (W) = 40 BJ 41Koefisien Angin C = 0.9 fu 4100

fy 25001. PERHITUNGAN IKATAN ANGINA. Perhitungan Tinggi Kolom Gevel

= 5 + 0 = 5 + 0 = 5 m

= 5 + 1 x 1.425 x sin 18 = 5 + 0.44 = 5.44 m

= 5 + 2 x 1.425 x sin 18 = 5 + 0.881 = 5.881 m

= 5 + 3 x 1.425 x sin 18 = 5 + 1.321 = 6.321 m

= 5 + 4 x 1.425 x sin 18 = 5 + 1.761 = 6.761 m

= 5 + 5 x 1.425 x sin 18 = 5 + 2.202 = 7.202 m

= 5 + 6 x 1.425 x sin 18 = 5 + 2.642 = 7.642 m

= 5 + 7 x 1.425 x sin 18 = 5 + 3.082 = 8.082 m

= 5 + 11 x tan 18 = 5 + 3.574 = 8.574 m

B. Perhitungan Luas Dinding Pikul

= = 7.07467054

= = 15.9396982

= = 18.3268411

= = 20.7139839

= = 11.2869484

C. Perhitungan Gaya Kerja Pada Ikatan Angin AtapGaya Yang Bekerja Pada Ikatan Angin Atap Akibat Beban Angin

R = 0.5 x C x W x A

= 0.5 x 0.9 x 40 x 7.07 = 127.3440696406 kg

= 0.5 x 0.9 x 40 x 15.9 = 286.9145678354 kg

= 0.5 x 0.9 x 40 x 18.3 = 329.8831392411 kg

= 0.5 x 0.9 x 40 x 20.7 = 372.8517106468 kg

= 0.5 x 0.9 x 40 x 11.3 = 203.1650706763 kg

= 1320.15855804 kg

D. Perhitungan Gaya Batang Ikatan AnginGaya Pada Ikatan Angin (Rangka Batang) --> Metode Keseimbangan Titik BuhulAsumsi Semua Gaya Batang Tarik

RTotal

f = = 25.41 °

Titik Buhul B

∑V =0 → + RTotal = 0

f → = -1320.15855804 Tekan

Titik Buhul A

∑V =0 → + + = 0

2.85 m → 127.34407 + 0.903 -1320.159 = 0

→ 0.9 = 1320.1586 + -127.3441

kg/m2

kg/cm2

kg/cm2

h1

h2

h3

h4

h5

h6

h7

h8

h9

A1 0.5 x (h1+h2) x 1.425 x cos 18° m2

A2 0.5 x (h2+h4) x 2 x 1.425 x cos 18° m2

A3 0.5 x (h4+h6) x 2 x 1.425 x cos 18° m2

A4 0.5 x (h6+h8) x 2 x 1.425 x cos 18° m2

A5 0.5 x (h8+h9) x 1.425 x cos 18° m2

R1

R2

R3

R4

R5

RTotal

tan-1(2.85/6.0)

V1 S1 V1

V1 kg →

R1 R2 R1 S1 cosf V1

S1 +

S1

6.0

m

A

B

→ = 1320.5405 Tarik

E. Perencanaan Dimensi Batang Tarik Ikatan Angin

Pu = = 1320.540457002 Tarik

a). Kontrol Leleh → f = 0.9

Pu ≤1320.5405 kg ≤ 0.9 x Ag x ###

Ag ≥ 0.58690687 → Menentukan

b). Kontrol Putus → f = 0.75 Ae = 0.75 Ag (Batang Bulat)

Pu ≤1320.5405 kg ≤ 0.75 x 0.75 Ag x ###

Ag ≥ 0.573 → Tidak Menentukan

Ag Perlu Max = 0.586907

Ag =

d =

d = 4 x 0.586906869779 / 3.14 0.5d = 0.864449556 cmd = 8.644 mm → Pakai Diameter (d) = 16 mm

F. Kontrol Dimensi Batang Tarik Ikatan Angin

Kontrol Kelangsingan → L/d ≤ 500

L = 2.85 + 6 0.5 = 6.6424769 m

L/d ≤ 500

6642.4769 mm / 16 mm ≤ 500

415.15481 ≤ 500 → OK

Jadi, Ikatan Angin dengan Diameter (d) = 16 mm dapat dipakai

S1 kg →

S1 kg →

f Ag fy

cm2 kg/cm2

cm2

f Ae fu

cm2 kg/cm2

cm2

cm2

0.25 x p x d2

(4 x Ag / p )0.5

Panjang Ikatan Angin →22

PERENCANAAN KOLOM GEVEL7 m 8 m 4.5 m 5

Modulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. MENENTUKAN KEBUTUHAN MOMEN INERSIA IxJarak Antar Kuda-Kuda (b) = 6 mBerat Penutup Atap (Asbes Gelombang) = 12.5Berat Gording = 6.63 kg/mBerat Dinding Seng Gelombang = 5Jarak Antar Regel = 2.5 mJumlah Penggantung Regel Melintang = 1 Buah

= 3 Buah

A. Beban Angin (W) C = 0.9 W = 40

= Bentang Gevel Max x C x W

= 2.8782988 m x 0.9 x 40= 103.61876 kg/m = 1.0361875823 kg/cm

h = Tinggi Kolom Gevel Max (Tengah) = 8.5741 m

=

= 0.125 x 103.62 kg/m x 8.574117= 952.1978 kgm

Lendutan Ijin (Syarat Kekakuan)

= h / 200

= 857.41167 cm / 200= 4.2870583 cm


y =

4.2870583 = 5 / 384 x 1.03618758 x 857.41 / 2000000 / Ix

Ix ≥ 850.44323

2. PERHITUNGAN KOLOM GEVELDATA PERENCANAANDirencanakan Menggunakan Profil Light Channel C 250 x 50 x 50 x 4

A = 250 mm Cy = 0.88 cm Zx = 108

B = 50 mm = 1050 Zy = 14

C = 0 mm = 23.3 BJ 41

t = 4 mm = 8.81 cm fu 4100

Ag = 13.47 = 1.32 cm fy 2500

W = 10.6 kg/m = 83.7 cm³

Cx = 0 cm = 5.66 cm³

A. Beban Mati (D)

Berat Sendiri Gevel =

= 10.6 kg/m x 8.5741 m = 90.8856 kg

Berat Dinding =

= 5 x 2.878298839603 m x 8.574= 123.3944 kg

Berat Regel Horizontal =

= 10.6 kg/m x 2.878298839603 m x 3

L1 = L2 = H1 = H2 =

kg/cm2

kg/m2

kg/m2

Jumlah Regel Horizontal (H2 = 5 m)

kg/m2

qw

kg/m2

Mw 0.125 x qw x h2

m2

ymax

5 / 384 x qw x h4 / (EIx)

cm4

Ix cm⁴

Iy cm⁴

ix kg/cm2

cm² iy kg/cm2

Sx

Sy

WGevel x Tinggi Kolom Gevel Max

WSeng 0.5mm x Bentang Gevel Max x Tinggi Kolom Gevel Max

kg/m2

WRegel x Bentang Gevel Max x Jumlah Regel

4

2

= 91.5299031 kg

Berat Total = Berat Sendiri Gevel + Berat Dinding + Berat Regel Horizontal= 90.8856 kg + 123.3944 kg + 91.5299030994= 305.8099 kg

Berat Alat-Alat Pengikat = 10% x Berat Total = 0.1 x 305.8099 kg = 30.580989 kg

= 336.3909 kg

B. Beban Hidup (L)

= 0 kg

C. Beban Angin (W)

= Bentang Gevel Max x C x W

= 2.8782988 m x 0.9 x 40= 103.61876 kg/m = 1.0361875823 kg/cm

=

= 0.125 x 103.62 kg/m x 8.574117= 952.1978 kgm

D. Beban Berfaktora). Beban Aksial Berfaktor

Nu =

= 1.2 x 336.39 kg + 1.6 x 0 kg= 403.66905 kg

b). Beban Momen Berfaktor Mux (Mntx) =

= 0.8 x 952.2 kgm= 761.75824 kgm

Muy (Mnty) = 0 kgm

E. Kontrol Koloma). Kontrol Penampang (Tekuk Lokal / Kelangsingan Elemen Penampang)Profil Light Channel C diatas tergolong PENAMPANG TIDAK KOMPAK

MASIH BOLEH DIPAKAI SEBAGAI KOLOM

b). Kontrol Kelangsingan Komponen Struktur (Tekuk Struktural) <Tidak Bergoyang>

Tekuk Terhadap Sumbu x → Faktor Tekuk kx = 1

λx = kx Lkx / ix → Panjang Tekuk Lkx = 8.574117 m= 1 x 857.41 cm / 8.81 cm= 97.32255 → Tidak Menentukan

Tekuk Terhadap Sumbu y → Faktor Tekuk ky = 1

λy = ky Lky / iy → Panjang Tekuk Lky = 2.5 m= 1 x 250 cm / 1.32 cm= 189.39394 → Menentukan

= 189.39394 ≤ 200 → OK

Beban Mati (ND)

Beban Hidup (NL)

qw

kg/m2

Mw 0.125 x qw x h2

m2

1.2 ND + 1.6 NL

1.2 MD + 1.6 ML + 0.8 MW

λmax

2

c). Kontrol Kekuatan → f = 0.85Parameter Kelangsingan

λc =

= 189.39394 / 3.1416 x 2500 / 2000000= 2.1314307 → λc ≥ 1.2 → ω=1.25λc2

→ Kolom Panjang (Tekuk Elastis)Faktor Tekuk (ω)

ω == 8.3167846

ω == 5.6787459

= 5.6787459

Kuat Nominal KolomNn =

= 13.47 x 2500 / 5.67874589077= 5930.0065 kg

= 403.66905 kg / 0.85 / 5930.006 kg

= 0.080085 ≤ 0.2 → Pakai Rumus Interaksi Beam Column 2

Rumus Interaksi Beam ColumnNu / 2φcNn + (Mux / φbMnx + Muy / φbMny)

F. Kontrol Baloka). Kolom Tidak Bergoyang (Gaya Tekan Kritis Euler Untuk Elemen Tidak Bergoyang <Ncrb>)


λx = kx Lkx / ix → Panjang Tekuk Lkx = 8.574117 m= 1 x 857.41 cm / 8.81 cm= 97.32255

Ncrbx =

= p x 2000000 x 13.47 /

= 28071.808 kg


λy = ky Lky / iy → Panjang Tekuk Lky = 2.5 m= 1 x 250 cm / 1.32 cm= 189.39394

Ncrby =

= p x 2000000 x 13.47 /

= 7412.5081 kg

b). Amplifikasi Momen Balok → Mu =

Cm = 1 (Komponen Struktur Tak Bergoyang, Beban Transversal, Ujung Sederhana)

= Cm / (1-(Nu/Ncrb)) → ≥ 1

Faktor Amplifikasi Momen Terhadap Sumbu x

λmax/p/ fy/E

kg/cm2 kg/cm2

1.43 / (1.6-0.67λc)

1.25 λc2

ωAktual

Ag fcr = Ag fy/ωcm2 kg/cm2

Nu / fNn

p2 E Ag / λx2

kg/cm2 cm2

p2 E Ag / λy2

kg/cm2 cm2

db Mnt

db db

2

2

= Cmx / (1-(Nu/Ncrbx))

= 1 / 1 - 403.67 kg / 28071.8 kg

= 1.0145897 ≥ 1 → = 1.01458967163

Faktor Amplifikasi Momen Terhadap Sumbu y

= Cmy / (1-(Nu/Ncrby))

= 1 / 1 - 403.67 kg / 7412.51 kg

= 1.0575943 ≥ 1 → = 1.05759428217

Momen Berfaktor Terhadap Sumbu x

Mux =

= 1.0145897 x 761.76 kgm= 772.87204 kgm

Momen Berfaktor Terhadap Sumbu y

Muy =

= 1.0575943 x 0 kgm= 0 kgm

c). Perhitungan Modulus Plastis Zx

Zx =Zx = 2 200 123 484 60.5Zx = 2 24600 29282Zx = 107764 = 107.764

Zy =Zy = 968 6.8 2 35.2 4.4 164.8 20.6Zy = 6582.4 2 154.88 3394.88Zy = 13681.92 = 13.68192

d). Kontrol Tekuk Lokal → Tegangan Residu (fr)= 70 MPa = 700

Profil Light Channel C diatas tergolong PENAMPANG TIDAK KOMPAK


Mpx = Zx fy

= 108 x 2500= 270000 kgcm

Mpy = Zy fy

= 14 x 2500= 35000 kgcm

Mrx = Sx ( fy - fr)

= 83.7 x 2500 - 700= 150660 kgcm

dbx

dbx

dby

dby

dbx Mntx

dby Mnty

2 x (Bt (0.5A - 0.5t) + (0.5At - t2) (0.25A - 0.5t))

mm3 cm3

(At - 2t2) (Cy - 0.5t) + 2 x [Cyt (0.5Cy) + (Bt - Cyt) (0.5B - 0.5Cy)]

mm3 cm3

cm3 kg/cm2

cm3 kg/cm2

cm3 kg/cm2 kg/cm2

x + x

x + x

xxxx +

x + x+ x +

Mry = Sy ( fy - fr)

= 5.66 x 2500 - 700= 10188 kgcm

λp = 1.76 E/fy = 1.76 x 2000000 / 2500

λp = 49.780317

= fy - fr = 2500 - 700 = 1800

Konstanta Puntir Torsi (J)J = =J = 0.5333333 + 0.1963J = 0.7296

h' = A - t = 250 - 4 = 246

= 0.25 x 23.3 x 24.6

= 3525.057

=

= p / 83.7 x 2000000 800000 0.73

= 105243.58

=

= 4 x 83.7 / 8E+05 x 0.73 x 3525

= 0.0000124443 = 1.24443296E-05

λr =

λr = 105243.58 / 2500 700 x 1 1 1.244433E-05

λr = 159.35294

λx = Lx / ix

λx = 857.41167 cm / 8.81 cm

λx = 97.32255

λy = Ly / iy

λy = 250 cm / 1.32 cm

λy = 189.39394

Mnx = Mpx - (Mpx - Mrx) x (λx-λp) / (λr-λp)= 270000 270000 150660 x 97.32255 49.78 / 159.4= 218219.82 kgcm

Mny = Mpy - (Mpy - Mry) x (λy-λp) / (λr-λp)= 35000 35000 10188 x 189.3939 49.78 / 159.4= 3385.4122 kgcm

e). Kontrol Tekuk Lateral

= 2.5 m = 250 cm

Lp = = 1.76 x 1.32 2000000 /

Lp = 65.710019 cm ≈ 66 cm

=

cm3 kg/cm2 kg/cm2

fL kg/cm2

∑ 1/3 bt3 1/3 At3 + 2/3 (B-t) t3

cm4



Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2

(x1/(fy-fr)) x 1+ 1+x2fL2


1.76 iy E/fy


2

xx x

2

- + + x

/

- - - -

- - - -

- + + x/

= 1.32 x 1E+05 2500 700 1 1 0.0000124443

= 210.34588 cm ≈ 210 cm

= 250 cm

Lp = 65.710019 cm → LB > LR → BENTANG PANJANG

= 210.34588 cm

f). Kekuatan Nominal Momen Lentur (Mn)

Mn = Mcr = 82.8950065806 kg/m

= 355.3757284171 kg

= 76175.82382673 kgcm =

Mx =

L = 8.5741167 m

= 2.5 m + 0.625 m = 3.125 m

= 2.5 m + 1.25 m = 3.75 m

= 2.5 m + 1.875 m = 4.375 m

= = 761.75823827 kgm

1110.54915 - 404.76077432 = 705.788376984 kgm

1332.65898 - 582.85551502 = 749.803466544 kgm

1554.76881 - 793.33111767 = 761.437694159 kgm

= 9521.978 / 1904 2117 2999 2284.313 = 1.02329 ≤


= = 83.7 x 2500 - 700

= 150660 kgcm = 1506.6 kgm

Mn =

Mn = 1.0232868 x 1506.6 2700 1506.6 210.3459 250 210.346 65.71

Mn = 1206.8759 kgm ≤ 2700 kgm

Mcr =

Mcr = 0.012859 27199488000000 + 631654681.6697 23.3

Mcr = 114350.94 kgcm = 1143.5094373 kgm ≤ 2700 kgm


Mnx = 218219.82 kgcm Mnx = McrMnx = 2182.1982 kgm Mnx = 1143.50943729 kgm

= 1143.50943729337 kgm

G. Kontrol Interaksi Beam Column → = 0.85 =

Nu+

Mux+

Muy≤ 1

LR

LR

LB

LR

qu

qu Vu

Mumax

Vux - 0.5qux2

XA

XB

XC

Mmax 0.125 x qu x L2

MA (x =3.125 m) =

MB (x =3.750 m) =

MC (x =4.375 m) =


Cb

cm3 kg/cm2


MR



MnxAktual

fc fb

- + + x/

+ ++

+ - x - -/

x x

+ + ≤ 1

403.6690545689 kg+

772.87204082 kgm+ 0 ≤ 1

10081.0110368 kg 1029.1584936 kgm

0.040042516876 + 0.750974748452541 + 0 ≤ 1

0.791017265329 ≤ 1 → OK

H. Kontrol Lendutan

Lendutan Ijin = h/200 = 857.41 / 200 = 4.287


y =

y = 5/384 1.04 857.41 2000000 1050

y = 3.4722855 cm ≤ = 4.287058329281 cm OK

I. Kontrol Kuat Geser → f = 0.9

Badan (Web) → h = A - 2t = 242 mm = 24.2

h/t ≤ → 242 / 4 ≤ 1100 / 250

→ 60.5 ≤ 69.57 → PLASTIS

≤ → ≤ ≤

Vn = 0.6 fy A t

Vn = 0.6 x 2500 x 25 cm x 0.4Vn = 15000 kg

≤ → 355.37572842 kg ≤ 0.9 x 15000

→ 355.37572842 kg ≤ 13500 kg OK

Jadi, Gevel dengan Profil Light Channel C 250 x 50 x 50 x 4 dapat dipakai

2fc Nn fb Mnx fb Mny

ymax

5 / 384 x qw x h4 / (EIx)

ymax

1100 / fy1/2


kg/cm2

Vu f Vn

4x x / x

PERENCANAAN KOLOM GEVELm

800000

cm³ 107.76

cm³ 13.682

m

Buah

kg/cm2

kg/cm2

kg/cm2

97.32

189

(Komponen Struktur Tak Bergoyang, Beban Transversal, Ujung Sederhana)

kg/cm2

2

2

GAMBAR BELUM BENAR

kg/cm2

mm

13.47 2

23.3

1800

49.78

49.78

2500

x

2

/

-

-

/

x 2

1800

762 kgm

2.3

≤ 2700

3525

TEKUK LATERAL

0.9

kg/cm2

x 2

x

cm

cm

cm

kg

f 0.6 fy A t

PERENCANAAN GORDING IKATAN ANGIN Modulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. DATA PENUTUP ATAPJenis = Asbes GelombangTebal = 6 mmBerat = 12.5Panjang = 305 cmLebar = 110 cmSudut Kemiringan Atap = 18 °Overlap = 20 cmJarak Gording Diagonal = 142.5 cm = 1.425 mJarak Gording Horizontal = 135.53 cm = 1.355256 mBentang Antar Kuda-Kuda = 6 mJumlah Penggantung Gording = 2 Buah2. PERHITUNGAN GORDING IKATAN ANGINDATA PERENCANAANDirencanakan Menggunakan Profil Double Light Lip Channel Back To Back Arrangement C 150 x 130 x 20 x 2.3

A = 150 mm Cy = 0 cm Zx = 79 cm³

B = 130 mm = 496 Zy = 31 cm³

C = 20 mm = 145 BJ 41

t = 2.3 mm = 5.95 cm fu 4100

Ag = 14.02 = 3.22 cm fy 2500

W = 11 kg/m = 66.1 cm³

Cx = 0 cm = 22.3 cm³

A. Beban Mati (D)

Berat Gording = = 11 kg/m

Berat Asbes =

= 12.5 x 1.15 x 1.425 m= 20.484375 kg/m

Berat Total = Berat Gording + Berat Asbes= 11 kg/m + 20.48437 kg/m= 31.484375 kg/m


34.6328125 kg/m

kg/cm2 kg/cm2

kg/m2

Ix cm⁴

Iy cm⁴

ix kg/cm2

cm² iy kg/cm2

Sx

Sy

WGording


kg/m2

Beban Mati (qD) =

A

C

=

= 0.125 x 34.6328125 kg/m x 0.951057 x 6= 148.21993 kgm

=

= 0.125 x 34.6328125 kg/m x 0.309017 x 2= 5.3510638 kgm


q = ≤ 20 = 40 - 0.8 x 18

q = 25.6 ≤ 20 → q = 20


= 20 x 1.3552555 m = 27.10511 kg/m

=

= 0.125 x 27.1051107144119 kg/m x 0.951057 x 6= 116.00321 kgm

=

= 0.125 x 27.1051107144119 kg/m x 0.309017 x 2= 4.1879699 kgm


=

= 0.25 x 100 kg x 0.951057 x 6 m= 142.65848 kgm

=

= 0.25 x 100 kg x 0.309017 x 2 m= 15.45085 kgm


Angin Tekan → = = 0.02 x 18 - 0.4

= -0.04 → Hisap


=


=


= → = 0 kgm

= 0.125 x 0 kg/m x 6= 0 kgm

MxD 1/8 x qD cosα x L2

m2

MyD 1/8 x qD sinα x (L/3)2

m2

40 - 0.8α kg/m2

kg/m2 kg/m2 kg/m2

qL

qL kg/m2

MxL1 1/8 x qL cosα x L2

m2

MyL1 1/8 x qL sinα x (L/3)2

m2

MxL2 1/4 x P cosα x L

MyL2 1/4 x P sinα x (L/3)

kg/m2

C1 0.02α-0.4

C1

C2


kg/m2


kg/m2

MxW 1/8 x qw1 x L2 MyW

m2

2

2

2

2

2


=

= 1.2 148 kgm 1.6 116.0032 kgm 0.8 0 kgm= 363.46906 kgm

=

= 1.2 5.35 kgm 1.6 4.18797 kgm 0.8 0 kgm= 13.122028 kgm


=

= 1.2 148 kgm 1.6 142.6585 kgm 0.8 0 kgm= 406.11748 kgm

=

= 1.2 5.35 kgm 1.6 15.45085 kgm 0.8 0 kgm= 31.142636 kgm

c). Beban Aksial (N) Dari Ikatan Angin

N = P = = 1320.1586 kg

d). Beban Berfaktor Untuk Kontrol Beam Column

= 363.46906 kgm

= 13.122028 kgm

= 406.11748 kgm

= 31.142636 kgm

Nu = 1056.1268 kg

E. Kontrol Koloma). Kontrol Penampang (Tekuk Lokal / Kelangsingan Elemen Penampang)Profil Double Light Lip Channel Back to Back Arrangement C diatas tergolong PENAMPANG TIDAK KOMPAK

MASIH BISA BERPERILAKU SEBAGAI KOLOM



λx = kx Lkx / ix → Panjang Tekuk Lkx = 6 m= 1 x 600 cm / 5.95 cm= 100.84034 → Menentukan


λy = ky Lky / iy → Panjang Tekuk Lky = 2 m= 1 x 200 cm / 3.22 cm= 62.111801 → Tidak Menentukan

= 100.84034 ≤ 200 → OK


λc =

Mux1 1.2 MxD + 1.6 MxL1 + 0.8 MxW

Muy1 1.2 MyD + 1.6 MyL1 + 0.8 MyW

Mux2 1.2 MxD + 1.6 MxL2 + 0.8 MxW

Muy2 1.2 MyD + 1.6 MyL2 + 0.8 MyW

RTotal Ikatan Angin

Mntx1

Mnty1

Mntx2

Mnty2

λmax

λmax/p/ fy/E

+x +x x

+x x

+x +x x

+x x

+x +x x

+ x

+ x

= 100.84034 / 3.1416 x 2500 / 2000000= 1.1348525 → 0.25 < λc < 1.2 → ω=1.43 / (1.6-0.67λc)

→ Kolom Menengah (Tekuk Inelastis)

Faktor Tekuk (ω)ω =

= 1.703093

ω == 1.6098627

= 1.703093


= 14.02 x 2500 / 1.703093= 20580.204 kg

= 1056.1268 kg / 0.85 / 20580.2 kg





λx = kx Lkx / ix → Panjang Tekuk Lkx = 6 m= 1 x 600 cm / 5.95 cm= 100.84034

Ncrbx =

= p x 2000000 x 14.02 / 100.8403

= 27215.053 kg


λy = ky Lky / iy → Panjang Tekuk Lky = 2 m= 1 x 200 cm / 3.22 cm= 62.111801

Ncrby =

= p x 2000000 x 14.02 / 62.1118

= 71734.736 kg


Cm = 1 (Komponen Struktur Tak Bergoyang, Beban Transversal, Ujung Sederhana)

= Cm / (1-(Nu/Ncrb)) → ≥ 1



= 1 / 1 - 1056.127 kg / 27215.05 kg

= 1.0403735 ≥ 1 → = 1.0403734785528

kg/cm2 kg/cm2

1.43 / (1.6-0.67λc)

1.25 λc2

ωAktual


Nu / fNn

p2 E Ag / λx2

kg/cm2 cm2

p2 E Ag / λy2

kg/cm2 cm2

db Mnt

db db

dbx

dbx

22

22



= 1 / 1 - 1056.127 kg / 71734.74 kg

= 1.0149427 ≥ 1 → = 1.01494266586666


=

= 1.0403735 x 363.47 kgm= 378.14357 kgm

=

= 1.0403735 x 406.12 kgm= 422.51385 kgm


=

= 1.0149427 x 13.122 kgm= 13.318107 kgm

=

= 1.0149427 x 31.143 kgm= 31.60799 kgm

c). Perhitungan Modulus Plastis Zx

Zx =Zx = 4 149.5 73.85 167.21 36.35 40.71 63.85Zx = 78871.968 = 78.871968

Zy =Zy = 4 149.5 32.5 167.21 1.15 40.71 63.85Zy = 30601.5 = 30.6015

d). Kontrol Tekuk Lokal → Tegangan Residu (fr)= 70 MPa = 700Profil Double Light Lip Channel Back to Back Arrangement C diatas tergolong PENAMPANG TIDAK KOMPAK


Mpx = Zx fy

= 79 x 2500= 197500 kgcm

Mpy = Zy fy

= 31 x 2500= 77500 kgcm

Mrx = Sx ( fy - fr)

= 66.1 x 2500 - 700= 118980 kgcm

Mry = Sy ( fy - fr)

= 22.3 x 2500 - 700

= 40140 kgcm

λp = 1.76 E/fy = 1.76 x 2000000 / 2500

λp = 49.780317

dby

dby

Mux1 dbx Mntx1

Mux2 dbx Mntx2

Muy1 dby Mnty1

Muy2 dby Mnty2

4 x (0.5Bt (0.5A - 0.5t) + (0.5At - t2) (0.25A - 0.5t) + (Ct - t2) (0.5A - 0.5C - 0.5t))

mm3 cm3

4 x (0.5Bt (0.25B) + (0.5At - t2) (0.5t) + (Ct - t2) (0.5B - 0.5t))

mm3 cm3

kg/cm2

cm3 kg/cm2

cm3 kg/cm2

cm3 kg/cm2 kg/cm2

cm3 kg/cm2 kg/cm2

x + x + x

x x x+

x

x +

= fy - fr = 2500 - 700 = 1800

Konstanta Puntir Torsi (J)J = =J = 2 x 0.0608 + 0.050858 + 0.014357J = 0.2521002

h' = A - t = 150 - 2.3 = 147.7 mm

= 0.25 x 145 x 14.77

= 7908.0426

=

= p / 66.1 2000000 800000 0.2521 14.02 2

= 79919.736

=

= 4 x 66.1 / 800000 x 0.2521 x 7908 145

= 0.0000234335 = 2.34335139321E-05

λr =

λr = 79919.736 / 2500 700 x 1 1 2.3433513932E-05 1800

λr = 138.78535

λx = Lx / ix

λx = 600 cm / 5.95 cm

λx = 100.84034

λy = Ly / iy

λy = 200 cm / 3.22 cm

λy = 62.111801

Mnx = Mpx - (Mpx - Mrx) x (λx-λp) / (λr-λp)= 197500 197500 118980 x 100.8403 49.78032 / 138.785 49.78032= 152454.99 kgcm

Mny = Mpy - (Mpy - Mry) x (λy-λp) / (λr-λp)= 77500 77500 40140 x 62.1118 49.78032 / 138.785 49.78032= 72323.841 kgcm

e). Kontrol Tekuk Lateral

= 0.38 m = 38 cm

Lp = = 1.76 x 3.22 2000000 / 2500

Lp = 160.29262 cm ≈ 160 cm

=

= 3.22 x 79920 2500 700 1 1 0.0000234335 1800

= 446.88883 cm ≈ 447 cm

= 38 cm


fL kg/cm2

∑ 1/3 bt3 2 x (1/3 At3 + 2/3 (0.5B-t) t3 + 2/3 (C-t) t3)

cm4



Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2

(x1/(fy-fr)) x 1+ 1+x2fL2


1.76 iy E/fy


LR

LR

LB

2

xx x

2

- + + x 2

/

- - - -

- - - -

/

- + + x 2/

x

= 446.88883 cm



Mny = 0.25 x 0.23 cm x 13 x 2500Mny = 24293.75 kgcm

f). Kekuatan Nominal Momen Lentur (Mn)TEKUK LOKAL TEKUK LATERAL



= 1524.549891 kgm

= 242.9375 kgm

= 0.85 = 0.9

Nu+

Mux+

Muy≤ 1

a). Beban Mati (D) + Beban Hidup (L) Terbagi Rata (Hujan) + Beban Angin (W)

Nu+ + ≤ 1

1056.12684643 kg+

378.143568690035 kgm+

13.3181065380775 kgm ≤34986.3463379 kg 1372.0949021386 kgm 218.64375 kgm

0.03018682878 + 0.275595783 + 0.060912358748318 ≤ 1

0.36669497037 ≤ 1 → OK

b). Beban Mati (D) + Beban Hidup (L) Terpusat (Manusia) + Beban Angin (W)Nu

+ + ≤ 1

1056.12684643 kg+

422.51385405981 kgm+

31.6079901311238 kgm ≤34986.3463379 kg 1372.0949021386 kgm 218.64375 kgm

0.03018682878 + 0.307933404 + 0.144563885915439 ≤ 1

0.48268411886 ≤ 1 → OK

J. Kontrol Lendutan

= L/180 = 600 / 180 = 3.33333 cm

Lendutan yang Terjadia). Beban Mati (D) + Beban Hidup (L) Terbagi Rata (Hujan) + Beban Angin (W)

=

= 5/384 0.35 0.2711 0.9510565 600 2000000 496

= 0.9988274 cm

LR

kg/cm2

(1/4 x t x B2) fy

cm2 kg/cm2

MnxAktual

MnyAktual

G. Kontrol Interaksi Beam Column → fc fb


Mux1 Muy1


Mux2 Muy2


Lendutan Ijin → fmax

fx 5 / 384 x (qD+qL) x cos α x Lx4 / (EIx)

fx

fx

2

4+x x x / x

=

= 5/384 0.35 0.2711 0.309017 200 2000000 145

= 0.0137055 cm

f =

f = 0.9976562 + 0.0002 0.5

f = 0.9989214 cm ≤ = 3.333333 cm OK

b). Beban Mati (D) + Beban Hidup (L) Terpusat (Manusia) + Beban Angin (W)

=

= 5/384 0.35 0.9511 600 100 0.951057 600 48 2E+06 496

= 0.991734 cm

=

= 5/384 0.35 0.309 200 100 0.309017 200 48 2E+06 145

= 0.0254479 cm

f =

f = 0.9835364 + 0.0006 0.5

f = 0.9920605 cm ≤ = 3.333333 cm OK

K. Kontrol Kuat Geser → f = 0.9 t = 2 x Tebal Badan Profil Light Lip Channel C

Badan (Web) → h = A - 2t = 145.4 mm = 14.54 cm

h/t ≤ → 145.4 / 4.6 ≤ 1100 / 250

→ 31.609 ≤ 69.57011 → PLASTIS

≤ → ≤ ≤

Vn = 0.6 fy A t

Vn = 0.6 x 2500 x 15 cm x 0.46 cmVn = 10350 kg

a). Beban Mati (D) + Beban Hidup (L) Terbagi Rata (Hujan) + Beban Angin (W)

=

= 0.5 x 34.633 kg/m x 6 m

= 103.89844 kg

=

= 0.5 x 27.105 kg/m x 6 m

= 81.315332 kg

=

= 1.2 x 103.9 kg + 1.6 x 81.31533 kg

= 254.78266 kg

≤ → 254.782656429177 kg ≤ 0.9 x 10350 kg

→ 254.782656429177 kg ≤ 9315 kg OK

b). Beban Mati (D) + Beban Hidup (L) Terpusat (Manusia) + Beban Angin (W)

=

fy 5 / 384 x (qD+qL) x sin α x Ly4 / (EIy)

fy

fy

(fx2 +fy2)0.5

fmax

fx 5 / 384 x qD x cos α x Lx4 / (EIx) + 1 / 48 x P x cos α x Lx3 / (EIx)

fx

fx

fy 5 / 384 x qD x sin α x Ly4 / (EIy) + 1 / 48 x P x sin α x Ly3 / (EIy)

fy

fy

(fx2 +fy2)0.5

fmax

1100 / fy1/2


kg/cm2

VD1 1/2 x qD x L

VD1

VD1

VL1 1/2 x qL x L

VL1

VL1

Vu1 1.2 VD1 + 1.6 VL1

Vu1

Vu1

Vu1 f Vn

VD2 1/2 x qD x L

4+x x x / x

4x x /xx + x 3 / x

4x x /xx + x 3 / x

= 0.5 x 34.633 kg/m x 6 m

= 103.89844 kg

= P

= 100 kg

=

= 1.2 x 103.9 kg + 1.6 x 100 kg

= 284.67813 kg

≤ → 284.678125 kg ≤ 0.9 x 10350 kg

→ 284.678125 kg ≤ 9315 kg OK

Jadi, Gording Ikatan Angin dengan Profil Double Light Lip Channel Back to Back Arrangement C 150 x 130 x 20 x 2.3 dapat dipakai

VD2

VD2

VL2

VL2

Vu2 1.2 VD2 + 1.6 VL2

Vu2

Vu2

Vu2 f Vn

78.87197

30.6015

Jadi, Gording Ikatan Angin dengan Profil Double Light Lip Channel Back to Back Arrangement C 150 x 130 x 20 x 2.3 dapat dipakai

PERENCANAAN PELAT LANTAI BONDEX1. PERHITUNGAN PELAT LANTAI BONDEX A → = 7 m

A. Beban Mati (D)Spesi 2 cm = = 0.02 m x 2100 = 42

Tegel 2 cm = = 0.02 m x 2400 = 48

Plafond = 11


= 108

B. Beban Hidup (L)


C. Beban Berguna

= + = 108 + 400 = 508 ≈ 500

Bentang Antar Balok Anak = 2.33 m ≈ 2.50 m → Tanpa Penyangga

Dari Tabel Perencanaan Praktis Pelat Bondex, untuk: Tebal Pelat Bondex = 0.75 mmBentang = 2.50 m Tanpa Penyangga → Bentang Menerus Dengan Tulangan NegatifBeban Berguna = 500 , didapat: Tebal Pelat = 9 cm

Spasi Tulangan Max = 2 Tebal Pelat = 180 mm Luas Tulangan Negatif = 2.86

8 mm

= = 0.25 x p x 8 x 8 = 50.27

= 0.503


Jumlah Tulangan Perlu n = As Negatif Perlu / As 8

Jumlah Tulangan Perlu n = 2.86 / 0.503 = 5.69 Buah/m → 6 Buah/mJarak Antar Tulangan S = 1 m / 6 = 0.16667 m = 167 mm ≈Jarak Antar Tulangan S ≈ 160 mm

→ 1000 / 160 x 0.503 ≥ 2.86

→ 3.142 ≥ 2.86

2. PERHITUNGAN PELAT LANTAI BONDEX B → = 8 m

A. Beban Mati (D)Spesi 2 cm = = 0.02 m x 2100 = 42

Tegel 2 cm = = 0.02 m x 2400 = 48

Plafond = 11


= 108

B. Beban Hidup (L)


C. Beban Berguna

= + = 108 + 400 = 508 ≈ 500

Bentang Antar Balok Anak = 2.0 m → Tanpa Penyangga

Dari Tabel Perencanaan Praktis Pelat Bondex, untuk: Tebal Pelat Bondex = 0.75 mmBentang = 2.00 m Tanpa Penyangga → Bentang Menerus Dengan Tulangan NegatifBeban Berguna = 500 , didapat: Tebal Pelat = 9 cm

Bentang Melintang L1



kg/m2

kg/m2

qD kg/m2

qL kg/m2


kg/m2

cm2/m


As 1 f 12 0.25 x p x f 2

As 1 f 12 cm2

cm2/m

1 f


cm2 cm2


Bentang Melintang L2



kg/m2

kg/m2

qD kg/m2

qL kg/m2


kg/m2

Spasi Tulangan Max = 2 Tebal Pelat = 180 mm Luas Tulangan Negatif = 1.79

8 mm

= = 0.25 x p x 8 x 8 = 50.27

= 0.503


Jumlah Tulangan Perlu n = As Negatif Perlu / As 8

Jumlah Tulangan Perlu n = 1.79 / 0.503 = 3.56 Buah/m → 4 Buah/mJarak Antar Tulangan S = 1 m / 4 = 0.25 m = 250 mm ≈Jarak Antar Tulangan S ≈ 180 mm

→ 1000 / 180 x 0.503 ≥ 1.79

→ 2.793 ≥ 1.79

cm2/m


As 1 f 12 0.25 x p x f 2

As 1 f 12 cm2

cm2/m

1 f


cm2 cm2


PERENCANAAN PELAT LANTAI BONDEX

+

160 mm

+

mm2

250 mm

mm2

PERENCANAAN BALOK ANAKPanjang Balok Anak (L) = 6 mModulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. MENENTUKAN KEBUTUHAN MOMEN INERSIA Ix DAN MODULUS PLASTIS ZxA. Berat Pelat LantaiPelat Bondex = 10.1

Pelat Beton = 0.09 m x 2400 = 216

Spesi 2 cm = 0.02 m x 2100 = 42

Tegel 2 cm = 0.02 m x 2400 = 48

Plafond = 11


= 334.1

B. Beban Mati (D) → Bentang Balok Anak Max = 2.33 mPerkiraan Berat Sendiri Balok Anak = 60.00 kg/mBerat Pelat Lantai = 334.1 x 2.3333 m = 779.57 kg/m +Berat Total = 839.57 kg/mBerat Alat-Alat Pengikat 10 % = 0.1 x 60.00 kg/m = 6 kg/m +

= 845.57 kg/m

C. Beban Hidup (L)

400 x 2.33 m = 933.3 kg/m


1.2 + 1.6

1.2 x 845.57 kg/m + 1.6 x 933.33 kg/m

1014.68 kg/m + 1493 kg/m 2508 kg/m

E. Kontrol Lendutan Ijin

= L/360 = 600 / 360 = 1.66667 cm


y =

1.66667 = 5/384 x 8.46 + 9.3333 x 600 / 2E+06 x IxIx ≥ 9005.6813

F. Beban Berfaktor dan Kontrol Kuat Rencana Momen Lentur → f = 0.9

L = 6 m

= 2508.0133 kg/m

= 0.5 x 2508 kg/m x 6 m

= 7524.04 kg

= 0.125 x 2508 kg/m x 6

= 11286.06 kgm

Mp = Zx x fy = 2500 Zx → Misal: Mn = 0.9 Mp = 2250 Zx≤

1128606 kgcm ≤ 0.9 x 2250 Zx

kg/cm2 kg/cm2

kg/m2

kg/m3 kg/m2

kg/m3 kg/m2

kg/m3 kg/m2

kg/m2

kg/m2

qLantai kg/m2

kg/m2

qD


qu = qD qL

qu =

qu = =

Lendutan Ijin ymax

5 / 384 x (qD+qL) x L4 / (EIx)

cm4

qu

qu

Vu 1/2 x qu x L =

Vu


Mumax

Mumax f Mn

kg/cm2 x cm3

4

2

Zx ≥ 557.3363

2. PERHITUNGAN BALOK ANAK → Profile Code = 26 → WF 300 x 200 x 8 x 12DATA PERENCANAANDirencanakan Menggunakan Profil WF 300 x 200 x 8 x 12

W = 56.8 kg/m h = = 160 cm³

d = 294 mm = 234 mm = 823 cm³ 822.6

= 200 mm = 11300 = 244 cm³ 244.32

= 8 mm = 1600 BJ 41

= 12 mm = 12.5 cm fu 4100

r = 18 mm = 4.71 cm fy 2500

Ag = 72.38 = 771 cm³

A. Berat Pelat LantaiPelat Bondex = 10.1

Pelat Beton = 0.09 m x 2400 = 216

Spesi 2 cm = 0.02 m x 2100 = 42

Tegel 2 cm = 0.02 m x 2400 = 48

Plafond = 11


= 334.1

B. Beban Mati (D)Berat Sendiri Balok Anak = 56.80 kg/mBerat Pelat Lantai = 334.1 x 2.3333 m = 779.57 kg/m +Berat Total = 836.37 kg/mBerat Alat-Alat Pengikat 10 % = 0.1 x 56.80 kg/m = 5.68 kg/m +

= 842.05 kg/m

C. Beban Hidup (L)

400 x 2.33 m = 933.3 kg/m


1.2 + 1.6

1.2 x 842.05 kg/m + 1.6 x 933.33 kg/m

1010.456 kg/m + 1493 kg/m 2503.8 kg/m

L = 6 m

= 2503.7893 kg/m

= 0.5 x 2504 kg/m x 6 m

= 7511.368 kg

= 0.125 x 2504 kg/m x 6

= 11267.052 kgm = 1126705.2 kgcm

cm3

d - 2(tf+r) Sy

Zx

bf Ix cm⁴ Zy

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx

kg/m2

kg/m3 kg/m2

kg/m3 kg/m2

kg/m3 kg/m2

kg/m2

kg/m2

qLantai kg/m2

kg/m2

qD


qu = qD qL

qu =

qu = =

qu

qu

Vu 1/2 x qu x L =

Vu


Mumax

2

E. Kontrol Penampang (Tekuk Lokal)Badan (Web)

≤ → 234 / 8 ≤ 1680 / 250

→ 29.25 ≤ 106.25 → OK

Sayap (Flange) 1≤ → 200 / 24 ≤ 170 / 250

→ 8.3333 ≤ 10.752 → OK


F. Kontrol Tekul Lateral → Tegangan Residu (fr)= 70 MPa = 700Bentang Antar Balok Anak = 2.3333 m

= 6 m = 600 cm

Lp = = 1.76 x 4.71 cm x 2E+06 / 2500

Lp = 234.46529 cm ≈ 234 cm

= fy - fr = 2500 - 700 = 1800


J = =

J = 0.3333333 x 27 x 0.8 + 0.67 x 20 x 1.2J = 4.608 + 23.04 = 27.648

h' = d - tf = 294 - 12 = 282 mm

= 0.25 x 1600 x 28.2

= 318096

=

= p / 771 x 2000000 800000 27.648 72.38 2

= 163035.3

=

= 4 x 771 / 8E+05 x 27.6 x 318096 / 1600

= 0.0000009663 = 9.662724E-07

=

= 4.71 x 163035 / 2500 700 x 1 1 1E-06 1800

= 742.89009 cm ≈ 743 cm

= 600 cm


= 742.89009 cm

G. Kekuatan Nominal Momen Lentur (Mn)


= 7511.368 kg

= 1126705.2 kgcm

= 11267.052 kgm

Mx =

L = 6 m

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

kg/cm2

kg/cm2


1.76 iy E/fy

fL kg/cm2


3

cm4 cm4 cm4



Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2


LR

LR

LB

LR

qu

qu Vu

Mumax

Vux - 0.5qux2

3 3

2

/xx x

2

- + + x 2

= 1.5 m

= 3 m

= 4.5 m

= = 11267.052 kgm

11267.1 - 2816.763 = 8450.289 kgm

22534.1 - 11267.052 = 11267.052 kgm

33801.2 - 25350.867 = 8450.289 kgm

= 140838.15 / 28168 25351 45068 25351 = 1.1364 ≤ 2.3


= = 771 x 2500 - 700

= 1387800 kgcm = 13878 kgm

Mn =

Mn = 1.1363636 x 13878 20575 13878 742.9 600 742.89 234.465 ≤ 20575

Mn = 17909.269 kgm ≤ 20575 kgm

Mcr =

Mcr = 0.00595 70778880000000000 + 109662271 1600 318096

Mcr = 2116990.4 kgcm = 21169.903975 kgm ≤ 20575 kgm


Mn = 2057500 kgcm Mn = Cb[MR + (Mp-MR) x (LR-LB) / (LR-Lp)Mn = 20575 kgm Mn = 17909.26852 kgm

= 17909.2685168645 kgm

H. Kontrol Kuat Momen Lentur → f = 0.9

≤ → 11267.052 kgm ≤ 0.9 x 17909.2685169 kgm

→ 11267.052 kgm ≤ 16118.34167 kgm OK

I. Kontrol Lendutan

= L/360 = 600 / 360 = 1.66667 cm


y =

y = 5/384 x 8.42 + 9.3333 x 600 / 2E+06 x 11300

y = 1.3256433 cm ≤ = 1.66666667 cm OK

J. Kontrol Kuat Geser → f = 0.9

Badan (Web)≤ → 234 / 8 ≤ 1100 / 250

→ 29.25 ≤ 69.57 → PLASTIS

≤ → ≤ → ≤→ 7511.368 kg ≤ 0.9 0.6 2500 29.4 0.8

XA

XB

XC

Mmax 0.125 x qu x L2

MA (x =1.5 m) =

MB (x =3.0 m) =

MC (x =4.5 m) =


Cb

cm3 kg/cm2


MR



MnAktual

Mumax f Mn

Lendutan Ijin ymax

5 / 384 x (qD+qL) x L4 / (EIx)

ymax

h/tw 1100 / fy1/2


+ ++

+ - x - -/

x x x x

x x

4

→ 7511.368 kg ≤ 31752 kg OK

Jadi, Balok Anak dengan Profil WF 300 x 200 x 8 x 12 dapat dipakai

x x x x

PEMBEBANAN PORTAL GEDUNGJarak Antar Kuda-Kuda (b) = 6 m Jumlah Lapangan Memanjang (n) =Modulus Elastisitas (E) = 2E+06 Modulus Geser (G) = 8000001. PEMBEBANAN PORTAL MELINTANG (TENGAH)A. Beban Atap

Berat Gording =

= 8.815 kg/m / 1.355 m x 6 m= 39.0259 kg/m

Berat Asbes =

= 12.5 x 1.15 x 6 m / cos

= 90.6886 kg/m

Berat Kuda-Kuda WF 400 x 200 x 8 x 13 (Perkiraan) → W = 66 kg/m

Berat Kuda-Kuda =

= 66 kg/m / cos 18= 69.3965 kg/m

Berat Total = Berat Gording + Berat Asbes + Berat Kuda-Kuda= 39.0259 kg/m + 90.69 kg/m + 69.4 kg/m= 199.1109741946 kg/m


= 219.0220716141 kg/m

Beban Hujan q = 20Beban Hidup (qL) = q x Bentang Antar Kuda-Kuda

= 20 x 6 m= 120 kg/m

c). Beban Angin (W) → Tekanan Tiup Angin (W) Dekat Pantai = 40

Angin Tekan → = = 0.02 x 18 -

= -0.04 → Hisap


=

= -0.04 x 40 x 6 m

= -9.6 kg/m → Di Pihak Angin

=

= -0.4 x 40 x 6 m= -96 kg/m → Di Belakang Angin

B. Beban Lantai → Berat per satuan Luas Pelat Bondex = 10.1a). Beban Terbagi Rata Mati Lantai

Bentang Melintang = 7 m

= Berat Profil Balok Induk Melintang Tepi WF 400 x 200 x 8 x 13 (Perkiraan)

kg/cm2 kg/cm2

a). Beban Mati Atap (qD)

WGording / Jarak Horizontal Gording x Bentang Antar Kuda-Kuda

WAsbes x Koefisien Impitan x Bentang Antar Kuda-Kuda / cos αkg/m2

WKuda-Kuda / cos α

Beban Mati (qD)

b). Beban Hidup Atap (qL)

kg/m2

kg/m2

kg/m2

C1 0.02α-0.4

C1

C2

qW1 C1 x W x Bentang Antar Kuda-Kuda

kg/m2


kg/m2

kg/m2

L1

qD

= 66 kg/m


= Berat Profil Balok Induk Melintang Tengah WF 600 x 200 x 11 x 17 (Perkiraan)

= 106 kg/m→ Koefisien Reduksi Beban Hidup (r) Gudang

b). Beban Terpusat Lantai → Untuk Perencanaan Balok Induk dan Portal & Gempa (r) =

→ = 7 m

→ = 2.333

Beban Pelat Bondex = Jarak Antar Balok Anak x Berat per satuan Luas Pelat Bondex

= 2.33333 m x 10.1

= 23.5667 kg/m

Beban Pelat Beton = Jarak Antar Balok Anak x Tebal Pelat Beton x Berat Volume Beton

= 2.33333 m x 0.09 m x 2400= 504 kg/m

Beban Spesi 2 cm = Jarak Antar Balok Anak x Tebal Spesi x Berat Volume Spesi= 2.33333 m x 0.02 m x 2100= 98 kg/m

Beban Tegel 2 cm = Jarak Antar Balok Anak x Tebal Tegel x Berat Volume Tegel Beton= 2.33333 m x 0.02 m x 2400= 112 kg/m

Beban Plafond + Penggantung Plafond = Jarak Antar Balok Anak x Beban Hidup Plafond= 2.333 m x 11 + 7= 42 kg/m

Beban Profil Balok Anak WF 300 x 200 x 8 x 12 = 56.8 kg/m

= 23.57 + 504 + 98 + 112 + 42 +

= 836.4 kg/m

= qTotal + Berat Alat Pengikat 10 % x Beban Profil Balok Anak

= 836.4 kg/m + 0.1 x 56.8 kg/m

= 842 kg/m

=

= 0.5 x 842 kg/m x 6 m x 2

= 5052 kg

=

= 2.333 m x 400 x 0.8

= 746.7 kg/m

=

= 0.5 x 746.67 kg/m x 6 m x 2

= 4480 kg

qD

L2

qD

qD

I. Beban Terpusat P1 Balok Anak pada Bentang Melintang L1

i). P1 Mati (P1D) Jarak Antar Balok Anak pada Bentang Melintang L1

kg/m2

kg/m3

kg/m3

kg/m3

kg/m2

qTotal

qTotal

q1D

q1D

q1D

P1D 1/2 x q1D x L x 2

P1D

P1D

ii). P1 Hidup (P1L)

q1L Jarak Antar Balok Anak x Beban Hidup Lantai Gudang x Koefisien Reduksi Beban Hidup Gudang

q1L kg/m2

q1L

P1L 1/2 x q1L x L x 2

P1L

P1L

→ = 8 m

→ = 2


= 2 m x 10.1

= 20.2 kg/m


= 2 m x 0.09 m x 2400= 432 kg/m

Beban Spesi 2 cm = Jarak Antar Balok Anak x Tebal Spesi x Berat Volume Spesi= 2 m x 0.02 m x 2100= 84 kg/m

Beban Tegel 2 cm = Jarak Antar Balok Anak x Tebal Tegel x Berat Volume Tegel Beton= 2 m x 0.02 m x 2400= 96 kg/m

Beban Plafond + Penggantung Plafond = Jarak Antar Balok Anak x Beban Hidup Plafond= 2 m x 11 + 7= 36 kg/m


= 20.2 + 432 + 84 + 96 + 36 +

= 725 kg/m


= 725 kg/m + 0.1 x 56.8 kg/m

= 730.68 kg/m

=

= 0.5 x 730.68 kg/m x 6 m x 2

= 4384 kg

=

= 2 m x 400 x 0.8

= 640 kg/m

=

= 0.5 x 640 kg/m x 6 m x 2

= 3840 kg

→

→ Bentang Pelat Yang Dipikul Balok Induk Memanjang Tengah =

Beban Pelat Bondex = Bentang Pelat Yang Dipikul x Berat per satuan Luas Pelat Bondex

= 2.16667 m x 10.1

II. Beban Terpusat P2 Balok Anak pada Bentang Melintang L2

i). P2 Mati (P2D) Jarak Antar Balok Anak pada Bentang Melintang L2

kg/m2

kg/m3

kg/m3

kg/m3

kg/m2

qTotal

qTotal

q2D

q2D

q2D

P2D 1/2 x q2D x L x 2

P2D

P2D

ii). P2 Hidup (P2L)

q2L Jarak Antar Balok Anak x Beban Hidup Lantai Gudang x Koefisien Reduksi Beban Hidup Gudang

q2L kg/m2

q2L

P2L 1/2 x q2L x L x 2

P2L

P2L

III. Beban Terpusat P3 Balok Induk Memanjang Tengah antara Bentang Melintang L1 dan L2

i). P3 Mati (P3D)

kg/m2

= 21.8833 kg/m

Beban Pelat Beton = Bentang Pelat Yang Dipikul x Tebal Pelat Beton x Berat Volume Beton

= 2.16667 m x 0.09 m x 2400= 468 kg/m

Beban Spesi 2 cm = Bentang Pelat Yang Dipikul x Tebal Spesi x Berat Volume Spesi= 2.16667 m x 0.02 m x 2100= 91 kg/m

Beban Tegel 2 cm = Bentang Pelat Yang Dipikul x Tebal Tegel x Berat Volume Tegel Beton= 2.16667 m x 0.02 m x 2400= 104 kg/m

Beban Plafond + Penggantung Plafond = Bentang Pelat Yang Dipikul x Beban Hidup Plafond= 2.167 m x 11 + 7= 39 kg/m

Beban Profil Balok Induk Memanjang WF 300 x 200 x 8 x 12 = 56.8 kg/m

= 21.88 + 468 + 91 + 104 + 39 +

= 780.7 kg/m

= qTotal + Berat Alat Pengikat 10 % x Beban Profil Balok Induk Memanjang

= 780.7 kg/m + 0.1 x 56.8 kg/m

= 786.4 kg/m

=

= 0.5 x 786.36 kg/m x 6 m x 2

= 4718 kg

=

= 2.167 m x 400 x 0.8

= 693.3 kg/m

=

= 0.5 x 693.33 kg/m x 6 m x 2

= 4160 kg

→ 7

→ Bentang Pelat Yang Dipikul Balok Induk Memanjang Tepi = 1.167


= 1.16667 m x 10.1

= 11.7833 kg/m


= 1.16667 m x 0.09 m x 2400= 252 kg/m


kg/m3

kg/m3

kg/m3

kg/m2

qTotal

qTotal

q3D

q3D

q3D

P3D 1/2 x q3D x L x 2

P3D

P3D

ii). P3 Hidup (P3L)

q3L Bentang Pelat Yang Dipikul x Beban Hidup Lantai Gudang x Koefisien Reduksi Beban Hidup Gudang

q3L kg/m2

q3L

P3L 1/2 x q3L x L x 2

P3L

P3L

IV. Beban Terpusat P4 Balok Induk Memanjang Tepi pada Bentang Melintang L1 =

i). P4 Mati (P4D)

kg/m2

kg/m3

kg/m3



Beban Profil Balok Induk Memanjang WF 300 x 200 x 8 x 12 = 56.8 kg/m

= 11.78 + 252 + 49 + 56 + 21 +

= 446.6 kg/m


= 446.6 kg/m + 0.1 x 56.8 kg/m

= 452.3 kg/m

=

= 0.5 x 452.26 kg/m x 6 m x 2

= 2714 kg

=

= 1.167 m x 400 x 0.8

= 373.3 kg/m

=

= 0.5 x 373.33 kg/m x 6 m x 2

= 2240 kg

C. Beban Angin Dinding → Tekanan Tiup Angin (W) Dekat Pantai = 40

= → = 0.9

= 0.9 x 40 x 6 m

= 216 kg/m

= → = -0.4

= -0.4 x 40 x 6 m

= -96 kg/m

D. Beban Horizontal Akibat Gempa (E) Arah Melintanga). Beban Atap → Bentang Atap Horizontal Melintang = 26 m

= 219.022071614 kg/m

=

= 219 kg/m x 26 m

= 5695 kg

= 120 kg/m

kg/m3

kg/m2

qTotal

qTotal

q4D

q4D

q4D

P4D 1/2 x q4D x L x 2

P4D

P4D

ii). P4 Hidup (P4L)

q4L Bentang Pelat Yang Dipikul x Beban Hidup Lantai Gudang x Koefisien Reduksi Beban Hidup Gudang

q4L kg/m2

q4L

P4L 1/2 x q4L x L x 2

P4L

P4L

kg/m2

qW1 C1 x W x Jarak Antar Kuda-Kuda C1

qW1 kg/m2

qW1


qW2 kg/m2

qW2

qDAtap

W1 qDAtap x Bentang Atap Horizontal

W1

W1

qLAtap

=

= 120 kg/m x 26 m

= 3120 kg

Berat Kolom → Coba Profil Kolom WF 350 x 350 x 13 x 13 → W = 106

= Berat Kolom x (1/2 x Tinggi Kolom Lantai 1) x 2 Kolom (Kanan Kiri)

= 106 kg/m x 2.5 m x 2 Kolom

= 530 kg

= + +

= 5695 kg + 3120 kg + 530 kg

= 9345 kg

b). Beban Lantai 1 → Bentang Antar Kolom Utama = 22 m

Berat Profil Balok Induk Melintang Tepi WF 400 x 200 x 8 x 13 (Perkiraan) = 66 kg/mBerat Profil Balok Induk Melintang Tengah WF 600 x 200 x 11 x 17 (Perkiraan) = 106 kg/m

=

= 66 kg/m x 14 m + 106 kg/m x 8

= 1772 kg

4.5 m 5Berat Kolom → Coba Profil Kolom WF 350 x 350 x 13 x 13 → W = 106

= Berat Kolom x (1/2x(Tinggi Kolom <Lantai Dasar+Lantai 1>)) x 2 Kolom (Kanan Kiri)

= 106 kg/m x 4.75 m x 2 Kolom

= 1007 kg

Berat Balok dan Pelat Memanjang

=

= 4 5052.28 4480 3 4384 3840 2 4718 4160 2714

= 38129 kg + 24672 kg + 27664 kg

= 90464.88 kg

= + +

= 1772 kg + 1007 kg + 90465 kg

= 93243.88 kg

c). Berat Total Gedung

=

= 9345 kg + 93243.88 kg

= 102588.4539 kg

Wihi

Atap →→ 9345 kg x 9.5 m = 88773

Lantai 1 →→ 93243.88 kg x 4.5 m = 419597.46

∑Wihi =

∑Wihi = 88773 + 419597.46∑Wihi = 508370.911689

W2 qLAtap x Bentang Atap Horizontal

W2

W2

W3

W3

W3

Berat Atap Total (WAtapTotal) W1 W2 W3

Berat Atap Total (WAtapTotal)


→ 2L1 + L2

W1 Berat Balok Induk Melintang Tepi x 2 x L1 + Berat Balok Induk Melintang Tengah x L2

W1

W1

Tinggi Kolom Lantai Dasar H1 = Tinggi Kolom Lantai 1 H2 =

W2

W2

W2

W3 4 (P1D+P1L) + 3 (P2D+P2L) + 2 (P3D+P3L) + 2 (P4D+P4L)

W3

W3

W3

Berat Lantai 1 Total (WLantai1Total) W1 W2 W3

Berat Lantai 1 Total (WLantai1Total)

Berat Lantai 1 Total (WLantai1Total)

Berat Total Gedung (WGedungTotal) WAtapTotal + WLantaiTotal

Berat Total Gedung (WGedungTotal)


WAtapTotal x Tinggi Kolom Total

WLantai1Total x Tinggi Kolom Lantai Dasar

WihiAtap + WihiLantai1

x + ++ ++x x+ +

d). Waktu Getar Alami Fundamental Struktur Gedung (T) → z = 0.19 →

T = → n = Jumlah Tingkat Struktur Gedung

T = 0.19 x 2 TingkatT = 0.38 detik

e). Gaya Geser Dasar → Surabaya → Zona Gempa 2, Tanah Lunak

V =

C = Respons Spektrum Gempa Rencana = 0.5I = Faktor Keutamaan Bangunan = 1 → Gedung Perniagaan

R = Faktor Reduksi Gempa Maksimum = 4.5 → SRPMB Baja

= Berat Struktur Bangunan = 102588.45386 kgV = 0.5 x 1 x 102588.45386 kg / 4.5V = 11399 kg

f). Gaya Geser Tiap LantaiFi =

Atap → = 88773 / 508370.91169 x 11399 kg

→ = 1990 kg

Lantai 1 → = 419597.46 / 508370.91169 x 11399 kg

→ = 9408 kg

2. PEMBEBANAN PORTAL MEMANJANG (TEPI)A. Beban Atap → Sosoran Atap Melintang dan Memanjang = 2 ma). Beban Terpusat Atap → Jumlah Gording Total = 22 Buah

Berat Gording =

= 8.815 kg/m x 5 m x 11 Buah= 484.825 kg

Berat Asbes =

= 12.5 x 1.15 x 65 / cos= 982.46 kg


Berat Kuda-Kuda =

= 66 kg/m / cos 18 x 13 m

= 902.155 kg

Berat Total = Berat Gording + Berat Asbes + Berat Kuda-Kuda= 484.825 kg + 982 kg + 902.2 kg= 2369.439604169 kg

Berat Alat-Alat Pengikat = 10% x Berat Total= 0.1 x 2369.439604 kg = 236.943960417 kg

= 2606.38356459 kg

Beban Hujan q = 20

= Beban Hujan q x Luas Atap Pengaruh

= 20 x 65

z n

CIWt / R

Wt

(Wihi / ∑Wihi) x V

F1

F1

F2

F2

I. Beban Terpusat P5 (Kolom Tepi)

i). P5 Mati (P5D)

WGording x Bentang Gording x Jumlah Gording 1/2 Lapangan

WAsbes x Koefisien Impitan x Luas Asbes Horizontal / cos αkg/m2 m2

WKuda-Kuda / cos α x Panjang Kuda-Kuda 1/2 Lapangan

P5D

ii). P5 Hidup (P5L)

kg/m2

P5L

P5L kg/m2 m2

= 1300 kg

Berat Gording =

= 8.815 kg/m x 6 m x 11 Buah= 581.79 kg

Berat Asbes =

= 12.5 x 1.15 x 78 / cos= 1179 kg


Berat Kuda-Kuda =

= 66 kg/m / cos 18 x 13 m= 902.155 kg



= 2929.18626806 kg

Beban Hujan q = 20


= 20 x 78

= 1560 kg

B. Beban Lantai → Berat per satuan Luas Pelat Bondex = 10.1Koefisien Reduksi Beban Hidup (r) Gudang

I. Beban Mati (D) → Untuk Perencanaan Balok Induk dan Portal & Gempa (r) =

7 m

Bentang Pelat Yang Dipikul Balok Induk Memanjang Tepi = 1.1667 m

= Berat Profil Balok Induk Memanjang WF 300 x 200 x 8 x 12 (Perkiraan)

= 56.8 kg/m


= 1.16667 m x 10.1

= 11.7833 kg/m


= 1.16667 m x 0.09 m x 2400= 252 kg/m


P5L

II. Beban Terpusat P6 (Kolom Tengah)

i). P6 Mati (P6D)


WAsbes x Koefisien Impitan x Luas Asbes / cos αkg/m2 m2


P6D

ii). P6 Hidup (P6L)

kg/m2

P6L

P6L kg/m2 m2

P6L

kg/m2

a). Beban Terbagi Rata Lantai →

Balok Induk Memanjang Tepi pada Bentang Melintang L1 =

qProfil

qProfil

kg/m2

kg/m3

kg/m3



= 11.78 + 252 + 49 + 56 + 21 +

= 446.6 kg/m


= 446.6 kg/m + 0.1 x 56.8 kg/m

= 452.3 kg/mII. Beban Hidup (L)

=

= 1.167 m x 400 x 0.8

= 373.3 kg/m

b). Beban Terpusat Lantai



= 66 kg/m

= 5052 kg = 4480 kg

→ Balok Anak Yang Menumpu Hanya 1 Sisi Saja

=

= 0.5 x 66 kg/m x 7 m + 2526.14 kg

= 231 kg + 2526 kg

= 2757 kg

=

= 0.5 x 4480 kg

= 2240 kg

→ Balok Anak Yang Menumpu Berasal Dari 2 Sisi (Kanan Kiri)

=

= 0.5 x 66 kg/m x 7 m + 5052.28 kg

= 231 kg + 5052 kg

= 5283 kg

=

= 1 x 4480 kg

kg/m3

kg/m2

qTotal

qTotal

qD

qD

qD

qL Bentang Pelat Yang Dipikul x Beban Hidup Lantai Gudang x Koefisien Reduksi Beban Hidup Gudang

qL kg/m2

qL

L1

qD

qD

Beban Terpusat dari Balok Anak pada Bentang Melintang Tepi L1 (Portal Melintang Tengah)

P1D P1L


i). P7 Mati (P7D) → Beban Terpusat Balok Anak Yang Masuk Kolom Portal Memanjang Tinggal 1/2 PTotal

P7D 1/2 x qD x L1 + 0.5 P1D

P7D

P7D

P7D

ii). P7 Hidup (P7L)

P7L 0.5 P1L

P7L

P7L


i). P8 Mati (P8D) → Beban Terpusat Balok Anak Yang Masuk Kolom Portal Memanjang adalah 1 PTotal

P8D 1/2 x qD x L1 + P1D

P8D

P8D

P8D

ii). P8 Hidup (P8L)

P8L 1 P1L

P8L

= 4480 kg


= =

= 0.9 x 40 x 3.5 m

= 126 kg/m

= =

= -0.4 x 40 x 3.5 m

= -56 kg/m

D. Beban Horizontal Akibat Gempa (E) Arah Memanjanga). Beban Atap → Bentang Atap Horizontal Memanjang = 52 m

= 56.8 kg/m → Berat Profil Balok Induk Memanjang WF 300 x 200 x 8 x 12

= +

= 56.8 kg/m x 52 m + 25717 kg

= 28671 kg

=

= 2 x 1300 kg + 7 x 1560 kg

= 13520 kg


= Berat Kolom x (1/2 x Tinggi Kolom Lantai 1) x Jumlah Kolom

= 106 kg/m x 2.5 m x 9 Kolom

= 2385 kg

= + +

= 28671 kg + 13520 kg + 2385 kg

= 44576 kg

b). Beban Lantai → Bentang Antar Kolom Utama = 48 mBerat Sendiri Balok Induk Memanjang WF 300 x 200 x 8 x 12 = 56.8 kg/m

=

= 56.8 kg/m x 48 m

= 2726.4 kg


= Berat Kolom x (1/2x(Tinggi Kolom <Lantai Dasar+Lantai 1>)) x Jumlah Kolom

= 106 kg/m x 4.75 m x 9 Kolom

= 4531.5 kg

Berat Balok dan Pelat Melintang

=

= 2 2757.14 2240 7 5283 4480

P8L

kg/m2

qW1 C1 x W x 1/2 Jarak Melintang Antar Kolom Utama (L1) → C1

qW1 kg/m2

qW1


qW2 kg/m2

qW2

qDAtap

W1 qDAtap x Bentang Atap Horizontal ∑PD (2P5D + 7P6D)

W1

W1

W2 ∑PL (2P5L + 7P6L)

W2

W2

W3

W3

W3




→ n x b

W1 W x Bentang Antar Kolom Utama

W1

W1

W2

W2

W2

W3 2 (P7D+P7L) + 7 (P8D+P8L)

W3 x + +x+

= 9994 kg + 68342.96 kg

= 78337.24 kg

= + +

= 2726.4 kg + 4531.5 kg + 78337 kg

= 85595.14 kg


=

= 44576 kg + 85595.14 kg

= 130170.811 kg

Wihi

Atap →→ 44576 kg x 9.5 m = 423468.8746 kg

Lantai 1 →→ 85595.14 kg x 4.5 m = 385178.13 kg

∑Wihi =

∑Wihi = 423468.874553 kg + 385178.13 kg∑Wihi = 808647.004553 kg





V =



= Berat Struktur Bangunan = 130170.81101 kg

V = 0.5 x 1 x 130170.81101 kg / 4.5V = 14463 kg


Atap → = 423468.87455 / 808647.00455 x 14463 kg

→ = 7574 kg

Lantai 1 → = 385178.13 / 808647.00455 x 14463 kg

→ = 6889 kg

3. PEMBEBANAN PORTAL MEMANJANG (TENGAH) → Tanpa Beban AtapA. Beban Lantai → Berat per satuan Luas Pelat Bondex = 10.1

Koefisien Reduksi Beban Hidup (r) GudangI. Beban Mati (D) → Untuk Perencanaan Balok Induk dan Portal & Gempa (r) =Balok Induk Memanjang Tengah pada Bentang Melintang Gabungan = 7.5 m

W3

W3

Berat Lantai Total (WLantaiTotal) W1 W2 W3

Berat Lantai Total (WLantaiTotal)






WLantaiTotal x Tinggi Kolom Lantai Dasar


z n

CIWt / R

Wt


F1

F1

F2

F2

kg/m2

a). Beban Terbagi Rata Lantai →

x + +x+

Bentang Pelat Yang Dipikul Balok Induk Memanjang Tengah = 2.1667 m


= 56.8 kg/m


= 2.16667 m x 10.1

= 21.8833 kg/m


= 2.16667 m x 0.09 m x 2400= 468 kg/m




= 21.88 + 468 + 91 + 104 + 39 +

= 780.7 kg/m


= 780.7 kg/m + 0.1 x 56.8 kg/m

= 786.4 kg/mII. Beban Hidup (L)

=

= 2.167 m x 400 x 0.8

= 693.3 kg/m


Bentang Melintang = 7 m = 8 m


= 66 kg/m

= Berat Profil Balok Induk Melintang Tengah WF 600 x 200 x 11 x 17 (Perkiraan)

= 106 kg/m

= 5052 kg = 4480 kg

= 4384 kg = 3840 kg

→ Balok Anak Yang Menumpu Hanya 1 Sisi Saja

qProfil

qProfil

kg/m2

kg/m3

kg/m3

kg/m3

kg/m2

qTotal

qTotal

qD

qD

qD

qL Bentang Pelat Yang Dipikul x Beban Hidup Lantai Gudang x Koefisien Reduksi Beban Hidup Gudang

qL kg/m2

qL

L1 L2

qDTepi

qDTepi

qDTengah

qDTengah

Beban Terpusat dari Balok Anak pada Bentang Melintang Tepi L1 (Portal Melintang Tengah)

P1D P1L

Beban Terpusat dari Balok Anak pada Bentang Melintang Tengah L2 (Portal Melintang Tengah)

P2D P2L


=

= 0.5 66 7 106 8 + 2526 + 3288.06

= 655 kg + 2526 kg + 3288 kg

= 6469.2 kg

=

= 0.5 x 4480 kg + 0.75 x 3840 kg

= 5120 kg

→ Balok Anak Yang Menumpu Berasal Dari 2 Sisi (Kanan Kiri)

=

= 0.5 66 7 106 8 + 5052 + 6576.12

= 655 kg + 5052 kg + 6576 kg

= 12283 kg

=

= 1 x 4480 kg + 1.5 x 3840 kg

= 10240 kg


= = 0.9

= 0.9 x 40 x 7.5 m

= 270 kg/m

= = -0.4

= -0.4 x 40 x 7.5 m

= -120 kg/m

D. Beban Horizontal Akibat Gempa (E) Arah Memanjanga). Beban Lantai → Bentang Antar Kolom Utama = 48 m

Berat Sendiri Balok Induk Memanjang WF 300 x 200 x 8 x 12 = 56.8 kg/m

=

= 56.8 kg/m x 48 m

= 2726.4 kg


= Berat Kolom x (1/2x(Tinggi Kolom <Lantai Dasar+Lantai 1>)) x Jumlah Kolom

= 106 kg/m x 4.75 m x 9 Kolom

= 4531.5 kg


=

i). P9 Mati (P9D) → Beban Terpusat Balok Anak Yang Masuk Kolom Portal Memanjang Tinggal 1/2 PTotal

P9D 1/2 x qDTepi x L1 + 1/2 qDTengah x L2 + 0.5 P1D + 0.75 P2D

P9D

P9D

P9D

ii). P9 Hidup (P9L)

P9L 0.5 P1L + 0.75 P2L

P9L

P9L


i). P10 Mati (P10D) → Beban Terpusat Balok Anak Yang Masuk Kolom Portal Memanjang adalah 1 PTotal

P10D 1/2 x qDTepi x L1 + 1/2 qDTengah x L2 + 1 P1D + 1.5 P2D

P10D

P10D

P10D

ii). P10 Hidup (P10L)

P10L 1 P1L + 1.5 P2L

P10L

P10L

kg/m2

qW1 C1 x W x 1/2 Jarak Melintang Antar Kolom Utama → C1

qW1 kg/m2

qW1

qW2 C2 x W x 1/2 Jarak Melintang Antar Kolom Utama → C2

qW2 kg/m2

qW2

→ n x b


W1

W1

W2

W2

W2

W3 2 (P9D+P9L) + 7 (P10D+P10L)

x + +x+

x x x+

x x x+

= 2 6469.2 5120 7 ### 10240

= 23178 kg + 157663.8 kg

= 180842.2 kg

= + +

= 2726.4 kg + 4531.5 kg + 180842 kg

= 188100.1 kg


=

= 188100.1 kg

Wihi

Lantai 1 →→ 188100.1 kg x 4.5 m = 846450.45 kg

∑Wihi =

∑Wihi = 846450.45 kg





V =




V = 0.5 x 1 x 188100.1 kg / 4.5V = 20900 kg


Lantai 1 → = 846450.45 / 846450.45 x 20900 kg

→ = 20900 kg

DAFTAR PEMBEBANAN PORTAL GEDUNG UNTUK INPUT PROGRAM ANALISIS STRUKTUR (SAP 2000)1. PEMBEBANAN PORTAL MELINTANG (TENGAH)

= 219.022071614 kg/m

= 120 kg/m

= -9.6 kg/m

= -96 kg/m

= 66 kg/m

= 106 kg/m

= 5052.28 kg

= 4480 kg

= 4384.08 kg

= 3840 kg

W3

W3

W3




Berat Total Gedung (WGedungTotal) WLantaiTotal


WLantaiTotal x Tinggi Kolom Lantai Dasar

WihiLantai1

z n

CIWt / R

Wt


F1

F1

qDAtap

qLAtap

qWAtap1

qWAtap2

qDTepi

qDTengah

P1D

P1L

P2D

P2L

x + +x+

= 4718.18 kg

= 4160 kg

= 2713.58 kg

= 2240 kg

= 216 kg/m

= -96 kg/m

= 1990.48261446 kg

= 9408.23448131 kg2. PEMBEBANAN PORTAL MEMANJANG (TEPI)

= 2606.38356459 kg

= 1300 kg

= 2929.18626806 kg

= 1560 kg

= 452.263333333 kg/m

= 373.333333333 kg/m

= 2757.14 kg

= 2240 kg

= 5283.28 kg

= 4480 kg

= 126 kg/m

= -56 kg/m

= 7574.14497794 kg

= 6889.27846712 kg3. PEMBEBANAN PORTAL MEMANJANG (TENGAH)

= 786.363333333 kg/m

= 693.333333333 kg/m

= 6469.2 kg

= 5120 kg

= 12283.4 kg

= 10240 kg

= 270 kg/m

= -120 kg/m

= 20900.0111111 kg

P3D

P3L

P4D

P4L

qWDinding1

qWDinding2

F1Atap

F2Lantai1

P5DAtap

P5LAtap

P6DAtap

P6Latap

qDLantai

qLLantai

P7DLantai

P7LLantai

P8DLantai

P8LLantai

qWDinding1

qWDinding2

F1Atap

F2Lantai1

qDLantai

qLLantai

P9DLantai

P9LLantai

P10DLantai

P10LLantai

qWDinding1

qWDinding2

F1Lantai1

PEMBEBANAN PORTAL GEDUNG8

18

0.4

kg/cm2

0.8

m

56.8

Jarak Antar Balok Anak x Beban Hidup Lantai Gudang x Koefisien Reduksi Beban Hidup Gudang

m

56.8

2.167 m

Jarak Antar Balok Anak x Beban Hidup Lantai Gudang x Koefisien Reduksi Beban Hidup Gudang

56.8

m

m

Bentang Pelat Yang Dipikul x Beban Hidup Lantai Gudang x Koefisien Reduksi Beban Hidup Gudang

56.8


kg/m

m

mkg/m

Berat Kolom x (1/2x(Tinggi Kolom <Lantai Dasar+Lantai 1>)) x 2 Kolom (Kanan Kiri)

2240

+ Berat Balok Induk Melintang Tengah x L2

+

Wilayah Gempa 2

Jumlah Tingkat Struktur Gedung = 2

18

18

0.8

56.8

Balok Anak Yang Menumpu Berasal Dari 2 Sisi (Kanan Kiri)


Beban Terpusat Balok Anak Yang Masuk Kolom Portal Memanjang Tinggal 1/2 PTotal

Beban Terpusat Balok Anak Yang Masuk Kolom Portal Memanjang adalah 1 PTotal

0.9

-0.4

Berat Profil Balok Induk Memanjang WF 300 x 200 x 8 x 12

kg/m

kg/m

Wilayah Gempa 2


0.8

56.8


Balok Anak Yang Menumpu Berasal Dari 2 Sisi (Kanan Kiri)

kg/m

Beban Terpusat Balok Anak Yang Masuk Kolom Portal Memanjang Tinggal 1/2 PTotal

Beban Terpusat Balok Anak Yang Masuk Kolom Portal Memanjang adalah 1 PTotal

Wilayah Gempa 2


DAFTAR PEMBEBANAN PORTAL GEDUNG UNTUK INPUT PROGRAM ANALISIS STRUKTUR (SAP 2000)

KONTROL PROFIL BALOK INDUK MELINTANG TEPI7 m

Modulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. PERHITUNGAN BALOK INDUK MELINTANG TEPI → Profile Code = 44DATA PERENCANAAN → WF 400 x 200 x 8 x 13Direncanakan Menggunakan Profil WF 400 x 200 x 8 x 13

W = 66 kg/m h = = 174 cm³

d = 400 mm = 342 mm = 1286 cm³

= 200 mm = 23700 = 266 cm³

= 8 mm = 1740 BJ 41

= 13 mm = 16.8 cm fu 4100

r = 16 mm = 4.54 cm fy 2500

Ag = 84.12 = 1190 cm³

A. Beban Berfaktor Untuk Beam ColumnDari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Melintang Tepi denganMomen Maksimum pada Batang 7 Portal Melintang Tengah dengan Beban Kombinasi 1.2D + 1.6L + 0.8W

= 16335.22 kgm MOMEN NEGATIF


= 10941.15 kgm MOMEN POSITIF

= 12342.71 kg

= 26990.98 kg

Nu = 6083.46 kg TEKAN

B. Beban Layan Untuk Kontrol LendutanDari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Melintang Tepi denganMomen Kondisi Layan pada Batang 7 Portal Melintang Tengah dengan Beban Kombinasi 1.0D + 1.0L




C. Kontrol Kolom

a). Kontrol Penampang (Tekuk Lokal / Kelangsingan Elemen Penampang)Badan (Web)

≤ → 342 / 8 ≤ 665 / 250

→ 42.75 ≥ 42.058 → NOT OKSayap (Flange) 0

≤ → 200 / 26 ≤ 250 / 250

→ 7.6923 ≤ 15.811 → OKProfil WF diatas tergolong PENAMPANG TIDAK KOMPAKMASIH BISA BERPERILAKU SEBAGAI KOLOM


Tekuk Terhadap Sumbu x → Faktor Tekuk kx 1

λx = kx Lkx / ix → Panjang Tekuk Lkx = 7 m= 1 x 700 cm / 16.8 cm= 41.666667 → Tidak Menentukan

Panjang Balok Induk Melintang Tepi (L1) =

kg/cm2 kg/cm2

d - 2(tf+r) Sy

Zx

bf Ix cm⁴ Zy

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx

MA

MB

MS

VA

VB

MA

MB

MS

h/tw 665 / fy1/2

bf/2tf 250 / fy1/2

Tekuk Terhadap Sumbu y → Faktor Tekuk ky 1

λy = ky Lky / iy → Panjang Tekuk Lky = 2.333 m= 1 x 233.33 cm / 4.54 cm= 51.395007 → Menentukan

= 51.395007 ≤ 200 → OK


λc =

= 51.395007 / 3.1416 x 2500 / 2000000= 0.578397 → 0.25 < λc < 1.2 → ω=1.43 / (1.6-0.67λc)



= 1.1794068

ω == 0.4181789

= 1.1794068


= 84.12 x 2500 / 1.17940675= 178309.98 kg

= 6083.46 kg / 0.85 / 178309.9845 kg



D. Kontrol Baloka). Kolom Tidak Bergoyang (Gaya Tekan Kritis Euler Untuk Elemen Tidak Bergoyang <Ncrb>)


λx = kx Lkx / ix → Panjang Tekuk Lkx = 7 m= 1 x 700 cm / 16.8 cm= 41.666667

Ncrbx =

= p x 2000000 x 84.12 / 41.667

= 956426.25 kg


λy = ky Lky / iy → Panjang Tekuk Lky = 2.333 m= 1 x 233.33 cm / 4.54 cm= 51.395007

Ncrby =

= p x 2000000 x 84.12 / 51.395

λmax

λmax/p/ fy/E

kg/cm2 kg/cm2

1.43 / (1.6-0.67λc)

1.25 λc2

ωAktual


Nu / fNn

p2 E Ag / λx2

kg/cm2 cm2

p2 E Ag / λy2

kg/cm2 cm2

22

22

= 628618.47 kg


Cm = 1 (Komponen Struktur Tak Bergoyang, Beban Transversal)

= Cm / (1-(Nu/Ncrb)) → ≥ 1



= 1 / 1 - 6083 kg / 956426.2528 kg

= 1.0064013 ≥ 1 → = 1.00640133



= 1 / 1 - 6083 kg / 628618.47424 kg

= 1.0097721 ≥ 1 → = 1.00977208


Mux =

= 1.0064013 x 24491.81 kgm= 24648.59 kgm


Muy =

= 1.0097721 x 0 kgm= 0 kgm

c). Kontrol Tekuk LokalBadan (Web)

≤ → 342 / 8 ≤ 1680 / 250

→ 42.75 ≤ 106.25 → OKSayap (Flange) 1

≤ → 200 / 26 ≤ 170 / 250

→ 7.6923 ≤ 10.752 → OKProfil WF diatas tergolong PENAMPANG KOMPAK


d). Kontrol Tekuk Lateral → Tegangan Residu (fr)= 70 MPa = 700

2.3333 m

Lp = = 1.76 x 4.54 cm x 2000000 / 2500

Lp = 226.00264 cm ≈ 226 cm

= fy - fr = 2500 - 700 = 1800


J = =

J = 0.3333333 x 37.4 x 0.8 + 0.667 x 20 x 1.3J = 6.3829333 + 29.293 = 35.6762667

h' = d - tf = 400 - 13 = 387 mm

= 0.25 x 1740 x 38.7

= 651495.15

db Mnt

db db

dbx

dbx

dby

dby

dbx Mntx

dby Mnty

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

kg/cm2

kg/cm2

Bentang Antar Balok Anak (LB) =

1.76 iy E/fy

fL kg/cm2


3

cm4 cm4 cm4



Iw cm6

3 3

2

=

= p / 1190 x 2000000 800000 35.6763 84.12 2

= 129356.2

=

= 4 x 1190 / 8E+05 x 35.68 x 651495 / 1740

= 0.0000026036 = 2.603617E-06

=

= 4.54 x 129356 / 2500 700 x 1 1 3E-06 1800

= 658.35744 cm ≈ 658 cm

= 233.33333 cm


= 658.35744 cm

→Mnx = Cb(MR+(Mp-MR) x (LR-LB)/(LR-Lp) = 79.2 kg/m

= VB = 26990.98 kg

= MB = 24491.81 kgm

= P3D = 4718.18 kg

= P3L = 4160 kg

L = 7 m = 12317.816 kg

= 0.5833333 m Mx =

= 1.1666667 m

= 1.75 m

8559.35 - 13.475 - 24491.81 = -15945.9393333 kgm

17118.7 - 53.9 - 24491.81 = -7427.01866667 kgm

25678 - 121.28 - 24491.81 = 1064.952 kgm

= 306147.63 / 61230 47838 29708 3195 = 2.16 ≤ 2.3

Mpx = Zx fy = 1286 x 2500Mpx = 3215000 kgcm = 32150 kgm

= = 1190 x 2500 - 700

= 2142000 kgcm = 21420 kgm

Mnx =

Mnx = 2.1564206 x 21420 32150 21420 658.4 233.3 658 226.003 ≤ 32150

Mnx = 68936.605 kgm ≥ 32150 kgm → Mn = 32150 kgm

Mcr =

Mcr = 0.029034 99322726400000000 + 725113792.733096 1740 651495.15

Mcr = 27868278 kgcm = 278682.775842 kgm ≤ 32150 kgm

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2


LR

LR

LB

LR

Balok Induk Melintang Tepi KIRI

qu

qu Vu

Mumax

PD

PL

Pu

XA (Vu-Pu)x - 0.5qux2 - Mu

XB

XC

MA (x =0.583 m) =

MB (x =1.167 m) =

MC (x =1.750 m) =


Cb

cm3 kg/cm2

MRx Sx (fy-fr) cm3 kg/cm2

MRx



/xx x

2

- + + x 2

+ ++

+ - x - -/

x x


Mnx = 3215000 kgcm Mnx = Cb[MR + (Mp-MR) x (LR-LB) / (LR-Lp)Mnx = 32150 kgm Mnx = 32150 kgm

= 32150 kgm

E. Kontrol Interaksi Beam Column → = 0.85 = 0.9

Nu+

Mux+

Muy≤ 1

6083.46 kg+

24648.590214 kgm+ 0 ≤ 1

303126.9736 kg 28935 kgm

0.020069016 + 0.851860729706 + 0 ≤ 1

0.871929745 ≤ 1 → OK

F. Kontrol Lendutan

= L/360 = 700 / 360 = 1.94444 cm


y =

y = (5 / 48) 700 760688 0.1 1310593 1696918 2000000 x 23700

y = 0.4952731 cm ≤ = 1.944444444 cm OK

G. Kontrol Kuat Geser → f = 0.9

Badan (Web)≤ → 342 / 8 ≤ 1100 / 250

→ 42.75 ≤ 69.57 → PLASTIS

≤ → ≤ → ≤→ 26990.98 kg ≤ 0.9 0.6 2500 40 0.8

→ 26990.98 kg ≤ 43200 kg OK

Jadi, Balok Induk Melintang Tepi dengan Profil WF 400 x 200 x 8 x 13 dapat dipakai

MnxAktual

fc fb


Lendutan Ijin ymax

5 / 48 L2 (MS - 0.1 (MA + MB)) / (EIx)

ymax

h/tw 1100 / fy1/2


2x x - x + /

x x x x

1285.95

265.984

Dari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Melintang Tepi dengan

Dari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Melintang Tepi dengan

GAMBAR BELUM BENAR

2

KONTROL PROFIL BALOK INDUK MELINTANG TENGAH8 m

Modulus Elastisitas (E) = 2E+06 Modulus Geser (G) = 8000001. PERHITUNGAN BALOK INDUK MELINTANG TENGAH → Profile Code = 67DATA PERENCANAAN → WF 600 x 200 x 11 x 17Direncanakan Menggunakan Profil WF 600 x 200 x 11 x 17

W = 106 kg/m h = = 228 cm³

d = 600 mm = 522 mm = 2863 cm³

= 200 mm = 77600 = 357 cm³

= 11 mm = 2280 BJ 41

= 17 mm = 24 cm fu 4100

r = 22 mm = 4.12 cm fy 2500

Ag = 134.4 = 2590 cm³

A. Beban Berfaktor Untuk Beam ColumnDari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Melintang Tengah denganMomen Maksimum pada Batang 8 Portal Melintang Tengah dengan Beban Kombinasi 1.2D + 1.6L + 0.8W




= 17383.29 kg

= 17849 kg


B. Beban Layan Untuk Kontrol LendutanDari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Melintang Tengah denganMomen Kondisi Layan pada Batang 8 Portal Melintang Tengah dengan Beban Kombinasi 1.0D + 1.0L




C. Kontrol Kolom


≤ → 522 / 11 ≤ 665 / 250


≤ → 200 / 34 ≤ 250 / 250




λx = kx Lkx / ix → Panjang Tekuk Lkx = 8 m= 1 x 800 cm / 24 cm= 33.333333 → Tidak Menentukan

Panjang Balok Induk Melintang Tengah (L2) =

kg/cm2 kg/cm2

d - 2(tf+r) Sy

Zx

bf Ix cm⁴ Zy

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx

MA

MB

MS

VA

VB

MA

MB

MS

h/tw 665 / fy1/2

bf/2tf 250 / fy1/2


λy = ky Lky / iy → Panjang Tekuk Lky = 2 m= 1 x 200 cm / 4.12 cm= 48.543689 → Menentukan

= 48.543689 ≤ 200 → OK


λc =

= 48.543689 / 3.14159 x 2500 / 2000000= 0.5463084 → 0.25 < λc < 1.2 → ω=1.43 / (1.6-0.67λc)



= 1.1588581

ω == 0.3730661

= 1.1588581


= 134.4 x 2500 / 1.1588581= 289940.59 kg

= 3104.7 kg / 0.85 / 289940.5895 kg





λx = kx Lkx / ix → Panjang Tekuk Lkx = 8 m= 1 x 800 cm / 24 cm= 33.333333

Ncrbx =

= p x 2000000 x 134.4 / 33.333

= 2387654.7 kg



Ncrby =

= p x 2000000 x 134.4 / 48.544

λmax

λmax/p/ fy/E

kg/cm2 kg/cm2

1.43 / (1.6-0.67λc)

1.25 λc2

ωAktual


Nu / fNn

p2 E Ag / λx2

kg/cm2 cm2

p2 E Ag / λy2

kg/cm2 cm2

22

22

= 1125805.7 kg



= Cm / (1-(Nu/Ncrb)) → ≥ 1



= 1 / 1 - 3104.7 kg / 2387654.6967 kg

= 1.001302 ≥ 1 → = 1.00130201



= 1 / 1 - 3104.7 kg / 1125805.719 kg

= 1.0027654 ≥ 1 → = 1.00276538


Mux =

= 1.001302 x 27642.95 kgm= 27678.941 kgm


Muy =

= 1.0027654 x 0 kgm= 0 kgm


≤ → 522 / 11 ≤ 1680 / 250

→ 47.4545 ≤ 106.25 → OKSayap (Flange) 1

≤ → 200 / 34 ≤ 170 / 250




2 m

Lp = = 1.76 x 4.12 cm x 2000000 / 2500

Lp = 205.09491 cm ≈ 205 cm

= fy - fr = 2500 - 700 = 1800


J = =

J = 0.3333333 x 56.6 x 1.1 + 0.667 x 20 x 1.7J = 25.111533 + 65.507 = 90.6182

h' = d - tf = 600 - 17 = 583 mm

= 0.25 x 2280 x 58.3

= 1937367.3

=

db Mnt

db db

dbx

dbx

dby

dby

dbx Mntx

dby Mnty

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

kg/cm2

kg/cm2

Bentang Antar Balok Anak (LB) =

1.76 iy E/fy

fL kg/cm2


3

cm4 cm4 cm4



Iw cm6

x1 p/Sx EGJA/2

3 3

2

/xx x

= p / 2590 x 2000000 800000 90.6182 134.4 2

= 119729.85

=

= 4 x 2590 / 8E+05 x 90.62 x 1937367 / 2280

= 0.0000043384 = 4.338362E-06

=

= 4.12 x 119730 / 2500 700 x 1 1 4E-06 1800

= 605.40769 cm ≈ 605 cm

= 200 cm


= 605.40769 cm

e). Kekuatan Nominal Momen Lentur (Mn) → Balok Induk Melintang Tengah

Mnx = Mp = Zx fy = 127.2 kg/m

= VB = 17849 kg

= MB = 27642.95 kgm

= P3D = 4718.18 kg

= P3L = 4160 kg

L = 8 m = 12317.816 kg

= 0.5 m Mx =

= 1 m

= 1.5 m

2765.59 - 15.9 - 27642.95 = -24893.258 kgm

5531.18 - 63.6 - 27642.95 = -22175.366 kgm

8296.78 - 143.1 - 27642.95 = -19489.274 kgm

= 345536.88 / 69107.4 74680 88701 58468 = 1.19 ≤ 2.3


= = 2590 x 2500 - 700

= 4662000 kgcm = 46620 kgm

Mnx =

Mnx = 1.1875897 x 46620 71575 46620 605.4 200 605 205.095 ≤ 71575

Mnx = 85378.926 kgm ≥ 71575 kgm → Mn = 71575 kgm

Mcr =

Mcr = 0.0186546 330575193600000000 + 986960440.108936 2280 1937367.3

Mcr = 40399948 kgcm = 403999.478031 kgm ≤ 71575 kgm

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2


LR

LR

LB

LR

qu

qu Vu

Mumax

PD

PL

Pu


XB

XC

MA (x =0.5 m) =

MB (x =1.0 m) =

MC (x =1.5 m) =


Cb

cm3 kg/cm2


MRx



/xx x

2

- + + x 2

+ ++

+ - x - -/

x x


Mnx = 7157500 kgcm Mnx = Mp = Zx fyMnx = 71575 kgm Mnx = 71575 kgm

= 71575 kgm


Nu+

Mux+

Muy≤ 1

3104.7 kg+

27678.94130527 kgm+ 0 ≤ 1

492899.002 kg 64417.5 kgm

0.00629886 + 0.4296804642414 + 0 ≤ 1

0.43597932 ≤ 1 → OK

F. Kontrol Lendutan

= L/360 = 800 / 360 = 2.22222 cm


y =

y = (5 / 48) 800 1447084 0.1 1927348 1927348 2000000 x 77600

y = 0.4560199 cm ≤ = 2.222222222 cm OK


Badan (Web)≤ → 522 / 11 ≤ 1100 / 250

→ 47.4545 ≤ 69.57 → PLASTIS

≤ → ≤ → ≤→ 17849 kg ≤ 0.9 0.6 2500 60 1.1

→ 17849 kg ≤ 89100 kg OK

Jadi, Balok Induk Melintang Tengah dengan Profil WF 600 x 200 x 11 x 17 dapat dipakai

MnxAktual

fc fb


Lendutan Ijin ymax

5 / 48 L2 (MS - 0.1 (MA + MB)) / (EIx)

ymax

h/tw 1100 / fy1/2


2x x - x + /

x x x x

2863.18

357.122

Dari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Melintang Tengah dengan

Dari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Melintang Tengah dengan

GAMBAR BELUM BENAR

2

KONTROL PROFIL BALOK INDUK MEMANJANG6 m

Modulus Elastisitas (E) = 2E+06 Modulus Geser (G) = 8000001. PERHITUNGAN BALOK INDUK MEMANJANG → Profile Code = 26DATA PERENCANAAN → WF 300 x 200 x 8 x 12Direncanakan Menggunakan Profil WF 300 x 200 x 8 x 12

W = 56.8 kg/m h = = 160 cm³

d = 294 mm = 234 mm = 823 cm³

= 200 mm = 11300 = 244 cm³

= 8 mm = 1600 BJ 41

= 12 mm = 12.5 cm fu 4100

r = 18 mm = 4.71 cm fy 2500

Ag = 72.38 = 771 cm³

A. Beban Berfaktor Untuk Beam ColumnDari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Memanjang denganMomen Maksimum pada Batang 10 Portal Memanjang Tengah dengan Beban Kombinasi 1.2D + 1.0L + 1.0E




= 3584.92 kg

= 6236.9 kg


B. Beban Layan Untuk Kontrol LendutanDari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Memanjang denganMomen Kondisi Layan pada Batang 10 Portal Memanjang Tengah dengan Beban Kombinasi 1.0D + 1.0L




C. Kontrol Kolom


≤ → 234 / 8 ≤ 665 / 250

→ 29.25 ≤ 42.058 → OKSayap (Flange) 1

≤ → 200 / 24 ≤ 250 / 250

→ 8.33333 ≤ 15.811 → OKProfil WF diatas tergolong PENAMPANG KOMPAKSANGAT BOLEH DIPAKAI SEBAGAI KOLOM



λx = kx Lkx / ix → Panjang Tekuk Lkx = 6 m= 1 x 600 cm / 12.5 cm= 48 → Tidak Menentukan

Panjang Balok Induk Memanjang (L) =kg/cm2 kg/cm2

d - 2(tf+r) Sy

Zx

bf Ix cm⁴ Zy

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx

MA

MB

MS

VA

VB

MA

MB

MS

h/tw 665 / fy1/2

bf/2tf 250 / fy1/2


λy = ky Lky / iy → Panjang Tekuk Lky = 6 m= 1 x 600 cm / 4.71 cm= 127.38854 → Menentukan

= 127.38854 ≤ 200 → OK


λc =

= 127.38854 / 3.14159 x 2500 / 2000000= 1.4336247 → λc ≥ 1.2 → ω=1.25λc2

→ Kolom Panjang (Tekuk Elastis)


= 2.2362218

ω == 2.5690998

= 2.5690998


= 72.38 x 2500 / 2.56909975= 70433.232 kg

= 19554.43 kg / 0.85 / 70433.23245 kg

= 0.3266244 ≥ 0.2 → Pakai Rumus Interaksi Beam Column 1

Rumus Interaksi Beam ColumnNu / φcNn + 8/9 x (Mux / φbMnx + Muy / φbMny)



λx = kx Lkx / ix → Panjang Tekuk Lkx = 6 m= 1 x 600 cm / 12.5 cm= 48

Ncrbx =

= p x 2000000 x 72.38 / 48

= 620105.87 kg



Ncrby =

= p x 2000000 x 72.38 / 127.39

λmax

λmax/p/ fy/E

kg/cm2 kg/cm2

1.43 / (1.6-0.67λc)

1.25 λc2

ωAktual


Nu / fNn

p2 E Ag / λx2

kg/cm2 cm2

p2 E Ag / λy2

kg/cm2 cm2

22

22

= 88041.541 kg



= Cm / (1-(Nu/Ncrb)) → ≥ 1



= 1 / 1 - 19554 kg / 620105.87374 kg

= 1.0325608 ≥ 1 → = 1.03256079



= 1 / 1 - 19554 kg / 88041.540568 kg

= 1.2855199 ≥ 1 → = 1.28551986


Mux =

= 1.0325608 x 8301.4 kgm= 8571.7001 kgm


Muy =

= 1.2855199 x 0 kgm= 0 kgm


≤ → 234 / 8 ≤ 1680 / 250

→ 29.25 ≤ 106.25 → OKSayap (Flange) 1

≤ → 200 / 24 ≤ 170 / 250




6 m

Lp = = 1.76 x 4.71 cm x 2000000 / 2500

Lp = 234.46529 cm ≈ 234 cm

= fy - fr = 2500 - 700 = 1800


J = =

J = 0.3333333 x 27 x 0.8 + 0.667 x 20 x 1.2J = 4.608 + 23.04 = 27.648

h' = d - tf = 294 - 12 = 282 mm

= 0.25 x 1600 x 28.2

= 318096

=

db Mnt

db db

dbx

dbx

dby

dby

dbx Mntx

dby Mnty

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

kg/cm2

kg/cm2

Bentang Balok Induk Memanjang (LB) =

1.76 iy E/fy

fL kg/cm2


3

cm4 cm4 cm4



Iw cm6

x1 p/Sx EGJA/2

3 3

2

/xx x

= p / 771 x 2000000 800000 27.648 72.38 2

= 163035.3

=

= 4 x 771 / 8E+05 x 27.65 x 318096 / 1600

= 0.0000009663 = 9.662724E-07

=

= 4.71 x 163035 / 2500 700 x 1 1 1E-06 1800

= 742.89009 cm ≈ 743 cm

= 600 cm


= 742.89009 cm

e). Kekuatan Nominal Momen Lentur (Mn) → Balok Induk Memanjang

Mnx = Cb(MR+(Mp-MR) x (LR-LB)/(LR-Lp) = 2052.96933 kg/m

= VB = 6236.9 kg

= MB = 8301.4 kgm

= P10D = 12283.4 kg

= P10L = 10240 kg

L = 6 m = 0 kg

= 1.5 m Mx =

= 3 m

= 4.5 m

9355.35 - 2309.6 - 8301.4 = -1255.6405 kgm

18710.7 - 9238.4 - 8301.4 = 1170.938 kgm

28066 - 20786 - 8301.4 = -1021.6645 kgm

= 103767.5 / 20753.5 3766.9 4684 3065 = 3.22 ≤ 2.3


= = 771 x 2500 - 700

= 1387800 kgcm = 13878 kgm

Mnx =

Mnx = 2.3 x 13878 20575 13878 742.9 600 743 234.465 ≤ 20575

Mnx = 36248.359 kgm ≥ 20575 kgm → Mn = 20575 kgm

Mcr =

Mcr = 0.0168373 70778880000000000 + 109662271.123215 1600 318096

Mcr = 5990666.7 kgcm = 59906.6672363 kgm ≤ 20575 kgm

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2


LR

LR

LB

LR

qu

qu Vu

Mumax

PD

PL

Pu


XB

XC

MA (x =1.5 m) =

MB (x =3.0 m) =

MC (x =4.5 m) =


Cb

cm3 kg/cm2


MRx



/xx x

2

- + + x 2

+ ++

+ - x - -/

x x


Mnx = 2057500 kgcm Mnx = Cb[MR + (Mp-MR) x (LR-LB) / (LR-Lp)Mnx = 20575 kgm Mnx = 20575 kgm

= 20575 kgm


Nu+

8 Mux+

Muy≤ 1

9

19554.43 kg+

8 8571.70014979 kgm+ 0 ≤ 1

59868.2476 kg 9 18517.5 kgm

0.32662439 + 0.4114642377231 + 0 ≤ 1

0.73808863 ≤ 1 → OK

F. Kontrol Lendutan

= L/360 = 600 / 360 = 1.66667 cm


y =

y = (5 / 48) 600 252609 0.1 342564 483944 2000000 x 11300

y = 0.2820103 cm ≤ = 1.666666667 cm OK


Badan (Web)≤ → 234 / 8 ≤ 1100 / 250

→ 29.25 ≤ 69.57 → PLASTIS

≤ → ≤ → ≤→ 6236.9 kg ≤ 0.9 0.6 2500 29.4 0.8

→ 6236.9 kg ≤ 31752 kg OK

Jadi, Balok Induk Memanjang dengan Profil WF 300 x 200 x 8 x 12 dapat dipakai

MnxAktual

fc fb

fc Nn fb Mnx fb Mny

Lendutan Ijin ymax

5 / 48 L2 (MS - 0.1 (MA + MB)) / (EIx)

ymax

h/tw 1100 / fy1/2


2x x - x + /

x x x x

x

x

822.6

244.32

GAMBAR BELUM BENAR

2

KONTROL PROFIL BALOK ATAP KUDA-KUDA11.5661 m

Modulus Elastisitas (E) = 2E+06 Modulus Geser (G) = 8000001. PERHITUNGAN BALOK ATAP KUDA-KUDA (KANAN) → Profile Code = 44DATA PERENCANAAN → WF 400 x 200 x 8 x 13Direncanakan Menggunakan Profil WF 400 x 200 x 8 x 13

W = 66 kg/m h = = 174 cm³

d = 400 mm = 342 mm = 1286 cm³

= 200 mm = 23700 = 266 cm³

= 8 mm = 1740 BJ 41

= 13 mm = 16.8 cm fu 4100

r = 16 mm = 4.54 cm fy 2500

Ag = 84.12 = 1190 cm³

A. Beban Berfaktor Untuk Beam ColumnDari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Atap Kuda-Kuda denganMomen Maksimum pada Batang 11 Portal Melintang Tengah dengan Beban Kombinasi 1.2D + 1.6L

= -1227.81 kgm MOMEN NEGATIF



= 1429.68 kg

= 3573.41 kg


B. Beban Layan Untuk Kontrol LendutanDari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Atap Kuda-Kuda denganMomen Kondisi Layan pada Batang 11 Portal Melintang Tengah dengan Beban Kombinasi 1.0D + 1.0L

= -950.99 kgm MOMEN NEGATIF



C. Kontrol Kolom


≤ → 342 / 8 ≤ 665 / 250


≤ → 200 / 26 ≤ 250 / 250




λx = kx Lkx / ix → Panjang Tekuk Lkx = 11.57 m= 1 x 1156.61 cm / 16.8 cm= 68.8457409 → Menentukan

Panjang Balok Atap Kuda-Kuda =kg/cm2 kg/cm2

d - 2(tf+r) Sy

Zx

bf Ix cm⁴ Zy

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx

MA

MB

MS

VA

VB

MA

MB

MS

h/tw 665 / fy1/2

bf/2tf 250 / fy1/2


λy = ky Lky / iy → Panjang Tekuk Lky = 1.425 m= 1 x 142.5 cm / 4.54 cm= 31.3876652 → Tidak Menentukan

= 68.8457409 ≤ 200 → OK


λc =

= 68.8457409 / 3.14159 x 2500 / 2000000= 0.7747868 → 0.25 < λc < 1.2 → ω=1.43 / (1.6-0.67λc)



= 1.32298035

ω == 0.75036823

= 1.32298035


= 84.12 x 2500 / 1.323= 158959.277 kg

= 6025.72 kg / 0.85 / 158959.2765 kg





λx = kx Lkx / ix → Panjang Tekuk Lkx = 11.57 m= 1 x 1156.61 cm / 16.8 cm= 68.8457409

Ncrbx =

= p x 2000000 x 84.12 / 68.846

= 350328 kg


λy = ky Lky / iy → Panjang Tekuk Lky = 1.425 m= 1 x 142.5 cm / 4.54 cm= 31.3876652

Ncrby =

= p x 2000000 x 84.12 / 31.388

λmax

λmax/p/ fy/E

kg/cm2 kg/cm2

1.43 / (1.6-0.67λc)

1.25 λc2

ωAktual


Nu / fNn

p2 E Ag / λx2

kg/cm2 cm2

p2 E Ag / λy2

kg/cm2 cm2

22

22

= 1685431 kg



= Cm / (1-(Nu/Ncrb)) → ≥ 1



= 1 / 1 - 6026 kg / 350327.99964 kg

= 1.01750125 ≥ 1 → = 1.01750125



= 1 / 1 - 6026 kg / 1685431.0174 kg

= 1.00358801 ≥ 1 → = 1.00358801


Mux =

= 1.01750125 x 11169.43 kgm= 11364.909 kgm


Muy =

= 1.00358801 x 0 kgm= 0 kgm


≤ → 342 / 8 ≤ 1680 / 250

→ 42.75 ≤ 106.25 → OKSayap (Flange) 1

≤ → 200 / 26 ≤ 170 / 250




1.425 m

Lp = = 1.76 x 4.54 cm x 2000000 /

Lp = 226.002641 cm ≈ 226 cm

= fy - fr = 2500 - 700 = 1800


J = =

J = 0.33333333 x 37.4 x 0.8 + 0.667 x 20 xJ = 6.38293333 + 29.293 = 35.6762667

h' = d - tf = 400 - 13 = 387 mm

= 0.25 x 1740 x 38.7

= 651495.15

=

db Mnt

db db

dbx

dbx

dby

dby

dbx Mntx

dby Mnty

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

kg/cm2

Bentang Diagonal Antar Gording (LB) =

1.76 iy E/fy

fL kg/cm2


3

cm4 cm4 cm4



Iw cm6

x1 p/Sx EGJA/2

3

2

/xx x

= p / 1190 x 2000000 800000 35.6763 84.12

= 129356.197

=

= 4 x 1190 / 8E+05 x 35.68 x 651495 /

= 0.0000026036 = 2.603617E-06

=

= 4.54 x 129356 / 2500 700 x 1 1 3E-06

= 658.357444 cm ≈ 658 cm

= 142.5 cm


= 658.357444 cm

e). Kekuatan Nominal Momen Lentur (Mn) → Balok Atap Kuda-Kuda

Mnx = Mp = Zx fy = 454.826486 kg/m

= VB = 3573.41 kg

= MB = 11169.43 kgm

= P3D = 4718.18 kg

= P3L = 4160 kg

L = 11.5660845 m = 96 kg/m

= 0.33881388 m Mx =

= 0.67762777 m

= 1.01644165 m

1178.19 - 10.639 - 11169.43 = -10001.8739852

2356.39 - 73.489 - 11169.43 = -8886.52970775

3534.58 - 188.55 - 11169.43 = -7823.39716756

= 139617.875 / 27923.6 30006 35546 23470 = 1.19 ≤ 2.3


= = 1190 x 2500 - 700

= 2142000 kgcm = 21420 kgm

Mnx =

Mnx = 1.19387122 x 21420 32150 21420 658.4 142.5 658 226.003 ≤Mnx = 40857.0589 kgm ≥ 32150 kgm → Mn = 32150 kgm

Mcr =

Mcr = 0.0263204 99322726400000000 + 1944151067.00437 1740 651495.15

Mcr = 39944756 kgcm = 399447.556474 kgm ≤ 32150 kgm

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2


LR

LR

LB

LR

qu

qu Vu

Mumax

PD

PL

qW


XB

XC

MA (x =0.3388 m) =

MB (x =0.6776 m) =

MC (x =1.016 m) =


Cb

cm3 kg/cm2


MRx



/xx x

2

- + + x

+ ++

+ - x - -/

x x


Mnx = 3215000 kgcm Mnx = Mp = Zx fyMnx = 32150 kgm Mnx = 32150 kgm

= 32150 kgm

E. Kontrol Interaksi Beam Column → = 0.85 =

Nu+

Mux+

Muy≤ 1

6025.72 kg+

11364.90897798 kgm+ 0 ≤ 1

270230.77 kg 28935 kgm

0.02229842 + 0.392773768031 + 0 ≤ 1

0.41507219 ≤ 1 → OK

F. Kontrol Lendutan

= L/360 = 1157 / 360 = 3.2128 cm


y =

y = (5 / 48) 1156.61 169104 0.1 -95099 835210 2000000 x

y = 0.27955756 cm ≤ = 3.212801241 cm OK


Badan (Web)≤ → 342 / 8 ≤ 1100 / 250

→ 42.75 ≤ 69.57 → PLASTIS

≤ → ≤ → ≤→ 3573.41 kg ≤ 0.9 0.6 2500 40

→ 3573.41 kg ≤ 43200 kg OK

Jadi, Balok Atap Kuda-Kuda dengan Profil WF 400 x 200 x 8 x 13 dapat dipakai

MnxAktual

fc fb


Lendutan Ijin ymax

5 / 48 L2 (MS - 0.1 (MA + MB)) / (EIx)

ymax

h/tw 1100 / fy1/2


2x - x + /

x x x x

x

KONTROL PROFIL BALOK ATAP KUDA-KUDA

1285.95

265.984

Dari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Atap Kuda-Kuda dengan

Dari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Atap Kuda-Kuda denganMomen Kondisi Layan pada Batang 11 Portal Melintang Tengah dengan Beban Kombinasi 1.0D + 1.0L

0

2500

1.3

kg/cm2

3

/

2

1740

1800

GAMBAR BELUM BENAR

kgm

kgm

kgm

32150

651495.15

/

x 2

TEKUK LATERAL

0.9

23700

0.8

f 0.6 fy d tw

x

KONTROL PROFIL KOLOM4.5 m 5 m

Modulus Elastisitas (E) = 2E+06 Modulus Geser (G) = 8000001. PERHITUNGAN KOLOM TENGAH (INTERIOR) LANTAI DASAR → Profile Code = 37DATA PERENCANAAN → WF 350 x 350 x 13 x 13Direncanakan Menggunakan Profil WF 350 x 350 x 13 x 13

W = 106 kg/m h = = 534 cm³

d = 338 mm = 272 mm = 1799 cm³

= 351 mm = 28200 = 814 cm³

= 13 mm = 9380 BJ 41

= 13 mm = 14.4 cm fu 4100

r = 20 mm = 8.33 cm fy 2500

Ag = 135.3 = 1670 cm³

A. Beban Berfaktor Untuk Beam Column Akibat Beban Gravitasi (Kolom Tak Bergoyang) <D+L>

a). Beban Arah Sumbu x (Portal Melintang)

= 3140.77 kgm

= 1486.78 kgm


b). Beban Arah Sumbu y (Portal Memanjang)

= 497.67 kgm

= 363.08 kgm


B. Beban Berfaktor Untuk Beam Column Akibat Beban Gempa (Kolom Bergoyang) <E>

a). Beban Arah Sumbu x (Portal Melintang)

= 8075.29 kgm

= 9128.68 kgm

Nu = 90.96 kg TEKAN

b). Beban Arah Sumbu y (Portal Memanjang)

= 5394.39 kgm

= 8295.2 kgm


C. Daftar Gaya Normal Berfaktor untuk Perhitungan Amplifikasi Momen (Lantai Dasar) <D+L>a). Beban Arah Sumbu x (Portal Melintang) b). Beban Arah Sumbu y (Portal Memanjang)

= 26739.55 kg = 21787.02 kg

= 44224.07 kg = 43778.26 kg

= 44224.07 kg = 43420.46 kg

= 26739.55 kg = 43450.91 kg

= 141927.24 kg = 43447.84 kg

= 43450.91 kg

= 43420.46 kg

= 43778.26 kg

= 21787.02 kg

Tinggi Kolom Lantai Dasar (H1) = Tinggi Kolom Lantai 1 (H2) =

kg/cm2 kg/cm2

d - 2(tf+r) Sy

Zx

bf Ix cm⁴ Zy

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx

Mntx1

Mntx1

Mntx2

Mntx2

Mnty1

Mnty1

Mnty2

Mnty2

Mltx1

Mltx1

Mltx2

Mltx2

Mlty1

Mlty1

Mlty2

Mlty2

Nux1 Nuy1

Nux2 Nuy2

Nux3 Nuy3

Nux4 Nuy4

∑Nux Nuy5

Nuy6

Nuy7

Nuy8

Nuy9

= 348321.14 kg

D. Penentuan Faktor Panjang Tekuk (kc) Kolom Tengah dan Tepi Arah x dan y Menggunakan Nomograma). Komponen Struktur Arah Sumbu x (Portal Melintang)I). Kolom Tengah → 1 Kolom 2 Balok

=

= 28200 450 / 23700 700 + 77600 800= 0.47889374 ≈ 0.479

= 1 → Kolom dengan Perletakan Jepit (Kaku)

kcx = 0.72 (Tak Bergoyang) (Dari Pembacaan Nomogram Komponen Struktur Tak Bergoyang)kcx = 1.23 (Bergoyang) (Dari Pembacaan Nomogram Komponen Struktur Bergoyang)

II). Kolom Tepi → 2 Kolom 1 Balok

=

= 28200 450 + 28200 500 / 23700 700= 3.51673699 ≈ 3.517


kcx = 0.84 (Tak Bergoyang) (Dari Pembacaan Nomogram Komponen Struktur Tak Bergoyang)kcx = 1.6 (Bergoyang) (Dari Pembacaan Nomogram Komponen Struktur Bergoyang)

b). Komponen Struktur Arah Sumbu y (Portal Memanjang)I). Kolom Tengah → 1 Kolom 2 Balok

=

= 9380 450 / 11300 600 + 11300 600= 0.55339233 ≈ 0.553


kcy = 0.76 (Tak Bergoyang) (Dari Pembacaan Nomogram Komponen Struktur Tak Bergoyang)kcy = 1.25 (Bergoyang) (Dari Pembacaan Nomogram Komponen Struktur Bergoyang)

II). Kolom Tepi → 1 Kolom 1 Balok

=

= 9380 450 / 11300 600= 1.10678466 ≈ 1.107


kcy = 0.77 (Tak Bergoyang) (Dari Pembacaan Nomogram Komponen Struktur Tak Bergoyang)kcy = 1.33 (Bergoyang) (Dari Pembacaan Nomogram Komponen Struktur Bergoyang)

E. Kontrol Kolom (Kolom Tengah Lantai Dasar)a). Kontrol Penampang (Tekuk Lokal / Kelangsingan Elemen Penampang)Badan (Web)

≤ → 272 / 13 ≤ 665 / 250

→ 20.9231 ≤ 42.058 → OKSayap (Flange) 1

≤ → 351 / 26 ≤ 250 / 250

→ 13.5 ≤ 15.811 → OKProfil WF diatas tergolong PENAMPANG KOMPAKSANGAT BOLEH DIPAKAI SEBAGAI KOLOM

∑Nuy

GA ∑(Ix / L)Column / ∑(Ix / L)Beam

GB

GA ∑(Ix / L)Column / ∑(Ix / L)Beam

GB

GA ∑(Iy / L)Column / ∑(Ix / L)Beam

GB

GA ∑(Iy / L)Column / ∑(Ix / L)Beam

GB

h/tw 665 / fy1/2

bf/2tf 250 / fy1/2

/ / /

/ / /

/ / /

/ /

b). Kontrol Kelangsingan Komponen Struktur (Tekuk Struktural) I). Portal Tidak Bergoyangi.) Kolom Tengah

Tekuk Terhadap Sumbu x → Faktor Tekuk kcx = 0.72

λx = kcx Lkx / ix → Panjang Tekuk Lkx = 4.5 m= 0.72 x 450 cm / 14.4 cm= 22.5

Tekuk Terhadap Sumbu y → Faktor Tekuk kcy = 0.76

λy = kcy Lky / iy → Panjang Tekuk Lky = 4.5 m= 0.76 x 450 cm / 8.33 cm= 41.0564226

ii.) Kolom Tepi


λx = kcx Lkx / ix → Panjang Tekuk Lkx = 4.5 m= 0.84 x 450 cm / 14.4 cm= 26.25



II). Portal Bergoyang (Menentukan Kekuatan Kolom)i.) Kolom Tengah


λx = kcx Lkx / ix → Panjang Tekuk Lkx = 4.5 m= 1.23 x 450 cm / 14.4 cm= 38.4375 → Tidak Menentukan


λy = kcy Lky / iy → Panjang Tekuk Lky = 4.5 m= 1.25 x 450 cm / 8.33 cm= 67.5270108 → Menentukan

ii.) Kolom Tepi


λx = kcx Lkx / ix → Panjang Tekuk Lkx = 4.5 m= 1.6 x 450 cm / 14.4 cm= 50



= 67.5270108 ≤ 200 → OK

c). Kontrol Kekuatan → f = 0.85

λmax

Parameter Kelangsingan

λc =

= 67.5270108 / 3.14159 x 2500 / 2000000= 0.75994587 → 0.25 < λc < 1.2 → ω=1.43 / (1.6-0.67λc)

→ Kolom Menengah (Tekuk Inelastis)Faktor Tekuk (ω)

ω == 1.31092084

ω == 0.72189716

= 1.31092084


= 135.3 x 2500 / 1.31092084= 258024.732 kg

= 44224.07 kg / 0.85 / 258024.7324 kg

= 0.20164082 ≥ 0.2 → Pakai Rumus Interaksi Beam Column 1

Rumus Interaksi Beam ColumnNu / φcNn + 8/9 x (Mux / φbMnx + Muy / φbMny)


Ncrbx =

= p x 2000000 x 135.3 / 22.5

= 5275486.32 kg

Ncrby =

= p x 2000000 x 135.3 / 41.06

= 1584401.8 kg

b). Kolom Bergoyang (Gaya Tekan Kritis Euler Untuk Elemen Bergoyang <Ncrs>)I.) Kolom Tengah

Ncrsx =

= p x 2000000 x 135.3 / 38.44

= 1807662.18 kg

Ncrsy =

= p x 2000000 x 135.3 / 67.53

= 585696.309 kg

II.) Kolom Tepi

Ncrsx =

= p x 2000000 x 135.3 / 50

= 1068285.98 kg

Ncrsy =

= p x 2000000 x 135.3 / 71.85

= 517355.691 kg

λmax/p/ fy/E

kg/cm2 kg/cm2

1.43 / (1.6-0.67λc)

1.25 λc2

ωAktual


Nu / fNn

p2 E Ag / λx2

kg/cm2 cm2

p2 E Ag / λy2

kg/cm2 cm2

p2 E Ag / λx2

kg/cm2 cm2

p2 E Ag / λy2

kg/cm2 cm2

p2 E Ag / λx2

kg/cm2 cm2

p2 E Ag / λy2

kg/cm2 cm2

22

22

22

22

22

22

= 2 Ncrsx Kolom Tengah + 2 Ncrsx Kolom Tepi

= 2 x 1807662.178437 kg + 2 x 1068285.98037 kg= 5751896.32 kg

= 7 Ncrsy Kolom Tengah + 2 Ncrsy Kolom Tepi

= 7 x 585696.3086504 kg + 2 x 517355.691258 kg= 5134585.54 kg

c). Amplifikasi Momen Balok → Mu =

= Cm / (1-(Nu/Ncrb)) → ≥ 1

= → ≥ 1

Cm = → <

Faktor Amplifikasi Momen Terhadap Sumbu xCmx = 0.6 - 0.4 x 1487 kgm / 3140.77 kgm

= 0.41064771


= 0.41064771 / 1 - 44224 kg / 5275486.3228 kg

= 0.41411924 ≤ 1 → = 1

= 141927.24 kg = 5751896.32 kg

=

= 1 / 1 - 141927.24 kg / 5751896.31762 kg

= 1.02529911 ≥ 1 → = 1.02529911

Faktor Amplifikasi Momen Terhadap Sumbu yCmy = 0.6 - 0.4 x 363.08 kgm / 497.67 kgm

= 0.3081761


= 0.3081761 / 1 - 44224 kg / 1584401.8045 kg

= 0.31702495 ≤ 1 → = 1

= 348321.14 kg = 5134585.54 kg

=

= 1 / 1 - 348321.14 kg / 5134585.54307 kg

= 1.07277516 ≥ 1 → = 1.07277516


Mux =

= 1 x 3140.77 kgm + 1.025 x 9128.68 kgm= 12500.3975 kgm


Muy =

= 1 x 497.67 kgm + 1.073 x 8295.2 kgm= 9396.55448 kgm

c). Kontrol Tekuk Lokal

∑Ncrsx

∑Ncrsy

db Mnt + ds Mlt

db db

ds 1 / (1-(∑Nu/∑Ncrs)) ds

0.6 - 0.4 (M1/M2) M1 M2

dbx

dbx

∑Nux ∑Ncrsx

dsx 1 / (1-(∑Nux/∑Ncrsx))

dsx

dby

dby

∑Nuy ∑Ncrsy

dsy 1 / (1-(∑Nuy/∑Ncrsy))

dsy

dbx Mntx + dsx Mltx

dby Mnty + dsy Mlty

Badan (Web)≤ → 272 / 13 ≤ 1680 / 250

→ 20.9231 ≤ 106.25 → OKSayap (Flange)

≤ → 351 / 26 ≤ 170 / 250

→ 13.5 ≥ 10.752 → NOT OK 0

≤ → 351 / 26 ≤ 370 / 180

→ 13.5 ≤ 27.578 → OK

Profil WF diatas tergolong PENAMPANG TIDAK KOMPAK → λp < λ < λrMn = Mp - (Mp - Mr) x (λ-λp) / (λr-λp)

Mpx = Zx fy

= 1799 x 2500= 4497500 kgcm

Mpy = Zy fy

= 814 x 2500= 2035000 kgcm

Mrx = Sx ( fy - fr)

= 1670 x 2500 - 700= 3006000 kgcm

Mry = Sy ( fy - fr)

= 534 x 2500 - 700= 961200 kgcm

λp == 170 / 250= 10.751744

λr == 370 / 250 - 70= 27.5781717

Mnx = Mpx - (Mpx - Mrx) x (λ-λp) / (λr-λp)= 4497500 4497500 3006000 x 13.5 10.8 / 27.58 10.752= 4253893.7 kgcm

Mny = Mpy - (Mpy - Mry) x (λ-λp) / (λr-λp)= 2035000 2035000 961200 x 13.5 10.75 / 27.5782 10.75= 1859616.53 kgcm


4.5 m = 450 cm

Lp = = 1.76 x 8.33 cm x 2000000 / 2500

Lp = 414.670044 cm ≈ 415 cm

= fy - fr = 2500 - 700 = 1800


h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

bf/2tf 370 / (fy-fr)1/2

cm3 kg/cm2

cm3 kg/cm2

cm3 kg/cm2 kg/cm2

cm3 kg/cm2 kg/cm2

170 / fy1/2

370 / (fy-fr)1/2

kg/cm2

Tinggi Kolom (LB) =

1.76 iy E/fy

fL kg/cm2

- - - -

- - - -

J = =

J = 0.33333333 x 31.2 x 1.3 + 0.667 x 35.1 x 1.3J = 22.8488 + 51.41 = 74.2586

h' = d - tf = 338 - 13 = 325 mm

= 0.25 x 9380 x 32.5

= 2476906.25

=

= p / 1670 x 2000000 800000 74.2586 135.3 2

= 168655.563

=

= 4 x 1670 / 8E+05 x 74.26 x 2476906 / 9380

= 0.0000008347 = 8.346919E-07

=

= 8.33 x 168656 / 2500 700 x 1 1 8E-07 1800

= 1334.78867 cm ≈ 1335 cm

= 450 cm


= 1334.78867 cm

e). Kekuatan Nominal Momen Lentur (Mn)Mnx = Cb(MR+(Mp-MR) x (LR-LB)/(LR-Lp)

= 9128.68 kgm

Hu = 3823.104444 kg

= 1.125 m Mx =

= 2.25 m

= 3.375 m

3823.1 x 1.125 - 9128.68 = -4827.6875 kgm

3823.1 x 2.25 - 9128.68 = -526.695 kgm

3823.1 x 3.375 - 9128.68 = 3774.2975 kgm

3823.1 x 4.5 - 9128.68 = 8075.29 kgm

= 114108.5 / 22821.7 14483 2107 11323 = 2.25 ≤ 2.3


= = 1670 x 2500 - 700

= 3006000 kgcm = 30060 kgm


3

cm4 cm4 cm4



Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2


LR

LR

LB

LR

Mumax

XA Hux - Mu

XB

XC

MA (x =1.125 m) =

MB (x =2.250 m) =

MC (x =3.375 m) =

Mumin (x =4.5 m) =


Cb

cm3 kg/cm2


MRx

3 3

2

/xx x

2

- + + x 2

+ ++

Mnx =

Mnx = 2.24913316 x 30060 44975 30060 1335 450 1335 414.67 ≤ 44975

Mnx = 99866.699 kgm ≥ 44975 kgm → Mn = 44975 kgm

Mcr =

Mcr = 0.01570191 1114473068800000000 + 194955148.663493 9380 2476906.25

Mcr = 37302997.4 kgcm = 373029.974464 kgm ≥ 44975 kgm


Mnx = 4253893.7 kgcm Mnx = Cb[MR + (Mp-MR) x (LR-LB) / (LR-Lp)Mnx = 42538.937 kgm Mnx = 44975 kgm

= 42538.9370186116 kgm

= 18596.1653171875 kgm

G. Kontrol Interaksi Beam Column → = 0.85 = 0.9

Nu+

8 Mux+

Muy≤ 1

9

44224.07 kg+

8 12500.3975042 kgm+

9396.5544765 kgm ≤219321.023 kg 9 38285.0433168 kgm 16736.548785 kgm

0.20164082 + 0.2902299040455 + 0.4991 ≤ 1

0.99092778 ≤ 1 → OK

Jadi, Kolom dengan Profil WF 350 x 350 x 13 x 13 dapat dipakai



MnxAktual

MnyAktual

fc fb

fc Nn fb Mnx fb Mny

+ - x - -/

x x

x

x

1799.34

813.989

D. Penentuan Faktor Panjang Tekuk (kc) Kolom Tengah dan Tepi Arah x dan y Menggunakan Nomogram

(Dari Pembacaan Nomogram Komponen Struktur Tak Bergoyang)




CEK

2

PERENCANAAN SAMBUNGAN BAUT DAN LAS1. PERHITUNGAN SAMBUNGAN KUDA-KUDA (DETAIL A)A. Beban Berfaktor Untuk Perencanaan Sambungan Baut Dan Las

Mu = 1227.81 kgmPu = 1503.25451 kg

B. Kontrol Sambungan Baut → f = 0.75

Mutu Baut = BJ 50 → = 5000

Mutu Pelat = BJ 41 → fu = 4100Tebal Pelat Penyambung (tp) = 13 mmDiameter Baut (db) = 12 mm

Luas 1 Buah Baut (Ab) = = 0.25 3.142 1.2 = 1.13097

= 0.5 → Tidak Ada Ulir Pada Bidang Geser = 1.5

m = 1 → Baut Bertampang SatuTinggi End Plate Tambahan = 200 mm

a). Kuat Geser Baut

Vd =

= 0.75 x 0.5 x 5000 x 1.131 x= 2120.57504 kg → Menentukan

b). Kuat Tumpu Baut

Rd =

= 0.75 x 2.4 x 1.2 cm x 1.3 cm x= 11512.8 kg → Tidak Menentukan

c). Kuat Tarik (Ulir) Baut

Td =

= 0.75 x 0.75 x 5000 x 1.131= 3180.86256 kg → Tidak Menentukan

= 2120.57504 kg

d). Kontrol Kekuatan Sambungan Baut (Momen Lentur) <Cara Pendekatan Titik Putar>I). Akibat Beban Sentris Pu → Jumlah Baut (n) = 6 Baut

Vu = Pu / n= 1503.25451 kg / 6 Baut= 250.542419 kg

Vu ≤ → 250.542 kg ≤ 2120.5750412 kg → OK

II). Akibat Beban Momen Lentur Mu

=

= 122781 kgcm 40 cm 2 20 40= 1227.81 kg

≤ Td → 1227.81 kg ≤ 3180.8625618 kg → OK

e). Kontrol Interaksi Geser Dan Tarik

fuv = Vu / Ab ≤= 250.542419 kg / 1.131 ≤ 0.75 0.5 5000

fub kg/cm2

kg/cm2

0.25 x p x db2 cm2

r1 r2

f r1 fub Ab m

kg/cm2 cm2

f 2.4 db tp fu

f 0.75 fub Ab

kg/cm2 cm2

Vnmin

Vnmin

Beban Tarik Max (Tumax)

Tumax Mu dmax / ∑d2

Tumax

f r1 fub m

cm2 kg/cm2

2x x

x / x + 22

x x x

= 221.528136 ≤ 1875

ft = ≤= 1.3 x 5000 - 1.5 x 221.528135583

= 6167.7078 ≥ = 5000

= 5000

Td =

= 0.75 x 5000 x 1.1309733553= 4241.15008 kg

≤ Td → 1227.81 kg ≤ 4241.1500823 kg → OK

Jadi, Sambungan Baut dengan Spesifikasi diatas Memenuhi Kriteria Kuat Geser, Kuat Tumpu, Kuat Tarik (Ulir), Interaksi Geser dan Tarik

C. Kontrol Sambungan Las → f = 0.75 → 1 ksi = 70.3

Mutu Las = → = 70 ksi = 4921

Mutu Pelat = BJ 41 → fu = 4100Tebal Pelat Penyambung (tp) = 13 mmTinggi End Plate Tambahan = 200 mm

a). Kontrol Tebal Las Maksimum Untuk Kekuatan Berimbang

=

= 0.707 x 4100 x 8 mm / 4921= 4.71238 mm

=

= 1.414 x 4100 x 13 mm / 4921= 15.3152 mm

b). Luas Sambungan Las → Misal: te = 1 cm

= 2 x 10 + 2 x 10 + 8 x

= 20 + 20 + 40

= 80

c). Garis Netral Las

ya =

= 374 + 974 + 374 + 387 + 574

= 2683 / 80= 33.5375 cm

yb =

= 40 cm + 20 cm - 2.6 cm - 33.5375= 23.8625 cm

d). Momen Inersia Las (Ix)Ix =

= 166.666667 166.667 11247.6 149.19 266.51 5694 4403 4598= 26691.9208

e). Modulus Elastis Las (Sx)

Sx =

kg/cm2 kg/cm2

1.3 fub - r2 fuv fub

kg/cm2 kg/cm2

kg/cm2 fub kg/cm2

ftAktual kg/cm2

f ft Ab

kg/cm2 cm2

Tumax

FE70XX E70XX kg/cm2

kg/cm2

aeffmax Badan 0.707 fu tw / E70XX

kg/cm2

aeffmax Sayap 1.414 fu tf / E70XX

kg/cm2

ALas

cm2 cm2 cm2

cm2

(2 x 10 x 18.7 + 2 x 10 x 48.7 + 2 x 5 x 37.4 + 2 x 5 x 38.7 + 2 x 5 x 57.4) / ALas

cm3

cm3 cm2

yTotal - ya

2 x (1/12) x 1 x 103 + 2 x (1/12) x 1 x 103 + 2 x 5 x ya2 + 2 x 5 x (37.4-ya)2 + 2 x 5 x (38.7-ya)

cm4

Ix / ymax

+ + +++ + +

x x x

= 26691.9208 / 33.5375 cm

= 795.882843

= f x 0.6 x

= 0.75 x 0.6 x 70 x 70.3

= 2214.45

g). Kontrol Kekuatan Sambungan Las (Momen Lentur) I). Akibat Beban Sentris Pu

fv =

= 1503.25451 kg / 80

= 18.7906814

II). Akibat Beban Momen Lentur Mufh = Mu / Sx

= 122781 kgcm / 795.88

= 154.270193

III). Tegangan Total Yang Terjadi Pada Sambungan Las

=

= 353.089708 + 23799.3= 155.410367

IV). Tebal Las Sudut Yang Diperlukan≥≥ 155.410367 / 2214.5 x 1 cm

≥ 0.07018012 cm

≥≥ 0.07018012 cm / 0.707

≥ 0.09926467 cm

= 0.10 cm = 1 mm

h). Kontrol Tebal Minimum dan Maksimum Las SudutTebal Pelat Penyambung (tp) = 13 mmTebal Pelat Paling Tebal (t) = 13 mm

= 5 mm → 10 < t ≤ 15

= 11.4 mm → tp ≥ 6.4 → a ≤ (tp - 1.6)

= 5 mm

Jadi, Sambungan Las dengan Spesifikasi diatas Memenuhi Kriteria Tebal Las Sudut Minimum dan Maksimum sehingga Dapat Dipakai

cm4

cm3

f). Kuat Rencana Las ( f fn)

f fn E70XX

kg/cm2

kg/cm2

Pu / ALas

cm2

kg/cm2

cm3

kg/cm2

fTotal (fv2 + fh2)1/2

kg/cm2

teperlu fTotal / f fn

teperlu kg/cm2 kg/cm2

teperlu

aperlu teperlu / 0.707

aperlu

aperlu

aperlu

amin Las

amax Las

aPakai

2. PERHITUNGAN SAMBUNGAN KUDA-KUDA DENGAN KOLOM (DETAIL B)A. Beban Berfaktor Untuk Perencanaan Sambungan Baut Dan Las

Mu = 11169.43 kgmPu = 3757.30563 kg


Mutu Baut = BJ 50 → = 5000


Luas 1 Buah Baut (Ab) = = 0.25 3.142 1.9 = 2.83529



a). Kuat Geser Baut

Vd =


b). Kuat Tumpu Baut

Rd =



Td =


= 5316.16382 kg


Vu = Pu / n= 3757.30563 kg / 10 Baut= 375.730563 kg

Vu ≤ → 375.731 kg ≤ 5316.1638185 kg → OK


=

= 1116943 kgcm 45 cm 2 10 25 35 45= 6322.31887 kg

≤ Td → 6322.32 kg ≤ 7974.2457277 kg → OK


fub kg/cm2

kg/cm2

0.25 x p x db2 cm2

r1 r2

f r1 fub Ab m

kg/cm2 cm2

f 2.4 db tp fu

f 0.75 fub Ab

kg/cm2 cm2

Vnmin

Vnmin



Tumax

2x x

x / x + 22 + 2 2+

fuv = Vu / Ab ≤= 375.730563 kg / 2.8353 ≤ 0.75 0.5 5000

= 132.519394 ≤ 1875

ft = ≤= 1.3 x 5000 - 1.5 x 132.519393507

= 6301.22091 ≥ = 5000

= 5000

Td =

= 0.75 x 5000 x 2.8352873699= 10632.3276 kg

≤ Td → 6322.32 kg ≤ 10632.327637 kg → OK

Jadi, Sambungan Baut diatas dapat dipakai


Mutu Las = → = 70 ksi = 4921



=

= 0.707 x 4100 x 8 mm / 4921= 4.71238 mm

=

= 1.414 x 4100 x 13 mm / 4921= 15.3152 mm


= 2 x 25 + 2 x 15 + 8 x

= 50 + 30 + 64

= 144


ya =

= 935 + 1461 + 598.4 + 619.2 + 918.4

= 4532 / 144= 31.4722222 cm

yb =

= 40 cm + 20 cm - 2.6 cm - 31.47222222= 25.9277778 cm

d). Momen Inersia Las (Ix)

f r1 fub m

cm2 kg/cm2

kg/cm2 kg/cm2


kg/cm2 kg/cm2

kg/cm2 fub kg/cm2

ftAktual kg/cm2

f ft Ab

kg/cm2 cm2

Tumax

FE70XX E70XX kg/cm2

kg/cm2


kg/cm2


kg/cm2

ALas

cm2 cm2 cm2

cm2

(2 x 25 x 18.7 + 2 x 15 x 48.7 + 2 x 8 x 37.4 + 2 x 8 x 38.7 + 2 x 8 x 57.4) / ALas

cm3

cm3 cm2

yTotal - ya

x x x

Ix == 2604.16667 562.5 15848 562.22 835.85 10756 8156 8904= 48229.1156


Sx =

= 48229.1156 / 31.4722222222 cm

= 1532.43439

= f x 0.6 x

= 0.75 x 0.6 x 70 x 70.3

= 2214.45


fv =

= 3757.30563 kg / 144

= 26.0924002


= 1116943 kgcm / 1532.4

= 728.8684


=

= 680.813347 + 531249= 729.335285


≥ 0.32935279 cm

≥≥ 0.32935279 cm / 0.707

≥ 0.46584553 cm

= 0.47 cm = 4.7 mm

h). Kontrol Tebal Minimum dan Maksimum Las SudutTebal Pelat Penyambung (tp) = 13 mmTebal Pelat Paling Tebal = 13 mm

= 5 mm → 10 < t ≤ 15

= 11.4 mm → tp ≥ 6.4 → a ≤ (tp - 1.6)

= 5 mm

Jadi, Sambungan Las diatas dapat dipakai

2 x (1/12) x 1 x 253 + 2 x (1/12) x 1 x 153 + 2 x 8 x ya2 + 2 x 8 x (37.4-ya)2 + 2 x 8 x (38.7-ya)

cm4

Ix / ymax

cm4

cm3


f fn E70XX

kg/cm2

kg/cm2

Pu / ALas

cm2

kg/cm2

cm3

kg/cm2


kg/cm2



teperlu


aperlu

aperlu

aperlu

amin Las

amax Las

aPakai

+ + +++ + +

3. PERHITUNGAN SAMBUNGAN KOLOM DENGAN OVERSTECK (DETAIL C)A. Beban Berfaktor Untuk Perencanaan Sambungan Baut Dan Las

Mu = 956.47 kgmPu = 956.462612 kg


Mutu Baut = BJ 50 → = 5000


Luas 1 Buah Baut (Ab) = = 0.25 3.142 1.2 = 1.13097



a). Kuat Geser Baut

Vd =


b). Kuat Tumpu Baut

Rd =



Td =


= 2120.57504 kg


Vu = Pu / n= 956.462612 kg / 4 Baut= 239.115653 kg

Vu ≤ → 239.116 kg ≤ 2120.5750412 kg → OK


fub kg/cm2

kg/cm2

0.25 x p x db2 cm2

r1 r2

f r1 fub Ab m

kg/cm2 cm2

f 2.4 db tp fu

f 0.75 fub Ab

kg/cm2 cm2

Vnmin

Vnmin


2x x

=

= 95647 kgcm 16 cm 2 16 0= 2988.96875 kg

≤ Td → 2988.97 kg ≤ 3180.8625618 kg → OK


fuv = Vu / Ab ≤= 239.115653 kg / 1.131 ≤ 0.75 0.5 5000

= 211.424656 ≤ 1875

ft = ≤= 1.3 x 5000 - 1.5 x 211.424656426

= 6182.86302 ≥ = 5000

= 5000

Td =

= 0.75 x 5000 x 1.1309733553= 4241.15008 kg

≤ Td → 2988.97 kg ≤ 4241.1500823 kg → OK



Mutu Las = → = 70 ksi = 4921



=

= 0.707 x 4100 x 6 mm / 4921= 3.53428 mm

=

= 1.414 x 4100 x 8 mm / 4921= 9.42475 mm


= 2 x 10.9 + 2 x 10.9 + 8 x

= 21.8 + 21.8 + 21.6

= 65.2


ya =

= 118.81 + 373.87 + 58.86 + 63.18 + 122.04

= 736.76 / 65.2


Tumax

f r1 fub m

cm2 kg/cm2

kg/cm2 kg/cm2


kg/cm2 kg/cm2

kg/cm2 fub kg/cm2

ftAktual kg/cm2

f ft Ab

kg/cm2 cm2

Tumax

FE70XX E70XX kg/cm2

kg/cm2


kg/cm2


kg/cm2

ALas

cm2 cm2 cm2

cm2

(2 x 10.9 x 5.45 + 2 x 10.9 x 17.15 + 2 x 2.7 x 10.9 + 2 x 2.7 x 11.7 + 2 x 2.7 x 22.6) / ALas

cm3

cm3 cm2

x / x + 22

x x x

= 11.3 cm

yb =

= 12.5 cm + 11.7 cm - 1.6 cm - 11.3= 11.3 cm


= 215.838167 215.838 689.526 0.864 0.864 689.5 746.1 746.1= 3304.55733


Sx =

= 3304.55733 / 11.3 cm

= 292.438702

= f x 0.6 x

= 0.75 x 0.6 x 70 x 70.3

= 2214.45


fv =

= 956.462612 kg / 65.2

= 14.669672


= 95647 kgcm / 292.44

= 327.066833


=

= 215.199276 + 106973= 327.395651


≥ 0.14784513 cm

≥≥ 0.14784513 cm / 0.707

≥ 0.20911617 cm

= 0.21 cm = 2.1 mm


yTotal - ya

2 x (1/12) x 1 x 10.93 + 2 x (1/12) x 1 x 10.93 + 2 x 2.7 x ya2 + 2 x 2.7 x (ya-10.9)2 + 2 x 2.7 x (11.7-ya)

cm4

Ix / ymax

cm4

cm3


f fn E70XX

kg/cm2

kg/cm2

Pu / ALas

cm2

kg/cm2

cm3

kg/cm2


kg/cm2



teperlu


aperlu

aperlu

aperlu

+ + +++ + +

= 5 mm → 10 < t ≤ 15

= 11.4 mm → tp ≥ 6.4 → a ≤ (tp - 1.6)

= 5 mm


4. PERHITUNGAN SAMBUNGAN BALOK INDUK MELINTANG TEPI DENGAN KOLOM TEPI (DETAIL D)A. Beban Berfaktor Untuk Perencanaan Sambungan Baut Dan Las

Mu = 53047.5 kgm Ry = 1.5 fy = 2500

Pu = 15705.7143 kg = 1286


Mutu Baut = BJ 50 → = 5000


Luas 1 Buah Baut (Ab) = = 0.25 3.142 3 = 7.06858



a). Kuat Geser Baut

Vd =


b). Kuat Tumpu Baut

Rd =

= 0.75 x 2.4 x 3 cm x 1.3 cm x= 28782 kg → Tidak Menentukan


Td =


= 13253.594 kg


Vu = Pu / n

amin Las

amax Las

aPakai

Zx cm3

fub kg/cm2

kg/cm2

0.25 x p x db2 cm2

r1 r2

f r1 fub Ab m

kg/cm2 cm2

f 2.4 db tp fu

f 0.75 fub Ab

kg/cm2 cm2

Vnmin

2x x

= 15705.7143 kg / 10 Baut= 1570.57143 kg

Vu ≤ → 1570.57 kg ≤ 13253.594007 kg → OK


=

= 5304750 kgcm 60 cm 2 0 20 40 50 60= 19647.2222 kg

≤ Td → 19647.2 kg ≤ 19880.391011 kg → OK


fuv = Vu / Ab ≤= 1570.57143 kg / 7.0686 ≤ 0.75 0.5 5000

= 222.190406 ≤ 1875

ft = ≤= 1.3 x 5000 - 1.5 x 222.190405632

= 6166.71439 ≥ = 5000

= 5000

Td =

= 0.75 x 5000 x 7.0685834706= 26507.188 kg

≤ Td → 19647.2 kg ≤ 26507.188015 kg → OK



Mutu Las = → = 70 ksi = 4921



=

= 0.707 x 4100 x 8 mm / 4921= 4.71238 mm

=

= 1.414 x 4100 x 13 mm / 4921= 15.3152 mm


= 2 x 37.4 + 2 x 37.4 + 8 x

= 74.8 + 74.8 + 76.8

Vnmin



Tumax

f r1 fub m

cm2 kg/cm2

kg/cm2 kg/cm2


kg/cm2 kg/cm2

kg/cm2 fub kg/cm2

ftAktual kg/cm2

f ft Ab

kg/cm2 cm2

Tumax

FE70XX E70XX kg/cm2

kg/cm2


kg/cm2


kg/cm2

ALas

cm2 cm2 cm2

x / x + 22 + 2 2+ 2+

x x x

= 226.4


ya =

= 1398.76 + 4293.52 + 718.08 + 743.04 + 1461.12

= 8614.52 / 226.4= 38.05 cm

yb =

= 40 cm + 38.7 cm - 2.6 cm - 38.05= 38.05 cm


= 8718.93733 8718.94 27797.8 8.112 8.112 27798 28007 28006.803= 129063.321


Sx =

= 129063.321 / 38.05 cm

= 3391.9401

= f x 0.6 x

= 0.75 x 0.6 x 70 x 70.3

= 2214.45


fv =

= 15705.7143 kg / 226.4

= 69.3715295


= 5304750 kgcm / 3391.9

= 1563.92797


=

= 4812.40911 + 2445871= 1565.46578


≥ 0.7069321 cm

≥≥ 0.7069321 cm / 0.707

≥ 0.99990396 cm

cm2

(2 x 37.4 x 18.7 + 2 x 37.4 x 57.4 + 2 x 9.6 x 37.4 + 2 x 9.6 x 38.7 + 2 x 9.6 x 76.1) / ALas

cm3

cm3 cm2

yTotal - ya


cm4

Ix / ymax

cm4

cm3


f fn E70XX

kg/cm2

kg/cm2

Pu / ALas

cm2

kg/cm2

cm3

kg/cm2


kg/cm2



teperlu


aperlu

aperlu

+ + +++ + +

= 1.00 cm = 10 mm


= 5 mm → 10 < t ≤ 15

= 11.4 mm → tp ≥ 6.4 → a ≤ (tp - 1.6)

= 10 mm


5. PERHITUNGAN SAMBUNGAN BALOK INDUK MELINTANG TENGAH DENGAN KOLOM TENGAH (DETAIL E)A. Beban Berfaktor Untuk Perencanaan Sambungan Baut Dan Las

Mu = 118098.75 kgm Ry = 1.5 fy = 2500

Pu = 26042.4838 kg = 2863


Mutu Baut = BJ 50 → = 5000


Luas 1 Buah Baut (Ab) = = 0.25 3.142 3 = 7.06858



a). Kuat Geser Baut

Vd =


b). Kuat Tumpu Baut

Rd =



Td =


aperlu

amin Las

amax Las

aPakai

Zx cm3

fub kg/cm2

kg/cm2

0.25 x p x db2 cm2

r1 r2

f r1 fub Ab m

kg/cm2 cm2

f 2.4 db tp fu

f 0.75 fub Ab

kg/cm2 cm2

2x x

= 13253.594 kg


Vu = Pu / n= 26042.4838 kg / 16 Baut= 1627.65523 kg

Vu ≤ → 1627.66 kg ≤ 13253.594007 kg → OK


=

= 11809875 kgcm 100 cm 2 10 20 0 40 60= 19552.7732 kg

≤ Td → 19552.8 kg ≤ 19880.391011 kg → OK


fuv = Vu / Ab ≤= 1627.65523 kg / 7.0686 ≤ 0.75 0.5 5000

= 230.266112 ≤ 1875

ft = ≤= 1.3 x 5000 - 1.5 x 230.266112178

= 6154.60083 ≥ = 5000

= 5000

Td =

= 0.75 x 5000 x 7.0685834706= 26507.188 kg

≤ Td → 19552.8 kg ≤ 26507.188015 kg → OK



Mutu Las = → = 70 ksi = 4921



=

= 0.707 x 4100 x 11 mm / 4921= 6.47952 mm

=

= 1.414 x 4100 x 17 mm / 4921

Vnmin

Vnmin



Tumax

f r1 fub m

cm2 kg/cm2

kg/cm2 kg/cm2


kg/cm2 kg/cm2

kg/cm2 fub kg/cm2

ftAktual kg/cm2

f ft Ab

kg/cm2 cm2

Tumax

FE70XX E70XX kg/cm2

kg/cm2


kg/cm2


kg/cm2

x / x + 22 + 2 2+ 2+ +

x x x

= 20.0276 mm


= 2 x 56.6 + 2 x 56.6 + 8 x

= 113.2 + 113.2 + 76.8

= 303.2


ya =

= 3203.56 + 9803.12 + 1087 + 1119 + 2206.08

= 17418.84 / 303.2= 57.45 cm

yb =

= 60 cm + 58.3 cm - 3.4 cm - 57.45= 57.45 cm


= 30220.2493 30220 63369.6 13.872 13.872 63370 96189 96188.587= 379584.713


Sx =

= 379584.713 / 57.45 cm

= 6607.21867

= f x 0.6 x

= 0.75 x 0.6 x 70 x 70.3

= 2214.45


fv =

= 26042.4838 kg / 303.2

= 85.8920968


= 11809875 kgcm / 6607.2

= 1787.42003


=

= 7377.45229 + 3194870= 1789.48256


≥ 0.80809346 cm

ALas

cm2 cm2 cm2

cm2

(2 x 56.6 x 28.3 + 2 x 56.6 x 86.6 + 2 x 9.6 x 56.6 + 2 x 9.6 x 58.3 + 2 x 9.6 x 114.9) / ALas

cm3

cm3 cm2

yTotal - ya


cm4

Ix / ymax

cm4

cm3


f fn E70XX

kg/cm2

kg/cm2

Pu / ALas

cm2

kg/cm2

cm3

kg/cm2


kg/cm2



teperlu

+ + +++ + +

≥≥ 0.80809346 cm / 0.707

≥ 1.14298933 cm

= 1.15 cm = 11.5 mm


= 6 mm → t > 15

= 15.4 mm → tp ≥ 6.4 → a ≤ (tp - 1.6)

= 12 mm


6. PERHITUNGAN SAMBUNGAN BALOK INDUK MELINTANG TENGAH DENGAN BALOK ANAK (SIMPLE CONNECTION)A. Beban Berfaktor Untuk Perencanaan Sambungan Baut Dan Las

Mu = 0 kgmPu = 17361.74 kg


Mutu Baut = BJ 50 → = 5000

Mutu Pelat = BJ 41 → fu = 4100Siku Penyambung Profil L 50 x 50 x 5 → b = d = 50 mm t = 5Diameter Baut (db) = 20 mm → BOR

Luas 1 Buah Baut (Ab) = = 0.25 3.142 2 = 3.14159


m = 2 → Baut Bertampang Dua

a). Kuat Geser Baut Pada Badan Balok Induk Melintang Tengah

Vd =

= 0.75 x 0.5 x 5000 x 3.142 x= 11780.9725 kg → Tidak Menentukan

b). Kuat Tumpu Baut Pada Badan Balok Induk Melintang Tengah

Rd =

= 0.75 x 2.4 x 2 cm x 1 cm x= 14760 kg → Tidak Menentukan


aperlu

aperlu

aperlu

amin Las

amax Las

aPakai

fub kg/cm2

kg/cm2

0.25 x p x db2 cm2

r1 r2

f r1 fub Ab m

kg/cm2 cm2

f 2.4 db tp fu

2x x

c). Kuat Tarik (Ulir) Baut Pada Badan Balok Induk Melintang Tengah

Td =

= 0.75 x 0.75 x 5000 x 3.142= 8835.72934 kg → Menentukan

= 8835.72934 kg

d). Kuat Geser Baut Pada Badan Balok Anak

Vd =

= 0.75 x 0.5 x 5000 x 3.142 x= 11780.9725 kg → Tidak Menentukan

e). Kuat Tumpu Baut Pada Badan Balok Anak

Rd =


f). Kuat Tarik (Ulir) Baut Pada Badan Balok Anak

Td =

= 0.75 x 0.75 x 5000 x 3.142= 8835.72934 kg → Menentukan

= 8835.72934 kg

g). Menghitung Jumlah Kebutuhan Baut Pada Badan Balok Induk Melintang Tengah (n)

n = 2 x Pu /

= 2 x 17361.7 kg / 8836 kg= 3.92989403 Buah → n = 4 Buah

h). Menghitung Jumlah Kebutuhan Baut Pada Badan Balok Anak (n)

n = Pu /

= 17361.74 kg / 8835.7 kg= 1.96494702 Buah → n = 2 Buah

i). Kontrol Block Shear Pada Siku Penyambung Badan Balok Induk Melintang Tengah Dan Balok Anak

Anv = =

= 15 cm - 2 x 2.15 cm x 0.5 cm= 5.35

=

= 2 x 0.75 x 0.6 x 4100 x 5.35

= 19741.5 kg ≥ Pu = 17361.74 kg → OK

f 0.75 fub Ab

kg/cm2 cm2

f Rnmin

f r1 fub Ab m

kg/cm2 cm2

f 2.4 db tp fu

f 0.75 fub Ab

kg/cm2 cm2

f Rnmin

f Rnmin

f Rnmin

Lnv x tL (L - n fp) x tL

cm2

2 f Pn 2 x f x (0.6 x fu x Anv)

kg/cm2

PERENCANAAN SAMBUNGAN BAUT DAN LAS

1

4100

1

kg/cm2

x

Jadi, Sambungan Baut dengan Spesifikasi diatas Memenuhi Kriteria Kuat Geser, Kuat Tumpu, Kuat Tarik (Ulir), Interaksi Geser dan Tarik

5

/ 80

33.5375 cm

kg/cm2

kg/cm2

kg/cm2

kg/cm2

cm2

+ 2 x 5 x (38.7-ya)2 + 2 x 5 x yb2 + 2 x 10 x (ya-18.7)2 + 2 x 10 x (48.7-ya)2

x

Jadi, Sambungan Las dengan Spesifikasi diatas Memenuhi Kriteria Tebal Las Sudut Minimum dan Maksimum sehingga Dapat Dipakai

1

4100 kg/cm2

1

Jadi, Sambungan Baut dengan Spesifikasi diatas Memenuhi Kriteria Kuat Geser, Kuat Tumpu, Kuat Tarik (Ulir), Interaksi Geser dan TarikJadi, Sambungan Las dengan Spesifikasi diatas Memenuhi Kriteria Tebal Las Sudut Minimum dan Maksimum sehingga Dapat Dipakai

8

/ 144

31.47222222 cm

kg/cm2

kg/cm2

kg/cm2

kg/cm2

cm2

x

+ 2 x 8 x (38.7-ya)2 + 2 x 8 x yb2 + 2 x 25 x (ya-18.7)2 + 2 x 15 x (48.7-ya)2

1

4100 kg/cm2

1

2.7

/ 65.2

kg/cm2

kg/cm2

kg/cm2

kg/cm2

(2 x 10.9 x 5.45 + 2 x 10.9 x 17.15 + 2 x 2.7 x 10.9 + 2 x 2.7 x 11.7 + 2 x 2.7 x 22.6) / ALas

cm2

x

11.3 cm

+ 2 x 2.7 x (ya-10.9)2 + 2 x 2.7 x (11.7-ya)2 + 2 x 2.7 x yb2 + 2 x 10.9 x (ya-5.45)2 + 2 x 10.9 x (17.15-ya)2

4. PERHITUNGAN SAMBUNGAN BALOK INDUK MELINTANG TEPI DENGAN KOLOM TEPI (DETAIL D)

1

4100

kg/cm2

kg/cm2

1

9.6

kg/cm2

kg/cm2

kg/cm2

kg/cm2

2

x

/ 226.4

38.05 cm

cm2


5. PERHITUNGAN SAMBUNGAN BALOK INDUK MELINTANG TENGAH DENGAN KOLOM TENGAH (DETAIL E)

1

4100

kg/cm2

kg/cm2

0 80 90 100

1

kg/cm2

kg/cm2

kg/cm2

kg/cm2

2 2 2 2+ ++ + 2

x

9.6

/ 303.2

57.45 cm

(2 x 56.6 x 28.3 + 2 x 56.6 x 86.6 + 2 x 9.6 x 56.6 + 2 x 9.6 x 58.3 + 2 x 9.6 x 114.9) / ALas

cm2


6. PERHITUNGAN SAMBUNGAN BALOK INDUK MELINTANG TENGAH DENGAN BALOK ANAK (SIMPLE CONNECTION)

mm

2

4100 kg/cm2

2

4100

i). Kontrol Block Shear Pada Siku Penyambung Badan Balok Induk Melintang Tengah Dan Balok Anak → = 20 mm + 1.5 mm

= 21.5 mm

Agt = 2.5 cm x 0.5 cm

= 1.25

Ant = 2.5 cm - 0.5 x 2.15 cm x

= 0.7125

Agv = 11 cm x 0.5 cm= 5.5

Anv = 11 cm - 1.5 x 2.15 cm x= 3.8875

fu Ant = 4100 x 0.7125= 2921.25 kg

kg/cm2

fp

fp

cm2

cm2

cm2

cm2

cm2

kg/cm2 cm2

0.6 fu Anv = 0.6 x 4100 x 3.8875= 9563.25 kg

=

=

kg/cm2 cm2

f Rn

Jadi, Sambungan Baut dengan Spesifikasi diatas Memenuhi Kriteria Kuat Geser, Kuat Tumpu, Kuat Tarik (Ulir), Interaksi Geser dan TarikJadi, Sambungan Las dengan Spesifikasi diatas Memenuhi Kriteria Tebal Las Sudut Minimum dan Maksimum sehingga Dapat Dipakai

0.5 cm

0.5 cm

PERENCANAAN PELAT LANDASAN KOLOM 1. PERHITUNGAN SAMBUNGAN KOLOM PONDASI <COLUMN BASE PLATE> (DETAIL 5)A. Beban Berfaktor Untuk Perencanaan Pelat Landasan Kolom

Mu = 10079.7 kgmPu = 35494 kgHu = 4536.8 kg

B. Kontrol Sambungan Kolom Pondasi (Column Base Plate)

Kolom Profil WF 350 x 350 x 13 x 13 → d = 338 mm = 351

Mutu Beton (f'c) = 25 MPa = 250

Diameter Baut Angker Perkiraan = 16 mm → Ab = 2.011We = 22.225 mm → We = 27.5 mm

= 25.4 mm → = 27.5 mm

h' = = 55 mm

a). Dimensi Pelat Landasan Kolom

H = 448 mm

h = H - We = 420.5 mm

B = 461 mm

b). Dimensi Pondasi BetonPanjang (P) = H + 2S = 448 mm + 2 x 50 mm = 548

Lebar (L) = B + 2S = 461 mm + 2 x 50 mm = 561

P/H = 548 mm / 448 mm = 1.223L/B = 561 mm / 461 mm = 1.217

Mutu Beton Modifikasi (f'cu) =

f'cu = 0.85 x 1.21692 x 250 = 258.595444685

Eksentrisitas (e) = Mu / Pu = 10080 kgm / 35494 kge = 0.2839832084 m = 28.398 cm ≥ H/6 = 7.46667 cm

c). Tebal Pelat Baja → Bm = B = 461 mmI). Arah H

≥≥ 5.5 cm 1.333 258.6 461 / 2500 461

≥ 2.0422936875 cm → ≥ 2.1 cm = 21 mm

II). Arah B

b = = 461 - 0.8 x 351 / 2

= 90.1 mm= 9.01 cm

≥≥ 9.01 cm 1.333 258.6 461 / 2500 461

≥ 3.3456483862 cm → ≥ 3.4 cm = 34 mm

III). Pengaruh Tu → f = 0.6

Garis Netral

a =

bf

kg/cm2

cm2

C1 C1

We + C1

d + 2 (We +C1) =

bf + 2 (We +C1) =

0.85 f'c (A2/A1)0.5

kg/cm2 kg/cm

t1 h' (1.333 f'cu Bm / (fy B))0.5

t10.5 '

t1 t1

(B - 0.8 bf) / 2

t2 b (1.333 f'cu Bm / (fy B))0.5

t20.5 '

t2 t2

h - (h2 - (Pu (2h - H) + 2 Mu) / (f f'cu B))0.5

x x x

x x x

2 x x - + x

= 42.05 - 42.05 - 35494 2 42.05 44.8 2 1007970= 6.1147337299 cm

Cu =

= 0.6 x 258.595 x 46.1 cm x 6.11473 cm= 43737.161686 kg

Tu = Cu - Pu= 43737.161686 kg - 35494 kg= 8243.1616862 kg

Kuat Tarik (Ulir) Baut → f = 0.75

Mutu Baut = BJ 50 → = 5000Diameter Baut (db) = 16 mm

Luas 1 Buah Baut (Ab) = = 0.25 3.142 1.6 = 2.01062


m = 1 → Baut Bertampang Satu

Td =

= 0.75 x 0.75 x 5000 x 2.011= 5654.8667765 kg

Kebutuhan Jumlah Baut (n) = Tu / Td = 8243.161686216 kg / 5654.866776 kgn = 1.4577110323 ≈ 2 Baut / Sisi = 4 Baut

Kontrol Interaksi Geser Dan Tarik (Badan Baut)

fuv = Hu / n Ab ≤= 4536.8 kg / 4 / 2.011 ≤ 0.75 0.5

= 564.10480142 ≤ 1875 → Kuat Geser Memenuhi

ft = ≤= 1.3 x 5000 - 1.5 x 564.104801421

= 5653.8427979 ≥ = 5000

= 5000

Td =

= 0.75 x 5000 x 2.0106192983= 7539.8223686 kg

Kebutuhan Jumlah Baut (n) = Tu / Td = 8243.161686216 kg / 7539.822369 kgn = 1.0932832742 ≈ 2 Baut / Sisi = 4 Baut

= 5654.8667765 kg → Kebutuhan Jumlah Baut (n) = 2 Baut / Sisi

≥≥ 2.108 x 8243.16 kg 5.5 2.75 cm / 2500 46.1

≥ 0.9348962024 cm → ≥ 1 cm = 10 mm

= 21 mm

= 34 mm

= 10 mm

f f'cu B a

kg/cm2

fub kg/cm2

0.25 x p x db2 cm2

r1 r2

f 0.75 fub Ab

kg/cm2 cm2

f r1 fub m

cm2

kg/cm2 kg/cm2


kg/cm2 kg/cm

kg/cm2 fub kg/cm2

ftAktual kg/cm2

f ft Ab

kg/cm2 cm2

Tdmin

t3 2.108 (Tu (h' - We) / (fy B))0.5

t3

t3 t3

t1

t2

t3

2 x x - + x

2x x

x x

- xx

= 34 mm

=

= 8243.1616862 kg / 2 Angker / 3.142 1.6 cm /= 51.858979115 cm ≈ 52 cm

60 cm ≥ 52 cm → OK

tpakai

d). Panjang Penyaluran Angker (LAngker)

LAngker Tu (1 Angker) / (Keliling Baut x (f'c)1/2)

LAngkerpakai =

x

PERENCANAAN PELAT LANDASAN KOLOM

mm

mmmm

→ e > H/6

kg/cm2

0.5/ x x

1007970 0.6 259 46.1

5000 1

= 4 Baut

kg/cm2

kg/cm2

0.5 '

0.5/ x x

x x

250 0.5 '

PERENCANAAN TANGGA

Modulus Elastisitas (E) = 2E+06 Modulus Geser (G) = 800000I. RENCANA TANJAKAN DAN INJAKANSyarat-Syarat Perencanaan Tanjakan, Injakan dan Kemiringan Tangga

60 cm ≤ 2t+i ≤ 65 cm → Syarat 1

25° ≤ α ≤ 40° → Syarat 2

t = Tinggi Tanjakani = Lebar Injakan

α = Kemiringan TanggaTinggi Tangga = 225 cm = 2.25 mTinggi Tanjakan (t) = 15 cm = 0.15 mLebar Injakan (i) = 30 cm = 0.3 mJumlah Tanjakan = 225 / 15 = 15 BuahJumlah Injakan = 15 - 1 = 14 BuahPanjang Bordes = 330 cm = 3.3 mLebar Bordes = 150 cm = 1.5 m

arctan 225 / 420 = 28.18 °

Cek Syarat TanggaSyarat 1 → 60 cm ≤ 2t+i ≤ 65 cm

→ 60 cm ≤ 60 ≤ 65 cm → OK

Syarat 2 → 25° ≤ α ≤ 40°

→ 25 ≤ 28.18 ≤ 40 → OK

II. PERENCANAAN PELAT ANAK TANGGA

= 7850 BJ 41

Lebar Tangga = 1.5 m fu 4100

Tebal Pelat Anak Tangga Rencana = 3 mm = 0.003 m fy 2500

A. Beban Mati (D)

= Berat Sendiri Pelat Anak Tangga

=

= 0.003 m x 1.5 m x 7850

= 35.325 kg/m = 0.353 kg/cm

kg/cm2 kg/cm2

Kemiringan Tangga (α) =

gBaja kg/m3

kg/cm2

kg/cm2

qD

qD Tebal Pelat x Lebar Tangga x gBaja

qD kg/m3

qD

B. Beban Hidup (L) → Beban Hidup Lantai Tangga = 300

= Beban Hidup Lantai Tangga x Lebar Tangga

= 300 x 1.5 m

= 450 kg/m = 4.5 kg/cm

C. Beban Kombinasi (Ultimate)

1.2 + 1.6

1.2 x 35.33 kg/m + 1.6 x 450 kg/m

762.39 kg/m

= 0.125 x 762.39 kg/m x 0.3

= 8.5768875 kgm

Vu = = 0.5 x 762.39 kg/m x 0.3 m

Vu = 114.3585 kg

E. Kontrol Tekul Lateral → Tegangan Residu (fr)= 70 MPa = 700

= 0.3 m = 30 cm

= fy - fr = 2500 - 700 = 1800

Luas Penampang Pelat Anak Tangga (A)A = Lebar Tangga x Tebal Pelat Anak TanggaA = 150 cm x 0.3 cm = 45

Konstanta Puntir Torsi (J)J = =

J = 0.3333333 x 150 cm x 0.3

J = 1.35

= = 0.083 x 150 cm x 0.3

= 0.3375

= = 0.083 x 0.3 cm x 150

= 84375

= = 84375 / 45

= 43.30127 cm

= = 0.167 x 150 cm x 0.3

= 2.25

= = 0.25 x 150 cm x 0.3

= 3.375

Mp = Zx fy = 3.375 x 2500 = 8437.5 kgcm

Lp =

Lp = 0.13 2E+06 43.3 1.35 45 8437.5Lp = 10400 cm

= Sx (fy-fr) = 2.25 x 1800 = 4050 kgcm

kg/m2

qL

qL kg/m2

qL

qu = qD qL

qu =

qu =


Mumax

1/2 x qu x L

kg/cm2

Panjang Anak Tangga (LB)

fL kg/cm2

cm2

∑ 1/3 bt3 1/3 x b x t3

cm3

cm4

Momen Inersia Sumbu x (Ix)

Ix 1/12bt3 cm3

Ix cm4

Momen Inersia Sumbu y (Iy)

Iy 1/12tb3 cm3

Iy cm4

Jari-Jari Girasi Sumbu y (iy)

iy Iy / A cm4 cm2

iy

Modulus Elastis Sumbu x (Sx)

Sx 1/6bt2 cm2

Sx cm3

Modulus Plastis Sumbu x (Zx)

Zx 1/4bt2 cm2

Zx cm3

cm3 kg/cm2

0.13 Eiy JA / Mp

MR cm3 kg/cm2

2

3

3

3

2

2

x xx x /

=

= 2 2E+06 43.3 1.35 45 4050

= 333333.33 cm

= 30 cm

Lp = 10400 cm → LB < Lp → BENTANG PENDEK

= 333333.33 cm

F. Kekuatan Nominal Momen Lentur (Mn)TEKUK LATERAL

Mn = Mp = Zx fyMn = 8437.5 kgcmMn = 84.375 kgm

G. Kontrol Kuat Momen Lentur → f = 0.9

≤ → 8.5768875 kgm ≤ 0.9 x 84.375 kgm

→ 8.5768875 kgm ≤ 75.9375 kgm OK

H. Kontrol Lendutan

= L/240 = 30 / 240 = 0.125 cm


y =

y = 5/384 4.85 30 / 2000000 0.337

y = 0.075832 cm ≤ = 0.125 cm OK

LR 2 Eiy JA / MR

LR

LR

LB

LR

Mumax f Mn

Lendutan Ijin → ymax

5 / 384 x (qD+qL) x L4 / (EIx)

ymax

x xx x /

4x x x

Lp = =0.13 Eiy JA / Mp

PERHITUNGAN TANGGAV. Balok Pembagi → Berat Volume Baja = 7850

→ Panjang Pelat Tangga (L) = 1.5 m1. PERHITUNGAN BALOK PEMBAGI → Lebar Pelat Tangga = 0.3 mDATA PERENCANAAN → Tebal Pelat Tangga = 0.003 mDirencanakan Menggunakan Profil WF 100 x 50 x 5 x 7

W = 9.3 kg/m h = 5.91 cm³

d = 100 mm = 70 mm 42 cm³

= 50 mm = 187 9 cm³

= 5 mm = 14.8 BJ 41

= 7 mm = 3.98 cm fu 4100

r = 8 mm = 1.12 cm fy 2500

Ag = 11.85 = 37.5 cm³

A. Beban Mati (D)Berat Sendiri Balok Pembagi = 9.30 kg/m = 0.093 kg/cm

Berat Pelat = Lebar Pelat x Tebal Pelat x Berat Volume Baja= 0.3 m x 0.003 m x 7850= 7.065 kg/m

Berat Total = Berat Sendiri Balok Pembagi + Berat Pelat= 9.30 kg/m + 7.065 kg/m= 16.37 kg/m

Berat Alat-Alat Pengikat 10 % = 0.1 x 16.37 kg/m = 1.637 kg/m

= 16.37 kg/m + 1.6365 kg/m

= 18.00 kg/m

B. Beban Hidup (L)

Beban Hidup Lantai Tangga = 300

= Beban Hidup Lantai Tangga x Lebar Pelat Tangga

= 300 x 0.3 m= 90 kg/m

kg/m3

d - 2(tf+r) Sy =

Zx =

bf Ix cm⁴ Zy =

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx

kg/m3

Beban Mati (qD)

kg/m2

Beban Hidup (qL)

kg/m2

25 cm

50 cm

150 cm

150 cm

Balok Pembagi


1.2 + 1.6

1.2 x 18.00 kg/m + 1.6 x 90 kg/m

21.6018 kg/m + 144 kg/m 165.6 kg/m

L = 1.5 m

= 165.6018 kg/m

= 0.5 x 165.6 kg/m x 1.5 m

= 124.20135 kg

= 0.125 x 165.6 kg/m x 1.5

= 46.575506 kgm = 4657.550625 kgcm

D. Kontrol Penampang (Tekuk Lokal)Badan (Web)

≤ → 70 / 5 ≤ 1680 / 250

→ 14 ≤ 106.25 → OK

Sayap (Flange) 1≤ → 50 / 14 ≤ 170 / 250

→ 3.571 ≤ 10.752 → OK


E. Kontrol Tekul Lateral → Tegangan Residu (fr)= 70 MPa = 700Bentang Balok Pembagi = 0.3 mPanjang Balok Pembagi = 1.5 m = 150 cm

Lp = = 1.76 x 1.12 cm x 2E+06 / 2500

Lp = 55.753955 cm ≈ 56 cm

= fy - fr = 2500 - 700 = 1800

= 0 cm



F. Kekuatan Nominal Momen Lentur (Mn)TEKUK LOKAL TEKUK LATERAL

Mn = 105000 kgcm Mn = Mp = Zx fyMn = 1050 kgm Mn = 1050 kgm

= 1050 kgm


≤ → 46.57550625 kgm ≤ 0.9 x 1050 kgm

→ 46.57550625 kgm ≤ 945 kgm OK

qu = qD qL

qu =

qu = =

qu

qu

Vu 1/2 x qu x L =

Vu


Mumax

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

kg/cm2

kg/cm2

1.76 iy E/fy

fL kg/cm2

LB

cm3 kg/cm2

MnAktual

Mumax f Mn

2

H. Kontrol Lendutan

= L/360 = 150 / 360 = 0.4167 cm


y =

y = 5/384 0.18 0.9 x 150 / 2000000 x 187

y = 0.0190354 cm ≤ = 0.416666667 cm OK

Lendutan Ijin → ymax

5 / 384 x (qD+qL) x L4 / (EIx)

ymax

4x +

41.8

9.288

kg/cm2

kg/cm2

PEMBEBANAN PORTAL GEDUNGJarak Antar Kuda-Kuda (b) = 6 mModulus Elastisitas (E) = 2E+06 Modulus Geser (G) = 8000001. PEMBEBANAN PORTAL MELINTANGA. Beban Atap

Berat Gording =

= 8.815 kg/m / 1.355 m x 6 m= 39.0259 kg/m

Berat Asbes =

= 12.5 x 1.15 x 6 m / cos 18

= 90.6886 kg/m

Berat Kuda-Kuda WF 300 x 200 x 8 x 12 (Perkiraan) → W = 56.6 kg/m

Berat Kuda-Kuda =

= 56.6 kg/m / cos 18= 59.5128 kg/m

Berat Total = Berat Gording + Berat Asbes + Berat Kuda-Kuda= 39.0259 kg/m + 90.69 kg/m + 59.51 kg/m= 189.2272292868 kg/m


= 208.1499522155 kg/m

Beban Hujan q = 20Beban Hidup (qL) = q x Bentang Antar Kuda-Kuda

= 20 x 6 m= 120 kg/m

c). Beban Angin (W) → Tekanan Tiup Angin (W) Dekat Pantai = 40

Angin Tekan → = = 0.02 x 18 - 0.4

= -0.04 → Hisap


=

= -0.04 x 40 x 6 m

= -9.6 kg/m → Di Pihak Angin

=

= -0.4 x 40 x 6 m= -96 kg/m → Di Belakang Angin

kg/cm2 kg/cm2

a). Beban Mati Atap (qD)

WGording / Jarak Horizontal Gording x Bentang Antar Kuda-Kuda

WAsbes x Koefisien Impitan x Bentang Antar Kuda-Kuda / cos αkg/m2

WKuda-Kuda / cos α

Beban Mati (qD)

b). Beban Hidup Atap (qL)

kg/m2

kg/m2

kg/m2

C1 0.02α-0.4

C1

C2


kg/m2


kg/m2


= Berat Profil Balok Induk Melintang WF 500 x 200 x 9 x 14

= 79.5 kg/m



= 2.33333 m x 10.1

= 23.5667 kg/m


= 2.33333 m x 0.09 m x 2400= 504 kg/m

Beban Spesi 2 cm = Jarak Antar Balok Anak x Tebal Spesi x Berat Volume Spesi

= 2.33333 m x 0.02 m x 2100= 98 kg/m

Beban Tegel 2 cm = Jarak Antar Balok Anak x Tebal Tegel x Berat Volume Tegel Beton= 2.33333 m x 0.02 m x 2400= 112 kg/m

Beban Plafond + Penggantung Plafond = Jarak Antar Balok Anak x Beban Hidup Plafond= 2.333 m x 11 + 7= 42 kg/m


= 23.57 + 504 + 98 + 112 + 42 + 56.8

= 836.4 kg/m


= 836.4 kg/m + 0.1 x 56.8 kg/m

= 842 kg/m

=

= 0.5 x 842 kg/m x 6 m x 2

= 5052 kg

= Jarak Antar Balok Anak x Beban Hidup Lantai Gudang

= 2.333 m x 400

= 933.3 kg/m

=

= 0.5 x 933.33 kg/m x 6 m x 2

= 5600 kg

kg/m2

qD

qD

I. Beban Terpusat P1

i). P1 Mati (P1D)

kg/m2

kg/m3

kg/m3

kg/m3

kg/m2

qTotal

qTotal

q1D

q1D

q1D

P1D 1/2 x q1D x L x 2

P1D

P1D

ii). P1 Hidup (P1L)

q1L

q1L kg/m2

q1L

P1L 1/2 x q1L x L x 2

P1L

P1L

→ Jarak Antar Balok Anak Bentang 8 m = 2 m


= 2 m x 10.1

= 20.2 kg/m


= 2 m x 0.09 m x 2400= 432 kg/m

Beban Spesi 2 cm = Jarak Antar Balok Anak x Tebal Spesi x Berat Volume Spesi= 2 m x 0.02 m x 2100= 84 kg/m

Beban Tegel 2 cm = Jarak Antar Balok Anak x Tebal Tegel x Berat Volume Tegel Beton= 2 m x 0.02 m x 2400= 96 kg/m

Beban Plafond + Penggantung Plafond = Jarak Antar Balok Anak x Beban Hidup Plafond= 2 m x 11 + 7= 36 kg/m


= 20.2 + 432 + 84 + 96 + 36 + 56.8

= 725 kg/m


= 725 kg/m + 0.1 x 56.8 kg/m

= 730.68 kg/m

=

= 0.5 x 730.68 kg/m x 6 m x 2

= 4384 kg

= Jarak Antar Balok Anak x Beban Hidup Lantai Gudang

= 2 m x 400

= 800 kg/m

=

= 0.5 x 800 kg/m x 6 m x 2

= 4800 kg

→ Bentang Pelat Yang Dipikul = 2.167 m


= 2.16667 m x 10.1

II. Beban Terpusat P2

i). P2 Mati (P2D)

kg/m2

kg/m3

kg/m3

kg/m3

kg/m2

qTotal

qTotal

q2D

q2D

q2D

P2D 1/2 x q2D x L x 2

P2D

P2D

ii). P2 Hidup (P2L)

q2L

q2L kg/m2

q2L

P2L 1/2 x q2L x L x 2

P2L

P2L

III. Beban Terpusat P3

i). P3 Mati (P3D)

kg/m2

= 21.8833 kg/m


= 2.16667 m x 0.09 m x 2400= 468 kg/m




Beban Profil Balok Induk Memanjang WF 400 x 200 x 8 x 13 = 66 kg/m

= 21.88 + 468 + 91 + 104 + 39 + 66

= 789.9 kg/m


= 789.9 kg/m + 0.1 x 66 kg/m

= 796.5 kg/m

=

= 0.5 x 796.48 kg/m x 6 m x 2

= 4778.9 kg

= Bentang Pelat Yang Dipikul x Beban Hidup Lantai Gudang

= 2.167 m x 400

= 866.7 kg/m

=

= 0.5 x 866.67 kg/m x 6 m x 2

= 5200 kg

Beban Pelat Bondex = 1/2 Jarak Antar Balok Anak x Berat per satuan Luas Pelat Bondex

= 1.16667 m x 10.1

= 11.7833 kg/m

Beban Pelat Beton = 1/2 Jarak Antar Balok Anak x Tebal Pelat Beton x Berat Volume Beton

= 1.16667 m x 0.09 m x 2400= 252 kg/m

Beban Spesi 2 cm = 1/2 Jarak Antar Balok Anak x Tebal Spesi x Berat Volume Spesi= 1.16667 m x 0.02 m x 2100= 49 kg/m

kg/m3

kg/m3

kg/m3

kg/m2

qTotal

qTotal

q3D

q3D

q3D

P3D 1/2 x q3D x L x 2

P3D

P3D

ii). P3 Hidup (P3L)

q3L

q3L kg/m2

q3L

P3L 1/2 x q3L x L x 2

P3L

P3L

IV. Beban Terpusat P4

i). P4 Mati (P4D)

kg/m2

kg/m3

kg/m3

Beban Tegel 2 cm = 1/2 Jarak Antar Balok Anak x Tebal Tegel x Berat Volume Tegel Beton= 1.16667 m x 0.02 m x 2400= 56 kg/m

Beban Plafond + Penggantung Plafond = 1/2 Jarak Antar Balok Anak x Beban Hidup Plafond= 1.167 m x 11 + 7= 21 kg/m

Beban Profil Balok Induk Memanjang WF 400 x 200 x 8 x 13 = 66 kg/m

= 11.78 + 252 + 49 + 56 + 21 + 66

= 455.8 kg/m


= 455.8 kg/m + 0.1 x 66 kg/m

= 462.4 kg/m

=

= 0.5 x 462.38 kg/m x 6 m x 2

= 2774.3 kg

= 1/2 Jarak Antar Balok Anak x Beban Hidup Lantai Gudang

= 1.167 m x 400

= 466.7 kg/m

=

= 0.5 x 466.67 kg/m x 6 m x 2

= 2800 kg


= → = 0.9

= 0.9 x 40 x 6 m

= 216 kg/m

= → = -0.4

= -0.4 x 40 x 6 m

= -96 kg/m

D. Beban Horizontal Akibat Gempa (E) Arah Melintanga). Beban Atap → Bentang Atap Horizontal = 26 m

= 208.149952215 kg/m

=

= 208.1 kg/m x 26 m

= 5412 kg

= 120 kg/m

kg/m3

kg/m2

qTotal

qTotal

q4D

q4D

q4D

P4D 1/2 x q4D x L x 2

P4D

P4D

ii). P4 Hidup (P4L)

q4L

q4L kg/m2

q4L

P4L 1/2 x q4L x L x 2

P4L

P4L

kg/m2


qW1 kg/m2

qW1


qW2 kg/m2

qW2

qDAtap

W1 qDAtap x Bentang Atap Horizontal

W1

W1

qLAtap

=

= 120 kg/m x 26 m

= 3120 kg

Berat Kolom → Coba Profil Kolom WF 250 x 250 x 14 x 14 → W = 82.2 kg/m

= Berat Kolom x (1/2 x Tinggi Kolom Lantai 1) x 2 Kolom (Kanan Kiri)

= 82.2 kg/m x 2.5 m x 2 Kolom

= 411 kg

= + +

= 5412 kg + 3120 kg + 411 kg

= 8943 kg

b). Beban Lantai → Bentang Antar Kolom Utama = 22 mBerat Sendiri Balok Induk Melintang WF 500 x 200 x 9 x 14 = 79.5 kg/m

=

= 79.5 kg/m x 22 m

= 1749 kg


= Berat Kolom x (1/2x(Tinggi Kolom <Lantai Dasar+Lantai 1)) x 2 Kolom (Kanan Kiri)

= 82.2 kg/m x 4.75 m x 2 Kolom

= 780.9 kg

Berat Balok dan Pelat Memanjang

=

= 4 5052.28 5600 3 4384 4800 2 4779 5200 2774 2800

= 42609 kg + 27552 kg + 31106 kg

= 101267.76 kg

= + +

= 1749 kg + 780.9 kg + 101268 kg

= 103797.66 kg


=

= 8943 kg + 103797.66 kg

= 112740.5588 kg

Wihi

Atap →→ 8943 kg x 9.5 m = 84958 kg

Lantai 1 →→ 103797.66 kg x 4.5 m = 467089.47 kg

∑Wihi =

∑Wihi = 84958 kg + 467089.47 kg∑Wihi = 552047.008197 kg

d). Periode Getar Bangunan (T) → Ct = 0.085 Untuk Rangka Momen BajaT = hn = Tinggi Struktur Bangunan = 9.5 m

W2 qLAtap x Bentang Atap Horizontal

W2

W2

W3

W3

W3





W1

W1

W2

W2

W2

W3 4 (P1D+P1L) + 3 (P2D+P2L) + 2 (P3D+P3L) + 2 (P4D+P4L)

W3

W3

W3








WLantaiTotal x Tinggi Kolom Lantai Dasar =


Ct(hn)3/4

x + ++ ++x x+ +

3/4

T = 0.085 x 9.5T = 0.46 detik


V =




V = 0.5 x 1 x 112740.55876 kg / 4.5V = 12527 kg


Atap → = 84958 / 552047.0082 x 12527 kg

→ = 1928 kg

Lantai 1 → = 467089.47 / 552047.0082 x 12527 kg

→ = 10599 kg

2. PEMBEBANAN PORTAL MEMANJANGA. Beban Atap → Sosoran Atap Melintang dan Memanjang = 2 ma). Beban Terpusat Atap → Jumlah Gording Total = 22 Buah

Berat Gording =

= 9.255 kg/m x 5 m x 11 Buah= 509.025 kg

Berat Asbes =

= 12.5 x 1.15 x 65 / cos 18= 982.46 kg


Berat Kuda-Kuda =

= 56.6 kg/m / cos 18 x 13 m= 773.666 kg



= 2491.6660124 kg

Beban Hujan q = 20


= 20 x 65

= 1300 kg

CIWt / R

Wt


F1

F1

F2

F2


i). P5 Mati (P5D)




P5D

ii). P5 Hidup (P5L)

kg/m2

P5L

P5L kg/m2 m2

P5L

3/4

Berat Gording =

= 9.255 kg/m x 6 m x 11 Buah= 610.83 kg

Berat Asbes =

= 12.5 x 1.15 x 78 / cos 18= 1179 kg


Berat Kuda-Kuda =

= 56.6 kg/m / cos 18 x 13 m= 773.666 kg



= 2819.79271587 kg

Beban Hujan q = 20


= 20 x 78

= 1560 kg



= 66 kg/m


Beban Pelat Bondex = 1/2 Bentang Kuda-Kuda x Berat per satuan Luas Pelat Bondex

= 3 m x 10.1

= 30.3 kg/m

Beban Pelat Beton = 1/2 Bentang Kuda-Kuda x Tebal Pelat Beton x Berat Volume Beton

= 3 m x 0.09 m x 2400= 648 kg/m

Beban Spesi 2 cm = 1/2 Bentang Kuda-Kuda x Tebal Spesi x Berat Volume Spesi= 3 m x 0.02 m x 2100= 126 kg/m

Beban Tegel 2 cm = 1/2 Bentang Kuda-Kuda x Tebal Tegel x Berat Volume Tegel Beton


i). P6 Mati (P6D)




P6D

ii). P6 Hidup (P6L)

kg/m2

P6L

P6L kg/m2 m2

P6L

kg/m2

qD

qD

I. Beban Terpusat P7

i). P7 Mati (P7D)

kg/m2

kg/m3

kg/m3

= 3 m x 0.02 m x 2400= 144 kg/m

Beban Plafond + Penggantung Plafond = 1/2 Bentang Kuda-Kuda x Beban Hidup Plafond= 3 m x 11 + 7= 54 kg/m

Beban Profil Balok Induk Melintang WF 500 x 200 x 9 x 14 = 106 kg/m

= 30.3 + 648 + 126 + 144 + 54 + 106

= 1108.3 kg/m

= qTotal + Berat Alat Pengikat 10 % x Beban Profil Balok Induk Melintang

= 1108.3 kg/m + 0.1 x 106 kg/m

= 1118.9 kg/m


= 1/2 x Beban Profil Balok Anak x Bentang Kuda-Kuda

= 0.5 x 56.8 kg/m x 6 m

= 170.4 kg

=

= 0.5 x 1118.9 kg/m x 7 m + 170.4 kg

= 4087 kg

= 1/2 Bentang Kuda-Kuda x Beban Hidup Lantai Gudang

= 3 m x 400

= 1200 kg/m

=

= 0.5 x 1200 kg/m x 7 m

= 4200 kg

Beban Pelat Bondex = Bentang Kuda-Kuda x Berat per satuan Luas Pelat Bondex

= 6 m x 10.1

= 60.6 kg/m

Beban Pelat Beton = Bentang Kuda-Kuda x Tebal Pelat Beton x Berat Volume Beton

= 6 m x 0.09 m x 2400= 1296 kg/m

Beban Spesi 2 cm = Bentang Kuda-Kuda x Tebal Spesi x Berat Volume Spesi= 6 m x 0.02 m x 2100= 252 kg/m

Beban Tegel 2 cm = Bentang Kuda-Kuda x Tebal Tegel x Berat Volume Tegel Beton

kg/m3

kg/m2

qTotal

qTotal

q7D

q7D

q7D

PBalok Anak

PBalok Anak

PBalok Anak

P7D 1/2 x q7D x L + PBalokAnak

P7D

P7D

ii). P7 Hidup (P7L)

q7L

q7L kg/m2

q7L

P7L 1/2 x q7L x L

P7L

P7L

II. Beban Terpusat P8

i). P8 Mati (P8D)

kg/m2

kg/m3

kg/m3

= 6 m x 0.02 m x 2400= 288 kg/m

Beban Plafond + Penggantung Plafond = Bentang Kuda-Kuda x Beban Hidup Plafond= 6 m x 11 + 7= 108 kg/m

Beban Profil Balok Induk Melintang WF 500 x 200 x 9 x 14 = 106 kg/m

= 60.6 + 1296 + 252 + 288 + 108 + 106

= 2110.6 kg/m

= qTotal + Berat Alat Pengikat 10 % x Beban Profil Balok Induk Melintang

= 2110.6 kg/m + 0.1 x 106 kg/m

= 2121.2 kg/m


= Beban Profil Balok Anak x Bentang Kuda-Kuda

= 56.8 kg/m x 6 m

= 340.8 kg

=

= 0.5 x 2121.2 kg/m x 7 m + 340.8 kg

= 7765 kg

= Bentang Kuda-Kuda x Beban Hidup Lantai Gudang

= 6 m x 400

= 2400 kg/m

=

= 0.5 x 2400 kg/m x 7 m

= 8400 kg


= = 0.9

= 0.9 x 40 x 3.5 m

= 126 kg/m

= = -0.4

= -0.4 x 40 x 3.5 m

= -56 kg/m

D. Beban Horizontal Akibat Gempa (E) Arah Memanjanga). Beban Atap → Bentang Atap Horizontal = 52 m

= 66 kg/m

= +

= 66 kg/m x 52 m + 24722 kg

= 28154 kg

kg/m3

kg/m2

qTotal

qTotal

q8D

q8D

q8D

PBalok Anak

PBalok Anak

PBalok Anak

P8D 1/2 x q8D x L + PBalokAnak

P8D

P8D

ii). P8 Hidup (P8L)

q8L

q8L kg/m2

q8L

P8L 1/2 x q8L x L

P8L

P8L

kg/m2


qW1 kg/m2

qW1


qW2 kg/m2

qW2

qDAtap

W1 qDAtap x Bentang Atap Horizontal ∑PD (2P5D + 7P6D)

W1

W1

=

= 2 x 1300 kg + 7 x 1560 kg

= 13520 kg


= Berat Kolom x (1/2 x Tinggi Kolom Lantai 1) x Jumlah Kolom

= 82.2 kg/m x 2.5 m x 9 Kolom

= 1849.5 kg

= + +

= 28154 kg + 13520 kg + 1849.5 kg

= 43523 kg

b). Beban Lantai → Bentang Antar Kolom Utama = 48 mBerat Sendiri Balok Induk Memanjang WF 400 x 200 x 8 x 13 = 66 kg/m

=

= 66 kg/m x 48 m

= 3168 kg


= Berat Kolom x (1/2x(Tinggi Kolom <Lantai Dasar+Lantai 1)) x Jumlah Kolom

= 82.2 kg/m x 4.75 m x 9 Kolom

= 3514 kg


=

= 2 4086.55 4200 7 7765 8400

= 16573 kg + 113155 kg

= 129728.1 kg

= + +

= 3168 kg + 3514 kg + 129728 kg

= 136410.15 kg


=

= 43523 kg + 136410.15 kg

= 179933.531 kg

Wihi

Atap →→ 43523 kg x 9.5 m = 413472.1198 kg

Lantai 1 →→ 136410.15 kg x 4.5 m = 613845.675 kg

∑Wihi =

W2 ∑PL (2P5L + 7P6L)

W2

W2

W3

W3

W3





W1

W1

W2

W2

W2

W3 2 (P7D+P7L) + 7 (P8D+P8L)

W3

W3

W3








WLantaiTotal x Tinggi Kolom Lantai Dasar =


x + +x+

∑Wihi = 413472.119841 kg + 613845.675 kg∑Wihi = 1027317.79484 kg

d). Periode Getar Bangunan (T) → Ct = 0.085 Untuk Rangka Momen BajaT = hn = Tinggi Struktur Bangunan = 9.5 mT = 0.085 x 9.5T = 0.46 detik


V =




V = 0.5 x 1 x 179933.53104 kg / 4.5V = 19993 kg


Atap → = 413472.11984 / 1027317.7948 x 19993 kg

→ = 8047 kg

Lantai 1 → = 613845.675 / 1027317.7948 x 19993 kg

→ = 11946 kg

Ct(hn)3/4

CIWt / R

Wt


F1

F1

F2

F2

3/4

KONTROL PROFIL BALOK INDUK MELINTANGPanjang Balok Induk Melintang (L) = 7 mModulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001.DATA-DATA PROFIL BALOK INDUK MELINTANGDATA PERENCANAANDirencanakan Menggunakan Profil WF 500 x 200 x 9 x 14

W = 79.5 kg/m h = 185 cm³

d = 496 mm = 428 mm 1836 cm³

= 199 mm = 41900 287 cm³

= 9 mm = 1840 BJ 41

= 14 mm = 20.3 cm fu 4100

r = 20 mm = 4.27 cm fy 2500

Ag = 101.3 = 1690 cm³

Dari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Balok Induk Melintang denganMomen Maksimum pada Batang 8 Portal Melintang dengan Beban Kombinasi 6 (1.2D + 1.6L + 0.8W)

= 31153.66 kgm

= 31550.94 kgm

= 30000 kgm

Vu = 25000 kg

A. Kontrol Penampang (Tekuk Lokal)Badan (Web)

≤ → 428 / 9 ≤ 1680 / 250

→ 47.556 ≤ 106.25 → OK

Sayap (Flange) 1≤ → 199 / 28 ≤ 170 / 250

→ 7.1071 ≤ 10.752 → OK

Profil WF diatas tergolong PENAMPANG KOMPAK

kg/cm2 kg/cm2

d - 2(tf+r) Sy =

Zx =

bf Ix cm⁴ Zy =

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx

MA

MB

MS

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

Mn = Mp = Zx fyMn = 1836 cm³ x 2500 = 4590000 kgcm

B. Kontrol Tekul Lateral → Tegangan Residu (fr)= 70 MPa = 700Bentang Antar Balok Anak = 2.3333 m

= 7 m = 700 cm

Lp = = 1.76 x 4.27 cm x 2000000 / 2500

Lp = 212.56196 cm ≈ 213 cm

= fy - fr = 2500 - 700 = 1800


J = =

J = 0.3333333 x 46.8 x 0.9 + 0.667 x 19.9 x 1.4J = 11.3724 + 36.4 = 47.7761333

h' = d - tf = 496 - 14 = 482 mm

= 0.25 x 1840 x 48.2

= 1068690.4

=

= p / 1690 x 2000000 800000 47.7761 101.3 2

= 115669.37

=

= 4 x 1690 / 8E+05 x 47.78 x 1068690 / 1840

= 0.0000045422 = 4.542192E-06

=

= 4.27 x 115669 / 2500 700 x 1 1 5E-06 1800

= 611.37607 cm ≈ 611 cm

= 233.33333 cm


= 611.37607 cm

C. Kekuatan Nominal Momen Lentur (Mn)


= 30000 kg

= #REF! kg

= #REF! kgcm = #REF! kgm

= = #REF! kgm

L = 7 m

1.75 #REF! 0.5 31551 1.75 1.75 30000

= = #REF! kgm

= #REF! / #REF! #REF! #REF! #REF! = ### ≤ 2.3


= = 1690 x 2500 - 700

= 3042000 kgcm = 30420 kgm

Mn =

kg/cm2

kg/cm2


1.76 iy E/fy

fL kg/cm2


3

cm4 cm4 cm4



Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2


LR

LR

LB

LR

qu

qu Pu

Vu

Mumax

Mu(x=3.5m) = Mumax MB

Mu(x=1.75m)=Mu(x=5.25m) =

Mu(x=1.75m)=Mu(x=5.25m) = MA MC


Cb

cm3 kg/cm2


MR


3 3

2

/xx x

2

- + + x 2

x x x 2

+ ++

+ - x - -/

Pu PuPu Pu

- - x

Mn = #REF! x 30420 45900 30420 611.4 233.3 611 212.562 ≤ 45900

Mn = #REF! kgm ≤ 45900 kgm → Mn = #REF! kgm

Mcr =

Mcr = #REF! 140652936533333000 + 725113792.733096 1840 1068690.4

Mcr = #REF! kgcm = #REF! kgm ≤ 45900 kgm


Mn = 4590000 kgcm Mn = Cb[MR + (Mp-MR) x (LR-LB) / (LR-Lp)Mn = 45900 kgm Mn = #REF! kgm

= #REF! kgm


≤ → #REF! kgm ≤ 0.9 x #REF! kgm

→ #REF! kgm ≤ #REF! kgm #REF!

H. Kontrol Lendutan

= L/360 = 700 / 360 = 1.94444 cm


y =

y = 5/384 315.51 700 23/648 0 0 700 ### 41900

y = 11.770617 cm ≤ = 1.944 cm NOT OK


Badan (Web)≤ → 428 / 9 ≤ 1100 / 250

→ 47.556 ≤ 69.57 → PLASTIS

≤ → ≤ → ≤→ #REF! kg ≤ 0.9 0.6 2500 49.6 0.9

→ #REF! kg ≤ 60264 kg #REF!


MnAktual

Mumax f Mn

Lendutan Ijin ymax

5 / 384 x qD x L4 / (EIx) + 23 / 648 x (PD+PL) L3 / (EIx)

ymax

h/tw 1100 / fy1/2


+ - x - -/

3

x x x x

4 /x x + x + x x

1835.66

286.684

KONTROL PROFIL KOLOMJarak Antar Kuda-Kuda (b) = 6 mModulus Elastisitas (E) = 2000000 Modulus Geser (G) = 800000

4.5 m

5 m1.DATA-DATA PROFIL KOLOMDATA PERENCANAANDirencanakan Menggunakan Profil WF 250 x 250 x 14 x 14

W = 82.2 kg/m h = 304 cm³

d = 250 mm = 190 mm 1015 cm³

= 255 mm = 11500 466 cm³

= 14 mm = 3880 BJ 41

= 14 mm = 10.5 cm fu 4100

r = 16 mm = 6.09 cm fy 2500

Ag = 104.7 = 919 cm³

Dari Hasil Analisa Struktur dengan menggunakan Program SAP 2000 v.14, Kolom denganMomen Maksimum pada Batang 8 Portal Melintang dan pada Batang 10 Portal Memanjang dengan Beban Kombinasi 6 (1.2D + 1.6L + 0.8W)

A. Gaya-Gaya yang Bekerja Pada Kolom → Kolom Tengah Lantai DasarPu = 25000 kg

kg/cm2 kg/cm2

Tinggi Kolom Lantai Dasar (H1) =

Tinggi Kolom Lantai 1 (H2)=

d - 2(tf+r) Sy =

Zx =

bf Ix cm⁴ Zy =

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx

1015.014

466.053

kg/cm2

kg/cm2

PERENCANAAN BALOK INDUK MELINTANGPanjang Balok Induk Melintang (L) = 7 mModulus Elastisitas (E) = 2000000 Modulus Geser (G) 8000001. MENENTUKAN KEBUTUHAN MOMEN INERSIA Ix DAN MODULUS PLASTIS ZxA. Beban Mati (D)

Perkiraan Berat Sendiri Balok Induk Melintang = 80.00 kg/m = 0.8 kg/cm

=

= 2 x 0.5 x 920.00 kg/m x 6 m

= 5520.02 kg

B. Beban Hidup (L) →

=

= 2 x 0.5 x 933.33 kg/m x 6 m

= 5600 kg


1.2 1.2 x 80.00 kg/m 96 kg/m

1.2 + 1.6

= 1.2 x 5520 kg + 1.6 x 5600 kg

= 15584.024 kg

D. Kontrol Lendutan Ijin

= L/360 = 700 / 360 = 1.94444 cm


y =

1.94444 = 5/384 0.80 700 23/648 5520 5600 700 2000000 IxIx ≥ 35454.965

E. Beban Berfaktor dan Kontrol Kuat Rencana Momen Lentur → f = 0.9

L = 6 m

= 96 kg/m

= 15584.024 kg

= = 0.5 96 7 + 2 15584

= 31504.048 kg

= 0.125 96 7 + 0.33 15584 7

= 36950.723 kgm


4E+06 kgcm ≤ 0.9 x 2250 x Zx

Zx ≥ 1824.727

kg/cm2 kg/cm2

a) Beban Mati Terbagi Rata (qD)

b) Beban Mati Terpusat Akibat Balok Anak (PD)

PD 2 x 0.5 x qDBalokAnak x L

PD

PD

Beban Hidup Terpusat Akibat Balok Anak (PL)

PL 2 x 0.5 x qLBalokAnak x L

PL

PL

qu = qD = =

Pu = PD PL

Pu

Pu

Lendutan Ijin ymax


cm4

qu

qu

Pu

Vu 1/2 x qu x L + 4/2 Pu

Vu

Mumax 1/8 x qu x L2 + 1/3 x Pu x L=

Mumax

Mumax f Mn

kg/cm2 cm3

cm3

4 3/x x + x + x x

Pu Pu

x xx

2 xxx x

Pu Pu

2. PERHITUNGAN BALOK INDUK MELINTANGDATA PERENCANAANDirencanakan Menggunakan Profil WF 500 x 200 x 9 x 14

W = 79.5 kg/m h = 185 cm³

d = 496 mm = 428 mm 1836 cm³

= 199 mm = 41900 287 cm³

= 9 mm = 1840 BJ 41

= 14 mm = 20.3 cm fu 4100

r = 20 mm = 4.27 cm fy 2500

Ag = 101.3 = 1690 cm³

A. Beban Mati (D)

Berat Sendiri Balok Induk Melintang = 79.50 kg/m = 0.795 kg/cm

=

= 2 x 0.5 x 920.00 kg/m x 6 m

= 5520.02 kg


=

= 2 x 0.5 x 933.33 kg/m x 6 m

= 5600 kg


1.2 1.2 x 79.50 kg/m 95.4 kg/m

1.2 + 1.6

= 1.2 x 5520 kg + 1.6 x 5600 kg

= 15584.024 kg

L = 7 m

= 95.4 kg/m

= 15584.024 kg

= = 0.5 95.4 7 + 2 15584

= 31501.948 kg

= 0.125 95.4 7 + 0.33 15584 7

= 36947.048 kgm = 3694704.7667 kgcm

d - 2(tf+r) Sy =

Zx =

bf Ix cm⁴ Zy =

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx




PD

PD



PL

PL

qu = qD = =

Pu = PD PL

Pu

Pu

qu

qu

Pu

Vu 1/2 x qu x L + 4/2 Pu

Vu


Mumax

Pu Pu

x xx

2 xxx x

Pu Pu


≤ → 428 / 9 ≤ 1680 / 250

→ 47.556 ≤ 106.25 → OK

Sayap (Flange) 1≤ → 199 / 28 ≤ 170 / 250

→ 7.1071 ≤ 10.752 → OK


E. Kontrol Tekul Lateral → Tegangan Residu (fr)= 70 MPa = 700Bentang Antar Balok Anak = 2.3333 m

= 7 m = 700 cm

Lp = = 1.76 x 4.27 cm x 2000000 / 2500

Lp = 212.56196 cm ≈ 213 cm

= fy - fr = 2500 - 700 = 1800


J = =

J = 0.3333333 x 46.8 x 0.9 + 0.667 x 19.9 x 1.4J = 11.3724 + 36.4 = 47.7761333

h' = d - tf = 496 - 14 = 482 mm

= 0.25 x 1840 x 48.2

= 1068690.4

=

= p / 1690 x 2000000 800000 47.7761 101.3 2

= 115669.37

=

= 4 x 1690 / 8E+05 x 47.78 x 1068690 / 1840

= 0.0000045422 = 4.542192E-06

=

= 4.27 x 115669 / 2500 700 x 1 1 5E-06 1800

= 611.37607 cm ≈ 611 cm

= 233.33333 cm


= 611.37607 cm

F. Kekuatan Nominal Momen Lentur (Mn)


= 15584.024 kg

= 31501.948 kg

= 3694704.767 kgcm = 36947 kgm

= = 36947 kgm

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

kg/cm2

kg/cm2


1.76 iy E/fy

fL kg/cm2


3

cm4 cm4 cm4



Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2


LR

LR

LB

LR

qu

qu Pu

Vu

Mumax

Mu(x=3.5m) = Mumax MB

3 3

2

/xx x

2

- + + x 2

Pu PuPu Pu

L = 7 m

1.75 31502 0.5 95.4 1.75 1.75 15584

= = 27710.2857 kgm

= 461838.1 / 92368 83131 1E+05 83131 = 1.14 ≤ 2.3


= = 1690 x 2500 - 700

= 3042000 kgcm = 30420 kgm

Mn =

Mn = 1.1363636 x 30420 45900 30420 611.4 233.3 611 212.562 ≤ 45900

Mn = 51242.906 kgm ≤ 45900 kgm → Mn = 45900 kgm

Mcr =

Mcr = 0.0153 140652936533333000 + 725113792.733096 1840 1068690.4

Mcr = 19149479 kgcm = 191494.78806 kgm ≤ 45900 kgm


Mn = 4590000 kgcm Mn = Cb[MR + (Mp-MR) x (LR-LB) / (LR-Lp)Mn = 45900 kgm Mn = 45900 kgm

= 45900 kgm


≤ → 36947.0476667 kgm ≤ 0.9 x 45900 kgm

→ 36947.0476667 kgm ≤ 41310 kgm OK

H. Kontrol Lendutan

= L/360 = 700 / 360 = 1.94444 cm


y =

y = 5/384 0.95 700 23/648 5520 5600 700 2000000 41900

y = 1.6510963 cm ≤ = 1.944444444 cm OK


Badan (Web)≤ → 428 / 9 ≤ 1100 / 250

→ 47.556 ≤ 69.57 → PLASTIS

≤ → ≤ → ≤→ 31501.948 kg ≤ 0.9 0.6 2500 49.6 0.9

→ 31501.948 kg ≤ 60264 kg OK

Mu(x=1.75m)=Mu(x=5.25m) =

Mu(x=1.75m)=Mu(x=5.25m) = MA MC


Cb

cm3 kg/cm2


MR



MnAktual

Mumax f Mn

Lendutan Ijin ymax


ymax

h/tw 1100 / fy1/2


x x x 2

+ ++

+ - x - -/

3

x x x x

- - x

4 /x x + x + x x

1835.66

286.684

PERENCANAAN BALOK INDUK MELINTANGPanjang Balok Induk Melintang Tengah (L) = 8 m

Modulus Elastisitas (E) = 2000000 Modulus Geser (G) = 8000001. MENENTUKAN KEBUTUHAN MOMEN INERSIA Ix DAN MODULUS PLASTIS ZxA. Beban Mati (D)

Perkiraan Berat Sendiri Balok Induk Melintang = 120.00 kg/m = 1.2 kg/cm

→ Bentang Balok Anak = 2.1667 m

=

= 2 x 0.5 x 786.36 kg/m x 6 m

= 4718.18 kg


=

= 2 x 0.5 x 933.33 kg/m x 6 m

= 5600 kg


1.2 1.2 x 120 kg/m 144 kg/m

1.2 + 1.6

= 1.2 x 4718.2 kg + 1.6 x 5600 kg

= 14621.816 kg

D. Kontrol Lendutan Ijin

= L/360 = 800 / 360 = 2.22222 cm


y =

2.22222 = 5/384 1.20 800 23/648 4718 5600 800 2000000 IxIx ≥ 43629.892

E. Beban Berfaktor dan Kontrol Kuat Rencana Momen Lentur → f = 0.9

L = 8 m

= 144 kg/m

= 14621.816 kg

= = 0.5 144 8 + 2.5 14621.8

= 37130.54 kg

= 0.125 144 8 + 0.5 14621.8 8

= 59639.264 kgm


6E+06 kgcm ≤ 0.9 x 2375 x Zx

Zx ≥ 2790.141

kg/cm2 kg/cm2




PD

PD



PL

PL

qu = qD = =

Pu = PD PL

Pu

Pu

Lendutan Ijin ymax


cm4

qu

qu

Pu

Vu 1/2 x qu x L + 5/2 Pu

Vu


Mumax

Mumax f Mn

kg/cm2 cm3

cm3

4 3/x x + x + x x

Pu Pu

x xx

2 xxx x

Pu PuPu

2. PERHITUNGAN BALOK INDUK MELINTANGDATA PERENCANAANDirencanakan Menggunakan Profil WF 600 x 200 x 11 x 17

W = 106 kg/m h = 228 cm³

d = 600 mm = 522 mm 2863 cm³

= 200 mm = 77600 357 cm³

= 11 mm = 2280 BJ 41

= 17 mm = 24 cm fu 4100

r = 22 mm = 4.12 cm fy 2500

Ag = 134.4 = 2590 cm³

A. Beban Mati (D)

Berat Sendiri Balok Induk Melintang = 106.00 kg/m = 1.06 kg/cm

=

= 2 x 0.5 x 786.36 kg/m x 6 m

= 4718.18 kg


=

= 2 x 0.5 x 933.33 kg/m x 6 m

= 5600 kg


1.2 1.2 x 106 kg/m 127.2 kg/m

1.2 + 1.6

= 1.2 x 4718.2 kg + 1.6 x 5600 kg

= 14621.816 kg

L = 8 m

= 127.2 kg/m

= 14621.816 kg

= = 0.5 127.2 8 + 2.5 14621.8

= 37063.34 kg

= 0.125 127.2 8 + 0.5 14621.8 8

= 59504.864 kgm = 5950486.4 kgcm

d - 2(tf+r) Sy =

Zx =

bf Ix cm⁴ Zy =

tw Iy cm⁴

tf ix kg/cm2

iy kg/cm2

cm² Sx




PD

PD



PL

PL

qu = qD = =

Pu = PD PL

Pu

Pu

qu

qu

Pu

Vu 1/2 x qu x L + 5/2 Pu

Vu


Mumax

Pu Pu

x xx

2 xxx x

Pu Pu


≤ → 522 / 11 ≤ 1680 / 250

→ 47.455 ≤ 106.25 → OK

Sayap (Flange) 1≤ → 200 / 34 ≤ 170 / 250

→ 5.8824 ≤ 10.752 → OK


E. Kontrol Tekul Lateral → Tegangan Residu (fr)= 70 MPa = 700Bentang Antar Balok Anak = 2 m

= 8 m = 800 cm

Lp = = 1.76 x 4.12 cm x 2000000 / 2500

Lp = 205.09491 cm ≈ 205 cm

= fy - fr = 2500 - 700 = 1800


J = =

J = 0.3333333 x 56.6 x 1.1 + 0.667 x 20 x 1.7J = 25.111533 + 65.51 = 90.6182

h' = d - tf = 600 - 17 = 583 mm

= 0.25 x 2280 x 58.3

= 1937367.3

=

= p / 2590 x 2000000 800000 90.6182 134.4 2

= 119729.85

=

= 4 x 2590 / 8E+05 x 90.62 x 1937367 / 2280

= 0.0000043384 = 4.338362E-06

=

= 4.12 x 119730 / 2500 700 x 1 1 4E-06 1800

= 605.40769 cm ≈ 605 cm

= 200 cm


= 605.40769 cm

F. Kekuatan Nominal Momen Lentur (Mn)

Mn = Mp = Zx fy = 127.2 kg/m

= 14621.816 kg

= 37063.34 kg

= 5950486.4 kgcm = 59505 kgm

= = 59505 kgm

h/tw 1680 / fy1/2

bf/2tf 170 / fy1/2

kg/cm2

kg/cm2


1.76 iy E/fy

fL kg/cm2


3

cm4 cm4 cm4



Iw cm6

x1 p/Sx EGJA/2

x1

x1 kg/cm2

x2 4 (Sx / (GJ))2 x Iw / Iy

x2

x2 cm4/kg2 cm4/kg2


LR

LR

LB

LR

qu

qu Pu

Vu

Mumax

Mu(x=4m) = Mumax MB

3 3

2

/xx x

2

- + + x 2

Pu PuPu Pu

L = 8 m

2.9167 22442 0.5 127.2 2.917 0.58 14621.8

= = 56384 kgm

= 743810.8 / 148762 2E+05 2E+05 169152.032 = 1.02582 ≤ 2.3


= = 2590 x 2500 - 700

= 4662000 kgcm = 46620 kgm

Mn =

Mn = 1.0258247 x 46620 71575 46620 605.4 200 605 205.095 ≤ 71575

Mn = 73749.215 kgm ≤ 71575 kgm → Mn = 71575 kgm

Mcr =

Mcr = 0.0161136 330575193600000000 + 986960440 2280 2E+06

Mcr = 34896954 kgcm = 3E+05 kgm ≤ 71575 kgm


Mn = 7157500 kgcm Mn = Mp = Zx fyMn = 71575 kgm Mn = 71575 kgm

= 71575 kgm


≤ → 59505 kgm ≤ 0.9 x 71575 kgm

→ 59505 kgm ≤ 64418 kgm OK

H. Kontrol Lendutan

= L/360 = 800 / 360 = 2.22222 cm


y =

y = 5/384 1.27 800 23/648 4718 5600 800 2000000 77600

y = 1.2518984 cm ≤ = 2.222 cm OK


Badan (Web)≤ → 522 / 11 ≤ 1100 / 250

→ 47.455 ≤ 69.57 → PLASTIS

≤ → ≤ → ≤→ 37063.34 kg ≤ 0.9 0.6 2500 60 1.1

→ 37063.34 kg ≤ 89100 kg OK

Mu(x=2.91667m)=Mu(x=4.0833m) =

Mu(x=2.91667m)=Mu(x=4.0833m) = MA MC


Cb

cm3 kg/cm2


MR



MnAktual

Mumax f Mn

Lendutan Ijin ymax


ymax

h/tw 1100 / fy1/2


x x x2

+ ++

+ - x - -/

3

x x x x

- - x

4 /x x + x + x x

x x

2863.18

357.122

Program Bangunan Baja

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