Question

1. The results of a sieve analysis on a soil sample were: Sieve Size (mm) Mass Retained (g) 0.0 5.5 25.7 23.1 22.0 17.3 12.7 6.9 0.6 0.3 0.15 0.063 2.3 grams passed the 63 um (or 0.063 mm) sieve. Plot the particle size distribution curve and estimate the following: (a) (b) (C) percentage of sand percentage of gravel percentage of cobble effective size uniformity coefficient coefficient of curvature. (e)

Answer 1

sieve size	Mass Retained	cumulative mass retained	percentage mass retained	% finer than
mm	g	g	%
10	0	0	0.00	100
6	5.5	5.5	4.76	95.24
2	25.7	31.2	27.01	72.99
1	23.1	54.3	47.01	52.99
0.6	22	76.3	66.06	33.94
0.3	17.3	93.6	81.04	18.96
0.15	12.7	106.3	92.03	7.97
0.063	6.9	113.2	98.01	1.99
PAN	2.3	115.5	100.00	0.00

Partical size distribution curve 100 95.24 72.99 % finer than 6052.99 33.94 40 18.96 7.97 1.99 0.01 0.1 sieve size mm, in log scale

A.

% sand ( 0.063 mm to 2 mm ) = 98.01-27.01% = 71%

B.

% Gravel (2 mm to 4 mm ) =

4.76	6
?	4
27.01	2
15.885

% Gravel (2 mm to 4 mm ) = 27.01-15.88 % = 11.13%

C. % cobbles (64 mm to 256 mm) = 0%

d,e,f

Effective size = D50

	For D10			For D30			For D60			For D50
Cu= D60/D10, Cc= D30^2/D60/D10		0.3	18.96		0.6	33.94		2	72.99		1	52.99
	D10	?	10		?	30		?	60		?	50
		0.15	7.97		0.3	18.96		1	52.99		0.6	33.94
	D10	0.178		D30	0.521		D60	1.351		D60	0.937
		Cu	7.600		Cc	1.131

Effective size = D50 = 0.937 mm

Cu uniformity coefficient

7.600

Cc coefficient of curvature

1.131

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