The document provides calculations for determining the required reinforcement of a concrete beam (balok) with the following information: - Concrete compressive strength is 20 MPa - Steel yield strength is 400 MPa - Beam dimensions are 25cm x 40cm - Loads include wall weight, floor finish weight, and live loads from balconies Bending moments are calculated at different points along the beam due to the varying loads. Required steel reinforcement is then determined based on the bending moment values and reinforcement ratios from code tables. Reinforcement amounts are provided for three sections of the beam labeled A-B, B-C, and C-D.

1. 93
100 300 400 400 200 100
200500500400400100200
200250500400500200
400 400 300 100 200200
VOID
VOID
A
B
C
D
1
A
B
B'
C
D
E
F
F'
2 2' 3 4 5 5' 6
1 2 3 4 5 5' 64'
250
2100
E
F
F'
A
B
C
D
E'
3.2.3. Perhitungan Balok
Data yang ada :
 Mutu Beton ( f’c ) = 20 Mpa
 Mutu Baja ( f’y ) = 400 Mpa (ulir)
 Berat Dinding = 2,5 kN/m2
 Berat Jenis Beton = 2400 kg/m3
 WL = 2 kN/m2
, 3 kN/m2
 Penutup = 1 kN/m2
 Tinggi dinding = 4 m

2. 94
200 500 400 500 250 250
A B C D E E' F
+
BALOK AS - 3
 Dimensi Balok
Diambil bentang terpanjang yaitu 5 m
h =
12
1
 =
12
1
. 500 = 41,66 ~ 40 cm
b =
3
2
h =
3
2
. 40 = 26,6 ~ 25 cm
Diambil dimensi balok = 25/40
1. BALOK AS 3 ( A-B)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 3,88kN/m
Beban hidup (WL)
– WL untuk balkon = 300 . 1 . 10 = 3,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(3)
=9,456 kN/m

3. 95
+
2. Pembebanan BALOK A-B
Beban mati(WD)
– q ekuivalen plat beban segitiga
=
3
1
.WU plat . lx
=
3
1
. 9,456 . 2 = 6,304kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m
– Tembok t=1m
2,5 kN/m2
. 1 = 2,500kN/m
q AB = 11,204kN/m
3. Menghitung Momen
MtA =
16
1
. q . l2
=
16
1
. 11,204 . 22
= 2,801 kNm
MlAB =
14
1
. q . l 2
=
14
1
. 11,204 . 22
= 3,201 kNm

4. 96
MtB =
10
1
. q . l2
=
10
1
. 11,204 . 22
= 4,482 kNm
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 400 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m
5. Mencari tulangan
>> MtA = 2,801 KNm
22
346,025,0
801,2


db
Mu
= 93,588 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
ASTA =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2

5. 97
>> MlAB = 3,021 KNm
22
346,025,0
021,3


db
Mu
= 106,953 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
ASlAB =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2
>> MtB = 4,482 KNm
22
346,025,0
482,4


db
Mu
= 149,754 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
ASlAB =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2
6. Pilih Tulangan
AS A = 2 D 16 (402 mm2
)
AS A-B = 2 D 16 (402 mm2
)
AS B = 2 D 16 (402 mm2
)

6. 98
+
2. BALOK AS 3 (B-C)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 3,88kN/m
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK B-C
Beban mati(WD)
– q ekuivalen plat beban segitiga
=
3
1
.WU plat . lx. 1
=
3
1
. 7,856 . 2,5 . 1 = 6,546kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m

7. 99
+
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
– Pembebanan P (diratakan)
> q ekuivalen plat trapesium
=
6
1
. WU . lx . (3-(
ly
lx
)2
)
=
6
1
. 7,856 . 2,5 . (3-(
4
5,2
)2
)
= 17,083 kN/m
>bs balok anak
=0,2 . 0,3 . 24
=1,44 kN/m
>q total = 17,083 kN/m + 1,44 kN/m
= 18,523 kN/m
>beban terpusat
=q . l = 18,523 . 4 m = 74,092 Kn
>dibagi di 2 tumpuan, tiap tumpuan menerima:
=74,092 : 2 = 37,046 kN
Diasumsikan diratakan di balok B-C
37,046 : 5 = 7,409 kN/m = 7,409kN/m
qBC = 26,355kN/m

8. 100
3. Menghitung Momen
MtB =
10
1
. q . l2
=
10
1
. 26,355 . 52
= 65,887 kNm
MlBC =
16
1
. q . l 2
=
16
1
. 26,355 . 52
= 41,719 kNm
MtC =
11
1
. q . l2
=
11
1
. 26,355 . 52
= 59,897 kNm
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 400 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

9. 101
5. Mencari tulangan
>> MtB = 65,887 KNm
22
346,025,0
65,887


db
Mu
= 2201,44 KN/m2
 interpolasi =
22002300
22002201


=
0075,00079,0
0075,0


= 0,0004 = 100 x – 0,75
= 0,0075
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0075
ASTB =  . b . d
= 0,0075 . 250 . 346
= 648,7 mm2
>> MlBC = 41,719 KNm
22
346,025,0
41,719


db
Mu
= 1375,88 KN/m2
 interpolasi =
13001400
13001376


=
0043,00046,0
0043,0


= 0,0228 = 100 x – 0,43
= 0,0045
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0045

10. 102
ASlBC =  . b . d
= 0,0045 . 250 . 346
= 389,25 mm2
>> MtC = 59,897 KNm
22
346,025,0
59,897


db
Mu
= 2001 KN/m2
 interpolasi =
20002100
20002001


=
0068,00072,0
0068,0


= 0,0004 = 100 x – 0,68
= 0,0068
 interpolasi >  min (0,0017)
Dipakai  min 0,0068
AStC =  . b . d
= 0,0068 . 250 . 346
= 588,2mm2
6. Pilih Tulangan
AS B = 4 D 16 (804 mm2
)
AS B-C = 2 D 16 (402 mm2
)
AS C = 3 D 16 (603 mm2
)

11. 103
+
3. BALOK AS 3 (C-D)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 3,88kN/m
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK B-C
Beban mati(WD)
– q ekuivalen plat beban segitiga
=
3
1
.WU plat . lx. 2
=
3
1
. 7,856 . 2 . 2 = 10,501kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m

12. 104
+
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
– Pembebanan P (diratakan)
> q ekuivalen plat trapesium
=
6
1
. WU . lx . (3-(
ly
lx
)2
)
=
6
1
. 7,856 . 2 . (3-(
4
2
)2
)
= 7,201 kN/m x 4
= 28,805 kN/m
>bs balok anak
=0,2 . 0,3 . 24 . 2
=2,88 kN/m
>q total = 28,805 kN/m + 2,88 kN/m
= 31,685 kN/m
>beban terpusat
=q . l = 31,685 . 4 m = 126,623 kN
Diasumsikan diratakan di balok C-D
126,623 : 4 = 31,658 kN/m = 31,658kN/m
qBC = 54,559kN/m

13. 105
3. Menghitung Momen
MtC =
11
1
. q . l2
=
11
1
. 54,559 . 42
= 79,358 kNm
MlCD =
16
1
. q . l 2
=
16
1
. 54,559 . 42
= 54,559 kNm
MtD =
11
1
. q . l2
=
11
1
. 54,559 . 42
= 79,358 kNm
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 400 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

14. 106
5. Mencari tulangan
>> MtC = 79,358 KNm
22
346,025,0
79,358


db
Mu
= 2651,54 KN/m2
 interpolasi =
26002700
26002651


=
0091,00095,0
0091,0


= 0,0204 = 100 x – 0,91
= 0,0093
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0093
ASTC =  . b . d
= 0,0093 . 250 . 346
= 804 mm2
>> MlCD = 54,559 KNm
22
346,025,0
54,559


db
Mu
= 1822,94 KN/m2
 interpolasi =
18001900
18001822


=
0061,00064,0
0061,0


= 0,0066 = 100 x – 0,61
= 0,0062
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0062

15. 107
ASlCD =  . b . d
= 0,0062 . 250 . 346
= 536,3 mm2
>> MtD = 79,358 KNm
22
346,025,0
79,358


db
Mu
= 2651,54 KN/m2
 interpolasi =
26002700
26002651


=
0091,00095,0
0091,0


= 0,0204 = 100 x – 0,91
= 0,0093
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0093
ASTD =  . b . d
= 0,0093 . 250 . 346
= 804 mm2
6. Pilih Tulangan
AS C = 4 D 16 (804 mm2
)
AS C-D = 3 D 16 (603 mm2
)
AS D = 4 D 16 (804 mm2
)

16. 108
+
4. BALOK AS 3 (D-E)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 3,88kN/m
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK D-E
Beban mati(WD)
– q ekuivalen plat beban segitiga
=
3
1
.WU plat . lx. 1
=
3
1
. 7,856 . 2,5 . 1 = 6,546kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m

17. 109
+
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
– Pembebanan P (diratakan)
> q ekuivalen plat trapesium
=
6
1
. WU . lx . (3-(
ly
lx
)2
)
=
6
1
. 7,856 . 2,5 . (3-(
4
5,2
)2
)
= 17,083 kN/m
>bs balok anak
=0,2 . 0,3 . 24
=1,44 kN/m
>q total = 17,083 kN/m + 1,44 kN/m
= 18,523 kN/m
>beban terpusat
=q . l = 18,523 . 4 m = 74,092 kN
>dibagi di 2 tumpuan, tiap tumpuan menerima:
=74,092 : 2 = 37,046 kN
Diasumsikan diratakan di balok D-E
37,046 : 5 = 7,409 kN/m = 7,409kN/m
qDE = 26,355kN/m

18. 110
3. Menghitung Momen
MtD =
11
1
. q . l2
=
11
1
. 26,355 . 52
= 59,897 kNm
MlDE =
16
1
. q . l 2
=
16
1
. 26,355 . 52
= 41,719 kNm
MtE =
11
1
. q . l2
=
11
1
. 26,355 . 52
= 59,897 kNm
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 400 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

19. 111
5. Mencari tulangan
>> MtD = 59,897 KNm
22
346,025,0
59,897


db
Mu
= 2001,3 KN/m2
 interpolasi =
20002100
20002001


=
0068,00072,0
0068,0


= 0,0004 = 100 x – 0,68
= 0,0068
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0068
ASTD =  . b . d
= 0,0068 . 250 . 346
= 588,2 mm2
>> MlDE = 41,719 KNm
22
346,025,0
41,719


db
Mu
= 1375,88 KN/m2
 interpolasi =
13001400
13001376


=
0043,00046,0
0043,0


= 0,0228 = 100 x – 0,43
= 0,0045
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0045

20. 112
ASlDE =  . b . d
= 0,0045 . 250 . 346
= 389,25 mm2
>> MtE = 59,897 KNm
22
346,025,0
59,897


db
Mu
= 2001,3 KN/m2
 interpolasi =
20002100
20002001


=
0068,00072,0
0068,0


= 0,0004 = 100 x – 0,68
= 0,0068
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0068
ASTE =  . b . d
= 0,0068 . 250 . 346
= 588,2 mm2
6. Pilih Tulangan
AS D = 3 D 16 (603 mm2
)
AS D-E = 2 D 16 (402 mm2
)
AS E = 3 D 16 (603 mm2
)

21. 113
+
5. BALOK AS 3 ( E-E’)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 3,88kN/m
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK A-B
Beban mati(WD)
– q ekuivalen plat beban segitiga
=
3
1
.WU plat . lx
=
3
1
. 7,856 . 2,5 . 2 = 13,093kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m

22. 114
+
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
q EE’ = 25,493kN/m
3. Menghitung Momen
MtE =
11
1
. q . l2
=
11
1
. 25,493 . 2,52
= 14,484 kNm
MlEE’ =
16
1
. q . l 2
=
16
1
. 25,493 . 2,52
= 9,958 kNm
MtE’ =
10
1
. q . l2
=
10
1
. 25,493 . 2,52
= 15,933 kNm
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 400 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm

23. 115
d = h – p – Ø sengkang – ½ Ø utama
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m
5. Mencari tulangan
>> MtE = 14,484 KNm
22
346,025,0
14,484


db
Mu
= 467,77 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
ASTE =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2
>> MlEE’ = 9,958 KNm
22
346,025,0
9,958


db
Mu
= 332,720 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
AslEE’ =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2

24. 116
+
>> MtE’ = 15,933 KNm
22
346,025,0
15,933


db
Mu
= 532,359 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
AstE’ =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2
6. Pilih Tulangan
ASE = 2 D 16 (402 mm2
)
AS E-E’ = 2 D 16 (402 mm2
)
AS E’ = 2 D 16 (402 mm2
)
6. BALOK AS 3 ( E’-F)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 3,88kN/m

25. 117
+
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK A-B
Beban mati(WD)
– q ekuivalen plat beban segitiga
=
3
1
.WU plat . lx
=
3
1
. 7,856 . 2,5 . 2 = 13,093kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
q E’F = 25,493kN/m

26. 118
3. Menghitung Momen
MtE’ =
10
1
. q . l2
=
10
1
. 25,493 . 2,52
= 15,933 kNm
MlE’F =
14
1
. q . l 2
=
14
1
. 25,493 . 2,52
= 11,380 kNm
MtF =
16
1
. q . l2
=
16
1
. 25,493 . 2,52
= 9,958 kNm
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 400 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

27. 119
5. Mencari tulangan
>> MtE’ = 15,933 KNm
22
346,025,0
15,933


db
Mu
= 532,359 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
ASTE’ =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2
>> MlE’F = 11,380 KNm
22
346,025,0
11,380


db
Mu
= 380,323 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
AslE’F =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2
>> MtF = 9,958 KNm
22
346,025,0
9,958


db
Mu
= 332,720 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017

28. 120
AstF =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2
6. Pilih Tulangan
ASE = 2 D 16 (402 mm2
)
AS E-E’ = 2 D 16 (402 mm2
)
AS E’ = 2 D 16 (402 mm2
)

29. 121
200 400 400 300 100 200
1 2 3 4 5'54'
+
BALOK AS-F
 Dimensi Balok
Diambil bentang terpanjang yaitu 4 m
h =
12
1
 =
12
1
. 400 = 33,3 ~ 35 cm
b =
3
2
h =
3
2
. 35 = 23,3 ~ 20 cm
Diambil dimensi balok = 20/35
1. BALOK AS F ( 1-2)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 3,88kN/m
Beban hidup (WL)
– WL untuk balkon = 300 . 1 . 10 = 3,00kN/m

30. 122
+
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(3)
=9,456 kN/m
2. Pembebanan BALOK 1-2
Beban mati(WD)
– q ekuivalen plat beban segitiga
=
3
1
.WU plat . lx
=
3
1
. 9,456 . 2 = 6,304kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,2 . 0,35 . 24 kN/m2
= 1,680kN/m
– Tembok t=1m
2,5 kN/m2
. 1 = 2,500kN/m
q 1-2 = 10,484kN/m
3. Menghitung Momen
Mt1 =
16
1
. q . l2
=
16
1
. 10,484 . 22
= 2,621 kNm # 5,11 kNm (hasil Beamax)

31. 123
Ml1-2 =
14
1
. q . l 2
=
14
1
. 10,484 . 22
= 2,995 kNm
MtB =
10
1
. q . l2
=
10
1
. 10,484 . 22
= 4,193 kNm # 20,7 kNm (hasil beamax)
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 350 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 350 – 40 – 8 – 6
= 296 mm = 0,296 m
5. Mencari tulangan
>> Mt1 = 5,11 KNm
22
296,020,0
5,11


db
Mu
= 291,613 KN/m2
 interpolasi <  min (0,0017)

32. 124
Dipakai  min 0,0017
AST1 =  . b . d
= 0,0017 . 200 . 296
= 100,64 mm2
>> Ml1-2 = 2,995 KNm
22
296,020,0
2,995


db
Mu
= 170,91 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
Asl1-2 =  . b . d
= 0,0017 . 200 . 296
= 100,64 mm2
>> Mt2 = 20,7 KNm
22
296,020,0
20,7


db
Mu
= 1181,291 KN/m2
 interpolasi =
11001200
11001181


=
0036,00039,0
0036,0


= 0,0243 = 100 x – 0,36
= 0,0038
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0038

33. 125
+
Ast2 =  . b . d
= 0,0038 . 200 . 296
= 224,96mm2
6. Pilih Tulangan
AS1 = 2 D 13 (265 mm2
)
AS 1-2 = 2 D 13 (265 mm2
)
AS 2 = 2 D 13 (265 mm2
)
2. BALOK AS F ( 2-3)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 2,88kN/m
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m

34. 126
+
2. Pembebanan BALOK 2-3
Beban mati(WD)
– q ekuivalen plat beban trapesium
=
6
1
. WU . lx . (3-(
ly
lx
)2
)
=
6
1
. 7,856 . 2,5 . (3-(
4
5,2
)2
) = 8,451kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,2 . 0,35 . 24 kN/m2
= 1,680kN/m
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
q 2-3 = 20,221kN/m
3. Menghitung Momen
Mt2 =
10
1
. q . l2
=
10
1
. 20,221 . 42
= 32,353 kNm
Ml2-3 =
16
1
. q . l 2
=
16
1
. 20,221 . 42
= 20,221 kNm

35. 127
Mt3 =
11
1
. q . l2
=
11
1
. 20,221 . 42
= 29,412 kNm
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 350 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 350 – 40 – 8 – 6
= 296 mm = 0,296 m
5. Mencari tulangan
>> Mt2 = 32,353 KNm
22
296,020,0
32,353


db
Mu
= 1846,295 KN/m2
 interpolasi =
18001900
18001846


=
0061,00064,0
0061,0


= 0,0138 = 100 x – 0,61
= 0,0062
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0062

36. 128
AST2 =  . b . d
= 0,0062 . 200 . 296
= 367,04 mm2
>> Ml2-3 = 20,221 KNm
22
296,020,0
20,221


db
Mu
= 1153,955 KN/m2
 interpolasi =
11001200
11001154


=
0036,00039,0
0036,0


= 0,0162 = 100 x – 0,36
= 0,0037
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0037
Asl2-3 =  . b . d
= 0,0037 . 200 . 296
= 219,04 mm2
>> Mt3 = 29,412 KNm
22
296,020,0
29,412


db
Mu
= 1678,46 KN/m2
 interpolasi =
16001700
16001678


=
0053,00057,0
0053,0


= 0,0312 = 100 x – 0,53
= 0,0056

37. 129
+
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0056
Ast3 =  . b . d
= 0,0056 . 200 . 296
= 331,52 mm2
6. Pilih Tulangan
AS2 = 3 D 13 (398 mm2
)
AS 2-3 = 2 D 13 (265 mm2
)
AS 3 = 3 D 13 (398 mm2
)
3. BALOK AS F ( 3-4)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 2,88kN/m
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m

38. 130
+
2. Pembebanan BALOK 3-4
Beban mati(WD)
– q ekuivalen plat beban trapesium
=
6
1
. WU . lx . (3-(
ly
lx
)2
)
=
6
1
. 7,856 . 2,5 . (3-(
4
5,2
)2
) = 8,451kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,2 . 0,35 . 24 kN/m2
= 1,680kN/m
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
q 3-4 = 20,221kN/m
3. Menghitung Momen
Mt3 =
11
1
. q . l2
=
11
1
. 20,221 . 42
= 29,412 kNm
Ml3-4 =
16
1
. q . l 2
=
16
1
. 20,221 . 42
= 20,221 kNm

39. 131
Mt4 =
11
1
. q . l2
=
11
1
. 20,221 . 42
= 29,412 kNm # 34,76 kNm (hasil beamax)
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 350 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 350 – 40 – 8 – 6
= 296 mm = 0,296 m
5. Mencari tulangan
>> Mt3 = 29,412 KNm
22
296,020,0
29,412


db
Mu
= 1678,46 KN/m2
 interpolasi =
16001700
16001678


=
0053,00057,0
0053,0


= 0,0312 = 100 x – 0,53
= 0,0056
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0056

40. 132
Ast3 =  . b . d
= 0,0056 . 200 . 296
= 331,52 mm2
>> Ml3-4 = 20,221 KNm
22
296,020,0
20,221


db
Mu
= 1153,955 KN/m2
 interpolasi =
11001200
11001154


=
0036,00039,0
0036,0


= 0,0162 = 100 x – 0,36
= 0,0037
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0037
Asl3-4 =  . b . d
= 0,0037 . 200 . 296
= 219,04 mm2
>> Mt4 = 34,76 KNm
22
296,020,0
34,76


db
Mu
= 1983,65 KN/m2
 interpolasi =
19002000
19001983


=
0064,00068,0
0064,0


= 0,0162 = 100 x – 0,64
= 0,0067

41. 133
+
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0067
Ast4 =  . b . d
= 0,0067 . 200 . 296
= 396,64 mm2
6. Pilih Tulangan
AS3 = 3 D 13 (398 mm2
)
AS 3-4 = 2 D 13 (265 mm2
)
AS 4 = 3 D 13 (398 mm2
)
4. BALOK AS F ( 4-4’)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 2,88kN/m
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m

42. 134
2. Pembebanan BALOK 4-4’
Beban mati(WD)
– q ekuivalen plat beban trapesium
=
6
1
. WU . lx . (3-(
ly
lx
)2
)
=
6
1
. 7,856 . 2 . (3-(
3
2
)2
) = 6,692kN/m
– ekuivalen plat beban segitiga
=
3
1
.WU plat . lx
=
3
1
. 7,856 . 2 . 1,5 = 7,856kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,2 . 0,35 . 24 kN/m2
= 1,680kN/m
– Tembok t=4m
=2,5 kN/m2
. 4 = 10,000kN/m
– pembebanan beban p (diratakan)
>q ekuivalen plat trapesium
=
6
1
. WU . lx . (3-(
ly
lx
)2
)
=
6
1
. 7,856 . 2 . (3-(
5,2
2
)2
) x 4
= 24,720 kN/m

43. 135
+
>bs balok anak
=0,2 . 0,3 . 24
=1,44 kN/m
>bs tembok t=4m
=2,5 . 4 = 10,000 kN/m
>q total = 17,083 kN/m + 1,44 kN/m+10,00 kN/m
= 34,720 kN/m
>beban terpusat
=q . l = 34,720 . 5 m = 173,601 kN
>dibagi di 2 tumpuan, tiap tumpuan menerima:
=173,61 : 2 = 86,805 kN
Diasumsikan diratakan di balok 4-4’
86,805 : 3 = 28,933 kN/m = 28,933kN/m
q 4-4’ = 55,080kN/m
3. Menghitung Momen
Mt4 =
11
1
. q . l2
=
11
1
. 55,080 . 32
= 45,065 kNm

44. 136
Ml4-4’ =
16
1
. q . l 2
=
16
1
. 55,080 . 32
= 30,982 kNm # 41,61 kNm (hasil beamax)
Mt4 =
11
1
. q . l2
=
11
1
. 55,080 . 32
= 45,065 kNm # 46,36 kNm (hasil beamax)
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 350 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 350 – 40 – 8 – 6
= 296 mm = 0,296 m
5. Mencari tulangan
>> Mt4 = 45,065 KNm
22
296,020,0
45,065


db
Mu
= 2571,733 KN/m2

45. 137
 interpolasi =
25002600
25002572


=
0087,00091,0
0087,0


= 0,0288 = 100 x – 0,87
= 0,0089
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0089
Ast4 =  . b . d
= 0,0089 . 200 . 296
= 526,88 mm2
>> Ml4-4’ = 41,61 KNm
22
296,020,0
41,61


db
Mu
= 2374,5 KN/m2
 interpolasi =
23002400
23002375


=
0079,00083,0
0079,0


= 0,0300 = 100 x – 0,79
= 0,0082
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0082
Asl4-4’ =  . b . d
= 0,0082 . 200 . 296
= 485,44 mm2

46. 138
+
>> Mt4’ = 46,36 KNm
22
296,020,0
46,36


db
Mu
= 2645,63 KN/m2
 interpolasi =
16002700
26002646


=
0091,00095,0
0091,0


= 0,0184 = 100 x – 0,91
= 0,0092
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0092
Ast4’ =  . b . d
= 0,0092 . 200 . 296
= 544,64 mm2
6. Pilih Tulangan
AS4 = 4 D 13 (531 mm2
)
AS 4-4’ = 3 D 13 (398 mm2
)
AS 4‘ = 5 D 13 (664 mm2
)
5. BALOK AS 3 ( 4’-5)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 3,88kN/m

47. 139
+
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK 4’-5
Beban mati(WD)
– q ekuivalen plat beban segitiga
=
3
1
.WU plat . lx
=
3
1
. 7,856 . 2 .
2
1
= 2,618kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,20 . 0,35 . 24 kN/2
= 1,680kN/m
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
q 4’-5 = 14,298kN/m

48. 140
3. Menghitung Momen
Mt4’ =
11
1
. q . l2
=
11
1
. 14,298 . 12
= 1,299 kNm # 46,36 kNm (hasil beamax)
Ml4’5 =
16
1
. q . l 2
=
16
1
. 14,298 . 12
= 0,893 kNm
Mt5 =
10
1
. q . l2
=
10
1
. 14,298 . 12
= 1,429 kNm # 5,17 kNm (hasil beamax)
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 400 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

49. 141
5. Mencari tulangan
>> Mt4’ = 46,36 KNm
22
296,020,0
46,36


db
Mu
= 2645,63 KN/m2
 interpolasi =
16002700
26002646


=
0091,00095,0
0091,0


= 0,0184 = 100 x – 0,91
= 0,0092
 interpolasi >  min (0,0017)
Dipakai  interpolasi 0,0092
AST4’ =  . b . d
= 0,0092 . 200 . 296
= 544,64 mm2
>> Ml4’5 = 0,893 KNm
22
296,020,0
0,893


db
Mu
= 50,961 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
Asl4’5 =  . b . d
= 0,0017 . 200 . 296
= 100,64 mm2

50. 142
+
>> Mt5 = 5,17 KNm
22
296,020,0
5,17


db
Mu
= 295,03 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
AstF =  . b . d
= 0,0017 . 200 . 296
= 100,64 mm2
6. Pilih Tulangan
AS4‘ = 5 D 13 (664 mm2
)
AS 4’-5 = 2 D 13 (265 mm2
)
AS 5 = 2 D 13 (265 mm2
)
6. BALOK AS 3 ( 5-5`)
1. Beban yang bekerja pada plat lantai
Beban mati(WD)
– B.s. plat = 0,12 . 2400 . 1 . 10 = 2880 N/m = 2,88kN/m
– Finishing =100 . 1 . 10 = 1000 N/m = 1,00kN/m
– WD TOTAL = 3,88kN/m

51. 143
+
Beban hidup (WL)
– WL untuk balkon = 200 . 1 . 10 = 2,00kN/m
Beban Ultimate (WU) =1,2 (WD)+1,6(WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK 5-5’
Beban mati(WD)
– q ekuivalen plat beban segitiga
=
3
1
.WU plat . lx
=
3
1
. 7,856 . 2 = 5,237kN/m
– Berat sendiri Balok
= b . h . 24 kN/m2
= 0,20 . 0,35 . 24 kN/2
= 1,680kN/m
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
q 5-5’ = 16,917kN/m

52. 144
3. Menghitung Momen
Mt5 =
10
1
. q . l2
=
10
1
. 16,917 . 22
= 6,766 kNm
Ml5-5’ =
14
1
. q . l 2
=
14
1
. 16,917 . 22
= 4,833 kNm # 7,47 kNm (hasil beamax)
Mt5’ =
16
1
. q . l2
=
16
1
. 16,917 . 22
= 4,229 kNm # 11,04 kNm (hasil beamax)
4. Menghitung luas tulangan perlu
Tinggi balok ( h ) = 400 mm
Selimut beton (p) = 40 mm (tabel 3)
Perkiraan diameter utama = Ø 12 mm
Perkiraan diameter sengkang = Ø 8 mm
d = h – p – Ø sengkang – ½ Ø utama
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

53. 145
5. Mencari tulangan
>> Mt5 = 6,766 KNm
22
296,020,0
6,766


db
Mu
= 386,116 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
AST5 =  . b . d
= 0,0017 . 200 . 296
= 100,64 mm2
>> Ml5-5’ = 7,47 KNm
22
296,020,0
7,47


db
Mu
= 426,292 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017
Asl5-5’ =  . b . d
= 0,0017 . 200 . 296
= 100,64 mm2
>> Mt5’ = 11,04 KNm
22
296,020,0
11,04


db
Mu
= 652,84 KN/m2
 interpolasi <  min (0,0017)
Dipakai  min 0,0017

54. 146
Ast5’ =  . b . d
= 0,0017 . 200 . 296
= 100,64 mm2
6. Pilih Tulangan
AS 5 = 2 D 13 (265 mm2
)
AS 5-5’ = 2 D 13 (265 mm2
)
AS 5’ = 2 D 13 (265 mm2
)