Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod todetermine its unit weight. The following data are obtained:Volume of bucket½ ft3Weight of empty bucketWeight of bucket filled with dry rodded coarse aggregate20.3 lbTrial 1 =69.6 lbTrial 2 = 68.2 lbTrial 3 = 71.6 lba. Calculate the average dry-rodded unit weightb. If the bulk dry specific gravity of the aggregate is 2.620, calculate the per-cent voids between aggregate particles for each trial.

The given data is: Volume of bucket = 1/2 ft3 Weight of Empty Bucket = 20.3 lb Weight of bucket…

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 = 1½ ft³ Weight of empty bucket Weight of bucket filled with dry rodded coarse aggregate 20.3 lb %3D 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.

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 = 1½ ft³ Weight of empty bucket Weight of bucket filled with dry rodded coarse aggregate 20.3 lb %3D 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.

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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