1. Find: wc wdish=50.00[g] wdish+wet soil=58.15[g]
wdish+dry soil=57.28[g] 4% 8%
12%
D) 16%
Find the water content. [pause] In this problem,

2. Find: wc wdish=50.00[g] wdish+wet soil=58.15[g]
wdish+dry soil=57.28[g] 4% 8%
12%
D) 16%
we’ve been given some laboratory test data from a water content test. Water content is often referred to as moisture content.

3. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
wdish+dry soil=57.28[g] 4% 8%
12%
D) 16%
Briefly, the procedure for this test is as follows.

4. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
wdish+dry soil=57.28[g] 4% 8%
12%
D) 16%
First, a dry, empty dish is weighed out on a scale.

5. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
wdish+dry soil=57.28[g] 4% 8%
12%
D) 16%
For this problem, the weight of the dish is grams.

6. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g] 4% 8%
12%
D) 16%
The second step is to add a sample of soil to the dish,

7. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil 4% 8%
12%
D) 16%
and the next step is to weight the dish and soil together, which for this problem was grams.

8. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil 4% 8%
12%
D) 16%
Then, the soil is cooked in the oven in order to remove all the water from the soil.

9. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil 4% 8%
12%
D) 16%
oven
---

10. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil 4% 8%
12%
D) 16%
oven
It’s typical to cook the soil overnight at about 100 degrees Fahrenheit.
100˚F

11. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
oven
The last step is to remove the dish from the oven, and weight on a scale, once again.
100˚F

12. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
oven
Now, the sample only weighs ---
100˚F

13. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
oven
57.28 grams, since the soil is dry.

14. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
The problem asked us to find the water content of the soil sample,

15. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
wwater
wc=
wsolid
which is defined as the weight of the water in the sample divided by the weight of the solid material in the sample.

16. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
wwater
wdish+wet soil-wdish+dry soil
wc= =
wsolid
wdish+dry soil-wdish
The weight of the water is equivalent to the weight of the dish with the wet soil, minus, the weight of the dish with the dry soil.

17. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
wwater
wdish+wet soil-wdish+dry soil
wc= =
wsolid
wdish+dry soil-wdish
The weight of the solid material is equivalent to the weight of the dish with the dry soil, minus, the weight, of the empty dish.

18. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
wwater
wdish+wet soil-wdish+dry soil
wc= =
wsolid
wdish+dry soil-wdish
Once we plug in our known variables,

19. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
wwater
wdish+wet soil-wdish+dry soil
wc= =
wsolid
wdish+dry soil-wdish
We compute the water content to be 0.12, or,
wc=0.12

20. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
wwater
wdish+wet soil-wdish+dry soil
wc= =
wsolid
wdish+dry soil-wdish
12 percent.
wc=0.12=12%

21. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
wwater
wdish+wet soil-wdish+dry soil
wc= =
wsolid
wdish+dry soil-wdish
Compared with the possible solutions,
wc=0.12=12%

22. Find: wc Lab Procedure: wdish=50.00[g] wdish+wet soil=58.15[g]
1.weigh the dish
wdish+wet soil=58.15[g]
2.add wet soil
wdish+dry soil=57.28[g]
3.weigh dish+wet soil
4.cook the soil
5.weigh dish+dry soil 4% 8%
12%
D) 16%
wwater
wdish+wet soil-wdish+dry soil
wc= =
wsolid
wdish+dry soil-wdish
the answer is C.
wc=0.12=12%
AnswerC

24. ( ) ? τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal
Find: σ’v ρc d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ φ γ Δ d ˚ d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
φ=α1-α2
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
τ [lb/ft2]
(5 [cm])2 * π/4
( )
H*C σfinal
ρcn=
1+e σinitial
log ‘ φ
c=0
400
1,400 σ3
Sand σ1