1 perhitungan-balok

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

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

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

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

97
>> MlAB = 3,021 KNm
22
346,025,0
021,3


db
Mu
= 106,953 KN/m2
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
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
)

98
+
2. BALOK AS 3 (B-C)
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
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK B-C
Beban mati(WD)
=
3
1
.WU plat . lx. 1
=
3
1
. 7,856 . 2,5 . 1 = 6,546kN/m
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m

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

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
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

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

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
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
)

103
+
3. BALOK AS 3 (C-D)
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
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK B-C
Beban mati(WD)
=
3
1
.WU plat . lx. 2
=
3
1
. 7,856 . 2 . 2 = 10,501kN/m
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m

104
+
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
=
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

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
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

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
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

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
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
)

108
+
4. BALOK AS 3 (D-E)
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
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK D-E
Beban mati(WD)
=
3
1
.WU plat . lx. 1
=
3
1
. 7,856 . 2,5 . 1 = 6,546kN/m
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m

109
+
– Tembok t=4m
2,5 kN/m2
. 4 = 10,000kN/m
=
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
=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

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
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

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
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

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
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
)

113
+
5. BALOK AS 3 ( E-E’)
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
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
Beban mati(WD)
=
3
1
.WU plat . lx
=
3
1
. 7,856 . 2,5 . 2 = 13,093kN/m
= b . h . 24 kN/m2
= 0,25 . 0,4 . 24 kN/2
= 2,400kN/m

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

115
= 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
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
AslEE’ =  . b . d
= 0,0017 . 250 . 346
= 147,05 mm2

116
+
>> MtE’ = 15,933 KNm
22
346,025,0
15,933


db
Mu
= 532,359 KN/m2
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)
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

117
+
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
Beban mati(WD)
=
3
1
.WU plat . lx
=
3
1
. 7,856 . 2,5 . 2 = 13,093kN/m
= 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

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
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

119
5. Mencari tulangan
>> MtE’ = 15,933 KNm
22
346,025,0
15,933


db
Mu
= 532,359 KN/m2
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
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

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
)

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)
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
Beban hidup (WL)

122
+
=1,2 (3,88)+1,6(3)
=9,456 kN/m
2. Pembebanan BALOK 1-2
Beban mati(WD)
=
3
1
.WU plat . lx
=
3
1
. 9,456 . 2 = 6,304kN/m
= 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)

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)
= 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

124
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
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

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
)
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
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m

126
+
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
= 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

127
Mt3 =
11
1
. q . l2
=
11
1
. 20,221 . 42
= 29,412 kNm
= 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

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
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

129
+
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
)
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
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m

130
+
Beban mati(WD)
=
6
1
. WU . lx . (3-(
ly
lx
)2
)
=
6
1
. 7,856 . 2,5 . (3-(
4
5,2
)2
) = 8,451kN/m
= 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

131
Mt4 =
11
1
. q . l2
=
11
1
. 20,221 . 42
= 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

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
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

133
+
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’)
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
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m

134
2. Pembebanan BALOK 4-4’
Beban mati(WD)
=
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
= 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

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
=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

136
Ml4-4’ =
16
1
. q . l 2
=
16
1
. 55,080 . 32
Mt4 =
11
1
. q . l2
=
11
1
. 55,080 . 32
= 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

137
 interpolasi =
25002600
25002572


=
0087,00091,0
0087,0


= 0,0288 = 100 x – 0,87
= 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
Asl4-4’ =  . b . d
= 0,0082 . 200 . 296
= 485,44 mm2

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
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)
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

139
+
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK 4’-5
Beban mati(WD)
=
3
1
.WU plat . lx
=
3
1
. 7,856 . 2 .
2
1
= 2,618kN/m
= 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

140
3. Menghitung Momen
Mt4’ =
11
1
. q . l2
=
11
1
. 14,298 . 12
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
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

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
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
Asl4’5 =  . b . d
= 0,0017 . 200 . 296
= 100,64 mm2

142
+
>> Mt5 = 5,17 KNm
22
296,020,0
5,17


db
Mu
= 295,03 KN/m2
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`)
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

143
+
Beban hidup (WL)
=1,2 (3,88)+1,6(2)
=7,856 kN/m
2. Pembebanan BALOK 5-5’
Beban mati(WD)
=
3
1
.WU plat . lx
=
3
1
. 7,856 . 2 = 5,237kN/m
= 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

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
Mt5’ =
16
1
. q . l2
=
16
1
. 16,917 . 22
= 400 – 40 – 8 – 6
= 346 mm = 0,346 m

145
5. Mencari tulangan
>> Mt5 = 6,766 KNm
22
296,020,0
6,766


db
Mu
= 386,116 KN/m2
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
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

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
)

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