BAB III Stabilitas Bendung

(1)

BAB III

ANALISA STABILITAS BENDUNG

Gaya-gaya yang bekerja pada tubuh bendung, akibat: Gaya-gaya yang bekerja pada tubuh bendung, akibat:

1.

1. Tekanan airTekanan air 2.

2. Tekanan lumpurTekanan lumpur 3.

3. Tekanan berat sendiri bendungTekanan berat sendiri bendung 4.

4. Gaya gempaGaya gempa 5.

5. Gaya angkat (uplift pressure)Gaya angkat (uplift pressure)

3.1.

3.1. Tekanan AirTekanan Air 3.1.1

3.1.1 Tekanan Air NormalTekanan Air Normal

Gambar 3.1 Diagram tekanan akibat air normal Gambar 3.1 Diagram tekanan akibat air normal

air air γ γ = 1 ton/m= 1 ton/m33 Pa Pa11 == ..γγ .h.h .. 2 2 1 1 ₂₂ air air == ..



 

11..

 

3,853,85 .. 2 2 1 1 22 = = 7,411 7,411 tonton

(2)

Tabel 3.1

Tabel 3.1 Perhitungan Tekanan Perhitungan Tekanan Akibat Air NormalAkibat Air Normal

Bagian

Bagian Gaya Gaya (t) (t) Lengan Lengan (m) (m) Momen Momen (tm)(tm)

V H X Y M V H X Y Mrr MM00 Pa Pa11 - - 7,411 7,411 - - 5,905 5,905 - - 43,76343,763 JUMLAH JUMLAH 7,411 7,411 43,76343,763 3.1.2

3.1.2 Tekanan Air BanjirTekanan Air Banjir

Gambar 3.2 Diagram tekanan akibat air banjir Gambar 3.2 Diagram tekanan akibat air banjir

Pf Pf 11 == 2 2 air air ..hh γ γ .. 2 2 1 1 = = ..



 

11..

 

3,853,85 22 2 2 1 1 = = 7,411 7,411 tonton Pf

Pf 22 = b = b . h .. h .γγair air = = 1,851.(3,85).(1) 1,851.(3,85).(1) = = 7,126 7,126 tonton

Pf Pf 33 == 2 2 air air..hh γ γ .. 2 2 1 1



==



 





2 2 2,634 2,634 .. 1 1 .. 2 2 1 1



= = - - 3,469 3,469 tonton 1 1



 





22 1 1

(3)

Tabel 3.2 Perhitungan Tekanan Akibat Air Banjir

Bagian Berat (ton) Lengan (m) Momen ™

V H x y Mr Mo Pf 1 7,411 5,663 41,972 Pf 2 7,411 6,305 46,728 Pf 3 -3,469 4,756 -16,498 Pf 4 3,469 0,892 3,094 JUMLAH 3,469 11,354 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)

(4)

PL1 = 2 1 . K a . lumpur  . h2 = 2 1 .(0,333).(0,6).(3,85)2 = 1,481 ton

Tabel 3.3 Perhitugan Tekanan Akibat Lumpur

Bagian Gaya (t) Lengan (m) Momen (tm)

V H x Y Mr M0

PL1 - 0,888 - 5,663 - 5,031

JUMLAH 0,888 5,301

3.3. Tekanan Berat Sendiri Bendung

Berat volume pasangan batu γ pas batu = 2,2 t/m2

Ditinjau 1 m lebar bendung

Gambar 3.4 Diagram tekanan berat sendiri bendung

W1 = b . h .  _{pasang an} = 2,0 . 2,0 . 2,2 = 8,800 ton

(5)

W5 = b . h .  _{pasang an} = 2,5 . 3,0 . 2,2 = 16,500 ton

W6 = 2/3 . b. h .  _{pasang an} = 2/3 . 1,5 . 1,5 . 2,2 = 3,300 ton

W7 = 1/2 . b. h .  _{pasang an} = 1/2 . 3,0 . 2,5 . 2,2 = 8,250 ton

Tabel 3.4 Perhitungan Tekanan Berat Sendiri Bendung

Bagian Gaya (ton) Vertikal Lengan (m) Momen (tm) 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

(6)

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





= 



 

_Σ.Δ







= 



 

_29,77.3,2







= 



 0,107 



Keterangan :

(7)

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

(8)

1.3800 3.0000 4.7680 A B c U1 U2 R L

 =



_

1+2

 = 21+2

_{31+2}

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)

A - B H = U U xH 2 2 1 = - 3 2 768 , 4 380 , 1 x  = - 9,222 t y = b a b a h















2

3

=















768 , 4 380 , 1 768 , 4 ) 380 , 1 2 ( 3 3 x = 1,224 m Ytotal= 1,224 m

(9)

B C 2.5000 5.0910 4.7680 C D 5.0910 2.0000 3.3500 D E 1.5000 3.3500 3.5440 B - C V = U U xH 2 2 1 V = - 2,5 2 091 . 5 768 , 4 x  = -12,324 t x = c b c b h















2 3 =















091 , 5 768 , 4 091 , 5 ) 768 , 4 2 ( 3 5 , 2 x = 1,236 m X total= 3 – 1,236 = 1,764 m C – D H = U U xH 2 2 1 H = 2,0 2 350 , 3 091 , 5 x  = 8,441 t y = d c d c h















2 3 =















350 , 3 091 , 5 350 , 3 ) 091 , 5 2 ( 3 0 , 2 x = 1,069 m Ytotal= 1,069 m D – E V = U U xH 2 2 1 V = - 1,5 2 544 , 3 350 , 3 x  = - 5,170 t x = e d e d h















2 3 =















544 ,

3

350 ,

3

544 ,

3 )

350 ,

3

2 (

3

5 ,

1

x = 0,743 m X total= (2,5 – 0,743) + 1,5 = 3,257 m

(10)

F G 2.6730 2.8030 1.0000 G 1.0000 2.8030 3.9320 H E F 3.5440 1.0000 2.6730 E – F H = U U xH 2 2 1 H = 1 2 673 , 2 544 , 3 x  = 3,109 t y =

f

e

f

e

h















2 3 =















673 , 2 544 , 3 673 , 2 ) 544 , 3 2 ( 3 1 x = 0,523 m Ytotal= 0,523 + 1,0 = 1,523 m F – G V = U U xH 2 2 1 V = - 1 2 803 , 2 673 , 2 x  = - 2,738 t x =

g

f

g

f

h















2 3 =















803 , 2 673 , 2 803 , 2 ) 673 , 2 2 ( 3 1 x =0,496 m X total= (1 - 0,496) + 2,5 + 1,5 = 4,504 m G - H H = U U xH 2 2 1 H = - 1,0 2 932 , 3 803 , 2 x  = - 3,367 t y =

h

g

h

g

h















2 3 =















932 , 3 803 , 2 932 , 3 ) 803 , 2 2 ( 3 0 , 1 x = 0,472 m Ytotal= 0,472 + 1,0 = 1,472 m

(11)

2.0000 3.9320 4.1910 H I I J 2.0000 2.4490 4.1910 H - I V = U U xH 2 2 1 V = - 2 2 191 , 4 932 , 3 x  = - 8,122 t x =

h

g

h

g

h















2 3 =















191 , 4 932 , 3 191 , 4 ) 932 , 3 2 ( 3 2 x = 0,989 m Xtotal = (2 – 0,989) + 2,5 + 1,5 + 1 = 6,011 m I - J H = U U xH 2 2 1 H = 2 2 449 , 2 191 , 4 x  = 6,640 t y =

h

g

h

g

h















2 3 =















449 , 2 191 , 4 449 , 2 ) 191 , 4 2 ( 3 2 x = 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₎

Uplift Force (t) Lengan (m) Lengan (m) Momen

V H x y x (total) y (total) Mr Mo A _5.23 _29.77 1.380 -9.222 1.224 1.224 11.292 B _8.23 _26.77 4.768 -12.324 1.236 1.764 21.740 C _8.23 _24.27 5.091 8.441 1.069 1.069 9.022 D _6.23 _22.27 3.350 -5.170 0.743 3.257 16.840 E _6.23 _20.77 3.544

(12)

3.109 0.523 1.523 4.734 F _5.23 _19.77 2.673 -2.738 0.496 4.504 12.332 G _5.23 _18.77 2.803 -3.367 0.472 1.472 4.957 H _6.23 _17.77 3.932 -8.122 0.989 6.011 48.824 I _6.23 _15.77 4.191 6.640 1.087 2.087 13.857 J _4.23 _13.77 2.449 Σ (JUMLAH) - 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 : f u = 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





= 



 

_{Σ.Δ=}



 

_{29,77.5,701=}



 0,192 



(13)

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 7.08 29.77 1.380 B 10.08 26.77 4.995 C 10.08 24.27 5.433 D 8.08 22.27 3.816 E 8.08 20.27 4.104 F 7.08 19.77 3.295 G 7.08 18.77 3.487 H 8.08 17.77 4.678 I 8.08 15.77 5.061 J 6.08 13.77 3.444

(14)

A 1.3800 3.0000 4.9550 B 2.5000 4.9550 5.4330 B C 3.8160 2.0000 D

Tabel 3.9 Perhitungan Uplift Pressure Akibat Air Banjir

Bagian Gambar Gaya angkat per 1 m panjang (t)

A-B H = U U xH 2 2 1 = - 3,0 2 955 , 4 380 , 1 x  = - 9,502 t y = b a b a h















2 3 =















955 , 4 380 , 1 955 , 4 ) 380 , 1 2 ( 3 0 , 3 x = 1,218 m Ytotal= 1,218 m B-C V = U U xH 2 2 1 V = - 2,5 2 433 , 5 955 , 4 x  = - 12,985 t x = c b c b h















2

3

=















433 , 5 955 , 4 433 , 5 ) 955 , 4 2 ( 3 5 , 2 x = 1,231 m X total= 2,5 – 1,231 = 1,269 m C-D H = U U xH 2 2 1 H = 2,0 2 816 , 3 433 , 5 x  = 9,250 t y = d c d c h















2 3

(15)

1.5000 3.8160 4.1040 D E 4.1040 E F 1.0000 3.2950 1.0000 3.2950 3.4870 F G D-E V = U U xH 2 2 1 V = - 1,5 2 104 , 4 816 , 3 x  = - 5,940 t x = e d e d h















2

3

=















104 , 4 816 , 3 104 , 4 ) 816 , 3 2 ( 3 5 , 1 x = 0,741 m X total= (1,5 - 0,741) + 2,5 = 3,259 m E-F H = U U xH 2 2 1 H = 1 2 295 , 3 104 , 4 x  = 3,699 t y =

f

e

f

e

h















2 3 =















295 , 3 104 , 4 295 , 3 ) 104 , 4 2 ( 3 1 x = 0,518 m Ytotal= 0,518 + 1,0 = 1,518 m F-G V = U U xH 2 2 1 V = - 1,0 2 487 , 3 295 , 3 x  = - 3,391 t x =

g

f

g

f

h















2 3 =















487 , 3 295 , 3 487 , 3 ) 295 , 3 2 ( 3 0 , 1 x = 0,495 m X total= (1,0-0,495)+2,5+1,5 = 4,505 m

(16)

1.0000 4.6780 3.4870 G H 2.0000 4.6780 5.0610 H I 2.0000 5.0610 3.4440 I J G-H H = U U xH 2 2 1 H = - 1,0 2 678 , 4 487 , 3 x  = - 4,082 t y =

h

g

h

g

h















2 3 =















678 , 4 487 , 3 678 , 4 ) 487 , 3 2 ( 3 0 , 1 x = 0,476 m Ytotal= 0,476 + 1,0 = 1,476 m H-I V = U U xH 2 2 1 V = - 2,0 2 061 , 5 678 , 4 x  = - 9,739 t x =

h

g

h

g

h















2 3 =















061 ,

5

678 ,

4

061 ,

5 )

678 ,

4

2 (

3

0 ,

2

x = 0,987 m Xtotal= (2 – 0,987) + 2,5 + 1,5 + 1,0 = 6,013 m I-J H = U U xH 2 2 1 H = 2,0 2 444 , 3 061 , 5 x  = 8,505 t y =

h

g

h

g

h















2 3

(17)

Tabel 3.10 Gaya Angkat Akibat Air Banjir Titik Hx (m) Lx (m) Ux (t/m2₎

Uplift Force (t) Lengan (m) Lengan (m) Momen

V H x y x (total) y (total) Mr Mo A _7.08 _29.77 1.380 -9.502 1.214 1.218 11.572 B _10.08 _26.77 4.955 -12.985 1.231 1.269 16.478 C _10.08 _24.27 5.433 9.250 1.058 1.058 9.789 D _8.08 _22.27 3.816 -5.940 0.741 3.259 19.358 E _8.08 _20.77 4.104 3.699 0.518 1.518 5.615 F _7.08 _19.77 3.295 -3.391 0.495 4.505 15.275 G 7.08 18.77 3.487 -4.082 0.476 1.476 6.025 H _8.08 _17.77 4.678 -9.739 0.987 6.031 58.736 I _8.08 _15.77 5.061 8.505 1.063 3.063 26.051 J _6.08 _13.77 3.444 Σ (JUMLAH) - 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

(18)

Tabel 3.11 Akumulasi Beban

_–

Beban Pada Bendung

3.6. 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

No. Keterangan Gaya (ton) Momen (ton.meter) Vertikal Horizontal Mr Mo 1 2 3 4 5 6 Tekanan Air a Air Normal 0.000 7.411 0.000 43.763 b Air Banjir 3.469 11.354 3.094 72.202 c Tekanan Lumpur 0.000 0.888 0.000 5.031 d Berat Sendiri Bendung 71.160 - 257.174 -Gaya Gempa e Gempa Horisontal - 7.116 27.259 27.259 f Gempa Vertikal 3.558 - 12.859 12.859 Gaya Angkat g Air Normal -14.177 2.800 15.053 56.746 h Air Banjir -16.027 3.935 21.824 62.626

(19)

b. Terhadap Geser (Sliding)

 = f .Σ

_{Σ =}

0,7 . 56,983

_{11,100 =3,594 ≥1,2 }

Dimana, f = koefisien geser (diambil f= 0,7) c. Terhadap Daya Dukung Tanah (Over Stressing)

Resultante beban vertikal bekerja sejarak a dari titik O

 = ΣΣ

_Σ

= 272,226105,540

_56,983

= 2,925 

Resultante beban vertikal bekerja sejarak e dari pusat berat bendung

 = 2 =

7 2  2,925 = 0,575 

Jarak e masih terletak didalam “Bidang Kern”

 < 6 →0,575<

7

6 0,575  < 1,167 

…….. (OK!)

Gambar 3.7 Jarak e Dalam Bidang Kern

 Tegangan yang terjadi pada tanah akibat beban – beban pada bendung:

σ

= y I x . M A V   = y 3 x x y x _._b _._b 12 1 b . 0,5 . e . V b . b V   

(20)

= y 2 x y x b . b e . V . 6 b . b V    =

_















x y b e . 6 1 b . x b V

Tegangan izin tanah dasar (σ’) = 2,0 kg/cm2 = 20 t/m2

 Tegangan tanah dikontrol per 1 meter panjang bendung :

σ =

_















x y b e . 6 1 b . x b V σmin =

 





_













7,0 0,575 . 6 1 1 7,0 983 , 56 = 4,130 t/m2 > 0 (OK!) σmax =

 





_













7,0 0,575 . 6 1 1 7,0. 983 , 56 = 12,151 t/m2 < σ’= 20 t/m2 (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)

 = Σ

_{Σ =}

282,091

_{139,858 =2,017 ≥1,5 }

 = f .Σ

_{Σ =}

0,7 . 58,602

_{16,177 =2,536 ≥1,2 }

(21)

 =



_

  =



_



2,427 = 1,073

 <



_

→1,073<



_

1,073  < 1,167 

Tegangan tanah dikontrol per 1 m panjang bendung:





=

,

_{ .}

1 +

 .−,

_

 = 16,071 

⁄



< 20  

⁄ →







= 58,602

_{7 .1 (1 }

6 .1,073

_{7 ) = 0,673  }

⁄ >0→



3.6.2 Dengan Pengaruh Gempa Horizontal

1. Keadaan Normal dengan Uplift Pressure

Tegangan ijin tanah (dengan gempa) σ’= 20 t/m2 x 1,3 = 26 t/m2

 ΣH = a(4) + c(4) + e(4) + g(4) = 7,411 + 0,888 + 7,116 + 2,800 = 18,216 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) + e(6) + g(6) = 43,763 + 5,031 + 27,259 + 56,746 = 132,799 tm Kontrol:

 =

_



=

,

_{,}

=2,050 ≥1,5 

 =

 .

_

=

, . ,

_{,}

=2,190 ≥1,2 

f = koefisien geser (diambil f= 0,7)

c. Terhadap Daya Dukung Tanah (Over Stressing)

 =

−

_

=

,−,

_{,}

= 2,447 

(22)

 <



_

→1,053<



_

1,053  < 1,167 





=

,

_{ .}

1 +

 .−,

_

 = 15,489 

⁄



< 26  

⁄ →







= 56,983

_{7 .1 (1 }

6 .1,053

_{7 ) = 0,792  }

⁄ >0→



2. Keadaan Normal tanpa Uplift Pressure

 ΣH = a(4) + c(4) + e(4) = 7,411 + 0,888 + 7,116 = 15,416 t  ΣV = a(3) + c(3) + d(3) = 0,000 + 0,000 + 71,160 = 71,160 t  ΣMr = a(5) + c(5) + d(5) = 0,000 + 0,000 + 257,174 = 257,174 tm  ΣM0= a(6) + c(6) + e(6) = 43,763 + 5,031 + 27,259 = 76,054 tm Kontrol:

 =

_



=

,

_{,}

=3,381 ≥1,5 

 =

 .

_

=

, . ,

_{,}

=3,231 ≥1,2 

 =

−

_

=

,−,

_{,}

= 2,545 

 =



_

  =



_

 2,545 = 0,955 

(23)





= 71,160

_{7 .1 (1 +}

6 .0,955

_{7 ) = 18,485 }

⁄



< 26  

⁄ →







= 71,160

_{7 .1 (1 }

6 .0,955

_{7 ) = 1,846  }

⁄ >0→



 ΣH = b(4) + c(4) + e(4) + h(4) = 11,354 + 0,888 + 7,116 + 3,935 = 23,293 t  ΣV = b(3) + c(3) + d(3) + h(3) = 3,469 + 0,000 + 71,160 – 16,027 = 58,602 t  ΣMr = b(5) + c(5) + d(5) + h(5) = 3,094 + 0,000 + 257,174 + 21,824 = 282,091 tm  ΣM0= b(6) + c(6) + e(6) + h(6) = 72,202 + 5,031 + 27,259 + 62,626 = 167,117 tm Kontrol:

 =

_



=

,

_{,}

=1,688 ≥1,5 

 =

 .

_

=

, . ,

_{,}

=1,761 ≥1,2 

 =

−

_

=

,−,

_{,}

= 1,962 

 =



_

  =



_

 1,962 = 1,538 

 <



_

→1,538<



_

1,538  < 1,167 





=

,

_{ .}

1 +

 .,

_

 = 19,408 

⁄



< 26  

⁄ →



(24)

4. Keadaan Banjir tanpa Uplift Pressure  ΣH = b(4) + c(4) + e(4) = 11,354 + 0,888 + 7,116 = 19,358 t  ΣV = b(3) + c(3) + d(3) = 3,469+ 0,000 + 71,160 = 74,629 t  ΣMr = b(5) + c(5) + d(5) + e(5) = 3,054 + 0,000 + 257,174 + 27,259 = 287,526 tm  ΣM0= b(6) + c(6) + e(6) = 72,202 + 5,031 + 27,259 = 104,492 tm Kontrol:

 =

_



=

,

_{,}

=2,752 ≥1,5 

 =

 .

_

=

, .,

_{,}

=1,047 ≥1,2 

 =

−

_

=

,−,

_{,}

= 2,453 

 =



_

  =



_

 2,453 = 1,047 

 <



_

→1,047<



_

1,047 < 1,167 





= 74,629

_{7 .1 (1 +}

6 .1,047

_{7 ) = 20,233 }

⁄



< 20  

⁄ →







= 74,629

_{7 .1 (1 }

6 .1,047

_{7 ) = 1,090  }

⁄ >0→



(25)

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) + f(3) + g(3) = 0,000 + 0,000 + 71,160 + 3,558 – 14,177= 60,541 t  ΣMr = a(5) + c(5) + d(5) + f(5) + g(5) = 0,000 + 0,000 + 257,174 + 12,859 + 15,053= 285,085 tm  ΣM0= a(6) + c(6) + f(6) + g(6) = 43,763 + 5,031 + 12,859 + 56,746 = 118,399 tm Kontrol:

 =

_



=

,

_{,}

=2,408 ≥1,5 

 = f .Σ

_{Σ =}

0,7 . 60,541

_{11,100 =3,818 ≥1,2 }

 =

−

_

=

,−,

_{,}

= 2,753 

 =



_

  =



_

 2,753 = 0,747 

 <



_

→0,747<



_

0,747  < 1,167 





=

,

_{ .}

1 +

 .−,

_

 = 14,184  

⁄



< 20  

⁄ →







= 60,541

_{7 .1 (1 }

6 .0,747

_{7 ) = 3,113  }

⁄ >0→



2. Keadaan Normal tanpa Uplift Pressure

 ΣH = a(4) + c(4)

= 7,411 + 0,888 = 8,300 t

(26)

= 0,000 + 0,000 + 71,160 + 3,558 = 74,718 t  ΣMr = a(5) + c(5) + d(5) + f(5) = 0,000 + 0,000 + 257,174 + 12,859 = 270,032 tm  ΣM0= a(6) + c(6) = 43,763 + 5,031 = 48,795 tm Kontrol:

 =

_



=

,

_{,}

=5,534 ≥1,5 

 =

 .

_

=

, . ,

_{,}

=6,3017 ≥1,2 

 =

−

_

=

,−,

_{,}

= 2,961 

 =



_

  =



_

 2,961 = 0,539 

 <



_

→0,539<



_

0,539  < 1,167 





= 74,718

_{7 .1 (1 +}

6 .0,539

_{7 ) = 15,606 }

⁄



< 20  

⁄ →







= 74,718

_{7 .1 (1 }

6 .0,539

_{7 ) = 5,742  }

⁄ >0→



 ΣH = b(4) + c(4) + h(4)

= 11,354 + 0,888 + 3,935 = 16,177 t

 ΣV = b(3) + c(3) + d(3) + f(3) + h(3)

(27)

= 72,202 + 5,031 + 12,859 + 62,626 = 152,717 tm Kontrol:

 =

_



=

,

_{,}

=1,847 ≥1,5 

 =

 .

_

=

, . ,

_{,}

=2,690 ≥1,2 

 =

−

_

=

,−,

_{,}

= 2,081 

 =



_

  =



_

 2,081 = 1,419 

 <



_

→1,419 <



_

1,419  < 1,167 





=

,

_{ .}

1 +

 .,

_

 = 19,678 

⁄



< 20  

⁄ →







= 62,160

_{7 .1 (1 }

6 .1,419

_{7 ) = 1,918  }

⁄ >0→



4. Keadaan Banjir tanpa Uplift Pressure

 ΣH = b(4) + c(4) = 11,354 + 0,888 = 12,242 t  ΣV = b(3) + c(3) + d(3) + f(3) = 3,469 + 0,000 + 71,160 + 3,558 = 78,187 t  ΣMr = b(5) + c(5) + d(5) + f(5) = 3,094 + 0,000 + 257,174 + 12,859 = 273,126 tm  ΣM0= b(6) + c(6) = 72,202 + 5,031 = 77,233 tm Kontrol:

(28)

 =

 .

_

=

, .,

_{,}

=4,471 ≥1,2 

 = ΣΣ

_Σ

= 273,12677,233

_78,187

= 2,505 

 =



_

  =



_

 2,505 = 0,995 

 <



_

→0,995<



_

0,995 < 1,167 





= 78,187

_{7 .1 (1 +}

6 .0,995

_{7 ) = 20,691  }

⁄



< 20  

⁄ →







= 78,187

_{7 .1 (1 }

6 .0,995

_{7 ) = 1,648  }

⁄ >0→



Tabel 3.12 Akumulasi Kombinasi Gaya-Gaya Yang Bekerja Pada Tubuh Bendung

No. Kombinasi Gaya – Gaya

SF Tegangan Tanah

Tanpa Gempa Dengan Gempa

pada Tubuh Bendung Guling Geser Max Min Max Min

≥1,50 ≥1,20 < 20 t/m2 _{> 0} _{< 26 t/m}2 _{> 0}

1

Tanpa Gempa

a. Air Normal + Gaya Angkat 2,579 3,594 12,151 4,130 - - b. Air Banjir + Gaya Angkat 2,017 2,536 16,071 0,673 -

-2

Dengan Gempa Horizontal

a. Air Normal + Gaya Angkat 2,050 2,190 - - 15,489 0,792

b. Air Normal 3,381 3,231 - - 18,485 1,846

c. Air Banjir + Gaya Angkat 1,688 1,761 - - 19,408 -2,665

d. Air Banjir 2,752 2,699 - - 20,233 1,090

3

Dengan Gempa Vertikal

a. Air Normal + Gaya Angkat 2,408 3,818 - - 14,184 3,113

b. Air Normal 5,534 6,302 - - 15,606 5,742