FAKULTAS DESAIN dan TEKNIK PERENCANAAN UJIAN TENGAH SEMESTER Periode Genap Tahun Akademik 2007/2008 Jurusan Kd. Kls./MK Mata Ujian Dosen
: Teknik Sipil : A / TS13404 : Struktur Baja 1 : Wiryanto Dewobroto
Hari / Tanggal Waktu Semester Sifat Ujian
: : : IV : Open Personal Notes Only
SOAL
1.
tidak hanya bisa menghitung, tetapi juga memahami memahami knowledge tentang Teori (30%). Menjadi engineer tidak materi yang dihitung. Jika topik yang akan dibahas adalah tentang sambungan struktur baja dengan baut mutu tinggi (A325 atau A490), maka jelaskan hal tersebut jika dikaitkan dengan : a. b. c.
2.
Slip Fatique (kelelahan logam) Perilaku sambungan jika digabung dengan sistem sambungan las.
SAMBUNGAN BAUT (75%) . Jika diketahui :
Spesifikasi Bahan : Baja karbon A36 E (modulus elastisitas) = 200 000 MPa Fy tegangan leleh minimum 36 ksi ≈ 250 MPa Fu tegangan tarik 58-80 ksi (mean 69 ksi) ≈ 475 MPa
HTB (High Tension Bolts) A325 or A325M Tegangan geser nominal tumpu, F nv = 330 Mpa (ulir di bidang geser) Diameter baut = 16 mm Diameter lubang = 18 mm (standard)
Adapun konfigurasi sambungan adalah sebagai berikut : 28
150
total 6 M16 A325 bolts
Plate 200x28 100 19 44
P 100
19
P 25
50
106
L 150x150x19 Area = 53.38 cm2
25
Potongan
25
45
45
45
Tampak Depan 25
Pertanyaan hitung Pu maksimum yang dapat diberikan pada sambungan tersebut (hitung pakai cara LRFD).
Selamat bekerja !
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JAWABAN NO.2 ********************************* *** Kekuatan Pelat penyambung *** ********************************* Plate 200x28
B 61
C 72
P 50
72 16
A
25
45
45
45 25
*** Keruntuhan tumpu Tinjau bagian pelat tepi A: Rn = 1.5 Lc tF u ≤ 3.0dtF u deformasi layan tidak masalah (AISC-LRFD J3-2b) Rn = 1.5 *16 * 28 * 475 ≤ 3.0 *16 * 28 * 475 Rn = 319,200 ≤ 638,400 ∴ RnA = 319.200 kN
Tinjau bagian pelat tengah B : Rn = 1.5 Lc tF u ≤ 3.0dtF u Rn = 1.5 * 61* 28 * 475 ≤ 3.0 *16 * 28 * 475 Rn = 1,216,950 ≤ 638,400 ∴ RnB
= 638.400 kN
Tinjau bagian pelat tengah C : Rn = 1.5 Lc tF u ≤ 3.0dtF u Rn = 1.5 * 72 * 28 * 475 ≤ 3.0 * 16 * 28 * 475 Rn = 1,436,400 ≤ 638,400 ∴ RnB = 638.400 kN
Jadi kekuatan ijin pelat terhadap bahaya tumpu Pu = φ ( RnA + RnB + RnC ) = 0.75(319.2 + 5 * 638.4) = 2633.4 kN *** Keruntuhan tarik 2
Ag = 200 * 28 = 5600 mm : leleh Pu = φ Ag F y = 0.9 * 5600 * 250 / 1000 = 1260 kN
Lurus
An = (200 − 2 *18) * 28 = 4592 mm 2 : fraktur Pu = φ An F u = 0.75 * 4592 * 475 / 1000 = 1636 kN
Staggers wn = wg −
∑ ∑ d +
s2 4g
= 200 − 3 * 16 + 2
45 2 4 * 25
Jadi kekuatan ijin pelat terhadap bahaya tarik Pu = 1260 kN **************************************** *** Kekuatan Profil Siku yang disambung *** **************************************** 16
A
61
72
B
C
P
= 192.5 mm
An = 192.5 * 28 = 5390 mm
2
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Tinjau bagian profil siku tepi A: Rn = 1.5 Lc tF u ≤ 3.0dtF u deformasi layan tidak masalah (AISC-LRFD J3-2b) Rn = 1.5 *16 *19 * 475 ≤ 3.0 *16 *19 * 475 Rn = 216,600 ≤ 433,200 ∴ RnA = 216.6 kN
Tinjau bagian profil siku tengah B : Rn = 1.5 Lc tF u ≤ 3.0dtF u Rn = 1.5 * 61 *19 * 475 ≤ 3.0 *16 *19 * 475 Rn = 825,788 ≤ 433,200 ∴ RnB = 433 kN
Tinjau bagian profil siku tengah C : Rn = 1.5 Lc tF u ≤ 3.0dtF u Rn = 1.5 * 72 *19 * 475 ≤ 3.0 *16 *19 * 475 Rn = 974,700 ≤ 433,200 ∴ RnC = 433 kN
Jadi kekuatan ijin pelat terhadap bahaya tumpu Pu = φ ( RnA + RnB + RnC ) = 0.75(2 * 216.6 + 4 * 433) = 1624 kN *** Keruntuhan tarik 2
Ag = 5338 mm : leleh Pu = φ Ag F y = 0.9 * 5338 * 250 / 1E 3 = 1201 kN An = 5338 − 2 *18 *19 = 4654 mm
2
Karena profil siku hanya tersambung pada satu sisi saja maka SHEAR-LAG harus ditinjau ⎛ x ⎞ ⎛ 44 ⎞ U = 1 − ⎜ ⎟ = 1 − ⎜ ⎟ = 0.67 ≤ 0.9 ⎝ L ⎠ ⎝ 135 ⎠ Ae = UAn = 0.67 * 4654 = 3118 mm
Akibat fraktur
2
Pu = φ Ae F u = 0.75 * 3118 * 475 / 1E 3 = 1110.8 kN (**)
Jadi kekuatan ijin pelat terhadap bahaya tarik ditentukan oleh keruntuhan fraktur. *** Keruntuhan blok geser
P 50 81
L 150x150x19 Area = 53.38 cm2
25 25
45
45
45
Tampak Depan
Diameter lobang baut M16 standar adalah 18 mm Luas Geser Agv = 160 *19 = 3040 mm2 Anv = (160 − 2.5 * 18) *19 = 2185 mm 2
Luas Tarik
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F u Ant = 475 * 1368 / 1E 3 = 650 kN
Kuat Geser-Blok dimana F u Ant > 0.6 F u Anv maka (AISC-LRFD J4-3a) φ Rn = φ ⎡⎣0.6 Fy Agv + Fu Ant ⎤⎦ ≤ φ ⎡⎣0.6 Fu Anv + Fu Ant ⎤⎦ Mekanisme leleh leleh geser dan putus tarik tarik φ Rn = 0.75[0.6 * 250 * 3040 + 475 * 1368 ] ≤ 0.75[0.6 * 475 * 2185 + 475 * 1368] φ Rn = 829,350 ≤ 954,393 Jadi kekuatan profil terhadap bahaya geser blok Pu = 829.35 kN
********************************* *** Kekuatan Baut Penyambung *** ********************************* Digunakan baut mutu tinggi A325M dimana kuat nominal terhadap geser 330 MPa, dan dianggap bidang geser kritis terletak pada daerah ulir-drat (kondisi umum)..
Kuat geser pada buat (satu sisi) adalah: Pu = φ nAb F v Pu = 0.75 * 6 *
1 π 16 2 4
* 330
Pu = 298.6 *10 3 N
Rangkuman perhitungan sambungan
1 2 3
Komponen yang dievaluasi Pelat Profil siku Baut mutu tinggi
Gaya maks , Pu (kN) 1260 829 299
Keterangan Leleh tarik Geser blok Geser satu bidang
Dari ketiga komponen tersebut maka kekuatan baut paling menentukan (akan pecah terlebih dahulu). Jadi kekuatan sambungan tersebut adalah 299 kN.
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FAKULTAS DESAIN dan TEKNIK PERENCANAAN UJIAN TENGAH SEMESTER Periode Genap Tahun Akademik 2009/2010 Jurusan Kode Kelas Mata Ujian Dosen
: Teknik Sipil : ACB : Struktur Baja 1 : Dr. Ir. Wiryanto Dewobroto, MT
Hari / Tanggal Waktu Semester Sifat Ujian
: Kamis / 4 Maret 2010 : 07.15 – 09.00 : IV : Open handnote (1 sheet)
Pengetahuan Soal Teori ( 40%) 1. Pada konstruksi baja dengan profil hot-rolled dikenal dikenal dua macam sistem sambungan, yaitu [a] las (weld (weld ) dan [b] baut. Jelaskan kedua macam alat sambung tersebut, adakah yang satu lebih baik dibanding yang lain, dan kapankah alat-alat sambung tersebut dipakai. Bisakah dalam satu sambungan ke dua alat sambung tersebut digabungkan secara bersama-sama. 2. Alat sambung baut mutu tinggi : dari sisi orientasi pemasangan baut mutu tinggi pada sambungan maka baut tersebut dapat dianggap mengalami gaya-gaya [a] tarik ; ; [b] geser; [c] tarik dan geser dan [d] tekan. Berikan dalam bentuk gambar bagaimana bentuk sambungan dan penempatan baut yang dimaksud. Gambarkan selengkap mungkin, minimal dapat dilihat dari dua dua sisi termasuk dimensi dimensi (ukuran), dan gambarkan secara proposional. 3. Sambungan dengan gaya tarik aksial, gambarkan bentuk sambungan yang menyebabkan gaya pada baut adalah [a] tarik dan dan [b] geser. Apa kelebihan dan kekurangan masingmasing bentuk tersebut. 4. Pada penggunaan baut mutu tinggi dikenal dua mekanisme kerja sambungan, yaitu [a] slipkritis; [b] bearing (tumpu). Pertanyaan : a. Pada baut dengan diorientasikan seperti apakah (lihat soal no.2) dua mekanisme tersebut dapat bekerja. b. Adakah syarat khusus agar mekanisme-mekanisme tersebut bekerja. c. Apakah dalam suatu sambungan dapat bekerja satu mekanisme saja, jika dapat terjadi mekanisme yang mana. d. Gambarkan dengan diagram gaya-deformasi keberadaan dua mekanisme tersebut. e. Jika ada dua mekanisme yang bekerja, apakah itu dapat terjadi bersama-sama. Jika tidak, mekanisme yang mana dulu yang dapat terjadi. Pengetahuan Design ( 60%) : konfigurasi sambungan baut mutu tinggi Jawaban : Lihat catatan kuliah dan penjelasan yang diberikan. Tidak dijawab agar tidak dijadikan hapalan.
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P
P
50 10 75
70
70
75
50 18
16
18
16
18
16
70 70
70
70
40
60.6
60.6 70 50 40
10M16 A325M
5 50
290
5
P
P
Masalah SAMBUNGAN BAUT pada baja konstruksi. Jika diketahui : Spesifikasi Bahan : Baja karbon A36 E (modulus elastisitas) = 200 000 MPa Fy leleh minimum 36 ksi ≈ 250 MPa Fu tarik 58-80 ksi (mean 69 ksi) ≈ 475 MPa
HTB (High Tension Bolts) A325 or A325M Tegangan geser nominal tumpu, F nv = 330 Mpa (ulir di bidang geser) Diameter baut = 16 mm Diameter lubang = 18 mm (standard)
Pertanyaan : a. Jika tebal pelat t = 5 mm (sisi luar) dan t = 10 mm (sisi dalam), berapa P maksimum yang dapat diberikan. b. Jika digunakan konfigurasi dan jumlah baut yang sama seperti di atas, berapa tebal pelat agar keruntuhan ditentukan dari kekuatan baut yang ada. c. Jika digunakan sambungan slip-kritis dengan memberikan Pretensioning min 70%
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Jawab : Kekuatan PELAT, pada bagian dalam t = 10 mm, sedang sisi luar t = 2 * 5 ≈ 10 mm, sama. Oleh karena itu pada perhitungan ditinjau pelat bagian dalam saja. 50
50
60.6
60.6
50
50
A 102 40
B G
75 18 70
35
18
C
70 103
P 290
41
H
P 70
D I
70 70
E
75
F
40
*** Kekuatan Tumpu sekitar Lubang Baut *** Tinjau bagian pelat tepi (paling kanan): Rn = 1.5 Lc tF u ≤ 3.0dtF u deformasi di lubang baut tidak menjadi masalah........................ (J3-6b) Rn = 1.5 * 41 *10 * 475 ≤ 3.0 *16 *10 * 475 Rn = 292125 ≤ 228000 ∴ Rn = 228 kN per baut (ada 3 baut)
Tinjau bagian pelat tengah : Rn = 1.5 Lc tF u ≤ 3.0dtF u deformasi di lubang baut tidak menjadi masalah........................ (J3-6b) Rn = 1.5 *102 *10 * 475 ≤ 3.0 *16 *10 * 475 Rn = 726750 ≤ 228000 ∴ Rn = 228 kN per baut (ada 4 baut)
Tinjau bagian pelat tepi (paling kiri):
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Potongan netto A-B-G-C-H-D-I-E-F s 2 wn = w g − ∑ d + ∑ 4 g wn = 290 − 18 * 7 + 6
602 4 * 35
= 318.286 >>> w g = 290 mm tidak menentukan
Potongan netto A-B-G-H-I-E-F wn = w g − ∑ d + ∑
s 2 4 g
wn = 290 − 18 * 5 + 2
60 2
= 251.43 mm 4 * 35 An = wnt = 251.43 *10 = 2514 mm 2 ; Ae = An : fraktur
P Ae F u u = φ
........................... ........................... (D2-2)
P u = φ Ae F u = 0.75 * 2514 * 475 / 1000 = 896 kN Kekuatan BAUT Kuat geser pada baut (dua sisi) adalah: ................................................................... .............................................. .............................................. ......................... .. (J3-1 ) P u = φ nAb F v ............................................
P u = 0.75 * 2 * 10 * 14 π 16 2 * 330 / 1000 P u = 995 kN Catatan : kondisi kritis ditentukan oleh kekuatan leleh (pelat utuh), selanjutnya potongan kritis adalah pada lubang (baut 4 baris ditengah), yang menjadi lintasan terpendek dan tidak dimungkinkan ada keruntuhan blok.
SAMBUNGAN FRIKSI dengan BAUT MUTU TINGGI (SLIP-CRITICAL) ................................................................... .............................................. .................................. ........... (J3-4) Rn = μ Du h scT b N s ............................................ dimana
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Rangkuman perhitungan sambungan menurut ANSI/AISC 360-05 Komponen yang Pu maks Efisiensi Keterangan dievaluasi (kN) sambungan 1 Pelat 652.5 (100 % pelat Sambungan utuh) Tumpu 2 Baut HTB 995 (153% pelat utuh) 3 HTB - Full 611. 611.8 8 (94% (94% pela pelatt utuh utuh)) Samb Sambun unga gan n Pretensioning Friksi
Applicabl e ya ya
JAWABAN PENGETAHUAN DESIGN (BOBOT 60%) a. Karena pelat sisi luar ada dua, masing-masing t = 5 mm dan pelat dalam hanya satu yaitu t = 10 mm maka pada prinsipnya kekuatan pelat sama. Hitungan yang disusun di atas didasarkan kekuatan satu pelat tunggal t = 10 mm. Adapun P maksimum yang dapat diberikan ditentukan kekuatan yielding pelat yaitu P = 652.5 kN. b. Jika memakai konfigurasi dan jumlah baut yang sama seperti di atas, dan kekuatan ditentukan oleh kekuatan baut maka ketebalan pelat yang diperlukan adalah 995 kN 995000 t = = = 16 mm untuk pelat dalam 8 mm untuk pelat luar. φ F y b pelat 0.9 * 240 * 290
c. P maksimum yang dapat diberikan sebelum terjadi slip = 612 kN
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Wiryanto Dewobroto ------------------------------------------------------------------------ Jurusan Teknik Sipil - Universitas Pelita Harapan Harapan,, Karawaci
F A K U L T A S D E S A IN I N d a n T E K N IK IK P E R E N C A N A A N U J IA N T E N G A H S E M E S T E R ( U T S ) G E N A P T A H U N A K A D E M IK 2010 / 2011 Jurusan
: TEKNIK SIP IL
Kd. Kel as Mat a Ujian Dosen
: AIJ : Struktur Baja 1 : Dr. Ir. Wiryanto Dewobroto, MT.
Hari / Ta Tanggal gal Waktu SKS Sifat Uji an
: Jumat / 4 Maret 201 2011 : 07:15 – 09:00 :2 : note tulisan tangan 1 lbr
Teori : (Bobot 25%) 1. Berd Berdas asar ark kan orientasi baut terhadap bekerjanya gaya yang dialihkan pada sambungan, berikan contoh bentuk-bentuk sambungannya (gambar / sketch). Dari segi efisiensi penggunaan baut mana yang lebih lebih baik, mengapa ? shear lag, pada kondisi apa terjadi. Apakah bisa hilang. Berikan 2. Apa Apa yan yang g dim dimaks aksud shear contoh sketch sambungan dengan shear lag dan yang tidak. 3. Jelask Jelaskan an apa yang yang dima dimaksu ksud d dengan dengan mekanisme slip kritis dan mekanisme tumpu. Kenapa bisa terjadi mekanisme baut slip kritis, dan faktor-faktor yang mempengaruhi. Bilamana suatu sambungan harus memakai mekanisme-mekanisme tersebut. 4. Tentang Tentang sambungan sambungan sistem sistem geser baut mutu mutu tinggi dan sistem sistem las, dapatkah dapatkah keduanya keduanya disatukan dengan cara kumulatif, argumentasi apa yang mendasari jawabanmu. 5. Apa yang dimaks dimaksud ud deng dengan an prying force, jelaskan. Hitungan (Bobot 75%) Diketahui profil siku tunggal L120x120x12 dan gusset plate tebal 12 mm, mutu ASTM A36 atau setara, disambung dengan baut mutu tinggi ASTM A325 , M20 (diameter (diameter 20 mm) . Tampak elemen dan potongan potongan sambungan sebagai berikut berikut (gambar bautnya bautnya belum ada) :
12
120
12
0 2 1
P
P
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Wiryanto Dewobroto ------------------------------------------------------------------------ Jurusan Teknik Sipil - Universitas Pelita Harapan Harapan,, Karawaci
Lampiran :
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Wiryanto Dewobroto ------------------------------------------------------------------------ Jurusan Teknik Sipil - Universitas Pelita Harapan Harapan,, Karawaci
Jawaban tentang soal HITUNGAN: (versi (versi 11 Maret 2011 pk 09:00) Hanya diketahui baut M20 (diameter 20mm) A STM A3235 dan gambar berikut:
12
120
12
0 2 1
P
P
profil siku L120x120x12 gusset plate t = 12 mm
Diminta untuk merencanakan sambungan secara optimal, yaitu aman tetapi menggunakan bahan yang sekecil mungkin. mungkin. Gaya P dari gambar di atas alihkan dari [1] pelat (gusset p late 12mm) ke [2] profil siku L120 melalui [3] baut dengan orientasi geser. J adi ada tiga material yang harus diperiksa. ************************** *********************************** *********
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Wiryanto Dewobroto ------------------------------------------------------------------------ Jurusan Teknik Sipil - Universitas Pelita Harapan Harapan,, Karawaci
Luas penampang netto (dengan lobang) 2
Tinjau potongan vertikal : A n = Ag – (20 + 2)*12 = 2472 mm Tinjau potongan zigzag : A n = Ag – (20 + 2)*12*2 + (s 2/4g) 2 An = 2208 +(36^2/(4*48) = 2215 mm ** govern ** Untuk memperhitungkan shear-lag perlu tahu jumlah baut dan panjang pa njang penyambungan penyambungan (L), untuk itu anggap bahwa yang menentukan adalah keruntuhan leleh. Baut geser Fnv = 330 MPa (lihat Tabel J3.2) 2 2 2 Luas baut M20 adalah A bolt = ¼ ¼ D D = 0.25 0.25 *20 = 314 mm Pu baut = = P Pn baut = 0.75 *314* 330/1000 = 77.7 kN / b aut (geser tunggal). n baut = P u siku / Pu baut = 615.6 / 77.7 = 7.922 digunakan 8 M20 A325. Jadi detail penempatan baut menjadi. Titik berat siku x = (108*12*6 + 120*12*60) 120*12 *60) / A g = 34.4 mm profil siku L120x120x12 0 3
108 2 1
0 2 1
12
8 4
0 6
6 0
posisi pinggir sambungan
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Wiryanto Dewobroto ------------------------------------------------------------------------ Jurusan Teknik Sipil - Universitas Pelita Harapan Harapan,, Karawaci
Hitung ulang A n dan Ae sebagai berikut: Luas penampang netto (dengan lobang) Tinjau potongan vertikal : A n = Ag – (20 + 2)*12 = 2472 mm 2 2 Tinjau potongan zigzag : A n = Ag – (20 + 2)*12*2 + (s /4g) An = 2208 +(60^2/(4*48) = 2227 mm 2 ** govern ** Ternyata penempatan baut zigzag sangat mempengaruhi A n (luas penampang netto). Oleh karena itu konfigurasinya perlu diubah lagi menjadi cukup satu garis garis saja, sbb :
30
60
60
posisi pinggir sambungan
profil siku L120x120x12
2 Ø 2 60
60
60
60
60
30
38 420 19
Luas penampang netto (dengan lobang) Tinjau potongan vertikal : A n = Ag – (20 + 2)*12 = 2472 mm 2 Faktor untuk memperhitungkan shear lag adalah U 1
x L
1 2
Luas potongan efektif A e = U * A n = 0.92 * 2472 = 2274 mm .
34.4 420
0.92
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Wiryanto Dewobroto ------------------------------------------------------------------------ Jurusan Teknik Sipil - Universitas Pelita Harapan Harapan,, Karawaci
**************************************** ** Keru Keruntuhan ntuhan dengan mekanisme geser blok. blok. ** **************************************** Karena gusset plate u kurannya kurannya tidak terbatas (karena tidak ada penjelasan pada soal) maka keruntuhan geser blok tidak terjadi pada gusset-plate. Sedangkan Sedangkan profil siku yang disambung tidak menyeluruh, yaitu hanya satu sisi saja (o leh karena itu ada pengaruh shear-lag ) maka dimungkinkan akan terjadi geser blok pada profil siku. Ini yang akan kita check sbb :
2 2 Ø
posisi pinggir sambungan
profil siku L120x120x12 450
0 6
Diameter lobang baut M20 standar adalah 22 mm Luas Geser Agv 450 *12 5400 mm
2
Anv 450 7.5 * 22 *12 3420 mm
2
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FAKULTAS DESAIN dan TEKNIK PERENCANAAN UJIAN AKHIR SEMESTER SEMESTER GENAP TA 2007/2008 Jurusan Mata Kuliah Dosen
: Teknik Sipil : Struktur Baja I : Ir. Wiryanto Dewobroto, MT.
Hari / Tanggal Waktu Semester Sifat Ujian
: Senin, 26-05-2008 : 14.10 – 15.50 : IV : catatan tangan 1 lembar
Ada dua rangka, sebut saja Frame 1 dan Frame 2, tampak samping pada gambar di bawah (kanan) Keduanya dijajarkan pada bagian kolom dipasang bracing, tampak depan gambar bawah kiri. Pada bagian balok (batang ABC) pada kedua frame pada prinsipnya juga ada bracing (tidak terlihat). Dengan penggabungan tersebut dihasilkan konstruksi yang mampu memikul pembebanan q secara stabil dan mandiri (sebagai struktur 3D). 2m q (t/m')
B2
10m
q (t/m')
q (t/m')
A
B1
B
m 3
D2
g i n c a ) b r y p t (
C
balok menerus menerus ABC m 6
m 3
D1
kolom BD
pin D
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Jawab :
Langkah pertama adalah mendapatkan gaya-gaya internal pada kolom, adapun konfigurasi struktur adalah 10m
2m q (t/m')
B2
q (t/m')
q (t/m')
A
B1
B
m 3
D2
g i n c a ) b r y p t (
C
balok meneru meneruss ABC m 6
m 3
D1
kolom BD
pin D
z
z
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Dari soal diketahu bahwa beban yang bekerja q (t/m) adalah beban hidup, jadi load faktornya = 1.6, sehingga PU = 1.6*R B = 1.6*72 t ≈ 1152 kN Jika pakai profil tunggal WF200.200.8.12 , maka : Dimension (mm)
Label
H B WH-200 200 200
t1 8
t2 12
Mass Momen of Inertia Radius of Gyration Section Modulus A per meter cm4 cm cm3 cm2 (kg/m) axis x-x axis y-y axis x-x axis y-y axis x-x axis y-y r 13 63.53 49.9 4720 1600 8.62 5.02 472 160
Baja ST37 ≈ BJ37 Fy = 240 Mpa Pakai cara AISC 1999 Sumbu kuat r = r 1−1 = r x = 86.2 mm λ 1−1 =
kL1−1 1 r x
F y
π E
=
1 * 6000 1 86.2 π
240 200000
= 0.7675 < 1.5
Untuk λ1-1 ≤ 1.5 λ 12−1
F cr = 0.658
F y = 0.658
0.76752
* 240 = 187.56 MPa
Pn = F cr Ag = 187.56 * 6353 / 1E 3 = 1191.6 kN
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Pakai cara AISC AISC 1999 atau 2005 tetapi
= 0.85
Modulus Potonga A n Hx B t r 1 r 2 Cx=Cy Ix= Iy max Iu min Iv r x=r y max r u min r v Sx=Sy mm mm mm mm cm2 kg/m cm cm4 cm4 cm4 cm cm cm cm3 130 x 130 12 14 7 30.0 23.6 3.64 472 750 194 3.97 5.00 2.54 50.4 Dimensi Standar Profil Siku Sama Kaki
berat satua n
cg
Momen Inersia Inersia
Radius Girasi
1
2
a
a
a
6000
Kolom Profil Siku Ganda 2 L 130 x 130 x 12 Jawab :
10.0
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Memakai persamaan persamaan E3-2 atau E3-3 untuk menghitung menghitung tekuk lentur dari batang tekan dimana dimana KL ⎛ KL ⎞ =⎜ ⎟ = 106.6 , sehingga r ⎝ r ⎠ m F e =
π 2 E 2
=
π 2 * 200000
⎛ KL ⎞ ⎜ ⎟ ⎝ r ⎠ m
(106.6)2
= 173.7 MPa >> 0.44F y = 105.6 MPa
E ⎛ KL ⎞ = 135.97 ⎜ ⎟ = 106.6 < 4.71 F y ⎝ r ⎠ m
Karena
KL r
≤ 4.71
F y ⎡ F ⎢ F cr = 0.658 e ⎢ ⎢
E F y
atau F e ≥ 0.44 F y maka digunakan
⎤ 240 ⎤ ⎡ ⎥ F = ⎢0.658173.7 ⎥ * 240 = 134.6 MPa ⎥ y ⎢ ⎥⎦ ⎣ ⎥
F y ⎡ F F cr = ⎢0.658 e ⎢ ⎢⎣
⎤ ⎥ F ⎥ y ⎥⎦
(AISC-LRFD E3-2)
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FAKULTAS DESAIN dan TEKNIK PERENCANAAN
Ujian Akhir Semester Periode Genap Tahun Akademik 2009/2010 Jurusan Kode Kelas Mata Ujian Dosen
: Teknik Sipil : ACB : Struktur Baja 1 : Dr. Ir. Wiryanto Dewobroto, MT
Hari / Tanggal Waktu Semester Sifat Ujian
Soal 1. (70%) 4 5 3
Pu bracing
A
Pu 3500
bracing
A
WF 150x75x5x7
: Senin, 17 Mei 2010 : 07.15 – 09.00 : IV : Open notes
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Soal 2 (30%)
Soal berikut didasarkan pada pengertian-pengertian yang disampaikan secara lesan di kelas. Adapun pertanyaannya adalah sebagai berikut : a. Jika beban P U diubah arahnya, dari tekan (seperti kondisi soal no.1) menjadi tarik (arah beban diputar 180o), elemen mana yang paling kritis, batang AB atau batang AC. Berapa gaya P maksimum tersebut.
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Tinjau batang AB profil WF 150 x 75 x 5 x 7 (AISC LRFD 2005)
Dimension (mm)
Label H
B
tw
tf
r
WF-150 150 75
5
7
8
A 2 cm 17.85
Momen of Inertia Radius of Gyration Section Modulus Mass cm4 cm cm3 per meter (kg/m) axis x-x axis y-y axis x-x axis y-y axis x-x axis y-y 14.0
666
49.5
E (modulus elastisitas) = 200000 MPa , G (modulus geser) = 80000 MPa
6.11
μ = 0.25
1.66
88.8
13.2
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Tinjau batang AB profil siku ganda L70x70 (AISC LRFD 2005)
Label HxB L70x70
t mm 6
r 1 mm 8.5
r 2 mm 4
y
x
A cm2 8.127
q kg/m 6.38
Cx=Cy cm 1.81
Ix=Iy cm4 29.4
Imax cm4 46.6
Imin cm4 12.2
r x=r y cm 1.98
r max max cm 2.49
r min min cm 1.27
Zx=Zy cm3 6.26
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Menghitung apakah tekuk torsi lentur menentukan ! Ini hanya di penampang T atau double siku yang memang dapat terjadi keruntuhan terhadaptekuk-torsi-lentur. Tekuk lentur ( flexural flexural torsional buckling ) akan terjadi pada sumbu y-y profil gabungan, yaitu sumbu kuatnya, dimana pelat kopel berpengaruh sehingga perlu memodifikasi kondisi tekuknya, lihat AISC LRFD E6.2.
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F crz = F cry
GJ A g r o2
=
80000 * 20160 1625.4 *1796.4
= 552.35 MPa
+ F crz = 120.4 + 552.35 = 672.75 MPa
H = 1 −
xo2 + yo2
sehingga
r o2
=1−
0 + 18.12 1796.4
= 0.8176
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