PERHITUNGAN BALOK ANGGAPUTRA

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93 100 300 400 400 200 100 200 500 500 400 400 100 200 200 250 500 400 500 200 400 400 300 100 200 200 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 4' 5 5' 6 250 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

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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 = 2,88 kN/m – Finishing =100 . 1 . 10 = 1000 N/m = 1,00 kN/m – WD TOTAL = 3,88 kN/m Beban hidup (WL) – WL untuk balkon = 300 . 1 . 10 = 3,00 kN/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 = 6,304 kN/m

– Berat sendiri Balok = b . h . 24 kN/m2 = 0,25 . 0,4 . 24 kN/2 = 2,400 kN/m – Tembok t=1m 2,5 kN/m2 . 1 = 2,500 kN/m q AB = 11,204 kN/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 2 2 346 , 0 25 , 0 801 , 2    d b 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 2 2 346 , 0 25 , 0 021 , 3    d b 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 2 2 346 , 0 25 , 0 482 , 4    d b 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)

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98

+

2. BALOK AS 3 (B-C)

=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 = 6,546 kN/m

– Berat sendiri Balok = b . h . 24 kN/m2

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99

+

– Tembok t=4m

2,5 kN/m2 . 4 = 10,000 kN/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 = 7,409 kN/m

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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

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

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101 5. Mencari tulangan >> MtB = 65,887 KNm 2 2 346 , 0 25 , 0 65,887    d b Mu = 2201,44 KN/m2  interpolasi = 2200 2300 2200 2201   = 0075 , 0 0079 , 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 2 2 346 , 0 25 , 0 41,719    d b Mu = 1375,88 KN/m2  interpolasi = 1300 1400 1300 1376   = 0043 , 0 0046 , 0 0043 , 0    = 0,0228 = 100 x – 0,43 = 0,0045  interpolasi >  min (0,0017) Dipakai  interpolasi 0,0045

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102 ASlBC =  . b . d = 0,0045 . 250 . 346 = 389,25 mm2 >> MtC = 59,897 KNm 2 2 346 , 0 25 , 0 59,897    d b Mu = 2001 KN/m2  interpolasi = 2000 2100 2000 2001   = 0068 , 0 0072 , 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)

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103

+

3. BALOK AS 3 (C-D)

=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 = 10,501 kN/m

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104

+

– Tembok t=4m

2,5 kN/m2 . 4 = 10,000 kN/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 = 31,658 kN/m

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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

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106 5. Mencari tulangan >> MtC = 79,358 KNm 2 2 346 , 0 25 , 0 79,358    d b Mu = 2651,54 KN/m2  interpolasi = 2600 2700 2600 2651   = 0091 , 0 0095 , 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 2 2 346 , 0 25 , 0 54,559    d b Mu = 1822,94 KN/m2  interpolasi = 1800 1900 1800 1822   = 0061 , 0 0064 , 0 0061 , 0    = 0,0066 = 100 x – 0,61 = 0,0062  interpolasi >  min (0,0017) Dipakai  interpolasi 0,0062

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107 ASlCD =  . b . d = 0,0062 . 250 . 346 = 536,3 mm2 >> MtD = 79,358 KNm 2 2 346 , 0 25 , 0 79,358    d b Mu = 2651,54 KN/m2  interpolasi = 2600 2700 2600 2651   = 0091 , 0 0095 , 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)

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108

+

4. BALOK AS 3 (D-E)

=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 = 6,546 kN/m

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109

+

– Tembok t=4m

2,5 kN/m2 . 4 = 10,000 kN/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

Diasumsikan diratakan di balok D-E

37,046 : 5 = 7,409 kN/m = 7,409 kN/m

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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

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111 5. Mencari tulangan >> MtD = 59,897 KNm 2 2 346 , 0 25 , 0 59,897    d b Mu = 2001,3 KN/m2  interpolasi = 2000 2100 2000 2001   = 0068 , 0 0072 , 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 2 2 346 , 0 25 , 0 41,719    d b Mu = 1375,88 KN/m2  interpolasi = 1300 1400 1300 1376   = 0043 , 0 0046 , 0 0043 , 0    = 0,0228 = 100 x – 0,43 = 0,0045  interpolasi >  min (0,0017) Dipakai  interpolasi 0,0045

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112 ASlDE =  . b . d = 0,0045 . 250 . 346 = 389,25 mm2 >> MtE = 59,897 KNm 2 2 346 , 0 25 , 0 59,897    d b Mu = 2001,3 KN/m2  interpolasi = 2000 2100 2000 2001   = 0068 , 0 0072 , 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)

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113

+

5. BALOK AS 3 ( E-E’)

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

– q ekuivalen plat beban segitiga = 3 1 .WU plat . lx = 3 1 . 7,856 . 2,5 . 2 = 13,093 kN/m

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114

+

– Tembok t=4m 2,5 kN/m2 . 4 = 10,000 kN/m q EE’ = 25,493 kN/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

Selimut beton (p) = 40 mm (tabel 3) Perkiraan diameter utama = Ø 12 mm Perkiraan diameter sengkang = Ø 8 mm

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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 2 2 346 , 0 25 , 0 14,484    d b 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 2 2 346 , 0 25 , 0 9,958    d b Mu = 332,720 KN/m2  interpolasi <  min (0,0017) Dipakai  min 0,0017 AslEE’ =  . b . d = 0,0017 . 250 . 346 = 147,05 mm2

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116

+

>> MtE’ = 15,933 KNm 2 2 346 , 0 25 , 0 15,933    d b 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 = 2,88 kN/m – Finishing =100 . 1 . 10 = 1000 N/m = 1,00 kN/m

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117

+

Beban hidup (WL)

– WL untuk balkon = 200 . 1 . 10 = 2,00 kN/m Beban Ultimate (WU) =1,2 (WD)+1,6(WL)

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

– q ekuivalen plat beban segitiga = 3 1 .WU plat . lx = 3 1 . 7,856 . 2,5 . 2 = 13,093 kN/m

= 0,25 . 0,4 . 24 kN/2 = 2,400 kN/m

– Tembok t=4m

2,5 kN/m2 . 4 = 10,000 kN/m

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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

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119 5. Mencari tulangan >> MtE’ = 15,933 KNm 2 2 346 , 0 25 , 0 15,933    d b 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 2 2 346 , 0 25 , 0 11,380    d b 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 2 2 346 , 0 25 , 0 9,958    d b Mu = 332,720 KN/m2  interpolasi <  min (0,0017) Dipakai  min 0,0017

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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)

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121 200 400 400 300 100 200 1 2 3 4 4' 5 5'

+

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 = 2,88 kN/m – Finishing =100 . 1 . 10 = 1000 N/m = 1,00 kN/m – WD TOTAL = 3,88 kN/m Beban hidup (WL) – WL untuk balkon = 300 . 1 . 10 = 3,00 kN/m

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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)

– Berat sendiri Balok = b . h . 24 kN/m2 = 0,2 . 0,35 . 24 kN/m2 = 1,680 kN/m – Tembok t=1m 2,5 kN/m2 . 1 = 2,500 kN/m q 1-2 = 10,484 kN/m 3. Menghitung Momen Mt1 = 16 1 . q . l2 = 16 1 . 10,484 . 22 = 2,621 kNm # 5,11 kNm (hasil Beamax)

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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)

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 2 2 296 , 0 20 , 0 5,11    d b Mu = 291,613 KN/m2  interpolasi <  min (0,0017)

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124 Dipakai  min 0,0017 AST1 =  . b . d = 0,0017 . 200 . 296 = 100,64 mm2 >> Ml1-2 = 2,995 KNm 2 2 296 , 0 20 , 0 2,995    d b 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 2 2 296 , 0 20 , 0 20,7    d b Mu = 1181,291 KN/m2  interpolasi = 1100 1200 1100 1181   = 0036 , 0 0039 , 0 0036 , 0    = 0,0243 = 100 x – 0,36 = 0,0038  interpolasi >  min (0,0017) Dipakai  interpolasi 0,0038

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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)

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

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126

+

– q ekuivalen plat beban trapesium = 6 1 . WU . lx . (3-( ly lx )2) = 6 1 . 7,856 . 2,5 . (3-( 4 5 , 2 )2) = 8,451 kN/m

– Berat sendiri Balok = b . h . 24 kN/m2 = 0,2 . 0,35 . 24 kN/m2 = 1,680 kN/m – Tembok t=4m 2,5 kN/m2 . 4 = 10,000 kN/m q 2-3 = 20,221 kN/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

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

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 2 2 296 , 0 20 , 0 32,353    d b Mu = 1846,295 KN/m2  interpolasi = 1800 1900 1800 1846   = 0061 , 0 0064 , 0 0061 , 0    = 0,0138 = 100 x – 0,61 = 0,0062  interpolasi >  min (0,0017) Dipakai  interpolasi 0,0062

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128 AST2 =  . b . d = 0,0062 . 200 . 296 = 367,04 mm2 >> Ml2-3 = 20,221 KNm 2 2 296 , 0 20 , 0 20,221    d b Mu = 1153,955 KN/m2  interpolasi = 1100 1200 1100 1154   = 0036 , 0 0039 , 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 2 2 296 , 0 20 , 0 29,412    d b Mu = 1678,46 KN/m2  interpolasi = 1600 1700 1600 1678   = 0053 , 0 0057 , 0 0053 , 0    = 0,0312 = 100 x – 0,53 = 0,0056

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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)

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

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130

+

– q ekuivalen plat beban trapesium = 6 1 . WU . lx . (3-( ly lx )2) = 6 1 . 7,856 . 2,5 . (3-( 4 5 , 2 )2) = 8,451 kN/m

– Berat sendiri Balok = b . h . 24 kN/m2 = 0,2 . 0,35 . 24 kN/m2 = 1,680 kN/m – Tembok t=4m 2,5 kN/m2 . 4 = 10,000 kN/m q 3-4 = 20,221 kN/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

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131 Mt4 = 11 1 . q . l2 = 11 1 . 20,221 . 42 = 29,412 kNm # 34,76 kNm (hasil beamax)

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 2 2 296 , 0 20 , 0 29,412    d b Mu = 1678,46 KN/m2  interpolasi = 1600 1700 1600 1678   = 0053 , 0 0057 , 0 0053 , 0    = 0,0312 = 100 x – 0,53 = 0,0056  interpolasi >  min (0,0017) Dipakai  interpolasi 0,0056

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132 Ast3 =  . b . d = 0,0056 . 200 . 296 = 331,52 mm2 >> Ml3-4 = 20,221 KNm 2 2 296 , 0 20 , 0 20,221    d b Mu = 1153,955 KN/m2  interpolasi = 1100 1200 1100 1154   = 0036 , 0 0039 , 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 2 2 296 , 0 20 , 0 34,76    d b Mu = 1983,65 KN/m2  interpolasi = 1900 2000 1900 1983   = 0064 , 0 0068 , 0 0064 , 0    = 0,0162 = 100 x – 0,64 = 0,0067

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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’)

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

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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) = 6,692 kN/m

– ekuivalen plat beban segitiga = 3 1 .WU plat . lx = 3 1 . 7,856 . 2 . 1,5 = 7,856 kN/m

= 0,2 . 0,35 . 24 kN/m2 = 1,680 kN/m

– Tembok t=4m

=2,5 kN/m2 . 4 = 10,000 kN/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

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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

Diasumsikan diratakan di balok 4-4’

86,805 : 3 = 28,933 kN/m = 28,933 kN/m q 4-4’ = 55,080 kN/m 3. Menghitung Momen Mt4 = 11 1 . q . l2 = 11 1 . 55,080 . 32 = 45,065 kNm

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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)

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 2 2 296 , 0 20 , 0 45,065    d b Mu = 2571,733 KN/m2

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137  interpolasi = 2500 2600 2500 2572   = 0087 , 0 0091 , 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 2 2 296 , 0 20 , 0 41,61    d b Mu = 2374,5 KN/m2  interpolasi = 2300 2400 2300 2375   = 0079 , 0 0083 , 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

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138

+

>> Mt4’ = 46,36 KNm 2 2 296 , 0 20 , 0 46,36    d b Mu = 2645,63 KN/m2  interpolasi = 1600 2700 2600 2646   = 0091 , 0 0095 , 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)

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139

+

Beban hidup (WL)

– WL untuk balkon = 200 . 1 . 10 = 2,00 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 = 2,618 kN/m

= 0,20 . 0,35 . 24 kN/2 = 1,680 kN/m

– Tembok t=4m

2,5 kN/m2 . 4 = 10,000 kN/m

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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)

= 400 – 40 – 8 – 6 = 346 mm = 0,346 m

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141 5. Mencari tulangan >> Mt4’ = 46,36 KNm 2 2 296 , 0 20 , 0 46,36    d b Mu = 2645,63 KN/m2  interpolasi = 1600 2700 2600 2646   = 0091 , 0 0095 , 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 2 2 296 , 0 20 , 0 0,893    d b Mu = 50,961 KN/m2  interpolasi <  min (0,0017) Dipakai  min 0,0017 Asl4’5 =  . b . d = 0,0017 . 200 . 296 = 100,64 mm2

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142

+

>> Mt5 = 5,17 KNm 2 2 296 , 0 20 , 0 5,17    d b 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`)

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143

+

Beban hidup (WL)

– WL untuk balkon = 200 . 1 . 10 = 2,00 kN/m

2. Pembebanan BALOK 5-5’ Beban mati(WD)

= 0,20 . 0,35 . 24 kN/2 = 1,680 kN/m

– Tembok t=4m

2,5 kN/m2 . 4 = 10,000 kN/m

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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)

= 400 – 40 – 8 – 6 = 346 mm = 0,346 m

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145 5. Mencari tulangan >> Mt5 = 6,766 KNm 2 2 296 , 0 20 , 0 6,766    d b 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 2 2 296 , 0 20 , 0 7,47    d b 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 2 2 296 , 0 20 , 0 11,04    d b Mu = 652,84 KN/m2  interpolasi <  min (0,0017) Dipakai  min 0,0017

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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)