CE404 Seepage Theories

Seepage and Uplift Pressure

Hydraulic Structures CE404http://hyd.uod.ac/

Faculty of Engineering and Applied ScienceSchool of Engineering

Civil Engineering Department

A number of methods are available to analysis the problemofseepage and uplift pressures but the most useful and easily adopted are:

1. Flow nets (graphical or experimental).

2. Bligh's Creep Theory.

3. Lane's weighted Creep Theory.

4. Khosla's Method

Seepage and Uplift Pressure

Flow Nets

In any hydraulic structure on permeable foundations, water flow from a region of high level (high pressure) to the region of low level (low pressure), beneath and around the structure.

High Pressure Low

pressure

In Fig. (a), a hydraulic structure having an upstreampiled cut off wall only, thestreamlines are compressed around the toes of the pile. In this case, the uplift pressureunderneath the floor varies from15 to 28% of H.

Flow Nets

In Fig (b) the structure having downstreamcutoff wall only, the uplift

pressure varies from70 to 85% of H.

Flow Nets

In Fig. c the structure having upstreamand downstreamcutoffs the upliftpressure varies from45 to 55% of H.

Flow Nets

Bligh's Creep Theory

According to Bligh’s theory, water creeps along the bottom contour of the structure.

(CREEP PATH)

Bligh's Creep Theory

The length of the path of water is called the length of creep and the loss of head is proportional to the length of creep.

Bligh's Creep Theory

If HL is the total head loss between upstream and downstream and L is the length of the creep, then the loss of head per unit of creep length (i.e. ) is called the hydraulic gradient. Bligh’s theory makes no discrimination between horizontal and vertical creeps.

Bligh's Creep Theory

Consider a section as shown in the figure below. LetH be the difference of waterlevels between upstreamand downstreamends (no water is shown in thedownstreamend). Water starts percolating atA and emerges atB.

Bligh's Creep Theory

Total creep length (L)

( ) ( )1 1 2 2 3

1 2 1 2 3

2 2 2

2

d L d L d

L L d d d

= + + + += + + + +

Bligh's Creep Theory

Head loss per unit length (hydraulic gradient)

( )1 2 32

H H

L b d d d= =

+ + + (1.1)

Bligh's Creep Theory

Hydraulic gradient drop at upstream cutoff

1

1

2

2

CH H

dH H H

L

Hd

L

= −

= − −

=

( )

( ) ( )

2 3

322 3

2

22

E

HH L d

LdL H H

H H L dL L L L

= +

= + = +

Safety against Piping or Undermining

Safety against piping can be ensured by providing sufficient creep length given by

L C H=

where,

C = Bligh’s coefficient for the soil.

1H L C=

Safety against Piping or Undermining

No. Type of soil Value of C Safe exit gradient less than

1 Fine sand 15 1/15 2 Coarse grained sand 12 1/12 3 Sand mixed with boulders and gravel 5 to 9 1/5 to 1/9 4 Light sand and mud 8 1/8

Hydraulic gradient 1H L C< for safety against piping.

Bligh’s coefficient for different types of soil

Safety against Uplift Pressure

If the uplift head at any point is H1 (meter of water) then uplift head has to be counterbalanced by the weight of floor thickness.

Uplift pressure 1w Hγ= , wγ = Unit weight of water gρ= .

Downward pressure( )w c c w cG t tγ γ= − , where cG is the specific gravity

of the floor material.

Safety against Uplift Pressure

For equilibrium,

( )( ) ( )

1

1 1

w w c c w c

c c c c c

H G t t

H G t t t G

γ γ γ= −

= − = −

Example

Find the hydraulic gradient and uplift pressure at a point 15 m from the upstream end of the floor in the figure below.

Example

Water percolates at point A and emerges at point B,

Total creep length 2 6 10 2 3 20 2 8 64m= × + + × + + × =

Head of water on structure= 6 m

Hydraulic gradient6 1

64 10.66= =

Example

According to Bligh’s theory, the structure would be safe on sand mixed with boulders

No. Type of soil Value of C Safe exit gradient less than

1 Fine sand 15 1/15 2 Coarse grained sand 12 1/12 3 Sand mixed with boulders and gravel 5 to 9 1/5 to 1/9 4 Light sand and mud 8 1/8

Hydraulic gradient 1H L C< for safety against piping.

Example

Creep length up to point C= 1 2 6 2 3 15 33L m= × + × + =

( )664 33 2.91

64CH m= − =

1

2.912.076 of concrete

2.4 1

C w Cc

c w c w

H Ht

G G

m

γγ γ

= =− −

= =−

Lane’s Weighted Creep Theory

From the analysis of 200 dams all over the world, Lane’s concluded that horizontal creep is less effective in reducing uplift than vertical creep. Therefore, he suggested a factor of 1/3 for horizontal creep against 1 for the vertical creep.

Lane’s Weighted Creep Theory

For the structure in the figure

( ) ( )

( )

1 1 2 2 3

1 2 1 2 3

1 2 3

1 12 2 2

3 31

23

23

Horizonals 3 Verticals

L d L d L d

L L d d d

bd d d

L

= + + + +

= + + + +

= + + +

= +∑ ∑

Lane’s Weighted Creep Theory

L C H>

Hydraulic gradient H

Lshould be less than

1

C

1H

L C<

Slopes steeper than 45º are taken as verticals.

Lane’s Weighted Creep Theory

Lane’s coefficient for different types of soil

Example

Find C for the following structure & the uplift pressure at point A.

H.W.Consider C = 5, Check if the following structure is safe using Lane'smethod. If it is safe find the thickness at A.

The floor thickness t to resist uplift is: ( )( )wc

wu yPt

γγγ

−−

= …………………………………………………….. (4)

Where: Pu = uplift pressure in meters of water. y = depth of water on floor (m) γc = density of concrete. γw = density of water.

Khosla's Method

It is used to find uplift pressure at the key points in a barrage or a weir. Inthis method a composite barrage or weir section is split up into a number ofsimple standard forms of known analytical solution, these are:

a – A straight horizontal floor of negligible thickness witha sheet pile at eitherend.

b – A straight horizontal floor depressed belowthe bed but with novertical cut off.

c – A straight horizontal floor of negligible thickness with sheet pile atsome intermediate position.

Ex: find the pressure at the key points for the structure below:

Sol.

5 m

10 m

w.s.

50 m

D

C E

Ex: find the pressure at the key points for the structure below:

φE1 = 100% of H

2.050

101 ===b

d

α

φ D = 26% of H φ D1 = 100 – 26 = 74% of H φ E = 40% φ C1 = 100 – 40 = 60% of H

Ex: find the pressure at the key points for the structure below:

Sol:

25.020

51 ===b

d

α

φ D' = 18% of H φ D'1 = 100 – 18 = 82% of H

Ex: Find the pressure percentage for the intermediate pile shown in the figurebelow:

45.2

10 ===b

dα

4.010

41 ==b

b

φ C = 42% of H To find φE

Read φC for the base ratio )1( 1

b

b− for the value of α and subtract

from 100.

φ C for )1( 1

b

b− = 0.6 and α = 4 = 29% of H

φ E = 100 – 29 = 71% of H

How to find φ D

To get value φ D for values of b

b1 less than 0.5.

b

b1 = 0.4

Read φ D (for )1( 1

b

b− = 0.6 and α = 4) = 44.8% of H

φ D = 100 – 44.8 = 55.2% of H

The percentage pressure observed from the curves for the simple form into which the profile has been Brocken up is valid for the profile as a whole if corrected for:-

Correction for interference of piles Cp

Cp = correction in a percentage.

b' = the distance between two piles.

b = total floor length

d = depth of pile on which the effect is to be taken.

D = depth of the pile line, the influence of which has to be determined on the

neighboring pile of depth of is to be measured bellow the level.

This correction is positive for points in the rear of back waterand subtractive for points forward in the direction of flow.

Effective of d.s. pile on u.s. pile (+ve).

Effective of u.s. pile on d.s. pile (–ve).

2 – Correction for the floor thickness Ct

The thickness of floor is assumed to be negligible and the pressure isfound at point E, C fromthe curves. The pressure at point E', C' are interpolatedby assuming straight line variation.

The correction either to be positive or negative.At point E is negative while it is positive at C.

3 – Correction due to slope CsC

b

bC S

S ′±=

Slop V: H C 1:1 11.2 1:2 6.5 1:3 4.5 1:4 3.3 1:5 2.8 1:6 2.5 1:7 2.3 1:8 1

The correction being plus for the downstream slope and minus for the upslope following the direction of water.

b'

bs

Example:Determine the correction percentage pressure at the key points.

Solution 1 – Upstream Pile b = 57 m d = 150 – 145 = 5 m

087.057

51 ===b

d

α

φ D1 = 100 – φ D = 100 – 18 = 82% of H φ C = 100 – φ E = 100 – 27 = 73% of H

2 – Intermediate Pile

b1 = 15.5 + 0.75 = 16.25 m b = 57 m d =150 – 145 = 5 =m

4.115

57===b

dα

285.057

25.161 ==b

b

For α = 11.4 and b

b1 =0.285; φ C1=58% of H

For α =11.4 and (1 – b

b1 =0.7155); φ D1=100 – φ D = 100 – 36 = 64% of H

For α =11.4 and (1 – b

b1 =0.7155); φ E1=100 – φ C = 100 – 30 = 70% of H

Correction for E1

a – ( ) %2.1)6470(5

111 −=−−=−−= DEt d

tC φφ

b – %35.157

44

5.15

419 −=

+−=pC

φ E1 corrected = 70 – 1.2 – 1.35 = 67.45% of H

Correction at C1

a – ( ) %2.1)5864(5

1

5

111 +=−−=−+= CDtC φφ

b – D = 149 – 141 = 8 m

%78.157

48

40

819 +=

++=pC

c – Cb

bC S

S ′−= %325.05.6*

40

2 −=−−=

φ C1 corrected = 58 + 1.2 + 1.78 – 0.325 = 60.65% of H

3 – Downstream Pile

b

d=α1

d = 147 – 141 = 6 m b = 57 m

105.057

61 ==α

For b

d=α1 =0.105, φ E2 = 29% of H

φ D2 = 20% of H φ C2 = 0% of H

Correction for E2

a – ( ) %25.2)2029(5

5.1

6

5.14514722 −=−−=−−−= DEtC φφ

b – %18.057

5.04

40

5.019 −=

+−=pC

φ E2 corrected = 29 – 2.25 – 0.18 = 26.57% of H

φE φD φC φE1 φD1 φC1 φE2 φD2 φC2 96.4 82 76.15 67.45 64 60.65 26.57 20 5

Correction for φC2= ( )226

5.1CDtC φφ −+=

( ) %50206

5.1 +=−+= Of H