Download - (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 =
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 2
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 =
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 3
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 =
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 4
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 5
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 6
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 7
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 8
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
→ Dipilih yang terkecil, s = 209.33 ≈ 250 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 9
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 10
s =As perlu
= 0.25 x 3.14 x 10 1000554.167
= 141.654 mm
s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm
→ Dipilih yang terkecil, s = 141.654 ≈ 140 mm
As ada =s
= 0.25 x 3.14 x 10 1000140
= 560.999 > As perlu
→ dipakai tulangan pokok Ф 10 - 140
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
→ Dipilih yang terkecil, s = 209.440 ≈ 200 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 11
= 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
= 0.00238→ 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
→ Dipilih yang terkecil, s = 158.400 ≈ 160 mm
As ada =s
= 0.25 x 3.14 x 10 1000160
= 490.874 > As perlu
→ dipakai tulangan pokok Ф 10 - 160
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 12
→ Dipilih yang terkecil, s = 209.440 ≈ 200 mm
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
a. Menentukan momenKx = 42 (PBI 1971)Ky = 37 (PBI 1971)
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
b. Menentukan tulangan arah lapangan x = tumpuan xMu = 4.910 kNm = 4.910 x 10 NmmRn = Mu = 4910000
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 13
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
= 0.00287→ 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
→ Dipilih yang terkecil, s = 141.654 ≈ 140 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
→ Dipilih yang terkecil, s = 209.333 ≈ 210 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 14
c. Menentukan tulangan arah lapangan y = tumpuan yMu = 4.326 kNm = 4.325 x 10 NmmRn = Mu = 4325000
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
→ Dipilih yang terkecil, s = 158.400 ≈ 160 mm
As ada =s
= 0.25 x 3.14 x 10 1000160
= 490.874 > As perlu→ dipakai tulangan pokok Ф 10 - 160
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 15
s =As bagi
= 0.25 x 3.1416 x 8 1000240
= 209.44 mms ≤ 5 h = 5 x 120 = 600s ≤ 450 mm
→ Dipilih yang terkecil, s = 209.440 ≈ 210 mm
As ada =s
= 0.25 x 3.14 x 8 1000210
= 239.359 > As bagi→ dipakai tulangan pokok Ф 8 - 210 mm
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 16
= 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
→ Dipilih yang terkecil, s = 141.654 ≈ 180 mm
As ada =s
= 0.25 x 3.14 x 10 1000180.000
= 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 1000
240= 209.33 mm
s ≤ 5 h = 5 x 120 = 600s ≤ 450 mm
→ Dipilih yang terkecil, s = 209.333 ≈ 250 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 17
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 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
b. Menentukan tulangan arah lapangan x = tumpuan xMu = 4.945 kNm = 4.944 x 10 NmmRn = Mu = 4944000
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 18
s =As perlu
= 0.25 x 3.14 x 10 1000554.167
= 141.654 mm
s ≤ 2 h = 2 x 120 = 240s ≤ 450 mm
→ Dipilih yang terkecil, s = 141.654 ≈ 140 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 3.1416 x 8 1000
240= 209.44 mm
s ≤ 5 h = 5 x 3.14 = 15.71s ≤ 450 mm
→ Dipilih yang terkecil, s = 209.440 ≈ 200 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 = 4.085 kNm = 4.084 x 10 NmmRn = Mu = 4084000
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 19
= 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
= 0.00238→ 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
→ Dipilih yang terkecil, s = 158.400 ≈ 160 mm
As ada =s
= 0.25 x 3.14 x 10 1000160
= 490.874 > As perlu→ dipakai tulangan pokok Ф 10 - 160
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 20
→ Dipilih yang terkecil, s = 209.440 ≈ 200 mm
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
a. Menentukan momenKx = 42 (PBI 1971)Ky = 37 (PBI 1971)
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
b. Menentukan tulangan arah lapangan x = tumpuan xMu = 6.145 kNm = 6.145 x 10 NmmRn = Mu = 6145000
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 21
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
→ Dipilih yang terkecil, s = 141.654 ≈ 140 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 3.1416 x 8 1000
240= 209.44 mm
s ≤ 5 h = 5 x 120 = 600s ≤ 450 mm
→ Dipilih yang terkecil, s = 209.440 ≈ 200 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 22
c. Menentukan tulangan arah lapangan y = tumpuan yMu = 5.414 kNm = 5.413 x 10 NmmRn = Mu = 5413000
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
→ Dipilih yang terkecil, s = 158.400 ≈ 160 mm
As ada =s
= 0.25 x 3.14 x 10 1000160
= 490.874 > As perlu→ dipakai tulangan pokok Ф 10 - 160
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
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 23
= 0.25 x 3.1416 x 8 1000240
= 209.44 mms ≤ 5 h = 5 x 120 = 600s ≤ 450 mm
→ Dipilih yang terkecil, s = 209.440 ≈ 200 mm
As ada =s
= 0.25 x 3.14 x 8 1000200
= 251.327 > As bagi→ dipakai tulangan pokok Ф 8 - 200 mm
2 x
1/4 . π . Ф2 .b
2 x
mm2
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 24
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 25
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 26
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)
B.S. Balok = 0.2 0.3 - 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 400 kg/m² = 90.6 kg/m
Pembebanan Lantai 3• 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
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 27
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
Momen Tumpuan :Mu1 = Mu10 = 1/16 x qu x = 0.0625 x 1867.3 x 12.25 = 1429.643 kgmMu2 = Mu9 = 1/10 x qu x = 0.10 x 1867.3 x 12.25 = 2287.428 kgmMu3 - Mu8 = 1/11 x qu x = 0.09 x 1867.3 x 12.25 = 2079.48 kgm
Momen Lapangan :Mu1-2 = Mu9-10 = 1/14 x qu x = 0.0714 x 1867.3 x 12.25 = 1633.877 kgmMu2-3 → Mu8-9 = 1/16 x qu x = 0.06 x 1867.3 x 12.25 = 1429.643 kgm
Gaya Geser :Qu = 1/2 x qu x L
= 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 28
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
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
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 29
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 . ز 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 30
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 31
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
ds1 = p + Ø tul. begel + (½ x Ø tul. pokok)= 40 + 8 +( ½ x 16 )= 56 mm
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 32
→- 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
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 33
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 . ز ¼ x π x 16 ²
Digunakan tulangan tarik As = 4 Ø 16 = 803.840 > Asu OKtulangan tekan As' = 2 Ø 16 = 401.920 (ditambahkan)
Jumlah tulangan maksimal pada 1 baris :
m =b - 2 x d'
+ 1Ø + Sn
=200 - 2 x 56
+ 116 + 25
= 3.146341 batang → maksimal 3 batang
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
Diambil nilai yang terbesar Asu
mm²mm²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 34
B. Hitung Momen Rencana1). Hitung tinggi blok tegangan beton tekan persegi ekuivalen a
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
ρ min < ρ < ρ maks OK
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)
Mn = As x fy x ( d - a/2)= 803.84 x 400 x 216.79= 69707304.35 Nmm
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
Kontrol semua tulangan tarik harus sudah leleh ( a ≤ a maks leleh )
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 35
Mr = Ø x Mn= 0.8 x 69707304.35= 55765843.48 Nmm > Mu = 15863057.14 Nmm → aman
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
Kontrol εc' harus ≤ εcu' = 0,003
εyβ1
Ø.Vc/2
Ø.Vc
( Vu - Ø. Vc / 2)/x = Vu / ( Vu - Ø. Vc / 2) / Vu
karena
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 36
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
→ Untuk daerah sepanjang x = 0.95 m,
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
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
mm²
dp²
karena
mm²
dp²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 37
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 38
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
Pembebanan Lantai 2• Beban Mati (qD)
B.S. Balok = 0.2 0.3 - 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 400 kg/m² = 160.0 kg/m
Pembebanan Lantai 3• 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
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 39
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
Momen Tumpuan :Mu1 = Mu10 = 1/16 x qu x = 0.0625 x 2202.9 x 12.25 = 1686.58 kgmMu2 = Mu9 = 1/10 x qu x = 0.10 x 2202.9 x 12.25 = 2698.528 kgmMu3 - Mu8 = 1/11 x qu x = 0.09 x 2202.9 x 12.25 = 2453.207 kgm
Momen Lapangan :Mu1-2 = Mu9-10 = 1/14 x qu x = 0.0714 x 2202.9 x 12.25 = 1927.52 kgmMu2-3 → Mu8-9 = 1/16 x qu x = 0.06 x 2202.9 x 12.25 = 1686.58 kgm
Gaya Geser :Qu = 1/2 x qu x L
= 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 = 38.5504 kN
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
Momen Tumpuan :Mu1 = Mu10 = 1/16 x qu x = 0.0625 x 2298.9 x 12.25 = 1760.08 kgmMu2 = Mu9 = 1/10 x qu x = 0.10 x 2298.9 x 12.25 = 2816.128 kgmMu3 - Mu8 = 1/11 x qu x = 0.09 x 2298.9 x 12.25 = 2560.116 kgm
Momen Lapangan :Mu1-2 = Mu9-10 = 1/14 x qu x = 0.0714 x 2298.9 x 12.25 = 2011.52 kgmMu2-3 → Mu8-9 = 1/16 x qu x = 0.06 x 2298.9 x 12.25 = 1760.08 kgm
Gaya Geser :Qu = 1/2 x qu x L
= 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 40
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
Momen Tumpuan :Mu1 = Mu10 = 1/16 x qu x = 0.0625 x 2202.9 x 12.25 = 1686.58 kgmMu2 = Mu9 = 1/10 x qu x = 0.10 x 2202.9 x 12.25 = 2698.528 kgmMu3 - Mu8 = 1/11 x qu x = 0.09 x 2202.9 x 12.25 = 2453.207 kgm
Momen Lapangan :Mu1-2 = Mu9-10 = 1/14 x qu x = 0.0714 x 2202.9 x 12.25 = 1927.52 kgmMu2-3 → Mu8-9 = 1/16 x qu x = 0.06 x 2202.9 x 12.25 = 1686.58 kgm
Gaya Geser :Qu = 1/2 x qu x L
= 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
ds1 = p + Ø tul. begel + (½ x Ø tul. pokok)= 40 + 8 +( ½ x 16 )= 56 mm
ds2 = Ø + Sn ( jarak minimum tulangan )= 16 + 25= 41 mm ≈ 40 mm
d = h - ds1= 300 - 56.0= 244.0 mm
5640
300244
200
L²L²L²
L²L²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 41
Mu tump (-) = 28.1613 kNm = 28161280.00 NmmMn = Mu / Ф
= 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 )
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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
→ Hitung luas tulangan perlu (As,u)
= 0.85 x fc x a x bfy
= 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 . ز 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 42
Digunakan tulangan tarik As = 4 Ø 16 = 803.840 > 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
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
B. Hitung Momen Rencana1). Hitung tinggi blok tegangan beton tekan persegi ekuivalen a
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
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 43
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 )
4). Hitung momen nominal (Mn)dan momen rencana (Mr)
Mn = As x fy x ( d - a/2)= 803.84 x 240 x 205.09= 39565541.95 Nmm
Mr = Ø x Mn= 0.8 x 39565541.95= 31652433.56 Nmm > Mu = 28161280.00 Nmm → aman
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
2. Perhitungan tulangan lapangan :A. Hitung Tulangan
ds1 = p + Ø tul. begel + (½ x Ø tul. pokok)= 40 + 8 +( ½ x 16 )= 56 mm
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 44
→- 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. (+) = 19.2752 kNm = 19275200.00 NmmMn = Mu / Ф
= 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
Karena K < Kmaks maka dipakai tulangan tunggal
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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
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
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 45
# Hitung luas tulangan perlu (As,u)
= 0.85 x fc x a x befy
= 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
# Hitung jumlah tulangan
n =As,u
=1245.417
= 6.20 ≈ 7 buah1/4 . π . ز ¼ x π x 16 ²
Digunakan tulangan tarik As = 7 Ø 16 = 1406.720 > Asu OKtulangan tekan As' = 2 Ø 16 = 401.920 (ditambahkan)
Jumlah tulangan maksimal pada 1 baris :
m =b - 2 x d'
+ 1Ø + Sn
=200 - 2 x 56
+ 116 + 25
= 3.146341 batang → maksimal 3 batang
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
B. Hitung Momen Rencana1). Hitung tinggi blok tegangan beton tekan persegi ekuivalen a
a =As x fy
0.85 x fc' x be
=1406.72 x 240
= 15.13 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
Asu
mm²
Diambil nilai yang terbesar Asu
mm²mm²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 46
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
ρ min < ρ < ρ maks OK
OKOK
3).
a maks leleh = 600 x β1 x dd600 + fy
= 600 x 0.85 x 224600 + 240
= 136 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)
Mn = As x fy x ( d - a/2)= 1406.72 x 240 x 176.43= 59566527.69 Nmm
Mr = Ø x Mn= 0.8 x 59566527.69= 47653222.15 Nmm > Mu = 19275200.00 Nmm → aman
5).
εy = fy / Es= 240 / 200000= 0.00120
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
Kontrol semua tulangan tarik harus sudah leleh ( a ≤ a maks leleh )
Kontrol εc' harus ≤ εcu' = 0,003
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 47
εc' =a
x =15.13
x 0.00120x d - a 0.85 x 224 - 15.13
= 0.000102273 < εcu' = 0.003 → aman
2. Perhitungan tulangan geser :Vu = 38.5504 kN = 38550.40 N
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
→ Untuk daerah sepanjang x = 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
Vs < 2/3 x fc' x b x d= 0.67 x 30 x 200 x 224= 163586.47 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
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
εyβ1
Ø.Vc/2
Ø.Vc
( Vu - Ø. Vc / 2)/x = Vu / ( Vu - Ø. Vc / 2) / Vu
karena
mm²
mm²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 48
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 = 224 / 2 = 112 mmDipilih spasi yang kecil , yaitu s = 112 mm ≈ 110 mmJadi dipakai begel Ø 8 - 110 mm
→ Untuk daerah sepanjang x = 0.75 m,
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
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
Vs < 2/3 x fc' x b x d= 0.67 x 30 x 200 x 224= 163586.47 N jadi syarat spasi begel s <= d/2 dan s <= 600 mm
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
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 = 224 / 2 = 112 mmDipilih spasi yang kecil , yaitu s = 112 mm ≈ 110 mmJadi dipakai begel Ø 8 - 110 mm
dp²
karena
mm²
mm²
dp²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 49
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 50
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 51
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 52
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 53
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 54
Gambar 4.1 Potongan Portal
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 55
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
ht segitiga = 1.75 ml = 4.00 m
4.00
Ra =( 4.00 + 0.5 ) x 1.75
x 0.52
= 1.96875
T1 = a x t x 0.5 segitiga= 1.75 x 1.75 x 0.5= 1.53125
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
ht segitiga = 1.5 ml = 3.50 m
3.50
2/3 x h
1/8 x h x L2
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 56
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 ( 0.333 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 = Mmax21.73438 = 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
ht segitiga = 0.5 ml = 3.50 m
3.50
Ra =( 3.50 + 2.5 ) x 0.5
x 0.52
= 0.75
T1 = a x t x 0.5 segitiga= 0.5 x 0.5 x 0.5= 0.125
1/8 x h x L2
2/3 x h
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 57
T2 = 2.5 x ( 0.5 / 2 )= 0.625
Mmax 1 = Ra x 1/2 x L - T1 x (1/3 x a + b/2) - T2 x b/4= 0.75 x 0.5 x 3.5 - 0.125 x ( 0.333 x 0.5 + 2.5 / 2 )
- 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 58
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 +
= 3.318 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 3.318
= 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 +
= 5.748 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 5.748
= 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 +
= 5.833 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 5.833
= 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 +
= 2.917 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 2.917
= 0.875 kN/m
Beban joint pada ABeban mati titik A = 3.50 x 4.839 = 16.94 kN
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 59
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
= 0.667 kN/mReduksi beban hidup = 0.3Beban hidup q1 = 0.3 x 0.667
= 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 60
= 6.517 kN/mReduksi beban hidup = 0.3Beban hidup q1 = 0.3 x 6.517
= 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 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 = 0.300Beban hidup q1 = 0.30 x 0.486
= 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 +
= 5.017 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 5.017
= 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 +
Beban mati qd = 21.703 kN/m
kN/m3
kN/m2
kN/m2
kN/m1
kN/m2
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 61
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 +
= 9.197 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 9.197
= 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 +
= 9.333 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 9.333
= 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 +
Beban mati qd = 16.470 kN/mbeban hidup :Berat lantai type a = 4.00 x 1.167 = 4.667 kN/m +
= 4.667 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 4.667
= 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 joint pada BBeban mati titikB = 3.50 x 18.560 = 64.96 kNBeban hidup titik B = 3.50 x 1.505 = 5.27 kN
Beban joint pada CBeban mati titik C = 3.50 x 21.703 = 75.96 kNBeban hidup titik C = 3.50 x 2.759 = 9.66 kN
Beban joint pada DBeban mati titik D = 3.50 x 21.860 = 76.51 kNBeban hidup titik D = 3.50 x 2.800 = 9.80 kN
Beban joint pada EBeban mati titik E = 3.50 x 16.470 = 57.64 kNBeban hidup titik E = 3.50 x 1.400 = 4.90 kN
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 62
Beban Yang Bekerja Pada Portal Line 4Beban merata line 4 bagian A-BBeban matiBerat sendiri balok = 0.1 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
= 0.667 kN/mReduksi beban hidup = 0.3Beban hidup q1 = 0.3 x 0.667
= 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 = 9.864 kN/mBeban hidupBerat lantai tipe d = 1.000 x 4.00 x 2 = 8.000 kN/m
= 8.000 kN/mReduksi beban hidup = 0.3Beban hidup q1 = 0.3 x 8
= 2.40 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 = 16.9153 kN/mBeban hidupBerat lantai tipe b = 1.303 x 4.00 x 2 = 10.427 kN/m
= 10.427 kN/mReduksi beban hidup = 0.3Beban hidup q1 = 0.3 x 10.43
= 3.128 kN/m
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HENDRICH DENY .T (12.21.916) 63
4.5 Pembebanan Plat AtapBj. Beton = 24.00Berat dinding = 2.50B. Plat Mati = 3.37B. Hidup lantai = 1.00
Beban merata line ABerat lantai type e = 3.37 x 0.486 = 1.637 kN/mBerat sendiri lisplank = 0.12 x ( 1.00 - 0.10 ) x 24.0 = 2.592 kN/m +
Beban mati qd = 4.229 kN/mbeban hidup :Berat lantai type e = 1.00 x 0.486 = 0.486 kN/m
= 0.486 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 0.486
= 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 +
= 1.619 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 1.619
= 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 +
= 2.299 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 2.299
= 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 +
Beban mati qd = 8.934 kN/m
kN/m3
kN/m2
kN/m2
kN/m2
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 64
beban hidup :Berat lantai type a = 1.00 x 1.167 x 2 = 2.333 kN/m +
= 2.333 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 2.333
= 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 +
Beban mati qd = 5.007 kN/mbeban hidup :Berat lantai type a = 1.00 x 1.167 = 1.167 kN/m +
= 1.167 kN/mReduksi beban hidup = 0.300Beban hidup q1 = 0.30 x 1.167
= 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 joint pada BBeban mati titikB = 3.50 x 6.530 = 22.85 kNBeban hidup titik B = 3.50 x 0.486 = 1.70 kN
Beban joint pada CBeban mati titik C = 3.50 x 8.820 = 30.87 kNBeban hidup titik C = 3.50 x 0.690 = 2.41 kN
Beban joint pada DBeban mati titik D = 3.50 x 8.934 = 31.27 kNBeban hidup titik D = 3.50 x 0.700 = 2.45 kN
Beban joint pada EBeban mati titik E = 3.50 x 5.007 = 17.52 kNBeban hidup titik E = 3.50 x 0.350 = 1.22 kN
Beban Yang Bekerja Pada Portal Line 4Beban merata line 4 bagian A-BBeban matiBerat sendiri balok = 0.1 x ( 0.15 - 0.12 ) x 24.00 = 0.072 kN/mBerat lantai tipe f = 0.333 x 3.37 x 2 = 2.244 kN/m
Beban mati qd = 2.316 kN/mBeban hidupBerat lantai tipe f = 0.333 x 1.00 x 2 = 0.667 kN/m
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 65
= 0.667 kN/mReduksi beban hidup = 0.3Beban hidup q1 = 0.3 x 0.667
= 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 3.37 x 2 = 6.732 kN/m
Beban mati qd = 7.36 kN/mBeban hidupBerat lantai tipe d = 1.000 x 1.00 x 2 = 2.000 kN/m
= 2.000 kN/mReduksi beban hidup = 0.3Beban hidup q1 = 0.3 x 2
= 0.60 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 3.37 x 2 = 8.774 kN/m
Beban mati qd = 13.65 kN/mBeban hidupBerat lantai tipe b = 1.303 x 1.00 x 2 = 2.607 kN/m
= 2.607 kN/mReduksi beban hidup = 0.3Beban hidup q1 = 0.3 x 2.607
= 0.782 kN/m
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 66
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 67
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
Total beban lt 2 W lt2 = 5068.84 + 453.6 = 5522.44 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 68
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
Total beban lt 1 W lt1 = 5099.47 + 283.5 = 5382.97 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 69
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)
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 70
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 71
BAB VIPORTAL MELINTANG
BALOK INDUK 35/70 ( LINE 4 )
A. Data Perencanaan
Ø tul.pokok = 26 mmØ tul. begel = 8 mmp = 50 mmfc' = 30 Mpafy = 400 Mpafy tul. Geser = 240 Mpaβ1 = 0.85BJ beton = 2400 kg/m³dimensi balok → b = 0.35 m 350 mm
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
ds1 = p + Ø tul. begel + (½ x Ø tul. pokok)= 50 + 8 +( ½ x 26 )= 71 mm
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 72
Mu tump (-) = 1177.76 kNm = 1177760000.00 NmmMn = Mu / Ф
= 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
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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
→ Hitung luas tulangan perlu (As,u)
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 73
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 . ز 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 . ز ¼ x π x 26 ²
Digunakan tulangan tarik As' = 8 Ø 26 = 4245.280 > Asu OK
Jumlah tulangan maksimal pada 1 baris :
m =b - 2 x ds1
+ 1Ø + Sn
=350 - 2 x 71
+ 126 + 25
= 5.0784314 batang → maksimal 5 batang
7151
70051
578.05171
350
d²
mm²
mm²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 74
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)
4). Hitung momen nominal (Mn)dan momen rencana (Mr)
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 )
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 75
Mr = Ø x Mn= 0.8 x 1596079775.11= 1276863820.1 Nmm > Mu = 1177760000.00 Nmm → aman
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
2. Perhitungan tulangan lapangan :A. Hitung Tulangan
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
Diambil nilai yang terkeci yaitu be = 2000 mmJarak bersih antar tulangan Sn = 25 mm 26 mm)
2000
700603.5
5171
350
Kontrol εc' harus ≤ εcu' = 0,003
εyβ1
Mencari lebar pelat efektif (be) Bentang balok (as-as) λ =Bentang bersih pelat λn =
be
be
be
( > Ø =
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 76
Mu lap. (+) = 203.3140 kNm = 203314000.00 NmmMn = Mu / Ф
= 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
Karena K < Kmaks maka dipakai tulangan tunggal
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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 fc x a x befy
= 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
# Hitung jumlah tulangan
n =As,u
=4224.500
= 7.96 ≈ 8 buah1/4 . π . ز ¼ x π x 26 ²
Ø x be x d²
382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²
Asu
mm²
Diambil nilai yang terbesar Asu
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 77
Digunakan tulangan tarik As = 8 Ø 26 = 4245.280 > Asu OKtulangan tekan As' = 2 Ø 26 = 1061.320 (ditambahkan)
Jumlah tulangan maksimal pada 1 baris :
m =b - 2 x d'
+ 1Ø + Sn
=350 - 2 x 71
+ 126 + 25
= 5.0784314 batang → maksimal 5 batang
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
B. Hitung Momen Rencana1). Hitung tinggi blok tegangan beton tekan persegi ekuivalen a
a =As x fy
0.85 x fc' x be
=4245.28 x 400
= 33.30 mm0.85 x 30 x 2000
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 = 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
Kontrol rasio tulangan ρ min ≤ ρ ≤ ρ maks
mm²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 78
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
3).
a maks leleh = 600 x β1 x dd600 + fy
= 600 x 0.85 x 604600 + 400
= 307.79 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)
Mn = As x fy x ( d - a/2)= 4245.28 x 400 x 586.85= 996540157.05 Nmm
Mr = Ø x Mn= 0.8 x 996540157.05= 797232125.64 Nmm > Mu = 203314000.00 Nmm → aman
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
Jika ρ < ρ min → balok diperkecilJika ρ > ρ maks → balok diperbesar
Kontrol semua tulangan tarik harus sudah leleh ( a ≤ a maks leleh )
Kontrol εc' harus ≤ εcu' = 0,003
εyβ1
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 79
2. Perhitungan tulangan geser :Vu = 387.595 kN = 387595.00 N
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
→ Untuk daerah sepanjang x = 3.20 m,
Vu > Ø.Vc maka gaya geser ditahan begelVu > Ø.Vc/2 NO
Ø.Vc/2 > Vu > Ø.Vc NO
Av,u =b x S
=350 x 1000
= 486.11113 x fy 3 x 240
Av,u =75 x fc' x b x S
=75 x 30 x 350 x 1000
1200 x fy 1200 x 240= 499.22629 mm²
Dipilih yang besar , jadi Av,u = 2394.172 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
2394.172
s = 41.96858 mm
Ø.Vc/2
Ø.Vc
( Vu - Ø. Vc / 2)/x = Vu / ( Vu - Ø. Vc / 2) / Vu
karena
mm²
dp²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 80
n = Jumlah kaki begeldp² = Diameter begel
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
→ Untuk daerah sepanjang x = 0.80 m,
Vud > Ø.Vc maka gaya geser ditahan begelVud > Ø.Vc/2 NO
Ø.Vc/2 > Vud > Ø.Vc NO
Luas begel / m :
Av,u =b x S
=350 x 1000
= 486.11113 x fy 3 x 240
Av,u =75 x fc' x b x S
=75 x 30 x 350 x 1000
1200 x fy 1200 x 240= 499.22629 mm²
Dipilih yang besar , jadi Av,u = 499.226 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
499.226
s = 201.27145 mm
n = Jumlah kaki begeldp² = Diameter begel
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 81
PORTAL MELINTANGBALOK INDUK 25/30 ( LINE 4 )
A. Data Perencanaan
Ø tul.pokok = 26 mmØ tul. begel = 8 mmp = 50 mmfc' = 30 Mpafy = 400 Mpafy tul. Geser = 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
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
ds1 = p + Ø tul. begel + (½ x Ø tul. pokok)= 50 + 8 +( ½ x 26 )= 71 mm
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 82
Mu tump (-) = 82.33 kNm = 82330000.00 NmmMn = Mu / Ф
= 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
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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
→ Hitung luas tulangan perlu (As,u)
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 83
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 . ز 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 . ز ¼ x π x 26 ²
Digunakan tulangan tarik As' = 2 Ø 26 = 1061.320 > Asu OK
Jumlah tulangan maksimal pada 1 baris :
m =b - 2 x ds1
+ 1Ø + Sn
=250 - 2 x 71
+ 126 + 25
= 3.1176471 batang → maksimal 3 batang
7151
300
203.5
71
250
d²
mm²
mm²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 84
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)
4). Hitung momen nominal (Mn)dan momen rencana (Mr)
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 )
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 85
Mr = Ø x Mn= 0.8 x 128506190.53= 102804952.4 Nmm > Mu = 82330000.00 Nmm → aman
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
2. Perhitungan tulangan lapangan :A. Hitung Tulangan
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
Diambil nilai yang terkeci yaitu be = 750 mmJarak bersih antar tulangan Sn = 25 mm 26 mm)
750
300229.0
5171
250
Kontrol εc' harus ≤ εcu' = 0,003
εyβ1
Mencari lebar pelat efektif (be) Bentang balok (as-as) λ =Bentang bersih pelat λn =
be
be
be
( > Ø =
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 86
Mu lap. (+) = 4.0050 kNm = 4005000.00 NmmMn = Mu / Ф
= 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
Karena K < Kmaks maka dipakai tulangan tunggal
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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
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 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
# Hitung jumlah tulangan
n =As,u
=601.125
= 1.13 ≈ 2 buah1/4 . π . ز ¼ x π x 26 ²
Ø x be x d²
382,5.β1.fc'.(600 + fy - 225.β1)(600 + fy)²
Asu
mm²
Diambil nilai yang terbesar Asu
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 87
Digunakan tulangan tarik As = 2 Ø 26 = 1061.320 > Asu OKtulangan tekan As' = 2 Ø 26 = 1061.320 (ditambahkan)
Jumlah tulangan maksimal pada 1 baris :
m =b - 2 x d'
+ 1Ø + Sn
=250 - 2 x 71
+ 126 + 25
= 3.1176471 batang → maksimal 3 batang
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
B. Hitung Momen Rencana1). Hitung tinggi blok tegangan beton tekan persegi ekuivalen a
a =As x fy
0.85 x fc' x be
=1061.32 x 400
= 22.20 mm0.85 x 30 x 750
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 = 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
Kontrol rasio tulangan ρ min ≤ ρ ≤ ρ maks
mm²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 88
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
3).
a maks leleh = 600 x β1 x dd600 + fy
= 600 x 0.85 x 204600 + 400
= 103.79 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)
Mn = As x fy x ( d - a/2)= 1061.32 x 400 x 192.40= 81679708.84 Nmm
Mr = Ø x Mn= 0.8 x 81679708.84= 65343767.07 Nmm > Mu = 4005000.00 Nmm → aman
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
Jika ρ < ρ min → balok diperkecilJika ρ > ρ maks → balok diperbesar
Kontrol semua tulangan tarik harus sudah leleh ( a ≤ a maks leleh )
Kontrol εc' harus ≤ εcu' = 0,003
εyβ1
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 89
2. Perhitungan tulangan geser :Vu = 66.715 kN = 66715.00 N
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
→ Untuk daerah sepanjang x = 1.10 m,
Vu > Ø.Vc maka gaya geser ditahan begelVu > Ø.Vc/2 NO
Ø.Vc/2 > Vu > Ø.Vc NO
Av,u =b x S
=250 x 1000
= 347.22223 x fy 3 x 240
Av,u =75 x fc' x b x S
=75 x 30 x 250 x 1000
1200 x fy 1200 x 240= 356.59021 mm²
Dipilih yang besar , jadi Av,u = 870.414 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
870.414
s = 115.43931 mm
Ø.Vc/2
Ø.Vc
( Vu - Ø. Vc / 2)/x = Vu / ( Vu - Ø. Vc / 2) / Vu
karena
mm²
dp²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 90
n = Jumlah kaki begeldp² = Diameter begel
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
→ Untuk daerah sepanjang x = 0.40 m,
Vud > Ø.Vc maka gaya geser ditahan begelVud > Ø.Vc/2 NO
Ø.Vc/2 > Vud > Ø.Vc NO
Luas begel / m :
Av,u =b x S
=250 x 1000
= 347.22223 x fy 3 x 240
Av,u =75 x fc' x b x S
=75 x 30 x 250 x 1000
1200 x fy 1200 x 240= 356.59021 mm²
Dipilih yang besar , jadi Av,u = 356.590 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
356.590
s = 281.78003 mm
n = Jumlah kaki begeldp² = Diameter begel
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
PERANCANGAN STRUKTUR BETON BERTULANG
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY . T (12.21.916) 92
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 Desak x > xbx 300 mma = β1*x 255.000 mmfs' = (x-d')*600 / x 458.000 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 1950.750 kNCs = As'*(fs'-0,85*fc') / 1000 854.158 kNfs = (d-x)*600 / x 58.000 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 58.000 MpaTs = (As*fs) / 1000 132.286 kNPnb = Cc + Cs - Ts 2672.622 kNy = h/2 200 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 268.681 kNme = (Mnb / Pnb)*1000 100.531 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY . T (12.21.916) 93
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY . T (12.21.916) 94
--- 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
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4500
Diagram Interaksi
Kolom 30/40, 6D22Pu, Mu
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 95
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 %
Beban SentrisPo Po = (0,85*fc'*(Ag-Ast)+Ast*fy) / 1000 12597.294 kNPn Pn = 0,8*Po 10077.835 kN
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 96
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 Desak x > xbx 400 mma = β1*x 340.000 mmfs' = (x-d')*600 / x 488.250 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 4335.000 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-x)*600 / x 188.250 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 188.250 MpaTs = (As*fs) / 1000 1243.429 kNPnb = Cc + Cs - Ts 5565.218 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1401.751 kNme = (Mnb / Pnb)*1000 251.877 mm
Kondisi Patah Desak x > xbx 450 mma = β1*x 382.500 mmfs' = (x-d')*600 / x 500.667 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 4876.875 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-x)*600 / x 100.667 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 100.667 MpaTs = (As*fs) / 1000 664.923 kNPnb = Cc + Cs - Ts 6685.599 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1238.108 kNme = (Mnb / Pnb)*1000 185.190 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 97
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
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 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 98
--- 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
Rekap (Urutan P dari terkecil)Pnb Mnb
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
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Diagram Interaksi
Kolom 50/60, 10D29Pu, Mu
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 99
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 %
Beban SentrisPo Po = (0,85*fc'*(Ag-Ast)+Ast*fy) / 1000 12597.294 kNPn Pn = 0,8*Po 10077.835 kN
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 100
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 Desak x > xbx 400 mma = β1*x 340.000 mmfs' = (x-d')*600 / x 488.250 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 4335.000 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-x)*600 / x 188.250 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 188.250 MpaTs = (As*fs) / 1000 1243.429 kNPnb = Cc + Cs - Ts 5565.218 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1401.751 kNme = (Mnb / Pnb)*1000 251.877 mm
Kondisi Patah Desak x > xbx 450 mma = β1*x 382.500 mmfs' = (x-d')*600 / x 500.667 Mpa--- fs' > fy ,kondisi tul.tekan leleh--- maka selanjutnya dipakai fs' = fy = 400 MpaCc = (0,85*fc'*b*a) / 1000 4876.875 kNCs = As'*(fs'-0,85*fc') / 1000 2473.647 kNfs = (d-x)*600 / x 100.667 Mpa--- fs < fy ,kondisi tul.tarik belum leleh--- maka selanjutnya dipakai fs = 100.667 MpaTs = (As*fs) / 1000 664.923 kNPnb = Cc + Cs - Ts 6685.599 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1238.108 kNme = (Mnb / Pnb)*1000 185.190 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 101
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
Kondisi Patah Tarik x < xbx 150 mma = β1*x 127.500 mmfs' = (x-d')*600 / x 302.000 Mpa--- fs' < fy ,kondisi tul.tekan belum leleh--- maka selanjutnya dipakai fs' = 302 MpaCc = (0,85*fc'*b*a) / 1000 1625.625 kNCs = As'*(fs'-0,85*fc') / 1000 1826.337 kNfs = (d-x)*600 / x 1502.000 Mpa--- fs > fy ,kondisi tul.tarik leleh--- maka selanjutnya dipakai fs = fy = 400 MpaTs = (As*fs) / 1000 2642.079 kNPnb = Cc + Cs - Ts 809.883 kNy = h/2 300 mmMnb = ((Cc*(y-a/2)+Cs*(y-d')+Ts*(d-y))/1000 1391.682 kNme = (Mnb / Pnb)*1000 1718.374 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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 102
--- 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
Rekap (Urutan P dari terkecil)Pnb Mnb
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
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Diagram Interaksi
Kolom 40/50, 3D22Pu, Mu
PERANCANGAN STRUKTU BETON BERTULANG
HENDRICH DENY .T (12.21.916) 103
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 104
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
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 105
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
Gaya geser yang ditahan beton (Ø.Vc) :
σ min + ( B - a ) x ( σ maks - σ min )
) x (
a x B x ( σ maks + σ a )
cf '
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 106
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
Jadi, Vu = 607.396 kN < Ø.Vc = 1732.425 kN → AMAN
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 '
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 107
= 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
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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
→ Hitung luas tulangan perlu (As,u)
= 0.85 x fc x a x bfy
= 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
Diambil nilai yang terbesar Asu
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 108
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 '
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 109
PONDASI BUJUR SANGKAR TENGAH BE
A. Data Perencanaan
P = 1473.22 kNMu = 926.169 kNmDf = 1.50 mØ tul = 29 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 = 0.60 m = 600 mm
dimensi kolom → bk = 0.50 m = 500 mmhk = 0.60 m = 600 mmDf-ts = 0.90 m = 900 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 0.60 m x 2400 kg/m³ x 1.2 = 6912 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 0.90 m ) - vol kolom ) x x 1600 kg/m³ x 1.2 = 6394 kg
= 14381 kg= 143.8 kN
• Beban terpusat (P) = = 1473.2 kNPu = 1617.0 kN
C. Daya Dukung Ultimit Tanah
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
karena σ maks = 1098.88 kN/m² < qs = 2000 kN/m² → AMAN
kN/m²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 110
D. Kekuatan Geser pondasi
ds = p + (½ x Ø tul)= 75 +( ½ x 29 )= 89.5 mm
d = ts - ds= 600 - 89.5= 510.5 mm
gaya geser 1 araha = ½ B– ½ hk – d
= ( 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 yang ditahan beton (Ø.Vc) :
Ø.Vc = Ø x ⅙ x x B x d= 0.75 x ⅙ x 30 x 2000 x 510.5= 699030.91 N= 699.031 kN
Jadi, Vu = 391.533 kN < Ø.Vc = 699.031 kN → AMAN
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
Vu = ((B² ) - (B1 x B2)) x (σ maks - σ min
)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
Gaya geser yang ditahan beton (Ø.Vc) :
σ min + ( B - a ) x ( σ maks - σ min )
) x (
a x B x ( σ maks + σ a )
cf '
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 111
Vc = ( 1 + 2
) ( ⅙ x x bo x d )βc
= ( 1 +2
) ( ⅙ x 30 x 4242 x 510.5 )1.20= 5271625.13 N= 5271.625 kN
Vc = ( 2 + αs x d
) (1
x x bo x d )→ αs = 40 ( Kolom dalam )
bo 12 → αs = 30 ( Kolom tepi )
= ( 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
Dipilih Vc yang terkecil, yaitu Vc = 3953.719 kNØ.Vc = 0.75 x 3953.719
= 2965.289 kN
Jadi, Vu = 1163.388 kN < Ø.Vc = 2965.289 kN → AMAN
E. Penulangan Pondasi
ds = p + (½ x Ø tul) + Ø tul= 75 +( ½ x 29 )+ 29= 119 mm
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 =
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 '
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 112
= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 )
= 7.8883 Mpa
Syarat K = 1.23746 < Kmaks = 7.888 → AMAN
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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
→ Hitung luas tulangan perlu (As,u)
= 0.85 x fc x a x bfy
= 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
s pakai dipilih yang terkecil yaitu s = 391.9418 mm ≈ 350 mmmaka digunakan tulangan dengan Ø 29 - 350 mm
Asu
mm²
Asu
Diambil nilai yang terbesar Asu
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 113
panjang penyaluran dan penyambungan tulangan dipakai tulangan Ø 29
= 4 x
=29 x 400
= 529.47 mm4 x 30
= 0,04 x db x fy= 0.04 x 29 x 400 = 464 mm < 529 mm → AMAN
jadi dipakai panjang penyaluran = 529.5 mm ≈ 500 mmuntuk memermudah pemasangan makan di gunakan Ø 29 - 500
E. Kuat Dukung Pondasi
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
tidak boleh kurang dari λdb
cf '
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 114
PONDASI BUJUR SANGKAR TEPI CF
A. Data Perencanaan
P = 1401.24 kNMu = 905.037 kNmDf = 1.50 mØ tul = 29 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 = 0.50 m = 500 mmhk = 0.60 m = 600 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 = 7104 kg
= 13939 kg= 139.4 kN
• Beban terpusat (P) = = 1401.2 kNPu = 1540.6 kN
C. Daya Dukung Ultimit Tanah
σ =Pu
±Mu
B² . B³⅙
=1540.6
±905.037
2.00 ² ( )⅙ x 2.00 ³
σ maks = 1063.935 kN/m²σ min = -293.620 kN/m²
karena σ maks = 1063.94 kN/m² < qs = 2000 kN/m² → AMAN
kN/m²
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 115
D. Kekuatan Geser pondasi
ds = p + (½ x Ø tul)= 75 +( ½ x 29 )= 89.5 mm
d = ts - ds= 500 - 89.5= 410.5 mm
gaya geser 1 araha = ½ B– ½ hk – d
= ( 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 yang ditahan beton (Ø.Vc) :
Ø.Vc = Ø x ⅙ x x B x d= 0.75 x ⅙ x 30 x 2000 x 410.5= 562100.27 N= 562.100 kN
Jadi, Vu = 559.130 kN < Ø.Vc = 562.100 kN → AMAN
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
Vu = ((B² ) - (B1 x B2)) x (σ maks - σ min
)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
Gaya geser yang ditahan beton (Ø.Vc) :
σ min + ( B - a ) x ( σ maks - σ min )
) x (
a x B x ( σ maks + σ a )
cf '
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 116
Vc = ( 1 + 2
) ( ⅙ x x bo x d )βc
= ( 1 +2
) ( ⅙ x 30 x 3842 x 410.5 )1.20= 3839269.79 N= 3839.270 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 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
Dipilih Vc yang terkecil, yaitu Vc = 2879.452 kNØ.Vc = 0.75 x 2879.452
= 2159.589 kN
Jadi, Vu = 1186.262 kN < Ø.Vc = 2159.589 kN → AMAN
E. Penulangan Pondasi
ds = p + (½ x Ø tul) + Ø tul= 75 +( ½ x 29 )+ 29= 119 mm
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 =
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 '
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 117
= 382.5 x 0.85 x 30 x ( 600 + 400 - 225 x 0.85 )( 600 + 400 )
= 7.8883 Mpa
Syarat K = 1.90547 < Kmaks = 7.888 → AMAN
→ Mencari tinggi blok tegangan beton tekon ekuivalen a
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
→ Hitung luas tulangan perlu (As,u)
= 0.85 x fc x a x bfy
= 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
s pakai dipilih yang terkecil yaitu s = 349.3259 mm ≈ 350 mmmaka digunakan tulangan dengan Ø 29 - 350 mm
Asu
mm²
Asu
Diambil nilai yang terbesar Asu
PERANCANGAN STRUKTUR BETON BERTULANG
HENDRICH DENY .T (12.21.916) 118
panjang penyaluran dan penyambungan tulangan dipakai tulangan Ø 29
= 4 x
=29 x 400
= 529.47 mm4 x 30
= 0,04 x db x fy= 0.04 x 29 x 400 = 464 mm < 529 mm → AMAN
jadi dipakai panjang penyaluran = 529.5 mm ≈ 500 mmuntuk memermudah pemasangan makan di gunakan Ø 29 - 500
E. Kuat Dukung Pondasi
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
tidak boleh kurang dari λdb
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
dimensi kolom → bk = 0.50 m = 500 mmhk = 0.60 m = 600 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/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
C. Daya Dukung Ultimit Tanah
σ =Pu
±Mu . B³⅙
=1612.61
±926.169
2.00 ² ( )⅙ x 2.00 ³
σ maks = 1097.780 kN/m²σ min = -291.474 kN/m²
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
gaya geser 1 araha = ½ B– ½ hk – d
= ( 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
Jadi, Vu = 532.129 kN < Ø.Vc = 596.333 kN → AMAN
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
σ min + ( B - a ) x ( σ maks - σ min )
) x (
a x B x ( σ maks + σ a )
Gaya geser yang ditahan beton (Ø.Vc) :
Ø.Vc
((B² ) - (B1 x B2))
Gaya geser yang ditahan beton (Ø.Vc) :
cf '
Vc = ( 1 + 2
) ( ⅙ x x bo x d )βc
= ( 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 )
bo 12 → αs = 30 ( Kolom tepi )
= ( 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
Dipilih Vc yang terkecil, yaitu Vc = 3134.326 kNØ.Vc = 0.75 x 3134.326
= 2350.744 kN
Jadi, Vu = 1222.074 kN < Ø.Vc = 2350.744 kN → AMAN
E. Penulangan Pondasi
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
σ min + ( B - x ) x ( σ maks - σ min )
) 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
s pakai dipilih yang terkecil yaitu s = 160.762 mm ≈ 160 mmmaka digunakan tulangan dengan Ø 29 - 160 mm
panjang penyaluran dan penyambungan tulangan dipakai tulangan Ø 29
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
dimensi pondasi → B = 2.00 m = 2000 mmL = 2.00 m = 2000 mm ts = 0.50 m = 500 mm
dimensi kolom → bk = 0.50 m = 500 mmhk = 0.70 m = 700 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 0.5 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 = 7008 kg
= 13843 kg= 138.4 kN
• Beban terpusat (N) = = 1401.2 kNPu = 1539.7 kN
C. Daya Dukung Ultimit Tanah
σ =Pu
±Mu
B² . B³⅙
=1539.7
±905.037
2.00 ² ( )⅙ x 2.00 ³
σ maks = 1063.695 kN/m²σ min = -293.860 kN/m²
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
gaya geser 1 araha = ½ B– ½ hk – d
= ( 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 yang ditahan beton (Ø.Vc) :
Ø.Vc = 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
Jadi, Vu = 425.095 kN < Ø.Vc = 596.333 kN → AMAN
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
Vu = ((B² ) - (B1 x B2)) x (σ maks - σ min
)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
Gaya geser yang ditahan beton (Ø.Vc) :
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 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 )
bo 12 → αs = 30 ( Kolom tepi )
= ( 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
Dipilih Vc yang terkecil, yaitu Vc = 3293.348 kNØ.Vc = 0.75 x 3293.348
= 2470.011 kN
Jadi, Vu = 1130.787 kN < Ø.Vc = 2470.011 kN → AMAN
E. Penulangan Pondasi
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
)
σ min + ( B - x ) x ( σ maks - σ min )
) 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
s pakai dipilih yang terkecil yaitu s = 333.199 mm ≈ 300 mmmaka digunakan tulangan dengan Ø 29 - 300 mm
panjang penyaluran dan penyambungan tulangan dipakai tulangan Ø 29
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