Note: Give me right solution according to the question.Graph of the Grain Size Distribution Curve SIEVE ANALYSISSeive No.Sieve + Soil (g)Empty Sieve (g)3/4"NA1/2"NA3/8"NA# 4505.4497.8#10502.1473.6#20670.2633.1#40405.2367.3#60402.3339.1#100360.3322.2#200373.4356.1PAN362357.4

Answered: SIEVE ANALYSIS Seive No. Sieve + Soil…

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 following table shows the lab results of a sieve analysis and Atterberg limits for an inorganic soil sample. The Grain Size graph is attached. What is the soil USCS Designation symbol? Sieve Sample Passing Sieve Size Size/No. (mm) Percentages 3" 75 100% 3/4" 19 100% No. 4 4.75 45% No. 10 2 35% No. 20 0.850 0.630 No. 40 0.425 22% No, 60 0.250 18% No. 100 0.150 7% No. 200 0.075 3% 0.020 0.002 20% LL 1% PL
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.
Find the Cu of the given sample. identify the answer in this choices. Weight Percentage Cumulative Percentage Sieve retained finer S.NO Weight retained (gm) percentage retained (C)N = (100 - C) Select one: size (gm) {W,/ W,}x 100 O a. 8.98 1 75mm 100 O b. 18.33 63mm 100 О с. 2.92 3 50mm 100 O d. 6.22 4 37.5mm 454 22.7 22.7 77.30 28mm 285 14.25 36.95 63.05 20mm 207 10.35 47.30 52.70 7 14mm 318 15.90 63.20 36.80 8 10mm 206 10.30 73.50 26.50 9 6.3mm 118 5.90 79.40 20.60 10 5.0mm 102 5.10 84.50 15.50 11 3.35mm 64 3.20 87.70 12.30 12 2.0mm 88 4.40 92.10 7.90 13 1.18mm 35 1.75 93.85 6.15 14 600µ 42 2.10 95.15 4.85 15 300µ 33 1.65 97.60 2.40 16 150µ 18 0.90 98.50 1.50 17 75µ 22 1.10 99.6 0.40 18 Pan 8 0.40 100 In the pan we have = 100 W,= 2000 -99.6 = 0.4 % of material 2. 5.
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 88 20 0.850 70 40 0.425 50 60 0.250 21 100 0.150 13 200 0.075 5 Pan --- 0 Determine the value of the coefficient of uniformity. Round off to three decimal places.
Correct answer please.
Complete the table. Show handwritten solution for each value in the table. Sieve Number 4 8 16 30 50 80 100 200 Pan Mass of Empty Sieve 515 480 424 401 372 347 341 340 364 Mass of Sieve + Soil Retained 637 612 606 664 611 670 679 600 693 Soil Retained Percent Retained Percent Passing
ASAP PLS ANSWER CORRECTLY. do not round off in initial values, round off final ans only plz ty
Explain LAP SPLICING. Illustrate a sample figure about it.
From a particle size distribution curve, it is found that 60% of particles are finer than 4.5 mm and the coefficient of uniformity is 6. It is also known that the coefficient of curvature is 2. Then 30% particles are finer than which size. 1 decimal place in mm) (up to
Given below is the result of a sieve analysis done on a soil sample. Percent Passing % 96 83 67 46 Sieve No. 4 10 20 40 60 100 200 Pan Diameter 4.750 2.000 0.850 0.425 0.250 0.150 0.075 26 13 8 0 Determine the value of the coefficient of uniformity.
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.

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SIEVE ANALYSIS Seive No. Sieve + Soil (g) Empty Sieve (g) 3/4" NA 1/2" NA 3/8" NA # 4 505.4 497.8 #10 502.1 473.6 #20 670.2 633.1 #40 405.2 367.3 #60 402.3 339.1 #100 360.3 322.2 #200 373.4 356.1 PAN 362 357.4