Calculating Water Flow And Soil Properties In An Essay.

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

1. Hydraulic conductivity

K_v =

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.

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 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

K_v = 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

K_v = 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

K_v = 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

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 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

1. Settlement of clay layer due to one dimensional consolidation

S_c-clay =

C_c 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, , C_c = 0.36

S_c-clay =

 =

= 0.096 m

Calculation of the settlement of peat layer

S_c-clay =

C_c 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, , C_c = 6.6, H = 2 m

S_c-clay =

 =

= 0.136 m

S_c = S_c-clay + S_c-peat

S_c = 0.096 + 0.136

= 0.232 m

Total consolidation settlement = 0.232 m

1. b)

calculation of time for 99% primary consolidation using the formula

t_99-clay =

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. T₉₉=1.781.

Substitute 200cm for H_dr. 1.781 for T₉₉. And 0.003cm²/sec for C_v.

t_99-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

t_99-clay =

for the peat layer assume a single drainage

Substitute 200cm for H_dr. 1.781 for T₉₉. And 0.025 cm²/sec for C_v.

t_99-peat=

2,849,600s*

=33 days

Hence,the time for 99% primary consolidation for clay is 33 days

Secondary compression in clay

S_c-clay = C’_a*Hlog[t₂/t₁]

C’_a secondary compression index , t₁ = initial time, t₂ = final time

_primary = c_C*log[

C_c = 0.36, ₀ = 60.76 kN/m²,

_primary =0.36*log[

= 0.36 *0.1407

= 0.0506

Computation of void ratio of clay

e_p = e₀ - _primary

substitution e₀ 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, t₁ = 275 days , t₂ = 365 day

S_s-clay = 0.0146 * 4*log[

= 0.0247

Computation of secondary compression in peat layer

S_s-peat = C’_a*H*log[t₂/t₁]

C’_a secondary compression index , t₁ = initial time, t₂ = final time

_primary = c_C*log[

C_c = 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

e_p = e₀ - _primary

substitution e₀ 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, t₁ = 365 days , t₂ = 33 day

S_s-clay = 0.0408 * 2*log[

= 0.109

d)

S = S_c + S_c-clay + S_c-peat

S_c = 0.232, S_c-clay = 0.0247 and S_c-peat = 0.109

S= 0.232 + 0.0247 + 0.109

= 0.365 m

1. 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 cm²/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)u₀

 = (1 – 0.85)*23.25

= 0.15 * 23.25

= 3.4875 kN/m²

Calculate the increase in effective stress after 3 months

₀ - u_z

_z = 3.4875 kN/m² and u₀ = 23.25 kN/m²

₀ - u_z

   = 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

S_s =

H = 3 m

C_c = 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

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 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, G_s = 2.68, e = 0.58

S = *100

= 69.31%