A sample of coarse aggregate has an oven dry weight of 1034.0 g and a moisture content of 4.0 %. The saturated surface dry weight is 1048.9g and the weight of the aggregate in water is 675.6 g. Determine using phase volume relationships: a) Apparent Specific Gravity (GA) b) Bulk Specific Gravity (GB) c) Bulk Specific Gravity SSD (GB (SSD)) d) Absorption, % e) Bulk Volume

Question

23-01-2020
Engineering

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A sample of coarse aggregate has an oven dry weight of 1034.0 g and a moisture content of 4.0 %. The saturated surface dry weight is 1048.9g and the weight of the aggregate in water is 675.6 g. Determine using phase volume relationships: a) Apparent Specific Gravity (GA) b) Bulk Specific Gravity (GB) c) Bulk Specific Gravity SSD (GB (SSD)) d) Absorption, % e) Bulk Volume

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DeniceSandidge DeniceSandidge · Answer 1 · 2020-01-24T12:04:41+01:00

Answer:

Apparent Specific Gravity = 2.88

bulk specific gravity = 2.76

Bulk Specific Gravity SSD = 2.80

absorption = 1.44%

bulk volume = 373.3

Explanation:

given data

oven dry weight A = 1034.0 g

moisture content = 4.0 %

saturated surface dry weight B = 1048.9 g

weight of the aggregate in water C = 675.6 g

solution

we get here Apparent Specific Gravity that is express as

Apparent Specific Gravity = [tex]\frac{A}{A-C}[/tex] ..........1

put here value

Apparent Specific Gravity = [tex]\frac{1034}{1034-675.6}[/tex]

Apparent Specific Gravity = 2.88

and

now we get bulk specific gravity that is

bulk specific gravity = [tex]\frac{A}{B-C}[/tex] ...................2

put here value

bulk specific gravity = [tex]\frac{1034}{1048.9-675.6}[/tex]

bulk specific gravity = 2.76

and

now we get Bulk Specific Gravity SSD

Bulk Specific Gravity SSD = [tex]\frac{B}{B-C}[/tex] ............3

Bulk Specific Gravity SSD = [tex]\frac{1048.9}{1048.9-675.6}[/tex]

Bulk Specific Gravity SSD = 2.80

and

now absorption will be here as

absorption = [tex]\frac{B-A}{A}[/tex] × 100% ................4

absorption = [tex]\frac{1048.9-1034}{1034}[/tex] × 100%

absorption = 1.44%

and

last we get bulk volume that is

bulk volume = [tex]\frac{weight\ displce\ water}{density\ water }[/tex]

bulk volume = [tex]\frac{1048.9-675.6}{1}[/tex]

bulk volume = 373.3