(C).Workout flakiness & elongation index from the following Flakiness Index Weights of aggregates passing through corresponding slot=w2=200 grams Weight of aggregates retained on a particular sieve =w1=700 grams Workout flakiness index? Elongation Index Weights of aggregates Retained on gauge length =w2=190 grams Weight of aggregates retained on a particular sieve =w1=540 grams Workout elongation index? Are these flakiness & elongation index within permissible range? Also mention bad effects of greater amount of flaky & elongated particles.

(C).Workout flakiness & elongation index from the following Flakiness Index Weights of aggregates passing through corresponding slot=w2=200 grams Weight of aggregates retained on a particular sieve =w1=700 grams Workout flakiness index? Elongation Index Weights of aggregates Retained on gauge length =w2=190 grams Weight of aggregates retained on a particular sieve =w1=540 grams Workout elongation index? Are these flakiness & elongation index within permissible range? Also mention bad effects of greater amount of flaky & elongated particles.

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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Give me right answer according to the given choices. Urgent please
A. Given properties of cement, sand and gravel is presented in the table below: Specific Density, Materials Gravity, kg/m3 Gs Cement 3.22 1510 Sand 2.75 1688 Gravel 2.58 1529 Solve for the number of bags, volume of sand, gravel and water for a 35m³ of concrete. B. Cement having a specific gravity of 3.35 and aggregates having a specific gravity of 2.65 and 2.67 for sand and gravel, respectively. Compute for the mass required for cement, sand, gravel and water. 3. Use 25 L of water and 1:2:4 design mix for problems 2A and 2B.
Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: (Note: Write answer with 2 decimal places) Volume of bucket = 1/2 cu.ft Weight of empty bucket = 20.3 Ib Weight of bucket filled with dry rodded coarse aggregate: Trial 1 = 76.6 Ib Trial 2 = 75.1 Ib Trial 3 = 78.8 Ib 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. {Note: Bulk Dry specific gravity must be multiplied to the unit weight of water to obtain percent voids)
Can you please write how did you come up with the answer
express your answers in the nearest three decimal numbers refer to MM = 08 and DD = 26
Please solve with a step by step solution and a clear understanding so I can understand
KINDLY ANSWER IT CORRECTLY AND CLEARLY THANKS
I need answer for this please.
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 = 14 LWeight of empty bucket = 9.21 kgWeight of bucket filled with dry rodded coarse aggregate:Trial 1 = 34.75 kgTrial 2 = 34.06 kgMeasureSampleA B CWet Mass (g) 521.0 522.4 523.4Dry Mass (g) 491.6 491.7 492.1Absorption (%) 2.5 2.4 2.3Table P 5 . 9 determine the amount of water required to increase the moisture content of aggregate to reach absorption?5.9 Three samples of fine aggregate have the properties shown in Table Trial 3 = 35.74 kga. Calculate the average dry-rodded unit weightb. If the bulk dry specific gravity of the aggregate is 2.620, calculate thepercent voids between aggregate particles for each trial.
Void Content—Calculate the void content in the aggregate using the bulk densitydetermined by either the rodding, jigging or shoveling procedure, as follows:
if total sample mass of aggregate is 164.9 grams, then the percent passing from sieve size #30 only is equal to ………………
Calculate the fineness modulus.