1. Find: CC Lab Test Data e 0.03 0.07 C) 0.38 D) 0.50 Load [kPa] 0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
Find the compression index, also referred to as the virgin compression index, C sub c. In this problem,

2. Find: CC Lab Test Data e 0.03 0.07 C) 0.38 D) 0.50 Load [kPa] 0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
We’ve been provided laboratory data from a consolidation test.

3. Find: CC Lab Test Data e Load [kPa] 0.919 6.25 0.900 0.890 12.5 25
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
The compression index can be determined by plotting ---

4. Find: CC e Lab Test Data e Load [kPa] Load [kPa] 0.919 6.25 0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
the laboratory test data on a semi-log graph, where ---
Load [kPa]

5. Find: CC e Lab Test Data e Load [kPa] Load [kPa] 0.919 6.25 0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
the vertical axis is the void ratio,
Load [kPa]

6. Find: CC e Lab Test Data e Load [kPa] Load [kPa] 0.919 6.25 0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
And the horizontal axis is the load. This horizontal axis is plotted on a log scale, and the units of load for this problem, are in kilopascals.
Load [kPa]

7. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
So we plot our data points.
0.5
0.4 1 10
100
1,000
Load [kPa]

8. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
---
0.5
0.4 1 10
100
1,000
Load [kPa]

9. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
Keep in mind, this drawing is not to scale
0.5
0.4 1 10
100
1,000
Load [kPa]

10. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
---
0.5
0.4 1 10
100
1,000
Load [kPa]

11. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
---
0.5
0.4 1 10
100
1,000
Load [kPa]

12. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
---
0.5
0.4 1 10
100
1,000
Load [kPa]

13. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
---
0.5
0.4 1 10
100
1,000
Load [kPa]

14. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
---
0.5
0.4 1 10
100
1,000
Load [kPa]

15. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
Once we’ve plotted all the points, we can compute the compression index,
0.5
0.4 1 10
100
1,000
Load [kPa]

16. Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
which is the slope of the best fit line formed by the data points, at loading stresses greater than the preconsolidation stress.
0.5
0.4 1 10
100
1,000
Load [kPa]

17. σ’p Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
So, if the preconsolidation stress is somewhere in here,
0.5
0.4 1 10
100
1,000
Load [kPa]

18. σ’p Find: CC e log 10 CC Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
log 10
0.8
0.7 CC
0.6
we’re interested in the magnitude of the slope of the best fit line along these last three, or four, data points. Where log base 10 of 10, is equal to 1.
0.5
0.4 1 10
100
1,000
Load [kPa]

19. σ’p Find: CC e 1 CC Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9 1
0.8
0.7 CC
0.6
Suppose the best fit line passes most closely ---
0.5
0.4 1 10
100
1,000
Load [kPa]

20. σ’p Find: CC e 1 CC Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9 1
0.8
0.7 CC
0.6
through the two data points at loads of 200 and 400 kilopascals.
0.5
0.4 1 10
100
1,000
Load [kPa]

21. σ’p Find: CC e 1 CC e 0.9 0.8 0.7 Load [kPa] 0.6 0.5 0.4 1 10 100
0.687
0.572
Load [kPa]
200
400
0.6
We can compute the compression index,
0.5
0.4 1 10
100
1,000
Load [kPa]

22. σ’p Find: CC CC= e 1 CC e e400-e200 400 log 200 0.9 0.8 0.7 Load [kPa]
0.687
0.572
Load [kPa]
200
400
0.6
by computing the slope of the line through these two points.
0.5
0.4 1 10
100
1,000
Load [kPa]

23. σ’p Find: CC CC= e 1 CC=0.382 CC e e400-e200 400 log 200 0.9 0.8 0.7
0.687
0.572
Load [kPa]
200
400
0.6
The compression index is [pause]
0.5
0.4 1 10
100
1,000
Load [kPa]

24. σ’p Find: CC CC= e 1 CC=0.382 CC e e400-e200 400 0.03 0.07 C) 0.38
log
200
0.9 1
0.8
CC=0.382
0.7 CC e
0.687
0.572
Load [kPa]
200
400
0.6
Returning to our possible solutions,
0.5
0.4 1 10
100
1,000
Load [kPa]

25. σ’p Find: CC CC= e 1 CC=0.382 AnswerC CC e e400-e200 400 0.03 0.07
log
200
0.9 1
0.8
CC=0.382
AnswerC
0.7 CC e
0.687
0.572
Load [kPa]
200
400
0.6
the answer is C. [pause] Now, if the problem happened to ask for the recompression index,
0.5
0.4 1 10
100
1,000
Load [kPa]

26. σ’p Find: CR e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
not the compression index, the same method can be used.
0.5
0.4 1 10
100
1,000
Load [kPa]

27. σ’p Find: CR e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800
0.9
0.8
0.7
0.6
the only difference would be, we’d solve for the magnitude of the slope ---
0.5
0.4 1 10
100
1,000
Load [kPa]

28. σ’p Find: CR e 1 CR Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4
0.919
0.900
0.890
0.868
0.825
0.771
0.687
0.572
0.463
Load [kPa]
6.25
12.5 25 50
100
200
400
800 1
0.9 CR
0.8
0.7
0.6
of the best line passing through the data points with stress values less than the preconcolidation stress.
0.5
0.4 1 10
100
1,000
Load [kPa]

29. σ’p Find: CR e 1 CR Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4
0.919
0.900
0.890
0.868
Load [kPa]
6.25
12.5 25 1
0.9 CR
0.8
0.7
0.6
If we assume a line passing through the two data points at loads of 6.25 and 12.5 kilopascals is most parallel the best fit line,
0.5
0.4 1 10
100
1,000
Load [kPa]

30. σ’p Find: CR e 1 CR e12.5-e6.25 CR= Lab Test Data e 12.5 log 6.25
0.919
0.900
0.890
0.868
Load [kPa]
6.25
12.5 25 1
0.9 CR
0.8
0.7
e12.5-e6.25
CR=
0.6
we can plug the appropriate numbers into our equation,
12.5
log
0.5
6.25
0.4 1 10
100
1,000
Load [kPa]

31. σ’p Find: CR e 1 CR e12.5-e6.25 CR= Lab Test Data e 12.5 log 6.25
0.919
0.900
0.890
0.868
Load [kPa]
6.25
12.5 25 1
0.9 CR
0.8
0.7
e12.5-e6.25
CR=
0.6
And solve for a recompression index value of
12.5
log
0.5
6.25
0.4
CR=0.033 1 10
100
1,000
Load [kPa]

32. ( ) ? γclay=53.1[lb/ft3] Index σ’v = Σ γ d H*C σfinal ρcn= log
Find: σ’v ρc d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4
( )
H*C σfinal
ρcn=
1+e σinitial
log ‘