Materials for Civil and Construction Engineers (4th Edition)
4th Edition
ISBN: 9780134320533
Author: Michael S. Mamlouk, John P. 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.
ProCoarse aggregate is placed in a rigid bucketand rodded with a tamping rod to determine itsunit weight. The following data are obtained:
Volume of bucket = 1/3cuftWeight of empty bucket is 18.5 lbWeight of bucket filled with dry rodded
coarse aggregate = 55.9lb
a. Calculate the dry-rodded unit weightb. If the bulk dry specific gravity of theaggregate is 2.63, calculate the percent voids inthe aggregate.
Chapter 5 Solutions
Materials for Civil and Construction Engineers (4th Edition)
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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- 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?arrow_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 = 0.481 %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_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
- 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.arrow_forward5.14 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_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_forwardH.W. Fine aggregate sample was tested to find its physical properties. The test results were as follows: Bulk specific gravity= 2.45, Apparent specific gravity= 2.5, mass of the vessel full of water = 1300 gm, mass of the vessel plus sand and topped up with water= 1500 gm, mass in wet condition 518 gm. Calculate 1- absorption% 2- moisture content% 3- Bulk density 4- Apparent density А-D % Absorption = *100 D D Apparent Specific Gravity = D- (B-C) SSD Aggregate (A) OD Aggregate (D) Container with H,O Container with H2O SSD Bulk Sp.Gr. = (C) and with A-(В-C) Aggregate (В) Weight of Agg. (WA) Sp.Gr.= Weight of an equal volume of water (V,*p) WA PA Density of Agg. VA*Pw Pw Density of Water Aarrow_forwardH.W. Fine aggregate sample was tested to find its physical properties. The test results were as follows: Bulk specific gravity= 2.45, Apparent specific gravity= 2.5, mass of the vessel full of water = 1300 gm, mass of the vessel plus sand and topped up with water= 1500 gm, mass in wet condition 518 gm. Calculate 1- absorption% 2- moisture content% 3- Bulk density 4- Apparent densityarrow_forward
- H.W. Fine aggregate sample was tested to find its physical properties. The test results were as follows: Bulk specific gravity= 2.45, Apparent specific gravity 2.5, mass of the vessel full of water = 1300 gm, mass of the vessel plus sand and topped up with water= 1500 gm, mass in wet condition 518 gm. Calculate 1- absorption% 2- moisture content% 3- Bulk density 4- Apparent density A-D % Absorption = D D- (B-C) Apparent Specific Gravity = Container SSD Aggregate (A) OD Aggregate (D) Container SSD Bulk Sp.Gr. = with H,O (C) with H,O and with A-(B-C) Aggregate (B) Weight of Agg. (W,) Sp.Gr.= Weight of an equal volume of water (V,*P) PA Density of Agg. Pw Density of Waterarrow_forwardPlease help, thanks. 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.arrow_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_forward
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Aggregates: Properties; Author: nptelhrd;https://www.youtube.com/watch?v=49yGZYeokKM;License: Standard YouTube License, CC-BY