A Standard Proctor Test was completed according to Method A described in Chapter 12. The following laboratory data was obtained: Mass of mold = Mass of moist specimen and mold = Volume of compaction mold = Mass of container= 51.5g Mass of container plus moist soil= 261.85g Mass of container plus oven-dried soil= 231.83g 1.98 kg 3.718 kg 0.0009438m3 K Determine the Moist density of compacted specimen. (expressed in Mg/m³, 3 decimal places) Your Answer:

A Standard Proctor Test was completed according to Method A described in Chapter 12. The following laboratory data was obtained: Mass of mold = Mass of moist specimen and mold = Volume of compaction mold = Mass of container= 51.5g Mass of container plus moist soil= 261.85g Mass of container plus oven-dried soil= 231.83g 1.98 kg 3.718 kg 0.0009438m3 K Determine the Moist density of compacted specimen. (expressed in Mg/m³, 3 decimal places) Your Answer:

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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In a liquid limit test using Fall cone apparatus, the following readings were recorded. While in the plastic limit test, the following data were obtained: Weight of soil + container = 22.12 grams Weight of dry soil + container = 20.42 grams Weight of container = 13.07 grams Determine the ff. (SHOW COMPLETE SOLUTION AND GRAPH) A. Find the liquid limit B. Compute the plastic limit. C. Find the liquid limit. D. Find the plastic limit. 40% 45% 50% 58% Water Content 12 16 21 35 Penetration of cone(mm) -PLS SHOW COMPLETE SOLUTION AND GRAPH. SHOW YOUR ANSWER IN 4 DECIMAL PLACES. I'LL RATE, THANKS!
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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
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