The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Water Moist Weight in Mold, gf Content, % 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (d) What is y, at wopt? (e) What is the range of water content if the relative compaction (R.C.) is required to be 95% of the modified Proctor Ya.max?

The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Water Moist Weight in Mold, gf Content, % 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (d) What is y, at wopt? (e) What is the range of water content if the relative compaction (R.C.) is required to be 95% of the modified Proctor Ya.max?

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter4: Plasticity And Structure Of Soil

Section: Chapter Questions

Problem 4.3P

See similar textbooks

Similar questions

The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Moist Weight in Mold, gf Water Content, % 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (a) Plot the Ya versus w relation. (b) Determine Ya.max and wopt• (c) Calculate S and e at the maximum dry unit weight point. (d) What is Y at w. (e) What is the range of water content if the relative compaction (R.C.) is required to be 95% of the modified Proctor Ya.max? pt? opr?
Show solution for prob 7
. Problem 3- From the following data of the grain size analysis Sieve size system #4 #8 # 10 #25 #60 # 100 #200 Pan Sieve opening (mm) 4.76 2.36 2.00 0.71 0.25 0.149 0.074 Determine; a) Cu and Cc % Finer (by weight) 80 56 51 35 18 13 7 I b) classify the soil using the unified soil classification
AASHTO SOIL CLASSIFICATION (by AASHTO) AASHTO = American Association of State Highway and Transport Officials Situation 2 Classification of Highway Subgrade Materials % FINER SOIL B SIEVE General classification Granular materials (35% or less of total sample passing No. 200) SOIL A SOIL C 10 83 100 90 A-1 А-2 Procedure: 40 48 92 76 Group classification A-1-a А-1-b А-3 A-2-4 A-2-5 A-2-6 A-2-7 O Group Classification with required test data in mind, proceed from left to right, in chart. The correct group will be found by process of elimination. The first group from the left consistent with the test data is the corect classification. 86 34 Sieve analysis (percentage passing) No. 10 200 20 LL 20 70 37 50 max. 30 max. No. 40 No. 200 Characteristics of fraction PI 32 18 50 max. 51 min. 15 max. 25 max. 10 max. 35 max. 35 max. 35 max. 35 max. Using AASHT0 method: 1. Determine the group symbol and index for soil A. 2. Determine the group symbol and index for soil B. 3. Determine the group symbol…
For a soil specimen, given the following: % passing sieve #4-92 %, sieve #10 -81%, sieve # 40-78%, LL-48%, PL-16%, GI-18.2 classify the soil by AASHTO classification system. GI = (F200-35)[0.2 + 0.005 (LL - 40)] +0.01 (F20015)(PI – 10)
No need to input units in final answer
A standard Proctor Test has yielded the values shown below. Determine: (a) The maximum dry unit weight and its OMC at Volume of 1/30 cubic ft. (b) The moisture for 93% of maximum dry unit weight. range STANDARD PROCTOR TEST RESULTS TRIAL WEIGHT OF SOIL (Ibs.) MOISTURE (%) |8.24 3.26 4.15 1 10.20 4.67 12.30 4. 4.02 14.60 3.36 16.80 23
Classify soil A depicted in the table according to the AASHTO classification system. Sieve analysis- Percent finer No. 10 Liquid limit Plasticity index Soil No. 40 No. 200 A 98 80 72 52 21 Select one: a. A-7-5 b. A-7-6 C. A-7 d. A-6
Answer ALL Questions Time: 2 Hours Q3: Classify the following soil by using the Unified soil classification system. Give the group symbols and the group names. Table 1. Sample 1 Grain Size Results (ASTM D422). Table 2. Atterberg Limits Results (ASTM D4318). Sleve No. Dia. (mm) % Passing Sample PL LL % in 12.7 100 G0 40 14 4.75 97 #10 2 94 #20 0.85 84 # 40 0.425 57 #60 0.25 32 140 0.106 15 #200 0.075 100 90 Atterberg Limit Results: LL = 60%, PL = 40% E 80 * 70- 60 - 50 40 30- a 20 10 1 30 a40 100 0.1200 10 80 0.01 Particle Diameter (mm) Answer ALL Questions Time: 2 Hours 70 CH or OH 60 A-line PI = 0.73(LL - 20) Or 50 40 30 CL MH OL 20 or CL-ML OH ML 10 - or OL 10 16 20 30 40 50 60 70 80 90 100 Liquid limit Figure 5.3 Plasticity chart Percent (%) Finer by Weight Plasticity index U-line PI = 0.(LL - 8) O Cengage Learning 2014
Please kindly answer all. C and d answers mandatory. Please answer all and help me.. thank you
- Once answered correctly will UPVOTE!!
• Problem 2--- From the following data of the grain size analysis Sieve size #4 #8 #10 #25 #60 #100 # 200 Pan Sieve opening (mm) 4.76 2.36 2.00 0.71 0.25 0.149 0.074 Determine; a) Cu and Cc system % Finer (by weight) 80 56 51 35 18 13 7 ——— b) classify the soil using the unified soil classification