tugas besar struktur baja 2 final ( ferdinand )

GAMBAR POTONGAN & DENAH KUDA-KUDA

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σy = 290 Mpa 2400 kg/cm2 < dilihat dari mutu baja BJ37

σy = ∂ ySF

=24001,5

=¿1600 kg/cm2

Perhitungan Sudut

Tan α= 16,524,25 = 0,6804

α = arch-1 0,6804 = 34,23°

Pembebanan Pada Gording

α = 34,23°

Coba profil Canal C = 200 . 75 . 8,5 berat sendiri = 29,4 kg/m (Try & Error )

Jarak antar Gording = 1 m

Beban Mati

Beban Atap (ᶐatap ) Atap yang dipilih Bluescope Lysaght TrimdekColorbound = 4,93 kg/mTebal = 0.48 mm

ᶐba = 4,93 x (0,5 x 1,65 x 2 ) x 1

= 8,15 kg/m

Berat Sendiri (ᶐbs ) = 25,3 kg/m Berat Sendiri (ᶐbm) = ᶐba + ᶐbs = 8,15 + 25,30 = 33,45 kg/m Beban Aksesoria (ᶐbA) = 25 % ᶐbm = 25 % x 33,45 = 8,36 kg/m Beban Mati total = ᶐbm + ᶐbA = 33,45 + 8,36 = 41,68 kg/m

ᶐbm Total = 41,68cos34,23 °= 48,68 kg/m

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Beban Hidup

Beban Orang = 100 kg Refernsi dari PBI ( Peratuan Pembebanan Indonesia)

Beban Hujan (W) = 20 kg

P = w x Jarak Kuda-kuda x Jarak antar gording= 20 x 6,8 x 2= 272 kg

Karena beban hujan ≥ beban orang maka, PBH = 272 kg

Px = Px . Sin α

= 272 Sin 34,23 = 153,004

Py = Py . Cos α

= 272 Cos 34,23 = 224

Pengaruh kemiringan Atap

ᶐbm x = ᶐbm x sin α

= 48,68 sin 34,23= 24,93 kg/m

ᶐbm y = ᶐbm x cos α

= 48,68 cos 34,23= 41,81 kg/m

Mbm x= ⅛ x ᶐbmx x (jarak kuda-kuda)2

= ⅛ x 24,93 x 6,82

= 144,09 kg/m

Mbm y= ⅛ x ᶐbmy x (jarak kuda-kuda)2 = ⅛ x 41,81 x 6,82

=271,43 kg/m

Mbh x =⅟4 x(Px) x (jarakkuda-kuda)

= ⅟4 x 153,004 x 6,8= 260,11 kg/m

Mbh y= ⅟4 x (P y) x (jarak kuda-kuda)

= ⅟4 x 224,89 x 6,8= 382,30 kg/m

Tegangan Lentur akibat Beban Atap

Mbt x = Mbmx + Mbhx

= 144,09 + 260,11

= 404,20 kg/m

Mbt y = Mbmy + Mbhy

= 271,43 + 382,30

= 653,73 kg/m

Tegangan Beban Atap Wx= 191 , Wy = 27

σbt x = Mbtx

Wy( table)= 404,20

27=1497

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σbt y = Mbty

Wx(table)=653,73

191=342

σ Total Atap = √ (δbt x )2+( δbt y )2

= √ (1497 )2+(342 )2

= 1535 ≤ σ’= 1600 kg/m ok

jadi profil yang dipakai adalah “C” 200 . 75 . 8,5

Tegangan Geser Tetap

Lx tetap = ½ . ᶐbm x . Ɩ + ½Px

= (½ . 24,93) x 6,8 + (½ . 153,004)

= 84,76 + 76,502

= 161,26

Ly tetap = ½ . ᶐbm y . Ɩ + ½Py

= (½ . 41,81) x 6,8 + (½ . 224,89)

= 142,15 + 112,45

= 254,59

d= 8,5 r=6 Wx = 191

Ix = 1910 Lx = 161,26 Wy = 27

Iy = 148 Ly = 254,59

Ʈ x =Lx .Wxd

10× Iy

= 161,26× 191

8,510

×148

= 244,84

Ʈy =Lx .Wxd

10× Iy

= 254,59× 278,510

×1910

= 5,99

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Ʈ Total = √ (δbs x )2+( δbs y )2

= √ (244,84 )2+(5,99 )2

= 244,91 ≤ Ʈ=0,6 . 1600

= 244,91 ≤ Ʈ=960

Periksa tegangan ijin ( Efek Huber Hangky) dalam keadaan tetap

σi = √ (δ )²+σ (Ʈ )²

= √ (1535,57 ) ²+3(244,91) ²

= 1584,212 ≤ δ = 1600 Kg/cm² ok

Jadi profil yang digunakan adalah canal “C” 200 . 75 . 8,5

Beban Sementara

Ci = 0,02 (α)-0,4

= 0,02 (34,32)- 0,4 = 0,2846 --> jika (-) tidak perlu dihitung

Beban angin = w angina x C x Jarak gording= 45 x 0,2846 x 6,8= 12,81 kg/m

Beban angin terpusat

Pa = ᶐa x Jarak kuda−kudacosα

= 12,81 x 6,8cos34.23 = 101,39

Max = 0May = ⅟4 x Pa x jarak kuda-kuda

= ⅟4 x 101,39 x 6,8

= 172,37

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Mbs x = Mbt y+ May= 404,20 + 0= 404,20

Mbs y = Mbt y+ May= 653,37 + 127,37= 826,10

Tegangan lentur akibat beban sementara

σbs x = Mbs x

Wy( table)= 404,20

27=14,97 x 100 = 1497

σbs y = Mbs y

Wy( table)=826,10

191=4,33 x 100= 433

σ Total sementara = √ (δbs x )2+( δbs y )2

= √ (1497 )2+(433 )2

= 1558 ≤ σ’ x 1,25

= 1558 ≤ 2000 ok

Tegangan geser sementara

Lx sementara= ½ . ᶐbm x . (jarak kuda-kuda) + ½Px= ½ . 24,93 . ( 6,8 ) + ½ 153,004= 161,264

Ly sementara= ½ . ᶐbm y . (jarak kuda-kuda) + Pa y . cosα= ½ . 41,81 . ( 6,8 ) +101,395 x 153,004 x cos34,23

= 229,24

Ʈ x =Lx .Wxd

10× Iy

= 161,26× 191

8,510

×148

= 244,84

Ʈy =Lx .Wxd

10× Iy

= 229,24 ×278,510

× 1910

= 5,40

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Ʈ Total = √ (δbs x )2+( δbs y )2

= √ (244,84 )2+(5,40 )2

= 244,89 ≤ Ʈ=0,6 . 1600

= 244,89 ≤ Ʈ=960

Periksa tegangan ijin ( Efek Huber Hangky) dalam keadaan tetap

σi = √ (δ )²+σ (Ʈ )²

= √ (1696,20 ) ²+3(244,89) ²

= 1748 ≤ δ = 0,6 . 1600 Kg/cm²

= 1748 ≤ δ = 2000 Kg/cm² ok

Periksa Kekuatan lendutan

Lendutan izin (δ’) = jarak kuda−kudaFy

=6,8× 100240

=2,833 cm

Beban terbagi rata qx= qbm x+qa x

= 24,93+0=24,93

qy= qbm y+qa y

= 41,81+172,37=214,18

Beban terpusat Gording

Px=153,004 Py=224,885

Lendutan pada Gording

δx = 5. qx

100.l ⁴

384. E . Iy+ 1. Px . l ³

48. E . Iy

=5(24 .10 ¯ ²)(6,8 .10) ⁴384.(2,1.10 ¯ 6) .1910

+ 1.224,805 .(680) ³48.(2,1. 10 ¯ 6). 1910

= 5,45

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δy = 5. qy

100.l ⁴

384. E . Ix+ 1. Py . l ³

48. E . Ix

=5(214,18 .10 ¯ ²)(680) ⁴384.(2,1. 10 ¯ 6) .1910

+1. 224,805 .(680) ³

48.(2,1. 10¯ 6).1910

= 1,86

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δ Total = √δx ²+δy ²

= √5 , 45²+1 ,86²

= 5,76 ≥ 2,83 Not ok

Karena Lendutan total lebih besar dari lendutan yang di izinkan, maka ada alternative lain untuk mengatasinya yaitu dengan menambahkan Track Stang. Ini lebih efisien dibandingkan dengan mengganti profil nya.

Jarak Track Stang

Jarak kuda-kuda = (JK . 100 )+103

=(6,8 .100 )+10

3=236,67

Lendutan Izin δ '=236,67240

=0,99

Lendutan pada Gording

δx = 5. qx

100. l4

384. E . Iy+ 1 . Px . l3

48. E . Iy

= 5(24 /100)(236,67) ⁴384.(2,1 .10 ¯ 6) .148

+ 1 .224,805.(236,67)³48.(2,1 .10 ¯ 6) .148

= 0,168

δy = 5. qx

100. l4

384. E . Ix+ 1. Py . l3

48. E . Ix

= 5(24/100)(236,67) ⁴384.(2,1. 10 ¯ 6).1910

+1 . 224,805 .(236,67) ³48.(2,1 . 10¯ 6) . 1910

= 0,037

δ Total = √δx ²+δy ²

= √0 , 168²+0 , 037²

= 0,172 ≤ δ’= 0,99 Ok

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PEMBEBANAN PADA KUDA – KUDA RANGKA BAJA

BEBAN ATAP

q atap = Berat atap x jarak kuda-kuda

q atap= 8.154 x1.25cos(34.23)

x 6.8=83.91

Pa 1 = Pa 11= ½ bentang x q atap= ½ x 3.25 x 83.91 = 136.35 kg

Pa 2 = Pa 10= {(½ bentang) + (½ bentang)} x q atap= {(½ x 3.255)+ (½ x 6)} x 83.91 = 388.08 kg

Pa 3 = Pa 4 = Pa 8 = Pa 9= {(½ bentang) x 2} x q atap= {(½ x 6) x 2} x 83.91 = 503.46 kg

Pa 5 = Pa 7= {(½ bentang + ½ bentang)} x q atap= {(½ x 6 + ½ x 3)} x 83.91 = 377.59 kg

Pa 6 = {(½ atap) x 2} x q atap= {(½ x 3) x 2} x 83.91 = 251.73 kg

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BEBAN GORDING

G = qbs gording (berat canal) x jarak kuda-kuda

G = 25.3 x 6.8

G = 172.04 kg

PG 1 = PG 11= (2 x ½ ) x G = 172.04 kg

PG 2 = PG 10= {(2 x ½) + (3 x ½)} x G = 430.1 kg

PG 3 = PG 9= {(3 x ½) + (4 x ½)} x G = 602.14 kg

PG 4 = PG 8={(4 x ½) + (4 x ½)} x G = 688.16 kg

PG 5 = PG 7= {(4 x ½) + (2 x ½)} x G = 516.12 kg

PG 6 = {(2 x ½) + (2 x ½)} x G = 344.08 kg

BEBAN HIDUP / BEBAN HUJAN

qh = W hujan x Jarak kuda – kuda

= 20 x 6.8

qh = 136 kg

Ph 1 = Ph 11= qh x (3.25 x ½) = 221 kg

Ph 2 = Ph 10= {(3.25 x ½) + (6 x ½)} x qh = 629 kg

Ph 3 = Ph 4 = Ph 9 = Ph 8= {(6 x ½) + (6 x ½)} x qh = 816 kg

Ph 5 = Ph 7= {(6 x ½) + (3 x ½)} x qh = 612 kg

Ph 6 = {(3 x ½) + (3 x ½)} x qh = 408 kg

BEBAN TOTAL / BEBAN TETAP TOTAL

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PBT = Pa + Pg + Ph

- PBT 1 = PBT 11 = Pa + Pg + Ph = 136.3538 + 172.04 + 221 = 529.3938 kg

- PBT 2 = PBT 10 = Pa + Pg + Ph = 388.0838 + 430.1 + 629 = 1447.1838 kg

- PBT 3 = PBT 9 = Pa + Pg + Ph = 503.46 + 602.14 + 816 = 1921.6 kg

- PBT 4 = PBT 8 = Pa + Pg + Ph = 503.46 + 688.16 + 816 = 2007.62 kg

- PBT 5 = PBT 7 = Pa + Pg + Ph = 377.595 + 516.12 + 612 = 1505.715 kg

- PBT 6 = Pa + Pg + Ph = 251.73 + 34.08 + 408 = 1003.81 kg

GAMBAR PEMBEBANAN UNTUK PEMBEBANAN BEBAN TETAP KUDA-KUDA

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MENCARI BEBAN SENDIRI

Menentukan Profil Awalo Periksa Angka Kelangsingan (λ)

Kestabilan = x < 200

X = Limin ; imin = L

λ ( Batang terpanjangketentuan kestabilan

)

imin = 1002.63200

=5.01315

Karena yang dipakai menggunakan profil ganda semua, maka imin = ix = iy dan dicoba menggunakan profil siku 180.180.16

Ix = Iy = 5.51 ≤ imin OK

Panjang Batang Luar + Panjang Batang Dalamo Panjang Batang Luar = Panjang (∑A + ∑B)

= a1+a2+a3+a4+a5+a6+a7+a8+a9+a10+b1+b2+b3+b4+b5+b6

= 393.10 + (725.72 x 3) + (361.86 x 2) + (725.72 x 3) + 393.10 + (647.06 x 2) + (621.18 x 6)

= 10887.44100 = 108.874 m

Panjang Batang Luar = 108.874 x 2 = 217.748 m

o Panjang Batang Dalam = d1+d2+d3+d4+d5+d6+d7+d8+d9+d10+d11+d12+d13+d14

= 304.75 + 550.74 + 424.99 + 770.11 + 625.14 + 1002.63 + 850.54 + 850.54 + 1002.63 + 625.14 + 770.11 + 424.99 + 550.74 + 304.75 + 1000

= 10057.8100 = 100.578 m

Panjang Batang Dalam = 100.578 x 2 = 201.156 m

Panjang Total = 217.748 + 201.156 = 418.904 m

qbs = Panjang total xberat profil x berat aksesorispanjang batang

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= 418.904 x 43.5 x1.2548.5 = 469.647 kg/m

PBS = qbs x L (Panjang batang x ½)

- PBS 1 = PBS 9 = qbs x L= 469.647 x (6.25 x ½ ) = 1467.6468 kg

- PBS 2 = PBS 8 = qbs x L= 469.647 x {(6.25 x ½ )+(6 x ½ )} = 2876.5878 kg

- PBS 3 = PBS 7 = qbs x L= 469.647 x {(6 x ½ )+(6 x ½ )} = 2817.882 kg

- PBS 4 = PBS 6 = qbs x L= 469.647 x {(6 x ½ )+(6 x ½ )} = 2817.882 kg

- PBS 5 = qbs x L= 469.647 x x {(6 x ½ )+(6 x ½ )} = 2817.882 kg

GAMBAR PEMBEBANAN UNTUK PEMBEBANAN BEBAN SEMENTARA KUDA-KUDA

Berat Sendiri Kuda-Kuda

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qbs = Panjang total xberat profil x berat aksesorispanjang batang

= 418.904 x 43.5 x1.2548.5 = 469.647 kg/m

Pemilihan Profil

Batang Terpanjang (lk) = 10.02 m = 1002.63i min

≤ 200

= 1002.63200

=5.01315≤ i

(i yang di gunakan = 5.51) = 1002.63240

=4.177 ≤ i

BEBAN AKIBAT ANGIN KIRI

MUKA

q angin = W angin x c x Jarak kuda−kudacos α ; α = 34.23O ; C = 0.2846

= 45 x o.2846 x 6.8cos 34.23°

q angin = 105.3326 kg/m

o P 1 = q angin x ½ L= 105.3326 x (3.931 x ½ ) = 207.031 kg

o P 2 = q angin x (½ L + ½ L)= 105.3326 x {(3.931 x ½) + (7.257 x ½)} = 589.230 kg

o P4 & P3 = q angin x 2 x ½L = 105.3326 x 2 x (7.252 x ½) = 764.3986 kg

o P5 = q angin x (½ L + ½ L)= 105.3326 x {(7.257 x ½) + (3.628 x ½)} = 573.5726 kg

o P6 = q angin x ½ L= 105.3326 x (3.628 x ½) = 191.0733 kg

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BELAKANG


= 45 x o.4 x6.8cos34.23°

q angin = 148.0431 kg/m

o P6 = q angin x ½ L= 148.0431 x (3.628 x ½) = 268.550 kg

o P7 = q angin x (½ L + ½ L)

= 148.0431 x {(7.257 x ½) + (3.628 x ½)} = 805.724 kgo P8 & P9 = q angin x (½ L + ½ L)

= 148.0431 x {(7.257 x ½) + (7.257 x ½)} = 1074.3488 kgo P10 = q angin x (½ L + ½ L)

= 148.0431 x {(7.257 x ½) + (3.931 x ½)} = 788.1815 kgo P11 = q angin x ½ L

= 148.0431 x (3.931 x ½) = 290.9787 kg

GAMBAR PEMBEBANAN UNTUK PEMBEBANAN BEBAN ANGIN KIRI

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BEBAN AKIBAT ANGIN KANAN

MUKA


= 45 x o.2846 x 6.8cos 34.23°

q angin = 105.3326 kg/m

o P 11 = q angin x ½ L= 105.3326 x (3.931 x ½ ) = 207.031 kg

o P 10 = q angin x (½ L + ½ L)= 105.3326 x {(3.931 x ½) + (7.257 x ½)} = 589.230 kg

o P8 & P9 = q angin x 2 x ½L = 105.3326 x 2 x (7.252 x ½) = 764.3986 kg


o P6 = q angin x ½ L= 105.3326 x (3.628 x ½) = 191.0733 kg

BELAKANG


= 45 x o.4 x6.8cos34.23°

q angin = 148.0431 kg/m

o P6 = q angin x ½ L= 148.0431 x (3.628 x ½) = 268.550 kg


o P3 & P4 = q angin x (½ L + ½ L)= 148.0431 x {(7.257 x ½) + (7.257 x ½)} = 1074.3488 kg


o P1 = q angin x ½ L

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= 148.0431 x (3.931 x ½) = 290.9787 kg

GAMBAR PEMBEBANAN UNTUK PEMBEBANAN BEBAN ANGIN KANAN

KONTROL MOMEN

1. Beban Total / Beban Tetap TotalKontrol ∑M = 0 , terhadap beban tetap (PBT)

o ∑MA = 0-Vb x 48.50 + P1 x 0 + P2 x 3.25 + P3 x 9.25 + P4 x 15.25 + P5 x 21.25 + P6 x 24.25 + P7 x 27.25 + P8 x 33.25 + P9 x 39.25 + P10 x 45.25 + P11 x 48.5 = 0

-Vb x 48.50 = -383809.6

Vb = 383821.648.5

Vb = 7913.8 kg

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o ∑MB = 0Va x 48.50 - P11 x 0 - P10 x 3.25 - P9 x 9.25 - P8 x 15.25 - P7 x 21.25 - P6 x 24.25 - P5 x 27.25 - P4 x 33.25 - P3 x 39.25 - P2 x 45.25 - P1 x 48.5 = 0

Va x 48.50 = 383765.9

Va = 383765.948.5

Va = 7912.7 kg

Va + Vb = P1 + P2 + P3 + P4 + P5 + P6 + P7 + P8 + P9 + P10 + P1115826.5 = 15826.8 = 0.3 ≈ 0 !OK!

2. Beban Sendirio ∑MA = 0-Vb x 48.50 + P1 x 0 + P2 x 6.25 + P3 x 12.25 + P4 x 18.25 + P5 x 24.25 + P6 x 30.25 + P7 x 36.25 + P8 x 42.25 + P9 x 48.25 = 0

-Vb x 48.50 = -552363.57

Vb = 552363.5748.5

Vb = 11388.94 kg

o ∑MB = 0Va x 48.50 - P9 x 0 - P8 x 6.25 - P7 x 12.25 - P6 x 18.25 - P5 x 24.25 - P4 x 30.25 - P3 x 36.25 - P2 x 42.25 - P1 x 48.25 = 0

Vb x 48.50 = 552363.57

Va = 552363.5748.5

Va = 11388.94 kg

Va + Vb = P1 + P2 + P3 + P4 + P5 + P6 + P7 + P8 + P9 + P10 + P11

22777.88 = 22777.87 = 0.01 ≈ 0 !OK!

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PERHITUNGAN GAYA BATANG RANGKA KUDA-KUDA

1. Gambar Rangka Kuda Kuda Baja

2. Pendimensian Rangka batang dalam Perhitungan SAP 2000

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3. Pembebanan pada Rangka Kuda Kuda (Pada Joint)

3.1 Pembebanan PBT (Pembebanan Beban Tetap)

3.2 Pembebanan PBS (Pembebanan Beban Sementara)

3.3 Pembebanan Angin Kiri

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3.4 Pembebanan Angin Kanan

4. Reaksi Batang yang ditimbulkan yang terinput di SAP 2000

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4.1 Akibat Load Dead

4.2 Akibat Load Live

4.3 Akibat Left Load Wind

4.4 Akibat Right Load Wind

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4.5 Akibat Beban SDL (Kombinasi )

Kombinasi 1,2, dan 3

Kombinasi 1

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Kombinasi 2

Kombinasi 3

Dari tabel combo d SAP dicari batang tekan & tarik, yg terbesar, profil ganda

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DESAIN BATANG TARIK

Dari tabel kombinasi pembebanan didapat

27

No Batang Station (m) OutputCase Gaya Tekan (-) Gaya Tarik

(+)A1 3.93585 Comb 2 -15672.71A2 7.22578 Comb 1 -19096.07A3 7.22578 Comb 1 -16561.68A4 7.25578 COMB1 -13928.74A5 3.62789 COMB1 -11085.66A6 3.62789 COMB1 -11085.66A7 7.25578 COMB1 -13928.74A8 7.25578 COMB1 -16561.68A9 7.25578 COMB1 -19096.07A10 3.93585 COMB3 -21163.73B1 6.47185 COMB1 18029.59B2 6.20455 COMB1 15670.37B3 6.2201 COMB1 13346.11B4 6.16117 COMB1 10677.28B5 6.16117 COMB1 10677.28B6 6.2201 COMB1 13346.11B7 6.20455 COMB1 15670.37B8 6.47185 COMB1 18029.59D1 3.04821 COMB1 -1646.54D2 5.50858 COMB2 1239.01D3 4.27102 COMB2 -2194.59D4 7.72622 COMB2 2253.65D5 6.24743 COMB1 -2823.45D6 10.01966 COMB2 3893.66D7 8.694 COMB2 -3734.23D8 8.694 COMB3 -5066.18D9 10.01966 COMB3 5160.34D10 6.24743 COMB3 -4197.07D11 7.72622 COMB3 3224.57D12 4.27102 COMB3 -3060.73D13 5.50858 COMB3 1696.63D14 3.04821 COMB3 -2489.86V1 10.2 COMB1 11548.76

585

586

587

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589

590

591

592

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596

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598

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600

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602

603

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606

607

608609610611612613614615616617

Desain batang tarik (profil ganda)Beban terbesar Batang B1 = 18.029,6 kg

Panjang batang = 6,50 m

Data-data profil → L 18 0. 18 0.1 6 A = 55,4 IX = IY = 1680 cm4

imin = iη = 3,5 cm ix = iy = 5,51 cm ex = ey = 50,2 mm = 5,02 cm = 0,0502 m

a = 2ey + t

= (2 x 5,02) + 1.9

= 11.94 cm

Sumbu bahan → sb.x= 5,51 → ix

Sumbu bebas bahan → sb.y = iy = √ ( iy )2+(12

a)2

= √ (5,51 )2+ (5,97 )2

= 8,12 cm

Diambil → ix = 5,51

Periksa Kestabilan

λx = LKimin ≤ 240 ; λy =

LKimin ≤ 240

λx = 6505,51 ≤ 240 ; λy =

6508.12 ≤ 240

= 117,9 ≤ 240 ….. ok ; = 80.05 ≤ 240 …. ok

28

618619

620621622623624625626627628

629

630

631

632

633

634

635

636

637

638

639

640

641

642

643

644

645

646

Periksa Kekuatan

tr = P

2x An ≤ (dimana = 1600 kg/cm2)

Dimana → An = 80% x 55,4 = 44,32

tr = 18.029,6

88,64 ≤

= 203.4 ≤ 1600 ….. ok

BATANG TEKAN

Dari tabel kombinasi pembebanan didapat

Desain batang tekan (profil ganda)Beban terbesar Batang A10 = -21085.9 kg

Panjang batang = 3.93 m

Data-data profil → L 18 0. 18 0.1 6 A = 55,4 IX = IY = 1680 cm4

imin = iη = 3,50 cm ix = iy = 5,51 cm ex = ey = 502 mm = 5,02 cm = 0,0502 m Iη = 679 cm2

a = 2ey + t

= (2 x 5,02) + 1.9

= 11.94 cm

Periksa Kekuatan

P . ω

A ≤

→ = = 1,21

Karena = 1.21 maka

29

…Ok.

647

648

649

650

651

652

653

654

655656

657658659660661662663664665666

667

668

669

670

671

672

673

674

= 1,25 (1,21)2

= 1,83

Periksa Kestabilan

λx = LKimin ≤ 200

λx = 3933.5 ≤ 200

= 112.3 ≤ 200 ….. ok

Sumbu bebas bahan → sb.y = iy = √ ( iy )2+(12

a)2

= √ (5,51 )2+ (5,97 )2

= 8,12

→ y = lkiy ≤ 200

= 3935,51 ≤ 200

= 71.32 ≤ 200 ….. ok

Desain Plat koppel

w² = x²

→ x² = I² +y²

I = √ (112.3)2−(71.32)²

I = 86.7

ii = 3,5 cm

i1 = I . i1

= 86.7 x 3,5 = 303.45 cm

30

675

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678

679

680

681

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683

684

685

686

687

688

689

690

691

692

693

694

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696

697

698

Jumlah medan

n 393

303.45

n = 1,29 pembulatan menjadi 3

lI = 3933 = 131 cm

Periksa angka kelangsingan terhadap sumbu bebas bahan

1 = l 1imin

= 1313,5 = 37,43

w = iy = √ ( y )2+1²

= √ (71.32 )2+ (37,43 )2

= 80,54

Hitung lebar plat Koppel ; (dicoba t = 1/4 “ = 0.6 cm)

112 . t . h3

10 . I .2al i

112 . t . b3

10 .679 .(2 ×5,97)131

b3 618,874 x12

0.6

b 23, 131

H = (2 x 18) + 1.6 = 37.6

ukuran plat Koppel

H x b x t = 37,6 x 23, 131 x 0,6

31

699

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701

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718

719

720

721

Periksa kekuatan plat Koppel

ukuran plat Koppel : 37,5 x 20.04 x 0,6

lI = 131 cm

w = 80,54

Mencari besarnya Q

Q = 2% x 21085.9 kg = 421.7 kg

Atau → Q = ww131 x 21085.9

Ww …..?

c = w❑ √ fy

E

= 80 ,54❑ √ 2400

2,1 x10⁶ = 0.867

Jadi : 0,25 c 1,2

ww= 1,43

1,6−0,6 c

ww= 1,43

1,6−0,6 ( 0.91 ) = 1.3243

jadi →Q = ww131 x 21085.9

Q = 1.3243

131 x 21085.9 = 213.163 kg

Tentukan gaya geser (T) atau (L)

T = Q2a

= 145.9 x 131

2 x5,97 = 2463 kg

32

722

723

724

725

726

727

728

729

730

731

732

733

734

735

736

737

738

739

740

741

742

743

= Lsxbix

= 32 x

Tb xh (1600 x 0,6 = 960)

= 32 x

246320.04 x37,5 ( 960)

= 4.82 ….. ok

Periksa kekuatan batang tekan

w x

90.76 112.3….. ok

Periksa kritis (c)

w = 90.76 → c = w❑ √ FY

E

= 90.76❑ √ 2100

2,1 .10⁶ = 0.91

Jadi : 0,25 c 1,2

Sehingga → ω = 1,43

1,6−0,67 c

ω = 1,43

1,6−0,67 (0.91) = 1,44

cek= ρ . w2x A (1600)

= 21085.9 x 1,44

2× 55,4 1600

= 274.04 1600….. ok

Jadi profil 10.200.18 bisa dipakai

33

744

745

746

747748

749

750

751

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753

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757

758

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760

761

762

763

764

765

Detail SambunganPada sambungan desain kuda-kuda ini menggunakan jenis sambungan Las yang digunakan sebagai pengikat antara profil kuda-kuda dan plat kopel yang terdapat di desain kuda kuda ini.

P = P4 ; P = gaya batang

A=Px √sin α+¿¿¿¿ ; α = sudut

a= t√2

; t = 1.6 ; a = 1.13

Ln = Aa

Lbr = Ln + 3a

Lbr total = Lbr x 4

DETAIL BATANG A1

P = 15672.71

4=3918.18

A = 3918.18 x√sin 34.23+¿¿¿¿ = 3.77

ln = 3.771.13

=3.33

lbr = 3.33 + (3 x 1.13) = 6.72

Lbr total Batang A1 = 6.72 x 4 = 26.88 ~ 27

DETAIL BATANG A2

34

766

767

768

769

770

771772

773

774775776

777

778

779

780

781782783

784

785

786

787

788789790

791

P = 19096.07

4=4774.02

A = 4774.02 x √sin 34.23+¿¿¿¿ = 4.59

ln = 4.591.13

=4.06

lbr = 4.06+ (3 x 1.13) = 7.45

Lbr total Batang A2 = 7.45 x 4 = 29.8 ~ 30

DETAIL BATANG B1

P = 18029.59

4=4507.40

A = 4507.40 x √sin 19.23+¿¿¿¿ = 4.70

ln = 4.701.13

=4.15

lbr = 4.15+ (3 x 1.13) = 7.55

Lbr total Batang B1 = 7.55 x 4 = 30.2 ~ 31

DETAIL BATANG B2

P = 15670.37

4=3917.59

A = 3917.59 x√sin 19.23+¿¿¿¿ = 4.08

ln = 4.081.13

=3.61

lbr = 3.61+ (3 x 1.13) = 7.00

Lbr total Batang B2 = 7.00 x 4 = 28

35

792

793

794

795796

797798

799

800

801

802803

804

805

806

807

808

809810

811812

TABEL DATA HASIL PERHITUNGAN SAMBUNGAN LAS

36

αno. Batang

A1A2

813814815816817818819

α 56.99no. Batang gaya batang P bar Á ln lbr

D2 1239.01 309.75 0.33 0.29 3.69 4.00D13 1696.63 424.16 0.45 0.40 3.80 4.00


D3 2194.59 548.65 0.59 0.52 3.91 4.00D12 3060.73 765.18 0.82 0.72 4.12 5.00


D4 2194.59 548.65 0.59 0.52 3.91 4.00D11 3060.73 765.18 0.82 0.72 4.12 5.00


D5 2823.45 705.86 0.76 0.67 4.06 5.00D10 2823.45 705.86 0.76 0.67 4.06 5.00


D6 3893.66 973.42 1.04 0.92 4.32 5.00D9 5160.34 1290.09 1.38 1.22 4.62 5.00


D7 3734.23 933.56 1.00 0.88 4.28 5.00D8 3567.29 891.82 0.96 0.84 4.24 5.00


V1 11548.76 2887.19 3.09 2.73 6.13 7.00

37

820

822

GAMBAR KUDA – KUDA BAJA TYPE SCISSORS BENTANG 48,5 M

38

35

823824825826827828829830831832833834835836837838839840841842843844845846847

39

36848849850851852853854855856857858859860861862863864865866867868869870871872

40

37873874875876877878879880881882883884885886887888889890891892893894895896897

41

38

898899900901902903904905906907908909910911912913914915916917918919920921922

tugas besar struktur baja 2 final ( ferdinand )

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