Perencanaan Beton
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PERENCANAAN BETON
Dimensi Balok sudah diketahui,Hitung M.max yang mampu dipikul Balok tersebut…………….???????Hitungan :
fc = 25 Mpa (fc : mutu beton )fy = 320 Mpa
h = 400 mm ( diketahui )b = 200 mm ( diketahui )d = 370 mmn = 3 bh
= 16 mmπ = 3.14a =
Hitung Mmax……..???
As = Luas Penampangn = Jumlah Tulangan
As = → (d² : diameter kwadrat)= 602.88
a = As x fy0,85 x fc x b
= 45.3933 mm
Mn = Momen NominalMn = As x fy ( d - a/2)
= 67,002,316.27 Nm= 6700.23 Kg.M
M max == 5360 Kg.m
5360 Kg.m
(fy : mutu baja,240 besi polos,320 besi ulir )
d/Ø
1/4 x π x d² x nmm²
Ø Mn
» Jadi M max yang mampu dipikul Balok tersebut adalah
PERENCANAAN BETON
Balok menerima Beban :M = 12000 Kg.m' = Nmmfc = 25 Mpa (fc : mutu beton )fy = 320 Mpa
Dimensi Tulangan :Ø 6 mmØ 8 mmØ 10 mmØ 12 mmØ 16 mmØ 19 mm
ρ min = 1,4/fy= 0.0044
Misalρ = 0.005 ∞ ρ = 0.07 (misal)
h = ³√2 x 1,2 x 10⁸ → m = fyØρ x fy(1-1/2ρm) 0,85 x fc
= 15.06= 580 mm
b = h/2= 290 mm → 300 mm
h = 600 mmd = 570 mm
As = ρ x b x d → A == 855 = 201
N → 4.25 →
ρ = Rasio TulanganMu = Momen Ultimate Beban
Mu = Nmm = 120000000 Nmm
fc = 25 MPa (fc : mutu beton )fy = 320 MPa
ρ min = 1,4/fy m = 15.06= 0.0044
1,2 x 10⁸
(fy : mutu baja,240 besi polos,320 besi ulir )
π/4 x d²
( 5 Ø 16)
1,2 x 10⁸
(fy : mutu baja,240 besi polos,320 besi ulir )
300
600 570
30
ρ max = x 600fy 600 + fy
= 0.0276
Rn = MuØ b x d²
= 1.54
ρ == 0.005
As = ρ x b x d → A == 855 = 201
N → 4.25 →
Tentukan Dimensi Balok dan Penulangannya,jika diharapkan rasio tulangan sebesar 0,007 ???
Balok menerima Beban :M = 12000 Kg.m' = = 120000000 Nmm
Momen akibat Beban Mati = 9000 Kg.mMomen akibat Beban Hidup = 11300 Kg.mBentang (L) = 4 m ( tinggi balok biasanya 1/12 dari bentang )
fc = 25 Mpa (fc : mutu beton )fy = 320 Mpa
Dimensi Tulangan :Ø 6 mmØ 8 mmØ 10 mm
Ø 12 mmØ 16 mmØ 19 mm
Mencari momen :
q = ( 1,2 x beban mati ) + ( 1,6 x beban hidup )= 28880 Kg.m
M == 57760 Kg.m= 577600000 Nmm
Misal
0,75 x 0,85 x β x fc
1/m(1-√1-(2.m.Rn/fy))
π/4 x d²mm²
( 5 Ø 16)
1,2 x 10⁸( jika dimensi sudah diketahui maka berat sendiri dihitung dan ditambahkan ke beban mati),Rumus berat sendiri = luas penampang x 2400 (berat beton per m³)
(fy : mutu baja,240 besi polos,320 besi ulir )
1/8 x q x L²
ρ = 0.007
h = ³√2 x 1,2 x 10⁸ → m = fyØρ x fy(1-1/2ρm) 0,85 x fc
= 15.06= 880 mm
b = h/2= 440 mm → 450 mm
h = 900 mmd = 870 mm
As = ρ x b x d → A == 2739 = 201
N → 13.63 →
π/4 x d²mm²
( 14 Ø 16)
→ 0.333333 mh = 33.33 cm → 40 cmb = 20 cm
h = 0.4 mb = 0.2 m
Misal berat sendiri balok :→ 192 Kg.m
Mencari momen :
q = ( 1,2 x (beban mati+berat sendiri balok ) + ( 1,6 x beban hidup )= 29110 Kg.m
M == 58221 Kg.m
= 582208000 Nmm
Misal
( jika dimensi sudah diketahui maka berat sendiri dihitung dan ditambahkan ke beban mati),Rumus berat sendiri = luas penampang x 2400 (berat beton per m³)
1/8 x q x L²
ρ = 0.007
h = ³√2 x 1,2 x 10⁸ → m = fyØρ x fy(1-1/2ρm) 0,85 x fc
= 15.06= 882 mm
b = h/2= 441 mm → 450 mm
h = 900 mmd = 870 mm
As = ρ x b x d → A == 2740 = 201
N → 13.64 →
π/4 x d²mm²
( 14 Ø 16)
Perhitungan Plat
→ Plat dihitung per 1 m
ρ = → As = ρ x b x d
→ m = fy0,85 x fc
Rn = MuØ b x d²
→ Hasil tulangan yang dipilih Ø 10 - 150
As = → 1000 adalah Plat dihitung per 1 m=
Perhitungan Penulangan pada Plat
Lx = 3 →Ly = 3.5
Ly/Lx = 1.17 Dibulatkan ∞ 1.2 ( Cari di Tabel Momen )
Ly/Lx > 2 = 2 Arah< 2 = 1 Arah
Mu =
Tulangan Geser
Gaya geser di tahan oleh :- Beton- Sengkang
Vn = VuØ geser
BetonVc = ⅙ √fc.b.d Ø = 0.75 → Faktor Reduksi Kekuatan Geser
Ø = 0.8 → Faktor Reduksi Kekuatan LenturSengkang
1/m(1-√1-(2.m.Rn/fy))
π/4 x d² x 1000/150π/4 x 10² x 1000/150
0,001 x q x Lx² x Cx
Vs = Vu - VcØ geser
Vs = Av.fy.d/s → Av = Luas Tulangan Geser (π/4 x d² x 2 ).....2 adlh jumlah sengkang kanan kiri s = Jarak Sengkang
Diameter ditentukan :Av = π/4 x d² x 2 → Ø 10/8
s = Av.fy.dVs
Contoh Perhitungan Tulangan Geser
Sengkang Ø 10 mm
d = h-selimut-sengkang-1/2 Ø = 555
Diketahui :Beban Hidup ( qL ) = 1,500 Kg/mBeban Mati ( qD ) = 2,000 Kg/mPanjang ( L ) = 10 MLebar ( b ) = 400 mmTinggi ( h ) = 600 mmTinggi - Selimut ( d ) = 555 mmMutu Beton ( fc ) = 25 MpaMutu Baja ( fy ) = 240 Mpa ( Besi Polos )
Hitungan :
q total = ( 1,2 x beban mati ) + ( 1,6 x beban hidup )= 4,800 Kg/m
M = ⅛ x q total x L²= 60,000 Kg/m
Vu = ⅟₂ x q total x L= 24,000 Kg/m= 240,000 N
Gaya Geser Beton Max.Vc = ⅙ √fc.b.d
= 185,000 N
Vs = Vu - VcØ geser
= 135,000 N
Ø 10 → Av = π/4 x d² x 2 = 157
s = Av.fy.dVs
= 154.906666666667 mm
Diperoleh sengkang Ø 10 - 150 mm
Tulangan Lentur
Mutu Beton ( fc ) = 25 MpaMutu Baja ( fy ) = 320 Mpa ( Besi Polos )
Mu = 60,000 Kg.m'= 6 x 10⁸ Nmm= 600,000,000 Nmm
m = fy Rn = Mu0,85 x fc Ø b x d²
= 15.06 = 6.09
ρ min = 1,4/fy= 0.0044
ρ max = x 600fy 600 + fy
= 0.0325Ø 19
ρ = Ø 29= 0.02301 Ø 32
As = ρ x b x d → A == 5107.85 = 660
N → 7.74 →
Plat Lantai
Di ket :Tebal Plat = 12 cm = 120 mmBeban Hidup = 250 Kg/m²Mutu Beto ( fc ) = 25 MpaMutu Baja ( fy ) = 240 Mpa
mm²
0,75 x 0,85 x β x fc
1/m(1-√1-(2.m.Rn/fy))
π/4 x d²mm²
( 8 Ø 29)
Ly = 4.75 MLx = 3 M
Beban Mati :BS. Plat 0.12 x 2400 = 288 Kg/m²Spasi 0.02 x 2200 = 44 Kg/m²Keramik 0.01 x 2400 = 24 Kg/m²Plafond + Penggantung = 18 Kg/m²
= 374 Kg/m²
q = ( 1,2 x beban mati ) + ( 1,6 x beban hidup )= 849 Kg/m²
Ly = 1.58Lx
= 1.60 → Ctx = 79 ( Dari Tabel Momen )Cty = 57
Mu == 603.50 Kg.m= 6.03 Nmm= 6,034,968 Nmm
b = 1000 mm ( Hitungan Plat Per 1 m )h = 120 mmd = 90 mm
Rn = MuØ b x d²
= 0.93m = fy
0,85 x fc= 11.29
ρ min = 1,4/fy= 0.0058
ρ perlu == 0.00397
Di pakai ρ min = 0.0058
-0,001 x q x Lx² x Ctx
x 10⁶
1/m(1-√1-(2.m.Rn/fy))
As = ρ x b x d= 525
→
Balok Jembatan
Di ket :Beban Terpusat ( P1 ) = 10 ton = 10000 KgBeban Terpusat ( P2 ) = 10 ton = 10000 KgBeban Merata ( q ) = 2400 Kg/mLebar ( L ) = 15 mVA/RA (q.L/2)+ P = 28,000 Kg = 280,000 N →VB/RB = 28,000 Kg = 280,000 Nb = 400 mmh = 900 mmd = 855 mm → Sengkang Ø 10fc = 25 Mpa d = h-selimut-sengkang-1/2 Ø fy ( untuk tul. Lentur ) = 320 Mpa = 855fy ( untuk tul. geser ) = 240 Mpa
M = → 5 adalah jarak= 117,500 Kg.m= 1.175 Nmm= 1,175,000,000 Nmm
m = fy Rn = Mu0,85 x fc Ø b x d²
= 15.06 = 5.02
ρ min = 1,4/fy= 0.0044
ρ max = x 600fy 600 + fy
= 0.0276Ø 19
ρ = Ø 29= 0.0182 Ø 32
As = ρ x b x d → A == 6219.99 = 804
N → 7.74 →
mm²
( Ø10 - 125)
1/8 x q x L² + ( P x 5 )
x 10⁹
0,75 x 0,85 x β x fc
1/m(1-√1-(2.m.Rn/fy))
π/4 x d²mm²
( 8 Ø 32 )
Tulangan Geser
Gaya Geser Beton Max. Ø = 0.75 → Faktor Reduksi Kekuatan GeserVc = ⅙ √fc.b.d Ø = 0.8 → Faktor Reduksi Kekuatan Lentur
= 285,000 N
Vs = Vu - VcØ geser
= 88,333 N
Ø 12 → Av = π/4 x d² x 2 = 226.08
s = Av.fy.dVs
= 525 mm
s max = d/2= 427.5 mm
Diperoleh sengkang Ø 12 - 430 mmatau
Ø 12 - 500 mm
mm²