The bulk unit weight of the dry-rodded aggregate inside a rigid container is 2800 kg/m3. 25 percent of the volume is occupied by the voids. What is the bulk specific gravity of the aggregate?
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The bulk unit weight of the dry-rodded aggregate inside a rigid container is 2800 kg/m3. 25 percent of the volume is occupied by the voids. What is the bulk specific gravity of the aggregate?
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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 = 1/3 ft3Weight of empty bucket = 18.5 lbWeight of bucket filled with dry rodded coarse aggregate = 55.9 lba. Calculate the dry-rodded unit weightb. If the bulk dry specific gravity of the aggregate is 2.630, calculate the percent voids in the aggregate.Coarse aggregates from two stockpiles having coarse aggregate angularity (crushed faces) of 40% and 90% were blended at a ratio of 30:70 by weight, respectively. What is the percent of crushed faces of the aggregate blend?The bulk unit weight of the dry-rodded aggregate inside a rigid container is 25 kN/m3. 35 percent of the volume is occupied by the voids. What is the bulk specific gravity of the aggregate?aaaaaaaaaaaaaaaaaaaaaaaa
- A sample of wet aggregate weighed 297.2 N. After drying in an oven, this sample weighed 281.5 N. Theabsorption of this aggregate is 2.5%. Determine the saturated surface dry weight of the aggregates.Coarse aggregate is placed in a rigid bucket and compacted with a tamping rod to determineits unit weight. The following data are obtained:• Volume of bucket = 0.5 ft3• Weight of empty bucket = 20.3 lb• Weight of bucket filled with dry compacted coarse aggregate = 76.8 lba) Calculate the dry-compacted unit weight of the aggregate.b) If the bulk dry specific gravity of the aggregate is 2.620, calculate the percent voids.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 = 0.3 cuft Weight of empty bucket is 19.8 lb Weight of bucket filled with dry rodded coarse aggregate = 57.6 lb If the bulk dry specific gravity of the aggregate is 2.63 and unit weight of water = to 62.4lb/ft^3, calculate the percent voids in the aggregate (one decimal only). Input magnitude only.
- We have four aggregates with specific gravity values of 2.46, 2.82, and 2.78 in equal proportions, e.g. 1/3 each. What is the specific gravity of the aggregate blend?Calculate the percent voids between aggregate particles that have been compacted by rodding, if the bulk dry-rodded unit weight is 90 lb/ft3 and the bulk dry specific gravity is 2.70.500 gm oven dry fine aggregate was tested to find its physical properties. The output from the test was as follows: SSD = 510 gm, mass of the vessel full of water = 1200 gm, mass of vessel plus sand and topped up with water= 1500 gm, mass in wet condition 518 gm, Apparent specific gravity= 2.5 . The grading of the fine aggregate is in the table below. Answer the followings:
- 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 = 12 ft3Weight of empty bucket = 20.3 lbWeight of bucket filled with dry rodded coarse aggregate:Trial 1 = 76.6 lbTrial 2 = 75.1 lb Trial 3 = 78.8 lba. Calculate the average dry-rodded unit weightb. If the bulk dry specific gravity of the aggregate is 2.620, calculate the percent voids between aggregate particles for each trial.The mass of aggregate B has been reduced and weighed to conduct an aggregate test shown in Table 2. (a) Calculate the: (i) dry bulk specific gravity (ii) SSD bulk specific gravity,Aggregates from three sources having bulk specific gravities of 2.753, 2.649, and 2.689 were blended at a ratio of 70:20:10 by weight, respectively. What is the bulk specific gravity of the aggregate blend?