bab 5 perencanaan portal.docx
TRANSCRIPT
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
dengan Ag = 21,90 cm2,
Zx= 76,5 cm3
Ag total = 11,85 + 21,90 = 33,75 cm2
Zx total = 37,5+ 76,5 = 114 cm3
Digunakan profil WF 150x150
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
Digunakan profil WF 250x175
dengan Ag = 56,24 cm2
Zx = 502 cm3
Akibat beban P
Ag=
PFy
= 27700250 = 110,8 mm2
Digunakan profil WF100x50
dengan Ag = 11,85 cm2,
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
Digunakan profil WF 100x50
dengan Ag = 11,85 cm2
Zx = 37,5 cm3
2. Akibat momen
Zx=
87440000250 = 572578,1 mm3 = 572,5781 cm3
Digunakan profil WF 200x200x8x12
dengan Ag = 63,53 cm2
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
Lr = π×r ts×√ E
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 .5Mmax2. 5Mmax+3Ma+4Mb+3Mc
=
(12 .5x 4 ,02 )(2 .5 x 4 ,02)+(3. 2 ,59)+( 4 x1 ,15)+(3x 0 ,29)
= 2,1576 < 2,3 (memenuhi)
Mn=Cb(Mp−(Mp−Mr ) Lr−LbLr−Lp )
=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
2. Kontrol Terhadap Aksial
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
3. Kontrol batas Penampang Kompak Profil
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
λ < λp = 22,2857 < 106,2525→Penampang Kompak
Tekuk Sayap
λ= bf2tf
=3522 x22
=8
λp=170√Fy
=170√250
=10 ,7517
λr=370√(Fy−Fr )
=370√(250−123 )
=32 ,8322
λ < λp = 8 < 10,7517→Penampang Kompak
4. Kontrol Kekuatan Geser
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
Vn = 0,6. fy. Aw .Cv
= 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!)
Sehingga balok dapat menahan beban
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
Lr = π×r ts×√ E
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 .5Mmax2. 5Mmax+3Ma+4Mb+3Mc
=
(12 .5 x 13 ,82 )(2 .5 x13 ,82 )+(3 x3 ,34 )+(4 x 9 ,08 )+(3 x3 ,39)
= 1,8971< 2,3 (memenuhi)
Mn=Cb(Mp−(Mp−Mr ) Lr−LbLr−Lp )
=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
2. Kontrol Terhadap Aksial
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
Sehingga balok dapat menahan beban
3. Kontrol batas Penampang Kompak Profil
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
λ < λp = 22,2857 < 106,2525→Penampang Kompak
Tekuk Sayap
λ= bf2tf
=3522 x22
=8
λp=170√Fy
=170√250
=10 ,7517
λr=370√(Fy−Fr )
=370√(250−123 )
=32 ,8322
λ < λp = 8 < 10,7517→Penampang Kompak
4. Kontrol Kekuatan Geser
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
Vn = 0,6. fy. Aw .Cv
= 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)