MEKANIKA TANAH 1

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TEKANAN REMBESAN DAN UPLIFT (Pertemuan 12)

Disusun oleh:

Tim KBK Geoteknik Prodi Teknik Sipil FT UNS

Lab. Mekanika Tanah FT UNS, Jl Ir Sutami 36 a Surakarta

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TEKANAN REMBESAN

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A. TEKANAN REMBESAN

• Air keadaan statis  tekanan hidrostatis

• Air mengalir  tekanan partikel hidrodinamis  arahnya sesuai aliran

• Besarnya tekanan rembesan  f(i)

• Besarnya gaya rembesan pada pias (dP)

dP = g

_w

dh dA

• Besarnya gaya rembesan persatuan volume

D = g

_w

i

• Kondisi kritis D = g’= g

_w

i

_c

 g

_eff

= g ’ – D < 0

• Gradien hidroulik kritis i

_c

i £ i

_c

i

_c

= G

_s

- 1 SF

1 + e

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UPLIFT

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Uplift pressure on Hydraulic Structures

21 m

6 m

Impervious layer

k k

k

_x



_z



a d

30 m

42 m

a b c d e f

N_f = 2 N_d = 7

m 7 3

21 





N

d

h H

Point a 

p

_a

   21  6   3  g

_w

 24 g

_w

Point b 

p

_b

   21  6   2  3  g

_w

 21 g

_w

Point f 

p

_f

   21  6   6  3  g

_w

 9 g

_w

Method 1

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21 m

6 m

Impervious layer

k k

k

_x



_z



a d

30 m

42 m

a b c d e f

N_f = 2 N_d = 7

m 7 3

21 





N

d

h H

Total head at point a (from datum) 

21 18 m 7

6  



 H

N h n

d d a

Uplift pressure at point a, by using Bernoulli Equation 

Datum

n_d = 0

1 3 2

5 4 6

7



_a _a



_w _w

w

a

h z

p  g   g ( 18  (  6 ))  24 g

Method 2

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21 m

6 m

Impervious layer

k k

k

_x



_z



a d

30 m

42 m

a b c d e f

N_f = 2 N_d = 7

a b c d e f

kN/m2

24g_w

kN/m2

21g_w

kN/m2

18g_w

kN/m2

15g_w

kN/m2

12g_w

kN/m2

9g_w

42 m

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2. (a) Calculate the seepage under the hydraulic structure (dam) (b) Draw uplift pressure distribution at the bottom of the dam

det /

,

k k

k

_x _z

m 10

5

2 

^⁵



5 m

4 m 10 m

5 m

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9. REMBESAN PADA

BENDUNGAN

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Seepage through an Earth Dam

Methods to calculate the rate of flow (debit rembesan) 1. Dupuit ’ s solution

2. Schaffernak ’ s solution 3. Casagrande ’ s solution

Drawing the seepage line 1. Isotropic

2. Anisotropic

3. Inflow, outflow and transfer condition

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Impervious layer

H₁

H₂ x

z

A

B dx

dz

d

1. Dupuit ’ s solution

dx i dz

kiA q



   

¹

0 2

H

H d

dz kz dx

q 

2²



2

2 H

1

H d

q  k 

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Impervious layer

H

x

A z

B dx

dz

d

2. Schaffernak ’ s solution













sin 1

BD

tan a A

dx i dz

kiA

q  





^d

cos a H

sin a

dx a

dz

z sin tan



 tan sin

ka q 

D C

a



 tan sin

dx ka kz dz

q  

 

 





 

 

²₂ ₂²

sin cos

cos

H d

a d

(13)

Impervious layer

H

x

A z

B dx

dz

d

3. Casagrande ’ s solution













sin 1

BF

sin a A

ds i dz

kiA

q    



s

a H

sin a

dx a

dz

z sin

²

sin

²



 ka q

F C

a



sin

²





 ka

dx kz dz q

       

 d

²

H

²

d

²

H

²

ctg

²

a

ds A

’ D

E

0,3(AD)

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Graphical solution (Taylor, 1948) HCH 1-3353-p239

Impervious layer

H

x

A z

B

d

F C

 A

’ D

0,3(AD)

sin mH a





d/H

(degree)

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Impervious layer

Example: Calculate the rate of flow through an earth dam by using the methods: (a) Dupuit ; (b) Shaffernak and (c) Casagrande

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Drawing the seepage line

Impervious layer

+ x

z

E 1

n

x=d

F V

 B

D₁

z=H

0,3(AD)

D C

1 m

B₁

B₂

R

S A

H

p p

 

 d H d 

p 

²



²



1 2

p p x z

4

²

2





A

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Impervious layer

+ x

z

E 1

n

x=d

F V

 B

D₁

z=H

0,3(AD)

D C

1 m

B₁

B₂

R

S A

H

p p

a c a

a RF

RS 





 

30







A

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HCH MT1-3354-p.265

 a a  c

a    





 a sin Z

_s

parabolic parabolic

parabolic

Earth dam drainage

Earth dam

drainage

drainage

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30





 HCH MT1-3354-p.245

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(2)

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(5)

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Example

sec / mm 4 ,

 0

 _z

x k k

Draw flow net

AB = equipotential line AC = flow line

BD  using Casagrande’s solution

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 



^d ^H ^d



p  ²  ²  1 2

p p x z

4 4 ²

2 



(2)

(3)

) 16 , 3 ( 4

) 16 , 3 (

4 ²

2 

 z x

N_f = 3 N_d = 16

(5)

width dam

per

/day m

46267

³



d f

N kH N q

Drainage

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Seepage through anisotropic earth dam k x

x k

x z

t





 



 

 k

_x

k

_z



'

k  

d f

N H N ' k q 

Inflow, outflow, and transfer condition

HCH MT1-3357-p.253

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Filter

 

  D

¹⁵₈₅ _s^f

 4 ~ 5

D

 

  D

¹⁵₁₅ _s^f

 4 ~ 5

D

    D

⁵⁰₅₀ _s^f

 25

D

To prevent from piping

Hydraulic conductivity of the filter

is high enough for drainage

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Example

sec /

m 10

5 ,

4 

^⁸

x

 k

sec /

m 10

6 ,

1 

^⁸

z

 k

x , k x

x k

x z

t

  0 6



 



 

  2 , 7 10 m/sec

'  k

_x

 k

_z

 

^⁸

k

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DAFTAR PUSTAKA

• Fathani, T.F., Slide kuliah Mekanika Tanah

• Hardiyatmo. 2006. Mekanika Tanah 1. Gadjah

Mada University Press, Yogyakarta

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SEKIAN DAN TERIMAKASIH

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Tim KBK Geoteknik Prodi Teknik Sipil UNS