1. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
2.25
2.50
C) 2.75
D) 3.00
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
Find the settlement from consolidation in inches from a load after 2 years. In this problem,
Clay
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
γT=118 [lb/ft3]
Gravel
4[ft]

2. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
2.25
2.50
C) 2.75
D) 3.00
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
A load of 400 pounds per square feet is applied to a soil profile consisting of ---
Clay
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
γT=118 [lb/ft3]
Gravel
4[ft]

3. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
2.25
2.50
C) 2.75
D) 3.00
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
a 6 foot thick layer of sand,
Clay
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
γT=118 [lb/ft3]
Gravel
4[ft]

4. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
2.25
2.50
C) 2.75
D) 3.00
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
a 20 foot thick layer of clay
Clay
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
γT=118 [lb/ft3]
Gravel
4[ft]

5. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
2.25
2.50
C) 2.75
D) 3.00
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
and a 4 foot thick layer of gravel.
Clay
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
γT=118 [lb/ft3]
Gravel
4[ft]

6. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
2.25
2.50
C) 2.75
D) 3.00
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
We’ve also been provided various soil properties. [pause]
Clay
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
γT=118 [lb/ft3]
Gravel
4[ft]

7. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
2.25
2.50
C) 2.75
D) 3.00
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
In this problem, we assume the clay is the only soil type which consolidates.
Clay
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
γT=118 [lb/ft3]
Gravel
4[ft]

8. ( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
Clay
So we begin by solving for the ultimate consolidation in the clay layer.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

9. ( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
Clay
The height of the clay layer is 20 feet.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

10. ( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
Clay
Since the clay is normally consolidated,
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

11. ( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
Clay
we’ll use 0.40 for our consolidation index.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

12. ( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3] ?
2 [ft]
γT=110 [lb/ft3]
20 [ft]
Clay
Next, we’ll solve for the initial void ratio.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

13. ( ) ? Find: ρc [in] from load after 2 years γT= H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3] ?
2 [ft]
(SG+e) * γW
γT=110 [lb/ft3]
20 [ft]
γT=
Clay
Since the clay layer is saturated, we know the total unit weight can be determined from the specific gravity, void ratio and unit weight of water.
1+e
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

14. ( ) ? Find: ρc [in] from load after 2 years γT - γW γT= H*C σfinal ρc=
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3] ?
2 [ft]
(SG+e) * γW
γT=110 [lb/ft3]
20 [ft]
γT=
Clay
After solving for the void ratio ---
1+e
CR=0.04
CC=0.40
normally consolidated
SG * γW - γT
CV=5 [ft2/year] e=
γT - γW
Grav
γT=118 [lb/ft3]
4[ft]

15. ( ) ? Find: ρc [in] from load after 2 years γT - γW γT= H*C σfinal ρc=
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3] ?
2 [ft]
(SG+e) * γW
γT=110 [lb/ft3]
20 [ft]
γT=
Clay
and plugging in our known values, assuming the specific gravity is 2.70,
1+e
CR=0.04
CC=0.40
2.70
normally consolidated
SG * γW - γT
CV=5 [ft2/year] e=
62.4 [lb/ft3]
γT - γW
Grav
γT=118 [lb/ft3]
4[ft]

16. ( ) ? Find: ρc [in] from load after 2 years γT - γW γT= H*C σfinal ρc=
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3] ?
2 [ft]
(SG+e) * γW
γT=110 [lb/ft3]
20 [ft]
γT=
Clay
we find the initial void ratio to be 1.23.
1+e
CR=0.04
CC=0.40
2.70
normally consolidated
SG * γW - γT
CV=5 [ft2/year] e=
62.4 [lb/ft3]
γT - γW
Grav
γT=118 [lb/ft3]
4[ft]
e=1.23

17. ( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
1.23
γT=110 [lb/ft3]
20 [ft]
Clay
Now, all we have to solve for ---
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

18. ( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft] ?
1.23
γT=110 [lb/ft3]
20 [ft]
Clay
is the initial and final vertical effective stress values.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

19. ( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft] ?
1.23
σ’i = Σ γ’ * d
γT=110 [lb/ft3]
20 [ft]
Clay
Let’s evaluate a point in the middle of the clay layer, which would be at a total depth of 16 feet.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

20. ( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft] ?
1.23
γT=110 [lb/ft3]
σ’i,= 4 [ft] * 106 [lb/ft3]
20 [ft]
We add the overburden stress from the unsaturated sand,
CR=0.04
CC=0.40 +…
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

21. ( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft] ?
1.23
γT=110 [lb/ft3]
σ’i,= 4 [ft] * 106 [lb/ft3]
20 [ft]
from the saturated sand,
CR=0.04
CC=0.40
+ 2 [ft] * ( ) [lb/ft3]
normally consolidated +…
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

22. ( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft] ?
1.23
γT=110 [lb/ft3]
σ’i,= 4 [ft] * 106 [lb/ft3]
20 [ft]
Clay
and from the top 10 feet of the clay later.
CR=0.04
CC=0.40
+ 2 [ft] * ( ) [lb/ft3]
normally consolidated
+(10 [ft]) * ( ) [lb/ft3]
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

23. ( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft] ?
1.23
γT=110 [lb/ft3]
σ’i,= 4 [ft] * 106 [lb/ft3]
20 [ft]
Clay
Our initial stress is 999 pounds per square feet,
CR=0.04
CC=0.40
+ 2 [ft] * ( ) [lb/ft3]
normally consolidated
+(10 [ft]) * ( ) [lb/ft3]
CV=5 [ft2/year]
σ’i,clay= 999 [lb/ft2]
Grav
γT=118 [lb/ft3]
4[ft]

24. ( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft] ?
1.23
γT=110 [lb/ft3]
σ’i,= 4 [ft] * 106 [lb/ft3]
20 [ft]
Clay
and after adding the load of 400 pounds per square feet,
CR=0.04
CC=0.40
+ 2 [ft] * ( ) [lb/ft3]
normally consolidated
+(10 [ft]) * ( ) [lb/ft3]
CV=5 [ft2/year]
σ’i,clay= 999 [lb/ft2]
Grav
γT=118 [lb/ft3]
4[ft]

25. ( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft] ?
1.23
γT=110 [lb/ft3]
σ’i,= 4 [ft] * 106 [lb/ft3]
20 [ft]
Clay
our final stress is 1,399 pounds per square feet,
CR=0.04
CC=0.40
+ 2 [ft] * ( ) [lb/ft3]
normally consolidated
+(10 [ft]) * ( ) [lb/ft3]
CV=5 [ft2/year]
σ’i,clay= 999 [lb/ft2]
Grav
γT=118 [lb/ft3]
σ’f,clay= 1,399 [lb/ft2]
4[ft]

26. ( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
1,399 [lb/ft2]
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
1.23
999 [lb/ft2]
γT=110 [lb/ft3]
20 [ft]
Clay
We update our settlement equation ---
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

27. ( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
1,399 [lb/ft2]
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
1.23
999 [lb/ft2]
γT=110 [lb/ft3]
ρc= 6.30 [in]
20 [ft]
Clay
to find the consolidation settlement is 6.30 inches. This is the maximum consolidation settlement which would occur,
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

28. ( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
1,399 [lb/ft2]
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
1.23
999 [lb/ft2]
γT=110 [lb/ft3]
ρc= 6.30 [in]
20 [ft]
Clay
if the soil had an infinite amount of time to consolidate.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Ultimate consolidation
(given infinite time)
Grav
γT=118 [lb/ft3]
4[ft]

29. ( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log
Load=400 [lb/ft2]
20 [ft]
0.40
1,399 [lb/ft2]
( )
H*C σfinal
ρc=
1+e σinitial
log ‘
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
1.23
999 [lb/ft2]
γT=110 [lb/ft3]
ρc= 6.30 [in]
20 [ft]
Clay
We’ll call this the ultimate consolidation. This is the consolidation which is typically referred to in consolidation settlement problems, when time is not mentioned.
CR=0.04
CC=0.40
ρult= 6.30 [in]
normally consolidated
CV=5 [ft2/year]
Ultimate consolidation
(given infinite time)
Grav
γT=118 [lb/ft3]
4[ft]

30. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
Clay
However, in this problem, were asked to find the consolidation settlement after 2 years. So, we already know the answer is between 0 and 6.30 inches.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

31. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
γT=110 [lb/ft3]
20 [ft]
Clay
The consolidation settlement after a certain time, equals the ultimate consolidation times the percent of consolidation which has occurred, U.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

32. <100% Find: ρc [in] from load after 2 years ρult= 6.30 [in]
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
Average degree
of consolidation
0% <
<100%
γT=110 [lb/ft3]
20 [ft]
Clay
We’ll call this ‘U’ value the ‘Average Degree of Consolidation’ which ranges from 0 and 100 percent.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

33. <100% Find: ρc [in] from load after 2 years ρult= 6.30 [in]
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
Average degree
of consolidation
0% <
<100%
γT=110 [lb/ft3]
20 [ft]
Clay
U is a function of big T, the ‘Time Factor,’
CR=0.04
CC=0.40
U = f ( T )
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

34. <100% Find: ρc [in] from load after 2 years ρult= 6.30 [in]
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
Average degree
of consolidation
0% <
<100%
γT=110 [lb/ft3]
20 [ft]
Clay
And the Time Factor, is equal to,
CR=0.04
CC=0.40
U = f ( T )
normally consolidated
CV=5 [ft2/year]
cv * t T=
Grav
γT=118 [lb/ft3]
4[ft]
(HD)2

35. <100% Find: ρc [in] from load after 2 years ρult= 6.30 [in]
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
Average degree
of consolidation
0% <
<100%
γT=110 [lb/ft3]
20 [ft]
Clay
the time rate of consolidation, which is sometimes called the consolidation coefficient,
CR=0.04
CC=0.40
U = f ( T )
5 [ft2/yr]
normally consolidated
CV=5 [ft2/year]
cv * t T=
Grav
γT=118 [lb/ft3]
4[ft]
(HD)2

36. <100% Find: ρc [in] from load after 2 years ρult= 6.30 [in]
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
Average degree
of consolidation
0% <
<100%
γT=110 [lb/ft3]
20 [ft]
Clay
times the time the soil has to consolidate,
CR=0.04
CC=0.40
U = f ( T )
5 [ft2/yr]
normally consolidated
2 [yr]
CV=5 [ft2/year]
cv * t T=
Grav
γT=118 [lb/ft3]
4[ft]
(HD)2

37. <100% ? Find: ρc [in] from load after 2 years ρult= 6.30 [in]
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
Average degree
of consolidation
0% <
<100%
γT=110 [lb/ft3]
20 [ft]
Clay
divided by the drainage distance, squared.
CR=0.04
CC=0.40
U = f ( T )
5 [ft2/yr]
normally consolidated
2 [yr]
CV=5 [ft2/year]
cv * t T=
Grav
γT=118 [lb/ft3]
4[ft]
(HD)2 ?

38. <100% ? Find: ρc [in] from load after 2 years ρult= 6.30 [in]
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
Average degree
of consolidation
0% <
<100%
γT=110 [lb/ft3]
20 [ft]
Clay
The drainage distance is the furthest distance a drop of water would have to travel to exit the consolidating layer.
CR=0.04
CC=0.40
U = f ( T )
5 [ft2/yr]
normally consolidated
2 [yr]
CV=5 [ft2/year]
cv * t T=
Grav
γT=118 [lb/ft3]
4[ft]
(HD)2 ?

39. <100% Find: ρc [in] from load after 2 years ρult= 6.30 [in]
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
Average degree
of consolidation
0% <
<100%
γT=110 [lb/ft3]
20 [ft]
10 [ft]
Clay
Since our clay layer is doubly drained, the drainage distance is half the layer layer thickness, or, 10 feet.
CR=0.04
CC=0.40
U = f ( T )
5 [ft2/yr]
normally consolidated
10 [ft]
2 [yr]
CV=5 [ft2/year]
cv * t T=
Grav
γT=118 [lb/ft3]
4[ft]
(HD)2
10 [ft]

40. <100% Find: ρc [in] from load after 2 years ρult= 6.30 [in]
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
γT=112 [lb/ft3]
2 [ft]
Average degree
of consolidation
0% <
<100%
γT=110 [lb/ft3]
20 [ft]
Clay
Which makes our Time Factor equal to [pause]
CR=0.04
CC=0.40
U = f ( T )
5 [ft2/yr]
normally consolidated
2 [yr]
CV=5 [ft2/year]
cv * t T=
=0.10
Grav
γT=118 [lb/ft3]
4[ft]
(HD)2
10 [ft]

41. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
T=0.10
γT=112 [lb/ft3]
2 [ft]
U = f ( T )
γT=110 [lb/ft3]
20 [ft]
Clay
Next we solve for our average degree of consolidation.
CR=0.04
CC=0.40
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

42. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
T=0.10
γT=112 [lb/ft3]
2 [ft]
U = f ( T )
γT=110 [lb/ft3]
20 [ft]
If T<0.285±
Clay
When T is less than about 0.285, the average degree of consolidation is equal to the square root of 4 times T, divided by PI, all times 100%.
CR=0.04
CC=0.40
Then 4T U≈
*100% π
normally consolidated
CV=5 [ft2/year]
Grav
γT=118 [lb/ft3]
4[ft]

43. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
T=0.10
γT=112 [lb/ft3]
2 [ft]
U = f ( T )
γT=110 [lb/ft3]
20 [ft]
If T<0.285±
Clay
When T is greater than about 0.285, the average degree of consolidation is this second equation.
CR=0.04
CC=0.40
Then 4T U≈
*100% π
normally consolidated
CV=5 [ft2/year]
If T>0.285±
Then
γT=118 [lb/ft3]
U≈100-10{(1.781-T)/0.933}

44. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
T=0.10
γT=112 [lb/ft3]
2 [ft]
U = f ( T )
γT=110 [lb/ft3]
20 [ft]
If T<0.285±
Clay
For our problem, T equals 0.10, so we’ll use the first equation to solve for ‘U.’
CR=0.04
CC=0.40
Then 4T U≈
*100% π
normally consolidated
CV=5 [ft2/year]
If T>0.285±
Then
γT=118 [lb/ft3]
U≈100-10{(1.781-T)/0.933}

45. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
T=0.10
γT=112 [lb/ft3]
2 [ft]
U = f ( T )
γT=110 [lb/ft3]
20 [ft]
If T<0.285±
Clay
After 2 years, this 20 foot thick clay layer has consolidated 35.7% of it’s ultimate consolidation settlement.
CR=0.04
CC=0.40
Then 4T U≈
*100% π
normally consolidated
CV=5 [ft2/year]
U=35.7%
Grav
γT=118 [lb/ft3]

46. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
T=0.10
γT=112 [lb/ft3]
2 [ft]
U = f ( T )
γT=110 [lb/ft3]
20 [ft]
If T<0.285±
Clay
---
CR=0.04
CC=0.40
Then 4T U≈
*100% π
normally consolidated
CV=5 [ft2/year]
U=35.7%
Grav
γT=118 [lb/ft3]

47. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
T=0.10
γT=112 [lb/ft3]
2 [ft]
U = f ( T )
γT=110 [lb/ft3]
20 [ft]
If T<0.285±
Clay
Or, 2.25 inches of settlement.
CR=0.04
CC=0.40
Then 4T U≈
*100% π
normally consolidated
CV=5 [ft2/year]
U=35.7%
Grav
γT=118 [lb/ft3]
ρ2yr = 2.25 [in]

48. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
2.25
2.50
C) 2.75
D) 3.00
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
T=0.10
γT=112 [lb/ft3]
2 [ft]
U = f ( T )
γT=110 [lb/ft3]
20 [ft]
If T<0.285±
Clay
Compared with our possible solutions,
CR=0.04
CC=0.40
Then 4T U≈
*100% π
normally consolidated
CV=5 [ft2/year]
U=35.7%
Grav
γT=118 [lb/ft3]
ρ2yr = 2.25 [in]

49. Find: ρc [in] from load after 2 years
Load=400 [lb/ft2]
ρult= 6.30 [in]
2.25
2.50
C) 2.75
D) 3.00
ρ2yr = U * ρult
γT=106 [lb/ft3]
4 [ft]
Sand
T=0.10
γT=112 [lb/ft3]
2 [ft]
U = f ( T )
γT=110 [lb/ft3]
20 [ft]
If T<0.285±
Clay
the answer is A.
CR=0.04
CC=0.40
Then 4T U≈
*100% π
normally consolidated
CV=5 [ft2/year]
U=35.7%
Answer  A
ρ2yr = 2.25 [in]

50. ( ) ? 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 ‘