Sieve Analysis & Gradation Flashcards
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According to ASTM C136, what is the correct order of sieves when stacking them for a sieve analysis?
Answer: Largest opening at the top, pan at the bottom
Sieves are stacked with the largest opening at the top and progressively smaller openings below, with the pan at the bottom to catch material passing the finest sieve.
What is the fineness modulus (FM) of a fine aggregate if the cumulative percentages retained on the standard sieves are: No.100=95, No.50=80, No.30=55, No.16=30, No.8=10, No.4=2?
Answer: 2.72
FM = (95+80+55+30+10+2) / 100 = 272/100 = 2.72; fineness modulus sums cumulative percent retained on the six standard sieves divided by 100.
When performing a washed sieve analysis per ASTM C117, after washing the sample over the No. 200 sieve, the material retained must be:
Answer: Dried to constant mass in an oven at 230±9°F (110±5°C)
After washing, the retained material is oven-dried to constant mass at 110±5°C (230±9°F) before performing the dry sieve analysis.
The nominal maximum size of an aggregate is defined as:
Answer: The sieve size one step larger than the first sieve to retain more than 10%
Nominal maximum size is the smallest sieve opening through which the entire amount of aggregate is permitted to pass, which is one sieve size larger than the maximum size.
A sieve is considered overloaded during a sieve analysis when:
Answer: The mass of material retained exceeds the limits specified in ASTM C136 for that sieve size
ASTM C136 specifies maximum mass limits per unit area for each sieve size; exceeding these limits causes particles to stack and prevents accurate separation.
Which gradation type is characterized by a gap or break in the particle size distribution, meaning certain intermediate sizes are absent or underrepresented?
Answer: Gap-graded
Gap-graded aggregate has a discontinuous particle size distribution where one or more intermediate sieve sizes have little or no material retained on them.
If a coarse aggregate sample has an original dry mass of 5,000 g, and after washing and drying the mass is 4,925 g, what is the percent passing the No. 200 sieve by washing?
Answer: 1.50%
Percent passing No. 200 = ((5000 - 4925) / 5000) × 100 = (75/5000) × 100 = 1.50%.