Geotechnical questions 1 of 2 HW 1 Principle of Geotechnical Engineering Due Wed

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Geotechnical questions 1 of 2 HW 1 Principle of Geotechnical Engineering Due Weds Sep 20 2.1 For a soil with Ds-0.42 mm, Dy-0.21 mm, and Do -0.16 mm, calculate the uniformity coefficient and the coefficient of gradation. The following are the results of a sieve analysis: U.S. sieve no Mass of soill retained (g) 2.3 28 10 20 128 221 100 24 a. Determine the percent finer than each sieve and plot a grain-size distribution curve b. Determine Dio- Do and D from the grain-size distribution curve. e. Calculate the uniformity coefficient, C d. Calculate the coefficient of gradation, Ce 2.7 The following are the results of a sieve and hydrometer analysis. Sieve numberl grain sire finer than 100 170 80 0.04 mm 002 mm 001 mm 0.005 mm 0.0015 mm 26 15 a. Draw the grain-size distribution curve. b. Determine the percentages of gravel, sand, silt, and clay according to the MIT c. Repeat Part b according to the USDA system. d. Repeat Purt b according to the AASHTO system. 3.8 when the moisture content of a soil is 26%, the degree of saturation is 72%, and the moist unit weight is 108 lb/it. Determine: a. Specific gravity of soil solids b. Void ratio c. Saturated unit weight 3.9 For a given soil, the following are known: G,-2.74, moist unit weight, -20.6 Kumo, and moisture content, w-16.6%. Determine: a. Dry unit weight

Explanation / Answer

Solutions:

ans 2.1-

coefficient of uniformity= D60 / D10

=0.42/ 0.16

= 2.625

coefficient of gradation= D302 / D60*D10

=0.212 / 0.42* 0.16

=0.65625

Ans 3.8= Given : ( all symbols according to convention)

w=26% , S=72% , Y moist= 108 lb/ft3

we know,

Y = (G + e* S) * Yw / (1+e)

e= void ratio

G= specific gravity

Yw = unit weight of water= 62.4

also G*w= e*S

putting the values we get,

108 = (G+G*0.26) 62.4 / 1 +(G*0.26 / 0.72)

or, 1.7307= 1.26 G / (1 + 0.361 G)

G= 2.725

thus, G*w= e*S

2.725* 0.26 = e* 0.72

e=0.984=1

Ysat = (G + e)*Yw / 1+e

=116.654 lb/ft3

Ans 3.9- Solution:

G=2.74

w=16.6% = 0.166

Y(moist)= 20.6 kN/m3

Y= (G+G*w)*Yw / 1+e

20.6= (2.74 + 2.74* 0.166) * 9.81 / 1+e

e=0.52

porosity =n= e/ (1+e)

0.52/ 1.52

=0.3427= 34.27%

G*w= e*S

2.74*0.166= 0.52*S

S= 87.469%

dry unit weight =Yd = G*Yw / (1+e)

=17.68 kN/ m3

Ans 3.10=

a. Y= ( G + e*S) Yw / 1+e

= (2.74 + 0.52*0.9) 9.81 / (1 +0.52)

=20.704 kN/m3

previously, given unit weight= 20.6 kN/m3

thus, water to be added= 20.7- 20.6= 0.1 kN

for S=100%

Y= (2.74 +0.52)* 9.81 / 1+0.52

=21.04 kN/m3

water added= 21.04-20.6= 0.44 kN

Ans 3.11=

w=23%=0.23

G=2.73

Y(moist)=1750 kg/m3

Yw = 1000kg/m3 ( for water)

dry density = Y/ 1+ w

1750/ ( 1+ 0.23)

=1422.76 kg/ m3

again,

1750= (G + Gw)Yw / (1+e)

e= 0.9188=0.92

n= e/ 1+e

=0.4788=47.88%

G*w =e*S

S=68.25%

Y(saturated)= (G+e)*Yw / (1+e)

=1901.04 kg/m3

thus , mass of water to be added= 1901.04- 1750= 151.04 kg

For the graphical questions , all table data not legible properly. It first needs to be plotted in a semi -log graph paper and then D60, D30 and D10 needs to be pointed and then uniformity and gradation coefficient are to be found out according to formula given in first question.

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