PERHITUNGAN BALOK ANGGAPUTRA

3.2.3.

Perhitungan Balok

1

2

3

200

400

4 400

4' 300

5 100

5'

6

200

200

F' 200

F'

F

500

250

F

250

E'

C

500

E

500

E

VOID

2100

D

D

400

400

D VOID

C

C

400

A 500

100

B'

B

B 200

200

B

A

A 100

1

2

300

2'

400

3

400

4

200

5

100

5'

6

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

=4m

93

BALOK AS - 3

200

A

500

B

400

500

C

D

250

E

250

E'

F

 Dimensi Balok

Diambil bentang terpanjang yaitu 5 m h=

1 1 . 500 = 41,66 ~ 40 cm  = 12 12

b=

2 2 h= . 40 = 26,6 ~ 25 cm 3 3

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

=

3,00 kN/m

+

Beban hidup (WL) –

WL untuk balkon = 300 . 1 . 10

Beban Ultimate (WU)

=1,2 (WD)+1,6(WL) =1,2 (3,88)+1,6(3) =9,456 kN/m

94

2.

Pembebanan BALOK A-B Beban mati(WD) –

q ekuivalen plat beban segitiga 1 = .WU plat . lx 3

= –

1 . 9,456 . 2 3

=

6,304 kN/m

=

2,400 kN/m

2,5 kN/m2 . 1

=

2,500 kN/m

q AB

=

11,204 kN/m

Berat sendiri Balok = b . h . 24 kN/m2 = 0,25 . 0,4 . 24 kN/2

–

3.

Tembok t=1m

+

Menghitung Momen MtA

=

1 . q . l2 16

=

1 . 11,204 . 22 16

= 2,801 kNm MlAB

=

1 .q.l2 14

=

1 . 11,204 . 22 14

= 3,201 kNm

95

MtB

=

1 . q . l2 10

=

1 . 11,204 . 22 10

= 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

Mu 2,801  = 93,588 KN/m2 2 2 bd 0,25  0,346  interpolasi <  min (0,0017) Dipakai  min 0,0017 ASTA

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

96

>> MlAB = 3,021 KNm

Mu 3,021  = 106,953 KN/m2 2 2 bd 0,25  0,346  interpolasi <  min (0,0017) Dipakai  min 0,0017 ASlAB

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

>> MtB = 4,482 KNm

Mu 4,482  = 149,754 KN/m2 2 bd 0,25  0,346 2  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)

97

2.

BALOK AS 3 (B-C)

1.


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

=

2,00 kN/m

=

6,546 kN/m

=

2,400 kN/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 1 = .WU plat . lx. 1 3

= –

1 . 7,856 . 2,5 . 1 3


98

–

Tembok t=4m 2,5 kN/m2 . 4

–

=

10,000 kN/m

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

=

7,409 kN/m

qBC

=

26,355 kN/m

Pembebanan P (diratakan) > q ekuivalen plat trapesium =

lx 1 . WU . lx . (3-( )2) 6 ly

=

1 2,5 2 . 7,856 . 2,5 . (3-( )) 6 4

= 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

+

99

3.

Menghitung Momen MtB

=

1 . q . l2 10

=

1 . 26,355 . 52 10

= 65,887 kNm MlBC

=

1 .q.l2 16

=

1 . 26,355 . 52 16

= 41,719 kNm MtC

=

1 . q . l2 11

=

1 . 26,355 . 52 11

= 59,897 kNm

4.


= 400 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


100

5.

Mencari tulangan >> MtB = 65,887 KNm

Mu 65,887  = 2201,44 KN/m2 2 2 bd 0,25  0,346  interpolasi

=

  0,0075 2201  2200 = 2300  2200 0,0079  0,0075

= 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


=

  0,0043 1376  1300 = 1400  1300 0,0046  0,0043

= 0,0228 = 100 x – 0,43 = 0,0045  interpolasi >  min (0,0017) Dipakai  interpolasi 0,0045

101

ASlBC

=.b.d = 0,0045 . 250 . 346 = 389,25 mm2

>> MtC = 59,897 KNm

Mu 59,897  = 2001 KN/m2 2 2 bd 0,25  0,346  interpolasi

=

  0,0068 2001  2000 = 2100  2000 0,0072  0,0068

= 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)

102

3.

BALOK AS 3 (C-D)

1.


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

=

2,00 kN/m

=

10,501 kN/m

=

2,400 kN/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) –


= –

1 . 7,856 . 2 . 2 3


103

–


–

=

10,000 kN/m

126,623 : 4 = 31,658 kN/m

=

31,658 kN/m

qBC

=

54,559 kN/m


lx 1 . WU . lx . (3-( )2) 6 ly

=

1 2 . 7,856 . 2 . (3-( )2) 6 4

= 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

+

104

3.

Menghitung Momen MtC

=

1 . q . l2 11

=

1 . 54,559 . 42 11

= 79,358 kNm MlCD

=

1 .q.l2 16

=

1 . 54,559 . 42 16

= 54,559 kNm MtD

=

1 . q . l2 11

=

1 . 54,559 . 42 11

= 79,358 kNm

4.


= 400 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


105

5.

Mencari tulangan >> MtC = 79,358 KNm


=

  0,0091 2651  2600 = 2700  2600 0,0095  0,0091

= 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


=

  0,0061 1822  1800 = 1900  1800 0,0064  0,0061


106

ASlCD

=.b.d = 0,0062 . 250 . 346 = 536,3 mm2

>> MtD = 79,358 KNm


=

  0,0091 2651  2600 = 2700  2600 0,0095  0,0091

= 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)

107

4.

BALOK AS 3 (D-E)

1.


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

=

2,00 kN/m

=

6,546 kN/m

=

2,400 kN/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) –


= –

1 . 7,856 . 2,5 . 1 3


108

–


–

=

10,000 kN/m

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

=

7,409 kN/m

qDE

=

26,355 kN/m


lx 1 . WU . lx . (3-( )2) 6 ly

=

1 2,5 2 . 7,856 . 2,5 . (3-( )) 6 4

= 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

+

109

3.

Menghitung Momen MtD

=

1 . q . l2 11

=

1 . 26,355 . 52 11

= 59,897 kNm MlDE

=

1 .q.l2 16

=

1 . 26,355 . 52 16

= 41,719 kNm MtE

=

1 . q . l2 11

=

1 . 26,355 . 52 11

= 59,897 kNm

4.


= 400 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


110

5.

Mencari tulangan >> MtD = 59,897 KNm


=

  0,0068 2001  2000 = 2100  2000 0,0072  0,0068

= 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


=

  0,0043 1376  1300 = 1400  1300 0,0046  0,0043


111

ASlDE

=.b.d = 0,0045 . 250 . 346 = 389,25 mm2

>> MtE = 59,897 KNm


=

  0,0068 2001  2000 = 2100  2000 0,0072  0,0068

= 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)

112

5.

BALOK AS 3 ( E-E’)

1.


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

=

2,00 kN/m

=

13,093 kN/m

=

2,400 kN/m

+



Beban Ultimate (WU)

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

2.



= –

1 . 7,856 . 2,5 . 2 3


113

–

3.


=

10,000 kN/m

q EE’

=

25,493 kN/m

+

Menghitung Momen MtE

=

1 . q . l2 11

=

1 . 25,493 . 2,52 11

= 14,484 kNm MlEE’

=

1 .q.l2 16

=

1 . 25,493 . 2,52 16

= 9,958 kNm MtE’

=

1 . q . l2 10

=

1 . 25,493 . 2,52 10

= 15,933 kNm

4.


= 400 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm 114


5.

Mencari tulangan >> MtE = 14,484 KNm

Mu 14,484  = 467,77 KN/m2 2 bd 0,25  0,346 2  interpolasi <  min (0,0017) Dipakai  min 0,0017 ASTE

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

>> MlEE’ = 9,958 KNm

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

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

115

>> MtE’ = 15,933 KNm

Mu 15,933  = 532,359 KN/m2 2 2 bd 0,25  0,346  interpolasi <  min (0,0017) Dipakai  min 0,0017 AstE’

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

6.

6.

Pilih Tulangan ASE

= 2 D 16 (402 mm2)

AS E-E’

= 2 D 16 (402 mm2)

AS E’

= 2 D 16 (402 mm2)

BALOK AS 3 ( E’-F)

1.


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

+

116



=

2,00 kN/m

=

13,093 kN/m

=

2,400 kN/m

2,5 kN/m2 . 4

=

10,000 kN/m

q E’F

=

25,493 kN/m

Beban Ultimate (WU)

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

2.



= –

1 . 7,856 . 2,5 . 2 3


–

Tembok t=4m

+

117

3.

Menghitung Momen MtE’

=

1 . q . l2 10

=

1 . 25,493 . 2,52 10

= 15,933 kNm MlE’F

=

1 .q.l2 14

=

1 . 25,493 . 2,52 14

= 11,380 kNm MtF

=

1 . q . l2 16

=

1 . 25,493 . 2,52 16

= 9,958 kNm

4.


= 400 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


118

5.

Mencari tulangan >> MtE’ = 15,933 KNm

Mu 15,933  = 532,359 KN/m2 2 2 bd 0,25  0,346  interpolasi <  min (0,0017) Dipakai  min 0,0017 ASTE’

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

>> MlE’F = 11,380 KNm

Mu 11,380  = 380,323 KN/m2 2 2 bd 0,25  0,346  interpolasi <  min (0,0017) Dipakai  min 0,0017 AslE’F

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

>> MtF = 9,958 KNm

Mu 9,958  = 332,720 KN/m2 2 2 bd 0,25  0,346  interpolasi <  min (0,0017) Dipakai  min 0,0017

119

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)

120

BALOK AS-F

200

400

1

400

2

3

300

4

100

4'

200

5

5'

 Dimensi Balok

Diambil bentang terpanjang yaitu 4 m h=

1 1 . 400 = 33,3 ~ 35 cm  = 12 12

b=

2 2 h= . 35 = 23,3 ~ 20 cm 3 3

Diambil dimensi balok = 20/35

1.

BALOK AS F ( 1-2)

1.


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

=

3,00 kN/m

+



121

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) –


= –

1 . 9,456 . 2 3

=

6,304 kN/m

=

1,680 kN/m

2,5 kN/m2 . 1

=

2,500 kN/m

q 1-2

=

10,484 kN/m

Berat sendiri Balok = b . h . 24 kN/m2 = 0,2 . 0,35 . 24 kN/m2

–

3.

Tembok t=1m

+

Menghitung Momen Mt1

=

1 . q . l2 16

=

1 . 10,484 . 22 16

= 2,621 kNm # 5,11 kNm (hasil Beamax)

122

Ml1-2

=

1 .q.l2 14

=

1 . 10,484 . 22 14

= 2,995 kNm MtB

=

1 . q . l2 10

=

1 . 10,484 . 22 10

= 4,193 kNm # 20,7 kNm (hasil beamax)

4.


= 350 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


5.

Mencari tulangan >> Mt1 = 5,11 KNm

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

123

Dipakai  min 0,0017 AST1

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

>> Ml1-2 = 2,995 KNm

Mu 2,995  = 170,91 KN/m2 2 bd 0,20  0,296 2  interpolasi <  min (0,0017) Dipakai  min 0,0017 Asl1-2

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

>> Mt2 = 20,7 KNm


=

  0,0036 1181  1100 = 1200  1100 0,0039  0,0036


124

Ast2

=.b.d = 0,0038 . 200 . 296 = 224,96mm2

6.

2.

Pilih Tulangan AS1

= 2 D 13 (265 mm2)

AS 1-2

= 2 D 13 (265 mm2)

AS 2

= 2 D 13 (265 mm2)

BALOK AS F ( 2-3)

1.


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

=

2,88 kN/m

=

2,00 kN/m

+



Beban Ultimate (WU)

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

125

2.


q ekuivalen plat beban trapesium

lx 1 = . WU . lx . (3-( )2) 6 ly = –

1 2,5 2 . 7,856 . 2,5 . (3-( )) 6 4

=

8,451 kN/m

=

1,680 kN/m

2,5 kN/m2 . 4

=

10,000 kN/m

q 2-3

=

20,221 kN/m


–

3.

Tembok t=4m

+


=

1 . q . l2 10

=

1 . 20,221 . 42 10

= 32,353 kNm Ml2-3

=

1 .q.l2 16

=

1 . 20,221 . 42 16

= 20,221 kNm

126

Mt3

=

1 . q . l2 11

=

1 . 20,221 . 42 11

= 29,412 kNm

4.


= 350 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


5.



=

  0,0061 1846  1800 = 1900  1800 0,0064  0,0061

= 0,0138 = 100 x – 0,61 = 0,0062  interpolasi >  min (0,0017) Dipakai  interpolasi 0,0062 127

AST2

=.b.d = 0,0062 . 200 . 296 = 367,04 mm2

>> Ml2-3 = 20,221 KNm


=

  0,0036 1154  1100 = 1200  1100 0,0039  0,0036

= 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


=

  0,0053 1678  1600 = 1700  1600 0,0057  0,0053

= 0,0312 = 100 x – 0,53 = 0,0056

128

 interpolasi >  min (0,0017) Dipakai  interpolasi 0,0056 Ast3

=.b.d = 0,0056 . 200 . 296 = 331,52 mm2

6.

3.

Pilih Tulangan AS2

= 3 D 13 (398 mm2)

AS 2-3

= 2 D 13 (265 mm2)

AS 3

= 3 D 13 (398 mm2)

BALOK AS F ( 3-4)

1.


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

=

2,88 kN/m

=

2,00 kN/m

+



Beban Ultimate (WU)

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

129

2.



lx 1 = . WU . lx . (3-( )2) 6 ly = –

1 2,5 2 . 7,856 . 2,5 . (3-( )) 6 4

=

8,451 kN/m

=

1,680 kN/m

2,5 kN/m2 . 4

=

10,000 kN/m

q 3-4

=

20,221 kN/m


–

3.

Tembok t=4m

+


=

1 . q . l2 11

=

1 . 20,221 . 42 11

= 29,412 kNm Ml3-4

=

1 .q.l2 16

=

1 . 20,221 . 42 16

= 20,221 kNm

130

Mt4

=

1 . q . l2 11

=

1 . 20,221 . 42 11


4.


= 350 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


5.



=

  0,0053 1678  1600 = 1700  1600 0,0057  0,0053

= 0,0312 = 100 x – 0,53 = 0,0056  interpolasi >  min (0,0017) Dipakai  interpolasi 0,0056 131

Ast3

=.b.d = 0,0056 . 200 . 296 = 331,52 mm2

>> Ml3-4 = 20,221 KNm


=

  0,0036 1154  1100 = 1200  1100 0,0039  0,0036

= 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


=

  0,0064 1983  1900 = 2000  1900 0,0068  0,0064

= 0,0162 = 100 x – 0,64 = 0,0067

132

 interpolasi >  min (0,0017) Dipakai  interpolasi 0,0067 Ast4

=.b.d = 0,0067 . 200 . 296 = 396,64 mm2

6.

4.

Pilih Tulangan AS3

= 3 D 13 (398 mm2)

AS 3-4

= 2 D 13 (265 mm2)

AS 4

= 3 D 13 (398 mm2)

BALOK AS F ( 4-4’)

1.


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

=

2,88 kN/m

=

2,00 kN/m

+



Beban Ultimate (WU)

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

133

2.

Pembebanan BALOK 4-4’ Beban mati(WD) –


lx 1 = . WU . lx . (3-( )2) 6 ly = –

1 2 . 7,856 . 2 . (3-( )2) 6 3

=

6,692 kN/m

=

7,856 kN/m

=

1,680 kN/m

=

10,000 kN/m

ekuivalen plat beban segitiga 1 = .WU plat . lx 3

= –

1 . 7,856 . 2 . 1,5 3


–

Tembok t=4m =2,5 kN/m2 . 4

–

pembebanan beban p (diratakan) >q ekuivalen plat trapesium =

lx 1 . WU . lx . (3-( )2) 6 ly

=

2 2 1 . 7,856 . 2 . (3-( ) )x4 2,5 6

= 24,720 kN/m

134

>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’

3.

86,805 : 3 = 28,933 kN/m

=

28,933 kN/m

q 4-4’

=

55,080 kN/m


=

1 . q . l2 11

=

1 . 55,080 . 32 11

= 45,065 kNm

135

+

Ml4-4’

=

1 .q.l2 16

=

1 . 55,080 . 32 16

= 30,982 kNm # 41,61 kNm (hasil beamax) Mt4

=

1 . q . l2 11

=

1 . 55,080 . 32 11


4.


= 350 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


5.


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

136

 interpolasi

=

  0,0087 2572  2500 = 2600  2500 0,0091  0,0087

= 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

Mu 41,61  = 2374,5 KN/m2 2 bd 0,20  0,296 2  interpolasi

=

  0,0079 2375  2300 = 2400  2300 0,0083  0,0079

= 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

137

>> Mt4’ = 46,36 KNm


=

  0,0091 2646  2600 = 2700  1600 0,0095  0,0091

= 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.

5.

Pilih Tulangan AS4

= 4 D 13 (531 mm2)

AS 4-4’

= 3 D 13 (398 mm2)

AS 4‘

= 5 D 13 (664 mm2)

BALOK AS 3 ( 4’-5)

1.


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

+

138



=

2,00 kN/m

=

2,618 kN/m

=

1,680 kN/m

2,5 kN/m2 . 4

=

10,000 kN/m

q 4’-5

=

14,298 kN/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) –


= –

1 1 . 7,856 . 2 . 3 2


–

Tembok t=4m

+

139

3.

Menghitung Momen Mt4’

=

1 . q . l2 11

=

1 . 14,298 . 12 11

= 1,299 kNm # 46,36 kNm (hasil beamax) Ml4’5

=

1 .q.l2 16

=

1 . 14,298 . 12 16

= 0,893 kNm Mt5

=

1 . q . l2 10

=

1 . 14,298 . 12 10


4.


= 400 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


140

5.

Mencari tulangan >> Mt4’ = 46,36 KNm


=

  0,0091 2646  2600 = 2700  1600 0,0095  0,0091

= 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

Mu 0,893  = 50,961 KN/m2 2 2 bd 0,20  0,296  interpolasi <  min (0,0017) Dipakai  min 0,0017 Asl4’5

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

141

>> Mt5 = 5,17 KNm

Mu 5,17  = 295,03 KN/m2 2 2 bd 0,20  0,296  interpolasi <  min (0,0017) Dipakai  min 0,0017 AstF

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

6.

6.

Pilih Tulangan AS4 ‘

= 5 D 13 (664 mm2)

AS 4’-5

= 2 D 13 (265 mm2)

AS 5

= 2 D 13 (265 mm2)

BALOK AS 3 ( 5-5`)

1.


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

+

142



=

2,00 kN/m

=

5,237 kN/m

=

1,680 kN/m

2,5 kN/m2 . 4

=

10,000 kN/m

q 5-5’

=

16,917 kN/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) –


= –

1 . 7,856 . 2 3


–

Tembok t=4m

+

143

3.


=

1 . q . l2 10

=

1 . 16,917 . 22 10

= 6,766 kNm Ml5-5’

=

1 .q.l2 14

=

1 . 16,917 . 22 14

= 4,833 kNm # 7,47 kNm (hasil beamax) Mt5’

=

1 . q . l2 16

=

1 . 16,917 . 22 16


4.


= 400 mm

Selimut beton (p)

= 40 mm (tabel 3)


= Ø 12 mm


= Ø 8 mm


144

5.


Mu 6,766  = 386,116 KN/m2 2 2 bd 0,20  0,296  interpolasi <  min (0,0017) Dipakai  min 0,0017 AST5

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

>> Ml5-5’ = 7,47 KNm

Mu 7,47  = 426,292 KN/m2 2 2 bd 0,20  0,296  interpolasi <  min (0,0017) Dipakai  min 0,0017 Asl5-5’

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

>> Mt5’ = 11,04 KNm

Mu 11,04  = 652,84 KN/m2 2 2 bd 0,20  0,296  interpolasi <  min (0,0017) Dipakai  min 0,0017

145

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)

146

PERHITUNGAN BALOK ANGGAPUTRA

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