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

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

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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The bulk unit weight of the dry-rodded aggregate inside a rigid container is 2800 kg/m3. 25 percent of the volume is occupied by the voids. What is the bulk specific gravity of the aggregate?
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
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
VFB=??
Use the following data:Oven Dry weight - 4.98 kgWeight under water - 3.02 kgSSD Weight - 5.07 kg1. Determine the apparent specific gravity of the coarse aggregates.2.Determine the bulk specific gravity SSD of the coarse aggregates3.Determine the bulk specific gravity (Dry) of the coarse aggregates4.
what is the requirement volume of fine aggregates (s.g. = 2..4) if the mass nedded for the concrete mix is 850 kg?
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
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 = 0.481 %3D Weight of empty bucket = 22.107 Ib %3! Weight of bucket filled with dry rodded coarse aggregate Trial 1 = 69.259 lb %3D Trial 2 = 70.567 lb Trial 3 = 69.708 lb • If the bulk dry specific gravity of the aggregate is 2.727, calculate the percent voids for trial 2. where unit weight of water is 62.40
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.
Given the following information from an asphalt mix: Gsb = 2.705 Gmm = 2.534 Gmb = 2.432 Gb = 1.014 Pb = 5.7% Unit weight of water = 62.4 Ib/ft3 Assume 1 ft for total volume. Draw mass/volume relationship and determine Mass of Mix (Mm) Ib., Mass of Binder (Mb) Ib., Volume of Binder (Vb) pcf, Volume of Void-less Mix (Vmm) Volume of Air (Va) pcf, Volume of Absorbed Asphalt (Vba) pcf, Volume of Effective Asphalt (Vbe) pcf. In addition, calculate Air Voids (VA) %, Voids in Mineral Aggregate (VMA) %, and Voids Filled with Asphalt (VFA)
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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