EFFECTIVE STRESS AND PORE WATER PRESSURE
5.5 EFFECTIV E PRESSURE DUE TO CAPILLAR Y WATER RIS E IN SOIL The term wate r level, wate r table an d phreatic surfac e designate th e locus o f the levels to which
water rises in observation wells in free communication with the voids of the soil at a site. The water table can also be defined as the surface at which the neutral stress uw in the soil is equal to zero.
If the water contained in the soil were subjected to no force other than gravity, the soil above the wate r table woul d be perfectly dry . In reality, every soil i n the fiel d i s completely saturate d above this level up to a certain height. The water that occupies th e voids of the soil located abov e the water table constitutes soil moisture.
If the lower part of the mass o f dry soil comes int o contact with water, the water rises in the voids to a certain height above the free water surface. The upward flow into the voids of the soil is attributed to the surface tension of the water. The height to which water rises above the water table against th e force of gravity is called capillary rise. The height of capillary ris e is greatest fo r very fine graine d soil materials. The water that rises abov e th e water table attains the maximu m heigh t hc onl y i n th e smalle r voids . A fe w larg e void s ma y effectivel y sto p capillary rise in certain parts of the soil mass. As a consequence, only a portion o f the capillary zone abov e th e fre e wate r surfac e remain s full y saturate d an d th e remainde r i s partiall y saturated.
The seat of the surface tension is located a t the boundary between air and water. Within the boundary zone the water is in a state of tension comparable to that in a stretched rubber membrane attached t o th e wall s o f th e void s o f a soil . However , i n contras t t o th e tensio n i n a stretched membrane, th e surfac e tension in the boundary film o f water is entirely unaffecte d b y eithe r the contraction or stretching of the film. The water held in the pores of soil above the free water surface is retained in a state of reduced pressure. This reduced pressure is called capillary pressure o r soil moisture suction pressure.
The existence o f surface tension can be demonstrated a s follows:
2 TeL cos a
Figure 5. 5 Needl e smeare d wit h grease floating on wate r
A grease d sewin g needle , Fig . 5.5 , ca n b e mad e t o floa t o n wate r becaus e wate r ha s n o affinity t o grease, and , therefore, the water surface curves down under the needle until the upward component o f the surface tension is large enough to support the weight of the needle. I n Fig. 5.5 , 7^ is the surface tension per unit length of the needle an d Wn the weight of the needle. The upward vertical force due to surface tension is 2TL co s a, wher e L is the length of the needle. Th e needle floats whe n this vertical force i s greater than the weight of the needle Wn acting downwards.
Rise o f Wate r i n Capillar y Tube s
The phenomenon o f capillary rise can be demonstrated b y immersing the lower end of a very small diameter glass tube into water. Such a tube is known as capillary tube . As soon as the lower end of the tube comes into contact wit h water, the attraction betwee n th e glass and the water molecule s combined wit h the surface tension of the water pulls the water up into the tube to a height hc above the water level as shown in Fig. 5.6(a). The height hc is known as the height of capillary rise . The upper surface of water assumes the shape of a cup, called the 'meniscus' tha t joins the walls of the tube at an angle a known as the contact angle.
On the other hand, if the tube is dipped int o mercury a depression o f the surface develops i n the tube below th e surfac e o f the mercury, with the formation of a convex meniscu s a s shown in Fig. 5.6(b). The reason for the difference between th e behavior of water and mercury resides in the different affinit y betwee n th e molecule s o f th e soli d an d wate r o r mercury . I f ther e i s a stron g affinity betwee n the molecules of the solid and the liquid, the surface of the liquid will climb up on the wal l of the solid unti l a definite contact angl e a i s established. Th e contact angl e betwee n a clean mois t glas s surfac e an d wate r i s zero , tha t is, th e wate r surfac e touche s th e glas s surfac e tangentially. For the case of a dry glass surfac e and water, a is not a constant. It may be as high as 45° at first then gradually reducing to much smaller values. Probably the inevitable contaminatio n of surface s cleane d b y ordinar y methods , an d th e humidit y o f ai r ar e responsibl e fo r suc h variations. Fig . 5.6(c ) show s th e contac t angle s betwee n wate r an d th e surface s unde r differen t conditions.
Surface Tensio n
Surface tension is a force that exists at the surface of the meniscus. Along the line of contact between the meniscus in a tube and the walls of the tube itself, the surface tension, Ts, is expressed a s the force per unit length acting in the direction of the tangent as shown in Fig. 5.7(a). The components o f this force along the wall and perpendicular to the wall are
Along the wall = T co s a pe r unit length of wall
Effective Stres s an d Por e Water Pressur e 151
Water (a)
Meniscus T,
Glass tube
Meniscus
Glass tube Convex meniscu s
\
Mercury(b)
a = 0 0 < a < 45° a > 90°
Moist glas s Dr y glass Greas y glas s surface surfac e surfac e (c)
Figure 5.6 Capillar y ris e and meniscus Normal to the wall = Ts si n a pe r unit length of wall.
The force normal to the wall tries to pull the walls of the tube together and the one along the wall produces a compressive forc e in the tube below the line of contact.
The meniscu s ca n b e visualize d a s a suspensio n bridg e i n thre e dimension s whic h i s supported o n th e wall s o f th e tube . Th e colum n o f wate r o f heigh t hc belo w th e meniscu s i s suspended fro m thi s bridg e b y mean s o f the molecula r attractio n o f the wate r molecules . I f th e meniscus ha s stoppe d movin g upwar d in th e tube, the n ther e mus t be equilibriu m betwee n th e weight of the column of water suspended from the meniscus and the force with which the meniscus is clinging to the wall of the tube. We can write the following equation of equilibrium
TidT cos « = or h = 4T co s a
(5.21) The surface tension Ts for water at 20 °C can be taken as equal to 75 x 10~8 kN per cm. The surface tensions of some of the common liquid s are given in Table 5.2.
Equation (5.21) can be simplified by assuming a = 0 for moist glass and by substituting for Ts. Therefore, for the case of water, the capillary height hc can be written as
h = 47" 4x75xlO- 8xl06 03 d dyw dx9.Sl
In Eq. (5.22) h an d d are expressed in cm, and, v = 9.81 kN/m3.
(5.22)
Table 5. 2 Surfac e tension o f som e liquid s at 2 0 ° C
Liquids 7" kN/c m x 1CT 8
Ethyl Alcohol Benzene
Carbon Tetra Chloride Mercury
Petroleum Water
22.03 28.90 26.80 573.00 26.00 75.00
Stress Distributio n i n Water Belo w th e Meniscu s
Figure 5.7(b ) show s a capillar y tub e wit h it s botto m en d immerse d i n water . Th e pressur e i s atmospheric at points A and B. Since point C is at the same leve l a s A, according to the laws of hydraulics, the pressure at C is also atmospheric. Since the point D which is just below the meniscus is higher than point C by the head hc, the pressure at D must be less than atmospheric by the amount hcyw. Therefore, th e pressure at any point in water between C and D is less than atmospheric. That means, the water above point C is in tension if we refer to atmospheric pressure a s zero pressure . The tension in water at any height h above C is given by hyw. By contrast, the pressure in the water below th e fre e surfac e A i s abov e atmospheri c an d therefor e i s i n compression . Th e stres s distribution in water is given in Fig. 5.7(b).
Tr si n a -* -a
Water
-*- Ts sin a a
- Glass tube
(a) Forces due to surface tension
Tension Stress distribution
\
(b)
uc = 4 Tsld
Water \ Capillar y tube wall under compressio n
(c)
Figure 5. 7 Capillar y pressure
Effective Stres s an d Por e Water Pressur e
Thus th e tension uw in water immediately below the meniscus is given by 47 cos a
153
(5.23) If rm is the radius of the meniscus, Fig. 5.7(a) , we can write,
r = d or d = 2r co s a
m 2cos a
Substituting for d in Eq. (5.23), w e have u = —4Ts cos a
2r cos a 2Ts
r (5.24)
It may be noted here that at the level of the meniscus the magnitude o f the capillary pressur e u tha t compresses th e wall of the tube is also equal to the capillary tension in the water just below the meniscus. The magnitude of the capillary pressure uc remains constant with depth as shown in Fig. 5.7(c ) wherea s th e capillar y tension , uw, in wate r varie s fro m a maximu m o f hcYw a t th e meniscus level to zero at the free wate r surface level as shown in Fig. 5.7(b).
Capillary Ris e o f Wate r i n Soils
In contrast to capillary tubes the continuous voids in soils have a variable width. They communicate with each other in all directions and constitute an intricate network of voids. When water rises into the network from below, the lower part of the network becomes completel y saturated . In the upper part, however, the water occupies onl y the narrowest voids and the wider areas remain filled wit h air. Fig. 5.8(a) shows a glass tube filled with fine sand. Sand would remain fully saturated only up to a heigh t h' whic h i s considerabl y smalle r tha n hc. A fe w larg e void s ma y effectivel y sto p capillary rise in certain parts . The water would rise, therefore, to a height of hc only in the smalle r voids. The zone between the depths (hc - h'J wil l remain partially saturated .
sampleSoil s
V~
**,"/:•• :-v
ft
•V°-j :!••'£
/ 1r
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)ry zone f TCMf 15 0cm
T
Partially - ewg 10 0 hc - h'c saturate d &zone = 3
c , '§ ,
?? s n
<+- ,, Saturate d 2
"c £i zon e O B
i 3 y
A /)
/ ^
"10 1. 0 0. 1 0.0 1 0.00 1 Grain size (mm) log scale (a) Height of capillary ris e (b) Rate of capillary ris e in soil consisting of
uniform quartz powde r Figure 5.8 Capillar y ris e in soil s
Capillary rise of water
/~z£r\
_ T,
////////SS/////S/S/////////////////////////S//7///////S/7////S//S/7/S//S
Figure 5.9 Capillar y siphonin g
The heigh t of the capillary rise i s greatest fo r very fine graine d soil s materials , bu t the rate of ris e i n suc h material s i s slo w becaus e o f thei r lo w permeability . Fig . 5.8(b ) show s th e relationship between th e height of capillary rise in 24 hours and the grain size of a uniform quartz powder. This clearly show s that the rise is a maximum for materials falling in the category o f silts and fine sands .
As the effective grai n siz e decreases, the size o f the voids also decreases , an d the height of capillary rise increases. A rough estimation of the height of capillary ris e can be determined fro m the equation,
C
h^^D~eLJ\Q (5 '25)
in which e is the void ratio, DIQ is Hazen's effective diameter i n centimeters, an d C is an empirical constant which can have a value between 0.1 and 0.5 sq. cm.
Capillary Siphonin g
Capillary force s ar e able to raise water against the force of gravity not only into capillary tubes or the void s in columns o f dry soil , bu t also int o narrow ope n channel s or V-shaped grooves . I f the highest point of the groove is located below the level to which the surface tension can lift the water, the capillary forces will pull the water into the descending part of the groove an d will slowly empty the vessel. This proces s i s known as capillary siphoning. Th e same process ma y also occur in the voids of soil. For example, water may flow ove r the crest of an impermeable core in a dam in spite of the fac t tha t th e elevation o f the free wate r surfac e i s below th e crest of the core as show n in Fig. 5.9.
Capillary Pressur e i n Soil s
The tension uw in water just below the meniscus is given by Eq. (5.23) as 4T cost f
Since thi s pressure i s below atmospheri c pressure , i t draws the grains of soils closer to each other a t al l points wher e th e menisc i touc h the soi l grains . Intergranular pressure o f thi s type is called capillary pressure. The effective o r intergranular pressure a t any point in a soil mass can be expressed b y
of = a-u, (5.26 )
Effective Stres s an d Por e Water Pressure Capillary fring e
155
(b) (c ) (d ) (e )
Figure 5.1 0 Effec t o f capillar y pressur e uc on soil vertical stres s diagra m where ot is the total pressure, t f i s the effective o r the intergranular pressure an d uw is the por e water pressure. Whe n the water is in compression uw is positive, an d when it is in tension uw is negative. Since uw is negative in the capillary zone, the intergranular pressure i s increased b y uw. The equation, therefore, can be written as
of = at-(-uw) = at + uw (5.27 )
The increas e i n th e intergranula r pressur e du e t o capillar y pressur e actin g o n th e grain s leads to greater strengt h of the soil mass .
Stress Conditio n i n Soil due to Surfac e Tension Force s
It is to be assumed here that the soil above the ground water table remains dry prior to the rise of capillary water. The stress condition in the dry soil mass changes due to the rise of capillary water.
Now conside r th e soi l profile given in Fig. 5.10(a). When a dry soil mass above th e GWT comes in contact with water, water rises by capillary action. Let the height of rise be hc and assume that the soil withi n this zone becomes saturate d due to capillary water . Assume that the menisci formed a t heigh t hc coincid e wit h th e groun d surface . Th e plan e o f th e menisc i i s calle d th e capillary fringe.
The vertical stress distribution of the dry soil mass is shown in Fig 5.10(b). The vertical stres s distribution o f th e saturate d mas s o f soi l i s give n i n Fi g 5.10(d) . Th e tensio n i n th e wate r i s maximum at the menisci level, say equal to uw and zero at the GWT level as shown in Fig. 5.10(e).
Prior to capillary rise the maximum pressure of the dry mass, rfd, a t the GWT level is
where, yd = dry unit weight of soil.
After th e capillary rise , th e maximum pressure of the saturate d weight of soil a t the GWT level is
Since th e por e wate r pressur e a t the GW T leve l i s zero , i t i s obviou s that th e differenc e between th e two pressures o/sat and tf d represent s th e increase i n pressure du e to capillary ris e which is actually the capillary pressure, which may be expressed a s
Mr ~~ i-Wsa t ~ ~ I A) (.3. )
By substitutin g for
^, an d l + e
in Eq . (a) , w e have, after simplifyin g
l + e ct (5.28)
where, e = void ratio, n = porosity
It is clear fro m Eq . (5.28 ) tha t the capillary pressure fo r soi l i s directly proportional t o the porosity o f the soil and this pressure is very much less than h./ whic h is used only for a fine bore and uniform diameter capillary tube.
The distributio n of capillar y pressur e uc (constan t wit h depth) i s give n i n Fig . 5.10(c) . Th e following equation for the pressure at any depth z may be written as per Fig. 5.10
(5.29)
Example 5. 1
The depth of water in a well is 3 m. Below the bottom of the well lies a layer of sand 5 meters thick overlying a clay deposit. Th e specific gravity of the solids o f sand and clay are respectively 2.6 4 and 2.70. Their water contents are respectively 25 and 20 percent. Compute the total, intergranular and pore water pressures a t points A and B shown in Fig. Ex. 5.1.
Solution
The formula for the submerged uni t weight is
l + e
Since the soil is saturated,
3m 5 m
2m 2 m
Sand
Clay Figure Ex. 5. 1
Effective Stres s an d Por e Water Pressur e 15 7
_ . 9.81(2.64-1 ) 3
For sand, y, = = 9.7 kN/mJ
* 1 + 0.25x2.64
For clay, y , = 9'81(2JO " ^ = 10.83 kN/m3
* 1 + 0.20x2.70 Pressure at point A
(i) Tota l pressure = 3 x 9.7 (sand) + 6 x 9.81 = 29.1 + 58.9 = 88 kN/m2 (ii) Effectiv e pressur e = 3 x 9.7 = 29.1 kN/m2
(iii) Por e wate r pressure = 6 x 9.81 = 58.9 kN/m2
Pressure at point B
(i) Tota l pressur e = 5 x 9.7 + 2 x 10.8 3 + 10 x 9.81 = 168. 3 kN/m2 (ii) Intergranula r pressure = 5 x 9.7 + 2 x 10.8 3 = 70.2 kN/m2 (iii) Por e wate r pressure =10x9.81=98.1 kN/m2
Example 5. 2
If water in the well in example 5.1 is pumped out up to the bottom of the well, estimate the change in the pressures a t points A and B given in Fig. Ex. 5.1.
Solution
Change in pressure a t points A and B
(i) Chang e in total pressure = decreas e in water pressure du e to pumping
= 3x9.81=29.4 3 kN/m2 (ii) Chang e in effective pressure = 0
(iii) Chang e in pore water pressure = decreas e i n water pressure due to pumping
= 3x9.81=29.4 3 kN/m2
Example 5. 3
A trench is excavated i n fine sand for a building foundation, u p to a depth of 1 3 ft. The excavation was carried out by providing the necessary side supports for pumping water. The water levels at the sides an d the bottom of the trench are as given Fig. Ex . 5.3. Examin e whether the bottom o f the trench is subjected to a quick condition if Gs = 2.64 and e = 0.7. If so, what is the remedy?
Solution
As per Fig. Ex. 5.3 the depth of the water table above the bottom of the trench = 10 ft. The sheeting is taken 6.5 ft below the bottom of the trench to increase th e seepage path.
G - 1 The equation for the critical gradient is / =
l + e
If the trench is to be stable, th e hydraulic gradient, / , prevailing at the bottom shoul d be less than i . The hydraulic gradient i is
//A\V/A\\
3 f t //A\V/A\\
10ft
6.5ft
_L
Figure Ex . 5. 3
There wil l be n o quick condition if,
L l + e
From the given data 2.64-1 = 1.6 4
1 + 0.7 1. 7
* = - * = , .54 L 6. 5
It is obvious tha t h/L > ic. There will be quick condition.
Remedy:
(i) Increas e L t o at least a 1 3 ft depth belo w th e botto m o f trench s o that h/L = 0.77 whic h gives a margin o f factor of safety.
or (ii ) Kee p the water table outside the trench at a low level by pumping out water. This reduce s the head h .
or (iii ) D o not pump water up to the bottom leve l of the trench. Arrange th e work in such a way that the work may be carried ou t with some water in the trench.
Any suggestio n given above shoul d be considered b y keeping i n view the site conditions and other practical considerations.
Example 5. 4
A clay layer 3.6 6 m thick rests beneath a deposit o f submerged san d 7.92 m thick. The top of the sand is located 3.0 5 m below th e surface of a lake. The saturated unit weight of the sand is 19.6 2 kN/m3 an d of the clay is 18.36 kN/m3.
Compute (a ) th e tota l vertica l pressure , (b ) th e por e wate r pressure , an d (c ) th e effectiv e vertical pressure a t mid height of the clay layer (Refer to Fig. Ex. 5.4) .
Solution
(a) Total pressure
The total pressure cr , over the midpoint of the clay is due to the saturated weights of clay and sand layers plus the weight of water over the bed of sand, that is
Effective Stres s an d Por e Water Pressur e 159
3.05m
T '
Lake/psxXK^v&tfSS^^
7.92m.'•'••'•'• . ' '••':••':''•. Submerged sand :: ' ...'••'[•' • ' . ' / •
3.66m'
I ! '6--''v'f' '/•'••?. J'''--/*'.'"'•}.•••'•'.•• '•£•'•: .v:'-;:'*'.'''*Vj'l' :C *•••
_I_ ' • '•'•:-'-'- '* ••"''••''-."•..* •• ' ';'•":•. "v •."-'•.•'•."•.. ' •• ' ':'•":•. •* • ; '•..' •
cr, = 3.66
Figure Ex. 5. 4
x 18.36 +7.92 x 19.62 + 3.05 x 9.81 = 33.6 + 155.4 + 29.9 = 218.9 kN/m2 (b) Pore water pressure is due to the total water column above the midpoint.
That is
u.. =3.66 x 9.81 + 7.92 x 9.81 + 3.05 x 9.81 = 125.6 kN/m2 (c) Effective vertical pressure
a-u = &' = 218.9-125.6 = 93.3 kN/m2
Example 5. 5
The surfac e of a saturated clay deposit i s located permanentl y below a body o f water as shown in Fig. Ex. 5.5. Laborator y test s hav e indicated tha t the average natura l water content of the clay is 41% and that the specific gravity of the solid matter is 2.74. What is the vertical effective pressure at a depth of 37 ft below the top of the clay.
Solution
To find th e effective pressure, w e have to find first the submerged uni t weight of soil expressed a s
Yh = l + e
Now fro m Eq. (3.14a) , e = *-wG, = wGs sinc e 5 = 1
or e = 0.47 x 2.74 = 1.29 Therefore,
(2.74-1.00)x62.4=47411b/ft3
* 1 + 1.29
Lake
/A\VA\VA\VA\\//>, \VA\VA\VA\\
Clay deposit 37ft
D
/A\VA\\
A\VA\VA\VW\\
Figure Ex. 5. 5
Effective pressure , a' = 37 x 47.41 = 1754 lb/ft:
Example 5. 6
If the water level in Ex. 5.5 remains unchanged and an excavation is made by dredging, what depth of clay must be removed to reduce the effective pressure at point A at a depth of 37 ft by 1000 lb/ft2? (Fig. Ex. 5.5)
Solution
As in Ex. 5.5, yb = 47.41 lb/ft3, let the depth of excavation be D. The effective depth over the point A i s (3 7 - D ) ft . Th e dept h o f D mus t b e suc h whic h give s a n effectiv e pressur e o f (1754 - 1000 ) lb/ft3 = 754 lb/ft2
or (3 7 -D)x 47.41 =75 4
^ 37x47.41-75 4 _ 1 i r or D = = 21.1 ft
47.41 Example 5. 7
The wate r table is lowered fro m a depth of 1 0 ft to a depth of 20 ft in a deposit o f silt. All the silt is saturated even after the water table is lowered. Its water content is 26%. Estimate the increase in the effective pressur e at a depth of 34 ft on account of lowering the water table. Assume Gs = 2.7.
Solution
Effective pressur e befor e lowering the water table.
The wate r table is at a depth of 1 0 ft and the soil above this depth remains saturate d but not submerged. Th e soi l fro m 1 0 ft t o 2 0 f t remains submerged. Therefore , th e effectiv e pressur e a t 34 ft depth is
(34-10)^