SOIL PERMEABILITY AND SEEPAGE
4.21 PIPIN G FAILUR E
Soil Permeability an d Seepage 13 1
Procedures hav e therefor e bee n develope d t o sketc h th e downstrea m divergence a s explained below. A s show n i n Fig . 4.23(a), E i s th e poin t a t whic h th e basi c parabol a intersect s th e discharge face. Let the distance ED be designated as Aa and the distance DF as a. The values of Aa and a + Aa vary with the angle, j3, made by the discharge fac e with the horizontal measured clockwise. The angl e may vary from 30 ° to 180° . The discharge face is horizontal as shown in Fig. 4.22(e). Casagrande (1937) determined th e ratios of Aa / (a + Aa) for a number of discharge slopes varyin g from 30° to 180 ° an d the relationship is shown in a graphical form Fig. 4.23(b).
The distanc e ( a + Aa) can be determined by constructing the basic parabol a wit h F as the focus. Wit h th e know n ( a + Aa ) an d th e discharg e fac e angl e j3 , A a ca n b e determine d fro m Fig. 4.23(b). The point D may therefore be marked out at a distance of Aa from E. With the point D known, the divergence DS may be sketched by eye.
It should be noted that the discharge length a, is neither an equipotential nor a flow line, since it is at atmospheri c pressure . I t is a boundary along which the head a t any point i s equal to the elevation.
Analytical Solutions fo r Determinin g a an d q
Casagrande (1937) proposed th e following equation for determining a for j8 < 30°
(4.61) cos/? ^jcos 2/? sin 2/?
L. Casagrande (1932 ) gave the following equation for a when {$ lies between 30° and 90°.
(4.62) The discharg e q per unit length through any cross-section o f the dam may be expressed a s follows:
For/?<30°, a = fcasin/?tan/ ? (4.63 )
For30°</?<90°, a = fca sin2/? (4.64) .
(a)
1
D(b)
Sheet pile wall
D/2 d'
\c
Figure 4.24 Pipin g failure Factor o f Safet y Against Heave
The prism aocd in Fig. 4.24(b) subjected to the possible uplift has a depth of D and width D/2.
The average uplift pressure on the base of prism is equal to Ywha- The total uplift force per unit length of wall is
_i
The submerged weigh t of the prism aocd is 1
(4.65)
where yb is the submerged uni t weight of the material. The factor of safety with respect t o piping can therefore be expressed a s
F = (4.66)
Soil Permeabilit y an d Seepag e 133
If it is not economical t o drive the sheet piles deeply enough to prevent heave, the factor of safety ca n be increased b y placing a weighted filte r over the prism aocd a s shown by the pris m aa'd'd. If the weight of such a filter is W(, the new factor of safety can be written as
F =
u
(4.67)Filter Requirement s t o Contro l Pipin g
Filter drain s are require d o n the downstrea m sides o f hydraulic structures and aroun d drainage pipes. A properly grade d filter prevents the erosion o f soil in contact with it due to seepage forces.
To prevent the movement of erodible soils into or through filters, the pore spaces between the filter particles should be small enough to hold some of the protected material s in place. Taylor (1948 ) shows that if three perfect spheres have diameters greater than 6.5 times the diameter o f a small sphere, th e smal l sphere s ca n mov e throug h th e large r a s show n i n Fig . 4.25(a). Soil s an d aggregates are always composed of ranges of particle sizes, and if pore spaces in filters are small enough to hold the 85 per cent size (D85) of the protected soil in place, the finer particles will also be held in place as exhibited schematically in Fig. 4.25(b).
The requirement s o f a filte r t o kee p th e protecte d soi l particle s fro m invadin g th e filte r significantly ar e based o n particle size . These requirement s were developed fro m test s by Terzaghi which were later extended by the U.S. Army Corps of Engineers (1953). The resulting filter specifica- tions relate the grading of the protective filter to that of the soil being protected b y the following;
D
5 filter <4 . s filter ^50 filter
85 soil D15 soil D <25
50 soil (4.68)
(a) Size of smallest spherical particl e which just fits the space between large r spheres Soil which has migrated into
filter and is held by D85 size soil particles
(b) Condition of the boundary betwee n protecte d soil and the filter materia l Figure 4.2 5 Requirement s o f a filte r
10 1.0
Grain size D mm 0.1 = 0.015 mm0.01
Figure 4.26 Grai n size distribution curves for grade d filter and protected material s
The criteria ma y be explained a s follows:
1. Th e 1 5 per cent size (D15) o f filter material must be less than 4 times the 85 per cent size (D85) of a protected soil. The ratio of D15 of a filter to D85 of a soil is called the piping ratio.
2. Th e 1 5 per cent size (D15) o f a filter material should be at least 4 times the 1 5 per cent size (D]5) of a protected soi l but not more than 20 times of the latter.
3. Th e 50 per cent size (D5Q) o f filter material should be less than 25 times the 50 per cent size (D50) of protected soil .
Experience indicate s that if the basic filter criteria mentioned above are satisfied in every part of a filter, piping cannot occur under even extremely sever e conditions.
A typica l grai n siz e distributio n curve o f a protecte d soi l an d th e limitin g size s o f filte r materials fo r constructing a graded filter is given in Fig. 4.26. Th e size of filter materials mus t fal l within the two curves C2 and C3 to satisfy th e requirements.
Example 4.1 6
Fig. Ex . 4.1 6 give s th e sectio n o f a homogeneou s da m wit h a hydrauli c conductivit y k = 7.87 4 x 10"5 in/sec. Draw the phreatic line and compute the seepage loss per foot length of the dam.
Soil Permeabilit y and Seepage 135 13.12ft
d = 68.9 ft Figure Ex. 4.1 6 Solution
The depth h of water on upstream side = 32.81 ft.
The projected length of slope A 'B on the water surface = 32.81 ft.
The point A on the water level is a point on the basic parabola. Therefore AA' = 0.3x32.81=9.84 ft.
F is the focus of the parabola. The distance of the directrix from th e focus F is v0 = 4d2 +h2 - d
where d = 68. 9 ft, h = 32.81 ft. Therefore y0 = V(68.9)2+(32.81)2-68.9 = 7.413 ft
The distance of the vertex of the parabola from F is
FV = a - . =
0 2 2 = 3J06 ft
The (jc , y) coordinates of the basic parabola may be obtained from Eq. (4.58) as
2yQv 2x7.413 14.83
Given below ar e values of y for various values of x
jt(ft) 0 15 30 45 68.9
y(ft) 7.41 6 16.6 5 22.3 6 26.8 8 32.8 1
The parabola has been constructed with the above coordinates as shown in Fig. Ex. 4.16.
From Fig. Ex. 4.16 A a + a = 24.6 ft From Fig. 4.23, for a slope angle )3 = 45°
-^--035
a + Aaor Aa = 0.35 ( a + Aa) = 0.35 x 24.6 = 8.61 ft From Eq . (4.60)
q = kyQ
where k = 7.874 x 10~5 in/sec or 6.56 x 10"6 ft/sec an d yQ = 7.413 f t q = 6.56 x 10-6 x 7.413 = 48.63 x 10"6 ft3/sec pe r ft length of dam.
Example 4.1 7
An earth dam which is anisotropic is given in Fig. Ex . 4.17(a). The hydraulic conductivities kx and kz i n th e horizonta l an d vertica l direction s ar e respectivel y 4. 5 x 10~8 m/ s an d 1. 6 x 10~8 m/s . Construct th e flo w ne t an d determine th e quantity of seepage through the dam. What i s the por e pressure a t point PI
Solution
The transforme d sectio n i s obtained b y multiplyin g the horizonta l distance s b y ^Jkz I kx an d b y keeping th e vertical dimensions unaltered. Fig. Ex . 4.17(a) is a natural section of the dam. The scale factor for transformation i n the horizontal directio n i s
Scale factor = P - = JL6xl°"8B = 0.6 ]kx V4.5X10- 8
The transforme d sectio n o f th e da m i s give n i n Fig. Ex. 4.17(b) . The isotropi c equivalen t coefficient o f permeability is
k =e
Confocal parabolas can be constructed with the focus of the parabola a t A. The basic parabol a passes throug h point G such that
GC=0.3 H C =0.3x2 7 = 8.10m The coordinates of G are:
x = +40.80 m, z = +18.0 m
72- 4 f l2
As per Eq. (4.58) x = 9 . (a )
Substituting for x and z, we get, 40.8 0 = Simplifying w e have, 4a2 + 163.2aQ - 324 = 0 Solving, aQ = 1.9 m
Substituting for aQ in Eq. (a) above, we can write
Soil Permeability and Seepage 137 15.0m,
h = 18.0m Blanket drain
B S
(b) Transformed section
Figure Ex. 4.17
Z2-14.4
7.6 (b)
By using Eq. (b), the coordinates o f a number of points on the basic parabola may be calculated . 7(m) -1. 9 0. 0 I o 10. 0 20. 0 30. 0
z(m) 0. 0 3. 8 7.2 4 9.5 1 12. 9 15.5 7 The basic parabola is shown in Fig. Ex. 4.17(b).
The flowne t i s complete d b y makin g the entry corrections b y ensurin g that the potential drops ar e equal between the successive equipotentia l lines at the top seepage lin e level .
As per Fig. Ex. 4.17(b), there are 3.8 flow channels and 18 equipotential drops. The seepag e per unit length of dam is
Nf 3 8
-^- = (2.7xlO-8)xl8x—= lxlO-7 m3/s
N 18
The quantity of seepage across section Az can also be calculated without the flownet by using Eq. (4.60 )
q = k^Q = 2keaQ = 2x2.7 x 1Q-8 x 1. 9 « 1 x 10~7 m3/sec per meter Pore pressur e at P
Let RS be the equipotential line passing through P. The number of equipotential drops up to point P equals 2.4
Total head loss = h = 18m, number of drops =18
Head loss per drop = — = 1 m.18 18
Therefore th e head at point P = 18 - 2.4(A/z ) = 18 - 2.4(1 ) = 15. 6 m
Assuming the base o f the dam as the datum, the elevation head of point P = 5.50 m.
Therefore th e pressure head at P = 15.6 - 5. 5 = 10.1 m.
The pore pressure at P is, therefore, uw = 10.1 x 9.81 = 99 kN/m2
Example 4.18
A sheet pile wall was driven across a river to a depth of 6 m below the river bed. It retains a head of water of 12.0 m. The soil below the river bed is silty sand and extends up to a depth of 12.0 m where it meets an impermeable stratum of clay. Flow net analysis gave A/,= 6 and Nd - 12 . The hydraulic conductivity of the sub-soil is k = 8 x 10~5 m /min. The average uplift pressure head ha at the bottom of the pile is 3.5 m. The saturate d unit weight of the soil ysat = 19. 5 kN/m3. Determine:
(a) The seepage less per meter length of pile per day.
(b) The factor of safety against heave on the downstream side of the pile.
Solution (a) Seepage loss,
The loss of head h = 12 m
Nf 6
q = kh—^- = (8x 10~5)x 12x — = 48x 10~5 m3/min = 69.12x 10~2 m3/day
Nd 1 2
(b) The Fs a s per Eq. (4.67) is (Ref. Fig 4.24 )
F W » + W> D7 » V h Jw ha = 3.5 m
Yb = Xsat -Yw= 19-5-9.8 1 = 9.69 kN/m3 6 x 9.69