BAB III Metodologi Beton

BAB III

PAGE

BAB IIIANALISA PERHITUNGAN STRUKTUR3.1 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

225D. Menghitung tebal plat atap

Diketahui data material :

fy = 300 Mpa

fc = 22,5 Mpa

lebar plat bw = 1000 mm = 1 m

tulangan = 8 mm

SHAPE \* MERGEFORMAT


Nilai banding panjang terhadap lebar bentang bersih =

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 cm3.1.2 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 tul = 120 25 0,5 . 8 = 91 mm

dy = h p 0,5 tul - tul = 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 STRUKTURa. 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 = mt = 0,663KNm

mtiy = mt = -0,663 KNmb. 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 = mt = -7,741 KNmc. 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 KNm3.1.5 PERHITUNGAN TULANGANb =

max = 0,75 x b = 0,75 x 0,0425 = 0,0319

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

dy = h p 0,5 tul - tul = 120 25 0,5 . 8 - 8 = 83 mmPlat Type ADaerah Lapangan x

D = dx = 91 mm

min = 0,0047Mu = 0,585 kNm max = 0,0319

Mn = kNm = 731000 NmmMenentukan yang diperlukan :

=

min = 0,0047 > perlu = 0,0003min = 0,0047luas tulangan perlu

Aslx = . bw . d = 0,0047 . 1000 . 91 = 427,7 mm2Maka digunakan 8 100 = 503 Kontrol jarak penulangan plat

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

Kontrol kekuatan plat

a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509 kNm > Mu = 0,585 kNm .....................OK!Daerah Lapangan ydy = 83 mm

min = 0,0047max = 0,0319

Mu = 0,585 kNm

Mn = kNm = 731000 Menentukan yang diperlukan :

=

min = 0,0047 > perlu = 0,0003 min = 0,0047luas tulangan perlu


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


a = mm

Mn = . 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

min = 0,0047max = 0,0319

Mu=1,327kNm

kNm = 1659000 Nmm

Menentukan yang diperlukan :

=

min = 0,0047 > perlu = 0,0007min = 0,0047

luas tulangan perlu

Aslx = . bw . d = 0,0047. 1000 . 91 = 427,7 mm2Maka digunakan 8 100 = 503 Kontrol jarak penulangan plat

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


a = mm

Mn = . 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

min = 0,0047max = 0,0319

Mu=1,327kNm


=



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


a = mm

Mn = . 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

min = 0,0047max = 0,0319Mu = 0,663kNm

Mn = kNm = 287000 Nmm


=



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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509 kNm > Mu = 0,663kNm .....................OK!Daerah tumpuan jepit iyD = dx = 83 mm

min = 0,0047max = 0,0319Mu = 0,663kNm

Mn = kNm = 287000 Nmm


=



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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (83-7,89/2) = 9,543 kNm > Mu = 0,663kNm .....................OK!Plat Type BDaerah Lapangan x

D = dx = 91 mm

min = 0,0047 max = 0,0319Mu = 7,865 kNmMn = kNm = 9831250 NmmMenentukan yang diperlukan :

=



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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509kNm > Mu = 7,865kNm .....................OK!Daerah Lapangan yD = dy = 83 mm

min = 0,0047max = 0,0319Mu = 1,997 kNm

Mn = kNm = 2496250 Nmm


=



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


a = mm


D = dx = 91 mm

min = 0,0047max = 0,0319Mu = 15,482 kNm

Mn = kNm = 19352500 Nmm


=

min = 0,0047 < perlu = 0,0082perlu = 0,0082luas tulangan perlu


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


a = mm

Mn = . 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

min = 0,0047max = 0,0319Mu = 8,864 kNm

Mn = kNm = 11080000 Nmm


=

min = 0,0047 < perlu = 0,0055 perlu = 0,0055luas tulangan perlu


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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (83-7,89/2) = 10,509 kNm > Mu = 8,864 kNm .......................OK!

Daerah tumpuan jepit ix

D = dx = 91 mm

min = 0,0047max = 0,0319Mu = 7,741 kNm

Mn = kNm = 9676250 Nmm


=



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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509 kNm > Mu = 7,741 kNm .....................OK!Plat Type CDaerah Lapangan x

D = dx = 91 mm

min = 0,0047 max = 0,0319Mu = 3,121 kNmMn = kNm = 390000 NmmMenentukan yang diperlukan :

=



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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509kNm > Mu = 3,121 kNm .....................OK!Daerah Lapangan yD = dy = 83 mm

min = 0,0047max = 0,0319Mu = 3,121 kNm

Mn = kNm = 390000 Nmm


=



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


a = mm


D = dx = 91 mm

min = 0,0047max = 0,0319Mu = 6,368 kNm

Mn = kNm = 796000 Nmm


=

min = 0,0047 > perlu = 0,0003min = 0,0047 luas tulangan perlu


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


a = mm

Mn = . 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

min = 0,0047max = 0,0319Mu = 6,368 kNm

Mn = kNm = 796000 Nmm


=

min = 0,0047 < perlu = 0,0004min = 0,0047luas tulangan perlu


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


a = mm

Mn = . 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 lantaiDiketahui data material :

fy = 300 Mpa

fc = 22,5 Mpa

lebar plat bw = 1000 mm = 1 m

tulangan = 8 mm




Nilai banding panjang terhadap lebar bentang bersih =

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 cm3.2.2 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/m2 += 3,585 kN/m2Maka, Wu = 1,2 (3,585) + 1,6 (2,5) = 8,302 kN/m2Perhitungan Geser :

Dx = h - tulangan = 120 25 10 = 90 mm

Dy = h - tulangan - tul = 120 25 10 10 = 80 mm

28,64 kN

Vc = 1/6.bw .d =0,8 x 1/6. 1000 . 90 = 56920,9 N = 56,92 kN

Syarat Vc = 46,92 kN > Vu = 28,64 kN OK!!!3.2.3 PEMODELAN STRUKTURa. Type A

= = 1,00

b. Type B

= = 3c. Type C

= = 1

d. Type D

= = 3

e.TypeE

= = 1,2

3.2.4 MENENTUKAN MOMENa. 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 = mt = -1,005KNm

mtiy = mt = -1,005 KNmb. 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 = mt = -6,72 KNmc. 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 KNmd. 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 5,752= 17,292 KNm

mly = 0,001 wu . Lx2. 16 = 0,016 8,302 5,752= 4,392 KNm

mt = -0,001 wu . Lx2. 124 = -0,124 8,302 5,752= -34,036 KNm

mty = -0,001 wu . Lx2. 71 = -0,071 8,302 5,752= -19,488 KNm

mtix = mt = -17,018 KNme. Menentukan momen plat atap type E dengan ly/l = 1,2

ml = 0,001 wu . Lx2. 34 = 0,034 8,302 5,752= 9,332 KNm

mly = 0,001 wu . Lx2. 22 = 0,022 8,302 5,752= 6,039 KNm

mt = -0,001 wu . Lx2. 63 = -0,063 8,302 5752= -17,292 KNm

mty = -0,001 wu . Lx2. 54 = -0,054 8,302 5752= -14,822 KNm3.2.5 PERHITUNGAN TULANGANb =

max = 0,75 x b = 0,75 x 0,051 = 0,0382Dx = h - tulangan = 120 25 10 = 90 mmDy = h - tulangan - tul = 120 25 10 10 = 80 mm

Plat Type ADaerah Lapangan x

min = 0,0047Mu = 0,887 kNm max = 0,0382

Dx = h - tulangan = 120 25 10 = 90 mmMn = kNm = 1109000 NmmMenentukan yang diperlukan :

=


Aslx = . bw . d = 0,0047 . 1000 . 90 = 423 mm2Maka digunakan 8 100 = 503

Kontrol jarak penulangan plat

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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509 kNm > Mu = 0,887 kNm .....................OK!Daerah Lapangan ymax = 0,0382

Mu = 0,887 kNm

min = 0,0047Dy = h - tulangan - tul = 120 25 10 10 = 80 mm


=




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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2) = 7,7 kNm > Mu = 0,887 kNm.......................OK!Daerah tumpuan x

min = 0,0047max = 0,0382

Mu=2,011kNm Dx = h - tulangan = 120 25 10 = 90 mm

kNm = 2513000 Nmm


=

min = 0,0047 > perlu = 0,001min = 0,0047

luas tulangan perlu

Aslx = . bw . d = 0,0047. 1000 . 90 = 423 mm2Maka digunakan 8 100 = 503


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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,9/2) = 10,509 kNm > Mu = 2,011kNm .....................OK!Daerah tumpuan y

min = 0,0047max = 0,0382

Mu=2,011kNm Dy = h - tulangan - tul = 120 25 10 10 = 80 mm


=




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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2) = 7,7 kNm > Mu = 2,011kNm .......................OK!Daerah tumpuan jepit ix

min = 0,0047max = 0,0382Mu = 1,005kNm

Dx = h - tulangan = 120 25 10 = 90 mmMn = kNm = 1256000 Nmm Menentukan yang diperlukan :

=




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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509 kNm > Mu = 1,005 kNm .....................OK!Daerah tumpuan jepit iymin = 0,0047max = 0,0382Mu = 1,005kNm


Mn = kNm = 1256000 Nmm


=




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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (83-7,89/2) = 9,543 kNm > Mu = 1,005kNm .....................OK!Plat Type BDaerah Lapangan x

min = 0,0047 max = 0,0382Mu = 11,925kNmDx = h - tulangan = 120 25 10 = 90 mmMn = kNm = 14906000 NmmMenentukan yang diperlukan :

=

min = 0,0047< perlu = 0,006perlu = 0,006luas tulangan perlu



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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509kNm > Mu = 11,925kNm .....................OK!Daerah Lapangan ymin = 0,0047max = 0,0382Mu = 3,029 kNm


Mn = kNm = 3786000 Nmm


=




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


a = mm


min = 0,0047max = 0,0382Mu = 17,292 kNm

Dx = h - tulangan = 120 25 10 = 90 mmMn = kNm = 23340000 Nmm


=




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


a = mm


min = 0,0047max = 0,0382Mu = 13,440 kNm


Mn = kNm = 16800000 Nmm


=




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


a = mm



min = 0,0047max = 0,0382Mu = 6,72 kNm



=




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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509 kNm > Mu = 6,72 kNm .....................OK!Plat Type CDaerah Lapangan x

min = 0,0047 max = 0,0382Mu = 4,732 kNmDx = h - tulangan = 120 25 10 = 90 mmMn = kNm = 5915000 NmmMenentukan yang diperlukan :

=




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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509kNm > Mu = 4,732 kNm .....................OK!Daerah Lapangan ymin = 0,0047max = 0,0382Mu = 9,654 kNm


Mn = kNm = 12067000 Nmm


=




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


a = mm


min = 0,0047max = 0,0382Mu = 4,732 kNm



=

min = 0,0047 > perlu = 0,002min = 0,0047 luas tulangan perlu



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


a = mm


min = 0,0047max = 0,0382Mu = 4,732 kNm


Mn = kNm = 5915000 Nmm


=




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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2) = 7,704 kNm > Mu = 4,732 kNm .......................OK!Plat Type DDaerah Lapangan xmin = 0,0047 max = 0,0382Mu = 17,292 kNmDx = h - tulangan = 120 25 10 = 90 mmMn = kNm = 21615000 Nmm Menentukan yang diperlukan :

=

min = 0,0047 < perlu 0,009perlu = 0,009luas tulangan perlu



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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509kNm > Mu = 17,292kNm .....................OK!Daerah Lapangan yD = dy = 83 mm

min = 0,0047max = 0,0382Mu = 4,392kNm


Mn = kNm = 5490000 Nmm


=




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


a = mm


min = 0,0047max = 0,0382Mu = 34,036 kNm


=

min = 0,0047 < perlu = 0,02

perlu = 0,02luas tulangan perlu



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


a = mm


min = 0,0047max = 0,0382Mu = 19,488 kNm


Mn = kNm = 24360000 Nmm


=

min = 0,0047 < perlu = 0,01




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


a = mm



min = 0,0047max = 0,0382Mu = 17,018 kNm


=

min = 0,0047< perlu = 0,009




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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509 kNm > Mu = 17,018 kNm .....................OK!Plat Type EDaerah Lapangan x

min = 0,0047 max = 0,0382Mu = 9,332 kNmDx = h - tulangan = 120 25 10 = 90 mmMn = kNm = 11665000 Nmm Menentukan yang diperlukan :

=

min = 0,0047 > perlu = 0,005perlu = 0,005luas tulangan perlu



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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 503 . 300 (91-7,89/2) = 10,509kNm > Mu = 9,332 kNm .....................OK!Daerah Lapangan ymin = 0,0047max = 0,0382Mu = 6,039kNm

Dy = h - tulangan - tul = 120 25 10 10 = 80 mmMn = kNm = 7550000 Nmm


=




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


a = mm


min = 0,0047max = 0,0382Mu = 17,292 kNm



=

min = 0,0047 < perlu = 0,009min = 0,009 luas tulangan perlu



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


a = mm


min = 0,0047max = 0,0382Mu = 14,822 kNm

Dy = h - tulangan - tul = 120 25 10 10 = 80 mmMn = kNm = 18527000 Nmm


=


Aslx = . bw . d = 0,01 . 1000 . 80= 800 mm2Maka digunakan 8 125 = 402


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


a = mm

Mn = . As . fy . (d-a/2) = 0,8 . 402 . 300 (83-6,3/2) = 7,704 kNm > Mu = 14,822 kNm .......................OK! Resume Plat Hasil Perhitungan

PLAT ATAP

Tipe A

Aslx = 8 250 = 201 mm2Asly= 8 250 = 201 mm2Astx= 8 150 = 201 mm2Asty= 8 150 = 201 mm2

Tipe B1


Tipe B2


Tipe C


Tipe DAslx = 8 250 = 201 mm2Asly= 8 250 = 201 mm2Astx= 8 250 = 201 mm2Asty= 8 250 = 201 mm2

Astix = 8 250 = 201 mm2

Tipe E


Astix = 8 250 = 201 mm2

PLAT LANTAI

Tipe A


Tipe B1


Tipe B2


Tipe C


Tipe D


Astix = 8 250 = 201 mm2

Tipe E


Astix = 8 250 = 201 mm2

3.3 PERHITUNGAN PORTAL

3.3.1 Dimensi Portal

Balok atap direncanakan 20/35Balok lantai direncanakan 20/40

Kolom direncanakan 30/40 ( untuk kolom lantai 1 dan lantai 2 )

Equivalen beban plat q plat

bidang C dan F ( dua buah segitiga )

luas =

berat q plat( IJ dan EF) per m =

bidang A ( dua buah segitiga )

luas =

bidang B

luas =

berat q plat( HI dan DE) per m =

bidang D luas =

bidang E

luas =

berat q plat( FG dan JK) per m =

3.3.2 Pembebanan Portal beban balok HI ( 20/35 )

berat sendiri balok = 0,2 x 0,35 x 24

= 1,68 KNm

berat q plat

=

= 15.551 KNm

total = 17,231 KNm beban balok IJ (20/35)


= 1,68 KNm

berat q plat

=

= 14.164 KNm

total = 15,844 KNm beban balok JK (20/35)


= 1,68 KNm

berat q plat

=

= 14.8912 KNm

total = 16,571 KNm

beban balok DE ( 20/40 )


= 1,92 KNm

berat q plat

=

= 17,772 KNm

total = 19,692 KNm

beban balok EF ( 20/40 )


= 1,92 KNm

berat q plat

=

= 16,189 KNm

total = 18,109 KNm

beban balok FG ( 20/40 )


= 1,92 KNm

berat q plat

=

= 17,019 KNm

total = 18,919 KNm

3.3.2 Perhitungan Cross

MENENTUKAN NILAI t2 Balok T dengan bentang 4,3 m

t2

tf t3

bw

Bw = 20 cm

h = 35 cm

tf = 10 cm

Mengingat plat dan balok monolit, maka balok diperbolehkan dianggap sebagai balok T, dengan ketentuan :t2 = diperoleh dari nilai terkecil :

panjang balok = x 4,3 m = 1,075 m

Jarak antar portal = 4,3 m

16 tf + bw = 16 . 10 + 20 = 180 cm = 1,8 m

Maka t2 atau bm untuk balok dengan bentang 4,3 meter adalah 1,075 m

Balok T dengan bentang 3,9 m

t2

tf

t3

bw

Bw = 20 cm

h = 35 cm

tf = 10 cm

Mengingat plat dan balok monolit, maka balok diperbolehkan dianggap sebagai balok T, dengan ketentuan :

t2 = diperoleh dari nilai terkecil :



16 tf + bw = 16 . 10 + 20 = 180 cm = 1,8 m


Balok T dengan bentang 2,3m

t2

tf

t3

bw

Bw = 20 cm

h = 35 cm

tf = 10 cm

Mengingat plat dan balok monolit, maka balok diperbolehkan dianggap sebagai balok T, dengan ketentuan :

t2 = diperoleh dari nilai terkecil :



16 tf + bw = 16 . 10 + 20 = 180 cm = 1,8 m


MOMEN INERSIA

Momen inersia balok atap berukuran 20/35 didapat dari SAP dan cross yaitu 0,00139514 cm4 (HI), 0,00135385 cm4 (IJ), 0,00112737 cm4 (JK)

Momen inersia balok lantai berukuran 20/40 didapat dari SAP dan cross yaitu 0,00208261 cm4 (DE), 0,00202118 cm4 (EF), 0,00168515 cm4 (FG)

Momen inersia kolom ukuran 30/40 = 1/12 b h3 = 1/12 . 40 . 303 = 90000 cm4 = 0,0009 m4KEKAKUAN

Atap

Kekakuan HI

Inersia = 0,00139514 cm4L= 4,3 meter

KHI ==295,560724

Kekakuan IJ


KIJ ==277,712589

Kekakuan JK ( dengan bentang L = 2,3 m )

KJK = 0

Lantai

Kekakuan DE


KDE ==387,462812

Kekakuan EF


KEF ==414,601018

Kekakuan FG ( dengan bentang L = 2,3 m )

KFG = 0

Kolom

Inersia = 0,0009 m4 Kekakuan HD = IE = JF ( dengan L = 3,8 m )

KHD = KIE = KJF = = 189,473864 Kekakuan AD = BE = CF ( dengan L = 4 m )

KHD = KIE = KJF = = 180DISTRIBUSI KEKAKUAN

Titik Kumpul H

Titik kumpul I

Titik kumpul J

Titik Kumpul D

Titik kumpul E

Titik kumpul F

MOMEN PRIMER

Atap

MHI = -1/12. Q . L2 = -1/12 x 17,231 x 4,32 = -26,549

MHI = 1/12. Q . L2 = -1/12 x 17,231 x 4,32 = 26,549

MIJ = -1/12. Q . L2 = -1/12 x 15,844 x 3,92 = -20,077

MIJ = 1/12. Q . L2 = -1/12 x 15,844 x 3,92 = 20,077

MJK = -1/12. Q . L2 = -1/12 x 16,571 x 2,32 = -43,83Lantai

MDE = -1/12. Q . L2 = -1/12 x 19,692 x 4,32 = -30,342MED = 1/12. Q . L2 = 1/12 x 19,692 x 4,32 = 30,342 MEF = -1/12. Q . L2 = -1/12 x 18,109 x 3,92 = -22,953MFE = 1/12. Q . L2 = 1/12 x 18,109 x 3,92 = 22,953 MFG = -1/12. Q . L2 = -1/12 x 18,919 x 2,32 = - 50,041Lanjutan Perhitungan Cross ada pada Lampiran3.4 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:

Fc = 25 Mpa

Fy = 400 Mpa

1 = 0,85

lentur = 0,8

sengkang = 0,6

Untuk perencanaan tulangan selanjutnya dapat diketahui nilai-nilai rasio tulangan maksimum dan minimum sebagai berikut :

min =

b =

max = 0,75 x b = 0,75 x 0,027 = 0,0203

3.4.1 Dimensionering Balok

BALOK ATAPBALOK LANTAI

Tinggi balok h = 350 mm = 35 cmTinggi balok h = 400mm = 40 cm

Lebar balok bw = 200 mm =20 cmLebar balok bw = 200 mm =20 cm

Lapisan tulangan lp = 1 lapis

pokok = 16 mm

sengkang = 8 mm

Degging P = 40 mm

FREE BODY BALOK HIQ = 17,231 KNm

L = 4,3 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 = -17,231 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, min = 0,0035, 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 216 = 402 mm2Kontrol kekuatan balok:

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

Mn > Mu = 35,386 KNm > 14,716 KNm OK TUMPUAN IH

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

Mu = 27,123 KNm, Mn =

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

Luas tulangan perlu

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

Maka digunakan 216 = 402 mm2Kontrol kekuatan balok

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

Mn > Mu = 35,386 KNm > 27,123 KNm OK LAPANGAN HI

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

Mu = 18,903 KNm, MT plat = 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 KNmMu =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


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

Mn > Mu = 35,386 KNm > 18,903 KNm OK

PERHITUNGAN TULANGAN SENGKANGKuat 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 SEGMEN1. 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 / = 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

4. Segmen 4 ( 1290-1720 mm )

Lihat lampiran

5. Segmen 5 ( 1720-2150 mm )

Lihat lampiran

6. Segmen 6 ( 2150-2580 mm )

Lihat lampiran

7. Segmen 7 ( 2580-3010 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 KNmVI = 28,218 KN

VJ= 33,357 KN


TUMPUAN IJd = 294 mm, Bw = 200 mm, min = 0,0035, max = 0,0203

Mu = 20,418 KNm , Mn =



Luas tulangan perlu

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


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

Mn > Mu = 35,386 KNm > 20,418 KNm OK

TUMPUAN JId = 294 mm, Bw =200 mm, min = 0,0035, max = 0,0203

Mu = 30,836 KNm, Mn =


Luas tulangan perlu

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


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

Mn > Mu = 35,386 KNm > 30,836 KNm OK

LAPANGAN IJd = 294 mm, Bw = 200 mm, min = 0,0035,

Mu = 4,494 KNm,

MT plat = 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 KNmMu =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


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

Mn > Mu = 35,386 KNm > 4,494 KNm OK

PERHITUNGAN TULANGAN SENGKANG

Kuat geser beton dapat dihitung sebagai berikut :



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


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



PERHITUNGAN PER SEGMEN

1. Segmen 1 ( 0 - 390 mm )


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

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

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

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

Mu = 43,83 KNm , Mn =


Karena = 0,0067 > = 0,0067 maka yang diambil min untuk menghitung luas penampang tulangan, jadi Luas tulangan perlu As = . bw . d = 0,0067 x 200 x 294 = 393,96 mm2


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

Mn > Mu = 35,386 KNm > 43,83 KNm OK

PERHITUNGAN PER SEGMEN

1. Segmen 1 ( 0-230 mm )


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

Vn = Vu / = 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

2. Segmen 2 ( 230-460 mm )

Lihat lampiran

3. Segmen 3( 460-690 mm )

Lihat lampiran

4. Segmen 4 ( 690-920 mm )

Lihat lampiran

5. Segmen 5 ( 920-1150 mm )

Lihat lampiran

6. Segmen 6 ( 1150-1380 mm )

Lihat lampiran

7. Segmen 7 ( 1380-1610 mm )

Lihat lampiran




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

d = 3444 mm, Bw = 200 mm, min = 0,0035, max = 0,0203

Mu = 19,19 KNm , Mn =



Luas tulangan perlu

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


Mn = . as .Fy .( d a/2 ) = 0,8 x 402 x 400 ( 344-37,8/2) = 41,81 KNm

Mn > Mu = 41,81 KNm > 19,19 KNm OK

TUMPUAN ED

d = 344 mm, Bw =200 mm, min = 0,0035, max = 0,0203

Mu = 32,749 KNm, Mn =


Luas tulangan perlu

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


Mn = . as .Fy .( d a/2 ) = 0,8 x 402 x 400 ( 344-37,8/2) = 41,81 KNm

Mn > Mu = 41,81 KNm > 32,749 KNm OK

LAPANGAN DE

d = 344 mm, Bw = 200 mm, min = 0,0035, max = 0,0203

Mu = 20,269 KNm, Mn =



Luas tulangan perlu

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


Mn = . as .Fy .( d a/2 ) = 0,8 x 402 x 400 ( 294-37,8/2) = 41,81 KNm

Mn > Mu = 41,81 KNm > 20,269 KNm OK

PERHITUNGAN TULANGAN SENGKANGKuat geser beton dapat dihitung sebagai berikut :



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


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



PERHITUNGAN PER SEGMEN1. Segmen 1 (0-430 mm)


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

Vn = Vu / = 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

2. Segmen 2 ( 430-860 mm )

Lihat lampiran

3. Segmen 3( 860-1290 mm )

Lihat lampiran

4. Segmen 4 ( 1290-1720 mm )

Lihat lampiran

5. Segmen 5 ( 1720-2150 mm )

Lihat lampiran

6. Segmen 6 ( 2150-2580 mm )

Lihat lampiran

7. Segmen 7 ( 2580-3010 mm )

Lihat lampiran




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

d = 344 mm, Bw = 200 mm, min = 0,0035, max = 0,0203

Mu = 50,041 KNm , Mn =


Karena = 0,0071 > = 0,0071 maka yang diambil min untuk menghitung luas penampang tulangan, jadi luas tulangan perlu

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


Mn = . as .Fy .( d a/2 ) = 0,8 x 402 x 400 ( 344-56,8/2) = 60,9 KNm

Mn > Mu = 60,9 KNm > 50,041 KNm OK

PERHITUNGAN PER SEGMEN1. Segmen 1 ( 0-330 mm )


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

Vn = Vu / = 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

2. Segmen 2 ( 230-460 mm )

Lihat lampiran

3. Segmen 3( 460-690 mm )

Lihat lampiran

4. Segmen 4 ( 690-920 mm )

Lihat lampiran

5. Segmen 5 ( 920-1150 mm )

Lihat lampiran

6. Segmen 6 ( 1150-1380 mm )

Lihat lampiran

7. Segmen 7 ( 1380-1610 mm )

Lihat lampiran




LAMPIRAN FG

DETAIL PENULANGAN POKOK DAN SENGKANG BALOK FG

TABEL PENULANGAN BALOK

NONAMA BALOKTUMPUANLAPANGANTUMPUAN

1DE2 16 = 402 mm22 16 = 402 mm22 16 = 402 mm2

2EF2 16 = 402 mm22 16 = 402 mm22 16 = 402 mm2

3FG3 16 = 402 mm22 16 = 402 mm22 16 = 402 mm2

4HI2 16 = 402 mm22 16 = 402 mm22 16 = 402 mm2

5IJ2 16 = 402 mm22 16 = 402 mm22 16 = 402 mm2

6JK2 16 = 402 mm22 16 = 402 mm22 16 = 402 mm2

3.5 PERHITUNGAN TULANGAN KOLOM

Hasil analisa gaya dalam yang dipakai untuk penulangan kolom adalah hasil perhitungan Cross.Data material :Fc = 25 Mpa

hFy = 400 Mpa

1 = 0,85 d lentur = 0,8

bw sengkang = 0,6

Untuk perencanaan tulangan selanjutnya dapat diketahui nilai-nilai rasio tulangan maksimum dan minimum sebagai berikut :

min =

b =

max = 0,75 x b = 0,75 x 0,027 = 0,0203

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

lapis tulangan Lp = 1 lapis

pokok = 16 mm, 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 DAd/h =56/300 = 0,19

e = Mu / Pu=6,612 / 73,681 = 89,738e/h = 89,738 / 300 = 0,299

Setelah di cari pada grafik interaksi kolom didapatkan r = 0,003jadi = r . = 0,003 . 1 = 0,003 min = 0,0035< = 0,003 < max = 0,0203

perlu = 0,0035

luas tulangan perlu :As = . bw . d . = 0,0035 x 400 x 300 =420 mm2

Maka digunakan 3 16 = 603 mm2 TUMPUAN ADd/h =56/300 = 0,19

e = Mu / Pu= 3,306 / 73,681 = 44,869e/h = 44,869 / 300 = 0,15

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

jadi = r . = 0,003 . 1 = 0,003

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

perlu = 0,0035

luas tulangan perlu :

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

Maka digunakan 3 16 = 603 mm2 PERHITUNGAN TULANGAN SENGKANGKuat geser beton dapat dihitung sebagai berikut :



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 )


Vu = VA = 2,479 KN

Vn = Vu / = 2,479 / 0,6 = 4,13 KNm

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

Smax = 122 mm120 mm

2. Segmen 2 ( 440-800mm )

Lihat lampiran

3. Segmen 3( 880-1200 mm )

Lihat lampiran

4. Segmen 4 ( 1200-1600mm )

Lihat lampiran

5. Segmen 5 ( 1600-2000 mm )

Lihat lampiran

6. Segmen 6 ( 2000-2400 mm )

Lihat lampiran

7. Segmen 7 ( 2400-2800 mm )

Lihat lampiran



10. Segmen 10 ( 3600-4000 mm ) Lihat lampiran

LAMPIRAN AD

FREE BODY KOLOM CFMCF = 0,567 KNm

MFC = 1,135 KNm

VC = -0,425 KNm

VF = 0,425 KNm

PFC = 108,228 KN

TUMPUAN CFd/h =56/300 = 0,19

e = Mu / Pu= 0,567 / 108,228 = 5,239e/h = 5,239 / 300 = 0,017


perlu = 0,004luas tulangan perlu :

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

Maka digunakan 3 16 = 603 mm2 TUMPUAN FCd/h =56/300 = 0,19

e = Mu / Pu= 1,135 / 108,228 = 10,49e/h = 10,49 / 300 = 0,365



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

Maka digunakan 3 16 = 603 mm2 PERHITUNGAN TULANGAN SENGKANG




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


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




1. Segmen 1 ( 0- 400 mm )


Vu = VA = 0,425 KN

Vn = Vu / = 0,425 / 0,6 =0,71KNm

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

Smax = 122 mm120 mm

2. Segmen 2 ( 440-800mm )

Lihat lampiran

3. Segmen 3( 880-1200 mm )

Lihat lampiran

4. Segmen 4 ( 1200-1600mm )

Lihat lampiran

5. Segmen 5 ( 1600-2000 mm )

Lihat lampiran

6. Segmen 6 ( 2000-2400 mm )

Lihat lampiran

7. Segmen 7 ( 2400-2800 mm )

Lihat lampiran



10. Segmen 10 ( 3600-4000 mm ) Lihat lampiran

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 = r . = 0 . 1 = 0 min = 0,0035< = 0 < max = 0,0203


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

Maka digunakan 3 16 = 603 mm2 TUMPUAN HDd/h =56/300 = 0,19

e = Mu / Pu= 14,262 / 33,62 = 423,68e/h = 423,68 / 300 = 1,412


jadi = r . = 0. 1 = 0

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

perlu = 0


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





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


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




1. Segmen 1 ( 0- 380 mm )


Vu = VA = 6,973 KN

Vn = Vu / = 6,973 / 0,6 = 11,62KNm

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

Smax = 122 mm120 mm

2. Segmen 2 (380-760mm )

Lihat lampiran

3. Segmen 3(760-1140 mm )

Lihat lampiran

4. Segmen 4 (1140-1520 mm )

Lihat lampiran

5. Segmen 5 (1520-1900 mm )

Lihat lampiran

6. Segmen 6 (1900-2280 mm )

Lihat lampiran

7. Segmen 7 (2280-2660 mm )

Lihat lampiran




LAMPIRAN DH

GAMBAR PENULANGAN SKEMATIS KOLOM DH

FREE BODY KOLOM FJMFJ = 2,359 KNm

MJF = 2,926 KNm

VF = -1,391 KNm

VJ= 1,391 KNm

PJF = 50,631 KN

TUMPUAN FJd/h =56/300 = 0,19

e = Mu / Pu= 2,359 / 50,631 = 46,592e/h = 46,592 / 300 = 0,155


jadi = r . = 0 . 1 = 0

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

perlu = 0,0035


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

Maka digunakan 3 16 = 603 mm2 TUMPUAN JFd/h =56/300 = 0,19

e = Mu / Pu= 2,926 / 33,62 = 57,791e/h = 57,791 / 300 = 0,193


jadi = r . = 0. 1 = 0

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

perlu = 0


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





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


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




11. Segmen 1 ( 0- 380 mm )


Vu = VA = 1,391 KN

Vn = Vu / = 1,391 / 0,6 = 2,32 KNm

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

Smax = 122 mm120 mm

12. Segmen 2 (380-760mm )

Lihat lampiran

13. Segmen 3(760-1140 mm )

Lihat lampiran

14. Segmen 4 (1140-1520 mm )

Lihat lampiran

15. Segmen 5 (1520-1900 mm )

Lihat lampiran

16. Segmen 6 (1900-2280 mm )

Lihat lampiran

17. Segmen 7 (2280-2660 mm )

Lihat lampiran




LAMPIRAN FJ

GAMBAR PENULANGAN SKEMATIS KOLOM FJ

TABEL PENULANGAN KOLOM

NONAMA KOLOMTUMPUANTUMPUAN

1AD3 16 = 603 mm23 16 = 603 mm2

2BE3 16 = 603 mm23 16 = 603 mm2

3CF3 16 = 603 mm23 16 = 603 mm2

4DH3 16 = 603 mm23 16 = 603 mm2

5EI3 16 = 603 mm23 16 = 603 mm2

6FJ3 16 = 603 mm23 16 = 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, = tg -1 . (200/330) = 31,2184 310136.24

3.6.2 Pembebanan TanggaPembebanan tangga :

beban mati

pelat tangga = 0,12 (1/cos 310136.24). 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 = 14 mm

tebal selimut beton = 20 mm

tinggi efektif pelat = h p 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 : min < < maks

pmin=

=

= 0,0035

b =

== 0,027

maks = 0,75 x b = 0,75 . 0,027= 0,02

menghitung perlu :

Rn=

=

= 3,616

=

=

= 0,031

= 0,01149 > min = 0,0035

diambil = 0,031

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

= 5766 mm2Dipakai tulangan 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

Mn= 0,8 . 148,001688 = 118,406 KNm

Mn= 118,406 KNm > Mu = 50,041 KNm..OK

Tulangan bagi = 0,002 . b . h

= 0,002 . 2000 . 120 = 480 mm2

Digunakan tulangan 10 150 (A= 524 mm2)

Skema penulangan plat tangga

b. penulangan pelat bordes

tebal pelat tangga ; h = 120 mm

tulangan yang digunakan = 14 mm

tebal selimut beton = 20 mm

tinggi efektif plat (dx)

= h p tulnagan

= 120 20 . 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

=

=

= 0,0135

= 0,0135 > min = 0,0035

diambil = 0,0135

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

= 2511 mm2Dipakai tulangan 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

Mn= 0,8 . 71,39 = 57,112 KNm

Mn= 57,112 KNm > Mu = 22,825 KNm

Tulangan bagi = 0,002 . b . h

= 0,002 . 1200 . 120 = 288 mm2

Digunakan tulangan 10 250 (A= 314 mm2)

Skema Penulangan Plat Bordes

3.7 PERHITUNGAN PONDASI73,681 kN 145,981kN 153,13 kN

2,479 kNA

-1,071 kN B2,188 kNC

3,306 kN-1,431 kN2,917 kN

FREE BODY PONDASI A FREEBODY PONDASI B FREEBODY PONDASI C

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

NoKedalaman (m)Konus (kg/cm2)

10,212

20,415

30,618

40,822

51,026

61,230

71,415

8 1,68

= 0 ( sudut gesek dalam tanah )

= 20 KN/m3 ( berat volume tanah )

C = 40 KN/m3 (kohesi )

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

Nc = 5,41

Nq = 1

N = 0

3.7.1 Perhitungan Kapasitas Daya DukungReaksi 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 ) = 166,954Sudut arah gaya yang di bentuk Pv dan reaksi horicontal ( RH ) dari pondasi :

0Mengingat 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 +/2 ) =1 + 0,2 .0,98. tg2 ( 45 + 0/ 2 ) = 1,197

Sq = 1

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

Dq = 1

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

Po = D = 1,4 x 20 = 28

Kapasitas daya dukung tanah , dengan N = 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 = 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 PondasiFc = 25 Mpa

Tebal pondasi direncanakan setebal = 200 mm

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

Dimensi rencana dipakai bujursangkar B = 1,4 m

3.7.3 Stabilitas PondasiKeliling 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 = 425 . 1,66 . 0,13(1000) = 4316

Vu =174,27 KN 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 Vc = 0,6 . 151,67 = 91 KN.................OK

3.7.4 Penulangan PondasiPerhitungan tulangan lentur dilakukan seperti halnya pada tulngan balok yang mana harus diketahui M, Bw, dan d

Panjang pembebanan lentur =

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

Bw = 1,4 m

d = 0,13 m

min = 0,0035, max = 0,0203

Mn = = = 25,7625

=0,0064

Luas tulangan

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

Digunakan 4 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 EMBED AutoCAD.Drawing.17

14 - 40

10 - 150

10 - 150

10 - 250

14 - 80

EMBED AutoCAD.Drawing.17

EMBED AutoCAD.Drawing.17

PAGE

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BAB III Metodologi Beton

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Transcript of BAB III Metodologi Beton