Perencanaan Jembatan Kayu

Konstruksi Jembatan Kayu

Spesifikasi Kayu dan Jembatan

1. Jenis kayu : Sonokeling

- Kelas kuat II

- Kelas awet I

- Mutu kayu A

2. BJ : 0,9

3. E : 100.000 Kg / cm2

4. Tegangan Izin

- σlt : 100 Kg/cm2

- σtr// / σtk// : 85 Kg/cm2

- σtk : 25 Kg/cm2

- τ : 12 Kg/cm2

5. Konstruksi menerima beban tetap dan

beban tidak tetap

- Faktor kali 5/4

6. Konstruksi tidak terlindung

- Faktor kali 5/6

Pembebanan

1. Beban tetap

Beban konstruksi : 21.600 Kg

2. Beban tidak tetap

- Beban orang

- Untuk setiap 1 m X 1,5 m dapat

berisikan 4 orang

- Maka, untuk 6 m X 1,5 m

berisikan 24 orang

- Diasumsikan beban 1 orang 100

Kg

- Maka, beban 24 orang 2.400

Kg

3. Ptotal

Ptotal = Beban tetap + Beban tidak

tetap

= 21.600 + 2.400

= 24.000 Kg

4. P

P = P total12

= 24.00012

Analisa Struktur

Sin α = 0,64

Cos α = 0,77

a. Kestabilan Struktur

m = 2j – 3

25 = 2 . 14 – 3

25 = 25

b. Reaksi Perletakan

RAV = RBV = 12P

2

= 1224.000

2 = 6.000 Kg

RBH = 0

ΣMJ = 0

RAV . 1 – 12

P . 1 + V1 . 1 = 0

V1 = 1.000 – 6000

V1 = - 5.000 Kg

ΣMC = 0

-B1 . 1,2 = 0

ΣMJ = 0

-V1 . 1 + A1 . 1,2 =

0

A1 = −5.0001,2

A1 = -4.166,67 Kg

ΣMD = 0

-V1 . 1 – D1V . 1 = 0

D1 = 5.000cos α

D1 = 6.508,54 Kg

ΣME = 0

V2 . 1 = 0

ΣMJ = 0

-A1 . 1,2 + A2 . 1,2 = 0

A2 = -4.166,67 Kg

ΣMK = 0

D1V . 1 + V2 . 1 + D2V . 1 – P . 1 = 0

D2 = 2.000−5.000cos α

D2 = -3.905,12 Kg

ΣME = 0

B1 . 1,2 + D1H . 1,2 + D1V . 1 + V2 . 1 – B2 . 1,2 – P . 1 = 0

B2 = 0 .1,2+4.166,67 .1,2+5.000 .1+0 .1−2.000 .1

1,2

B2 = 6.666,67 Kg

∑ML = 0

V3 . 1 – P . 1

= 0

V3 = 2.000 Kg

∑ME = 0

B2 . 1,2 – B3 . 1,2 = 0

B3 = 6.666,67 Kg

∑ML = 0

-A2 . 1,2 – D2H . 1,2 – D2V . 1 – V3 . 1 + A3 . 1,2 = 0

A3 =

−4.166,67 .1,2−2.500 .1,2– 3.000 .1+2.000.11,2

A3 = -7.500 Kg

∑MF = 0

-D2V . 1 – V3 . 1 – D3V . 1 = 0

D3 = 3.000 .1−2.000 .1

cos α

D3 = -1.301,71 Kg

∑ME = 0

V4 . 1 = 0

TABEL HASIL PERHITUNGAN GAYA BATANG

No. Nama Batang

Panjang (m)

Gaya Batang (Kg)Tarik (+) Tekan(-)

1 V1 1,20 5.0002 A1 1,00 4.166,673 D1 1,56 6.508,544 B1 1,00 0 05 V2 1,20 0 06 A2 1,00 4.166,677 D2 1,56 3.905,128 B2 1,00 6.666,679 V3 1,20 2.000

10 A3 1,00 7.50011 D3 1,56 1.301,7112 B3 1,00 6.666,6713 V4 1,20 0 014 A4 1,00 7.50015 D4 1,56 1.301,7116 B4 1,00 6.666,6717 V5 1,20 2.00018 A5 1,00 4.166,6719 D5 1,56 3.905,1220 B5 1,00 6.666,6721 V6 1,20 0 022 A6 1,00 4.166,6723 D6 1,56 6.508,5424 B6 1,00 0 025 V7 1,20 5.000

Dimensi Kayu

1. Batang tarik

BJ = 0,9

P = 6.666,67 Kg

Sambungan baut C = 80%

σtr// = 150 . BJ . 5/4 . 5/6

= 150 . 0,9 . 5/4 . 5/6

σtr// = 140,625 Kg/cm2 ≥ σtr// = PAnetto

σtr// = PAnetto

Anetto = Pσ tr /¿

= 6.666,67140,625

= 47,407cm2

Anetto = 0,8 Abruto

Abruto = Anetto0,8

Abruto = 47,4070,8

Abruto = 59,259 cm2 dimensi kayu 8/10 Abruto = 64 cm2 ≥ Abruto syarat ok!

2. Batang tekan

BJ = 0,9

σtk// = 150 . BJ . 5/4 . 5/6

= 150 . 0,9 . 5/4 . 5/6

σtk// = 140,625 Kg/cm2

a. Lk = 1 m

P = 7,5 T

Imin = 50 . Ptk . lk2

= 50 . 7,5 . 12

= 375 cm4

I = 112

. b . h3

=112

. 8 . 103

= 666,67 cm4 ≥ Imin

ok!

Cek kekuatan

imin = √ IminAbruto

imin = √ 37580imin = 2,165 cm

λ = lkimin

λ = 1002,165

λ = 46,188 ω = 1,44

σtk// = P .ωAbruto

=7.500 .1,44

80

= 135 Kg/cm2 ≤ σtk//

ok!

b. Lk = 1,2 m

P = 5 T


= 50 . 5 . 1,22

= 360 cm4

I = 112

. b . h3

=112

. 8 . 103

= 666,67 cm4 ≥ Imin

ok!

Cek kekuatan


imin = √ 36080imin = 2,121 cm

λ = lkimin

λ = 1202,121

λ = 56,568 ω = 1,60


=5.000 .1,60

80

= 100 Kg/cm2 ≤ σtk//

ok!

c. Lk = 1,56 m

P = 3,90512 T


= 50 . 3,90512 . 1,562

= 305 cm4

I = 112

. b . h3

=112

. 8 . 103

= 666,67 cm4 ≥ Imin

ok!

Cek kekuatan


imin = √ 30580imin = 1,952cm

λ = lkimin

λ = 1001,952

λ = 80 ω = 2,14


=3.905,12.2,14

80

= 104,462Kg/cm2 ≤ σtk//

ok!

Perencanaan Sambungan

Gambar Layout Sambungan

- Sambungan baut dengan pelat simpul dari baja sambungan tampang dua

- Kelas kuat kayu II golongan II

- λb = 4,3

- S = 100 . d . b3 ( 1 – 0.6 sin α )

- S = 200 . d . b1 ( 1 – 0.6 sin α )

- S = 430 . d2 ( 1 – 0.35 sin α )

- Diameter baut 18 mm- Tebal plat = 0.3 d = 0.3(10)= 3 mm asumsi kayu penyambung 4/10- S ditambah 25 % karena menerima beban tetap dan beban tidak tetap- S dikalikan 5/6 karena konstruksi tidak terlindung- Semua Gaya batang dibagi 2 untuk 2 sisi

a. Sambungan - A

1. batang V1 = -2.500 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 90o)

S1 = 100 (1,8) (8) (0,4) = 576 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 90o )

S2 =200 . 1,8 . 4 (0,4)

S2 =576 kg (diambil S yang terkecil)

c. S3 = 430 d2 . (1 – 0,35 sin 90o)

S3 = 430 (1,8)2 (0,65)= 905,58 kg

d. Sr = 1,25 . S . 56

Sr = 1,25 . 576 . 56

Sr = 600 kg

e. η= P/S = 2.500/ 600 = 4,17 5 Baut

2. batang B1 = 0 kg 8/10

3. Jarak – jarak baut

2d = 2 . (1,8) = 3,6 5 cm

5d = 5 . (1,8) = 9 cm

7d = 7 . (1,8) = 12,6 ≈ 13 cm

3,5d = 3,5 . (1,8) = 6,3 ≈ 7 cm

b. Sambungan – C

1. batang A1 = -2.083,33 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 00)

S1 = 100 (1,8) (8) (1) = 1440 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 00 )

S2 =200 . 1,8 . 4 = 1440 kg

c. S3 = 430 d2 . (1 – 0,35 sin 00)

S3 = 430 (1,8)2

S3 = 1393,2 kg (diambil S yang terkecil)

d. Sr = 1,25 . S . 56

Sr = 1,25 . 1393,2 . 56

Sr = 1451,25 kg

e. η= P/S = 2.083,33 / 1451,25 = 1,435 2 Baut

2. batang D1 = 3254,27 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin (90 - α))

S1 = 100 (1,8) (8) (0,539)

S1 = 776,256 kg = 776,256 kg (diambil S yang terkecil)

b. S2 = 200 d.b1 (1 – 0,6 sin (90 - α))

S2 =200 . 1,8 . 4 (0,539) = 776,256 kg

c. S3 = 430 d2 . (1 – 0,35 sin (90 - α))

S3 = 430 (1,8)2 (0,731)= 1.018,599 kg

d. Sr = 1,25 . S . 56

Sr = 1,25 . 776,256 . 56

Sr = 808,6 kg

e. η = P/S = 3254,27 / 808,6 = 4,024 5 Baut

3. batang V1 = -2.500 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 90o)

S1 = 100 (1,8) (8) (0,4) = 576 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 90o )

S2 =200 . 1,8 . 4 (0,4)


c. S3 = 430 d2 . (1 – 0,35 sin 90o)

S3 = 430 (1,8)2 (0,65)= 905,58 kg

d. Sr = 1,25 . S . 56

Sr = 1,25 . 576 . 56

Sr = 600 kg

e. η= P/S = 2.500/ 600 = 4,17 5 Baut


2d = 2 . (1,8) = 3,6 5 cm

5d = 5 . (1,8) = 9 cm

7d = 7 . (1,8) = 12,6 ≈ 13 cm

3,5d = 3,5 . (1,8) = 6,3 ≈ 7 cm

c. Sambungan – D

1. batang A1 = A2= -2.083,33 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 00)

S1 = 100 (1,8) (8) (1) = 1440 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 00 )

S2 =200 . 1,8 . 4 = 1440 kg

c. S3 = 430 d2 . (1 – 0,35 sin 00)

S3 = 430 (1,8)2


d. Sr = 1,25 . S . 56

Sr = 1,25 . 1393,2 . 56

Sr = 1451,25 kg

e. ηA1=A2 = P/S = 2.083,33 / 1451,25 = 1,435 2 Baut

2. batang V2 = 0 kg 8/10


2d = 2 . (1,8) = 3,6 5 cm

5d = 5 . (1,8) = 9 cm

7d = 7 . (1,8) = 12,6 ≈ 13 cm

3,5d = 3,5 . (1,8) = 6,3 ≈ 7 cm

d. Sambungan – J

1. batang B1 = 0 kg 8/10

batang B2 = 3.333,33 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 00)

S1 = 100 (1,8) (8) (1) = 1440 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 00 )

S2 =200 . 1,8 . 4 = 1440 kg

c. S3 = 430 d2 . (1 – 0,35 sin 00)

S3 = 430 (1,8)2


d. Sr = 1,25 . S . 56

Sr = 1,25 . 1393,2 . 56

Sr = 1451,25 kg

e. ηB2= P/S = 3.333,33 / 1451,25 = 2,297 3 Baut

2. batang D1 = 3254,27 kg 8/10

batang D2 = -1952,53 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin (90 - α))

S1 = 100 (1,8) (8) (0,539)


b. S2 = 200 d.b1 (1 – 0,6 sin (90 - α))

S2 =200 . 1,8 . 4 (0,539) = 776,256 kg

c. S3 = 430 d2 . (1 – 0,35 sin (90 - α))

S3 = 430 (1,8)2 (0,731)= 1.018,599 kg

d. Sr = 1,25 . S . 56

Sr = 1,25 . 776,256 . 56

Sr = 808,6 kg

f. ΗD1 = P/S = 3254,27 / 808,6 = 4,024 5 Baut

ηD2= P/S = 1952,56 / 808,6 = 2,415 3 Baut

3. batang V2 = 0 kg 8/10


2d = 2 . (1,8) = 3,6 5 cm

5d = 5 . (1,8) = 9 cm

7d = 7 . (1,8) = 12,6 ≈ 13 cm

3,5d = 3,5 . (1,8) = 6,3 ≈ 7 cm

e. Sambungan – K

1. batang B2 = B3 = 3.333,33 kg

8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 00)

S1 = 100 (1,8) (8) (1) = 1440 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 00 )

S2 =200 . 1,8 . 4 = 1440 kg

c. S3 = 430 d2 . (1 – 0,35 sin 00)

S3 = 430 (1,8)2


d. Sr = 1,25 . S . 56

Sr = 1,25 . 1393,2 . 56

Sr = 1451,25 kg

e. ηB2=B3 = P/S = 3.333,33 / 1451,25 = 2,297 3 Baut

2. batang V3 = 1.000 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 90o)

S1 = 100 (1,8) (8) (0,4) = 576 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 90o )

S2 =200 . 1,8 . 4 (0,4)


c. S3 = 430 d2 . (1 – 0,35 sin 90o)

S3 = 430 (1,8)2 (0,65)= 905,58 kg

d. Sr = 1,25 . S . 56

Sr = 1,25 . 576 . 56

Sr = 600 kg

e. η= P/S = 1.000 / 600 = 1,67 2 Baut


2d = 2 . (1,8) = 3,6 5 cm

5d = 5 . (1,8) = 9 cm

7d = 7 . (1,8) = 12,6 ≈ 13 cm

3,5d = 3,5 . (1,8) = 6,3 ≈ 7 cm

f. Sambungan – E

1. batang A2 = -2.083,33 kg 8/10

batang A3 = -3.750 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 00)

S1 = 100 (1,8) (8) (1) = 1440 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 00 )

S2 =200 . 1,8 . 4 = 1440 kg

c. S3 = 430 d2 . (1 – 0,35 sin 00)

S3 = 430 (1,8)2


d. Sr = 1,25 . S . 56

Sr = 1,25 . 1393,2 . 56

Sr = 1451,25 kg

e. ηA2 = P/S = 2.083,33 / 1451,25 = 1,435 2 Baut

ηA3 = P/S = 3750 / 1451,25 = 2,584 3 Baut

2. batang D2 = -1952,56 kg 8/10

batang D3 = -650,854 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin (90 - α))

S1 = 100 (1,8) (8) (0,539)


b. S2 = 200 d.b1 (1 – 0,6 sin (90 - α))

S2 =200 . 1,8 . 4 (0,539) = 776,256 kg

c. S3 = 430 d2 . (1 – 0,35 sin (90 - α))

S3 = 430 (1,8)2 (0,731)= 1.018,599 kg

d. Sr = 1,25 . S . 56

Sr = 1,25 . 776,256 . 56

Sr = 808,6 kg

e. ηD2= P/S = 1952,56 / 808,6 = 2,415 3 Baut

ηD3= P/S = 650,854 / 808,6 = 0,805 1 Baut

3. batang V3 = 1.000 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 90o)

S1 = 100 (1,8) (8) (0,4) = 576 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 90o )

S2 =200 . 1,8 . 4 (0,4)


c. S3 = 430 d2 . (1 – 0,35 sin 90o)

S3 = 430 (1,8)2 (0,65)= 905,58 kg

d. Sr = 1,25 . S . 56

Sr = 1,25 . 576 . 56

Sr = 600 kg

e. η= P/S = 1.000/ 600 = 1,67 2 Baut


2d = 2 . (1,8) = 3,6 5 cm

5d = 5 . (1,8) = 9 cm

7d = 7 . (1,8) = 12,6 ≈ 13 cm

3,5d = 3,5 . (1,8) = 6,3 ≈ 7 cm

g. Sambungan - F

1. batang A3 = A4 = -3750 kg

8/10

a. S1 = 100 d b3 . (1 – 0,6 sin

00)

S1 = 100 (1,8) (8) (1) = 1440 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 00 )

S2 =200 . 1,8 . 4 = 1440 kg

c. S3 = 430 d2 . (1 – 0,35 sin 00)

S3 = 430 (1,8)2


d. Sr = 1,25 . S . 56

Sr = 1,25 . 1393,2 . 56

Sr = 1451,25 kg

e. ηA3=A4 = P/S = 3750 / 1451,25 = 2,584 3 Baut

2. batang V4 = 0 8/10


2d = 2 . (1,8) = 3,6 5 cm

5d = 5 . (1,8) = 9 cm

7d = 7 . (1,8) = 12,6 ≈ 13 cm

3,5d = 3,5 . (1,8) = 6,3 ≈ 7 cm

h. Sambungan – L

1. batang B3 = B4 = 3.333,33 kg

8/10

a. S1 = 100 d b3 . (1 – 0,6 sin 00)

S1 = 100 (1,8) (8) (1) = 1440 kg

b. S2 = 200d.b1 ( 1 – 0.6 sin 00 )

S2 =200 . 1,8 . 4 = 1440 kg

c. S3 = 430 d2 . (1 – 0,35 sin 00)

S3 = 430 (1,8)2


d. Sr = 1,25 . S . 56

Sr = 1,25 . 1393,2 . 56

Sr = 1451,25 kg

e. ηB3=B4 = P/S = 3.333,33 / 1451,25 = 2,297 3 Baut

2. batang D3 = D4 = -650,854 kg 8/10

a. S1 = 100 d b3 . (1 – 0,6 sin (90 - α))

S1 = 100 (1,8) (8) (0,539)


b. S2 = 200 d.b1 (1 – 0,6 sin (90 - α))

S2 =200 . 1,8 . 4 (0,539) = 776,256 kg

c. S3 = 430 d2 . (1 – 0,35 sin (90 - α))

S3 = 430 (1,8)2 (0,731)= 1.018,599 kg

d. Sr = 1,25 . S . 56

Sr = 1,25 . 776,256 . 56

Sr = 808,6 kg

e. ηD3= P/S = 650,854 / 808,6 = 0,805 1 Baut

3. batang V4 = 0 8/10


2d = 2 . (1,8) = 3,6 5 cm

5d = 5 . (1,8) = 9 cm

7d = 7 . (1,8) = 12,6 ≈ 13 cm

3,5d = 3,5 . (1,8) = 6,3 ≈ 7 cm

Tugas Besar Konstruksi Kayu

Tugas Besar Ini Disusun Sebagai Persyaratan Nilai Semester 3

Oleh

Nama : Oki Baihaqqi

N.I.M : 1111020040

Kelas : 2 – Sipil – 1 Pagi

Jurusan Teknik SipilPoliteknik Negeri Jakarta

2012

Documents

Perencanaan Jembatan Kayu