Tugas Beton terbaru

BAB III

ANALISA PERHITUNGAN STRUKTUR

3.1 PERHITUNGAN PLAT ATAP

3.1.1 PERHITUNGAN TEBAL PLAT ATAP

A. Perkiraan tinggi balok

a. Balok arah X, dengan bentang Lx = 5 m = 5000 mm

hx =

b. Balok arah Y, dengan bentang Ly = 5 m = 5000 mm

hy =

B. Perkiraan lebar balok

a. Balok arah X, dengan hx = 450 mm

b =

b. Balok arah Y, dengan hy= 450mm

b =

C. Gambar penampang balok plat atap

a. Balok arah X b. Balok arah Y

450 450

225 225

D. Menghitung tebal plat atap

Diketahui data material :fy = 300 Mpafc = 22,5 Mpalebar plat bw = 1000 mm = 1 m

Ø tulangan = 8 mm

Nilai banding panjang terhadap lebar bentang bersih β =

Persamaan perhitungan tebal plat sebagai berikut

Batas minimun tebal plat atap

h min = =

Batas maximum tebal plat atap

h max =

syarat h min ≤ h ≤ hmax → 10,61≤ h ≤ 13,26 cm

diambil tebal plat atap h = 12 cm

3.1.2 PEMBEBANAN PLAT ATAP

Wu = 1,2 WD + 1,6 WL

Beban hidup hidup/bekerja (WL ) untuk plat atap = 1 kN/m2

WD didapat dari

Berat plat = 0,12 x 24 = 2,88 kN/m2

Lapisan aspal = 0,005 x 14 = 0,07 kN/m2

Plafond & Penggantung = 0,07 + 0,11= 0,18 kN/m2

Berat genangan air hujan = 0,05 x 10 = 0,5 kN/m2 +

= 3.63 kN/m2

Maka, Wu = 1,2 (3,63) + 1,6 (1) = 5,956 kN/m2

Perhitungan Geser :

dx = h – p – 0,5 Фtul = 120 – 25 – 0,5 . 8 = 91 mm

dy = h – p – 0,5 Фtul - Фtul = 120 – 25 – 0,5 . 8 - 8 = 83 mm

= 0,6 (reduksi geser)

Vc = 45,5 kN > Vu = 16,353 kN ……. OK (tidak memerlukan tulangan geser)

3.1.3 PEMODELAN STRUKTUR

a. Type A

= = 1,00

b. Type B

= = 2,53 ≈ 2,5

c. Type C

= = 1,00

3.1.4 MENENTUKAN MOMEN

a. Menentukan momen plat atap type A dengan ly/l×= 1,00

ml× = 0,001 wu . Lx2. 30 = 0,03 × 5,956 × 1,88752 = 0,585 KNm

mly = 0,001 wu . Lx2. 30 = 0,03 × 5,956 × 1,88752 = 0,585 KNm

mt× = -0,001 wu . Lx2. 68 = -0,068 × 5,956 × 1,88752 = -1,327 KNm

mty = -0,001 wu . Lx2. 68 = -0,068 × 5,956 × 1,88752 = -1,327 KNm

mtix = ½ mt× = 0,663KNm

mtiy = ½ mt× = -0,663 KNm

b. Menentukan momen plat atap type B dengan ly/l× = 2,5 > 2(analisa plat satu arah)

ml× = 0,001 wu . Lx2. 63 = 0,063 × 5,956 × 4,7752 = 7,865 KNm

mly = 0,001 wu . Lx2. 16 = 0,016 × 5,956 × 4,7752 = 1,997 KNm

mt× = -0,001 wu . Lx2. 124 = -0,124 × 5,956 × 4,7752 = -15,482 KNm

mty = -0,001 wu . Lx2. 71 = -0,071 × 5,956 × 4,7752 = -8,864 KNm

mtix = ½ mt× = -7,741 KNm

c. Menentukan momen plat atap type C dengan ly/l× = 1,00

ml× = 0,001 wu . Lx2. 25 = 0,025 × 5,956 × 4,7752 = 3,121 KNm

mly = 0,001 wu . Lx2. 25 = 0,025 × 5,956 × 4,7752 = 3,121 KNm

mt× = -0,001 wu . Lx2. 51 = -0,051 × 5,956 × 4,7752 = -6,368 KNm

mty = -0,001 wu . Lx2. 51 = -0,051 × 5,956 × 4,7752 = -6,368 KNm

3.1.5 PERHITUNGAN TULANGAN

ρb =

ρmax = 0,75 x ρb = 0,75 x 0,0425 = 0,0319

dx = h – p – 0,5 Фtul = 120 – 25 – 0,5 . 8 = 91 mm

dy = h – p – 0,5 Фtul - Фtul = 120 – 25 – 0,5 . 8 - 8 = 83 mm

Plat Type A

Daerah Lapangan x

D = dx = 91 mm

ρmin = 0,0047

Mu = 0,585 kNm

ρmax = 0,0319

Mn = kNm = 731000 Nmm

Menentukan ρ yang diperlukan :

ρ =

ρmin = 0,0047 > ρperlu = 0,0003

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 91 = 427,7 mm2

Maka digunakan ø 8 – 100 = 503

Kontrol jarak penulangan plat

S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm

Kontrol kekuatan plat

a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509 kNm > Mu = 0,585 kNm .....................OK!

Daerah Lapangan y

dy = 83 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 0,585 kNm

Mn = kNm = 731000


ρ =

ρmin = 0,0047 > ρperlu = 0,0003

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 83 = 390,1 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,7 kNm > Mu = 0,585 kNm .......................OK!

Daerah tumpuan x

D = dx = 91 mm

ρmin = 0,0047

ρmax = 0,0319

Mu=1,327kNm

kNm = 1659000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0007

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 91 = 427,7 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,9/2)

= 10,509 kNm > Mu = 1,327kNm .....................OK!

Daerah tumpuan y

dy = 83 mm ρmin = 0,0047

ρmax = 0,0319 Mu=1,327kNm

Mn = kNm = 1659000


ρ =

ρmin = 0,0047 > ρperlu = 0,0007

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 83 = 390,1 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,7 kNm > Mu = 1,327kNm .......................OK!

Daerah tumpuan jepit ix

D = dx = 91 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 0,663kNm

Mn = kNm = 287000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0001

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 91 = 427,7 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509 kNm > Mu = 0,663kNm .....................OK!

Daerah tumpuan jepit iy

D = dx = 83 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 0,663kNm

Mn = kNm = 287000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0001

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 83 = 390,1 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (83-7,89/2)

= 9,543 kNm > Mu = 0,663kNm .....................OK!

Plat Type B

Daerah Lapangan x

D = dx = 91 mm ρmin = 0,0047

ρmax = 0,0319 Mu = 7,865 kNm

Mn = kNm = 9831250 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0040

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 91 = 427,7 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509kNm > Mu = 7,865kNm .....................OK!

Daerah Lapangan y

D = dy = 83 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 1,997 kNm

Mn = kNm = 2496250 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,001

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 83 = 390,1 mm2



S = 125 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,704 kNm > Mu = 1,997

kNm .......................OK!

Daerah tumpuan x

D = dx = 91 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 15,482 kNm

Mn = kNm = 19352500 Nmm


ρ =

ρmin = 0,0047 < ρperlu = 0,0082

ρperlu = 0,0082

luas tulangan perlu

Aslx = ρ . bw . d = 0,0082. 1000 . 91 = 746,2 mm2



S = 50 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 1005 . 300 (91-15,76/2)

= 20,048 kNm > Mu = 15,482

kNm .....................OK!

Daerah tumpuan y

D = dy = 83 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 8,864 kNm

Mn = kNm = 11080000 Nmm


ρ =

ρmin = 0,0047 < ρperlu = 0,0055

ρperlu = 0,0055

luas tulangan perlu

Aslx = ρ . bw . d = 0,0055 . 1000 . 83 = 456,5 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (83-7,89/2)

= 10,509 kNm > Mu = 8,864

kNm .......................OK!


D = dx = 91 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 7,741 kNm

Mn = kNm = 9676250 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0039

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 91 = 427,7 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509 kNm > Mu = 7,741 kNm .....................OK!

Plat Type C

Daerah Lapangan x

D = dx = 91 mm ρmin = 0,0047

ρmax = 0,0319 Mu = 3,121 kNm

Mn = kNm = 390000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,00016

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 91 = 427,7 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509kNm > Mu = 3,121 kNm .....................OK!

Daerah Lapangan y

D = dy = 83 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 3,121 kNm

Mn = kNm = 390000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0002

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 83 = 390,1 mm2



S = 125 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,704 kNm > Mu = 3,121 kNm .......................OK!

Daerah tumpuan x

D = dx = 91 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 6,368 kNm

Mn = kNm = 796000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0003

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 91 = 427,7 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509kNm > Mu = 6,368 kNm .....................OK!

Daerah tumpuan y

D = dy = 83 mm

ρmin = 0,0047

ρmax = 0,0319

Mu = 6,368 kNm

Mn = kNm = 796000 Nmm


ρ =

ρmin = 0,0047 < ρperlu = 0,0004

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 83 = 390,1 mm2



S = 125 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,704 kNm > Mu = 6,368 kNm .......................OK!

3.2 PERHITUNGAN PLAT LANTAI

3.2.1 PERHITUNGAN TEBAL PLAT LANTAI

A. Perkiraan tinggi balok

a. Balok arah X, dengan bentang Lx = 6 m = 6000 mm

hx =

b. Balok arah Y, dengan bentang Lx = 5 m = 5000 mm

hy =

B. Perkiraan lebar balok

a. Balok arah X, dengan hx = 50 cm

b =

b. Balok arah Y, dengan hy= 45 cm

b =

C. Gambar penampang balok plat Lantai

a. Balok arah X b. Balok arah Y

500 450

250 225

D. Menghitung tebal plat lantai

Diketahui data material :

fy = 300 Mpa

fc = 22,5 Mpa

lebar plat bw = 1000 mm = 1 m

Ø tulangan = 12 mm

Nilai banding panjang terhadap lebar bentang bersih β =

Persamaan perhitungan tebal plat sebagai berikut

Batas minimun tebal plat atap

h min = =

Batas maximum tebal plat atap

h max =

syarat h min ≤ h ≤ hmax → 10,2≤ h ≤ 15,97 cm

diambil tebal plat atap h = 12 cm

3.2.2 PEMBEBANAN PLAT LANTAI

Wu = 1,2 WD + 1,6 WL

Beban hidup hidup/bekerja (WL ) untuk plat atap = 2,5 kN/m2

WD didapat dari

- berat plat = 0,12 . 24 = 2,88 kN/m2

- adukan 2 cm = 0,022 . 21 = 0,42 kN/m2

- tegel keramik 0,5 = 0,005. 21 = 0,105 kN/m2

- plafond = 0,11 + 0,007 = 0,18 kN/m 2 +

= 3,585 kN/m2

Maka, Wu = 1,2 (3,585) + 1,6 (2,5) = 8,302 kN/m2

Perhitungan Geser :

Dx = h – ρ - ½Ø tulangan = 120 – 25 – ½ 10 = 90 mm

Dy = h – ρ - ½Ø tulangan - Ø tul = 120 – 25 – ½ 10 – 10 = 80 mm

28,64 kN

Vc = ф 1/6 .bw .dx = 0,8x 1/6 . 1000 . 90 = 56920,998 N = 56,92 kN

Syarat Vc = 56,92 kN > Vu = 28,64 kN ……OK!!!

3.2.3 PEMODELAN STRUKTUR

a. Type A

= = 1,00

b. Type B

= = 3

c. Type C

= = 1

d. Type D

= = 3

e. Type E

= = 1,2

3.2.4 MENENTUKAN MOMEN

a. Menentukan momen plat atap type A dengan ly/l×= 1,00

ml× = 0,001 wu . Lx2. 30 = 0,03 × 8,302 × 1,88752 = 0,887 KNm

mly = 0,001 wu . Lx2. 30 = 0,03 × 8,302 × 1,88752 = 0,887 KNm

mt× = -0,001 wu . Lx2. 68 = -0,068 × 8,302 × 1,88752 = -2,011 KNm

mty = -0,001 wu . Lx2. 68 = -0,068 × 8,302 × 1,88752 = -2,011 KNm

mtix = ½ mt× = -1,005KNm

mtiy = ½ mt× = -1,005 KNm

b. Menentukan momen plat atap type B dengan ly/l× = 3 > 2 (analisa plat satu arah)

ml× = 0,001 wu . Lx2. 63 = 0,063 × 8,302 × 4,7752 = 11,925 KNm

mly = 0,001 wu . Lx2. 16 = 0,016 × 8,302 × 4,7752 = 3,029 KNm

mt× = -0,001 wu . Lx2. 124 = -0,124 × 8,302 × 4,7752 = -23,472 KNm

mty = -0,001 wu . Lx2. 71 = -0,071 × 8,302 × 4,7752 = -13,440 KNm

mtix = ½ mt× = -6,72 KNm

c. Menentukan momen plat atap type C dengan ly/l× = 1,00

ml× = 0,001 wu . Lx2. 25 = 0,025 × 8,302 × 4,7752 = 4,732 KNm

mly = 0,001 wu . Lx2. 25 = 0,025 × 8,302 × 4,7752 = 4,732 KNm

mt× = -0,001 wu . Lx2. 51 = -0,051 × 8,302 × 4,7752 = -9,654 KNm

mty = -0,001 wu . Lx2. 51 = -0,051 × 8,302 × 4,7752 = -9,654 KNm

d. Menentukan momen plat atap type D dengan ly/l× = 3 > 2(analisa plat satu arah)

ml× = 0,001 wu . Lx2. 63 = 0,063 × 8,302 × 1,88752 = 1,863 KNm

mly = 0,001 wu . Lx2. 16 = 0,016 × 8,302 × 1,88752 = 0,473 KNm

mt× = -0,001 wu . Lx2. 124 = -0,124 × 8,302 × 1,88752 = -3,667 KNm

mty = -0,001 wu . Lx2. 71 = -0,071 × 8,302 × 1,88752 = -2,1 KNm

mtix = ½ mt× = -1,833 KNm

e. Menentukan momen plat atap type E dengan ly/l× = 1,2

ml× = 0,001 wu . Lx2. 34 = 0,034 × 8,302 × 4,7752 = 6,436 KNm

mly = 0,001 wu . Lx2. 22 = 0,022 × 8,302 × 4,7752 = 4,164 KNm

mt× = -0,001 wu . Lx2. 63 = -0,063 × 8,302 × 4,7752 = -11,925 KNm

mty = -0,001 wu . Lx2. 54 = -0,054 × 8,302 × 4,7752 = -10,222 KNm

3.2.5 PERHITUNGAN TULANGAN

ρb =

ρmax = 0,75 x ρb = 0,75 x 0,051 = 0,0382



Plat Type A

Daerah Lapangan x

ρmin = 0,0047 Mu = 0,887 kNm

ρmax = 0,0382


Mn = kNm = 1109000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0004

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 90 = 423 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509 kNm > Mu = 0,887 kNm .....................OK!

Daerah Lapangan y

ρmax = 0,0382 Mu = 0,887 kNm

ρmin = 0,0047


Mn = kNm = 1109000


ρ =

ρmin = 0,0047 > ρperlu = 0,0003

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 80 = 376 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,7 kNm > Mu = 0,887 kNm.......................OK!

Daerah tumpuan x

ρmin = 0,0047

ρmax = 0,0382

Mu=2,011kNm


kNm = 2513000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,001

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 90 = 423 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,9/2)

= 10,509 kNm > Mu = 2,011kNm .....................OK!

Daerah tumpuan y

ρmin = 0,0047

ρmax = 0,0382 Mu=2,011kNm


Mn = kNm = 2513000


ρ =

ρmin = 0,0047 > ρperlu = 0,0012

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 80 = 376 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,7 kNm > Mu = 2,011kNm .......................OK!


ρmin = 0,0047

ρmax = 0,0382 Mu = 1,005kNm


Mn = kNm = 1256000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0005

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 90 = 423 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509 kNm > Mu = 1,005 kNm .....................OK!

Daerah tumpuan jepit iy

ρmin = 0,0047

ρmax = 0,0382

Mu = 1,005kNm


Mn = kNm = 1256000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0006

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 80 = 376 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (83-7,89/2)

= 9,543 kNm > Mu = 1,005kNm .....................OK!

Plat Type B

Daerah Lapangan x

ρmin = 0,0047

ρmax = 0,0382 Mu = 11,925kNm


Mn = kNm = 14906000 Nmm


ρ =

ρmin = 0,0047< ρperlu = 0,006

ρperlu = 0,006

luas tulangan perlu

Aslx = ρ . bw . d = 0,006 . 1000 . 90 = 540 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509kNm > Mu = 11,925kNm .....................OK!

Daerah Lapangan y

ρmin = 0,0047

ρmax = 0,0382

Mu = 3,029 kNm


Mn = kNm = 3786000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,002

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 80 = 376 mm2



S = 125 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,704 kNm > Mu = 3,029

kNm .......................OK!

Daerah tumpuan x

ρmin = 0,0047ρmax = 0,0382

Mu = 23,472 kNm


Mn = kNm = 29340000 Nmm


ρ =

ρmin = 0,0047 < ρperlu = 0,01

ρperlu = 0,01

luas tulangan perlu

Aslx = ρ . bw . d = 0,01. 1000 . 90 = 900 mm2



S = 50 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 1005 . 300 (91-15,76/2)

= 20,048 kNm > Mu = 15,482

kNm .....................OK!

Daerah tumpuan y

ρmin = 0,0047

ρmax = 0,0382

Mu = 13,440 kNm


Mn = kNm = 16800000 Nmm


ρ =

ρmin = 0,0047 < ρperlu = 0,009

ρperlu = 0,009

luas tulangan perlu

Aslx = ρ . bw . d = 0,009 . 1000 . 80 = 720 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (83-7,89/2)

= 10,509 kNm > Mu = 13,440

kNm .......................OK!


ρmin = 0,0047

ρmax = 0,0382

Mu = 6,72 kNm


Mn = kNm = 84000000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,003

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 90 = 423 mm2

Maka digunakan ø 10– 175 = 449


S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509 kNm > Mu = 6,72 kNm .....................OK!

Plat Type C

Daerah Lapangan x

ρmin = 0,0047

ρmax = 0,0382 Mu = 4,732 kNm


Mn = kNm = 5915000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,002

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 90 = 423 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509kNm > Mu = 4,732 kNm .....................OK!

Daerah Lapangan y

ρmin = 0,0047

ρmax = 0,0382

Mu = 9,654 kNm


Mn = kNm = 12067000 Nmm


ρ =

ρmin = 0,0047 < ρperlu = 0,006

ρperlu = 0,006

luas tulangan perlu

Aslx = ρ . bw . d = 0,006 . 1000 . 80 = 480 mm2



S = 125 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,704 kNm > Mu = 9,654 kNm .......................OK!

Daerah tumpuan x

ρmin = 0,0047

ρmax = 0,0382

Mu = 4,732 kNm


Mn = kNm = 5915000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,002

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 90 = 423 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,5cvv222nb09kNm > Mu = 4,732

kNm .....................OK!

Daerah tumpuan y

ρmin = 0,0047

ρmax = 0,0382

Mu = 4,732 kNm


Mn = kNm = 5915000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,003

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 80 = 376 mm2



S = 125 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,704 kNm > Mu = 4,732 kNm .......................OK!

Plat Type D

Daerah Lapangan x

ρmin = 0,0047

ρmax = 0,0382 Mu = 1,863 kNm


Mn = kNm = 2329000 Nmm


ρ =

ρmin = 0,0047 > ρperlu 0,001

ρperlu = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 90 = 423 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509kNm > Mu = 1,863kNm .....................OK!

Daerah Lapangan y

D = dy = 83 mm

ρmin = 0,0047

ρmax = 0,0382

Mu = 0,473kNm


Mn = kNm = 591000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,0003

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 80 = 376 mm2



S = 125 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,704 kNm > Mu = 0,473

kNm .......................OK!

Daerah tumpuan x

ρmin = 0,0047

ρmax = 0,0382

Mu = 3,667 kNm


Mn = kNm = 4584000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,002

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 90 = 423 mm2



S = 50 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 1005 . 300 (91-15,76/2)

= 20,048 kNm > Mu = 3,667 kNm .....................OK!

Daerah tumpuan y

ρmin = 0,0047

ρmax = 0,0382

Mu = 2,1 kNm


Mn = kNm = 2625000 Nmm


ρ =

ρmin = 0,0047 < ρperlu = 0,001

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 80 = 376 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (83-7,89/2)

= 10,509 kNm > Mu = 2,1 kNm .......................OK!


ρmin = 0,0047

ρmax = 0,0382

Mu = 1,833 kNm


Mn = kNm = 2291000 Nmm


ρ =

ρmin = 0,0047< ρperlu = 0,0009

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047. 1000 . 90 = 423 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509 kNm > Mu = 1,833 kNm .....................OK!

Plat Type E

Daerah Lapangan x

ρmin = 0,0047

ρmax = 0,0382 Mu = 6,436 kNm


Mn = kNm = 8045000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,003

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 90 = 423 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509kNm > Mu = 6,436 kNm .....................OK!

Daerah Lapangan y

ρmin = 0,0047

ρmax = 0,0382

Mu = 4,164kNm


Mn = kNm = 5205000 Nmm


ρ =

ρmin = 0,0047 > ρperlu = 0,003

ρmin = 0,0047

luas tulangan perlu

Aslx = ρ . bw . d = 0,0047 . 1000 . 80 = 376 mm2



S = 125 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,704 kNm > Mu = 4,614

kNm .......................OK!

Daerah tumpuan x

ρmin = 0,0047

ρmax = 0,0382

Mu = 11, 925 kNm


Mn = kNm = 14906000 Nmm


ρ =

ρmin = 0,0047 < ρperlu = 0,006

ρperlu = 0,006

luas tulangan perlu

Aslx = ρ . bw . d = 0,006. 1000 . 90 = 540 mm2



S = 100 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2)

= 10,509kNm > Mu = 11,925 kNm .....................OK!

Daerah tumpuan y

ρmin = 0,0047

ρmax = 0,0382

Mu = 10,222 kNm


Mn = kNm = 12777000 Nmm


ρ =

ρmin = 0,0047 < ρperlu = 0,007

ρperlu = 0,007

luas tulangan perlu

Aslx = ρ . bw . d = 0,007 . 1000 . 80= 560 mm2



S = 125 mm < S max = 3 . h = 3 . 120 = 360 mm


a = mm

ΦMn = Φ . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2)

= 7,704 kNm > Mu = 10,222 kNm .......................OK!

Resume Plat Hasil Perhitungan

PLAT ATAP

Tipe A

Aslx = Ǿ8 – 100 = 201 mm2

Asly = Ǿ8 – 125 = 201 mm2

Astx = Ǿ8 – 100 = 201 mm2

Asty = Ǿ8 – 125 = 201 mm2

Astix = Ǿ8 – 100 = 201 mm2

Astiy = Ǿ8 – 100 = 201 mm2

Tipe B

Aslx = Ǿ8 – 100 = 201 mm2

Asly = Ǿ8 – 125 = 201 mm2

Astx = Ǿ8 – 50 = 201 mm2

Asty = Ǿ8 – 100 = 201 mm2

Astix = Ǿ8 – 100 = 201 mm2

Tipe C

Aslx = Ǿ8 – 100 = 201 mm2

Asly = Ǿ8 – 125 = 201 mm2

Astx = Ǿ8 – 100 = 201 mm2

Asty = Ǿ8 – 125 = 201 mm2

PLAT LANTAI

Tipe A

Aslx = Ǿ10 – 175 = 175 mm2

Asly = Ǿ10 – 200 = 200 mm2

Astx = Ǿ10 – 175 = 175 mm2

Asty = Ǿ10 – 150 = 200 mm2

Astix = Ǿ10 – 250 = 175 mm2

AstiY = Ǿ10 – 250 = 200 mm2

Tipe B1

Aslx = Ǿ10 – 250 = 125 mm2

Asly = Ǿ10 – 250 = 200 mm2

Astx = Ǿ10 – 175 = 75 mm2

Asty = Ǿ10 – 150 = 100 mm2

Astix = Ǿ10 – 250 = 175 mm2

Tipe B2

Aslx = Ǿ10 – 250 = 201 mm2

Asly = Ǿ10 – 250 = 201 mm2

Astx = Ǿ10 – 175 = 201 mm2

Asty = Ǿ10 – 150 = 201 mm2

Tipe C

Aslx = Ǿ10 – 250 = 201 mm2

Asly = Ǿ10 – 250 = 201 mm2

Astx = Ǿ10 – 175 = 201 mm2

Asty = Ǿ10 – 150 = 201 mm2

Tipe D

Aslx = Ǿ10 – 250 = 175 mm2

Asly = Ǿ10 – 250 = 200 mm2

Astx = Ǿ10 – 250 = 175 mm2

Asty = Ǿ10 – 250 = 200 mm2

Astix = Ǿ10 – 250 = 175 mm2

Tipe E

Aslx = Ǿ10 – 250 = 175 mm2

Asly = Ǿ10 – 250 = 200 mm2

Astx = Ǿ10 – 250 = 125 mm2

Asty = Ǿ10 – 250 = 125 mm2

3.3 PERHITUNGAN PORTAL

3.3.1 Dimensi Portal

Balok atap direncanakan 22,5/45

h = 45 cm ≥ 1/12 (500) = 41,67 cmBalok lantai direncanakan 25/50

h = 50 cm ≥1/12 (600) = 50 cmKolom direncanakan 30/40 ( untuk kolom lantai 1 dan lantai 2 )

Equivalen beban plat q plat

bidang 1

bidang K dan P ( dua buah segitiga )

luas =

berat q plat( EF dan KL) per m =

bidang 2

bidang L

luas =

bidang Q

luas =

berat q plat( FG ) per m =

bidang 3

bidang M

luas =

bidang R

luas =

berat q plat(GH dan LM) per m =

bidang 4

bidang N

luas =

bidang S

luas =

berat q plat(HI dan MN) per m =

bidang 5

bidang O dan T ( dua buah segitiga )

luas =

berat q plat( IJ dan NO) per m =

3.3.2 Pembebanan Portal

beban balok KL ( 22,5/45)berat sendiri balok = 0,225 . 0,45 . 24 = 2,43 KN/m

berat q plat = = 11,912 KN/m

total = 14,342 KN/m beban balok LM (22,5/45)

berat sendiri balok = 0,225 . 0,45 . 24 = 2,43 KN/m


total = 17,022 KN/m beban balok MN (22,5/45)

berat sendiri balok = 0,225 . 0,45 . 24 = 2,43 KN/m


total = 17,022 KN/mbeban balok NO (22,5/45)berat sendiri balok = 0,225 . 0,45 . 24 = 2,43 KN/m


total = 14,342 KN/m beban balok EF ( 25/50)

berat sendiri balok = 0,25 . 0,5 . 24 = 3 KN/m


total = 19,604 KN/m beban balok FG (25/50)



total = 31,711 KN/m beban balok GH (25/50)



total = 23,34 KN/m beban balok HI (25/50)



total = 23,34 KN/m beban balok IJ (25/50)berat sendiri balok = 0,25 . 0,5 . 24 = 3 KN/m


total = 19,604 KN/m

3.3.2 Perhitungan Cross

MENENTUKAN NILAI t2

Balok T dengan bentang 5 m

t2

tf

t3

bw

Bw = 22,5 cm

h = 45 cm

tf = 12 cm

Mengingat plat dan balok monolit, maka balok diperbolehkan dianggap sebagai balok

T, dengan ketentuan :

t2 = diperoleh dari nilai terkecil :

¼ panjang balok = ¼ x 5 m = 1,25 m

Jarak antar portal = 5 m

16 tf + bw = 16 . 12 + 22,5 = 214,5 cm = 2,145 m

Maka t2 atau bm untuk balok dengan bentang 5 meter adalah 1,25 m

Balok T dengan bentang 6 m

t2

tf

t3

bw

Bw = 25 cm

h = 50 cm

tf = 12 cm






16 tf + bw = 16 . 12 + 25 = 217 cm = 2,17 m


Balok T dengan bentang 2m

t2

tf

t3

bw

Bw = 22,5 cm

h = 45 cm

tf = 12 cm






16 tf + bw = 16 . 12 + 22,5 = 214,5 cm = 2,145 m


MOMEN INERSIA

Momen inersia balok atap berukuran 22,5/45 didapat dari SAP dan cross yaitu 0,00242578cm4 (KL), 0,00336814 cm4 (LM), 0,00336814 cm4 (MN), 0,00242578 cm4

(NO) Momen inersia balok lantai berukuran 25/50 didapat dari SAP dan cross yaitu

0,00351355cm4 (EF), 0,00524553cm4 (FG), 0,00495837cm4 (GH), 0,00336814 cm4

0,00495837 (HI), 0,00351355cm4 (IJ) Momen inersia kolom ukuran 30/40 = 1/12 b h3 = 1/12 . 30 . 403 = 160000 cm4 =

0,0016 m4

KEKAKUAN

Atap Kekakuan KL( dengan bentang L = 2 m)

KKL = 0 Kekakuan LM

Inersia = 0,00336814 cm4

L= 5 meter

KLM = = 538,902

Kekakuan MNInersia = 0,00495837 cm4

L= 5 meter

KMN = =538,902

Kekakuan NO ( dengan bentang L = 2 m )KNO = 0

Lantai Kekakuan EF (dengan bentang L = 2)

KEF = 0 Kekakuan FG

Inersia = 0,00524553 cm4

L= 6 meter

KFG = = 699,404

Kekakuan GHInersia = 0,00495837 cm4

L= 5 meter

KGH = = 793,3393

Kekakuan HIInersia = 0,00933644 cm4

L= 5 meter

KHI = = 793,3393

Kekakuan IJ (dengan bentang L = 2)KIJ = 0

KolomInersia = 0,0009 m4

Kekakuan AF = BG = CH = DI ( dengan L = 5 m )

KAF = KBG = KCH = KDI = = 256

Kekakuan GL = HM = IN ( dengan L = 5 m )

KGL = KHM = KIN = = 256

DISTRIBUSI KEKAKUAN Titik Kumpul L

Titik Kumpul M

Titik Kumpul N

Titik Kumpul F

Titik Kumpul G

Titik Kumpul H

Titik Kumpul I

MOMEN PRIMERAtap

MLK = 1/12. Q . L2 = 1/12 x 14,342 x 22 = 4,780667 MML = 1/12. Q . L2 = 1/12 x 17,022 x 52 = 35,4625

MLM = -1/12. Q . L2 = -1/12 x 17,022 x 52 = -35,4625 MNM = 1/12. Q . L2 = 1/12 x 17,022x 52 = 35,4625

MMN = -1/12. Q . L2 = -1/12 x 17,0227 x 52 = -35,4625

MNO = -1/12. Q . L2 = -1/12 x 14,342 x 22 = -4,780667

Lantai MFE = 1/12. Q . L2 = 1/12 x 19,604 x 22 = 4,534667 MGF = 1/12. Q . L2 = 1/12 x 31,711 x 62 = 95,133

MFG = -1/12. Q . L2 = -1/12 x 31,711 x 62 = -95,133 MHG = 1/12. Q . L2 = 1/12 x 22,34x 52 = 48,625

MGH = -1/12. Q . L2 = -1/12 x 22,34 x52 = -48,625 MIH = 1/12. Q . L2 = 1/12 x 22,34 x 52 = 48,625

MHI = -1/12. Q . L2 = -1/12 x 22,34 x52 = -48,625

MJI = -1/12. Q . L2 = -1/12 x 19,604 x 22 = -4,534667

Lanjutan Perhitungan Cross ada pada Lampiran

3.4 PERHITUNGAN TULANGAN BALOK

Hasil analisa gaya dalam yang dipakai untuk penulangan balok dan kolom adalah

hasil perhitungan Cross. Berikut perhitungan penulangan untuk balok HI,JK dan DE,FG.

h

d

bw

Data material:

F’c = 22,5 Mpa F’y = 300 Mpa

β1 = 0,85 ф lentur = 0,8

ф sengkang = 0,6

Untuk perencanaan tulangan selanjutnya dapat diketahui nilai-nilai rasio tulangan

maksimum dan minimum sebagai berikut :

ρ min =

ρb =

ρmax = 0,75 x ρb = 0,75 x 0,036 = 0,027

3.4.1 Dimensionering Balok

BALOK ATAP BALOK LANTAI

Tinggi balok h = 450 mm = 45 cm Tinggi balok h = 500mm = 50 cm

Lebar balok bw = 225 mm =22,5 cm Lebar balok bw = 250 mm =25 cm

Lapisan tulangan lp = 1 lapis

Ǿ pokok = 10 mm

Ǿ sengkang = 8 mm

Degging P = 40 mm

FREE BODY BALOK HI

Q = 23,34 KNm L = 5 M

MHI = 14,716 KNm MIH = 27,123 KNm

VH = - 34,159 KN VI= 39,39 KN

Daerah momen tumpuan HI dan IH dapat

dihitung dengan rumus abc:

a= - q = -23,34 KNm

b = VH = 34,159 KN

c = - MHI = - 14,716 KNm

jarak

Jarak H = = 0,491m = 491 mm

Jarak I = = 3,473 m = 3473 mm

momen tumpuan HI berjarak 0 – 491 mm

momen tumpuan IH berjarak 3473 mm – 4300 mm

momen lapangan direncanakan berada di tengah-tengah bentang HI

dapat di hitung Mlap

Mlap = - MHI + VH L/2 – Q L/2 ( L/2 )/2

= -14,716 + (34,159. 4,3/2) –( 17,231 x 4,3/2 x (4,3/2) /2 )

= 18,903 KNm dengan jarak Mlap yaitu 491 mm – 3473 mm

PERHITUNGAN TULANGAN LENTUR

TUMPUAN HI

d = 294 mm, Bw = 200 mm, ρ min = 0,0035, ρ max = 0,0203

Mu = 14,716 KNm , Mn =


Karena = 0,0027 < = 0,0035 maka yang diambil min untuk menghitung luas

penampang tulangan, jadi

Luas tulangan perlu

As = . bw . d = 0,0035 x 200 x 294 = 205,8 mm2

Maka digunakan 2Ǿ16 = 402 mm2

Kontrol kekuatan balok:

φMn = φ . as .Fy .( d – a/2 ) = 0,8 x 402 x 400 ( 294-37,8/2) = 35,386 KNm

φMn > Mu = 35,386 KNm > 14,716 KNm OK

TUMPUAN IH

d = 294 mm, Bw =200 mm, ρ min = 0,0035, ρ max = 0,0203

Mu = 27,123 KNm, Mn =

Karena = 0,0052 > = 0,0052 maka yang diambil min untuk menghitung luas


Luas tulangan perlu

As = . bw . d = 0,0052 x 200 x 294 = 305,76 mm2


Kontrol kekuatan balok


φMn > Mu = 35,386 KNm > 27,123 KNm OK

LAPANGAN HI

d = 294 mm, Bw = 200 mm, ρ min = 0,0035,

Mu = 18,903 KNm,

MT plat = Ф 0,85 . f c’ . b . hf (d – hf/2 )

= 0,8 x 0,85 x 25 x 825 x 100 (294-100/2 )

= 342210000 Nmm = 342,21 KNm

Mu =18,903 < Mtplat = 342,21 KNm ...analisa balok T persegi b = 825 mm


Mn =



Luas tulangan perlu

As = . bw . d = 0,0035 x 200 x 294 = 205,8 mm2




φMn > Mu = 35,386 KNm > 18,903 KNm OK

PERHITUNGAN TULANGAN SENGKANG

Kuat geser beton dapat dihitung sebagai berikut :

Menghitung jarak ( Smax ), diambil dari nilai di bawah ini :

Jika , maka nilai Smax ditentukan dengan rumus : ( tulangan geser tak

dihitung )

Smax = d/2 = 294/2 = 147 mm

Jika , maka nilai Smac ditentukan dengan rumus :

Smax = d/4 = 249 / 4 = 73.5 mm

Dari kedua nilai Smax diatas, tidak boleh lebih besar dari

Smax Yang Terjadi = 147 mm, maka Smax = 140 mm

PERHITUNGAN PER SEGMEN

1. Segmen 1 ( 0-430 mm )

Menghitung Vu per-segmen dengan Rumus ( diambil nilai Vu terbesar yaitu pada x =

0 m )

Vu = VH – qx = 34,159 – (17,231 x 0 ) = 34,159 KN

Vn = Vu / Ф = 34,159 / 0,6 = 56,93 KNm

Vs = [ vn ] – vc = 56,93 – 49 = 7,93 KNm

Kerena S=1490,85 > Smax=147, digunakan S = 147 mm →140 mm

2. Segmen 2 ( 430-860 mm ) Lihat lampiran

3. Segmen 3( 860-1290 mm ) Lihat lampiran





8. Segmen 8 ( 3010 – 3440 mm ) Lihat lampiran


10. Segmen 10 ( 3870 - 4300mm ) Lihat lampiran

LAMPIRAN HI

DETAIL PENULANGAN POKOK DAN SENGKANG BALOK HI

FREE BODY BALOK IJ

Q = 15,844 KNm L = 3,9 M

MIJ = 20,418 KNm MJI = 30,836 KNm

Mlap = 4,494 KNm

VI = 28,218 KN VJ= 33,357 KN


TUMPUAN IJ


Mu = 20,418 KNm , Mn =




Luas tulangan perlu

As = . bw . d = 0,0038 x 200 x 294 = 223,44 mm2




φMn > Mu = 35,386 KNm > 20,418 KNm OK

TUMPUAN JI


Mu = 30,836 KNm, Mn =



Luas tulangan perlu

As = . bw . d = 0,0059 x 200 x 294 = 346,92 mm2




φMn > Mu = 35,386 KNm > 30,836 KNm OK

LAPANGAN IJ

d = 294 mm, Bw = 200 mm, ρ min = 0,0035,

Mu = 4,494 KNm,

MT plat = Ф 0,85 . f c’ . b . hf (d – hf/2 )

= 0,8 x 0,85 x 25 x 825 x 100 (294-100/2 )

= 342210000 Nmm = 342,21 KNm

Mu =4,494 < Mtplat = 342,21 KNm ...analisa balok T persegi b = 825 mm


Mn =



Luas tulangan perlu

As = . bw . d = 0,0035 x 200 x 294 = 205,8 mm2




φMn > Mu = 35,386 KNm > 4,494 KNm OK





dihitung )

Smax = d/2 = 294/2 = 147 mm


Smax = d/4 = 249 / 4 = 73.5 mm




1. Segmen 1 ( 0 - 390 mm )


0 m )

Vu = VH – qx = 20,418 – (15,844 x 0 ) = 20,418 KN

Vn = Vu / Ф = 20,418 / 0,6 = 25,5225 KNm

Vs = [ vn ] – vc = 25,5225 – 49 = -23.4775 KNm < 0...........tidak dihitung

digunakan S = 147 mm →140 mm

2. Segmen 2 ( 390 - 780 mm ) Lihat lampiran

3. Segmen 3( 860 - 1170 mm ) Lihat lampiran

4. Segmen 4 (1170 - 1560 mm ) Lihat lampiran




8. Segmen 8 (2730 – 3120 mm ) Lihat lampiran



LAMPIRAN HI

DETAIL PENULANGAN POKOK DAN SENGKANG BALOK IJ

FREE BODY BALOK JK

Q = 15,844 KNm L = 2,3 M

MJK = 43,83 KNm MKJ = 0 KNm

Mlap = 10,958 KNm

VJ = 38,113 KN VK= 0 KN

TUMPUAN JK


Mu = 43,83 KNm , Mn =



penampang tulangan, jadi Luas tulangan perlu

As = . bw . d = 0,0067 x 200 x 294 = 393,96 mm2




φMn > Mu = 35,386 KNm > 43,83 KNm OK


1. Segmen 1 ( 0-230 mm )


0 m )

Vu = VJ – qx = 43,83 – (9,1056 x 0 ) = 43,83 KN

Vn = Vu / Ф = 43,83 / 0,6 = 73,05 KNm

Vs = [ vn ] – vc = 73,05 – 49 = 24,05 Knm

Karena S = 814,217 mm > Smax = 142 mm, digunakan S = 147 mm → s = 140 mm










DETAIL PENULANGAN POKOK DAN SENGKANG BALOK JK

FREE BODY BALOK DE

Q = 19,692 KNm L = 4,3 M

MDE = 19,19 KNm MED = 32,749 KNm

Mlap = 20,269 KNm

VD = 39,522 KN VE= 45,154 KN


TUMPUAN DE


Mu = 19,19 KNm , Mn =




Luas tulangan perlu

As = . bw . d = 0,0035 x 200 x 344 = 204,8 mm2




φMn > Mu = 41,81 KNm > 19,19 KNm OK

TUMPUAN ED


Mu = 32,749 KNm, Mn =



Luas tulangan perlu

As = . bw . d = 0,0045 x 200 x 344 = 309,6 mm2




φMn > Mu = 41,81 KNm > 32,749 KNm OK

LAPANGAN DE


Mu = 20,269 KNm, Mn =




Luas tulangan perlu

As = . bw . d = 0,0035 x 200 x 344 = 204,8 mm2




φMn > Mu = 41,81 KNm > 20,269 KNm OK





dihitung )

Smax = d/2 = 344/2 = 172 mm


Smax = d/4 = 344 / 4 = 86 mm




1. Segmen 1 (0-430 mm)


0 m )

Vu = VD – qx = 39,522 – (15,5787 x 0 ) = 39,522 KN

Vn = Vu / Ф = 39,522 / 0,6 = 65,87 KNm

Vs = [ vn ] – vc = 65,87– 57,33 = 8,54 KNm

Karena S = 1619,796 mm > Smax = 172 mm, digunakan S = 172 mm → 170 mm










LAMPIRAN DE

DETAIL PENULANGAN POKOK DAN SENGKANG BALOK DE

FREE BODY BALOK FG

Q = 18,919 KNm L = 2,3 M

MFG = 50,041 KNm MGF = 0 KNm

Mlap = 12,51 KNm

VF = 34,514 KN VG= 0 KN

TUMPUAN FG


Mu = 50,041 KNm , Mn =



penampang tulangan, jadi luas tulangan perlu

As = . bw . d = 0,0071 x 200 x 344 = 488,48 mm2




φMn > Mu = 60,9 KNm > 50,041 KNm OK


1. Segmen 1 ( 0-330 mm )


0 m )

Vu = VF– qx = 34,514 – (15,5787 x 0 ) = 34,514 KN

Vn = Vu / Ф = 34,514 / 0,6 = 72,52 KNm

Vs = [ vn ] – vc = 72,52 – 57,333 = 15,19 Knm

Karena S = 910,668mm > Smax = 172 mm, digunakan S = 172 mm → s = 170 mm










LAMPIRAN FG

DETAIL PENULANGAN POKOK DAN SENGKANG BALOK FG

TABEL PENULANGAN BALOK

NO NAMA BALOK TUMPUAN LAPANGAN TUMPUAN

1 DE 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2

2 EF 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2

3 FG 3 Ø16 = 402 mm2 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2

4 HI 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2

5 IJ 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2

6 JK 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2 2 Ø16 = 402 mm2

3.5 PERHITUNGAN TULANGAN KOLOM

Hasil analisa gaya dalam yang dipakai untuk penulangan kolom adalah hasil

perhitungan Cross.

Data material :

F’c = 25 Mpa h

F’y = 400 Mpa

β1 = 0,85 d’

ф lentur = 0,8 bw

ф sengkang = 0,6

Untuk perencanaan tulangan selanjutnya dapat diketahui nilai-nilai rasio tulangan

maksimum dan minimum sebagai berikut :

ρ min =

ρb =

ρmax = 0,75 x ρb = 0,75 x 0,027 = 0,0203

tinggi kolom h = 300 mm , lebar kolom bw = 400 mm

lapis tulangan Lp = 1 lapis

Ǿ pokok = 16 mm , Ǿ sengkang = 8 mm , Degging P = 40 mm

Perhitungan tulangan lentur


FREE BODY KOLOM AD

MAD = 3,306 KNm

MDA = 6,612 KNm

VA = -2,479 KNm

VD = 2,479 KNm

PDA = 73,681 KN

TUMPUAN DA

d’/h =56/300 = 0,19

e = Mu / Pu=6,612 / 73,681 = 89,738

e/h = 89,738 / 300 = 0,299

Setelah di cari pada grafik interaksi kolom didapatkan r = 0,003

jadi ρ = r . β = 0,003 . 1 = 0,003

ρ min = 0,0035< ρ = 0,003 < ρ max = 0,0203

ρ perlu = 0,0035

luas tulangan perlu :

As = . bw . d . = 0,0035 x 400 x 300 =420 mm2

Maka digunakan 3 Ǿ 16 = 603 mm2

TUMPUAN AD

d’/h =56/300 = 0,19

e = Mu / Pu= 3,306 / 73,681 = 44,869

e/h = 44,869 / 300 = 0,15


jadi ρ = r . β = 0,003 . 1 = 0,003

ρ min = 0,0035< ρ = 0,003 < ρ max = 0,0203

ρ perlu = 0,0035


As = . bw . d . = 0,0035 x 400 x 300 =420 mm2






dihitung )

Smax = d/2 = 244/2 = 122 mm

Jika , maka nilai Smax ditentukan dengan rumus :

Smax = d/4 = 244 / 4 = 61 mm


Smax yang terjadi = 122 mm, maka Smax = 120 mm

PERHITUNGAN PER SEGMEN ( tinggi kolom 4 m = 4400 mm )

1. Segmen 1 ( 0- 400 mm )


0 m )

Vu = VA = 2,479 KN

Vn = Vu / Ф = 2,479 / 0,6 = 4,13 KNm

Vs = [ vn ] – vc = 4,13 – 81,333 = - 77,203 Knm < 0 ..... tidak perlu dihitung

Smax = 122 mm→120 mm

2. Segmen 2 ( 440-800mm ) Lihat lampiran









LAMPIRAN AD

FREE BODY KOLOM CF

MCF = 0,567 KNm

MFC = 1,135 KNm

VC = -0,425 KNm

VF = 0,425 KNm

PFC = 108,228 KN

TUMPUAN CF

d’/h =56/300 = 0,19

e = Mu / Pu= 0,567 / 108,228 = 5,239

e/h = 5,239 / 300 = 0,017


jadi ρ = r . β = 0,004 . 1 = 0,004

ρ min = 0,0035< ρ = 0,004 < ρ max = 0,0203

ρ perlu = 0,004


As = . bw . d . = 0,004 x 400 x 300 = 480 mm2


TUMPUAN FC

d’/h =56/300 = 0,19

e = Mu / Pu= 1,135 / 108,228 = 10,49

e/h = 10,49 / 300 = 0,365


jadi ρ = r . β = 0,004 . 1 = 0,004

ρ min = 0,0035< ρ = 0,004 < ρ max = 0,0203

ρ perlu = 0,004


As = . bw . d . = 0,004 x 400 x 300 = 480 mm2






dihitung )

Smax = d/2 = 244/2 = 122 mm


Smax = d/4 = 244 / 4 = 61 mm




1. Segmen 1 ( 0- 400 mm )


0 m )

Vu = VA = 0,425 KN

Vn = Vu / Ф = 0,425 / 0,6 =0,71KNm

Vs = [ vn ] – vc = 0,71 – 108,228 = - 107,518 Knm < 0 ..... tidak perlu dihitung











LAMPIRAN CF

GAMBAR PENULANGAN SKEMATIS KOLOM CF

FREE BODY KOLOM DH

MDH = 12,235 KNm

MHD = 14,262 KNm

VD = -6,973 KNm

VH= 6,973 KNm

PHD = 33,662 KN

TUMPUAN DH

d’/h =56/300 = 0,19

e = Mu / Pu= 0,567 / 33,662 = 363,47

e/h = 363,47 / 300 = 1,212

Setelah di cari pada grafik interaksi kolom didapatkan r = 0

jadi ρ = r . β = 0 . 1 = 0

ρ min = 0,0035< ρ = 0 < ρ max = 0,0203

ρ perlu = 0,0035


As = . bw . d . = 0,0035 x 400 x 300 = 420 mm2


TUMPUAN HD

d’/h =56/300 = 0,19

e = Mu / Pu= 14,262 / 33,62 = 423,68

e/h = 423,68 / 300 = 1,412


jadi ρ = r . β = 0. 1 = 0

ρ min = 0,0035< ρ = 0 < ρ max = 0,0203

ρ perlu = 0


As = . bw . d . = 0,0035 x 400 x 300 = 420 mm2






dihitung )

Smax = d/2 = 244/2 = 122 mm


Smax = d/4 = 244 / 4 = 61 mm




1. Segmen 1 ( 0- 380 mm )


0 m )

Vu = VA = 6,973 KN

Vn = Vu / Ф = 6,973 / 0,6 = 11,62KNm

Vs = [ vn ] – vc = 11,62 – 81,333 = - -69,71 Knm < 0 ..... tidak perlu dihitung


2. Segmen 2 (380-760mm ) Lihat lampiran

3. Segmen 3(760-1140 mm ) Lihat lampiran

4. Segmen 4 (1140-1520 mm ) Lihat lampiran







LAMPIRAN DH

GAMBAR PENULANGAN SKEMATIS KOLOM DH

FREE BODY KOLOM FJ

MFJ = 2,359 KNm

MJF = 2,926 KNm

VF = -1,391 KNm

VJ= 1,391 KNm

PJF = 50,631 KN

TUMPUAN FJ

d’/h =56/300 = 0,19

e = Mu / Pu= 2,359 / 50,631 = 46,592

e/h = 46,592 / 300 = 0,155


jadi ρ = r . β = 0 . 1 = 0

ρ min = 0,0035< ρ = 0 < ρ max = 0,0203

ρ perlu = 0,0035


As = . bw . d . = 0,0035 x 400 x 300 = 420 mm2


TUMPUAN JF

d’/h =56/300 = 0,19

e = Mu / Pu= 2,926 / 33,62 = 57,791

e/h = 57,791 / 300 = 0,193


jadi ρ = r . β = 0. 1 = 0

ρ min = 0,0035< ρ = 0 < ρ max = 0,0203

ρ perlu = 0


As = . bw . d . = 0,0035 x 400 x 300 = 420 mm2






dihitung )

Smax = d/2 = 244/2 = 122 mm


Smax = d/4 = 244 / 4 = 61 mm




11. Segmen 1 ( 0- 380 mm )


0 m )

Vu = VA = 1,391 KN

Vn = Vu / Ф = 1,391 / 0,6 = 2,32 KNm

Vs = [ vn ] – vc = 2,32 – 81,333 = - -79,013 Knm < 0 ..... tidak perlu dihitung


12. Segmen 2 (380-760mm ) Lihat lampiran

13. Segmen 3(760-1140 mm ) Lihat lampiran








LAMPIRAN FJ

GAMBAR PENULANGAN SKEMATIS KOLOM FJ

TABEL PENULANGAN KOLOM

NO NAMA KOLOM TUMPUAN TUMPUAN

1 AD 3 Ø16 = 603 mm2 3 Ø16 = 603 mm2

2 BE 3 Ø16 = 603 mm2 3 Ø16 = 603 mm2

3 CF 3 Ø16 = 603 mm2 3 Ø16 = 603 mm2

4 DH 3 Ø16 = 603 mm2 3 Ø16 = 603 mm2

5 EI 3 Ø16 = 603 mm2 3 Ø16 = 603 mm2

6 FJ 3 Ø16 = 603 mm2 3 Ø16 = 603 mm2

3.6 PERHITUNGAN TANGGA

3.6.1 Ketentuan-ketentuan tangga:

- perbedaan ketinggian antara dua lantai = 400 cm

- tinggi optrede = 17,5 cm

- menentukan antrede sesuai dengan syarat tangga yaitu 55 – 65 cm

2 x optrede + antrede = 64 cm

2 x 17,5 + antrede = 64 cm

Antrede = 64 – ( 2 x 17,5) = 30 cm

- jumlah optrede ((nopt) = (h/17,5) = 400 / 17,5 = 22,8 ~ 23 buah

- panjang miring anak tangga ( A) = = = 0,5 m

- tebal pelat tangga dan bordes = 12 cm

- lebar tangga = 2 x 200 = 400 cm

- panjang datar tangga = 330 cm

- tinggi bordes (ho) = 400/2 = 200 cm

- panjang miring tangga AB = = 388,5 cm = 3,858 m

- kemiringan tangga, α = tg -1 . (200/330) = 31,2184 31013’6.24’’

3.6.2 Pembebanan Tangga

Pembebanan tangga :

- beban mati

pelat tangga = 0,12 (1/cos 31013’6.24”). 2,00 . 24 = 6,73 kN/m

anak tangga = (1/2. 0,175. 0,3 . 2,00 . 24

= 1,260 kNm

Untuk 1 m terdapat anak tangga 100/30 = 3,33

Maka = 3,33 x 1,260 = 4,196 kN/m +

= 10,926 kN/m

- beban hidup = 2,5 . 2 = 5,00 kN/m

- beban tangga terfaktor (Wu1) = 1,2 WDLI + 1,6 WLL

= 1,2 . 10,926 + 1,6 . 5,00

= 21,111 kN/m

Pembebanan bordes

- beban mati

pelat bordes = 0,12 . 1,20. 24 = 3,456 KN/m

tegel 0,5 cm = 0,05 . 1,20 . 17 = 0,102 KN/m

spesi 1,5 cm = 0,015 . 1,20 . 21 = 0,378 KN/m +

= 3,936 KN/m

- beban hidup = 2,5 . 1,2 = 1,8 KN/m

- beban bordes terfaktor (WU2) = 1,2 WDLI + 1,6 WLL

= 1,2 . 3,939 + 1,6 . 1,8

= 7,6068 kN/m

3.6.3 Perhitungan Penulangan Konstruksi Tangga

a. penulangan pelat tangga

tebal pelat lantai = 120 mm , tulangan yang digunakan = Ф 14 mm

tebal selimut beton = 20 mm

tinggi efektif pelat = h – p – ½ Ф tulangan = 120 – 21 – 1/2/ 14 = 93 mm

lebar pelat = 2,00 m ~ 2000 mm

Perhitungan momen dilakukan dengan menggunakan program SAP 2000, adapunn

hasilnya adalah sebagai berikut

Mu = Mmaks = 50,041 KNm

Syarat untuk momen : ρmin < ρ < ρmaks

pmin = = = 0,0035

ρb = = = 0,027

ρmaks = 0,75 x ρb = 0,75 . 0,027 = 0,02

menghitung ρperlu :

Rn = = = 3,616

ρ =

Ф 14 - 40

Ф 10 - 150

Ф 10 - 150

ρ =

ρ = 0,031

ρ = 0,01149 > ρ min = 0,0035

diambil ρ = 0,031

luas tulangan perlu = As perlu = ρ.b.d = 0,031 x 2000 x 93

= 5766 mm2

Dipakai tulangan Ф 14 – 40 (A= 6157 mm2 )

Control kekuatan pelat tangga :

As = 5766 mm2 > Amin = 0,002 . 50 . 2000 = 200 mm2

a = = = 57,66 mm

Mn = As.fy.(d-a.2)

Mn = 5766.400.(93-57,66/2)

Mn = 1480016880Nmm = 148,001688 KNm

ФMn = 0,8 . 148,001688 = 118,406 KNm

ФMn = 118,406 KNm > Mu = 50,041 KNm………..OK

Tulangan bagi = 0,002 . b . h

= 0,002 . 2000 . 120 = 480 mm2

Digunakan tulangan Ф 10 – 150 (A= 524 mm2)

Skema penulangan plat tangga

b. penulangan pelat bordes

tebal pelat tangga ; h = 120 mm

tulangan yang digunakan = Ф 14 mm

tebal selimut beton = 20 mm

tinggi efektif plat (dx) = h – p – ½ Ф tulnagan

= 120 – 20 – ½ . 14 = 93 mm

Lebar plat = 1,2 m

Perhitungan momen dilakukan dengan menggunakan program SAP 2000, adapunn

hasilnya adalah sebagai berikut

Mu = Mmaks = 22,825 KNm

Mn =

pmin = 0,0035

Rn = = = 2,749

ρ =

ρ =

ρ = 0,0135

ρ = 0,0135 > ρ min = 0,0035

diambil ρ = 0,0135

luas tulangan perlu = As perlu = ρ.b.d = 0,0135 x 2000 x 93

= 2511 mm2

Dipakai tulangan Ф 14 – 80 (A= 2770 mm2 )

Control kekuatan pelat :

As = 2511 mm2 > Amin = 0,002 .120 . 1200 = 288 mm2

a = = = 41,85 mm

Mn = As.fy.(d-a.2)

Mn = 2511.400.(93-41,85/2)

Mn = 71,39 KNm

ФMn = 0,8 . 71,39 = 57,112 KNm

ФMn = 57,112 KNm > Mu = 22,825 KNm

Tulangan bagi = 0,002 . b . h

= 0,002 . 1200 . 120 = 288 mm2

Digunakan tulangan Ф 10 – 250 (A= 314 mm2)

Skema Penulangan Plat Bordes

Ф 10 - 250

Ф 14 - 80

3.7 PERHITUNGAN PONDASI

73,681 kN 145,981kN 153,13 kN

2,479 kN A -1,071 kN B 2,188 kN C

3,306 kN -1,431 kN 2,917 kN

FREE BODY PONDASI A FREEBODY PONDASI B FREEBODY PONDASI C

Sebagai data awal dari penyelidikan tanah, diperoleh data sebagai berikut :

No Kedalaman (m) Konus (kg/cm2)

1 0,2 12

2 0,4 15

3 0,6 18

4 0,8 22

5 1,0 26

6 1,2 30

7 1,4 15

8 1,6 8

Φ = 0 ( sudut gesek dalam tanah )

Γ = 20 KN/m3 ( berat volume tanah )

C = 40 KN/m3 (kohesi )

Didapatkan dari tabel kapasitas daya dukung meyerhorf ( 1963 ) dengan φ = 0 , maka :

Nc = 5,41

Nq = 1

Nγ = 0

3.7.1 Perhitungan Kapasitas Daya Dukung

Reaksi horizontal = 2,479 KN

Reaksi vertikal = 153,13 KN

Momen = 3,306 KN

Perhitungan daya dukung tanah

Direncanakan kedalaman tanah D = 1,4 m

Beban Vertikal total Pv = reaksi vertikal + berat kolom ( 30/40 – 8,6 meter )

= 153,13+ 0,3. 0,4. 7,8 ( 24 ) = 166,954

Sudut α arah gaya yang di bentuk Pv dan reaksi horicontal ( RH ) dari pondasi :

0

Mengingat banyaknya parameter perhitungan untuk mendesain suatu pondasi maka untuk

perhitungan di tentukan terlebih dahulu dimensi pondasi bujur sangkar yaitu :

B x B = 1,4 m x 1,4 m

Luas pondasi = 1,4 x 1,4 = 1,96 m2

e ( eksentrisitas ) =

karena terdapat eksentrisitas maka salah satu lebar pondasi B di revisi menjadi B’

B’ = B – 2e = 1,4 – 2 (0,02) = 1,37

Jadi letak posisi beban vertikal

Sc = 1 + 0,2 B’/B tg2 ( 45 +φ/2 ) =1 + 0,2 .0,98. tg2 ( 45 + 0/ 2 ) = 1,197

Sq = 1

Dc = 1 + 0,2 D/B’ Tg2 (45+φ/2) = 1 + 0,2 . 1,02 tg2 ( 45 + 0/2 ) = 1,2

Dq = 1

Ic = iq = 1 – α/90 = 1 – 0,75 / 90 = 0,99

Po = D γ = 1,4 x 20 = 28

Kapasitas daya dukung tanah , dengan Nγ = 0

qu = Sc.dc.ic.c.Nc + Dq . dq . iq .Po. Nq

= 1,197 .1,20 . 0,99 .40 . 5,41 + 1.1. 0,99 . 24 . 1

= 320,09

qun = qu – D γ = 320,09 – 24 = 292,09

dengan menggunakan angka keamanan = 3, maka

q safe = qun/3 = 292,09/ 3 = 97,36

Psafe = q safe .B . B = 97,36 x 1,4 x 1,4 = 190,82

Kontrol Pv =166,954KN ≤ P safe = 190,82 KN ........................OK

3.7.2 Dimensionering Pondasi

Fc’ = 25 Mpa

Tebal pondasi direncanakan setebal = 200 mm

D = h – (1/2 tul pokok + degging ) = 200 – ( ½ 20 +60 ) = 130 mm = 0,13 m

Dimensi rencana dipakai bujursangkar B = 1,4 m

3.7.3 Stabilitas Pondasi

Keliling kritis bo = 2 (( 0,3 + 0,13/2 ) + ( 0,4 + 0,13/2 )) = 1,66 m

Luas kritis = ( 0,3 + 0,13/2 ) . (0,4 + 0,13/2 ) = 0,17 m2

Luas pondasi = 1,4 x 1,4 = 1,96 m2

Luas yang terbebani geser = 1,96 – 0,17 = 1,79

Gaya geser Vu = q safe . luas yang terbebani geser = 97,36 . 1,79 = 174,27

Vc = 4 √fc’ . bo . d = 4√25 . 1,66 . 0,13(1000) = 4316

Vu =174,27 KN ≤ φ Vc =0,6 . 4316 = 2589,6 KN..............OK

Untuk geser pons cukup diantisipasi oleh kapasitas beton saja

Kontrol geser satu arah

t = 0,2m 1,4 m

Daerah pembabanan geser satu arah

1,5 m

470 130 300

Panjang pembebanan geser =

Gaya geser Vu = q safe . 1,4 . 0,42 = 97,36 . 1,4 . 0,42 = 57,248

Vc = 1/6 √Fc’ . bw. D = 1/6 √25. 1,4 . 0,13(1000) = 151,67

Vu = 57,248KN ≤ φ Vc = 0,6 . 151,67 = 91 KN.................OK

3.7.4 Penulangan Pondasi

Perhitungan tulangan lentur dilakukan seperti halnya pada tulngan balok yang mana harus

diketahui M, Bw, dan d

Panjang pembebanan lentur =

M = q safe . ½ . panjang pemb lentur . bw . 4 = 97,36 . ½ .0,552. 1,4 = 20,61

Bw = 1,4 m

d = 0,13 m

ρ min = 0,0035, ρ max = 0,0203

Mn = = = 25,7625

=0,0064

Luas tulangan

As = ρ . bw . d =0,0064 . 1400. 130 = 1164,8

Digunakan 4 Ǿ 20 = 1256,63 mm2 , untuk kepraktisan penulangan digunakan system poor

dengan tulanagan tersebar masing-masing sisi = 4 buah.

Karena pada perhitungan pondasi ini menerima gaya vertical, gaya horizontal, momen yang

paling besar maka dimensi pondasi untuk A, B, dan C disamakan dengan perhitungan.

Skema penulangan pondasi

Rencana Penulangan Pondasi Telapak

Tugas Beton terbaru

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