Faculty & Staff Profiles

Gradation Solution

Geology and Soil Mechanics, UW-Stout

Gradation Curve Solution

To perform this analysis, one needs to calculate the percent by weight passing each sieve.  (Note:  When the units of grams or kilograms are "called" weights, it is technically incorrect.  It should be mass retained.  You may encounter this incorrect usage in the field and this is why I'm exposing you to it.)

  1. Calculate the total weight of the sample:  W = 43 + 195 + 281 + 127 + 44 + 25 + 135 = 850 g

  2. Calculate the percent passing each sieve (example for sieve No. 4)
    weight that got through sieve #4 = 850 - 195 - 43 = 612 g
    percent by weight that got through sieve #4 = (612/850)x100% = 72%

 

U.S Sieve Size

Weight Retained (g)

% passing

3/8 in.

43

95

No. 4

195

72

No. 10

281

39

No. 40

127

24

No. 100

44

19

No. 200

25

16

pan (<0.075mm)

135

(N/A)

The gradation curve parameter D50 refers to the average sieve opening that would allow 50% by weight to pass through.  Thus, after drawing a smooth curve that best fits the data points on the gradation graph (shown below), follow the 50% passing horizontal line (red) over to the right until it intersects the curve.  At this point on the graph drop vertically down and read the sieve diameter off of the horizontal axis.

grad-soln-graph

This indicates that D50 has a value of about 2.8 mm. 
Similarly, D10 = 0.03 mm, D60 = 3.4 mm, D30 = 1.2 mm and
CU = D60/D10 = 3.4/0.03 = 113
CC = (D30)2/(D60.D10) = (1.2)2/( (3.4)(0.03) ) = 14

It should be noted that this soil sample is about 16 percent fines.  The portion of the curve to the right of No. 200 sieve is an "educated guess" or extrapolation  The same with the part of the curve that is left of 3/8 in sieve.