Void Content—Calculate the void content in the aggregate using the bulk densitydetermined by either the rodding, jigging or shoveling procedure, as follows: % Voids = 100[(S X W)- MY(S × W)where:M= bulk density of the aggregate, lb/ft [ kg/m'],= bulk specific gravity (dry basis) as determined inaccordance with Test Method C 127 or Test MethodSC 128, anddensity of water, 62.3 lb/ft [998 kg/m³].W I. DATA TO BE GATHERED:Calibration:1. Mass of the measure plus plate, kg (Mm+p)11.69 kg2. Mass of the measure plus plate and water, kg (Mm+p+w)15.47 kg%3D3. Mass of the water to fill the measure, kg (Mw)3.78 kg4. Temperature of water in the measure, °C (Tw)28°Bulk Density:1. Mass of aggregate plus measure, kg (G)14.54 kg%3D2. Mass of the measure, kg (T)9.14 kg3. Volume of the measure, m³ (V)0.00379 m3

Answered: Image /qna-images/answer/c5b09c9d-3c0c-4fcb-9814-92e21ded64a1.jpg

Calibration: 1. Mass of the measure plus plate, kg (Mm+p) 11.69 kg 15.47 kg 2. Mass of the measure plus plate and water, kg (Mm+p+w) 3. Mass of the water to fill the measure, kg (Mw) 3.78 kg %3D 4. Temperature of water in the measure, °C (Tw) 28° Bulk Density: 1. Mass of aggregate plus measure, kg (G) 14.54 kg 9.14 kg %3D 2. Mass of the measure, kg (T) %3D 3. Volume of the measure, m³ (V) 0.00379 m³ %3D II II||

Calibration: 1. Mass of the measure plus plate, kg (Mm+p) 11.69 kg 15.47 kg 2. Mass of the measure plus plate and water, kg (Mm+p+w) 3. Mass of the water to fill the measure, kg (Mw) 3.78 kg %3D 4. Temperature of water in the measure, °C (Tw) 28° Bulk Density: 1. Mass of aggregate plus measure, kg (G) 14.54 kg 9.14 kg %3D 2. Mass of the measure, kg (T) %3D 3. Volume of the measure, m³ (V) 0.00379 m³ %3D II II||

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

If Slump=120 Maximum size of aggregate=25 How is the amount of water interpolated ?
Determine Admixtures in ml/m3 Fine aggregate requirements (dry), in kg/m3 Net Water Content, in kg/m3
A box measuring 50 cm x 50 cm x 50 cm is filled to the top with dry coarse aggregate of mass 187.5 kg. The water absorption and specific gravity of the aggregate are 0.5% and 2.5, respectively. The maximum quantity of water (in kg, round off to 2 decimal places) required to fill the box completely is bris
Choose one answer 1.target mean strength fm (mpa) 1. 28.12 2. 33.12 3. 40.12 4. 25.12 2.w/c ratio 1. 0.58 2. 0.61 3. 0.51 4. 0.48 3 .the free water content is (kg/m3) 1. 180 2. 205 3. 190 4 .160 4.total aggregate content is (kg/m3) 1. 1889.7 2. 1869.7 3. 1899.7 4. 1879.7 5.the coarse aggregate is (kg.m3) 1. 1150.51 2. 1170.51 3. 1130.51 4. 1190.51
Strength requirements, in MPa – 30.68 Mpa Water cement ratio, in decimal – 0.45 Coarse aggregate requirements (dry), in kg/m3 – 992.kg/m3 Air content, in percent – 3.5% Water content, kg/m3 – 184 kg/m3 Cement, in kg/m3 – 408.89 kg/m3 Determine the Admixtures in ml/m3 Fine aggregate requirements (dry), in kg/m3 Net Water Content, in kg/m3
Given the following data: Slump Requirement = 4" ements Max. Aggregate Size = 1.0" nts Water/cement Ratio = 0.55 Mass of Cement = 350.91 kg/m³ Mass of the Coarse Aggregates = 1072 kg/m³ Total Mass of the Concrete = 2375 kg/m³ Compute for the mass of Fine Aggregates.
Compute for the concrete mix proportion for a 6" diameter concrete cylinder mold with 12" height and 6"x6"x21" concrete beam mold Slump = 25mm to 50mm fc' = 3000psi at 28 days Maximum size of Coarse Aggregates (gravel) = 3/4" Unit weight of Coarse aggregate= 1650 kg/m³ Cement Type I Specific Gravity=3.15 Specific Gravity for Coarse Aggregates=2.73 Specific Gravity for Fine Aggregate=2.66 Fineness Modulus (FM)=2.63 Interpolate values that are not present/exact in each table hint: compute for the mix proportion of 1cubic meter of concrete after getting the mix proportions relate it to the volume of the structure above (concrete cylinder and concrete beam) to get the batch weights of the materials.
Problem 1: 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.
determine the effective specific gravity of aggregates. DATA: Specific Gravity: Fine Agg = 2.603 Coarse Agg = 2.740 Asphalt Cement = 1.048 Percent By Weight: Fine Agg = 47.90 Coarse Agg= 47.78 Asphalt Cement= 100-FA-CA
Standard Test Method for Ductility of Bituminous Materials 7) What is the effect of improper level of briquette placement ?8) What is the effect of rate of pulling on ductility value ?9) What is the effect of dimensions of the briquette on ductility value
Given the following data:slump requirement = 4" max aggregate size = 1" mass pf water = .193 kg/m3 water/cement ration = .48 mass of coarse aggregates = 1105.5 kg/m3 total mass of the concrete = 2375 kg/m3 compute for the mass of fine aggregate
Situation 2. Given the following data for a concrete mix:Target Concrete Strength = 40 MPaSlump = 75 mm to 150 mmWater – Cement Ratio = 0.48Unit Weight of Concrete = 23.4 KN/m 3Water Content = 180 Kg/m 3Entrapped Air = 1%Fine Aggregate = 33% of the total volume of the aggregatesSpecific Gravity of Materials:Cement = 3.15Fine Aggregate = 2.68Coarse Aggregate = 2.64 4. Calculate the required weight (kg) of cement and water per cubic meter of concrete. A. 530 C. 632 B. 587 D. 555 5. Calculate the required weight (kg) of fine aggregates per cubic meter of concrete. A. 611 C. 562 B. 745 D. 452 6. Calculate the required weight (kg) of coarse aggregates per cubic meter of concrete. A. 1222 C. 904 B. 1124 D. 1490