Beban Kuda-Kuda

3.6 PERHITUNGAN BEBAN PADA KUDA-KUDA

3.6.2 Beban Kuda-Kuda

51 Cek Kapasitas Penampang Pada Kondisi Patah

Ae = (85% × A gab) × 0,9 = (85% × 38 cm²) × 0,9 = 29,07 cm²

Pn = 0,75 × Ae × Fu

= 0,75 × 29,07 cm² × 5098,58 kg/cm² = 111162 > 271,32 (OK)

Batang Diagonal dengan profil siku 100 x 100 x 10 dengan t = 10 mm (OK)

52 – Berat batang tegak digunakan profil Double L ukuran 100 x 100 x 10 – Berat batang diagonal digunakan profil Double L ukuran 100 x 100 x 10 – Berat batang tepi digunakan profil Double L ukuran 150 x 150 x 19

No Batang Berat/Panjang Panjang Jumlah Berat Total

kg/m m buah Kg

1 Batang Tepi Atas 54,6 1,18 18 1159,704

2 Batang Tepi Bawah 54,6 1,18 18 1159,704

3 Batang Tegak 29,8 0.75 19 424,650

4 Batang Diagonal 29,8 1,16 18 622,224

Σ 3366,282

– Berat sambungan 5% kuda-kuda = 5% × 3366,282 kg

= 168,314 kg

– Berat Total kuda – kuda = 3366,282 kg + 168,282 kg

= 3534,596 kg

– Jumlah titik buhul = 38

– Beban yang diterima tiap titik buhul : 𝑝 =𝑏𝑒𝑟𝑎𝑡 𝑡𝑜𝑡𝑎𝑙 𝑘𝑢𝑑𝑎 − 𝑘𝑢𝑑𝑎

𝑗𝑢𝑚𝑙𝑎ℎ 𝑡𝑖𝑡𝑖𝑘 𝑏𝑢ℎ𝑢𝑙 𝑝 =3534,596 kg

38 𝑝 = 93,018 𝑘𝑔 Beban Hidup

RL = 2 × 67,8 kg = 135,6 kg Beban Mati

RD = 2 × 37,530 = 75,060 kg R = RL + RD

= (135,6 kg + 75,060 kg) = 210,660 kg

Bagian Tengah = 210,660 kg

53 Bagian Tepi =

2× 210,660 kg = 105,330 𝑘𝑔

Beban Angin

Angin Datang (Tekan) : Rw = 15,30 kg Angin Pergi (Hisap) :

Rw = 22,950 kg Beban Kombinasi 1 : 1,4D

Beban Kombinasi 2 : 0,9D + 1,0W

Beban Kombinasi 3 : 1,2D + 1,6L + 0,5W

Beban Komninasi 4 : 1,2D + 1,3W + 0,5(La atau H)

Gambar 3. 7 Nodal

Gambar 3. 8 Beam

54 Gambar 3. 9 Beban Mati

Gambar 3. 10 Beban Hidup

Gambar 3. 11 Beban Angin

55 Batang Tepi Atas

Gambar 3. 12 Gaya Tekan Batang Tepi Atas

Fy = 290 Mpa = 2957,18 kg/cm² Fu = 500 Mpa = 5098,58 kg/cm² p profil = 1,800 m = 118 cm

½ p profil = 0,59 cm = 59 cm

Dicoba Batang Diagonal dengan profil siku dobel 150 x 150, t = 12 mm

– A = 53,38 cm²

– T = 12 cm

– Lx = 1730 cm⁴

– Ly = 451 cm⁴

– Cx = cy = 4,52 cm – Ix max = 5,69 cm – Ix min = 2,91 cm – Ix = Iy = 4,52 cm

56 Untuk Profil Ganda :

A gab = 2 × 53,38 cm² = 106,76 cm² Ix gab = 2 × 1730 cm⁴ = 3460 cm⁴

Iy gab = (2 × Ly) + (A gab × (Cx=Xy + 0,75)²) = (2 × 451) + (106,76 × (4,52 + 0,75)²)

= 12717,172cm⁴

ix gab =

√

^{𝐼𝑥̅ 𝑔𝑎𝑏}_{𝐴 𝑔𝑎𝑏}

=

√

³⁴⁶⁰

106,76 = 5,693 cm iy gab =

√

^{𝐼𝑦 𝑔𝑎𝑏}

𝐴 𝑔𝑎𝑏

=

√

^21886,7883

106,76 = 14,318 cm

• Cek Penampang Batang Tekan Diambil Sampel Terbesar

Pu = 18422 kg

L = 1,18 m = 118 cm

• Cek Kelangsingan λ = ^L

i min < 200

= ¹¹⁸

14,318 < 200

= 8,24 < 200

• Perencanaan Plat Kopel Coba dengan 2 daerah L = ¹¹⁸

2 = 59 cm λ bat tunggal

L

ix min= ⁵⁹

2,91 = 20,27 < 40 (OK)

λ gab

L

iy gab = ¹¹⁸

14,318 = 8,24 < 200 (OK)

57 Flens

λr = 0,56 ×√²⁰⁰⁰⁰⁰

290 = 14,71 b/tf = ¹⁵⁰

12

= 12.5 < 14,71 (OK) (Penampang Langsing)

Web

λr = 1,49 ×√²⁰⁰⁰⁰⁰

290 = 39,13 b/tf = ¹⁵⁰⁻¹²

12

= 11.5 < 39,13 (OK)

(Penampang Tak Langsing)

λ < 4,71 × √²⁰⁰⁰⁰⁰

290

8,24 < 123,691 Fe = ^𝜋

2×𝐸 λ𝑔𝑎𝑏² = ^3,14

2×200000

8,45² = 29033,4569

Fcr = 0,658^(Fy/Fe) × Fy = 0,658(290/29033,4569) × 290

= 288,79 MPa → 2944,79 kg/cm²

∅pn = 0,9 × Fcr × A gab

= 0,9 × 2944,79 × 106,76

= 282947,49 > 18422 (OK)

Profil Siku Double 150 x 150 dengan t = 12 OK

58 Batang Tepi Bawah

Gambar 3. 13 Gaya Tarik dan Tekan Batang Tepi Bawah

Fy = 290 Mpa = 2957,18 kg/cm² Fu = 500 Mpa = 5098,58 kg/cm² p profil = 1,18 m = 118 cm

½ p profil = 0,59 cm = 59 cm

Dicoba Batang Diagonal dengan profil siku dobel 150 x 150, t = 12 mm

– A = 53,38 cm²

– T = 19 cm

– Lx = 1730 cm⁴

– Ly = 451 cm⁴

– Cx = cy = 4,52 cm – Ix max = 5,69 cm – Ix min = 2,91 cm – Ix = Iy = 4,52 cm

59 Untuk Profil Ganda :

A gab = 2 × 53,38 cm² = 106,76 cm² Ix gab = 2 × 1730 cm⁴ = 3460 cm⁴

Iy gab = (2 × Ly) + (A gab × (Cx=Xy + 0,75)²) = (2 × 451) + (106,76 × (4,52 + 0,75)²)

= 21886,7883 cm⁴

ix gab =

√

^{𝐼𝑥̅ 𝑔𝑎𝑏}_{𝐴 𝑔𝑎𝑏}

=

√

³⁴⁶⁰

106,76 = 5,693 cm iy gab =

√

^{𝐼𝑦 𝑔𝑎𝑏}

𝐴 𝑔𝑎𝑏

=

√

21886,7883

106,76 = 14,318 cm

• Cek Penampang Batang Tarik Diambil Sampel Terbesar

L = 1,18 m = 118cm Pu = 17585 kg

• Cek Kelangsingan λ = ¹¹⁸

14,318 = 8,241 < 300 (OK)

• Cek Kapasitas Penampang Pada Kondisi Leleh pn = 0,9 × Fy × A gab

= 0,9 × 2957,18 × 106,76

= 284137,683 > 17585 (OK)

• Cek Kapasitas Penampang Pada Kondisi Patah Ae = (85% × Agab) × 0,9

= (85% × 106,76) × 0,9 = 81,6714

Pn = 0,75 × Ae × Fu

= 0,75 × 81,6714 × 5098,58

60 Pn = 312306,125> 17585 (OK)

Profil Siku Double 150 x 150 dengan t = 19 OK Cek Penampang Batang Tekan

Diambil Sampel Terbesar Pu = 3406,4 kg

L = 1,18 m = 118 cm

• Cek Kelangsingan λ = ^L

i min < 200

= ¹¹⁸

14,318 < 200

= 8,24 < 200

• Perencanaan Plat Kopel Coba dengan 2 daerah L = ¹¹⁸

2 = 59 cm λ bat tunggal

L

ix min= ⁵⁹

2,91 = 20,27 < 40 (OK)

λ gab

L

iy gab = ¹¹⁸

14,318 = 8,24 < 200 (OK) Flens

λr = 0,56 ×√²⁰⁰⁰⁰⁰

290 = 14,71 b/tf = ¹⁵⁰

12

= 12.5 < 14,71 (OK) (Penampang Langsing)

Web

λr = 1,49 ×√²⁰⁰⁰⁰⁰

290 = 39,13 b/tf = ¹⁵⁰⁻¹²

12

= 11.5 < 39,13 (OK)

(Penampang Tak Langsing)

λ < 4,71 × √²⁰⁰⁰⁰⁰

290

61 Fe = ^𝜋

2×𝐸 λ𝑔𝑎𝑏² = ^3,14

2×200000

8,45² = 29033,4569

Fcr = 0,658^(Fy/Fe) × Fy = 0,658(290/29033,4569) × 290

= 288,79 MPa → 2944,79 kg/cm²

∅pn = 0,9 × Fcr × A gab

= 0,9 × 2944,79 × 106,76

= 282947,49 > 3406,4 (OK)

Profil Siku Double 150 x 150 dengan t = 12 OK Batang Diagonal

Gambar 3. 14 Gaya Tekan dan Tarik Batang Diagonal

Fy = 290 Mpa = 2957,18 kg/cm² Fu = 500 Mpa = 5098,58kg/cm² p profil = 1,16 m = 116cm

½ p profil = 0,58 m = 58 cm

Dicoba Batang Diagonal dengan profil siku dobel 100 x 100 dengan t = 10 mm

– A = 19 cm²

– T = 10 cm

– Lx = 278 cm⁴

– Ly = 72 cm⁴

– Cx = cy = 2,82 cm – Ix max = 2,83 cm – Ix min = 1,95 cm – Ix = Iy = 3,04 cm

62 A gab = 2 × 19 cm² = 38 cm²

Ix gab = 2 × 278 cm⁴ = 556 cm⁴

Iy gab = (2 × Ly) + (A gab × (Cx=Xy + 0,75)²)

= (2 × 72) + (38 × (2,82 +0,75 )²)

= 147,02cm⁴ ix gab =

√

^{𝐼𝑥̅ 𝑔𝑎𝑏}

𝐴 𝑔𝑎𝑏

=

√

⁵⁵⁶

38 = 3,8251 cm iy gab =

√

^{𝐼𝑦 𝑔𝑎𝑏}_{𝐴 𝑔𝑎𝑏}

=

√

^147,02

38 = 1,967 cm

• Cek Penampang Batang Tekan Diambil Sampel Terbesar

Pu = 2980,6

L = 1,16 m = 116 cm

• Cek Kelangsingan λ = ^L

iy gab < 300

= ¹¹⁶

1,967 < 300

= 58,97 < 300 (OK)

• Perencanaan Plat Kopel Coba dengan 2 daerah L = ¹¹⁶

2 = 58 cm λ bat tunggal

L

ix min= ⁶⁰

1,95 = 29,74 < 40 (OK) λ gab

Lk

ix gab = ^29,74

3,825 = 7,78 < 200 (OK)

63 λr = 0,56 ×√²⁰⁰⁰⁰⁰

290 = 14,706 b/tf = ¹⁰⁰

10

= 10 <

14,706

(OK) (Penampang tak Langsing)

Web

λr = 1,49 ×√²⁰⁰⁰⁰⁰

290 = 12,868 b/tf = ¹⁰⁰⁻¹⁰

10

= 9 <

12,868

(OK) (Penampang tak Langsing)

λ = 4,47 × √²⁰⁰⁰⁰⁰

290

= 117,388 Fe = ^𝜋

2×𝐸 λ𝑔𝑎𝑏² = ^3,14

2×200000

7,78² = 32613,2 Fcr = 0,658^(Fy/Fe) × Fy

= 0,658(290/32613,2) × 290

= 288,923 = 2944,793 kg/cm3

∅pn = 0,9 × Fcr × A gab

= 0,9 × 2944,793 × 38

= 1001711,9 > 196,3768 (OK)

Profil Siku Double 100 x 100 dengan t = 10 OK

• Cek Penampang Batang Tarik Diambil Sampel Terbesar

L = 1,16 m = 116 cm Pu = 4732,6 kg

• Cek Kelangsingan Λ = ¹¹⁶

1,967 = 58,974 < 300 (OK)

• Cek Kapasitas Penampang Pada Kondisi Leleh pn = 0,9 × Fy × A gab

= 0,9 × 2957,18 × 38

= 101135,6 > 4732,6 (OK)

64 Ae = (85% × Agab) × 0,9

= (85% × 38) × 0,9

= 29,07

Pn = 0,75 × Ae × Fu

= 0,75 × 29,07 × 5098,58

= 111161,8

Pn = 111161,8 > 4732,6 (OK)

Profil Siku Double 100 x 100 dengan t = 10 OK Batang Tegak

Gambar 3. 15 Gaya Tekan dan Tarik Batang Tegak

Mutu Baja BJ50 :

Fy = 290 Mpa = 2957,18 kg/cm² Fu = 500 Mpa = 5098,59 kg/cm² p profil = 0.75 m = 75 cm

½ p profil = 0,375 cm = 3,75 cm

Dicoba Batang Diagonal dengan profil siku dobel 100 x 100, t = 10 mm

– A = 19 cm²

– T = 10 cm

– Lx = 278 cm⁴

– Ly = 72 cm⁴

– Cx = cy = 2,82 cm – Ix max = 2,83 cm – Ix min = 1,95 cm – Ix = Iy = 3,04 cm

65 A gab = 2 × 19 cm² = 38 cm²

Ix gab = 2 × 278 cm⁴ = 556 cm⁴

Iy gab = (2 × Ly) + (A gab × (Cx=Xy + 0,75)²)

= (2 × 278) + (38× (2,82+ 0,75)²)

= 147,022 cm⁴ ix gab =

√

^{𝐼𝑥̅ 𝑔𝑎𝑏}

𝐴 𝑔𝑎𝑏

=

√

⁵⁵⁶

38 = 3,825 cm iy gab =

√

^{𝐼𝑦 𝑔𝑎𝑏}_{𝐴 𝑔𝑎𝑏}

=

√

^147,022

38 = 1,967 cm

• Cek Penampang Batang Tarik Diambil Sampel Terbesar L = 0,75 m = 75 cm Pu = 11355 kg

• Cek Kelangsingan λ = ⁷⁵

1,967 = 38,13 < 300 (OK)

• Cek Kapasitas Penampang Pada Kondisi Leleh Pn = 0,9 × Fy × A gab

= 0,9 × 2957,18 × 38

= 101135,6 > 11355 (OK)

• Cek Kapasitas Penampang Pada Kondisi Patah Ae = (85% × Agab) × 0,9

= (85% × 38) × 0,9

= 29,07

Pn = 0,75 × Ae × Fu

= 0,75 × 29,07 × 5098,6

= 111162

Pn = 111162 > 11355 (OK)

Profil Siku Dobel 100 x 100 dengan t = 10 mm (OK)

66 Diambil Sampel Terbesar

Pu = 2599 kg

L = 0,75 m = 75 cm

• Cek Kelangsingan λ = ^L

iy gab < 300

= ⁷⁵

1,967 < 300

= 38,13 < 300 (OK)

• Perencanaan Plat Kopel Coba dengan 2 daerah L = ⁷⁵

2 = 37,5 cm λ bat tunggal

L

ix min = ^37,5

1,95 = 19,23 < 40 (OK) λ gab

Lk

ix gab = ^19,23

3,825 = 5,0275 < 200 (OK) Flens

λr = 0,56 ×√²⁰⁰⁰⁰⁰

290 = 14,706 b/tf = ¹⁰⁰

10

= 10 <

14,706

(OK) (Penampang tak Langsing)

Web

λr = 1,49 ×√²⁰⁰⁰⁰⁰

290 = 12,868 b/tf = ¹⁰⁰⁻¹⁰

10

= 9 <

12,868

(OK) (Penampang tak Langsing)

λ = 4,47 × √²⁰⁰⁰⁰⁰

290

= 117,388

67 Fe =

λ𝑔𝑎𝑏² =

5,0275² = 78018 Fcr = 0,658^(Fy/Fe) × Fy

= 0,658(290/78018) × 290

= 289,55 Mpa = 2952,5 kg/cm3

∅pn = 0,9 × Fcr × A gab

= 0,9 × 2952,5 × 38

= 100975,5 > 2599 (OK)

Profil Siku Double 100 x 100 dengan t = 10

68 Data Kolom

L = 5,5 m = 550 cm E = 200000 Mpa

Fy = 290 Mpa = 2957,18 kg/cm²

Direncanakan kolom dengan profil WF 250 x 250 x 9 x 14 B = 250 mm = 25 cm

H =250 mm = 25 cm A = 119,8 cm²

Ix = 10800 cm⁴ Iy = 3650 cm⁴ ix = 10,8 cm iy = 6,29 cm t1 = 9 mm t2 = 14 mm r = 16 mm q = 72,4 kg/m zx = 967 cm³ zy = 292 cm³