In the table below, find the missing values for pure sand sample and without fine materials ote (weight of sample 1000g). Sieve opening (mm) Retained weight Retained weight% Cumulative percentage Passing% 4.75 2.36 100 10 90 1.18 0.6 250 25 50 0.3 200 20 30 0.15 10 Pan

In the table below, find the missing values for pure sand sample and without fine materials ote (weight of sample 1000g). Sieve opening (mm) Retained weight Retained weight% Cumulative percentage Passing% 4.75 2.36 100 10 90 1.18 0.6 250 25 50 0.3 200 20 30 0.15 10 Pan

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: . Using the sieve results of a batch of sand and gravel shown in the table below,calculate the…

Q: 2- The following are the results of a sieve analysis: U.S. sieve no. Mass of soil retained (g) 4 10…

A: Given,

Q: QI: The following are the results of a laboratory consolidation test on a soil specimen obtained…

A: Given: Dry mass of specimen, MS = 128 g Height of specimen at beginning of test = 2.54 cm Specific…

Q: The sieve analysis has been 0/4 conducted on a soil sample and it was found that the percentage of…

A: The given figures are shown below:

Q: A liquid limit test performed on a soil sample has the

A: here water content is given to us and also N = 20

Q: 7. Given are the results from the liquid limit and plastic limit test for a soil. Calculate: a.…

Q: Q1: The results of a standard Proctor test are given in the table. The volume of the proctor mold…

A: Standard proctor test is a laboratory method used to determine the moisture content at which soil…

Q: Let us assume that a soil sample taken in the field is taken and it must be identified by the SUCS…

Q: Following are the results of a compaction test: Mass of mould 2925 3095 3150 3125 3070 + Wet soil…

Q: A sample of sand has the following properties: Moist mass = 625.2 g Dry mass = 589.9 g absorption =…

Q: A Sample of soil was tested in the laboratory with the following results Liquid limit =30% Plastic…

A: we have Liquid limit =30% Plastic limit =12% U.S. Sieve Size Percentage Passing…

Q: From the following data, solve for the degree of compaction. If this is for Item 201, Aggregate Base…

Q: Q/ What the purpose from sand content measurement, what is the acceptable percentage and size…

A: Sand content measurement is the sand volume which includes void spaces between the grains which are…

Q: A liquid limit test on a clay was performed with the following results. The natural water content of…

Q: The following are the data obtained in the shrinkage limit test: The initial weight of saturated…

Q: A wet sand sample has a volume of 4.64 x 10-4 m3 and weighs 8 N. After oven-drying, the weight…

A: Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: Given: Weight of wet sand sample (Ww) = 500 g Weight of Dry sand sample (Wd)…

Q: The grain-size characteristics of soil re given in the following table. Use Table 2.3 for the…

A: OPTION - A is correct

Q: A Standard Proctor test has yielded the values shown below. Determine the maximum dry unit weight…

A: Given Standard proctor test Volume V = 1/30 ft3

Q: From the table shows a summary of a liquid limit and plastic limit test. Compute the liquidity…

Q: A saturated soil sample, weighing 178 g has a volume of 96 cc. If the specific gravity of soil…

Q: The following results were obtained from a liquid limit test on a clay using the Casagrande cup…

A: Given data: Natural water content (wn) =38% Plastic limit (wp) =21%

Q: Hydrometer and sieve analysis were conducted on a soil sample. Find D10 Deo uniformity coefficient…

A: Given that: Grain size distribution curve:

Q: Question 19 A water content test was conducted and the data were recorded as follows: The mass of…

Q: From the tabulated data and result of determining consistency in reference to our laboratory…

Q: 4. The following data were obtained from the liquid and plastic limit tests for a soil: Liquid Limit…

Q: The following data were obtained from laboratory tests for a cohesive soil: moisture content =…

Q: -) The following are the results of a sieve analysis: US. Sieve No. Mass of Soil Retained on each…

A: Sieve Mass % mass Cumulative % Cumulative % No Sieve size Retained Retained Retained Passing…

Q: Volumes Weight Va= అదికలయకు Vina L.Quid WwE Wm= Vs= Salid

A: Given volume dry sand sample fills = 200 cm 3 weight of solids, =260 g (Ws), Specific gravity of…

Q: 10. Sieve analysis and hydrometer analysis were performed on a soil sample. The table below…

Q: Sieve Size Sieve opening (mm) Percent Finer 3 in. 76.2 100% 2 in. 50.8 90% 1 in. 25.4 80% #4 4.75…

A: Given the sieve sizes and percentage finer soil in those sieves.

Q: Two soil samples A and B were found to have the physical properties shown in Table 1. Check the…

A: Given :

Q: 737 grams of dry sand completely fills a 500 cm³ permeameter tube. The specific gravity of the sand…

Q: A liquid limit test conducted on a soil sample in the cup device gave the following results: | 27 N…

Q: 7.17 The grain-size analysis data for a sand is given in the following table. Estimate the hydraulic…

Q: If the weight of sample of sand in sieve analysis test was 420 gm and the weight retained on sieve…

Q: Following is the result from the liquid limit test for a soil. y = -18.67In(x) + 111.39 60 55 50 45…

A: Liquid limit is defined as water content corresponding to 25 number of drops.

Q: Complete the blanks in the following sieve analysis table. Use only numbers with no percentage or…

Q: Determine the Fineness Modulus of the sand given its sieve analysis in the following table. round to…

Q: The results of a sieve analysis are reported in table 1. The total weight of the sample was 500 g.…

A: Due to lack of graph I am using interpolation method…

Q: Two samples of sand are mixed together to get a mixture with fineness modulus = 2.8. Find the mix…

A: We have to determine the value of mix proportion by the data we have samples of sand are mixed…

Q: No. Weight Retained cumulative Passing of (mm) Retained Retained sieve (gm) 4.75 70 80 0.6 71 0.425…

A: Solution:- No. of sieve Dia. (mm) Weight retained (gm) Retained % Cumulative retained % Passing %…

Q: A constant-head permeability test is performed in a laboratory where the soil sample is 25 cm in…

Q: The following results were obtained from a liquid limit test on a clay using the Casagrande cup…

A: Given data Number of blows against water content Natural water content According to guidelines…

Q: Q2. Using the sieve results of a batch of sand and gravel shown in the table below and if the P sand…

A: Percentage of sand = 38 % Finness modulus is the sum of cumulative percentage retained on each…

Q: We made sieve analysis for a soil sample, we had these results, complete the table. Sieve opening…

Q: Question 4 Comment the four types of sieve analysis diagrams, which appear in the figure below:…

Q: 9. Soil sample for liquid limit test should pass through the number sieve. А. 10 В. 20 С. 40 D. 50…

A: 9) ans -- number 40 seive.. In Astm test the atterberg limits were performed on the soil finer than…

Q: SITUATION 3. Following are the results of a sieve analysis. SIEVE NO. OPENING (mm) RETAINED 4. 4.75…

A: ASKED: To determine- effective particle size uniformity coefficient coefficient of curvature…

Concept explainers

Weight-Volume relationships

Weight is defined as the quantity of matter contained in a body. Weight can also be defined as the force acting on a body due to gravity. Mathematically, weight can be defined as-

Question

In the table below, find the missing values for pure sand sample and without fine materials ote (weight
of sample 1000g).
Sieve opening (mm) Retained weight Retained weight% Cumulative percentage Passing%
4.75
2.36
100
10
90
1.18
X
X
0.6
250
25
50
0.3
200
20
30
0.15
10
Pan

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

A sample of a dry coarse grained material of 500 grams was shaken through a nest of sieves and the following results are obtained: 20 | 200 40 0.425 Sieve No. 10 2.00 4 60 100 Pan Opening (mm) % Passing 4.75 0.85 0.25 0.15 0.075 -- 94 60 30 10 7 5 3 31. Determine the uniformity coefficient. b. 4.424 32. Determine the coefficient of curvature. a. 4.706 C. 1.250 d. 0.85 а. 4.706 b. 4.424 C. 1.250 d. 0.85 33. Given that the soil is non-plastic, classify the soil by using the Unified Soil Classification System. a. GW b. GP C. Sw d. SP
a. what is the effective size diameter? b. what is the uniformity coefficient? he following are the results of a sieve analysis. Sieve No. Opening Size (mm) Mass Retained (g) 4 4.750 28 10 2.000 42 20 0.850 48 40 0.425 128 60 0.250 221 100 0.150 86 200 0.075 40 Pan 24
SHOW COMPLETE SOLUTION PLS VALUES:X = 425Y = 25 A = 5
1. A sample of dry coarse-grained material was shaken through a nest of sieves and the following results were obtained. Sieve No. % Finer a. Determine Effective Size b. Determine the Uniformity Coefficient c. Determine the Coefficient of Curvature 4 100 10 97 20 77 40 59 100 23 200
SITUATION 1 A sample of dry coarse- grained materials was taken through a layer of sieves and the following results were obtained. Make the necessary calculations of the following questions. For Situation 1 (Sieve Analysis) SIZE (mm) WEIGHT RETAINED 18.96 gms. 0.25 0.149 33.18 0.074 45.03 0.052 54.51 0.02 42.66 0.01 11.85 0.004 4.74 0.001 4.74 PAN 21.33 Draw the particle distribution curve (manually) Your answer
Problem A sample of a dry coarse grained material of 500 grams was shaken through a nest of sieves and the following results are obtained: Sieve No. Opening (mm) % Passing 4 10 20 40 60 100 200 Pan 0.425 0.25 0.075 2.00 60 4.75 0.85 0.15 94 30 10 7 3 12. Determine the uniformity coefficient. a. 4.706 b. 4.424 C. 1.250 d. 0.85 13. Determine the coefficient of curvature. a. 4.706 b. 4.424 1.250 d. 0.85 C.
Q2: The dial readings recorded during a consolidation test at a certain load increment are given below. Determine c, by both the square root of time and log of time methods. The thickness of the sample = 2 cm and the sample is drained both sides. Time Dial Reading Time min Dial Reading cm x 10-4 min cm x 10-4 0 240 15 622 0.10 318 H 30 738 C₁=0.848- 0.25 340 60 842 0.50 360 120 930 1.00 385 240 975 2.00 415 1200 1070 C,=0.197 4.00 464 8.00 530 90 H² tso
Q1)- A) compare by drawing between the Clay minerals of 1/1 and clay minerals of 1/2
The mass of the container plus sand inside is 590 g. During the sand specimen preparation, you pluviated the sand into the shear box and then measured the mass of the container plus sand afterward recorded 406 g. Assume the specimen height is 1.5 inches. What is the unit weight (lb/ft³) of the sand specimen? Assume a diameter of 2.5in.
SITUATION 3. Following are the results of a sieve analysis. SIEVE NO. OPENING (mm) RETAINED 4 4.75 10 2.0 40 20 0.85 60 40 0.425 89 60 0.25 140 80 0.18 122 100 0.15 210 200 0.075 56 PAN 12 a. Find the effective particle size. b. Determine the uniformity coefficient. c. Find the coefficient of curvature.
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
Three samples of aggregates were taken weighed in moist and dry condition. Moist Dry Unit Unit Absorption Volume Aggregate Wt. Wt. (%) (m³) (kN/m) (kN/m³) Sample 1 23.425 20.018 5.14 0.366 Sample 2 22.7 18.757 3.87 0.695 Sample 3 20.292 18.387 2.62 0.237 Where; Unit weight of water = 9.81 kN/m3 Determine the average free moisture content of the three samples.