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GUNAKAN SEBAIK" NYA PASTI JAYA HUMENDRUUNIVERSITAS MERCUBUANA
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PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK0.85 Fc'
Sc
As' - aub Cc
55 51
Cub
As +
4 ST
30
Mu = 39.8 Tm} d = 0.5
Fc' = 250
Fy = 3200
Keadaan Balance
Persamaan Deformasi/Kompatibilitas :
1.=
Cub
d - Cub0.003
=Cub
0.0016 51-Cub 0.153 = 0.0016 Cub + 0.003 CubCub = 33.26 cmaub = 0.85 x 33.26 = 28.272 cm
Persamaan statika : bajaSc = As'.Fs' = 0.5 As . Fy = 1600 AsCc =0.85 Fc'. Aub . B = 180234St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 180234 = 3200 As
180234.0 = 1600 As
Asb = 112.65
As'b = 56.33
Jadi , Sc = 56.33 x 3200 = 180256.0 kgCc =0.85 x 250 x 28.272 x 30 = 180234.0 kgSt = 112.65 x 3200 = 360480.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (180256 x (51 - 4)) + (180234 x (51 - (0.5 x28.272)))Mnb = 15116178.176 kgcm
Mub = 0.8 x 15116178.176 kgcm= 120.929 Tm > ... Mu, dimensi balok kuat
ec' = 0.003
es'
es
dia e dia s
kg/cm2
kg/cm2
ec'
es
S H = 0
cm2
cm2
S M = 0
Design Tulangan (Under-reinforced)Cu = 10 cmAu = 0.85Cu = 0.85 x 10 = 8.5 cm
Persamaan Deformasi/Kompatibilitas :
1.=
Cu=
10
Cu - d' 10 - 40.003
=10
6
0.018 = 10
0.0018 > 0.0016
tulangan tekan sudah leleh,maka Fs' = Fy = 3200
Persamaan statika : bajaSc = As'.Fs' = 0.5 As x = 1600 AsCc = 0.85 Fc'. Au . B = 54187.5St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 54187.5 = 3200 As
54187.5 = 1600 As
As = 33.87
As' = 16.94
Jadi , Sc = 16.94 x 3200 = 54208.0 kgCc = 0.85 x 250 x 8.5 x 30 = 54187.5 kgSt = 33.87 x 3200 = 108384.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMn = (54208 x (51 - 4)) + (54187.5 x (51 - (0.5 x8.5))) 2547776Mn = 5081041.63 kgcm
Mu = 0.8 x 5081041.63 kgcm= 40.65 Tm > Mu, dimensi balok kuat
Keadaan Under Reinforce = 0.7 x Asb (max). 84.488
diambil As = 33.87
As' = 16.935
Tulangan diameter 25 , Luas = 4.91Tulangan tarik 7 buahTulangan tekan 4 buah
Jadi Dipakai As = 34.37
As' = 19.64
0.02246 OK, r min < r OK, r < r max
0.012837 OK, r min < r OK, r < r max
0.00455 = As min = 6.96
0.03035 = As max = 46.44
Dimensi & tulangan terpasang
ec' ec'
es' es'
es'
es'
es' =
kg/cm2
kg/cm2
S H = 0
cm2
cm2
S M = 0
cm2
cm2
cm2
cm2
cm2
r =
r ' =
r tulangan min = cm2
r tulangan max = cm2
4d25 7d25
55 55
7d25 4d25
30 30
Tumpuan Lapangan
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK0.85 Fc'
Sc
As' - aub Cc
50 46
Cub
As +
4 ST
25
Mu = 39.8 Tm} d = 0.5
Fc' = 250
Fy = 3200
Keadaan Balance
Persamaan Deformasi/Kompatibilitas :
1.=
Cub
d - Cub0.003
=Cub
0.0016 46-Cub 0.138 = 0.0016 Cub + 0.003 CubCub = 30 cmaub = 0.85 x 30 = 25.500 cm
Persamaan statika : bajaSc = As'.Fs' = 0.5 As . Fy = 1600 AsCc =0.85 Fc'. Aub . B = 135469St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 135468.75 = 3200 As
135468.8 = 1600 As
Asb = 84.67
As'b = 42.34
Jadi , Sc = 42.34 x 3200 = 135488.0 kgCc =0.85 x 250 x 25.5 x 25 = 135468.8 kgSt = 84.67 x 3200 = 270944.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (135488 x (46 - 4)) + (135468.75 x (46 - (0.5 x25.5)))Mnb = 10194831.938 kgcm
Mub = 0.8 x 10194831.938 kgcm= 81.559 Tm > ... Mu, dimensi balok kuat
ec' = 0.003
es'
es
dia e dia s
kg/cm2
kg/cm2
ec'
es
S H = 0
cm2
cm2
S M = 0
Design Tulangan (Under-reinforced)Cu = 12 cmAu = 0.85Cu = 0.85 x 12 = 10.2 cm
Persamaan Deformasi/Kompatibilitas :
1.=
Cu=
12
Cu - d' 12 - 40.003
=12
8
0.024 = 12
0.0020 > 0.0016
tulangan tekan sudah leleh,maka Fs' = Fy = 3200
Persamaan statika : bajaSc = As'.Fs' = 0.5 As x = 1600 AsCc = 0.85 Fc'. Au . B = 54187.5St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 54187.5 = 3200 As
54187.5 = 1600 As
As = 33.87
As' = 16.94
Jadi , Sc = 16.94 x 3200 = 54208.0 kgCc = 0.85 x 250 x 10.2 x 25 = 54187.5 kgSt = 33.87 x 3200 = 108384.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMn = (54208 x (46 - 4)) + (54187.5 x (46 - (0.5 x10.2))) 2276736Mn = 4493004.75 kgcm
Mu = 0.8 x 4493004.75 kgcm= 35.94 Tm NOT OK
Keadaan Under Reinforce = 0.7 x Asb (max). 63.503
diambil As = 33.87
As' = 16.935
Tulangan diameter 25 , Luas = 4.91Tulangan tarik 7 buahTulangan tekan 4 buah
Jadi Dipakai As = 34.37
As' = 19.64
0.02989 OK, r min < r OK, r < r max
0.017078 OK, r min < r OK, r < r max
0.00455 = As min = 5.23
0.03035 = As max = 34.90
Dimensi & tulangan terpasang
ec' ec'
es' es'
es'
es'
es' =
kg/cm2
kg/cm2
S H = 0
cm2
cm2
S M = 0
cm2
cm2
cm2
cm2
cm2
r =
r ' =
r tulangan min = cm2
r tulangan max = cm2
4d25 7d25
50 50
7d25 4d25
25 25
Tumpuan Lapangan
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK0.85 Fc'
Sc
As' - aub Cc
60 56
Cub
As +
4 ST
30
Mu = 39.8 Tm} d = 0.5
Fc' = 250
Fy = 3200
Keadaan Balance
Persamaan Deformasi/Kompatibilitas :
1.=
Cub
d - Cub0.003
=Cub
0.0016 56-Cub 0.168 = 0.0016 Cub + 0.003 CubCub = 36.52 cmaub = 0.85 x 36.52 = 31.043 cm
Persamaan statika : bajaSc = As'.Fs' = 0.5 As . Fy = 1600 AsCc =0.85 Fc'. Aub . B = 197899St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 197899.125 = 3200 As
197899.1 = 1600 As
Asb = 123.69
As'b = 61.85
Jadi , Sc = 61.85 x 3200 = 197920.0 kgCc =0.85 x 250 x 31.043 x 30 = 197899.1 kgSt = 123.69 x 3200 = 395808.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (197920 x (56 - 4)) + (197899.125 x (56 - (0.5 x31.043)))Mnb = 18302499.731 kgcm
Mub = 0.8 x 18302499.731 kgcm= 146.420 Tm > ... Mu, dimensi balok kuat
ec' = 0.003
es'
es
dia e dia s
kg/cm2
kg/cm2
ec'
es
S H = 0
cm2
cm2
S M = 0
Design Tulangan (Under-reinforced)Cu = 9 cmAu = 0.85Cu = 0.85 x 9 = 7.65 cm
Persamaan Deformasi/Kompatibilitas :
1.=
Cu=
9
Cu - d' 9 - 40.003
=9
5
0.015 = 9
0.0017 > 0.0016
tulangan tekan sudah leleh,maka Fs' = Fy = 3200
Persamaan statika : bajaSc = As'.Fs' = 0.5 As x = 1600 AsCc = 0.85 Fc'. Au . B = 48768.8St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 48768.75 = 3200 As
48768.75 = 1600 As
As = 30.48
As' = 15.24
Jadi , Sc = 15.24 x 3200 = 48768.0 kgCc = 0.85 x 250 x 7.65 x 30 = 48768.8 kgSt = 30.48 x 3200 = 97536.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMn = (48768 x (56 - 4)) + (48768.75 x (56 - (0.5 x7.65)) 2535936Mn = 5080445.53 kgcm
Mu = 0.8 x 5080445.53 kgcm= 40.64 Tm > Mu, dimensi balok kuat
Keadaan Under Reinforce = 0.7 x Asb (max). 92.768
diambil As = 30.48
As' = 15.24
Tulangan diameter 25 , Luas = 4.91Tulangan tarik 7 buahTulangan tekan 4 buah
Jadi Dipakai As = 34.37
As' = 19.64
0.02046 OK, r min < r OK, r < r max
0.01169 OK, r min < r OK, r < r max
0.00455 = As min = 7.64
0.03035 = As max = 50.99
Dimensi & tulangan terpasang
ec' ec'
es' es'
es'
es'
es' =
kg/cm2
kg/cm2
S H = 0
cm2
cm2
S M = 0
cm2
cm2
cm2
cm2
cm2
r =
r ' =
r tulangan min = cm2
r tulangan max = cm2
4d25 7d25
60 60
7d25 4d25
30 30
Tumpuan Lapangan
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK
document.xls Page 13 Of 32 Balok 25x50
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK
Nama : SHOLIK KRISTIANTONIM : 41107120046
0.85 Fc'Sc
As' - aub Cc
40 35
Cub
As +
5 ST
30
Mu = 34.56 Tm} d = 0.5
Fc' = 250
Fy = 2400
Keadaan Balance
Persamaan Deformasi/Kompatibilitas :
1.=
Cub
d - Cub0.003
=Cub
0.0012 35-Cub 0.105 = 0.0012 Cub + 0.003 CubCub = 25 cmaub = 0.85 x 25 = 21.250 cm
Persamaan statika : bajaSc = As'.Fs' = 0.5 As . Fy = 1200 AsCc =0.85 Fc'. Aub . B = 135469St = As. Fy = 2400 As
1. Sc + Cc = ST1200 As + 135468.75 = 2400 As
135468.8 = 1200 As
Asb = 112.89
As'b = 56.45
Jadi , Sc = 56.45 x 2400 = 135480.0 kgCc =0.85 x 250 x 21.25 x 30 = 135468.8 kgSt = 112.89 x 2400 = 270936.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (135480 x (35 - 5)) + (135468.75 x (35 - (0.5 x21.25)))Mnb = 7366450.781 kgcm
Mub = 0.8 x 7366450.781 kgcm= 58.932 Tm > ... Mu, dimensi balok kuat
ec' = 0.003
es'
es
dia e dia s
kg/cm2
kg/cm2
ec'
es
S H = 0
cm2
cm2
S M = 0
document.xls Page 14 Of 32 Balok 25x50
Design Tulangan (Under-reinforced)Cu = 8 cmAu = 0.85Cu = 0.85 x 8 = 6.8 cm
Persamaan Deformasi/Kompatibilitas :
1.=
Cu=
8
Cu - d' 8 - 50.003
=8
3
0.009 = 8
0.0011 > 0.0012
karena tulangan tekan belum leleh,maka Fs'= 4000
Persamaan statika : bajaSc = As'.Fs' = 0.5 As x = 2000 AsCc = 0.85 Fc'. Au . B = 43350St = As. Fy = 2400 As
1. Sc + Cc = ST2000 As + 43350 = 2400 As
43350 = 400 As
As = 108.38
As' = 54.19
Jadi , Sc = 54.19 x 4000 = 216760.0 kgCc = 0.85 x 250 x 6.8 x 30 = 43350.0 kgSt = 108.38 x 2400 = 260112.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMn = (216760 x (35 - 5)) + (43350 x (35 - (0.5 x6.8))) 6502800Mn = 7872660 kgcm
Mu = 0.8 x 7872660 kgcm= 62.98 Tm > Mu, dimensi balok kuat
84.668
diambil As = 108.38
As' = 54.19
Tulangan diameter 25 , Luas = 4.91Tulangan tarik 23 buahTulangan tekan 12 buah
Jadi Dipakai As = 112.93
As' = 58.92
0.10755 OK, r min < r NOT OK
0.056114 OK, r min < r ' NOT OK
0.002100.02440
ec' ec'
es' es'
es'
es'
es' =
kg/cm2
kg/cm2
S H = 0
cm2
cm2
S M = 0
Keadaan Under Reinforce : As.maks < 0.7 x Asb (max).
cm2
cm2
cm2
cm2
cm2
r =
r ' =
r tulangan min =r tulangan max =
document.xls Page 15 Of 32 Balok 25x50
Dimensi & tulangan terpasang
12d25
40
23d25
30
Lapangan Tumpuan
document.xls Page 16 Of 32 Balok 30x55
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK
Nama : SHOLIK KRISTIANTONIM : 41107120046
0.85 Fc'Sc
As' - aub Cc
55 51
Cub
As +
4 ST
30
Mu = 39.8 Tm} d = 0.5
Fc' = 250
Fy = 3200
Keadaan Balance
Persamaan Deformasi/Kompatibilitas :
1.=
Cub
d - Cub0.003
=Cub
0.0016 51-Cub 0.153 = 0.0016 Cub + 0.003 CubCub = 33.26 cmaub = 0.85 x 33.26 = 28.272 cm
Persamaan statika : bajaSc = As'.Fs' = 0.5 As . Fy = 1600 AsCc =0.85 Fc'. Aub . B = 180234St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 180234 = 3200 As
180234.0 = 1600 As
Asb = 112.65
As'b = 56.33
Jadi , Sc = 56.33 x 3200 = 180256.0 kgCc =0.85 x 250 x 28.272 x 30 = 180234.0 kgSt = 112.65 x 3200 = 360480.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (180256 x (51 - 4)) + (180234 x (51 - (0.5 x28.272)))Mnb = 15116178.176 kgcm
Mub = 0.8 x 15116178.176 kgcm= 120.929 Tm > ... Mu, dimensi balok kuat
ec' = 0.003
es'
es
dia e dia s
kg/cm2
kg/cm2
ec'
es
S H = 0
cm2
cm2
S M = 0
document.xls Page 17 Of 32 Balok 30x55
Design Tulangan (Under-reinforced)Cu = 10 cmAu = 0.85Cu = 0.85 x 10 = 8.5 cm
Persamaan Deformasi/Kompatibilitas :
1.=
Cu=
10
Cu - d' 10 - 40.003
=10
6
0.018 = 10
0.0018 > 0.0016
tulangan tekan sudah leleh,maka Fs' = Fy = 3200
Persamaan statika : bajaSc = As'.Fs' = 0.5 As x = 1600 AsCc = 0.85 Fc'. Au . B = 54187.5St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 54187.5 = 3200 As
54187.5 = 1600 As
As = 33.87
As' = 16.94
Jadi , Sc = 16.94 x 3200 = 54208.0 kgCc = 0.85 x 250 x 8.5 x 30 = 54187.5 kgSt = 33.87 x 3200 = 108384.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMn = (54208 x (51 - 4)) + (54187.5 x (51 - (0.5 x8.5))) 2547776Mn = 5081041.63 kgcm
Mu = 0.8 x 5081041.63 kgcm= 40.65 Tm > Mu, dimensi balok kuat
Keadaan Under Reinforce = 0.7 x Asb (max). 84.488
diambil As = 33.87
As' = 16.935
Tulangan diameter 25 , Luas = 4.91Tulangan tarik 7 buahTulangan tekan 4 buah
Jadi Dipakai As = 34.37
As' = 19.64
0.02246 OK, r min < r OK, r < r max
0.012837 OK, r min < r ' OK, r '< r max
0.004550.03035
ec' ec'
es' es'
es'
es'
es' =
kg/cm2
kg/cm2
S H = 0
cm2
cm2
S M = 0
cm2
cm2
cm2
cm2
cm2
r =
r ' =
r tulangan min =r tulangan max =
document.xls Page 18 Of 32 Balok 30x55
Dimensi & tulangan terpasang
4d25
55
7d25
30
Tumpuan Lapangan
document.xls Page 19 Of 32 Balok 30x60
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK
Nama : SHOLIK KRISTIANTONIM : 41107120046
0.85 Fc'Sc
As' - aub Cc
60 56
Cub
As +
4 ST
30
Mu = 39.8 Tm} d = 0.5
Fc' = 250
Fy = 3200
Keadaan Balance
Persamaan Deformasi/Kompatibilitas :
1.=
Cub
d - Cub0.003
=Cub
0.0016 56-Cub 0.168 = 0.0016 Cub + 0.003 CubCub = 36.52 cmaub = 0.85 x 36.52 = 31.043 cm
Persamaan statika : bajaSc = As'.Fs' = 0.5 As . Fy = 1600 AsCc =0.85 Fc'. Aub . B = 197899St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 197899.125 = 3200 As
197899.1 = 1600 As
Asb = 123.69
As'b = 61.85
Jadi , Sc = 61.85 x 3200 = 197920.0 kgCc =0.85 x 250 x 31.043 x 30 = 197899.1 kgSt = 123.69 x 3200 = 395808.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (197920 x (56 - 4)) + (197899.125 x (56 - (0.5 x31.043)))Mnb = 18302499.731 kgcm
Mub = 0.8 x 18302499.731 kgcm= 146.420 Tm > ... Mu, dimensi balok kuat
ec' = 0.003
es'
es
dia e dia s
kg/cm2
kg/cm2
ec'
es
S H = 0
cm2
cm2
S M = 0
document.xls Page 20 Of 32 Balok 30x60
Design Tulangan (Under-reinforced)Cu = 9 cmAu = 0.85Cu = 0.85 x 9 = 7.65 cm
Persamaan Deformasi/Kompatibilitas :
1.=
Cu=
9
Cu - d' 9 - 40.003
=9
5
0.015 = 9
0.0017 > 0.0016
tulangan tekan sudah leleh,maka Fs' = Fy = 3200
Persamaan statika : bajaSc = As'.Fs' = 0.5 As x = 1600 AsCc = 0.85 Fc'. Au . B = 48768.8St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 48768.75 = 3200 As
48768.75 = 1600 As
As = 30.48
As' = 15.24
Jadi , Sc = 15.24 x 3200 = 48768.0 kgCc = 0.85 x 250 x 7.65 x 30 = 48768.8 kgSt = 30.48 x 3200 = 97536.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMn = (48768 x (56 - 4)) + (48768.75 x (56 - (0.5 x7.65)) 2535936Mn = 5080445.53 kgcm
Mu = 0.8 x 5080445.53 kgcm= 40.64 Tm > Mu, dimensi balok kuat
Keadaan Under Reinforce = 0.7 x Asb (max). 92.768
diambil As = 30.48
As' = 15.24
Tulangan diameter 25 , Luas = 4.91Tulangan tarik 7 buahTulangan tekan 4 buah
Jadi Dipakai As = 34.37
As' = 19.64
0.02046 OK, r min < r OK, r < r max
0.01169 OK, r min < r ' OK, r '< r max
0.004550.03035
ec' ec'
es' es'
es'
es'
es' =
kg/cm2
kg/cm2
S H = 0
cm2
cm2
S M = 0
cm2
cm2
cm2
cm2
cm2
r =
r ' =
r tulangan min =r tulangan max =
document.xls Page 21 Of 32 Balok 30x60
Dimensi & tulangan terpasang
4d25
60
7d25
30
Tumpuan Lapangan
document.xls Page 22 Of 32 Balok 35x70
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK
Nama : SHOLIK KRISTIANTONIM : 41107120046
0.85 Fc'Sc
As' - aub Cc
50 45
Cub
As +
5 ST
25
Mu = 34.56 Tm} d = 0.5
Fc' = 300
Fy = 4000
Keadaan Balance
Persamaan Deformasi/Kompatibilitas :
1.=
Cub
d - Cub0.003
=Cub
0.0020 45-Cub 0.135 = 0.002 Cub + 0.003 CubCub = 27 cmaub = 0.85 x 27 = 22.950 cm
Persamaan statika : bajaSc = As'.Fs' = 0.5 As . Fy = 2000 AsCc =0.85 Fc'. Aub . B = 146306St = As. Fy = 4000 As
1. Sc + Cc = ST2000 As + 146306.25 = 4000 As
146306.3 = 2000 As
Asb = 73.15
As'b = 36.58
Jadi , Sc = 36.58 x 4000 = 146320.0 kgCc =0.85 x 300 x 22.95 x 25 = 146306.3 kgSt = 73.15 x 4000 = 292600.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (146320 x (45 - 5)) + (146306.25 x (45 - (0.5 x22.95)))Mnb = 10757717.031 kgcm
Mub = 0.8 x 10757717.031 kgcm= 86.062 Tm > ... Mu, dimensi balok kuat
ec' = 0.003
es'
es
dia e dia s
kg/cm2
kg/cm2
ec'
es
S H = 0
cm2
cm2
S M = 0
document.xls Page 23 Of 32 Balok 35x70
Design Tulangan (Under-reinforced)Cu = 20 cmAu = 0.85Cu = 0.85 x 20 = 17 cm
Persamaan Deformasi/Kompatibilitas :
1.=
Cu=
20
Cu - d' 20 - 50.003
=20
15
0.045 = 20
0.0023 > 0.002
tulangan tekan sudah leleh,maka Fs' = Fy = 2600
Persamaan statika : bajaSc = As'.Fs' = 0.5 As x = 1300 AsCc = 0.85 Fc'. Au . B = 108375St = As. Fy = 4000 As
1. Sc + Cc = ST1300 As + 108375 = 4000 As
108375 = 2700 As
As = 40.14
As' = 20.07
Jadi , Sc = 20.07 x 2600 = 52182.0 kgCc = 0.85 x 300 x 17 x 25 = 108375.0 kgSt = 40.14 x 4000 = 160560.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMn = (52182 x (45 - 5)) + (108375 x (45 - (0.5 x17))) 2087280Mn = 6042967.5 kgcm
Mu = 0.8 x 6042967.5 kgcm= 48.34 Tm > Mu, dimensi balok kuat
Keadaan Under Reinforce = 0.7 x Asb (max). 54.863
diambil As = 40.14
As' = 20.07
Tulangan diameter 22 , Luas = 3.8Tulangan tarik 11 buahTulangan tekan 6 buah
Jadi Dipakai As = 41.8
As' = 22.8
0.03716 OK, r min < r NOT OK
0.020267 OK, r min < r ' OK, r '< r max
0.002100.02440
ec' ec'
es' es'
es'
es'
es' =
kg/cm2
kg/cm2
S H = 0
cm2
cm2
S M = 0
cm2
cm2
cm2
cm2
cm2
r =
r ' =
r tulangan min =r tulangan max =
document.xls Page 24 Of 32 Balok 35x70
Dimensi & tulangan terpasang
6d22
50
11d22
25
Tumpuan Lapangan
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK0.85 Fc'
Sc
As' - aub Cc
60 56
Cub
As +
4 ST
30
Mu = 39.8 Tm} d = 0.5
Fc' = 250
Fy = 3200
Keadaan Balance
Persamaan Deformasi/Kompatibilitas :
1.=
Cub
d - Cub0.003
=Cub
0.0016 56-Cub 0.168 = 0.0016 Cub + 0.003 CubCub = 36.52 cmaub = 0.85 x 36.52 = 31.043 cm
Persamaan statika : bajaSc = As'.Fs' = 0.5 As . Fy = 1600 AsCc =0.85 Fc'. Aub . B = 197899St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 197899.125 = 3200 As
197899.125 = 1600 As
Asb = 123.69
As'b = 61.85
Jadi , Sc = 61.85 x 3200 = 197920.0 kgCc =0.85 x 250 x 31.043 x 30 = 197899.1 kgSt = 123.69 x 3200 = 395808.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (197920 x (56 - 4)) + (197899.125 x (56 - (0.5 x31.043)))Mnb = 18302499.731 kgcm
Mub = 0.8 x 18302499.731 kgcm= 146.420 Tm > ... Mu, dimensi balok kuat
ec' = 0.003
es'
es
dia e dia s
kg/cm2
kg/cm2
ec'
es
S H = 0
cm2
cm2
S M = 0
Design Tulangan (Under-reinforced)Cu = 9 cmAu = 0.85Cu = 0.85 x 9 = 7.65 cm
Persamaan Deformasi/Kompatibilitas :
1.=
Cu=
9
Cu - d' 9 - 40.003
=9
5
0.015 = 9
0.0017 > 0.0016
tulangan tekan sudah leleh,maka Fs' = Fy = 3200
Persamaan statika : bajaSc = As'.Fs' = 0.5 As x = 1600 AsCc = 0.85 Fc'. Au . B = 48768.8St = As. Fy = 3200 As
1. Sc + Cc = ST1600 As + 48768.75 = 3200 As
48768.75 = 1600 As
As = 30.48
As' = 15.24
Jadi , Sc = 15.24 x 3200 = 48768.0 kgCc = 0.85 x 250 x 7.65 x 30 = 48768.8 kgSt = 30.48 x 3200 = 97536.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMn = (48768 x (56 - 4)) + (48768.75 x (56 - (0.5 x7.65)) 2535936Mn = 5080445.53 kgcm
Mu = 0.8 x 5080445.53 kgcm= 40.64 Tm > Mu, dimensi balok kuat
Keadaan Under Reinforce = 0.7 x Asb (max). 92.768
diambil As = 30.48
As' = 15.24
Tulangan diameter 25 , Luas = 4.91Tulangan tarik 7 buahTulangan tekan 4 buah
Jadi Dipakai As = 34.37
As' = 19.64
0.02046 OK, r min < r < r max
0.01169 OK, r min < r < r max
0.00455 = As min = 7.64
0.03035 = As max = 50.99
Dimensi & tulangan terpasang
ec' ec'
es' es'
es'
es'
es' =
kg/cm2
kg/cm2
S H = 0
cm2
cm2
S M = 0
cm2
cm2
cm2
cm2
cm2
r =
r ' =
r tulangan min = cm2
r tulangan max = cm2
4d25 7d25
60 60
7d25 4d25
30 30
Tumpuan Lapangan
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK0.85 Fc'
Sc
As' - aub Cc
45 40
Cub
As + Mnb
5 ST
30
Mu = 32.52 Tm} d = 0.5
Fc' = 250
Fy = 4000
Keadaan Balance
Persamaan Deformasi/Kompatibilitas :
1.=
Cub
d - Cub0.003
=Cub
0.0020 40-Cub 0.12 = 0.002 Cub + 0.003 CubCub = 24 cmaub = 0.85 x 24 = 20.400 cm
Persamaan statika : Sc = As'.Fs' = 0.5 As . Fy = 2000 As (tul. Tekan sudah leleh)Cc =0.85 Fc'. Aub . B = 130050St = As. Fy = 4000 As
1. Sc + Cc = ST2000 As + 130050 = 4000 As
130050 = 2000 As
Asb = 65.03
As'b = 32.52
Jadi , Sc = 32.52 x 4000 = 130080.0 kgCc =0.85 x 250 x 20.4 x 30 = 130050.0 kgSt = 65.03 x 4000 = 260120.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (130080 x (40 - 5)) + (130050 x (40 - (0.5 x20.4)))Mnb = 8428290 kgcm
Mub = 0.8 x 8428290 kgcm= 67.426 Tm > ... Mu, dimensi balok kuat >
ec' = 0.003
es'
es
dia e dia s
kg/cm2
kg/cm2
ec'
es
S H = 0
cm2
cm2
S M 2= 0
Design Tulangan (Under-reinforced) COBA 1: Cu = 16 cmAu = 0.85Cu = 0.85 x 16 = 13.6 cm
Persamaan Deformasi/Kompatibilitas :
1.=
Cu=
16
Cu - d' 16 - 50.003
=16
11
0.033 = 16
0.0021 > 0.002
tulangan tekan sudah leleh,maka Fs' = Fy = 2400
Persamaan statika : Sc = As'.Fs' = 0.5 As x fy = 1200 AsCc = 0.85 Fc'. Au . B = 86700St = As. Fy = 4000 As
1. Sc + Cc = ST1200 As + 86700 = 4000 As
86700 = 2800 As
As = 30.96
As' = 15.48
Jadi , Sc = 15.48 x 2400 = 37152.0 kgCc = 0.85 x 250 x 13.6 x 30 = 86700.0 kgSt = 30.96 x 4000 = 123840.0 kg
2. Sc (d - d') + Cc (d - 0.5au) = MnMn = (37152 x (40 - 5)) + (86700 x (40 - (0.5 x13.6))) 1300320Mn = 4178760 kgcm
Mu = 0.8 x 4178760 kgcm= 33.43 Tm > Mu, dimensi balok kuat
Keadaan Under Reinforce = 0.7 x Asb (max). 48.773
diambil As = 30.96 OK, As max < As < As min
As' = 15.48Perhitungan tulangan di bawah ini tidak terpakai.
Sc = 15.5 x 2400 = 37152.00 kgCc = 0.85 x 250 x 13.6 x 30 = 86700.00 kgSt = 31 x 4000 = 123840.00 kg
PERSAMAAN :1. Sc (d - d') + Cc (d - 0.5au) = Mn
Mn = 4178760.00 kg cmMu = 3343008.00 kg cm
= 33.43 Tm > M = 32.52 TmOK, dimensi & tulangan cukup
Tulangan diameter 22 , Luas = 3.8Tulangan tarik 9 buahTulangan tekan 5 buah
0.022933
ec' ec'
es' es'
es'
es'
es' =
kg/cm2
kg/cm2
S H = 0
cm2
cm2
S M = 0
cm2
cm2
S M = 0
Mu = f Mn
cm2
r =
0.0035 = As min = 4.73
0.03035 = As max = 40.97
Dimensi & tulangan terpasang
5d22 9d22
45 45
9d22 5d22
30 30
Tumpuan Lapangan
r tulangan min = cm2
r tulangan max = cm2
PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK
Mnb
