Please help I'm in hurry. We learned that there are limitations on the quantity of material which may beretained on a particular sieve. This is to ensure that all particles will haveopportunity to reach sieve openings a number of times during the sievingoperation. What is the allowed quantity of material that can be retained on a25.0 mm sieve? Assume that the sieve has a 12-in diameter.O 2.9 kg6.3 kg3.2 kg4.2 kgWhich of the following is true? *OSurface moisture content is equal to the combined total evaporable moisture contentand the absorptionAbsorption is equal to difference between surface moisture content and surfacemoistureOSurface moisture content is equal to the difference between the total evaporablemoisture content and the absorptionTotal evaporable moisture content is equal to difference between surface moisturecontent and absorptionIt is very clear in the ASTM D75/D75M-19 the minimum size of field samplerequired for each aggregate size in the fine and coarse aggregate. However,instances arise when a unit of material consists of combined coarse and fineaggregates. What shall be used as the minimum size of field sample in thisinstance?O coarse aggregate's mnimum + 10 kg.O fine aggregate's mnimum + 10 kgfine aggregate's mnimumcoarse aggregate's mnimumWhich of the following sources of heat wasn't suitable for drying of aggregates *based on the standard?O Ventilated microwave overO Electric heat lampsElectric or gas hot plateHair blower*In performing ASTM C566 and when a hot plate is used in drying, it can beexpedited by adding what substance to cover the moisture content. This will beignited and allowed to burn during drying over the hot plateO Anhydrous Denatured AlcoholEthyl Alcohol000OGasolineIcopropyl Alcohol

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We learned that there are limitations on the quantity of material which may be retained on a particular sieve. This is to ensure that all particles will have opportunity to reach sieve openings a number of times during the sieving operation. What is the allowed quantity of material that can be retained on a 25.0 mm sieve? Assume that the sieve has a 12-in diameter.

We learned that there are limitations on the quantity of material which may be retained on a particular sieve. This is to ensure that all particles will have opportunity to reach sieve openings a number of times during the sieving operation. What is the allowed quantity of material that can be retained on a 25.0 mm sieve? Assume that the sieve has a 12-in diameter.

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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Question 4 Comment the four types of sieve analysis diagrams, which appear in the figure below: Continuum, Uniform, Gaps, Open. % Passing 100 O Continuum A Gaps Sieve size Open Uniform
Q.2) Answer the following question. 1- The sieve analysis below was done for which type of aggregate. Weight of sample = 2000gm Sieve size in. (mm) 3/8" 9.5 mm No.4 4.75 mm No.8 2.36 mm No.16 1.18 mm No.30 0.6 mm No.50 0.3 mm No.100 0.15 mm Pan Percent Weight retained (g) retained 0 2000*100 94 Klcc→ -2000 4032000X100 →0 358420000 360 2000 X 492-20000 72 2000 x 6² 118 982 Percent cumulative @ 4.7 > 95 20 i7-75 18 2212000x100 · 100 29 24.614 57 39 3 Percent passing 0 5 25 уз 61 86 97 317 ASTM Limit 100 95_100 80_100 40 85 25_60 10_30 2_10
4. MECHANICAL SEPARATION A sieve analysis gives the following results: Sieve size (mm) Wt. retained (g) 1.000 0 0.500 64 0.250 0.125 0.063 324 240 48 24 Through 0.063 Plot a cumulative size analysis and a size-distribution analysis, and estimate the weights, per 1000 kg of powder, which would lie in the size ranges 0.150 to 0.200 mm and 0.250 to 0.350 mm.
Q.2) Answer the following question. 1- The sieve analysis below was done for which type of aggregate. Weight of sample = 2000gm Sieve size in. (mm) 3/8" 9.5 mm No.4 4.75 mm No.8 2.36 mm No.16 1.18 mm No.30 0.6 mm No.50 0.3 mm No.100 0.15 mm Pan Percent Weight retained (g) retained 02000*100 94 →0 Klcc→ +-2000 4032000X100 72 Percent cumulative @ 2212000x100112 4.7 > 95 358420000²18 30 / 75 982 57 360 2000 Alco isies 39 4924200009 24.614 · 100 3 Percent passing 0 5 25 уз 61 86 97 317 ASTM Limit 100 95_100 SO_100 40 85 25_60 10_30 2_10
If the particle passing on sieve No.200 is 72%, passing on sieve No. 4 is 90%; what is the % of gravel. (provide the answer in percentage, value only; e.g. 20)
An aggregate sample has a compacted dry unit weight of 1615 kg/m3 and a dry specific gravity of 2.15 If this aggregate is placed in a container and compacted with the same procedure used in the determination of compacted unit weight, what will be the amount of pores, in percent of the container volume, between the aggregate particles? (density of water = 1000 kg/m3)
A saturated aggregate sample with an absolute volume of 5.05x10^-5 m^3 has a mass of 178 g before drying and 139 g after drying in an oven. Assume the water absorption by volume to be 59.18%. Determine the following: a. Specific gravity of the solid particles b. Mass specific gravity c. Void Ratio d. Porosity
From a Fine Aggregate sample, the ff data are given below to determine only the SPECIFIC GRAVITIES and ABSORPTION OF THE SAMPLE. MASS of oven dried aggregate 428.3g MASS of PYC filled with water =652.5g MASS of aggregate and water =785.3g MASS of SSD Aggregate =459.1g MASS OF AN EMPTY PYCNOMETER =125.0g MASS of oven dried aggregate 428.3g 652.5g 785.3g 459.1g 125.0g MASS of PYC filled with water MASS of aggregate and water MASS of SSD Aggregate MASS OF AN EMPTY PYCNOMETER IL || || |I
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 = 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.
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Q1: C: A sieve analysis test was performed on a sample of aggregate and produced the results shown in the Table below, Calculate the percent passing through each sieve. Plot the percent passing versus sieve size. Sieve size, mm Amount Retained, g 25 19 376.7 12.5 888.4 9.5 506.2 4.75 1038.4 2.35 900.1 1.18 891.5 0.6 712.6 0.3 625.2 0.15 581.5 0.075 242.9 Pan 44.9
Sand particles does have liquid limit ? Justify your answer