DISAIN JEMBATAN RANGKA BAJA
2 SIPIL 1 PAGI UNTUK MELENGKAPI TUGAS BAJA
OLEH: GILANG ADITYA.P DANIEL HASIBUAN SADDAM FIRGIAWAN
DEPOK,15 JUNI 2010
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
*Gaya-Gaya Batang
BUHUL A
∑ V = 0 −
1 2
P− S1.0, 6 + 3 P= 0
− S 1 =
−2,5 2, 5 P
0,6 0, 6
∑ H = 0
S1SIN 53 + S 2 = 0 S 2 = −4,17 P .0, 8 S 2 = −3, 33P
S1 − 4,17 P
BUHUL B
∑ V = 0
S1COS 53 + S 3 + S 6SIN1 4 = 0 4,17 P.0, 6 + S 3 + 3, 44.0, 24 = 0
S 3 = −3, 33P
∑ H = 0 − S1SIN 53 + S 6COS1 4 = 0 −4,17.0, 8 P + S 6.0, 97 = 0
S6 =
3,33 P 0,97
= 3, 44 P
BUHUL C ∑ V = 0 − P− S3 − S5 SIN2 6, 6 = 0 − P+ 3, 33 P− S50 , 45 = 0
S 5 =
−2,33 P −0,45
∑ H = 0 − S 2 + S 4 + S 5COS 2 6, 6 = 0
3, 33 P+ S4 + 5,18 P.0, 89 = 0
S 4 = −7, 96 P
S 5 = 5,18 P
BUHUL D ∑ V = 0
S 7 + S 5COS 63, 4 − S 6COS 7 6 = 0 S 7 + 5,18 P.0, 45 − 3, 44 P .0, 24 = 0 S 7 = −1, 5 P
∑ H = 0
S10 − S 5SIN 63, 4 − S 6 SIN 7 6 = 0 S10 − 5,18 P.0, 89 − 3, 44 P .0, 97 = 0 S10 = 7, 97 P
BUHUL E
∑ V = 0
∑ H = 0
− P− S7 − S9 SIN2 6, 6 = 0
− S 4 + S 8 + S 9COS 2 6, 6 = 0
− P+ 1, 5 P− S9. 0, 45 = 0
7, 96 P+ S8 + 1,11 P.0, 89 = 0
S9 =
0,5 0, 5 0,45
= 1,11P
BUHUL F ∑ V = 0 − P − S 11 = 0
S 11 = − P
S 8 = −9 P
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
GAYA YANG BEKERJA
GAYA AXIAL
GAYA GESER
MOMEN
TABEL BEBAN BERJALAN
GRAFIK BEBAN BERJALAN
TOTAL GAYA YANG BEKERJA DISETIAP BATANG
PERENCANAAN RANGKA UTAMA
Tebal trotoar Lebar trotoar Panjang segmen Tinggi rangka Lebar jembatan Beban jalur Beban garis
= 20 cm = 100 cm =4m =3m =8m = 9 kN/m = 49 kN/m
PEMBEBANAN PADA DIAFRAGMA
Beban mati mati ∴ Beban 4 x0, 5 = 24 kN/ m − Berat perker asan= 24 x −
Berat aspal
= 22 4x 0x, 05 = 4, 4
/ m kN
qdl = 28, 4kN / m M max =
1
x28, 4 x82 = 227, 2 kN .m
8 Mu = 1, 4 xM max = 1, 4 x227, 2 = 318,1kN .m
BEBAN MATI
B.mati +diafragma = RA x factor beban = 113.6 x 1.1 B.trotoar = Bj beton x t x I x λ x λ x factor beban = 24 x 0,2 x 1 x 4 x 1,3 PDl
=124,96 kN =24,96kN = 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 φ tarik φ tarik φ fraktur φ fraktur
= 4014,11 kN (s10) = 0,9 = 0,75
Fy= 290 Mpa= 290 N/ mm2 = 0, 29 kN/ mm2 = 29 kN/ cm2 Fu= 500 Mpa= 500 N/ mm2 = 0, 5 kN/ mm2 = 50 kN/ cm2 φ baut = 50mm = 5cm Preliminary disain ;
Pu≤ φ Pn Pu= φ . Fy. Ag Ag =
Pu
=
4380,74
φ Fy 0, 9 2x9
A=g167,8
2
cm
Dicoba menggunakan profil IWF 400x400 Tw = 13 mm = 21 mm Tf 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
+
Cek kuat leleh tarik
Pu≤ φ Pn 4380, 74 ≤ 0, 9 xFy. Ag 4380, 74 74 ≤ 0, 9 x29 x218, 7 4380, 74 74 ≤ 5708, 07 07 kN.........( OK ) Cek kuat leleh fraktur
Pu≤ φ Pn φ Pn= 0, 75 xF.u Ag
An= Ag− 4( luaslub ang) =2 A1n870 − 10( 21 ( 20 +x1)) =n17460 A
2 m m
φ Pn= φ FuxAe φ Pn= 0, 75 75 x0, 5 1x5714 φ
=P5892,75 n
Pu≤ φ Pn 4380, 74 74 ≤ 5892, 75 75 kN.............( OK )
Ae= AnxU A=e17460 0,x9 = 15714
DISAIN BATANG TEKAN
08kN .......( s8) u = −4942, 08 φ Tekan = 0,85
Fy= 290 Mpa= 290 N/ mm2 = 0, 29 kN/ mm2 = 29 kN/ cm2 Fu= 500 Mpa= 500 N/ mm2 = 0, 5 kN/ mm2 = 50 kN/ cm2 kc = 0, 7 L= 4 m= 400 cm 200.000 =E
Mpa
F= r 70 Mpa Dicoba menggunakan profil IWF 400x400
kN
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
Cek kelangsingan penampang
kc.L
≤ 140 r min 0, 7 x 400 ≤ 140 10,1
27, 7 ≤ 140.........( OK ) Cek local buckling
λ =
b 2.tf
λ p =
=
400 2 x 21
170
=
170
= 9,98
290
Fy 370
λ r =
= 9,52
370
=
Fy− Fr
290 − 70
∴ λ < λ ........ ⇒ p
= 24,9
Penampang kompak
Cek lateral buckling
21 − 2 x22 H− 2 tf − 2 r 400 − 2 x21
λ =
=
λ p =
tw 1680 1680 =
Fy 2550
λ r =
290
13 = 98,65
2550
=
Fy− Fr
= 24,15
290 − 70
∴ λ < λ .....p ..... ⇒
= 171,9
Penampaknogm pak
Cek kuat tekan nominal
λ c = λ c =
kc.L
Fy
r min.π
E
0, 7 x 400
290
10,1 x3,14
200000
λ c = 8,83 ,83 x0,038 ,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, 66 0,33 )218, 7 x29 kN 4942, 08kN ≤ 5152, 45 45 kN...........( OK )
DISAIN SAMBUNGAN DENGAN BAUT BUHUL
A
= 2289380 .N 38 kN ....( 1)s P= u 2289, 38
Dbau=t 20 mm 2 m m
Aba=u1256 t Jumlah baut : S1 = 40mm S 2 = 140mm S = 40mm t. pelat = 20mm tf = 21mm φ f = 0.75
F=u 410
Mpa
Sambungan tipe tumpu Rd = 2, 4.φ f . d. tp. Fu. n Rd = 2, 4 x0, 75 x20 x( 20 x2) x410. n n=
2289380 590400
BUHUL
= 3, 8 ⇒ 4 Baut
B
....( 4s) = 4376500 .N P= u 4376, 5 kN
Dbau=t 20 mm Aba= ut1256
m2 m
Jumlah baut : S1 = 40mm S 2 = 140mm S = 40mm t. pelat = 20mm tf = 21mm φ f = 0.75
F=u 410
Mpa
Sambungan tipe tumpu Rd = 2, 4.φ f . d. tp. Fu. n Rd = 2, 4 x0, 75 x20 x(20 x2) x410. n n=
4376500 590400
= 7, 4 ⇒ 8 Baut
BUHUL C
08 kN ....( 4s) = 4942080 .N P= u 4942, 08
Dbau=t 20 mm Aba= ut1256
m2 m
Jumlah baut : S1 = 40mm S 2 = 140mm S = 40mm t. pelat = 20mm tf = 21mm φ f = 0.75
F=u 410
Mpa
Sambungan tipe tumpu Rd = 2, 4.φ f . d. tp. Fu. n Rd = 2, 4 x0, 75 x20 x(20 x2) x410. n n=
4942080 590400
BUHUL
= 8, 4 ⇒ 10 Baut
D
= 4942080 .N 08 kN ....( 8)s P= u 4942, 08
Dbau=t 20 mm Aba= ut1256
m2 m
Jumlah baut : S1 = 40mm S 2 = 140mm S = 40mm t. pelat = 20mm tf = 21mm φ f = 0.75
F=u 410
Mpa
Sambungan tipe tumpu Rd = 2, 4.φ f . d. tp. Fu. n Rd = 2, 4 x0, 75 x20 x(20 x2) x410. n n=
4942080 590400
= 8, 4 ⇒ 10 Baut
BUHUL
H
38 kN ....( 1)s = 2289380 .N P= u 2289, 38
Dbau=t 20 mm Aba= ut1256
m2 m
Jumlah baut : S1 = 40mm S 2 = 140mm S = 40mm t. pelat = 20mm tf = 21mm φ f = 0.75
F=u 410
Mpa
Sambungan tipe tumpu Rd = 2, 4.φ f . d. tp. Fu. n Rd = 2, 4 x0, 75 x20 x(20 x2) x410. n n=
2289380 590400
BUHUL
= 3, 8 ⇒ 4 Baut
I
= 4380620 .N 62 kN ....( 10s) P= u 4380, 62
Dbau=t 20 mm Aba= ut1256
m2 m
Jumlah baut : S1 = 40mm S 2 = 140mm S = 40mm t. pelat = 20mm tf = 21mm φ f = 0.75
F=u 410
Mpa
Sambungan tipe tumpu Rd = 2, 4.φ f . d. tp. Fu. n Rd = 2, 4 x0, 75 x20 x(20 x2) x410. n n=
4380620 590400
= 7, 4 ⇒ 8 Baut
BUHUL
J
62 kN ....( 10s) = 4380620 .N P= u 4380, 62
Dbau=t 20 mm m2 m
Aba= ut1256 Jumlah baut : S1 = 40mm S 2 = 140mm S = 40mm t. pelat = 20mm tf = 21mm φ f = 0.75
F=u 410
Mpa
Sambungan tipe tumpu Rd = 2, 4.φ f . d. tp. Fu. n Rd = 2, 4 x0, 75 x20 x(20 x2) x410. n n=
4380620 590400
= 7, 4 ⇒ 8 Baut
• CEK BLOCK SHEAR
Ag=t 4. . S t=f 4.40.21 = 3360 An=t 4. .S t− f4
2
mm
Dlub ang
tf 2 21 x 21x − 4 21 = 2478 = 4 40 2
). + 3 2S Ag=s4( 1 S
m2m
= tf 4(40 + 3.140).21 = 38640
m2m
3, 5 Dlub ang. tf Ans= 4( S1 + 3 S2). tf t f − 4 x3, 2
3, 5 x21 x21 = 32466 mm = 38640 − 4 x
FuAn=t500 24 x 78 = 1239000 Mpa = 1239 kN 0, 6
FuA= ns0, 6 50x0 32x466 = 9739800
Mp=a9739, 8
kN
0, 6 FuAns> FuAn.t...... ⇒ RETAK GESER− LE LELEH TARIK φ
P=n φ ( FyAg+t 0, 6 FuAn) s ,75(0,29 x3360 + 9739,8) ,8) = 0,75( = 0,75 10714 07 ,2 x 14,2
kN
65kN ................( OK ) = 8035, 65
Cek kuat leleh fraktur
Pu≤ φ Pn φ Pn= 0, 75 xF.u Ag
An= Ag Ag− 10( luaslub ang) =2 A1n870 − 10( 21 ( 20 +x1)) =n17460 A
2
mm
Ae= AnxU A=e17460 0,x9 = 15714 φ Pn= φ FuxAe φ Pn= 0, 75 75 0 x, 5 1x5714 φ
n =P5892,75
kN
Pu≤ φ Pn 4380, 74 74 ≤ 5892, 75 75 kN.............( OK )
LAMPIRAN 1
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