1. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
Sand 17 26
C) 42
D) 53
γT=100 [lb/ft3]
10 [ft]
Clay
25 [ft]
OCR=1.0
Find the time, in years, it will take the given soil profile to settle 90% of it’s ultimate consolidation. In this problem,
γT=105 [lb/ft3]
cV =10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

2. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
Sand 17 26
C) 42
D) 53
γT=100 [lb/ft3]
10 [ft]
Clay
25 [ft]
OCR=1.0
a 10 foot layer of sand sits above ---
γT=105 [lb/ft3]
cV =10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

3. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
Sand 17 26
C) 42
D) 53
γT=100 [lb/ft3]
10 [ft]
Clay
25 [ft]
OCR=1.0
a 25 foot layer of clay,
γT=105 [lb/ft3]
cV =10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

4. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
Sand 17 26
C) 42
D) 53
γT=100 [lb/ft3]
10 [ft]
Clay
25 [ft]
OCR=1.0
which is above an impermeable bedrock.
γT=105 [lb/ft3]
cV =10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

5. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
Sand 17 26
C) 42
D) 53
γT=100 [lb/ft3]
10 [ft]
Clay
25 [ft]
OCR=1.0
It is assumed the ground water table is not present.
γT=105 [lb/ft3]
cV =10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

6. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
Sand 17 26
C) 42
D) 53
γT=100 [lb/ft3]
10 [ft]
Clay
25 [ft]
OCR=1.0
We’ve also been given a number of the soil properties.
γT=105 [lb/ft3]
cV =10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

7. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2
Clay
25 [ft]
OCR=1.0
We begin with our time rate of consolidation equation, and solve for our time, little t.
γT=105 [lb/ft3]
cV =10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

8. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
We’ve been given the consolidation coefficient for the clay layer, cv
γT=105 [lb/ft3]
cV=10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

9. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
which is 10 feet squared per year. cv
γT=105 [lb/ft3]
cV=10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

10. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
We also know the drainage depth is 25 feet. This is because the impermeable bedrock would prevent water from leaving the clay layer in the downward direction. cv
γT=105 [lb/ft3]
cV=10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

11. ? Find: Time [yr] for 90% consolidation to occur cv * t T= (HD)2
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2 ?
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
But we don’t yet know the time factor, big ‘T.’ cv
γT=105 [lb/ft3]
cV=10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

12. ? Find: Time [yr] for 90% consolidation to occur cv * t T= (HD)2
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2 ?
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
When a soil’s average degree of consolidation, ‘U’ is less than 60% of its ultimate consolidation, cv
γT=105 [lb/ft3]
If U<60%
cV=10 [ft2/yr]
CC=0.35
CR=0.05
Impermeable Bedrock

13. ? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T=
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2 ?
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
the time factor, ‘T,’ is equal to PI times ‘U’ squared, all divided by 4. Where U is a decimal. cv
γT=105 [lb/ft3]
π * U2
If U<60%
then T=
cV=10 [ft2/yr] 4
CC=0.35
CR=0.05
Impermeable Bedrock

14. ? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T=
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2 ?
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
When a soil’s average degree of consolidation is greater than 60%, of its ultimate consolidation, cv
γT=105 [lb/ft3]
π * U2
If U<60%
then T=
cV=10 [ft2/yr] 4
CC=0.35
CR=0.05
If U>60%
Impermeable Bedrock

15. ? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T=
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2 ?
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
the time factor, is equal to minus times the log of 100 percent minus ‘U.’ Where ‘U’ is a percent. cv
γT=105 [lb/ft3]
π * U2
If U<60%
then T=
cV=10 [ft2/yr] 4
CC=0.35
CR=0.05
then T= *log(100%-U)
If U>60%
Impermeable Bedrock

16. ? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T=
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2 ?
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
In our problem, the average degree of consolidation is 90%, cv
γT=105 [lb/ft3]
π * U2
If U<60%
then T=
cV=10 [ft2/yr] 4
CC=0.35
CR=0.05
then T= *log(100%-U)
If U>60%
Impermeable Bedrock

17. ? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T=
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2 ?
T * (HD)2
Clay t=
25 [ft]
OCR=1.0
we we plug this into our equation, and solve for T. cv
γT=105 [lb/ft3]
π * U2
If U<60%
then T=
cV=10 [ft2/yr] 4
CC=0.35
CR=0.05
then T= *log(100%-U)
If U>60%
Impermeable Bedrock

18. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2
Clay
T * (HD)2 t=
25 [ft]
25 [ft]
OCR=1.0
The time factor is We plug this value into our time rate of consolidation equation, and solve for time. cv
cV=10 [ft2/yr]
γT=105 [lb/ft3]
π * U2
If U<60%
then T=
cV=10 [ft2/yr] 4
CC=0.35
CR=0.05
then T= *log(100%-U)
If U>60%
T=0.848

19. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2]
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2
Clay
T * (HD)2 t=
25 [ft]
25 [ft]
OCR=1.0
It takes the soil 53 years to consolidate 90% of it’s ultimate consolidation settlement. cv
cV=10 [ft2/yr]
γT=105 [lb/ft3]
π * U2
If U<60%
then T=
cV=10 [ft2/yr]
t=53[yr] 4
CC=0.35
CR=0.05
then T= *log(100%-U)
If U<60%
T=0.848

20. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2] 17 26
C) 42
D) 53
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2
Clay
T * (HD)2 t=
25 [ft]
25 [ft]
OCR=1.0
Looking back at our possible solutions, cv
cV=10 [ft2/yr]
γT=105 [lb/ft3]
π * U2
If U<60%
then T=
cV=10 [ft2/yr]
t=53[yr] 4
CC=0.35
CR=0.05
then T= *log(100%-U)
If U<60%
T=0.848

21. Find: Time [yr] for 90% consolidation to occur
Load=300 [lb/ft2] 17 26
C) 42
D) 53
cv * t
Sand T=
γT=100 [lb/ft3]
10 [ft]
(HD)2
Clay
T * (HD)2 t=
25 [ft]
25 [ft]
OCR=1.0
The answer is D. cv
cV=10 [ft2/yr]
γT=105 [lb/ft3]
π * U2
If U<60%
then T=
cV=10 [ft2/yr]
t=53[yr] 4
CC=0.35
CR=0.05
then T= *log(100%-U)
If U<60%
AnswerD
T=0.848

22. ( ) ? 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 ‘