A sample of dry coarse-grained material was taken through a nest of sieves and the following results were obtained: Sieve no. Opening (mm) Mass retained (g) 4 4.75 10 2.00 14.8 20 0.85 98.0 40 0.425 90.1 100 0.15 181.9 200 0.075 108.8 Pan 6.1 Using the particle size distribution curve, find the uniformity coefficient. 5.9 O 2.9 O 4.2 O 3.3 O.

A sample of dry coarse-grained material was taken through a nest of sieves and the following results were obtained: Sieve no. Opening (mm) Mass retained (g) 4 4.75 10 2.00 14.8 20 0.85 98.0 40 0.425 90.1 100 0.15 181.9 200 0.075 108.8 Pan 6.1 Using the particle size distribution curve, find the uniformity coefficient. 5.9 O 2.9 O 4.2 O 3.3 O.

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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Given below is the result of a sieve analysis done on a soil sample. Sieve No. Diameter Percent Passing % 4 4.750 95 10 2.000 89 20 0.850 71 40 0.425 47 60 0.250 21 100 0.150 14 200 0.075 7 Pan --- 0 Determine the value of the coefficient of uniformity. Round off to three decimal places.
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Situation 3. Following are the results of a sieve analysis. Determine: a. Cu b. Cc c. So Sieve No. 4 6 10 20 40 60 100 200 Pan Diameter (mm) 4.760 2.380 2.000 0.840 0.420 0.250 0.149 0.074 Soil Sample Mass Retained (g) 100 180 86 160 105 134 110 65 45
Problem 1: A sieve analysis of 1,135 grams of fine rock yields the following data in the table. The size of a sieve Restrained in each a Through a sieve(gr) sieve (gr) % in 1135 ½ in 56 1079 % in. 171 908 NO.4 262 646 NO.8 203 443 NO.30 182 261 NO.50 68 193 NO.100 57 136 NO.200 51 85 Pan 85 The total weight of example 1135 grams
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Given below is the result of a sieve analysis done on a soil sample.Sieve No. Diameter Percent Passing %4 4.750 9510 2.000 8220 0.850 7340 0.425 5360 0.250 23100 0.150 14200 0.075 7Pan --- 0Determine the value of the coefficient of uniformity
a. determine the effective size diameter d. determine the coefficient of uniformity c. determing the coefficient curvature The results of sieve analysis are shown below. Sieve No. Opening Size (mm) Mass Retained (g) 4 4.750 10 2.000 18.5 20 0.850 53.2 40 0.425 90.5 60 0.250 81.8 100 0.150 92.2 200 0.075 58.5 Pan 26.5
The figure and the table below, shows the results of consolidation test for a soil sample, Find the cc value. TI Applied pressure, kPa 25 50 100 200 400 800 Voids ratio 0.634 0.619 0.583 0.523 0456 0383 0.65 0.6 0.55 0.5 0.45 0.4 of 0.35 10 100 1000 PRESSURE 0.15 0.21 0.12 no one 0.30 VOIDS RATIO
Given below is the result of a sieve analysis done on a soil sample. Sieve No. Diameter Percent Passing % 4 4.750 96 10 2.000 81 20 0.850 72 40 0.425 50 60 0.250 27 100 0.150 14 200 0.075 5 Pan --- 0 Determine the value of the coefficient of gradation/curvature.
Given below is the result of a sieve analysis done on a soil sample. Sieve No. Diameter Percent Passing % 4 4.750 95 10 2.000 83 20 0.850 69 40 0.425 50 60 0.250 27 100 0.150 13 200 0.075 7 Pan --- 0 Determine the value of the coefficient of uniformity. ROUND OFF TO THREE DECIMAL PLACES. USE STORED VALUES.
Given below is the result of a sieve analysis done on a soil sample. Sieve No. Diameter Percent Passing % 4 4.750 97 10 2.000 82 20 0.850 71 40 0.425 42 60 0.250 27 100 0.150 13 200 0.075 8 Pan --- 0 Determine the value of the coefficient of gradation/curvature. Round off to three decimal places.