bab 5 perencanaan portal.docx

BAB 5

PERENCANAAN PORTAL

Tampak Samping

Potongan A-A

Tampak Atas

2,02

1,63

4

4

1

PEMBEBANAN

1. Beban Atap

Ra = 2,77 ton

2. Pembebanan Balok Induk

Penentuan profil berdasarkan bentang :

d >

L20 ; dimana

L20

=350020

=175mm

Dicoba menggunakan profil WF 250 x 250 x 11 x 11

Data profil W 250x250x11x11

A = 82,06 cm² tf =11 mm

d = 244 mm bf = 252 mm

tw = 11 mm Wt = 64,4 kg/m

h0 = d – 2tf = 244 – 22 = 222 mm

(Tabel Profil Konstruksi Baja, Ir. Rudy Gunawan dengan petunjuk Ir. Morisco)

Beban mati (qd)

Berat pelat = 0,15 m x 2400 kg/m3 x 1 = 360 kg/m2

Beban Spesi = 0,02 m x 2100 kg/m3 x 1 = 42 kg/m2

Beban Keramik = 0,01 m x 1700 kg/m3 x 1 = 17 kg/m2

Beban Plafon + penggantung = 11 kg/m2 + 7 kg/m2 = 18 kg/m2

Berat Pasir = 0,02 m x 1800 kg/m3 x 1 = 36 kg/m2

Beban instalasi listrik = 7 kg/m 2 +

∑qd = 480 kg/m2

Berat dinding = 0,15 m x 4 m x 1700 kg/m3 = 1020 kg/m

Berat sendiri balok induk = 64,4 kg/m

Berat sendiri balok anak = 31,5 kg/m

Ptembok = 0,5.Ra + Beban tembok

= 0,5.2770 + 1115,9 = 2500,9 kg/m

Beban hidup (ql)

ql = 250 kg/m2

Beban hidup fungsi gedung untuk pasar berdasarkan PPIUG untuk gedung

1983, harus diambil minimum sebesar 250 kg/m2

o Potongan Vertikal

1. Tipe 1

Lx = 3,5 m

Ly = 3,5 m

sehingga,

Leq = 13Lx=

13.3,5= 1,1667 m

Pembebanan

1. Beban Mati

qd = qd . 2Leq + berat dinding + Berat balok induk

= 480 kg/m2 . 2. 1,1667 m + 1020 kg/m + 64,4 kg/m

= 2204,4 kg/m

2. Beban Hidup (ql)

ql = ql1 . 2Leq

= 250 kg/m2 .2 .1,1667 m

= 583,33 kg/m

3. Beban Berfaktor (qu)

qu = 1,2 qd + 1,6 ql

= (1,2 . 2204,4 kg/m) + (1,6 . 583,33 kg/m)

= 3578,613 kg/m

Digunakan qu yang terbesar qu = 3578,613 kg/m’

Lx

Leq ½ Lx

3. Beban akibat angin

V = 20km/jam = 5,556m/dtk

P = 1/16 V2 = 1/16 . 202 = 25 km/jam = 6,945 m/dt

Pangin = 25 kg/m2

Koefisien angin tekan = 0,6

Bidang luar berupa dinding vertikal yang berada di pihak angin (PPIUG hal 23)

qw= Koefisien angin tekan x p x b

= 0,6 x 25 x 4 = 60 kg/m

I. Mw = ½ qw.h2= ½ .60. (8)2 = 1920 kg.m

II. Mw = W1.l1 + W2. (l1 + l2)

= W1.4,5 + W2( 4,5 + 4,5)

Mw I = Mw II

2733,75 = W1.4 + W2(4,5 + 4,5)

W 1 =M .h1

∑ h2 =1920 x 442+( 4+4 ) ¿

¿ 2 ¿¿¿¿

= 96 kg

W 2 =M . (h1+h2)∑ h2 =

1920 x (4+4 )42+ (4+4 )¿

¿ 2 ¿¿¿¿

= 192 kg

Pra Design

Pra Design Kolom Atas ( Batang 2 – 3 dan 9 - 10 )

a. Akibat Beban Aksial

Ag =

PFy

= 27700250 = 110,8 mm2

Digunakan profil WF 100x50

dengan Ag = 11,85 cm2,

Zx= 37,5 cm3

b. Akibat Momen

M2-3 = 768 kgm

Pa = 2770 kg

W2 = 192 kg

Mlt = W2 . h2

= 192 kg/m . 4 m

= 768 kgm

Zx =

MFy

= 7680000 250 = 30720 mm3

Digunakan profil WF100x100


Zx= 76,5 cm3

Ag total = 11,85 + 21,90 = 33,75 cm2

Zx total = 37,5+ 76,5 = 114 cm3


denganAg = 40,14 cm2

Zx = 219 cm3

Wt = 31,5 kg/m

Pra Design Balok

Gambar 5. Pembebanan pada portal

Hasil dari SAP 2000

Gambar 5. Momen 3-3 Diagram SAP 2000

Dari SAP di dapat nilai

M tumpuan : 13,85 ton m

M Lapangan : 8,36 ton m

Pradesign Balok – Kolom

Zx=

MFy

= 90800000 250 = 334400 mm3 = 334,4 cm3


dengan Ag = 56,24 cm2

Zx = 502 cm3

Akibat beban P

Ag=

PFy

= 27700250 = 110,8 mm2

Digunakan profil WF100x50


Zx = 37,5 cm3

Ag total = 56,24 + 11,85 = 68,09 cm2

Zx total = 502 + 37,5 = 539,5 cm3

Digunakan profil WF 350x350x14x22

Dengan Ag= 202 cm2

Zx = 2670 cm3

Wt = 159 kg/m

Pra Design Kolom Bawah ( Batang1-2 )

Mlt = W1 . h1 .

= 96 . 4 .

= 384 kgm

M total = Mlt + Mu balok

= 384 + 8360

= 8744 kg m

Ptotal = Pplat

= 3578,613 . 4

= 14314,45 kg

Jadi, design harus mampu menahan

Ptotal = 14314,45 kg

Mtotal = 8744 kgm

W1 = 0,1215 t

1

2

P

Balok – Kolom

1. Akibat beban P

Ag =

PFy

= 143144,5250 = 572,5781 mm2 = 5,7258 cm2



Zx = 37,5 cm3

2. Akibat momen

Zx=

87440000250 = 572578,1 mm3 = 572,5781 cm3

Digunakan profil WF 200x200x8x12


Zx = 472 cm3

Ag total = 11,85 + 63,53 = 75,38 cm2

Zx total = 37,5 + 472 = 509,5 cm3

Digunakan profil WF 250 x 250x11x11

dengan Ag = 202 cm2

Zx = 2670 cm3

Pra Design Kolom Bawah ( Batang 4 – 5 , 6 - 7,8 –9)

Digunakan profil W 350 x 350x14x22

dengan Ag = 202 cm2

Zx = 2670 cm3

Pembebanan Akhir dan Pra Design Profil

Gambar 5. Shear Force 2-2 Diagram SAP 2000 v10

qu=3,73 t/m

Gambar 5. Axial Force Diagram SAP 2000 v14

Gambar 5. Momen Force 3-3 Diagram SAP 2000 v14

Perhitungan Kapasitas PortalPerhitungan Kolom Atas (Batang 2-3 ) Efek Kolom Terhadap Aksial

Dicoba Profil W150x150

Data – data Profil W150x150:

A = 40,14 cm²d = 150 mm

tw = 7 mm

tf = 10 mm

Ix = 1640 cm4

Sx = 219 cm³Sy = 75,1 cm³rx = 6,39 cm

ry = 3,75 cm

Iy = 563 cm4

1. Kontrol Stabilitas Momen

Lp= 1 ,76 (r y )√ E

Fy =1 ,76 (37 ,5 )√200000

250 = 1866,7619 mm = 1,8668 m

rts = √ Iy×hoSx

=√5630000×(150−2 x10 )219000

= 57,8101

Lr = π×r ts×√ E

fy− fr

=3 ,14×57 ,8101×√200000

250−123

h2 = 4 mPa = 2,77 tonVu = 0,27 tonMlt = W2 x H2 = 0,192 x 4 = 0,768 tonm∆ H = 0,000471 ftH2= 4 m

W2 = 0,27 tPa = 2,77 t

= 7207,2026 mm = 7,2072 m

Didapatkan :

Lb = 4 m

Lp =1,8668 m

Lr = 7,2072 m

Karena Lp< Lb<Lr

Zx= (b.tf)(H-tf)+(tw.(h/2-tf)2

= (150 x 10) (150 - 10) + ( 7 x ( (150/2) - 10)2 )

= 239575 mm3

Mp = Zx.Fy = 239575 x 250 = 59893750 Nmm = 59,8938 kNm

Mr = Sx ( fy-fr) = 219000 (250-123) = 27813000 Nmm = 27,813 kNm

Dari perhitungan SAP 2000 v14 diperoleh

Mmaks = 710 kgm

MA = ¾ L = 532,5 kgm

MB = ½ L = 355 kgm

MC = ¼ L = 177,5 kgm

Cb =

12 .5Mmax2. 5Mmax+3Ma+4Mb+3Mc

=

(12 .5 x710 )(2 .5 x710 )+(3 x532 ,5)+(4 x 355)+(3 x 177 ,5 )

= 1,67< 2,3 (memenuhi)

Mn=Cb(Mp−(Mp−Mr ) Lr−LbLr−Lp )

=1 ,67(59,8938 - (59,8938-27,813 ) 7 ,2072−4,5

7 ,2072−1 ,8668 ) = 72,8640 kNm > Mp = 59,8938 kNm

Maka diambil Mn = 59,8938 kNm

ØMn = 0,85 x 59,8938 = 50,9097 kNm

ØMn > Mu

5,0909 tm > 0,71 tm

Kontrol :

MuØMn

=710 5090,97

=0 ,1395<1

Sehingga kolom dapat menahan momen

2. Kontrol Terhadap Aksial

K= 1 (jepit-jepit)

k . Lry

= 1 .4003 ,75 = 106,67

λC =

k . Lry √ Fy

π2 . E = 106,67. √250π2 . 200000 = 1,2004

Untuk λc ¿1,2

ωx =1,25λc2

=1,25 . 1,20042= 1,8013

Nnx= Ag x

f yωx

= 4014 x

2501,8013

= 557108 N

Ø Pn = Ø Nnx = 0,85 x 557108= 473542 N

Ø Pn > Pmax

47,35 ton> 2,77 ton ……………(OK!)

PuφC Pn

= 2,77 47 ,35 = 0,058

Maka digunakan,

Pu2xPn

+ 8Mu9.φbMn

= 2 ,772x 47 ,35

+ 8x 0 ,719 x 5 ,0909

= 0,1532 < 1

Sehingga balok dapat menahan beban

3. Kontrol batas Penampang Kompak Profil

Tekuk Badan

λ=h0

tw=

(150−2x 10)7

=18 ,5714

λp=1680√Fy

=1680√250

=106 ,2525

λr=2550√Fy

=2550√250

=161 ,2762

λ < λp =18,5714 < 106,2525→Penampang Kompak

Tekuk Sayap

λ= bf2tf

=1502 x10

=7,5

λp=170√Fy

=170√250

=10 ,7517

λr=370√(Fy−Fr )

=370√(250−123 )

=32 ,8322

λ < λp = 7,5 < 10,7517→Penampang Kompak

4. Kontrol Kekuatan Geser

Vu = 270 kg

h0 = d – (2 x tf )

= 150 – 2x10= 130 mm

λ=h0

tw=130

7=18 ,5714

2 ,24√E /Fy=2,24 √200000 /250=63 ,3568h0

tw<2 ,24√E /Fy→18 ,5714<63 ,3568→Cv=1

Aw = (d – 2.tf) tw

Pu = 22,37 t

W2 = 0,192 t

= ( 150 – 2.10 ) 7

= 910 mm2

Vn = 0,6. fy. Aw .Cv

= 0,6 x 250 x 910 x 1

= 136500 N

Ø Vn = 0,9 x 136500 N= 122850 N

12285 kg > Vu = 270 kg............... (Aman)

Perhitungan Kolom Bawah ( batang4-5 ) Efek Kolom Terhadap Aksial

Dicoba Profil W350x350x14x22

Data – data Profil W350x350x14x22

Ag = 202 cm²d = 356 mm

tw = 14 mm

tf = 22 mm

bf = 352 mm

Ix = 47600 cm4

Sx = 2670 cm³Sy = 909 cm³rx = 15,3 cm

ry = 8,9 cm

Iy = 16000 cm4

L = 4,5 m

GA= 1 ( jepit )

Pu = 22,37 tonW2 = 192 kgVu = 1,28 tonMu = 4,02 tonm

4 m

GB =

(47600450 )

(47600450

+47600900 ) = 0,6667

k = 0,74 (dari monogram LRFD hal 59)

k . Lry

= 0 ,74 ×2508,9 = 20,7865

λC =

k . Lry √ Fy

π2 . E

= 20,7865√250π2 . 200000 = 0,2339

Untuk 0 ,25>λcω = 1

Nnx = Ag x

f yωx

4,02 tm

1,73 tm

= 20200 x 250

1

= 5096891,401 N

Ø Pn = Ø Nnx = 0,85 x 5096891,401 = 4332357,691 N

Ø Pn > Pmax

433,2358 ton > 22,37 ton ……………(OK!)

Sehingga kolom dapat menahan beban

1. Lateral Torsional Backling (LTB)

Lp= 1 ,76 (r y )√ E

Fy =1 ,76 (89)√200000

250 = 4430,4482 mm = 4,4304 m

rts = √ Iy×hoSx

=√160000000×(356−2 x22 )2670000

= 136,7356


fy− fr

=3 ,14×136 ,7356×√200000

250−123

= 17046,8685 mm = 17,0469 m

Didapatkan :

Lb = 4,5 m

Lp = 3,4596 m

Lr = 11,3398 m

Karena Lp< Lb<Lr

Zx= (b.tf)(H-tf)+(tw.(h/2-tf)2

= (352 x 22) (356 - 22) + ( 14 x ( (356/2) - 22)2 )

= 2927200 mm3

Mp = Zx.Fy = 2927200 x 250 = 731800000 Nmm = 731,8 kNm

Mr = Sx ( fy-fr) = 2670000 (250-123) = 339090000 Nmm = 339,09kNm

Dari perhitungan SAP 2000 v14 diperoleh

Mmaks = 4,02 tonm

MA = ¼ L = 2,59 tonm

MB = ½ L = 1,15 tonm

MC = ¾ L = 0,29 tonm

Cb =


=

(12 .5x 4 ,02 )(2 .5 x 4 ,02)+(3. 2 ,59)+( 4 x1 ,15)+(3x 0 ,29)

= 2,1576 < 2,3 (memenuhi)


=2 ,1576 (1200,1520 - (1200,1520-766,29 ) 4,5−3 ,4596

11 ,3398−3 ,4596 ) = 2465,8573 kNm > Mp = 731,8 kNm

Maka diambil Mn = 731m8 kNm

ØMn = 0,85 x 731,8 = 622,03 kNm

ØMn > Mu

6220,3 kg m > 4020 kg m

Kontrol :

MuØMn

=4020 6220,3

=0 ,6463<1

Sehingga kolom dapat menahan momen


k . Lry

= 0,7 ×2508,9 = 19,6629

λC =

k . Lry √ Fy

π2 . E

= 19,6629√250π2 . 200000 = 0,2213

Untuk λc<0 ,25 ω =1

Nnx= Ag x

f yωx

= 20200 x 250

1

= 5050000 N

Ø Pn = Ø Nnx = 0,85 x 5050000 = 4292500 N

Ø Pn > Pmax

429,25 ton > 25,2982 ton ……………(OK!)

PuφC Pn

= 25,2982429 ,25 = 0,0589 <0,2

Maka digunakan,

Pu2xPn

+ Mu.φbMn

= 25 ,29822. 429 ,25

+ 4 ,026 ,22

= 0,67577 < 1

Sehingga kolom dapat menahan beban


Tekuk Badan

λ=h0

tw=

(356−2 x22)14

=22,2857

λp=1680√Fy

=1680√250

=106 ,2525

λr=2550√Fy

=2550√250

=161 ,2762


Tekuk Sayap

λ= bf2tf

=3522 x22

=8

λp=170√Fy

=170√250

=10 ,7517


=370√(250−123 )

=32 ,8322

λ < λp = 8 < 10,7517→Penampang Kompak


Vu = 1,28 ton

h0 = d – (2 x tf) = 356 - (2x22) = 312

λ=h0

tw=312

14=22 ,2857

2 ,24√E /Fy=2,24 √200000/250=63 ,3568h0

tw<2 ,24√E /Fy→22 ,2857<63 ,3568→Cv=1

Aw = (d – 2.tf) tw

= (356 – 2.22)14

= 4368 mm2


= 0,6 x 250 x 4368 x 1

= 655200 N

Ø Vn = 0,9 x 655200 N = 589680 N

58,968 ton> Vu =1,28 ton............... (Aman)

Perhitungan Balok ( Batang 5-7 )

Dicoba Profil W350x350x14x22

Data – data Profil W350x350x14x22

Ag = 202 cm²d = 356 mm

tw =14 mm

tf = 22 mm

bf = 352 mm

Ix = 47600 cm4

Sx = 2670 cm³Sy = 909 cm³rx = 15,3 cm

ry = 8,9 cm

Iy = 16000 cm4

Pu = 1,93 ton

Vu = 13,82 ton

L = 7 m

k = 1 (jepit-jepit)

k . Lry

= 1×7008,9 = 78,6517

λC =

k . Lry √ Fy

π2 . E = 78,6517. √250π2 . 200000 = 0,8851

Untuk λc<1,2 ; ω=1

Nnx = Ag x

f yωx

= 20200 x 250

1

= 5050000 N

Ø Pn = Ø Nnx = 0,75 x 5050000 = 3787500 N

Ø Pn > Pmax

378,750 ton > 2,8097 ton ……………(OK!)


1. Kontrol Stabilitas Momen

Lp= 1 ,76 (r y )√ E

Fy =1 ,76 (37 ,5 )√200000

250 = 1866,7619 mm = 1,8668 m

rts = √ Iy×hoSx

=√160000000×(356−2 x22 )2670000

= 136,7356


fy− fr

=3 ,14×57 ,8101×√200000

250−123

= 7207,2026 mm = 7,2072 m

Didapatkan :

Lb = 7 m

Lp = 3,4596 m

Lr = 11,3398 m

Karena Lp < Lb <Lr

Zx =(b.tf)(H-tf)+(tw.(h/2-tf)2

=(352x22)(356-22)+(14x(356/2-22)2)

= 2927200 mm3

Mp= Zx.Fy = 2927200 x 250 = 731800000 Nmm = 731,8 kNm

Mr = Sx ( fy-fr) = 2670000 (250-123) = 339090000 Nmm = 339,09 kNm

Mmax = 13,82 tonm

MA = ¼ L = 3,34 tonm

MB = ½ L = 9,08 tonm

MC = ¾ L = 3,39 tonm

Cb =


=

(12 .5 x 13 ,82 )(2 .5 x13 ,82 )+(3 x3 ,34 )+(4 x 9 ,08 )+(3 x3 ,39)

= 1,8971< 2,3 (memenuhi)


=1 ,8971(731,8 - (731,8 -339,09 )( 7−3 ,4596

11 ,3398−3 ,4596 ))= 1042,8374 kNm

Mn = 1042,8374 kNm < Mp = 1200,1520 kNm

Maka Mn diambil 1042,8374 kNm

ØMn = 0,85 x 1042,8374 kNm = 886,4118 kNm

Kontrol

Mn > Mu

88,64 tonm > 13,82 tonm

Sehingga balok dapat menahan momen


k . Lry

= 1×7008,9 = 78,6517

λC =

k . Lry √ Fy

π2 . E = 78,6517. √250π2 . 200000 = 0,8851

Untuk 0 ,25<λc<1,2

ω =1.43/(1.6-0.67λcx )

=1.43/(1.6-0.67 x 0,8851) = 1,4201

Nnx = Ag x

f yωx

= 20200 x

2501,4201

= 3556087,599 N

Ø Pn = Ø Nnx = 0,85 x 3556087,599 = 3022674,46 N

Ø Pn > Pmax

302,2674 ton > 1,93 ton ……………(OK!)

PuφC Pn

= 1,93302 ,2674 = 0,0064 < 0,2

Maka digunakan,

Pu2xPn

+ Mu.φbMn

= 1 ,932. 302 ,2674

+13 ,8288 ,64

= 0,1591 < 1



Tekuk Badan

λ=h0

tw=

(356−2 x22)14

=22,2857

λp=1680√Fy

=1680√250

=106 ,2525

λr=2550√Fy

=2550√250

=161 ,2762


Tekuk Sayap

λ= bf2tf

=3522 x22

=8

λp=170√Fy

=170√250

=10 ,7517


=370√(250−123 )

=32 ,8322

λ < λp = 8 < 10,7517→Penampang Kompak


Vu= 13,82 ton

h0 = d – 2 x tf = 356 – (2 x 22) = 312

λ=h0

tw=312

14=22 ,2857

2 ,24√E /Fy=2,24 √200000 /250=63 ,3568h0

tw<2 ,24√E /Fy→22 ,2857<63 ,3568→Cv=1

Aw = (d – 2.tf) tw

= (356 – 2.22)14

= 4368 mm2


= 0,6 x 250 x 4368 x 1

= 655200 N

Ø Vn = 0,9 x 655200 N = 589680 N

58,968 ton > Vu = 15,0056 ton............... (Aman)

bab 5 perencanaan portal.docx

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