Classification of Soils for Engineering Purposes - Soil Properties: Testing, Measurement, and Evaluation - page 159

(2.00 mm), and D 10 corresponds to a No. 100 sieve (0.150 mm). Sub-
stituting these data into Eqs. (11-2) and (11-3) gives

9.5
0.150 63.3

C u

2.00

2

1 2

C c

2.8

0.150 9.5

1 21 2

Because C u (63.3) is greater than 4 and C c (2.8) is between 1 and 3,
this sample meets both criteria for a well-graded gravel. Hence,
from Table 11-3, the soil is classified GW (i.e., well-graded gravel),
according to the Unified Soil Classification System.

Example 11-4
Given:
A sample of inorganic soil was tested in the laboratory with the fol-
lowing results:
1. Liquid limit

42.3%

2. Plastic limit 15.8%

3. Sieve analysis data:

U.S. Sieve Size Percentage Passing

No. 4 100

No. 10 93.2

No. 40 81.0

No. 200 60.2

Required:
Classify the soil sample by the Unified Soil Classification System.
Solution:
Because the percentage passing the No. 200 sieve is 60.2%, which
is greater than 50%, go to the lower block (labeled “Fine-Grained
Soils”) in Table 11-3. The liquid limit is 42.3%, which is less than
50%, so go to the block labeled “Silts and Clays, Liquid limit less
than 50.” Now, because the sample is an inorganic soil and the plas-
ticity index is 42.3 - 15.8, or 26.5%, which is greater than 7, refer
next to the Plasticity Chart (Figure 11-3). With a liquid limit of
42.3% and plasticity index of 26.5%, the sample is located above the
A-line. Return to Table 11-3 and go to the block labeled “CL.” Thus
the soil is classified CL according to the Unified Soil Classification
System.
Figure 11-8 lists a BASIC microcomputer program, prepared by
Stevens [5], that classifies soils by the Unified Soil Classification