EM 1110-2-1100 (Part III)
30 Apr 02
EXAMPLE PROBLEM III-1-1
FIND:
The statistics of the sediment size distribution shown in Figure III-1-2 and their qualitative
descriptions.
GIVEN:
The needed phi values are: φ05 = 0.56, φ16 = 0.80, φ25 = 0.93, φ50 = 1.37, φ75 = 1.87, φ84 = 2.08,
and φ95 = 2.48.
SOLUTION:
In phi units, the median grain size is given as:
φ50 = Mdφ = 1.37φ
From Equation 1-1b, the median grain size in millimeters is found as:
Md = 2-1.37 = 0.39 mm
From Equation 1-2, the mean grain size is found in phi units as:
Mφ = (0.80 + 1.37 + 2.08) / 3 = 1.42φ
From Equation 1-1b, this is converted to millimeters as:
D = 2-1.42 = 0.37 mm
From Equation 1-3, the standard deviation is found as:
σφ = (2.08 - 0.80)/4 + (2.48 -0.56)/6.0 = 0.32 + 0.32 = 0.64 φ
From Equation 1-4, the coefficient of skewness is found as:
αφ = 0.055 + 0.078 = 0.13
From Equation 1-5, the coefficient of kurtosis is found as:
βφ = 1.92 / 2.29 = 0.84
Thus, using Tables III-1-2 and III-1-3, the sediment is a medium sand (Wentworth classification),
it is moderately well sorted, and the distribution is fine-skewed and platykurtic.
Coastal Sediment Properties
III-1-13
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