1. Find: Saturation, S 11% d=2.8 [in] 17% 23% 83% L=5.5 [in] clay
Wtotal=2.58 [lb]
SG=2.75
wc=15%
Find the saturation of the sample of clay. In this problem,

2. Find: Saturation, S 11% d=2.8 [in] 17% 23% 83% L=5.5 [in] clay
Wtotal=2.58 [lb]
SG=2.75
wc=15%
a cylindrical sample of clay is provided with it’s known dimensions,

3. Find: Saturation, S 11% d=2.8 [in] 17% 23% 83% L=5.5 [in] clay
Wtotal=2.58 [lb]
SG=2.75
wc=15%
as well as it’s total weight, specific gravity and water content.

4. Find: Saturation, S VW S= VV d=2.8 [in] L=5.5 [in] clay
Wtotal=2.58 [lb]
SG=2.75
wc=15%
Saturation is defined as the volume of water divided by the volume of voids. To solve for these two variables,

5. Find: Saturation, S clay A W S Wtotal=2.58 [lb] SG=2.75 wc=15%
V [ft3]
W [lb]
we’ll need to set up a block block diagram for the sample of clay.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

6. Find: Saturation, S clay A W S 2.58 Wtotal=2.58 [lb] SG=2.75 wc=15%
V [ft3]
W [lb]
2.58
We were given the total weight is 2.58 pounds.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

7. Find: Saturation, S clay π * d2 VT = L * A 4 W S 2.58 Wtotal=2.58 [lb]
V [ft3]
W [lb]
π * d2
VT = L * 4
2.58
and, we can calculate the total volume by knowing, the cylinder’s length and diameter.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

8. Find: Saturation, S clay π * d2 VT = L * A 4 W S VT = 0.0196 [ft3]
V [ft3]
W [lb]
π * d2
VT = L * 4
VT = [ft3]
0.0196
2.58
The total volume of the sample is cubic feet.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

9. Find: Saturation, S clay A W S 0.0196 2.58 Wtotal=2.58 [lb]
V [ft3]
W [lb]
0.0196
2.58
Since we know the weight of air is zero, and we know the total weight,
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

10. Find: Saturation, S clay wc= A W S WW WS 0.0196 2.58 Wtotal=2.58 [lb]
V [ft3]
W [lb]
wc= WW WS WW WS
0.0196
2.58
we can use the water content to solve for the weight of water and weight of solid.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

11. Find: Saturation, S clay wc= A W WW=wc * WS S WW WS 0.0196 2.58
V [ft3]
W [lb]
wc= WW WS
WW=wc * WS WW WS
0.0196
2.58
The weight of water is 15% the weight of the solid.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

12. Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS WW WS
V [ft3]
W [lb]
wc= WW WS
WW=wc * WS WW
WW=0.15 * WS WS
0.0196
2.58
We also know ---
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

13. Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS
V [ft3]
W [lb]
wc= WW WS
WW=wc * WS WW
WW=0.15 * WS WS
WT = WW + WS
0.0196
2.58
the total weight equals the weight of water plus the weight of solid.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

14. Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS
V [ft3]
W [lb]
wc= WW WS
WW=wc * WS WW
WW=0.15 * WS WS
WT = WW + WS
0.0196
2.58
WS = WT - WW
Rearranging the equation,
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

15. Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS
V [ft3]
W [lb]
wc= WW WS
WW=wc * WS WW
WW=0.15 * WS WS
WT = WW + WS
0.0196
2.58
WS = WT - WW
the weight of solid is 2.58 pounds minus the weight of water.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

16. Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS
V [ft3]
W [lb]
wc= WW WS
WW=wc * WS WW
WW=0.15 * WS WS
WT = WW + WS
0.0196
2.58
WS = WT - WW
If we plug this into a previous equation and skip a couple steps of algebra,
WS = 2.58 [lb] - WW
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

17. Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS
V [ft3]
W [lb]
wc= WW WS
WW=wc * WS
0.337
WW=0.15 * WS WS
WT = WW + WS
0.0196
2.58
WS = WT - WW
we find the weight of water is pounds.
WS = 2.58 [lb] - WW
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
1.15 * WW = [lb]
WW = [lb]
L=5.5 [in] d=2.8 [in]

18. Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS
V [ft3]
W [lb]
wc= WW WS
WW=wc * WS
0.337
WW=0.15 * WS
2.243
WT = WW + WS
0.0196
2.58
WS = WT - WW
And by quick subtraction, the weight of solids is pounds. [pause]
WS = 2.58 [lb] - WW
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
1.15 * WW = [lb]
WW = [lb]
WS = [lb]
L=5.5 [in] d=2.8 [in]

19. γS Find: Saturation, S clay A W S WS VS= VA VW 0.337 VS 2.243 0.0196
V [ft3]
W [lb] WS
VS= γS VA VW
0.337 VS
2.243
0.0196
2.58
Next, we solve for the volumes of air, water and solid.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

20. γS γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 VW
V [ft3]
W [lb] WS
2.243 [lb]
VS= γS = VA lb
62.4
*2.75
ft3 γW VW
0.337 VS
2.243
0.0196
2.58
The volume of solid is the weight of solid divided the unit weigh of solid,
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

21. γS γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 VW
V [ft3]
W [lb] WS
2.243 [lb]
VS= γS = VA lb
62.4
*2.75
ft3 γW VW
0.337
VS= [ft3] VS
2.243
0.0196
2.58
and equals cubic feet.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

22. γS γW γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 VW
V [ft3]
W [lb] WS
2.243 [lb]
VS= γS = VA lb
62.4
*2.75
ft3 γW VW
0.337
VS= [ft3]
0.0131
2.243 WW
0.337 [lb]
VW= γW =
0.0196
2.58 lb
62.4
The volume of water is solved much the same,
ft3
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

23. γS γW γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 VW
V [ft3]
W [lb] WS
2.243 [lb]
VS= γS = VA lb
62.4
*2.75
ft3 γW VW
0.337
VS= [ft3]
0.0131
2.243 WW
0.337 [lb]
VW= γW =
0.0196
2.58 lb
62.4
and equals cubic feet.
ft3
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
VW= [ft3]
L=5.5 [in] d=2.8 [in]

24. γS γW γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4
V [ft3]
W [lb] WS
2.243 [lb]
VS= γS = VA lb
62.4
*2.75
ft3 γW
0.0054
0.337
VS= [ft3]
0.0131
2.243 WW
0.337 [lb]
VW= γW =
0.0196
2.58 lb
62.4
The volume of air is the total volume minus the volume of solid minus the volume of water,
ft3
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
VW= [ft3]
VA = VT - VS - VW
L=5.5 [in] d=2.8 [in]

25. γS γW γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4
V [ft3]
W [lb] WS
2.243 [lb]
VS= γS = VA lb
62.4
*2.75
ft3 γW
0.0054
0.337
VS= [ft3]
0.0131
2.243 WW
0.337 [lb]
VW= γW =
0.0196
2.58 lb
62.4
and equals cubic feet.
ft3
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
VW= [ft3]
VA = VT - VS - VW
VA = [ft3]
L=5.5 [in] d=2.8 [in]

26. Find: Saturation, S VW S= VV clay A W S 0.0011 0.0054 0.337 0.0131
V [ft3]
W [lb] VW S=
0.0011 VV
0.0054
0.337
0.0131
2.243
0.0196
2.58
With our block diagram solved, we can compute the saturation of the soil.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

27. Find:Saturation, S VW VW S= = VV VW+VA clay A W S 0.0011 0.0054 0.337
V [ft3]
W [lb] VW VW S= =
0.0011 VV
VW+VA
0.0054
0.337
0.0131
2.243
0.0196
2.58
After plugging in the appropriate values,
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

28. Find:Saturation, S VW VW S= = VV VW+VA S = 83.1% clay A W S 0.0011
V [ft3]
W [lb] VW VW S= =
0.0011 VV
VW+VA
0.0054
0.337
S = 83.1%
0.0131
2.243
0.0196
2.58
we find the saturation is 83.1 percent.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

29. Find:Saturation, S VW VW S= = VV VW+VA S = 83.1% clay A W S 0.0011
V [ft3]
W [lb] VW VW S= =
0.0011 VV
VW+VA
0.0054
0.337
S = 83.1%
0.0131
2.243
0.0196
2.58
11%
17%
23%
83%
When reviewing the possible solutions,
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

30. Find:Saturation, S VW VW S= = VV VW+VA S = 83.1% AnswerD clay A W S
V [ft3]
W [lb] VW VW S= =
0.0011 VV
VW+VA
0.0054
0.337
S = 83.1%
0.0131
2.243
AnswerD
0.0196
2.58
11%
17%
23%
83%
the answer is D.
clay
Wtotal=2.58 [lb]
SG=2.75 wc=15%
L=5.5 [in] d=2.8 [in]

31. ( ) ? τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal
wc= WW WS 1
Index
γclay=53.1[lb/ft3]
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]
62.4 lb
ft3 27
yd3
ft3
(5 [cm])2 * π/4
( )
H*C σfinal
ρcn=
1+e σinitial
log ‘ φ
size[mm]
% passing 0%
20%
40%
60%
80%
100% 10 1
0.1
0.01
c=0
400
1,400 σ3
Sand σ1