≤ contoh 1 dua plat akan disambung dengan las lumer bentuk tumpul, ukuran plat 1 = ukuran plat 2 yaitu 10/60. Menerima gaya beban sebesar P= 3000 kg
≠ = 1400 kg/ cm
2
Kontrol kekuatan las. Penyelesaian : Flas
= l net . a
a
= 10 mm
l br br
= 60 mm
l net
= l br – 3a 3a = 60- 30 mm= 3 cm br –
= = × = 10 1000 00 kg/c kg/cmm < = 1400 kg/ cm
2
Contoh 2 : Sebuah pelat tebal (s) = 8 mm, l = = 300 mm, disambungkan pada tiang baja profil DIN dengan las tumpul. Sambungan tersebut menahan gaya P = 1000 kg sejauh e = 150 mm dari las
= 1400 kg/ cm ̅̅ = 0,6 = 840 kg/ cm 2
2
Kontrol kekuatan las Penyelesain Tebal las (a) = s = 0,8 cm L br
= 300 mm = 30 cm
l net
= l br – – 3a 3a = 30- (3
×0,8)=27,6 cm
Pindahkan gaya P pada kedudukan las maka timbuk momen M = P. e dan gaya geser P
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M = P. e = 1000 kg
=
W las
= 1/6
×15 cm = 1500 kg cm
= ×0,8×30 = 12 1200 kg cm =125/ <1400/ = 15000 120 3
. = . . . Dimana : D = P;S = Sxlas b = alas ; I = Ix las
×1/4 l I = 1⁄12 . ℎ
Sx = (1/2.l.a (1/2.l.a)) x
Dimana : b = a h = l
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= 121 × 0,8 × 30 0 = 180 8000 . . 1000 10 00 × × 90 = . . = 0,8 × 18 1800 00 = 694 94,4,444 / / < ̅̅ Jadi las cukup menahan beban tersebut b. las Sudut arah gaya P / / las
sambungan las sudut
Bidang retak
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Bidang retak untuk las sambung dibebani gaya P dengan arah / / las, akan membuat sudut 450 dengan kaki las. Contoh : Las dipasang pada dua bagian Gaya P akan ditahan oleh las atas dan bawah sebesar P1 dan P2.
= = ⁄2 = + ×= 1⁄2 = + × = 1⁄2 = ≤ ̅̅ = ≤ ̅̅ ̅ = 0,6 =. =,707.. =. =,707.. = 3 Dimana :
= panjang las neto
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panjang las minimum = 40 mm panjang las maksimum = 40 a = 28,28 s contoh : Diketahui : Gaya P = 2000 kg Las dipasang pada dua muka.
= 1400 kg/ cm ; 0,6=840 kg/cm 2
2
Hitung panjang las Penyelesaian : s = 10 mm = 1 cm
×1=0.707 cm
a= 0,707s = 0.707
b= 200 mm = 20 cm
1⁄2 b = 10 cm P = P =1⁄2 P = 1000 kg e1 = e2 = 1
2
l1 = l2 = l br
=. =,707. ̅̅ = ⁄ maka = ⁄ =1,19 ² = /² 0,707 = 1,19 19 ²² = 1,683 cm = + 3a
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Las dipasang pada tiga bagian
= . ̅̅ = . . ̅ = 3=3 ∑ = 0 ( ) + . = 0 . × 1⁄2 × × 1⁄2 = 0 = 1⁄2 1⁄2 ∑ = 0 ( ) + . = 0 . × 1⁄2 × × 1⁄2 = 0 = 1⁄2 1⁄2 Contoh :
s = 10 mm = 1 cm; a = 0,707s = 0,707 cm b = 100 mm = 10 cm; e1 =e2 = ½
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l 3n = l br - 3a br = 10 cm - 3 x 0,707 cm= 7,879 cm
= ̅ = . = 7,879 cm× 0,70 ,707 × 84 8400 / / ²² = 4680 kg ∑ = 0 = 0 = = (−. ) = 2.660 660 = ambil = ̅ = 840 / /² = = . = 3,167 167 =. =0,707. 3,167 cm² = 0,707 cm. = ,,² =4,479 ⊥
Arah gaya P las
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⊥
Bidang retak utk las sudut yang dibebani gaya P dengan arah las, akan membentuk sudut 67½o dengan kaki las. Untuk mempermudah dalam perhitungan, diambil sudut 450.
⊥
Gaya P las,akan ditahan oleh las sudut atas dan bawah sebesar P1 dan p2. Untuk s1 = s1, maka P1= P= P2= = ½P. gaya p/2 bekerja pada titik berat bidang retak ½P // P.Gaya geser D = ½P sin 450 Gaya normal N = Gaya ½P sin 450 D=N
= ≤ ̅̅ dengan ̅̅ = 0,6 = ≤ = √ 3 ≤ = luas bidang geser = luas bidang retak
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= ≤ ̅̅ F = l a = 3 → = 0,707 707 ≤ ̅̅ = . = 3 → = 0,707 707 ≤ F = l a ≤ F = l a = 3 = 3 gs1
1n. 1
tr1
1n. 1
tr1
1n. 1
Contoh soal : Diketahui Konstruksi seperti pada gambar gaya P = 5000 kg; las sudut selebar plat =1400 kg/cm2 = = 0,6 = 840 kg/cm2 Penyelesaian : Tebal plat tidak sama S1 = 8 mm = 0,8 cm a1= 0,707 cm S2 = 10 cm = 1 cm a2 = 0,707 1 = 1,70 7 cm b = 100 mm = 10 cm l1br = = l2br = b= 10 cm l1n = l1br -3a -3a1 =10-3 0,566 = 8,302 cm l1n = l1br -3a -3a1 =10-3 0,707 = 7,879 cm
̅̅
→ ×0,8=0,566 → × ≥ → × × 0,8 × 50 = . = 0.8 = 2.222 .222 1 50000 = 0 → = 5.00 .000 2.222 222 277 7788
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= 49 4999 / /² ²
= 3 = 499 499 3(499 499)) = 998 ≤ Tegangan. geser dari las sudut sepanjang L yang menerima beban P yakni
= , Kapasitas geser las adalah R n dimana
× 0.707× a Lw 0.75´ f ×0.707× a × L dimana ϕ = 0,75 ´
R n = f w ϕR n =
w
w
Dimana f w = teg.geser ultimit electroda = 0,6 x kuat tarik electroda las (tergantung pada electrode yang digunakan pada proses SMAW) •
Kuat tarik dari electroda electroda las antara lain ; 413, 482, 551, 620, 688, 758, atau 827 MPa.
•
Terminologi standarelectrode las seterusnya.
yang
dipakai adalah E60XX,E70XX, E80XX, dan
E – electrode electrode 70 – 70 – tensile tensile strength of electrode (ksi) = 482 MPa XX – XX – type type of coating Kekuatan dari elektroda diperhiutngkan dari base metal dipakai :
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Tegangan geser las sudut =beban/ luas bidang patahan geser Kondisi batas las sudut ditentukan oleh': (1) Patahan geser pada lintasan kritis atau kuat geser las
= Untuk kaki las sudut yang sama (equal):
= (0,707) Misalkan elektroda las E70XX, mempunyai tegangan geser las sebesar f w = 0.60 FEXX
f = 0.75 x 0.60 x 482 = 217 MPa w
(2) Kemampuan geser base metal atau pelat : fR n = 0.9 x 0.6 Fy x Luas base metal yang menerima geser dimana Fy = Teg.leleh pada base metal. Contoh kuat las geser menerima beban adalah: Vn = 0.75 x (0.707 a) x L w x f w > T
Batasan dimensi sambungan las Minimum size (amin) Tergantung pada tebal pelat yang paling tipis pada sambungan Maximum size (a
)
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Jika Lw < 100 a, a, maka panjang las efektif (Lw eff ) = Lw Jika Lw < 300 a, maka panjang las efektif (Lw-eff ) = Lw (1.2 – (1.2 – 0.002 0.002 Lw/a) Jika Lw > 300 a, maka panjang las efektif (Lw-eff ) = 0.6 Lw
Pemutusan las
Lap joint – joint – pemutusan pemutusan las sudut pada jarak > a dari ujung. Pengait las pada daerah sudut harus > 2 a
PEDOMAN PERENCANAAN LAS SUDUT
Dua tipe las sudut yang digunakan
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F w = 0.6 F 0.6 F Exx(1+ 0.5sin15 )
= 0.75
Maksimun ukuran las, t-max
≤ 6 } _ ≤ {6 ≥ 6 2 − ≤ Minimu ukuran las, t-min
Contoh : Perencanaan kekuatan sambungan Las
Tentukan kemampuan batang tarik pada sistem sambungan seperti tergambar. Batang tarik adalah pelat persegi ukuran 100x10 mm dilas pada pelat sambung tebal 15 mm dengan memakai elektroda E70XX.
Pertimbangkan batang tarik tersebut pada kondisi leleh dan patah. Periksa juga kuat geser las dan base metal metal disekitar disekitar pelat dengan Fy = 573 MPa. MPa.
Ada gambar nanti dipoer point Langkah pertama : periksa batasan dimensi dari las t
= 10 mm (batang tarik) tarik) t
= 15
(pelat sambung) oleh karena itu,
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dimana ; Ae = U A (luas penampang tarik efektif) Ae = Ag = 100 x 10 = 1000 mm Fu = teg.tarik ultimit pelat = 597 MPa Maka : fR n = o,75 x 1000 x 448 /1000 = 336 kN Beban tarik yang dapat ditaham oleh sistem sambungan tersebut adalah sebesar 230 kN. Las kait pada pojok sambungan tidak termasuk dalam hitungan ini.
SAMBUNGAN LAS MENERIMA MOMEN Diasumsikan bahwa rotasi pada bidang patahan sambungan las terjadi disekitar pusat elastis las. Perbedaan dengan sambungan baut adalah pada sambungan las kekuatannya dihitung sebagai per-satuan panjang las. Tegangan geser pada las akibat momen, M adalah
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Contoh soal Tentukan ukuran las yang diizinkan untuk sambungan “bracket” seperti tergambar. Beban mati yang bekerja 50 kN dan beban hidup 120 kN secara secara terpusat (lihat gambar). Digunakan mutu baja A36 untuk bracket dan mutu baja A992 untuk kolom. Hitung untuk tebal las t eff = 25 mm Step 1 :Hitung beban ultimit: Pu = 1.2D + 1.6L = 1.2(50)+1.6(120) = 252 2 52 kN Step II:Hitung tegangan las akibat gaya geser :
× =360N/mm = ++ Step III: Hitung titik berat sambungan las:
̅̅ (700 700)) =200( =200(100 100)()(22) or ̅̅ =57,1 =57,1 mm Step IV: Hitung momen lentur pada titik berat las:
e = 250+ 200 – 57.1 57.1 = 392.9 M = Pe = 252(392.9)=99011 kN-mm.
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ΦRn= (0.9)0.6Fyt = 0.9(0.6)(248 0.9(0.6)( 248)) (15) = 2009 2009 N/mm Kontrol kekuatan pada sambungan las 2009 N/mm > 1703 N/mm Step VIII: Hitung tebal kaki las , asumsi Fw asumsi Fw = 0.6FEXX
= 1703 = (,707) 11,1,1 ,707) 0,75(5(0,707) 07)(0,6×482) = 11 Sehingga dapat dipakai tebal kaki las sudut 12 mm dengan E70XX