220
300 450
2 16
8-100 4
16
300 450
4 16
8-100 2
16
14,14 19,7 ok Dipakai sengkang Ø 8 – 100 tumpuan dan Ø 8 – 140 lapangan
• Cek Terhadap Pengaruh Geser Lentur
h b
V .
. 7
. 8
= τ
= 45
30 7
5874,821 8
x x
x = 4,973 kgcm
2
Syarat :
bm
τ τ ≤
4,973 ≤ 1,35 √
’bk
= 1,35 x √225
4,973 ≤ 20,25 kgcm
2
aman
Lapangan Tumpuan
Gambar 5.50 Penulangan Balok Memanjang Lantai Gerbang B
5.7.3 Perhitungan Pelat Lantai dan Balok Lantai Kamar
5.7.3.1 Perhitungan Pelat lantai
Gambar 5.51 Tampak Samping Kamar
+17,00 +16,00
+10,38
1685
221
Gambar 5.52 Potongan Melintang Kamar
1. Pembebanan Pelat Lantai f’c = 225 kgcm
2
fy = 2400 kgcm
2
β 9
36 1500
8 ,
ln
min
+ +
= fy
h ln = Iy = 7240 mm
Ix = 2700 mm diambil sama dengan jarak antar counterfort dinding kamar atau gerbang B
= lylx = 2,681 681
, 2
9 36
1500 240
8 ,
7240
min
x h
+ +
=
591 ,
115
min
= h
mm 36
1500 8
, ln
fy h
maks
+ =
36 1500
240 8
, 7240
+ =
maks
h 067
, 193
=
maks
h mm
Syarat : IyIx 2,0
h
min
= 120 mm IyIx
≥2,0 h
min
= 90 mm
1,75 2,12
2,75 0,58
2,4 0,6
2,4 2,4
0,6 2,4
7,24
0,20
222 Berdasarkan Tabel 10 Dasar-Dasar Perencanaan Beton Bertulang-Seri
Beton 1, h
min
= 127.lx = 127 x 2700 = 100 mm Dipilih tebal pelat = 200 mm
¾ Kondisi 1 Gerbang Kosong H = tinggi muka air tanah dari dasar pelat
H = +13,00 – +10,38 = 2,62 m, dengan pembebanan : - Berat sendiri pelat = 0,24 x 2400 = -480 kgm
2
- Gaya Uplift = 2,62 x 1000 = 2620 kgm
2
q
total
= -480 + 2620 = 2140 kgm
2
ke atas ¾ Kondisi 2 Gerbang Penuh Air
H = tinggi muka air dalam gerbang H = +16,00 – +10,38 = 5,62 m, dengan pembebanan :
- Berat sendiri pelat = 0,20 x 2400 = -480 kgm
2
- Gaya Uplift = 2,62 x 1000 = 2620 kgm
2
- Berat Air = 5,62 x 1000 = -5620 kgm
2
q
total
= -480 + 2620 + -5620 = -3480 kgm
2
ke bawah
Pembebanan pada saat gerbang terisi air lebih besar daripada saat gerbang kosong, sehingga untuk perhitungan pelat lantai didasarkan pada pembebanan
kondisi 2 dengan arah pembebanan ke bawah.
2. Penulangan Pelat Lantai q = 3480 kgm
2
IyIx = 2,681 Untuk kondisi terjepit di keempat sisinya :
Iy Ix
223 M
Ix
= 0,001.q
u
.Ix
2
.x M
Iy
= 0,001.q
u
.Ix
2
.x M
tx
= -0,001.q
u
.Ix
2
.x M
ty
= -0,001.q
u
.Ix
2
.x Dimana nilai x berturut-turut berdasarkan perbandingan lylx = 2,681
adalah 63,1; 14; 83 dan 50,3 Tabel Hal 26 Grafik dan Tabel
Perhitungan Beton Bertulang, Seri Beton- 4, W.C. Vis, Gideon Kusuma
M
Ix
= 0,001 x 3480 x 2,700
2
x 63,1 = 1600,797 kgm M
Iy
= 0,001 x 3480 x 2,700
2
x 14 = 355,169 kgm M
tx
= -0,001 x 3480 x 2,700
2
x 83 = -2105,644 kgm M
ty
= -0,001 x 3480 x 2,700
2
x 50,3 = -1276,071 kgm Diambil momen yang maksimum
Tulangan Lapangan Arah X Mu = 1600,797 kgm = 1,601 x 10
7
Nmm Mn = Mu
φ = 1,601 x 10
7
0,8 = 2,001 x 10
7
Nmm h = 200 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba menggunakan tulangan Ø 16 mm d’ = tebal selimut beton = 40 mm
d = h – d’ – Ø2 = 200 – 40 – 162 = 152 mm k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 2,001 x 10
7
1000 x 152
2
x 19,125 = 0,045 F = 1 -
k 2
1 −
= 1 - 045
, 2
1 −
= 0,046 Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 0,046 x 1000 x 152 x 19,125240 = 557,175 mm
2
ρ = Asb.d = 557,1751000 x 152 = 3,666 x 10
-3
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
224 ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton- 1, W.C. Vis, Gideon Kusuma
atau ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
atau ρ
maks
= fy
R F
1 max
. =
240 125
, 19
455 ,
x =0,03626
ρ ρ
min
dipakai ρ
min
As = ρ.b.d = 5,83 x 10
-3
x 1000 x 152 = 886,160 mm
2
Tulangan terpasang = Ø 16 – 225 As = 894 mm
2
Cek terhadap rasio tulangan: ρ = As
terpasang
b.d = 8941000x152 = 5,882 x 10
-3
ρ
min
ρ ρ
maks
ok Tulangan Lapangan Arah Y
Mu = 355,169 kgm = 3,552 x 10
6
Nmm Mn = Mu
φ = 3,552 x 10
6
0,8 = 4,440 x 10
6
Nmm h = 200 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 16 mm d’ = tebal selimut beton = 40 mm
d = h4 – d’ – Ø tulangan arah X - Ø2 = 200 – 40 – 16 – 162 = 136 mm k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 4,440 x 10
6
1000 x 136
2
x 19,125 = 0,013 F = 1 -
k 2
1 −
= 1 - 013
, 2
1 −
= 0,013 Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 0,013 x 1000 x 136 x 19,125240 = 140,888 mm
2
ρ = Asb.d = 140,8881000 x 136 = 1,036 x 10
-3
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton- 1, W.C. Vis, Gideon Kusuma
225 atau
ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
ρ ρ
min
dipakai ρ
min
As = ρ.b.d = 5,83 x 10
-3
x 1000 x 136 = 792,880 mm
2
Tulangan terpasang = Ø 16 – 250 As = 804 mm
2
Cek terhadap rasio tulangan : ρ = As
terpasang
b.d = 8041000x136 = 5,912 x 10
-3
ρ
min
ρ ρ
maks
ok Tulangan Tumpuan Arah X
Mu = 2105,644 kgm = 2,106 x 10
7
Nmm Mn = Mu
φ = 2,106 x 10
7
0,8 = 2,633 x 10
7
Nmm h = 200 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba menggunakan tulangan Ø 16 mm d’ = tebal selimut beton = 40 mm
d = h – d’ – Ø2 = 200 – 40 – 162 = 152 mm k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 2,633 x 10
7
1000 x 152
2
x 19,125 = 0,059 F = 1 -
k 2
1 −
= 1 - 059
, 2
1 −
= 0,061 Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 0,061x 1000 x 152 x 19,125240 = 738,863 mm
2
ρ = Asb.d = 738,8631000 x 152 = 4,861 x 10
-3
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton- 1, W.C. Vis, Gideon Kusuma
atau ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
atau ρ
maks
= fy
R F
1 max
. =
240 125
, 19
455 ,
x =0,03626
226 ρ ρ
min
dipakai ρ
min
As = ρ.b.d = 5,83 x 10
-3
x 1000 x 152 = 886,160 mm
2
Tulangan terpasang = Ø 16 – 225 As = 894 mm
2
Cek terhadap rasio tulangan: ρ = As
terpasang
b.d = 8941000x152 = 5,882 x 10
-3
ρ
min
ρ ρ
maks
ok Tulangan Tumpuan Arah Y
Mu = 1276,071 kgm = 1,276 x 10
7
Nmm Mn = Mu
φ = 1,276 x 10
7
0,8 = 1,595 x 10
7
Nmm h = 200 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 16 mm d’ = tebal selimut beton = 40 mm
d = h4 – d’ – Ø tulangan arah X - Ø2 = 200 – 40 – 16 – 162 = 136 mm k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 1,595 x 10
7
1000 x 136
2
x 19,125 = 0,045 F = 1 -
k 2
1 −
= 1 - 045
, 2
1 −
= 0,046 Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 0,046 x 1000 x 136 x 19,125240 = 498,525 mm
2
ρ = Asb.d = 498,525 1000 x 136 = 3,666 x 10
-3
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton- 1, W.C. Vis, Gideon Kusuma
atau ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
ρ ρ
min
dipakai ρ
min
As = ρ.b.d = 5,83 x 10
-3
x 1000 x 136 = 792,880 mm
2
Tulangan terpasang = Ø 16 – 250 As = 804 mm
2
Cek terhadap rasio tulangan :
227 ρ = As
terpasang
b.d = 8041000x136 = 5,912 x 10
-3
ρ
min
ρ ρ
maks
ok
5.7.3.2 Perhitungan Balok Lantai
