What is the percent passing on 4.75 mm sieve? b. What is the fineness modulus of the fine aggregate sample?
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a. What is the percent passing on 4.75 mm sieve?
b. What is the fineness modulus of the fine aggregate sample?
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- 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-CACoarse 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.If Slump=120 Maximum size of aggregate=25 How is the amount of water interpolated ?
- 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.Is it correct to say that the fineness modulus of aggregate B is 0.44 asumming aggregate B is a fine aggregate and also that the percent passing the 1.18mm, No. 16 sieve is 56%?A sieve analysis test was performed on a sample of fine aggregate and produced the following results:Sieve, mm 4.75 2.36 2.00 1.18 0.60 0.30 0.15 0.075 panAmount retained, g 0 33.2 56.9 83.1 151.4 40.4 72.0 58.3 15.6Calculate the percent passing each sieve, and draw a 0.45 power gradation chart with the use of a spreadsheet program. The specific steps or functions selected in the following description can vary somewhat depending on the version of the spreadsheet program being used.
- 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 kg 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.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.An aggregate sample was collected in a container and weighed as 528.7 g. The sample was oven dried and theoven dry weight was measured as 506.4 g. If the empty weight of the container was 79.6 g, determine the MC of theaggregates
- 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 = 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.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 =14L Weight of empty bucket = 9.21kg Weight of bucket filled with dry rodded coarse aggregate: Trial 1 =34,75 kg Trial 2 =34.06 kg Trial 3 =35.74 kg 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.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 2.325