Pearson eText for Materials for Civil and Construction Engineers -- Instant Access (Pearson+)
4th Edition
ISBN: 9780137505586
Author: Michael Mamlouk, John Zaniewski
Publisher: PEARSON+
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Textbook Question
Chapter 5, Problem 5.13QP
Calculate the percent voids between aggregate particles that have been compacted by rodding, if the dry-rodded unit weight is 72.5 lb/ft3 and the bulk dry specific gravity is 2.639.
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Calculate the percent voids between aggregate particles that have been compacted by rodding, if the dry-rodded unit weight is 88.0 lb/cu ft and the bulk dry specific gravity is 2.701.
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.
Laboratory specific gravity and absorption tests are run on two
coarse aggregate sizes, which have to be blended. The results
are as follows:
Aggregate A: Buck specific gravity = 2.604; absorption
0.544%
Aggregate B: Buck specific gravity = 2; absorption = 8%
a. What is the unit weight of Aggregate A, kn/m?
Chapter 5 Solutions
Pearson eText for Materials for Civil and Construction Engineers -- Instant Access (Pearson+)
Ch. 5 - Prob. 5.1QPCh. 5 - Discuss five different desirable characteristics...Ch. 5 - Discuss five different desirable characteristics...Ch. 5 - The shape and surface texture of aggregate...Ch. 5 - Define the following terms: a. Saturated...Ch. 5 - Three samples of fine aggregate have the...Ch. 5 - A sample of wet aggregate weighed 297.2 N. After...Ch. 5 - 46.5 kg (102.3 lb) of fine aggregate is mixed with...Ch. 5 - Samples of coarse aggregate from a stockpile are...Ch. 5 - Base course aggregate has a target dry density of...
Ch. 5 - Calculate the percent voids between aggregate...Ch. 5 - Calculate the percent voids between aggregate...Ch. 5 - Coarse aggregate is placed in a rigid bucket and...Ch. 5 - The following laboratory tests are performed on...Ch. 5 - Students in the materials lab performed the...Ch. 5 - The specific gravity and absorption test (ASTM...Ch. 5 - Prob. 5.18QPCh. 5 - Calculate the sieve analysis shown in Table P5.19...Ch. 5 - Calculate the sieve analysis shown in Table P5.20,...Ch. 5 - A sieve analysis test was performed on a sample of...Ch. 5 - A sieve analysis test was performed on a sample of...Ch. 5 - Draw a graph to show the cumulative percent...Ch. 5 - Referring to Table 5.6, plot the specification...Ch. 5 - Referring to the aggregate gradations A, B, and C...Ch. 5 - Table P5.26 shows the grain size distributions of...Ch. 5 - Table P5.27 shows the grain size distributions of...Ch. 5 - Three aggregates are to be mixed together in the...Ch. 5 - Table P5.29 shows the grain size distribution for...Ch. 5 - Laboratory specific gravity and absorption tests...Ch. 5 - Table P5.31 shows the grain size distribution for...Ch. 5 - Prob. 5.32QPCh. 5 - Laboratory specific gravity and absorption tests...Ch. 5 - Prob. 5.34QPCh. 5 - Define the fineness modulus of aggregate. What is...Ch. 5 - Calculate the fineness modulus of aggregate A in...Ch. 5 - Calculate the fineness modulus of aggregate B in...Ch. 5 - A portland cement concrete mix requires mixing...Ch. 5 - Discuss the effect of the amount of material...Ch. 5 - Aggregates from three sources having the...Ch. 5 - Aggregates from three sources having the...Ch. 5 - A contractor is considering using three stockpiles...Ch. 5 - Prob. 5.43QPCh. 5 - What are the typical deleterious substances in...Ch. 5 - Review ASTM D75 and summarize the following: a....
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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 = 0.481 %3D Weight of empty bucket = 22.107 Ib %3! Weight of bucket filled with dry rodded coarse aggregate Trial 1 = 69.259 lb %3D Trial 2 = 70.567 lb Trial 3 = 69.708 lb • If the bulk dry specific gravity of the aggregate is 2.727, calculate the percent voids for trial 2. where unit weight of water is 62.40arrow_forwardCoarse 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.arrow_forwardCalculate the percent voids between aggregate particles that have been compacted by rodding, if the dry-rodded unit weight is 1161 kg/m3and the bulkdry specific gravity is 2.639.arrow_forward
- 5.14 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 = ¹/2 ft³ Weight of empty bucket = 20.3 lb Weight of bucket filled with dry rodded coarse aggregate: Trial 176.6 lb Trial 275.1 lb Questions and Problems 217 Trial 3 78.8 lb a. Calculate the average dry-rodded unit weight b. If the bulk dry specific gravity of the aggregate is 2.620, calculate the percent voids between aggregate particles for each trial.arrow_forwardPls solve fast and correct do not solve wrongarrow_forwardCoarse 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.arrow_forward
- 2.638, 2.514, 2.437 O 2.651, 2.472, 2.364arrow_forwardIn specific gravity test for coarse aggregate we found the following results: dry weight A= 3008 grams SSD weight B = 3037 grams %3D Submerged weight C=1907 grams The SSD specific gravity Select one: O a. 2.456 O b. 2.877 O c. 2.688 O d. 2.899arrow_forwardQиestion no.08 The result of specific gravity test is given below Weight of saturated aggregate in water =618gm Weight of oven dry sample = 982 gm Weight of saturated surface dry aggregate = 990 gm (i) Calculate the specific gravity of aggregate?arrow_forward
- formed? Q1: B: The specific gravity and absorption test was performed on fine aggregate and the following data were obtained: Mass of SSD sand = 500.0 g Mass of pycnometer with water only = 623.0 g Mass of pycnometer with sand and water= 938.2 g Mass of dry sand = 495.5 g Calculate the specific gravity values (dry bulk, SSD, and apparent) and the absorption of the fine aggregate.arrow_forwardPlease show me all steps I need to be able to calculate on my own I have the answer but I want to understand how to work the problem.arrow_forwardLaboratory specific gravity and absorption tests are run on two coarse aggregate sizes, which have to be blended. The results are as follows: Aggregate A: Buck specific gravity = 2.395; absorption = 4.69% Aggregate B: Buck specific gravity = 2.601; absorption = 0.51% What is the specific gravity of a mixture of 50]% aggregate A and 50% aggregate B by weight?arrow_forward
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Aggregates: Properties; Author: nptelhrd;https://www.youtube.com/watch?v=49yGZYeokKM;License: Standard YouTube License, CC-BY