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PERANCANGAN IRIGASI BANGUNAN AIR
BAB III ANALISA STABILITAS BENDUNG Gaya-gaya yang bekerja pada tubuh bendung, akibat: 1. Tekanan air 2. Tekanan lumpur 3. Tekanan berat sendiri bendung 4. Gaya gempa 5. Gaya angkat (uplift pressure)
3.1.
Tekanan Air 3.1.1 Tekanan Air Normal
Gambar 3.1 Diagram tekanan akibat air normal γ air = 1 ton/m3
Pa1 =
1 .γair.h 2 . 2
=
1 2 .1 . 3,85 . 2
= 7,411 ton
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PERANCANGAN IRIGASI BANGUNAN AIR Tabel 3.1 Perhitungan Tekanan Akibat Air Normal Gaya (t)
Bagian Pa1
Momen (tm)
V
H
X
Y
Mr
M0
-
7,411
-
5,905
-
43,763
JUMLAH
3.1.2
Lengan (m)
7,411
43,763
Tekanan Air Banjir
Gambar 3.2 Diagram tekanan akibat air banjir
Pf1 =
1 . γair . h 2 2
Pf2 = b . h . γ air
1 2 .1 . 3,85 2
= 7,411 ton
= 1,851.(3,85).(1)
= 7,126 ton
=
1 1 2 . 2,634 Pf3 = . γair . h 2 = .1 2 2
Pf4 =
1 . γair . h 2 2
=
1 2 .1 . 2,634 2
= - 3,469 ton = 3,469 ton
46
PERANCANGAN IRIGASI BANGUNAN AIR Tabel 3.2 Perhitungan Tekanan Akibat Air Banjir Bagian
Berat (ton) V
H
Lengan (m) x
y
Momen ™ Mr
Mo
Pf1
7,411
5,663
41,972
Pf2
7,411
6,305
46,728
Pf3
-3,469
4,756
-16,498
Pf4
3,469
JUMLAH
3,469
0,892 11,354
3,094 3,904
72,202
3.2. Tekanan Lumpur
lumpur
= 0,6 ton/m3
θ
= 300
Ka
= tan2 (450 – θ/2) = tan2 (450 – 30o/2) = 0,333
Keterangan : γlumpur = berat volume lumpur (t/m3) θ
= sudut gesek dalam
Ka
= tekanan lumpur aktif (0,333)
Gambar 3.3 Diagram tekanan akibat lumpur
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PERANCANGAN IRIGASI BANGUNAN AIR
PL1 = =
1 . Ka . lumpur . h2 2 1 .(0,333).(0,6).(3,85)2 2
= 1,481 ton
Tabel 3.3 Perhitugan Tekanan Akibat Lumpur Gaya (t) Bagian PL1
Lengan (m)
Momen (tm)
V
H
x
Y
Mr
M0
-
0,888
-
5,663
-
5,031
0,888
JUMLAH
5,301
3.3. Tekanan Berat Sendiri Bendung Berat volume pasangan batu γpas batu = 2,2 t/m2 Ditinjau 1 m lebar bendung
Gambar 3.4 Diagram tekanan berat sendiri bendung
W1
= b . h . pasangan
= 2,0 . 2,0 . 2,2
W2
= b . h . pasangan
= 1,649 . 3,848 . 2,2 = 13,960 ton
W3
= b . h . pasangan
= 1,0 . 4,0 . 2,2
= 8,800 ton
W4
= b . h . pasangan
= 1,5 . 3,5 . 2,2
= 11,550 ton
= 8,800 ton
48
PERANCANGAN IRIGASI BANGUNAN AIR W5
= b . h . pasangan
W6
= 2/3 . b. h . pasangan = 2/3 . 1,5 . 1,5 . 2,2 = 3,300 ton
W7
= 1/2 . b. h . pasangan = 1/2 . 3,0 . 2,5 . 2,2 = 8,250 ton
= 2,5 . 3,0 . 2,2
= 16,500 ton
Tabel 3.4 Perhitungan Tekanan Berat Sendiri Bendung Bagian
Lengan (m)
Gaya (ton)
Momen (tm)
Vertikal
x
y
Mr
M0
W1
8,800
6,000
3,000
52,800
26,400
W2
13,960
5,825
5,924
81,311
82,701
W3
8,800
4,500
5,000
39,600
44,000
W4
11,550
3,250
3,750
37,538
43,313
W5
16,500
1,250
1,500
20,625
24,750
W6
3,300
3,500
6,000
11,550
19,800
W7
8,250
1,667
3,833
13,750
31,625
∑
71,160
257,174
272,588
Pada badan bendung yang berbentuk parabola, luas penampang digunakan pendekatan : A = 2/3 . L . H
Didapat: ΣW
= 71,160 ton
ΣMo
= 272,588 tm
ΣMr
= 257,174 tm
3.4. Gaya Gempa 3.4.1 Gempa Horisontal Gaya Gempa Horisontal (H)
= Kh. ΣW = 0,10. 71,160 = 7,116 ton
Momen akibat H = Kh. ΣMo = 0,10. 272,588 = 27,259 tm
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PERANCANGAN IRIGASI BANGUNAN AIR Keterangan: Kh = Koefisien gempa horisontal (diambil: Kh = 0,10) ΣW = Total berat sendiri bendung (t) Mo = Momen guling akibat berat sendiri bendung (tm)
3.4.2 Gempa Vertikal Gaya Gempa Vertikal (V) = Kv. ΣW = 0,05. 71,160 = 3,558 ton Momen akibat V = Kv. ΣMr = 0,05. 257,174 = 12,859 tm Keterangan : Kv = Koefisien gempa vertikal (diambil: Kv = 0,005) Mr = Momen tahanan akibat berat sendiri
3.5. Gaya Angkat (Uplift Pressure) 3.5.1 Tekanan Air Normal ΣL
= Lh + Lv = 19,15 + 10,62 = 29,77 m
ΔH (air normal) = elev. MAN – elev. Dasar sungai = 168,85 – 165,00 = 3,85 m 𝐿𝑋 . Δ𝐻 Σ𝐿 𝐿𝑋 𝑈𝑋 = 𝐻𝑋 − . 3,2 29,77 𝑈𝑋 = 𝐻𝑋 −
𝑈𝑋 = 𝐻𝑋 − 0,107 𝐿𝑋 Keterangan : Hx = tinggi muka air dari titik yang dicari (m) Lx
= panjang rayapan (m)
ΣL = total rayapan (m)
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PERANCANGAN IRIGASI BANGUNAN AIR ΔH = tinggi muka air normal dari lantai dasar bendung (m) Ux = uplift pressure di titik x (t/m2)
Gambar 3.5 Rayapan gaya angkat akibat muka air normal
Tabel 3.5 Perhitungan Tinggi Air Normal Terhadap Muka Bendung Titik
Hx (m)
Lx (m)
Ux (t/m2)
A
5.23
29..77
1.380
B
8.23
26.77
4.768
C
8.23
24.27
5.091
D
6.23
22.27
3.350
E
6.23
20.77
3.544
F
5.23
19.77
2.673
G
5.23
18.77
2.803
H
6.23
17.77
3.932
I
6.23
15.77
4.191
J
4.23
13.77
2.449
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PERANCANGAN IRIGASI BANGUNAN AIR U1
𝑐
𝑅 = (𝑈1 + 𝑈2) 2
𝐿=
𝑐(2𝑥𝑈1 + 𝑈2) 3(𝑈1 + 𝑈2)
c
R L U2
Keterangan: R
= Resultante Uplift Pressure (ton)
L
= Lengan momen (m)
U1
= Uplift Pressure terendah di setiap bagian (ton)
U2
= Uplift Pressure tertinggi di setiap bagian (ton)
Tabel 3.6 Perhitungan Uplift Pressure Akibat Air Normal Bagian
Gambar
Gaya angkat per 1 m panjang (t) H=
1.3800
U1 U 2 xH 2
=-
A
1,380 4,768 x3 2
= - 9,222 t A-B
h 2a b 3 a b
y =
3.0000
B
4.7680
=
3 (2 x1,380 ) 4,768 3 1,380 4,768
= 1,224 m Ytotal = 1,224 m
52
PERANCANGAN IRIGASI BANGUNAN AIR
V= C
2.5000
U1 U 2 xH 2
B
V =-
4,768 5.091 x 2,5 2
= -12,324 t B-C
h 2b c 3 bc
x =
4.7680
5.0910
=
2,5 (2 x 4,768) 5,091 3 4,768 5,091
= 1,236 m X total = 3 – 1,236 = 1,764 m H=
U1 U 2 xH 2
H=
5,091 3,350 x 2,0 2
3.3500
D
= 8,441 t C–D
2.0000
h 2c d 3 cd
y=
C 5.0910
=
2,0 (2 x5,091) 3,350 3 5,091 3,350
= 1,069 m Ytotal = 1,069 m V= E
1.5000
D
U1 U 2 xH 2
V =D–E
3.5440
3.3500
3,350 3,544 x1,5 = - 5,170 t 2
h 2d e 3 d e
x =
=
1,5 (2 x3,350) 3,544 = 0,743 m 3 3,350 3,544
X total = (2,5 – 0,743) + 1,5 = 3,257 m
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PERANCANGAN IRIGASI BANGUNAN AIR
2.6730
F
E–F
1.0000
H =
U1 U 2 xH 2
H =
3,544 2,673 x1 2
= 3,109 t
h 2e f 3 e f
y =
E
3.5440
=
1 (2 x3,544 ) 2,673 3 3,544 2,673
= 0,523 m Ytotal = 0,523 + 1,0 = 1,523 m V =
G
1.0000
F
U1 U 2 xH 2
V =-
2,673 2,803 x1 2
= - 2,738 t F–G
2.6730
2.8030
h2f g 3 f g
x =
=
1 (2 x 2,673) 2,803 = 0,496 m 3 2,673 2,803
X total = (1 - 0,496) + 2,5 + 1,5 = 4,504 m
U1 U 2 xH 2 2,803 3,932 x1,0 H =2 = - 3,367 t
H=
G
G-H
2.8030
h 2g h 3 g h 1,0 (2 x 2,803) 3,932 = 3 2,803 3,932
y =
1.0000
H
3.9320
= 0,472 m Ytotal = 0,472 + 1,0 = 1,472 m
54
PERANCANGAN IRIGASI BANGUNAN AIR U1 U 2 xH 2 3,932 4,191 V =x2 2 = - 8,122 t
V=
2.0000
H
I
H-I
h 2g h 3 g h 2 (2 x3,932 ) 4,191 = 3 3,932 4,191
x =
3.9320
4.1910
= 0,989 m Xtotal = (2 – 0,989) + 2,5 + 1,5 + 1 = 6,011 m U1 U 2 H = xH 2 H = 2.4490
J
4,191 2,449 x2 2
= 6,640 t I-J
2.0000
h 2g h 3 g h
y =
I
4.1910
=
2 (2 x 4,191) 2,449 3 4,191 2,449
= 1,087 m Ytotal = 1,087 + 1 = 2,087 m
Tabel 3.7 Gaya Angkat Air Normal Titik
Hx (m)
Lx (m)
Ux (t/m2)
A
5.23
29.77
1.380
Uplift Force (t) V
H
Lengan (m) x
-9.222 B
8.23
26.77
8.23
24.27
1.224 1.236
6.23
22.27
6.23
20.77
Mo
11.292
1.764
21.740
1.069
1.069
9.022
3.350 -5.170
E
Mr
5.091 8.441
D
1.224
Momen
4.768 -12.324
C
y
Lengan (m) x y (total) (total)
0.743
3.257
16.840
3.544
55
PERANCANGAN IRIGASI BANGUNAN AIR 3.109 F
5.23
19.77
5.23
18.77
0.496
6.23
17.77
6.23
15.77
0.472 0.989
4.23 13.77 Σ (JUMLAH)
12.332 1.472
4.957
6.011
48.824
4.191 6.640
J
4.504
3.932 -8.122
I
4.734
2.803 -3.367
H
1.523
2.673 -2.738
G
0.523
1.087
2.087
13.857
2.449 - 28.355
5.600
30.106
113.491
Catatan : Searah jarum jam (+) Berlawanan arah jarum jam (-)
Gaya Angkat Akibat Air Normal : 1. Tekanan Vertikal V = fu x ΣV = 0,5 x (- 28,355) = - 14,178 ton 2. Tekanan Horizontal H = fu x ΣH = 0,5 x 5,600
= 2,800 ton
3. Momen Mr = 0.5 x ΣMr = 0,5 x (30,106) = 15,053 t.m Mo = 0.5 x ΣMo = 0,5 x (113,491) = 56,746 t.m Dimana :
fu = koefisien reduksi untuk jenis tanah keras (50 %)
3.5.2 Tekanan Air Banjir ΣL
= 29,77 m
ΔH (air banjir) = elev. M.A.B – elev. Dasar sungai = 170,701 – 165,000 = 5,701 m 𝑈𝑋 = 𝐻𝑋 −
𝐿𝑋 𝐿𝑋 . Δ𝐻 = 𝐻𝑋 − . 5,701 = 𝐻𝑋 − 0,192 𝐿𝑋 Σ𝐿 29,77
Keterangan: Hx = tinggi muka air dari titik yang dicari (m) Lx = panjang rayapan (m) ΣL = total rayapan (m)
56
PERANCANGAN IRIGASI BANGUNAN AIR ΔH = tinggi muka air banjir dari lantai dasar bendung (m) Ux = uplift pressure di titik x (t/m2)
Gambar 3.6 Rayapan gaya angkat akibat muka air banjir
Tabel 3.8 Perhitungan Tinggi Air Banjir Terhadap Muka Bendung Titik
Hx (m)
Lx (m)
Ux (tm²)
A B C D E F G H I J
7.08 10.08 10.08 8.08 8.08 7.08 7.08 8.08 8.08 6.08
29.77 26.77 24.27 22.27 20.27 19.77 18.77 17.77 15.77 13.77
1.380 4.995 5.433 3.816 4.104 3.295 3.487 4.678 5.061 3.444
57
PERANCANGAN IRIGASI BANGUNAN AIR Tabel 3.9 Perhitungan Uplift Pressure Akibat Air Banjir Bagian
Gambar
Gaya angkat per 1 m panjang (t) U1 U 2 xH 2
H= A
1.3800
=-
1,380 4,955 x3,0 2
= - 9,502 t
3.0000
h 2a b 3 a b
A-B
y = B 4.9550
3,0 (2 x1,380 ) 4,955 3 1,380 4,955
=
= 1,218 m Ytotal = 1,218 m U1 U 2 xH 2
V= 2.5000
B
C
V =-
4,955 5,433 x 2,5 2
= - 12,985 t B-C
h 2b c 3 bc
4.9550
x =
5.4330
2,5 (2 x 4,955) 5,433 3 4,955 5,433
=
= 1,231 m X total = 2,5 – 1,231 = 1,269 m H=
U1 U 2 xH 2
H =
5,433 3,816 x 2,0 = 9,250 t 2
3.8160
D
2.0000
C-D
h 2c d 3 cd
y= C 5.4330
=
2,0 (2 x5,433) 3,816 = 1,058 m 3 5,433 3,816
Ytotal = 1,058 m
58
PERANCANGAN IRIGASI BANGUNAN AIR
E
D-E
1.5000
U1 U 2 xH 2 3,816 4,104 V =x1,5 2 = - 5,940 t
V= D
h 2d e 3 d e 1,5 (2 x3,816 ) 4,104 = 3 3,816 4,104
x =
3.8160
4.1040
= 0,741 m X total = (1,5 - 0,741) + 2,5 = 3,259 m
3.2950
F 1.0000
E-F
H=
U1 U 2 xH 2
H =
4,104 3,295 x1 2
= 3,699 t
h 2e f 3 e f
y =
E
4.1040
=
1 (2 x 4,104 ) 3,295 3 4,104 3,295
= 0,518 m Ytotal = 0,518 + 1,0 = 1,518 m V=
G
1.0000
F
U1 U 2 xH 2
V =-
3,295 3,487 x1,0 2
= - 3,391 t
h2f g 3 f g
x =
F-G 3.4870
3.2950
=
1,0 (2 x3,295) 3,487 3 3,295 3,487
= 0,495 m X total = (1,0-0,495)+2,5+1,5 = 4,505 m
59
PERANCANGAN IRIGASI BANGUNAN AIR
H=
U1 U 2 xH 2
H =3.4870
G
G-H
3,487 4,678 x1,0 2
= - 4,082 t
h 2g h 3 g h
y =
1.0000
H
4.6780
=
1,0 (2 x3,487 ) 4,678 3 3,487 4,678
= 0,476 m Ytotal = 0,476 + 1,0 = 1,476 m V= I
2.0000
H
U1 U 2 xH 2
V =-
4,678 5,061 x 2,0 2
= - 9,739 t H-I
h 2g h 3 g h
x =
4.6780
5.0610
=
2,0 (2 x 4,678) 5,061 3 4,678 5,061
= 0,987 m Xtotal = (2 – 0,987) + 2,5 + 1,5 + 1,0 = 6,013 m
3.4440
J
2.0000
I-J 5.0610
I
H=
U1 U 2 xH 2
H=
5,061 3,444 x 2,0 2
= 8,505 t
h 2g h 3 g h
y =
=
2,0 (2 x5,061) 3,444 = 1,063 m 3 5,061 3,444
Ytotal = 1,063 + 2 = 3,063 m
60
PERANCANGAN IRIGASI BANGUNAN AIR Tabel 3.10 Gaya Angkat Akibat Air Banjir Titik
Hx (m)
Lx (m)
Ux (t/m2)
A
7.08
29.77
1.380
Uplift Force (t) V
H
Lengan (m) x
-9.502 B
10.08
26.77
10.08
24.27
1.214 1.231
8.08
22.27
8.08
20.77
7.08
19.77
0.741
7.08
18.77
8.08
17.77
8.08
15.77
6.08 13.77 Σ (JUMLAH)
3.259
0.495
19.358 1.518
5.615
4.505
15.275
3.487 0.476
1.476
6.025
4.678 0.987
6.031
58.736
5.061 8.505
J
9.789
3.295
-9.739 I
1.058
0.518
-4.082 H
16.478
4.104
-3.391 G
11.572
1.269 1.058
3.699 F
Mo
3.816 -5.940
E
Mr
5.433 9.250
D
1.218
Momen
4.955 -12.985
C
y
Lengan (m) x y (total) (total)
1.063
3.063
26.051
3.444 - 32.054
7.870
43.648
125.251
Catatan : Searah jarum jam (+) Berlawanan arah jarum jam (-)
Gaya Angkat Akibat Air Banjir : 1. Tekanan Vertikal V = fu x ΣV = 0,5 x (-32.054) = - 16.027 ton 2. Tekanan Horizontal H = fu x ΣH = 0,5 x (7.870) = 3.935 ton 3. Momen Mr = 0.5 x ΣMr = 0,5 x (43.648) = 21.824 t.m Mo = 0.5 x ΣMo = 0,5 x (125.251) = 62.626 t.m Dimana :
fu = koefisien reduksi untuk jenis tanah keras (50 %)
61
PERANCANGAN IRIGASI BANGUNAN AIR Tabel 3.11 Akumulasi Beban – Beban Pada Bendung Gaya (ton) Momen (ton.meter) Keterangan Vertikal Horizontal Mr Mo 2 3 4 5 6
No. 1
Tekanan Air a Air Normal b Air Banjir c Tekanan Lumpur Berat Sendiri d Bendung Gaya Gempa e Gempa Horisontal f Gempa Vertikal Gaya Angkat g Air Normal h Air Banjir 3.6.
0.000 3.469 0.000
7.411 11.354 0.888
0.000 3.094 0.000
43.763 72.202 5.031
71.160
-
257.174
-
3.558
7.116 -
27.259 12.859
27.259 12.859
-14.177 -16.027
2.800 3.935
15.053 21.824
56.746 62.626
Kontrol Stabilitas Bendung Ketentuan : 1. Tegangan tanah dasar yang diijinkan (σ’)
= 2,0 kg/cm2 = 20 t/m2
2. Over Turning safety factor (guling)
= 1,5 kg/cm2
3. Sliding safety factor (geser)
= 1,2 kg/cm2
Kombinasi gaya – gaya yang bekerja pada bendung : 3.6.1 Tanpa Pengaruh Gempa 1. Keadaan Normal dengan Uplift Pressure
ΣH = a(4) + c(4) + g(4) = 7,411 + 0,888 + 2,800 = 11,100 t
ΣV = a(3) + c(3) + d(3) + g(3) = 0,000 + 0,000 + 71,160 – 14,177 = 56,983 t
ΣMr = a(5) + c(5) + d(5) + g(5) = 0,000 + 0,000 + 257,174 + 15,053 = 272,226 tm
ΣM0 = a(6) + c(6) + g(6) = 43,763 + 5,031 + 56,746 = 105,540 tm
Kontrol: a. Terhadap Guling (Over Turning) 𝑆𝐹 =
Σ𝑀𝑟 272,226 = = 2,579 ≥ 1,5 (𝑂𝐾) Σ𝑀𝑜 105,540
62
PERANCANGAN IRIGASI BANGUNAN AIR b. Terhadap Geser (Sliding) 𝑆𝐹 =
f . Σ𝑉 0,7 . 56,983 = = 3,594 ≥ 1,2 (𝑂𝐾) Σ𝐻 11,100
Dimana, f = koefisien geser (diambil f= 0,7) c. Terhadap Daya Dukung Tanah (Over Stressing) Resultante beban vertikal bekerja sejarak a dari titik O 𝑎=
Σ𝑀𝑟 − Σ𝑀𝑜 272,226 − 105,540 = = 2,925 𝑚 Σ𝑉 56,983
Resultante beban vertikal bekerja sejarak e dari pusat berat bendung 𝑒=
𝐵 7 − 𝑎 = − 2,925 = 0,575 𝑚 2 2
Jarak e masih terletak didalam “Bidang Kern” 𝑒
0
56,983 6 . 0,575 1 = 12,151 t/m2 < σ’= 20 t/m2 7,0. 1 7,0
(OK!)
(OK!)
2. Keadaan Banjir dengan Uplift Pressure
ΣH = b(4) + c(4) + h(4) = 11,354 + 0,888 + 3,935 = 16,177 t
ΣV = b(3) + c(3) + d(3) + h(3) = 0,000 + 0,000 + 71,160 – 16,027 = 58,602 t
ΣMr = b(5) + c(5) + d(5) + h(5) = 0,000 + 0,000 + 257,174 + 21,824 = 282,091 tm
ΣM0 = b(6) + c(6) + h(6) = 72,202 + 5,031 + 62,626 = 139,858 tm
Kontrol: a. Terhadap Guling (Over Turning) 𝑆𝐹 =
Σ𝑀𝑟 282,091 = = 2,017 ≥ 1,5 (𝑂𝐾) Σ𝑀𝑜 139,858
b. Terhadap Geser (Sliding) 𝑆𝐹 = f
f . Σ𝑉 0,7 . 58,602 = = 2,536 ≥ 1,2 (𝑂𝐾) Σ𝐻 16,177 = koefisien geser (diambil f= 0,7)
c. Terhadap Daya Dukung Tanah (Over Stressing) Resultante beban vertikal bekerja sejarak a dari titik O
𝑎=
Σ𝑀𝑟−Σ𝑀𝑜 Σ𝑉
=
282,091−139,858 58,602
= 2,427 m
64
PERANCANGAN IRIGASI BANGUNAN AIR Resultante beban vertikal bekerja sejarak e dari pusat berat bendung 𝐵
7
2
2
𝑒 = − 𝑎 = − 2,427 = 1,073 𝑒
0
Guling ≥1,50
Geser ≥1,20
2,579 2,017
3,594 2,536
12,151 16,071
4,130 0,673
-
-
2,050 3,381 1,688 2,752
2,190 3,231 1,761 2,699
-
-
15,489 18,485 19,408 20,233
0,792 1,846 -2,665 1,090
2,408 5,534 1,847 3,536
3,818 6,302 2,690 4,471
-
-
14,184 15,606 19,678
3,113 5,742 -1,918 20,691
1,648
72
