Tutorial UTS Struktur Baja
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![Page 1: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/1.jpg)
Aozora Insan Kamil, (hampir).S.T
15011086Sumber: -Salmon, Steel Structure Design and Behavior
-SNI-03-1729-2002 Tata Cara PerencanaanStruktur Baja Untuk Bangunan Gedung
- Brockenbrough, Structural Steel Designer Handbook
- Agus Setiawan, LRFD
*Tutorial UTS Struktur Baja
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*Ingat!
≠
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*Tension Member
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*Contoh 1
*Sebuah profil siku 40x40x4 memiliki tegangan leleh sebesar 410 Mpa. Berapakah Besar tahanan tarik nominalnya?
Jawab:
*
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*Hole Effect
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2 mm untuk SI
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*Holes?
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*Holes in an
angle?
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*Effective Area
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*Contoh 2
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*Contoh 2
Tampak samping
Tampak atas
Profil: L 40x40 t=4 Fu: 520 MPaD baut: 8 mm, jarak antar baut 4 cm
Tentukan Tn akibat fraktur pada lubang baut!
T
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Jawab:
1. Hitung
< penentu
Dalam hal ini =0
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2. Hitung
X=11,7
Maka
3. Hitung
Disebut juga keruntuhan Fraktur
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*Block Shear Formula
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*Contoh 3
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*Apabila diketahui bahwa mutu baja BJ 37, tentukan kekuatan tarik ultimit!
T
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JAWAB:
*1. Yielding Gross Section (leleh)
Untuk BJ 37:
Berdasarkan Tabel:
*2. Fracture of Nett Section (Fraktur)
Menentukan terkecil antara
A-B-C-D-E dan A-B-D-E
![Page 21: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/21.jpg)
Untuk A-B-C-D-E
Untuk A-B-D-E
Maka keruntuhan terjadi pada bidang A-B-C-D-E
Maka
![Page 22: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/22.jpg)
Sehingga
*Combination of Shear and Tension (Blok Geser)
Besarnya luasan bidang geser kotor
dan untuk luasan bidang geser bersih
Cek
Maka syarat terpenuhi -> Shear Yielding, Tension Fracture
![Page 23: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/23.jpg)
Jenis Keruntuhan Batas (ton)
Leleh 18,6
Fraktur 19,8
Blok Geser 35,7
PENENTU!
![Page 24: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/24.jpg)
*Compression Member
![Page 25: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/25.jpg)
*Kekompakan
![Page 26: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/26.jpg)
*Residual Stresses
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*Contoh 4
*BJ 52
*Hitung Beban Tekan Ultimate!
*Diasumsikan penampang kompak
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Jawab:
*dikarenakan adanya pengekang lateral pada arah x, maka tidak dapat dipastikan bahwa keadaan tumpuan jepit-jepit pada arah y memiliki panjang efektif yang lebih pendek.
Cek panjang Efektif:
-untuk panjang efektif x tanpa pengekang
-sedangkan untuk arah y (desain)
Untuk panjang efektif x dengan pengekang , sehingga arah y menjadi lebih rentan mengalami tekuk gunakan perhitungan untuk arah y.
x y
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*Panjang Efektif
![Page 36: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/36.jpg)
Sehingga beban ultimate yang dapat dipikul adalah
![Page 37: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/37.jpg)
*Tekuk Lentur Torsi
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*\
![Page 39: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/39.jpg)
*PenampangGabungan
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*Syarat-syarat
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*Contoh 5
*Sebuah penampang gabungan yang terdiri atas 2 buah profil Lip Channel 150x50x20,t=3,2 mm.
*BJ 37
*Diasumsukan bahwa seluruh persyaratan penampang gabungan terpenuhi.
*Tumpuan jepit-sendi, L=4 m
*Digunakan 10 buah pelat kopel
*Hitunglah kapasitas penampang di samping!
![Page 44: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/44.jpg)
*Properti Penampang
![Page 45: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/45.jpg)
Jawab:
1. Cek kekompakan Penampang
Maka kekompakan flens OK!
Maka kekompakan web OK!
![Page 46: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/46.jpg)
Untuk arah x
Pengecekan kekuatan terhadap tekuk global
![Page 47: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/47.jpg)
Untuk arah y
Karena m=2
Pengecekan kekuatan terhadap tekuk global
![Page 48: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/48.jpg)
Untuk pengecekan terhadap tekuk lentur torsi, ingat kembali persamaan berikut
Hitung Momen inersia gabungan
Maka
![Page 49: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/49.jpg)
Maka kuat tekan adalah
Kriteria Pu (ton)
Tekuk Lentur Arah x 7,4
Tekuk Lentur Arah y 0,7
Tekuk Lentur Torsi 1,4
PENENTU
![Page 50: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/50.jpg)
*Elemen Lentur
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*Terkekang Lateral
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*Tidak Terkekang Lateral
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*Ringkasan Z dan Cb
Section Z
2
1,7
1,6
1,5
1,27
1,12
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*Contoh 6
Sebuah balok dengan properti di bawah ini sepanjang 6 m terkekang lateral di tengah bentang. Apabila mutu baja adalah BJ37, tentukan Momen ultimate yang dapat dipikul oleh balok! (beban terdistribusi merata)
![Page 65: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/65.jpg)
Jawab:
Pengecekan kekompakan
Web
FlensWeb
Flens
![Page 66: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/66.jpg)
Hitung besarnya λ penampang
Maka flens: tak kompak
Web: kompak diambil tak kompak
![Page 67: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/67.jpg)
Pengecekan tekuk torsi lateral
![Page 68: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/68.jpg)
Karena balok terkekang lateral di tengah bentang, maka panjang terkekang menjadi ½ L=3000 mm, sehingga balok termasuk ke dalam bentang menengah.
Maka diambil
![Page 69: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/69.jpg)
Kriteria Mu (kN.m)
Kekompakan 442,29
Torsi Lateral 462,1
MENENTUKAN
![Page 70: Tutorial UTS Struktur Baja](https://reader033.fdocuments.us/reader033/viewer/2022061513/55cf925d550346f57b95d88f/html5/thumbnails/70.jpg)
SEKIAN DAN TERIMAKASIH