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.
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.
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.
 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
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
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


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.
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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