Chapter 2 1 .For a soil, suppose that D10 = 0.08 mm, D30 = 0.22 mm, and D60 = 0.41 mm. Calculate the uniformity coeffici
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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 D60 = 0.78 mm.
=
0.18 mm, D30 = 0.32 mm, and
Cu = D60/ D10 = 0.78/0.18 = 1.8 Cz =
D2 30/ (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 =
D 2 30/ (D60) (D10) = 0,172 /(0.30)(0.09) = 1.0703
4, Following are the results of a sieve analysis. Make the necessary calculations and draw a particle-size distribution curve
Solution The 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 a From Figure 2.28, D10 = 0.15 mm D30 = 0.17 mm D60 = 0.27 mm
Part b Cu = D60/ D10 = 0,27/.15 = 1.8 Cz =
D
2
(0.27)(0.15) = 0.71
30/ (D60) (D10) =
0.17
2
/
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. Solution From 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%