Nama
Nama : M. ALFIANSYAH SYUKUR: M. ALFIANSYAH SYUKUR NIM
NIM : D111 10 902: D111 10 902
JURUSAN SIPIL FAKULTAS TEKNIK UNIVERSITAS HASANUDDIN JURUSAN SIPIL FAKULTAS TEKNIK UNIVERSITAS HASANUDDIN
Dik
Diketaetahui hui suasuatu tu ranrangka gka kakap p dendengan gan datdata a sebsebagaagai i berberikuikut! t! (P (P daldalam am tonton)) Keterangan:Keterangan:
((L L ddaallaam m mmeetteerr) ) :: Mohon diisi Mohon diisi Beban (P)
Beban (P) :: maks. 58 tonmaks. 58 ton
P
P3 3 = = 1122..5 5 ttoonn Panjang per segmenPanjang per segmen :: maks. 14.5 meter maks. 14.5 meter Profil yang digunakan
Profil yang digunakan :: Profil L sama kaki Profil L sama kaki Mutu baja yg tersedia
Mutu baja yg tersedia :: BJ 33BJ 33 BJ 37 BJ 37 BJ 44 BJ 44 P P2 2 = = 1122..5 5 ttoon n PP4 4 = = 1122..5 5 ttoonn BJ 52BJ 52 P P1 1 = = 1122..5 5 ttoon n PP5 5 = = 1122..5 5 ttoonn 30 30 2 2..5 5 m m 22..5 5 m m 22..5 5 m m 22..5 5 mm
Menghitung Reaksi Perletakan Menghitung Reaksi Perletakan
∑MA ∑MA = = 00 --RRBBV V 110 0 + + P5 P5 110 0 + + PP4 4 77..5 5 + + PP3 3 5 5 + + PP2 2 22..5 5 = = 00 --RRBBV V 110 0 + + 1122..5 5 110 0 + + 1122..5 5 7..5 75 + + 1122..5 5 5 5 + + 1122..5 5 22..5 5 = = 00 --RRBBVV 1100 ++ 112255 ++ 9933..7755 ++ 6622..55 ++ 3311..2255 == 00 --RRBBV V 110 0 = = - - 331122..55 R RBBV V = = 3311..225 5 ttoonn ∑MB ∑MB = = 00 R RAAVV 1100 -- PP11 1100 -- PP22 77..55 -- PP33 55 -- PP44 22..55 == 00 R RAAV V 110 0 - - 1122..5 5 110 0 - - 1122..5 5 7..5 75 - - 1122..5 5 5 5 - - 1122..5 5 22..5 5 = = 00 R RAAVV 1100 -- 121255 -- 9933..7755 -- 6622..55 -- 3311..2255 == 00 R RAAV V 110 0 = = 331122..55 R RAAV V = = 3311..225 5 ttoonn KONTROL KONTROL ∑V ∑V = = 00
TUGAS FINAL STRUKTUR BAJA I
TUGAS FINAL STRUKTUR BAJA I
LL11 LL22 LL33 LL44 A A BB C C D D E E F F G G HH S1 S1 S2 S2 S3S3 S4 S4 S5 S6 S7 S8 S5 S6 S7 S8 S9 S9 S10 S10 S12S12 S11 S11 S13 S13
RAV + RBV - P1 - P2 - P3 - P4 - P5 = 0
31.25 + 31.25 - 12.5 - 12.5 - 12.5 - 12.5 - 12.5 = 0
0 = 0 OK
Menghitung Gaya-Gaya Batang
TITIK BUHUL A ∑V = 0 RAV - P1 + S1 sin 30 = 0 P1 = 12.5 ton 31.25 - 12.5 + S1 0.5 = 0 S1 0.5 = -18.75 30 S1 = -37.5 ton ∑H = 0
RAV = 31.25 ton S1 cos 30 + S5 = 0
-37.5 0.866 + S5 = 0 -32.476 + S5 = 0 S5 = 32.476 ton TITIK BUHUL F ∑V = 0 S9 = 0 ∑H = 0 -S5 + S6 = 0 S5 = 3 2.4 76 to n -32.476 + S6 = 0 S6 = 32.476 ton TITIK BUHUL C ∑V = 0
S1 sin 30 - P2 + S2 sin 30 - S10 sin 30 = 0
P2= 12.5 ton 37.5 0.5 - 12.5 + S2 0.5 - S10 0.5 = 0
18.75 - 12.5 + S2 0.5 - S10 0.5 = 0
S2 0.5 - S10 0.5 = -6.25
S1 = 37.5 ton ∑H = 0
S1 cos 30 + S2 cos 30 + S10 cos 30 = 0
37.5 0.866 + S2 0.866 + S10 0.866 = 0 30 32.476 + S2 0.866 + S10 0.866 = 0 S2 0.866 + S10 0.866 = -32.476 S1 S5 F S6 S9 C S2 S9 S10 A
METODE ELEMINASI S2 0.866 + S10 0.866 = -32.476 x 1 S2 0.866 + S10 0.866 = -32.476 S2 0.5 - S10 0.5 = -6.25 x 1.732 S2 0.866 - S10 0.866 = 10.825 -S10 1.732 = -21.651 S10 = -12.5 ton S2 0.5 - S10 0.5 = -6.25 S2 0.5 - -12.5 0.5 = -6.25 S2 0.5 + 6.25 = -6.25 S2 0.5 = -12.5 S2 = -25 ton TITIK BUHUL D P3 = 12.5 ton ∑V = 0 S2 sin 30 - P3 - S11 - S3 sin 30 = 0 25 0.5 - 12.5 - S11 - S3 0.5 = 0 12.5 - 12.5 - S11 - S3 0.5 = 0 S2 = 25 ton - S11 - S3 0.5 = 0 ∑H = 0 30 S2 cos 30 + S3 cos 30 = 0 25 0.866 + S3 0.866 = 0 21.651 + S3 0.866 = 0 S3 0.866 = -21.651 S3 = -25 ton - S11 - S3 0.5 = 0 - S11 - -25 0.5 = 0 - S11 + 12.5 = 0 - S11 = -12.5 S11 = 12.5 ton D S3 S11
TITIK BUHUL G S11 = 12.5 ton ∑V = 0 -S10 sin 30 + S11 + S12 sin 30 = 0 -12.5 0.5 + 12.5 + S12 0.5 = 0 -6.25 + 12.5 + S12 0.5 = 0 S12 0.5 = -6.25 S10 = 12.5 ton S12 = -12.5 ton ∑H = 0 -S6 + S10 cos 30 + S12 cos 30 + S7 = 0 -32.476 + 12.5 0.866 + -12.5 0.866 + S7 = 0 30 -32.476 + 10.825 + -10.825 + S7 = 0 S6 = 32.476 ton S7 = 32.476 ton
REKAPITULASI GAYA-GAYA BATANG
Nama Tekan Tarik Panjang Batang (ton) (ton) (meter)
S1 37.5 2.887 S2 25 2.887 S3 25 2.887 S4 37.5 2.887 S5 32.476 2.500 S6 32.476 2.500 S7 32.476 2.500 S8 32.476 2.500 S9 0 0 1.443 S10 12.5 2.887 S11 12.5 2.887 S12 12.5 2.887 S13 0 0 1.443 S7 S12
BATANG BAWAH (S5, S6, S7, S8) BATANG TEGAK (S9, S11, S13)
Direncanakan menggunakan baja BJ 33 Silakan pilih Direncanakan menggunakan baja BJ 33 Silakan pilih
fy = 200 MPa = 2000 kg/cm² fy = 200 MPa = 2000 kg/cm²
P max = 32.48 ton = 32476 kg P max = 12.50 ton = 12500 kg
Lk = 2.50 meter = 250.00 cm Lk = 2.89 meter = 288.68 cm
32476 12500
2000 2000
= 16.24 cm² = 6.25 cm²
Digunakan profil L sama kaki 75.75.12 Digunakan profil L sama kaki 55.55.6
Luas penampang profil (Ag) 16.7 cm² lihat Buku Teknik Sipil, Hal 255-258 Luas penampang profil (Ag) 6.31 cm² lihat Buku Teknik Sipil, Hal 255-258
Jari-jari inersia (ix) 2.22 cm Jari-jari inersia (ix) 1.66 cm
Kontrol Kelangsingan Batang Kontrol Kelangsingan Batang
λ < 240 λ < 240 Lk Lk ix ix 250 288.675 2.22 1.66 112.6126 < 240 OK 173.9007 < 240 OK A
PERHITUNGAN BATANG TARIK
= P max = fy A = = < 240 P max fy < 240 < 240 < 240 33 33
BATANG ATAS (S1, S2, S3, S4) BATANG DIAGONAL TENGAH (S10, S12)
Direncanakan menggunakan baja BJ 33 Silakan pilih Direncanakan menggunakan baja BJ 33 Silakan pilih
fy = 200 MPa = 2000 kg/cm² fy = 200 MPa = 2000 kg/cm²
P max = 37.50 ton = 37500 kg P max = 12.50 ton = 12500 kg
Lk = 2.89 meter = 288.68 cm Lk = 2.89 meter = 288.68 cm
Imin = 1.69 x P max x Lk² Imin = 1.69 x P max x Lk²
= 1.69 x 37.50 x 2.89 ² = 1.69 x 12.50 x 2.89 ²
= 528.13 cm⁴ = 176.04 cm⁴
Digunakan profil L sama kaki 130.130.14 Digunakan profil L sama kaki 100.100.10
Momen Inersia (IX) 540 cm⁴ lihat Buku Teknik Sipil, Hal 255-258 Momen Inersia (IX) 177 cm⁴ lihat Buku Teknik Sipil, Hal 255-258
Luas Penampang Profil (Ag) 34.7 cm² Luas Penampang Profil (Ag) 19.2 cm²
Jari-Jari Inersia (ix) 3.94 cm Jari-Jari Inersia (ix) 3.04 cm
Menentukan faktor tekuk (ω) Menentukan faktor tekuk (ω)
Lk Lk
ix ix
288.68 288.68
3.94 3.04
= 73.27 ≈ 74 = 94.96 ≈ 95
Berdasarkan nilai λ di atas, diperoleh faktor tekuk (ω) = 1.432 Berdasarkan nilai λ di atas, diperoleh faktor tekuk (ω) = 1.736
(lihat Buku Teknik Sipil, Hal 321-325) (lihat Buku Teknik Sipil, Hal 321-325)
Kontrol Kontrol ω P max ω P max 1.432 37500 1.736 12500 ≤ 2000 OK ≤ 2000 OK 1547.55 λ 1130.21 λ = = Ag ≤ fy ≤ 2000 19.2 ≤ 2000 34.7
PERHITUNGAN BATANG TEKAN
= =
≤ fy
Ag
