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1. i
TUGAS BESAR STRUKTUR BAJA I
Tugas besar ini dibuat untuk memenuhi salah satu tugas dalam mata kuliah Struktur
Baja I yang diampu oleh Ibu Ririt Aprilin Sumarsono ST.M.Sc.Eng.
Dibuat Oleh :
Nousseva Renna
5415164015
PRODI PENDIDIKAN TEKNIK BANGUNAN
FAKULTAS TEKNIK
UNIVERSITAS NEGERI JAKARTA
2018

2. ii
KATA PENGANTAR
Puji syukur penulis panjatkan kehadirat Tuhan Yang Maha Esa atas rahmat dan karunia-
Nya sehingga penulis dapat menyelesaikan tugas besar Struktur Baja ini. Makalah ini dibuat
sebagai salah satu tugas besar dalam mata kuliah Struktur Baja.
Makalah ini tidak mungkin selesai tanpa adanya bantuan dari berbagai pihak. Oleh karena
itu, selayaknya penulis mengucapkan terima kasih kepada:
1. Ibu Ririt Aprilin Sumarsono ST.M.Sc.Eng selaku dosen pembimbing mata kuliah
Struktur Baja;
2. Orang tua saya tercinta atas doanya yang tidak pernah putus;
3. Rekan-rekan prodi Pendidikan Teknik Bangunan.
Penulis menyadari bahwa makalah ini masih memiliki banyak kekurangan, untuk itu
penulis mengharapkan saran dan kritik dari pembaca agar makalah ini bisa menjadi lebih baik
lagi.
Jakarta, 6 Januari 2018
Nousseva Renna

3. iii
DAFTAR ISI
HALAMAN JUDUL................................................................................................................ i
KATA PENGANTAR............................................................................................................... ii
DAFTAR ISI........................................................................................................................... iii
Kontrol Kestabilan............................................................................................................... 1
Reaksi Perletakan................................................................................................................ 1
Mencari Sudut – Sudut Segitiga.......................................................................................... 2
Mencari Gaya Perbatang (Metode Titik Buhul) ................................................................ 2
Mendesain Batang Tarik ..................................................................................................... 4
Mendesain Batang Tekan ................................................................................................... 7
Mendesain Batang 0 ........................................................................................................... 9
Mendesain Baut.................................................................................................................. 10
Detail Sambungan............................................................................................................... 17

4. 1
2s – m – r = 0
2(8) – 13 – 3 =0
 𝜮𝑴 𝑨 = 𝟎
𝑃1(3,2) + 𝑃2(6) − 𝑃3(4,5) + 𝑃4(2,4) − 𝑉𝐸(12) = 0
𝑉𝐸 =
140(3,2) + 450(6) − 25(4,5) + 250(2,4)
12
= 302,96 𝑘𝑁
 𝜮𝑴 𝑬 = 𝟎
𝑃1(3,2) − 𝑃2(6) − 𝑃3(4,5) − 𝑃4(9,6) + 𝑉𝐴(12) = 0
𝑉𝐴 =
−140(3,2) + 450(6) + 25(4,5) + 250(9,6)
12
= 397,04 𝑘𝑁
 𝜮𝑯 = 𝟎
𝐻𝐴 + 𝑃1 − 𝑃3 = 0
𝐻𝐴 = 𝑃3 − 𝑃1
𝐻𝐴 = 25 − 140 = −115
 𝜮𝑽 = 𝟎
𝑉𝐸 + 𝑉𝐴 − 𝑃2 − 𝑃4 = 0
302,96 + 397,04 − 450 − 250 = 0
Reaksi Perletakan
1,3 m
3,2 m
H
G
F
E
3,6 m 3,6 m 2,4 m2,4 m
C DB
A
𝑃3 = 25 𝑘𝑁
𝑃2 = 450 𝑘𝑁
𝑃1 = 140 𝑘𝑁
𝑃4 = 250 𝑘𝑁
Konstrol Kestabilan

5. 2
F X = √2,42 + 3,22 = 4𝑚
X = √3,62 + 4,52 = 5,76 𝑚
G
Titik Buhul 𝜮𝑽 = 𝟎 𝜮𝑯 = 𝟎
Titik A SAF Sin(53,13) + 𝑉𝐴= 0
𝑆𝐴𝐹 =
−397,04
𝑆𝑖𝑛(53,13)
= −496,3 𝑘𝑁 (Tekan)
-𝐻𝐴 - SAF Cos(53,13) + SAB
SAB = 115 + 496,3 Cos(53,13)
= 412,8 kN (Tarik)
Titik F SAF Cos(36,87)– 465,46 Sin(19,86) –SFB
= 0
SFB = 496,3(Cos 36,87)-465,5 Sin(19,86)
= 239 kN (Tarik)
140+SAF Sin(36,87) + SFG (Cos
19,86)=0
𝑆𝐹𝐺 =
−140 − 496,3(𝑆𝑖𝑛 36,87)
𝐶𝑜𝑠(19,86)
= −465,5 𝑘𝑁 (Tekan)
𝑆𝐹𝐺
𝑆𝐴𝐹
𝛽
ℎ1140
𝑆𝐹𝐵
𝛼
𝑆𝐴𝐹
115
𝑉𝐴
𝑖
ℎ
𝑔
3,2m1,3m
3,6 m
4,5m
3,6 m
𝑒
𝑑 𝑓
𝑐
𝛽
𝛼
B
A
2,4 m
3,2 m
𝐶𝑜𝑠 𝛼 =
𝐴𝐵2
+ 𝐴𝐹2
− 𝐵𝐹2
2(𝐴𝐵)(𝐴𝐹)
=
2,42
+ 42
− 3,22
2(2,4)(4)
= 0,6
𝛼 = 53,13°
𝐶𝑜𝑠 𝛽 =
𝐵𝐹2
+ 𝐴𝐹2
− 𝐵𝐹2
2(𝐵𝐹)(𝐴𝐹)
=
3,22
+ 42
− 2,42
2(3,2)(4)
= 0,8 𝛽 = 36,87°
𝐶𝑜𝑠 𝑐 =
𝐴𝐵2
+ 𝐵𝐹2
− 𝐴𝐹2
2(𝐴𝐵)(𝐵𝐹)
=
2,42
+ 3,22
− 42
2(2,4)(3,2)
= 0 𝑐 = 90°
G
CB
𝐶𝑜𝑠 𝑑 =
𝐵𝐶2
+ 𝐵𝐺2
− 𝐶𝐺2
2(𝐵𝐶)(𝐵𝐺)
=
3,62
+ 5,762
− 4,52
2(3,6)(5,76)
= 0,624
𝐶𝑜𝑠 𝑒 =
𝐵𝐺2
+ 𝐶𝐺2
− 𝐵𝐶2
2(𝐵𝐺)(𝐶𝐺)
=
5,762
+ 4,52
− 3,62
2(5,76)(4,5)
= 0,781
𝐶𝑜𝑠 𝑓 =
𝐵𝐶2
+ 𝐶𝐺2
− 𝐵𝐺2
2(𝐵𝐶)(𝐶𝐺)
=
3,62
+ 4,52
− 5,762
2(3,6)(4,5)
= 0
𝑑 = 51,34°
𝑒 = 38,66°
𝑓 = 90°
F
B
X = √3,62 + 1,32 = 3,83 𝑚
𝑔 = 180° − 90° − 51,34° = 38,66°
𝐶𝑜𝑠 ℎ =
𝐵𝐹2
+ 𝐹𝐺2
− 𝐵𝐺2
2(𝐵𝐹)(𝐹𝐺)
=
3,22
+ 3,832
− 5,762
2(3,2)(3,83)
= 0,337
𝐶𝑜𝑠 𝑖 =
𝐵𝐺2
+ 𝐹𝐺2
− 𝐵𝐹2
2(𝐵𝐺)(𝐹𝐺)
=
5,762
+ 3,832
− 3,22
2(5,76)(3,83)
= 0,852
𝑔 = 38,66°
ℎ = 109,86°
𝑖 = 31,48°
Mencari sudut –sudut Segitiga
Mencari Gaya Perbatang (Metode Titik Buhul)

6. 3
Titik B SFB – 250 + SBG Sin(51,34) = 0
𝑆𝐵𝐺 =
250 − 239
𝑆𝑖𝑛 (51,34)
= 14,1 𝑘𝑁 (Tarik)
-SAB + SBG Cos(51,34) + SBC = 0
SBC = 412,8 – 14,1 Cos(51,34)
= 403,9 kN (Tarik)
Titik C SGC = 0 -SBC + SCD = 0
SCD = +SBC
= 403,9 kN (Tarik)
Titik E SHE Sin (53,13) + 302,96 = 0
𝑆𝐻𝐸 =
−302,96
𝑆𝑖𝑛 (53,13)
= −378,7 𝑘𝑁 (Tekan)
-SDE + SHE Cos (53,13) = 0
SDE = 378,7 Cos (53,13)
= 227,2 kN (Tarik)
Titik H -SHG Sin (19,86)-SHD+SHE Cos(36,87) =
0
SHD = -241,58 Sin(19,86) + 378,69
Cos(36,87) = 0
= 220,9 kN (Tarik)
-SHG Cos 19,86 – SHE Sin 36,87 = 0
𝑆𝐻𝐺 =
−378,69 𝑆𝑖𝑛 36,87
𝐶𝑜𝑠 (19,86)
= −241,6 𝑘𝑁 (Tekan)
Titik D SHD + SDG Sin (51,34) = 0
𝑆𝐷𝐺 =
−220,9
𝑆𝑖𝑛 (51,34)
= −282,9 𝑘𝑁 (Tekan)
-SCD + SDG Cos(51,34) + SDE = 0
-403,9 + 176,7 + 227,2 = 0
0 = 0
Titik G -450 + SFG Cos(70,14) – SBG
Cos(38,66) + SGD Cos (38,66) + SGH
Cos(70,14) = 0
-450 + 465,5 Cos(70,14) – 14,1
Cos(38,66) + 282,9 Cos(38,66) +
241,6 Cos(70,14) = 0
-450 + 158 – 11 + 221 + 82 = 0
-25 + SFG Cos(19,86) – SBG
Cos(51,34) – SGD Cos(51,34) –
SGH Cos(19,86) = 0
-25 + 465,5 Cos(19,86) – 14,1
Cos(51,34) – 282,9 Cos(51,34) –
241,6 Cos(19,86) = 0
-25 + 437,8 – 8,8 – 176,8 – 227,2
= 0
No. Batang Tarik (kN) Tekan (kN) Panjang Bentang (mm)
1. SAF 496,3 4000
2. SAB 412,8 2400
3. SFG 465,5 3830
4. SFB 239 3200
5. SBG 14,1 5760
6. SBC 403,9 3600
7. SGC 0 0 4500
8. SCD 403,9 3600
9. SHE 378.7 4000
10. SDE 227,2 2400
11. SHG 241,6 3830
12. SDH 220,9 3200
13. SDG 282,9 5760
𝑆𝐷𝐸𝑆𝐶𝐷
𝑆𝐷𝐺
𝑑
𝑆𝐻𝐷
𝑆𝐻𝐺
𝑆𝐻𝐸
25
𝛽
ℎ1
𝑆𝐻𝐸
𝑆𝐶𝐷
𝛼
𝑆𝐵𝐺
250
𝑆𝐵𝐶𝑆𝐴𝐵
SFB
SGC
𝑉𝐸
450
𝑆𝐵𝐺
𝑆𝐹𝐺
𝑆𝐺𝐷
𝑆𝐺𝐻
𝑆𝐺𝐶
𝑆𝐷𝐸
𝑆𝐵𝐶
𝑆𝐻𝐷

7. 4
 SAB
P = 412,8 kN
L = 2,4 m = 2400mm
Mutu baja A 500 Grade A :
Fy = 269 MPa
Fu = 310 MPa
Py = 0,9.Fy.Ag
𝐴𝑔 =
𝑃𝑦
0,9. 𝐹𝑦
=
412,8 𝑥 103
0,9.269
= 1705,08 𝑚𝑚2
= 17,0508 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm²
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
Cek Kondisi Leleh
Py = 0,9.Fy.Ag
= 0,9 x 269 x 5121
= 1239794,1 N
= 1239,7941 kN
Py = 1239,7941 kN > P = 412,8 kN (OK!)
Cek Kondisi Fraktur
Pu = 0,75.Fu.Ae
= 0,75 x 310 x (0,85x5121)
= 1012037,6N
= 1012,0376 kN
Pu = 1012,0376 kN > P = 412,8 kN (OK!)
Cek Kelangsingan Batang
𝜆 =
𝐿
𝑖 𝑦
=
2400
43,8
= 54,8
𝜆 = 54,8 < 240 (OK!)
 SFB
P = 239 kN
L = 3,2 m = 3200mm
Mutu baja A 500 Grade A :
Fy = 269 MPa
Fu = 310 MPa
Py = 0,9.Fy.Ag
𝐴𝑔 =
𝑃𝑦
0,9. 𝐹𝑦
=
239 𝑥 103
0,9.269
= 987,2 𝑚𝑚2
= 9,87 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm²
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
Cek Kondisi Leleh
Py = 0,9.Fy.Ag
= 0,9 x 269 x 5121
= 1239794,1 N
= 1239,7941 kN
Py = 1239,7941 kN > P = 239 kN (OK!)
Cek Kondisi Fraktur
Pu = 0,75.Fu.Ae
= 0,75 x 310 x (0,85x5121)
= 1012037,6N
= 1012,0376 kN
Pu = 1012,0376 kN > P = 239 kN (OK!)
Cek Kelangsingan Batang
𝜆 =
𝐿
𝑖 𝑦
=
3200
43,8
= 73,06
𝜆 = 73,06 < 240 (OK!)
 SBG
P = 14,1 Kn P = 403,9 kN
L = 5,76 m = 5760 mm L = 3,6 m = 3600 mm
Mutu baja A 500 Grade A : Mutu baja A 500 Grade A :
Fy = 269 MPa Fy = 269 MPa
Fu = 310 MPa Fu = 310 MPa
Py = 0,9.Fy.Ag Py = 0,9.Fy.Ag
 SBC
Mendesain Batang Tarik

8. 5
𝐴𝑔 =
𝑃𝑦
0,9.𝐹𝑦
𝐴𝑔 =
𝑃𝑦
0,9.𝐹𝑦
=
14,1 𝑥 103
0,9.269
=
403,9 𝑥 103
0,9.269
= 58,2 𝑚𝑚2
= 0,5824 𝑐𝑚2
Pilih profil WF 125 x 125 x 6,5 x 9
Ag = 30,31 cm² = 3031 mm²
𝑖 𝑥 = 5,29 𝑐𝑚 = 52,9 𝑚𝑚
𝑖 𝑦 = 3,11 𝑐𝑚 = 31,1 𝑚𝑚
Cek Kondisi Leleh
Py = 0,9.Fy.Ag
= 0,9 x 269 x 3031
= 733805,1N
= 733,8051 kN
Py = 733,8051 kN > P = 14,1 kN (OK!)
Cek Kondisi Fraktur
Pu = 0,75.Fu.Ae
= 0,75 x 310 x (0,85x3031)
= 599001,4 N
= 599,0014 kN
Pu = 599,0014 kN > P = 14,1 kN (OK!)
Cek Kelangsingan Batang
𝜆 =
𝐿
𝑖 𝑦
=
5760
31,1
= 233,2
𝜆 = 185,21 < 240 (OK!)
= 1668,32 𝑚𝑚2
= 16,6832 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm²
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
Cek Kondisi Leleh
Py = 0,9.Fy.Ag
= 0,9 x 269 x 5121
= 1239794,1 N
= 1239,7941 kN
Py = 1239,7941 kN > P = 403,9 kN (OK!)
Cek Kondisi Fraktur
Pu = 0,75.Fu.Ae
= 0,75 x 310 x (0,85x5121)
= 1012037,6N
= 1012,0376 kN
Pu = 1012,0376 kN > P = 403,9 kN (OK!)
Cek Kelangsingan Batang
𝜆 =
𝐿
𝑖 𝑦
=
3600
43,8
= 82,2
𝜆 = 82,2 < 240 (OK!)
 SCD
P = 403,9 Kn P = 227,2 kN
L = 3,6 m = 3600 mm L = 2,4 m = 2400 mm
Mutu baja A 500 Grade A : Mutu baja A 500 Grade A :
Fy = 269 MPa Fy = 269 MPa
Fu = 310 MPa Fu = 310 MPa
Py = 0,9.Fy.Ag Py = 0,9.Fy.Ag
𝐴𝑔 =
𝑃𝑦
0,9.𝐹𝑦
𝐴𝑔 =
𝑃𝑦
0,9.𝐹𝑦
=
403,9 𝑥 103
0,9.269
=
227,2 𝑥 103
0,9.269
= 1668,32 𝑚𝑚2
= 938,46 𝑚𝑚2
= 16,6832 𝑐𝑚2
= 9,3846 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11 Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm² Ag = 51,21 cm² = 5121 mm²
 SDE

9. 6
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
Cek Kondisi Leleh
Py = 0,9.Fy.Ag
= 0,9 x 269 x 5121
= 1239794,1 N
= 1239,7941 kN
Py = 1239,7941 kN > P = 403,9 kN (OK!)
Cek Kondisi Fraktur
Pu = 0,75.Fu.Ae
= 0,75 x 310 x (0,85x5121)
= 1012037,6N
= 1012,0376 kN
Pu = 1012,0376 kN > P = 403,9 kN (OK!)
Cek Kelangsingan Batang
𝜆 =
𝐿
𝑖 𝑦
=
3600
43,8
= 82,2
𝜆 = 82,2 < 240 (OK!)
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
Cek Kondisi Leleh
Py = 0,9.Fy.Ag
= 0,9 x 269 x 5121
= 1239794,1 N
= 1239,7941 kN
Py = 1239,7941 kN > P = 227,2 kN (OK!)
Cek Kondisi Fraktur
Pu = 0,75.Fu.Ae
= 0,75 x 310 x (0,85x5121)
= 1012037,6N
= 1012,0376 kN
Pu = 1012,0376 kN > P = 227,2 kN (OK!)
Cek Kelangsingan Batang
𝜆 =
𝐿
𝑖 𝑦
=
2400
43,8
= 54,8
𝜆 = 54,8 < 240 (OK!)
 SDH
P = 220,9 kN
L = 3,2 m = 3200 mm
Mutu baja A 500 Grade A :
Fy = 269 MPa
Py = 0,9.Fy.Ag
𝐴𝑔 =
𝑃𝑦
0,9. 𝐹𝑦
=
220,9 𝑥 103
0,9.269
= 912,43 𝑚𝑚2
= 9,1243 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm²
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
Cek Kondisi Leleh
Py = 0,9.Fy.Ag = 0,9 x 269 x 5121 = 1239794,1 N = 1239,7941 kN
Py = 1239,7941 kN > P = 220,9 kN (OK!)
Cek Kondisi Fraktur
Pu = 0,75.Fu.Ae = 0,75 x 310 x (0,85x5121) = 1012037,6N = 1012,0376 kN
Pu = 1012,0376 kN > P = 220,9 kN (OK!)
Cek Kelangsingan Batang
𝜆 =
𝐿
𝑖 𝑦
=
3200
43,8
= 73,06
𝜆 = 73,06 < 240 (OK!)

10. 7
 SAF
P = 496,3 kN
L = 4 m = 4000 mm
Mutu baja A 500 Grade A :
Fy = 269 MPa
Fu = 310 MPa
𝑃𝑈 = ∅𝑃𝑛
496,3 = 0,85 . 𝐴𝑔 .
𝐹𝑦
𝜔
𝐴𝑔 =
496,3 𝑥 103
𝑥1
0,85 𝑥 269
= 2170,57 𝑚𝑚2
= 21,7057 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm²
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
𝜆 𝑐 =
𝑘. 𝐿
𝑟
√
𝐹𝑦
𝜋2 𝐸
=
4000
43,8
√
269
𝜋22𝑥105
= 1,066
𝜔 =
1,43
1,6 − 0,67𝜆 𝑐
=
1,43
1,6 − 0,67(1,066)
= 1,614
Cek
𝑃𝑈 = ∅𝑃𝑛 = 0,85 𝑥 5121 𝑥
269
1,614
= 725475𝑁
𝑃𝑈 = 725,475𝑘𝑁 > 𝑃 = 496,3 kN (OK!)
Cek Tekuk
a. Web b.Flange
ℎ
𝑡 𝑤
≤ 1,49√
𝐸
𝐹𝑦
𝑏
2𝑡 𝑓
≤ 0,56√
𝐸
𝐹𝑦
175
7,5
≤ 1,49√
2𝑥105
269
175
2𝑥11
≤ 0,56√
2𝑥105
269
23,33 ≤ 40,63(OK!) 7,95 ≤ 15,27(OK!)
 SFG
P = 465,5 kN
L = 3,83 m = 3830 mm
Mutu baja A 500 Grade A :
Fy = 269 MPa
Fu = 310 MPa
𝑃𝑈 = ∅𝑃𝑛
465,5 = 0,85 . 𝐴𝑔 .
𝐹𝑦
𝜔
𝐴𝑔 =
465,5 𝑥 103
𝑥1
0,85 𝑥 269
= 2035,86 𝑚𝑚2
= 20,3586 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm²
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
𝜆 𝑐 =
𝑘. 𝐿
𝑟
√
𝐹𝑦
𝜋2 𝐸
=
3830
43,8
√
269
𝜋22𝑥105
= 1,02
𝜔 =
1,43
1,6 − 0,67𝜆 𝑐
=
1,43
1,6 − 0,67(1,02)
= 1,56
Cek
𝑃𝑈 = ∅𝑃𝑛 = 0,85 𝑥 5121 𝑥
269
1,56
= 750587,6 𝑁
𝑃𝑈 = 750,587𝑘𝑁 > 𝑃 = 465,5kN(OK!)
Cek Tekuk
a. Web b.Flange
ℎ
𝑡 𝑤
≤ 1,49√
𝐸
𝐹𝑦
𝑏
2𝑡 𝑓
≤ 0,56√
𝐸
𝐹𝑦
175
7,5
≤ 1,49√
2𝑥105
269
175
2𝑥11
≤ 0,56√
2𝑥105
269
23,33 ≤ 40,63(OK!) 7,95 ≤ 15,27(OK!)
Asumsi 𝜔 = 1
Mendesain Batang Tekan
Asumsi 𝜔 = 1

11. 8
 SHE
P = 378,7 kN
L = 4 m = 4000 mm
Mutu baja A 500 Grade A :
Fy = 269 MPa
Fu = 310 MPa
𝑃𝑈 = ∅𝑃𝑛
378,7 = 0,85 . 𝐴𝑔 .
𝐹𝑦
𝜔
𝐴𝑔 =
378,7 𝑥 103
𝑥1
0,85 𝑥 269
= 1656,24 𝑚𝑚2
= 16,5624 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm²
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
𝜆 𝑐 =
𝑘. 𝐿
𝑟
√
𝐹𝑦
𝜋2 𝐸
=
4000
43,8
√
269
𝜋22𝑥105
= 1,066
𝜔 =
1,43
1,6 − 0,67𝜆 𝑐
=
1,43
1,6 − 0,67(1,066)
= 1,614
Cek
𝑃𝑈 = ∅𝑃𝑛 = 0,85 𝑥 5121 𝑥
269
1,614
= 725475𝑁
𝑃𝑈 = 725,475 𝑘𝑁 > 𝑃 = 378,7 kN(OK!)
Cek Tekuk
a. Web b. Flange
ℎ
𝑡 𝑤
≤ 1,49√
𝐸
𝐹𝑦
𝑏
2𝑡 𝑓
≤ 0,56√
𝐸
𝐹𝑦
175
7,5
≤ 1,49√
2𝑥105
269
175
2𝑥11
≤ 0,56√
2𝑥105
269
23,33 ≤ 40,63 (OK!) 7,95 ≤ 15,27(OK!)
 SHG
P = 241,6 kN
L = 3,83 m = 3830 mm
Mutu baja A 500 Grade A :
Fy = 269 MPa
Fu = 310 MPa
𝑃𝑈 = ∅𝑃𝑛
465,5 = 0,85 . 𝐴𝑔 .
𝐹𝑦
𝜔
𝐴𝑔 =
241,6 𝑥 103
𝑥1
0,85 𝑥 269
= 1056,64 𝑚𝑚2
= 10,5664 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm²
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
𝜆 𝑐 =
𝑘. 𝐿
𝑟
√
𝐹𝑦
𝜋2 𝐸
=
3830
43,8
√
269
𝜋22𝑥105
= 1,02
𝜔 =
1,43
1,6 − 0,67𝜆 𝑐
=
1,43
1,6 − 0,67(1,02)
= 1,56
Cek
𝑃𝑈 = ∅𝑃𝑛 = 0,85 𝑥 5121 𝑥
269
1,56
= 750587,6 𝑁
𝑃𝑈 = 750,587 𝑘𝑁 > 𝑃 = 241,6 kN(OK!)
Cek Tekuk
a. Web b. Flange
ℎ
𝑡 𝑤
≤ 1,49√
𝐸
𝐹𝑦
𝑏
2𝑡 𝑓
≤ 0,56√
𝐸
𝐹𝑦
175
7,5
≤ 1,49√
2𝑥105
269
175
2𝑥11
≤ 0,56√
2𝑥105
269
23,33 ≤ 40,63(OK!) 7,95 ≤ 15,27(OK!)
Asumsi 𝜔 = 1
Asumsi 𝜔 = 1

12. 9
 SDG
P = 282,9 kN
L = 5,76 m = 5760 mm
Mutu baja A 500 Grade A :
Fy = 269 MPa
Fu = 310 MPa
𝑃𝑈 = ∅𝑃𝑛
282,9 = 0,85 . 𝐴𝑔 .
𝐹𝑦
𝜔
𝐴𝑔 =
282,9 𝑥 103
𝑥1
0,85 𝑥 269
= 1237,26 𝑚𝑚2
= 12,3726 𝑐𝑚2
Pilih profil WF 175 x 175 x 7,5 x 11
Ag = 51,21 cm² = 5121 mm²
𝑖 𝑥 = 7,5 𝑐𝑚 = 75 𝑚𝑚
𝑖 𝑦 = 4,38 𝑐𝑚 = 43,8 𝑚𝑚
𝜆 𝑐 =
𝑘. 𝐿
𝑟
√
𝐹𝑦
𝜋2 𝐸
=
5760
43,8
√
269
𝜋22𝑥105
= 1,54
𝜔 = 1,25 𝑥 𝜆 𝑐
2
= 1,25 𝑥 1,542
= 2,96
Cek
𝑃𝑈 = ∅𝑃𝑛 = 0,85 𝑥 5121 𝑥
269
2,9
= 395579,9𝑁
𝑃𝑈 = 395,5799 𝑘𝑁 > 𝑃 = 282,9 kN (OK!)
Cek Tekuk
b. Web b. Flange
ℎ
𝑡 𝑤
≤ 1,49√
𝐸
𝐹𝑦
𝑏
2𝑡 𝑓
≤ 0,56√
𝐸
𝐹𝑦
175
7,5
≤ 1,49√
2𝑥105
269
175
2𝑥11
≤ 0,56√
2𝑥105
269
23,33 ≤ 40,63 (OK!) 7,95 ≤ 15,27 (OK!)
SGC Profil yang digunakan adalah profil WF 175 x 175 x 7,5 x 11
Asumsi 𝜔 = 1
Mendesain Batang 0

13. 10
Batang Tampak Depan Tampak Samping
Batang AB 175 x 175 x 7,5 x 11
Batang AF 175 x 175 x 7,5 x 11
Batang FB 175 x 175 x 7,5 x 11
Batang BC 175 x 175 x 7,5 x 11
Batang BG 100 x 100 x 6 x 8
Batang FG 175 x 175 x 7,5 x 11
Batang GC 175 x 175 x 7,5 x 11
Batang CD 175 x 175 x 7,5 x 11
Batang DG 175 x 175 x 7,5 x 11
Batang DH 175 x 175 x 7,5 x 11
Batang HG 175 x 175 x 7,5 x 11
Batang DE 175 x 175 x 7,5 x 11
Batang EH 175 x 175 x 7,5 x 11
2400
4000
3200
3600
5760
3830
4500
3600
5760
3200
3830
2400
4000
PROFIL YANG DIGUNAKAN
Batang Profil WF
SAB 175 x 175 x 7,5 x 11
SFB 175 x 175 x 7,5 x 11
SBG 125 x 125 x 6,5 x 9
SBC 175 x 175 x 7,5 x 11
SCD 175 x 175 x 7,5 x 11
SDE 175 x 175 x 7,5 x 11
SDH 175 x 175 x 7,5 x 11
SAF 175 x 175 x 7,5 x 11
SFG 175 x 175 x 7,5 x 11
SHE 175 x 175 x 7,5 x 11
SHG 175 x 175 x 7,5 x 11
SDG 175 x 175 x 7,5 x 11
Mutu baja A 500 Grade A :
Fy = 269 MPa
Fu = 310 MPa
Edge-Distance (Le)
Minimum Le Maximum Le
1,25 d (given by manufacture)
12 t (or 150 mm)1,5 d (cut by machine)
1,75 d (cut by hands)
Bolts Spacing (s)
Minimum S Maximum S
3 d 15 t (or 200mm)
2 ⅔ d 32 t (or 300mm)
4 t + 100 mm (or 200 mm)
Mendesain Baut
Le S Le
HPlate
LeLe
B Plate

14. 11
25 150 150 25
2512525
19,1
175
350
 SAB
𝑃𝑦 = 412,8 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
412,8
50,285
= 8,21
≈ 10 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
 SFB
𝑃𝑦 = 239 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N = 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Bauat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
239
50,285
= 4,75
≈ 6 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
25 75 75 75 75 252512525
19,1
175
350

15. 12
 SBG
𝑃𝑦 = 14,1 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 125 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 250 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
14,1
50,285
= 0,4
≈ 6 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
 SBC
𝑃𝑦 = 403,9 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
403,9
50,285
= 8,03
≈ 10 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
25 100 100 25
257525
19,1
125
250
25 75 75 75 75 25
2512525
19,1
175
350

16. 13
 SCD
𝑃𝑦 = 403,9 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
403,9
50,285
= 8,03
≈ 6 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
 SDE
𝑃𝑦 = 227,2 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
227,2
50,285
= 4,5
≈ 6 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
25 75 75 75 75 25
2512525
19,1
175
350
25 150 150 25
2512525
19,1
175
350

17. 14
 SDH
𝑃𝑦 = 220,9 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
220,9
50,285
= 4,39
≈ 6 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
 SAF
𝑃𝑦 = 496,3 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
496,3
50,285
= 9,87
≈ 10 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
25 150 150 25
2512525
19,1
175
350
25 75 75 75 75 25
2512525
19,1
175
350

18. 15
 SFG
𝑃𝑦 = 465,5 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
465,5
50,285
= 9,26
≈ 10 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
 SHE
𝑃𝑦 = 378,7 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
378,7
50,285
= 7,53
≈ 8 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
25 75 75 75 75 25
2512525
19,1
175
350
25 100 100 100 25
2512525
19,1
175
350

19. 16
 SHG
𝑃𝑦 = 241,6 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
241,6
50,285
= 4,8
≈ 6 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
 SDG
𝑃𝑦 = 282,9 kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
282,9
50,285
= 5,63
≈ 6 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
25 150 150 252512525
19,1
350
175
25 150 150 25
2512525
19,1
175
350

20. 17
 SGC
𝑃𝑦 = 0kN
𝑑 𝑏 = 19,1 mm
𝑡 𝑝𝑙𝑎𝑡 = 5mm
ℎ 𝑝𝑙𝑎𝑡 = 175 mm
𝑏 𝑝𝑙𝑎𝑡 = 2 x ℎ 𝑝𝑙𝑎𝑡 = 350 mm
𝑓𝑢
𝑏
= 585 MPa
Tahanan Geser Baut Dengan Ulir
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,4 𝑓𝑢
𝑏
)m . 𝐴 𝑏
= 0,75 (0,4 . 585)1.
1
4
𝜋19,1²
= 50285 N
= 50,285 kN
Tahanan Tarik Baut
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (0,75. 𝑓𝑢
𝑏
.𝐴 𝑏)
= 0,75 (0,4 . 585 .
1
4
𝜋19,1²)
= 94283,4 N
= 94,2834 Kn
Tahanan Tumpu Buat
𝑅 𝑢 = ∅𝑅 𝑛
= 0,75 (2,4. 𝑑 𝑏.t.𝑓𝑢)
= 0,75 (2,4 . 19,1 . 5 . 310)
= 53289 N
= 53,289 kN
Jumlah Baut
𝑛 =
𝑃𝑦
𝑅 𝑢
=
0
50,285
= 0
≈ 6 𝑏𝑎𝑢𝑡 (2 𝑙𝑖𝑛𝑒𝑠)
𝐿 𝑒 min = 1,25 𝑑 = 23,875
𝐿 𝑒 max = 12 𝑡 = 60
𝑆 min = 3 𝑑 = 57,3
𝑆 max = 32 𝑥 𝑡 = 160
Titik A
25 150 150 25
2512525
19,1
175
350
Detail Sambungan

21. 18
Titik B
Titik C

22. 19
Titik D
Titik E

23. 20
Titik F
Titik G

24. 21
Titik H