Soil-HW51

DEPARTMENT OF CIVIL ENGINEERING

CEG133/ V77 GEOTECHNICAL ENGINEERING

ASSIGNMENT NO. : 3

Weight-Volume Relationship TOPIC

PREPARED BY:

Reyes, Reycelyn D. DATE SUBMITTED:

_____________

SUBMITTED TO: __ ENGR. GERARDO DB. ABESTILLA____

3.1 For a given soil, the following are given: Gs = 2.67; moist unit weight, γ = 17.61 KN/m3, and moisture content, w = 10.8%. Determine: a.) Dry unit weight b.) Void ratio c.) Porosity d.) Degree of saturation

Solution: γ

a) γd =

=

1+𝑤 γ𝑤 𝐺𝑠

b) γd =

17.61 1.108

= 15.89 kN/m3

1+𝑒 (2.67)(9.81)

15.89 =

1+𝑒

e = 0.65 c) n =

e 1+𝑒

=

0.65 1.65

= 0.39

d) Se = wGs S =

(0.108)(2.67) 0.65

x 100%

= 44.50% 3.2 Refer to Problem 3.1. Determine the weight of water, in KN, to be added per cubic meter of soil for: a.) 80% DEGREE OF SATURATION b.) 100% DEGREE OF SATURATION

Solution: a. ) S =

v𝑤 𝑣𝑣

∆S = 80 – 44.50 = 35.5% n =

v𝑤

0.39

=

𝑣𝑇

; VT = 1m3 v𝑣 1

Vv = 0.39 0.355 =

v𝑤 0.39

;

Vw = (0.355)(0.39) Vw = 0.140 m3 Ww = 0.140 x 9.81 = 1.37 Kn

b. ) ∆S = 100 - 44.5 = 55.5 S =

v𝑤 𝑣𝑣

; 0.555 =

Vw = 0.218 m3 Ww = 0.138 x 9.81 = 2.14 kN

v𝑤 0.39

3.3 A 0.4-m3 moist of soil sample has the following: Moist mass = 711.2 kg Dry mass = 623.9 kg Specific gravity of soil solids = 2.68 Estimate: a. Moisture content b. Moist density c. Dry density d. Void ratio e. Porosity

Solution: a. ) w =

711.2−623.9 623.9

( 100)

w = 13.99% 711.2

b. ) ρ =

0.4 ρ

c.) ρd =

= 1778 kg/m3

1+ w 1778

ρd =

1+ .1399

ρd = 1559.79 kg/m3 G𝑠 ρ𝑤

d. ) ρd =

1+e

e = 0.7182 e. ) n =

e 1+ e

n = 0.4180 3.4 For a given sand, the maximum and minimum void ratios are 0.78 and 0.43, respectively. Given Gs = 2.67, determine the dry unit weight of the soil in kN/m3 when the relative density is 65%.

Solution: Dr =

e𝑚𝑎𝑥 − e 𝑒𝑚𝑎𝑥 −e𝑚𝑖𝑛

; 0.65 =

0.78− e 0.78−0.43

e = 0.5525 γd =

(2.67)(9.81) 1.5525

= 16.87 kN/m3

3.5 In its natural state, a moist soil has a volume of 9.35 x 10-3 m3 and weighs 177.6 x 10-3 kN. The oven-dry weight of the soil is 153.6 x 10-3 kN. If Gs = 2.67, calculate the moisture content, moist unit weight, dry unit weight, void ratio, porosity, and degree of saturation.

Solution: 177.6𝑥10−3

γm =

9.35x10−3

γm = 18.99 kN/m3 w =

(177.6x10−3)−(153.6x10−3) 153.6x10−3

x 100

w = 15.625% 153.6x10−3

γd =

9.35 𝑥 10−3

γ = 16.423Kn/m3 G𝑠 γ𝑤

γd =

1+e

16.428 =

(2.67)(9.81) 1+e

e = 0.5944 e

n =n+ e = 0.3728 S =

𝑤G𝑠 e

=

S =

G 𝑠 γ𝑤 1+e 0.15625(2.17) 0.5944

x 100

S = 70.19 % 3.6 For a given sandy soil, emax = 0.75, emin = 0.46, and Gs = 2.68. What will be the moist unit weight of compaction (kN/m3) in the field if Dr = 78% and w = 9%?

Solution: Dr =

e𝑚𝑎𝑥 − e e𝑚𝑎𝑥 − e𝑚𝑖𝑛

; 0.78 =

0.75 − 𝑒 0.75−0.46

E = 0.5238 γd =

G 𝑠 γ𝑤 1+𝑒

=

(2.68)(9.81) 1.5238

= 17.25 kN/m3 γd =

γ𝑚𝑜𝑖𝑠𝑡 1+𝑤

; 17.25 =

γ𝑚𝑜𝑖𝑠𝑡 1+𝑤

γmoist = 18.81kN/m3

3.7 The moist weight of 5.66 x 10-3 of a soil is 102.3 x 10-3 kN. The moisture content and the specific gravity of the soil solids are determined in the laboratory to be 11% and 2.7, respectively. Calculate the following: a. Moist unit weight (kN/m3) b. Dry unit weight (kN/m3) c. Void Ratio d. Porosity e. Degree of saturation (%) f. Volume occupied by water (m3)

Solution: V𝑚𝑜𝑖𝑠𝑡

a. ) γmoist =

=

𝑉

102.3 x 10−3 5.66 𝑥 10−3

γmoist = 18.07 kN/m3 γ𝑚𝑜𝑖𝑠𝑡

b.) γd =

=

1+ 𝑤

18.07 1.11

γd = 16.28 kN/m3 G 𝑠 γ𝑤

c. ) γd =

; 16.28 =

1+𝑒

(2.7)(9.81) 1+𝑒

e = 0.63 d. ) h =

𝑒 1+𝑒

0.63

=

= 0.39

1.63

e. ) Se = wGs; S(0.63) = (0.11)(2.7) S = 47.39% f. ) s =

V𝑤 V𝑣

0.39 =

; n =

V𝑣 V

V𝑣 5.66x10−3 V𝑤

0.4739 =

; Vv = 2.180 x10-3

2.180 x 10−3 -3

Vw = 1.03 x 10

;

m3

3.8 For a given sandy soil, the maximum and minimum dry unit weights are 16.89 kN/m3 and 14.46 kN/m3, respectively. Given Gs = 2.65, determine the moist unit weight of this soil when the relative density is 60% and the moisture content is 8%.

γd max = 16.98 kN/m3

0.60 =

γd(min) = 14.46 kN/m3

1 1 − 14.46 γ𝑑 1 1 − 14.46 16.98

Gs = 2.65

γd

= 15.87 kN/m3

Dr = 60%

γm = γd(1 + w)

w = 8%

γm = 15.87 ( 1+0.08) γm = 17.14 kN/m3

3.9 The saturated unit weight of a soil is 19.8 kN/m3. The moisture content of the soil is 17.1%. Determine the following: a. Dry unit weight b. Specific gravity of soil solids c. Void ratio

Solution: γsat =19.8 kN/m3 w = 17.1% wGs = Se 0.171Gs = e (G𝑠 +𝑒)(γ𝑤 )

γsat =

1+ e

; 19.8 =

(G𝑠 +0.171G𝑠 )(9.81) 1+ 0.171G𝑠

Gs = 2.44 (G𝑠 γ𝑤 )

γd =

1+ e

γsat = γd (1 + w) 19.8 = γd (1+0.171) γd = 16.91 kN/m3 e = 0.171 (2.44) e = 0.4172 3.11 The unit weight of a soil is 14.94 kN/m3. The moisture content of this soil is 19.2% when the degree of saturation is 60%. Determine: a. Void ratio b. Specific gravity of soil solids c. Saturated unit weight

Solution: a. ) γ

=

Se

(G𝑠 +𝑒)(γ𝑤 )

( +Se)γ𝑤 w

1+ e

1+ e

; 14.94 =

14.94 =

(0.6𝑒) +0.6𝑒)(9.81) 0.192

(

1+ e

e = 0.69 b. ) Se = wGs; (0.6)(0.69) = (0.192)Gs Gs = 2.16 c. ) γsat =

(G𝑠 +𝑒)(γ𝑤 ) 1+ e

=

(2.16+0.69)(9.81) 1.69

γsat = 16.54 kN/m3

3.12 A saturated soil has the following characteristics: initial volume(Vi) = 24.6 cm3, final volume (Vf) = 15.9 cm3, mass of wet sol (m1) = 44 g, and mass of dry soil (m2) = 30.1 g. Determine the shrinkage limit.

Solution: SL = SL =

m1 −m2

(100)-

V 𝑖 −V𝑓

(ρw)(100)

m2 44−30.1

m2 24.6−15.9

30.1

30.1

(100)-

(1g/cm3)(100)

SL = 17.28% 3.13 The moist unit weights and degrees of saturation of a soil are given in the table. γ (kN/m3) S(%) 16.62 50 17.71 75 Determine: a. e c. Degree of saturation b. Gs d. Mass of water, in kg/m3, to be added to reach full saturation

Solution: a. ) w = 18% Gs = 2.73 1680

ρd =

1+0.18

ρd = 1423.73 kg/m3 (1+0.18)(2.73)(1000)

b. ) 1680 =

1+e

e = 0.9175 n =

e 1+e

= 0.4785

c.) Se = wGs S =

0.18(2.73) 0.9175

(100)

S = 53.56% d. ) ρsat =

(G𝑠 +𝑒)(ρ𝑤 ) 1+e

(100)

ρsat = 1902.22kg/m3 Mass of water ρsat – ρ = 1902.22 – 1680 = 222.22 kN/m3

3.15 Refer to Problem 3.14. Determine the weight of the water, in kN, that will be in 0.0708m3 of the soil when it is saturated.

Solution: n = n =

e 1+e V𝑣 V

0.8

=

= 0.44

1.8

; 0.44 =

V𝑣 0.0708

Vv = 0.0315 m3 S =

V𝑤 V𝑣

; 1 =

V𝑤 V𝑣

; Vw = Vv = 0.0315 m3

Ww = Vwγw =0.0315 x 9.81 Ww = 0.309 kN

3.16 The dry density of a soil is 1780 kg/m3. Given Gs = 2.68, what would be the moisture content of the soil when saturated?

γd =

1780 x 9.81

γd =

G 𝑠 γ𝑤

1000

1+e

= 17.46 kN/m3

; 17.46 =

2.68 x 9.81 1+e

e = 0.5056 Se = wGs ; (1)(0.5056) = w(2.68) w = 18.87%

3.17

The porosity of a soil is 0.35. Given Gs = 2.69, calculate: a.Saturated unit weight ( kN/m3) b. Moisture content when moist unit weight = 17.5 kN/m3

Solution: G 𝑠 γ𝑤

γsat =

1+e (2.69+0.54)(9.81)

=

1.54

= 20.59 kN/m3 n

e =

1−n 0.35

=

1−0.35

= 0.54 b.) γmoist = 17.54 kN/m3 γ =

(G𝑠 +𝑆𝑒)γ𝑤 1+e

17.5 =

(2.69+5(0.54))(9.81) 1.54

S = 10.11% Se = wGs (0.1011)(0.54) = w(2.69) w = 2.02%