127 CHAPTER 5 EARTH DAMS
5.1
INTRODUCTION Earth dams for the storage of water for irrigation have been built since the earliest times. These dams were however, limited in height but not necessarily in extent. Earth dams are now being built to unprecedented heights. Sites which have hitherto been considered unfit for the construction of darns are now being exploited. Development of soil mechanics, study of behavior of earth dams, and the development of better construction techniques have all been helpful in creating confidence to build higher dams with improved designs and more ingenious details. The result is that the highest dam in the world today is an earth dam. The highest earth/rockfill dams in the world are Roguni U.S.S.R ( 335 m ) Nurek, U.S.S.R. ( 300 m ); (Fig. 5.1) Mica, Tehri India ( 260 m ) Canada ( 244 m ) and Oroville, U.S.A. ( 235 m ).
Fig. 5.1 Nurek dam U.S.S.R. In spite of these developments it is difficult to establish mathematical solutions to the problems of design, and many of its components are still guided by experience or judgment. For a realistic design of an earth dam it. is necessary that the foundation conditions and materials of construction are thoroughly investigated. It is also necessary that controlled methods of construction are used to achieve necessary degree of compaction at predetermined moisture. The discussion in the text is limited to design procedure for earth dams which are rolled fill type of construction. This type of construction is now being used almost entirely for the construction of earth dams to the exclusion of hydraulic and semi-hydraulic fills. In this type, the major portion of the embankment is constructed in successive mechanically compacted layers of 150 mm to 220 mm thickness.
128 5.2
Foundation for Earth Dams
The essential requirements of a foundation for an. earth dam are (i) that it provides stable support for the embankment under all conditions of saturation and loading, and (ii) that it provides sufficient resistance to seepage to prevent piping and excessive loss of water. In general foundations may be grouped into three main classes according to their predominant characteristics. 1. Rock foundations, 2. Pervious foundations, and 3. Impervious foundations. Rock Foundations These foundations, including shale generally do not present any problem of bearing strength. The principal considerations are erosive leakage, excessive loss of water through joints, fissures, crevices, permeable strata and along fault planes, etc. Grouting is usually done to treat this type of foundation. Shale may however present foundation problems specially if they contain joints, faults, seams filled with soft material and weak layers. Such defects and excess pore water pressure may control the overall strength of foundation.
FOUNDATION FOR EARTH DAMS Pervious Foundations Often the foundations for earth dams consist of recent alluvial deposits composed of relatively pervious sand and gravels overlying impervious geological formations like rock or clay. There are two basic problems with which these types of foundations are associated viz. (i) excessive amount of under seepage, and (ii) piping and boils caused by forces exerted due to seepage. The treatment which may be provided to control these problems is governed by the thickness of pervious strata viz, whether the pervious foundation extend to a moderate depth or to an infinite depth. Loose fine sand or coarse silt deposits in a foundation present one of the most difficult problems. The difficulty arises not only due to low strength or high compressibility of the loose sand, but also through a phenomenon known as liquefaction. A certain fine uniform sand in a loose condition when subjected to sudden applications of shock (as in earth quake) loses all its shear strength and behaves as though it were a heavy viscous fluid. This phenomenon is exhibited by uniform sands which are very fine and consists of rounded grains and their relative density is less than 50%. Fortunately no failures of this type have occurred. At Obra dam in U.P., the foundations were thick loose sands and its susceptibility to liquefaction was carefully evaluated by field and laboratory investigations of the foundation material and by actual blasting tests. It was concluded that the foundations were not likely to liquefy.
129 Impervious Foundations Foundations of silt and clay extending to large depths are sufficiently impermeable to preclude the necessity of providing treatment for under seepage and piping. The main problem with these, foundations may be excessive pore water pressure and significant deformations. Where the embankments are constructed on foundations consisting of brittle, highly plastic or over-consolidated clays, serious investigations are required as their presence may cause excessive deformations. The embankment design in such cases would be controlled by likely strains in the foundations. If there is silt and clay to large depths, then there is not much necessity of providing treatment for under seepage and piping. The main problem with these foundations is of stability for which generally the slopes of the embankments are made flatter or berms on either side are provided. If the structure crosses swampy or similar area where the foundation material will be of plastic nature, the matter would require serious investigations as plastic clays are very deficient in shear strength. An approximate method to determine the safety of foundation material against the shear stress is as follows : (i) Determine the total horizontal shear under the slope of dam by the formula, φ h 12 − h 22 S= γ tan 2 45 − 1 2 2 where, h1
=
h2 γ φ1
= = =
tan φ1 =
(5.1)
vertical distance from top of dam down to the rigid boundary such as rock, the strength of which is great as compared to the overlying material. vertical distance from base of dam to the rigid boundary. effective weight per cubic metre of material. equivalent angle of internal friction given by,
C + γ h 1 tan φ γ h1
(ii) Calculate the average unit shear by the formula Sa = S/b where Sa = average horizontal foundation shear per sq. m. b = horizontal distance along base from top shoulder of slope to the toe of dam.
Fig. 5.2 Location of maximum Shear-Definition sketch
(5.2)
130 (iii) The maximum unit shear may be obtained by multiplying the average unit stress by 1.4 i.e. Smax =1.4 Sa. The location of the point of maximum shear may be taken 0.4 b from upper shoulder of the slope as shown in fig. 5.2. 5.3
Causes of Failures of Earth Dams
Like most other damages to engineering structures, earth dam failures are caused by improper design frequently based on insufficient investigations, and lack of care in construction and maintenance. Failures of earth darns may be grouped into the following basic causes (a) Hydraulic failures (b) Seepage failures and (c) Structural failures Hydraulic Failures They account for about one third of the failure of dams and are produced by surface erosion of the dam by water. They include wash-outs from overtopping (Fig. l-5-3a), wave erosion of upstream face, scour from the discharge of the spillway etc. and erosion from rainfall. Seepage Failures Seepage of water through the foundation or embankment has been responsible for more than one third of earth dam failures. Seepage is inevitable in all earth dams and ordinarily it does no harm. Uncontrolled seepage; may however, cause erosion within the embankment or in the foundation which may lead to piping (Fig. 5.3 b). Piping is the progressive erosion which develops through under the dam. It begins at a point of concentrated seepage where the gradients are sufficiently high to produce erosive velocities. If forces resisting erosion i.e. cohesion, inter-locking effect, weight of soil particles, action of downstream filter etc. are less than those’ which tend to cause, the soil particles are washed away causing piping failure. Seepage failures are generally caused by (a) pervious foundations, (b) leakage through embankments, (c) conduit leakage and (d) sloughing. Pervious Foundations Presence of strata and lenses of sand or gravel of high permeability or cavities and fissures in the foundation may permit concentrated flow of water from the reservoir causing piping. Presence of buried channels under the seat of dam have also been responsible for this type of failure.
131 Leakage Through Embankrnents The following are the common causes of embankment leaks which lead to piping: (i) Poor construction control which includes insufficient compaction adjacent to outlet conduits and poor bond between embankment and the foundation or between the successive layers of the embankment. (ii) Cracking in the embankment or in the conduits caused by foundation settlement (Fig. 5.3 c), (iii) Animal burrows (iv) Shrinkage and dry cracks (Fig. 5.3 d) (v) Presence of roots, pockets of gravel or boulders in the embankment. Conduit Leakage Conduits through the dam have been responsible for nearly one third of the seepage failure and more than one eighth of all failures. Failures are of two types (i) contact seepage along the outside of the conduit which develops into piping and (ii) seepage through leaks in the conduit which may also develop into piping. Contact seepage along the conduit wall is caused either by a zone of poorly compacted soil or small gap between the conduit and remainder of the embankment. Seepage through poorly compacted zones soon develops into piping. Conduit cracking is caused by differential settlement or by overloading from embankment. Sloughing Failure due to sloughing takes place where downstream portion of the dam becomes saturated either due to choking of filter toe drain, or due to the presence of highly pervious layer in the body of the dam. The process begins when a small amount of material at the downstream toe is eroded and produces a small slide. It leaves a relatively steep face which becomes saturated by seepage from the reservoir and slumps again, forming a higher and more unstable face. This process is continued until the remaining portion of the dam is too thin to withstand the water pressure and complete failure occurs. Structural Failures Structural failures of the embankment or its foundation account for about one fifth of the total number of failures. Structural failures may result in slides in foundation or embankment due to various causes as explained below. Foundation Failures (Fig. 5.3 e). Faults and seams of weathered rocks, shale, soft clay strata are responsible for the foundation failure in which the top of the embankment cracks and subsides and the lower slope moves outward and large mud waves are formed beyond the toe. Another form of foundation failure occurs because of excessive pore water pressure in confined seams of silt or sand. Pore water pressure in the confined cohesionless seams, artesian pressure in the abutments or consolidation of clays interbedded with the sands or silt, reduces
132 the strength of the soil to the extent that it may not be able to resist the shear stresses induced by the embankment. The movement develops very rapidly without warning. Excess settlement of foundation may also cause cracking of the embankment (Fig.5.3 c).
Fig. 5.3 Types of failures—Earth dams Slides in Embankment (Fig 5.3 f and g) An embankment is subjected to shear stresses imposed by pool fluctuations, seepage or earthquake forces. Embankment slides may occur when the slopes are too steep for the shear strength of the embankment material to resist the stresses imposed. Usually the movement develops slowly and is preceded by cracks on the top or the slope near the top. Failure of this type are usually due to faulty design and construction. In high dams slope failure may occur during dissipation of pore pressure just after construction. The upstream slope failure may occur due to sudden drawdown as shown in fig. 5.3 f. The downstream slope is critical during steady seepage condition. 5.4 Design Criteria of Earth Dams Based on the experience of failures, the following main design criteria may be laid down for the safety of an earth dam. 1. To prevent hydraulic failures the dam must be so designed that erosion of the embankment is prevented. This implies that the following conditions are satisfied. (a) Spillway capacity is sufficient to pass the peak flow. (b) Overtopping by wave action at maximum water level is prevented. (c) The original height of structure is sufficient to maintain the minimum safe freeboard after settlement has occurred. (d) Erosion of the embankment due to wave action and surface run-off does not occur. (e) The crest should be wide enough to withstand wave action and earthquake shock.
133 2. To prevent the seepage failures, the flow of water through the body of the dam and its foundation must not be sufficiently large in quantity to defeat the purpose of the structure nor at a pressure sufficiently high to cause piping. This implies that (a) (b) (c)
(d)
Quantity of seepage water through the dam section and foundation should be limited. The seepage line should be well within the downstream face of the dam to prevent sloughing. Seepage water through the dam or foundation should not remove any particle or in other words cause piping. The driving force depends upon the pressure gradient while the resisting force depends upon the strength characteristics of the boundary material. There should not be any leakage of water from the upstream to downstream face. Such leakage may occur through conduits, at joints between earth and concrete sections or through holes made by aquatic animals.
3. To prevent structural failures, the embankment and its foundation must be stable under all conditions. This implies that (i) The upstream and downstream slopes of the embankment should be stable under all loading conditions to which they may be subjected including earthquake. (ii) The foundation shear stresses should be within the permissible limits of shear strength of the material. 5.5 Prevention of Erosion-Embankment Details Spillway Capacity It must be calculated and fixed by relevant hydrological studies and flood routing such that sufficient freeboard is available between the maximum flood level and top of the dam. Freeboard for wave action The required allowance for waves is based on the effect of wind of maximum velocity blowing down the reservoir and setting up a wave splash on the dam face. Various empirical formulae depending on wind velocity and reservoir fetch have been suggested for computing wave heights. The MolitorStevenson formulas are normally used which are hw = 0.032
FV + 0.763 − 0.271 4 F where F< 32 km.
hw = 0.032
FV , where F > 32 km.
(5.4) (5.5)
134 where hw = wave height in meters measured between trough and crest. V = wind velocity in km per hour F = fetch in kilometers. On a sloping surface the wave rides along the slope upto a vertical height of 1.5 times the wave height above the reservoir level hence 1.5 hw is provided as freeboard. According to U.S.B.R. criteria distinction is to be made between the normal and minimum freeboards. Normal freeboard is defined as the difference in elevation between the crest of the dam and normal reservoir water surface. According to the fetch of reservoir the freeboard may be provided as given in Table 5.1. Table 5.1 Recommended Values of Freeboards Fetch in km
Less than
Normal freeboard in meter 1.5 1.5 4.0 8.0 15.0
1.25 1.50 1.80 2.50 3.00
Minimum freeboard in meter 1.00 1.25 1.50 1.80 2.20
It is also recommended that freeboard shown in, Table 5.1 be increased by 50 percent if a smooth pavement is provided as protection on the upstream slope. Settlement allowance Settlement of an embankment will be caused by consolidation in the foundation and in the fill. A settlement allowance of 2% is considered adequate and is generally provided. However, in case of dams of more than 30m height, an extra 1% allowance is provided to account for the settlement due to earthquake. Upstream slope protection Surface protection of upstream slope is meant to prevent the destructive wave action. Usual type of surface protection for the upstream slope is stone rip-rap either dry dumped or hand placed. When a thin layer is used, hand placed rip-rap may be more economical than dumped rip-rap. There are several empirical methods to find out the thickness of rip-rap. These methods take into account the wave height, embankment slope, weight of average size rock and its specific gravity.
135 Hand placed rip-rap The hand placed rip-rap for upstream protection is very common in this country. The size of stones used for hand placed rip-rap may be determined with the following formula,
dm = 2.23 C. hw
where
γw 1+ s2 . γ − γ w s ( s + 2)
(5.6)
dm = diameter of stone brought to form a ball, in metre, in the zone of maximum blow of the wave. γw = unit weight of water in t/m3 γ = unit weight of stones in t/m3 s = slope of embankment. hw = height of wave in metre. C = factor depending on the type of protection
For hand placed rip-rap C = 0.54 For rock fill or dumped rip-rap C = 0.80 Average size of stone required dav = dm/O.85 (average shape) The average weight of the stone can be found out by the formula
Wav =
π (d av ) 3 .γ 6
(5.7)
Filter below rip-rap A layer of filter material consisting of gravel Or crushed rock is always required under rip-rap to prevent waves from eroding the underlying embankment material. No definite rule can be given for the minimum necessary thickness of the filter layer. Most filters are constructed with thickness ranging from 20 cm to 75 cm. The following factors govern the thickness of the filter layer (i) Wave action The less the wave action, the less the need for a thick filter under the rip-rap. (ii)
Gradation of rip-rap If the rip-rap is well graded with plenty of quarry fines to fill the larger voids there is less stress on the filter.
(iii) Plasticity and gradation of embankment materials If the embankment material is well graded granular soil with a tough clay hinder it needs less protection against erosion than if it is fine silty sand.
136 (iv)
The cost of filter If the material for the filter is obtained without washing or screening as a pit run natural gravel and quarry wash, and is not expensive as compared to the cost of average embankment material, there is no reason not to use maximum filter thickness. The following limits are recommended to satisfy stability criteria for graded uniform filters:
(1)
D15 of filter = 5 to 40, provided that the filter does not contain more than 5 D15 of base material percent of material finer than 0.075 mm ( I.S. sieve no.-minus 75 micron ).
(2) (3)
D15 of filter D 85 of base
= 5 or less
D 85 of filter = 2 or more Max. opening of pipe drain
(4) The grain size curve of filter should be roughly parallel to that of the base material. D15 or D85 is the size at which 15 percent and 85 percent (respectively) of the total soil particles are smaller. The percentage is by weight as determined by mechanical analysis. If more than one filter layer is required; the same criterion is followed; the finer filter is considered as the base material for selection of the coarser filter. Horizontal filter layers can safely be made thinner than steeply inclined or vertical filters. Minimum thickness of each horizontal layer is 15 cms for sand and 30 cms for gravel. If the D of filter filter contains excessive fines or coarse material such that 15 > 4 but < 5 or D15 of base
D15 of filter > 5 but < 6, the thickness of filter layer may be increased by 50 percent. D15 of base Design Criteria The following criteria of upstream protection is being adopted by Central Design Directorate of Irrigation Department, Uttar Pradesh. (i) Upto 5 m height of bound No pitching (ii) Between 5 m and 10 m 0.25 m stone pitching over 0.15 m graded shingle or spalls. (iii) Above l0m and upto 15m 0.30 m stone pitching over 0.15 m graded shingle or spalls. (iv) Above 15 m and upto 25 m 0.5 m stone pitching over 0.25 m graded shingle or spalls. (v) Above 25 m and upto 50m 0.5 m stone pitching over 0.3 m grade shingle or spalls. (vi) Above 50 m and upto 75 0.75 m stone pitching over 0.5 m graded shingle or spalls. (vii) Above 75 m 1.0 m stone pitching over 0.75 m graded shingle. The entire upstream face should be pitched right upto the top of bund.
137 Downstream Slope Protection The problem of erosion of downstream slopes due to surface runoff may be effectively controlled by turfing. In area too deficient in rainfall during parts of the year to maintain a proper cover, berms and other erosion control be applied. Crest Width The crest or top width of an earth dam should not be less than 4 m for maintenance purposes. However, the width depends on several considerations such as (i) nature of embankment material and minimum allowable percolation distance through the embankment at normal reservoir water level (ii) height and importance of structure (iii) required width to provide embankment mass for resistance to earthquake shock and (iv) roadway requirements. Common practice is fairly well represented by the formula, Bt = 5/3.
H
5.6
Seepage Through Dams Phreatic or Seepage line The two dimensional flow of fluid through porous soil can be expressed by Laplace’s equation
∂ 2φ ∂ 2φ + =0 ∂x 2 ∂z 2 Graphically, the equation can be represented by two sets of curves that intersect at right angles. The combined representation of two sets of lines is called a flow net (Fig. 5.4). With the help of a flow net, the seepage problems can be analyzed at any point within the section of the embankment. The seepage or phreatic line may be defined as the line within a dam section below which there are positive hydrostatic pressures in the dam. On the line itself, the hydrostatic pressure is zero. Above the line, there will be a zone of capillary situation. The phreatic line represents the top flow line or the boundary condition for drawing the flow net.
Fig. 5.4
Flow net through earth dam
The location of the phreatic line is necessary in order to draw accurately the flow net. It is also useful in analyzing stability of the dam. It may be noted that the location of the seepage line is dependent only on the cross section of the dam. Its position is not influenced by the permeability of the material composing the dam so long as the material is homogeneous.
138 According to A. Cassagrande the phreatic line for the homogeneous fill section is a basic parabola except at the ingress and egress points. The presence of a pervious stratum below the dam does not influence the position of the phreatic line. For graphical construction of the phreatic line the following procedure may be adopted. Stepwise procedure for locating phreatic line 1. With horizontal drainage filter (Fig.5.5) (i) The horizontal distance between upstream toe A and the point ‘B’ where water surface meets the upstream face is calculated or measured (say L). The point B0 is then located on the water surface at a distance 0.3 L from B. (ii) The basic parabola has to pass through B0 and have its focus at F which is the starting point of the horizontal
Fig. 5.5 Phreatic line with horizontal filter drainage. With these points known the basic parabola may be constructed graphically. (iii) With centre B0 and radius B0F, draw an arc to meet the water line at C. Draw the vertical line CD which is the directrix. Let FD, the focal distance = yo. Bisect the distance FD to get the point E, the vertex of the parabola. Draw FG parallel to CD and equal to yo. Knowing Bo,G and E the basic parabola can be drawn. The focal distance y0 can also be determined on the consideration that if (x y) is one point on the parabola,
x 2 + y 2 = x+yo
(5.9)
Since the point B0 of coordinates d, h lies on the equation. yo = d 2 + h 2 − d (5.10) (iv) The ingress portion of phreatic line is joined to the base parabola from point B, keeping the starting end normal to the upstream face. 2. With inclined discharge face (Fig. 5.6) For embankments with no drainage measures the base parabola cuts the discharge face at point G0 at a distance (a + ∆a) along the discharge face from point F, and extends beyond the limits of the embankments. The actual seepage line meets the discharge face (at point G) at a distance a below the point G0. The value of ‘a’ can be worked out from the curve (5.7) after A.
139 ∆a Cassagrande giving the value of a + ∆a
for different angles (α) of discharging face. The
value of (a+ ∆a) can either be measured directly on the face when the parabola has been drawn or its value determined from the equation, yo (5.11) a + ∆a = 1 - cos α 3. With rock toe (Fig. 5.8) The basic parabola may be drawn in a similar way taking F
Fig. 5.6 Phreatic line with inclined discharge face
as focus. As already shown this parabola it self is the seepage line for a horizontal filter. (For a horizontal filter α = 180°). For a rock toe, an appropriate value of α, measured clockwise from the horizontal base ∆a should be taken and the value of a + ∆a is corrected at the egress point.
Fig. 5.7
read from the curve given in fig. 5.7. The parabola
Fig. 5.8
For values of α between 30° to 180°, the distance ‘a’ measured along the slope from the toe may be determined as explained above. However, if it is less than 30°, the distance ‘a’ or the point of emergence of the phreatic line at the downstream slope may be determined with the help of Schaffernak’s equation.
140 d d2 h − − 2 cosα cos sin 2 α
(5.12)
a = d 2 + h 2 − d 2 − h 2 cot 2 α
(5.13)
a=
or
where d and h are the coordinates of the initial point B, as explained in fig. 5.5 Quantity of Seepage Consider earth embankment of homogeneous material given in fig. 5.4. Flow net through the dam section has been drawn by trial and error method. If h is the total hydraulic head and Nd is the number of potential drops (9 in fig. 5.4) the potential drop ∆h = h Consider a field of length l, the field being an approximate square its width is also equal Nd ∆h . l The discharge through the field is given by h ∆h ∆q = K . .l = K∆h = K l Nd
to l, the hydraulic gradient across the field
If Nf is the total number of flow channels ( Nf = 3 in fig. 5.4 ) the seepage per unit width of embankment is q = Kh
Nf
(5.14)
Nd
The discharge through a homogeneous embankment with horizontal filter may also be calculated with the help of equation 5.9. The equation of base parabola under steady seepage condition may be written as y=
2.x. y 0 + y 02
For a unit length, y represents the area of flow. dy q = K . .y dx dy d = 2 xy 0 + y 02 dx dx
[
]
1/ 2
= y 0 / 2 xy 0 + y 02
Since discharge passing through any vertical plane is the same, at x = 0 we have
dy = 1 and dx
y = y0. ∴q = K y0 (5.15) The equation although is applicable to embankments with horizontal filters but hold good for determining approximate discharge in all other cases.
141 Example 5.1 For the homogeneous dam section shown in fig. 5.9, draw the pheratic line if a horizontal filter of length 10 m is provided. Also determined discharge per m length of dam if K 10-5 m / sec. Solution The angle α of discharge face = tan-1/1.5 = 33.69' From Fig. 5.7,
∆a = 0.35 a + ∆a
Fig. 5.9 Design example 15.2 — Dam section h = 22, L = 22 x 2.5 = 55, 0.3 L = 16.5m d = l6.5 + (2.5 + 1.5) 3 + 6 + 1.5 x 22 -10 = 57.5 m. Equation 5.10
y0 = d 2 + h 2 − d
= 57.5 2 + 22 2 − 57.5 = 4.06 Equation of the parabola is
x 2 + y 2 = x + 4.06 with the help of this equation, the pheratic line can be drawn. For example at x = 10, y = 9.88 at x = 20, y = 13.37 Discharge per metre length q = K yo ∴ q = 4.06 x 10-5 m2 / sec.
142 Flow net in Anisotropic Soil Natural deposits made by flowing water and rolled earth dam sections are horizontally stratified. Such stratifications display much higher horizontal permeability (Kh) than vertical permeability (Kv). When the horizontal and vertical permeability differ in a dam section, the flow net may be drawn in the usual manner in a transformed section. To make a transformed section the horizontal dimensions are multiplied by
Kv while vertical dimensions remain unchanged. Kh
After drawing the flow net in the transformed section, the flow net is restored back in the original section. The procedure is illustrated in .Fig. 5.10 in case when the horizontal permeability is 4 times the vertical permeability. It may be noted that while the fields are approximate squares in the transformed section, they get distorted when restored back in the original section. The higher the ratio between the horizontal and vertical permeability the more would be the distortion in the flow net. The flow lines tend to come near the downstream face of the dam, and for a certain ratio the flow lines may even intersect the section making it unsafe. The discharges, seepage force, pore water pressure can be determined with help of the flow net in the original section.
Fig 5.10 Transformation of section having different permeability in horizontal and vertical directions. The quantity of seepage in an anisotropic section can also be determined by using an effective permeability coefficient (K') in equation 5.15, K' is given by K' = 5.7
Kv Kh
(5.16)
Control of Seepage Through Foundations Different methods to control seepage of water through foundations are described below. The suitability of the method of treatment depends primarily on the nature of foundation. 1. Grouting and grout curtain 2. Cut off trenches. 3. Partial cut-off
143 4. 5. 6. 7. 8.
Sheet piling cutoff Cement bound curtain cutoff Cast in situ concrete diaphragm Upstream blanket Pressure relief wells.
Grouting and grout curtain Certain materials when infected as grout in the foundation strata acts as a binder and fills the voids, thus reducing the permeability and increasing its stability. Cement, asphalt, clay and various chemicals are used as grouting materials. The choice of grouting material, pattern, depth and sequence of grouting are related to the foundation conditions, type and height of the dam and objective desired. Cement grouting is extensively’ used in rock foundations. In pervious foundations the choice of a suitable grout depends primarily upon the grain size of the material and its permeability. Table 5.2. indicates approximate range of grain sizes that can normally be grouted by different types of grout materials and mixtures. Table 5.2 Type of grout
Cement Clay-cement bentomite Clay-chemical, bentomite-chemical Chemical
Diameter of the material that can be grouted mm 0.5 0.3 0.2 0.1
to to to to
1.4 0.5 0.4 0.2
Blanket grouting is done to a depth of 5 to 10 m through holes spaced 3 to 5 m to prevent piping. Curtain grouting is done to much greater depth to reduce seepage through foundation. The number of lines and spacing of holes depends upon the nature of foundation and width of grout curtain which is usually 1/3 to 1/5 of the water head. A single line of grout holes has been adopted in a number of dams. It is however desirable to have two lines of holes. The pervious zones should be grouted first with coarse grouts with holes spaced apart followed by finer grouts in further stages with holes spaced closely. Cut off trenches (Fig. 5.11 a) The cut-off trenches with side sloping or vertical are excavated below the dams and filled with well compacted impervious material. These trenches are located well upstream of the centre line of the dam but within a point where the cover of the impervious embankment above the trenches is sufficient to resist the percolation at least to the extent offered by the trench itself. The centre line of the trench is kept parallel to the centre line of the dam. The trench should be provided up to bed rock or other impervious strata. The bottom width of the
144 trench is governed by the space required for treatment of foundation and type of equipment used for rolling. Usually minimum 5 m width is adopted. Other method of cut-off may be economical as compared to deep cutoff trenches. The maximum depth of trench is governed by economical considerations. For moderate pervious foundations positive cut-off up to hard stratum is provided. The cut-off may be (i) sheet pile (ii) cement bound curtain, (iii) concrete diaphragm. Partial cut-off trench (Fig. 5.11 c) A partial cut off trench is effective in stratified foundation by intersecting more impervious layer in the foundation and by increasing the vertical path of seepage. In a uniform pervious strata the role of partial cut off in reducing percolation is very limited. A cut off going to 80% of the total depth of pervious
strata reduces the seepage discharge by only 50%. Thus with a partial cut off the reliance is primarily on the length of the seepage path. Therefore for treatment of deep pervious foundation where it is not economically possible to provide a positive cut-off, partial cut-off along with upstream blanket is provided to reduce the discharge and seepage pressures.
145 Sheet piling cut-off ( Fig. 5.11 b ) Steel sheet piling cut off can be used in silty, sandy and fine
Fig 5.11 Typical foundation treatment and trenches details
gravel foundations. If the foundation strata contains boulders the sheet piles will not easily penetrate. It is very difficult to make sheet pile cut off water tight. There are always chances of leakage through the joints and at contact with bed rock. These disadvantages and limitations of sheet pile cut off have been overcome by use of cellular sheet pile wall back filled with concrete. Another method of sealing the interlocks is to drill arrow of boles and fill them with bentonite before driving the sheet pipe line across the holes. Cement bound curtain cut off This type of cut off is used in pervious foundations which do not contain large cobbles and boulders. The grout is pumped through hollow rotating drill rod which is fixed with a mixing head on the end. The grout is forced down with vanes provided for in the mixing head which results in the formation of cylindrical elements of cement impregnated sand and gravel. A continuous curtain is formed by successive over lapping of such cylinders.
146 Cast in-situ concrete diaphragm In this process a trench 5 to 10 m long and up to 1.2 m wide is made by a percussion tool of the required diameter connected to a rigid tube through which drilling mud is circulated from the bottom to the top. The mud retains the walls of the trench by hydrostatic pressure as .well as by forming a cake of bentonite along the side of the trench. The excavating tool and the pipes travel horizontally to and fro over the length of the panels. The vertical percussion motion of the tool along with the horizontal movement of the assembly excavates a trench equal in length to the horizontal movement and the excavated material is sucked out through the rigid tube along with the drilling mud. This drilling mud is then led into large tanks where the excavated material settles down. The mud is replanished with bentonite and is returned to the trench. After excavating the trench to the full depth, it is backfilled by tremic concrete. Alternate panels are first completed, and the intervening panels are subsequently excavated and backfilled with concrete. This process is suitable for making of cut-offs in sand material and it is for the first time that this process has been adopted in Obra dam in India (Fig. 5.12)
Fig. 5.12 Cross section of Obra Dam U.P.
Fig. 5.13 Upstream blanket Upstream Blanket (i) Advantages An impervious clay blanket placed upstream of a dam and connected to the impervious section is a convenient way of effecting moderate reduction in the amount of seepage (Fig. 5.13). The quantity of seepage is some what less than inversely proportional to the total length of impervious material. The effectiveness of a blanket depends upon the proportionate increase in number of equipotential drop that results from its addition to the dam. A blanket is advantageous only when an appreciable length of the path of flow can be affected by a blanket of reasonable length.
147 (ii)
Length of blanket Usually the material of the blanket is so tight in relation to pervious stratum that it is not necessary to consider flow through the blanket in determining the blanket length. The length of the blanket is given by (Fig. 5.13). K .h.d − p.q.b L= (5.17) pq in which
L = K= h = d = p =
length of upstream blanket in metres. mean horizontal permeability coefficient of the pervious stratum. gross head in metres on impervious upstream blanket. depth of pervious stratum in metres percentage (stated as a decimal) of flow under the dam without a blanket, to which it is desired to reduce the seepage by means of the blanket. For example, if the seepage is required to be reduced to 25% of its original value, then p = 0.25. b = length of impervious portion of base of dam in metres. q = flow under dam, without a blanket, per metre of dam K. (h/b). d.
(iii)
Thickness of blanket The thickness of the blanket is a function of the relative permeability of the blanket, permeability of the foundation material and its depth. Theoretically it should vary in thickness from zero at its upper edge to the maximum where it joins the impervious section of the dam. The thickness of blanket ‘t’ at distance ‘h’ from upstream toe of blanket may be determined from the equation, t = ( K2/K1 ) x b x ( B/d ) (5.18) where t = thickness of blanket at point under consideration, b = distance from point under consideration to upstream toe of blanket, K1 = average permeability of stratum, K2 = permeability of blanket, B = length of blanket from upstream toe to impervious section d = depth of pervious stratum in meter. For normal condition the thickness of upstream blanket is kept between 1.5 to 3.0 m and length about 10 times the head of ponded water. In case of fine sand or silty foundations, the length of blanket is kept 15 times the head.
Pressure Relief Wells The primary purpose of relief wells is to reduce artesian pressures which otherwise would cause formation of sand boils and piping. Relief wells also intercept and provide controlled outlets for seepage which otherwise would emerge uncontrolled downstream of the dam. Theoretically, piping occurs when the uplift pressure at a point at some level in the foundation near downstream toe reaches, is greater than the combined weight of soil and water
148 above it. If the thickness of impervious layer is equal to the reservoir head, h1, the uplift pressure beneath the layer cannot exceed the weight of layer, because the saturated weight of soils equals approximately twice of water. Thus, if the thickness of the top impervious structure is equal to the reservoir head, there is no danger against piping. In this situation no further treatment of foundation is required. If the thickness of the top impervious stratum is less than reservoir head, h, but is too thick for treatment by drainage trenches or if the pervious foundation is stratified, pressure relief wells are required. Relief wells should be designed to penetrate into the principal pervious strata to obtain efficient pressure relief, especially where the foundation is stratified. Whereas full penetration is desirable in case of shallow depth of pervious aquifers (6 m to 9 meter thickness), at least 0.50 percent penetration should be ensured into thicker aquifers. Generally depths of wells equal to the height of the dam is satisfactory. Wells should be spaced sufficiently close together (generally 15 metre apart) to intercept seepage and reduce uplift pressures between wells. The wells must offer little resistance to infiltrator of seepage and discharge there of. The wells must be so designed that they do not become ineffective due to clogging or corrosion. Where no control measures are present, relief wells should be so designed that pressure gradients between wells or downstream from the wells do not exceed 0.5 to 0.6. Where downstream berms are provided pressure gradient between wells should not exceed 0.6 to 0.7. The slots in the well screen must have adequate area and at the same time be of such size as to prevent movement of filter through the screen after development of the well and shall satisfy the following criteria. D85 filter ≥ 1.2 slotwidth MinD85 filter ≥ 1.0 or Holediameter The gradation of the filter must also comply with the following criteria ( Max) D85 filter ≤ 5.0 ( Min) D85 sand and
( Min) D15 filter ≤ 4.0 ( Max) D15 sand
The well screen consists of G.I. pipe 10 cm to 15 cm diameter slotted with 4.75 mm to 6 mm by 50 mm size opening covering about 10 percent of the circumferential area of the pipe. Vertical slotting is preferable to horizontal slotting. The pipe should be coated with anticorrosive paint or regularized after slotting. A type design of pressure relief well is illustrated in fig. 5.14.
149
Fig. 5.14 Type design of relief well
In fig. 5.11 appropriate treatments for the following foundation conditions are illustrated. 1. Pervious foundation for shallow depth. 2. pervious foundation extending to a moderate depth. 3. Pervious foundation extending to great depth. 4. Thin impervious layer underlain by pervious foundation. 5. Impervious layer of thickness greater than 3 m and less than the height of the dam. 5.8
Drainage In Earth Dams Drainage in earth dams is primarily provided to bring the phreatic line well within the downstream face. A proper drainage system also helps in avoiding heaving and piping by arresting the soil particles, which may otherwise move by seepage discharge. The drainage system also reduces the pore water pressure in the downstream portion of the dam and thus the stability of the downstream face is increased. Design of drainage arrangement is governed mainly by the height of the dam, availability of pervious material, and the permeability of the foundation. Different types of arrangements are shown in fig. 5.15 and short description of each follows.
150 Horizontal drainage blanket The horizontal blanket is provided over that portion of the foundation downstream from the impervious zone of the dam where high upward seepage forces exist. The blanket must be pervious to drain off effectively and its design should fulfill the usual filter criteria, in order to prevent the movement of particles of the foundation or embankment by seepage discharge. They may be provided on homogeneous pervious foundations overlain by thin impervious layers, in order to stabilize the foundation and relieve pressures that may break through the impervious layer. The requirement of the length of blanket may be determined theoretically by means of flow net, provided the horizontal and vertical permeability of the foundation are known. Generally a length three times the height of dam is sufficient. The horizontal blanket is generally combined with a rock toe (Fig. 5.15 a). Chimney Drains The main disadvantage of the horizontal drainage blankets is to make the earth dam embankment stratified and consequently more pervious in the horizontal direction. The inclined or vertical chimney drains are thus provided in many homogeneous dams to intercept seepage water before it reaches the downstream slope. A number of dams with chimney arrangements have recently been constructed in this country.( Fig. 5.15)
Fig.5.15 Types of drainage arrangements
Rock Toe The rock toe consists of sizes usually varying from 15 cm to 20 cm ( Fig. 5.16 ). Transition filter is required to be provided between the homogeneous fill and the rock toe. The filter usually comprises
151 30 cm thick layer of fine sand (15% size = 0.09 mm) 45 cm thick layer of coarse sand (15% size 0.81 mm) 60 cm thick layer of gravel (15% size = 7.3 mm).
Fig. 5.16 Rock toe and filter
The capacity of the filter should as a rule be kept twice the discharge calculated by the Darcy’s formula. The minimum required height of rock toe is governed by two factors (a) The minimum allowable cover on the phreatic line. (b) Downstream water level. A minimum cover of 1 m between the seepage line and downstream slope is considered adequate. The height of rock toe is generally kept between 1/3 to 1/4 the height of the dam. The top of rock toe must be sufficiently higher than the tail water level to prevent any wave splashes on the downstream face. Sometimes it may not be economical to raise the level of rock toe for this purpose. In such cases stone pitching extending to a minimum vertical height of 1.5 m above the downstream-H.F.L. is provided in continuation of the rock toe. Generally the discharge face (inward face) of rock toe is kept at slope of 1:1 and the outer face is a continuation of the downstream slope. In case of high dams, considerable economy can be affected by providing a berm 3 m to 6 m wide at top of the rock toe and reducing its outer slope to 1.5:1 to 2:1. Toe Drains and Drainage Trenches The purpose of these drains is to collect the seepage from the horizontal blanket which discharges into the spillway stilling basin or into the river channel below the dam. Toe drains are also used on impervious foundations to ensure that any seepage that may come through the foundation or the embankment is collected and the ground water is kept below the surface to avoid the creation of boggy areas below the dam. The toe drains ( Fig. 5.17 ) may be of vitrified clay or perforated asphalt dipped corrugated metal pipe. Drainage trenches can be used to control seepage where the top stratum is thin and pervious foundation is shallow so that the trench can be built to penetrate the aquifer substantially. Where the pervious foundation is deep, a drainage trench of any reasonable depth would attract only a small portion of under
152 seepage, its effect would be local and detrimental because the under seepage would bypass the trench. A typical design of drainage trench is shown in fig. 5.18
Fig 5.17 Typical toe drain
Fig. 5.18 Details of drainage trench The filter comprising the drainage layers should ‘be designed in accordance with the filter criteria. 5.9 Stability of Slopes Methods of Investigating Stability of Slopes Every soil mass which has slope at its end is subjected to shear stresses on internal surfaces in the soil mass, near the slope. This is due to the force of gravity which tries to pull down the portions of’ the soil mass, adjoining the slope. If, however, the shearing resistance of the soil is greater than the shearing stress induced along the most severely stressed or critical internal surface, the slope will remain stable; and if on the other hand the shearing resistance of the soil, at any time after the construction of the slope becomes less than the induced shearing stress, the portion of soil mass between the slope and the critical internal surface will slide down along this surface, until, the new slope formed by the sliding mass makes the shearing stress less than the shearing strength of the soil. (The stability of slopes of earth-structures thus depends on the shear resistance or strength of the soil.) The well known method of investigating stability of slope is Swedish method, devised by Swedish Engineers in 1922 which is simple, and therefore, is more commonly used. In this method, the curved slip surface is taken to be an arc of circle with a certain centre. There will be a number of such likely slip circles’ with their respective centres. It is necessary to pick up the most dangerous of critical slip circle i.e. that circle along which the
153 soil has the least shear resistance. The centre of this circle is located by trial and error. Below are shown methods of locating the critical slip circle in the case of various types of soils, assuming that the soil slope is of homogeneous structure, i.e., it consists of one type of soil only. (i)
Cohesive soil Let PHV ( Fig.5.19 ) be any one slip circle with center O1, x is the distance of the centroid G of area DVHP from the centre O1. The position of G can be obtained in the same manner as the centroid of an irregular plane area. Actuating moment which may = W.x kg metres cause a slip along surface PHV
Fig. 5.19
Slip circle method
where, W = weight of soil mass of area DVHP and length one metre. For cohesive soil φ = 0 and, its shearing resistance depends on cohesion only and is the same along the entire surface VHP. The maximum resisting moment mobilized by the soil to prevent the likely Occurrence of a slip. = cohesive shear resistance developed along VHP = (L x 1) x c x R kg metres = cLR kg metres where L = length of arc VHP in meters c = unit cohesion in kg/m2 R = radius of likely slip circle in metres But, L = R α where α is the angle of arc VHP, in radians ∴ Maximum resisting moment = c R α R kg metre = c R2 α kg metre
154 Factor of safety against slipping = along surface VHP
Maximum resisting moment actuating moment
cR 2α W .x Various other slip circles like the above said one, e.g. QH1V, SH2V, TH3V, etc. are drawn to scale and the factor of safety is found, as shown above, for each such slip circle. The slip circle giving the minimum factor of safety is the critical slip circle, because, the failure, if it occurs, will occur along this, slip circle. This critical slip circle is located as shown below. The values of the factor of safety worked out for various slip circles say four, are plotted as shown in fig. 5.19, and a curve is drawn passing through the tops of the ordinates representing the four values of the factors of safety corresponding to four slip circles. The lowest point on this curve is noted and a vertical line is drawn through it, meeting the top of soil mass at a point say Z; V and Z then mark the two points through which the required critical slip circle will pass. =
c- φ Soil The shear strength of such a soil due to cohesion and internal friction varies from point to point along the slip circle. The equation for shear strength of this soil is, S = c + pn tan φ, kg/m2 where, c is constant along the slip circle but pn which is the pressure normal to the slip circle due to weight of soil varies from point to point along the slip circle. The method of locating the critical slip circle is as follows: The whole soil mass adjoining the slope, is divided into vertical strips or slices as shown in fig. 5.20 (i). The fifth strip from the left in the figure it taken for the study and is shown on a bigger scale in fig. Fig. 5.20 (ii). Taking one running metre of the strip, its equilibrium is practically due to two forces only viz; its weight W5 and its shearing strength S5 along the curved surface of length l5 in at the bottom of the strip. The weight W5 is equivalent to two forces, (ii)
Fig. 5.20 Swedish method for c - φ soil
155 namely, Wn5 which is normal to the curved surface and, the tangential force Wt5 which is tangential to the curved surface as shown in fig. 5.20. We will, therefore have the equation, S5 = ( c. l5 + Wn5 tan φ ) kg where, c.l5 = cohesive strength per metre length of the strip and Wn5 tan φ = frictional strength per metre length of the strip. The moment of this shear strength is = R. S5, kg metres = R ( c.15 + Wn5 tan φ ) kg metres and the moment of shear strength along the entire slip surface VHP = Σ R ( c.15 + Wn5 tan φ ) = Σ R ( ct5 + W n5 tan φ ) = Σ R (c.L.+ tan φ Wn5) kg.m where, L = length of the whole slip surface VHP. The actuating moment along slip surface at bottom of the strip R.. Wt5 kg metres ∴ The actuating moment along the entire slip surface PHV = Σ R Wt5 kg metres The factor of safety, against shear = resisting moment failure, along slip surface PHV actuating moment c.L + tan φ ∑ Wn or, in general, the factor of safety = ∑ Wt The weight of each strip one metre long, is proportional to its area and this latter can be found approximately by the trapezoidal formula, or, accurately by a planimeter. Wn and Wt of each strip can be found graphically by drawing the triangle of forces for each strip as shown in fig. 5.20 (i). The factor of safety is found for each of other slip circles and the critical circle is located as usual. It should be carefully noted, that, the tangential components of the weights of the first few strips near the toe of the slope, will resist the slipping tendency and these components must therefore, be taken with their proper sign. Location of critical slip surface The location of critical slip surface by trial and error entails much loss of time. To save time in locating the centre of critical slip circle for homogeneous sections, Fellenius construction, for locating the line on which the centre of the critical circle is most likely to lie, is generally adopted. Various circles with center on this line are tried until the one with minimum factor of safety is found. According to the Fellenius construction, the position of the line on which the centre of the critical circle lies depends only
156
Fig . 5.21
Fellenius method of locating line of center for critical slip circle
on the height and slopes of the embankment. Referring to fig. 5.21, the values of α , and β are taken from Table 5.3 for the embankment slope. From point A and B these angles are drawn to meet at point 01. Point 02 is then located at horizontal distance of 4.5 H, starting directly below point A, and at a depth equal to H, the height of the embankment. The line joining 01 and 02 points is the line on which the centre of the critical circle lies. The maximum depth to which the rupture can occur is limited by the presence of hard stratum underneath. Table 5.3 Recommended values of α and β- Fellenius construction Slope hor: vertical
α
β
1:1 2:1.
27.5° 25°
37° 32°
3:1 4:1 5:1
25° 25° 25°
35° 36° 37°
Pore Water Pressure When moist soil mass is loaded without permitting air/or water to escape, part of the load causes the soil grains to deform elastically or to undergo non-elastic rearrangement but without significant change in their solid volume. This part of the load is carried on the soil skeleton as effective stress. The remaining portion of the load is carried by stress in the air and water contained in the voids and is known as pore-water pressure. The pore-water pressures below the phreatic line reduces the shearing strength of the soil mass in accordance with Coulomb’s equation. S = c + ( pn – u ) tan φ
157 The practice of accounting for pore pressures in drawdown and steady-seepage conditions is different from that of end of construction conditions. Pore Pressures in drawdown and steady conditions: Pore pressures on any slice of a failure are taken to act normally on the arc surface and are equivalent to the weight of water transmitted by that slice. Consider a slice ( Fig. 5.22 ) below phreatic line and above dead storage level or tail water level. If h is the average height and d its width, force due to pore pressure. u = h.w x arc length = h x w x d sec θ. The resisting force due to pore pressure will, be reduced to u tan φ. = ( h d.w. sec θ ) tan φ = (h x arc length) x w tan φ
Fig. 5.22 Pore water pressure If are length for a small slice be taken approximately equal to the width of the slice, the contribution of pore pressure becomes (h.b.w.) tan φ With the above approximation the effect of pore pressures will amount in reducing the saturated weight W of the soil by the weight of water of the same volume. This would mean that submerged weight may he taken for saturated weight of soil while calculating the resisting component. Hence a_steady seepage an drawdown conditions, weigh of soil below phreatic line and above dead storage level is taken as submerged for calculating resisting forces and saturated for actuating forces. Pore Pressure just after construction In this condition the pore pressures are maximum and hence their correct evaluation from proper consolidation test is advisable. It has been found that the pore pressure generally varies from 1.1 H to 1.25 H. In the absence of a reliable test it is reasonable to adopt a value of pore pressures equal to 1.25 H. The effect of pore pressures estimated on above principle is calculated separately as u tan φ u being 1.25 H sec θ , and is deducted from the value of pn tan φ.
158 Critical cases for analysis Reservoir drawdown The upstream slope of the dam is subjected to most adverse condition during sudden drawdown. When sudden drawdown takes place, the water pressure acting on the upstream slope is removed above the drawdown level while the saturation line remains higher. Since the drainage is not as rapid as the drawdown, this means that the resisting forces are reduced in comparison to actuating forces. To take account of this fact in stability computations, the resisting forces are calculated for submerged weight of the material below water surface, and the actuating forces are calculated for the saturated weight of the material below water surface. All materials below drawdown level are submerged, and therefore, resisting and actuating force below the drawdown level are calculated on the basis of the submerged weight of the materials. It has been observed for homogeneous section that drawdown to an intermediate pool level usually presents a more critical case than complete draw down. In all upstream slope stability analysis, therefore, such a condition should be tested. For each slip circle, the intermediate surface of the pool is assumed to intersect the embankment slope directly below the centre of the circle. This represents the worst condition for assumed failure arc. Steady Seepage Condition The effect of seepage through the embankment is to reduce embankment stability by increasing the actuating forces and decreasing the resisting forces. For the analysis of this condition, reservoir is assumed to be at the normal storage level in which case the phreatic line is assu1’riecl to have fully developed. The upstream slope will be subjected to water pressure from the reservoir while the rest of darn will be subjected to pore pressure from the established flow net. This condition is critical for the downstream slope only. ‘End of Construction’ Condition Stability at the end of construction is most critical for homogeneous embankments constructed of plastic materials. Immediately on completion of embankment there would be construction pore pressure due to consolidation of fill under the embankment load. There would be no water loads. Earthquake There is little experience on the performance of earth dams shaken by earthquakes in seismically active regions. Practice to account the horizontal acceleration caused by earthquake is, however, a common practice in stability computations for darns lying in seismic active regions. It is recommended that applicable values of seismic coefficient may be taken into account in stability computation for steady seepage conditions, while half of its values be adopted in case of sudden drawdown and end of construction conditions.
159 Factor of Safety Factor of safety in the present context is known as the ratio of forces resisting movement to these tending to produce movement. The minimum required factor of safety in earth dams never exceeds 1.5, which may seem to be quite a low figure. There are several reasons for the apparent lowness of the acceptable factor of safety in earth dam designs. (a) The figures used for the strength of earth materials are necessarily taken quite close to the minimum values obtained, and the strength which would be mobilized before disastrous failure could occur would usually be much greater. (b) Usually the factor of safety increases with the passage of time owing to consolidation etc. so that a factor of safety which was originally 1.5 may eventually become 2. (c) In most cases the forces tending to produce movement are taken at the upper possible limit but may actually be materially less than the assumed values. According to the U.S.B.R. practice a factor of safety of 1.5 is adopted for all conditions. The factor of safety as recommended by U.S.B.R. are on higher side, High dams have recently been designed with lower factor of safety up to 1.25 for reservoir drawdown and end of construction conditions. When under earthquake condition, factor of safety of unity is considered adequate. Tentative section For fixing tentative section of earth dams the upstream and downstream slopes may be taken from table 5.4 given by Terzaghi. Table 5.4 Recommended Dam Slopes-Terzaghi Type of material Homogeneous well graded material Homogeneous coarse silt Homogeneous silt clay or clay Height less than l5m or so Height more than 15m or so Sand or sand and gravel with clay core With R.C. core wall
U.S. slope
D.S. slope
h:v 2½ : 1
h:v 2:1
3:1 2½:1 3:1
2½:1 2:1 2½:1
3:1 2½:1
2½:1 2:1
160 5.10 Selection of Type of Earth Dams The selection of type of dam i.e. earth, rock fill, concrete, concrete gravity, concrete arch, buttress etc. When a particular site favours an earth dam, a further decision must be made as to the type of earth dam. The type of earth dam would be dictated essentially by the materials available at or near the site as also the foundations. In the interest of economy, the design of earth dam should be adopted to the full utilization of the available materials. Thus, if nothing but sand is available the design should utilize the sand in the natural state for the bulk of the dam, limiting the imported material like clay or silt fore providing impervious member to the minimum. Earth dams may be classified into three main types : 1. Homogeneous type 2. Zoned type and 3. Diaphragm type Homogeneous Type A purely homogeneous type of dam is composed entirely of a single type of material as shown in fig. 5.23 a. Since the action of seepage is not favourable in such a purely homogeneous section, the upstream slope should be, relatively flat for safety in rapid drawndown (when embankment is relatively impervious, and the downstream slope must also be flat to provide a slope sufficiently stable to resist the forces resulting from a high saturation level. For a completely homogeneous section, it is inevitable that seepage emergence will occur in the downstream slope approximately at about one third the water depth regardless of permeability of material or slope flatness. Though formerly very common, the pure homogenous section has fallen into disuse because of the development of a modified homogeneous section (Fig. 5.23 b) in which small ‘amount of carefully placed drainage materials control the action of seepage and thus permit much steeper slopes.
Fig . 5.23 Homogenous type earth dams
161 A fully homogeneous section might be found convenient where the slopes are required to be flat because of a weak foundation, although even in this case some drainage measures may be necessary. The modified homogeneous type of embankment is most suitable in localities where readily available soils show practically no variation in permeability. Zoned type It is the most common type of dam section in which a central impervious core is flanked by shells of materials considerably more pervious. The shells enclose and protect the core; the upstream shell affords stability against rapid drawdown and the downstream shell acts as a drain that controls the line of seepage. For most effective control of steady seepage or drawdown seepage, the section should have a progressive increase in permeability from the centre towards each slope ( Fig. 5.24 ). Innumerable modifications might be made of the zoned type section depending upon the arrangements of seepage control and drainage arrangements.
Fig. 5.24 Zone type earth dam section The width, of core in the zoned type can be chosen, within reasonable limits, to meet the best adjustment in the quantity and the cost of impervious soils available. The minimum width should be adequate to reduce seepage and permit ease of construction. The minimum base width should be equal to the height of the embankment. If the width is less than the height of the embankment, the dam is considered as diaphragm type. Similarly if the core is larger than the size shown in fig. 5.25 the dam may be considered as homogeneous type.
Fig. 5.25 Size range of impervious cores in zoned embankment-after U.S.B.R.
162 All transitions between materials, of different gradations subject to seepage must be checked for filter action. If the gradation range between adjacent zones of the dam or foundation is so great that a natural filter will not be established, a special filter zone will be required to provide the necessary filter action. Diaphragm type ( Fig. 5.26 ) In this type the entire dam is composed of pervious material i.e. sand, gravel or rock and thin diaphragm of impervious materials is provided to retain water. The diaphragm may consist of earth, cement concrete, or other material in the central or upstream face of the dam. The position of the impervious diaphragm may vary to extreme limits-upstream face on one side and central core on the other end. The intermediate cases are termed as burned blanket or inclined diaphragm type as shown in fig. 5.27.
Fig. 5.26
Fig. 5.27
Diaphragm type earth dam
Inclined diaphragm type earth dam
Internal diaphragm of rigid materials like concrete have the disadvantage of getting ruptured due to settlement of the dam.
Fig. 5.28 (a)
Nanak Sagar dam (Constructed after breach) U.P.
163
Fig. 5.28 (b) Section of the u.s. riprap
It is, therefore, preferable to have core of the earth. The width of the earth core at the base should be 0.3 to 0.5 times the height of dam. There are certain advantages and disadvantages in locating the core centrally or in an inclined position. However, the construction convenience dictates the position of the core. An inclined diaphragm provides slightly better stability against earthquake. 5.11 MAINTENANCE & TREATMENT OF COMMON TROUBLES IN EARTH DAMS Proper maintenance of an earth dam is very essential. Small defects arising from the negligence may cause breach which may lead to serious damage and devastation. In spite of good design, some problems may be subsequently met with during the operation of the reservoir. The following are the most common troubles which may need proper and timely treatment. 1. Horizontal piping through embankment. 2. Boils. 3. Boggy areas downstream of the dam, and 4. Cracks in the embankment. 15.12 DESIGN OF EARTH DAM Design Example 5.2 An earth dam has to be constructed at a place where soils have the following characteristics: (i) Sand mixed with gravel (a) Saturated unit wt 2 gm/cc (2 tIm3) (b) Sp. Gravity 2.65 (c) Angle of internal friction 33° (d) Cohesion nil (e) Dry unit wt. 1.6 gm/cc (1.6 t/m3) This material constitutes the foundation to a depth of 15.Om, below the base of the dam after which rock is available.
164 (ii) Silty clay (a) Saturated unit wt. 1.76 gm/cc (1.76 tim2) (b) Specific Gravity 2.55 (c) Angle of internal friction 12° (d) Cohesion 3.t/m2 The material has to be obtained from an average distance of 1.5 km. (iii) Levels (a) River bed level 103.0 m (h) Dead storage level 123.0 m (c) F.S.L. 155.0 m (d) H.F.L. 158.0 m Assume the material above phreatic surface to be 40% saturated. (e) Earthquake factor (i) For sudden drawdown α = 0.15 (ii) For steady seepage α = 0.10 (iii)For just after construction α = 0.05 Design a suitable section of the earth dam and indicate the position of seepage line. Also draw cross section of the dam. Design 1. Type of Dam Section As both pervious and impervious materials are available in the close proximity, the dam shall consist of a composite section. The shell of the dam will consist of pervious material and core of impervious material. A cutoff trench up to rock level with 1: 1 side slopes shall also be provided to check the seepage through the foundation material. 2. Tentative cross section of the Dam (i) Freeboard (a) Freeboard for wave action: hw = 0.032 VF + 0.763 - 0.27 1 4 F where hw = height of wave in metres V = wind velocity in km. p.h. F = fetch in kms. Assume V = 150 km. ph. and F = 20 kms. hw = 0.032 150x 20 + 0.763 - 0.271 4 20 =1.958 metres Along slopes, the waves rides to a height of 1.5 hw. ∴ 1.5 hw = 150 x 1958 = 2.94 metres (b) Freeboard due to settlement
Then
165 Settlement allowance of 2% is generally taken both for the foundation and the embankment. 2 ∴ Settlement = 60 x = 1.20 m (Assuming dam height to be 60 m) 100 For earth dam more than 30 m in height, an extra settlement allowance of 1% due to earthquake is also taken. 60x1 ∴ Settlement due to earthquake = = 0.6 m 100 ∴ Total freeboard due to settlement = 1.20 + 0.60 m = 1.80m. Thus total freeboard required both due to wave action and settlement = 2.94 + 1.80 m = 4.74 m. Provide a freeboard of 5.0 m above H.F.L. (ii) Top level and width (a) Top R.L. of dam = 158 + 5 = 163.0 m Height of dam above bed level = (163.0—103.0) = 60m (b) Crest width : 5 Bt = H 3 where Bt = Crest width in metres H = Height of dam in metre 5 ∴ Bt = 60 = 12.90 m 3 Provide crest width of dam as 13.0 m. (iii) Slopes : For 60.0 m height of the dam adopt 3.5: 1 slope in the upstream and 3 : I in the downstream; Adopt slopes of the core material 1 : 1. (iv) Pitching Upstream : To prevent destructive wave action, the upstream slope shall be provided with 0.75 m thick stone pitching over 0.5 m thick graded shingle right up to the top of the dam. Downstream The downstream slope shall be provided with 0.5 m stone pitching to check the erosion due to surface runoff. 3. Foundation (3) Shear stress in the foundation The horizontal shear under the slope of dam is given by h − h2 φ S=γ 1 tan 2 45o − 1 2 2 2
where
2
φ1 = Equivalent angle of internal friction for composite section. γ = Effective wt/cum of the composite material, and h1 and h2 are heights top
166 and toe of the dam respectively to the rock level. Here h1 = 75m, h2 =15m, b = 2l0m γa = Average effective wt. of the composite material 2 x150 + 1.76 x 60 = = 1.93 t / cum. 210 Equivalent angle of internal friction for C— φ soil is given by C + γ h1 tan φ tan φ1 = γ h1 For shell material, C = 0, For core material, C = 3 t /m2
∴φ1 = 33° φ = 12°
3 + 1.76 x 60 x tan 12 o = 0.239 1.76 x 60 ∴ φ1 = 13.5° (for core material). Since the dam is of composite section, the equivalent value of internal friction ∴ tan φ1 =
150 x33o + 60 x 13.5 o = = 27.4° 210 ∴ Total shear S =
(for composite section)
75 2 − 15 2 27.4 o x1.93 tan 2 45 o − 2 2
= 1926.5 t say 1900 tonnes. 1900 210 = 9.05 t/m2 Maximum unit shear = 1.4 Sa = 1.4 x 9.05= 12.67 t/m2 (ii) Factor of Safety against foundation shear : We should now work out the factor of safety in the foundation against shear at the point of maximum unit shear which occurs at 84 m (0.4 b) from shoulder. The mean effective unit wt. r at this location. 36 x 2 + 15 x1 = = 1.706 t /m3 51 Shear strength at point of maximum shear = c + γ h tan φ1 Substituting c = 0, φ1 = 33° r = 1.706t/m2and h =51 m, we get Shear strength = l.706 x 51 x tan 33° = 1.706 x 51 x O.65 = 56.5 t m2 Factor of safety against shear at point of maximum shear 56.5 = = 4.45 okay 12.67
Average unit shear in the foundation (Sa) =
167 4. Position of Seepage Line in the Composite Section The shells of the dam which consist of sand and gravel are several times as pervious as the central portion of the dam consisting of silty clay. The upstream pervious shell has practically no effect on the position of seepage line and the downstream shell will act as a drain. Hence in computing the position of the seepage line, the effect of the outer shell should be neglected. (a) Construction of base parabola: The focus of the base parabola is located at the downstream toe of the core denoted by F. Taking focus as origin, the equation of base parabola is given by y 2 − y0 x= 2 y0
2
where x and y are the points on the parabola and y0 is the focal distance which is given by y0 y0 =
h2 + d 2 - d
Fig.5.29 Design Problem Dimension
From figure 5.29, h = 52 m and d = 96.6 m ∴ Y0 = Thus
52 2 + 96.6 2 - 96.6 = 13.106 say 13.0 m
x=
y 2 − 13 2 2 x13
or
y=
26 x + 169
168 The values of y for different values of x are given below : x = 0 10 20 30 40 50 60 70 80 80 96.6 y = 13 20.7 26.2 30.8 348 38.3 41.6 44.6 47.4 50.1 52.0 Thus the base parabola is plotted for the above computed values of x & y. (b) Correction at ingress and egress points The base parabola intersects the downstream face at a distance a + ∆a along the face from the point A. y0 a + ∆a = 1 − cosα Substituting y0 = 13 m and α = 45°, we get 13 a + ∆a = = 44.4m 1 − cos 45 o ∆a From fig. 15.7, = 0.34 a+∆ a ∆a = 034 x 44.4 = 151 m. Thus the seepage line meets the downstream face tangentially at point C, 15.1 m along the face from the point Co. The seepage line is completed at the downstream end by sketching a short transition curve from point C to the base parabola. The upstream end of seepage line is sketched with short transition curve as shown in fig. 5.30. The upstream shell material is so excessively pervious as compared to the central section that it wi1I have no effect on the seepage line. Hence through this section the seepage is drawn as a straight line. As the foundation material is also very pervious, the seepage line will pass beneath the base of the dam (figure 5.29). or
5.
Stability Computation of Upstream and Downstream Slopes The stability of upstream slope has been computed for the following conditions (i) Sudden drawdown condition (ii) Just after construction condition Both the conditions shall be analysed with and without earthquake. The stability of the downstream slope shall be computed for the following conditions: (i) Steady seepage condition (ii) Just after construction condition Both the conditions shall be analysed with and without earthquake. For calculating the actuating forces in sudden drowdown condition, the saturated weight of soil is taken in drawdown range between f.s.l. and d.s.l. Below d.s.l. the submerged weight of soil is taken while above f.s.l., the weight of soil is takesn as 40% saturated. The unit weight of soil are being calculated hereinafter for different conditions. (a) Sand mixed with gravel Weight of fully saturated soil = 2 gm/cc (2 t /m3) ∴ Submerged weight = 2 - 1 = 1 gm/cc = (1t / m3) 40% saturated weight Let V be the volume of solids in unit volume of soil then
169 V x 2.65 +(l - V)1 = 2 or V = 0.606 ∴ Weight of 40% saturated soil 40 = 0.606 x 2.65 + 0.394 x = 1.76 100 say = 1.75 gm / cc (1.75 t/m 3) (b) Silty Clay Submerged weight = 1.76 - 1 = 0.76 gm/cc = 0.76 t/m2 Weight of 40% saturated clay Let V be the volume of solids in unit volume of soil then V x 2.55 + (l-V) x 1 = 1.76; V = 0.49 40 Weight of 40% saturated earth = 0.49 x 2.55 + 0.51 x 100 3 = 1.45 t / m For just after construction condition no pore pressure effect in sandy soil is taken into account. Thus an amount equivalent to the value corresponding to pore pressure is to be deducted from (ΣW tan φ + c.L.). The value of unit weight of dry sand i.e. of shell material is taken as 1.6. The values of unit weight of shell and core material under various conditions of stability analysis both for upstream and downstream slopes are given in Table 5.5 Table 5.5 Values of unit weight under different stability conditions For Calculation of N Condition U.S. Slopes (i) Sudden drawdown (a) Without Earthquake (b) With Earthquake (ii) Just after Const. Condition (a) Without Earthquake (b) With Earthquake
Moist Shell
Shell above d.s.l
Shell below d.s.l
Core above d.s.l
Core below d.s.l
Moist Core
Moist Shell
Shell above d.s.l
Shell below d.s.l
Core above d.s.l
Core below d.s.l
1.45
1.75
1.00
1.00
0.76
0.76
1.45
1.75
2.00
1.0
1.76
0.76
1.45
1.75
1.60
1.00
1.45
0.76
1.45
1.75
2.00
1.0
1.45
0.76
Condition
For Calculation of N Moist Core
D.S. Slope (i) Steady Seepage Condition (a) Without Earthquake (b) With Earthquake (ii) Just after Construction (a) Without Earthquake (b) With Earthquake
For Calculation of T
Moist Core
For Calculation of T
Shell above d.s.l
Shell below d.s.l
Core below d.s.l
Core above d.s.l
1.45
1.75
1.0
0.76
0.76
1.45
1.60
1.0
0.76
1.45
Moist Core
Shell above d.s.l
Shell below d.s.l
Core above d.s.l
Core below d.s.l
1.45
1.75
1.00
1.76
0.76
1.45
1.60
1.00
1.45
0.76
170 A — STABILITY COMPUTATION OF THE UPSTREAM SLOPES: (1) Sudden Drawdown Condition (a) Without Earthquake Adopting the above values of the weight of different types of soils under different conditions, the stability computations have been worked out in Table 5.6. A slip circle has been drawn with a radius of 150 m as shown in figure 5.31. The entire circle has been divided into nine slices. Actuating force ΣT = 1 [99.7 x 1.45 + 58.2 x 1.75 + 297 x l.76 + 757.6 x 2 + 92 x 0.76 + 119 x 1] = 1 [144.60 + 102.00 + 523.00 + 1515.20 + 69.90 + 119.00] = 2473.76 t say 2474 t. Resisting force ΣN = 1 [79.6 x 1.45 + 57.3 x 1.75 + 322 x 0.76 + 1694.8 x 1 + 140.5 x 0.76+3411 x 1] = 1 [115.50 + 100.00 + 245.00 + 1694.80 + 107.00 + 3411.OO] = 5673.3 t. Radius of slip circle = 150 m. Length of arc in the core material 35 = 2 π x 150 x =91.5m 360 ΣN . tan φ + c. L. Factor of safety = ΣT ΣN for shell material = 100 + 1694.8 + 3412 = 5205.8 say 5206 t ΣN for core material = 115.5 + 245.0 + 107 = 467.7 t φ for shell material is 33° and for core is 12 ° and C = 3 t / m2 •
5206 tan 33o + 467.5 tan 12 o + 3 x91.5 F.S. = 2474 5206 x 0.65 + 467.50 x0.2125 + 274.5 = 2474 3380 + 9.4 + 274.5 3749.0 = = 2474 2474 = 1.515 Okey.
171
S.No
Table 5.6 Stability analysis of upstream slope for sudden draw down condition Area x Are Slice No. Sin θ cos θ θ a sin θ
1
1
(Submerged shell below d.s.l.)
390
-20.5°
2
2
(Submerged shell below d.s.I
550
-12.5°
3 4 5 6
3 3 4 4
(Submerged shell below d.s 1.) (Submerged shell above d.s 1.) (Submerged shell below d.s 1.) (Submerged shell above d.s 1.)
953 72 930 406
7 8 9 10 11 12 13 14 15 16 17 18 19 20
5 5 6 6 6 7 7 7 7 8 8 8 8 9
(Submerged shell below d.s 1.) (Submerged shell above d.s 1.) (Submerged shell below d.s 1.) (Submerged shell above d.s 1.) (Submerged core below d.s 1.) (Moist shell), (Submerged shell above d.s 1.) (Submerged core below d.s 1.) (Submerged core above d.s 1.) (Moist shell), (Submerged shell above d.s 1.) (Moist core) (Submerged core above d.s 1.) (Moist core)
497 410 200 518 82 12 418 88 188 70 72 58 252 70
Area x cos θ
-0.35
0.936
-136.5
365.0
-0.208 -3.5° -0.061
0.978 0.998
-114.4
538.0
10.0° 0.174.
0.984
-58.0 -4.4 162.0 70.5
951.00 71.00 915.0 400.0
170.0 140.0 95.40 247.0 39.0 7.2 252.0 53.0 113.0 51.0 52.6 42.3 184.0 57.4
467.0 385.00 176.0 455.0 72.0 9.6 334.0 70.22 150.0 47.7 49.0 39.6 172.0 40.0
20.0°
0.342
0.939
28.5°
0.477
0.878
37.0°
0.602
0.798
47.0°
0.731
0.682
55.0°
0.819
0.573
Values of Area x cos θ Moist Core = 39.6 + 40 = 79.6 Moist shell = 47.7 + 9.6 = 57.3 Submerged shell above d.s.l. = 71.8 + 400.0 + 385.0 + 455.0 + 334.0 + 49.0 = 1694.8 Submerged shell below d.s.1. = 365 + 538 + 951 + 915 + 467 + 176 = 3412 Submerged core above d.s.l. = 150 + 172.0 = 322 Submerged core below d.s.1. = 70.22 + 72.0 = 142.22 Values of Area x sin θ Moist core = 42.3 + 57.4 = 99.7 Moist shell= 51.0 + 7.2 = 58.2 Submerged shell above d.s.l. = - 4.4 + 70.5 + 140.0 + 247.0 + 252.0 + 52.60 = - 4.4 + 762.6 = 757.7
172 Submerged shell below d.s,1. = - (136.5 + 114.4 + 5.8) + (162:0 + 170.0 + 95.4) = - 308.9 + 427.4 = 1 18.5 Submerged core above d.s.1. = 113 + 184 = 297 Submerged core below d.s.l. = 53.0 + 39.0 = 92.0
Fig . 5.31 Graphical method of stability computation – Design example
(b)
With Earthquake Earthquake factor for sudden drawdown condition, α = 0.05 Increase in ΣT = 0.05 x 5673.3 = 288.7 t. Reduction in ΣN tan φ ΣT for shell material =.102 + 1515.2 + 119 = 1736.2 t ΣT for core material = 144.8 + 523 + 69.9 = 737.7 t Reduction in ΣN tan φ = 0.05 (737.7 x 0.2125 + 1736.2 x 0.65) = 0.05 (157 + 1130) = 64.4t Factor of Safety with earthquake 3749.0 − 64.4 3684.60 = = 2474.0 + 283.7 2757.7 = 1.34 Okay.
173 (ii) Just After Construction Condition (a) Without Earthquake In this condition the shell material should be assumed to be dry and the clay core material as moist with unit weight as 1.45 tIm3. Actuating forces ΣT = 1 (99.7 x 1.45 + 58.2 x 1.75 + 297 x 1.45 + 757.6 x 1.6 + 92 x 0.76 + 119.x 1.0) = 1 (144.8 + 102.0 + 430 + 1210.0 + 70.0 + 119.0) = 2075.8 say 2076 t ΣN tan φ = (79.6 x 1.45 x 0.2125 + 57.3 x 1.75 x 0.65 + 322 x 1.45 x 0.2125 + 1694.8 x 1.6 x 0.65 + 140.5 x 0.76 x 0.2125 + 3411 x 1 x 0.65) = 1 (24.6 + 65.0 + 99.4 + 1760 + 22.7 + 2220.0) = 4191.7 t say 4192 t Pore Pressure For slice 6, head is 8 m; sec θ = 1.138, width of strip = 20 m ∴ h x 1 sec θ = 182.0 t For slice 7, head = 14m; sec θ = 1.252, width of strip = 20 m ∴ h . 1 sec θ = 351.0 t For slice 8, head = 18 m; sec θ = 1.743, width of strip = 10 m ∴ h . 1 sec θ = 498.0 t For slice 9, head = 10 m; sec θ = 1.743, width of strip = 10 m ∴ h . 1 sec θ = 174.3 t Σ h. 1 sec θ = 182.0 + 351.0 + 498.0 + 174.3 = 1205.3 t Now P = w h l sec θ = l x 1205.3 = 1205.3 ∴ P tan φ = 1205.3 x 0.2125 = 256 t 4192.0 + 274.5 − 256 4210.5 Factor of safety = = 2076 2076 = 2.03 Okey. (b) With Earthquake (α = 0.05) Increase in T = 0.05 x 1 (79.6 x 1.45 + 57.3 x 1.75 + 322 x 1.45 + 1694.8 x 1.6 + 140.5 x 0.76 + 34ll x l) = 0.05 (115.5 + 100.1 + 467.0 + 2710.0 + 106.8 + 3411.0) = 0.05 x 6910.4 = 345.5 t Decrease in ΣN tan φ = 0.05 x 1 (99.7 x 1.45 x 0.2125 + 58.2 x 1.75 x 0.65 + 297 x 1.45 x 0.2125 + 757.6 x 1.6 x 0.65 + 92 x 0.76 x 0.2125 + 119 x l x 0.65) = 0.05 {644.8 x 0.2125 + 1431 x 0.65} = 0.05 x 1069.5 = 53.5 t
F.S.=
174
4210.5 − 53.5 4157.0 = 1.72 O.K = 2076 + 345.5 2421.0
B STABILITY COMPUTATION OF DOWNSTREAM SLOPE (i) Steady Seepage Condition: (a) Without Earthquake: The stability computation have been worked out in Table 5.7. A slip circle has been drawn with a radius of 150 m passing near the toe and touching the impervious rock. The entire slip surface has been divided into 9 slices. Actuating forces ΣT = 1 x (90.8 x 1.45 + 817.5 x 1.75 + 127 x 1.76 + 17 x 0.76 + 376.3 x 1) = 1 x (132 +1430 + 224 + 12.9 + 376.3) = 2175.2t ΣN tan φ = 1 x {(71.3 x 1.45 + 138.3 x 0.76 + 30.6 x 0.76) 0.2125 + (l3229 x 1.75 + 3333.1 x l) 0.65} = {(103.3 + 105 + 23.2 ) 0.2125 + (2320+3330) x 0.65}
S.No.
Table 5.7 Stability analysis of D.S. slope under steady seepage condition
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Slice No. 1 (Submerged shell below d.s.l of fail water level) 2 (Submerged shell below d.s.l.) 3 (Submerged shell below d.s.l.) 4 (Submerged shell below d.s.l.) 4 (Moist shell) 5 (Submerged shell below d.s.l.) 5 (Moist shell below d.s.l.) 6 (Submerged shell below d.s.l.) 6 (Moist shell) 6 (Submerged shell below d.s.l.) 6 (Submerged shell below d.s.l.) 7 (Moist shell) 7 (Submerged core below p.l) 7 (Moist core above P.l.) 8 (Moist shell) 8 (Submerged core below p.l) 8 (Moist core above p.l.) 9 (Moist shell) 9 (Moist core)
Area x sin θ
Area x cos θ
sin θ
cos θ
260 -19°
-0.325
0.945
- 84.5
246
502 655 743 89 956 380 320 420 35 23 550 135 6 206 53 59 52 51
-10° - 2° + 7°
-0.174 -0.035 0.122
0.935 0.999 0.992
19°
0.325
0.945
0.485
0.875
0.643
0.766
0.755
0.656
0.835
0.550
-87.5 -23.0 90.5 10.8 311.0 213.5 155.0 204.0 17.0 14.8 290.0 87.0 3.8 155.6 40.0 44.5 43.5 42.5
495 654 737.5 88.3 903.0 359.0 280.0 367.0 30.6 17.6 345.0 103.5 4.6 135.0 34.8 38.7 28.6 28.0
Area
θ
29°
40°
49°
56°
Values of Area x sin θ Moist core = 42.5 + 44.5 + 3.8 = 90.8 Moist shell above d.s.l. = 10.8 + 123.5 + 290.0 + 155.6 + 43.5 = 817.5 Submerged shell = - (84.5 + 87.5 + 23.0)+ 90.5 + 311.0 + 155.0 +14.8 = -195.0 + 571.3 = 376.3 Saturated core = 87.0 + 40.0 = 127.0 Submerged core below d.s l. = 17.0
175
Values of Area x Cos θ Moist core = 4.6 + 38.7 + 28.0 = 71.3 Moist shell above d.s1. = 88.3 + 359.0 + 345.0 + 135.0 + 28.6 = 1320.9 Submerged core above d.s.l. or below p.1. = 103.5 + 34.8 = 138.3 Submerged core below d.s.1. = 30.6 Submerged shell below d.s.l. = 246 + 495.0 + 654.0 + 737.5 + 903.0 + 280.0 + 17.6 = 3333.1 = (231.5 x 0.2125 + 5653 x 0.65) = 3659.2 say 3659 t. Factor of Safety ΣN tanφ + c.L. ΣT Length of arc (L) 36 = 2 π x 137 x 360 = 86 m ∴ cL = 3 x 86 = 258t 3669 + 258 3916 F.S. = = 2175.2 2175.2 = 1.8 (b)With Earthquake (Earthquake factor α = 0.1) Increase in Σ T = 0.1 x 1 (71.3 x 1.45 + 138.3 x 0.76 + 30.6 x 0.76 + 1320.9 x 1.75+3333) = 0.1 (439 + 5653) = 609.3t Decrease in ΣN tan φ = 0.1 [(90.8 x 1.45 + 127.0 x 1.76 + 17 x 0.76) x 0.2125 + 0.65 (817.5 x 1.75 + 376.3 x 1)] = 0.1(368.9 x 0.2125 + 1806.3 x 0.65) = 125.2 t Factor of Safety = L =
K . h . d − p . q .b = 1.377 pq
(ii) Just after Construction Condition: (a) Without Earthquake Effect
Actuating forces Σ T = 1 (90.8 x l.45 + 817.5 x 1.6 + 127 x 1.45 +19 x 0.76 + 376.3 x 1) = (132 + 1313 + 184 +13 + 376) = 2015 t Σ N tan φ = 1 [(71.3 x 1.45 + 138.3 x 1.45 + 30.6 x 0.76) 0.2125 + (1321 x 1.6 + 3333 x 1) 0.65]. = (327.4 x 0.2 125 + 5448 x 0.65) = 3609.60 t Pore Pressure For slice 6, head = 2 m, sec θ = 1.14 Width of strip 1 = 20 m ∴ P = wh 1 sec θ = 1 x 2 x 2O x l.14 = 45.6t For slice 7, head = 8m, sec θ = 1.3 Width of strip 1 = 20 m ∴ P = wh l sec θ = l x 1.3 x 20 = 208t For slice 8, head = 10 m, Sec θ = 1.525 Width of strip 1 = 14 m ∴ P = wh l sec θ =1 x lO x 14 x l.525 = 213 t For slice 9, head = 4 m, Sec θ = 1.81 Width of strip 1 = 13 m ∴ P = wh l sec θ = l x 4 x 1.81 x 13 = 94 t ∴ Σ P = 45.6 + 208 + 213 + 94 = 560.6 t P tan φ = 560.0 x 0.2125 = 121 t 3609.6 + 258 − 121 3746.6 F.S. = = =1.86 Okay 2015 2015 (b) With Earthquake: Earthquake factor α = 0.05 Increase in Σ T = 0.05 (71.3 x 1.45 +1321 x l.6 + 1383 x 1.45 + 30.6 x 0.76 + 3333 x 1) = 0.05 (103.2 + 2115 + 201 + 23.2 + 3333) = 288.87 t Decrease in Σ tan φ = 0.05 {(90.8 x 1.45 + 123 x 1.45 + 17 x 0.76) 0.2125 + (817.5 x 1.6 + 376.3 x 1) x 0.65} = 0.05 (329 x 0.2125 + l686.3 x 0.65) = 58.5t 3746.6 − 5.85 3698.1 F. S. = = 2015 + 288.8 2303.8 = 1.6 O.K
176
177 The values of ‘Factor of Safety’ under different conditions for upstream and downstream are given in Table 5.8 The factor of safety in normal case for all conditions without earthquake is greater than 1.5 and that with earthquake is more than unity. Hence the assumed slopes of the dam section both in the upstream and downstream are safe under all conditions. But the slip circles as marked in fig. 5.30 may not be the critical, one; other slip circles should be drawn and for each condition the factor of safety be calculated. The critical circle is one with minimum factor of safety. The detailed design has been illustrated in figure 5.30 Table 5.8 Factor of Safety Under Different Conditions
SL. No. A
Conditions of Analysis UPSTREAM SLOPE (i ) Sudden drawdown
(ii) Just after construction B
DOWNSTREAM SLOPE (i ) Steady Seepage
(ii) Just after construction
Factor of Safety Without E. Q. With E. Q. 1.515 Safe > 1.5 2.03 Safe > 1.5
1.34 Safe > 1.0 1.72 Safe > 1.0
1.80 Safe > 1.5 1.86 Safe > 1.5
1.377 Safe > 10 1.60 Safe > 1.0
