cee 3400A_LAB5
.pdf
keyboard_arrow_up
School
Georgia State University *
*We aren’t endorsed by this school
Course
4600
Subject
Civil Engineering
Date
Apr 3, 2024
Type
Pages
7
Uploaded by Fkhairy
Georgia Institute of Technology School of Civil and Environmental Engineering Soil Mechanics Laboratory MEMORANDUM To: Ryann Khalil
Date: 3/04/2024 From: Fukhraj Khairy
Lab Partners: Jordan P.G
Benjamin G.M Darien R Subject
: Compaction
Sample Description Name:
Piedmont Soil Source:
North Georgia Condition:
N/A Visual Classification and Unified Symbol:
ML Remarks:
N/A Test Procedure The following ASTM standards and procedures were used to determine the maximum dry unit weight of compaction of soils that can be used for the specification of field compaction. Table 1
: ASTM Standards Used
In this laboratory, both variants of the Proctor Compaction Test were carried out. The procedures for these tests are essentially the same, but there are a few crucial distinctions. To begin with, secure a sample of Piedmont Soil that has been sieved through Sieve #4. Add water to the soil sample to reach a water content that is approximately 6% below the optimum level, which is around 12% for this type of soil. Make sure the soil is mixed well. To calculate the amount of water needed to achieve the desired water content, use the formula: Mw = ω * Ms, where ω is the water content and Ms is the mass of the soil. Next, measure the combined weight of the Proctor mold and its base plate. Once you have recorded the weight, attach the collar to the mold’s top. start adding the moist soil into the mold.
Take into account that the soil will be compacted, so the amount you need to add depends on the test method: 1.
For the Standard Proctor Test, fill the mold to slightly more than half of its height. 2.
For the Modified Proctor Test, fill the mold to slightly more than one-third of its height. After the soil is added, gently position the hammer inside the mold and initiate the hammer drops. After loading the soil, cautiously situate the hammer in the mold and start executing the 25 drops. Rotate the mold a little with each hit to guarantee even compaction throughout the sample. Post compaction, redo the steps for reloading the mold, and then compact 25 more times. For the standard method, carry out these steps five times until the mold is filled; for the modified test, also repeat five times. Once compacting concludes, the soil should slightly surpass the mold's rim. Detach the collar and employ a straight edge to scrape off surplus soil. Ascertain the mass of mold, base plate, and compacted soil. Remove the base plate, then employ a jack to eject the compacted soil cylinder. Get a moisture container, find out its mass. Extract a sample from the center of the compacted soil and put it in the container. Jot down the mass of moisture container with the wet soil sample.
Add extra water to the primary soil sample aiming to boost the water content by 4%. Redo all the preceding steps. Elevate the water content by an additional 4% twice and reiterate all steps. After completing all experiments (four in total), position the moisture containers in the oven to dry until weight stabilizes. The subsequent day, evaluate the mass of the moisture containers coupled with the dry soil samples. Test Results The equations used to calculate the Actual Water Content and the Dry Unit Weight are shown below. An example calculation is performed for Table A1, Test 1 of the Modified Proctor Test.
Mass of Water = (mass of container + wet soil) − (mass of container + dry soil) = 235g – 195.96g = 39.04g Mass of Dry Soil = (mass of container + dry soil) − (mass of container) = 195.96g – 8.5g = 187.46g Water content (%) ?𝑎??? ??????? =
?𝑎?? ?? ?𝑎???
?𝑎?? ?? ??𝑦 ??𝑖?
× 100
=
39.04
187.46
∗ 100 = 20.83
Sample calcs for Dry unit weight. 𝑉????
= 943.69 ??
^3 ?
????
= 4343.4 ??
?
?????? ????
= 5899 ??
?
??𝑖?
= ?
?????? ????
− ?
????
= 5899 − 4343.4 = 1555.6?? ∗ 1??/1000?? = 1.5556??
??
??𝑖?
= ?
??𝑖?
∗ 9.81
?
?
2
= 1.5556 ∗ 9.81 = 15.26 ?
𝑦
?𝑢??
=
??
??𝑖?
𝑉
????
=
15.26
943.69
= 0.01617 ∗ 1000 = 16.17
𝐾?
?
3
?𝐶 = 0.2083
𝑦
?
=
𝑦
?𝑢??
1 + ?𝐶
=
16.17
1 + 0.2083
= 13.38
??
?
3
Dry Unit Weight of Zero Air-Voids Line 𝑦
??𝑦=
𝑦
𝑚
𝜔+
1
𝐺
𝑠
=
9.81𝑘𝑁/𝑚^3
0.2083+(
1
2.65
)
=16.75 ?𝑁/?^3
Table 2
: Summarized Test Results STANDARD
Mass of Water (g) 39.04 21.32 35.83 73.16 Mass of dry soil (g) 187.46 77.38 113.77 203.94 Water content (%) 20.83 27.55 31.49 35.87 Bulk Unit Weight(kN/m^3) 16.17 18.45 17.86 17.34 Dry unit weight(kN/M^3) 13.38 14.46 13.58 12.76 Zero Air Voids kN/m^3 16.75 15.03 14.17 13.33 MODIFIED Mass of Water (g) 5.29 11.02 17.89 15.66 Mass of dry soil (g) 32.49 44.92 67.88 49.39 Water content (%) 16.28 24.53 26.36 31.71 Bulk Unit Weight(kN/m^3) 20.41 19.41 20 18.32 Dry unit weight(kN/M^3) 17.55 15.59 15.83 13.91 Zero Air Voids kN/m^3 18.16 15.76 15.31 14.13
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Related Questions
Classify soil A depicted in the table according to the AASHTO classification system.
Sieve analysis-
Percent finer
Liquid
limit
Plasticity
index
Soil
No. 10
No. 40
No. 200
A 92
30
68
42
18
Select one:
a. A-2-4
b. A-2-5
C. A-2-7
d. A-2-6
arrow_forward
Soil classification requires all of the results of the following tests:
Liquid limit and Plastic limit tests
Atterberg limit and Hydrometer tests
Atterberg limit and grain size analysis test
Sieve and Hydrometer analysis tests
arrow_forward
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
arrow_forward
liquid limit was 25.72%.
arrow_forward
• 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
arrow_forward
Do asap
arrow_forward
(1)Determine the liquid limit.
(2)Determine the plastic limit
(3)Allot group symbols and give soil description by USCS.
arrow_forward
A sample soil for proposed building-the engineer in-charge decided to under go soil testing. A 480 cubic centimeter (cc) soil sample taken from the site weighs 850.5 grams. After oven drying, it weighed 594.4 grams.
-If the specific gravity of solids is 2.72, what will be the result of the void ratio (e) of the soil?
-What will be result in saturation (S)?
arrow_forward
Please answer
arrow_forward
Calculate the uniformity coefficient and the coefficient of gradation of the soil with the following tabulated sieve analysis data and evaluate the grading of the soil
LL=25%
PI=15%
Cu=_____
Cc=_____
AASHTO CLASSIFICATION= ______
arrow_forward
5.5 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
(d)
What is y at wopi?
(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?
arrow_forward
For a soil specimen, the following are given:
% passing Sieve No. 4 = 92%
% passing Sieve No. 10 = 81%
% passing Sieve No. 40 = 78%
% passing Sieve No. 200 = 65%
LL = 48%
PI = 32%
Classify the soil by the AASHTO system.
O A-2-7 (15)
O A-7-6 (18)
O A-7-6 (32)
O A-2-5 (6)
arrow_forward
b) During lab investigation the soil fall under "CL". Comment on its type and usefulness.
arrow_forward
Hello, I have an exam and I need to answer all the options
arrow_forward
Soil Mechanics
Based on the sieve analysis a soil has % finer for the following sieves: No. 10- 90% ; No. 40- 74% and No. 200- 58%. The liquid limit is 38 and Plasticity index is 12. Classify the soil by AASHT0 classification and give also the group index.
a. A-6 (9)
b. A-7-5(8)
c. A-7-6(8)
d. A-6 (5)
arrow_forward
Q3- Deascribe the following with aids of sketches if necessary:
i-
How to determine the optimum moisture content of soil in lab.
How to determine the soil porosity in lab.
ii-
iii-
Why a pycnometr calibration has to be conducted for specific gravity test?
arrow_forward
Based on the sieve analysis a soil has % finer for the following sieves: No. 10- 100% ; No. 40- 78% and No. 200- 82%. The liquid limit is 32 and Plasticity index is 12. Classify the soil by AASHT0 classification and give also the group index.
a.A-6 (5)
b.A-6 (9)
c.A-7-5(8)
d.A-7-6(8)
arrow_forward
Which of the following Limit is having the lowest water content at which the soil is completely saturated?
Select one:
a. Plastic Limit
O b. Shrinkage Limit
O c. Consistency Limit
O d. Liquid Limit
arrow_forward
Given the test results that follow, classify the soil in both the Unified Soil Classification System (USCS) and the AASHTO system.
arrow_forward
complete solutions with particle size distribution thank you
arrow_forward
|Laboratory testing was performed on two soil samples (A and B).
(a) Determine the USCS classification symbol for Sample A.
(b) Determine the AASHTO classification for Sample B.
Sieve No.
Sieve Opening (mm)
A- Percent Passing
B - Percent Passing
3 inch
76.2
100
1.5 inch
38.1
98
0.75 inch
19.1
96
4
4.75
77
100
10
2.00
96
20
0.85
55
94
40
0.425
73
100
0.15
30
200
0.075
18
55
Liquid Limit
32
52
Plastic Limit
25
32
A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter
of 2.83 inches and weighed 1.71 lbs. (a) Calculate the wet density of the CL
sample. (b) A small piece of the CL sample had a wet weight of 140.9 grams and
a weight of 85.2 grams after drying. Compute the water content. (c) Compute the
dry density of the CL sample
A Standard Proctor test was performed on a soil with a specific gravity of 2.71.
For the water content and wet soil unit weight in the following table:
a) Plot the moisture-dry unit weight curve
b) Find the maximum dry density and optimum…
arrow_forward
I need the answer as soon as possible
arrow_forward
A sample soil for proposed building-the engineer in-charge decided to under go soil testing. A 480 cubic centimeter (cc) soil sample taken from the site weighs 850.5 grams. After oven drying, it weighed 594.4 grams. If the specific gravity of solids is 2.72, what will be the result of the void ratio (e) of the soil?
arrow_forward
Sieve analysis on a dry sample of mass 1000g showed that 980 g and 270 g of soil pass through the 4.75 mm and 0.075 mm sieve respectively. The liquid limit and plastic limits of the soil fraction passing through the 0.075 mm sieves are 40% and 18%. The soil may classify as (USCS): (Justify your answer)
a.SC-SM
b.SM
c.SC
d.GC
arrow_forward
A soil sample is having liquid limit of 60% and plastic limit of 25%. This soil can be
classified as
A CH
B MH Your answer is
C MI
D CI
arrow_forward
- Once answered correctly will UPVOTE!!
arrow_forward
SEE MORE QUESTIONS
Recommended textbooks for you
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305970939/9781305970939_smallCoverImage.gif)
Principles of Geotechnical Engineering (MindTap C...
Civil Engineering
ISBN:9781305970939
Author:Braja M. Das, Khaled Sobhan
Publisher:Cengage Learning
Related Questions
- Classify soil A depicted in the table according to the AASHTO classification system. Sieve analysis- Percent finer Liquid limit Plasticity index Soil No. 10 No. 40 No. 200 A 92 30 68 42 18 Select one: a. A-2-4 b. A-2-5 C. A-2-7 d. A-2-6arrow_forwardSoil classification requires all of the results of the following tests: Liquid limit and Plastic limit tests Atterberg limit and Hydrometer tests Atterberg limit and grain size analysis test Sieve and Hydrometer analysis testsarrow_forwardAnswer 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 2014arrow_forward
- liquid limit was 25.72%.arrow_forward• 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 classificationarrow_forwardDo asaparrow_forward
- (1)Determine the liquid limit. (2)Determine the plastic limit (3)Allot group symbols and give soil description by USCS.arrow_forwardA sample soil for proposed building-the engineer in-charge decided to under go soil testing. A 480 cubic centimeter (cc) soil sample taken from the site weighs 850.5 grams. After oven drying, it weighed 594.4 grams. -If the specific gravity of solids is 2.72, what will be the result of the void ratio (e) of the soil? -What will be result in saturation (S)?arrow_forwardPlease answerarrow_forward
- Calculate the uniformity coefficient and the coefficient of gradation of the soil with the following tabulated sieve analysis data and evaluate the grading of the soil LL=25% PI=15% Cu=_____ Cc=_____ AASHTO CLASSIFICATION= ______arrow_forward5.5 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 (d) What is y at wopi? (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?arrow_forwardFor a soil specimen, the following are given: % passing Sieve No. 4 = 92% % passing Sieve No. 10 = 81% % passing Sieve No. 40 = 78% % passing Sieve No. 200 = 65% LL = 48% PI = 32% Classify the soil by the AASHTO system. O A-2-7 (15) O A-7-6 (18) O A-7-6 (32) O A-2-5 (6)arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Principles of Geotechnical Engineering (MindTap C...Civil EngineeringISBN:9781305970939Author:Braja M. Das, Khaled SobhanPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305970939/9781305970939_smallCoverImage.gif)
Principles of Geotechnical Engineering (MindTap C...
Civil Engineering
ISBN:9781305970939
Author:Braja M. Das, Khaled Sobhan
Publisher:Cengage Learning