UJIAN AKHIR SEMESTER
JURUSAN TEKNIK KELAUTAN FAKULTAS TEKNOLOGI KELAUTAN ITS 2009
BAGUS RENGGA
4307 100 020
JURUSAN TEKNIK KELAUTAN FAKULTAS TEKNOLOGI KELAUTAN INSTITUT TEKNOLOGI SEPULUH NOPEMBER SURABAYA See more on : bagusrengga.wordpress.com Contact :
[email protected]
NOMOR 1.... FORMULASI MATRIX DAN DEFLEKSINYA
Diketahui : δb
δc
3 δD
P = 2 Kn A1 = 6 cm2 = 6 x 10-4 m2 A2 = 4 cm2 = 4 x 10-4 m2 A3 = 2 cm2 = 2 x 10-4 m2 E aluminiun = 75 Gpa = 75 x 106 Kn/m L = 1 m
Ditanya : • Formulasi Matriks [K] {δ} = {P} • Defleksi di titik D ?
Jawab Gunakan metode Potensial Energi Total (PET) Pet = Ut – Lp Ut = ∑1/2 Cn (∆l)2 Lp = ∑ Pδ Ut = ½ C1 (∆lb )2 + ½ C2 (∆lc )2 + ½ C3 (∆lD )2
Dengan : C1 = EA1/L = 6 x 10-4 E/L C2 = EA2/L = 4 x 10-4 E/L C3 = EA3/L = 2 x 10-4 E/L (∆lb ) = δb’ = δb (∆lc ) = δb’ + δc’ = δd (∆ld ) = δb’ + δc’ + δd’ = δd
Lp = δb (P1 + P2 + P2) + δc (P2 + P3) + δd (P3) = δb (3P) + δc (2P) + δd (P)
Maka PET = [ (½)(6x10-4(Eδb)2/L) + (½)(4x10-4(Eδc)2/L) + (½)(2x10-4(Eδd)2/L) ] – [3pδb + 2Pδc + Pδd] Untuk mendapatkan formulasi matriksnya, maka persamaan PET harus diturnkan terhadap defleksi pada masing-masing titik yang mengalami defleksi.
Sehingga ∂PET/∂δb = 0 =∂{[ (½)(6x10-4E(δb)2/L) + (½)(4x10-4E(δc)2/L) + (½)(2x10-4E(δd)2/L) ] – [3pδb + 2Pδc + Pδd] } / ∂δb = 0 =[ (½)(12x10-4Eδb/L) ] – [3p] = 0 = [ (½)(12x10-4Eδb/L) ] = [3p] ………… [E1]
∂PET/∂δc = 0 =∂{[ (½)(6x10-4E(δb)2/L) + (½)(4x10-4E(δc)2/L) + (½)(2x10-4E(δd)2/L) ] – [3pδb + 2Pδc + Pδd] } / ∂δc = 0 =[ (½)(8x10-4Eδc/L) ] – [2p] = 0 = [ (½)(8x10-4Eδc/L) ] = [2p] ………… [E2]
∂PET/∂δd = 0 =∂{[ (½)(6x10-4E(δb)2/L) + (½)(4x10-4E(δc)2/L) + (½)(2x10-4E(δd)2/L) ] – [3pδb + 2Pδc + Pδd] } / ∂δd = 0
=[ (½)(4x10-4Eδd/L) ] – [p] = 0 = [ (½)(8x10-4Eδd/L) ] = [p] ………… [E3]
Sehingga [ (½)(12x10-4Eδb/L) ] = [3p] ………… [E1] [ (½)(8x10-4Eδc/L) ] = [2p] ………… [E2] [ (½)(8x10-4Eδd/L) ] = [p] ………… [E3]
Maka formulasi matriksnya adalah : [K] {δ} = {P}
Mencari defleksi di tiap elemen batang : Dengan E aluminium = 75 x 106 KN/m P = 2 KN
[ (½)(12x10-4Eδb/L) ] = [3p] δb = 3LP/ 12x10-4E = 3(1)(2) / (12x10-4(75 x 106)) = 6.667 x 10-5 m [ (½)(8x10-4Eδc/L) ] = [2p] δc = 2LP/ 12x10-4E = 2(1)(2) / (8x10-4(75 x 106)) = 6.667 x 10-5 m [ (½)(8x10-4Eδc/L) ] = P δd = LP/ 8x10-4E = (1)(2) / (4x10-4(75 x 106)) = 6.667 x 10-5 m
Defleksi di titik D adalah δtotal
= δb + δc + δd
= 6.667 x 10-5
+ 6.667 x 10-5 + 6.667 x 10-5
= 20.001 x 10-5 m
NOMOR 2.... PENENTUAN NILAI LENDUTAN
Sebuah jacket sederhana dibuat dari baja dan dikenai gaya-gaya seperti gambar di bawah. Batang vertikal mempunyai diameter 60 cm dan tebal 5 cm. Batang horizontal berdiameter 50 cm dan tebal 3 cm. Batang diagonal memiliki diameter 40 cm dan tebal 2 cm. Tentukan: -Lendutan horizontal di titik H -Lendutan vertikal di titik H -Lendutan horizontal di titik D Asumsi awal tarik (+)
PENYELESAIAN Perhitungan Reaksi Perletakan ΣMB = 0 RA.10 + 2.10 + 2.20 + 4.30 + 2.10 + 2.20 + 4.30 – 20.10 = 0 RA.10 + 20 + 40 + 120 + 20 + 40 + 120 – 200 = 0 RA.10 + 40 + 80 + 240 – 200 = 0 RA = – 40 – 80 – 240 + 200 10 RA = 16 KN (tarik)
Perhitungan Reaksi Perletakan ΣMA = 0 -RB.10 + 2.10 + 2.20 + 4.30 + 2.10 + 2.20 + 4.30 + 20.10 = 0 -RB.10 + 40 + 80 + 240 + 200 = 0 RB = 40 + 80 + 240 + 200 10 RB = 56 KN (tekan)
Perhitungan Reaksi Perletakan ΣH = 0 RAH + 2 + 2 + 2 + 2 + 4 + 4 = 0 RAH + 16 = 0 RAH = - 16 KN (tekan)
PERHITUNGAN JOINT Joint B ΣV = 0 RB + S4 = 0
ΣH = 0 S2 = 0
56 + S4 = 0 S4 = - 56 KN (tarik) S4 = 56 KN (tekan)
S4
S2
Rb
PERHITUNGAN JOINT Joint A ΣV = 0 -RA + S1 + S3Sinα = 0 -16 + S1 + S3Sin45 = 0 S1 = 0
ΣH = 0 -RA + S2 + S3Cosα = 0 -16 + 0 + S3Cos45 = 0 S3 = 22,63 KN (tarik)
S4
S3
S2
RaH
Ra
PERHITUNGAN JOINT Joint D ΣV = 0 -S4 – S8 + S3Sinα = 0 -56 + S8 + 16 = 0 S8 = - 40 KN (tarik) S8 = 40 KN (tekan)
ΣH = 0 S6 - 2 + S3Cosα = 0 S6 - 2 + 16 = 0 S6 = -14 KN (tarik) S6 = 14 KN (tekan)
S8
P
S6
S3
S4
PERHITUNGAN JOINT Joint C ΣV = 0 -S1 – S5 + S7Sinα = 0 0 – S5 + 12 = 0 S5 = - 12 KN (tarik) S5 = 12 KN (tekan)
ΣH = 0 P + S6 + S7Cosα = 0 2 - 14 + S7Cosα = 0 S7 = 16,8 KN (tarik)
S5
P
S7
S6
S1
PERHITUNGAN JOINT Joint F ΣV = 0 S12 + S8 - S7Sinα = 0 S12 + 40 - 12 = 0 S12 = - 28 KN (tarik) S12 = 28 KN (tekan)
ΣH = 0 S10 - P + S7Cosα = 0 S10 - 2 + 12 = 0 S10 = - 10 KN (tarik) S10 = 10 KN (tekan)
S12
S10
S7
P
S8
PERHITUNGAN JOINT Joint E ΣV = 0 S9 + S11 + S11Sinα = 0 S9 + 8 + 12 = 0 S9 = - 20 KN (tarik) S9 = 20 KN (tekan)
ΣH = 0 -S10 + P + S11Cosα = 0 -S10 + 2 - 12 = 0 S11 = 11,31 KN (tarik)
S9
P
S11
S10
S5
PERHITUNGAN JOINT Joint H ΣV = 0 P - S12 + S11Sinα = 0 20 - 28 + 8 = 0 0=0 Terbukti !!
ΣH = 0 S13 - P + S11Cosα = 0 S10 - 4 + 8 = 0 S13 = - 4 KN (tarik) S13 = 4 KN (tekan)
P
S13
S11
P
S12
PERHITUNGAN JOINT Joint G ΣV = 0 S9 - P = 0 20 - 20 = 0 0=0 Terbukti!!
ΣH = 0 P – S13 = 0 4-4=0 0=0 Terbukti!!
P
P
S13
S9
PERHITUNGAN EA EA BATANG HORIZONTAL E baja = 83.106 KN/M2 EA = (83.106).1/4.3,14.(D2 - d2) EA = (83.106).(1/4).(3,14).(0,52 – 0,44d2) EA = 3,68.106 KN
EA BATANG VERTIKAL EA = (83.106).1/4.3,14.(D2 - d2) EA = (83.106).(1/4).(3,14).(0,62 – 0,5d2) EA = 7,138.106 KN
EA BATANG DIAGONAL EA = (83.106).1/4.3,14.(D2 - d2) EA = (83.106).(1/4).(3,14).(0,42 – 0,36d2) EA = 7,138.106 KN
PENYELESAIAN Pengaruh P di titik H
A) Menyebabkan δ Horizontal di titik H (δHH) Reaksi Perletakan di titik A ΣMB = 0 RA.10 + P.30 = 0 RA.10 + 1.30 = 0 RA = 30 KN (tarik), mengarah
kebawah ΣMA = 0
-RB.10 + P.30 = 0 -RB.10 + 1.30 = 0 RB = 3 KN (tekan),mengarah
keatas
ΣH = 0 RAH – P = 0 RAH = P RAH = 1 KN
PERHITUNGAN JOINT Joint B ΣV = 0 RB + S4 = 0
ΣH = 0 S2 = 0
3 + S4 = 0 S4 = - 3 KN (tarik) S4 = 3 KN (tekan)
S4
S2
Rb
PERHITUNGAN JOINT Joint A ΣV = 0 -RA + S1 + S3Sinα = 0 -3 + S1 + 1 = 0 S1 = 2 KN (tarik)
ΣH = 0 -RA + S2 + S3Cosα = 0 -1 + 0 + S3Cos45 = 0 S3 = 1,4 KN (tarik)
S1
S3
S2
RaH
Ra
PERHITUNGAN JOINT Joint D ΣV = 0 -S4 + S8 - S3Sinα = 0 3 + S8 - 1 = 0 S8 = - 2 KN (tarik) S8 = 2 KN (tekan)
ΣH = 0 S6 + S3Cosα = 0 S6 + 1 = 0 S6 = -1 KN (tarik) S6 = 1 KN (tekan)
S8
S6
S3
S4
PERHITUNGAN JOINT Joint C ΣV = 0 -S1 + S5 + S7Sinα = 0 -2 + S5 + 1 = 0 S5 = 1 KN (tarik)
ΣH = 0 S6 + S7Cosα = 0 -1 + S7.0,5.1,4 = 0 S7 = 1,4 KN (tarik)
S5
S7
S6
S1
PERHITUNGAN JOINT Joint F ΣV = 0 S12 - S8 - S7Sinα = 0 S12 + 2 - 1 = 0 S12 = - 1 KN (tarik) S12 = 1 KN (tekan)
ΣH = 0 S10 + S7Cosα = 0 S10 + 1 = 0 S10 = - 1 KN (tarik) S10 = 1 KN (tekan)
S12
S10
S7
S8
PERHITUNGAN JOINT Joint E ΣV = 0 S9 - S5 + S11Sinα = 0 S9 - 1 + 1 = 0 S9 = 0
ΣH = 0 S10 + S11Cosα = 0 1 – S11Cosα = 0 S11 = 1,4 KN (tarik)
S9
S11
S10
S5
PERHITUNGAN JOINT Joint H ΣH = 0 S13 - P + S11Cosα = 0 S10 - 1 + 1 = 0 S13 = 0
ΣV = 0 S12 + S11Sinα = 0 S12 + 1 = 0 S12 = -1 KN (tarik) S12 = 1 KN (tekan) S13
S11
P
S12
TABEL LENDUTAN HORIZONTAL Tabel Lendutan Horizontal di titik H
Batang AC AB AD BD CE CD CF DF EG EF EH FH GH
Ni 0 0 22,4 -56 -12 -14 16,8 -40 -20 -10 11,2 -28 -4
Ni^ 2 0 1,4 -3 1 -1 1,4 -2 0 -1 1,4 -1 0
EA 7,138.106 3,68.106 1,98.106 7,138.106 7,138.106 3,68.106 1,98.106 7,138.106 7,138.106 3,68.106 1,98.106 7,138.106 3,68.106
Li 10 10 14 10 10 10 14 10 10 10 14 10 10
Jadi Lendutan Horizontal di titik H
(Ni^.N.Li)/EA 0 0 2,28. 10-4 2,35.10-4 2,38. 10-5 3,8. 10-5 1,71. 10-6 1,12.10-4 0 2,7. 10-5 1,14. 10-4 -39,23.10-5 0 Σ = 9,88. 10-4 m
PENYELESAIAN Pengaruh P di titik H
B) Menyebabkan δ Vertikal di titik H (δHv) Reaksi Perletakan di titik A ΣMB = 0 RA.10 = 0 RA.10 = 0 RA = 0 ΣMA = 0 -RB.10 - P.10 = 0 -RB.10 - P.10 = 0 RB = -1 KN (tarik),mengarah
kebawah
PERHITUNGAN JOINT Joint B ΣV = 0 RB + S4 = 0
ΣH = 0 S2 = 0
-1 + S4 = 0 S4 = 1 KN (tarik)
S4
S2
Rb
PERHITUNGAN JOINT Joint A ΣV = 0 -RA + S1 + S3Sinα = 0 0 + S1 + 0 = 0 S1 = 0 KN
ΣH = 0 -RA + S2 + S3Cosα = 0 -0 + 0 + S3Cos45 = 0 S3 = 0 KN
S1
S3
S2
Ra
PERHITUNGAN JOINT Joint D ΣV = 0 S4 + S8 + S3Sinα = 0 -1 + S8 - 0 = 0 S8 = 1 KN (tarik)
ΣH = 0 S6 + S3Cosα = 0 S6 + 0 = 0 S6 = 0 KN
S8
S6
S3
S4
PERHITUNGAN JOINT Joint C ΣV = 0 -S1 + S5 + S7Sinα = 0 0 + S5 + 0 = 0 S5 = 0 KN
ΣH = 0 S6 + S7Cosα = 0 0 + S7Cosα = 0 S7 = 0 KN
S5
S7
S6
S1
PERHITUNGAN JOINT Joint F ΣV = 0 S12 - S8 - S7Sinα = 0 S12 - 1 + 0 = 0 S12 = 1 KN (tarik)
ΣH = 0 S10 + S7Cosα = 0 S10 + 0 = 0 S10 = 0 KN
S12
S10
S7
S8
PERHITUNGAN JOINT Joint E ΣV = 0 S9 - S5 + S11Sinα = 0 S9 - 0 + 0 = 0 S9 = 0
ΣH = 0 S10 + S11Cosα = 0 S11Cosα = 0 S11 = 0 KN
S9
S11
S10
S5
PERHITUNGAN JOINT Joint H ΣH = 0 S13 + S11Cosα = 0 S13 + 0 = 0 S13 = 0
ΣV = 0 P - S12 - S11Sinα = 0 S11Sinα - 1 + 1 = 0 S11 = 0
P
S13
S11
S12
TABEL LENDUTAN HORIZONTAL Tabel Lendutan Vertikal di titik H
Batang AC AB AD BD CE CD CF DF EG EF EH FH GH
Ni 0 0 22,6 -56 -12 -14 16,8 -40 -20 -10 11,2 -28 -4
Ni^ 0 0 0 1 0 0 0 1 0 0 0 1 0
EA 7,138.106 3,68.106 1,98.106 7,138.106 7,138.106 3,68.106 1,98.106 7,138.106 7,138.106 3,168.106 1,98.106 7,138.106 3,98.106
Li 10 10 14 10 10 10 14 10 10 10 10 10 14
Jadi Lendutan Vertikal di titik H
(Ni^.N.Li)/EA 0 0 0 -78,45.10-6 0 0 0 -56,04. 10-6 0 0 0 -39,23. 10-6 0 Σ = -17,372. 10-5 m
PENYELESAIAN Pengaruh P di titik H
C) Menyebabkan δ Horizontal di titik D (δDH) Reaksi Perletakan di titik A & B ΣMB = 0 RA.10 + P.10 = 0 RA = -1 KN (tarik), mengarah
kebawah
ΣMA = 0 -RB.10 + P.10 = 0 -RB.10 + P.10 = 0 RB = 1 KN (tekan), mengarah ke
atas
ΣH = 0 RAx – P = 0 RAx = P RAx = 1 KN
PERHITUNGAN JOINT Joint H ΣH = 0 S13 + S11Cosα = 0 0 + S11 = 0 S11 = 0
ΣV = 0 S11 + S12 = 0 0 + S12 = 0 S12 = 0
S13
S11
S12
PERHITUNGAN JOINT Joint E ΣV = 0 S9 - S5 + S11Sinα = 0 0 - S5 + 0 = 0 S5 = 0
ΣH = 0 S10 + S11Cosα = 0 S10 + 0 = 0 S10 = 0 KN
S9
S11
S10
S5
PERHITUNGAN JOINT Joint F ΣV = 0 S12 - S8 - S7Sinα = 0 0 - S8 + 0 = 0 S8 = 0 KN (tarik)
ΣH = 0 S10 + S7Cosα = 0 0 + S7 = 0 S7 = 0 KN
S12
S10
S7
S8
PERHITUNGAN JOINT Joint C ΣV = 0 -S1 + S5 + S7Sinα = 0 -S1 + 0 + 0 = 0 S1 = 0 KN
ΣH = 0 S6 + S7Cosα = 0 S6 + 0 = 0 S6 = 0 KN
S5
S7
S6
S1
PERHITUNGAN JOINT Joint D ΣV = 0 -S4 + S8 - S3Sinα = 0 -S4 + 0 - 1 = 0 S4 = -1 KN (tarik) S4 = 1 KN (tekan)
ΣH = 0 1 - S6 - S3Cosα = 0 1 - 0 + S3Cosα = 0 S3Cosα = 1 S3 = 1,4 KN
S8
P
S6
S3
S4
PERHITUNGAN JOINT Joint A ΣV = 0 -RA + S1 + S3Sinα = 0 -1 + o + 1 = 0 0 = 0 KN
ΣH = 0 -RA + S2 + S3Cosα = 0 -1 + S2 + 1 = 0 S2 = 0 KN
S1
RaX
S3
S2
Ra
TABEL LENDUTAN HORIZONTAL Tabel Lendutan Horizontal di titik D
Batang AC AB AD BD CE CD CF DF EG EF EH FH GH
Ni 0 0 22,4 -56 -12 -14 16,8 -40 -20 -10 11,2 -28 -4
Ni^ 0 0 1,4 -1 0 0 0 0 0 0 0 0 0
EA 7,138.106 3,68.106 1,98.106 7,138.106 7,138.106 3,68.106 1,98.106 7,138.106 7,138.106 3,168.106 1,98.106 7,138.106 3,68.106
Li 10 10 14 10 10 10 14 10 10 10 14 10 10
Jadi Lendutan Horizontal di titik D
(Ni^.N.Li)/EA 0 0 2,2. 10-4 7,84. 10-5 0 0 0 0 0 0 0 0 0 Σ = 3. 10-4 m
TERIMA KASIH ATAS PERHATIANNYA See more on : bagusrengga.wordpress.com Contact :
[email protected]
