laporan tugas ujian mid beton 1
TRANSCRIPT
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
1
LAPORAN TUGAS UJIAN MID
STRUKTUR BETON 1
Oleh :
Nama : Desi Ariyo Nurhidayat
NIM : 30201203259
FAKULTAS TEKNIK JURUSAN TEKNIK SIPIL
UNIVERSITAS ISLAM SULTAN AGUNG
SEMARANG
2014
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Struktur
Beton 1
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SOAL
1.
Balok Penahan Momen
Hitung luas tulangan tarik dan jumlah tulangan tarik balok persegi beton bertulang dengan data sebagai berikut :
fc'=200 kg/cm2
(20 MPa); fy=4000 kg/cm2 (400 MPa); d=1,9 cm (19 mm)
a. b=30cm; h=50cm; Mu=20 tm
b. b=35cm; h=50cm; Mu=20 tm
c. b=30cm; h=50cm; Mu=10 tm
d. b=40cm; h=70cm; Mu=40 tm
2. Balok Penahan Lintang
Hitung tegangan geser & kuat geser beton dengan data sebagai berikut :
a. Gaya Lintang (Vu) = 20 ton
Mutu Beton (fc’) = 250 kg/cm
2
Mutu Baja sengkang (fyv) = 2200 kg/cm2
Tulangan sengkang (ds) =
∅12 cmLebar (b) =35 cm
Tinggi (h) =65 cm
= 0,9*65
= 59 cm
b. Gaya Lintang (Vu) = 35 ton
Mutu Beton (fc’
) = 200 kg/cm2
Mutu Baja sengkang (fyv) = 2100 kg/cm2
Tulangan sengkang (ds) = ∅14 cm
Lebar (b) =30 cm
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Struktur
Beton 1
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Tinggi (h) =55 cm
= 0,9*55
= 50 cm
c. Gaya Lintang (Vu) = 20 ton
Mutu Beton (fc’) = 345 kg/cm
2
Mutu Baja sengkang (fyv) = 5000 kg/cm2
Tulangan sengkang (ds) = ∅12 cm
Lebar (b) =20 cm
Tinggi (h) =40 cm
= 0,9*40
= 36 cm
d. Gaya Lintang (Vu) = 22 ton
Mutu Beton (fc’) = 400 kg/cm2
Mutu Baja sengkang (fyv) = 5200 kg/cm2
Tulangan sengkang (ds) = ∅8 cm
Lebar (b) =20 cm
Tinggi (h) =35 cm
= 0,9*35
= 32 cm
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Struktur
Beton 1
4
3. Balok Penahan Kolom
Hitung dan gambar prosedur perencanaan tulangan pokok kolom beton bertulang 4 sisi dengan
ketentuan sebagai berikut :
a.
b = 50 cmh = 50 cm
fc’ = 200 kg/cm2
fy = 4000 kg/cm2
Mu = 40 tm
Pu = 100 t
Diameter Tulangan = 19 mm
b. b = 55 cm
h = 55 cm
fc’ = 250 kg/cm2
fy = 4000 kg/cm2
Mu = 28 tm
Pu = 200 t
Diameter Tulangan = 19 mm
c. b = 60 cm
h = 60 cm
fc’
= 150 kg/cm2
fy = 4000 kg/cm2
Mu = 10 tmPu = 250 t
Diameter Tulangan = 19 mm
d. b = 30 cm
h = 30 cm
fc’
= 150 kg/cm2
fy = 4000 kg/cm2
Mu = 8 tm
Pu = 40 t
Diameter Tulangan = 19 mm
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Struktur
Beton 1
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JAWABAN
1. Balok Penahan Momen
1.a Diketahui : fc’ = 200 kg/cm2
( 20 Mpa )
fy = 4000 kg/cm2
( 400 Mpa )
b = 30 cm
h = 50 cm
d = 0.9 x 50 = 45 cm
Mu = 20 tm = 2000000 kg/cm
Ditanya : Luas Tulangan (As)
Jumlah Tulangan (n)
Penyelesaian :
1. Koefisien penampang
Rn =2^***85.0*
' d b fc
Mu
=2^45*30*200*85.0*8.0
2000000
=8262000
2000000
= 0,24207
2. Indeks tulangan ( n )
n = 1- )*21( Rn
= 1 - )24207,0*21(
= 1 – 0,71823
= 0,28177
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Struktur
Beton 1
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3. Rasio tulangan ( )
= n fy
fc'*85.0
= 0,218774000
200*85.0
= 0,01198
ρ min =14
=14
4000
= 0,00350
ρ max =0,75 .0,85 .0,85 .fc ′
fy .6000 / (6000+fy)
=0,75 .0,85 .0,85 .200
4000 .6000 / (6000+4000)
= 0,01626
4. Luas Tulangan (As)
As = *b*d
= 0,001198 * 30 *45
= 16,1665 cm
2
5. Jumlah tulangan (nb)
Nb = As/Ab (dibulatkan keatas)
Dimana Ab = ¼*3.14*1,92
Misal dipakai diameter tulangan 19 mm
Ab = ¼*3.14*1,92
= 2,83385 cm
2
nb = As/Ab
= 16,1665 cm2
/ 2,83385 cm2
= 5,704 dibulatkan menjadi 6 batang
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Struktur
Beton 1
7
Perhitungan menggunakan dsgwin
DATA:
Beam Width, Bw = 30.00 cm
Beam Height, Ht = 50.00 cm
Factored Moment, Mu = 20000.00 kg.m
Factored Shear, Vu = 0.00 kg
Factored Normal, Pu = 0.00 kg
Factored Torsion, Tu = 0.00 kg.m
Concrete Compressive Strength fc1 = 200.00 kg/cm2
Concrete Crack Strength fcr = 150.00 kg/cm2
Concrete Initial Strength fci = 120.00 kg/cm2
Main Rebar Yield Strength fy = 4000.00 kg/cm2
Stirrups Rebar Yield Strength fyv = 2400.00 kg/cm2
Main rebar diameter dbm = 1.90 cm
2nd main rebar diameter dbn = 1.90 cm
Stirrups Rebar diameter dbv = 1.00 cm
Min concrete cover, Cover = 4.00 cm
Min rebar clear space, Minclrspc = 2.50 cm
RESULT:
Effective Depth, d = 44.05 cm
Distance, d1 = H - d = 5.95 cmds = 43.32 cm
ds1 = 5.95 cm
Minimum Rebar Ratio, romin = 0.00352 (Using Romin = 200/Fy)
Maximum Rebar Ratio, romax = 0.01874
Flexural Reinforcement: OK
Normal Direction:
Beta Factor, Beta = 0.850
Location of neutral axis, c = 9.6063 cm
Length of compression block, a = 8.1654 cm
Total Rebar, nb = 4/ 6 d19
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Struktur
Beton 1
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Total Rebar area, Ast = 17.0117 cm2 (ro = 0.01287)
Nominal Moment Capacity, Md = 20963.44 kg.m
Inverted Section Direction
Beta Factor, Beta = 0.850
Location of neutral axis, c1 = 7.5919 cm
Length of compression block, a1 = 6.4532 cm
Total Rebar, nb = 6/ 4 d19 Total Rebar area, As1 = 11.3411 cm2 (ro1 = 0.00858)
Nominal Moment Capacity, Md1 = 14471.02 kg.m
Shear Reinforcement: OK
Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2
Ag = bw*h = 0.00 cm2
pw = Ast/(bw*d) = 0.0000
x1 = bw-2*cv-2*dbv = 0.00 cm
y1 = h-2*cv-2*dbv = 0.00 cm
Vn = Vu/phi = 0.00 kg
Tn = Tu/phi = 0.00 kg.m
Vc = = 0.00 kg
Tc = = 0.00 kg.m
Vnc = Vn - Vc = 0.00 kg
Tnc = Tn - Tc = 0.00 kg.m
Avs = Vnc/(fy*d) = 0.00 cm2
Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2Atst = Avs + 2*Ats = 0.00 cm2
Additional Longitudinal Rebar, At = 0.00 cm2
Stirrups Spacing, spc = 2540.01 cm
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
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1.b Diketahui : fc’ = 200 kg/cm2
( 20 Mpa )
fy = 4000 kg/cm2
( 400 Mpa )
b = 35 cmh = 50 cm
d = 0.9 x 50 = 45 cm
Mu = 20 tm = 2000000 kg/cm
Ditanya : Luas Tulangan (As)
Jumlah Tulangan (n)
Penyelesaian :
1. Koefisien penampang
Rn =2^***85.0* ' d b fc
Mu
=2^54*35*200*85.0*8.0
2000000
=9639000
2000000
= 0,20749
2.
Indeks tulangan ( n )
n = 1- )*21( Rn
= 1 - )20749,0*21(
= 1 – 0,76488
= 0,23512
3. Rasio tulangan ( )
= n fy
fc'*85.0
= 0,235124000
200*85.0
= 0,00999
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
10
ρ min =14
=14
4000
= 0,00350
ρ max = 0,75 .0,85 .0,85 .fc ′ fy .6000
/ (6000+fy)
=0,75 .0,85 .0,85 .200
4000 .6000 / (6000+4000)
= 0,01626
4. 45Luas Tulangan (As)
As = *b*d
= 0,00999 * 30 * 45
= 15,7383 cm2
5.
Jumlah tulangan (nb)
Nb = As/Ab (dibulatkan keatas)
Dimana Ab = ¼*3.14*1,92
Misal dipakai diameter tulangan 19 mm
Ab = ¼*3.14*1,92
= 2,83385 cm2
nb = As/Ab
= 15,7383 cm2
/ 2,83385 cm2
= 5,55225 dibulatkan menjadi 6 batang
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Struktur
Beton 1
11
Perhitungan menggunakan dsgwin
DATA:
Beam Width, Bw = 35.00 cm
Beam Height, Ht = 50.00 cm
Factored Moment, Mu = 20000.00 kg.m
Factored Shear, Vu = 0.00 kg
Factored Normal, Pu = 0.00 kg
Factored Torsion, Tu = 0.00 kg.m
Concrete Compressive Strength fc1 = 200.00 kg/cm2Concrete Crack Strength fcr = 150.00 kg/cm2
Concrete Initial Strength fci = 120.00 kg/cm2
Main Rebar Yield Strength fy = 4000.00 kg/cm2
Stirrups Rebar Yield Strength fyv = 2400.00 kg/cm2
Main rebar diameter dbm = 1.90 cm
2nd main rebar diameter dbn = 1.90 cm
Stirrups Rebar diameter dbv = 1.00 cm
Min concrete cover, Cover = 4.00 cm
Min rebar clear space, Minclrspc = 2.50 cm
RESULT:
Effective Depth, d = 44.05 cm
Distance, d1 = H - d = 5.95 cm
ds = 44.05 cmds1 = 5.95 cm
Minimum Rebar Ratio, romin = 0.00352 (Using Romin = 200/Fy)
Maximum Rebar Ratio, romax = 0.01539
Flexural Reinforcement: OK
Normal Direction:
Beta Factor, Beta = 0.850
Location of neutral axis, c = 10.4874 cm
Length of compression block, a = 8.9143 cm
Total Rebar, nb = 2/ 6 d19
Total Rebar area, Ast = 15.9182 cm2 (ro = 0.01103)
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
12
Nominal Moment Capacity, Md = 21374.00 kg.m
Inverted Section Direction
Beta Factor, Beta = 0.850
Location of neutral axis, c1 = 5.6190 cm
Length of compression block, a1 = 4.7761 cm
Total Rebar, nb = 6/ 2 d19
Total Rebar area, As1 = 5.6706 cm2 (ro1 = 0.00368)Nominal Moment Capacity, Md1 = 7723.30 kg.m
Shear Reinforcement: OK
Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2
Ag = bw*h = 0.00 cm2
pw = Ast/(bw*d) = 0.0000
x1 = bw-2*cv-2*dbv = 0.00 cm
y1 = h-2*cv-2*dbv = 0.00 cm
Vn = Vu/phi = 0.00 kg
Tn = Tu/phi = 0.00 kg.m
Vc = = 0.00 kg
Tc = = 0.00 kg.m
Vnc = Vn - Vc = 0.00 kg
Tnc = Tn - Tc = 0.00 kg.m
Avs = Vnc/(fy*d) = 0.00 cm2
Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2
Atst = Avs + 2*Ats = 0.00 cm2
Additional Longitudinal Rebar, At = 0.00 cm2
Stirrups Spacing, spc = 2540.01 cm
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
13
1.c Diketahui : fc’ = 200 kg/cm2
( 20 Mpa )
fy = 4000 kg/cm2
( 400 Mpa )
b = 30 cm
h = 50 cm
d = 0.9 x 50 = 45 cmMu = 10 tm = 1000000 kg/cm
Ditanya : Luas Tulangan (As)
Jumlah Tulangan (n)
Penyelesaian :
1. Koefisien penampang
Rn =2^***85.0*8.0 ' d b fc
Mu
=2^45*30*200*85.0*8.0
1000000
=82617316
1000000
= 0,12104
2. Indeks tulangan ( n )
n = 1- )*21( Rn
= 1 - )12104,0*21(
= 1 – 0,87059
= 0,12941
3. Rasio tulangan ( )
= n fy
fc'*85.0
= 0,129414000
200*85.0
= 0,00550
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
14
ρ min =14
=14
4000
= 0,00350
ρ max = 0,75 .0,85 .0,85 .fc ′ fy .6000
/ (6000+fy)
=0,75 .0,85 .0,85 .200
4000 .6000 / (6000+4000)
= 0,01626
4. Luas Tulangan (As)
As = *b*d
= 0,00550 * 30 *45
= 7,42487 cm2
5.
Jumlah tulangan (nb)
Nb = As/Ab (dibulatkan keatas)
Dimana Ab = ¼*3.14*1,92
Misal dipakai diameter tulangan 19 mm
Ab = ¼*3.14*1,92
= 2,83385 cm2
nb = As/Ab
= 7,42487 cm2
/ 2,83385 cm2
= 2,62006 dibulatkan menjadi 3 batang
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Struktur
Beton 1
15
Perhitungan menggunakan dsgwin
DATA:
Beam Width, Bw = 30.00 cm
Beam Height, Ht = 50.00 cm
Factored Moment, Mu = 10000.00 kg.m
Factored Shear, Vu = 0.00 kg
Factored Normal, Pu = 0.00 kg
Factored Torsion, Tu = 0.00 kg.m
Concrete Compressive Strength fc1 = 200.00 kg/cm2
Concrete Crack Strength fcr = 150.00 kg/cm2
Concrete Initial Strength fci = 120.00 kg/cm2
Main Rebar Yield Strength fy = 4000.00 kg/cm2
Stirrups Rebar Yield Strength fyv = 2400.00 kg/cm2
Main rebar diameter dbm = 1.90 cm
2nd main rebar diameter dbn = 1.90 cm
Stirrups Rebar diameter dbv = 1.00 cm
Min concrete cover, Cover = 4.00 cm
Min rebar clear space, Minclrspc = 2.50 cm
RESULT:
Effective Depth, d = 44.05 cm
Distance, d1 = H - d = 5.95 cm
ds = 44.05 cm
ds1 = 5.95 cm
Minimum Rebar Ratio, romin = 0.00352 (Using Romin = 200/Fy)
Maximum Rebar Ratio, romax = 0.01595
Flexural Reinforcement: OK
Normal Direction:
Beta Factor, Beta = 0.850
Location of neutral axis, c = 6.8239 cm
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Struktur
Beton 1
16
Length of compression block, a = 5.8003 cm
Total Rebar, nb = 2/ 3 d19
Total Rebar area, Ast = 7.6684 cm2 (ro = 0.00644)
Nominal Moment Capacity, Md = 11092.03 kg.m
Inverted Section Direction
Beta Factor, Beta = 0.850
Location of neutral axis, c1 = 5.7023 cmLength of compression block, a1 = 4.8470 cm
Total Rebar, nb = 3/ 2 d19
Total Rebar area, As1 = 5.6706 cm2 (ro1 = 0.00429)
Nominal Moment Capacity, Md1 = 7610.91 kg.m
Shear Reinforcement: OK
Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2
Ag = bw*h = 0.00 cm2
pw = Ast/(bw*d) = 0.0000
x1 = bw-2*cv-2*dbv = 0.00 cm
y1 = h-2*cv-2*dbv = 0.00 cm
Vn = Vu/phi = 0.00 kg
Tn = Tu/phi = 0.00 kg.m
Vc = = 0.00 kg
Tc = = 0.00 kg.m
Vnc = Vn - Vc = 0.00 kg
Tnc = Tn - Tc = 0.00 kg.mAvs = Vnc/(fy*d) = 0.00 cm2
Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2
Atst = Avs + 2*Ats = 0.00 cm2
Additional Longitudinal Rebar, At = 0.00 cm2
Stirrups Spacing, spc = 2540.01 cm
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
17
1.d Diketahui : fc’ = 200 kg/cm2
( 20 Mpa )
fy = 4000 kg/cm2
( 400 Mpa )
b = 40 cm
h = 70 cm
d = 0,9 x 70 = 63 cmMu = 40 tm = 4000000 kg/cm
Ditanya : Luas Tulangan (As)
Jumlah Tulangan (n)
Penyelesaian :
1. Koefisien penampang
Rn =2^***85.0*8.0 ' d b fc
Mu
=2^63*40*200*85.0*8.0
4000000
=21591360
4000000
= 0,18525
2. Indeks tulangan ( n )
n = 1- )*21( Rn
= 1 - )18525,0*21(
= 1 – 0,7934
= 0,20660
3. Rasio tulangan ( )
= n fy
fc'*85,0
= 0,206604000
200*85,0
= 0,00878
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
18
ρ min =14
=14
4000
= 0,00350
ρ max = 0,75 .0,85 .0,85 .fc ′ fy .6000
/ (6000+fy)
=0,75 .0,85 .0,85 .200
4000 .6000 / (6000+4000)
= 0,01626
4. Luas Tulangan (As)
As = *b*d
= 0,00878 * 40 *63
= 22,12686 cm2
5.
Jumlah tulangan (nb)
Nb = As/Ab (dibulatkan keatas)
Dimana Ab = ¼*3.14*1,92
Misal dipakai diameter tulangan 19 mm
Ab = ¼*3.14*1,92
= 2,83385 cm2
nb = As/Ab
= 22,12686 cm2
/ 2,83385 cm2
= 7,80805 dibulatkan menjadi 8 batang
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
19
Perhitungan menggunakan dsgwin
DATA:
Beam Width, Bw = 40.00 cm
Beam Height, Ht = 70.00 cm
Factored Moment, Mu = 40000.00 kg.m
Factored Shear, Vu = 0.00 kg
Factored Normal, Pu = 0.00 kgFactored Torsion, Tu = 0.00 kg.m
Concrete Compressive Strength fc1 = 200.00 kg/cm2
Concrete Crack Strength fcr = 150.00 kg/cm2
Concrete Initial Strength fci = 120.00 kg/cm2
Main Rebar Yield Strength fy = 4000.00 kg/cm2
Stirrups Rebar Yield Strength fyv = 2400.00 kg/cm2
Main rebar diameter dbm = 1.90 cm
2nd main rebar diameter dbn = 1.90 cm
Stirrups Rebar diameter dbv = 1.00 cm
Min concrete cover, Cover = 4.00 cm
Min rebar clear space, Minclrspc = 2.50 cm
RESULT:
Effective Depth, d = 64.05 cm
Distance, d1 = H - d = 5.95 cm
ds = 63.50 cm
ds1 = 5.95 cm
Minimum Rebar Ratio, romin = 0.00352 (Using Romin = 200/Fy)
Maximum Rebar Ratio, romax = 0.01517
Flexural Reinforcement: OK
Normal Direction:
Beta Factor, Beta = 0.850
Location of neutral axis, c = 12.5429 cm
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Struktur
Beton 1
20
Length of compression block, a = 10.6614 cm
Total Rebar, nb = 2/ 8 d19
Total Rebar area, Ast = 21.5352 cm2 (ro = 0.00885)
Nominal Moment Capacity, Md = 42130.61 kg.m
Inverted Section Direction
Beta Factor, Beta = 0.850
Location of neutral axis, c1 = 5.9656 cmLength of compression block, a1 = 5.0708 cm
Total Rebar, nb = 8/ 2 d19
Total Rebar area, As1 = 5.6706 cm2 (ro1 = 0.00221)
Nominal Moment Capacity, Md1 = 11536.49 kg.m
Shear Reinforcement: OK
Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2
Ag = bw*h = 0.00 cm2
pw = Ast/(bw*d) = 0.0000
x1 = bw-2*cv-2*dbv = 0.00 cm
y1 = h-2*cv-2*dbv = 0.00 cm
Vn = Vu/phi = 0.00 kg
Tn = Tu/phi = 0.00 kg.m
Vc = = 0.00 kg
Tc = = 0.00 kg.m
Vnc = Vn - Vc = 0.00 kg
Tnc = Tn - Tc = 0.00 kg.mAvs = Vnc/(fy*d) = 0.00 cm2
Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2
Atst = Avs + 2*Ats = 0.00 cm2
Additional Longitudinal Rebar, At = 0.00 cm2
Stirrups Spacing, spc = 2540.01 cm
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
21
2. Balok Penahan Lintang
2.a. Diketahui : Vu = 20 ton
fc’
= 250 kg/cm2
fyv = 2200 kg/cm2
ds = ∅12 cm b = 35 cm
h = 65 cm
= 0.9 * 65
= 59 cm
Ditanya : Tegangan Geser (Vn)
Kuat Geser (Vc)
Penyelesaian :
a. Tegangan Geser ( Vn ) : Vn =Vu
ɸ.b.d=
20.000
0,75.35.59 =
20.000
1548,75= 12,91 kg/cm2
b. Kuat Geser ( Vc ) : vc = 0,53 √ fcI = 0,53√ 250 = 8,38 kg/cm2
c. Kontrol (Vc
Vn) :
Vn
vc
=12,91
8,38
= 1,5
: 3 <Vn
vc≤ 1
: s =Asv .fyv
b (Vn−vc) =
2.0,25.3,14.1,22 .2200
35 (12,91−8,38) =
4973,76
158,55 =
: 31,4 cmJadi s = 30 cm
∅12 – 300 mm
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
22
Gambar penulangan geser untuk setengah batang
Gambar detail
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
23
2.b Diketahui : Vu = 35 ton
fc’
= 200 kg/cm2
fyv = 2100 kg/cm2
ds = ∅14 cm b = 30 cm
h = 55 cm
= 0.9 * 55
= 50 cm
Ditanya : Tegangan Geser (Vn)
Kuat Geser (Vc)
Penyelesaian :
a. Tegangan Geser ( Vn ) : Vn =Vu
ɸ.b.d=
35.000
0,75.30.50 =
35.000
1125= 31,1 kg/cm2
b. Kuat Geser ( Vc ) : vc = 0,53 √ fcI = 0,53√ 200 = 7,5 kg/cm2
c.
Kontrol (VcVn ) : Vn
vc = 31.1
7.5 = 4.1
: 5 <Vn
vc≤ 3
: s =Asv .fyv
b (Vn−vc) =
2.0,25.3,14.1,22 .2100
35 (31.1−7.5) =
6462.12
708 =
: 9.1 cm
Jadi s = 8.5 cm
∅14 – 85 mm
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Struktur
Beton 1
24
Gambar penulangan geser untuk setengah batang
Gambar detail
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Struktur
Beton 1
25
2.c Diketahui : Vu = 20 ton
fc’
= 345 kg/cm2
fyv = 5000 kg/cm2
ds = ∅12 cm b = 20 cm
h = 40 cm
= 0.9 * 40
= 36 cm
Ditanya : Tegangan Geser (Vn)
Kuat Geser (Vc)
Penyelesaian :
a. Tegangan Geser ( Vn ) : Vn =Vu
ɸ.b.d=
20.000
0,75.20.36 =
20.000
540= 37 kg/cm
2
b. Kuat Geser ( Vc ) : vc = 0,53 √ fcI = 0,53√ 345 = 9,8 kg/cm2
c.
Kontrol (
Vc
Vn) :
Vn
vc
=37
9.8
= 3,8
: 5 <Vn
vc≤ 3
: s =Asv .fyv
b (Vn−vc ) =
2.0,25.3,14.1,22 .5000
20 (37−9.8) =
11304
544 =
: 20.8 cm
Jadi s = 19 cm
∅12 – 190 mm
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Struktur
Beton 1
26
Gambar penulangan geser untuk setengah batang
Gambar detail
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Struktur
Beton 1
27
2.d Diketahui : Vu = 39 ton
fc’
= 300 kg/cm2
fyv = 3000 kg/cm2
ds = ∅16 cm b = 25 cm
h = 55 cm
= 0,9 * 55
= 50 cm
Ditanya : Tegangan Geser (Vn)
Kuat Geser (Vc)
Penyelesaian :
a. Tegangan Geser ( Vn ) : Vn =Vu
ɸ.b.d=
39.000
0,75.25.50 =
39.000
937,5= 41,6 kg/cm2
b. Kuat Geser ( Vc ) : vc = 0,53 √ fcI = 0,53√ 300 = 9,2 kg/cm2
c. Kontrol (Vc
Vn) :
Vn
vc =
41,1
9,2 = 4,5
: 5 <Vn
vc≤ 3
: s =Asv .fyv
b (Vn−vc) =
2.0,25.3,14.1,22 .3000
25 (41,6−9,2) =
12057,6
810 =
: 14,8 cm
Jadi s = 13 cm
∅16 – 130 mm
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Struktur
Beton 1
28
Gambar penulangan geser untuk setengah batang
Gambar detail
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Struktur
Beton 1
29
3. Balok Penahan Kolom
3.a Diketahui :b = 50 cm
h = 50 cm
fc’
= 200 kg/cm2
fy = 4000 kg/cm2
Mu = 40 tm
Pu = 100 t
DiameterTulangan = 19 mm
Ditanya :Perencanaan Tulangan Kolom
Penyelesaian :
Sumbu x = Muɸ.b.ℎ2 .0,85.f ′
= 40x105
0,65.50.502 .0,85.200 = 4.000.000
13.812.500 = 0,29
Sumbu y =Pu
ɸ.b.ℎ .0,85.f ′ =
100x103
0,65.50.50.0,85.200 =
100.000
176.250 = 0,57
Dari Grafik: fc’ = 200 kg/cm2
= 20 Mpa = 0,80 r = 0,0422
t = r. = 0,0422 x 0,80
= 0,034
= 0,01
Ast = t x b x h
= 0,034 x 50 x 50
= 85 cm
db = ¼ . 3,14 . 1,92 = 2,83 cm
2
n =Ast
db
=85
2,83
= 30 ~32 D 19 mm
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Struktur
Beton 1
30
Gambar Penampang
Diagram Interaksi
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Struktur
Beton 1
31
Perhitungan menggunakan dsgwin
DATA:
Column Width, Bw = 50.00 cm
Column Height, Ht = 50.00 cm
Factored Moment, Mu = 40000.00 kg.m
Factored Shear, Vu = 0.00 kg
Factored Normal, Pu = 100000.00 kg
Factored Torsion, Tu = 0.00 kg.m
Concrete Compressive Strength fc1 = 200.00 kg/cm2
Concrete Crack Strength fcr = 200.00 kg/cm2
Concrete Initial Strength fci = 120.00 kg/cm2
Main Rebar Yield Strength fy = 4000.00 kg/cm2
Stirrups Rebar Yield Strength fyv = 2400.00 kg/cm2
Main rebar diameter dbm = 1.90 cm
2nd main rebar diameter dbn = 1.90 cm
Stirrups Rebar diameter dbv = 1.00 cm
Min concrete cover, Cover = 9.00 cm
Min rebar clear space, Minclrspc = 2.50 cm
RESULT:
Flexural Reinforcement: OK
Balanced neutral axis, cbal = 23.6102 cm
Balanced Compr. block, abal = 20.0686 cm
Balanced Moment Capacity, Mnb = 70844.83 kg.m
Balanced Normal Capacity, Pnb = 130202.40 kg
Balanced Eccentricity, ebal = 54.41 cm
Location of neutral axis, c = 24.7896 cm
Length of compression block, a = 21.0711 cm
Total Rebar (incl. side bar) nbt = 32 d19
Total Side Rebar nbs = 0 d19
Total Rebar area, Ast = 90.7292 cm2 = 3.63%
Nominal Moment Capacity, Md = 41408.85 kg.m
Nominal Normal Capacity, Pd = 100009.10 kg
Shear Reinforcement: OK
Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2
Ag = bw*h = 0.00 cm2
pw = Ast/(bw*d) = 0.0000
x1 = bw-2*cv-2*dbv = 0.00 cm
y1 = h-2*cv-2*dbv = 0.00 cm
Vn = Vu/phi = 0.00 kg
Tn = Tu/phi = 0.00 kg.m
Vc = = 0.00 kg
Tc = = 0.00 kg.m
Vnc = Vn - Vc = 0.00 kg
Tnc = Tn - Tc = 0.00 kg.m
Avs = Vnc/(fy*d) = 0.00 cm2
Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2
Atst = Avs + 2*Ats = 0.00 cm2
Additional Longitudinal Rebar, At = 0.00 cm2Stirrups Spacing, spc = 30.40 cm
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Struktur
Beton 1
32
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Struktur
Beton 1
33
3.b Diketahui :b = 55 cm
h = 55 cm
fc’
= 250 kg/cm2
fy = 4000 kg/cm2
Mu = 28 tm
Pu = 200 t
DiameterTulangan = 19 mm
Ditanya :Perencanaan Tulangan Kolom
Penyelesaian :
Sumbu x =
Mu
ɸ.b.ℎ2 .0,85.f =
28x105
0,65.55.552 .0,85.250 =
2.800.000
22.980.547 = 0,12
Sumbu y =Pu
ɸ.b.ℎ .0,85.f =
200x103
0,65.55.55.0,85.250 =
200.000
417.828,13 = 0,48
Dari Grafik: fc’ = 200 kg/cm2
= 20 Mpa = 1,0 r = 0,01
t = r.
= 0,01 x 1,0
= 0,01
= 0,01
Ast = t x b x h
= 0,01 x 55 x 55
= 30,25 cm
db = ¼ . 3,14 . 1,92 = 2,83 cm
2
n =Ast
db =
30,25
2,83 = 10,6 ~12 D 19 mm
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Struktur
Beton 1
34
Gambar Penampang
Gambar Diagram Interaksi
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Struktur
Beton 1
35
Perhitungan menggunakan dsgwin
DATA:
Column Width, Bw = 55.00 cm
Column Height, Ht = 55.00 cm
Factored Moment, Mu = 28000.00 kg.m
Factored Shear, Vu = 0.00 kg
Factored Normal, Pu = 200000.00 kg
Factored Torsion, Tu = 0.00 kg.m
Concrete Compressive Strength fc1 = 250.00 kg/cm2
Concrete Crack Strength fcr = 150.00 kg/cm2
Concrete Initial Strength fci = 120.00 kg/cm2
Main Rebar Yield Strength fy = 4000.00 kg/cm2
Stirrups Rebar Yield Strength fyv = 2400.00 kg/cm2
Main rebar diameter dbm = 1.90 cm
2nd main rebar diameter dbn = 1.90 cm
Stirrups Rebar diameter dbv = 1.00 cm
Min concrete cover, Cover = 5.00 cm
Min rebar clear space, Minclrspc = 2.50 cm
RESULT:
Flexural Reinforcement: OK
Balanced neutral axis, cbal = 29.0517 cm
Balanced Compr. block, abal = 24.6939 cm
Balanced Moment Capacity, Mnb = 70957.14 kg.m
Balanced Normal Capacity, Pnb = 284994.47 kg
Balanced Eccentricity, ebal = 24.90 cm
Location of neutral axis, c = 32.2265 cm
Length of compression block, a = 27.3926 cm
Total Rebar (incl. side bar) nbt = 12 d19
Total Side Rebar nbs = 0 d19
Total Rebar area, Ast = 34.0234 cm2 = 1.12%
Nominal Moment Capacity, Md = 40759.42 kg.m
Nominal Normal Capacity, Pd = 200093.86 kg
Shear Reinforcement: OK
Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2
Ag = bw*h = 0.00 cm2
pw = Ast/(bw*d) = 0.0000
x1 = bw-2*cv-2*dbv = 0.00 cm
y1 = h-2*cv-2*dbv = 0.00 cm
Vn = Vu/phi = 0.00 kg
Tn = Tu/phi = 0.00 kg.m
Vc = = 0.00 kg
Tc = = 0.00 kg.m
Vnc = Vn - Vc = 0.00 kg
Tnc = Tn - Tc = 0.00 kg.m
Avs = Vnc/(fy*d) = 0.00 cm2
Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2
Atst = Avs + 2*Ats = 0.00 cm2
Additional Longitudinal Rebar, At = 0.00 cm2Stirrups Spacing, spc = 30.40 cm
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Struktur
Beton 1
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Struktur
Beton 1
37
3.c Diketahui :b = 60 cm
h = 60 cm
fc ' = 150 kg/cm2
fy = 4000 kg/cm2
Mu = 10 tm
Pu = 250 t
DiameterTulangan = 19 mm
Ditanya :Perencanaan Tulangan Kolom
Penyelesaian :
Sumbu x =
Mu
ɸ.b.ℎ2 .0,85.f =
10x105
0,65.60.602 .0,85.150 =
1.000.000
17.901.000 = 0,05
Sumbu y =Pu
ɸ.b.ℎ .0,85.f =
250x103
0,65.60.60.0,85.150 =
250.000
298.350 = 0,84
Dari Grafik: fc’ = 150 kg/cm2
= 15 Mpa = 0,6 r = 0,01
t = r.
= 0,01 x 0,6
= 0,006
= 0,01
Ast = t x b x h
= 0,01 x 60 x 60
= 21,6 cm
db = ¼ . 3,14 . 1,92 = 2,83 cm
2
n =Ast
db =
21,6
2,83 = 7,6~8 D 19 mm
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
38
Gambar Penampang
Gambar Diagram Interaksi
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Struktur
Beton 1
39
Perhitungan menggunakan dsgwin
DATA:
Column Width, Bw = 60.00 cm
Column Height, Ht = 60.00 cm
Factored Moment, Mu = 10000.00 kg.m
Factored Shear, Vu = 0.00 kg
Factored Normal, Pu = 250000.00 kg
Factored Torsion, Tu = 0.00 kg.m
Concrete Compressive Strength fc1 = 150.00 kg/cm2
Concrete Crack Strength fcr = 120.00 kg/cm2
Concrete Initial Strength fci = 120.00 kg/cm2
Main Rebar Yield Strength fy = 4000.00 kg/cm2
Stirrups Rebar Yield Strength fyv = 2400.00 kg/cm2
Main rebar diameter dbm = 1.90 cm
2nd main rebar diameter dbn = 1.90 cm
Stirrups Rebar diameter dbv = 1.00 cm
Min concrete cover, Cover = 4.00 cm
Min rebar clear space, Minclrspc = 2.50 cm
RESULT:
Flexural Reinforcement: OK
CONCENTRIC LOADING CONDITION
Total Rebar (incl. side bar) nbt = 14 d19
Total Side Rebar nbs = 0 d19
Total Rebar area, Ast = 39.6940 cm2 = 1.10%
Nominal Moment Capacity, Md = 11764.10 kg.m
Nominal Normal Capacity, Pd = 294102.46 kg
Shear Reinforcement: OK
Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2
Ag = bw*h = 0.00 cm2
pw = Ast/(bw*d) = 0.0000
x1 = bw-2*cv-2*dbv = 0.00 cm
y1 = h-2*cv-2*dbv = 0.00 cmVn = Vu/phi = 0.00 kg
Tn = Tu/phi = 0.00 kg.m
Vc = = 0.00 kg
Tc = = 0.00 kg.m
Vnc = Vn - Vc = 0.00 kg
Tnc = Tn - Tc = 0.00 kg.m
Avs = Vnc/(fy*d) = 0.00 cm2
Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2
Atst = Avs + 2*Ats = 0.00 cm2
Additional Longitudinal Rebar, At = 0.00 cm2
Stirrups Spacing, spc = 30.40 cm
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
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Struktur
Beton 1
41
3.d Diketahui :b = 30 cm
h = 30 cm
fcI
= 150 kg/cm2
fy = 4000kg/cm
2
Mu = 8 tm
Pu = 40 t
DiameterTulangan = 19 mm
Ditanya :Perencanaan Tulangan Kolom
Penyelesaian :
Sumbu x =Mu
ɸ.b.ℎ2 .0,85.f =
8x105
0,65.30.30.0,85.150 =
800.000
2.237.625 = 0,36
Sumbu y =Pu
ɸ.b.ℎ .0,85.f =
40x103
0,65.30.30.0,85.150 =
40.000
74.587,5 = 0,54
Dari Grafik: fc’ = 150 kg/cm2
= 15 Mpa = 0,6 r = 0,0568
t = r.
= 0,0568 x 0,6
= 0,034 = 0,01
Ast = t x b x h
= 0,034 x 30 x 30
= 30,6 cm
db = ¼ . 3,14 . 1,92 = 2,83 cm
2
n =Ast
db =
30,6
2,83 = 10,8~12 D 19 mm
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8/18/2019 Laporan Tugas Ujian Mid Beton 1
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Struktur
Beton 1
42
Gambar Penampang
Gambar Diagram Interaksi
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Struktur
Beton 1
43
Perhitungan menggunakan dsgwin
DATA:
Column Width, Bw = 30.00 cm
Column Height, Ht = 30.00 cm
Factored Moment, Mu = 8000.00 kg.m
Factored Shear, Vu = 0.00 kg
Factored Normal, Pu = 40000.00 kg
Factored Torsion, Tu = 0.00 kg.m
Concrete Compressive Strength fc1 = 150.00 kg/cm2
Concrete Crack Strength fcr = 150.00 kg/cm2
Concrete Initial Strength fci = 120.00 kg/cm2
Main Rebar Yield Strength fy = 4000.00 kg/cm2
Stirrups Rebar Yield Strength fyv = 2400.00 kg/cm2
Main rebar diameter dbm = 1.90 cm
2nd main rebar diameter dbn = 1.90 cm
Stirrups Rebar diameter dbv = 1.00 cm
Min concrete cover, Cover = 4.00 cmMin rebar clear space, Minclrspc = 2.50 cm
RESULT:
Flexural Reinforcement: OK
Balanced neutral axis, cbal = 14.5410 cm
Balanced Compr. block, abal = 12.3598 cm
Balanced Moment Capacity, Mnb = 15695.33 kg.m
Balanced Normal Capacity, Pnb = 38537.27 kg
Balanced Eccentricity, ebal = 40.73 cm
Location of neutral axis, c = 16.2499 cm
Length of compression block, a = 13.8124 cm
Total Rebar (incl. side bar) nbt = 12 d19
Total Side Rebar nbs = 0 d19
Total Rebar area, Ast = 34.0234 cm2 = 3.78%
Nominal Moment Capacity, Md = 8741.35 kg.m
Nominal Normal Capacity, Pd = 40002.38 kg
Shear Reinforcement: OK
Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2
Ag = bw*h = 0.00 cm2
pw = Ast/(bw*d) = 0.0000
x1 = bw-2*cv-2*dbv = 0.00 cm
y1 = h-2*cv-2*dbv = 0.00 cm
Vn = Vu/phi = 0.00 kg
Tn = Tu/phi = 0.00 kg.m
Vc = = 0.00 kg
Tc = = 0.00 kg.m
Vnc = Vn - Vc = 0.00 kg
Tnc = Tn - Tc = 0.00 kg.m
Avs = Vnc/(fy*d) = 0.00 cm2
Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2
Atst = Avs + 2*Ats = 0.00 cm2
Additional Longitudinal Rebar, At = 0.00 cm2
Stirrups Spacing, spc = 30.00 cm
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Struktur
Beton 1