PlasticsLiquid limitlimit11Can number#7#8#9#10Mass of empty can (g)Mass of can + soil (wet) (g)1.71.81.71.720.423.226.36.3Mass of can + soil (dry) (g)14.716.217.55.3Number of blows3328231. Determine the liquid limit.2. Determine the plastic limit.3. Allot group symbols and give soil description by USCS

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Answered: Plastics Liquid limit limit 1 1 Can…

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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Hazen-Williams formula was used in the derivation. 2- Explain why the Aluminum sulfate (alum) is widely used as a c 1-What are the factors allecting coagulation. b-Prove that the value of the exponent x in the relation (h=kQ) is 1.85 when Q5: distinguish its occurrence? a-Explain why cavitation oecurred in pumps and how one can icoagulant.
The data set of the diameters of the metal cylinders manufactured an automatic machine has sample size of n=28, mean of x= 49.94 mm and std. of S = 0.15 mm. Expected diameter of metal cylinder is µ = 50.00 mm. Can we be 95% confident that machine calibrated properly? Solution: v=n-1=27 ; ta/2=? P[X for 2 S VR μ+ (a Sl Vi -1-a =0.95
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 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.
Calculate the B.O.D. of a reference sample given that 1ml of seed material was used in the reference sample. (300 mL BOD bottles were used) Seed Material Reference Sample Volume Used 5 mL 5 mL 8.6 mg/L 5.8 mg/L 8.7 mg/L 4.8 mg/L Initial D.O. Final D.O.
(Same given conditions as the previous problem) Convert one hr UH to 2 hr UH. What is the value of "C" in the table? time (hr) 0 0.5 1 1.5 2 2.5 3 3.5 1-hr UH, cfs 0 32 44 52 36 22 10 O 2-hr UH, cfs 0 A B CO С
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What is the moisture contect for 2
18. The results for sieve analysis is shown in the following table: Sieve No. 10 20 40 60 100 % Passing 69.3 59.1 48.3 38.5 28.4 19.8 4.5 Liquid Limit Plastic Limit Non-Plastic (NP) Non-Plastic (NP)
Results trom the Atterberg Limit Test. > Uguid Limit TEST NUMBER Number of Blows Weight of wet sol• container, gm Weight of dry sol- container, gm Weight of container, gm Weight of water, gm Weight of dry sol. gm Water content% 2 29 21 29 18 78 13 24 14 26 12 22 10 1327 38 2247 19 44 12 74 20 21 27 18.75 13 20 > Plastic Limit and Natural Water Corrent NATURAL WATER CONTENT PLASTIC LIMIT TEST NUMBER 2 Number of Blows Weight of wet soll• container, gm Weight of dry sol- container, gm Weight of contaner, gm Weight of water, gm Weight of dry sol m Water content% Average, % 23 20 20 42 12.90 2280 20 19 12 96 17 53 14.84 950 16 97 14 36 955 Determine the following: o) Plot the water content versus number of blows on a log scale graph (manually) b) Uquid limit of the sol c) Plastic limit of the soll d) Liquidity index of the soll
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A river water was taken for analysis. The results for anions and cations analysis are shown below, 165 mg/l 70 mg/l HCO, 2+ Ca Са 173 mg/l 30 mg/1 SO42 2+ Mg 202 mg/l 125 mg/l ci Na Determine and justify either the analysis is acceptable or not. II II II

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