6.balok
DESCRIPTION
bTRANSCRIPT
Type B1 pada daerah tumpuan
Mu= 596798 (kg.cm) Diameter tulangan 16 mm
h= 60 (cm) Diameter sengkang 8 mm
b= 30 (cm) Penutup beton d' 5 cm
fc' = 200 (kg/cm2) d= 53 (cm)
fy = 2400 (kg/cm2)
1 Koefisen penampang : Rn = Mu/(0.80*0.85*fc’*b*d^2)Rn= 0.051
2 Indeks tulangan : wn = 1-(1-2*Rn)^0.5w = 0.053
3 Rasio tulangan (r) :
a r = wn*0.85*fc'/fy = 0.0037
b r min = 14/fy = 0.0058c rmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy) = 0.0323
4 Luas tulangan (As) :
a r = 0.0037 rmin = 0.0058Apakah : { r< rmin} ? 0.0058
b r = 0.0037 rmin = 0.0058 rmax = 0.0323
Apakah : {rmin< r ≤ rmax}? Salah 0
c r = 0.0037 rmax = 0.0323Apakah :{ r > rmax }? 0
Luas tulangan (As) :r dipakai = 0.0058As= r dipakai*b*d 9.345 cm2
Bila : r > rmax dan diinginkan ditahan tulangan tekan (tulangan rangkap) :
r1= rmax ;
As1= r1*b*d; wn1= r1*fy/(0.85*fc);
Rn1=wn1*(1-0.5*wn1); M1= Rn1*0.85*fc’*b*d2;
M2= Mu/0.80 – M1; As2 = M2/{(d-d’)*(fy-0.85*fc’);
As= As1 + As2
As’=As2r1 = 0.0000 M1= kg.cm
As1= 0.0000 cm2 M2= kg.cmwn1= 0.0000 As2= cm2
Rn1= 0.0000 As= cm2As'= cm2
5 Pilihan jumlah tulangan (nb) :
db(cm) Ab(cm2) nb=As/Ab Jumlah
1.6 2.010 5 D 161.6 2.010 0 D 16
6 Susunan tulangan :
Tulangan tarik As (Atas) 5 D 16
Tulangan tekan As' (Bawah) 2 D 16
0.000
Salah
DESAIN TULANGAN BALOK
0
0
0.000
0.000
4.6500.000
Benar
7 Tulangan geser
Lebar = cm Jumlah kaki = 2 kaki
Tinggi = cm Selimut beton = 5.00 cm
Gaya geser beban terfaktor Vud = kg f'c = 200 kg/cm
2
Momen torsi beban terfaktor Tud= kg-m fy = 2400 kg/cm2
Diameter tulangan sengkang = mm Faktor reduksi f = 0.75
Perhitungan :
Av = cm2smak = 27.5 cm
d = cm Vsmin = 4825 kg
Vc = 1/6*fc’*b*d = kg fVsmin = 3619 kg
fVc = 0,75 * Vc = kg Vsmak = 49193 kg
Tn = kg-m fVsmak = 36895 kg
fTu = kg-m
fVs =
Penulangan :
s[cm] fVc [kg] Vuk [kg] Vud[kg]
200.00 9224 9721 6529
Type B1 pada daerah lapangan
Mu= 451261 (kg.cm) Diameter tulangan 16 mm
h= 60 (cm) Diameter sengkang 8 mm
b= 30 (cm) Penutup beton d' 5 cm
fc' = 200 (kg/cm2) d= 53 (cm)
fy = 2400 (kg/cm2)
1 Koefisen penampang : Rn = Mu/(0.80*0.85*fc’*b*d^2)Rn= 0.03879
2 Indeks tulangan : wn = 1-(1-2*Rn)^0.5wn = 0.03957
3 Rasio tulangan (r) :
a r = wn*0.85*fc'/fy = 0.00280
b r min = 14/fy = 0.00583c rmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy) = 0.03225
4 Luas tulangan (As) :
a r = 0.00280 rmin = 0.00583Apakah : { r< rmin} ? 0.00583
b r = 0.00280 rmin = 0.00583 rmax = 0.03225
Apakah : {rmin< r ≤ rmax}? Salah 0
c r = 0.00280 rmax = 0.03225Apakah :{ r > rmax }? 0
Luas tulangan (As) :r dipakai = 0.00583As= r dipakai*b*d 9.345 cm2
Bila : r > rmax dan diinginkan ditahan tulangan tekan (tulangan rangkap) :
r1= rmax ;
Benar
Salah
Keterangan
jarak sengkang memenuhi
498
498
fVs [kg]
1207
906
6529
0
1.01
55.00
12298
9224
8
30
60
As1= r1*b*d; wn1= r1*fy/(0.85*fc);
Rn1=wn1*(1-0.5*wn1); M1= Rn1*0.85*fc’*b*d2;
M2= Mu/0.80 – M1; As2 = M2/{(d-d’)*(fy-0.85*fc’);
As= As1 + As2
As’=As2r1 = 0.0000 M1= kg.cm
As1= 0.0000 cm2 M2= kg.cmwn1= 0.0000 As2= cm2
Rn1= 0.0000 As= cm2As'= cm2
5 Pilihan jumlah tulangan (nb) :
db(cm) Ab(cm2) nb=As/Ab Jumlah
1.6 2.010 0 D 161.6 2.010 5 D 16
6 Susunan tulangan :
Tulangan tarik As (Atas) 2 D 16
Tulangan tekan As' (Bawah) 5 D 16
7 Tulangan geser
Lebar = cm Jumlah kaki = 2 kaki
Tinggi = cm Selimut beton = 5.00 cm
Gaya geser beban terfaktor Vud = kg f'c = 200 kg/cm
2
Momen torsi beban terfaktor Tud= kg-m fy = 2400 kg/cm2
Diameter tulangan sengkang = mm Faktor reduksi f = 0.75
Perhitungan :
Av = cm2smak = 27.5 cm
d = cm Vsmin = 4825 kg
Vc = 1/6*fc’*b*d = kg fVsmin = 3619 kg
fVc = 0,75 * Vc = kg Vsmak = 49193 kg
Tn = kg-m fVsmak = 36895 kg
fTu = kg-m
fVs =
Penulangan :
s[cm] fVc [kg] Vuk [kg] Vud[kg]
200.00 9224 9721 125
8 Sket Balok
BALOK
UKURAN
LOKASI
ATAS 5 D 2 D 16
SAMPING 2 2 10
BAWAH 2 D 5 D 16
SENGKANG 8 - 2008 - 200
498
fVs [kg] Keterangan
498 jarak sengkang memenuhi
B1
300 x 600
0
8
LAPANGAN
1.01
55.00
12298
9224
1207
906
4.6500.000
30
60
125
0
0
0.000
10
0.0000.000
16
TUMPUAN
300 x 600
16
SECTION