HOMEWORK SOLUTION

HW #2

GEOTECHNICAL ENGINEERING I

(Question no. 3.1, 3.2, 3.5, 3.7, & 3.9 ..Al-Khafaji textbook)

3.1 Explain the main differences between free water and hydration water as it relates to soil particles.

Free water            : Water which occupies the void of the soil which has primary influence on soil behavior, for granular soil it has no effect/influence on the individual particle structure when its removed but for clay is vice versa.

Hydration water    : As a part of the crystal structure. Usually it cannot be removed by oven dry at 110o C except for gypsum & some tropical clay.

3.2 Determine the height of capillary rise above the free water table if the soil is silt with D10 = 0.055 mm. How long would it take for the water to achieve 50% of its maximum capillary rise and what would the corresponding pore water pressure be? Assume that k=10-6 cm/s.

Solution:

hc = 1.5/0.0055=272.73 cm

Z/hc =50%=0.5

See table 3.6 for uniform inorganic silt (assumed)

e max = 1.1

e min = 0.4

take the average           e avg    = (1.1+0.4)/2 = 0.75

h     = e / (1+e) = 0.43

t = h . hc/K [  z/hc + ln (1  z/hc) ]

= (0.43 . 272.73/10-6)  [ 0.5 + ln (1- 0.5) ]

= 22651123.14 second = 262.2 days

Pore water pressure

p = gw . h = 9.81 . (272.73/(2.100))

= 13.38 N/m2

3.5 A soil sample has a void ratio of 0.70 and a degree of saturation of 0.80. The specific gravity of solids is 2.70. Use SI units and assume gw = 9.81 kN/m3, then compute:

(a) Dry unit weight (gd )

(b) Bulk unit weight (gm)

(c) Saturated unit weight (gsat)

(d) Water content (w)

Solution:

(a)    gd  = Gs. gw/ (1+e)

= 2.7 x 9.81 / (1+0.7)

= 15.581 kN/m3

(b) gm = gd (1+w)                            where   w = e. Sr/Gs = 0.7 . 0.8/ 2.7  = 0.207

= 15.581 (1+ 0.207)

= 18.81 kN/m3

(c)gsat = (Gs+e). gw  /(1+e) = (2.7+0.7) . 9.81 / (1+0.7) = 19.62 kN/m3

(d) w= e. Sr/Gs = 0.7 . 0.8/ 2.7  = 0.207

3.7 A soil sample extracted from below the ground water table was determined to have a water content of 24% and a specific gravity of 2.75. Calculate the void ratio (e) and the saturated unit weight (gsat).

Solution:

e = w.Gs/Sr = 0.24 x 2.75 / 1 = 0.66

gsat = (Gs + e ) gw / (1+e) = (2.75+0.66) 9.81 / (1+0.66) = 20.15  kN/m3

or,

gsat = (Gs + e ) gw / (1+e) = (2.75+0.66) 62.4 / (1+0.66) = 128.18  pcf

3.9 For a saturated soil, given gd = 16 kN/m3 and w = 28%, determine:

(a) Saturated unit weight (gsat)

(b) Void ratio (e)

(c) Specific gravity of the solid (Gs)

(d) Water content at 70% saturation (w at Sr=70%)

Solution:

(a)    gsat = gd (1+w) = 16 (1+0.28) = 20.48  kN/m3

(b)  e = (gsat - gd )/ (gd - gb) = (gsat - gd )/ (gd  (gsat-gw)

=  (20.48  16)/(16  (20.48  9.81) = 4.48 / 5.33 = 0.84

(c) Gs = gd (1+e) / gw                or,   Gs = [gsat (1+0.84)/9.81]  - 0.84 =3

= 16(1+0.84)/9.81

= 3

(d) w  at Sr = 70%

w = Sr . e/Gs = 0.7 x 0.84 / 3 = 0.196 = 19.6%