Kv =
Substitution of H = 600 mm, H1 = H2 = H3 = 200 mm K1 = 5 * 10-3, K2 = 4.2 *10-2 and K3 = 3.96 * 10-4
Determine the quantity of water flowing through the sample per hour.
Kv =
= 1.076 * 10-3 cm/sec
Quantity of water flowing through the sample
q = Kv*i*A
= Kv*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 cm3/sec
= 536.04 cm3/hr
The quantity of water flowing through the sample per hour = 536.04 cm3/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
Kv = K1 * i1
(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
Kv = K1 * i1
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
Kv = K1 * i1
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
Vs = V/n
For soil I
V = 0.000843 and n = 0.5
Vs = 0.000843/0.5
= 0.00168 cm/sec
For soil II
V = 0.000843 and n = 0.6
Vs = 0.000843/0.6
= 0.0014 cm/sec
For soil III
V = 0.000843 and n = 0.33
Vs = 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 x1 direction
1*A*sec
Sum forces in the y1 direction
x1y1 *A*sec
xy = yx, simplified
1 = x*cos2 + y*sin2 + 2*xy*sin
x1y1 = -(x –y) sin + xy(cos2
Using the trigonometric identities
sin
1 =
1face, substitution of
[2 + xy2
Cos2
Sin2
Substituting for R and rearranging
Larger principles
Sc-clay =
Cc compression index
0 is the effective overburden pressure
Bearing capacity of clay
q =
where
q = 19 * 2 = 38 kN/m2
calculation of m1
m1 = L/B
L = B = 10 m
m1 = 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
n1 = z/b
b = 5m and z = 4
n1 = 4/5
= 0.8
Tabulate the value for different z and calculate the increase in effective stress
|
m1 |
Z (m) |
b = B/2 |
n1 = z/b |
q |
I4 |
|
|
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
e0 = 1.1, , Cc = 0.36
Sc-clay =
=
= 0.096 m
Calculation of the settlement of peat layer
Sc-clay =
Cc compression index
0 is the effective overburden pressure
Bearing capacity of clay
q =
where
q = 19 * 2 = 38 kN/m2
calculation of m1
m1 = L/B
L = B = 10 m
m1 = 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
n1 = z/b
b = 5m and z = 8
n1 = 8/5
= 1.6
Tabulate the value for different z and calculate the increase in effective stress
|
m1 |
Z (m) |
b = B/2 |
n1 = z/b |
q |
I4 |
|
|
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
e0 = 5.9, , Cc = 6.6, H = 2 m
Sc-clay =
=
= 0.136 m
Sc = Sc-clay + Sc-peat
Sc = 0.096 + 0.136
= 0.232 m
Total consolidation settlement = 0.232 m
calculation of time for 99% primary consolidation using the formula
t99-clay =
Here,the maximum drainage path is Hdr,the time factor is T99, and the coefficient of consolidation is cv.
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,
Hdr=
=2m
=200cm
For 99% degree of consolidation.
Variation of Tv with U. T99=1.781.
Substitute 200cm for Hdr. 1.781 for T99. And 0.003cm2/sec for Cv.
t99-peat=
23,746,666s*
=275 days
Hence,the time for 99% primary consolidation for clay is 275 days
Calculation of the time 99% primary consolidation
t99-clay =
for the peat layer assume a single drainage
Substitute 200cm for Hdr. 1.781 for T99. And 0.025 cm2/sec for Cv.
t99-peat=
2,849,600s*
=33 days
Hence,the time for 99% primary consolidation for clay is 33 days
Secondary compression in clay
Sc-clay = C’a*Hlog[t2/t1]
C’a secondary compression index , t1 = initial time, t2 = final time
primary = cC*log[
Cc = 0.36, 0 = 60.76 kN/m2,
primary =0.36*log[
= 0.36 *0.1407
= 0.0506
Computation of void ratio of clay
ep = e0 – primary
substitution e0 1.1, primary =0.0506
ep = 1.1 – 0.0506
= 1.049
Calculation of the primary compression index
C’a =
Ca = 003, ep = 1.049
C’a =
= 0.0146
Substitution of C’a = 0.0146, H = 4 m, t1 = 275 days , t2 = 365 day
Ss-clay = 0.0146 * 4*log[
= 0.0247
Computation of secondary compression in peat layer
Ss-peat = C’a*H*log[t2/t1]
C’a secondary compression index , t1 = initial time, t2 = final time
primary = cC*log[
Cc = 6.6, 0 = 83.33 kN/m2, , e0 = 5.9
primary =6.6*log[
= 6.6 *0.071
= 0.468
Computation of void ratio of clay
ep = e0 – primary
substitution e0 5.9, primary =0.468
ep = 5.9 – 0.468
= 5.432
Calculation of the primary compression index
C’a =
Ca = 0.263, ep = 5.432
C’a =
= 0.0408
Substitution of C’a = 0.0408, H = 2 m, t1 = 365 days , t2 = 33 day
Ss-clay = 0.0408 * 2*log[
= 0.109
d)
S = Sc + Sc-clay + Sc-peat
Sc = 0.232, Sc-clay = 0.0247 and Sc-peat = 0.109
S= 0.232 + 0.0247 + 0.109
= 0.365 m
Tv =
Time = t, Cv = coefficient of consolidation, Hdr = maximum drainage path
t = 3 months , Cv = 0.003 cm2/sec, , Hdr = 200 cm
Tv =
= 0.5832
Hdr = 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/Hdr
Hdr = 2 m, z = 3 m
z/Hdr = 3/2 = 1.5
Variation of Uz with Tv and z/Hdr
Uz = 0.85
Computation of excess pore water pressure at point A
uz = (1 – Uz)u0
= (1 – 0.85)*23.25
= 0.15 * 23.25
= 3.4875 kN/m2
Calculate the increase in effective stress after 3 months
0 – uz
z = 3.4875 kN/m2 and u0 = 23.25 kN/m2
0 – uz
= 23.25 – 3.4875
= 19.76 kN/m2
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/m2,
0 = 0 +
= 68.95 + 19.76
= 88.71 kN/m2
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/m2
Settlement
Ss =
H = 3 m
Cc = 0.009(LL-10)
= 0.009(40 -10) = 0.27
e0 = 1.0
0 = [(15.72 *2) + (18.87*3)+(17.3*2)]-[9.81 * 4.5]
= 78.505 kN/m2
Ss =
=
= 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 m3
29 kN/m3
Determination of dry unit weight = d =
= 15.73 kN/m3
Table of moist unit weight and dry unit weight values
|
V in cm3 |
Weight of moist soil in N |
Moisture content (W) (%) |
d in kN/m3
|
|
|
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/m3
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, Gs = 2.68, e = 0.58
S = *100
= 69.31%
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