Materials for Civil and Construction Engineers (4th Edition)
4th Edition
ISBN: 9780134320533
Author: Michael S. Mamlouk, John P. Zaniewski
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Textbook Question
Chapter 5, Problem 5.12QP
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.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
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.
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.
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.
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....
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.Similar questions
- 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_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 = 2.604; absorption 0.544% Aggregate B: Buck specific gravity = 2; absorption = 8% a. What is the unit weight of Aggregate A, kn/m?arrow_forwardCalculate 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.arrow_forward
- Calculate 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_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_forwardformed? 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_forward
- 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_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 = 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.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 = 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.arrow_forward
- choose true or false If you know that weights of coarse aggregate as follow :Weight of aggregate stock = 5350 gm, weight of aggregate in oven dry= 5200 gm, weight of aggregate submerged in water =3254 gm, weight of aggregate SSD= 5220 gm , According to weights above , the oven dry bulk specific gravity is 2.325arrow_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_forwarda. What is the percent passing on 4.75 mm sieve? b. What is the fineness modulus of the fine aggregate sample?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Structural Analysis (10th Edition)Civil EngineeringISBN:9780134610672Author:Russell C. HibbelerPublisher:PEARSONPrinciples of Foundation Engineering (MindTap Cou...Civil EngineeringISBN:9781337705028Author:Braja M. Das, Nagaratnam SivakuganPublisher:Cengage Learning
- Fundamentals of Structural AnalysisCivil EngineeringISBN:9780073398006Author:Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel LanningPublisher:McGraw-Hill EducationTraffic and Highway EngineeringCivil EngineeringISBN:9781305156241Author:Garber, Nicholas J.Publisher:Cengage Learning
Structural Analysis (10th Edition)
Civil Engineering
ISBN:9780134610672
Author:Russell C. Hibbeler
Publisher:PEARSON
Principles of Foundation Engineering (MindTap Cou...
Civil Engineering
ISBN:9781337705028
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Cengage Learning
Fundamentals of Structural Analysis
Civil Engineering
ISBN:9780073398006
Author:Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher:McGraw-Hill Education
Traffic and Highway Engineering
Civil Engineering
ISBN:9781305156241
Author:Garber, Nicholas J.
Publisher:Cengage Learning
Aggregates: Properties; Author: nptelhrd;https://www.youtube.com/watch?v=49yGZYeokKM;License: Standard YouTube License, CC-BY