perhitungan portal baja sederhana
DESCRIPTION
Perhitungan portal baja sederhanaTRANSCRIPT
1. Gambar Struktur PortalP5
P4 P6P3 P7
P2 P8P1 P9
5.000
1.25 11.000 1.25
1.25
6.00
6.00
6.00
6.00
6.00
6.00
1.25
1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50
1.25 12.000 1.25
0
0
0 00 00
0
0
0
0
0
QI Q2
0 00 00
00
00
00
00
00
00
00
00
00
6.000
24°
27°
1.5000
0
0
0
0
1.25 12.000 1.25
0 00 00
2. Perencanaan Gording● Direncanakan Menggunakan Atap Seng = 10.00 kg/m²
● = 13.40● Menggunakan Balok = Baja WF 300 x 300● Menggunakan Kolom = Baja WF 300 x 300● = 2400.00● Jumlah Medan (n) = 6 Medan● Tinggi kolom = 6.00 m● Koevisien Reduksi Beban Hidup = 0.50● panjang q1 = 2.00 m● panjang q2 = 1.50 m
● Beban Mati Atap (WLD)Beban Mati Atap Tepi = 10.00 x 0.5 x 1.50 + 1.25
= 20.00Beban Mati Atap Tengah = 10.00 x 0.5 x 1.50 + 0.50 x 1.50
= 15.00Beban (P1) = 20.00Beban (P2) = 15.00Beban (P3) = 15.00Beban (P4) = 15.00Beban (P5) = 15.00Beban (P6) = 15.00Beban (P7) = 15.00Beban (P8) = 15.00Beban (P9) = 20.00
Diambil beban terbesar = 20.00Akibat Sambungan (10% x Beban Mati) = 0.10 x 20.00
= 2.00
= 20.00 + 2.00= 22.00
= 15.00 + 2.00= 17.00
Beban Mati Pada (Q1) = x L= 22.00 x 0.5 x 6.00 + 0.50 x 0.00= 66.00 Kg
Beban Mati Pada (Q2) = x L= 17.00 x 0.5 x 6.00 + 0.50 x 0.00= 51.00 Kg
Beban Hidup (W LL)
● Beban Kebetulan = 100.00● Beban air Hujan Setebal = 0.05 m
● Koefisien Reduksi = 0.500
● Baban Hidup Atap (Q1) = Luas Atap x ql Atap x Koevisien Reduksi= 3.000 x 2.000 x 100.000 x 0.50 = 300.000 Kg
Baban Hidup Atap (Q2) = Luas Atap x ql Atap x Koevisien Reduksi= 3.000 x 1.500 x 100.000 x 0.50 = 225.000 Kg
● Baban Air Hujan (Q1) = 0.050 x 3.000 x 2.000 x 1000.00 = 300.000 KgBaban Air Hujan (Q2) = 0.050 x 3.000 x 1.500 x 1000.00 = 225.000 Kg
● Beban Hidup Atap (Q1) = Baban Hidup Atap (Q1) + Baban Air Hujan (Q1)= 300.000 + 300.000
Berat Gording Ditaksir [12 kg/m
Tegangan Leleh baja (Fy) Kg/m²
Kg/m³
kg/m
kg/mkg/mkg/mkg/mkg/mkg/mkg/mkg/mkg/mkg/mkg/m
kg/m
Maka qTotal (Q1)kg/m
Maka qTotal (Q2)kg/m
qTotal
qTotal
Kg/m2
Kg/m3
= 600.000 Kg
● Beban Hidup Atap (Q2) = Baban Hidup Atap (Q2) + Baban Air Hujan (Q2)= 225.000 + 225.000= 450.000 Kg
Faktor Beban dan Kombinasi Pembebanan
Sumber : SNI 03-1729-2002 /Perencanaan Struktur Baja dengan Metode LRFD (Halaman 11)
● Total Beban Mati Pada (Q1) = 1.200 x Beban Mati Pada (Q1)= 1.200 x 66.000= 79.200 Kg
● Total Beban Mati Pada (Q2) = 1.200 x Beban Mati Pada (Q2)= 1.200 x 51.000= 61.200 Kg
● Total Beban Hidup Pada (Q1) = 1.600 x Beban Hidup Atap (Q1)= 1.600 x 600.000= 960.000 Kg
● Total Beban Hidup Pada (Q2) = 1.600 x Beban Hidup Atap (Q2)= 1.600 x 450.000= 720.000 Kg
Beban Merata Pada Gording (Q1) =Beban Mati Atap (Q1) + Beban Mati Atap (Q1)
Panjang Gording
=79.200 + 960.000
3.000= 346.400 Kg/m
Beban Merata Pada Gording (Q2) =Beban Mati Atap (Q2) + Beban Mati Atap (Q2)
Panjang Gording
=61.200 + 720.000
3.000= 260.400 Kg/m
● Total Beban Atap Pada Q1 = Beban Mati Pada (Q1) + Beban Hidup Atap (Q1)= 79.200 + 960.000= 1039.200 Kg
● Total Beban Atap Pada Q2 = Beban Mati Pada (Q2) + Beban Hidup Atap (Q2)= 61.200 + 720.000= 781.200 Kg
● Dengan Jumlah Medan (n) = 6 maka di dapat nilai momen :M1 = 0.078M2 = 0.105M3 = 0.033 diambil = 0.105M4 = 0.079M5 = 0.046
M1 = 0.105 x 1039.20 M2 = 0.105 x 781.20= 109.116 Kg.m = 82.026 Kg.m= 10911.600 Kg.cm = 8202.600 Kg.cm
Perhitungan Dimensi Gording● Tegangan Leleh Baja (Fy) = 2400.00 kg/m²
● Tegangan Ijin =Fy
=2400.00
1.5 1.5= 1600.00 kg/m²
●
Wx = 41.20 cm³ b = 50.00 mmIx = 206.00 cm⁴ d/tw = 6.00 mmIy = 29.30 cm⁴ t/tf = 8.50 mm
Berat = 10.60 Kg/m ht = 64 mmh = 100.00 mm fr = 70.00Ø = 0.90
Dicoba Profil [10 dengan data sebagai berikut
Faktor Reduksi (Ø) Untuk Keadaan Kekuatan Batas
Sumber : SNI 03 – 1729 – 2002 (Halaman 18 Dari 184)
Perhitungan Momen Pada Gordingq = Berat Maksimum Beban Merata Atap + Berat Sendiri Gording
= 346.400 + 10.600
= 357.000 Kg/m
357.000 Kg/m
3.00
Mx/Mu =1
x 357.000 x 3.00 ²8
= 401.625 Kg.m= 0.401625 Ton.m
● KontrolSNI 03 – 1729 – 2002 (Halaman 35 Dari 184)
Mn≥ Mu
Ø
≥ 401.6250.90
≥ 446.250 Kg/m≥ 0.446 Ton.m
Pemeriksaan Terhadap Tekuk lokal
Flens λp =170
=170
= 3.470fy 2400.000
λr =370
=370
= 7.553fy 2400.000
λf =b
=50.000
= 2.9412 . tf 2 x 8.500
Baja [10
Mu ≤ Ø Mn
00
00
Web λp =1680
=1680
= 34.293fy 2400.000
λr =2550
=2550
= 52.052fy 2400.000
=h
=100.000
= 16.667tw 6.000
Kontrolλf= 2.941 < λf= 3.470 Penampang Kompak
λw =h
=64.000
= 10.667tw 6.000
Kontrolλw= 10.667 < λp= 34.293 Penampang Kompak
Modulus PlastisZx = b . tf . ( h - tf ) + 1/4 . tw . ( h - 2tf )²
= 50 x 8.5 x 100 - 8.5 + 0.25 x 6.00 x 100 - 2 . 9 2
= 49221.000 mm²
Mp = Zx x fy= 49221.000 x 2400.000= 118130400.000= 11.813 Ton.m
Mr = ( fy - fr ) x Sx = ( fy - fr ) . Ix /(h/2)
= 2400.00 - 70.00 x206.000 x 10⁴
100.000 : 2.00= 95996000.000= 9.600 Ton.m
Mn =λr - λ
xλ - λp
λr - λp λr - λp
=7.553 - 2.941
x 11.813 +2.941 - 3.470
x 9.6007.553 - 3.470 7.553 - 3.470
= 12.100 Ton.m● Kontrol
Mn= ( 12.09982) Ton.m > ( 0.44625) Ton.m OK
● Kontrol Tegangan
= Mx = 401.63 = 9.75 kg/m²Wx 41.20
= 9.75 kg/m² < = 1600.00 kg/m² OK
● Kontrol Lendutan
=1
x L400
=1
x 600.00400
= 1.500 Cm
fmax =K x Q x L³
I
λf
x Mp x Mr
Mu/Ǿ=
max
fijin
K = 3.22 (Tabel Potma)
fmax =K x Q x L³
Ix
=3.22 x 0.357 x 6.00
206.00
= 1.205 Cm
fmax == 1.205= 1.0978811 cm
Kontrol
fmax = 1.0978811 cm < = 1.500 cm Ok
= 10.60 Kg/m
3. Pembebanan Atap● Berat Gording
- Berat Gording Untuk (Q1) = Panjang Gording x Berat Gording/m= 0.50 x 6.00 + 0.5 x 0.00 x W= 3.00 x 10.60= 31.800 Kg
- Berat Gording Untuk (Q2) = Panjang Gording x Berat Gording/m= 0.50 x 6.00 + 0.5 x 0.00 x W= 3.00 x 10.60= 31.800 Kg
● Total Berat Penutup Atap- Berat Penutup Atap Untuk (Q1) = Total Beban Atap Pada (Q1) + Berat Gording (Q1)
= 1039.200 + 31.800= 1071 Kg
- Berat Penutup Atap Untuk (Q2) = Total Beban Atap Pada (Q2) + Berat Gording (Q2)= 781.200 + 31.800= 813 Kg
Diketahui beban Pada Atap:● Berat Beban (P1) = 1071.00 Kg● Berat Beban (P2) = 813.00 Kg● Berat Beban (P3) = 813.00 Kg● Berat Beban (P4) = 813.00 Kg● Berat Beban (P5) = 813.00 Kg● Berat Beban (P6) = 813.00 Kg● Berat Beban (P7) = 813.00 Kg● Berat Beban (P8) = 813.00 Kg● Berat Beban (P9) = 1071.00 Kg
Jumlah = 7833.000 Kg= 7.833 Ton
Perhitungan Reaksi(RA = RB) =ΣPD
=7833.000
2 2
= 3916.500 Kg
= 3.917 Ton
fmax²
fijin
Maka digunakan Gording dengan Profil [10 Dengan berat
4. Skema Pembebanan Pada Struktur
P1 P2 P3 P4 P5 P6 P7 P8 P9
6.000
RA = 3916.500 Kg RB = 3916.500 Kg
1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50
1.25 11.000 1.25
5. Perhitungan Dimensi BalokP1 P2 P3 P4 P5 P6 P7 P8 P9
11.00
qu = P1 + P2 + P3 + P4 + P5 + P6 + P7 +
= P8 + P9= 1071.000 + 813.000 + 813.000 + 813.000 + 813.000 + 813.000 + 813.000 +
813.000 + 1071.000
= 7833.00 Kg= 7.833 Ton
Di coba dengan menggunakan Profil WF 300 x 300Dengan Data Sebagai Berikut
● tf/t2 = 15.000 mm● tw/t1 = 10.000 mm● b = 300.000 mm
● d = 300.000 mm
● fy = 2400.000 Mpa
● = 18.000 mm
● fr = 70.000 Mpa
● Ix = 20400.000 Cm⁴
● Iy = 6750.000 Cm⁴● Berat = 94.00 Kg/m
r0
0
0
00
00
00
00
00
00
00
00
00
0 00 00
00
tftw d
Mu =1
x qu x8
=1
x 7.833 + 1.20 x 0.094 x 11.0002
8= 120.180 Ton.m
Mn =MuǾ
=120.180
0.900= 133.534 Ton.m
Zx =
Zy =
h =
Pemeriksaan Terhadap Tekuk lokal
Flens λp =170
=170
= 3.470fy 2400.000
λr =370
=370
= 7.553fy 2400.000
λf =b
=300.000
= 10.0002 . tf 2 x 15.000
Web λp =1680
=1680
= 34.293fy 2400.000
λr =2550
=2550
= 52.052fy 2400.000
=d
=300.000
= 30.000tw 10.000
Kontrolλf= 10.000 < λf= 3.470 Penampang Tidak kompak
h = d - 2 x + tf
= 300.00 - 2 x 18.000 + 15.000= 234.000 mm
λw =h
=234.000
= 23.400tw 10.000
Kontrolλw= 23.400 < λp= 34.293 Penampang Kompak
Modulus PlastisZx = b . tf . ( d - tf ) + 1/4 . tw . ( d - 2tf )²
= 300 x 15 x 300 - 15 + 0.25 x 10 x 300 - 2 . 15 2
= 1464750.000 mm²
L2
b . tf ( d - tf ) + 1/4 . tw . ( d - 2 tf )2
1/2 . b2 . tf + 1/4 . tw2 . ( d - 2 tf )2
d - 2 ( r0 + tf )
λf
r0
Mp = Zx x fy= 1464750.000 x 2400.000= 3515400000.000= 351.540 Ton.m
Mr = ( fy - fr ) x Sx = ( fy - fr ) . Ix /(d/2)
= 2400.00 - 70.00 x20400.000 x 10⁴
300.000 : 2.00= 3168800000.000= 316.880 Ton.m
Mn =λr - λ
xλ - λp
λr - λp λr - λp
=7.553 - 10.000
x 351.540 +10.000 - 3.470
x 316.8807.553 - 3.470 7.553 - 3.470
= 296.102 Ton.mKontrol
Mp= ( 351.540) Ton.m > ( 133.5) Ton.m OK
6. Perhitungan Dimensi Kolomd b tw tf
Dicoba profil WF 300 300 10.00 15.00
fy = 240 Mpa ro = 18 mm
fu = 370 Mpa h = d - 2 ( tf + ro )
Nu = 3.9165 ton = 234 mm
d = 300 mm r/ix = 131.0 mm
b = 300 mm ry/iy = 75.1 mm
= 10 mm A = 11980 mm²
= 15 mm L = 6000 mm
Berat = 94.00 Kg/m
Ix = 20400.000 Cm⁴
Iy = 6750.000 Cm⁴ Nu
Kondisi tumpuan Sendi - Sendi, k = 1.0
x Mp x Mr
Mu/Ǿ=
tw/t1
tf/t2
b
tf
dhtw
Sumber : SNI 03-1729-2002 /Perencanaan Struktur Baja dengan Metode LRFD (Halaman 57)
Periksa kelangsingan penampangFlens b / 2
=300
= 102 x 15.0
250=
250= 16.137
fy 240.00
b / 2= 10.000 < λr = 16.137 OK
Web h=
234= 23.400
10
665=
665= 42.926
fy 240.00h
= 23.400 < λr = 42.926 OK
Arah sumbu kuat ( sumbu x )
=k.Lx
=1.0 x 6000
= 45.802rx 131
=fy
=45.8 240
= 0.5050341π E 3.14 200000
0.25 < > 1.2 =1.43
1.6 - 0.67 λcx
=1.43
= 1.13352 - ( 0.67 x 0.505 )
= = = 11980 .240
= 253.67 ton1.1335
Nu=
4= 0.02 ton < 1 OK
Nn 0.85 x 253.666
Arah sumbu lemah ( sumbu y )
=k.Ly
=1 x 6000
= 79.893ry 75.1
=fy
=79.89 240
= 0.881π E 3.14 200000
0.25 < < 1.2 =1.43
1.6 - 0.67 λcy
=1.43
= 1.41622 - ( 0.67 x 0.88 )
= = = 11980 .240
= 203.03 ton1.4162
Nu=
4= 0.02 ton < 1 OK
Nn 0.85 x 203.03
tf
tf
tw
tw
λx
λcxλx
λcx ωx
ωx
Nn Ag x fcr Ag .fy
ωx
Øc.
λy
λcxλy
λcy ωy
ωy
Nn Ag . fcr Ag .fy
ωy
Øc.
7. Analisa Struktur Dengan Metode CrossP1 P2 P3 P4 P5 P6 P7 P8 P9
C D
Baja WF 300 x 300
Baja WF 300 x 300 Baja WF 300 x 3005.000
A B
RA = 3916.500 Kg RB = 3916.500 Kg
1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50
1.25 11.000 1.25
Perhitungan Momen Primer Batang C - DP1 P2 P3 P4 P5 P6 P7 P8 P9
94.000 Kg/m
C DBaja WF 300 x 300
1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50
11.00
Kombinasi Pembebanan● Berat Beban (P1) = 1.071 ton● Berat Beban (P2) = 0.813 ton● Berat Beban (P3) = 0.813 ton● Berat Beban (P4) = 0.813 ton● Berat Beban (P5) = 0.813 ton● Berat Beban (P6) = 0.813 ton● Berat Beban (P7) = 0.813 ton● Berat Beban (P8) = 0.813 ton● Berat Beban (P9) = 1.071 ton
0
0
00
00
00
00
00
00
00
00
00
0 00 00
00
00
00
00
00
00
00
00
00
00
Angka Kekakuan Batang● Ix = 20400.000 Cm⁴ = 0.000204 m⁴
● Iy = 6750.000 Cm⁴ = 0.000068 m⁴
MºCD = -P1 a b 2
-P2 a b 2
-P3 a b 2
L 2 L 2 L 2
-P4 a b 2
-P5 a b 2
-P6 a b 2
L 2 L 2 L 2
-P7 a b 2
-P8 a b 2
-P9 a b 2
L 2 L 2 L 2
-1
x q x L2
12
= -1.07 0.00 11.00 2
-0.81 1.50 9.50 2
-0.81 3.00 8.00 2
11.00 ² 11.00 ² 11.00 ²
-0.81 4.50 6.50 2
-0.81 6.00 5.00 2
-0.81 7.50 3.50 2
11.00 ² 11.00 ² 11.00 ²
-0.81 9.00 2.00 2
-0.81 10.50 0.50 2
-1.07 12.00 -1.00 2
11.00 ² 11.00 ² 11.00 ²
-1
x 0.094 x 11.00 ²12
= - 0 - 0.9095857 - 1.29004959 - 1.277451 - 1.0078512 - 0.6173089- 0.2418843 - 0.0176374 - 0.10621488 - 0.947833
= - 5.3617686 Ton.m
MºDC = +P1 a 2 b
+P2 a 2 b
+P3 a 2 b
L 2 L 2 L 2
+P4 a 2 b
+P5 a 2 b
+P6 a 2 b
L 2 L 2 L 2
+P7 a b
+P8 a 2 b
+P9 a 2 b
L 2 L 2 L 2
+1
x q x L2
12
= +1.07 0.00 2 11.00
+0.81 1.50 2 9.50
+0.81 3.00 2 8.00
11.00 ² 11.00 ² 11.00 ²
+0.81 4.50 2 6.50
+0.81 6.00 2 5.00
+0.81 7.50 2 3.50
11.00 ² 11.00 ² 11.00 ²
+0.81 9.00 2 2.00
+0.81 10.50 2 0.50
+1.07 12.00 2 -1.00
11.00 ² 11.00 ² 11.00 ²
+1
x 0.094 x 11.00 ²12
= + 0 + 0.1436188 + 0.4837686 + 0.884389 + 1.2094215 + 1.3228048+ 1.08847934 + 0.3703853 + -1.2745785 + 0.947833
= + 5.50286777 Ton.m
Perhitungan Momen Primer Batang A - C
94.000 Kg/m
A CBaja WF 300 x 300
5.00
Angka Kekakuan Batang● Ix = 20400.000 Cm⁴ = 0.000204 m⁴
● Iy = 6750.000 Cm⁴ = 0.000068 m⁴
MºAC = 0.000 Ton.m
MºCA = +1
x 0.0940 x 5.00 ²8
= + 0.2938 Ton.m
Perhitungan Momen Primer Batang D - B
94.000 Kg/m
D BBaja WF 300 x 300
5.00
Angka Kekakuan Batang● Ix = 20400.000 Cm⁴ = 0.000204 m⁴
● Iy = 6750.000 Cm⁴ = 0.000068 m⁴
MºDB = -1
x 0.0940 x 5.00 ²8
= - 0.2938 Ton.m
MºBD = 0.000 Ton.m
Perhitungan Angka Distribusi● Angka Kekakuan Perbatang di Join C
KCA : KCD =3.00 x EI
:4.00 x EI
L L
=3.00 x 0.000204
:4.00 x 0.000204
5.000 11.000= 0.0001224 : 7.4181818E-05
● Angka Kekakuan Join CKCA + KCD = 0.0001224 + 7.4181818E-05
= 0.0001965818
● Angka Kekakuan Batang Join C
KCA =0.0001224
= 0.622641509430.0001965818
= 1KCD =
7.418182E-05= 0.37735849057
0.000196581800
00
00
00
Perhitungan Angka Distribusi● Angka Kekakuan Perbatang di Join D
KDC : KDB =4.00 x EI
:3.00 x EI
L L
=4.00 x 0.000204
:3.00 x 0.000204
11.000 5.000= 7.418182E-05 : 0.0001224
● Angka Kekakuan Join DKDC + KDB = 7.418182E-05 + 0.0001224
= 0.0001965818
● Angka Kekakuan Batang Join D
KDC =7.418182E-05
= 0.377358490570.0001965818
= 1KDB =
0.0001224= 0.62264150943
0.0001965818
Tabel Momen Pada Metode CrossTITIK A C D B
BATANG AC CA CD DC DB BC
AD 0 0.62264151 0.37735849 0.37735849 0.62264151 0
M' 0 0.29375000 -5.36176860 5.50286777 -0.29375000 0
3.15555875 1.91245985 0.95622992
-1.16327315 -2.32654630 -3.83880139
0.72430215 0.43897100 0.21948550
-0.04141236 -0.08282472 -0.13666078
0.02578505 0.01562731 0.00781365
-0.00147427 -0.00294855 -0.00486510
0.00091794 0.00055633 0.00027816
-0.00005248 -0.00010497 -0.00017320
0.00003268 0.00001981 0.00000990
-0.00000187 -0.00000374 -0.00000617
0.00000116 0.00000071 0.00000035
-0.00000007 -0.00000013 -0.00000022
0.00000004 0.00000003 0.00000001
0.00000000 0.00000000 -0.00000001
0.00000000 0.00000000 0.00000000
0.00000000 0.00000000 0.00000000
0.00000000 0.00000000 0.00000000
0.00000000 0.00000000 0.00000000
0.00000000 0.00000000 0.00000000
0.00000000 4.20034778 -4.20034778 4.27425687 -4.27425687 0.00000000
8. Skema Pembebanan Pada StrukturMCD = -4.2003 ton.m MDC = 4.2743 ton.m
C D
MCA = 4.2003 ton.m MDB = -4.2743 ton.m5.000
A MAC = 0.0000 ton.m MDB = 0.0000 ton.m B
RA = 3916.500 Kg RB = 3916.500 Kg
11.000
9. Perhitungan Reaksi Perbatang● Reaksi Pada Batang A - C
94.000 Kg/m
A CBaja WF 300 x 300
5.00
RA RC
RA = RC =1
x q x L2
=1
x 94.000 x 5.0002
= 235.000 Kg= 0.235 Ton
● Reaksi Pada Batang D - B
94.000 Kg/m
D BBaja WF 300 x 300
5.00
RD RB
00
00
00
00
00
00
00
00
RD = RB =1
x q x L2
=1
x 94.000 x 5.0002
= 235.000 Kg= 0.235 Ton
Perhitungan Momen Primer Batang C - DP1 P2 P3 P4 P5 P6 P7 P8 P9
94.000 Kg/m
C DBaja WF 300 x 300
1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50
11.00
RC RD
RA = RB =1
x q x L + P1 + P2 + P3 +2
P4 + P5 + P6 + P7 + P8 +
P9
=1
x 94.00 X 11.000 + 1071.000 + 813.000 + 813.000 +2
813.000 + 813.000 + 813.000 + 813.000 + 813.000 +
1071.000
= 4433.500 Kg= 4.434 Ton
10. Free BodyP1 P2 P3 P4 P5 P6 P7 P8 P9
94.000 Kg/m
C D
1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50
11.00
RC RD4.43350 4.43350
0.38185 -0.38185
00
00
00
00
00
00
00
00
00
00
00
00
00
00
00
00
00
00
00
00
00
00
-0.38857 0.38857
RC RD
4.42678 4.44022
0.235 0.840 1.075 1.090 0.855 0.235RC RC RD RD
0.47 0.470
0.235 -0.840 -0.605 -0.620 -0.855 0.235RA RA RB RB
RA RB4.42678 4.44022
Reaksi Pada Join C dan D
RHC = 1.075 Ton RHD = 1.090 Ton
RVC = 4.427 Ton RVD = 4.440 Ton
11. Perhitungan Sambungan
Data Balok
H = 1.075 Ton ● tf/t2 = 15.000 mm● tw/t1 = 10.000 mm● b = 300.000 mm● d = 300.000 mm
M = 4.200 ton.m ● fy = 2400.000 Mpa
● = 18.000 mm
● fr = 70.000 Mpa● Ix = 20400.000 Cm⁴● Iy = 6750.000 Cm⁴
V = 4.427 Ton ● Berat = 94.00 Kg/m
Data Kolom● fy = 240 Mpa ● ro = 18 mm
● fu = 370 Mpa ● h = d - 2 ( tf + ro )
● Nu = 3.9165 ton = 234 mm
● d/h = 300 mm ● r/ix = 131.0 mm
● b = 300 mm ● ry/iy = 75.1 mm
r0
C D
● = 10 mm ● A = 11980 mm²
● = 15 mm ● L = 6000 mm
● Berat = 94.00 Kg/m
● Ix = 20400.000 Cm⁴
● Iy = 6750.000 Cm⁴
=Sin
=300
= 1300 ts Cos 300
θ = Arc tan 1θ = 45 º
300 ts = 300² + 300²= 424.264069 mm= 0.70710678= 0.70710678
Gaya yang Bekerja pada Flens Ratter
F =M
=4.2003
= 14001.159 Kgd balok 300.00
Komponen Horizontal yang di dukung oleh badan profil (Web) Sebagai Geser
=Fy
=2400
= 16001.5 1.5
Fw = x tw x dc
= 0.6 x 1600 x 1 x 30= 28800 Kg
Komponen Horizontal sebagai gaya tekan yang di dukung diagonal pengaku/Penegar= F - Fw= 14001.159 - 28800.00= -14798.841 Kg
Gaya penampang pada Penegar
Fs = tsθ xtsdc
= -14798.841 x424.26407300 mm
= -20928.72 Kg
Luas Penampang Penegar yang di perlukan
As =Fs
=-20928.72
= -13.080 cm²1600
Di coba Penegar = 2 Cm x 7 CmAs' = 2 x 2.00 x 7.00
= 28As' > As28 > -13.1 OK
tw/t1
tf/t2
tan θ
Sin θCos θ
Kg/cm²
tsθ
Syarat bs/ts ≤ 17
cm²
Kontrol
bs=
2.00 x 7.00
=7 < 17 OK
ts 2.0
Kontrol Tegangan
=F
=14001.159
1 x 30 + 2.5 x 28 x 0.7071 x 0.707 ²= 255.7348
<255.7348 < 960.00 OK
Tegangan Tekan Pada Diagonal
s =F
tw. Dc +
=14001.159
1 x 30+ 28 x 0.7071
2.5 x 28 x 0.7071 x 0.7071
= 677.82093821
s <677.8209 < 1600.00 OK
Perhitungan BautDiketahui :● Digunakan Baut Ø1/2" → fy = 2400.000 Kg/cm²
● =fy
=2400.000
= 1600.0001.500 1.500
● tp = 1.600 x = 1.600 x 1600.000 = 2560.000● = 0.600 x = 0.600 x 1600.000 = 960.000● Ø baut = 1/2 '' = 1.270 cm● Ø lubang untuk baut = 1.270 + 0.100 = 1.370 cm
N Geser = 2 x 1/4. x π d² x 0.000= 2 x 0.250 x 3.140 1.370 ² x 960.000= 2828.864 Kg
N Tumpu = d x 0.800 x tp= 1.370 x 0.800 x 2560.000= 2805.760 Kg
Dari hasil perhitungan N Geser dan N tumpu diambil yang terkecil yitu :N Tumpu = 2805.760 Kg
● Jumlah Baut
n1 =S1
=4426.781
= 1.578 ≈ 3.0 Buah BautN Tumpu 2805.760
Luas Ø Lubang Baut = 2 x 1/4. x π x n= 2 x 0.250 x 3.140 1.370 ² x 3= 8.840 Kg
● Kontrol :
τ =S1
=4426.781
= 500.756 Kg/cm2
tw. Dc + 2,5. As'. Sin θ. Cos²θ
Kg/cm²
As'. Cosθ2,5. As'. Sin θ. Cosθ
Kg/cm²
Kg/cm²
Kg/cm²Kg/cm²
d²
τ =A Ø Lbg Baut
=8.840
= 500.756 Kg/cm
= 500.756 Kg/cm2
< = 960.000 Kg/cm2
OK
Sumber : SNI 03 – 1729 – 2002 (Halaman 104 dari 184)
12. Perhitungan Plat KakiDirencanakan :● Panjang Plat Kaki (b) = 54.000 Cm
1.075 Ton● Lebar Plat kaki (h) = 30.000 Cm● Tebal Plat Kaki (t) = 2.000 Cm● Gaya Horizontal (H) = 1075.070 Kg● Beban Aksial ( P ) = 4426.781 Kg
● Perhitungan Momen 5.000M = Gaya x Jarak
= 1075.070 x 5.000 M= 5375.35 Kg.m
b = 30
h = 544.4268 ton
Kontrol Teganagan
W =1
x b x6
=1
x 30.000 x 54.00 ²6
= 14580.00 Cm³
h²
0
00 0
0
0
F = b x h= 30.00 x 54.00= 1620.00 cm²
σ =P
±M
F W
=4426.781
±1075.070
1620.000 14580.000= 2.733 ± 0.074
σmax = 2.733 + 0.074= 2.806 Kg/cm²
σmin = 2.733 - 0.074= 2.659 Kg/cm²
x=
σmin50 - x σmax
σmax x = σmin 50 - x2.806 x = 2.659 50 - x2.806 x = 132.942 - 2.659 x5.465 x = 132.942
x =132.942
5.465x = 24.325 Cm
σ1 = σmax = 2.806 Kg/cm²σ2 = σmin = 2.659 Kg/cm²
+
-
a bd
T
54 cm
=σ1
x L+ σ2
=2.806
x 542.806 + 2.659
= 27.729 cm
a2 =σ2
x L+ σ2
=2.659
x 542.806 + 2.659
= 26.271 cm
Gaya Tarik (T) = 0.500 x a1 x σmin x b= 0.500 x 27.729 x 2.659 x 30.000= 1105.889 Kg
a1 : a2 = σ1 : σ2 = a1σ1
σ1
0
0 0 0
0
Gaya Tekan (d) = 0.500 x a2 x σmax x b= 0.500 x 26.271 x 2.806 x 30.000= 1105.889 Kg
Menghitung Tebal PlatM = T x W
5375.348 = 1105.889 x 1/6. x b x5375.348 = 1105.889 x 1/6. x 30 x s²5375.348 = 5529.447
=5375.3485529.447
s = 0.972s = 0.986 cm » 2 cm
Gaya Angkur (Jangkar)Diketahui :● Digunakan angkur Ø 7/8 " → fy = 2400.000 Kg/cm²
Perhitungan Gaya Tarik yang terjadi pada Plat kaki τ = 0.500 x b x σmin x x
= 0.500 x 54 x 2.659 x 30.00= 2153.664 Kg
Tegangan Ijin Angkur
=fy
=2400.000
= 1600.0001.500 1.500
Penampang angkur
F Perlu =τ
=2153.664
= 1.346 cm²σ 1600.000
Dipakai Angkur Ø 7/8 " = 2.720 cm
F =1
x 3.140 x 2.720 ²4
= 5.808 cm²
Kontrol Paku :F > F Perlu
5.808 cm² > 1.346 cm² OK
Jumlah Angkur
n = F Perlu
=1.346
= 0.232 BuahF 5.808
≈ 2 BuahJadi Dipakai jangkar 2 Ø 7/8"
Luas 2 Buah Angkur = 2.000 x 2.720= 5.440 cm²
Kontrol :F 2 Buah Angkur > F Perlu
5.440 cm² > 1.346 cm² OK
Dengan Demikian menggunakan 2 Buah Angkur Memenuhi
s²
s²
s²
Kg/cm²
13. Perencanaan Pondasi
30 cm
54 cm
60
30 30 cm 150 cm
54 cm60
150 cm
48 54 cm 48
Pondasi Derencanakan dengan Data Sebagai Berikut :● fy = 320 Mpa● fc = 25 Mpa● Panjang (h) = 100 Cm● Panjang (a) = 20 Cm● Panjang (b) = 20 Cm● Panjang (c) = 10 Cm● Panjang (b1) = 150 Cm● Panjang (b2) = 170 Cm
V = Rv + (berat WF x tinggi kolom)= 4426.781 + 94.000 x 5.000= 4896.781 kg
0
00 0
0
000
0
000
0
0
0
0
0 0 0
Berat kolom beton = 1.000 x 0.540 x 0.300 x 2400= 388.800 kg
Berat plat pondasi =16.2 + 225.0
x 0.2 + 0.20 x 1.5 x 1.5 x 24002
= 58968.000 Kg
Vt = 4896.781 + 388.800 + 58968.000= 64253.581 kg
σ =Vt
=64253.581
= 28557.147 Kg/cm²A 1.5 x 1.5
Momen arah x
Mx = σ x B x 0.480 x 0.5 x 0.480= 28557.147 x 1.5 x 0.480 x 0.5 x 0.480= 4934.675 kg.m
Mu = Faktor beban mati x Mx= 1.2 x 4934.675= 5921.610 kg.m
Mn =Mu
=5921.610
= 7402.013 kg.m 7402.013 x 104
Ø 0.8
h = 540 mmtebal selimut beton (s 40 mm
dx = h - s - 0.5 x Ø= 540 - 40 - 0.5 x 10= 495.000 mm
As perlu = 0.85 x fc' x b x dxx 1- 1-
2 xfy 0.85 x fc' x b x
=0.85 x 25 x 1500 x 495
x 1 - 1 -2 x 74020125.302
320 0.850 x 25 x 1500 x 495 ²= 49306.641 x 1 - 0.990= 469.534 mm²
As min = 0.002 x b x h= 0.002 x 1500 x 540= 1620.00 mm²
As t =0.85 x fc' x β1
x600
x b x dxfy 600 + fy
=0.85 x 25 x 0.8
x600
x 1500 x 495320 600 + 320
= 25725.20 mm²
Mn x 10⁴d²
As max = 0.75 x As t= 0.75 x 25725.204= 19293.90 mm²
As perlu = 469.53 mm² > As min = 1620.00 mm²< As max = 19293.90 mm²
Dipakai Ø = 16 mm
Jarak tulangan (as) =0.25 x π x x b
Asperlu
=0.25 x 3.14 x 16 ² x 1500
469.534= 642.00 mm ~ 200 mm
As plat =0.25 x π x x b
As= 0.25 x 3.14 x 16 ² x 1500
200= 1507.20 mm² > As perlu = 469.53 mm² OK
KesimpulanAs + = Ø16-200As - = Ø10-200
Momen Arah yMy = σ x B x 0.600 x 0.5 x 0.600
= 28557.147 x 1.5 x 0.600 x 0.5 x 0.600= 7710.430 kg.m
y
x
Mu = faktor baban mati x My= 1.2 x 7710.430= 9252.516 kg.m
Mn =Mu dy = h - s - Ø x - 0.5 ØØ = 300 - 20 - 10 - 0.5 10
= 9252.516 = 265.000 mm0.8
= 11565.64 kg.m
As perlu = 0.85 x fc' x b x dxx 1- 1-
2 xfy 0.85 x fc' x b x
=0.85 x 25 x 1500 x 265
x 1 - 1 -2 x 115656445.785
320 0.850 x 25 x 1500 x 265 ²= 26396.484 x 1 - 0.947= 1401.055 mm²
As min = 0.002 x b x h= 0.002 x 1500 x 300
D²
D²
Mn x 10⁴d²
= 900.00 mm²
As t=
0.85 x fc x β1x
600x b x dy
fy 600 + fy
=0.85 x 25 x 0.8
x600
x 1500 x 265320 600 + 320
= 13772.08 mm²
As max = 0.75 x As t= 0.75 x 13772.079= 10329.06 mm²
As perlu = 1401.06 mm² > As min = 900.00 mm²< As max = 10329.06 mm²
Dipakai Ø = 16 mm
Jarak tulangan (as) =0.25 x π x x b
Asperlu
=0.25 x 3.14 x 16 ² x 1500
1401.055= 215.15 mm ~ 200 mm
As plat =0.25 x π x x b
As= 0.25 x 3.14 x 16 ² x 1500
200= 1507.20 mm² > As perlu = 1401.06 mm² OK
KesimpulanAs + = Ø16-200As - = Ø10-200
D²
D²