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Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: Volume of bucket = ½ ft3 Weight of empty bucket Weight of bucket filled with dry rodded coarse aggregate = 20.3 lb Trial 1 = 69.6 lb Trial 2 = 68.2 lb Trial 3 = 71.6 lb %3D a. Calculate the average dry-rodded unit weight b. If the bulk dry specific gravity of the aggregate is 2.620, calculate the per- cent voids between aggregate particles for each trial.

Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: Volume of bucket = ½ ft3 Weight of empty bucket Weight of bucket filled with dry rodded coarse aggregate = 20.3 lb Trial 1 = 69.6 lb Trial 2 = 68.2 lb Trial 3 = 71.6 lb %3D a. Calculate the average dry-rodded unit weight b. If the bulk dry specific gravity of the aggregate is 2.620, calculate the per- cent voids between aggregate particles for each trial.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: Volume of bucket = ½ ft³ Weight of empty bucket = 20.3 lb Weight of bucket filled with dry rodded coarse aggregate Trial 1 = 69.6 lb Trial 2 = 68.2 lb Trial 3 = 71.6 lb a. Calculate the average dry-rodded unit weight b. If the bulk dry specific gravity of the aggregate is 2.620, calculate the per- cent voids between aggregate particles for each trial.
Transcribed Image Text:Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: Volume of bucket = ½ ft³ Weight of empty bucket = 20.3 lb Weight of bucket filled with dry rodded coarse aggregate Trial 1 = 69.6 lb Trial 2 = 68.2 lb Trial 3 = 71.6 lb a. Calculate the average dry-rodded unit weight b. If the bulk dry specific gravity of the aggregate is 2.620, calculate the per- cent voids between aggregate particles for each trial.
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