Contents preface (VII) introduction 1—37


Critical Conditions for the Stability of an Embankment Dam



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15.4.1. Critical Conditions for the Stability of an Embankment Dam
The following conditions are considered critical for the stability of an embankment dam (12).
(i) During Construction With or Without Partial Pool
When an embankment is constructed, water is added to soil during compaction. This increases pore pressures and may cause failure of both upstream (if there is no water in the reservoir) and downstream slopes. During construction, the reservoir is usually empty and, hence, both the upstream and downstream slopes need to be analysed. The magnitude and the distribution of the construction pore pressures depend primarily on the construction moisture content in the embankment, natural moisture content in the foundation, soil properties, construction rate, dam height, and internal drainage. Since high construction pore pressures exist only during the first few years of the life of the dam (when reservoir may not have been filled completely), more conservative and expensive design is avoided and, instead, suitable steps are taken to reduce the construction pore pressures. These steps include:
(i) compacting impervious core at an average moisture content which is 1–3 per cent below the optimum moisture content,
(ii) making impervious core section thinner,
(iii) providing internal drains within impervious core section to accelerate pore pressure dissipation,
(iv) accepting a lower factor of safety on the plea that a slide during construction would not cause catastrophic failure and only delay the completion of the dam, and
(v) having longer construction period so that some pore pressure is dissipated from the already-constructed fill before laying another layer.
For analysing the stability of slopes during construction, one needs information about pore pressure. As per USBR’s simple approach, the pore pressure head during construction equals, approximately, 1.25 times the height of the fill. Alternatively, one can measure pore pressures in sealed laboratory specimens which have been compacted and subjected to increasing stresses simulating the field conditions in respect of the water content, densities and increasing stresses, and use them for analysis. However, there can be some difficulties in simulation. If pore pressures are allowed to dissipate during construction by causing break in construction, the pore pressures are estimated (12) by using the following Bishop’s method which uses Hilf’s equation.
Bishop’s method requires the knowledge of volume change – effective stress relationship [(∆V/V) v/s σ’] for the compacted fill. This is determined experimentally by testing representative specimens. The volume change – pore pressure relationship [(∆V/V) v/s ∆u ] is obtained for each stage of construction using Boyle’s law and Henry’s law (for solubility of air in water).
Any sample of compacted fill having a volume of Vo , Fig. 15.23 includes volume of soil grains Vs and volume of voids Vv which equals the sum of the volume of water Vw and volume of air Va occupying the voids. The volume of air dissolved per unit volume of water is defined as Henry’s coefficient of solubility, H whose approximate value is 0.02. The degree of saturation So is the ratio of volume of water and volume of voids. Thus,

Initial volume of free air = Va + dissolved air

(at initial pressure po) = (VvVw) + HVw

= VvVv So + HSoVv



= Vv(1 – So + SoH)

516 IRRIGATION AND WATER RESOURCES ENGINEERING















s¢, u, s (N/cm2)



















0

10

20

30

40

50

60




70

80




0


































1






















Va (Air)






















Vv

V




(Water)

























w



































































Vo







2




s













Vs (Soil)










u




























DV (%)































































Vo
















Block diagram







1

1

1
















3










of a soil







3.23

























1b

1b

1b










specimen

























4





































2













2







2







5

2b
















2b

s¢ 2b











































Fig.

15.23 Variation of u, σ′ and σ with ∆V/Vo














At a new pressure p, the volume of air, using Boyle’s law, is, therefore, given as


ppo Vv(1 – So + SoH)


Therefore, change in volume of air, ∆V = (po/p) Vv (1 – So + SoH) – Vv (1 – So + SoH)



p { V + Vv (1 So + So H)} = poVv (1 – So + SoH)








po

= 1 +



















V /Vv




















































p



































































1

S +

S H














































































































































o

o














































= 1 +










V /no Vo




in which no is the initial porosity of soil.
















1

S +

S H
















































































































o

o





































1 −




po




= –

























V /Vo























































no

(1 − So + So H)


































p





















































































p po = –






















( ∆V /Vo ) p








































no

( 1 − So + So H)






























































































= –




























( ∆V /Vo ) po











































n

o

( 1 − S

+

S H ) ( p / p)























































o




o







o








































= –











































(∆V/Vo ) po












































































F
















V/Vo




I







































































































no

(1 − So

+

So H) G

1 +



















J






















no

(1

So +



































































H




So H)K
















= –




























(∆V/Vo ) po











































V




+ n

(1 − S

+

S H)










































































































Vo




o










o

o
















































































































EMBANKMENT DAMS

517

If the initial pressure po is the atmospheric pressure, pa then change in pore pressure,



u = ppo = ppa.













F

V I



















G




J pa




























Therefore,

u = –







H

Vo K

(15.29)







V

+ n (1 − S + S H)


































o




oo
















Vo













Equation (15.29) is known as Hilf’s equation. This equation is applicable under the following conditions:
(i) There is only vertical compression and no lateral bulging of the fill.
(ii) Pressures in pore water and pore air are the same. Strictly speaking, Hilf’s equation gives pore air pressure (say, Ua) whereas the effective stress depends on pore water pressure (say, Uw) which equals the sum of Ua and capillary pressure Uc which is negative for unsaturated condition and zero for saturated condition. Since earth fills are usually placed at 80 to 90 per cent degree of saturation, the capillary pressure is usually small and, therefore, neglected.
(iii) Decrease in volume of fill is due to compression of pore air and dissolution of pore air in pore water.
(iv) There is no dissipation of pore pressure during construction. (v) Boyle’s and Henry’s laws are applicable.

Further, when all the air in pores has gone into solution,


V = Va = VvVw = Vv – VvSo = Vv (1 – So)




V

=




Vv

(1 – S )



















V




Vo








































o






















o











































Thus,




= no(1 – So) with negative sign, of course.




















































u =













n o ( 1 So ) pa






















n ( 1

S

+ S H ) − ( 1 − S )n

o



















o







o




o

o










=







(1 − So ) pa

=

( 1 − So )Vv pa

























So H







HV














































w
















and,

u =







pa Va



















(15.30)










H V





































w
























when all the air in pores has gone into solution, the soil is fully saturated and any further increase in load (∆σ) results in equal increase in pore pressure (∆u) as there is no air to get compressed.


In order to prepare total stress versus pore pressure relationship for an earthfill being constructed in stages, following steps are followed:
(i) Using experimental data, plot σ′ against ∆V/Vo, Fig. 15.23.
(ii) For different assumed values of ∆V/Vo, compute ∆u using Hilf’s equation, Eq. (15.29) and the total stress σ = σ’ + ∆u. Carry out these computations till either σ = ρgH1 or

518 IRRIGATION AND WATER RESOURCES ENGINEERING
[∆V/Vo] = no (1 – So) i.e., when all pore air has gone into solution. Here, H1 is the height of the dam (or earthfill) at the end of the relevant stage of construction.
(iii) Plot σ v/s ∆V/Vo and ∆u v/s ∆V/Vo, Fig. 15.23.
(iv) Let the pore pressure ∆u at the end of the first stage be dissipated by x per cent before the beginning of the second stage.
Pore pressure at the beginning of the second stage




u




=

F

1 −

x

I

u































1b




G
















J

1






















H










100K















σ′




= σ′







+




x




u







1b

1




100








































1








































F







x I







and

σ1b = σ′1b + G 1










J ∆u1

= ρgH1














































H




100K







Also,

pob = po + u1b (in absolute terms)




With absolute pore pressure pob, Hilf’s equation can again be used to compute rise in pore pressure with further loading for the second stage of loading provided that the values of So and no are recalculated at 1b. During the loading stage under undrained conditions, the degree of saturation S can be expressed as follows:




S =

Initial volume of water

=

So Vv




=

So













volume of voids




Vv + ∆V

1 +

V




















































V























































v







Since

V =

po

V (1 – S

o

+ S H) – V (1 – S

o

+ S H)










p



















v

F

o




I

v







o
















V













po


































Vv

= (1 – So + SoH) G

p

− 1J











































H




K




























For the beginning of the second stage, the degree of saturation Sob corresponding to new pore pressure (gauge) pob is, therefore, expressed as:













Sob =



















So

























F po










I




























1 + G







− 1J (1 −

So + So H)











































H pob










K













Similarly, new value of porosity at stage 1b,
















nob =

New volume of voids
















Initial volume



















Vv + ∆V1b







V
















=



















= no +




1b



















Vo





































V



















F ∆V I







o














































= no + G







J
































































H Vo




K 1b
















F

V I





































Here,

G




J

is the value of ∆V/Vo at 1b and is negative.













H

Vo K

1b





































EMBANKMENT DAMS

519

Here, it has been implicity assumed that there is no drainage during the construction (loading stage) and drainage is only during the periods when construction is stopped temporarily for dissipation of pore pressure. However, there will be some drainage during the construction which can be accounted for by applying a suitable dissipation factor at the end of each step of construction.


(v) Using now 1b as origin [ i.e.,u = 0 and (∆V/ Vo) = 0], plot ∆u v/s ∆V/Vo and σ v/s ∆V/Vo using steps (ii) and (iii) with new values of porosity nob and degree of saturation, Sob
computed in step iv.
(vi) Likewise, computations can be carried out for further stages of construction, if any. (vii) Plot σ versus u on a separate graph sheet from the above results,
Above steps of computations have been illustrated in the following example.

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