Lampiran Perhitungan Penulangan PERHITUNGAN BALOK B1

113

Lampiran Perhitungan Penulangan

PERHITUNGAN BALOK B1 LANTAI 2

1. Menghitung tulangan lentur balok

a. Data perencanaan

Data perencanaan yang diperlukan meliputi :

- Momen Ultimate = 75592434 Nmm

- Mutu beton (fc’) = 30 Mpa

- Mutu baja (fy) = 400 Mpa

- Dimensi balok (B1) = 200 x 400 mm

- Diameter tulangan longitudinal = 16 mm

- Diameter tulangan sengkang (øs)= 10 mm

- Tebal selimut beton (ts) = 40 mm

- d’ = ts + øs + (0,5 x D)

= 40 + 10 + (0,5 x 16)

= 58 mm

- d = h - (ts + øs + (0,5 x D))

= 400 – (40 + 10 + (0,5 x 16))

= 342 mm

b. Menghitung tulangan tumpuan

- Hitung momen nominal (Mn)

mmNMu

Mn .839915939,0

75592434

- Hitung jarak Xb, Xmax, Xmin, dan X rencana

Xb =

dfy

600

600

=

342400600

600

= 205.2 mm

Xmax = 0,75 x Xb

= 0,75 x 205.2

=153.9 mm

114

Xmin = 2

Dtsts

= 2

161040

= 58 mm

Xrencana = 100 mm (Xmin < Xrencana < Xmax)

- Hitung Cc, Cs, Mnc, Mns

Cc = 0,85 x f’c x b x β1 x Xrencana

= 0,85 x 30 x 200 x 0,85 x 100

= 433500 N

Mnc =

2

XrencanadCc

=

2

10085,0342 433500

= 129833250 N.mm

Mns = Mn-Mnc

= 83991593 – 129833250

= -45841657 N.mm

Mns (-) maka menggunakan tulangan tunggal

Perhitungan balok dengan tulangan tunggal

- Menghitung nilai ρ

ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.85

400 x

600

600 + 400

= 0.0325

ρ max = 0.75 x ρb

= 0.75 x 0.0325

= 0.0243

115

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035

- Menghitung ρ perlu

Rn = 𝑀𝑛

𝑏 𝑥 𝑑 𝑥 𝑑

= 83991593.23

200 𝑥 342 𝑥 342

= 3.59

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

ρ perlu =

fy

Rnm

m

211

1

400

6.316211

7.15

1

= 0.0097

Cek : ρ min < ρ perlu < ρ max

0.35 0.0097 < 0.0243 (OK )

- Menghitung jumlah tulangan

As perlu = ρ x b x d

= 0.0097 x 200 x 342

= 665 mm2

116

n = 𝐴𝑠 𝑝𝑒𝑟𝑙𝑢

𝐴𝑠 𝑏𝑎𝑗𝑎

= 665

201

= 3.3 ≈ 4 tulangan Tarik

As aktual = 4 x 201 = 804 mm²

- Kontrol spasi tulangan

S = 1

)()2()2(

n

amatulanganutngtulsengkantsb

= 13

)163()102()402(200

= 26 mm > 25 mm

- Menghitung kekuatan balok

T = As x fy

= 804 x 400

= 321536

a = 𝐴𝑠 𝑥 𝑓𝑦

0.85 𝑥 𝑓𝑐 𝑥 𝑏𝑤

= 803.84 𝑥 400

0.85 𝑥 30 𝑥 200 = 63

Mn = T x (d - 𝑎

2 )

= 321536 x ( 342 - 63

2 )

= 99829488.54 N.mm

Cek ø. Mn > Mu

0.9 x 99829489 > 75592433.91

89846540 > 75592433.91 (OK )

117

Pengecekan tulangan lentur SRPMK

- Pengecekan M(+) 4 – D 16

As terpasang = 803.84



= 803.84 𝑥 400

0.85 𝑥 30 𝑥 200 = 63 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 200 x 63

= 321536

ø. Mn = ø. x Cc x ( d - 𝑎

2 )

= 0.9 x 321536 x ( 319 - 63

2 )

= 83132868.01 Nmm

- Pengecekan M(-) 2 – D 16



= 401.92 𝑥 400

0.85 𝑥 30 𝑥 200

= 31.5 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 200 x 31.5

= 160768


2 )

= 0.9 x 160768 x ( 319 - 31.5

2 )

= 43846994.28 Nm

118

Tabel 4.13 Rekapitulasi Momen Terbesar pada setiap lantai

Tulangan As Terpasang ØMn

M(-) 2 - D 16 401.92 83132868

M(+) 4 - D 16 803.84 43846994

(Sumber: Perhitungan sedndiri)

Pengecekan dengan SNI 2847:2013 Pasal 21.5.2

- Pasal 21.5.2.1 Jumlah tulangan atas maupun bawah tidak boleh

kurang dari yang diberikan persamaan (10-3) tetapi tidak kurang

dari:

As min < As actual

1.4𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 401.92

𝑥 200 𝑥 319

400 < 401.92

223.16 < 401.92 (OK )

2. Pasal 21.5.2.2(1) Momen positif tidak boleh kurang dari setengah

momen negatif yang disediakan pada muka join tersebut.

M ( + ) > 0.5 M ( - )

43846994 > 0.5 x 83132868

43846994 > 41566434 (OK )

3. Pasal 21.5.2.2(2) Sedikitnya harus disediakan dua buah tulangan

menerus, baik di sisi atas maupun sisi bawah penampang.

Tulangan menerus atas = 4 ≥ 2 (OK )

Tulangan menerus bawah = 2 ≥ 2 (OK )

4. Pasal 21.5.2.3 Baik kuat lentur negatif maupun kuat lentur momen

positif pada setiap penampang disepanjang bentang tidak boleh

kurang dari 0.25 kuat lentur terbesar yang disediakan pada kedua

muka kolom tersebut.

119

M ( + ) > 0.25 M max

831322868 > 21233830 (OK )

M ( - ) > 0.25 M max

43846994.28 > 21233829.75 (OK )

c. Menghitung tulangan lapangan


mmNMu

Mn .629936959,0

56694325


Xb =

dfy

600

600

=

342400600

600

= 205.2 mm

Xmax = 0,75 x Xb

= 0,75 x 205.2

=153.9 mm

Xmin = 2

Dtsts

= 2

161040

= 58 mm




= 0,85 x 30 x 200 x 0,85 x 100

= 433500 N

120

Mnc =

2

XrencanadCc

=

2

10085,0342 433500

= 129833250 N.mm

Mns = Mn-Mnc

= 62993695 – 129833250

= -66839555 N.mm




ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.85

400 x

600

600 + 400

= 0.0325

ρ max = 0.75 x ρb

= 0.75 x 0.0325

= 0.0243

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035


Rn = 𝑀𝑛


= 62993694.93

200 𝑥 342 𝑥 342

= 2.69

121

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

ρ perlu =

fy

Rnm

m

211

1

400

7.216211

7.15

1

= 0.0071


0.35 < 0.0071 < 0.0243 (OK )



= 0.0071 x 200 x 342

= 488 mm2



= 488

201


As aktual = 3 x 201 = 603 mm²


S = 1

)()2()2(

n


= 13

)163()102()402(200

= 26 mm > 25 mm


122

T = As x fy

= 603 x 400

= 241152



= 602.88 𝑥 400

0.85 𝑥 30 𝑥 200

= 47.3

Mn = T x (d - 𝑎

2 )

= 241152 x ( 342 - 47.3

2 )

= 76772583.3 N.mm

Cek ø. Mn > Mu

.9 x 76772583 > 75592433.91

69095325 > 75592433.91 (OK )






= 602.88 𝑥 400

0.85 𝑥 30 𝑥 200

= 47.3 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 200 x 47.3

= 241152


2 )

= 0.9 x 241152 x ( 319 - 47.3

2 ) = 64060071.21 Nmm

123





= 401.92 𝑥 400

0.85 𝑥 30 𝑥 200

= 31.5 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 200 x 31.5

= 160768


2 )

= 0.9 x 160768 x ( 319 - 31.5

2 )

= 43846994.28 Nmm



M(-) 2 - D 16 401.92 64060071

M(+) 3 - D 16 602.88 43846994.28





dari:

As min < As actual

1.4 𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 602.88

1.4 𝑥 200 𝑥 319

400 < 602.88

223.16 < 602.88 (OK )

124

- Pasal 21.5.2.2(1) Momen positif tidak boleh kurang dari setengah


M ( + ) > 0.5 M ( - )

43846994.28 > 0.5 x 64060071

43846994.28 > 32030035.61 (OK )

- Pasal 21.5.2.2(2) Sedikitnya harus disediakan dua buah tulangan




- Pasal 21.5.2.3 Baik kuat lentur negatif maupun kuat lentur momen




M ( + ) > 0.25 M max

64060071 > 16015017.8 (OK )

M ( - ) > 0.25 M max

43846994.28 > 16015017.8 (OK )

- Menghitung tulangan geser balok

Berdasarkan SNI 2847-2013 pasal 23.3(4) gaya geser rencana Vc harus

ditentukan dari peninjauan gaya static pada bagian komponen struktur

antara dua muaka tumpuan. Mpr harus dihitung dari tulangan terpasang

dengan tegangan tarik 1,25fy.

a. Menghitung kebutuhan tulangan geser pada sendi plastis

- Menghitung nilai Mpr

Untuk tulangan terpasang 4 –D 16

As actual = 803.84 mm2

125

a = 𝐴𝑠 𝑎𝑘𝑡𝑢𝑎𝑙 𝑥 1.25 𝑓𝑦

0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 803.84 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 200

= 47.28

Mpr (-) = As x 1.25 x fy x ( d - 𝑎

2 )

= 803.84 x 1.25 x 240 x ( 342 - 47

2 )

= 76772583.3 Nmm

= 7677.25833 Kgmm

Untuk tulangan terpasang 3 – D 16



0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 401.92 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 200

= 24

Mpr (+) = As x 1.25 x fy x ( d - 𝑎

2 )

= 401.92 x 1.25 x 240 x ( 342 - 24

2 )

= 39811641.83 Nmm

= 3981.164 Kgmm

- Menghitung nilai Vs

Beban ekivalen

Keramik = 0.01 x 24 = 0.24 kg/m

Spesi = 0.01 x 21 = 0.21 kg/m

Pelat lantai = 0.12 x 2400 = 288 kg/m +

Total = 288.45 kg/m

126

Beban mati

Balok = 0.2 x 0.4 x 2400 = 192 kg/m

Dinding = 3.5 x 250 = 875 kg/m

Beban ekifalen = 2/3 x 288.45 = 192.30 kg/m +

Total (DL) = 1259.30 kg/m

Beban hidup

Beban ekifalen = 2/3 x 250 = 166.67 kg/m +

Total (QL) = 166.67 kg/m

Beban ultimate

Q ultimate = 1.2 DL + LL

= 1.2 (1259.30) + 166.67

= 1677 kg/m

Ln = 4 – 2(0,5 x lb)

= 8 – 2 (0,5 x 0.30)

= 3.7 m

Vu 1 = 𝑞 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑥 𝐿𝑛

2

= 1677.83 𝑥 3.7

2

= 3150.93 kg

Vu 2 = 𝑀𝑝𝑟 (−) + 𝑀𝑝𝑟 (+)

𝐿𝑛

= 7677.26 + 3981.16

3.7

= 3150.92 kg

Vu = Vu1 + Vu2

` = 3104 + 3151

= 6255 kg

Cek Vu2 > 0.5 x Vu

3151 > 0.5 x 6255

3151 > 3127 (OK )

127

Karena gaya geser akibat gempa saja lebih besar dari setengah total

gaya geser maka nilai dari Vc = 0.

Vs = 𝑉𝑢

∅

= 6255

0.75

= 8339 kg

= 83399. N

- Menghitung jarak sengkang(s)

Menggunakan sengkang tertutup 2 kaki dengan tulangan diameter 8

mm S rencana = 85 mm

Av sengkang = 100 mm2

Smin = 𝑓𝑦 𝑥 𝑑 𝑥 𝐴𝑣𝑠

𝑉𝑠

= 240 𝑥 342 𝑥 100

83399

= 99 mm

Berdasarkan Pasal 21.5.3.2. SNI 2847:2013 “Sengkang tertutup

pertama harus ditempatkan tidak lebih dari 50 mm dari muka

struktur penumpu. Spasi sengkang tertutup tidak boleh melebihi

yang terkecil dari:

d/4 ≥ S

85.5 mm ≥ 85 mm (OK )

6D ≥ S

132 mm ≥ 85 mm (OK )

150 mm ≥ S

150 mm ≥ 85 mm (OK )

128

Av = 𝑉𝑠 𝑥 𝑆𝑚𝑖𝑛

𝑓𝑦 𝑥 𝑑

= 83399 𝑥 85

240 𝑥 342

= 86,4 mm2

Berdasarkan Pasal 13.5.6.9 . SNI 2847:2013 syarat nilai Vs adalah

Vs max = 2

3 x b x d x √ f'c

= 2

3 x 200 x 342 x √ 30

= 249761 N > 83106 (OK )

Jadi penulangan geser pada daerah sendi plastis digunkan sengkang

2 kaki tertutup dengan diameter 8 mm dan jarak 85 mm.

b. Menghitung kebutuhan tulangan geser pada luar sendi plastis


Vu = Vu – 1.2 q ultimate

= 6255 – 1.2 (1677)

= 42415 kg

Vc = √𝑓𝑐 𝑥 𝑏 𝑥 𝑑

6

= √30 𝑥 200 𝑥 342

6

= 37464.22 N

= 3746 kg

Vs = 𝑉𝑢

∅ - Vc < 0.33 x b x d x √ f'c

= 42338

0.75 – 37464 < 0.33 x 200 x 342 x √ 30

18986 kg < 123632 kg (OK )

129



mm dan jarak 150 mm


S min = (Av x fy x d)/Vs

= (100 x 240 x 342)/83399

= 98.9 mm

Smax = 0.5 x d

= 0.5 x 342

= 171 mm

Cek : S min < S rencana < S max

99 < 150 < 171 (OK )

Jadi penulangan geser pada daerah luar sendi plastis digunkan

sengkang 2 kaki tertutup dengan diameter 8 mm dan jarak 150 mm.



a. Data Perencanaan







- Diameter tulangan sengkang (øs)= 10 mm


- d’ = ts + øs + (0,5 x D)

= 40 + 10 + (0,5 x 16)

= 58 mm

- d = h - (ts + øs + (0,5 x D))

= 400 – (40 + 10 + (0,5 x 16))

= 342 mm

130



mmNMu

Mn .747525189,0

67277266


Xb =

dfy

600

600

=

342400600

600

= 205.2 mm

Xmax = 0,75 x Xb

= 0,75 x 205.2

=153.9 mm

Xmin = 2

Dtsts

= 2

161040

= 58 mm




= 0,85 x 30 x 200 x 0,85 x 100

= 433500 N

131

Mnc =

2

XrencanadCc

=

2

10085,0342 433500

= 129833250 N.mm

Mns = Mn-Mnc

= 74752518 – 129833250

= -55080732 N.mm




ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.85

400 x

600

600 + 400

= 0.0325

ρ max = 0.75 x ρb

= 0.75 x 0.0325

= 0.0243

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035


Rn = 𝑀𝑛


= 74752517.98

200 𝑥 342 𝑥 342

= 3.19

132

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

ρ perlu =

fy

Rnm

m

211

1

400

2.316211

7.15

1

= 0.0086


0.0035 < 0.0086 < 0.0243 (OK)



= 0.0086 x 200 x 342

= 586 mm2



= 586

201


As aktual = 3 x 201 = 603 mm²


S = 1

)()2()2(

n


= 13

)163()102()402(200

= 26 mm > 25 mm

133


T = As x fy

= 603 x 400

= 241152



= 602.88 𝑥 400

0.85 𝑥 30 𝑥 200

= 47.3

Mn = T x (d - 𝑎

2 )

= 241152 x ( 342 - 47.3

2 )

= 76772583.3 N.mm

Cek : ø. Mn > Mu

0.9 x 76772583 > 67277266.18

69095325 > 67277266.18 (OK )






= 602.88 𝑥 400

0.85 𝑥 30 𝑥 200

= 47.3 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 200 x 47.3

= 241152


2 )

= 0.9 x 241152 x ( 319 - 47.3

2 )

= 64060071.21 Nmm

134





= 401.92 𝑥 400

0.85 𝑥 30 𝑥 200

= 31.5 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 200 x 31.5

= 160768


2 )

= 0.9 x 160768 x ( 319 - 31.5

2 )

= 43846994.28 Nmm



M(-) 2 - D 16 401.92 64060071

M(+) 3 - D 16 602.88 43846994

(Sumber: Perhitungan sendiri)




dari:

As min < As actual

1.4 𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 401.92

1.4 𝑥 200 𝑥 319

400 < 401.92

223.16 < 401.92 (OK )

135



M ( + ) > 0.5 M ( - )

43846994 > 0.5 x 64060071

43846994 > 32030035.6 (OK )









M ( + ) > 0.25 M max

64060071 > 21233830 (OK )

M ( - ) > 0.25 M max

43846994.28 > 21233829.75 (OK )



mmNMu

Mn .560643889,0

50457950


Xb =

dfy

600

600

=

342400600

600

= 205.2 mm

136

Xmax = 0,75 x Xb

= 0,75 x 205.2

=153.9 mm

Xmin = 2

Dtsts

= 2

161040

= 58 mm




= 0,85 x 30 x 200 x 0,85 x 100

= 433500 N

Mnc =

2

XrencanadCc

=

2

10085,0342 433500

= 129833250 N.mm

Mns = Mn-Mnc

= 56064388 – 129833250

= -73768862 N.mm


137



ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.85

400 x

600

600 + 400

= 0.0325

ρ max = 0.75 x ρb

= 0.75 x 0.0325

= 0.0243

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035


Rn = 𝑀𝑛


= 56064388.48

200 𝑥 342 𝑥 342

= 2.39

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

ρ perlu =

fy

Rnm

m

211

1

400

4.216211

7.15

1

= 0.0063

138


0.0035 < 0.0063 < 0.0243 (OK )



= 0.0063 x 200 x 342

= 4331 mm2



= 431

201

= 2.1 ≈ 3 tulangan tarik

As aktual = 3 x 201 = 603 mm²


S = 1

)()2()2(

n


= 13

)163()102()402(200

= 26 mm > 25 mm


T = As x fy

= 603 x 400

= 241152



= 602.88 𝑥 400

0.85 𝑥 30 𝑥 200

= 47.3

Mn = T x (d - 𝑎

2 )

= 241152 x ( 342 - 47.3

2 )

= 76772583.3 N.mm

139

Cek ø. Mn > Mu

0.9 x 76772583 > 67277266.18

69095325 > 67277266.18 (OK )






= 602.88 𝑥 400

0.85 𝑥 30 𝑥 200

= 47.3 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 200 x 47.3

= 241152


2 )

= 0.9 x 241152 x ( 319 - 47.3

2 )

= 64060071.21 Nmm





= 401.92 𝑥 400

0.85 𝑥 30 𝑥 200

= 31.5 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 200 x 31.5

= 160768

140


2 )

= 0.9 x 160768 x ( 319 - 31.5

2 )

= 43846994.28 Nmm



M(-) 2 - D 16 401.92 64060071

M(+) 3 - D 16 602.88 43846994.28





dari:

As min < As actual

1.4 𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 602.88

1.4 𝑥 200 𝑥 319

400 < 602.88

223.16 < 602.88 (OK )



M ( + ) > 0.5 M ( - )

43846994.28 > 0.5 x 64060071

43846994.28 > 32030035.61 (OK )







141



M ( + ) > 0.25 M max

64060071 > 16015017.8 (OK )

M ( - ) > 0.25 M max

43846994.28 > 16015017.8 (OK )

2. Menghitung tulangan geser balok










0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 803.84 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 200

= 47.28

Mpr (-) = As x 1.25 x fy x ( d - 𝑎

2 )

= 803.84 x 1.25 x 240 x ( 342 - 47

2 )

= 76772583.3 Nmm

= 7677.25833 Kgmm




0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 401.92 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 200

= 24

142

Mpr (+) = As x 1.25 x fy x ( d - 𝑎

2 )

= 401.92 x 1.25 x 240 x ( 342 - 24

2 )

= 39811641.83 Nmm

= 3981.164 Kgmm


Beban ekivalen

Keramik = 0.01 x 24 = 0.24 kg/m

Spesi = 0.01 x 21 = 0.21 kg/m


Total = 288.45 kg/m

Beban mati

Balok = 0.3 x 0.6 x 2400 = 432 kg/m

Dinding = 3.5 x 250 = 875 kg/m



Beban hidup



Beban ultimate


= 1.2 (1499.30) + 166.67

= 1665.97 kg/m

Ln = 4 – 2(0,5 x lb)

= 8 – 2 (0,5 x 0.30)

= 3.7 m


2

= 1665.97𝑥 3.7

2

= 3082.04 kg

143

Vu 2 = 𝑀𝑝𝑟 (−) + 𝑀𝑝𝑟 (+)

𝐿𝑛

= 7677.26 + 3981.16

3.7

= 3150.92 kg

Vu = Vu1 + Vu2

` = 3082 + 3151

= 6233 kg

Cek : Vu2 > 0.5 x Vu

3151 > 0.5 x 6233

3151 > 3116 (OK )



Vs = 𝑉𝑢

∅

= 6232.96

0.75

= 8310.6 kg

= 83106. N



mm



𝑉𝑠

= 240 𝑥 342 𝑥 100

83106

= 99 mm ≈ 100




yang terkecil dari:

144

d/4 ≥ S min

85.5 mm ≥ 100 mm (OK )

6D ≥ S min

132 mm ≥ 100 mm (OK )

150 mm ≥ S min

150 mm ≥ 100 mm (OK )


𝑓𝑦 𝑥 𝑑

= 83106.18 𝑥 100

240 𝑥 342

= 101 mm2


Vs max = 2

3 x b x d x √ f'c

= 2

3 x 200 x 342 x √ 30

= 249761 N > 83106 (OK )






= 6232.96 – 1.2 (1665.97)

= 42338.03 kg


6

= √30 𝑥 200 𝑥 342

6

= 37464.22 N

= 3746 kg

145

Vs = 𝑉𝑢

∅ - Vc < 0.33 x b x d x √ f'c

= 42338

0.75 – 37464 < 0.33 x 200 x 342 x √ 30

18986 kg < 123632 kg (OK )



mm dan jarak 250 mm


S min = 𝐴𝑣 𝑥 𝑓𝑦 𝑥 𝑑

𝑉𝑠

= 100 𝑥 240 𝑥 342

83106

= 99.2 mm

Smax = 0.5 x d

= 0.5 x 342

= 171 mm

Cek S min < S rencana < S max

99 < 150 < 171 (OK )





a. Data Perencanaan







- Diameter tulangan sengkang (øs) = 8 mm


146

- d’ = ts + øs + (0,5 x D)

= 20 + 8 + (0,5 x 10)

= 33 mm

- d = h - (ts + øs + (0,5 x D))

= 150 – (20 + 8 + (0,5 x 10))

= 117 mm



mmNMu

Mn .2.44206109,0

3978549


Xb =

dfy

600

600

= 117

400600

600

= 70.2 mm

Xmax = 0,75 x Xb

= 0,75 x 70.2

= 52.65 mm

Xmin = 2

Dtsts

= 2

10820

= 33 mm


147



= 0,85 x 30 x 110 x 0,85 x 50

= 119212.5 N

Mnc =

2

XrencanadCc

=

2

5085,0117 119212.5

= 11414597 N.mm

Mns = Mn-Mnc

= 4420610.2 – 11414597

= -6993987 N.mm




ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.85

400 x

600

600 + 400

= 0.0325

ρ max = 0.75 x ρb

= 0.75 x 0.0325

= 0.0243

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035

148


Rn = 𝑀𝑛


= 4420610.2

110 𝑥 117 𝑥 117

= 2.9

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

ρ perlu =

fy

Rnm

m

211

1

400

9.216211

7.15

1

= 0.0078


0.0035 < 0.0078 < 0.0243 (OK )



= 0.0078 x 110 x 117

= 101 mm2



= 101

78.5


As aktual = 2 x 78.5 = 157 mm²

149


S = 1

)()2()2(

n


= 12

)102()82()202(110

= 34 mm > 25 mm


T = As x fy

= 157 x 400

= 62800



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4

Mn = T x (d - 𝑎

2 )

= 62800 x ( 117 - 22.4

2 )

= 6644598.22 N.mm

Cek : ø. Mn > Mu

0.9 x 6644598 > 3978549.2

5980138 > 3978549.2 (OK )



As terpasang = 157

A = 𝐴𝑠 𝑥 𝑓𝑦


= 157 𝑥 400

0.85 𝑥 30 𝑥 110 = 22.4 mm

150

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 110 x 22.4

= 62800


2 )

= 0.9 x 62800 x ( 104 - 22.4

2 )

= 5234074.4 Nmm


As terpasang = 157



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 110 x 22.4

= 62800


2 )

= 0.9 x 62800 x ( 104 - 22.4

2 )

= 5234074.4 Nmm



M(-) 2 - D 16 157 5234074

M(+) 2 - D 16 157 5234074

(Sumber : Perhitungan Sendiri)

151




dari:

As min < As actual

1.4 𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 157

1.4 𝑥 110 𝑥 104

400 < 157

39.96 < 157 (OK )



M ( + ) > 0.5 M ( - )

5234074.4 > 0.5 x 5234074.4

5234074.4 > 2617037.2 (OK )









M ( + ) > 0.25 M max

5234074.4 > 1117570 (OK )

M ( - ) > 0.25 M max

5234074.24 > 1117570 (OK )

152



mmNMu

Mn .2.39785499,0

3580694.2


Xb =

dfy

600

600

=

117400600

600

= 70.2 mm

Xmax = 0,75 x Xb

= 0,75 x 70.2

=52.65 mm

Xmin = 2

Dtsts

= 2

10820

= 33 mm




= 0,85 x 30 x 110 x 0,85 x 50

= 119212.5 N

153

Mnc =

2

XrencanadCc

=

2

5085,0117 119212.5

= 11414597 N.mm

Mns = Mn-Mnc

= 3978549.2 – 11414597

= -7436048 N.mm




ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.85

400 x

600

600 + 400

= 0.0325

ρ max = 0.75 x ρb

= 0.75 x 0.0325

= 0.0243

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035


Rn = 𝑀𝑛


= 3978549.15

110 𝑥 117 𝑥 117 = 2.6

154

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

ρ perlu =

fy

Rnm

m

211

1

400

6.216211

7.15

1

= 0.0070


0.0035 < 0.0070 < 0.0243 (OK )


As perlu= ρ x b x d

= 0.0070 x 110 x 117

= 89.9 mm2



= 89.9

78.5


As aktual = 2 x 78.5 = 157 mm²


S = 1

)()2()2(

n


= 12

)102()82()202(110

= 34 mm > 25 mm

155


T = As x fy

= 157 x 400

= 62800



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4

Mn = T x (d - 𝑎

2 )

= 62800 x ( 117 - 22.4

2 )

= 6644598.22 N.mm

Cek : ø. Mn > Mu

0.9 x 6644598 > 3580694

5980138 > 3580694 (OK )



As terpasang = 157



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 110 x 22.4

= 62800

156


2 )

= 0.9 x 62800 x ( 104 - 22.4

2 )

= 5234074.4 Nmm


As terpasang = 157



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 110 x 22.4

= 62800


2 )

= 0.9 x 160768 x ( 319 - 31.5

2 )

= 5234074.4 Nmm



M(-) 2 - D 16 157 5234074.4

M(+) 2 - D 16 157 5234074.4





dari:

157

As min < As actual

1.4 𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 157

1.4 𝑥 110 𝑥 104

400 < 157

39.69 < 157 (OK )



M ( + ) > 0.5 M ( - )

5234074.4 > 0.5 x 5234074.4

5234074.4 > 2617037.2 (OK )









M ( + ) > 0.25 M max

5234074.4 > 1308518.6 (OK )

M ( - ) > 0.25 M max

5234074.4 > 1308518.6 (OK )





dengan tegangan tarik 1,25fy


158



As actual = 157 mm2


0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 157 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 110

= 16.8

Mpr (-) = As x 1.25 x fy x ( d - 𝑎

2 )

= 157 x 1.25 x 240 x ( 117 - 17

2 )

= 5115261.5 Nmm

= 511.52615 Kgmm


As actual = 157 mm2


0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 157 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 110

= 17

Mpr (+) = As x 1.25 x fy x ( d - 𝑎

2 )

= 157 x 1.25 x 240 x ( 117 - 17

2 )

= 5115261.5 Nmm

= 511.5 Kgmm


Beban ekivalen

Keramik = 0.01 x 24 = 0.24 kg/m

Spesi = 0.01 x 21 = 0.21 kg/m


Total = 288.45 kg/m

159

Beban mati

Balok = 0.11 x 0.15 x 2400 = 39.6 kg/m

Dinding = 3.5 x 250 = 875 kg/m



Beban hidup



Beban ultimate


= 1.2 (1106.90) + 166.67

= 1494.9 kg/m

Ln = 4 – 2(0,5 x lb)

= 8 – 2 (0,5 x 0.30)

= 3.7 m


2

= 1494.5𝑥 3.7

2

= 2765.7 kg

Vu 1 = 𝑀𝑝𝑟 (−) + 𝑀𝑝𝑟 (+)

𝐿𝑛

= 511.53 + 511.53

3.7

= 276.5 kg

Vu = Vu1 + Vu2

` = 277 + 2766 = 3042 kg

Cek Vu2 > 0.5 x Vu

2766 > 0.5 x 3042

2766 > 1521 (OK )



160

Vs = 𝑉𝑢

∅

= 3042.15

0.75

= 4056 kg

= 40562 N



mm



𝑉𝑠

= 240 𝑥 117 𝑥 100

40562

= 70 mm ≈ 85




yang terkecil dari:

6D ≥ S min

60 mm ≥ 85 mm (OK )

150 mm ≥ S min

150 mm ≥ 85 mm (OK )


𝑓𝑦 𝑥 𝑑

= 40562.03 𝑥 85

240 𝑥 117

= 123 mm2


161

Vs max = 2

3 x b x d x √ f'c

= 2

3 x 110 x 117 x √ 30

= 46995 N > 450562 (OK )






= 3042.15 – 1.2 (1494.95)

= 12482.16 kg


6

= √30𝑥 110 𝑥 117

6

= 7049.2 N

= 705 kg

Vs = 𝑉𝑢

∅ - Vc < 0.33 x b x d x √ f'c

= 12482

0.75 – 7049 < 0.33 x 110 x 117 x √ 30

= 9594 kg < 23262 kg (OK )



mm dan jarak 120 mm



𝑉𝑠

= 100 𝑥 400 𝑥 117

40562 = 116 mm ≈ 120

162

Cek : S rencana > S min

120 > 116 (OK )





a. Data Perencanaan









- d’ = ts + øs + (0,5 x D)

= 20 + 8 + (0,5 x 10)

= 33 mm

- d = h - (ts + øs + (0,5 x D))

= 150 – (20 + 8 + (0,5 x 10))

= 117 mm



mmNMu

Mn .1.39343439,0

3540909


Xb =

dfy

600

600

=

117400600

600

= 70.2 mm

163

Xmax = 0,75 x Xb

= 0,75 x 70.2

= 52.65 mm

Xmin = 2

Dtsts

= 2

10820

= 33 mm




= 0,85 x 30 x 110 x 0,85 x 50

= 119212.5 N

Mnc =

2

XrencanadCc

=

2

5085,0117 119212.5

= 11414597 N.mm

Mns = Mn-Mnc

= 3934343.1 – 11414597

= -7480254 N.mm



ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.85

400 x

600

600 + 400 = 0.0325

164

ρ max = 0.75 x ρb

= 0.75 x 0.0325

= 0.0243

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035


Rn = 𝑀𝑛


= 3934343.1

110 𝑥 117 𝑥 117

= 2.6

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

ρ perlu =

fy

Rnm

m

211

1

400

6.216211

7.15

1

= 0.0069

Cek ρ min < ρ perlu < ρ max

0.0035 < 0.0069 < 0.0243 (OK )



= 0.0078 x 110 x 117

= 101 mm2

165



= 88.9

78.5


As aktual = 2 x 78.5 = 157 mm²


S = 1

)()2()2(

n


= 12

)102()82()202(110

= 34 mm > 25 mm


T = As x fy

= 157 x 400

= 62800



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4

Mn = T x (d - 𝑎

2 )

= 62800 x ( 117 - 22.4

2 )

= 6644598.22 N.mm

Cek ø. Mn > Mu

0.9 x 6644598 > 3978549.2

5980138 > 3978549.2 (OK )

166



As terpasang = 157



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 110 x 22.4

= 62800


2 )

= 0.9 x 62800 x ( 104 - 22.4

2 )

= 5234074.4 Nmm


As terpasang = 157



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 110 x 22.4

= 62800


2 )

= 0.9 x 62800 x ( 104 - 22.4

2 )

= 5234074.4 Nmm

167

Tabel Rekapitulasi Momen Terbesar pada setiap lantai


M(-) 2 - D 10 157 5234074

M(+) 2 - D 10 157 5234074





dari:

As min < As actual

1.4 𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 157

1.4 𝑥 110 𝑥 104

400 < 157

39.96 < 157 (OK )



M ( + ) > 0.5 M ( - )

5234074.4 > 0.5 x 5234074.4

5234074.4 > 2617037.2 (OK )









M ( + ) > 0.25 M max

5234074.4 > 1117570 (OK )

168

M ( - ) > 0.25 M max

5234074.24 > 1117570 (OK )



mmNMu

Mn .3.29507579,0

2655682


Xb =

dfy

600

600

=

117400600

600

= 70.2 mm

Xmax = 0,75 x Xb

= 0,75 x 70.2

=52.65 mm

Xmin = 2

Dtsts

= 2

10820

= 33 mm




= 0,85 x 30 x 110 x 0,85 x 50

= 119212.5 N

169

Mnc =

2

XrencanadCc

=

2

5085,0117 119212.5

= 11414597 N.mm

Mns = Mn-Mnc

= 2950757.3 – 11414597

= -8463840 N.mm


Hitung Perhitungan balok dengan tulangan tunggal


ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.85

400 x

600

600 + 400

= 0.0325

ρ max = 0.75 x ρb

= 0.75 x 0.0325

= 0.0243

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035


Rn = 𝑀𝑛


= 2950757.29

110 𝑥 117 𝑥 117

= 1.95

170

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

ρ perlu =

fy

Rnm

m

211

1

400

216211

7.15

1

= 0.0051

Cek ρ min < ρ perlu < ρ max

0.0035 < 0.0051 < 0.0243 (OK )



= 0.0051 x 110 x 117

= 65.7 mm2



= 65.7

78.5


As aktual = 2 x 78.5 = 157 mm²


S = 1

)()2()2(

n


= 12

)102()82()202(110

= 34 mm > 25 mm

171


T = As x fy

= 157 x 400

= 62800



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4

Mn = T x (d - 𝑎

2 )

= 62800 x ( 117 - 22.4

2 )

= 6644598.22 N.mm

Cek ø. Mn > Mu

0.9 x 6644598 > 3580694

5980138 > 2655682 (OK )



As terpasang = 157



= 157 𝑥 400

0.85 𝑥 30 𝑥 110 = 22.4 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 110 x 22.4

= 62800


2 )

= 0.9 x 62800 x ( 104 - 22.4

2 )

= 5234074.4 Nmm


As terpasang = 157

172



= 157 𝑥 400

0.85 𝑥 30 𝑥 110

= 22.4 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 110 x 22.4

= 62800


2 )

= 0.9 x 160768 x ( 319 - 31.5

2 )

= 5234074.4 Nmm



M(-) 2 - D 10 157 5234074.4

M(+) 2 - D 10 157 5234074.4





dari:

As min < As actual

1.4 𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 157

1.4 𝑥 110 𝑥 104

400 < 157

39.69 < 157 (OK )

173



M ( + ) > 0.5 M ( - )

5234074.4 > 0.5 x 5234074.4

5234074.4 > 2617037.2 (OK )









M ( + ) > 0.25 M max

5234074.4 > 1308518.6 (OK )

M ( - ) > 0.25 M max

5234074.4 > 1308518.6 (OK )









As actual = 157 mm2


0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 157 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 110

= 16.8

174

Mpr (-) = As x 1.25 x fy x ( d - 𝑎

2 )

= 157 x 1.25 x 240 x ( 117 - 17

2 )

= 5115261.5 Nmm

= 511.52615 Kgmm


As actual = 157 mm2


0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 157 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 110

= 17

Mpr (+) = As x 1.25 x fy x ( d - 𝑎

2 )

= 157 x 1.25 x 240 x ( 117 - 17

2 )

= 5115261.5 Nmm

= 511.5 Kgmm


Beban ekivalen

Keramik = 0.01 x 24 = 0.24 kg/m

Spesi = 0.01 x 21 = 0.21 kg/m


Total = 288.45 kg/m

Beban mati

Balok = 0.11 x 0.15 x 2400 = 39.6 kg/m

Dinding = 3.5 x 250 = 875 kg/m



Beban hidup



Beban ultimate


= 1.2 (1106.90) + 166.67

= 1494.9 kg/m

175

Ln = 4 – 2(0,5 x lb)

= 8 – 2 (0,5 x 0.30)

= 3.7 m

Vu 1 = 𝑀𝑝𝑟 (−) + 𝑀𝑝𝑟 (+)

𝐿𝑛

= 511.53 + 511.53

3.7

= 276.5 kg


2

= 1494.5𝑥 3.7

2

= 2765.7 kg

Vu = Vu1 + Vu2

` = 277 + 2766

= 3042 kg

Cek Vu2 > 0.5 x Vu

2766 > 0.5 x 3042

2766 > 1521 (OK )



Vs = 𝑉𝑢

∅

= 3042.15

0.75

= 4056 kg

= 40562 N




𝑉𝑠

= 240 𝑥 117 𝑥 100

40562

= 70 mm ≈ 85

176




yang terkecil dari:

6D ≥ S min

60 mm ≥ 85 mm (OK )

150 mm ≥ S min

150 mm ≥ 85 mm (OK )


𝑓𝑦 𝑥 𝑑

= 40562.03 𝑥 85

240 𝑥 117

= 123 mm2


Vs max = 2

3 x b x d x √ f'c

= 2

3 x 110 x 117 x √ 30

= 46995 N > 450562 (OK )






= 3042.15 – 1.2 (1494.95)

= 12482.16 kg


6

= √30 𝑥 110 𝑥 117

6

= 7049.2 N

= 705 kg

177

Vs = 𝑉𝑢

∅ - Vc < 0.33 x b x d x √ f'c

= 12482

0.75 – 7049 < 0.33 x 110 x 117 x √ 30

= 9594 kg < 23262 kg (OK )



mm dan jarak 120 mm



𝑉𝑠

= 100 𝑥 400 𝑥 117

40562

= 116 mm ≈ 120

Cek S rencana > S min

120 > 116 (OK )



PERHITUNGAN BALOK G1 LANTAI 2


a. Data Perencanaan





- Dimensi balok (G1) = 300 x 600 mm




- d’ = ts + øs + (0,5 x D)

= 40 + 10 + (0,5 x 22) = 61 mm

- d = h - (ts + øs + (0,5 x D))

= 600 – (40 + 10 + (0,5 x 19))

= 539 mm

178



mmNMu

Mn .3965510819,0

356895973


Xb =

dfy

600

600

=

539400600

600

= 323.4 mm

Xmax = 0,75 x Xb

= 0,75 x 323.4

=242.55 mm

Xmin = 2

Dtsts

= 2

221040

= 61 mm




= 0,85 x 30 x 300 x 0,80 x 90

= 550800 N

Mnc =

2

XrencanadCc

=

2

9080,0539550800

= 277052400 N.mm

179

Mns = Mn-Mnc

= 396551081 – 277052400

= 119498681 N.mm

Mns (+) maka menggunakan tulangan rangkap

Cs = 'dd

Mns

= 61539

119498681

= 249997.24

- Hitung f’s

f’s = 600'

Xrencana

dXrencana

= 60090

6190

= 193 < 400 Mpa

f’s < fy ( Tulangan tekan tidak leleh)

- Hitung A’s dan As

A’s =

)'85,0(' cfsf

Cs

=

)3085,0(193

249997.24

= 1489.56 mm2

As =

fy

CsCc

=

400

249997.24550800

= 2001.99 mm2

180

- Hitung kebutuhan tulangan

Tulangan Tekan:

n =94.379

1489.56

= 3.92 ≈ 4 tulangan tekan

A’s actual = 4 x 379.94 = 1519.76 mm²

Tulangan Tarik:

n = 94.379

2001.99


As aktual = 6 x 379.94 = 2279.64 mm²


S = 1

)()2()2(

n


= 14

)224()82()402(300

= 38.7 mm > 25 mm

- Karena penyusunan menggunakan tulangan 2 lapis maka harus

dilakukan pehitungan ulang

Y = 𝑛1 𝑥 (𝐴𝑠𝐷 𝑥 𝑦1 )+(𝑛2 𝑥 𝐴𝑠𝐷 𝑥 𝑦2 )

(𝑛1+𝑛2)𝑥 𝐴𝑠𝐷

= 4 𝑥 (380 𝑥61 )+(2 𝑥 380 𝑥 122 )

( 4+2 ) 𝑥 380

= 81.3 mm

d = h – y

= 600 – 81.3

= 518.7 mm


= 0,85 x 30 x 300 x 0,80 x X

= 6120 X

181

Cs = (𝑥−𝑑′)

𝑥 x bw – ( β1 x f’c ) x As actual

= (𝑥−61)

𝑥 x 600 – ( 0.80 x 30 ) x 1520

= 875381.76 - 5339828.36

𝒙

T = As actual x fy

= 1519.76 x 400

= 607904

T = Cc + Cs

607904 = 6120 X + 875381.76 - 53398287.36

𝒙

Dengan rumus ABC didapat nilai

x1 = - 93.4 Mpa

x2 = 49.7 Mpa

maka x yang digunakan adalah 49.7 Mpa

hitung nilai f’s

f’s = (𝑥−𝑑′)

𝑥 x bw

f’s = (49.7−61)

49.7 x 600

= - 136.37

- Hitung Mn1 dan Mn2

Mn1 = (As x fy) – (A’s x f’s) x (d - 𝑎

2 )

= ( 2001.99 x 400 ) – ( 1489.56 x (-136.37) ) x (519 - 42

2 )

= 499496655.5 Nmm

Mn2 = (d – d’) x (A’s x f’s)

= ( 519 – 61 ) x ( 1489.56 x (-136.37) )

= - 92965764.53 Nmm

182

Mn total = Mn1 + Mn 2

= 499496655.5 Nmm + (-92965764.53) Nmm

= 406530891 Nmm

- Cek ø. Mn total > Mu

0.9 x 406530891 > 356895973

365877802 > 356895973 (OK )



mmNMu

Mn .7.2379306489,0

8.214137583


Xb = 600

600+𝑓𝑦 x d

= 600

600+400 x 539

= 323.4 mm

X max = 0.75 x Xb

= 0.75 x 323.4

= 242.55 mm

X min = ts + øs + 𝑑

2

= 40 + 10 + 22

2

= 61 mm

X rencana = 100 mm



= 0,85 x 30 x 300 x 0,80 x 100

= 612000 N

183

Mnc =

2

XrencanadCc

=

2

10080,0539612000

= 305388000 N.mm

Mns = Mn-Mnc

= 237930648.7 – 305388000

= - 67457351.33 N.mm

Mns ( - ) maka menggunakan tulangan tunggal



ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.80

400 x

600

600 + 400

= 0.0306

ρ max = 0.75 x ρb

= 0.75 x 0.0306

= 0.0229

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035


Rn = 𝑀𝑛


= 237930648.7

300 𝑥 539 𝑥 539

= 2.7

184

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

ρ perlu =

fy

Rnm

m

211

1

400

7.216211

7.15

1

= 0.0072


0.0035 < 0.0072 < 0.0229 (OK)



= 0.0072 x 300 x 539

= 1170 mm2



= 1170

379.94

= 3.1 ~ 4 tulangan tarik

As aktual = 4 x 379.94 = 1519.76 mm²


T = As x fy

= 1520 x 400

= 607904

185



= 1519.76 𝑥 400

0.85 𝑥 30 𝑥 300

= 79.5

Mn = T x (d - 𝑎

2 )

= 607904 x ( 518.67 - 79.5

2 )

= 291146124.8 N.mm

Cek ø. Mn > Mu

0.9 x 291146124.8 > 214137583.8

262031512 > 214137583.8 (OK )






= 1519.76 𝑥 400

0.85 𝑥 30 𝑥 300

= 79.5 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 300 x 79.5

= 607904


2 )

= 0.9 x 607904 x ( 519 - 79.5

2 )

= 262031512.3 Nmm

186





= 2279.64 𝑥 400

0.85 𝑥 30 𝑥 300

= 119 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 300 x 119

= 911856


2 )

= 0.9 x 911856 x ( 519 - 119

2 )

= 376743712.3 Nmm



M(-) 6 - D 22 2279.64 376743712

M(+) 4 - D 22 1519.76 262031512.3





dari:

As min < As actual

1.4 𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 1519.76

1.4 𝑥 300 𝑥 519

400 < 1519.76

544.60 < 1519.76 (OK )

187



M ( + ) > 0.5 M ( - )

262031512.3 > 0.5 x 376743712

262031512.3 > 188371856.1 (OK )









M ( + ) > 0.25 M max

376743712 > 94185928.07 (OK )

M ( - ) > 0.25 M max

262031512.3 > 94185928.07 (OK )


berdasarkan SNI 2847-2013 pasal 23.3(4) gaya geser rencana Vc harus






Untuk tulangan terpasang 6 –D22



0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 2279.64 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 300

= 89.39

188

Mpr (-) = As x 1.25 x fy x ( d - 𝑎

2 )

= 2279.64 x 1.25 x 240 x ( 519 - 89.39

2 )

= 324142816.2 Nmm

= 32414.28162 Kgmm




0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 1519.76 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 300

= 59.59

Mpr (-) = As x 1.25 x fy x ( d - 𝑎

2 )

= 1519.76 x 1.25 x 240 x ( 519 - 59.59

2 )

= 222888359.2 Nmm

= 22288.83592 Kgmm


Beban ekivalen

Keramik = 0.01 x 24 = 0.24 kg/m

Spesi = 0.01 x 21 = 0.21 kg/m


Total = 288.45 kg/m

Beban mati

Balok = 0.3 x 0.6 x 2400 = 432 kg/m

Dinding = 3.5 x 250 = 875 kg/m



Beban hidup



189

Beban ultimate


= 1.2 (1499.30) + 166.67

= 1815.89 kg/m

Ln = 8 – 2(0,5 x lb)

= 8 – 2 (0,5 x 0.30)

= 7.7 m


2

= 1815.89 𝑥 7.7

2

= 6991.2 kg

Vu 2 = 𝑀𝑝𝑟 (−) + 𝑀𝑝𝑟 (+)

𝐿𝑛

= 32414.282 + 22288.84

7.7

= 7104.3 kg

Vu = Vu1 + Vu2

` = 6991.2 + 7104.3

= 14095.5 kg

Cek Vu2 > 0.5 x Vu

7104.3 > 0.5 x 14095.5

7104.3 > 7047.75 (OK )



Vs = 𝑉𝑢

∅

= 14095.5

0.75

= 18794.00419 kg

= 187940.0419 N

190


Menggunakan sengkang tertutup 2 kaki dengan tulangan diameter

10mm

Av sengkang = 157mm2


𝑉𝑠

= 240 𝑥 519 𝑥 157

187940.0419

= 104 mm ≈ 110




yang terkecil dari:

d/4 ≥ S min

130 mm ≥ 110 mm (OK )

6D ≥ S min

132 mm ≥ 110 mm (OK )

150 mm ≥ S min

150 mm ≥ 110 mm (OK )


𝑓𝑦 𝑥 𝑑

= 187940.04 𝑥 110

240 𝑥 519

= 166 mm2


Vs max = 2

3 x b x d x √ f'c

= 2

3 x 300 x 519 x √ 30

= 568171 N > 187940 (OK )

191






= 14095.503 – 1.2 (1815.897)

= 119164.27 kg


6

= √30 𝑥 300 𝑥 519

6

= 85225.63 N

= 8523 kg

Vs = 𝑉𝑢

∅ - Vc < 0.33 x b x d x √ f'c

= 119164

0.75 – 85226 < 0.33 x 300 x 519 x √ 30

73660 kg < 281245 kg (OK )



mm dan jarak 250 mm



𝑉𝑠

= 100 𝑥 240 𝑥 519

73660

= 170 mm

Smax = 0.5 x d

= 0.5 x 519

= 259 mm

192

Cek : S min < S rencana < S max

170 < 250 < 259 (OK )



PERHITUNGAN BALOK G1 LANTAI 3


a. Data Perencanaan





- Dimensi balok (G1) = 300 x 600 mm




- d’ = ts + øs + (0,5 x D)

= 40 + 10 + (0,5 x 22)

= 61 mm

- d = h - (ts + øs + (0,5 x D))

= 600 – (40 + 10 + (0,5 x 19))

= 539 mm



mmNMu

Mn .4280050519,0

385204546


Xb =

dfy

600

600

=

539400600

600

= 323.4 mm

193

Xmax = 0,75 x Xb

= 0,75 x 323.4

=242.55 mm

Xmin = 2

Dtsts

= 2

221040

= 61 mm




= 0,85 x 30 x 300 x 0,80 x 100

= 612000 N

Mnc =

2

XrencanadCc

=

2

10080,0539612000

= 305388000 N.mm

Mns = Mn-Mnc

= 428005051 – 305388000

= 122617051 N.mm

Mns (+) maka menggunakan tulangan rangkap

Cs = 'dd

Mns

= 61539

122617051

= 256521.03

194

- Hitung f’s

f’s = 600'

Xrencana

dXrencana

= 600100

61100

= 234 < 400 Mpa

f’s < fy ( Tulangan tekan tidak leleh)

- Hitung A’s dan As

A’s =

)'85,0(' cfsf

Cs

=

)3085,0(234

256521.03

= 1230.32 mm2

As =

fy

CsCc

=

400

256521.03612000

= 2171.3 mm2

- Hitung kebutuhan tulangan

Tulangan Tekan :

n =94.379

1230.32

= 3.24 ≈ 4 tulangan tekan

A’s actual = 4 x 379.94 = 1519.76 mm²

Tulangan Tarik:

n = 94.379

2171.3


As aktual = 6 x 379.94 = 2279.64 mm²

195


S = 1

)()2()2(

n


= 14

)224()82()402(300

= 38.7 mm > 25 mm

- Karena penyusunan menggunakan tulangan 2 lapis maka harus

dilakukan pehitungan ulang

Y = 𝑛1 𝑥 (𝐴𝑠𝐷 𝑥 𝑦1 )+(𝑛2 𝑥 𝐴𝑠𝐷 𝑥 𝑦2 )

(𝑛1+𝑛2)𝑥 𝐴𝑠𝐷

= 4 𝑥 (380 𝑥61 )+(2 𝑥 380 𝑥 122 )

( 4+2 ) 𝑥 380

= 81.3 mm

d = h – y

= 600 – 81.3

= 518.7 mm


= 0,85 x 30 x 300 x 0,80 x X

= 6120 X

Cs = (𝑥−𝑑′)

𝑥 x bw – ( β1 x f’c ) x As actual

= (𝑥−61)

𝑥 x 600 – ( 0.80 x 30 ) x 1520

= 875381.76 - 5339828.36

𝒙

T = As actual x fy

= 1519.76 x 400

= 607904

T = Cc + Cs

196

607904 = 6120 X + 875381.76 - 53398287.36

𝒙

Dengan rumus ABC didapat nilai

x1 = - 93.4 Mpa

x2 = 49.7 Mpa

maka x yang digunakan adalah 49.7 Mpa

hitung nilai f’s

f’s = (𝑥−𝑑′)

𝑥 x bw

f’s = (49.7−61)

49.7 x 600

= - 136.37

- Hitung Mn1 dan Mn2

Mn1 = (As x fy) – (A’s x f’s) x (d - 𝑎

2 )

= ( 2171.3 x 400 ) – ( 1230.32 x (-136.37) ) x (519 - 42

2 )

= 515602818.2 Nmm

Mn2 = (d – d’) x (A’s x f’s)

= ( 519 – 61 ) x (1230.32 x (-136.37) )

= - 76786162.08 Nmm

Mn total = Mn1 + Mn 2

= 515602818.2 Nmm + (-76786162.08) Nmm

= 438816656.1 Nmm

- Cek : ø. Mn total > Mu

0.9 x 438816656.1 > 385204546

394934990 > 385204546 (OK )

197



mmNMu

Mn .2568030319,0

231122728


Xb = 600

600+𝑓𝑦 x d

= 600

600+400 x 539

= 323.4 mm

X max = 0.75 x Xb

= 0.75 x 323.4

= 242.55 mm

X min = ts + øs + 𝑑

2

= 40 + 10 + 22

2

= 61 mm

X rencana = 100 mm



= 0,85 x 30 x 300 x 0,80 x 100

= 612000 N

Mnc =

2

XrencanadCc

=

2

10080,0539612000

= 305388000 N.mm

198

Mns = Mn-Mnc

= 256803031 – 305388000

= - 48584969.33 N.mm

Mns ( - ) maka menggunakan tulangan tunggal



ρb = 0.85 𝑥 𝑓′𝑐 𝑥 β

𝑓𝑦 x

600

𝑓𝑦 + 600

= 0.85 𝑥 30 𝑥 0.80

400 x

600

600 + 400

= 0.0306

ρ max = 0.75 x ρb

= 0.75 x 0.0306

= 0.0229

ρ min = 1.4

𝑓𝑦

= 1.4

400

= 0.0035


Rn = 𝑀𝑛


= 256803030.7

300 𝑥 539 𝑥 539

= 2.95

M = 𝑓𝑦

0.85 𝑥 𝑓′𝑐

= 400

0.85 𝑥 30

= 15.69

199

ρ perlu =

fy

Rnm

m

211

1

400

95.216211

7.15

1

= 0.0078


0.0035 < 0.0078 < 0.0229 (OK )



= 0.0078 x 300 x 539

= 1269 mm2



= 1269

379.94

= 3.3 ~ 4 tulangan tarik

As aktual = 4 x 379.94 = 1519.76 mm²


T = As x fy

= 1519.76 x 400

= 607904



= 1519.76 𝑥 400

0.85 𝑥 30 𝑥 300

= 79.5

200

Mn = T x (d - 𝑎

2 )

= 607904 x ( 518.67 - 79.5

2 )

= 291146124.8 N.mm

Cek ø. Mn > Mu

0.9 x 291146124.8 > 231122727.6

262031512 > 231122727.6 (OK )






= 1519.76 𝑥 400

0.85 𝑥 30 𝑥 300

= 79.5 mm

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 300 x 79.5

= 607904


2 )

= 0.9 x 607904 x ( 519 - 79.5

2 )

= 262031512.3 Nmm





= 2279.64 𝑥 400

0.85 𝑥 30 𝑥 300

= 119 mm

201

Cc = 0.85 x f’c x b x a

= 0.85 x 30 x 300 x 119

= 911856


2 )

= 0.9 x 911856 x ( 519 - 119

2 )

= 376743712.3 Nmm



M(-) 6 - D 22 2279.64 376743712

M(+) 4 - D 22 1519.76 262031512.3





dari:

As min < As actual

1.4 𝑥 𝑏 𝑥 𝑑

𝑓𝑦 < 1519.76

1.4 𝑥 300 𝑥 519

400 < 1519.76

544.60 < 1519.76 (OK )



M ( + ) > 0.5 M ( - )

262031512.3 > 0.5 x 376743712

262031512.3 > 188371856.1 (OK )

202









M ( + ) > 0.25 M max

376743712 > 94185928.07 (OK )

M ( - ) > 0.25 M max

262031512.3 > 94185928.07 (OK )


berdasarkan SNI 2847-2013 pasal 23.3(4) gaya geser rencana Vc harus



dengan tegangan tarik 1,25fy

c. Menghitung kebutuhan tulangan geser pada sendi plastis





0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 2279.64 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 300

= 89.39

203

Mpr (-) = As x 1.25 x fy x ( d - 𝑎

2 )

= 2279.64 x 1.25 x 240 x ( 519 - 89.39

2 )

= 324142816.2 Nmm

= 32414.28162 Kgmm




0.85 𝑥 𝑓𝑐 𝑥 𝑏

= 1519.76 𝑥 1.25 (240)

0.85 𝑥 30 𝑥 300

= 59.59

Mpr (-) = As x 1.25 x fy x ( d - 𝑎

2 )

= 1519.76 x 1.25 x 240 x ( 519 - 59.59

2 )

= 222888359.2 Nmm

= 22288.83592 Kgmm


Beban ekivalen

Keramik = 0.01 x 24 = 0.24 kg/m

Spesi = 0.01 x 21 = 0.21 kg/m


Total = 288.45 kg/m

Beban mati

Balok = 0.3 x 0.6 x 2400 = 432 kg/m

Dinding = 3.5 x 250 = 875 kg/m



Beban hidup



204

Beban ultimate


= 1.2 (1499.30) + 166.67

= 1815.89 kg/m

Ln = 8 – 2(0,5 x lb)

= 8 – 2 (0,5 x 0.30)

= 7.7 m


2

= 1815.89 𝑥 7.7

2

= 6991.2 kg

Vu 2 = 𝑀𝑝𝑟 (−) + 𝑀𝑝𝑟 (+)

𝐿𝑛

= 32414.282 + 22288.84

7.7

= 7104.3 kg

Vu = Vu1 + Vu2

` = 6991.2 + 7104.3

= 14095.5 kg

Cek Vu2 > 0.5 x Vu

7104.3 > 0.5 x 14095.5

7104.3 > 7047.75 (OK )



Vs = 𝑉𝑢

∅

= 14095.5

0.75

= 18794.00419 kg = 187940.0419 N

205


Menggunakan sengkang tertutup 2 kaki dengan tulangan diameter

10 mm



𝑉𝑠

= 240 𝑥 519 𝑥 157

187940.0419

= 104 mm ≈ 110 mm




yang terkecil dari:

d/4 ≥ S min

130 mm ≥ 110 mm (OK )

6D ≥ S min

132 mm ≥ 110 mm (OK )

150 mm ≥ S min

150 mm ≥ 110 mm (OK )


𝑓𝑦 𝑥 𝑑

= 187940.04 𝑥 110

240 𝑥 519

= 166 mm2


Vs max = 2

3 x b x d x √ f'c

= 2

3 x 300 x 519 x √ 30

= 568171 N > 187940 (OK )

206



d. Menghitung kebutuhan tulangan geser pada luar sendi plastis



= 14095.503 – 1.2 (1815.897)

= 119164.27 kg


6

= √30 𝑥 300 𝑥 519

6

= 85225.63 N

= 8523 kg

Vs = 𝑉𝑢

∅ - Vc < 0.33 x b x d x √ f'c

= 119164

0.75 – 85226 < 0.33 x 300 x 519 x √ 30

73660 kg < 281245 kg (OK )



mm dan jarak 250 mm



𝑉𝑠

= 100 𝑥 240 𝑥 519

73660

= 170 mm

Smax = 0.5 x d

= 0.5 x 519

= 259 mm

Documents

Lampiran Perhitungan Penulangan PERHITUNGAN BALOK B1