Chapter 2

1 .For a soil, suppose that D10 = 0.08 mm, D30 = 0.22 mm, and D60 = 0.41 mm. Calculate the uniformity coefficient and the coefficient of gradation.

Solution:

2. Repeat Problem 1 with the following: D10 = 0.18 mm, D30 = 0.32 mm, andD60 = 0.78 mm.

Cu = D60/ D10 = 0.78/0.18 = 1.8

Cz = D230/ (D60) (D10) = 0,322/(0.78)(0.18) = 0.71

3.

a. Determine the percent finer than each sieve and plot a grain-size distribution curve.b. Determine D10, D30, and D60 from the grain-size distribution curve.c. Calculate the uniformity coefficient, Cu.d. Calculate the coefficient of gradation, Cc.

Solution:

c,) Cu = D60/ D10 = 0.30/0.09 = 3.3333

d.) Cz = D230/ (D60) (D10) = 0,172/(0.30)(0.09) = 1.0703

4, Following are the results of a sieve analysis. Make the necessary calculations anddraw a particle-size distribution curve

SolutionThe following table can now be prepared.

5. For the particle-size distribution curve shown in Figure 2.28 determine a. D10, D30, and D60 b. Uniformity coefficient, Cu c. Coefficient of gradation, Cz

Solution:

Part aFrom Figure 2.28,

D10 = 0.15 mmD30 = 0.17 mmD60 = 0.27 mm

Part bCu = D60/ D10 = 0,27/.15 = 1.8

Cz = D230/ (D60) (D10) = 0.172/(0.27)

(0.15)

= 0.71

6. For the particle-size distribution curve shown in Figure 2.28, determine the percentages of gravel, sand, silt, and clay-size particles present. Use the Unified Soil Classification System.SolutionFrom Figure 2.28, we can prepare the following table

7. The particle-size characteristics of a soil are given in this table. Draw the particlesize distribution curve.

a) According USCS % Gravel = 0% % Sand = % passing on 4.74mm- % passing on 0.075mm =100-93=7% % Silt& clay = % Fines=100%-% gravel-%sand = 100-0-7=93%

b) According AASHTO % Grave l = 0% % Sand = % passing on 4.74mm- % passing on 0.075mm

=100-93=7% % Silt = % passing on 0.075mm- % passing on 0.002mm = 93-40=53% % Clay = % passing on 0.002mm = 40%