K =
R .
d .
b Mn
1 2
→ R
1
= β
1
x f’c = 0,85 x 22,5 = 19,125 N mm
2
=
19,125 x
720 x
1000 ,063
1289399450
2
= 1,301 x 10
-2
F = 1 - √1 - 2k = 1 - √1 - 2 x 1,301 x 10
-2
= 1,309 x 10
- 2
F F
maks
→ tulangan tunggal under reinforced As =
y 1
f R
. d
. b
. F
= 240
,125 19
x 720
x 1000
x 10
x 1,309
-2
= 751,094 mm
2
ρ =
b.d As
=
720 x
1000 751,094
= 1,04 x 10
-4
ρ
min
= 5,833 x 10
-3
As
min
= ρ
min.
b.d = 5,833 x 10
-3
x 1000 x 720 = 4200 mm
2
Tulangan terpasang : Ø 20 - 50 As = 6284 mm
2
4.6.4.4. Perhitungan Bagian Perkuatan Dinding Counterfort
H = 10,25 m b
3
= 0,85 m b
4
= 3,95 m b
5
= 3,95 m
1. Perhitungan Titik Berat Counterfort Yz =
⅓.H = ⅓ x 10,25 = 3,417 m Xz =
⅓.B
5
= ⅓ x 3,95 = 1,317 m
a O
Xz Yz
a
L H
b5 b3
b4
2. Perhitungan tinggi balok : Tan
α =
1 3
5
B B
B H
→ α = 67,959
o
Sin a =
1 3
5
B B
B X
→ a = 3,661 m
Cos α =
L B
B B
1 3
5
→ L = 11,059 cm 3. Pembebanan Counterfort
Tabel 4.22
Pembebanan Counterfort P
Gaya ton
Titik Berat Y
m X
m Lengan
m Momen
tm Yz
Xz Pa
1
Pa
2
Pa
3
Pa
4
Pa
5
Pa
6
Q G
1
G
2
G
3
G
4
G
5
G
6
G
7
G
8
G
9
1,622 10,285
24,658 9,898
21,782 2,22
4,15 4,176
5,46 7,392
33,408 20,029
10,917 13,058
19,864 5,669
3,417 3,417
3,417 3,417
3,417 3,417
3,417 3,417
3,417 3,417
3,417 3,417
3,417 3,417
3,417 3,417
1,317 1,317
1,317 1,317
1,317 1,317
1,317 1,317
1,317 1,317
1,317 1,317
1,317 1,317
1,317 1,317
8,1 7,35
3,35 2,517
-0,375 -0,783
1,875 -0,5
-0,475 -0,4
-0,4 1,875
1,9 1,9
1,975 1,975
4,683 3,933
-0,067 -0,9
-3,792 -4,2
-0,558 1,817
1,792 1,717
1,717 -0,558
-0,583 -0,583
-0,658 -0,658
7,595 40,456
-1,644 -8,908
-82,591 -9,324
-2,317 7,586
9,783 12,69
57,35 -11,183
-6,368 -7,617
-13,077 -3,732
ΣM = -11,301 tm 4. Perhitungan Tulangan Lentur
Mu = 11,301 tm = 113012173,98 Nmm Mn =
Mu
= 8
, 98
113012173, = 141265217,475 Nmm
Tulangan direncanakan Ø 25 mm
d = a - tebal selimut - ½ Ø tulangan lentur
= 3661 - 50 - 252 = 3598,817 cm K =
R .
d .
b Mn
1 2
→ R
1
= β
1
x f’c = 0,85 x 22,5 = 19,125 N mm
2
=
19,125 x
3598,817 x
300 475
141265217,
2
= 1,9 x 10
-3
F = 1 - √1 - 2k = 1 - √1 – 2 x 8,19 x 10
-3
= 1,9 x 10
-3
F F
maks
→ tulangan tunggal under reinforced As =
y 1
f R
. d
. b
. F
= 240
,125 19
x 3598,817
x 300
x 10
x 1,9
-3
= 163,711 mm
2
ρ =
b.d As
= 3598,817
x 300
711 ,
163 = 1,5 x 10
-4
ρ
min
= 5,833 x 10
-3
As
min
= ρ
min
.b.d = 5,833 x 10
-3
x 300 x 3598,817 = 6297,930 mm
2
Tulangan terpasang : 21 Ø 20 As = 6596 mm
2
Pemasangan 3 baris, 7 tulangan 5. Perhitungan Tulangan Horisontal
Gaya horisontal yang diperhitungkan : ΣP = Pa
1
+ Pa
2
+ Pa
3
+ Pa
4
+ Pa
5
+ Pa
6
= 1,622 + 10,285 + 24,658 + 9,898 + 21,782 + 2,22 = 70,465 ton = 704653,541 N
f
y
=
As P
As = 704653,541 240 = 2936,056 mm
2
untuk 2 sisi As = 1468,028 mm
2
untuk 1 sisi Pemasangan tulangan untuk tiap 1 meter pias :
As = 1468,028 10,25 = 143,222 mm
2
Tulangan terpasang = Ø 12 - 250 As = 452 mm
2
6. Perhitungan Tulangan Vertikal Gaya vertikal yang diperhitungkan :
ΣG = Q + G
1
+ G
2
+ G
3
+ G
4
+ G
5
+ G
6
+ G
7
+ G
8
+G
9
= 4,15 + 4,176 + 5,46 + 7,392 + 33,408 + 20,029 + 10,917 + 13,058 + 19,864 + 5,669
= 124,123 ton = 1241231,34 N
f
y
=
As G
As = 1241231,34 240 = 5171,797 mm
2
untuk 2 sisi As = 2585,899 mm
2
untuk 1 sisi Pemasangan tulangan untuk 1 meter pias :
As = 2585,899 3,95 = 654,658 mm
2
Tulangan terpasang = Ø 12 - 250 As = 452 mm
2
4.6.5. Perhitungan Konstruksi Dinding Beda Elevasi A - C 4.6.5.1. Perhitungan Pembebanan Terhadap Dinding
b3 q
b5 b4
H
H4 H3
H2 b2
b1
A B
H1 -5,25
-9,5 -10,55
?2 = 1,676 tm C2 = 0,14 tm²
Ø2 = 21°
?3 = 1,688 tm C3 = 0,15 tm²
Ø3 = 23°
Gambar 4.47
Diagram Tegangan Tanah Beda Elevasi A - C
1. Perhitungan tegangan tanah a. Rencana dimensi dinding
Tabel 4.23
Dimensi Dinding Beda Elevasi A - C H
4
=
4 1
H -
12 1
H; diambil 0,3 m b
1
= 0,25 m H = 5 + 0,3 = 5,3 m
b
2
= 0,3 m H
1
= 1,2 m b
3
=
12 1
H -
10 1
H; diambil 0,35 m H
2
= 1,6 m b
4
=
3 1
H ; diambil 1,8 m H
3
= 2,2 m b
5
= 1,8 m q
= 2,5856 tm
2
B = 0,4 - 0,7 H ; diambil 3,95 m
b. Perhitungan koefisien tekanan tanah aktif Ka
=
Sin 1
Sin 1
atau Ka = tan
2
45° - φ2
Ka
1
= 14
Sin 1
14 Sin
1
Ka
1
= 0,6104 Ka
2
= 21
Sin 1
21 Sin
1
Ka
2
= 0.4724 Ka
3
= 23
Sin 1
23 Sin
1
Ka
3
= 0.438 Mencari nilai q :
Akibat beban pelat beton D = 0,12 x 2400 = 288 kgm
2
Akibat beban hidup L = diambil 1400 kgm
2
Kombinasi pembebanan, q = 1,2.D + 1,6.L
= 1 ,2 x 288 + 1,6 x 1400 = 2585,6 kgm
2
c. Perhitungan tegangan tanah aktif
s a4 s a3
s a2 s a1
Pa4 Pa3
Pa2 Pa1
-10,55 -9,5
- 5,25
?2 = 1,676 tm C2 = 0,14 tm²
Ø2 = 21°
?3 = 1,688 tm C3 = 0,15 tm²
Ø3 = 23°
Gambar 4.48
Tegangan Tanah Aktif σ
a1
= Ka
2
.q - 2.C
2
. √Ka
2
= 0,472 x 2,586 - 2 x 0,14 x √0,472
= 1,029 tm
2
σ
a2
= Ka
2
. γ
2
.h
1
= 0,472 x 1,676 x 4,25 = 3,365 tm
2
σ
a3
= σ
a1
+ σ
a2
= 1,029 + 3,365 = 4,394 tm
2
σ
a4
= Ka
3
.γ
3
.h
2
= 0,438 x 1,688 x 1,05 = 0,777 tm
2
d. Perhitungan tekanan tanah aktif per 1 m lebar Pa1 =
σ
a1
.h
1
= 1,029 x 4,25 = 4,373 t Pa2 = ½.
σ
a2
.h
1
= ½ x 3,365 x 4,25 = 7,151 t Pa3 =
σ
a3
.h
2
= 4,394 x 1,05 = 4,614 t Pa4 = ½.
σ
a4
.h
2
= ½ x 0,777 x 1,05 = 0,408 t e. Perhitungan gaya-gaya vertikal per 1 m lebar
G4 G3
G8 G7
G6 G5
G2 G1
Q
1,2
0,35 1,8
1,8 0,3
2,2 1,6
0,3 0,25
A
Gambar 4.49
Gaya-Gaya Vertikal pada Dinding Beda Elevasi A - C
1. Akibat beban merata : Q = q.B - b
4
- b
1
= 1 6,8 - 3 - 0,6 = 4,913 t 2. Akibat berat sendiri struktur :
G
1
= H
1
.b
1
. γ
c
= 1,2 x 0,25 x 2,4 = 0,72 t G
2
= H
2
.b
2
. γ
c
= 1,6 x 0,3 x 2,4 = 1,152 t G
3
= H
3
.b
3
. γ
c
= 2,2 x 0,35 x 2,4 = 1,848 t G
4
= H
4
.B. γ
c
= 0,3 x 3,95 x 2,4 = 2,844 t
3. Akibat berat tanah di atas struktur : G
5
= b
3
+ b
5
- b
1
.H
1
.γ
2
= 0,35 + 1,8 - 0,25 x 1,2 x 1,676 = 3,822 t
G
6
= b
5
+ b
3
- b
2
.H
2
.γ
2
= 1,8 + 0,35 - 0,3 x 2,4 x 1,676 = 4,962 t
G
7
= b
5
.H
4
+ H
3
- h
2
.γ
2
= 1,8x 0,3 + 2,2 - 1,05 x 1,676 = 4,375 t
G
8
= b
5
.h
2
- H
4
.γ
3
= 1,8 x 1,05 - 0,3 x 1,688 = 2,279 t
f. Perhitungan momen terhadap titik A 1. Momen aktif
Tabel 4.24
Momen Aktif Horizontal
P Gaya ton
Lengan m M
aktif
tm
Pa
1
Pa
2
Pa
3
Pa
4
4,373 7,151
4,614 0,408
3,175 2,467
0,525 0,35
13,883 17,639
2,422 0,143
ΣP = 16,545 ΣM
aktif
= 34,088
2. Momen pasif
Tabel 4.25
Momen Pasif Vertikal
G Gaya ton
Lengan m M
pasif
tm
Q G
1
G
2
G
3
G
4
G
5
G
6
G
7
G
8
4,913 0,72
1,152 1,848
2,844 3,822
4,962 4,375
2,279 3
1,925 1,95
1,814 1,975
3 3,025
3,05 3,05
14,738 1,386
2,246 3,352
5,617 11,466
15,01 13,344
6,952 ΣG = 26,915
ΣM
Pasif
= 74,111
2. Cek Stabilitas Struktur : a. Kontrol terhadap guling
SF =
aktif pasif
M M
≥ 2
SF =
088 ,
34 111
, 74
≥ 2 SF
= 2,174 ≥ 2 Aman
b. Kontrol terhadap geser SF
= P
P C
B tan
G
pasif 2
≥ 1,5
SF =
545 ,
16 14
, x
95 ,
3 23
tan 915
, 26
≥ 1,5 SF
= 0,724 1,5 Perlu Tiang Pancang c. Kontrol terhadap eksentrisitas
e = ½.B -
G M
M
aktif pasif
≤
6 1
B e
= ½.3,95 - 915
, 26
088 ,
34 111
, 74
≤
6 1
x 3,95 e
= 1,975 – 1,487 ≤ 0,658
e = 0,488
≤ 0,648 Aman d. Daya dukung tanah
Nc =
40 3
, 4
228
=
23 40
23 x
3 ,
4 228
= 19,229
Nq =
40 5
40
=
23 40
23 x
5 40
= 9,118
N γ =
40
6
=
23 40
23 x
6
= 8,118 q
ult
= C
3
.Nc + γ
3
.H
4
.Nq + ½.B. γ
3
N γ
= 0,15 x 19,229 + 1,688 x 0,3 x 9,118 + ½ x 3,95 x
1,688 x 8,118 = 34,572 tm
2
Daya dukung tanah yang diijinkan ditentukan dengan membagi q
ult
dengan suatu faktor keamanan SF yaitu :
q
all
=
SF q
ult
→ diambil SF = 3
q
all
=
3 572
, 34
q
all
= 11,524 tm
2
e. Tegangan tanah yang terjadi Kondisi yang harus diperhitungkan adalah pada saat kamar penuh,
dengan berat air yang mempengaruhi dinding pada bagian toe sepanjang 1,8 m.
W = H
air
.B. γ
w
= 5 - 1 x 3,95 x 1 = 15,8 tm Tegangan yang terjadi :
σ
maks,min
=
B e
6 1
x 1
x B
W G
=
95 ,
3 488
, x
6 1
x 1
x 95
, 3
8 ,
15 915
, 26
σ
maks
= 18,829 tm
2
q
all
= 11,524 tm
2
Perlu tiang pancang σ
min
= 2,799 tm
2
4.6.5.2. Perhitungan Bagian Tapak Dinding
