bab iii analisa stabilitas bendung rev

Tugas Irigasi dan

JURUSAN TEKNIK SIPILFAKULTAS TEKNIKUNIVERSITAS UDAYANA

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

γair = 1 ton/m3

Pa1 = 12

. γ air . h2

= 12

. (1 ) . (4,0 )2 = 8 ton

Pa2 = b . h .γair = (1,5).(4,0).(1) = 6 ton

47

Tugas Irigasi dan


Tabel. 3.1 Perhitungan Tekanan Air Normal

BagianGaya (t) Lengan (m) Momen (tm)

V H x y Mr M0

Pa1 - 8 - 4,170 - 33,36

Pa2 6 - 6,536 - 39,216 -

JUMLAH 6 8 39,216 33,36

3.1.2 Tekanan Air Banjir (Flood)

Gambar 3.2 Diagram Tekanan Akibat Air Banjir

Pf1 = 12

. γ air . h2

= 12

. (1 ) . (4,0 )2= 8 ton

Pf2 = b . h .γair = (1,609).(4,0).(1) = 6,436 ton

Pf3 = b . h .γair = (1,5).(4,0).(1) = 6 ton

Pf4 = b . h .γair = (2,513).(1,609).(1) = 4,043 ton

48

Tugas Irigasi dan


Pf5 =

12

. γ air . h2

=

12

. (1,95 )2 .1= 1,9 ton

Pf6 = −1

2. γ air . h2

= −1

2. (1 ) . (1,95 )2

= - 1,9 ton

Tabel.3.2 Perhitungan Tekanan Air Banjir


V H x y Mr M0

Pf1 - 8 - 4,170 - 33,36

Pf2 - 6,436 - 4,837 - 31,131

Pf3 6 - 7,286 - 43,716 -

Pf4 4,043 - 6,78 - 27,412 -

Pf5 1,9 - 0,65 - 1,235 -

Pf6 - -1,9 - 3,15 - -5,985

JUMLAH 11,943 12,536 72,363 58,506

3.2 Tekanan Lumpur

γlumpur = 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

49

Tugas Irigasi dan


Gambar 3.3 Diagram Tekanan Akibat Lumpur

PL1 = 12 .Ka.h2.γlumpur

= 12 .(0,333). (4,0)2.(0,6).

= 1,6 ton

PL2 = b.h.γlumpur

= (1,5).(4,0). 0,6

= 3,6 ton

Tabel. 3.3 Perhitungan Tekanan Lumpur


V H x y Mr M0

PL1 - 1,6 - 4,170 - 6,672

PL2 3,6 - 6,536 - 23,530 -

JUMLAH 3,6 1,6 23,530 6,672

50

Tugas Irigasi dan


3.3 Tekanan Berat Sendiri Bendung

Berat volume pasangan batuγ pasangan = 2,2 t/m2

Gambar 3.4 Diagram Tekanan Akibat Berat Sendiri Bendung

Pada badan bendung yang berbentuk parabola, luas penampang digunakan pendekatan :

A = 2/3 . L . H

51

Tugas Irigasi dan


W1 = b . h . γ pasangan = 1,472 . 1,84 . 2,2 = 5,959 ton






W7 = 2/3 . b. h . γ pasangan = 2/3 . 1,013 . 0,45 . 2,2 = 0,669 ton





Tabel 3.4 Perhitungan Tekanan Berat Sendiri Bendung

Bagian B (m) H (m)Gaya (ton) Lengan (m) Momen (tm)

Vertikal x y Mr M0

W1 1,472 1,84 5,959 7,264 1,92 43,284 11,441

W2 1,013 5,38 11,990 6,023 3,69 72,215 44,243

W3 1,015 5,61 12,527 5,008 3,805 62,736 47,666

W4 1,5 3,5 11,550 3,75 3,75 43,313 43,313

W5 1,5 3 9,900 2,25 2,5 22,275 24,750

W6 1,5 2,5 8,250 0,75 1,25 6,188 10,313

W7 1,013 0,45 0,669 5,895 6,549 3,941 4,379

W8 1,015 0,23 0,342 5,134 6,692 1,758 2,291

W9 1,5 1,1 2,420 3,938 5,913 9,530 14,309

W10 1,5 1,5 3,300 2,438 4,563 8,045 15,058

W11 1,5 1,5 3,300 0,938 3,063 3,095 10,108∑ 70,207 276,379 227,869

52

Tugas Irigasi dan


3.4 Gaya Gempa

3.4.1 Gempa Horizontal

Gaya Horizontal (H) = Kh . ΣV1

= 0,10 . 70,207

= 7,021 ton

Momen akibat gempa horizontal :

M0 = Mr = Kh . ΣM01

= 0,10 . 227,869 = 22,787 tm

Keterangan :

H = gaya gempa horizontal (t)

Kh = koefisien gempa horizontal, (Pondasi batu : Kh = 0,10)

V1 = berat sendiri bendung (t)

M01 = momen guling akibat berat sendiri (tm)

3.4.2 Gempa Vertikal

Gaya Vertikal (V) = Kv . ΣW

= 0,05 .( 70,207 )

= 3,51 ton

Momen akibat gempa vertikal :

Mr = Kv . ΣMr1

= 0,05. (276,379)

= 13,82 tm

Keterangan :

V = gaya gempa vertikal (t)

Kv = koefisien gempa vertikal, (Pondasi batu : Kv = 0,05)

Mr1 = momen tahanan akibat berat sendiri (tm)

53

Tugas Irigasi dan


3.5 Gaya Angkat (Uplift Pressure)

3.5.1 Air Normal

ΣL = Lh + Lv

= 15,85 + 13,47

= 29,32 m

ΔH (air normal) = elev. MAN – elev. Dasar sungai

= 45,50 – 41,50

= 4,00 m

Ux = Hx –

Lx

ΣL . ΔH

Ux = Hx –

Lx

29 , 32 .(4,00)

Ux = Hx – 0,136 Lx

Keterangan :

Hx = tinggi muka air dari titik yang dicari (m)

Lx = panjang rayapan (m)

ΣL = total rayapan (m)

ΔH = tinggi muka air normal (m)

Ux = uplift pressure di titik x (t/m2)

e = ( h

3 ) 2U 1+U 2U 1+U 2

R = U 1+U 2

2xH

Gambar 3.5 Rayapan Gaya Angkat

54

Tugas Irigasi dan


Gambar 3.6 Gaya Angkat Akibat Air Normal

Tabel 3.5 Perhitungan Tinggi air normal terhadap muka bendung

Titik Hx (m) Lx (m) Ux (t/m2)

a 4,337 29,190 0,747

b 6,837 26,690 3,554

c 6,837 25,190 3,739

d 5,837 24,190 2,862

e 5,837 22,690 3,046

f 4,837 21,690 2,169

g 4,837 20,190 2,354

h 5,837 19,190 3,477

i 5,837 15,690 3,907

j 4,837 14,690 3,030

55

Tugas Irigasi dan


Tabel 3.6 Perhitungan Gaya Angkat Akibat Air Normal

Bagian Gambar Gaya angkat per 1 m panjang (t)

a - b

H = U 1+U 2

2xH

= -

0 ,747+3 ,5542

x 2,5

= -5,376 t

y =( h

3 ) 2 a+ba+b

=

2,53 ( (2 x 0 ,747 )+3 , 554

0 ,747+3 ,554 ) = 0,978 m

Ytotal = 0,978 m

b - c

V = U 1+U 2

2xH

V =

3 ,554+3 ,7392

x1,5

= 5,470 t

x = ( h3 ) 2b+c

b+c

=

1,53 ( (2 x 3 ,554 )+3 ,739

3 ,554+3 ,739 ) = 0,744 m

X total = 1,50 – 0,744 = 0,756 m

56

Tugas Irigasi dan


c – d

H =

U 1+U 22

xH

H = 3 ,739+2 ,862

2x1,0

= 3,30 t

y = ( h3 ) 2c+d

c+d

=

1,03 ( (2 x 3 ,739 )+2 ,862

3 ,739+2 ,862 ) = 0,522 m

Ytotal = 0,522 m

d – e

V =

U 1+U 22

xH

V = 2, 862+3 ,046

2x 1,5

= 4,431 t

x = ( h

3 ) 2d+ed+e

=

1,53 ( (2x 2,862 )+3 , 046

2,862+3 , 046 ) = 0,742 m

X total = (1,5 – 0,742) + 1,5 = 2,258 m

e – f

H =

U 1+U 22

xH

H = 3 ,046+2 ,169

2x 1

= 2,608 t

y = ( h

3 ) 2e+ fe+ f

=

13 ( (2 x 3 ,046 )+2 , 169

3 ,046+2 ,169 )= 0,528 m

57

Tugas Irigasi dan


Ytotal = 0,528 + 1,0 = 1,528 m

f – g

V = U 1+U 2

2xH

V =

2,169+2 , 3542

x 1,5

= 3,392 t

x = ( h

3 ) 2 f +gf +g

=

1,53 ( (2 x 2, 169 )+2 , 354

2, 169+2 , 354 )= 0,740 m

X total = (1,5-0,740)+1,5+1,5 = 3,76 m

g - h

H = U 1+U 2

2xH

H = -

2, 354+3 , 4772

x 1,0

= -2,915 t

y =( h

3 ) 2 g+hg+h

=

1,03 ( (2 x 2,354 )+3 , 477

2,354+3 , 477 ) = 0,468 m

Ytotal = 0,468 + 1,0 = 1,468 m

h - i

V =

U 1+U 22

xH

V = 3 , 477+3 , 907

2x 3,5

= 12,922 t

x =( h

3 ) 2 g+hg+h

58

Tugas Irigasi dan


=

3,53 ( (2x 3 ,477)+3 ,907

3 ,477+3 ,907 )= 1,716 m

Xtotal =(3,5-1,716)+1,5+1,5+1,5=6,284 m

i - j

H =

U 1+U 22

xH

H = 3 ,907+3 , 030

2x1,0

= 3,469 t

y =( h

3 ) 2 g+hg+h

=

1,03 ( (2 x 3 ,907 )+3 , 030

3 ,907+3 , 030 ) = 0,521 m

Ytotal = 0,521 + 1 = 1,521 m

Tabel 3.7 Gaya Angkat Akibat Air Normal

59

Tugas Irigasi dan


V H x y x (total) y (total) Mr Moa 4,337 29,190 0,747

-5,376 0,978 0,978 -5,258b 6,837 26,690 3,554

5,470 0,744 0,756 4,135c 6,837 25,190 3,739

3,300 0,522 0,522 1,723d 5,837 24,190 2,862

4,431 0,742 2,258 3,289e 5,837 22,690 3,046

2,608 0,528 1,531 1,377f 4,837 21,690 2,169

3,392 0,740 3,760 2,509g 4,837 20,190 2,354

-2,915 0,468 1,468 -1,364h 5,837 19,190 3,477

12,922 1,716 6,284 22,173i 5,837 15,690 3,907

3,469 0,521 1,521 1,807j 4,837 14,690 3,030

26,214 1,085 -1,714 32,106

Lengan (m) Momen

Σ (JUMLAH)

Titik Hx (m) Lx (m) Ux (t/m2)Uplift Force (t) Lengan (m)

Catatan : Searah jarum jam (+)

Berlawanan arah harum jam (-)

Gaya Angkat Akibat Air Normal :

1. Tekanan Vertikal

V = fu x ΣV = 0,5 x (26,214) = 13,107 ton

2. Tekanan Horizontal

H = fu x ΣH = 0,5 x (1,085) = 0,543 ton

3. Momen

Mr = 0.5 x ΣMr = 0,5 x (-1,714) = -0,857 tm

Mo = 0.5 x ΣMo = 0,5 x (32,106) = 16,053 tm

Dimana : fu = koefisien reduksi untuk jenis tanah keras (50 %)

3.5.2 Air Banjir

ΣL = 29,19 m

60

Tugas Irigasi dan


ΔH (air banjir) = elev. M.A.B – elev. Dasar sungai

= 47,535 – 41,50

= 6,035 m

Ux = Hx -

Lx

ΣL . ΔH

Ux = Hx -

Lx

29,19 . 6,035

Ux = Hx - 0,207 Lx

Keterangan :

Hx = tinggi muka air banjir dari titik yang dicari (m)

Lx = panjang rayapan (m)

ΣL = total rayapan (m)

ΔH = beda tinggi M.A.B dengan muka air di hilir (m)

Ux = uplift pressure di titik x (t/m2)

Gambar 3.7 Gaya Angkat Akibat Air Banjir

61

Tugas Irigasi dan


Tabel 3.8 Perhitungan Tinggi Air Banjir Terhadap Muka Bendung

Titik Hx (m) Lx (m) Ux (t/m2)

a 6,372 29,190 0,943

b 8,872 26,690 3,908

c 8,872 25,190 4,187

d 7,872 24,190 3,373

e 7,872 22,690 3,652

f 6,872 21,690 2,838

g 6,872 20,190 3,117

h 7,872 19,190 4,303

i 7,872 15,690 4,954

j 6,872 14,690 4,140Tabel 3.9 Perhitungan Gaya Angkat Akibat Air Banjir

Bagian Gambar Gaya angkat per 1 m panjang (t)

a - b

H = U 1+U 2

2xH

= -

0 , 943+3 , 9082

x 2,5

= -6,063 t

y =( h

3 ) 2 a+ba+b

=

2,53 ( (2 x 0 , 943)+3 ,908

0 , 943+3 ,908 )

62

Tugas Irigasi dan


= 0,995 m

Ytotal = 0,995 m

b - c

V = U 1+U 2

2xH

V =

3 ,908+4 ,1872

x1,5

= 6,071 t

x = ( h3 ) 2b+c

b+c

=

1,53 ( (2x 3 ,908 )+4 ,187

3 ,908+4 , 187 ) = 0,741 m

X total = 1,50 – 0,741 = 0,759 m

c – d

H = U 1+U 2

2xH

H = 4 ,187+3 , 373

2x 1,0

= 3,78 t

y = ( h3 ) 2c+d

c+d

=

1,03 ( (2 x 4 , 187 )+3 ,373

4 , 187+3 , 373 ) = 0,518 m

Ytotal = 0,518 m

63

Tugas Irigasi dan


d – e

V =

U 1+U 22

xH

V = 3 ,373+3 ,652

2x 1,5

= 5,268 t

x = ( h

3 ) 2d+ed+e

=

1,53 ( (2x 3 ,373 )+3 ,652

3 ,373+3 , 652 ) = 0,740 m

X total = (1,5 – 0,740) + 1,5 = 2,26 m

e – f

H =

U 1+U 22

xH

H = 3 ,652+2 , 838

2x 1

= 3,245 t

y = ( h

3 ) 2e+ fe+ f

=

13 ( (2x 3 ,652 )+2, 838

3 ,652+2 , 838 )= 0,521 m

Ytotal = 0,521 + 1,0 = 1,521 m

f – g

V =

U 1+U 22

xH

V = 2, 838+3 ,117

2x 1,5

= 4,466 t

x = ( h

3 ) 2 f +gf +g

=

1,53 ( (2 x 2, 838 )+3 ,117

2,838+3 , 117 )= 0,738 m

64

Tugas Irigasi dan


X total = (1,5-0,738)+1,5+1,5 = 3,762 m

g - h

H = U 1+U 2

2xH

H = -

3 ,117+4 ,3032

x1,0

= -3,710 t

y =( h

3 ) 2 g+hg+h

=

1,03 ( (2 x 3 ,711)+4 ,303

3 ,711+4 , 303 ) = 0,473 m

Ytotal = 0,473 + 1,0 = 1,473 m

h - i

V = U 1+U 2

2xH

V =

4 ,303+4 , 9542

x 3,5

= 16,199 t

x =( h

3 ) 2 g+hg+h

=

3,53 ( (2x 4 , 303 )+4 , 954

4 , 303+4 , 954 )= 1,709 m

Xtotal =(3,5-1,709)+1,5+1,5+1,5=6,291 m

i - j

H = U 1+U 2

2xH

H =

4 ,954+4 ,1402

x1,0

= 4,547 t

y =( h

3 ) 2 g+hg+h

65

Tugas Irigasi dan


=

1,03 ( (2x 4 ,954 )+4 , 140

4 , 954+4 ,140 ) = 0,515 m

Ytotal = 0,515 + 1 = 1,515 m

Tabel 3.10 Gaya Angkat Akibat Air Banjir

V H x y x (total) y (total) Mr Moa 6,372 29,190 0,943

-6,063 0,995 0,933 -5,657b 8,872 26,690 3,908

6,071 0,741 0,759 4,608c 8,872 25,190 4,187

3,780 0,518 0,518 1,958d 7,872 24,190 3,373

5,268 0,740 2,260 11,906e 7,872 22,690 3,652

3,245 0,521 1,521 4,935f 6,872 21,690 2,838

4,466 0,738 3,762 16,800g 6,872 20,190 3,117

-3,710 0,473 1,473 -5,464h 7,872 19,190 4,303

16,199 1,709 6,291 101,905i 7,872 15,690 4,954

4,547 0,515 1,515 6,888j 6,872 14,690 4,140

32,003 1,798 2,660 135,219

Lengan (m) Momen

Σ (JUMLAH)

Titik Hx (m) Lx (m) Ux (t/m2)Uplift Force (t) Lengan (m)

Catatan : Searah jarum jam (+)

Berlawanan arah harum jam (-)

Gaya Angkat Akibat Air Banjir :

1. Tekanan Vertikal

V = fu x ΣV = 0,5 x (32,003) = 16,002 ton

2. Tekanan Horizontal

H = fu x ΣH = 0,5 x (1,798) = 0,899 ton

3. Momen

66

Tugas Irigasi dan


Mr = 0.5 x ΣMr = 0,5 x (2,660) = 1,33 tm

Mo = 0.5 x ΣMo = 0,5 x (135,219) = 67,610 tm

Dimana : fu = koefisien reduksi untuk jenis tanah keras (50 %)

Tabel 3.11 Akumulasi Beban-Beban pada Bendung

No Keterangan Gaya (t) Momen (ton meter)Vertikal Horisontal Mr Mo

1 2 3 4 5 6

Tekanan Aira Air Normal 6,000 8,000 39,216 33,36b Air Banjir 11,943 12,536 72,363 58,506c Tekanan Lumpur 3,6 1,6 23,53 6,672d Berat Sendiri Bendung 70,207 - 276,379 -

Gaya Gempae Gempa Horisontal - 7,021 22,787 22,787f Gempa Vertikal 3,51 - 13,82 13,820

Gaya Angkatg Air Normal 13,107 0,543 -0,857 16,053h Air Banjir 16,002 0,899 1,33 67,61

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 Gempa

Tegangan ijin tanah σ’= 20 t/m2

1. Keadaan Air Normal dengan Uplift Pressure

ΣH = a(4) + c(4) + g(4)

= 8 + 1,6 + 0,543 = 10,143 t

67

Tugas Irigasi dan


ΣV = a(3) + c(3) + d(3) + g(3)

= 6 + 3,6 + 70,207 + 13,107 = 92,914 t

ΣMr = a(5) + c(5) + d(5) + g(5)

= 39,216 + 23,53 + 276,379 – 0,857 = 338,268 tm

ΣM0 = a(6) + c(6) + g(6)

= 33,36 + 6,672 + 16,053 = 56,085 tm

Kontrol :

a) Terhadap guling (over turning)

SF =

∑ M r

∑ M 0 =

338,268 56,085

=6 ,031.............. ≥ 1,50 (OK!)

b) Terhadap geser (sliding)

SF =

f ∑V

∑ H =

0 ,70 . (92 , 914 )10 , 143

=6 , 412.......≥ 1,20 (OK!)

keterangan : f = koefisien geser

c) Terhadap daya dukung tanah (over stressing)

68

Tugas Irigasi dan


Gambar 3.11. Jarak e dalam bidang Kern

Resultante beban vertikal bekerja sejarak a dari titik O.

a =

∑ M r−∑ M 0

∑V =

338 , 268−56 ,08592 , 914

=3 ,037 m

Resultante beban vertikal bekerja sejarak e dari pusat berat bendung.

e =

B2−a

=

8,02

−3 , 037=0 , 963 m

Jarak e masih terletak di dalam ‘ Bidang Kern’

e < B6

e = 0,963 m < 8,06 0,889 m < 1,333 m (OK!)

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

σ =

∑ VA

±M . x

I y

=

∑ Vbx . b y

±∑ V . e . 0,5 . bx

112

. bx3 . b y

=

∑ Vbx . b y

±6.∑V . e

bx2 . b y

69

Tugas Irigasi dan


=

∑ Vbx . b y (1±6 . e

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

Tegangan tanah dikontrol per 1 meter panjang bendung :

σ =

∑ Vbx . b y (1±6 . e

bx )

σmin =

92 , 9148,0 . (1 ) (1− 6 . (0,963 )

8,0 ) = 3,226 t/m2 > 0 (OK!)

σmax =

92 , 9148,0 . (1 ) (1+

6 . (0,963 )8,0 )

= 19,718 t/m2 < σ’= 20 t/m2 (OK!)

2. Keadaan Banjir dengan Uplift Pressure

ΣH = b(4) + c(4) + h(4)

= 12,536 + 1,6 + 0,899 = 15,062 t

ΣV = b(3) + c(3) + d(3) + h(3)

= 11,943 + 3,6 + 70,207 + 16,002 = 101,752 t

ΣMr = b(5) + c(5) + d(5) + h(5)

= 72,363 + 23,53 + 276,379 +1,33 = 373,602 tm

ΣM0 = b(6) + c(6) + h(6)

= 58,506 + 6,672 + 67,61 = 132,788 tm

Kontrol :

a) Terhadap guling (over turning)

SF =

∑ M r

∑ M 0 =

373 , 602132 ,788

=2 ,814 ≥ 1,50 (OK !)

b) Terhadap geser (sliding)

SF =

f ∑V

∑ H =

0,7 . (101,752 )15 ,062

=4 ,729 ≥ 1,20 (OK !)


70

Tugas Irigasi dan


c) Terhadap daya dukung tanah (over stressing)


a =

∑ M r−∑ M 0

∑V =

373 ,602−132 ,788101 ,752

=2 ,867 m


e =

B2−a

=

8,02

−2 , 867= 1,133 m

Jarak e masih terletak di dalam ‘ Bidang Kern’

e < B6

e = 1,133 m < 8,06 1,133 m < 1,333 m (OK!)

Tegangan pada tanah dasar

σ =

∑ Vbx . b y (1±6 . e

bx )

σmin =

101 ,7528,0 . (1 ) (1− 6 . (1,133 )

8,0 )= 1,911 t/m2 > 0 (OK!)

σmax =

101 ,7528,0 . (1 ) (1+ 6 . (1,133 )

8,0 )= 19,973 t/m2 < σ’= 20 t/m2 (OK !)

3.6.2 Dengan Gempa Horizontal

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


ΣH = a(4) + c(4) + e(4) + g(4)

= 8 + 1,6 + 7,021 + 0,543 = 17,164 t

ΣV = a(3) + c(3) + d(3) + g(3)

= 6 + 3,6 + 70,207 +13,107 = 92,914 t

ΣMr = a(5) + c(5) + d(5) +g(5)

= 39,216 + 23,53 + 276,379 – 0,857 = 338,268 tm

71

Tugas Irigasi dan


ΣM0 = a(6) + c(6) + e(6) + g(6)

= 33,36 + 6,672 + 22,787 + 16,053 = 78,872 tm

Kontrol :

a). Terhadap guling (over turning)

SF =

∑ M r

∑ M 0 =

338 , 26878 , 872

=4 , 289 ≥ 1,50 (OK !)

b).Terhadap geser (sliding)

SF =

f ∑V

∑ H =

0,7 . (92,914 )17 , 164

=3,8≥ 1,20 (OK !)


c). Terhadap daya dukung tanah (over stressing)


a =

∑ M r−∑ M0

∑V =

338 , 268−78 , 87292 , 914

=2, 791 m


e = B2−a

= 8,02

−2 ,79=1 ,21 m <

B6 = 1,333 m


σ =

∑ Vbx . b y (1±6 . e

bx )

σmin =

92 , 9148,0 . (1 ) (1−6 . (1,21 )

8,0 )= 1,074 t/m2 > 0 (OK !)

σmax =

92 ,9148,0 . (1 ) (1+

6 . (1,21 )8,0 )

= 22,154 t/m2 < σ’= 26 t/m2 (OK !)

2. Keadaan Air Normal tanpa Uplift Pressure

ΣH = a(4) + c(4) + e(4)

= 8 + 1,6 + 7,021 = 16,621 t

ΣV = a(3) + c(3) + d(3)

= 6 + 3,6 + 70,207 = 79,807 t

72

Tugas Irigasi dan


ΣMr = a(5) + c(5) + d(5)

= 39,216 + 23,53 + 276,379 = 339,125 tm

ΣM0 = a(6) + c(6) + e(6)

= 33,36 + 6,672 + 22,787 = 62,819 tm

Kontrol :


SF =

∑ M r

∑ M 0 =

339 , 12562 ,819

=5 ,398 ≥ 1,50 (OK !)


SF =

f ∑V

∑ H =

0,7 . (79,807 )16 , 621

=3 ,361 ≥ 1,20 (OK !)




a =

∑ M r−∑ M 0

∑V =

339 , 125−62 , 81979 , 807

=3 , 462 m


e = B2−a

= 8,02

−3 , 462=0 ,538 m <

B6 = 1,333 m


σ =

∑ Vbx . b y (1±6 . e

bx )

σmin =

79 , 8078,0 . (1 ) (1−

6. (0,538 )8,0 )

= 5,95 t/m2 > 0 (OK !)

σmax =

79 , 8078,0 . (1 ) (1+

6 . (0,538 )8,0 )

= 14,001 t/m2 < σ’= 26 t/m2 (OK !)

73

Tugas Irigasi dan


3. Keadaan Air Banjir dengan Uplift Pressure

ΣH = b(4) + c(4) + e(4) + h(4)

= 12,536 + 1,6 + 7,021 + 0,899 = 22,056 t

ΣV = b(3) + c(3) + d(3) + h(3)

= 11,943 + 3,6 + 70,207 + 16,002 = 101,752 t

ΣMr = b(5) + c(5) + d(5) + h(5)

= 72,363 + 23,53 + 276,379 + 1,33 = 373,602 tm

ΣM0 = b(6) + c(6) + e(6) + h(6)

= 58,506 + 6,672 + 22,787 + 67,61 = 155,575tm

Kontrol :


SF =

∑ M r

∑ M 0 =

373 ,602155 ,575

=2 ,401 ≥ 1,50 (OK !)


SF =

f ∑V

∑ H =

0,7 . (101,752 )22 ,056

=3 ,23 ≥ 1,20 (OK !)




a =

∑ M r−∑ M 0

∑V =

373 , 602−155 ,575101 , 056

=2 , 725m


e = B2−a

= 8,02

−2 ,725=1 ,275 m <

B6 = 1,333 m


σ =

∑ Vbx . b y (1±6 . e

bx )74

Tugas Irigasi dan


σmin =

101 ,7528,0 . (1 ) (1− 6 . (1,275 )

8,0 )= 0,556 t/m2 > 0 (OK !)

σmax =

101 ,7528,0 . (1 ) (1+ 6 . (1,275 )

8,0 )= 24,881 t/m2 < σ’= 26 t/m2 (OK !)

4. Keadaan Air Banjir tanpa Uplift Pressure

ΣH = b(4) + c(4) + e(4)

= 12,536 + 1,6 + 7,021 = 21,157 t

ΣV = b(3) + c(3) + d(3)

= 11,943 + 3,6 + 70,207 = 85,75 t

ΣMr = b(5) + c(5) + d(5) + e(5)

= 72,363 + 23,53 + 276,379 + 22,787 = 396,059 tm

ΣM0 = b(6) + c(6) + e(6)

= 58,506 + 6,672 + 22,787 = 87,965 tm

Kontrol :


SF =

∑ M r

∑ M 0 =

396 , 05987 ,965

=4 ,502≥ 1,50 (OK !)


SF =

f ∑V

∑ H =

0,7 . (85,75 )21 ,157

=2 ,837 ≥ 1,20 (OK !)




a =

∑ M r−∑ M 0

∑V =

396 , 059−87 , 96585 , 75

=3 ,593 m


e = B2−a

= 8,02

−3 ,593=0 , 407 m <

B6 = 1,333 m


75

Tugas Irigasi dan


σ =

∑ Vbx . b y (1±6 . e

bx )

σmin =

85 , 758,0 . (1 ) (1− 6 . (0,407 )

8,0 )= 10,173 > 0 (OK !)

σmax =

85 , 758,0 . (1 ) (1+ 6 . (0,407 )

8,0 )= 11,264 t/m2 < σ’= 26 t/m2 (OK !)

3.6.3 Dengan Gempa Vertikal

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


ΣH = a(4) + c(4) + g(4)

= 8 + 1,6 + 0,543 = 10,143 t

ΣV = a(3) + c(3) + d(3) + f(3) – g(3)

= 6 + 3,6 + 70,207 + 3,51 – 13,107 = 70,21 t

ΣMr = a(5) + c(5) + d(5) + f(5) + g(5)

= 39,216 + 23,53 + 276,379 + 13,82 -0,857 = 352,088 tm

ΣM0 = a(6) +c(6) + f(6) + g(6)

= 33,36 + 6,672 + 13,820 + 16,053 = 69,905 tm

Kontrol :


SF =

∑ M r

∑ M 0 =

352 , 08869 , 905

=19 , 342 ≥ 1,50 (OK !)


SF =

f ∑V

∑ H =

0,7 . (70,21 )10 ,143

=4 , 845 ≥ 1,20 (OK !)




a =

∑ M r−∑ M0

∑V =

352 , 088−69 , 90570 ,21

=4 , 019 m

76

Tugas Irigasi dan



e =

B2−a

=

8,02

−4 , 019=−0 , 019 m <

B6 = 1,333 m


σ =

∑ Vbx . b y (1±6 . e

bx )

σmin =

70 ,218,0 . (1 ) (1−6 . (−0 ,019 )

8,0 )= 8,797 t/m2 > 0 (OK !)

σmax =

70 ,218,0 . (1 ) (1+ 6 . (-0,019 )

8,0 ) = 8,755 t/m2 < σ’= 26 t/m2 (OK !)

2. Keadaan Air Normal tanpa Uplift Pressure

ΣH = a(4) + c(4)

= 8 + 1,6 = 9,6 t

ΣV = a(3) + c(3) + d(3) + f(3)

= 6 + 3,6 + 70,207 + 3,51 = 83,371 t

ΣMr = a(5) + c(5) + d(5) + f(5)

= 39,216 + 23,53 + 276,379 + 13,82 = 352,945 tm

ΣM0 = a(6) + c(6)

= 33,36 + 6,672 = 40,032 tm

Kontrol :


SF =

∑ M r

∑ M 0 =

352 , 94540 ,032

=8 ,817 ≥ 1,50 (OK !)


SF =

f ∑V

∑ H =

0,7 . (83,371 )9,6

=6 ,079 ≥ 1,20 (OK !)


77

Tugas Irigasi dan




a =

∑ M r−∑ M 0

∑V =

352 ,945−40 ,03283 ,371

=3 , 753 m


e =

B2−a

=

8,02

−3 , 753=0 , 247 m <

B6 = 1,333 m


σ =

∑ Vbx . b y (1±6 . e

bx )

σmin =

83 , 3718,0 . (1 ) (1− 6 . (0,247 )

8,0 ) = 8,491 t/m2 > 0 (OK !)

σmax =

83 ,3718,0 . (1 ) (1+

6 . (0,247 )8,0 )

= 12,352 t/m2 < σ’= 26 t/m2 (OK !)

3. Keadaan Air Banjir dengan Uplift Pressure

ΣH = b(4) + c(4) + h(4)

= 12,536 + 1,6 + 0,899 = 15,035 t

ΣV = b(3) + c(3) + d(3) + f(3) – h(3)

= 11,943 + 3,6+ 70,207 + 3,51 – 16,002 = 73,258 t

ΣMr = b(5) + c(5) + d(5) + h(5)

= 72,363 + 23,53 + 276,379 + 1,33 = 373,602 tm

ΣM0 = b(6) + c(6) + f(6) + h(6)

= 58,506 + 6.672 + 13,820 + 67,61 = 146,608 tm

Kontrol :


SF =

∑ M r

∑ M 0 =

373 ,602146 ,608

=2,548 ≥ 1,50 (OK !)

78

Tugas Irigasi dan



SF =

f ∑V

∑ H =

0,7 . (73,258 )15 , 035

=3 ,411 ≥ 1,20 (OK !)




a =

∑ M r−∑ M 0

∑V =

373 , 602−146 , 60873 ,258

=3 ,099 m


e =

B2−a

=

8,02

−3 , 099=0 ,901 m <

B6 = 1,333 m


σ =

∑ Vbx . b y (1±6 . e

bx )

σmin =

73 , 2588,0 . (1 ) (1−

6 . (0,901 )8,0 )

= 2,969 t/m2 > 0 (OK !)

σmax =

73 , 2588,0 . (1 ) (1+

6 . (0,901 )8,0 )

= 15,345 t/m2 < σ’= 26 t/m2 (OK !)

4. Keadaan Air Banjir tanpa Uplift Pressure

ΣH = b(4) + c(4)

= 12,536 + 1,6 = 14,136 t

ΣV = b(3) + c(3) + d(3) + f(3)

= 11,943 + 3,6 + 70,207 + 3,51 = 89,26 t

ΣMr = b(5) + c(5) + d(5) + f(5)

= 72,363 + 23,53 + 276,379 + 13,82 = 386,092 tm

ΣM0 = b(6) + c(6)

= 58,506 + 6,672 = 65,178 tm

Kontrol :

79

Tugas Irigasi dan



SF =

∑ M r

∑ M 0 =

386 ,09265 , 178

=5 ,9240 ≥ 1,50 (OK !)


SF =

f ∑V

∑ H =

0,7 . (89,26 )14 ,136

=4 , 42 ≥ 1,20 (OK !)




a =

∑ M r−∑ M 0

∑V =

386 , 092−65 , 17889 , 26

=3 ,595 m


e = B2−a

= 8,02

−3 ,595=0 ,405 m <

B6 = 1,333 m


σ =

∑ Vbx . b y (1±6 . e

bx )

σmin =

89 , 268,0 . (1 ) (1− 6 . (0,405 )

8,0 )= 7,768 t/m2 > 0 (OK !!)

σmax =

89 , 268,0 . (1 ) (1+ 6 . (0,405 )

8,0 ) = 14,547 t/m2 < σ’= 26 t/m2 (OK !!)

Tabel 3.12 Akumulasi Kombinasi Gaya-Gaya yang Bekerja pada Tubuh Bendung

80

Tugas Irigasi dan


Guling Geser Max Min Max Min≥1,50 ≥1,20 < 20 t/m2 > 0 < 26 t/m2 > 0

a. Air Normal + gaya angkat 6,031 6,412 19,718 3,226 - -b. Air Banjir + gaya angkat 2,814 4,792 19,973 1,911 - -

a. Air Normal + gaya angkat 4,289 3,800 - - 22,154 1,074b. Air Normal 5,398 3,361 - - 14,001 5,950c. Air Banjir + gaya angkat 2,401 3,230 - - 24,881 0,556d. Air Banjir 4,502 2,837 - - 11,264 10,173

a. Air Normal + gaya angkat 19,342 4,845 - - 8,755 8,797b. Air Normal 8,817 6,079 - - 12,352 8,491c. Air Banjir + gaya angkat 2,548 3,411 - - 15,345 2,969d. Air Banjir 5,924 4,420 - - 14,547 7,768

1Tanpa Gempa

2

Dengan Gempa Horizontal

3

Dengan Gempa Vertikal

Tegangan TanahTanpa Gempa Dengan GempaKombinasi gaya – gaya pada tubuh

bendung

SF

81

bab iii analisa stabilitas bendung rev

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