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)

300

600 570

30

→ 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

Luas Tulangan Geser (π/4 x d² x 2 ).....2 adlh jumlah sengkang kanan kiri

Karena Simetris

mmh-selimut-sengkang-1/2 Ø

Faktor Reduksi Kekuatan GeserFaktor Reduksi Kekuatan Lentur