77567225-perhitungan

x = 10.866 2.309401

tan 30° = 0.577 3.059401 z14.589102 z2

No.Batang z3

Batang A (m) Batang B (m) Batang V (m) Batang D (m) z41 1.155 1.155 0.750 1.662 z52 1.155 1.155 0.750 1.662 z63 1.155 1.155 0.750 1.662 z74 1.155 1.155 0.750 1.662 z85 1.155 1.155 0.750 1.662 z96 1.155 1.155 0.750 1.662 z107 1.155 1.155 0.750 1.662 z118 1.155 1.155 0.750 1.662 z129 0.750

Total 9.238 9.238 6.750 13.295

cos 30° =

03.5

77.577

7.758.1558.3278.7328.9059.3099.482

10.059

BEBAN MATIBeban Atap

Produk = Union DeckAtap = Spandek

Lebar = 680 mmPanjang = 9400 mm (max lenght 12000 mm)Tebal = 0.4 mm

4.53

Jarak gording = 1.155 mJarak kuda2 = 4 m

Titik Buhul Berat (kg)10 10.46211 20.92312 20.92313 20.92314 20.92315 20.92316 20.92317 20.92318 10.462

BEBAN TOTAL

Titik Buhul B. Mati (kg) B. Hidup (kg) B. Pekerja (kg)B. Angin (kg)

Hujan (kg)X Y miring

1 6.075 0.000 0 0 0 0 02 8.869 0.000 0 0 0 0 03 8.869 0.000 0 0 0 0 04 8.869 0.000 0 0 0 0 05 6.609 0.000 0 0 0 0 06 8.869 0.000 0 0 0 0 07 8.869 0.000 0 0 0 0 08 8.869 0.000 0 0 0 0 09 6.075 0.000 0 0 0 0 0

10 41.076 0.000 100 -5.774 -10.000 -11.547 36.95011 56.592 0.000 100 -11.547 -20.000 -23.094 73.90112 56.592 0.000 100 -11.547 -20.000 -23.094 73.90113 56.592 0.000 100 -11.547 -20.000 -23.094 73.90114 58.852 0.000 100 -11.547 -20.000 -23.094 73.90115 56.592 0.000 100 -11.547 -20.000 -23.094 73.90116 56.592 0.000 100 -11.547 -20.000 -23.094 73.90117 56.592 0.000 100 -11.547 -20.000 -23.094 73.90118 41.076 0.000 100 -5.774 -10.000 -11.547 36.950

Berat/m2 = kg/m2

Berat sendiri kuda-kuda baja Berat akibat gording

Profil = Double angle Profil = U ChannelMutu Baja = 37 Mpa Mutu Baja = 37

Batang Atas & Batang Bawah d = 1252L 40x40x4 bf = 65Berat per unit = 4.84 kg/m tw = 6

Batang Diagonal & Batang Vertikal tf = 8

2L 30x30x3 A = 17.11Berat per unit = 2.72 kg/m w = 13.4

Titik Buhul Berat (kg) Titik Buhul Berat (kg) Berat Gording = 26.81 6.075 10 3.8142 8.869 11 8.8693 8.869 12 8.8694 8.869 13 8.8695 6.609 14 11.1296 8.869 15 8.8697 8.869 16 8.8698 8.869 17 8.8699 6.075 18 3.814

KOMBINASI PEMBEBANANKombinasi yang digunakan:1,4D 1,2D+1,6L+0,5La

Titik Buhul Beban (kg) Titik Buhul Beban (kg) Titik Buhul1 8.504 10 57.506 12 12.416 11 79.229 23 12.416 12 79.229 34 12.416 13 79.229 45 9.252 14 82.393 56 12.416 15 79.229 67 12.416 16 79.229 78 12.416 17 79.229 89 8.504 18 57.506 9

1,2D+1,6La+0,8W 1,2D+1,3W+0,5LaTitik Buhul Beban (kg) Titik Buhul Beban (kg) Titik Buhul

1 7.289 10 201.291 12 10.643 11 211.911 23 10.643 12 211.911 34 10.643 13 211.911 45 7.931 14 214.623 56 10.643 15 211.911 67 10.643 16 211.911 78 10.643 17 211.911 89 7.289 18 201.291 9

Beban akibat air hujan Beban akibat pekerja

Beban Air Hujan = 16Mpa

mm Titik Buhul Berat (kg) Titik Buhul Berat (kg)

mm 10 36.950 10 100mm 11 73.901 11 100

mm 12 73.901 12 100

13 73.901 13 100

kg/m 14 73.901 14 10015 73.901 15 100

kg 16 73.901 16 10017 73.901 17 10018 36.950 18 100

1,2D+1,6L+0,5LaBeban (kg) Titik Buhul Beban (kg)

7.289 10 99.29110.643 11 117.91110.643 12 117.91110.643 13 117.9117.931 14 120.623

10.643 15 117.91110.643 16 117.91110.643 17 117.9117.289 18 99.291

1,2D+1,3W+0,5LaBeban (kg) Titik Buhul Beban (kg)

7.289 10 86.29110.643 11 91.91110.643 12 91.91110.643 13 91.9117.931 14 94.623

10.643 15 91.91110.643 16 91.91110.643 17 91.9117.289 18 86.291

kg/m2

cm2

BEBAN ANGINAngin Kiri

Angin tekan Angin hisapKoefisien = 0.2 Koefisien = -0.4Tekanan angin minimum = 25 Tekanan angin minimum = 25

Titik Buhul Berat (kg) Titik Buhul Berat (kg)10 11.547 14 -23.094

11 23.094 15 -46.18812 23.094 16 -46.18813 23.094 17 -46.18814 11.547 18 -23.094

0.5000.866

Titik BuhulBerat

Titik BuhulBerat

x y x10 5.774 10.000 14 -11.54711 11.547 20.000 15 -23.09412 11.547 20.000 16 -23.09413 11.547 20.000 17 -23.09414 5.774 10.000 18 -11.547

kg/m2

sin 30° =cos 30° =

Angin Kanan

Angin tekanKoefisien = 0.2Tekanan angin minimum = 25

Titik Buhul Berat (kg)14 11.547

15 23.09416 23.09417 23.09418 11.547

BeratTitik Buhul

Beraty x y

-20.000 14 5.774 10.000-40.000 15 11.547 20.000-40.000 16 11.547 20.000-40.000 17 11.547 20.000-20.000 18 5.774 10.000

kg/m2 kg/m2

Angin hisapKoefisien = -0.4Tekanan angin minimum = 25

Titik Buhul Berat (kg)10 -23.094

11 -46.18812 -46.18813 -46.18814 -23.094

Titik BuhulBerat

x y10 -11.547 -20.00011 -23.094 -40.00012 -23.094 -40.00013 -23.094 -40.00014 -11.547 -20.000

kg/m2

1.3.1 Perencanaan Profil C untuk Gording Dengan ukuran :

C

d = 125 mm w = 13.4 kg/m iy = 1.96

bf = 65 mm Cy = 1.94 cm Zx = 68

tw = 6 mm Ix = 425 Zy = 14.4

tf = 8 mm Iy = 65.5 α = 30

A = 17.11 ix = 4.99 cm Jarak kuda-kuda = 4

Mutu Baja = BJ 37fu = 3700 370 Mpafy = 2400 240 Mpa

1.3.2 Perencanaan Pembebanan

1.3.2.1 Perhitungan Beban

Beban Mati

Berat Gording =Berat Spandex = w x l

= 4.53 x 1.155 =Berat Total =

alat Pengikat dll 10 % = 0.1 x 18.631 ==

Beban Hidup

40 - 24 =q =

= 1.000 x 16.00 =

=

Beban Angin

Tekanan Angin =Angin Tekan =Angin Hisap =

1.000 x 10.00 =

Beban Mati + Beban Hidup > dari Beban Angin Hisap : 20.494 +Beban Angin Hisap tidak perlu diperhitungkan ==> 5

1.3.2.2 Perhitungan Momen Akibat Beban thp Sbx dan Sby

Beban Mati

0.125 x ( 20.494 x 0.866 x0.125 x ( 20.494 x 0.5 x

Beban Hidup Terbagi Rata

0.125 x ( 16 x 0.866 x0.125 x ( 16 x 0.5 x

cm4

cm4

cm2

kg/cm2 =

kg/cm2 =

qD

Beban Terbagi Rata = (40 - 0.8 a) =

qL = jarak gording horisontal x q

Beban Hidup Terpusat, PL

q = jrk gording horisontal x angin hisap =

qw =

MXD = 1/8 (qD x cosa) L2 =

MYD = 1/8(qDxsina xL/3)2 =

MXLD = 1/8 (qL x cosa) L2 =

MYL = 1/8(qLxsinaxL/3)2 =

Beban Hidup Terpusat

0.25 x ( 100 x 0.866 x0.25 x ( 100 x 0.5 x

Beban Angin Terbagi Rata

= 0.125 x 5 x 16

* Mu Beban Mati ,Beban Angin dan Beban Hidup Terbagi Rata

Sumbu X Sumbu Y35.496 kgm 2.2771 kgm27.713 kgm 1.7778 kgm

Mw = 10 kgm

1.2 x 35.496 + 1.6 x 27.7128 + 0.81.2 x 2.2771 + 1.6 x 1.778 + 0.8

* Mu Beban Mati, Beban Angin dan Beban Hidup Terpusat

Sumbu X Sumbu Y35.4964 kgm 2.2771 kgm86.6025 kgm 16.667 kgm

Mw = 10 kgm

1.2 x 35.496 + 1.6 x 86.6025 + 0.81.2 x 2.2771 + 1.6 x 16.6667 + 0.8

1.3.3 Kontrol Kekuatan Profil

1.3.3.1 Penampang Profil

Untuk Sayap Untuk Badanb

≤250 h

≤665

fy tw fy65 ≤ 250 125 ≤ 6658 240 6 240

8.13 ≤ 16.14 20.8 ≤ 42.9OK OK

Penampang Profil Kompak, maka Mnx = Mpx

1.3.3.2 Kontrol Lateral Buckling

500 mm =

1.76 x x Efy

= 1.76 x 1.96 x 200000 = 99.58 cm

MXL = 1/4 (qL x cosa) L =

MYL = 1/4(qL x sina)(L/3) =

MXW = 1/8 x qw x L2

1.3.3.3 Besar Momen Berfaktor ( Mu = 1.2 MD + 1.6 ML + 0.8 MW )

MD = MD =ML = ML =

MUX =MUY =

MD = MD =ML = ML =

MUX =MUY =

tf

Jarak Baut Pengikat / pengaku lateral = LB =

LP = iY

xx

xx

xx

= 1.76 x 1.96 x240

= 99.58 cm

Ternyata : < maka : Mnx = Mpx

Mnx = Mpx = Zx . Fy = 68 x 2400 = 1632 KgmMny = Zy ( satu sayap ) * fy

== 0.25 x 0.8 x 6.5 2 x 2400 == 202.8 kgm

1.3.3.3 Persamaan Iterasi

Mux + Muy ≤ 1

Beban Mati , Beban Angin dan Beban Hidup Terbagi Rata

94.936 + 5.577 ≤ 10.9 x 1632 0.9 x 202.8

0.065 + 0.031 ≤ 10.095 ≤ 1 OK

Beban Mati , Beban Angin dan Beban Hidup Terpusat

189.160 + 29.399 ≤ 10.9 x 1632 0.9 x 202.8

0.116 + 0.161 ≤ 10.277 ≤ 1 OK

1.3.3.4 Kontrol Lendutan Profil

Lendutan Ijin f = L = 400 = 2.222 cm180 180

Lendutan Akibat Beban Merata (1)

fx = 5 x = 5 x 0.3649384 E x Ix 384 x

= 0.124 cm

fy = 5 x = 5 x 0.3649384 E x Iy 384 x

= 0.006 cm

Lendutan Akibat Beban Terpusat (2)

fx = 1 x P = 1 x 10048 E x Ix 48 x

= 0.136 cm

fy = 1 x P = 1 x 10048 E x Iy 48 x

= 0.019 cm

LB LP

1/4 x tf x bf2 x fy

fb . Mnx fb . Mny

qD + L cos a L4

qD + L sin a (L/3)4

cos a L3

sin a L3

Lendutan Akibat Beban Angin merata (3)

fx = 5 x = 5 x 0.05384 E x Ix 384 x

= 0.01698 cm

fy = 5 x = 5 x 0.05384 E x Iy 384 x

= 0.001 cm

Lendutan total yang terjadi

= = ( 0.124 + 0.136 + 0.017 0.006 + 0.019

0.278 cm < 2.2222 cm OK

qW cos a L4

qW sin a (L/3)4

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

) 2 + (

ftot = fijin =

cm

°

m

13.4 kg/m

5.231 kg/m

18.631 kg/m

1.863 kg/m

20.494 kg/m

161616 kg/m

100 kg

255

10 (menentukan = q)10 kg/m

16 > 10kg/m

16 ) = 35.496 kgm1.78 ) = 2.277 kgm

16 ) = 27.713 kgm1.78 ) = 1.778 kgm

cm3

cm3

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

4 ) = 86.603 kgm1.33 ) = 16.667 kgm

= 10 kgm

x 10 = 94.936 kgmx 0 = 5.577 kgm

x 10 = 189.160 kgmx 0 = 29.399 kgm

50 cm

20280 kgcm

x 0.866 x 400 4

2000000 x 425

x 0.5 x 133.333 4

2000000 x 65.5

x 0.866 x 400 3

2000000 x 425

x 0.5 x 133.333 3

2000000 x 65.5

x 0.866 x 400 4

2000000 x 425

x 0.5 x 133.333 4

2000000 x 65.5

+ 0.001 ) 2

cos 30 = 0.866 53.01 D2* 0.602 + B2* 0.866 =sin 30 = 0.500 sin α2 = 0.799 D2* 0.799 + B2* 0.500 =tan 30 = 0.577 cos α2 = 0.602 D2* 0.481 + B2* 0.692 =

D2* 0.481 + B2* 0.301 =R1v = 987.517 0,75 + tan 30 = 1.327 B2* 0.391 =R9v = 987.517 B2 =

No A B V D D3* 0.602 + B3* 0.866 =1 0.000 1199.251 -201.291 -1726.002 D3* 0.799 + B3* 0.500 =2 -1199.251 2055.8584 567.026 -1232.859 D3* 0.481 + B3* 0.692 =3 -2055.858 2569.823 344.472 -739.715 D3* 0.481 + B3* 0.301 =4 -2569.823 2741.1445 121.919 -246.572 B3* 0.391 =5 -2569.823 2741.1445 2749.075 -246.572 B3 =6 -2055.858 2569.823 121.919 -739.7157 -1199.251 2055.8584 344.472 -1232.859 D4* 0.602 + B4* 0.866 =8 0.000 1199.251 567.026 -1726.002 D4* 0.799 + B4* 0.500 =9 -201.291 D4* 0.481 + B4* 0.692 =

D4* 0.481 + B4* 0.301 =B4* 0.391 =

B4 =

α2 =

1038.5816 *0,799 D1* 0.602 + B1* 0.866 = 043.242 *0,602 D1* 0.799 + B1* 0.500 = -778.936

829.517 D1* 0.481 + B1* 0.692 = 026.020 - D1* 0.481 + B1* 0.301 = -468.7067

803.497 B1* 0.391 = 468.7072055.858 B1 = 1199.251

1780.426 *0,799694.099 *0,602

1422.030417.658 -

1004.3712569.823

2225.532 *0,7991173.6349 *0,602

1777.537706.207 -

1071.3302741.1445

*0,799*0,602

-

Batang Atas (tekan)Profil : Double Angle 40.40 Kondisi Tumpuan Jepit-jepit k =BJ 37 fy = 240 Mpa BATANG A

fu = 370 Mpa arah sumbu kuat (sumbu x):

b = 40 mm = (k.Lx)/ix

t = 4 mm = 62.029T = 9 mm

r1 = 6 mm =

r2 = 3 mm = 0.684

Ag = 616

w = 4.84 kg/m 0,25 ˂ λcx ˂ 1,2 =

Ix = 8.96 =

Iy = 24.14

ix = 1.21 =

iy = 1.98 = 118038.25 N

Sx = 3.11

Sy = 5.43 = -2569.823

Periksa kelangsingan penampang = 0.256 ˂b/t ≤10 ≤ 12.910 arah sumbu lemah (sumbu y):

b/t ≤ OK!!!!! =

= 37.907

=

= 0.418

0,25 ˂ λcx ˂ 1,2 =

=

=

= 136462.69 N

= -2569.823

= 0.222 ˂

Batang Bawah (batang tarik)

BATANG BPeriksa syarat kelangsingan b. Tarik

λ =λ = 95.430 ˂ 240 OK!!!

Kondisi Leleh

λx

λx

λcx (λx/π)*((fy/E)^0,5)

λcx

mm2

ωx

cm4 ωx

cm4

cm Nn Ag*(fy/ωx)

cm Nn

cm3

cm3 Nu trbsr

Nu /(ɸc*Nn)200/√fy

λr λy (k.Ly)/iy

λy

λcy (λy/π)*((fy/E)^0,5)

λcy

ωy

ωy

Nn Ag*(fy/ωy)

Nn

Nu trbsr

Nu /(ɸc*Nn)

L/imin

=

= 133056 N 13305.6 kg

= 2741.145

˃13305.6 ˃ 2741.1445 OK!!!!!

ɸTn ɸ*fy*Ag

ɸTn

Tu trbsr

ɸTn Tu

Batang Diagonal, V1 & V9 (tekan)0.65 Profil : Double Angle 30.30 Kondisi Tumpuan Jepit-jepit

BJ 37 fy = 240 Mpa BATANG Dfu = 370 Mpa arah sumbu kuat (sumbu x):

b = 30 mm

t = 3 mmT = 9 mm

r1 = 4 mm

r2 = 2 mm

Ag = 346

w = 2.72 kg/m λcx >1,2

1.252 Ix = 2.84

Iy = 8.6

ix = 0.91

= 11803.825 kg iy = 1.58

Sx = 1.31

Sy = 2.49

1 OK!!!!! Periksa kelangsingan penampangb/t ≤10 ≤ 12.910 arah sumbu lemah (sumbu y):

b/t ≤ OK!!!!!

0,25 ˂ λcx ˂ 1,2

1.083

= 13646.269 kg

1 OK!!!!!Batang vertikal (batang tarik)

BATANG Vertikal (kec.V1&V9)Periksa syarat kelangsingan b. Tarik

λ =λ = 82.418 ˂ 240 OK!!!

Kondisi Leleh

λx

λx

λcx

λcx

mm2

1,43/(1,6-0,67*λc)

cm4

cm4

cm Nn

cm Nn

cm3

cm3 Nu trbsr

Nu /(ɸc*Nn)200/√fy

λr λy

λy

λcy

λcy

1,43/(1,6-0,67*λcy)

Nn

Nn

Nu trbsr

Nu /(ɸc*Nn)

L/imin

=

= 74736 N 7473.6 kg

= 2749.075

˃7473.6 ˃ 2749.075 OK!!!!!

ɸTn ɸ*fy*Ag

ɸTn

Tu trbsr

ɸTn Tu

Kondisi Tumpuan Jepit-jepit k = 0.65BATANG V1 & V9

arah sumbu kuat (sumbu x): arah sumbu kuat (sumbu x):= (k.Lx)/ix = (k.Lx)/ix

= 118.706007 118.706 = 53.571

= =

= 1.309 = 0.591

= 0,25 ˂ λcx ˂ 1,2

= 2.142

= =

= 38774.88 N = 3877.488 kg = 69929.24

= -1726.0022 = -201.291

= 0.524 ˂ 1 OK!!!!! = 0.034

arah sumbu lemah (sumbu y): arah sumbu lemah (sumbu y):= =

= 68.369 = 30.854

= =

= 0.754 = 0.340

0,25 ˂ λcx ˂ 1,2 = 0,25 ˂ λcx ˂ 1,2

= 1.306

= =

= 63581.0916 N = 6358.109 kg = 79675.046

= -1726.0022 = -201.29115

= 0.319 ˂ 1 OK!!!!! = 0.030

λx

λx

(λx/π)*((fy/E)^0,5) λcx (λx/π)*((fy/E)^0,5)

λcx

ωx 1,25*(λc^2) ωx

ωx ωx

Ag*(fy/ωx) Nn Ag*(fy/ωx)

Nn

Nu trbsr

Nu /(ɸc*Nn)

(k.Ly)/iy λy (k.Ly)/iy

λy

(λy/π)*((fy/E)^0,5) λcy (λy/π)*((fy/E)^0,5)

λcy

ωy 1,43/(1,6-0,67*λcy) ωy

ωy ωy

Ag*(fy/ωy) Nn Ag*(fy/ωy)

Nn

Nu trbsr

Nu /(ɸc*Nn)

=

= 1.187

N = 6992.924 kg

˂ 1 OK!!!!!

=

= 1.042

N = 7967.505 kg

˂ 1 OK!!!!!

(λx/π)*((fy/E)^0,5)

1,43/(1,6-0,67*λc)

(λy/π)*((fy/E)^0,5)

1,43/(1,6-0,67*λcy)

Pre - Eliminary Design

3Perencanaan Bondex dan Balok Anak3.1 Data - Data perencanaan

Beban Hidup : 400 Kg/m2Beban Finishing : 0 Kg/m2Beban Berguna : 400 Kg/m3

Berat Beton Kering : 2400 kg/m3Panjang Bentang Beban Bondex yang Dipikul Oleh Balok Anak : 4 mPanjang Balok Anak : 4 m

3.2 Perencanaan Pelat Lantai Bondex

3.2.1 Data Perencanaan

Berat Sendiri Beton = 2400 kg/m3Berat Sendiri Bondex = 0 kg/m2Berat Spesi per cm Tebal = 0 kg/m2Berat Tegel = 0 kg/m2


Beban Mati

Berat Beton = 2400 * 0.12 = 288 Kg/m2Berat Bondex = 0 Kg/m2Berat Spesi 2 Cm = 0 * 2 = 0 Kg/m2Berat Tegel 2 Cm = 0 * 2 = 0 Kg/m2

qD = 288 Kg/m2Beban Hidup

Beban Hidup Lantai gudang = 400 Kg/m2Beban Finishing = 0 Kg/m2

qL = 400 Kg/m2

3.2.3 Perencanaan Tebal Lantai Beton dan Tulangan Negatif

3.2.3.1 Perencanaan Tebal Lantai

qL = 400 kg/m2

Beban Berguna yang Dipakai = 500 kg/m2Jarak Antar Balok = 400 cmJarak Kuda - Kuda = 400 cm

Dari Tabel Brosur ( Bentang Menerus dengan Tulangan Negatif ),didapat :t = 12 mmA = 3.57 cm2/m

3.2.3.2 Perencanaan Tulangan Negatif

10 mmDirencanakan Tulangan Dengan φ =

As = 0.785 mm2

Banyaknya Tulangan Yang diperlukan Tiap 1 m = A = 3.57As 0.785

= 4.547771 Buah= 5 Buah

Jarak Tulangan Tarik = 200 cm

3.3 Perencanaan Dimensi Balok Anak


Beban Mati ( D )

Bondex = 4 0 = 0 kg/mPlat Beton = 4 0.12 2400 = 1152 kg/mTegel + Spesi = 4 0 = 0 kg/m

qD = 1152 kg/m

Beban Hidup ( L )

qL = 4 400 = 1600 kg/m

3.3.3 Perhitungan qU , Mu Max dan Du Max

qU = 1.2 qD + 1.6 qLqU = 1.2 1152 1.6 1600 = 3942.4 Kg/m

= 0.125 3942.4 16 = 7884.8 Kgm

= 0.5 3942.4 4 = 7884.8 Kg

3.3.4 Perhitungan Ix Profil Yang Diperlukan

Y = L = 400 = 1.111111360 360

Ix > 5 ( 11.52 16 ) 2.56E+10384 2100000 1.111111

Ix > 3931.429 cm4

3.3.5 Perencanaan Profil WF untuk Balok Anak

250 x 250 x 9 x 14

Pasang Tulangan Tarik φ10 - 200

2

8

1max lqMu u

lqDu u2

1max

EY

lqLqDIx

4*)(

384

5

A = 92.18 cm2 tf = 14 mm Zx = 867 cm3W = 72.4 kg/m Ix = 10800 cm4 Zy = 292 cm3a = 250 mm Iy = 3650 cm4 h = 190 mm

bf = 250 mm tw = 9 mm r = 16 mmiy = 6.29 cm ix = 10.8 cm

Mutu Baja = BJ 37fu = 3700 kg/cm2fy = 2400 kg/cm2

3.3.6 Perencanaan Pembebanan + Beban Profil

Beban Mati ( D )

Bondex = 4 0 = 0 kg/mPlat Beton = 4 0.12 2400 = 1152 kg/mTegel + Spesi = 4 0 = 0 kg/mBerat Profil = = 72.4 kg/m

qD = 1224.4 kg/m

Beban Hidup ( L )

qL = 4 400 = 1600

3.3.7 Perhitungan qU , Mu Max dan Du Max ( Berat Profil Dimasukkan )

qU = 1.2 qD + 1.6 qLqU = 1.2 1224.4 1.6 1600 = 4029.28 Kg/m

= 0.125 4029.28 16 = 8058.56 Kgm

= 0.5 4029.28 4 = 8058.56 Kg

3.3.8 Kontrol Lendutan Balok

Y = L = 400 = 1.111111360 360

= 5 ( 12.244 16 ) 2.56E+10384 2100000 10800

= 0.415 < 1.111111

OK

3.3.9 Kontrol Kuat Rencana Momen Lentur

3.3.9.1 Kontrol Penampang

2

8

1max lqMu u

lqDu u2

1max

EIx

lqLqDY

4*)(

384

5max

untuk Sayap untuk Badan

250 170 190 168028 15.49193 9 15.49193

8.929 10.973 21.111 108.444OK OK


Mp = fy * Zx= 2400 * 867= 2080800 kgcm= 20808 kgm


Jrk Pengikat Lateral : 1000 mm = 100 cm

Lp = 319.575 cm

Ternyata Lp > Lb maka Mnx = Mpx

Mnx = Mpx = Zx. Fy = 867 * 2400 = 20808 KgmMny = Zy ( 1 flen ) * fy

== 0.25 1.4 625 2400 = 525000 kgcm= 5250 kgm

0.9 Mp = 0.9 * 20808 = 18727.2 kgm

0.9 Mp > Mu18727.2 > 8058.56

OK

3.3.9.3 Kontrol Kuat Rencana Geser

190 < 11009 15.49193

fytf

b 170

2

fyt

h 1680

fy

EiyLp *76.1

fybftf *)2**4/1(x x x

fytw

h 1100

21.11111 < 71.00469Plastis

Vn = 0.6 fy Aw= 0.6 2400 0.9 25= 32400 Kg

Vu < ФVn

8058.56 < 0.9 324008058.56 < 29160

OK









Beban Mati




qL = 400 Kg/m2



qL = 400 kg/m2





As = 0.785 mm2


= 4.547771 Buah= 5 Buah




Beban Mati ( D )


qD = 576 kg/m

Beban Hidup ( L )

qL = 2 400 = 800 kg/m


qU = 1.2 qD + 1.6 qLqU = 1.2 576 1.6 800 = 1971.2 Kg/m

Mu=(1/24)*q*L^2 = 0.041667 1971.2 16 = 1314.133 Kgm

= 0.5 1971.2 4 = 3942.4 Kg


Y = L = 400 = 1.111111360 360

Ix > 5 ( 5.76 8 ) 2.56E+10384 2100000 1.111111

Ix > 1965.714 cm4


200 x 200 x 8 x 12


lqDu u2

1max

EY

lqLqDIx

4*)(

384

5





Beban Mati ( D )


qD = 625.9 kg/m

Beban Hidup ( L )

qL = 2 400 = 800


qU = 1.2 qD + 1.6 qLqU = 1.2 625.9 1.6 800 = 2031.08 Kg/m

= 0.041667 2031.08 16 = 1354.053 Kgm

= 0.5 2031.08 4 = 4062.16 Kg


Y = L = 400 = 1.111111360 360

= 5 ( 6.259 8 ) 2.56E+10384 2100000 4720

= 0.480 < 1.111111

OK



2

8

1max lqMu u

lqDu u2

1max

EIx

lqLqDY

4*)(

384

5max


200 170 150 168024 15.49193 8 15.49193

8.333333 10.97345 18.75 108.4435OK OK


Mp = fy * Zx= 2400 * 472= 1132800 kgcm= 11328 kgm



Lp = 255.0503 cm



== 0.25 1.2 400 2400 = 288000 kgcm= 2880 kgm

0.9 Mp = 0.9 * 11328 = 10195.2 kgm

0.9 Mp > Mu10195.2 > 1354.053

OK


150 < 11008 15.49193

fytf

b 170

2

fyt

h 1680

fy

EiyLp *76.1


fytw

h 1100

18.75 < 71.00469Plastis

Vn = 0.6 fy Aw= 0.6 2400 0.8 20= 23040 Kg

Vu < ФVn

4062.16 < 0.9 230404062.16 < 20736

OK









Beban Mati




qL = 400 Kg/m2



qL = 400 kg/m2





As = 0.785 mm2


= 4.547771 Buah= 5 Buah




Beban Mati ( D )


qD = 1152 kg/m

Beban Hidup ( L )

qL = 4 400 = 1600 kg/m


qU = 1.2 qD + 1.6 qLqU = 1.2 1152 1.6 1600 = 3942.4 Kg/m

Mu=(1/24)*q*L^2 = 0.041667 3942.4 16 = 2628.267 Kgm

= 0.5 3942.4 4 = 7884.8 Kg


Y = L = 400 = 1.111111360 360

Ix > 5 ( 11.52 16 ) 2.56E+10384 2100000 1.111111

Ix > 3931.429 cm4


200 x 200 x 8 x 12


lqDu u2

1max

EY

lqLqDIx

4*)(

384

5





Beban Mati ( D )


qD = 1561.9 kg/m

Beban Hidup ( L )

qL = 4 400 = 1600


qU = 1.2 qD + 1.6 qLqU = 1.2 1561.9 1.6 1600 = 4434.28 Kg/m

= 0.041667 4434.28 16 = 2956.187 Kgm

= 0.5 4434.28 4 = 8868.56 Kg


Y = L = 400 = 1.111111360 360

= 5 ( 15.619 16 ) 2.56E+10384 2100000 4720

= 1.063324 < 1.111111

OK



2

8

1max lqMu u

lqDu u2

1max

EIx

lqLqDY

4*)(

384

5max


200 170 150 168024 15.49193 8 15.49193

8.333333 10.97345 18.75 108.4435OK OK


Mp = fy * Zx= 2400 * 472= 1132800 kgcm= 11328 kgm



Lp = 255.0503 cm



== 0.25 1.2 400 2400 = 288000 kgcm= 2880 kgm

0.9 Mp = 0.9 * 11328 = 10195.2 kgm

0.9 Mp > Mu10195.2 > 2956.187

OK


150 < 11008 15.49193

fytf

b 170

2

fyt

h 1680

fy

EiyLp *76.1


fytw

h 1100

18.75 < 71.00469Plastis

Vn = 0.6 fy Aw= 0.6 2400 0.8 20= 23040 Kg

Vu < ФVn

8868.56 < 0.9 230408868.56 < 20736

OK

Nu = 780.58 kgL = 4 m

WF 250 x 250 x 9 x 14




kondisi tumpuan jepit-jepit k = 0.653.3.9.1 Kontrol Penampang


250 170 190 168028 15.491933385 9 15.49193

8.928571 10.973452814 21.11111 108.4435OK OK

arah sumbu kuat (sumbu x)k*Lx

= 0.65 4000 = 24.07407ix 108

(fy/E)^0,5 = 24.07407 15.49193 = 0.265455π 3.141593 447.2136

1.43 = 1.0055231,6 - 0,67λcx

Nn = Ag*fcr = Ag*(fy*ωx) = 220016.8302 kg

Nu=

780.58= 0.004174 < 1 OK

0,85* 220016.8302

arah sumbu lemah (sumbu y)k*Ly

= 0.65 4000 = 41.33545iy 62.9

λy(fy/E)^0,5 = 41.33545 15.49193 = 0.455789

λx =

λcx = λx

0,25<λcx<1,2 ωx =

ɸc*Nn

λy =

λcy =

fytf

b 170

2

fyt

h 1680

π(fy/E)^0,5 =

3.141593 447.2136= 0.455789

1.43 = 1.104571,6 - 0,67λcy

Nn = Ag*fcr = Ag*(fy*ωy) = 200287.9266 kg

Nu=

780.58= 0.004585 < 1 OK

0,85* 200287.9266

0,25<λcy<1,2 ωy =

ɸc*Nn

BUHUL 1Diasumsikan:

Tebal pelat (tp) = 5 mm

jumlah bidang geser (m) = 1

Baut tanpa ulir pada bidang geser (r1) = 0.5

digunakan bD 8 Tension = 37 kg/cm2

370 Mpa

Mutu Baja = BJ 37fu = 3700 kg/cm2 = 370 Mpa

fy = 2400 kg/cm2 = 240 Mpa

a. Tahanan geser baut

Ab =1

π = 0.25 3.1415927 644

= 50.2655 mm2 untuk satu irisandirencanakan pasangan baut dengan dua irisan, maka:Ab = 2 50.265482 = 100.530965 mm2Rn = m r1 Ab

= 1 0.5 370 100.530965 = 18598.229 N/baut

0.75 18598.229 = 13948.6714 N/baut

b. Tahanan tumpu baut

Rn = 2.4 db tp Fu

= 2.4 8 5 370 = 35520 N/baut

0.75 35520 = 26640 N/baut

Fub =

D2

Fub

ɸRn =

ɸRn =

Keterangan Tahanan geser baut Tahanan tumpu bautTitik Buhul Batang m r1 d Ab Rn ɸRn d tp fu

1B1 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V1 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370D1 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

2

B1 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370B2 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V2 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370D2 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

3

B2 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370B3 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V3 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370D3 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

4

B3 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370B4 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V4 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370D4 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

5B4 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V5 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370B5 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

10A1 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V1 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

11

A1 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370A2 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370D1 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V2 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

12

A2 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370A3 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370D2 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V3 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

13

A3 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370A4 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370D3 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V4 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

14

A4 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370A5 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370D4 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370D5 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370V5 1 0.5 370 8 100.531 18598.23 13948.67 8 5 370

Fub

Tahanan tumpu bautTu n n dipakai

Rn ɸRn35520 26640 11992.51 0.85976 235520 26640 2012.912 0.144308 235520 26640 17260.02 1.237395 235520 26640 11992.51 0.85976 235520 26640 20558.58 1.473874 235520 26640 5670.257 0.406509 235520 26640 12328.59 0.883854 235520 26640 20558.58 1.473874 235520 26640 25698.23 1.842342 235520 26640 3444.725 0.246957 235520 26640 7397.152 0.530312 235520 26640 25698.23 1.842342 235520 26640 27411.45 1.965165 235520 26640 1219.193 0.087406 235520 26640 2465.717 0.176771 235520 26640 27411.45 1.965165 235520 26640 27490.75 1.970851 235520 26640 27411.45 1.965165 235520 26640 0 0 235520 26640 2012.912 0.144308 235520 26640 0 0 235520 26640 11992.51 0.85976 235520 26640 17260.02 1.237395 235520 26640 5670.257 0.406509 235520 26640 11992.51 0.85976 235520 26640 20558.58 1.473874 235520 26640 12328.59 0.883854 235520 26640 3444.725 0.246957 235520 26640 20558.58 1.473874 235520 26640 25698.23 1.842342 235520 26640 7397.152 0.530312 235520 26640 1219.193 0.087406 235520 26640 25698.23 1.842342 235520 26640 25698.23 1.842342 235520 26640 2465.717 0.176771 235520 26640 2465.717 0.176771 235520 26640 27490.75 1.970851 2

KOLOM TEPIP = 7186.26 KG (DARI SAP)Mutu Baja = BJ 41

fu = 4100 kg/cm2 = 410 Mpa

fy = 2500 kg/cm2 = 250 Mpa

WF 300 x 300 x 10 x 15

A = 119.8 cm2 tf = 15 mm Zx = 1360 cm3W = 94 kg/m Ix = 13100 cm4 Zy = 450 cm3a = 300 mm Iy = 7510 cm4 h = 234 mm


a. Menentukan ukuran pelat kakiL = 28.5 cm 29 cmB = 1,7*L = 48.45 cm 49 cma = 0,35*L = 9.975 cm 10 cm

Kenyataan B = 490 mma = B-L = 490 - 300 = 95 mm

2 2

P = 7186.26 KG (DARI SAP)B = 47.5 cmPondasi Beton : fyc digunakan= 2500 kg/cm2

F = 1,5*L*B*fyc L = F/(1,5*B*fyc) = 0.060516 cm 0.605159 mm

fyc = F = F = 1666.667 kg/cm2LB 1,5*L*B

< 2500 OK

b. Menentukan tebal pelat kakiMb = a^2*fyc

= 75208.3333333 kg.cm2

W = s^2*L6

fy = Mb=

75208.3333333 6W s^2

2500 = 75208.3333333 6s^2

s = 13.435 cmatau

s = a*(3fyc/fy)^0,5 s = 13.435 cm

KOLOM TENGAHP = 1905.39 KG (DARI SAP)Mutu Baja = BJ 41

fu = 4100 kg/cm2 = 410 Mpa

fy = 2500 kg/cm2 = 250 Mpa

WF 300 x 300 x 10 x 15



a. Menentukan ukuran pelat kakiL = 28.5 cm 29 cmB = 1,7*L = 48.45 cm 49 cma = 0,35*L = 9.975 cm 10 cm

Kenyataan B = 490 mma = B-L = 490 - 300 = 95 mm

2 2

P = 1905.39 KG (DARI SAP)B = 47.5 cmPondasi Beton : fyc digunakan= 2500 kg/cm2

F = 1,5*L*B*fyc L = F/(1,5*B*fyc) = 0.016045 cm 0.160454 mm

fyc = F = F = 1666.667 kg/cm2LB 1,5*L*B

< 2500 OK

b. Menentukan tebal pelat kakiMb = a^2*fyc

= 75208.3333333 kg.cm2

W = s^2*L6

fy = Mb=

75208.3333333 6W s^2

2500 = 75208.3333333 6s^2

s = 13.435 cmatau

s = a*(3fyc/fy)^0,5 s = 13.435 cm

5.3 Perencanaan Beban Akibat Plat Lantai, Kolom Memanjang dan Melintang

Lx = 4 m

Ly = 4 m

A ( Balok Induk Melintang )

350 x 350 x 12 x 19



B ( Balok Anak )

250 x 250 x 9 x 14



C ( Balok Induk Memanjang)

300 x 300 x 10 x 15

A = 119.8 cm2 tf = 15 mm Zx = 1360 cm3W = 94 kg/m Ix = 20400 cm4 Zy = 450 cm3

a = 300 mm Iy = 6750 cm4 h = 234 mmbf = 300 mm tw = 10 mm r = 18 mmiy = 7.51 cm ix = 13.1 cm

Beban Mati

- Bondex = 10.1 kg/m2

- Beton = 0.12 x 2400 = 288 kg/m2

- Beban Finishing = 90 kg/m2

qM = 388.1 kg/m2

Beban Hidup

qL = 400 kg/m2

5.3.1 Perencanaan Pembebanan Portal Melintang

5.3.1.1 Beban P1

Beban Mati

Berat Pelat = 388.1 4 4 = 6209.6 kg

Balok induk melintang = 137 4 = 548 kg

Balok anak = 72.4 4 = 289.6 kg

Pm1 = 7047.2 kg

Beban Hidup

Ph1 = 400 4 4 = 6400 kg

5.3.1.4 Beban P2

Beban Mati

Berat Pelat = 388.1 2 4 = 3104.8 kg

Balok induk memanjang = 94 4 = 376 kg


Pm2 = 3754.8 kg

Beban Hidup

Ph2 = 400 2 4 = 3200 kg

5.3.2 Perencanaan Pembebanan Portal Memanjang

5.3.2.1 Beban P3

Beban Mati

Berat Pelat 388.1 4 2 = 3104.8 kg


Balok induk memanjang = 94 2 = 188 kg

Pm3 = 3840.8 kg

Beban Hidup

Ph3 = 400 4 2 = 3200 kg

x xx

x

x x

x x

x xxx

x x

x xxx

x

5.3.2.2 Beban P4

Beban Mati

- Berat Pelat = 388.1 4 2 = 3104.8 kg

- Balok induk memanjang = 94 4 = 376 kg

- Balok induk melintang = 137 4 = 548 kg

Pm4 = 4028.8 kg

Beban Hidup

Ph4 = 400 4 2 = 3200 kg

5.3.3 Beban Portal - Portal Melintang

P1 Pm1 = 7047.2 kg

Ph1 = 6400 kg

P2 Pm2 = 3754.8 kg

Ph2 = 3200 kg

5.3.4 Beban - beban Portal Memanjang

P3 Pm3 = 3840.8 kg

Ph3 = 3200 kg

P4 Pm4 = 4028.8 kg

Ph4 = 3200 kg

x x

x xxx

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