Question

In: Civil Engineering

In a laboratory work a rigid bucket with a diameter of 1 ft and height of...

In a laboratory work a rigid bucket with a diameter of 1 ft and height of 1.25 ft was filled with loose aggregate. The weight of aggregate was recorded to be 110 lb and its specific gravity was found to be 2.68. Given the above data, calculate (a) the bulk density of the aggregate and (b) the void content of the aggregate sample.

Expert Solution

Ans a) We know,

Bulk Density = Weight / Volume

Given, Weight = 110 lb

Volume = (/4)()(1.25) = 0.98125

=> Bulk Density = 110 lb / 0.98125

=> Bulk Desnity = 112.1 lb/

Ans b) We know,

Volume of void = Total Volume - Volume of solid

Total volume = (/4)()(1.25) = 0.98125

Volume of solid = Weight / Density of solid

Density of solid = Specific gravity x 62.4

Specific gravity = 2.68

=> Density of soild = 2.68 x 62.4 = 167.23 lb/

=> Volume of soild = 110 lb / 167.23 lb/ = 0.65776

Hence,

=> Volume of void = 0.98125 - 0.65776 = 0.3235

Hence, void content (e) = volume of void / volume of solid

=> e = 0.3235 / 0.65776 = 0.491

Ally Wells answered 8 months ago

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