BAB III Stabilitas Bendung

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 =

Pa1 =

1 ton/m 3 1 2

2

.γair .h .

=

1 2

.1.3,85 . 2

= 7,411 ton

45

Tabel 3.1 Perhitungan 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

Pf 1 =

1 2

. γair . h

2

=

Pf 2 = b . h . γ air Pf 3 =  Pf 4 =

1 2

1 2

. γair . h

. γair . h

2

1 2

.1.3,85

2

= 7,411 ton

= 1,851.(3,85).(1) 2

=  =

1 2

1 2

.1.2,634 

.1.2,634 

2

2

= 7,126 ton = - 3,469 ton = 3,469 ton

46

Tabel 3.2 Perhitungan Tekanan Akibat Air Banjir Bagian

Berat (ton) V

H

Lengan (m) x

y

Momen ™

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

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/m 3) θ

= sudut gesek dalam

Ka

= tekanan lumpur aktif (0,333)

Gambar 3.3 Diagram tekanan akibat lumpur

47

PL1 = =

1 2 1 2

. K a .  lumpur . h2 .(0,333).(0,6).(3,85)2

= 1,481 ton

Tabel 3.3 Perhitugan Tekanan Akibat Lumpur Bagian

PL1

Gaya (t)

Momen (tm)

V

H

x

Y

Mr

M0

-

0,888

-

5,663

-

5,031

0,888

JUMLAH

3.3.

Lengan (m)

5,301

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

W2

=b.h.

 pasang an

= 1,649 . 3,848 . 2,2

= 13,960 ton

W3

=b.h.

 pasang an

= 1,0 . 4,0 . 2,2

= 8,800 ton

W4

=b.h.

 pasang an

= 1,5 . 3,5 . 2,2

= 11,550 ton

48

W5

=b.h.

= 2,5 . 3,0 . 2,2

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

 pasang an

= 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

3.4.

= 71,160 ton

ΣMo

= 272,588 tm

ΣMr

= 257,174 tm

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

49

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

 .Δ  =   Σ  .3,2  =   29,77  =   0,107  Keterangan : Hx = tinggi muka air dari titik yang dicari (m) Lx

= panjang rayapan (m)

ΣL = total rayapan (m)

50

ΔH = tinggi muka air normal dari lantai dasar bendung (m) Ux = uplift pressure di titik x (t/m 2)

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

51

 =  1+2  = 21+2 31+2

U1

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

U 1  U 2

=-

A

xH

2

1,380  4,768 2

3

x

= - 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

V= C

2.5000

U 1  U 2

xH

2

B

V =-

4,768  5.091

x

2

2,5

= -12,324 t B-C

5.0910

 h  2b  c  3  b  c

x =  

4.7680

=

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= H=

U 1  U 2

xH

2

5,091  3,350

x

2

3.3500

D

2,0

= 8,441 t 2.0000

C – D

 h  2c  d  3  c  d

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

U 1  U 2

D

V =D – E

3.5440

3.3500

xH

2

3,350  3,544 2

1,5 = - 5,170 t

x

 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

53

H =

H = 2.6730

F

y = E

3.5440

xH

2

3,544  2,673

1

x

2

= 3,109 t

1.0000

E – F

U 1  U 2

=

 h  2e  f    3  e  f 1  ( 2 x3,544)  2,673 





3  3,544  2,673 

= 0,523 m Ytotal = 0,523 + 1,0 = 1,523 m V =

U 1  U 2

1.0000

F

G

xH

2

V =-

2,673  2,803 2

1

x

= - 2,738 t F – G

2.6730

2.8030

x =

=

 h  2 f  g    3  f  g 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

H=

U 1  U 2

H =G

G-H

2.8030

xH

2

2,803  3,932 2

1,0

x

= - 3,367 t

 h  2 g  h  3  g  h

y = 

1.0000

H

3.9320

=

1,0  (2 x 2,803)  3,932 





3  2,803  3,932  = 0,472 m Ytotal = 0,472 + 1,0 = 1,472 m

54

V=

2.0000

U 1  U 2

H

I

xH

2

V =-

3,932  4,191

x

2

2

= - 8,122 t H-I

 h  2 g  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 H =

H = 2.4490

U 1  U 2

xH

2

4,191  2,449 2

J

x

2

= 6,640 t I-J

2.0000

 h  2 g  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

Mr

1.224

11.292

Mo

1.764

21.740

1.069

1.069

9.022

3.350 -5.170

E

y (total)

5.091 8.441

D

x (total)

Momen

4.768 -12.324

C

y

Lengan (m)

0.743

3.257

16.840

3.544

55

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 :

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

 .Δ=   .5,701=  0,192   =   Σ 29,77 Keterangan: Hx = tinggi muka air dari titik yang dicari (m) Lx = panjang rayapan (m) ΣL = total rayapan (m)

56

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 C D E F

10.08 10.08 8.08 8.08 7.08

26.77 24.27 22.27 20.27 19.77

4.995 5.433 3.816 4.104 3.295

G H I J

7.08 8.08 8.08 6.08

18.77 17.77 15.77 13.77

3.487 4.678 5.061 3.444

57

Tabel 3.9 Perhitungan Uplift Pressure Akibat Air Banjir Bagian

Gambar

Gaya angkat per 1 m panjang (t) U 1  U 2

H=

xH

2

1.3800

A

=-

1,380  4,955

x

2

3,0

= - 9,502 t

3.0000

 h  2a  b  3  a  b

A-B

y =  B

3,0  (2 x1,380)  4,955 

4.9550



=



3  1,380  4,955 

= 1,218 m Ytotal = 1,218 m U 1  U 2

V= B

C

xH

2

2.5000

V =-

4,955  5,433

x

2

2,5

= - 12,985 t B-C

 h  2b  c    3  b  c

4.9550

x =

5.4330

2,5  (2 x4,955)  5,433 



=



3  4,955  5,433 

= 1,231 m X total = 2,5 – 1,231 = 1,269 m U 1  U 2

H=

xH

2

3.8160

D

H =

5,433  3,816 2

x

2,0 = 9,250 t

2.0000

C-D

 h  2c  d  3  c  d

y=   C 5.4330

=

2,0  ( 2 x5,433)  3,816 



 = 1,058 m

3  5,433  3,816 

Ytotal = 1,058 m

58

V=

1.5000

E

U 1  U 2

xH

2

D

V =-

3,816  4,104

1,5

x

2

= - 5,940 t D-E

 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

H= H = 3.2950

F 1.0000

E-F

xH

2

4,104  3,295

1

x

2

= 3,699 t

 h  2e  f  3  e  f

y =  

E

4.1040

U 1  U 2

=

1  (2 x 4,104 )  3,295 





3  4,104  3,295 

= 0,518 m Ytotal = 0,518 + 1,0 = 1,518 m V= 1.0000

G

F

U 1  U 2

xH

2

V=-

3,295  3,487 2

1,0

x

= - 3,391 t x =

F-G 3.4870

3.2950

=

 h  2 f  g    3  f  g 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

H=

U 1  U 2

H =3.4870

G

G-H

xH

2

3,487  4,678

1,0

x

2

= - 4,082 t

 h  2 g  h  3  g  h

1.0000

y = 

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

U 1  U 2

xH

2

V =-

4,678  5,061

x

2

2,0

= - 9,739 t H-I

 h  2 g  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 H= 3.4440

J

H= 2.0000

I

xH

2

5,061  3,444 2

x

2,0

= 8,505 t

I-J 5.0610

U 1  U 2

 h  2 g  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

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

16.478 1.058

0.518

-4.082 H

11.572

Mo

4.104

-3.391 G

1.218 1.269

1.058

3.699 F

Mr

3.816 -5.940

E

y (total)

5.433 9.250

D

x (total)

Momen

4.955 -12.985

C

y

Lengan (m)

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 :

f u = koefisien reduksi untuk jenis tanah keras (50 %)

61

Tabel 3.11 Akumulasi Beban – Beban Pada Bendung No.

Keterangan

1

2

Gaya (ton) Vertikal Horizontal 3 4

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.

Momen (ton.meter) Mr Mo 5 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)

Σ = 272,226 =2,579 ≥1,5   = Σ 105,540 62

b. Terhadap Geser (Sliding)

.Σ = 0,7 . 56,983 =3,594 ≥1,2   = f Σ 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

272,226105,540 = 2,925   = ΣΣ = Σ 56,983 Resultante beban vertikal bekerja sejarak e dari pusat berat bendung

 = 2   = 72  2,925 = 0,575  Jarak e masih terletak didalam “Bidang Kern”

 < 6 →0,575< 76 0,575  < 1,167 

…….. (OK!)

Gambar 3.7 Jarak e Dalam Bidang Kern



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

σ = =

 V A



V

b x . b y

M . x

Iy 

 V . e . 0,5 . b x

1 12

3

. b x . b y

63

=

V

b x . b y



6.V.e 2

b x . b y

 V  6 . e  1   b x  b x . b y 

=

Tegangan izin tanah dasar (σ’) = 2,0 kg/cm 2 = 20 t/m 2 

Tegangan tanah dikontrol per 1 meter panjang bendung : σ =

σmin =

σmax =

 V  6 . e  1   b x . b y  b x 

56,983  6 . 0,575   1   = 4,130 t/m2 7,0 1  7,0 

> 0

56,983  6 . 0,575   1   = 12,151 t/m 2 < 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)

Σ = 282,091 =2,017 ≥1,5   = Σ 139,858 b. Terhadap Geser (Sliding)

.Σ = 0,7 . 58,602 =2,536 ≥1,2   = f Σ 16,177 f

= koefisien geser (diambil f= 0,7)

c. Terhadap Daya Dukung Tanah (Over Stressing) Resultante beban vertikal bekerja sejarak a dari titik O

,−,  = − =  ,

= 2,427 m

64

Resultante beban vertikal bekerja sejarak e dari pusat berat bendung

 =    =    <  →1,073< 

2,427 = 1,073

1,073  < 1,167 

Tegangan tanah dikontrol per 1 m panjang bendung:

 .−,    = ,  . 1 +   = 16,071 ⁄ < 20 ⁄ →

6 .1,073) = 0,673 ⁄ >0→  = 58,602 (1  7 .1 7 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/m 2 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: a.

Terhadap Guling (Over Turning)

 = , =2,050 ≥1,5   =  , b.

Terhadap Geser (Sliding)

. , =2,190 ≥1,2   = . = ,, f c.


Terhadap Daya Dukung Tanah (Over Stressing) Resultante beban vertikal bekerja sejarak a dari titik O

,−, = 2,4 47   = − =  , Resultante beban vertikal bekerja sejarak e dari pusat berat bendung

 =    =   2,447 = 1,053  65

 <  →1,053<  1,053  < 1,167  Tegangan tanah dikontrol per 1 m panjang bendung:

 .−, = 15,489⁄ < 26 ⁄ →  = , 1 +  .  6 .1,053) = 0,792 ⁄ >0→  = 56,983 (1  7 .1 7 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: a.


 = , =3,381 ≥1,5   =  , b.


. , =3,231 ≥1,2   = . = ,, f = koefisien geser (diambil f= 0,7) c.


,−, = 2,5 45   = − =  , Resultante beban vertikal bekerja sejarak e dari pusat berat bendung

 =    =   2,545 = 0,955   <  →0,955<  0,955  < 1,167 

66

Tegangan tanah dikontrol per 1 m panjang bendung:

6 .0,955) = 18,485⁄ < 26 ⁄ →  = 71,160 (1 + 7 .1 7 6 .0,955) = 1,846 ⁄ >0→  = 71,160 (1  7 .1 7 3. Keadaan Banjir dengan Uplift Pressure 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: a.


 = , =1,688 ≥1,5   =  , b.


. , =1,761 ≥1,2   = . = ,, f c.



,−, = 1,9 62   = − =  , Resultante beban vertikal bekerja sejarak e dari pusat berat bendung

 =    =   1,962 = 1,538   <  →1,538<  1,538  < 1,167  Tegangan tanah dikontrol per 1 m panjang bendung:

 ., = 19,408⁄ < 26 ⁄ →  = , 1 +  .   ., = 2,665 ⁄ >0→  = , 1   .  67

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: a.


 = , =2,752 ≥1,5   =  , b.


., =1,047 ≥1,2   = . = ,, f c.



,−, = 2,4 53   = − =  , Resultante beban vertikal bekerja sejarak e dari pusat berat bendung

 =    =   2,453 = 1,047   <  →1,047<  1,047 < 1,167  Tegangan tanah dikontrol per 1 m panjang bendung:

6 .1,047) = 20,233⁄ < 20 ⁄ →  = 74,629 (1 + 7 .1 7 6 .1,047) = 1,090 ⁄ >0→  = 74,629 (1  7 .1 7 3.7.3

Dengan Pengaruh Gempa Vertikal Tegangan ijin tanah (dengan gempa) σ’= 20 t/m2 x 1,3 = 26 t/m 2

68

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: a.


 = , =2,408 ≥1,5   =  , b.


.Σ = 0,7 . 60,541 =3,818 ≥1,2   = f Σ 11,100 f

c.



,−, = 2,7 53   = − =  , Resultante beban vertikal bekerja sejarak e dari pusat berat bendung


 .−, = 14,184 ⁄ < 20 ⁄ →  = , 1 +  .  6 .0,747) = 3,113 ⁄ >0→  = 60,541 (1  7 .1 7 2. Keadaan Normal tanpa Uplift Pressure 

ΣH = a(4) + c(4) = 7,411 + 0,888 = 8,300 t



ΣV = a(3) + c(3) + d(3) + f(3)

69

= 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: a.


 = , =5,534 ≥1,5   =  , b.


. , =6,3017 ≥1,2   = . = ,, f c.



,−, = 2,9 61   = − =  , Resultante beban vertikal bekerja sejarak e dari pusat berat bendung


6 .0,539) = 15,606⁄ < 20 ⁄ →  = 74,718 (1 + 7 .1 7 6 .0,539) = 5,742 ⁄ >0→  = 74,718 (1  7 .1 7 3. 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) + f(3) + h(3) = 3,469 + 0,000 + 71,160 + 3,558 – 16,027 = 62,160 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) + f(6) + h(6)

70

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


 = , =1,847 ≥1,5   =  , b.


. , =2,690 ≥1,2   = . = ,, f c.



,−, = 2,0 81   = − =  , Resultante beban vertikal bekerja sejarak e dari pusat berat bendung

 =    =   2,081 = 1,419   <  →1,419 <  1,419  < 1,167  Tegangan tanah dikontrol per 1 m panjang bendung:

 .,  = 19,678⁄ < 20 ⁄ →  = , 1 +  .  6 .1,419 ) = 1,918 ⁄ >0→  = 62,160 (1  7 .1 7 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: a.


 = , =3,536 ≥1,5   =  , 71

b.


., =4,471 ≥1,2   = . = ,, f c.



273,12677,233 = 2,505   = ΣΣ = Σ 78,187 Resultante beban vertikal bekerja sejarak e dari pusat berat bendung

 =    =   2,505 = 0,995   <  →0,995<  0,995 < 1,167  Tegangan tanah dikontrol per 1 m panjang bendung:

6 .0,995) = 20,691 ⁄ < 20 ⁄ →  = 78,187 (1 + 7 .1 7 6 .0,995) = 1,648 ⁄ >0→  = 78,187 (1  7 .1 7 Tabel 3.12 Akumulasi Kombinasi Gaya-Gaya Yang Bekerja Pada Tubuh Bendung Tegangan Tanah

SF No.

1

2

3

Kombinasi Gaya – Gaya pada Tubuh Bendung Tanpa Gempa a. Air Normal + Gaya Angkat b. Air Banjir + Gaya Angkat Dengan Gempa Horizontal a. Air Normal + Gaya Angkat b. Air Normal c. Air Banjir + Gaya Angkat d. Air Banjir Dengan Gempa Vertikal a. Air Normal + Gaya Angkat b. Air Normal c. Air Banjir + Gaya Angkat d. Air Banjir

Tanpa Gempa

Dengan Gempa

Guling ≥1,50

Geser ≥1,20

Max < 20 t/m2

Min >0

Max < 26 t/m2

Min >0

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

BAB III Stabilitas Bendung

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