Foundation engineers are often challenged by the existence of soft compressible soils at the construction site. The a soil profile with a silty sand (?= 15 kN/m3; ?sat=17 kN/m3) underlain by high?plasticity clay (?sat=17 kN/m3) and a peat layer (?sat=16kN/m3), followed by dense sand. To expedite consolidation and minimize future settlement, an additional 2?m?thick fill material, compacted to a unit weight of 19 kN/m3, will be placed on top of the silty sand layer. The plan area of the fill is 10 m X 10 m. The fill load will be left in place for 2 years, after which construction will begin with the fill becoming part of the permanent foundation. Undisturbed samples collected from the clay and organic layers had the following properties:
1. Estimate the total consolidation settlement under the action of the fill load. Consider both the clay and peat layers to be normally consolidated.
2. Estimate the time for 99% primary consolidation in each layer. Are the layers singly or doubly drained? Explain.
3. Estimate the secondary compression in each layer.
4. What will be the total settlement after 2 years?
5. Determine the effective stress at point A three months after the application of the fill load.
Question 4:
Calculate the settlement of the 3?m thick clay layer (see the following figure) that will result from the load (Q=4000 kg) by a 2?m square footing. The clay is normally consolidated. Using the weight average method to calculate the average increase of effective pressure in the clay layer.
The weight average method is
Question 5:
The results of a standard proctor test are given in the following table.
1. Determine the maximum dry unit weight of compaction and the optimum moisture content. Given: mold volume=953.3 cm3.
2. Determine the void ratio and the degree of saturation at the optimum moisture content. Given: Gs=2.68.
Calculation of Water Flow
 Hydraulic conductivity
K_{v} =
Substitution of H = 600 mm, H_{1} = H_{2} = H_{3} = 200 mm K_{1} = 5 * 10^{3}, K_{2} = 4.2 *10^{2} and K_{3} = 3.96 * 10^{4}
Determine the quantity of water flowing through the sample per hour.
K_{v} =
= 1.076 * 10^{3} cm/sec
Quantity of water flowing through the sample
q = K_{v}*i*A
= K_{v}*i*
i is the hydraulic gradient, A = cross section area of soil
substitute 47/60 for I and d = 15 cm
q = 1.076 * 10^{3} * 47/60 *
= 0.1489 cm^{3}/sec
= 536.04 cm^{3}/hr
The quantity of water flowing through the sample per hour = 536.04 cm^{3}/hr
b)
Calculate when x = 0
Pressure head is zero, because above the datum line Y – Y there is no water pressure
The value of Z is 220 mm
Calculate the total head at entrance and exit of each layer using the equation
h =
Elevation head is Z, pressure head is
h = 0 + 470
= 470 mm
Therefore, the total head is 470 mm
Calculate when x = 200 mm
Pressure head is zero, because above the datum line Y – Y there is no water pressure
The value of Z is 220 mm
Calculate the total head at entrance and exit of each layer using the equation
h =
here, elevation head is Z, pressure head is
Equate the discharge velocities
K_{v} = K_{1} * i_{1}
(1.076 * 10^{3}) * 47/60 = 0.005*
0.00084 = 2.5 * 10^{5}
h = 470 – 33.7
= 436.29 mm
Calculate the value of
= 436.29 + 220
656.29 mm
the total head is 656.29 mm
Calculate when X = 400 mm
Pressure head is zero because the datum line Y – Y there is no water pressure
The value of Z is 220 mm
Equate the discharge velocities
K_{v} = K_{1} * i_{1}
1.076 * 10^{3} * 47/60 = 0.042 *
0.00084 = 2.1 * 10^{5}
h = 436.29 – 4
= 432.29 mm
Calculate the value of
= 432.29 + 220
652.29 mm
the total head is 652.29 mm
Calculate when x = 600 mm
Pressure head is zero because the datum line Y – Y there is no water pressure
The value of Z is 220 mm
Equate the discharge velocities
K_{v} = K_{1} * i_{1}
1.076 * 10^{3} * 47/60 = 0.00039 *
0.00084 = 1.95 * 10^{4}
h = 436.29 – 432.24
= 0 mm
Calculate the value of
= 0 + 220
200 mm
the total head is 200 mm
d) discharge velocity = ki
k = hydraulic conductivity of the soil and I hydraulic gradient
k = 0.001076
v = 0.001076 * 47/60
= 0.000843 cm/sec
Calculation of the seepage velocity by using the equation
V_{s} = V/n
For soil I
V = 0.000843 and n = 0.5
V_{s} = 0.000843/0.5
= 0.00168 cm/sec
For soil II
V = 0.000843 and n = 0.6
V_{s} = 0.000843/0.6
= 0.0014 cm/sec
For soil III
V = 0.000843 and n = 0.33
Calculation of Pressure Head at Different Depths
V_{s} = 0.000843/0.33
e)
calculation of the height in A
height = (pressure head at X)_{200 mm}
= 656.29 mm
Therefore, the height of water in A is 656.29 mm
Calculate the height of water in B
height = (pressure head at X)_{400 mm}
= 652.29 mm
Therefore, the height of water in A is 652.29 mm
Q2)
The plane stress in XY plane, _{z} = 0 and _{ZX} = _{XZ} =_{ZY} =_{YZ} = 0
Sum forces in the x_{1} direction
_{1}*A*sec
Sum forces in the y_{1} direction
_{x1y1} *A*sec
_{xy} = _{yx}, simplified
_{1} = _{x}*cos^{2}^{ }+ _{y}*sin^{2}^{ }+ 2*_{xy}*sin
_{x1y1} = (_{x} –_{y}) sin + _{xy}(cos^{2}
Using the trigonometric identities
sin
_{1} =
_{1}face, substitution of
[^{2} + _{xy}^{2}
Cos2
Sin2
Substituting for R and rearranging
Larger principles
 Settlement of clay layer due to one dimensional consolidation
S_{cclay} =
C_{c} compression index
_{0} is the effective overburden pressure
Bearing capacity of clay
q =
where
q = 19 * 2 = 38 kN/m^{2}
calculation of m_{1}
m_{1} = L/B
L = B = 10 m
m_{1} = 10/10 = 1
Calculation of value of b
b = B/2
B = 10 m
b = 10/2 = 5 m
consider a clay layer of different depths from fill load = z
let z = 4, 6, 8
for z = 4
n_{1} = z/b
b = 5m and z = 4
n_{1} = 4/5
= 0.8
Tabulate the value for different z and calculate the increase in effective stress
m_{1} 
Z (m) 
b = B/2 
n_{1} = z/b 
q 
I_{4} 

1 
4 
5 
0.8 
38 
0.8 
30.4 
1 
6 
5 
1.2 
38 
0.606 
23.02 
1 
8 
5 
1.6 
38 
0.449 
17.06 
_{0} = 2 * 15 + 2*17 + 4/2 *18 –[(2 +2) *9.81]
Calculate the settlement of clay layer due to one dimensional consolidation
e_{0} = 1.1, , C_{c} = 0.36
S_{cclay} =
=
= 0.096 m
Calculation of the settlement of peat layer
S_{cclay} =
C_{c} compression index
_{0} is the effective overburden pressure
Bearing capacity of clay
q =
where
q = 19 * 2 = 38 kN/m^{2}
calculation of m_{1}
m_{1} = L/B
L = B = 10 m
m_{1} = 10/10 = 1
Calculation of value of b
b = B/2
B = 10 m
b = 10/2 = 5 m
consider a clay layer of different depths from fill load = z
let z = 8, 9, 10
for z = 8
n_{1} = z/b
b = 5m and z = 8
n_{1} = 8/5
= 1.6
Tabulate the value for different z and calculate the increase in effective stress
m_{1} 
Z (m) 
b = B/2 
n_{1} = z/b 
q 
I_{4} 

1 
8 
5 
1.6 
38 
0.449 
17.06 
1 
9 
5 
1.8 
38 
0.388 
14.74 
1 
10 
5 
2.0 
38 
0.336 
12.76 
Calculate the settlement of peat layer due to one dimensional consolidation
e_{0} = 5.9, , C_{c} = 6.6, H = 2 m
S_{cclay} =
=
= 0.136 m
S_{c} = S_{cclay }+ S_{cpeat}
S_{c} = 0.096 + 0.136
= 0.232 m
Total consolidation settlement = 0.232 m
 b)
calculation of time for 99% primary consolidation using the formula
t_{99clay} =
Here,the maximum drainage path is H_{dr},the time factor is T_{99, }and the coefficient of consolidation is c_{v}.
For the clay layer, a drainage condition is assumed since the top is silty sand which has a high permeability and the bottom is a peat layer has a high void ratio and considerable permeability.
Calculation of Discharge Velocity
For double drainage condition,
H_{dr}=
=2m
=200cm
For 99% degree of consolidation.
Variation of T_{v} with U. T_{99}=1.781.
Substitute 200cm for H_{dr. }1.781 for T_{99}. And 0.003cm^{2}/sec for C_{v}.
t_{99peat}=
23,746,666s*
=275 days
Hence,the time for 99% primary consolidation for clay is 275 days
Calculation of the time 99% primary consolidation
t_{99clay} =
for the peat layer assume a single drainage
Substitute 200cm for H_{dr. }1.781 for T_{99}. And 0.025 cm^{2}/sec for C_{v}.
t_{99peat}=
2,849,600s*
=33 days
Hence,the time for 99% primary consolidation for clay is 33 days
Secondary compression in clay
S_{cclay} = C’_{a}*Hlog[t_{2}/t_{1}]
C’_{a} secondary compression index , t_{1} = initial time, t_{2} = final time
_{primary }= c_{C}*log[
C_{c} = 0.36, _{0} = 60.76 kN/m^{2},
_{primary }=0.36*log[
= 0.36 *0.1407
= 0.0506
Computation of void ratio of clay
e_{p} = e_{0}  _{primary }
substitution e_{0} 1.1, _{primary }=0.0506
e_{p} = 1.1 – 0.0506
= 1.049
Calculation of the primary compression index
C’_{a} =
C_{a} = 003, e_{p} = 1.049
C’_{a} =
= 0.0146
Substitution of C’_{a} = 0.0146, H = 4 m, t_{1} = 275 days , t_{2} = 365 day
S_{sclay} = 0.0146 * 4*log[
= 0.0247
Computation of secondary compression in peat layer
S_{speat} = C’_{a}*H*log[t_{2}/t_{1}]
C’_{a} secondary compression index , t_{1} = initial time, t_{2} = final time
_{primary }= c_{C}*log[
C_{c} = 6.6, _{0} = 83.33 kN/m^{2}, , e_{0} = 5.9
_{primary }=6.6*log[
= 6.6 *0.071
= 0.468
Computation of void ratio of clay
e_{p} = e_{0}  _{primary }
substitution e_{0} 5.9, _{primary }=0.468
e_{p} = 5.9 – 0.468
= 5.432
Calculation of the primary compression index
C’_{a} =
C_{a} = 0.263, e_{p} = 5.432
C’_{a} =
= 0.0408
Substitution of C’_{a} = 0.0408, H = 2 m, t_{1} = 365 days , t_{2} = 33 day
S_{sclay} = 0.0408 * 2*log[
= 0.109
d)
S = S_{c} + S_{cclay} + S_{cpeat}
S_{c} = 0.232, S_{cclay} = 0.0247 and S_{cpeat} = 0.109
S= 0.232 + 0.0247 + 0.109
= 0.365 m
 e) calculation of time factor in three months
T_{v} =
Time = t, C_{v} = coefficient of consolidation, H_{dr} = maximum drainage path
t = 3 months , C_{v} = 0.003 cm^{2}/sec^{, }, H_{dr} = 200 cm
T_{v} =
= 0.5832
H_{dr} = 4/2 = 2 m , since the c lay layer is a double drainage
= 200 cm
Calculation of degree of consolidation
Calculate the ratio of = z/H_{dr}
H_{dr} = 2 m, z = 3 m
z/H_{dr} = 3/2 = 1.5
Variation of U_{z} with T_{v} and z/H_{dr}
U_{z} = 0.85
Computation of excess pore water pressure at point A
u_{z} = (1 – U_{z})u_{0}
= (1 – 0.85)*23.25
= 0.15 * 23.25
= 3.4875 kN/m^{2}
Calculate the increase in effective stress after 3 months
_{0}  u_{z}
_{z} = 3.4875 kN/m^{2} and u_{0} = 23.25 kN/m^{2}
_{0}  u_{z}
= 23.25 – 3.4875
= 19.76 kN/m^{2}
Effective stress at point A on clay layer
_{0} = 2 *15 +2*17+3*18 – (2+3)9.81
Final effective stress after 3 month
_{0} = _{0} +
_{0} = 68.95 kN/m^{2},
_{0} = _{0} +
= 68.95 + 19.76
= 88.71 kN/m^{2}
Q4)
Q = 200 kips
10 ft = 3
100 pcf = 15.72
120 pcf = 18.87
110 pcf = 17.3
Z = from footing to centre of clay layer
= 2 + 3 +2 = 7 m
= 484.44 N/m^{2}
Settlement
S_{s} =
H = 3 m
C_{c} = 0.009(LL10)
= 0.009(40 10) = 0.27
e_{0} = 1.0
_{0} = [(15.72 *2) + (18.87*3)+(17.3*2)][9.81 * 4.5]
= 78.505 kN/m^{2}
S_{s} =
=
= 2.164 *10^{3} m
Q5)
Trial no. 
Weight of moist soil in N 
Moisture content (%) 
1 
16.48 
9.9 
2 
16.78 
10.6 
3 
17.36 
12.1 
4 
17.95 
13.8 
5 
18.25 
15.1 
6 
18.44 
17.4 
7 
18.34 
19.4 
8 
18.15 
21.2 
Moist unit weight
Substituting = 16.48 * 10^{3} kN for W and 953.3 *10^{6} m^{3}
29 kN/m^{3}
Determination of dry unit weight = _{d} =
= 15.73 kN/m^{3}
Table of moist unit weight and dry unit weight values
V in cm^{3} 
Weight of moist soil in N 
Moisture content (W) (%) 
_{d} in kN/m^{3} ^{ } 

953.3 
16.48 
17.29 
9.9 
15.73 
953.3 
16.78 
17.60 
10.6 
15.91 
953.3 
17.36 
18.21 
12.1 
16.24 
953.3 
17.95 
18.83 
13.8 
16.55 
953.3 
18.25 
19.14 
15.1 
16.63 
953.3 
18.44 
19.34 
17.4 
16.47 
953.3 
18.34 
19.24 
19.4 
16.11 
953.3 
18.15 
19.04 
21.2 
15.71 
From the graph
The maximum dry unit weight = 16.64 kN/m^{3}
Optimum moisture content = 15.0%
The void ratio € using the dry unit weight
_{s} is the specific gravity of soil
16.64 =
1 + e = 1.58
e = 0.58
finding the degree of saturation (S)
Substitute 15.0% for W, G_{s} = 2.68, e = 0.58
S = *100
= 69.31%
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