7308ENG Advanced Soil Mechanics: Seepage, Stress & Consolidation

This assignment delves into advanced soil mechanics, focusing on seepage analysis through layered soil specimens, stress distribution, and consolidation settlement calculations. It begins by determining the quantity of water flowing through a soil sample with varying hydraulic conductivities, followed by calculating and plotting the elevation head, pressure head, and total head at different points within the sample. The assignment further examines discharge and seepage velocities along the sample axis. Additionally, it covers the calculation of consolidation settlement for clay and peat layers under given loading conditions, including primary and secondary consolidation. The time required for primary consolidation is also computed. Desklib offers a wealth of similar solved assignments and study materials for students.

Solution
Q1)
i) Hydraulic conductivity
Kv =
H
H 1
K 1 + H 2
K 2 + H 3
K 3
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 =
60
20
5∗10−3 + 20
4.2∗10−2 + 20
3.96∗10−3
= 1.076 * 10-3 cm/sec
Quantity of water flowing through the sample
q = Kv*i*A
= Kv*i* π d2
4
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 * π∗152
4
= 0.1489 cm3/sec
=
0.1489 c m2
1
60∗60 hr
= 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 = H
γw + Z
Elevation head is Z, pressure head is u
γw

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 = u
γw + Z
here, elevation head is Z, pressure head is u
γw
Equate the discharge velocities
Kv = K1 * i1
(1.076 * 10-3) * 47/60 = 0.005* ∆ H
20
0.00084 = 2.5 * 10-5 ∆ H
∆ H =33.7 mm
calculate the total head
h = 470 – 33.7
= 436.29 mm
Calculate the value of u
γw
H
γw = 436.29 + 220
¿ 656.29 mm
Therefore , 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 * ∆ H
20
0.00084 = 2.1 * 10-5 ∆ H
∆ H =4 mm
calculate the total head
h = 436.29 – 4
= 432.29 mm

Calculate the value of u
γw
H
γw = 432.29 + 220
¿ 652.29 mm
Therefore , 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 * ∆ H
20
0.00084 = 1.95 * 10-4 ∆ H
∆ H =432.24 mm
calculate the total head
h = 436.29 – 432.24
= 0 mm
Calculate the value of u
γw
H
γw = 0 + 220
¿ 200 mm
Therefore , the total head is 200 mm

Part c
0 100 200 300 400 500 600 700
-400
-200
0
200
400
600
800
Total head
elevation
Series 3
Distance along the sample
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
= 0.00255 cm/sec
0 200 400 600
0
0.0005
0.001
0.0015
0.002
0.0025
0.003
seepage velocity
seepage velocity
Distance along the sample
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, σ z = 0 and τ ZX = τ XZ =τ ZY =τ YZ = 0

Sum forces in the x1 direction
σ 1*A*sec
θ− ( σx∗A )∗cosθ− ( τxy∗A )∗sinθ− ( σy∗A∗tanθ )∗sinθ− ( τyx∗A∗tanθ )∗cosθ=0
Sum forces in the y1 direction
τ x1y1 *A*secθ+ ( σx∗A ) sinθ− ( τxy∗A ) cosθ− ( σy∗Atanθ ) cosθ− ( τyx∗Atanθ ) sinθ=0
using τxy = τ yx, simplified
σ 1 = σ x*cos2
θ + σ y*sin2
θ + 2*τ xy*sinθ cosθ
τ x1y1 = -( σ x –σ y) sinθcosθ + τ xy(cos2
θ−sin2 θ ¿
Using the trigonometric identities
cos2 θ=1+ cos 2θ
2 ,
sin2 θ= 1−cos 2θ
2
sinθcosθ=sin 2 θ
2
σ 1 = σx+ σy
2 + σx−σy
2 cos 2θ +τxy∗sin 2 θ
for pressure at y1face, substitution of θ+ 900 for θ
σ 1 = σx+ σy
2 − σx−σy
2 cosθ−τxy∗sin 2θ
σ 1 = σx+ σy
2 + σx−σy
2 cos 2θ +τxy∗sin 2 θ
∂ σ 1
∂ θ = σx−σy
2 (−2 sinθ ) + τxy∗( 2 cos 2θ )=0
∂ σ 1
∂ θ =− ( σx−σy ) sin 2θ +2∗τxy∗( cos 2 θ )=0

tan2 θ= 2 τxy
σx−σy
σ 1 = σx+ σy
2 + σx−σy
2 cos 2θ +τxy∗sin 2 θ
R2 = [ σx−σy
2 ¿2 + τ xy2
Cos2θ= σx−σy
2 R
Sin2θ= τxy
R
σ 1 = σx+ σy
2 + σx−σy
2 [ σx−σy
2 R ]+ τxy∗τxy
R
Substituting for R and rearranging
Larger principles
σ 1 = σx+ σy
2 + √ [ σx +σy
2 ]2
+ τx y2
smaller principle ,
σ 3 = σx +σy−σ 1= σx+ σy
2 − √ [ σx+σy
2 ]2
+τx y2
Q3)
a) Settlement of clay layer due to one dimensional consolidation

Sc-clay = Cc∗H
1+ e 0 log [ σ 0+∆ σ
σ 0 ]
Cc compression index
∆ σ increase∈effective stress
σ 0 is the effective overburden pressure
H isthe height of the clay layer
Bearing capacity of clay
q = γ∗H
where γ=19 kN
m2 , H =2 m
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
∆ σ'=qI 4
q=38 kN
m2 ∧l=0.8
∆ σ' =38∗0.8=30.4 kN / m3
Tabulate the value for different z and calculate the increase in effective stress
m1 Z (m) b = B/2 n1 = z/b q I4 ∆ σ'=qI 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

∆ σ'= 30.4+4∗23.02∗17.06
6
¿ 23.25 kN /m2
effective overburden pressure up ¿ middle of clay layer
σ '0 = 2 * 15 + 2*17 + 4/2 *18 –[(2 +2) *9.81]
¿ 60.76 kN /m2
Calculate the settlement of clay layer due to one dimensional consolidation
e0 = 1.1, σ ' 0=60.76 kN
m2 for , ∆ σ'=23.25 kN / m2
, Cc = 0.36
Sc-clay = Cc∗H
1+ e 0 log [ σ 0+∆ σ
σ 0 ]
= 0.36∗4
1+1.1 log [ 60.76+23.23
60.76 ]
= 0.096 m
Calculation of the settlement of peat layer
Sc-clay = Cc∗H
1+ e 0 log [ σ 0+∆ σ
σ 0 ]
Cc compression index
∆ σ increase∈effective stress
σ 0 is the effective overburden pressure
H isthe height of the clay layer
Bearing capacity of clay
q = γ∗H
where γ=19 kN
m2 , H =2 m
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
∆ σ'=qI 4
q=38 kN
m2 ∧l=0.449
∆ σ'=38∗0.449=17.06 kN /m3
Tabulate the value for different z and calculate the increase in effective stress
m1 Z (m) b = B/2 n1 = z/b q I4 ∆ σ'=qI 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
∆ σ'=17.06 +4∗14.74∗12.76
6
¿ 14.8 kN /m2
Calculate the settlement of peat layer due to one dimensional consolidation
e0 = 5.9, σ ' 0=83.33 kN
m2 , ∆ σ ' =14.8 kN /m2
, Cc = 6.6, H = 2 m
Sc-clay = Cc∗H
1+ e 0 log [ σ 0+∆ σ
σ 0 ]
= 6.6∗2
1+ 5.9 log [ 83.33+14.8
83.33 ]
= 0.136 m
Sc = Sc-clay + Sc-peat

Sc = 0.096 + 0.136
= 0.232 m
Total consolidation settlement = 0.232 m
b)
calculation of time for 99% primary consolidation using the formula
t99-clay = H2 T 99
Cv
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= 4
2
=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= 2002∗1.781
0.003
23,746,666s* 1 day
86,400 s
=275 days
Hence,the time for 99% primary consolidation for clay is 275 days
Calculation of the time 99% primary consolidation