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1. DISAIN JEMBATAN RANGKA BAJA
2 SIPIL 1 PAGI
UNTUK MELENGKAPI TUGAS BAJA
OLEH:
GILANG ADITYA.P
DANIEL HASIBUAN
SADDAM FIRGIAWAN
DEPOK,15 JUNI 2010

2. PEMODELAN STRUKTUR
TAMPAK DEPAN
TAMPAK ATAS
SEPESIFIKASI JEMBATAN
-Jembatan rangka baja
-Panjang bentang 24 meter
-Lebar jembatan 8 meter
-Tinggi jembatan 3 meter
PERHITUNGAN GAYA-GAYA DALAM
*Reaksi Perletakan
= Ravx24 –
=
24Rav = 72P
Rav = 3P
Rav = Rbv maka Rbv = 3P

3. *Gaya-Gaya Batang
BUHUL A
∑V = 0
1 ∑H =0
− P − S1.0, 6 + 3P = 0
2 S1SIN 53 + S 2 = 0
−2,5P S 2 = −4,17 P.0,8
− S1 =
0, 6 S 2 = −3,33P
S1 − 4,17 P
BUHUL B
∑H =0
∑V = 0
− S1SIN 53 + S 6COS14 = 0
S1COS 53 + S 3 + S 6SIN14 = 0
−4,17.0,8P + S 6.0,97 = 0
4,17 P.0, 6 + S 3 + 3, 44.0, 24 = 0
3,33P
S 3 = −3,33P S6 = = 3, 44 P
0,97

4. BUHUL C
∑V = 0
∑H =0
− P − S 3 − S 5SIN 26,6 = 0
− S 2 + S 4 + S 5COS 26, 6 = 0
− P + 3,33P − S 50, 45 = 0
3,33P + S 4 + 5,18P.0,89 = 0
−2,33P
S5 = S 4 = −7,96 P
−0, 45
S 5 = 5,18 P
BUHUL D
∑V = 0
∑H =0
S 7 + S 5COS 63, 4 − S 6COS 76 = 0
S10 − S 5SIN 63, 4 − S 6SIN 76 = 0
S 7 + 5,18P.0, 45 − 3, 44 P.0, 24 = 0
S10 − 5,18P.0,89 − 3, 44 P.0,97 = 0
S 7 = −1,5 P
S10 = 7,97 P

5. BUHUL E
∑V = 0 ∑H =0
− P − S 7 − S 9SIN 26,6 = 0 − S 4 + S 8 + S 9COS 26, 6 = 0
− P + 1,5P − S 9.0, 45 = 0 7,96 P + S 8 + 1,11P.0,89 = 0
0,5 S 8 = −9 P
S9 = = 1,11P
0, 45
BUHUL F
∑V = 0
− P − S11 = 0
S11 = − P

6. TABEL GAYA BATANG
BEBAN MATI ,HIDUP DAN BERJALAN
TABEL BEBAN HIDUP DAN BEBAN MATI
NO.BATANG GAYA JENIS
S1 4,17P TARIK
S2 -3,33P TEKAN
S3 -3,33P TEKAN
S4 -7,96P TEKAN
S5 5,18P TARIK
S6 3,44P TARIK
S7 -1,5P TEKAN
S8 -9P TEKAN
S9 1,11P TARIK
S10 7,97P TARIK
S11 -P TEKAN

7. GAYA YANG BEKERJA
GAYA AXIAL
GAYA GESER
MOMEN

8. TABEL BEBAN BERJALAN

9. GRAFIK BEBAN BERJALAN

10. TOTAL GAYA YANG BEKERJA DISETIAP BATANG
PERENCANAAN RANGKA UTAMA
Tebal trotoar = 20 cm
Lebar trotoar = 100 cm
Panjang segmen =4m
Tinggi rangka =3m
Lebar jembatan =8m
Beban jalur = 9 kN/m
Beban garis = 49 kN/m
PEMBEBANAN PADA DIAFRAGMA
∴ Beban mati
− Berat per ker asan = 24 x 4 x0,5 = 24kN / m
− Berat aspal = 22 x 4 x0, 05 = 4, 4kN / m
qdl = 28, 4kN / m
1
M max = x 28, 4 x82 = 227, 2kN .m
8
Mu = 1, 4 xM max = 1, 4 x 227, 2 = 318,1kN .m

11. BEBAN MATI
B.mati +diafragma = RA x factor beban
= 113.6 x 1.1 =124,96 kN
B.trotoar = Bj beton x t x I x λ x factor beban
= 24 x 0,2 x 1 x 4 x 1,3 =24,96kN +
PDl = 149,92 kN
BEBAN HIDUP
B.lajur = 9 x 6 x 4 x 2 x 0,5 = 216 kN
B.pejalan kaki = 5 x 1 x 4 x2 = 40 kN
B.genangan air = 0,5 x 6 x 4 x 1 x 0,5 = 6 kN
+
qll = 262 kN
BEBAN BERJALAN
B.garis (KEL) = 49 x 6 x 1,4 x 2 x 0,5 = 411,6 kN
DISAIN BATANG TARIK
Pu = 4014,11 kN (s10)
φ tarik = 0,9
φ fraktur = 0,75
Fy = 290Mpa = 290 N / mm 2 = 0, 29kN / mm 2 = 29kN / cm 2
Fu = 500Mpa = 500 N / mm 2 = 0,5kN / mm2 = 50kN / cm2
φ baut = 50mm = 5cm
Preliminary disain ;
Pu ≤ φ Pn
Pu = φ .Fy. Ag
Pu 4380, 74
Ag = =
φ Fy 0,9 x 29
Ag = 167,8cm2
Dicoba menggunakan profil IWF 400x400
Tw = 13 mm
Tf = 21 mm
R = 22 mm
Ag = 218,7 cm2
Ix = 66600cm4
Iy = 22400 cm4
ix = 17,5 cm
iy= r min= 10,1cm
Sx =3330 cm3
Zx = 3672,35 cm3

12. Cek kuat leleh tarik
Pu ≤ φ Pn
4380,74 ≤ 0,9 xFy. Ag
4380,74 ≤ 0,9 x 29 x 218, 7
4380,74 ≤ 5708,07 kN .........(OK )
Cek kuat leleh fraktur
Pu ≤ φ Pn
φ Pn = 0, 75 xFu. Ag φ Pn = φ FuxAe
An = Ag − 4(luas lub ang ) φ Pn = 0, 75 x0,5 x15714
An = 21870 − 10(21x(20 + 1)) φ Pn = 5892,75kN
An = 17460mm 2 Pu ≤ φ Pn
Ae = AnxU 4380,74 ≤ 5892, 75kN .............(OK )
Ae = 17460 x0,9 = 15714
DISAIN BATANG TEKAN
Nu = −4942, 08kN .......( s8)
φTekan = 0,85
Fy = 290Mpa = 290 N / mm2 = 0, 29kN / mm 2 = 29kN / cm 2
Fu = 500Mpa = 500 N / mm2 = 0,5kN / mm 2 = 50kN / cm 2
kc = 0, 7
L = 4m = 400cm
E = 200.000 Mpa
Fr = 70Mpa
Dicoba menggunakan profil IWF 400x400
Tw = 13 mm
Tf = 21 mm
R = 22 mm
Ag = 218,7 cm2
Ix = 66600cm4
Iy = 22400 cm4
ix = 17,5 cm
iy= r min= 10,1cm
Sx =3330 cm3
Zx = 3672,35 cm3

13. Cek kelangsingan penampang
kc.L
≤ 140
r min
0, 7 x 400
≤ 140
10,1
27, 7 ≤ 140.........(OK )
Cek local buckling
b 400
λ= = = 9,52
2.tf 2 x 21
170 170
λp = = = 9,98
Fy 290
370 370
λr = = = 24,9
Fy − Fr 290 − 70
∴ λ < λ p........ ⇒ Penampang kompak
Cek lateral buckling
H − 2tf − 2r 400 − 2 x 21 − 2 x 22
λ= = = 24,15
tw 13
1680 1680
λp = = = 98,65
Fy 290
2550 2550
λr = = = 171,9
Fy − Fr 290 − 70
∴ λ < λ p.......... ⇒ Penampang kompak
Cek kuat tekan nominal
kc.L Fy
λc =
r min .π E
0, 7 x 400 290
λc =
10,1x3,14 200000
λ c = 8,83x0, 038
λ c = 0,33 < 1,5
Maka;
Nu ≤ φ Nn
2
4942, 08kN ≤ φ (0, 66λ c ) Ag.Fy
2
4942, 08kN ≤ 0,85(0, 660,33 )218, 7 x 29kN
4942, 08kN ≤ 5152, 45kN ...........(OK )

14. DISAIN SAMBUNGAN DENGAN BAUT
BUHUL A
Pu = 2289,38kN = 2289380 N .....( s1)
Dbaut = 20mm
Abaut = 1256mm2
Jumlah baut :
S1 = 40mm
S 2 = 140mm
S = 40mm
t. pelat = 20mm
tf = 21mm
φ f = 0.75
Fu = 410Mpa
Sambungan tipe tumpu
Rd = 2, 4.φ f .d .tp.Fu.n
Rd = 2, 4 x0,75 x 20 x(20 x 2) x 410.n
2289380
n= = 3,8 ⇒ 4 Baut
590400
BUHUL B
Pu = 4376,5kN = 4376500 N .....( s 4)
Dbaut = 20mm
Abaut = 1256mm 2
Jumlah baut :
S1 = 40mm
S 2 = 140mm
S = 40mm
t. pelat = 20mm
tf = 21mm
φ f = 0.75
Fu = 410Mpa
Sambungan tipe tumpu
Rd = 2, 4.φ f .d .tp.Fu.n
Rd = 2, 4 x0, 75 x 20 x(20 x 2) x 410.n
4376500
n= = 7, 4 ⇒ 8 Baut
590400

15. BUHUL C
Pu = 4942, 08kN = 4942080 N .....( s 4)
Dbaut = 20mm
Abaut = 1256mm 2
Jumlah baut :
S1 = 40mm
S 2 = 140mm
S = 40mm
t. pelat = 20mm
tf = 21mm
φ f = 0.75
Fu = 410Mpa
Sambungan tipe tumpu
Rd = 2, 4.φ f .d .tp.Fu.n
Rd = 2, 4 x0, 75 x 20 x(20 x 2) x 410.n
4942080
n= = 8, 4 ⇒ 10 Baut
590400
BUHUL D
Pu = 4942, 08kN = 4942080 N .....( s8)
Dbaut = 20mm
Abaut = 1256mm 2
Jumlah baut :
S1 = 40mm
S 2 = 140mm
S = 40mm
t. pelat = 20mm
tf = 21mm
φ f = 0.75
Fu = 410Mpa
Sambungan tipe tumpu
Rd = 2, 4.φ f .d .tp.Fu.n
Rd = 2, 4 x0, 75 x 20 x(20 x 2) x 410.n
4942080
n= = 8, 4 ⇒ 10 Baut
590400

16. BUHUL H
Pu = 2289,38kN = 2289380 N .....( s1)
Dbaut = 20mm
Abaut = 1256mm2
Jumlah baut :
S1 = 40mm
S 2 = 140mm
S = 40mm
t. pelat = 20mm
tf = 21mm
φ f = 0.75
Fu = 410Mpa
Sambungan tipe tumpu
Rd = 2, 4.φ f .d .tp.Fu.n
Rd = 2, 4 x0,75 x 20 x(20 x 2) x 410.n
2289380
n= = 3,8 ⇒ 4 Baut
590400
BUHUL I
Pu = 4380, 62kN = 4380620 N .....( s10)
Dbaut = 20mm
Abaut = 1256mm 2
Jumlah baut :
S1 = 40mm
S 2 = 140mm
S = 40mm
t. pelat = 20mm
tf = 21mm
φ f = 0.75
Fu = 410Mpa
Sambungan tipe tumpu
Rd = 2, 4.φ f .d .tp.Fu.n
Rd = 2, 4 x0, 75 x 20 x(20 x 2) x 410.n
4380620
n= = 7, 4 ⇒ 8 Baut
590400

17. BUHUL J
Pu = 4380, 62kN = 4380620 N .....( s10)
Dbaut = 20mm
Abaut = 1256mm 2
Jumlah baut :
S1 = 40mm
S 2 = 140mm
S = 40mm
t. pelat = 20mm
tf = 21mm
φ f = 0.75
Fu = 410Mpa
Sambungan tipe tumpu
Rd = 2, 4.φ f .d .tp.Fu.n
Rd = 2, 4 x0, 75 x 20 x(20 x 2) x 410.n
4380620
n= = 7, 4 ⇒ 8 Baut
590400
• CEK BLOCK SHEAR
Agt = 4.S .tf = 4.40.21 = 3360mm 2
D lub ang
Ant = 4.S .tf − 4 tf
2
21
= 4 x 40 x 21 − 4 21 = 2478mm 2
2
Ags = 4( S1 + 3S 2).tf = 4(40 + 3.140).21 = 38640mm 2
Ans = 4( S1 + 3S 2).tf − 4 x3,5 D lub ang.tf
= 38640 − 4 x3,5 x 21x 21 = 32466mm2
FuAnt = 500 x 2478 = 1239000 Mpa = 1239kN
0, 6 FuAns = 0,6 x500 x32466 = 9739800Mpa = 9739,8kN
0, 6 FuAns > FuAnt....... ⇒ RETAK GESER − LELEH TARIK
φ Pn = φ ( FyAgt + 0, 6 FuAns)
= 0, 75(0, 29 x3360 + 9739,8)
= 0, 75 x10714, 2kN
= 8035, 65kN ................(OK )

18. Cek kuat leleh fraktur
Pu ≤ φ Pn
φ Pn = 0, 75 xFu. Ag
An = Ag − 10(luas lub ang )
An = 21870 − 10(21x(20 + 1))
An = 17460mm 2
Ae = AnxU
Ae = 17460 x0,9 = 15714
φ Pn = φ FuxAe
φ Pn = 0, 75 x0,5 x15714
φ Pn = 5892,75kN
Pu ≤ φ Pn
4380,74 ≤ 5892, 75kN .............(OK )

19. LAMPIRAN 1
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