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7308ENG The Foundation Engineers

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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. 

Answer:

  1. Hydraulic conductivity

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

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 = -(xy) 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

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

  1. b)

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

  1. e) calculation of time factor in three months

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%

This problem has been solved.


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