Desain Penulangan.docx

PENULANGAN Hasil tulangan diperoleh dari luas tulangan atas (As Top) terbesar dan luas tulangan bawah (As Min) terbesar melalui program ETABS. Summary Balok B1 Balok B2 As Top As Bot At Shear As Top As Bot At Shear Story (mm2) (mm2) (mm2/m) (mm2) (mm2) (mm2/m) 1 2 3 4 5

323 312 321 321 104

Tulangan Balok B1

241 232 234 236 59

202.2 199.32 198.98 199.37 51.2

212 220 223 223 66

185 183 185 186 33

199.22 200.98 201.78 201.92 57.79

ETABS 2016 16.0.0

License #*1S8ZDXPNU9WFLC4

ETABS 2016 Concrete Frame Design ACI 318-14 Beam Section Design

Beam Elem ent Details (Sum m ary) Level

Elem ent

Unique Nam e

Section ID

Com bo ID

Station Loc

Length (m m )

LLRF

Type

Story1

B20

20

Balok B1

Envelope

250

5700

1

Sway Special

Section Properties b (m m )

h (m m )

bf (m m )

ds (m m )

dct (m m )

dcb (m m )

250

450

250

0

50

50

Material Properties Ec (MPa)

f' c (MPa)

Lt.Wt Factor (Unitless)

fy (MPa)

f ys (MPa)

26587.21

32

1

420

420

Design Code Param eters ΦT

ΦCTied

ΦCSpiral

ΦVns

ΦVs

ΦVj oint

0.9

0.65

0.75

0.75

0.6

0.85

Design Mom ent and Flexural Reinforcem ent for Mom ent, M u3 Design -Mom ent kN-m Top

(+2 Axis)

Design +Mom ent kN-m

-Mom ent Rebar mm²

-35.601

Bottom (-2 Axis)

17.8005

+Mom ent Rebar mm²

Minim um Rebar mm²

Required Rebar mm²

242

0

323

323

0

121

161

161

Shear Force and Reinforcem ent for Shear, V u2 Shear V u2 kN

Shear ΦV c kN

Shear ΦV s kN

Shear V p kN

Rebar Av /S m m ²/m

20.0578

0

20.0578

18.2128

202.02

Torsion Force and Torsion Reinforcem ent for Torsion, T u Φ*Tu kN-m

Tth kN-m

Tcr kN-m

Area Ao cm ²

Perim eter, p h mm

Rebar At /s m m ²/m

Rebar Al mm²

0.0067

3.1801

12.7204

494.5

1044.4

0

0

ETABS 2016 Concrete Frame Design ACI 318-14 Beam Section Design

Level

Element

Unique Name

Story1

B21

26

Beam Element Details (Summary) Section ID Combo ID Station Loc Balok B2

b (mm)

h (mm)

200

350

Envelope

Section Properties bf (mm) ds (mm) 200

Length (mm)

LLRF

Type

4700

1

Sway Special

4450

dct (mm)

dcb (mm)

50

50

0

Ec (MPa)

f'c (MPa)

Material Properties Lt.Wt Factor (Unitless)

fy (MPa)

fys (MPa)

26587.21

32

1

420

420

ΦT

ΦCTied

0.9

0.65

Design Code Parameters ΦCSpiral ΦVns 0.75

0.75

ΦVs

ΦVjoint

0.6

0.85

Design Moment and Flexural Reinforcement for Moment, Mu3 -Moment +Moment Minimum Design Design Rebar Rebar Rebar -Moment +Moment mm² mm² mm² kN-m kN-m Top

(+2 Axis)

-20.3347

Bottom (-2 Axis)

10.1673

Shear Vu2 kN 14.5443

186

0

201

201

0

93

124

124

Shear Force and Reinforcement for Shear, Vu2 Shear ΦVc Shear ΦVs Shear Vp kN kN kN 0

14.5443

Rebar Av /S mm²/m

11.8453

195.32

Φ*Tu kN-m

Torsion Force and Torsion Reinforcement for Torsion, Tu Area Ao Rebar At /s Tth Tcr Perimeter, ph cm² mm²/m kN-m kN-m mm

0.0107

1.5661

6.2644

246.6

Required Rebar mm²

744.4

0

Rebar Al mm² 0

Tulangan Balok B2



Tulangan Longitudinal Luas Story Balok Tulangan (mm2) A

B

B1 1 B2

C

Tulangan D16 (mm2)

Desain Jumlah tulangan

Keterangan

E

F=D/E

G

2 2 2 2

2D16 2D16 2D16 2D16

D

As Top As Bot As Top As Bot

323 241 201 184

201.062 201.062 201.062 201.062

Balok B1 Story 1 memiliki luas tulangan terbesar dan diperoleh jumlah tulangan sebanyak 2 buah (tulangan minimum), sehingga balok lain yang memilki luas tulangan yang dibutuhkan lebih kecil digunakan pula tulangan minimum sebanyak 2 buah. 

Tulangan Transversal Balok B1 Vu = 20057.8 N d’ = 50 mm d = h – d’ = 450 – 50 = 400 mm √f′c √32 = 250 × 400 × = 94280 N 6 6 Vu 20057.8 Vn = = = 26743.733N 0,75 0.75 Vc = b × d ×

jika Vc > Vn " gunakan Avmin " Coba dengan n = 2 kaki Ø10: 1 Avs = Avmin 2

1 b×s 2 × × π × D2 = 4 3 × fy 1 250. s 2 × × π × 102 = 4 3 × 420 s = 791.84 mm Syarat s ≤ d/2 791.84 mm ≤ 400/2 =200 mm :: digunakan s = 200 mm untuk tulangan tumpuan :: tulangan geser daerah lapangan digunakan: S = d/4 = 400/4 = 100 mm



Tulangan Transversal Balok B2 Vu = 14544.3 N d’ = 50 mm d = h – d’ = 450 – 50 = 400 mm √f′c √32 = 250 × 400 × = 94280 N 6 6 Vu 14544.3 Vn = = = 19392.4 N 0,75 0.75 Vc = b × d ×

jika Vc > Vn " gunakan Avmin " Coba dengan n = 2 kaki Ø10: 1 Avs = Avmin 2

1 b×s 2 × × π × D2 = 4 3 × fy 1 250. s 2 × × π × 102 = 4 3 × 420 s = 791.84 mm Syarat s ≤ d/2 791.84 mm ≤ 400/2 =200 mm :: digunakan s = 200 mm untuk tulangan tumpuan :: tulangan geser daerah lapangan digunakan: S = d/4 = 400/4 = 100 mm

Pengecekan Kolom Pengecekan kolom dilakukan menggunakan bantuan program PCAColumn. Gaya aksial diperoleh dari hasil ETABS. Berikut adalah gaya aksial yang telah diurutkan dari gaya yang terbesar. TABLE: Column Forces Story Column Unique Name Load Case/Combo Station m Story1 C11 251 QX 2 0 Story1 C11 251 QX 2 1.775 Story1 C11 251 QX 2 3.55 Story1 C11 251 QX 4 0 Story1 C11 251 QX 4 1.775

Story1 Story1 Story1 Story1 Story1 Story1

C14 C27 C27 C12 C12 C12

266 Envelope Min 331 Comb7 Min 331 Envelope Min 256 Comb7 Max 256 Comb7 Min 256 Envelope Min

0 0 0 0 0 0

P kN 254.2984 254.2984 254.2984 254.2984 254.2984

V2 kN -0.9185 -0.9185 -0.9185 -0.9185 -0.9185

V3 kN 70.6322 70.6322 70.6322 70.6322 70.6322

T kN-m -0.3194 -0.3194 -0.3194 -0.3194 -0.3194

M2 kN-m 183.3885 58.0163 -67.3559 183.3885 58.0163

M3 kN-m -2.0384 -0.408 1.2224 -2.0384 -0.408

-2627.9805 -2627.9995 -2627.9995 -2631.7547 -2631.8801 -2631.8801

0.5705 -1.6765 -1.6765 1.7204 1.4944 0.5983

-1.2021 1.0193 0.3702 1.1214 1.0491 0.3812

-0.001 -0.0007 -0.0014 0.0039 -0.0002 -0.0012

-1.7136 1.301 0.379 1.4974 1.3336 0.3885

0.6411 -2.3314 -2.3314 2.3622 1.8532 0.6788

P ( kN) 6000 (Pmax)

4000

2000

1 -450

-350

-250

-150

-50

50

150

250

350

450

M (91°) (k N -m)

(Pmin)

-2000

500

My (k N -m)

300

1 100 Mx (k N -m) -500

-300

-100

100 -100

-300

-500 P = 254 k N

300

500

Penulangan Pelat Lantai 

Data Perhitungan: Lebar pelat (b)

= 1000 mm

Tinggi pelat (h)

= 150 mm

Tebal Selimut Beton (d’) = 20 mm (asumsi) = h – d’ = 150 – 20 = 130 mm

Tinggi Eefektif (d)

Diperoleh gaya momen maksimum dan minimum dari ETABS M11 (kN) 21.546 -21.545

max min

ρmin =

M22 (kN) 30.976 -30.65

1,4 1,4 = =0,00333 fy 420

ρmax = 0,75 × ρbalance = 0,75 × (

0,85 × 𝛽1 × 𝑓′𝑐 600 × ) 𝑓𝑦 600 + 𝑓𝑦

0,85 × 0,834 × 32 600 ρmax = 0,75 × ( × ) = 0,0238 420 600 + 420

1. Desain Tulangan Arah X a. Tulangan Lapangan Arah X Mnx+ =

Mlx 30.976 = = 38.72 kNm = 38720000 Nmm ɸ 0,8

ρmin < ρ < ρmax

Mnx+

2=

ρ × fy × (1 - 0,59 ×

b×d 38720000

2=

1000 × 130

ρ × fy ) f'c

ρ × 420 × (1 - 0,59 ×

ρ × 420 ) 32

ρ = 0,0057 :: maka digunakan: ρ = 0,00571 As = ρ × b × d = 0,00571 × 1000 × 130 = 742.3 mm2

Coba tulangan ø10 D10 = ¼ × π × d2 = ¼ × π × 102 = 78,539 mm2 n=

742.3 = 9.451 ≈ 10 buah ø10 / meter 78,539

Jarak antar tulangan (s) =

b 1000 = = 100 mm n 10

b. Tulangan Tumpuan Arah X Mnx- =

Mtx 30.65 = = 38.3125 kNm = 38312500 Nmm ɸ 0,8

ρmin < ρ < ρmax

Mnx-

2=

b×d

ρ × fy × (1 - 0,59 ×

38312500

2=

1000 × 130

ρ × fy ) f'c

ρ × 420 × (1 - 0,59 ×

ρ × 420 ) 32

ρ = 0,00564

:: maka digunakan: ρ = 0,00564 As = ρ × b × d = 0,00564 × 1000 × 130 = 733.2 mm2 Coba tulangan ø10 D10 = ¼ × π × d2 = ¼ × π × 102 = 78,539 mm2 n=

733.2 = 9.335 ≈ 10 buah ø10 / meter 78,539

Jarak antar tulangan (s) =

b 1000 = = 100 mm n 10

2. Desain Tulangan Arah Y a. Tulangan Lapangan Arah Y Mny+ =

Mly 21.546 = = 26.325 kNm = 2632500 Nmm ɸ 0,8

ρmin < ρ < ρmax

Mny+

2=

ρ × fy × (1 - 0,59 ×

b×d 2632500

1000 × 130

2=

ρ × fy ) f'c

ρ × 420 × (1 - 0,59 ×

ρ × 420 ) 32

Didapat: ρ = 0,00372 ::bdigunakan: ρ = 0,00372 As = ρ × b × d = 0,00372 × 1000 × 130 = 483.6 mm2 D10 = ¼ × π × d2 = ¼ × π × 102 = 78,539 mm2 Coba tulangan ø10 n=

483.6 = 6.157 ≈ 7 buah D10 / meter 78,539

Jarak antar tulangan (s) =

b 1000 = = 142.85 mm ≈ 120 mm n 7

∷ karena jarak jadi 120mm maka tulangan menjadi : 8 buah ø10 / meter

b. Tulangan Tumpuan Arah Y Mny- =

Mly 21.545 = = 26.93125 kNm = 26931250 Nmm ɸ 0,8

ρmin < ρ < ρmax Mny-

2 = ρ × fy × (1 - 0,59 ×

b×d 26931250

ρ × fy ) f'c

2 = ρ × 420 × (1 - 0,59 ×

1000 × 130

ρ × 420 ) 32

Didapat: ρ = 0,00391 ::bdigunakan: ρ = 0,00391 As = ρ × b × d = 0,00391 × 1000 × 130 = 506.3 mm2 D10 = ¼ × π × d2 = ¼ × π × 102 = 78,539 mm2 Coba tulangan ø10 n=

508.3 = 6.471 ≈ 7 buah D10 / meter 78,539

Jarak antar tulangan (s) =

b 1000 = = 142.85 mm ≈ 120 mm n 7

∷ karena jarak jadi 120mm maka tulangan menjadi : 8 buah ø10 / meter