133 As = F.b.d.R
l
fy = 8,894x10
-3
x 1000 x 320 x 19,125240 = 226,789 mm
2
ρ = Asb.d = 226,7891000 x 320 = 7,087 x 10
-4
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
min
= ρ
min
.b.d = 5,83 x 10
-3
x 1000 x 320 = 1865,6 mm
2
Tulangan pokok terpasang = Ø 20 – 150 As = 2094 mm
2
Cek : ρ = As
terpasang
b.d = 20941000x320 = 6,544 x 10
-3
ρ
min
ρ ρ
maks
ok
5.6.1.4 Perhitungan Bagian Perkuatan Dinding Counterfort
¾ Perhitungan titik berat counterfort sebagai balok Xz = 13.b5 = 13 x 1,50 = 0,500 m
Yz = 13.H = 13 x 4,12 = 1,373 m
H L
o
Xz Yz
a
a
b
5
134 ¾ Perhitungan tinggi balok
Tan α = Hb5+b3-b1 = 4,121,5+0,4-0,3 = 2,575 α = 68,776
a = Sin α.b5 = Sin 68,776
x 1,50 = 1,398 m L = b5+b3-b1Cos
α = 1,5+0,4-0,3Cos 68,776 = 4,419 m
¾ Pembebanan counterfort
Tabel 5.8 Pembebanan Counterfort
Titik Berat Terhadap Titik O
P Gaya ton
Yz m Xz m
Y m X m
Lengan m
Momen tm
Pa1 Pa2
Pa3 Pa4
Pa5 Pa6
Paw Q
G1 G2
G3 G4
G5 G6
G7 G8
G9 G10
1,458 0,691
2,840 0,344
1,311 0,049
0,110 1,600
0,720 0,941
1,920 2,856
2,632 2,550
0,318 4,822
0,121 0,180
1,373 1,373
1,373 1,373
1,373 1,373
1,373 1,373
1,373 1,373
1,373 1,373
1,373 1,373
1,373 1,373
1,373 1,373
0,500 0,500
0,500 0,500
0,500 0,500
0,500 0,500
0,500 0,500
0,500 0,500
0,500 0,500
0,500 0,500
0,500 0,500
3,120 2,453
1,120 0,787
-0,115 -0,193
-0,193 -
- -
- -
- -
- -
- -
- -
- -
- -
- 0,700
-0,250 -0,225
-0,200 -0,200
0,700 0,725
0,725 0,750
0,750 0,750
1,747 1,080
-0,253 -0,586
-1,488 -1,566
-1,566 -0,200
0,750 0,725
0,700 0,700
-0,200 -0,225
-0,225 -0,250
-0,250 -0,250
2,547 0,746
-0,719 -0,202
-1,951 -0,077
-0,172 -0,320
0,540 0,682
1,344 1,999
-0,526 -0,574
-0,072 -1,206
-0,030 -0,045
Σ M = 1,964
135 ¾ Perhitungan tulangan lentur
Mu = 1,964 = 1,964 x 10
7
Nmm Mn = Mu
φ = 1,964 x 10
7
0,8 = 2,455 x 10
7
Nmm Tebal counterfort, syarat
≥ 200 m; diambil b = 300 m Jarak antar counterfort = 0,3 – 0,6 H
diambil = 2,675 m f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 20 mm d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan Beton
Bertulang, Seri Beton –1, W. C. Vis, Gideon Kusuma
a = tinggi counterfort sebagai balok = 1398 mm d = a – d’ – Ø2 = 1398 – 50 – 202 = 1338 mm
k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 2,455 x 10
7
300 x 1338
2
x 19,125 = 2,390 x 10
-3
F = 1 - k
2 1
− = 1 -
10 x
2,390 2
1
-3
− = 2,393 x 10
-3
Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 2,393 x 10
-3
x 300 x 1338 x 19,125240 = 76,543 mm
2
ρ = Asb.d = 76,543300 x 1338 = 1,907 x 10
-4
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
min
= ρ
min
.b.d = 5,83 x 10
-3
x 300 x 1338 = 2340,162 mm
2
Tulangan terpasang = 8 Ø 20 As = 2513 mm
2
pemasangan 2 baris 4 tulangan Cek :
ρ = As
terpasang
b.d = 2513300x1338 = 6,261 x 10
-3
ρ
min
ρ ρ
maks
ok
136 ¾ Perhitungan tulangan horisontal
∑P = Pa1 + Pa2 + Pa3 + Pa4 + Pa5 + Pa6 + Paw = 1,458 + 0,691 + 2,840 + 0,344 + 1,311 + 0,049 + 0,11
= 6,803 t = 68030 N
fy = ∑PAs
As = ∑Pfy
= 68030240
= 283,458 mm
2
untuk 2 sisi = 141,729 mm
2
untuk 1 sisi Pemasangan tulangan untuk tiap 1 meter pias :
As = 141,729 H = 141,729 4,12 = 34,400 mm
2
Tulangan terpasang = Ø 12 - 250 As = 452 mm
2
¾ Perhitungan tulangan vertikal ∑G= Q + G1 + G2 + G3 + G4 + G5 + G6 + G7 + G8 + G9 +G10
= 1,6 + 0,72 + 0,941 + 1,92 + 2,856 + 2,632 + 2,55 + 0,318 + 4,822 +0,121 + 0,18
= 18,66 t = 186600 N
fy = ∑GAs
As = ∑Gfy
= 186600240
= 777,500 mm
2
untuk 2 sisi = 388,750 mm
2
untuk 1 sisi Pemasangan tulangan untuk tiap 1 meter pias :
As = 388,750 b5 = 384,583 1,5 = 259,167 mm
2
Tulangan terpasang = Ø 12 - 250 As = 452 mm
2
137
5.6.2 Perhitungan Konstruksi Dinding Gerbang B
