200
300 450
2 16
8-100 4
16
300 450
4 16
8-100 2
16
7,651 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
362 ,
8725 8
x x
x = 7,387 kgcm
2
Syarat :
bm
τ τ ≤
7,387 ≤ 1,35 √
’bk
= 1,35 x √225
7,387 ≤ 20,25 kgcm
2
aman
Lapangan Tumpuan
Gambar 5.41 Penulangan Balok Memanjang Lantai Gerbang A
5.7.2 Perhitungan Pelat Lantai dan Balok Lantai Gerbang B
5.7.2.1 Perhitungan Pelat Lantai
Gambar 5.42 Tampak Samping Gerbang B
250 350
100 23,25
23,25 100
100 100
23,25 23,25
250
+17,00 +16,00
+10,38
+13,50
+10,88
1350 680,5
Pelat 1 669,5
Pelat 2
201
Gambar 5.43
Potongan Melintang Gerbang B
1. Pembebanan Pelat Lantai f’c = 225 kgcm
2
fy = 2400 kgcm
2
β 9
36 1500
8 ,
ln
min
+ +
+ =
fy h
ln = Iy = 2700 mm sama dengan jarak antar counterfort dinding gerbang B Ix = 1220
= lylx = 2,213 231
, 2
9 36
1500 240
8 ,
2700
min
x h
+ +
=
354 ,
46
min
= h
mm 36
1500 8
, ln
fy h
maks
+ =
36 1500
240 8
, 2700
+ =
maks
h 72
=
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 1,22
0,5 0,12
202 Berdasarkan Tabel 10 Dasar-Dasar Perencanaan Beton Bertulang-Seri
Beton 1, h
min
= 127.lx = 127 x 1220 = 45,185 mm Dipilih tebal pelat = 120 mm paling aman
Konstruksi Ambang Trap, direncanakan tidak terbuat dari material beton bertulang dan hanya digunakan untuk memperkuat stabilitas konstruksi pintu
gerbang, maka ketebalan pelat lantai sebenarnya tetap sebesar 120 mm. ¾ Kondisi 1 Gerbang Kosong
H = tinggi muka air tanah dari dasar pelat H = +13,00 – +10,38 = 2,62 m, dengan pembebanan :
- Beban akibat kostruksi ambang = ditentukan sebesar -900 kgm
2
- Berat sendiri pelat = 0,12 x 2400 = -288 kgm
2
- Gaya Uplift = 2,62 x 1000 = 2620 kgm
2
q
total
= -900 + -288 + 2620 = 1432 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 :
- Beban akibat kostruksi ambang = ditentukan sebesar -900 kgm
2
- Berat sendiri pelat = 0,12 x 2400 = -288 kgm
2
- Gaya Uplift = 2,62 x 1000 = 2620 kgm
2
- Berat Air = 5,62 x 1000 = -5620 kgm
2
q
total
= -900 + -288 + 2620 + -5620 = -4188 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 = 4188 kgm
2
IyIx = 2,213
203 Untuk kondisi terjepit di keempat sisinya :
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 = 2,213
adalah 59,7; 14,6; 82,4 dan 52,1 Tabel Hal 26 Grafik dan Tabel
Perhitungan Beton Bertulang, Seri Beton- 4, W.C. Vis, Gideon Kusuma
M
Ix
= 0,001 x 4188 x 1,220
2
x 59,7 = 372,135 kgm M
Iy
= 0,001 x 4188 x 1,220
2
x 14,6 = 91,008 kgm M
tx
= -0,001 x 4188 x 1,220
2
x 82,4 = -513,634 kgm M
ty
= -0,001 x 4188 x 1,220
2
x 52,1 = -324,761 kgm Diambil momen yang maksimum
Tulangan Lapangan Arah X Mu = 372,135 kgm = 3,721 x 10
6
Nmm Mn = Mu
φ = 3,721 x 10
6
0,8 = 4,651 x 10
6
Nmm h = 120 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba menggunakan tulangan Ø 12 mm d’ = tebal selimut beton = 40 mm
d = h – d’ – Ø2 = 120 – 40 – 122 = 74 mm k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 4,561 x 10
6
1000 x 74
2
x 19,125 = 0,044 F = 1 -
k 2
1 −
= 1 - 044
, 2
1 −
= 0,045
Iy Ix
204 Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 0,045 x 1000 x 74 x 19,125240 = 265,359 mm
2
ρ = Asb.d = 265,3591000 x 74 = 3,586 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
ρ ρ
min
dipakai ρ
min
As = ρ.b.d = 5,83 x 10
-3
x 1000 x 74 = 431,42 mm
2
Tulangan terpasang = Ø 12 – 250 As = 452 mm
2
Cek terhadap rasio tulangan: ρ = As
terpasang
b.d = 4521000x74 = 6,108 x 10
-3
ρ
min
ρ ρ
maks
ok Tulangan Lapangan Arah Y
Mu = 91,008 kgm = 9,101x 10
5
Nmm Mn = Mu
φ = 9,101x 10
5
0,8 = 1,138 x 10
6
Nmm h = 120 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 12 mm d’ = tebal selimut beton = 40 mm
d = h4 – d’ – Ø tulangan arah X - Ø2 = 120 – 40 – 12 – 122 = 62 mm k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 1,138 x 10
6
1000 x 62
2
x 19,125 = 0,015 F = 1 -
k 2
1 −
= 1 - 015
, 2
1 −
= 0,015 Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced
205 As = F.b.d.R
l
fy = 0,015 x 1000 x 62 x 19,125240 = 74,109 mm
2
ρ = Asb.d = 74,1091000 x 62 = 1,195 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 62 = 361,46 mm
2
Tulangan terpasang = Ø 12 – 250 As = 452 mm
2
Cek terhadap rasio tulangan : ρ = As
terpasang
b.d = 4521000x62 = 7,290 x 10
-3
ρ
min
ρ ρ
maks
ok Tulangan Tumpuan Arah X
Mu = 513,634 kgm = 5,136 x 10
6
Nmm Mn = Mu
φ = 5,136 x 10
6
0,8 = 6,420 x 10
6
Nmm h = 120 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba menggunakan tulangan Ø 12 mm d’ = tebal selimut beton = 40 mm
d = h – d’ – Ø2 = 120 – 40 – 122 = 74 mm k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 6,420 x 10
6
1000 x 74
2
x 19,125 = 0,061 F = 1 -
k 2
1 −
= 1 - 061
, 2
1 −
= 0,063 Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 0,063x 1000 x 74 x 19,125240 = 371,503 mm
2
ρ = Asb.d = 371,5031000 x 74 = 5,020 x 10
-3
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
206 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 74 = 431,42 mm
2
Tulangan terpasang = Ø 12 – 250 As = 452 mm
2
Cek terhadap rasio tulangan: ρ = As
terpasang
b.d = 4521000x74 = 6,108 x 10
-3
ρ
min
ρ ρ
maks
ok Tulangan Tumpuan Arah Y
Mu = 324,761 kgm = 3,248 x 10
6
Nmm Mn = Mu
φ = 3,248 x 10
6
0,8 = 4,060 x 10
6
Nmm h = 120 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 12 mm d’ = tebal selimut beton = 40 mm
d = h4 – d’ – Ø tulangan arah X - Ø2 = 120 – 40 – 12 – 122 = 62 mm k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 4,060 x 10
6
1000 x 62
2
x 19,125 = 0,055 F = 1 -
k 2
1 −
= 1 - 055
, 2
1 −
= 0,057 Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 0,057x 1000 x 62 x 19,125240 = 281,616 mm
2
ρ = Asb.d = 281,616 1000 x 62 = 4,542 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
207 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 62 = 361,46 mm
2
Tulangan terpasang = Ø 12 – 250 As = 452 mm
2
Cek terhadap rasio tulangan : ρ = As
terpasang
b.d = 4521000x62 = 7,290 x 10
-3
ρ
min
ρ ρ
maks
ok
5.7.2.2 Perhitungan Balok Lantai
