Hitungan_Balok_1_ok.xls
12
Mata Kuliah : STRUKTUR BETON I (Ir.H.Sumirin, MS.) Topik : Contoh Soal Disain Penampang Balok Menahan Momen Lentur ( Tulangan Tungga Diketahui : Mu= 50 5.00E+06 (kg.cm) h= 70 (cm) d= 63 (cm) b= 30 (cm) fc' = 200 fy = 4000 1 Koefisen penampang : Rn = Mu/(0.80*0.85*fc’*b*d^2) Rn= 0.30877 2 0.38156 3 a 0.01622 b 0.00350 c 0.01626 4 Luas tulangan (As) : a 0.01622 0.00350 Salah 0 b 0.01622 0.00350 0.01626 Apakah : Benar 0.01622 c 0.01622 0.01626 Salah 0 Luas tulangan (As) : 0.01622 30.6487 cm2 5 Pilihan jumlah tulangan (nb) : db(cm) Ab(cm2) nb=As/Ab Jumlah 1.3 1.327 23.102 24 D 13 1.6 2.010 15.251 16 D 16 (kg/cm 2 ) (kg/cm 2 ) Indeks tulangan : wn = 1-(1-2*Rn)^0.5 wn = Rasio tulangan (r) : r = wn*0.85*fc'/fy r = rmin = 14/fy rmin = rmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy) rmax = r = rmin = Apakah : { r< rmin} ? r = rmin = rmax = {rmin< r ≤ rmax}? r = rmax = Apakah :{ r > rmax }? r dipakai = As= rdipakai*b*d
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Transcript of Hitungan_Balok_1_ok.xls
Mata Kuliah : STRUKTUR BETON I (Ir.H.Sumirin, MS.)Topik : Contoh Soal Disain Penampang Balok
Menahan Momen Lentur ( Tulangan Tunggal )Diketahui :
Mu= 50 5.00E+06 (kg.cm)
h= 70 (cm)
d= 63 (cm)
b= 30 (cm)
fc' = 200
fy = 4000
1 Koefisen penampang : Rn = Mu/(0.80*0.85*fc’*b*d^2)Rn= 0.30877
20.38156
3a
0.01622b
0.00350c
0.016264 Luas tulangan (As) :
a 0.016220.00350
Salah 0b 0.01622
0.003500.01626
Apakah :Benar 0.016216
c 0.016220.01626
Salah 0
Luas tulangan (As) :0.01621630.64874 cm2
5 Pilihan jumlah tulangan (nb) :db(cm) Ab(cm2) nb=As/Ab Jumlah
1.3 1.327 23.102 24 D 131.6 2.010 15.251 16 D 16
(kg/cm2)
(kg/cm2)
Indeks tulangan : wn = 1-(1-2*Rn)^0.5wn = Rasio tulangan (r) : r = wn*0.85*fc'/fyr = rmin = 14/fyrmin =rmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)rmax =
r = rmin =Apakah : { r< rmin} ?r = rmin =rmax =
{rmin< r ≤ rmax}?r = rmax =Apakah :{ r > rmax }?
r dipakai = As= rdipakai*b*d
1.9 2.834 10.815 11 D 192.2 3.799 8.067 9 D 222.5 4.906 6.247 7 D 25
6 Perhitungan parameter penampang :c = jarak dari serat tekan terluar ke garis netral (cm)a = tinggi blok tegangan beton ekivalen (cm)Cc = gaya tekan blok tegangan beton (kg)Ts = gaya tarik baja tulangtan (kg)
a=As*fy/(0.85*fc'*b)
a= 24.038 (cm)
c= 28.280 (cm)Cc= a*b*0.85*fc’
Cc= 122595.0 (kg) = 122.6 (ton)Ts=As*fy
Ts= 122595.0 (kg) = 122.6 (ton)
Mata Kuliah : STRUKTUR BETON I (Ir.H.Sumirin, MS.)Topik : Contoh Soal Disain Penampang Balok
Menahan Momen Lentur ( Tulangan Rangkap)Diketahui :
Mu= 50 5.00E+06 (kg.cm)
h= 70 (cm)
d= 63 (cm)
b= 30 (cm)
fc' = 200
fy = 4000
1 Koefisen penampang : Rn = Mu/(0.80*0.85*fc’*b*d^2)Rn= 0.30877
20.38156
3a
c=a/b1 =a/0.85
(kg/cm2)
(kg/cm2)
Indeks tulangan : wn = 1-(1-2*Rn)^0.5wn = Rasio tulangan (r) : r = wn*0.85*fc'/fy
0.01622b
0.00350c
0.016264 Luas tulangan (As) :
a 0.016220.00350
Salah 0b 0.01622
0.003500.01626
Apakah :Benar 0.016216
c 0.016220.01626
Salah 0
Luas tulangan (As) :0.01621630.64874 cm2
5 Pilihan jumlah tulangan (nb) :db(cm) Ab(cm2) nb=As/Ab Jumlah
1.3 1.327 23.102 24 D 131.6 2.010 15.251 16 D 161.9 2.834 10.815 11 D 192.2 3.799 8.067 9 D 222.5 4.906 6.247 7 D 25
6 Perhitungan parameter penampang :c = jarak dari serat tekan terluar ke garis netral (cm)a = tinggi blok tegangan beton ekivalen (cm)Cc = gaya tekan blok tegangan beton (kg)Ts = gaya tarik baja tulangtan (kg)
a=As*fy/(0.85*fc'*b)
a= 24.038 (cm)
c= 28.280 (cm)Cc= a*b*0.85*fc’
Cc= 122595.0 (kg) = 122.6 (ton)Ts=As*fy
Ts= 122595.0 (kg) = 122.6 (ton)
r = rmin = 14/fyrmin =rmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)rmax =
r = rmin =Apakah : { r< rmin} ?r = rmin =rmax =
{rmin< r ≤ rmax}?r = rmax =Apakah :{ r > rmax }?
r dipakai = As= rdipakai*b*d
c=a/b1 =a/0.85
Mata Kuliah : STRUKTUR BETON I (Ir.H.Sumirin, MS.)Topik : Contoh Soal Disain Penampang Balok
Menahan Momen Lentur ( Tulangan Tunggal )Diketahui :
Mu= 10 1.00E+06 (kg.cm)
h= 70 (cm)
d= 63 (cm)
b= 30 (cm)
fc' = 200
fy = 4000
1 Koefisen penampang : Rn = Mu/(0.80*0.85*fc’*b*d^2)Rn= 0.06175
20.06379
3a
0.00271b
0.00350c
0.016264 Luas tulangan (As) :
a 0.002710.00350
Benar 0.0035b 0.00271
0.003500.01626
Apakah :Salah 0
c 0.002710.01626
Salah 0
Luas tulangan (As) :0.00356.615 cm2
5 Pilihan jumlah tulangan (nb) :db(cm) Ab(cm2) nb=As/Ab Jumlah
1.3 1.327 4.986 5 D 131.6 2.010 3.292 4 D 16
(kg/cm2)
(kg/cm2)
Indeks tulangan : wn = 1-(1-2*Rn)^0.5wn = Rasio tulangan (r) : r = wn*0.85*fc'/fyr = rmin = 14/fyrmin =rmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)rmax =
r = rmin =Apakah : { r< rmin} ?r = rmin =rmax =
{rmin< r ≤ rmax}?r = rmax =Apakah :{ r > rmax }?
r dipakai = As= rdipakai*b*d
1.9 2.834 2.334 3 D 192.2 3.799 1.741 2 D 222.5 4.906 1.348 2 D 25
6 Perhitungan parameter penampang :c = jarak dari serat tekan terluar ke garis netral (cm)a = tinggi blok tegangan beton ekivalen (cm)Cc = gaya tekan blok tegangan beton (kg)Ts = gaya tarik baja tulangtan (kg)
a=As*fy/(0.85*fc'*b)
a= 4.019 (cm)
c= 4.728 (cm)Cc= a*b*0.85*fc’
Cc= 20494.9 (kg) = 20.5 (ton)Ts=As*fy
Ts= 20494.9 (kg) = 20.5 (ton)
Mata Kuliah : STRUKTUR BETON I (Ir.H.Sumirin, MS.)Topik : Contoh Soal Disain Penampang Balok
Menahan Momen Lentur ( Tulangan Rangkap)Diketahui :
Mu= 10 1.00E+06 (kg.cm)
h= 70 (cm)
d= 63 (cm)
b= 30 (cm)
fc' = 200
fy = 4000
1 Koefisen penampang : Rn = Mu/(0.80*0.85*fc’*b*d^2)Rn= 0.06175
20.06379
3a
c=a/b1 =a/0.85
(kg/cm2)
(kg/cm2)
Indeks tulangan : wn = 1-(1-2*Rn)^0.5wn = Rasio tulangan (r) : r = wn*0.85*fc'/fy
0.00271b
0.00350c
0.016264 Luas tulangan (As) :
a 0.002710.00350
Benar 0.0035b 0.00271
0.003500.01626
Apakah :Salah 0
c 0.002710.01626
Salah 0
Luas tulangan (As) :0.00356.615 cm2
5 Pilihan jumlah tulangan (nb) :db(cm) Ab(cm2) nb=As/Ab Jumlah
1.3 1.327 4.986 5 D 131.6 2.010 3.292 4 D 161.9 2.834 2.334 3 D 192.2 3.799 1.741 2 D 222.5 4.906 1.348 2 D 25
6 Perhitungan parameter penampang :c = jarak dari serat tekan terluar ke garis netral (cm)a = tinggi blok tegangan beton ekivalen (cm)Cc = gaya tekan blok tegangan beton (kg)Ts = gaya tarik baja tulangtan (kg)
a=As*fy/(0.85*fc'*b)
a= 4.019 (cm)
c= 4.728 (cm)Cc= a*b*0.85*fc’
Cc= 20494.9 (kg) = 20.5 (ton)Ts=As*fy
Ts= 20494.9 (kg) = 20.5 (ton)
r = rmin = 14/fyrmin =rmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)rmax =
r = rmin =Apakah : { r< rmin} ?r = rmin =rmax =
{rmin< r ≤ rmax}?r = rmax =Apakah :{ r > rmax }?
r dipakai = As= rdipakai*b*d
c=a/b1 =a/0.85
Mata Kuliah : STRUKTUR BETON I (Ir.H.Sumirin, MS.)Topik : Contoh Soal Disain Penampang Balok
Menahan Momen Lentur ( Tulangan Tunggal )Diketahui :
Mu= 80 8.00E+06 (kg.cm)
h= 70 (cm)
d= 63 (cm)
b= 30 (cm)
fc' = 200
fy = 4000
1 Koefisen penampang : Rn = Mu/(0.80*0.85*fc’*b*d^2)Rn= 0.49402
20.89068
3a
0.03785b
0.00350c
0.016264 Luas tulangan (As) :
a 0.037850.00350
Salah 0b 0.03785
0.003500.01626
Apakah :Salah 0
c 0.037850.01626
Benar Perbesar Ukuran Balok !!!
Luas tulangan (As) :?????? cm2
5 Pilihan jumlah tulangan (nb) :db(cm) Ab(cm2) nb=As/Ab Jumlah
1.3 1.327 ??? ??? D 131.6 2.010 ??? ??? D 16
(kg/cm2)
(kg/cm2)
Indeks tulangan : wn = 1-(1-2*Rn)^0.5wn = Rasio tulangan (r) : r = wn*0.85*fc'/fyr = rmin = 14/fyrmin =rmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)rmax =
r = rmin =Apakah : { r< rmin} ?r = rmin =rmax =
{rmin< r ≤ rmax}?r = rmax =Apakah :{ r > rmax }?
r dipakai = As= rdipakai*b*d
1.9 2.834 ??? ??? D 192.2 3.799 ??? ??? D 222.5 4.906 ??? ??? D 25
6 Perhitungan parameter penampang :c = jarak dari serat tekan terluar ke garis netral (cm)a = tinggi blok tegangan beton ekivalen (cm)Cc = gaya tekan blok tegangan beton (kg)Ts = gaya tarik baja tulangtan (kg)
a=As*fy/(0.85*fc'*b)
a= 0.000 (cm)
c= 0.000 (cm)Cc= a*b*0.85*fc’
Cc= 0.0 (kg) = 0.0 (ton)Ts=As*fy
Ts= 0.0 (kg) = 0.0 (ton)
Mata Kuliah : STRUKTUR BETON I (Ir.H.Sumirin, MS.)Topik : Contoh Soal Disain Penampang Balok
Menahan Momen Lentur ( Tulangan Rangkap)Diketahui :
Mu= 80 8.00E+06 (kg.cm)
h= 70 (cm)
d= 63 (cm)
b= 30 (cm)
fc' = 200
fy = 4000
1 Koefisen penampang : Rn = Mu/(0.80*0.85*fc’*b*d^2)Rn= 0.49402
20.89068
3a
c=a/b1 =a/0.85
(kg/cm2)
(kg/cm2)
Indeks tulangan : wn = 1-(1-2*Rn)^0.5wn = Rasio tulangan (r) : r = wn*0.85*fc'/fy
0.03785b
0.00350c
0.016264 Luas tulangan (As) :
a 0.037850.00350
Salah 0b 0.03785
0.003500.01626
Apakah :Salah 0
c 0.037850.01626
Benar Perbesar Ukuran Balok !!!
Luas tulangan (As) :?????? cm2
5 Pilihan jumlah tulangan (nb) :db(cm) Ab(cm2) nb=As/Ab Jumlah
1.3 1.327 ??? ??? D 131.6 2.010 ??? ??? D 161.9 2.834 ??? ??? D 192.2 3.799 ??? ??? D 222.5 4.906 ??? ??? D 25
6 Perhitungan parameter penampang :c = jarak dari serat tekan terluar ke garis netral (cm)a = tinggi blok tegangan beton ekivalen (cm)Cc = gaya tekan blok tegangan beton (kg)Ts = gaya tarik baja tulangtan (kg)
a=As*fy/(0.85*fc'*b)
a= 0.000 (cm)
c= 0.000 (cm)Cc= a*b*0.85*fc’
Cc= 0.0 (kg) = 0.0 (ton)Ts=As*fy
Ts= 0.0 (kg) = 0.0 (ton)
r = rmin = 14/fyrmin =rmax = 0.75*0.85*0.85*fc'/fy * 6000/(6000+fy)rmax =
r = rmin =Apakah : { r< rmin} ?r = rmin =rmax =
{rmin< r ≤ rmax}?r = rmax =Apakah :{ r > rmax }?
r dipakai = As= rdipakai*b*d
c=a/b1 =a/0.85
