(HDT)Perancangan Beton Bertulang.xlsx

Jarak antar portal = 3.50 = P7Jumlah Trave = 9 = P8Kuat Tekan Beton ( f’c ) = 30 = P10Tegangan Leleh Tulangan ( fy )(tul.pelat, balok anak, sloof dan tul.geser) = 240 = P11Tegangan Leleh Tulangan ( fy )(tul.memanjang balok induk, kolom, tul.tangga dan pondasi) = 400 = P13Tegangan Ijin Tanah (σ) = 2 = P15

=Balok kantilever ( L = 100 h = 15 cm = M27

b = 10 cm = M29=

Balok induk melintang ( L = 300 h = 25 cm = N34b = 20 cm = N36

=Balok induk melintang ( L = 800 h = 70 cm = N41

b = 35 cm = N43=

Balok induk memanjang ( L = 350 h = 30 cm = N50b = 20 cm = N52

=Balok anak ( L = 350 h = 30 cm = N50

b = 20 cm = N52=

Ukuran kolom kiri h = 40 cm = N60b = 30 cm = K60

=Kolom tengah h = 60 cm = N63

b = 50 cm = K63=

Kolom kanan h = 60 cm = N66b = 50 cm = K66

Plat tipe A Ly = P = 350 cmLx = L = 100 cm

Plat tipe B Ly = P = 350 cmLx = L = 300 cm

Plat tipe C Ly = P = 400 cmLx = L = 350 cm

PERANCANGAN STRUKTUR BETON BERTULANG

HENDRICH DENY .T (12.21.916) 1

BAB IDATA PERENCANAAN

1.1Data Perencanaana. Fungsi Bangunan = Kantor b. Lokasi = NTTc. Jenis struktur = SRPMKd. Jarak antar portal = 3.50 me. Jumlah Trave = 9 buahf. Penampang Kolom = Persegi panjangg. Kuat Tekan Beton ( f’c ) = 30 Mpah. Tegangan Leleh Tulangan ( fy ) = 240 Mpa

(tul.pelat, balok anak, sloof dan tul.geser)i. Tegangan Leleh Tulangan ( fy ) = 400 Mpa

(tul.memanjang balok induk, kolom, tul.tangga dan pondasi)j. Tegangan Ijin Tanah (σ) = 2 Mpa

1.2Dimensi BalokPerkiraan dimensi balok dalam SNI 03-2847-2002 dinyatakan dengan rumus :

1 L (untuk balok diatas dua tumpuan)161 L (untuk balok kantilever)8

b = 2 h3

1. Balok kantilever ( L = 100 cm )1 x 100 = 12.5 ≈ 15 cm8

b = 2 x 15 = 10 ≈ 10 cm3

-> ukuran balok kantilever = 10 / 15 cm

2. Balok induk melintang ( L = 300 cm )1

x 300 = 25 ≈ 25 cm12

b = 2x 25 = 16.67 ≈ 20 cm

3-> ukuran Balok induk melintang = 20 / 25 cm

3. Balok induk melintang ( L = 800 cm )1

x 800 = 66.67 ≈ 70 cm12

b = 1x 70 = 35 ≈ 35 cm

2-> ukuran Balok induk melintang = 35 / 70 cm

hmin =

hmin =

hmin =

hmin =

hmin =



4. Balok induk memanjang ( L = 350 cm )1

x 350 = 29.17 ≈ 30 cm12

b = 2x 30 = 20 ≈ 20 cm

3-> ukuran Balok induk memanjang = 20 / 30 cm

5. Balok anak ( L = 350 cm )-> ukuran balok anak = 20 / 30 cm

1.3Dimensi Kolom1. Kolom kiri

Ukuran kolom kiri = 30 / 40 cm

2. Kolom tengahUkuran kolom tengah = 50 / 60 cm

3. Kolom kananUkuran kolom kanan = 50 / 60 cm

hmin =



BAB IIPERHITUNGAN PELAT

1.1Data Perencanaana. Fungsi Bangunan = Kantorb. Mutu Beton (f'c) = 30 Mpac. Mutu Baja (fy) = 240 Mpad. Beban Hidup = 250 (untuk gedung perkantoran)e. Beban Hidup = 400 (untuk ruang pertemuan / aula)

f. β1 = 0.85

1.2 Menentukan Tebal Plat1. Plat tipe A

a. Lx = 100 cmb. Ly = 350 cmc. Dimensi balok x (B1) = 10 / 15 cmd. Dimensi balok y (B4) = 20 / 30 cm

Ln = Ly – 2 (1/2 x Lebar balok x)= 350 - 2 ( 1/2 x 10 )= 340 cm

Sn = Lx – 2 (1/2 x Lebar balok y)= 100 - ( 1/2 x 20 )= 90 cm

β =Ln

=340

= 3.78 > 2 (Plat satu arah)Sn 90

Menurut SNI 03-2847-2002 tebal mininum plat satu arah :1 Lx (satu ujung menerus)

24= 1

x 100 = 4.17 ≈ 12 cm = 120 mm24

2. Plat tipe B

a. Lx = 300 cmb. Ly = 350 cmc. Dimensi balok x (B2) = 20 / 25 cmd. Dimensi balok y (B4) = 20 / 30 cm

kg/m2

kg/m2

hmin =



Ln = Ly – 2 (1/2 x Lebar balok x)= 350 - 2 ( 1/2 x 20 )= 330 cm = 3300 mm

Sn = Lx – 2 (1/2 x Lebar balok y)= 300 - 2 ( 1/2 x 20 )= 280 cm = 2800 mm

β =Ln

=3300

= 1.179 < 2 (Plat dua arah)Sn 2800

Menurut SNI 03-2847-2002 tebal mininum plat dua arah :

= =0.08 x 20 x 25

= 0.6030.08 x 300 x 12

= =0.08 x 20 x 30

= 0.8930.08 x 350 x 12

= =0.60 + 0.89

= 0.7482 2

karena αm 0.748 → αm > 2 maka:h

=Ln (0,8 + fy/1500)

=3300 x ( 0.8 + 240 / 1500 )

36 + 5 x 1.179= 75.621

syarat : h ≥ 90 mm → maka diambil h = 120 mm

3. Plat tipe C

a. Ly = 400 cmb. Lx = 350 cmc. Dimensi balok y (B3) = 35 / 70 cmd. Dimensi balok x (B4) = 20 / 30 cm

Ln = Ly – 2 (1/2 x Lebar balok x)= 350 - 2 ( 1/2 x 20 )= 330 cm = 3300 mm

Sn = Lx – 2 (1/2 x Lebar balok y)= 400 - 2 ( 1/2 x 35 )= 365 cm = 3650 mm

β =Ln

=3300

= 0.904 < 2 (Plat dua arah)Sn 3650

α1Eb1 x Ib1

3

Ep1 x Ip13

α2Eb2 x Ib2

3

Ep2 x Ip23

αmα1 x α2

36 + 5 β



Menurut SNI 03-2847-2002 tebal mininum plat dua arah :

= =0.08 x 20 x 30

= 0.8930.08 x 350 x 12

= =0.08 x 35 x 70

= 17.370.08 x 400 x 12

= =0.893 + 17.37

= 9.1312 2

karena αm 9.131 → αm > 2 maka:h

=Ln (0,8 + fy/1500)

=3300 x ( 0.8 + 240 / 1500 )

36 + 5 x 0.904= 78.183 mm

syarat : h ≥ 90 mm → maka diambil h = 120 mm

1.3 Pembebanan PlatPembebanan pelat tipe A = B = Ca.

Berat plat (lantai 1, 2 & 3)Berat sendiri plat = 0.12 x 2400 = 288Berat keramik per cm = 1 x 24 = 24Berat spesi per cm = 2 x 21 = 42Berat pasir = 0.03 x 1600 = 48Berat plafon = = 11Berat penggantung = = 7

= 420Berat aksesoris = 10% x 420 = 42

= 462

Berat plat (kantilever)Berat sendiri plat = 0.12 x 2400 = 288Berat plafon = = 11Berat penggantung = = 7

= 306Berat aksesoris = 10% x 306 = 30.6

= 337Berat plat (lantai atap)Berat sendiri plat = 0.12 x 2400 = 288Berat plafon = = 11Berat penggantung = = 7

= 306Berat aksesoris = 10% x 306 = 30.6

= 336.6

Berat Sandaran = 0.12 m x 1 m x 2400 = 288 kg/m

α1Eb1 x Ib1

3

Ep1 x Ip13

α2Eb2 x Ib2

3

Ep2 x Ip23

αmα1 x α2

36 + 5 β

Beban mati (qD)

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

kg/m2

Berat sandaran kantilever (PD)kg/m3



b.ᵒ Berdasarkan PMI 1970:16 untuk bangunan kantor :

Beban hidup Lantai 1 dan 3 = 250ᵒ Berdasarkan PMI 1970:16 untuk ruang pertemuan/aula :

Beban hidup Lantai 2 (aula saja) = 400ᵒ Berdasarkan PMI 1970:16 untuk ruang atap :

Beban hidup Lantai atap dan semua kantilever = 100

c. Beban ultimate (qu & pu)ᵒ Untuk lantai 1 dan 3,qu =

= 1.2 x 462 + 1.6 x 250= 954.4= 9.544

ᵒ Untuk lantai 2, qu == 1.2 x 462 + 1.6 x 400= 1194.4= 11.944

ᵒ Untuk kantilever, qu == 1.2 x 336.6 + 1.6 x 100= 563.92= 5.6392

ᵒ Untuk atap, qu == 1.2 x 336.6 + 1.6 x 100= 563.92= 5.6392

ᵒ sandaran kantilever, pu = 1.2 x= 1.2 x 288= 345.6 kg= 3.456 kN

1.4 Menghitung d efektif1). Plat tipe B & C

a. Tebal pelat (h) = 120 mmb. Ø Tulangan pokok = 10 mmc. Ø Tulangan bagi = 8 mmd. Tebal selimut beton (p) = 20 mm

dx = h – p – ½ Ø tul. Pokok= 120 - 20 - 0.5 x 10= 95 mm

dy = h – p – Ø tul. Pokok – ½ Ø tul. Pokok= 120 - 20 - 10 - 0.5 x 10= 85 mm

2). Plat tipe Aa. Tebal pelat (h) = 120 mmb. Ø Tulangan pokok = 10 mmc. Ø Tulangan bagi = 8 mmd. Tebal selimut beton (p) = 20 mm

Beban hidup (qL)

kg/m2

kg/m2

kg/m2

1,2. qD + 1,6. qL

kg/m2

kN/m2

1,2. qD + 1,6. qL

kg/m2

kN/m2

1,2. qD + 1,6. qL

kg/m2

kN/m2

1,2. qD + 1,6. qL

kg/m2

kN/m2

PD



dx = h – p – ½ Ø tul. Pokok= 120 - 20 - 0.5 x 10= 95 mm

dy = h – p – Ø tul. Pokok – ½ Ø tul. Pokok= 120 - 20 - 10 - 0.5 x 10= 85 mm

1.5 Perhitungan momen dan penulangan pelat Lt. 1 & 31. Plat tipe A

Ly=

3.50= 3.5 > 2 (pelat satu arah)

Lx 1.00

a. Menentukan MomenMu = 1

2= 1 5.64 x 1.00 + 3.456 x 1.00

2= 6.2756 kNm

b. Menentukan tulangan pokokMu = 6.28 kNm = 6275600 NmmRn = Mu = 6275600

0.8 x 1000 x 95= 0.869197 Mpa= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

= 0.75 x(0.85 x 30

0.85 600

)240 600 + 240

= 0.04838= 1 = 1.4 = 0.00583

fy 240= 0.85 x f'c

x( 1 - 1 -2 x Rn

)fy 0.85 x fc'

= 0.85 x 30x( 1 - 1 -

2 x 0.8692)

240 0.85 x 30= 0.0037

→ ρperlu < ρ min < ρmaks, maka dipakai ρ perlu = 0.0058

qu . l2 + pu . l

2

Ф . b . dx2 2

ρmax x ρb

β1

ρmin

ρ perlu



As perlu = ρ . b . dx = 0.0058 x 1000 x 95 = 554.167

s =As perlu

= 0.25 x 3.14 x 10 1000554.167

= 141.654 mm

s ≤ 3 h = 3 x 120 = 360s ≤ 450 mm

→ Dipilih yang terkecil, s = 141.65 ≈ 180 mm

As ada =s

= 0.25 x 3.14 x 10 1000180

= 436.111 > As perlu

→ dipakai tulangan pokok Ф 10 - 180

c. Menentukan tulangan bagiAs bagi = 20% As perlu = 20%x 554.167 = 110.833As bagi = 0,002 . b . h 0.002 x 1000 x 120 = 240→ Dipilih yang terbesar, As bagi = 240s =

As bagi= 0.25 x 3.14 x 8 x 1000

240= 209.33 mm

s ≤ 5 h = 5 x 120 = 600s ≤ 450 mm


As ada=s

= 0.25 x 3.14 x 8 1000250

= 201.062 > As perlu

→ dipakai tulangan pokok Ф 8 - 250 mm

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2

1/4 . π . Ф2 .b

2

1/4 . π . Ф2 .b

2 x

mm2



2. Plat tipe B

Ly=

3.50= 1.1667 < 2 (pelat dua arah)

Lx 3.00

a. Menentukan momenKx = 46 (PBI 1971)Ky = 38 (PBI 1971)

Mlx = -Mtx = 0.001= 0.001 x 9.54 x 3 x 46= 3.951 kNm

Mly = -Mty = 0.001= 0.001 x 9.54 x 3 x 38= 3.264 kNm

b. Menentukan tulangan arah lapangan x = tumpuan xMu = 3.951 kNm = 3.951 x 10 NmmRn = Mu = 3951000

0.8 x 1000 x 95= 0.547 Mpa= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

= 0.75 x(0.85 x 30

0.85600

)240 600 + 240

= 0.04838

= 1.4 = 1.4 = 0.00583fy 240

0.85 x f'cx( 1 - 1 -

2 x Rn)

fy 0.85 x fc'= 0.85 x 30

x( 1 - 1 -2 x 0.547

)240 0.85 x 30

= 0.00231→ ρperlu < ρmin < ρmax , maka dipakai ρ min = 0.00583

As perlu = ρ . b . dx= 0.00583 x 1000 x 95= 554.167

x Wu x lx2 x Kx 2

x Wu x lx2 x Ky2

6

Ф . b . dx2 2

ρmax x ρb

β1

ρmin

ρ perlu=

mm2



s =As perlu

= 0.25 x 3.14 x 10 1000554.167

= 141.654 mm

s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 10 1000140

= 560.999 > As perlu


c. Menentukan tulangan bagi tumpuan xAs bagi = 20% As perlu = 20%x 554.167 = 110.833As bagi = 0,002 . b . h 0.002 x 1000 x 120 = 240→ Dipilih yang terbesar, As bagi = 240s =

As bagi= 0.25 x 3.1416 x 8 1000

240= 209.44 mm

s ≤ 5 h = 5 x 120 = 600s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 8 1000200

= 251.327 > As bagi

→ dipakai tulangan pokok Ф 8 - 200 mm

c. Menentukan tulangan arah lapangan y = tumpuan yMu = 3.264 kNm = 3.264 x 10 NmmRn = Mu = 3264000

0.8 x 1000 x 85= 0.565 Mpa

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π .Ф2 .b

2 x

mm2

6

Ф . b . dx2 2



= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

=0.75 x(

0.85 x 300.85

600)

240 600 + 240= 0.04838

= 1.4 = 1.4 = 0.00583fy 240

= 0.85 x f'cx( 1 - 1 -

2 x Rnfy 0.85 x fc'

= 0.85 x 30x( 1 - 1 -

2 x 0.56240 0.85 x 30


As perlu = ρ . b . dy = 0.00583 x 1000 x 85 = 495.833

s =As perlu

= 0.25 x 3.1416 x 10 1000495.833

= 158.400 mm

s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 10 1000160

= 490.874 > As perlu


Menentukan tulangan bagi tumpuan yAs bagi = 20% As perlu = 20%x 495.833 = 99.167As bagi = 0,002 . b . h 0.002 x 1000 x 120 = 240→ Dipilih yang terbesar, As bagi = 240s =

As bagi= 0.25 x 3.1416 x 8 1000

240= 209.44 mm

s ≤ 5 h = 5 x 120 = 600s ≤ 450 mm

ρmax x ρb

β1

ρmin

ρ perlu

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2

1/4 . π . Ф2 .b

2 x




As ada =s

= 0.25 x 3.14 x 8 1000200

= 251.327 > As bagi→ dipakai tulangan pokok Ф 8 - 200 mm

3. Plat tipe C

Ly=

4.00= 1.1429 < 2 (pelat dua arah)

Lx 3.50


Mlx = -Mtx = 0.001= 0.001 x 9.54 x 3.50 x 42= 4.910 kNm

Mly = -Mty = 0.001= 0.001 x 9.54 x 3.5 x 37= 4.326 kNm


0.8 x 1000 x 95= 0.680 Mpa= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

= 0.75 x(0.85 x 30

0.85600

)240 600 + 240

= 0.04838

= 1.4 = 1.4 = 0.00583fy 240

1/4 . π . Ф2 .b

2 x

mm2

x Wu x lx2 x Kx 2

x Wu x lx2 x Ky2

6

Ф . b . dx2 2

ρmax x ρb

β1

ρmin



0.85 x f'cx( 1 - 1 -

2 x Rn)

fy 0.85 x fc'= 0.85 x 30

x( 1 - 1 -2 x 0.680

)240 0.85 x 30


As perlu = ρ . b . dx= 0.00583 x 1000 x 95= 554.167

s =As perlu

= 0.25 x 3.14 x 10 1000554.167

= 141.654 mm

s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 10 1000140

= 560.999 > As perlu→ dipakai tulangan pokok Ф 10 - 140

Menentukan tulangan bagi tumpuan xAs bagi = 20% As perlu = 20%x 554.167 = 110.833As bagi = 0,002 . b . h 0.002 x 1000 x 120 = 240→ Dipilih yang terbesar, As bagi = 240s =

As bagi= 0.25 x 3.14 x 8 1000

240= 209.33 mm

s ≤ 5 h = 5 x 120 = 600s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 8 1000210

= 239.238 > As bagi→ dipakai tulangan bagi Ф 8 - 210 mm

ρ perlu=

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π .Ф2 .b

2 x

mm2




0.8 x 1000 x 95= 0.599 Mpa= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

=0.75 x(

0.85 x 300.85

600)

240 600 + 240= 0.04838

= 1.4 = 1.4 = 0.00583fy 240

= 0.85 x f'cx( 1 - 1 -

2 x Rnfy 0.85 x fc'

= 0.85 x 30x( 1 - 1 -

2 x 0.6240 0.85 x 30

= 0.00253→ ρperlu < ρmin<ρmax , maka dipakai ρ min =0.00583

As perlu = ρ . b . dy = 0.00583 x 1000 x 85 = 495.833

s =As perlu

= 0.25 x 3.1416 x 10 1000495.833

= 158.400 mm

s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 10 1000160


Menentukan tulangan bagi tumpuan yAs bagi = 20% As perlu = 20%x 495.833 = 99.167As bagi = 0,002 . b . h 0.002 x 1000 x 120 = 240→ Dipilih yang terbesar, As bagi = 240

6

Ф . b . dx2 2

ρmax x ρb

β1

ρmin

ρ perlu

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2



s =As bagi

= 0.25 x 3.1416 x 8 1000240

= 209.44 mms ≤ 5 h = 5 x 120 = 600s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 8 1000210


1.6 Perhitungan momen dan penulangan pelat Lt. 21. Plat tipe A

Ly=

3.5= 3.5 > 2 (pelat satu arah)

Lx 1.00

a. Menentukan MomenMu = 1

2= 1 5.64 x 1.00 + 3.456 x 1.00

2= 6.2756 kNm

b. Menentukan tulangan pokokMu = 6.28 kNm = 6.28 x 10 NmmRn = Mu = 6275600

0.8 x 1000 x 95= 0.869197 Mpa= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

= 0.75 x(0.85 x 30

0.85 600

)240 600 + 240

= 0.04838= 1 = 1.4 = 0.00583

fy 240= 0.85 x f'c

x( 1 - 1 -2 x Rn

)fy 0.85 x fc'

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

qu . l2 + pu . l

2

6

Ф . b . dx2 2

ρmax x ρb

β1

ρmin

ρ perlu



= 0.85 x 30x( 1 - 1 -

2 x 0.87)

240 0.85 x 240= 0.0005

→ ρperlu < ρ min < ρmaks, maka dipakai ρ perlu = 0.0058

As perlu = ρ . b . dx = 0.0058 x 1000 x 95 = 554.167

s =As perlu

= 0.25 x 3.14 x 10 1000554.167

= 141.654 mm

s ≤ 3 h = 3 x 120 = 360s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 10 1000180.000


c. Menentukan tulangan bagiAs bagi = 20% As perlu = 20%x 554.167 = 110.833As bagi = 0,002 . b . h 0.002 x 1000 x 120 = 240→ Dipilih yang terbesar, As bagi = 240s =

As bagi= 0.25 x 3.14 x 8 1000

240= 209.33 mm

s ≤ 5 h = 5 x 120 = 600s ≤ 450 mm


As ada=s

= 0.25 x 3.14 x 8 1000250

= 201.062 > As perlu→ dipakai tulangan pokok Ф 8 - 250 mm

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2



2. Plat tipe B

Ly=

3.50= 1.1667 < 2 (pelat dua arah)

Lx 3.00


Mlx = -Mtx = 0.001= 0.001 x 11.9 x 3.00 x 46= 4.945 kNm

Mly = -Mty = 0.001= 0.001 x 11.9 x 3.00 x 38= 4.085 kNm


0.8 x 1000 x 95= 0.685 Mpa= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

= 0.75 x(0.85 x 30

0.85600

)240 600 + 240

= 0.04838

= 1.4 = 1.4 = 0.00583fy 240

0.85 x f'cx( 1 - 1 -

2 x Rn)

fy 0.85 x fc'= 0.85 x 30

x( 1 - 1 -2 x 0.685

)240 0.85 x 30

= 0.00289→ ρperlu < ρmin , maka dipakai ρ min = 0.00583

As perlu = ρ . b . dx= 0.00583 x 1000 x 95= 554.167

x Wu x lx2 x Kx 2

x Wu x lx2 x Ky2

6

Ф . b . dx2 2

ρmax x ρb

β1

ρmin

ρ perlu=

mm2



s =As perlu

= 0.25 x 3.14 x 10 1000554.167

= 141.654 mm

s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 10 1000140



As bagi= 0.25 x 3.1416 x 8 1000

240= 209.44 mm

s ≤ 5 h = 5 x 3.14 = 15.71s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 8 1000200



0.8 x 1000 x 95= 0.566 Mpa

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π .Ф2 .b

2 x

mm2

6

Ф . b . dx2 2



= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

=0.75 x(

0.85 x 300.85

600)

240 600 + 240= 0.04838

= 1.4 = 1.4 = 0.00583fy 240

= 0.85 x f'cx( 1 - 1 -

2 x Rnfy 0.85 x fc'

= 0.85 x 30x( 1 - 1 -

2 x 0.57240 0.85 x 30


As perlu = ρ . b . dy = 0.00583 x 1000 x 85 = 495.833

s =As perlu

= 0.25 x 3.1416 x 10 1000495.833

= 158.400 mm

s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 10 1000160



As bagi= 0.25 x 3.1416 x 8 1000

240= 209.44 mm

s ≤ 5 h = 5 x 120 = 600s ≤ 450 mm

ρmax x ρb

β1

ρmin

ρ perlu

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2

1/4 . π . Ф2 .b

2 x




As ada =s

= 0.25 x 3.14 x 8 1000200


3. Plat tipe C

Ly=

4.00= 1.1429 < 2 (pelat dua arah)

Lx 3.50


Mlx = -Mtx = 0.001= 0.001 x 11.9 x 3.50 x 42= 6.145 kNm

Mly = -Mty = 0.001= 0.001 x 11.9 x 3.5 x 37= 5.414 kNm


0.8 x 1000 x 95= 0.851 Mpa= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

= 0.75 x(0.85 x 30

0.85600

240 600 + 240= 0.04838

= 1.4 = 1.4 = 0.00583fy 240

1/4 . π . Ф2 .b

2 x

mm2

x Wu x lx2 x Kx 2

x Wu x lx2 x Ky2

6

Ф . b . dx2 2

ρmax x ρb

β1

ρmin



0.85 x f'cx( 1 - 1 -

2 x Rn)

fy 0.85 x fc'= 0.85 x 30

x( 1 - 1 -2 x 0.851

)240 0.85 x 30

= 0.00361→ ρperlu < ρmin < ρmax, maka dipakai ρ min = 0.00583

As perlu = ρ . b . dx= 0.00583 x 1000 x 95= 554.167

s =As perlu

= 0.25 x 3.14 x 10 1000554.167

= 141.654 mm

s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 10 1000140



As bagi= 0.25 x 3.1416 x 8 1000

240= 209.44 mm

s ≤ 5 h = 5 x 120 = 600s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 8 1000200

= 251.327 > As bagi→ dipakai tulangan bagi Ф 8 - 200 mm

ρ perlu=

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π .Ф2 .b

2 x

mm2




0.8 x 1000 x 95= 0.750 Mpa= 0.75

= 0.75 x(0.85 x fc' 600

)fy 600 + fy

=0.75 x(

0.85 x 300.85

600)

240 600 + 240= 0.04838

= 1.4 = 1.4 = 0.00583fy 240

= 0.85 x f'cx( 1 - 1 -

2 x Rnfy 0.85 x fc'

= 0.85 x 30x( 1 - 1 -

2 x 0.75240 0.85 x 30

= 0.00317→ ρperlu < ρmin <ρmax , maka dipakai ρ min = 0.00583

As perlu = ρ . b . dy = 0.00583 x 1000 x 85 = 495.833

s =As perlu

= 0.25 x 3.1416 x 10 1000495.833

= 158.400 mm

s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 10 1000160



As bagi

6

Ф . b . dx2 2

ρmax x ρb

β1

ρmin

ρ perlu

mm2

1/4 . π . Ф2 .b

2 x

1/4 . π . Ф2 .b

2 x

mm2

mm2

mm2

mm2

1/4 . π . Ф2 .b



= 0.25 x 3.1416 x 8 1000240

= 209.44 mms ≤ 5 h = 5 x 120 = 600s ≤ 450 mm


As ada =s

= 0.25 x 3.14 x 8 1000200


2 x

1/4 . π . Ф2 .b

2 x

mm2



BAB IIIBALOK INDUK LT. 1 & 3( LINE C )

A. Data Perencanaan

Ø tul.pokok = 16 mmØ tul. begel = 8 mmp = 40 mmfc' = 30 Mpafy = 400 Mpafy tul. Geser = 240 Mpaβ1 = 0.85BJ beton = 2400 kg/m³dimensi balok → b = 0.20 m 200 mm

h = 0.30 m 300 mmhf = 0.12 m 120 mm

fungsi lantai 1 & 3 = kantorfungsi lantai 2 = aulaReduksi beban hidup = 0.3Bentang L = 3.5 mtinggi lantai 1 = 4.0 m = 4000 mmtinggi lantai 2 & 3 = 4.0 m = 4000 mmbeban mati → beban plat lantai = 462 kg/m²

beban ½ bata = 250 kg/m²beban hidup → beban plat kantor = 250 kg/m²

beban plat aula = 400 kg/m²

Gambar 3.1 Area pembebanan balok induk



B. Perataan Beban

Perataan beban type Ca b a

a = 1.5 m

1.5b = 0.5 m

ht segitiga = 1.5 ml = 3.50 m

3.50

Ra =( 3.50 + 0.5 ) x 1.5

x 0.52

= 1.5

T1 = a x t x 0.5 segitiga= 1.5 x 1.5 x 0.5= 1.125

T2 = 0.5 x ( 1.5 / 2 )= 0.375

Mmax 1 = Ra x 1/2 x L - T1 x (1/3 x a + b/2) - T2 x b/4= 1.5 x 0.5 x 3.5 - 1.125 x ( 0.3333 x 1.5 + 0.5 / 2 )

- 0.375 x 0.5 / 4= 1.73438

Mmax 2 =

= 0.125 x h x 3.5 ²= 1.53125 h

Mmax1 = Mmax21.734375 = 1.5313 h

h =1.734375

= 1.133 m1.53125

1/8 x h x L2



Perataan beban type A

2 m Lx = 2.00 m = 2000 mmh

h = ²/з x Lx3.50 m = ²/з x 2.00 m

= 0.667 m

C. Pembebanan BalokPembebanan Lantai 1 • Beban Mati (qD)

B.S. Balok = 0.20 0.30 - 0.12 2400 kg/m³ = 86.4 kg/mBerat plat = 1.133 x 0.667 m x 462 kg/m² = 348.9 kg/mBerat dinding = 4.0 m x 250 kg/m² = 1000.0 kg/m

= 1435.3 kg/m

• Beban Hidup (qL)Berat plat = 0.3 x 1.133 x 0.667 m x 250 kg/m² = 56.6 kg/m

Pembebanan Lantai 2• Beban Mati (qD)


= 1435.3 kg/m




= 1435.3 kg/m


Beban Terfaktorqu lt.1 = 1.2 qd + 1.6 ql qu lt.3 = 1.2 qd + 1.6 ql

= 1.2 x 1435.3 + 1.6 x 56.6 = 1.2 x 1435.3 + 1.6 x 56.6= 1812.9 kg/m = 1812.9 kg/m

qu lt.2 = 1.2 qd + 1.6 ql= 1.2 x 1435.3 + 1.6 x 90.6= 1867.3 kg/m

m x ( ) m x

m x ( ) m x

m x ( ) m x



D. PERHITUNGAN STATIKA1. Balok induk Lantai 1

1/16 1/10 1/11 1/11 1/11 1/11 1/11 1/11 1/10 1/16

1 2 3 4 5 6 7 8 9 10

1/14 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/14

Momen Tumpuan :Mu1 = Mu10 = 1/16 x qu x = 0.0625 x 1812.9 x 12.25 = 1388.018 kgmMu2 = Mu9 = 1/10 x qu x = 0.10 x 1812.9 x 12.25 = 2220.828 kgmMu3 - Mu8 = 1/11 x qu x = 0.09 x 1812.9 x 12.25 = 2018.935 kgm

Momen Lapangan :Mu1-2 = Mu9-10 = 1/14 x qu x = 0.0714 x 1812.9 x 12.25 = 1586.306 kgmMu2-3 → Mu8-9 = 1/16 x qu x = 0.06 x 1812.9 x 12.25 = 1388.018 kgm

Gaya Geser :Qu = 1/2 x qu x L

= 0.5 x 1812.9 x 3.5= 3172.6114 Kg

Hasil :• Mu tump = 2220.83 kgm = 22.20828 kNm• Mu lap = 1586.31 kgm = 15.863057 kNm• Vu = 3172.61 kg = 31.726114 kN

2. Balok induk Lantai 2

1/16 1/10 1/11 1/11 1/11 1/11 1/11 1/11 1/10 1/16

1 2 3 4 5 6 7 8 9 10

1/14 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/14




= 0.5 x 1867.3 x 3.5= 3267.7543 Kg

• Mu tump = 2287.43 kgm = 22.87428 kNm• Mu lap = 1633.88 kgm = 16.338771 kNm• Vu = 3267.75 kg = 32.677543 kN

L²L²L²

L²L²

L²L²L²

L²L²



3. Balok induk Lantai 3

1/16 1/10 1/11 1/11 1/11 1/11 1/11 1/11 1/10 1/16

1 2 3 4 5 6 7 8 9 10

1/14 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/14




= 0.5 x 1812.9 x 3.5= 3172.6114 Kg


E. Perhitungan Penulangan Balok induk Lantai 1 & 31. Perhitungan tulangan tumpuan :A. Hitung Tulangan

ds1 = p + Ø tul. begel + (½ x Ø tul. pokok)= 40 + 8 +( ½ x 16 )= 56 mm

d = h - ds1= 300 - 56.0= 244.0 mm

300244

56

200

L²L²L²

L²L²



Mu tump (-) = 22.2083 kNm = 22208280.00 NmmMn = Mu / Ф

= 22208280 / 0.8= 27760350 Nmm

K =Mu

=22208280.00

= 2.331 Mpa0.8 x 200 x 244.0 ²

Kmaks =

= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 )

= 7.8883 Mpa

Karena K < Kmaks maka dipakai tulangan tunggalJika K > Kmaks ( maka dipakai tulangan rangkap )

→ Mencari tinggi blok tegangan beton tekon ekuivalen a

a = 1 -1 -

2 x Kx d

0.85 x fc'

= 1 -1 -

2 x 2.331x 244

0.85 x 30

= 23.43 mm

→ Hitung luas tulangan perlu (As,u)

= 0.85 x fc x a x bfy

= 0.85 x 30 x 23.43 x 200400

= 298.78

=1.4 x b x d

=1.4 x 200 x 244

= 170.80 mm²fy 400

=fc' x b x d

=30 x 200 x 244

= 167.06 mm²4 x fy 4 x 400

= 298.78 mm²

→ Hitung jumlah tulangan

n =As,u

=298.78

= 1.49 ≈ 2 buah1/4 . π . Ø² ¼ x π x 16 ²

Ø x b x d²

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²

Asu

mm²

Asu

Asu

Diambil nilai yang terbesar Asu



Digunakan tulangan tarik As = 2 Ø 16 = 401.920 > Asu OKtulangan tekan As' = 2 Ø 16 = 401.920 (ditambahkan)

Jumlah tulangan maksimal pada 1 baris :

m =b - 2 x ds1

+ 1Ø + Sn

=200 - 2 x 56

+ 116 + 25

= 3.146341 batang → maksimal 3 batang

56

300244

200

B. Hitung Momen Rencana1). Hitung tinggi blok tegangan beton tekan persegi ekuivalen a

a =As x fy

0.85 x fc' x b

=401.92 x 400

= 31.52 mm0.85 x 30 x 200

2).

As = 2 Ø 16 = 2 x (1/4) x π x 16 ² = 401.920

ρ =As

=401.92

= 0.00824be x d 200 x 244

ρ min =fc'

=30

= 0.003424 x fy 4 x 400

ρ maks =

= 0.75 x (0,85 . fc'

β1600

)fy 600 + fy

= 0.75 x (0.85 x 30

0.85600

)400 600 + 400

= 0.02438

ρ min < ρ < ρ maks OK

OKOK

mm²mm²

Kontrol rasio tulangan ρ min ≤ ρ ≤ ρ maks

mm²

0,75 x ρ b

Jika ρ < ρ min → balok diperkecilJika ρ > ρ maks → balok diperbesar



3).

a maks leleh = 600 x β1 x d600 + fy

= 600 x 0.85 x 244600 + 400

= 124.44 mm

a < a maks leleh OKjika a > a maks leleh ( ukuran balok kurang besar )

4). Hitung momen nominal (Mn)dan momen rencana (Mr)

Mn = As x fy x ( d - a/2)= 401.92 x 400 x 228.24= 36693436.13 Nmm

Mr = Ø x Mn= 0.8 x 36693436.13= 29354748.91 Nmm > Mu = 22208280.00 Nmm → aman

5).

εy = fy / Es= 400 / 200000= 0.00200

εc' =a

x =31.52

x 0.00200x d - a 0.85 x 244 - 31.52

= 0.00034908 < εcu' = 0.003 → aman

2. Perhitungan tulangan lapangan :A. Hitung Tulangan


ds2 = Ø + Sn ( jarak minimum tulangan )= 16 + 25= 41 mm ≈ 40 mm

d = h - ds1= 300 - 56.0= 244.0 mm

Kontrol semua tulangan tarik harus sudah leleh ( a ≤ a maks leleh )

Kontrol εc' harus ≤ εcu' = 0,003

εyβ1



→- 3500 mm- 3150 mm

≤ 1x λ =

1x 3500 = 875 mm

4 4

≤ 16 x hf + b = 16 x 120 + 200 = 2120 mm

≤ λn x b = 3150 x 200 = 630000 mm

Diambil nilai yang terkeci yaitu be = 875 mmJarak bersih antar tulangan Sn = 25 mm 16 mm)

875

300244

4056

200

Mu lap. (+) = 15.8631 kNm = 15863057.14 NmmMn = Mu / Ф

= 15863057 / 0.8= 19828821 Nmm

K =Mu

=15863057.14

= 0.381 Mpa0.8 x 875 x 244.0 ²

Kmaks =

= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 ) ²

= 7.8883 Mpa

Karena K < Kmaks maka dipakai tulangan tunggal


a = 1 -1 -

2 x Kx d

0.85 x fc'

= 1 -1 -

2 x 0.381x 244

0.85 x 30

= 3.67 mm

Mencari lebar pelat efektif (be) Bentang balok (as-as) λ =Bentang bersih pelat λn =

be

be

be

( > Ø =

Ø x be x d²

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²



karena a < hf maka beton tekan berada di sayap atau balok T palsu (hitung luas As,u )dihitung sebagai balok persegi panjang dengan lebar balok = be

# Hitung luas tulangan perlu (As,u)

= 0.85 x fc x a x befy

= 0.85 x 30 x 3.67 x 875400

= 204.70

Amin =1.4 x be x d

=1.4 x 875 x 244

= 747.25 mm²fy 400

Amin =fc' x be x d

=30 x 875 x 244

= 730.87 mm²4 x fy 4 x 400

= 747.25 mm²

# Hitung jumlah tulangan

n =As,u

=747.250

= 3.72 ≈ 4 buah1/4 . π . Ø² ¼ x π x 16 ²



m =b - 2 x d'

+ 1Ø + Sn

=200 - 2 x 56

+ 116 + 25


Karena n > m maka tulangan lebih dari 1 baris ds2 = ds1 + ( ds2 / 2 )

= 56 + ( 20 ) = 76 mmjadi d = h - ds2

= 300 - 76 = 224 mm

875

300224

4056

200

Asu

mm²


mm²mm²




a =As x fy

0.85 x fc' x be

=803.84 x 400

= 14.41 mm0.85 x 30 x 875

karena a < hf maka beton tekan berada di sayap atau balok T palsu (kontrol rasio tulangan )dihitung sebagai balok persegi panjang dengan lebar balok = be

2).

As = 4 Ø 16 = 4 x (1/4) x π x 16 ² = 803.840

ρ =As

=803.84

= 0.00410be x d 875 x 224

ρ min =fc'

=30

= 0.003424 x fy 4 x 400

ρ maks =

= 0.75 x (0,85 . fc'

β1600

)fy 600 + fy

= 0.75 x (0.85 x 30

0.85600

)400 600 + 400

= 0.02438


OKOK

3).

a maks leleh = 600 x β1 x dd600 + fy

= 600 x 0.85 x 244600 + 400

= 124.44 mm

dd = jarak antara titik berat tulangan tarik paling dalam dan tepi serat beton tekan

a < a maks leleh (OK)

4). Hitung momen nominal (Mn) dan momen rencana (Mr)



mm²

0,75 x ρ b






5).

εy = fy / Es= 400 / 200000= 0.00200

εc' =a

x =14.41

x 0.00200x d - a 0.85 x 244 - 14.41

= 0.000147686 < εcu' = 0.003 → aman

2. Perhitungan tulangan geser :Vu = 31.726114 kN = 31726.11 N

d = 244 mm = 0.24 m

0.24 1.51

Vu = 31726.11Vud = 27302.59 N

Vut = 0

0.80 0.951.75

Vud = Vut +x

x ( Vu - Vut )y

= 0 +1.51

x ( 31726.114 - 0 )1.75

= 27302.587 N

= Ø x 1/6 x fc' x b x d= 0.75 x 0.17 x 30 x 200 x 244= 33411.076 N

Ø.Vc/2 = 33411.076 / 2= 16705.538 N

1.75 → x = 1.75 x= 1.75 x ( 31726.114 - 16705.538 ) / 31726.11= 0.83 m ≈ 0.80 m

→ Untuk daerah sepanjang x = 0.80 m,

Vu < Ø.Vc NOVu > Ø.Vc/2 NO

Ø.Vc/2 > Vu < Ø.Vc dipakai luas begel perlu minimal per meter panjang balok (Av,u) yang besar


εyβ1

Ø.Vc/2

Ø.Vc

( Vu - Ø. Vc / 2)/x = Vu / ( Vu - Ø. Vc / 2) / Vu

karena



Av,u =b x S

=200 x 1000

= 277.77783 x fy 3 x 240

Av,u =75 x fc' x b x S

=75 x 30 x 200 x 1000

1200 x fy 1200 x 240= 285.27217 mm²

Dipilih yang besar , jadi Av,u = 285.272 mm²

Spasi begel :s =

n x 1/4 x π x x SAv,u

s =2 x 0.25 x 3.14 x 64 x 1000

285.272

s = 352.22504 mm

n = Jumlah kaki begeldp² = Diameter begel

Syarat spasi s = d / 2 = 244 / 2 = 122 mmDipilih spasi yang kecil , yaitu s = 122 mm ≈ 120 mmJadi dipakai begel Ø 8 - 120 mm


Vud < Ø.Vc NOVud > Ø.Vc/2 NO

Ø.Vc/2 > Vud < Ø.Vc dipakai luas begel perlu minimal per meter panjang balok (Av,u) yang besar

Luas begel / m :

Av,u =b x S

=200 x 1000

= 277.77783 x fy 3 x 240


=75 x 30 x 200 x 1000

1200 x fy 1200 x 240= 285.27217 mm²


Spasi begel :s =


s =2 x 0.25 x 3.14 x 64 x 1000

285.272

s = 352.22504 mm



mm²

dp²

karena

mm²

dp²



BALOK ANAK ( LINE D )

A. Data Perencanaan

Ø tul.pokok = 16 mmØ tul. begel = 8 mmp = 40 mmfc' = 30 Mpafy = 240 Mpaβ1 = 0.85BJ beton = 2400 kg/m³dimensi balok → b = 0.20 m 200 mm

h = 0.30 m 300 mmhf = 0.12 m 120 mm

fungsi lantai 1 & 3 = kantorfungsi lantai 2 = aulaReduksi beban hidup = 0.3Bentang L = 3.5 mtinggi lantai 1 = 4.0 m = 4000 mmtinggi lantai 2 & 3 = 4.0 m = 4000 mmbeban mati → beban plat lantai = 462 kg/m²

beban ½ bata = 250 kg/m²beban hidup → beban plat kantor = 250 kg/m²

beban plat aula = 400 kg/m²

Gambar 3.1 Area pembebanan balok anak



B. Perataan Beban

2 m Lx = 2.00 m = 2000 mmh

h = ²/з x Lx3.50 m = ²/з x 2.00 m

= 0.667 m

C. Pembebanan BalokPembebanan Lantai 1 • Beban Mati (qD)

B.S. Balok = 0.20 0.30 - 0.12 2400 kg/m³ = 86.4 kg/mBerat plat = 2 x 0.667 m x 462 kg/m² = 616.0 kg/mBerat dinding = 4.0 m x 250 kg/m² = 1000.0 kg/m

= 1702.4 kg/m

• Beban Hidup (qL)Berat plat = 0.3 x 2 x 0.667 m x 250 kg/m² = 100.0 kg/m



= 1702.4 kg/m




= 1702.4 kg/m


Beban Terfaktorqu lt.1 = 1.2 qd + 1.6 ql qu lt.3 = 1.2 qd + 1.6 ql

= 1.2 x 1702.4 + 1.6 x 100.0 = 1.2 x 1702.4 + 1.6 x 100.0= 2202.9 kg/m = 2202.9 kg/m

qu lt.2 = 1.2 qd + 1.6 ql= 1.2 x 1702.4 + 1.6 x 160.0= 2298.9 kg/m

m x ( ) m x

m x ( ) m x

m x ( ) m x



D. PERHITUNGAN STATIKA1. Balok anak Lantai 1

1/16 1/10 1/11 1/11 1/11 1/11 1/11 1/11 1/10 1/16

1 2 3 4 5 6 7 8 9 10

1/14 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/14




= 0.5 x 2202.9 x 3.5= 3855.04 Kg


2. Balok anak Lantai 2

1/16 1/10 1/11 1/11 1/11 1/11 1/11 1/11 1/10 1/16

1 2 3 4 5 6 7 8 9 10

1/14 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/14




= 0.5 x 2298.9 x 3.5= 4023.04 Kg

• Mu tump = 2816.13 kgm = 28.16128 kNm• Mu lap = 2011.52 kgm = 20.1152 kNm• Vu = 4023.04 kg

L²L²L²

L²L²

L²L²L²

L²L²



3. Balok anak Lantai 3

1/16 1/10 1/11 1/11 1/11 1/11 1/11 1/11 1/10 1/16

1 2 3 4 5 6 7 8 9 10

1/14 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/14




= 0.5 x 2202.9 x 3.5= 3855.04 Kg

Hasil :• Mu tump = 2698.53 kgm = 26.98528 kNm• Mu lap = 1927.52 kgm = 19.2752 kNm• Vu = 3855.04 kg

E. Perhitungan Penulangan Balok anak Lantai 21. Perhitungan tulangan tumpuan :A. Hitung Tulangan



d = h - ds1= 300 - 56.0= 244.0 mm

5640

300244

200

L²L²L²

L²L²




= 28161280 / 0.8= 35201600 Nmm

K =Mu

=28161280.00

= 2.956 Mpa0.8 x 200 x 244.0 ²

Kmaks =

= 382.5 x 0.85 x 30 x ( 600 + 240 - 225 x 0.85 )( 600 + 240 )

= 8.9679 Mpa

Karena K < Kmaks maka dipakai tulangan tunggalJika K > Kmaks ( maka dipakai tulangan rangkap )


a = 1 -1 -

2 x Kx d

0.85 x fc'

= 1 -1 -

2 x 2.956x 244

0.85 x 30

= 30.15 mm



= 0.85 x 30 x 30.15 x 200240

= 640.71

=1.4 x b x d

=1.4 x 200 x 244

= 284.67 mm²fy 240

=fc' x b x d

=30 x 200 x 244

= 278.43 mm²4 x fy 4 x 240

= 640.71 mm²

→ Hitung jumlah tulangan

n =As,u

=640.71

= 3.19 ≈ 4 buah1/4 . π . Ø² ¼ x π x 16 ²

Ø x b x d²

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²

Asu

mm²

Asu

Asu






m =b - 2 x ds1

+ 1Ø + Sn

=200 - 2 x 56

+ 116 + 25


Karena n > m maka tulangan lebih dari 1 baris ds2 = ds1 + ( ds2 / 2 )

= 56 + ( 20 ) = 76 mm

jadi d = h - ds2= 300 - 76 = 224 mm

5640

300224

200


a =As x fy

0.85 x fc' x b

=803.84 x 240

= 37.83 mm0.85 x 30 x 200

2).

As = 4 Ø 16 = 4 x (1/4) x π x 16 ² = 803.840

ρ =As

=803.84

= 0.01794be x d 200 x 224

ρ min =fc'

=30

= 0.005714 x fy 4 x 240

ρ maks =

= 0.75 x (0,85 . fc'

β1600

)fy 600 + fy

= 0.75 x (0.85 x 30

0.85600

)240 600 + 240

= 0.04838


OKOK

mm²mm²


mm²

0,75 x ρ b




3).

a maks leleh = 600 x β1 x d600 + fy

= 600 x 0.85 x 224600 + 240

= 136 mm

a < a maks leleh OKjika a > a maks leleh ( ukuran balok kurang besar )




5).

εy = fy / Es= 240 / 200000= 0.00120

εc' =a

x =37.83

x 0.00120x d - a 0.85 x 224 - 37.83

= 0.00028685 < εcu' = 0.003 → aman




d = h - ds1= 300 - 56.0= 244.0 mm



εyβ1



→- 3500 mm- 3150 mm

≤ 1x λ =

1x 3500 = 875 mm

4 4

≤ 16 x hf + b = 16 x 120 + 200 = 2120 mm

≤ λn x b = 3150 x 200 = 630000 mm


875

300244

4056

200


= 19275200 / 0.8= 24094000 Nmm

K =Mu

=19275200.00

= 0.463 Mpa0.8 x 875 x 244.0 ²

Kmaks =

= 382.5 x 0.85 x 30 x ( 600 + 240 - 225 x 0.85 )( 600 + 240 ) ²

= 8.9679 Mpa



a = 1 -1 -

2 x Kx d

0.85 x fc'

= 1 -1 -

2 x 0.463x 244

0.85 x 30

= 4.47 mm



be

be

be

( > Ø =

Ø x be x d²

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²





= 0.85 x 30 x 4.47 x 875240

= 415.24

Amin =1.4 x be x d

=1.4 x 875 x 244

= 1245.42 mm²fy 240

Amin =fc' x be x d

=30 x 875 x 244

= 1218.11 mm²4 x fy 4 x 240

= 1245.42 mm²


n =As,u

=1245.417

= 6.20 ≈ 7 buah1/4 . π . Ø² ¼ x π x 16 ²



m =b - 2 x d'

+ 1Ø + Sn

=200 - 2 x 56

+ 116 + 25


Karena n > m maka tulangan lebih dari 1 baris ds2 = ds1 + ( ds3 / 2 ) + ds2

= 56 + ( 20 ) = 116 mm

jadi d = h - ds2= 300 - 116 = 184 mm

875

300184

404056

200


a =As x fy

0.85 x fc' x be

=1406.72 x 240

= 15.13 mm0.85 x 30 x 875


Asu

mm²


mm²mm²



2).

As = 7 Ø 16 = 7 x (1/4) x π x 16 ² = 1406.720

ρ =As

=1406.72

= 0.00874be x d 875 x 184

ρ min =fc'

=30

= 0.005714 x fy 4 x 240

ρ maks =

= 0.75 x (0,85 . fc'

β1600

)fy 600 + fy

= 0.75 x (0.85 x 30

0.85600

)240 600 + 240

= 0.04838


OKOK

3).


= 600 x 0.85 x 224600 + 240

= 136 mm






5).

εy = fy / Es= 240 / 200000= 0.00120


mm²

0,75 x ρ b






εc' =a

x =15.13

x 0.00120x d - a 0.85 x 224 - 15.13

= 0.000102273 < εcu' = 0.003 → aman


d = 224 mm = 0.22 m

0.22 1.53

Vu = 38550.40Vud = 33615.95 N

Vut = 0

1.0 0.751.75

Vud = Vut +x

x ( Vu - Vut )y

= 0 +1.53

x ( 38550.4 - 0 )1.75

= 33615.949 N

= Ø x 1/6 x fc' x b x d= 0.75 x 0.17 x 30 x 200 x 224= 30672.463 N

Ø.Vc/2 = 30672.463 / 2= 15336.232 N

1.75 → x = 1.75 x= 1.75 x ( 38550.4 - 15336.232 ) / 38550.4= 1.05 m ≈ 1.00 m


Vu > Ø.Vc maka gaya geser ditahan begelVu > Ø.Vc/2 NO

Ø.Vc/2 > Vu > Ø.Vc NO

Vs = ( Vu - Ø.Vc ) / Ø= ( 38550.40 - 30672.463 ) / 0.75= 10503.916 N

Vs < 1/3 x fc' x b x d= 0.33 x 30 x 200 x 224= 81793.235 N jadi syarat spasi begel s <= d/2 dan s <= 600 mm


Luas begel / m :Av,u =

Vs x S=

10503.916 x 1000= 195.3853

fy x d 240 x 224

Av,u =b x S

=200 x 1000

= 277.77783 x fy 3 x 240


=75 x 30 x 200 x 1000

1200 x fy 1200 x 240= 285.27217 mm²

εyβ1

Ø.Vc/2

Ø.Vc


karena

mm²

mm²




Spasi begel :s =


s =2 x 0.25 x 3.14 x 64 x 1000

285.272

s = 352.22504 mm




Vud > Ø.Vc maka gaya geser ditahan begelVud > Ø.Vc/2 NO

Ø.Vc/2 > Vud > Ø.Vc NO

Vs = ( Vud - Ø.Vc ) / Ø= ( 33615.95 - 30672.463 ) / 0.75= 3924.6474 N



Luas begel / m :Av,u =

Vs x S=

3924.6474 x 1000= 73.00311

fy x d 240 x 224

Av,u =b x S

=200 x 1000

= 277.77783 x fy 3 x 240


=75 x 30 x 200 x 1000

1200 x fy 1200 x 240= 285.27217 mm²


Spasi begel :s =


s =2 x 0.25 x 3.14 x 64 x 1000

285.272

s = 352.22504 mm



dp²

karena

mm²

mm²

dp²



BAB IVTANGGA

A. Penentuan Dimensi Tangga

Ø tul.pokok = 12 mmØ tul. begel = 8 mmp = 40 mmfc' = 30 Mpafy = 240 Mpaβ1 = 0.85BJ beton = 2400 kg/m³dimensi balok → b = 0.25 m 250 mm

h = 0.30 m 300 mmhf = 0.12 m 120 mm

Reduksi beban hidup = 0.3tinggi lantai 1 = 4.5 m = 4500 mmtinggi lantai 2 & 3 = 4.0 m = 4000 mmbeban hidup → beban plat tangga = 300 kg/m²

beban plat bordes = 300 kg/m²

syarat kenyamanan tanggaT + 2R = 60Dimana :T = Tead (anak tangga dalam)R = Riser (anak tangga naik)

R direncanakan = 20 cm, maka :T = 60 - 2R

= 65 - ( 2 x 20 )= 25 cm

Jumlah langkah naik

n =200

= 10 buah20

Panjang langkah datar= (n-1) x T= ( 10 - 1 ) x 25= 225

Jumlah langkah datar

n =225

= 9 buah25

Panjang miring tanggaL = ( 225 + ( 200 )²

= 301.0399 cm

Sudut kemiringan tangga

θ = arc tan20

= 38.66 °25

)²



y = 25 sin θ= 25 sin 38.66 °= 15.62 cm

= ½ . y= ½ x 15.617= 7.809 cm = 78.09 mm

Tebal plat tanggaL = 3010.399 mmh = L

x( 0.4 +fy

) =3010

x( 0.4 +240

)20 700 20 700

= 112 mm

H = h += 112 + 78.0869= 190 mm

Bordes

H = 1500 x( 0.4 +240

)700

10= 111 mm

PembebananPembebanan tangga

Berat sendiri = 0.21 x 1.4 x 2400 = 705.6 kg/mBerat tegel = 1 x 24 x 2 = 48 kg/mBeban spesi = 1 x 21 x 3 = 63 kg/mBeban sandaran kayu = 50 x 1 = 50 kg/m +

qd = 866.6 kg/mBeban hidup (ql) = 300

Beban terfaktor (qu)qu = 1.2 qd + 1.6 ql

= 1.2 x 866.6 + 1.6 x 300= 1519.92 kg/m

Pembebanan bordes* Beban mati (qd)

Berat sendiri = 0.1 x 3 x 2400 = 720 kg/mBerat tegel = 2 x 24 x 1 = 48 kg/mBeban spesi = 3 x 21 x 1 = 63 kg/m

qd = 831 kg/mBeban hidup (ql) = 300

Beban terfaktor (qu)qu = 1.2 qd + 1.6 ql

= 1.2 x 831 + 1.6 x 300= 1477.2 kg/m

Perhitungan momen dilakukan melalui program STAAD PRO 2004Untuk itu hasilnya dapat dilihat pada tabel.

Δh

∆h

kg/m2

kg/m2



Penulangan plat tangga dan plat bordesData perencanaan :fc' = 30 Mpafy = 400 Mpad = h - Selimut beton - 1/2 tulangan pokok

= 210 - 40 - 0.5 x 12= 164 mm

l min = 0.002l max =

0.75 (0.85 x 30

x 0.85 x(600

)400 600 + 400

= 0.0244

Penulangan tumpuan arah tanggaMu = 5.497 kNm = 5.497 x 10 Nmm

Rn = MuФ x b x

= 5.4970.8 x 1000 x 94

= 0.778 Mpa

= 0.75 x

= 0.75 x(0.85 x f'c

x x600

)fy 600 + fy

= 0.75 x(0.85 x 30

x 0.85 x600

)400 600 + 400

= 0.0243844

= 1.4 = 1.4 = 0.00583fy 240

= 0.85 x f'cx( 1 - 1 -

2 x Rn)

fy 0.85 x fc

= 0.85 x 30x( 1 - 1 -

2 x 0.778)

400 0.85 x 30= 0.00197< , maka dupakai = = 0.00583

As perlu = ρ x b x dx = 0.00583 x 1000 x 94= 548.333

s = = 0.25 x 3.14 x 12 x 1000As perlu 548.333

= 206.152 mms ≤ 2 h = 2 x 120 = 240 mms ≤ 450 mm

Dipilih yang terkecil, s = 206.152 mm= 200 mm

As ada = = 0.25 x 3.14 x 12 x 1000s 200.00

= 565.200 > As perlu

6

dx2

2

ρ maks ρb

β1

ρ min

ρ perlu

ρ perlu ρ min ρ perlu ρ min

mm2

1/4 . Π . Ф 2 . B 2

1/4 . Π . Ф 2 . B 2

mm2



dipakai tulangan pokok Ф 12 - 200

Tulangan bagi tumpuan xAs bagi = 20 % x As perlu

= 20 % x 548.333= 109.66667

dipilih yang terbesar, As bagi = 240s = = 0.25 x 3.14 x 8 x 1000

As perlu 240.00= 209.333 mm

s ≤ 5 h = 5 x 120 = 600 mms ≤ 450 mmdipilih yang terkecil, s = 209.440 ~ 200 mm

As ada = = 0.25 x 3.14 x 8 x 1000s 200.00

= 251.200 > Asbagidipakai tulangan bagi Ф 8 - 200

mm2

mm2

1/4 . Π . Ф 2 . B 2

1/4 . Π . Ф 2 . B 2

mm2



BAB VPEMBEBANAN PORTAL

4.1 Data PerencanaanDimensi penampangTebal plat lantai : h = 0.12 mTebal plat kantilever : h = 0.10 mDimensi balok line A : b = 0.12 m h = 1.00 mDimensi balok line B : b = 0.25 m h = 0.30 mDimensi balok line C : b = 0.25 m h = 0.30 mDimensi balok line D : b = 0.25 m h = 0.30 mDimensi balok line E : b = 0.25 m h = 0.30 m

Dimensi balok A-B : b = 0.10 m h = 0.15 mDimensi balok B-C : b = 0.20 m h = 0.25 mDimensi balok C-D : b = 0.35 m h = 0.70 mDimensi balok D-E : b = 0.35 m h = 0.70 m

Perataan beban tipe a : h = 1.167 mPerataan beban tipe b : h = 1.303 mPerataan beban tipe c : h = 1.133 mPerataan beban tipe d : h = 1.000 mPerataan beban tipe e : h = 0.486 mPerataan beban tipe f : h = 0.333 m

Jarak portal = 3.50 mDimensi kolom a = 0.50 m b = 0.60 mDimensi kolom a = 0.30 m b = 0.40 m



Gambar 4.1 Potongan Portal



4.2 Perataan BebanA. Perataan beban type A

h =1.75 m = 0.667 x 1.75

h = 1.167 m

3.50 m

B. Perataan beban type Ba b a

a = 1.75 m

1.75b = 0.5 m


4.00

Ra =( 4.00 + 0.5 ) x 1.75

x 0.52

= 1.96875


T2 = 0.5 x ( 1.75 / 2 )= 0.4375

Mmax 1 = Ra x 1/2 x L - T1 x (1/3 x a + b/2) - T2 x b/4= 1.96875 x 0.5 x 4 - 1.53125 x ( 0.333 x 1.75 + 0.5 / 2 )

- 0.4375 x 0.5 / 4= 2.60677

Mmax 2 =

= 0.125 x h x 4 ²= 2 h

Mmax1 = Mmax22.60677 = 2 h

h =2.60677

= 1.303 m2

C. Perataan beban type Ca b a

a = 1.5 m

1.5b = 0.5 m


3.50

2/3 x h

1/8 x h x L2



Ra =( 3.50 + 0.5 ) x 1.5

x 0.52

= 1.5


T2 = 0.5 x ( 1.5 / 2 )= 0.375


- 0.375 x 0.5 / 4= 1.73438

Mmax 2 == 0.125 x h x 3.5 ²= 1.53125 h

Mmax1 = Mmax21.73438 = 1.531 h

h =1.73438

= 1.133 m1.53125

D. Perataan beban type D

h =1.5 m = 0.667 x 1.5

h = 1.000 m

3.00 m

E. Perataan beban type Ea b a

a = 0.5 m

0.5b = 2.5 m


3.50

Ra =( 3.50 + 2.5 ) x 0.5

x 0.52

= 0.75


1/8 x h x L2

2/3 x h



T2 = 2.5 x ( 0.5 / 2 )= 0.625


- 0.625 x 2.5 / 4= 0.74479

Mmax 2 == 0.125 x h x 3.5 ²= 1.53125 h

Mmax1 = Mmax20.74479 = 1.531 h

h =0.74479

= 0.486 m1.53125

F. Perataan beban type F

h =0.5 m = 0.667 x 0.5

h = 0.333 m

1.00 m

4.3 Pembebanan Lantai 1 dan 3Tinggi dinding = 4.00 mBj. Beton = 24.00Berat dinding = 2.50B. Mati lantai = 4.62B. Hidup lantai = 1.00 ( untuk kantilever)B. Hidup lantai = 2.50 ( untuk bangunan kantor)

Beban merata line ABerat lantai type e = 4.62 x 0.486 = 2.247 kN/mBerat sendiri lisplank = 0.12 x ( 1.00 - 0.10 ) x 24.0 = 2.592 kN/m +

Beban mati qd = 4.839 kN/mbeban hidup :Berat lantai type e = 1.00 x 0.486 = 0.486 kN/m +

= 0.486 kN/mReduksi beban hidup = 0.30Beban hidup q1 = 0.30 x 0.486

= 0.146 kN/m

Beban merata line Bbeban mati :Berat sendiri balok = 0.25 x ( 0.30 - 0.12 ) x 24.0 = 1.08 kN/mBerat dinding = 1.00 x 4.00 x 2.50 = 10.000 kN/mBerat lantai type c = 4.62 x 1.133 = 5.233 kN/mBerat lantai type e = 4.62 x 0.486 = 2.247 kN/m +

Beban mati qd = 18.560 kN/m

1/8 x h x L2

2/3 x h

kN/m3

kN/m2

kN/m2

kN/m1

kN/m2



beban hidup :Berat lantai type c = 2.50 x 1.133 = 2.832 kN/mBerat lantai type e = 1.00 x 0.486 = 0.486 kN/m +


= 0.995 kN/m

Beban merata line Cbeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.08 kN/mBerat dinding = 1.00 x 4.00 x 2.50 = 10.000 kN/mBerat lantai type a = 4.62 x 1.167 = 5.390 kN/mBerat lantai type c = 4.62 x 1.133 = 5.233 kN/m +

Beban mati qd = 21.703 kN/mbeban hidup :Berat lantai type a = 2.50 x 1.167 = 2.917 kN/mBerat lantai type c = 2.50 x 1.133 = 2.832 kN/m +


= 1.724 kN/m

Beban merata line Dbeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.08 kN/mBerat dinding = 1.00 x 4.00 x 2.50 = 10.000 kN/mBerat lantai type a = 4.62 x 1.167 x 2 = 10.780 kN/m +

Beban mati qd = 21.860 kN/mbeban hidup :Berat lantai type a = 2.50 x 1.167 x 2 = 5.833 kN/m +


= 1.750 kN/mBeban merata line Ebeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.08 kN/mBerat dinding = 1.00 x 4.00 x 2.50 = 10.000 kN/mBerat lantai type a = 4.62 x 1.167 = 5.390 kN/m +

Beban mati qd = 16.470 kN/mbeban hidup :Berat lantai type a = 2.50 x 1.167 = 2.917 kN/m +


= 0.875 kN/m

Beban joint pada ABeban mati titik A = 3.50 x 4.839 = 16.94 kN



Beban hidup titik A = 3.50 x 0.146 = 0.51 kN

Beban joint pada BBeban mati titikB = 3.50 x 18.560 = 64.96 kNBeban hidup titik B = 3.50 x 0.995 = 3.48 kN

Beban joint pada CBeban mati titik C = 3.50 x 21.703 = 75.96 kNBeban hidup titik C = 3.50 x 1.724 = 6.04 kN

Beban joint pada DBeban mati titik D = 3.50 x 21.860 = 76.51 kNBeban hidup titik D = 3.50 x 1.750 = 6.12 kN

Beban joint pada EBeban mati titik E = 3.50 x 16.470 = 57.64 kNBeban hidup titik E = 3.50 x 0.875 = 3.06 kN

Beban Yang Bekerja Pada Portal Line 4Beban merata line 4 bagian A-BBeban matiBerat sendiri balok = 0.10 x ( 0.15 - 0.10 ) x 24.00 = 0.12 kN/mBerat lantai tipe f = 0.333 x 4.62 x 2 = 3.08 kN/m

Beban mati qd = 3.20 kN/mBeban hidupberat lantai tipe f = 0.333 x 1.00 x 2 = 0.667 kN/m


= 0.20 kN/m

Beban merata line 4 bagian B-CBeban matiBerat sendiri balok = 0.2 x( 0.25 - 0.12 )x 24.0 = 0.624 kN/mBerat lantai tipe d = 1.000 x 4.62 x 2 = 9.24 kN/m

Beban mati qd = 9.86 kN/mBeban hidupBerat lantai tipe d = 1.000 x 2.50 x 2 = 5.000 kN/m

= 5.000 kN/mReduksi beban hidup = 0.3Beban hidup q1 = 0.3 x 5

= 1.50 kN/m

Beban merata line 4 bagian C-D = D-EBeban matiBerat sendiri balok = 0.35 x( 0.70 - 0.12 )x 24.0 = 4.872 kN/mBerat lantai tipe b = 1.303 x 4.62 x 2 = 12.0433 kN/m

Beban mati qd = 16.92 kN/mBeban hidupBerat lantai tipe b = 1.303 x 2.50 x 2 = 6.517 kN/m




= 1.955 kN/m

4.4 Pembebanan Lantai 2Tinggi dinding = 4.00 mBj. Beton = 24.00Berat dinding = 2.50B. Mati lantai = 4.62B. Hidup lantai = 1.00 ( untuk kantilever)B. Hidup lantai = 4.00 ( untuk aula)


Beban mati qd = 4.839 kN/mbeban hidup :Berat lantai type e = 1.00 x 0.486 = 0.486 kN/m


= 0.146 kN/m

Beban merata line Bbeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.08 kN/mBerat dinding = 1.00 x 4.00 x 2.50 = 10.000 kN/mBerat lantai type c = 4.62 x 1.133 = 5.233 kN/mBerat lantai type e = 4.62 x 0.486 = 2.247 kN/m +

Beban mati qd = 18.560 kN/mbeban hidup :Berat lantai type c = 4.00 x 1.133 = 4.531 kN/mBerat lantai type e = 1.00 x 0.486 = 0.486 kN/m +


= 1.505 kN/m

Beban merata line Cbeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.08 kN/mBerat dinding = 1.00 x 4.00 x 2.50 = 10.000 kN/mBerat lantai type a = 4.62 x 1.167 = 5.390 kN/mBerat lantai type c = 4.62 x 1.133 = 5.233 kN/m +


kN/m3

kN/m2

kN/m2

kN/m1

kN/m2



beban hidup :Berat lantai type a = 4.00 x 1.167 = 4.667 kN/mBerat lantai type c = 4.00 x 1.133 = 4.531 kN/m +


= 2.759 kN/mBeban merata line Dbeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.08 kN/mBerat dinding = 1.00 x 4.00 x 2.50 = 10.000 kN/mBerat lantai type a = 4.62 x 1.167 x 2 = 10.780 kN/m +

Beban mati qd = 21.860 kN/mbeban hidup :Berat lantai type a = 4.00 x 1.167 x 2 = 9.333 kN/m +


= 2.80 kN/mBeban merata line Ebeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.08 kN/mBerat dinding = 1.00 x 4.00 x 2.50 = 10.000 kN/mBerat lantai type a = 4.62 x 1.167 = 5.390 kN/m +



= 1.400 kN/m

Beban joint pada ABeban mati titik A = 3.50 x 4.839 = 16.94 kNBeban hidup titik A = 3.50 x 0.146 = 0.51 kN








Beban mati qd = 3.20 kN/mBeban hidupBerat lantai tipe f = 0.333 x 1.00 x 2 = 0.667 kN/m


= 0.20 kN/m




= 2.40 kN/m




= 3.128 kN/m



4.5 Pembebanan Plat AtapBj. Beton = 24.00Berat dinding = 2.50B. Plat Mati = 3.37B. Hidup lantai = 1.00


Beban mati qd = 4.229 kN/mbeban hidup :Berat lantai type e = 1.00 x 0.486 = 0.486 kN/m


= 0.146 kN/m

Beban merata line Bbeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.08 kN/mBerat lantai type c = 3.37 x 1.133 = 3.81 kN/mBerat lantai type e = 3.37 x 0.486 = 1.64 kN/m +

Beban mati qd = 6.530 kN/mbeban hidup :Berat lantai type c = 1.00 x 1.133 = 1.133 kN/mBerat lantai type e = 1.00 x 0.486 = 0.486 kN/m +


= 0.486 kN/m

Beban merata line Cbeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.080 kN/mBerat lantai type a = 3.37 x 1.167 = 3.93 kN/mBerat lantai type c = 3.37 x 1.133 = 3.813 kN/m +

Beban mati qd = 8.820 kN/mbeban hidup :Berat lantai type a = 1.00 x 1.167 = 1.17 kN/mBerat lantai type c = 1.00 x 1.133 = 1.133 kN/m +


= 0.690 kN/m

Beban merata line Dbeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.080 kN/mBerat lantai type a = 3.37 x 1.167 x 2 = 7.854 kN/m +


kN/m3

kN/m2

kN/m2

kN/m2



beban hidup :Berat lantai type a = 1.00 x 1.167 x 2 = 2.333 kN/m +


= 0.700 kN/m

Beban merata line Ebeban mati :Berat sendiri balok = 0.250 x ( 0.300 - 0.120 ) x 24.0 = 1.080 kN/mBerat lantai type a = 3.37 x 1.167 = 3.927 kN/m +



= 0.350 kN/m

Beban joint pada ABeban mati titik A = 3.50 x 4.229 = 14.80 kNBeban hidup titik A = 3.50 x 0.146 = 0.51 kN






Beban mati qd = 2.316 kN/mBeban hidupBerat lantai tipe f = 0.333 x 1.00 x 2 = 0.667 kN/m




= 0.20 kN/m




= 0.60 kN/m




= 0.782 kN/m



4.6 Beban GempaBeban mati atapBerat lantai = 12 x 31.5 x 3.37 = 1272.35 kNBerat kolom = 0.50 x 0.60 x 2 x 20 x 24 = 288.00 kNBerat kolom = 0.30 x 0.40 x 2 x 10 x 24 = 57.60 kNBerat (kolom praktis) = 0.15 x 0.15 x 2 x 11 x 24 = 11.88 kNBerat lisplank (Balok A) = 0.12 x 1.0 x 31.5 x 24 = 90.72 kNBerat balok B+C+D+E = 0.25 x 0.30 x 3.5 x 36 x 24 = 226.80 kNBerat balok A-B = 0.10 x 0.15 x 1 x 10 x 24 = 3.60 kNBerat balok B-C = 0.20 x 0.25 x 3 x 10 x 24 = 36.00 kNBerat balok C-D = 0.35 x 0.70 x 4 x 10 x 24 = 235.20 kNBerat balok D-E = 0.35 x 0.70 x 4 x 10 x 24 = 235.20 kNBerat dinding line B Lt 3 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line C Lt 3 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line D Lt 3 = 3.5 x 2 x 2.5 x 6 = 105.00 kNBerat dinding line E Lt 3 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line 1&10 Lt 3 = 11 x 2 x 2.5 x 2 = 110.00 kNBerat dinding line 2+3+5+6+8 Lt 3 = 7 x 2 x 2.5 x 5 = 175.00 kNBerat dinding line 4+7+9 Lt 3 = 3 x 2 x 2.5 x 3 = 45.00 kN

WD atap = 3364.85 kN

Beban hidupBeban hidup atap = 12 x 31.5 x 1.00 = 378 kNReduksi beban hidup = 0.3Beban hidup atap WL atap = 113.4 kN

Total beban atap Watap = 3364.85 + 113.4 = 3478.25 kN

Beban mati lantai 3Berat lantai = 12 x 31.5 x 4.620 = 1746.36 kNBerat kolom = 0.50 x 0.60 x 4 x 20 x 24 = 576.00 kNBerat kolom = 0.30 x 0.40 x 4 x 10 x 24 = 115.20 kNBerat (kolom praktis) = 0.15 x 0.15 x 4 x 11 x 24 = 23.76 kNBerat lisplank (Balok A) = 0.12 x 1.0 x 31.5 x 24 = 90.72 kNBerat balok B+C+D+E = 0.25 x 0.30 x 3.5 x 36 x 24 = 226.80 kNBerat balok A-B = 0.10 x 0.15 x 1 x 10 x 24 = 3.60 kNBerat balok B-C = 0.20 x 0.25 x 3 x 10 x 24 = 36.00 kNBerat balok C-D = 0.35 x 0.70 x 4 x 10 x 24 = 235.20 kNBerat balok D-E = 0.35 x 0.70 x 4 x 10 x 24 = 235.20 kNBerat dinding line B Lt 3 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line B Lt 2 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line C Lt 3 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line C Lt 2 = 3.5 x 2 x 2.5 x 5 = 87.50 kNBerat dinding line D Lt 3 = 3.5 x 2 x 2.5 x 6 = 105.00 kNBerat dinding line D Lt 2 = 3.5 x 2 x 2.5 x 6 = 105.00 kNBerat dinding line E Lt 3 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line E Lt 2 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line 1&10 Lt 3 = 11 x 2 x 2.5 x 2 = 110.00 kNBerat dinding line 1,2&10 Lt 2 = 11 x 2 x 2.5 x 3 = 165.00 kNBerat dinding line 2+3+5+6+8 Lt 3 = 7 x 2 x 2.5 x 5 = 175.00 kNBerat dinding line 4+7+9 Lt 3 = 3 x 2 x 2.5 x 3 = 45.00 kN



Berat dinding line 3+5 Lt 2 = 4 x 2 x 2.5 x 2 = 40.00 kNBerat dinding line 6+7+8+9 Lt 2 = 7 x 2 x 2.5 x 4 = 140.00 kN

WD lt.3 = 5048.84 kN

Beban hidupBeban hidup lt.3 = 12 x 31.5 x 2.50 = 945 kNReduksi beban hidup = 0.3Beban hidup lt.3 W Lt 3 = 283.5 kN

Total beban lt 3 W lt3 = 5048.84 + 283.5 = 5332.34 kN

Beban mati lantai 2Berat lantai = 12 x 31.5 x 4.620 = 1746.36 kNBerat kolom = 0.50 x 0.60 x 4 x 20 x 24 = 576.00 kNBerat kolom = 0.30 x 0.40 x 4 x 10 x 24 = 115.20 kNBerat (kolom praktis) = 0.15 x 0.15 x 4 x 11 x 24 = 23.76 kNBerat lisplank (Balok A) = 0.12 x 1.0 x 31.5 x 24 = 90.72 kNBerat balok B+C+D+E = 0.25 x 0.30 x 3.5 x 36 x 24 = 226.80 kNBerat balok A-B = 0.10 x 0.15 x 1 x 10 x 24 = 3.60 kNBerat balok B-C = 0.20 x 0.25 x 3 x 10 x 24 = 36.00 kNBerat balok C-D = 0.35 x 0.70 x 4 x 10 x 24 = 235.20 kNBerat balok D-E = 0.35 x 0.70 x 4 x 10 x 24 = 235.20 kNBerat dinding line B Lt 2 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line B Lt 1 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line C Lt 2 = 3.5 x 2 x 2.5 x 5 = 87.50 kNBerat dinding line C Lt 1 = 3.5 x 2 x 2.5 x 7 = 122.50 kNBerat dinding line D Lt 2 = 3.5 x 2 x 2.5 x 6 = 105.00 kNBerat dinding line D Lt 1 = 3.5 x 2 x 2.5 x 6 = 105.00 kNBerat dinding line E Lt 2 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line E Lt 1 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line 1,2&10 Lt 2 = 11 x 2 x 2.5 x 3 = 165.00 kNBerat dinding line 3&5 Lt 2 = 4 x 2 x 2.5 x 2 = 40.00 kNBerat dinding line 6+7+8+9 Lt 2 = 7 x 2 x 2.5 x 4 = 140.00 kNBerat dinding line 1,2&10 Lt 1 = 11 x 2 x 2.5 x 3 = 165.00 kNBerat dinding line 3&5 Lt 1 = 8 x 2 x 2.5 x 2 = 80.00 kNBerat dinding line 6+7+8+9 Lt 1 = 7 x 2 x 2.5 x 4 = 140.00 kN

WD lt.2 = 5068.84 kN

Beban hidupBeban hidup lt.2 = 12 x 31.5 x 4.00 = 1512 kNReduksi beban hidup = 0.3Beban hidup lt.2 W lt.2 = 453.6 kN


Beban mati lantai 1Berat lantai = 12 x 31.50 x 4.6 = 1746.36 kNBerat kolom = 0.50 x 0.60 x 4.0 x 20 x 24 = 576.00 kNBerat kolom = 0.30 x 0.40 x 4.0 x 10 x 24 = 115.20 kNBerat (kolom praktis) = 0.15 x 0.15 x 4.0 x 11 x 24 = 23.76 kNBerat lisplank (Balok A) = 0.12 x 1.00 x 31.5 x 24 = 90.72 kN



Berat balok B+C+D+E = 0.25 x 0.30 x 3.5 x 36 x 24 = 226.80 kNBerat balok A-B = 0.10 x 0.15 x 1.0 x 10 x 24 = 3.60 kNBerat balok B-C = 0.20 x 0.25 x 3.0 x 10 x 24 = 36.00 kNBerat balok C-D = 0.35 x 0.70 x 4.0 x 10 x 24 = 235.20 kNBerat balok D-E = 0.35 x 0.70 x 4.0 x 10 x 24 = 235.20 kNBerat dinding line B Lt 1 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line B Lt 0 = 3.5 x 2.25 x 2.5 x 9 = 177.19 kNBerat dinding line C Lt 1 = 3.5 x 2 x 2.5 x 7 = 122.50 kNBerat dinding line C Lt 0 = 3.5 x 2.25 x 2.5 x 6 = 118.13 kNBerat dinding line D Lt 1 = 3.5 x 2 x 2.5 x 6 = 105.00 kNBerat dinding line D Lt 0 = 3.5 x 2.25 x 2.5 x 2 = 39.38 kNBerat dinding line E Lt 1 = 3.5 x 2 x 2.5 x 9 = 157.50 kNBerat dinding line E Lt 0 = 3.5 x 2.25 x 2.5 x 9 = 177.19 kNBerat dinding line 1,2&10 Lt 1 = 11 x 2 x 2.5 x 3 = 165.00 kNBerat dinding line 3&5 Lt 1 = 8 x 2 x 2.5 x 2 = 80.00 kNBerat dinding line 6+7+8+9 Lt 1 = 7 x 2 x 2.5 x 4 = 140.00 kNBerat dinding line 1,2&10 Lt 0 = 11 x 2.25 x 2.5 x 3 = 185.63 kNBerat dinding line 3+5+6 Lt 0 = 8 x 2.25 x 2.5 x 3 = 135.00 kNBerat dinding line 7+8+9 Lt 0 = 3 x 2.25 x 2.5 x 3 = 50.63 kN

WD lt.1 = 5099.47 kN

Beban hidupBeban hidup lt.1 = 12 x 31.50 x 2.50 = 945 kNReduksi beban hidup = 0.3Beban hidup lt.1 W lt.1 = 283.5 kN


Wt = W atap + W3 + W2 + W1= 3478.25 + 5332.34 + 5522.44 + 5382.97= 19715.993 kN

Waktu getar bangunan (T)gengan rumus empiris :

T =

H = 16.50 mT = 0.060 x 16.5

= 0.491 dtk

Kontrol pembatasan waktu getar alami fundamental, T sesuai pasal 5.6

Dimana koefisien ditetapkan menurut tabel 8Koefisien yang membatasi waktu getar alami fundamental struktur gedung:

Wilayah Gempa ζ1 0.22 0.193 0.184 0.175 0.166 0.15

0,06 x H3/4

3/4

Syarat T1 < ζ. n



Dari tabel diatas NTT termasuk wilayah Gempa 5 maka:

ζ = 0.16 ( BUKU STANDAR PERENCANAAN KETAHANAN GEMPA UNTUK STRUKTUR BANGUNAN GEDUNG )

n = 3 (tingkat)

T = ζ x n= 0.16 x 3= 0.48 dtk < T empiris = 0.491 dtk

Maka dipakai T = 0.48 dtk

Koefisien gempa dasar untuk Wilayah Gempa 5 dan untuk tanah sedangDari gambar didapat nilai :

Dari gambar di dapat nilai C = 0.83 (SNI Pasal 4.7.6)

Faktor keutamaan I dan faktor reduksi gempa RI = 1.0R = 3.2

Gaya geser horizontal total akibat gempav = C x I

x WtR

= 0.830 x 1.0x 19715.99

3.2= 5113.84 kN

C

TGambar Respons Spectrum Gempa Rencana

Wilayah Gempa 5

C= 0.9/T (Tanah Lunak)

C= 0.4/T (Tanah sedang)

C= 0.35/T (Tanah keras)



Distribusi gaya geser horizontal total akibat gempa ke sepanjang tinggi gedunga. Arah X

H/B = 16.50 / 11 = 1.50 < 3b. Arah Y

H/A = 16.50 / 31.5 = 0.524 < 3

maka beban gempa nominal pada tiap lantai dihitung dengan rumus berikut :

dimana :Fi = gaya geser horizontalWi = berat lantai tingkat ke iZi = ketinggian lantai tingkat ke iV = gaya geser horizontal total akibat gempa

=3478.248 x 16.50

x 5113.83568192517.942

= 1524.474 kN

Perhitungan yang lain ditabelkan sbb :

TingkatZi Wi Wi x Zi Fix,y Untuk tiap portal

(m) (kN) (kNm) kN 1/8 Fix (kN) 1/3 Fiy (kN)Atap 16.50 3478.248 57391.092 1524.474 190.559 63.520

3 12.50 5332.340 66654.250 1770.530 221.316 73.7722 8.50 5522.440 46940.740 1246.882 155.860 51.9531 4.00 5382.965 21531.860 571.949 71.494 23.831

192517.942 5113.836 639.229 213.076

Sehingga didapat gaya gempa untuk arah X adalah :E = 1/8 Fi x

E atap = 190.56 kNE 3 = 221.32 kNE 2 = 155.86 kNE 1 = 71.49 kN

dan gaya gempa untuk arah Y adalah :E = 1/3 Fi y

E atap = 63.52 kNE 3 = 73.77 kNE 2 = 51.95 kNE 1 = 23.83 kN

Σwi . Zi

xVWixZi

WixZiFi

xVWixZi

WixZiFi



BAB VIPORTAL MELINTANG

BALOK INDUK 35/70 ( LINE 4 )

A. Data Perencanaan


h = 0.70 m 700 mmhf = 0.12 m 120 mm

fungsi lantai 1 & 3 = kantorfungsi lantai 2 = aulaReduksi beban hidup = 0.3Bentang L = 8 mtinggi lantai 1 = 4.0 m = 4000 mmtinggi lantai 2 & 3 = 4.0 m = 4000 mm

E. Perhitungan Penulangan Balok Induk 1. Perhitungan tulangan tumpuan : Balok 3 Joint 7-8 (35/70)A. Hitung Tulangan


ds = Ø + Sn ( jarak minimum tulangan )= 26 + 25= 51 mm ≈ 51 mm

ds' = 71 + ( 2 x 51 ) / 2= 122 mm

dd = 71 + ( 51 / 2 )= 97 mm

d = h - (ds1+ds2+ds/2)= 700 - 122.0= 578.0 mm

7151

70051

578.05171

350




= 1177760000 / 0.8= 1472200000 Nmm

K =Mu

=1177760000.00

= 12.591 Mpa0.8 x 350 x 578.0 ²

Kmaks =

= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 )

= 7.8883 Mpa

Karena K > Kmaks maka dipakai tulangan rangkap (K=0,8. Kmaks)

Diambil = 0.80 x Kmaks = 0.80 x 7.8883= 6.3107 Mpa


a = 1 -1 -

2 xx d

0.85 x fc'

= 1 -1 -

2 x 6.311x 578.0

0.85 x 30

= 167.24 mm

a maks leleh =600 x β1 x d

600 + fy

=600 x 0.85 x 578.0

600 + 400

= 294.78 mm


A1 = 0.85 x fc x a x bfy

= 0.85 x 30 x 167.24 x 350400

= 3731.44

Ø x b x d²

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²

K1

K1

mm²



A2 = ( K - K1 ) x b x( d - ds' ) x fy

= ( 12.591 - 6.311 ) x 350 x 578.0 ²( 578.00 - 122.00 ) x 400

= 4025.75 mm²

Asu = A1 + A2 = 3731.44 + 4025.75 = 7757.19 mm²

As'u = A2 = 4025.75 mm²

→ Hitung jumlah tulanganTulangan Tarik Asu

n =Asu

=7757.19

= 14.62 ≈ 15 buah1/4 . π . Ø² ¼ x π x 26 ²

Digunakan tulangan tarik As = 15 Ø 26 = 7959.900 > Asu OK

Tulangan Tekan As'u

n =As'u

=4025.75

= 7.59 ≈ 8 buah1/4 . π . Ø² ¼ x π x 26 ²

Digunakan tulangan tarik As' = 8 Ø 26 = 4245.280 > Asu OK


m =b - 2 x ds1

+ 1Ø + Sn

=350 - 2 x 71

+ 126 + 25


7151

70051

578.05171

350

d²

mm²

mm²



B. Hitung Momen Rencana

1).

As = 15 Ø 26 = 15 x (1/4) x π x 26 ² = 7959.900As' = 8 Ø 26 = 8 x (1/4) x π x 26 ² = 4245.280

ρ =(As - As')

=3714.62

= 0.01836be x d 350 x 578.0

ρ maks =

= 0.75 x (0,85 . fc'

β1600

)fy 600 + fy

= 0.75 x (0.85 x 30

0.85600

)400 600 + 400

= 0.02438

Syarat = ρ < ρ maks OK

2). Hitung tinggi blok tegangan beton tekan persegi ekuivalen a

a =(As - As') x fy

0.85 x fc' x b

=3714.62 x 400

= 166.48 mm0.85 x 30 x 350

3).

a min leleh =600 x β1 x dd

600 + fy

=600 x 0.85 x 97

600 - 400

= 246.08 mm

a < a min leleh Tulangan Tekan sudah Leleh (nilai a sudah betul)


Mnc = 0.85 x fc x a x b x (d-a/2)= 0.85 x 30 x 166.48 x 350 x 494.76= 735136991.11 Nmm

Mns = As' x fy x ( d - ds)= 4245.28 x 400 x 507.00= 860942784.00 Nmm

Mn = Mnc + Mns= 1596079775.11 Nmm

Kontrol rasio tulangan ρ ≤ ρ maks

mm²mm²

0,75 x ρ b

Kontrol semua tulangan tarik harus sudah leleh ( a ≤ a min leleh )




5).

εy = fy / Es= 400 / 200000= 0.00200

εc' =a

x =166.48

x 0.00200x d - a 0.85 x 578.00 - 166.48

= 0.00095189 < εcu' = 0.003 → aman


ds1 = p + Ø tul. begel + (½ x Ø tul. pokok)= 50 + 8 +( ½ x 26 )= 71.0 mm

ds = Ø + Sn ( jarak minimum tulangan )= 26 + 25= 51.0 mm ≈ 51 mm

d = h - (ds1+ds/2)= 700 - 96.5= 603.5 mm

→- 8000 mm- 7400 mm

≤ 1x λ =

1x 8000 = 2000 mm

4 4

≤ 16 x hf + b = 16 x 120 + 350 = 2270 mm

≤ λn x b = 7400 x 350 = 2590000 mm


2000

700603.5

5171

350


εyβ1


be

be

be

( > Ø =




= 203314000 / 0.8= 254142500 Nmm

K =Mu

=203314000.00

= 0.349 Mpa0.8 x 2000 x 603.5 ²

Kmaks =

= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 ) ²

= 7.8883 Mpa



a = 1 -1 -

2 x Kx d

0.85 x fc'

= 1 -1 -

2 x 0.349x 603.5

0.85 x 30

= 8.31 mm

karena a < hf maka beton tekan berada di sayap atau balok T palsu (hitung luas As,u )dihitung sebagai balok persegi panjang dengan lebar balok = be# Hitung luas tulangan perlu (As,u)


= 0.85 x 30 x 8.31 x 2000400

= 1060.09

Amin =1.4 x be x d

=1.4 x 2000 x 604

= 4224.50 mm²fy 400

Amin =fc' x be x d

=30 x 2000 x 604

= 4131.88 mm²4 x fy 4 x 400

= 4224.50 mm²


n =As,u

=4224.500

= 7.96 ≈ 8 buah1/4 . π . Ø² ¼ x π x 26 ²

Ø x be x d²

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²

Asu

mm²






m =b - 2 x d'

+ 1Ø + Sn

=350 - 2 x 71

+ 126 + 25


Karena n > m maka tulangan lebih dari 1 baris ds2 = ds1 + ( ds / 2 )

= 71 + ( 26 ) = 97 mm

jadi d = h - ds2= 700 - 97 = 603.5 mm

2000

603.5700

5171

350


a =As x fy

0.85 x fc' x be

=4245.28 x 400

= 33.30 mm0.85 x 30 x 2000


2).

As = 8 Ø 26 = 8 x (1/4) x π x 26 ² = 4245.280

ρ =As

=4245.28

= 0.00352be x d 2000 x 604

ρ min =fc'

=30

= 0.003424 x fy 4 x 400

mm²mm²


mm²



ρ maks =

= 0.75 x (0,85 . fc'

β1600

)fy 600 + fy

= 0.75 x (0.85 x 30

0.85600

)400 600 + 400

= 0.02438


OKOK

3).


= 600 x 0.85 x 604600 + 400

= 307.79 mm






5).εy = fy / Es

= 400 / 200000= 0.00200

εc' =a

x =33.30

x 0.00200x d - a 0.85 x 604 - 33.30

= 0.000137397 < εcu' = 0.003 → aman

0,75 x ρ b




εyβ1




d = 578.0 mm = 0.58 m

0.58 3.42

Vu = 387595Vud = 331587.5 N

Vut = 0

3.20 0.804.00

Vud = Vut +x

x ( Vu - Vut )y

= 0 +3.42

x ( 387595 - 0 )4.00

= 331587.52 N

= Ø x 1/6 x fc' x b x d= 0.75 x 0.17 x 30 x 350 x 578= 138505.34 N

Ø.Vc/2 = 138505.34 / 2= 69252.671 N

4.00 → x = 4.00 x= 4.00 x ( 387595 - 69252.671 ) / 387595= 3.29 m ≈ 3.20 m




Av,u =b x S

=350 x 1000

= 486.11113 x fy 3 x 240


=75 x 30 x 350 x 1000

1200 x fy 1200 x 240= 499.22629 mm²


Spasi begel :s =


s =2 x 0.25 x 3.14 x 64 x 1000

2394.172

s = 41.96858 mm

Ø.Vc/2

Ø.Vc


karena

mm²

dp²




Syarat spasi s = d / 2 = 578 / 2 = 289 mmDipilih spasi yang kecil , yaitu s = 41.969 mm ≈ 40 mmJadi dipakai begel Ø 8 - 40 mm




Luas begel / m :

Av,u =b x S

=350 x 1000

= 486.11113 x fy 3 x 240


=75 x 30 x 350 x 1000

1200 x fy 1200 x 240= 499.22629 mm²


Spasi begel :s =


s =2 x 0.25 x 3.14 x 64 x 1000

499.226

s = 201.27145 mm


Syarat spasi s = d / 2 = 578 / 2 = 289 mmDipilih spasi yang kecil , yaitu s = 201.27 mm ≈ 200 mmJadi dipakai begel Ø 8 - 200 mm

karena

mm²

dp²



PORTAL MELINTANGBALOK INDUK 25/30 ( LINE 4 )

A. Data Perencanaan


h = 0.30 m 300 mmhf = 0.12 m 120 mm

fungsi lantai 1 & 3 = kantorfungsi lantai 2 = aulaReduksi beban hidup = 0.3Bentang L = 3 mtinggi lantai 1 = 4.0 m = 4000 mmtinggi lantai 2 & 3 = 4.0 m = 4000 mm

E. Perhitungan Penulangan Balok Induk 1. Perhitungan tulangan tumpuan : Balok 2 Joint 6-7 (25/30)A. Hitung Tulangan


ds = Ø + Sn ( jarak minimum tulangan )= 26 + 25= 51 mm ≈ 51 mm

ds' = 71 + ( 1 x 51 ) / 2= 96.5 mm

dd = 71 mm

d = h - (ds1+ds2+ds/2)= 300 - 96.5= 203.5 mm

7151

300

203.5

71

250




= 82330000 / 0.8= 102912500 Nmm

K =Mu

=82330000.00

= 9.940 Mpa0.8 x 250 x 203.5 ²

Kmaks =

= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 )

= 7.8883 Mpa

Karena K > Kmaks maka dipakai tulangan rangkap (K=0,8. Kmaks)

Diambil = 0.80 x Kmaks = 0.80 x 7.8883= 6.3107 Mpa


a = 1 -1 -

2 xx d

0.85 x fc'

= 1 -1 -

2 x 6.311x 203.5

0.85 x 30

= 58.88 mm

a maks leleh =600 x β1 x d

600 + fy

=600 x 0.85 x 203.5

600 + 400

= 103.79 mm


A1 = 0.85 x fc x a x bfy

= 0.85 x 30 x 58.88 x 250400

= 938.39

Ø x b x d²

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²

K1

K1

mm²



A2 = ( K - K1 ) x b x( d - ds' ) x fy

= ( 9.940 - 6.311 ) x 250 x 203.5 ²( 203.50 - 96.50 ) x 400

= 877.98 mm²

Asu = A1 + A2 = 938.39 + 877.98 = 1816.38 mm²

As'u = A2 = 877.98 mm²

→ Hitung jumlah tulanganTulangan Tarik Asu

n =Asu

=1816.38

= 3.42 ≈ 4 buah1/4 . π . Ø² ¼ x π x 26 ²

Digunakan tulangan tarik As = 4 Ø 26 = 2122.640 > Asu OK

Tulangan Tekan As'u

n =As'u

=877.98

= 1.65 ≈ 2 buah1/4 . π . Ø² ¼ x π x 26 ²

Digunakan tulangan tarik As' = 2 Ø 26 = 1061.320 > Asu OK


m =b - 2 x ds1

+ 1Ø + Sn

=250 - 2 x 71

+ 126 + 25


7151

300

203.5

71

250

d²

mm²

mm²



B. Hitung Momen Rencana

1).

As = 4 Ø 26 = 4 x (1/4) x π x 26 ² = 2122.640As' = 2 Ø 26 = 2 x (1/4) x π x 26 ² = 1061.320

ρ =(As - As')

=1061.32

= 0.02086be x d 250 x 203.5

ρ maks =

= 0.75 x (0,85 . fc'

β1600

)fy 600 + fy

= 0.75 x (0.85 x 30

0.85600

)400 600 + 400

= 0.02438

Syarat = ρ < ρ maks OK

2). Hitung tinggi blok tegangan beton tekan persegi ekuivalen a

a =(As - As') x fy

0.85 x fc' x b

=1061.32 x 400

= 66.59 mm0.85 x 30 x 250

3).

a min leleh =600 x β1 x dd

600 + fy

=600 x 0.85 x 71

600 - 400

= 181.05 mm

a < a min leleh Tulangan Tekan sudah Leleh (nilai a sudah betul)


Mnc = 0.85 x fc x a x b x (d-a/2)= 0.85 x 30 x 66.59 x 250 x 170.20= 72256230.53 Nmm

Mns = As' x fy x ( d - ds)= 1061.32 x 400 x 132.50= 56249960.00 Nmm

Mn = Mnc + Mns= 128506190.53 Nmm

Kontrol rasio tulangan ρ ≤ ρ maks

mm²mm²

0,75 x ρ b

Kontrol semua tulangan tarik harus sudah leleh ( a ≤ a min leleh )




5).

εy = fy / Es= 400 / 200000= 0.00200

εc' =a

x =66.59

x 0.00200x d - a 0.85 x 203.50 - 66.59

= 0.00114449 < εcu' = 0.003 → aman


ds1 = p + Ø tul. begel + (½ x Ø tul. pokok)= 50 + 8 +( ½ x 26 )= 71.0 mm

ds = Ø + Sn ( jarak minimum tulangan )= 26 + 25= 51.0 mm ≈ 51 mm

d = h - (ds1+ds/2)= 300 - 71.0= 229.0 mm

→- 3000 mm- 2550 mm

≤ 1x λ =

1x 3000 = 750 mm

4 4

≤ 16 x hf + b = 16 x 120 + 250 = 2170 mm

≤ λn x b = 2550 x 250 = 637500 mm


750

300229.0

5171

250


εyβ1


be

be

be

( > Ø =




= 4005000 / 0.8= 5006250 Nmm

K =Mu

=4005000.00

= 0.127 Mpa0.8 x 750 x 229.0 ²

Kmaks =

= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 ) ²

= 7.8883 Mpa



a = 1 -1 -

2 x Kx d

0.85 x fc'

= 1 -1 -

2 x 0.127x 229.0

0.85 x 30

= 1.15 mm




= 0.85 x 30 x 1.15 x 750400

= 54.79

Amin =1.4 x be x d

=1.4 x 750 x 229

= 601.13 mm²fy 400

Amin =fc' x be x d

=30 x 750 x 229

= 587.95 mm²4 x fy 4 x 400

= 601.13 mm²


n =As,u

=601.125

= 1.13 ≈ 2 buah1/4 . π . Ø² ¼ x π x 26 ²

Ø x be x d²

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²

Asu

mm²






m =b - 2 x d'

+ 1Ø + Sn

=250 - 2 x 71

+ 126 + 25


Karena n < m maka tulangan 1 barisds2 = ds1 + ( ds / 2 )

= 71 + ( 26 ) = 97 mm

jadi d = h - ds2= 300 - 97 = 203.5 mm

750

203.5300

71

250


a =As x fy

0.85 x fc' x be

=1061.32 x 400

= 22.20 mm0.85 x 30 x 750


2).

As = 2 Ø 26 = 2 x (1/4) x π x 26 ² = 1061.320

ρ =As

=1061.32

= 0.00695be x d 750 x 204

ρ min =fc'

=30

= 0.003424 x fy 4 x 400

mm²mm²


mm²



ρ maks =

= 0.75 x (0,85 . fc'

β1600

)fy 600 + fy

= 0.75 x (0.85 x 30

0.85600

)400 600 + 400

= 0.02438


OKOK

3).


= 600 x 0.85 x 204600 + 400

= 103.79 mm






5).εy = fy / Es

= 400 / 200000= 0.00200

εc' =a

x =22.20

x 0.00200x d - a 0.85 x 204 - 22.20

= 0.000288079 < εcu' = 0.003 → aman

0,75 x ρ b




εyβ1




d = 203.5 mm = 0.20 m

0.20 1.30

Vu = 66715Vud = 57664 N

Vut = 0

1.10 0.401.50

Vud = Vut +x

x ( Vu - Vut )y

= 0 +1.30

x ( 66715 - 0 )1.50

= 57663.998 N

= Ø x 1/6 x fc' x b x d= 0.75 x 0.17 x 30 x 250 x 204= 34831.731 N

Ø.Vc/2 = 34831.731 / 2= 17415.866 N

1.50 → x = 1.50 x= 1.50 x ( 66715 - 17415.866 ) / 66715= 1.11 m ≈ 1.10 m




Av,u =b x S

=250 x 1000

= 347.22223 x fy 3 x 240


=75 x 30 x 250 x 1000

1200 x fy 1200 x 240= 356.59021 mm²


Spasi begel :s =


s =2 x 0.25 x 3.14 x 64 x 1000

870.414

s = 115.43931 mm

Ø.Vc/2

Ø.Vc


karena

mm²

dp²




Syarat spasi s = d / 2 = 204 / 2 = 101.75 mmDipilih spasi yang kecil , yaitu s = 101.75 mm ≈ 100 mmJadi dipakai begel Ø 8 - 100 mm




Luas begel / m :

Av,u =b x S

=250 x 1000

= 347.22223 x fy 3 x 240


=75 x 30 x 250 x 1000

1200 x fy 1200 x 240= 356.59021 mm²


Spasi begel :s =


s =2 x 0.25 x 3.14 x 64 x 1000

356.590

s = 281.78003 mm


Syarat spasi s = d / 2 = 204 / 2 = 101.75 mmDipilih spasi yang kecil , yaitu s = 101.75 mm ≈ 100 mmJadi dipakai begel Ø 8 - 100 mm

karena

mm²

dp²


HENDRICH DENY . T (12.21.916) 91

BAB VIIPERHITUNGAN DIAGRAM INTERAKSI KOLOM JOINT 2

Data PerencanaanMomen M 58.993 kNmAxial P 567.805 kNMomen Nominal Mn 90.76 kNmAxial Nominal Pn 873.55 kNLebar Kolom b 300 mmTinggi Kolom h 400 mmDiameter Tul.Tarik = Tul.Tekan D 22 mmDiameter Sengkang 10 mmJumlah Tul.Tarik = Tul.Tekan n 6 buahTebal Selimut Beton 50 mmTinggi Efektif Kolom d' = selimut+sengkang+1/2 D tul 71 mm

d = h-d' 329 mmJumlah Tul. Dalam 1 baris 3.54838709677419 3 buahTegangan Tekan Beton fc' 30 MpaTegangan Leleh Tulangan fy 400 MpaModulus Elastisitas Beton Ec = 4700*fc^0,5 25743 MpaModulus Elastisitas Tulangan Es 200000 MpaFaktor Tegangan Beton β1 0.85Faktor Reduksi Ø 0.65Regangan Leleh Tulangan εy = fy / Es 0.002Luas Tul.Tarik = Tul.Tekan As = As' = n*1/4*π*D² 2280.796 mm²Luas Total Tulangan Ast = 2*(n*1/4*π*D²) 4561.593 mm²Luas Total Kolom Ag = b*h 120000 mm²% Luas Tulangan ρg = (Ast/Ag)*100 3.801 %

Beban SentrisPo Po = (0,85*fc'*(Ag-Ast)+Ast*fy) / 1000 4768.316 kNPn Pn = 0,8*Po 3814.653 kN

Kondisi Seimbang x = xbxb = 600*d / (600+fy) 197.400 mmab = β1*xb 167.790 mmfs' = (xb-d')*600 / xb 384.195 Mpa--- fs' < fy ,kondisi tul.tekan belum leleh--- maka selanjutnya dipakai fs' = 384.195 MpaCc = (0,85*fc'*b*ab) / 1000 1283.594 kNCs = As'*(fs'-0,85*fc') / 1000 818.109 kNfs = (d-xb)*600 / xb 400 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 400 MpaTs = (As*fs) / 1000 912.319 kNPnb = Cc + Cs - Ts 1189.384 kNy = h/2 200 mmMnb = ((Cc*(y-ab/2)+Cs*(y-d')+Ts*(d-y))/1000 372.257 kNme = (Mnb / Pnb)*1000 312.983 mm

Kondisi Patah Desak x > xbx 200 mma = β1*x 170.000 mmfs' = (x-d')*600 / x 387.000 Mpa--- fs' < fy ,kondisi tul.tekan belum leleh--- maka selanjutnya dipakai fs' = 387 Mpa



Cc = (0,85*fc'*b*a) / 1000 1300.500 kNCs = As'*(fs'-0,85*fc') / 1000 824.508 kNfs = (d-x)*600 / x 387.000 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 387.000 MpaTs = (As*fs) / 1000 882.668 kNPnb = Cc + Cs - Ts 1242.340 kNy = h/2 200 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 369.783 kNme = (Mnb / Pnb)*1000 297.651 mm

Kondisi Patah Desak x > xbx 250 mma = β1*x 212.500 mmfs' = (x-d')*600 / x 429.600 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 1625.625 kNCs = As'*(fs'-0,85*fc') / 1000 854.158 kNfs = (d-x)*600 / x 189.600 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 189.600 MpaTs = (As*fs) / 1000 432.439 kNPnb = Cc + Cs - Ts 2047.344 kNy = h/2 200 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 318.373 kNme = (Mnb / Pnb)*1000 155.506 mm


Kondisi Patah Tarik x < xbx 175 mma = β1*x 148.750 mmfs' = (x-d')*600 / x 356.571 Mpa--- fs' < fy ,kondisi tul.tekan belum leleh--- maka selanjutnya dipakai fs' = 357 MpaCc = (0,85*fc'*b*a) / 1000 1137.938 kNCs = As'*(fs'-0,85*fc') / 1000 755.106 kNfs = (d-x)*600 / x 528.000 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400.000 Mpa



Ts = (As*fs) / 1000 912.319 kNPnb = Cc + Cs - Ts 980.725 kNy = h/2 200 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 358.051 kNme = (Mnb / Pnb)*1000 365.088 mm

Kondisi Patah Tarik x < xbx 150 mma = β1*x 127.500 mmfs' = (x-d')*600 / x 316.000 Mpa--- fs' < fy ,kondisi tul.tekan belum leleh--- maka selanjutnya dipakai fs' = 316 MpaCc = (0,85*fc'*b*a) / 1000 975.375 kNCs = As'*(fs'-0,85*fc') / 1000 662.571 kNfs = (d-x)*600 / x 716.000 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400.000 MpaTs = (As*fs) / 1000 912.319 kNPnb = Cc + Cs - Ts 725.628 kNy = h/2 200 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 336.056 kNme = (Mnb / Pnb)*1000 463.124 mm

Kondisi Patah Tarik x < xbx 125 mma = β1*x 106.250 mmfs' = (x-d')*600 / x 259.200 Mpa--- fs' < fy ,kondisi tul.tekan belum leleh--- maka selanjutnya dipakai fs' = 259 MpaCc = (0,85*fc'*b*a) / 1000 812.813 kNCs = As'*(fs'-0,85*fc') / 1000 533.022 kNfs = (d-x)*600 / x 979.200 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400 MpaTs = (As*fs) / 1000 912.319 kNPnb = Cc + Cs - Ts 433.516 kNy = h/2 200 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 305.831 kNme = (Mnb / Pnb)*1000 705.466 mm

Lentur Murnix = tinggi garis netrala = β1*xfs' = (x-d')*600 / xfs = (d-x)*600 / xCc = (0,85*fc'*b*β1*x)Cs = As'*(fs'-0,85*fc')Ts = (As*fs)Cc + Cs - Ts = 0--- setelah di uraikan menjadi persamaan ABC :A = 0,85*fc'*b*β1 6502.500B = 600*As'-0,85*As'-As*fy 454220.576C = - 600*As'*d' -97161920.953 mmx = (-B+sqrt(B^2-4*A*C))/2*A 92.204 mma = β1*x 78.373 mmfs' = (x-d')*600 / x 137.979 Mpa



--- fs' < fy ,kondisi tul.tekan belum leleh--- maka selanjutnya dipakai fs' = 137.979 MpaCc = (0,85*fc'*b*a) / 1000 599.554 kNCs = As'*(fs'-0,85*fc') / 1000 256.542 kNfs = (d-x)*600 / x 1540.913 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400 MpaTs = (As*fs) / 1000 912.319 kNy = h/2 200 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 247.199 kNm

Rekap (Urutan P dari terkecil)Pnb Mnb

0 247.199433.516 305.831725.628 336.056980.725 358.0511242.340 369.7831189.384 372.2572047.344 318.3732672.622 268.6813814.653 0

Pn Mn873.55 90.76

0.000 50.000 100.000 150.000 200.000 250.000 300.000 350.000 400.0000

500

1000

1500

2000

2500

3000

3500

4000

4500

Diagram Interaksi

Kolom 30/40, 6D22Pu, Mu



PERHITUNGAN DIAGRAM INTERAKSI KOLOM JOINT 3

Data PerencanaanMomen M 926.169 kNmAxial P 1473.221 kNMomen Nominal Mn 1424.88 kNmAxial Nominal Pn 2266.49 kNLebar Kolom b 500 mmTinggi Kolom h 600 mmDiameter Tul.Tarik = Tul.Tekan D 29 mmDiameter Sengkang 10 mmJumlah Tul.Tarik = Tul.Tekan n 10 buahTebal Selimut Beton 50 mmTinggi Efektif Kolom d' = selimut+sengkang+1/2 D tul 74.5 mm

d = h-d' 525.5 mmJumlah Tul. Dalam 1 baris 6.08695652173913 6 buahTegangan Tekan Beton fc' 30 MpaTegangan Leleh Tulangan fy 400 MpaModulus Elastisitas Beton Ec = 4700*fc^0,5 25743 MpaModulus Elastisitas Tulangan Es 200000 MpaFaktor Tegangan Beton β1 0.85Faktor Reduksi Ø 0.65Regangan Leleh Tulangan εy = fy / Es 0.002Luas Tul.Tarik = Tul.Tekan As = As' = n*1/4*π*D² 6605.199 mm²Luas Total Tulangan Ast = 2*(n*1/4*π*D²) 13210.397 mm²Luas Total Kolom Ag = b*h 300000 mm²% Luas Tulangan ρg = (Ast/Ag)*100 4.403 %


Kondisi Seimbang x = xbxb = 600*d / (600+fy) 315.300 mmab = β1*xb 268.005 mmfs' = (xb-d')*600 / xb 458.230 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*ab) / 1000 3417.064 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-xb)*600 / xb 400 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 400 MpaTs = (As*fs) / 1000 2642.079 kNPnb = Cc + Cs - Ts 3248.631 kNy = h/2 300 mmMnb = ((Cc*(y-ab/2)+Cs*(y-d')+Ts*(d-y))/1000 1720.820 kNme = (Mnb / Pnb)*1000 529.706 mm

Kondisi Patah Desak x > xbx 350 mma = β1*x 297.500 mmfs' = (x-d')*600 / x 472.286 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 Mpa



Cc = (0,85*fc'*b*a) / 1000 3793.125 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-x)*600 / x 300.857 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 300.857 MpaTs = (As*fs) / 1000 1987.221 kNPnb = Cc + Cs - Ts 4279.551 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1579.636 kNme = (Mnb / Pnb)*1000 369.113 mm



Kondisi Patah Tarik x < xbx 300 mma = β1*x 255.000 mmfs' = (x-d')*600 / x 451.000 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 3251.250 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-x)*600 / x 451.000 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400.000 Mpa



Ts = (As*fs) / 1000 2642.079 kNPnb = Cc + Cs - Ts 3082.817 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1714.437 kNme = (Mnb / Pnb)*1000 556.127 mm

Kondisi Patah Tarik x < xbx 250 mma = β1*x 212.500 mmfs' = (x-d')*600 / x 421.200 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 2709.375 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-x)*600 / x 661.200 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400.000 MpaTs = (As*fs) / 1000 2642.079 kNPnb = Cc + Cs - Ts 2540.942 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1678.538 kNme = (Mnb / Pnb)*1000 660.597 mm


Lentur Murnix = tinggi garis netrala = β1*xfs' = (x-d')*600 / xfs = (d-x)*600 / xCc = (0,85*fc'*b*β1*x)Cs = As'*(fs'-0,85*fc')Ts = (As*fs)Cc + Cs - Ts = 0--- setelah di uraikan menjadi persamaan ABC :A = 0,85*fc'*b*β1 10837.500B = 600*As'-0,85*As'-As*fy 1315425.292C = - 600*As'*d' -295252375.372 mmx = (-B+sqrt(B^2-4*A*C))/2*A 115.171 mma = β1*x 97.896 mmfs' = (x-d')*600 / x 211.882 Mpa



--- fs' < fy ,kondisi tul.tekan belum leleh--- maka selanjutnya dipakai fs' = 211.882 MpaCc = (0,85*fc'*b*a) / 1000 1248.169 kNCs = As'*(fs'-0,85*fc') / 1000 1231.093 kNfs = (d-x)*600 / x 2137.662 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400 MpaTs = (As*fs) / 1000 2642.079 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1186.756 kNm


0 1186.7561843.845 1584.6062540.942 1678.5383082.817 1714.4374279.551 1579.6363248.631 1720.8205565.218 1401.7516685.599 1238.10810077.835 0

Pu Mu2266.49 1424.88

0

2000

4000

6000

8000

10000

12000

Diagram Interaksi




PERHITUNGAN DIAGRAM INTERAKSI KOLOM JOINT 4

Data PerencanaanMomen M 905.037 kNmAxial P 1401.238 kNMomen Nominal Mn 1392.36 kNmAxial Nominal Pn 2155.75 kNLebar Kolom b 500 mmTinggi Kolom h 600 mmDiameter Tul.Tarik = Tul.Tekan D 29 mmDiameter Sengkang 10 mmJumlah Tul.Tarik = Tul.Tekan n 10 buahTebal Selimut Beton 50 mmTinggi Efektif Kolom d' = selimut+sengkang+1/2 D tul 74.5 mm

d = h-d' 525.5 mmTegangan Tekan Beton fc' 30 MpaTegangan Leleh Tulangan fy 400 MpaModulus Elastisitas Beton Ec = 4700*fc^0,5 25743 MpaModulus Elastisitas Tulangan Es 200000 MpaFaktor Tegangan Beton β1 0.85Faktor Reduksi Ø 0.65Regangan Leleh Tulangan εy = fy / Es 0.002Luas Tul.Tarik = Tul.Tekan As = As' = n*1/4*π*D² 6605.199 mm²Luas Total Tulangan Ast = 2*(n*1/4*π*D²) 13210.397 mm²Luas Total Kolom Ag = b*h 300000 mm²% Luas Tulangan ρg = (Ast/Ag)*100 4.403 %


Kondisi Seimbang x = xbxb = 600*d / (600+fy) 315.300 mmab = β1*xb 268.005 mmfs' = (xb-d')*600 / xb 458.230 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*ab) / 1000 3417.064 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-xb)*600 / xb 400 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 400 MpaTs = (As*fs) / 1000 2642.079 kNPnb = Cc + Cs - Ts 3248.631 kNy = h/2 300 mmMnb = ((Cc*(y-ab/2)+Cs*(y-d')+Ts*(d-y))/1000 1720.820 kNme = (Mnb / Pnb)*1000 529.706 mm

Kondisi Patah Desak x > xbx 350 mma = β1*x 297.500 mmfs' = (x-d')*600 / x 472.286 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 3793.125 kN



Cs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-x)*600 / x 300.857 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 300.857 MpaTs = (As*fs) / 1000 1987.221 kNPnb = Cc + Cs - Ts 4279.551 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1579.636 kNme = (Mnb / Pnb)*1000 369.113 mm



Kondisi Patah Tarik x < xbx 300 mma = β1*x 255.000 mmfs' = (x-d')*600 / x 451.000 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 3251.250 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-x)*600 / x 451.000 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400.000 MpaTs = (As*fs) / 1000 2642.079 kN



Pnb = Cc + Cs - Ts 3082.817 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1714.437 kNme = (Mnb / Pnb)*1000 556.127 mm

Kondisi Patah Tarik x < xbx 200 mma = β1*x 170.000 mmfs' = (x-d')*600 / x 376.500 Mpa--- fs' < fy ,kondisi tul.tekan belum leleh--- maka selanjutnya dipakai fs' = 377 MpaCc = (0,85*fc'*b*a) / 1000 2167.500 kNCs = As'*(fs'-0,85*fc') / 1000 2318.425 kNfs = (d-x)*600 / x 976.500 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400.000 MpaTs = (As*fs) / 1000 2642.079 kNPnb = Cc + Cs - Ts 1843.845 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1584.606 kNme = (Mnb / Pnb)*1000 859.403 mm


Lentur Murnix = tinggi garis netrala = β1*xfs' = (x-d')*600 / xfs = (d-x)*600 / xCc = (0,85*fc'*b*β1*x)Cs = As'*(fs'-0,85*fc')Ts = (As*fs)Cc + Cs - Ts = 0--- setelah di uraikan menjadi persamaan ABC :A = 0,85*fc'*b*β1 10837.500B = 600*As'-0,85*As'-As*fy 1315425.292C = - 600*As'*d' -295252375.372 mmx = (-B+sqrt(B^2-4*A*C))/2*A 115.171 mma = β1*x 97.896 mmfs' = (x-d')*600 / x 211.882 Mpa--- fs' < fy ,kondisi tul.tekan belum leleh



--- maka selanjutnya dipakai fs' = 211.882 MpaCc = (0,85*fc'*b*a) / 1000 1248.169 kNCs = As'*(fs'-0,85*fc') / 1000 1231.093 kNfs = (d-x)*600 / x 2137.662 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400 MpaTs = (As*fs) / 1000 2642.079 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1186.756 kNm


0 1186.756809.883 1391.6821843.845 1584.6063082.817 1714.4374279.551 1579.6363248.631 1720.8205565.218 1401.7516685.599 1238.10810077.835 0

Pn Mn2155.75 1392.36

0

2000

4000

6000

8000

10000

12000

Diagram Interaksi


PERANCANGAN STRUKTU BETON BERTULANG


BAB VIII PERHITUNGAN PONDASI

A. Data Perencanaan

> Tepi AD :P = 567.805 kNMu = 58.993 kNm

> Tepi BE :P = 1473.22 kNMu = 926.169 kNm

> Tepi CF :P = 1401.238 kNMu = 905.037 kNm

Gambar 8.1 Pondasi

A

D

CB

E F



PONDASI BUJUR SANGKAR TEPI AD

A. Data Perencanaan

P = 567.805 kNMu = 58.993 kNmDf = 1.50 mØ tul = 22 mmp = 75 mmfc' = 30 Mpafy = 400 Mpaγ = 1600 kg/m³σ = 2 Mpa = 2000

dimensi pondasi → B = 2.00 m = 2000 mmL = 2.00 m = 2000 mm ts = 0.50 m = 500 mm

dimensi kolom → bk = 0.30 m = 300 mmhk = 0.40 m = 400 mmDf-ts = 1.00 m = 1000 mm

B. Pembebanan Pondasi

Berat penutup lantai :• Keramik 1 cm = 1 cm x 24 kg/m²/cm = 24 kg/m²• Spesi 2 cm = 2 cm x 21 kg/m²/cm = 42 kg/m²• Pasir 3 cm = 0.03 m x 1600 kg/m³ = 48 kg/m²

= 114 kg/m²

• Berat penutup lantai =2.00 m x 2.00 m x 114 kg/m² x 1.2 = 547 kg

• Berat plat pondasi =2.00 m x 2.00 m x 0.50 m x 2400 kg/m³ x 1.2 = 5760 kg

• Berat lantai kerja =2.00 m x 2.00 m x 0.05 m x 2200 kg/m³ x 1.2 = 528 kg

• Berat urugan =(( 2.00 m x 2.00 m x 1.00 m ) - vol kolom ) x x 1600 kg/m³ x 1.2 = 7450 kg

= 14285 kg= 142.8 kN

• Beban terpusat (P) = = 567.8 kNPu = 710.7 kN

C. Daya Dukung Ultimit Tanah

σ =Pu

±Mu

B² . B³⅙

=710.7

±58.993

2.00 ² ( )⅙ x 2.00 ³

σ maks = 221.908 kN/m²σ min = 133.418 kN/m²

karena σ maks = 221.91 kN/m² < qs = 2000 kN/m² → AMAN

kN/m²



D. Kekuatan Geser pondasi

ds = p + (½ x Ø tul)= 75 +( ½ x 22 )= 86 mm

d = ts - ds= 500 - 86.0= 414.0 mm

gaya geser 1 araha = ½ B– ½ hk – d

= ( ½ x 2000 )-( ½ x 400 ) - 414= 386 mm= 0.386 m

σ a =B

= 133.42 + ( 2.00 - 0.386 221.908 - 133.42 )2.00

= 204.830 kN/m²

Vu =2

= 0.386 x 2.00 x ( 221.908 + 204.830 )2

= 164.7207 kN

Gaya geser yang ditahan beton (Ø.Vc) :

Ø.Vc = Ø x ⅙ x x B x d= 0.75 x ⅙ x 30 x 2000 x 414.0= 566892.85 N= 566.893 kN

Jadi, Vu = 164.721 kN < Ø.Vc = 566.893 kN → AMAN

gaya geser 2 arah

B1 = bk + d = 300 + 414.0 = 714 mm = 0.714 mB2 = hk + d = 400 + 414.0 = 814 mm = 0.814 m

Vu = ((B² ) - (B1 x B2)) x (σ maks - σ min

)2

= (( 2.00 ² ) - ( 0.714 x 0.814 )) x (221.908 + 133.42

)2

= 607.396 kN

βc =hk

=400

= 1.33bk 300

bo = 2 x (B1 + B2)= 2 x ( 714 + 814 )= 3056 mm²


σ min + ( B - a ) x ( σ maks - σ min )

) x (

a x B x ( σ maks + σ a )

cf '



Vc = ( 1 + 2

) ( ⅙ x x bo x d )βc

= ( 1 +2

) ( ⅙ x 30 x 3056 x 414.0 )1.33= 2887374.23 N= 2887.374 kN

Vc = ( 2 + αs x d

) (1

x x bo x d )→ αs = 40 ( Kolom dalam )

bo 12 → αs = 30 ( Kolom tepi )

= ( 1 +30 x 414

) (1

x 30 x 3056 x 414.0 )3056 12= 2924411.23 N= 2924.411 kN

Vc = ⅓ x x bo x d= ⅓ x 30 x 3056 x 414.0= 2309899.39 N= 2309.899 kN

Dipilih Vc yang terkecil, yaitu Vc = 2309.899 kNØ.Vc = 0.75 x 2309.899

= 1732.425 kN


E. Penulangan Pondasi

ds = p + (½ x Ø tul) + Ø tul= 75 +( ½ x 22 )+ 22= 108 mm

d = ts - ds= 500 - 108.0= 392.0 mm

x = ½ B– ½ hk= ( ½ x 2.00 )-( ½ x 0.40 )= 0.800 m

σ x =B

= 133.42 + ( 2.00 - 0.800 221.908 - 133.42 )2.00

= 186.512 kN/m²

Mu == ½ x 186.512 x 0.800 ² + ⅓ 221.908 - 186.512 0.800 ²= 67.24 kNm= 67235008.0 Nmm

K =Mu

=67235008.000

= 0.54693 MpaΦ x b x d² 0.8 x 1000 x 392.0 ²

Kmaks =

σ min + ( B - x ) x ( σ maks - σ min )

) x (

½ . σ x . x² + ⅓ . ( σ maks - σ x ) . x²x ( ) x

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²

cf '

cf '

cf '



= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 )

= 7.8883 Mpa

Syarat K = 0.546932 < Kmaks = 7.888 → AMAN


a = 1 -1 -

2 x Kx d

0.85 x fc'

= 1 -1 -

2 x 0.547x 392.0

0.85 x 30

= 8.50 mm



= 0.85 x 30 x 8.50 x 1000400

= 541.87

=1.4 x b x d

=1.4 x 1000 x 392

= 1372.00 mm²fy 400

= 1372.00 mm²

s =1/4 x π x D² x b

Asu

=0.25 x π x 22 ² x 1000

1372.00= 277.0647 mm

s ≤ ( 2 . ts ) = 2 x 500 = 1000 mm

s ≤ 450 mm

s pakai dipilih yang terkecil yaitu s = 277.0647 mm ≈ 260 mmmaka digunakan tulangan dengan Ø 22 - 260 mm

Asu

mm²

Asu




panjang penyaluran dan penyambungan tulangan dipakai tulangan Ø 22

= 4 x

=22 x 400

= 401.66 mm4 x 30

= 0,04 x db x fy= 0.04 x 22 x 400 = 352 mm < 402 mm → AMAN

jadi dipakai panjang penyaluran = 401.7 mm ≈ 400 mmuntuk memermudah pemasangan makan di gunakan Ø 22 - 400

E. Kuat Dukung Pondasi

Pu = Ø x 0.85 x fc' x A= 0.7 x 0.85 x 30 x 120000= 2142000.0 N = 2142 kN

Pu = 710.7 kN < Pu = 2142 kN → AMAN

λdbdb x fy

tidak boleh kurang dari λdb

cf '



PONDASI BUJUR SANGKAR TENGAH BE

A. Data Perencanaan






= 114 kg/m²





= 14381 kg= 143.8 kN



σ =Pu

±Mu

B² . B³⅙

=1617.0

±926.169

2.00 ² ( )⅙ x 2.00 ³

σ maks = 1098.884 kN/m²σ min = -290.370 kN/m²


kN/m²




ds = p + (½ x Ø tul)= 75 +( ½ x 29 )= 89.5 mm

d = ts - ds= 600 - 89.5= 510.5 mm


= ( ½ x 2000 )-( ½ x 600 ) - 510.5= 189.5 mm= 0.190 m

σ a =B

= -290.37 + ( 2.00 - 0.190 1098.884 - -290.37 )2.00

= 967.252 kN/m²

Vu =2

= 0.190 x 2.00 x ( 1098.884 + 967.252 )2

= 391.533 kN




gaya geser 2 arah

B1 = bk + d = 500 + 510.5 = 1011 mm = 1.011 mB2 = hk + d = 600 + 510.5 = 1111 mm = 1.111 m


)2

= (( 2.00 ² ) - ( 1.011 x 1.111 )) x (1098.884 + -290.37

)2

= 1163.388 kN

βc =hk

=600

= 1.20bk 500

bo = 2 x (B1 + B2)= 2 x ( 1011 + 1111 )= 4242 mm²



) x (


cf '



Vc = ( 1 + 2


= ( 1 +2

) ( ⅙ x 30 x 4242 x 510.5 )1.20= 5271625.13 N= 5271.625 kN

Vc = ( 2 + αs x d

) (1



= ( 1 +40 x 511

) (1

x 30 x 4242 x 510.5 )4242 12= 5746500.13 N= 5746.500 kN

Vc = ⅓ x x bo x d= ⅓ x 30 x 4242 x 510.5= 3953718.85 N= 3953.719 kN


= 2965.289 kN




d = ts - ds= 600 - 118.5= 481.5 mm

x = ½ B– ½ hk= ( ½ x 2.00 )-( ½ x 0.60 )= 0.700 m

σ x =B

= -290.37 + ( 2.00 - 0.700 1098.884 - -290.37 )2.00

= 612.645 kN/m²

Mu == ½ x 612.645 x 0.700 ² + ⅓ 1098.884 - 612.645 0.700 ²= 229.52 kNm= 229517084.1 Nmm

K =Mu

=229517084.125

= 1.23746 MpaΦ x b x d² 0.8 x 1000 x 481.5 ²

Kmaks =


) x (

½ . σ x . x² + ⅓ . ( σ maks - σ x ) . x²x ( ) x

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²

cf '

cf '

cf '



= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 )

= 7.8883 Mpa



a = 1 -1 -

2 x Kx d

0.85 x fc'

= 1 -1 -

2 x 1.237x 481.5

0.85 x 30

= 23.96 mm



= 0.85 x 30 x 23.96 x 1000400

= 1527.61

=1.4 x b x d

=1.4 x 1000 x 482

= 1685.25 mm²fy 400

= 1685.25 mm²

s =1/4 x π x D² x b

Asu

=0.25 x π x 29 ² x 1000

1685.25= 391.942 mm

s ≤ ( 2 . ts ) = 2 x 600 = 1200 mm

s ≤ 450 mm


Asu

mm²

Asu





= 4 x

=29 x 400

= 529.47 mm4 x 30




Pu= Ø x 0.85 x fc' x A= 0.7 x 0.85 x 30 x 300000= 5355000.0 N = 5355 kN

Pu = 1617.0 kN < Pu = 5355 kN → AMAN

λdbdb x fy


cf '



PONDASI BUJUR SANGKAR TEPI CF

A. Data Perencanaan






= 114 kg/m²





= 13939 kg= 139.4 kN



σ =Pu

±Mu

B² . B³⅙

=1540.6

±905.037

2.00 ² ( )⅙ x 2.00 ³



kN/m²




ds = p + (½ x Ø tul)= 75 +( ½ x 29 )= 89.5 mm

d = ts - ds= 500 - 89.5= 410.5 mm


= ( ½ x 2000 )-( ½ x 600 ) - 410.5= 289.5 mm= 0.290 m

σ a =B

= -293.62 + ( 2.00 - 0.290 1063.935 - -293.62 )2.00

= 867.429 kN/m²

Vu =2

= 0.290 x 2.00 x ( 1063.935 + 867.429 )2

= 559.13 kN




gaya geser 2 arah

B1 = bk + d = 500 + 410.5 = 911 mm = 0.911 mB2 = hk + d = 600 + 410.5 = 1011 mm = 1.011 m


)2

= (( 2.00 ² ) - ( 0.911 x 1.011 )) x (1063.935 + -293.62

)2

= 1186.262 kN

βc =hk

=600

= 1.20bk 500

bo = 2 x (B1 + B2)= 2 x ( 911 + 1011 )= 3842 mm²



) x (


cf '



Vc = ( 1 + 2


= ( 1 +2

) ( ⅙ x 30 x 3842 x 410.5 )1.20= 3839269.79 N= 3839.270 kN

Vc = ( 2 + αs x d

) (1



= ( 1 +30 x 411

) (1

x 30 x 3842 x 410.5 )3842 12= 3027284.71 N= 3027.285 kN

Vc = ⅓ x x bo x d= ⅓ x 30 x 3842 x 410.5= 2879452.34 N= 2879.452 kN


= 2159.589 kN




d = ts - ds= 500 - 118.5= 381.5 mm

x = ½ B– ½ hk= ( ½ x 2.00 )-( ½ x 0.60 )= 0.700 m

σ x =B

= -293.62 + ( 2.00 - 0.700 1063.935 - -293.62 )2.00

= 588.791 kN/m²

Mu == ½ x 588.791 x 0.700 ² + ⅓ 1063.935 - 588.791 0.700 ²= 221.86 kNm= 221860674.9 Nmm

K =Mu

=221860674.875

= 1.90547 MpaΦ x b x d² 0.8 x 1000 x 381.5 ²

Kmaks =


) x (

½ . σ x . x² + ⅓ . ( σ maks - σ x ) . x²x ( ) x

382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²

cf '

cf '

cf '



= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 )

= 7.8883 Mpa



a = 1 -1 -

2 x Kx d

0.85 x fc'

= 1 -1 -

2 x 1.905x 381.5

0.85 x 30

= 29.66 mm



= 0.85 x 30 x 29.66 x 1000400

= 1890.84

=1.4 x b x d

=1.4 x 1000 x 382

= 1335.25 mm²fy 400

= 1890.84 mm²

s =1/4 x π x D² x b

Asu

=0.25 x π x 29 ² x 1000

1890.84= 349.3259 mm

s ≤ ( 2 . ts ) = 2 x 500 = 1000 mm

s ≤ 450 mm


Asu

mm²

Asu





= 4 x

=29 x 400

= 529.47 mm4 x 30




Pu= Ø x 0.85 x fc' x A= 0.7 x 0.85 x 30 x 300000= 5355000.0 N = 5355 kN

Pu = 1540.6 kN < Pu = 5355 kN → AMAN

λdbdb x fy


cf '

PONDASI BUJUR SANGKAR TENGAH JOINT 3

A. Data Perencanaan

N = 1473.22 kNMu = 926.169 kNmDf = 1.50 m

= 29 mmp = 50 mmfc' = 30 Mpafy = 390 Mpaγ = 1600 kg/m³σ = 2 Mpa = 2000

dimensi pondasi → B = 2.00 m = 2000 mmL = 2.00 m = 2000 mmts = 0.50 m = 500 mm



Berat penutup lantai :Keramik 1 cm = 1 cm x 24 kg/m²/cm = 24 kg/m²Spesi 2 cm = 2 cm x 21 kg/m²/cm = 42 kg/m²Pasir 3 cm = 0.03 m x 1600 = 48 kg/m²

= 114 kg/m²

Berat penutup lantai =2.00 m x 2.00 m x 114 kg/m² x 1.2 = 547 kgBerat plat pondasi =2.00 m x 2.00 m x 0.5 m x 2400 kg/m³ x 1.2 = 5760 kgBerat lantai kerja =2.00 m x 2.00 m x 0.05 m x 2200 kg/m³ x 1.2 = 528 kgBerat urugan =(( 2.00 m x 2.00 m x 1.00 m ) - vol kolom ) x 1600 kg/m³ x 1.2 = 7104 kg

= 13939 kg= 139.4 kN

Beban terpusat (N) = = 1473.2 kNPu = 1612.6 kN


σ =Pu

±Mu . B³⅙

=1612.61

±926.169

2.00 ² ( )⅙ x 2.00 ³


karena σ maks = 1097.8 kN/m² < qs = 2000 kN/m² → AMAND. Kekuatan Geser pondasi

Ø tul

kN/m²

••• kg/m³

•

•

•

•

•

B²

ds = p + (½ x Ø tul)= 50 +( ½ x 29 )= 64.5 mm

d = ts - ds= 500 - 64.5= 435.5 mm


= ( ½ x 2000 )-( ½ x 600 ) - 435.5= 264.5 mm= 0.265 m

σ a =B

= -291.47 + ( 2.00 - 0.265 1097.780 - -291.47 )2.00

= 914.051 kN/m²

Vu =2

= 0.265 x 2.00 x ( 1097.780 + 914.051 )2

= 532.1294 kN

= Ø x ⅙ x x B x d= 0.75 x ⅙ x 30 x 2000 x 435.5= 596332.93 N= 596.333 kN


gaya geser 2 arah

B1 = bk + d = 500 + 435.5 = 936 mm = 0.936 mB2 = hk + d = 600 + 435.5 = 1036 mm = 1.036 m

Vu = x (σ maks - σ min

)2

= (( 2.00 ² ) - ( 0.936 x 1.036 )) x (1097.78 + -291.47

)2.00

= 1222.074 kN

βc =hk

=600

= 1.20bk 500

bo = 2 x (B1 + B2)= 2 x ( 936 + 1036 )= 3942 mm²


) x (



Ø.Vc

((B² ) - (B1 x B2))


cf '

Vc = ( 1 + 2


= ( 1 +2

) ( ⅙ x 30 x 3942 x 435.5 )1.20= 4179101.20 N= 4179.101 kN

Vc = ( 2 + ) (1

x x bo x d )→ αs = 40 ( Kolom tengah )


= ( 1 +40 x 435.5

) (1

x 30 x 3942 x 435.5 )3942 12= 4246288.05 N= 4246.288 kN

Vc = ⅓ x x bo x d= ⅓ x 30 x 3942 x 435.5= 3134325.90 N= 3134.326 kN


= 2350.744 kN



ds = p + (½ x Ø tul) + Ø tul= 50 +( ½ x 29 )+ 29= 93.5 mm

d = ts - ds= 500 - 93.5= 406.5 mm

x = ½ B– ½ hk= ( ½ x 2.00 )-( ½ x 0.60 )= 0.700 m

σ x =B

= -291.47 + ( 2.00 - 0.700 1097.780 - -291.47 )2.00

= 611.541 kN/m²

Mu == ½ x 611.541 x 0.700 1097.780 - 611.541 0.700 ²= 229.25 kNm= 229246604.1 Nmm

k =Mu

=229246604.125

= 1.734171 N/mm²Φ x b x d² 0.8 x 1000 x 406.5 ²

k = ρ x 0,8 x fy x (1-0,588 x ρ xfy

)fc'

1.7342 = ρ x 0.8 x 390 x ( 1 - 0.588 x ρ x390

)

αs x d


) x (

½ . σ x . x² + ⅓ . ( σ maks - σ x ) . x²² + ⅓ x ( ) x

cf '

cf '

cf '

1.7342 = ρ x 0.8 x 390 x ( 1 - 0.588 x ρ x30

)

0 = -2384.93 ρ² + 312 ρ + -1.7342

dengan rumus ABC didapatkan :

ρ₁₂ =- b b²-4ac

2a

=- 312 ± 312 ² - 4 x -2384.9 x -1.7342

2 x -2384.928

ρ₁ = 0.00582→ ρ pakai = 0.00582ρ₂ = 0.12500

ρ min =1.4

=1.4

= 0.00359fy 390

Jadi, ρ = 0.00582 > ρ min = 0.00359 → ρ = ρ min = 0.00582As = ρ x b x d

= 0.00582 x 1000 x 406.5= 2364.563 mm²

s =0.25 x π x D² x b

As

=0.25 x π x 22 ² x 1000

2364.56= 160.762 mm

s ≤ ( 2 . ts ) = 2 x 500 = 1000 mm

s ≤ 450 mm



ldb = 4 x

=29 x 390

= 516.23 mm4 x 30

tidak boleh kurang dari ldb = 0,04 x db x fy= 0.04 x 29 x 390 = 452.4 mm < 516 mm

jadi dipakai panjang penyaluran = 516 mm ≈ 390 mmuntuk memermudah pemasangan makan di gunakan Ø 29 - 390

db x fy

cf '

PONDASI BUJUR SANGKAR TENGAH JOINT 3

PONDASI BUJUR SANGKAR TEPI JOINT 4

A. Data Perencanaan

N = 1401.24 kNMu = 905.037 kNmDf = 1.50 mØ tul = 29 mmp = 50 mmfc' = 30 Mpafy = 390 Mpaγ = 1600 kg/m³σ = 2 Mpa = 2000





= 114 kg/m²





= 13843 kg= 138.4 kN

• Beban terpusat (N) = = 1401.2 kNPu = 1539.7 kN


σ =Pu

±Mu

B² . B³⅙

=1539.7

±905.037

2.00 ² ( )⅙ x 2.00 ³


karena σ maks = 1063.70 kN/m² < qs = 2000 kN/m² → AMAND. Kekuatan Geser pondasi

kN/m²

ds = p + (½ x Ø tul)= 50 +( ½ x 29 )= 64.5 mm

d = ts - ds= 500 - 64.5= 435.5 mm


= ( ½ x 2000 )-( ½ x 700 ) - 435.5= 214.5 mm= 0.215 m

σ a =B

= -293.86 + ( 2.00 - 0.215 1063.695 - -293.86 )2.00

= 918.097 kN/m²

Vu =2

= 0.215 x 2.00 x ( 1063.695 + 918.097 )2

= 425.0945 kN




gaya geser 2 arah

B1 = bk + d = 500 + 435.5 = 936 mm = 0.936 mB2 = hk + d = 700 + 435.5 = 1136 mm = 1.136 m


)2

= (( 2.00 ² ) - ( 0.936 x 1.136 )) x (1063.695 + -293.86

)2

= 1130.787 kN

βc =hk

=700

= 1.40bk 500

bo = 2 x (B1 + B2)= 2 x ( 936 + 1136 )= 4142 mm²



) x (


cf '

Vc = ( 1 + 2


= ( 1 +2

) ( ⅙ x 30 x 4142 x 435.5 )1.40= 3999065.45 N= 3999.065 kN

Vc = ( 2 + αs x d

) (1

x x bo x d )→ αs = 40 ( Kolom tengah )


= ( 1 +30 x 435.5

) (1

x 30 x 4142 x 435.5 )4142 12= 3420366.93 N= 3420.367 kN

Vc = ⅓ x x bo x d= ⅓ x 30 x 4142 x 435.5= 3293348.02 N= 3293.348 kN


= 2470.011 kN



ds = p + (½ x Ø tul) + Ø tul= 50 +( ½ x 29 )+ 29= 93.5 mm

d = ts - ds= 500 - 93.5= 406.5 mm

x = ½ B– ½ hk= ( ½ x 2.00 )-( ½ x 0.70 )= 0.650 m

σ x =B

= -293.86 + ( 2.00 - 0.650 1063.695 - -293.86 )2.00

= 622.490 kN/m²

Mu == ½ x 622.490 x 0.650 ² + ⅓ 1063.695 - 622.490 0.650 ²= 193.64 kNm= 193637398.3 Nmm

k =Mu

=193637398.297

= 1.4648 N/mm²Φ x b x d² 0.8 x 1000 x 406.5 ²

k = ρ x 0,8 x fy x (1-0,588 x ρ xfy

)fc'

1.4648 = ρ x 0.8 x 390 x ( 1 - 0.588 x ρ x390

)


) x (

½ . σ x . x² + ⅓ . ( σ maks - σ x ) . x²x ( ) x

cf '

cf '

cf '

1.4648 = ρ x 0.8 x 390 x ( 1 - 0.588 x ρ x30

)

0 = -2384.93 ρ² + 312 ρ + -1.4648

dengan rumus ABC didapatkan :

ρ₁₂ =- b b²-4ac

2a

=- 312 ± 312 ² - 4 x -2384.9 x -1.4648

2 x -2384.9

ρ₁ = 0.00488→ ρ pakai = 0.00488ρ₂ = 0.12594

ρ min =1.4

=1.4

= 0.00359fy 390

Jadi, ρ = 0.00488 > ρ min = 0.00359 → ρ = ρ min = 0.00488As = ρ x b x d

= 0.00488 x 1000 x 406.5= 1982.36 mm²

s =0.25 x π x D² x b

As

=0.25 x π x 29 ² x 1000

1982.36= 333.199 mm

s ≤ ( 2 . ts ) = 2 x 500 = 1000 mm

s ≤ 450 mm



ldb = 4 x

=29 x 390

= 516 mm4 x 30

tidak boleh kurang dari ldb = 0,04 x db x fy= 0.04 x 29 x 390 = 452.4 mm < 516 mm

jadi dipakai panjang penyaluran = 516 mm ≈ 500 mmuntuk memermudah pemasangan makan di gunakan Ø 29 - 500

db x fy

cf '

PONDASI BUJUR SANGKAR TEPI JOINT 4

(HDT)Perancangan Beton Bertulang.xlsx

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