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 bucket = ½ ft³Weight of empty bucket = 20.3 lbWeight of bucket filled with dry rodded coarse aggregateTrial 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.

aTrial-1γd1=W2-W1V=69.6-20.3 lb0.5ft3=98.6lbft3Trial-2γd2=W2-W1V=68.2-20.3…

Answered: Coarse aggregate is placed in a rigid…

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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Q-3: Two aggregates (A1 and A2) were obtained from the field. A sieve analysis was performed on each sample and the samples were each brought to the oven dry condition (OD) and weighed. Data was obtained for each of aggregates and is presented in tables below. a) Plot the gradation for each aggregate on the graph. Indicate which gradation curvebelongs to which aggregate.(Show all necessary calculation steps in a table)b) Determine the gradation type of each aggregate.c) Determine the fineness modulus (FM) for A1 and A2. Indicate which aggregate iscoarser and which is finer.d) Determine the total percent moisture content of each aggregate, A1 and A2.Describe the field moisture state of each aggregate as dry or wet by comparingwith its absorption capacities.e) Determine the mass of each sample in the SSD state. For each sample, how muchadd water or dry the sample to bring it to SSD state?f) If for A1 and A2 masses under water are 302.0 g and 319.0 g, respectively,determine oven dry…
Compressive Strength is conducted for each aggregate with 3 specimens for each no. of days of curing. The recorded maximum load when the specimen breaks are as follows 7 Days (kN) | 486.73 14 Days (kN) 28 Days (kN) 678.8 ... 746.3 3/8" Sample 513.14 694.2 754.7 360.71 679.8 759.8 7 Days (kN) 14 Days (kN) 28 Days (kN) 231.86 339.8 377.7 3/4" Sample 252.04 339.6 381.1 206.69 343.3 379.9 Compute the compressive strength, in psi, for each size of aggregate on each no. of days of curing. Please indicate if the sample passed the required strength for each no.of days of curing. Conversion: 1lb-f = 4.488 N Attach your solution on the box below, image type.
The table below presents the results of the sieve analysis conducted on a sample of aggregate. Given the bulk specific gravities of 2.65 for coarse aggregate, 2.69 for fine aggregate, and 2.73 for filler, determine the bulk density of the aggregate sample. Additionally, calculate the maximum aggregate size and nominal aggregate maximum size. Sieve size (mm) 37.5 % Passing 100 Specification limit 25 92 100 90-100 19 83 76-90 12.5 9.5 4.75 2.36 0.3 71 62 53 32 15 56-80 48-74 29-59 19-45 5-17 0.075 4 2-8
(14) IfC 400 kg/m: W 200 kg/m: Air content 4% and CA (SSD) = 840 kg/m. Using the volume method, the weight of fine aggregate (SSD) (kg/m) is (a) 855 (b) 890 (c) 794 (d) 870 (e) 750
What should be the observation and analysis for sieve analysis of fine aggregate if the value of fineness modulus is 3.7, percent error is 1.8%, coefficient of uniformity is 4.1 and Coefficient of curvature is 1.1? Also, what is the conclusion for this?
Samples of coarse aggregate from a stockpileare brought to the laboratory for determination of specific gravities. Thefollowing weights are found: (5 points) 1. Mass of moist aggregate sample as broughtto the laboratory: 5,298 g 2. Mass of oven dried aggregate: 5,216 g 3. Mass of aggregates submerged in water:3,295 g 4. Mass of SSD (Saturated Surface Dry)Aggregate: 5,227 g Find 1. The aggregate bulk dry specific gravity 2. The aggregate apparent specific gravity 3. Absorption (report as percent)
Thoroughly mix the aggregate sample and reduce it to an amount suitable for testing, using a sample splitter or by quartering. The minimum sample size should be as follows: Minimum Mass, kg Fine aggregate with at least 95% passing 2.36-mm sieve 0.1 Fine aggregate with at least 85% passing 4.75-mm sieve 0.5 Coarse aggregate with a nominal maximum size of 9.5 mm 1 Coarse aggregate of a nominal maximum size of 12.5 mm 2 Coarse aggregate of a nominal maximum size of 19.0 mm Coarse aggregate of a nominal maximum size of 25.0 mm 10 Coarse aggregate of a nominal maximum size of 37.5 mm 15
Analysis of Fine AggregatesFill in the table:
Give me right answer according to the given choices. Urgent please
2) How should the moisture contents of the three samples compare? 3) Why is it important to determine the moisture content of aggregates? 4) Describe the natural moisture content (in general terms rather than numerical terms) of aggregate stockpiles and how it is dealt with in the construction industry.
Fine aggregate sample was tested to find its physical properties. The test results were as follows: Bulk specific gravity= 2.42, Apparent specific gravity= 2.58, mass of the vessel full of water = 1350 gm, mass of the vessel plus sand and topped up with water= 1600 gm, mass in wet condition 541 gm. Calculate 1- absorption% 2- moisture content% 3- Bulk density 4- Apparent density
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