1 PERILAKU BALOK BETON MENAHAN MOMEN
Rumus-Rumus Prosedur Perhitungan
a.
Merencanakan Dimensi Penampang Balok (As =……)
Ø Koefisien penampang = Rn = Mu/(0.80*0.85*fc’*b*d^2)
Ø Indeks Tulangan = ω = 1-(1-2*Rn)^0.5
Ø Rasio Tulangan = ρ = wn*0.85*fc'/fy
ρmin = 14/fy
ρmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)
Syarat : ρmin <= ρ <= ρmax ρ dipakai = ρ terbesar antara
ρ & ρmin
Ø Luas Tulangan = As = ρ dipakai*b*d
Ø Jumlah Tulangan = nb = As/Ab = As/0.25*π*(d^2)
Buat
beberapa alternatif ukuran tulangan : contoh : D16, D19, D22, D25, D32
Pilih yang paling ekonomis dan sesuai
struktur
Contoh-Contoh Perhitungan
Soal 1
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Diketahui :
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Mu=
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22tm
=
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2.20E+06
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(kg.cm)
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h=
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80
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(cm)
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d=
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72
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(cm)
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b=
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35
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(cm)
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fc' =
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240
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(kg/cm2)
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fy =
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4000
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(kg/cm2)
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1
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Koefisen
penampang : Rn =
Mu/(0.80*0.85*fc’*b*d^2)
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Rn=
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0.07430
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2
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Indeks tulangan
: wn = 1-(1-2*Rn)^0.5
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wn =
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0.07728
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3
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Rasio tulangan
(r) :
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a
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r =
wn*0.85*fc'/fy
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r =
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0.00394
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b
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rmin = 14/fy
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rmin =
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0.00350
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c
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rmax =
0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)
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rmax =
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0.01951
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4
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Luas
tulangan (As) :
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a
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r =
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0.00394
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rmin =
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0.00350
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Apakah : {
r< rmin} ?
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0
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b
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r =
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0.00394
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rmin =
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0.00350
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rmax =
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0.01951
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Apakah :
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{rmin< r ≤
rmax}?
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0.00394
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c
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r =
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0.00394
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rmax =
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0.01951
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Apakah :{ r
> rmax }?
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0
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Luas
tulangan (As) :
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r dipakai =
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0.00394
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As=
rdipakai*b*d
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9.93241
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cm2
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5
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Pilihan jumlah
tulangan (nb) :
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db(cm)
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Ab(cm2)
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nb=As/Ab
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Jumlah
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1.3
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1.327
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7.487
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8
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D
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13
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1.6
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2.010
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4.942
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5
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D
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16
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1.9
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2.834
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3.505
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4
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D
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19
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2.2
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3.799
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2.614
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3
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D
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22
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2.5
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4.906
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2.024
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3
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D
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25
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-----> Jadi dipilih yang ekonomis menggunakan baja
tulangan 5D16
1.6 x 5 = 8.0
2.5 x 6 = 15.0
1.0 x 2 = 2.0
= 25 <= 35 (OK)
Soal 2
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Diketahui :
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Mu=
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46
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4.60E+06
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(kg.cm)
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h=
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80
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(cm)
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d=
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72
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(cm)
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b=
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30
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(cm)
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fc' =
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180
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(kg/cm2)
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fy =
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3200
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(kg/cm2)
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1
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Koefisen
penampang : Rn =
Mu/(0.80*0.85*fc’*b*d^2)
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Rn=
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0.24165
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2
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Indeks tulangan
: wn = 1-(1-2*Rn)^0.5
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wn =
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0.28118
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3
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Rasio tulangan
(r) :
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a
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r =
wn*0.85*fc'/fy
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r =
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0.01344
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b
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rmin = 14/fy
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rmin =
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0.00438
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c
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rmax =
0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)
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rmax =
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0.01988
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4
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Luas
tulangan (As) :
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a
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r =
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0.01344
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rmin =
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0.00438
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Apakah : {
r< rmin} ?
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0
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b
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r =
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0.01344
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rmin =
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0.00438
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rmax =
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0.01988
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Apakah :
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{rmin< r ≤
rmax}?
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0.01344
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c
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r =
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0.01344
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rmax =
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0.01988
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Apakah :{ r
> rmax }?
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0
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Luas
tulangan (As) :
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r dipakai =
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0.01344
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As=
rdipakai*b*d
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29.0393
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cm2
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5
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Pilihan jumlah
tulangan (nb) :
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db(cm)
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Ab(cm2)
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nb=As/Ab
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Jumlah
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1.3
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1.327
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21.889
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22
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D
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13
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1.6
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2.010
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14.450
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15
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D
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16
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1.9
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2.834
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10.247
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11
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D
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19
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2.2
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3.799
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7.643
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8
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D
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22
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2.5
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4.906
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5.919
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6
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D
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25
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-----> Jadi dipilih
yang ekonomis menggunakan baja tulangan 6D25
Soal 3
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Diketahui :
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Mu=
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10
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1.00E+06
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(kg.cm)
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h=
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70
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(cm)
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d=
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63
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(cm)
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b=
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30
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(cm)
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fc' =
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200
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(kg/cm2)
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fy =
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4000
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(kg/cm2)
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1
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Koefisen
penampang : Rn =
Mu/(0.80*0.85*fc’*b*d^2)
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|
||||
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Rn=
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0.06175
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2
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Indeks tulangan
: wn = 1-(1-2*Rn)^0.5
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wn =
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0.06379
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3
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Rasio tulangan
(r) :
|
|
|
|
|
|
|
|
a
|
r =
wn*0.85*fc'/fy
|
|
|||||
|
|
r =
|
0.00271
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|
||||
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b
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rmin = 14/fy
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|
|||||
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rmin =
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0.00350
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|
||||
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c
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rmax =
0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)
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|||||
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rmax =
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0.01626
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4
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Luas
tulangan (As) :
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|
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a
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r =
|
0.00271
|
|
||||
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rmin =
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0.00350
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|
||||
|
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Apakah : {
r< rmin} ?
|
0.0035
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b
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r =
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0.00271
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rmin =
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0.00350
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rmax =
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0.01626
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Apakah :
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|||||
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{rmin< r ≤
rmax}?
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0
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||||
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c
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r =
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0.00271
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rmax =
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0.01626
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||||
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Apakah :{ r
> rmax }?
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0
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||||||
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Luas
tulangan (As) :
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r dipakai =
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0.0035
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As=
rdipakai*b*d
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6.615
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cm2
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5
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Pilihan jumlah
tulangan (nb) :
|
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||
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db(cm)
|
Ab(cm2)
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nb=As/Ab
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Jumlah
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1.3
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1.327
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4.986
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5
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D
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13
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1.6
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2.010
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3.292
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4
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D
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16
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1.9
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2.834
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2.334
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3
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D
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19
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2.2
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3.799
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1.741
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2
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D
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22
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2.5
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4.906
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1.348
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2
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D
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25
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-----> Jadi dipilih
yang ekonomis menggunakan baja tulangan 5D13
b. Menganalisa Penampang
Balok (Mu = …….)
Ø Rasio tulangan = ρ=As/(b*d)
Ø Indeks tulangan = ω=(fy*ρ)/(0.85*fc')
Ø Koefisien penampang = Rn=ω*(1-(ω/2))
Ø Mu=Rn*Ø*0.85*fc'*b*d^2
Contoh Soal
:
Soal 1
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Diketahui :
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|||
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|||
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h=
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80
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(cm)
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d=
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72
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(cm)
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b=
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35
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(cm)
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|
|
fc' =
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240
|
(kg/cm2)
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|
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fy =
|
4000
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(kg/cm2)
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As=
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10
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cm²
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ρ=As/(b*d)
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ρ=
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0.0039683
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ω=(fy*ρ)/(0.85*fc')
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ω=
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0.0778089
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Rn=ω*(1-(ω/2))
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Rn=
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0.0747818
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Mu=Rn*Ø*0.85*fc'*b*d^2
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Mu=
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22.143641
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tm
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---------->
Jadi momen yang dapat ditahan oleh luas tulangan 20 cm2 yaitu 22,14 tm
Soal
2
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Diketahui :
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|||
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h=
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70
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(cm)
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d=
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63
|
(cm)
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b=
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30
|
(cm)
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|
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fc' =
|
240
|
(kg/cm2)
|
|
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fy =
|
4000
|
(kg/cm2)
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As=
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15
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cm²
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ρ=As/(b*d)
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ρ=
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0.0079365
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ω=(fy*ρ)/(0.85*fc')
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ω=
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0.1556178
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Rn=ω*(1-(ω/2))
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Rn=
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0.1435094
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Mu=Rn*Ø*0.85*fc'*b*d^2
|
|||
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Mu=
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27.887059
|
tm
|
|
---------->
Jadi momen yang dapat ditahan oleh luas tulangan 20 cm2 yaitu 27,89 tm
Soal 3
Diketahui :
| |||
h=
|
90
|
(cm)
| |
d=
|
81
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(cm)
| |
b=
|
40
|
(cm)
| |
fc' =
|
240
|
(kg/cm2)
| |
fy =
|
4000
|
(kg/cm2)
| |
As=
|
20
|
cm²
| |
ρ=As/(b*d)
| |||
ρ=
|
0.0061728
| ||
ω=(fy*ρ)/(0.85*fc')
| |||
ω=
|
0.1210361
| ||
Rn=ω*(1-(ω/2))
| |||
Rn=
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0.1137112
| ||
Mu=Rn*Ø*0.85*fc'*b*d^2
| |||
Mu=
|
48.702745
|
tm
| |
----------> Jadi momen yang dapat ditahan oleh luas tulangan 20 cm2 yaitu 48,7 tm
